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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Lie Algebras
  }
    \\ {\normalsize Lectures by Brian Boe. University of Georgia, Fall
2022} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Lie Algebras taught by Brian Boe at the
University of Georgia in Fall 2022. As such, any errors or inaccuracies
are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2022-12-27 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{finite-dimensional-semisimple-lie-algebras-over-mathbfc-wednesday-august-17}{%
\section{\texorpdfstring{Finite-dimensional Semisimple Lie Algebras over
\({\mathbf{C}}\) (Wednesday, August
17)}{Finite-dimensional Semisimple Lie Algebras over \{\textbackslash mathbf\{C\}\} (Wednesday, August 17)}}\label{finite-dimensional-semisimple-lie-algebras-over-mathbfc-wednesday-august-17}}

\hypertarget{humphreys-1.1}{%
\subsection{Humphreys 1.1}\label{humphreys-1.1}}

\begin{remark}

Main goal: understand semisimple finite-dimensional Lie algebras over
\({\mathbf{C}}\). These are extremely well-understood, but there are
open problems in infinite-dimensional representations, representations
over other fields, and Lie superalgebras.

\end{remark}

\begin{remark}

Recall that an associative algebras is a ring with the structure of a
\(k{\hbox{-}}\)vector space, and \emph{algebra} generally means a
non-associative algebra. Given any algebra, one can define a new
bilinear product
\begin{align*}
[{-}, {-}]: A\otimes_k A &\to A \\
a\otimes b &\mapsto ab-ba
\end{align*}
called the \textbf{commutator bracket}. This yields a new algebra
\(A_L\) which is an example of a Lie algebra.

\end{remark}

\begin{definition}[Lie algebra]

For \(L\in {}_{{ \mathbf{F} }}{\mathsf{Mod}}\) with an operation
\([{-}, {-}]: L\times L\to L\) (called the \textbf{bracket}) is a
\textbf{Lie algebra} if

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \([{-},{-}]\) is bilinear,
\item
  \([x, x] = 0\) for all \(x\in L\), and
\item
  the Jacobi identity holds: \([x[yz]] + [y[zx]] + [z[xy]] = 0\).
\end{enumerate}

\end{definition}

\begin{exercise}[?]

Check that \([ab]\coloneqq ab-ba\) satisfies the Jacobi identity.

\end{exercise}

\begin{remark}

\envlist

\begin{itemize}
\item
  Expanding \([x+y, x+y] = 0\) yields \([xy] = -[yx]\). Note that this
  is equivalent to axiom 2 when
  \(\operatorname{ch}{ \mathbf{F} }\neq 2\) (given axiom 1).
\item
  The Jacobi identity can be rewritten as
  \([x[yz]] = [[xy]z] + [y[xz]]\), where the second term is an error
  term measuring the failure of associativity. Note that this is
  essentially the Leibniz rule.
\end{itemize}

\end{remark}

\begin{definition}[Abelian Lie algebras]

A Lie algebra \(L\in\mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\) is
\textbf{abelian} if \([xy]=0\) for all \(x,y\in L\).

\end{definition}

\begin{definition}[Morphisms of Lie algebras]

A \textbf{morphism} in
\(\mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\) is a morphism
\(\phi\in {}_{{ \mathbf{F} }}{\mathsf{Mod}}(L, L')\) satisfying
\(\phi( [xy] ) = [ \phi(x) \phi(y) ]\).

\end{definition}

\begin{exercise}[?]

Check that if \(\phi\) has an inverse in
\({}_{{ \mathbf{F} }}{\mathsf{Mod}}\), then \(\phi\) automatically has
an inverse in \(\mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\).

\end{exercise}

\begin{definition}[Subobjects]

A vector subspace \(K\leq L\) is a \textbf{Lie subalgebra} if
\([xy]\in K\) for all \(x,y\in K\).

\end{definition}

\begin{remark}

\envlist

\begin{itemize}
\tightlist
\item
  Note that any nonzero \(x\in L\) determines a 1-dimensional Lie
  subalgebra \(K\coloneqq{ \mathbf{F} }\cdot x\), which is in fact
  abelian.
\item
  A big source of Lie algebras: left-invariant vector fields on a Lie
  group.
\item
  We'll restrict to finite-dimensional algebras for the remainder of the
  class.
\end{itemize}

\end{remark}

\hypertarget{humphreys-1.2-linear-lie-algebras}{%
\subsection{Humphreys 1.2: Linear Lie
Algebras}\label{humphreys-1.2-linear-lie-algebras}}

\begin{remark}

For \(V\in {}_{{ \mathbf{F} }}{\mathsf{Mod}}\), the endomorphisms
\(A\coloneqq{ \operatorname{End} }_{ \mathbf{F} }(V)\) is an associative
algebra over \({ \mathbf{F} }\). Thus it can be made into a Lie algebra
\({\mathfrak{gl}}(V) \coloneqq A_L\) by defining \([xy] = xy-yx\) as
above.

\end{remark}

\begin{definition}[Linear Lie algebras]

Any subalgebra \(K\leq {\mathfrak{gl}}(V)\) is a \textbf{linear Lie
algebra}.

\end{definition}

\begin{remark}

After picking a basis for \(V\), there is a noncanonical isomorphism
\({ \operatorname{End} }_{ \mathbf{F} }(V) \cong \operatorname{Mat}_{n\times n}({ \mathbf{F} })\)
where \(n\coloneqq\dim_{ \mathbf{F} }V\). The resulting Lie algebra is
\({\mathfrak{gl}}_n({ \mathbf{F} }) \coloneqq\operatorname{Mat}_{n\times n}({ \mathbf{F} })_L\).

\end{remark}

\begin{fact}

By Ado-Iwasawa, any finite-dimensional Lie algebra is isomorphic to some
linear Lie algebra.

\end{fact}

\begin{example}[?]

The upper triangular matrices form a subalgebra
\({\mathfrak{t}}_n({ \mathbf{F} }) \leq {\mathfrak{gl}}_n({ \mathbf{F} })\).\footnote{You
  get something interesting if you take the commutator bracket of two
  upper triangular matrices.} This is sometimes called the Borel and
denoted \({\mathfrak{b}}\). There is also a subalgebra
\({\mathfrak{n}}_n({ \mathbf{F} })\) of strictly upper triangular
matrices. The diagonal matrices form a maximal torus/Cartan subalgebra
\({\mathfrak{h}}_n({ \mathbf{F} })\) which is abelian.

\end{example}

\begin{example}[Classical Lie algebras]

\envlist

\begin{itemize}
\tightlist
\item
  Type \(A_n \leadsto {\mathfrak{sl}}_{n+1}({ \mathbf{F} })\) is the
  special linear Lie algebra, traceless matrices.
\item
  Type \(B_n \leadsto {\mathfrak{so}}_{2n+1}({ \mathbf{F} })\) is the
  odd orthogonal Lie algebra.
\item
  Type \(C_n \leadsto {\mathfrak{sp}}_{2n}({ \mathbf{F} })\) is the
  symplectic Lie algebra.
\item
  Type \(D_n \leadsto {\mathfrak{so}}_{2n}({ \mathbf{F} })\) is the even
  orthogonal Lie algebra.
\item
  The remaining 3 are defined by matrices satisfying \(sx = -x^t s\)
  where \(s\) is one of the following:

  \begin{itemize}
  \tightlist
  \item
    \({ \begin{bmatrix}  {1} & {0} & {0} \\  {0} & {0} & {I_n} \\  {0} & {I_n} & {0} \end{bmatrix} }\)
    corresponding to \({\mathfrak{so}}_{2n+1}\),
  \item
    \({ \begin{bmatrix}  {0} & {I_n} \\  {-I_n} & {0} \end{bmatrix} }\)
    corresponding to \({\mathfrak{sp}}_{2n}\),
  \item
    \({ \begin{bmatrix}  {0} & {I_n} \\  {I_n} & {0} \end{bmatrix} }\)
    corresponding to \({\mathfrak{so}}_{2n}\).
  \end{itemize}
\end{itemize}

These can be viewed as the matrices of a nodegenerate bilinear form:
writing \(N\) for the size of the matrices, the matrices act on
\(V \coloneqq{ \mathbf{F} }^N\) by a bilinear form
\(f: V\times V\to { \mathbf{F} }\) given by \(f(v, w) = v^t s w\). The
form will be symmetric for \({\mathfrak{so}}\) and skew-symmetric for
\({\mathfrak{sp}}\). The equation \(sx=-x^ts\) is a version of
preserving the bilinear form \(s\). Note that these are the Lie algebras
of the Lie groups
\(G = {\operatorname{SO}}_{2n+1}({ \mathbf{F} }), {\operatorname{Sp}}_{2n}({ \mathbf{F} }), {\operatorname{SO}}_{2n}({ \mathbf{F} })\)
defined by the condition \(f(gv, gw) = f(v, w)\) for all
\(v,w\in { \mathbf{F} }^N\) where
\(G = \left\{{g\in \operatorname{GL}_N({ \mathbf{F} }) {~\mathrel{\Big\vert}~}f(gv, gw) = f(v, w)}\right\}\).
This is equivalent to the condition that \(f(gv, w) = f(v, g^{-1}w)\).

\end{example}

\begin{remark}

Philosophy: \(G\to {\mathfrak{g}}\) sends products to sums.
\todo{Check, might have gotten this backward.}

\end{remark}

\begin{exercise}[?]

Check that the definitions of
\({\operatorname{SO}}_n({ \mathbf{F} }), {\operatorname{Sp}}_n({ \mathbf{F} })\)
yield Lie algebras.

\end{exercise}

\hypertarget{friday-august-19}{%
\section{Friday, August 19}\label{friday-august-19}}

\hypertarget{humphreys-1.3}{%
\subsection{Humphreys 1.3}\label{humphreys-1.3}}

\begin{definition}[Derivations]

Let \(A\in \mathsf{Alg}_{/ {{ \mathbf{F} }}}\), not necessarily
associative (e.g.~a Lie algebra). An
\textbf{\({ \mathbf{F} }{\hbox{-}}\)derivation} is a morphism
\(D: A \to A\) such that \(D(ab) = D(a)b + aD(b)\). Equipped with the
commutator bracket, this defines a Lie algebra
\(\mathop{\mathrm{Der}}_{ \mathbf{F} }(A) \leq {\mathfrak{gl}}_{ \mathbf{F} }(A)\).\footnote{The
  usual product somehow involves ``second-order terms'', while the
  commutator product cancels higher order terms to give something
  first-order.}

\end{definition}

\begin{warnings}

If \(D, D'\) are derivations, then the composition \(D\circ D'\) is
\emph{not} generally a derivation.

\end{warnings}

\begin{definition}[Adjoint]

If \(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\), for
\(x\in L\) fixed define the \textbf{adjoint operator}
\begin{align*}
 { \operatorname{ad}}_x: L &\to L \\
y &\mapsto [x, y]
.\end{align*}
Note that
\({ \operatorname{ad}}_x\in \mathop{\mathrm{Der}}_{ \mathbf{F} }(A)\) by
the Jacobi identity. Any derivation of this form is an \textbf{inner
derivation}, and all other derivations are \textbf{outer derivations}.

\end{definition}

\begin{remark}

Given \(x\in K\leq L\), note that \(K\) is a Lie subalgebra, and we'll
want to distinguish \({ \operatorname{ad}}_L x\) and
\({ \operatorname{ad}}_K x\) (which may differ). Note that
\({\mathfrak{gl}}_n({ \mathbf{F} }) \geq {\mathfrak{b}}= {\mathfrak{h}}\oplus {\mathfrak{n}}\),
where \({\mathfrak{b}}\) are upper triangular, \({\mathfrak{h}}\)
diagonal, and \({\mathfrak{n}}\) strictly upper triangular matrices. If
\(x\in {\mathfrak{h}}\) then note that
\({ \operatorname{ad}}_{\mathfrak{h}}x = 0\), but
\({ \operatorname{ad}}_{\mathfrak{g}}x {\mathfrak{h}}\neq 0\).

\end{remark}

\hypertarget{humphreys-2.1}{%
\subsection{Humphreys 2.1}\label{humphreys-2.1}}

\begin{remark}

Some notes:

\begin{itemize}
\tightlist
\item
  \(L\geq I\) is an \textbf{ideal} if \([\ell, i] \in I\). Note that
  this is like a left ideal in a ring, but since \(I\) is closed under
  scalar multiplication and \([i, \ell] = -[\ell, i]\), this is closed
  to a two-sided ideal.
\item
  If \(A, B \subseteq L\) define
  \([A, B] \coloneqq\mathop{\mathrm{span}}_{ \mathbf{F} }\left\{{[a,b] {~\mathrel{\Big\vert}~}a\in A,b\in B}\right\}\).
  \emph{Not} taking the span generally won't even yield a subalgebra.
\item
  For ideals \(I,J{~\trianglelefteq~}L\),
  \(I+J, [I,J] {~\trianglelefteq~}L\).
\item
  \(L/I \coloneqq\left\{{x+I {~\mathrel{\Big\vert}~}x\in L}\right\}\)
  with \([x+I, y+I] \coloneqq[x,y] + I\).\footnote{One should check that
    this is well-defined.}
\item
  Ideals are subalgebras (since this only requires closure under
  bracketing), but subalgebras are not necessarily ideals.
\item
  Centers:
  \(Z(L) = \left\{{x\in L {~\mathrel{\Big\vert}~}[xy] = 0 \forall y\in L}\right\} \leq L\).
\item
  Derived ideals
  \(L' \coloneqq L^1 \coloneqq[L, L] {~\trianglelefteq~}L\).
\item
  If
  \({\mathfrak{g}}\geq {\mathfrak{b}}= {\mathfrak{h}}\oplus {\mathfrak{n}}\),
  and \({\mathfrak{n}}{~\trianglelefteq~}{\mathfrak{b}}\) using the fact
  that products of upper-triangular matrices involve multiplying
  diagonals, and bracketing/subtracting cancels the diagonal off.
  Moreover \({\mathfrak{b}}/{\mathfrak{n}}\cong {\mathfrak{b}}\).
\item
  For \(K \subseteq L\) a subspace, the normalizer is
  \(N_L(K) \coloneqq\left\{{x\in L{~\mathrel{\Big\vert}~}[x, K] \subseteq K}\right\}\).
  If \(K = N_L\) then \(K\) is self-normalizing.
\item
  The centralizer of \(K\) in \(L\) is
  \(C_L(K) \coloneqq\left\{{x\in L{~\mathrel{\Big\vert}~}[x, K] = 0}\right\} \leq L\),
  which is a subalgebra by the Jacobi identity.
\end{itemize}

\end{remark}

\begin{exercise}[?]

Is \({\mathfrak{h}}{~\trianglelefteq~}{\mathfrak{b}}\)?

\end{exercise}

\begin{definition}[Simple Lie algebras]

A Lie algebra \(L\) is \textbf{simple} if \(L\neq 0\) and
\(\operatorname{Id}(L) = \left\{{0, L}\right\}\) and \([L, L] \neq 0\).
Note that \([LL] \neq 0\) only rules out the 1-dimensional Lie algebra,
since \([L, L] = 0\) and if \(0 < K < L\) then \(K{~\trianglelefteq~}L\)
since \([L,K] = 0\).

\end{definition}

\begin{example}[?]

Let \(L = {\mathfrak{sl}}_2({\mathbf{C}})\), so
\(\operatorname{tr}(x) = 0\). This has standard basis
\begin{align*}
x = {
\begin{bmatrix}
  {0} & {1} 
\\
  {0} & {0}
\end{bmatrix}
}, \qquad
y = {
\begin{bmatrix}
  {0} & {0} 
\\
  {1} & {0}
\end{bmatrix}
},\qquad
h = {
\begin{bmatrix}
  {1} & {0} 
\\
  {0} & {-1}
\end{bmatrix}
}. \\
[xy]=h,\quad [hx] = 2x,\quad [hy] = -2y
.\end{align*}

\end{example}

\begin{exercise}[?]

Prove that \({\mathfrak{sl}}_2({\mathbf{C}})\) is simple.

\end{exercise}

\begin{exercise}[?]

Show that for \(K\leq L\), the normalizer \(N_L(K)\) is the largest
subalgebra of \(L\) in which \(K\) is an ideal.

\end{exercise}

\begin{exercise}[?]

Show that
\({\mathfrak{h}}\subseteq {\mathfrak{g}}\coloneqq{\mathfrak{sl}}_n({\mathbf{C}})\)
is self-normalizing subalgebra of \({\mathfrak{g}}\).

\begin{quote}
Hint: use \([h, e_{ij}] = (h_i - h_j) e_{ij}\) where
\(h = \operatorname{diag}(h_1,\cdots, h_n)\). The standard basis is
\({\mathfrak{h}}= \left\langle{e_{11} - e_{22}, e_{22} - e_{33}, \cdots, e_{n-1, n-1} - e_{n,n} }\right\rangle\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

What is \(\dim {\mathfrak{sl}}_3({\mathbf{C}})\)? What is the basis for
\({\mathfrak{g}}\) and \({\mathfrak{h}}\)?

\end{exercise}

\hypertarget{humphreys-2.2}{%
\subsection{Humphreys 2.2}\label{humphreys-2.2}}

\begin{remark}

Notes:

\begin{itemize}
\tightlist
\item
  Let
  \(L, L'\in \mathsf{C} \coloneqq\mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}, \phi \in \mathsf{C}(L, L'), \ker \phi = \left\{{x\in L{~\mathrel{\Big\vert}~}\phi(x) = 0}\right\} {~\trianglelefteq~}L\).

  \begin{itemize}
  \tightlist
  \item
    Note that if \(x\in \ker \phi, y\in L\) then
    \(\phi([xy]) = [\phi(x)\phi(y)] = [0\phi(y)] = 0\).
  \end{itemize}
\item
  A \textbf{representation} of \(L\) is some
  \(\phi\in \mathsf{C}(L, {\mathfrak{gl}}(V))\) for some
  \(V\in { \mathsf{Vect}}_{/ {{ \mathbf{F} }}}\).
\item
  The usual 3 isomorphism theorems for groups hold for Lie algebras.
\item
  \({ \operatorname{ad}}: L\to {\mathfrak{gl}}(L)\) where
  \(x\mapsto { \operatorname{ad}}x\) is a representation.
\item
  \(\ker { \operatorname{ad}}= Z(L)\), so if \(L\) is simple then
  \(Z(L) = 0\) and \({ \operatorname{ad}}\) is injective. Thus any
  simple Lie algebra is linear.

  \begin{itemize}
  \tightlist
  \item
    Compare to: any finite dimensional Lie algebra is linear.
  \end{itemize}
\end{itemize}

\end{remark}

\hypertarget{monday-august-22}{%
\section{Monday, August 22}\label{monday-august-22}}

\hypertarget{humphreys-2.3-automorphisms}{%
\subsection{Humphreys 2.3:
Automorphisms}\label{humphreys-2.3-automorphisms}}

\begin{remark}

Let \(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\), then
\(\mathop{\mathrm{Aut}}(L)\) is the group of isomorphisms
\(L { \, \xrightarrow{\sim}\, }L\). Some important examples: if \(L\) is
linear and \(g\in \operatorname{GL}(V)\), if \(gLg^{-1}= L\) then
\(x\mapsto gxg^{-1}\) is an automorphism. This holds for example if
\(L = {\mathfrak{gl}}_n({ \mathbf{F} })\) or
\({\mathfrak{sl}}_n({ \mathbf{F} })\). Assume
\(\operatorname{ch}{ \mathbf{F} }= 0\) and let \(x\in L\) with
\({ \operatorname{ad}}x\) nilpotent, say
\(( { \operatorname{ad}}x)^k=0\). Then the power series expansion
\(e^{ { \operatorname{ad}}x} = \sum_{n\geq 0} ( { \operatorname{ad}}x)^n\)
is a polynomial.

\end{remark}

\begin{claim}

\(\exp^{ { \operatorname{ad}}x}\in \mathop{\mathrm{Aut}}(L)\) is an
automorphism. More generally, \(e^\delta\in \mathop{\mathrm{Aut}}(L)\)
for \(\delta\) any nilpotent derivation.

\end{claim}

\begin{lemma}[Generalized Leibniz rule]

\begin{align*}
\delta^n(xy) = \sum_{i=0}^n {n\choose i} \delta^{n-i}(x) \delta^{i}(y)
.\end{align*}

\end{lemma}

\begin{remark}

One can prove this by induction. Then check that
\(\exp(\delta(x))\exp(\delta(y)) = \exp(\delta(xy))\) and writing
\(\exp(\delta) = 1+\eta\) there is an inverse
\(1-\eta +\eta^2 +\cdots \pm \eta^{k-1}\). Automorphisms which are of
the form \(\exp(\delta)\) for \(\delta\) nilpotent derivation are called
\textbf{inner automorphisms}, and all others are \textbf{outer
automorphisms}.

\end{remark}

\hypertarget{solvable-and-nilpotent-lie-algebras-wednesday-august-24}{%
\section{Solvable and Nilpotent Lie Algebras (Wednesday, August
24)}\label{solvable-and-nilpotent-lie-algebras-wednesday-august-24}}

\hypertarget{humphreys-3.1}{%
\subsection{Humphreys 3.1}\label{humphreys-3.1}}

\begin{definition}[Derived series]

Recall that if \(L \subseteq {\mathfrak{g}}\) is any subset, the
\textbf{derived algebra} \([LL]\) is the span of \([xy]\) for
\(x,y\in L\). This is the analog of needing to take products of
commutators to generate a commutator subgroup for groups. Define the
\textbf{derived series} of \(L\) as
\begin{align*}
L^{(0)} = [LL], \quad L^{(1)} = [L^{(0)}, L^{(0)}], \cdots \quad L^{(i+1)} = [L^{(i)} L^{(i)}]
.\end{align*}

\end{definition}

\begin{proposition}[?]

These are all ideals.

\end{proposition}

\begin{proof}[?]

By induction on \(i\) -- it STS that \([x[ab]] \in L^{(i)}\) for
\(a,b \in L^{(i-1)}\) and \(x\in L\). Use the Jacobi identity and the
induction hypothesis that \(L^{(i-1)} {~\trianglelefteq~}L\):
\begin{align*}
[x,[ab]] = [[xa]b] + [a[xb]] \in L^{(i-1)} + L^{(i-1)} \subseteq L^{(i)}
.\end{align*}

\end{proof}

\begin{definition}[Solvable]

If \(L^{(n)} = 0\) for some \(n\geq 01\) then \(L\) is called
\textbf{solvable}.

\end{definition}

\begin{remark}

Note that

\begin{itemize}
\tightlist
\item
  \(L\) abelian implies solvable, since \(L^{(1)} = 0\).
\item
  \(L\) simple implies non-solvable, since this forces \(L^{(1)} = L\).
\end{itemize}

\end{remark}

\begin{exercise}[?]

Let \({\mathfrak{b}}\coloneqq{\mathfrak{b}}_n({ \mathbf{F} })\) be upper
triangular matrices, show that \({\mathfrak{b}}\) is solvable.

\begin{quote}
Use that \([{\mathfrak{b}}{\mathfrak{b}}] = {\mathfrak{n}}\) is strictly
upper triangular since diagonals cancel. More generally, bracketing
matrices with \(n\) diagonals of zeros yields matrices with about
\(2^n\) diagonals of zeros.
\end{quote}

\end{exercise}

\begin{proposition}[?]

Let \(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\), then

\begin{itemize}
\tightlist
\item
  \(L\) solvable implies solvability of all subalgebras and homomorphic
  images of \(L\).
\item
  If \(I{~\trianglelefteq~}L\) and \(L/I\) are solvable, then \(L\) is
  solvable.
\item
  If \(I,J{~\trianglelefteq~}L\) are solvable then \(I+J\) is
  solvable.\footnote{Use the third isomorphism theorem.}
\end{itemize}

\end{proposition}

\begin{exercise}[?]

Prove these.

\end{exercise}

\begin{definition}[Radical and semisimple]

Every
\(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}^{{\mathrm{fd}}}\)
has a unique maximal solvable ideal, the sum of all solvable ideals,
called the \textbf{radical} of \(L\), denote \(\sqrt{(}L)\). \(L\) is
\textbf{semisimple} if \(\sqrt{(}L) = 0\).

\end{definition}

\begin{exercise}[?]

Prove that any simple algebra is semisimple, and in general
\(L/\sqrt{(}L)\) is semisimple (if nonzero).

\end{exercise}

\begin{proposition}[?]

Assume \({\mathfrak{sl}}_n({\mathbf{C}})\) is simple, then
\(R\coloneqq\sqrt{(}{\mathfrak{gl}}_n({\mathbf{C}})) = Z({\mathfrak{g}}) \supseteq{\mathbf{C}}\operatorname{id}_n\)
for \({\mathfrak{g}}\coloneqq{\mathfrak{gl}}_n({\mathbf{C}})\).

\end{proposition}

\begin{proof}[?]

\(\supseteq\): Centers are always solvable ideals, since it's abelian
and brackets are ideals, and the radical is sum of all solvable ideals.

\(\subseteq\): Suppose \(Z\subsetneq R\) is proper, then there is a
non-scalar matrix \(x\in R\). Write \(x = aI_n + y\) for
\(a = {\mathrm{tr}}(x)/n\) and
\(0\neq y\in {\mathfrak{sl}}_n({\mathbf{C}})\) is traceless. Consider
\(I = \left\langle{x}\right\rangle {~\trianglelefteq~}{\mathfrak{gl}}_n({\mathbf{C}})\),
i.e.~the span of all brackets \([zx]\) for \(z\in {\mathfrak{g}}\) and
their iterated brackets containing \(x\), e.g.~\([z_1[z_2x]]\). Note
that \([zx]=[zy]\) since \(aI_n\) is central. Since
\({\mathfrak{sl}}_n({\mathbf{C}})\) is simple, so
\(\left\langle{y}\right\rangle_{{\mathfrak{sl}}_n({\mathbf{C}})} = {\mathfrak{sl}}_n({\mathbf{C}})\)
and thus \({\mathfrak{sl}}_n({\mathbf{C}}) \subseteq I\). This
containment must be proper, since
\(I \subseteq \sqrt{(}{\mathfrak{g}})\) and the latter is solvable, so
\(I\) must be solvable -- but \({\mathfrak{sl}}_n({\mathbf{C}})\) is not
solvable. We can thus choose \(x\in I\) such that \(x = aI_n + y\) with
\(a\neq 0\) and \(0\neq y \in {\mathfrak{sl}}_n({\mathbf{C}})\), so
\(x-y= aI \in I\) since \(y\in I\) because
\({\mathfrak{sl}}_n({\mathbf{C}}) \subseteq I\). Since \(a\neq 0\), we
must have \(I_n\in I\). Then \({\mathbf{C}}\cdot I_n \subseteq I\),
forcing \(I = {\mathfrak{g}}\) since every matrix in
\({\mathfrak{gl}}_n({\mathbf{C}})\) is a scalar multiple of the identity
plus a traceless matrix. This contradicts that \(I\) is solvable, since
\({\mathfrak{g}}^{(1)} \coloneqq[{\mathfrak{g}}{\mathfrak{g}}] = {\mathfrak{sl}}_n({\mathbf{C}})\).
But \({\mathfrak{g}}^{(1)} = {\mathfrak{sl}}_n({\mathbf{C}})\), so the
derived series never terminates. \(\contradiction\)

\end{proof}

\hypertarget{humphreys-3.2}{%
\subsection{Humphreys 3.2}\label{humphreys-3.2}}

\begin{definition}[Lower central series and nilpotent algebras]

The \textbf{descending/lower central series} of \(L\) is defined as
\begin{align*}
L^0 = L, \quad L^1 = [LL], \quad \cdots L^i = [L, L^{i-1}]
.\end{align*}
\(L\) is \textbf{nilpotent} if \(L^n=0\) for some \(n\).

\end{definition}

\begin{exercise}[?]

Check that \(L^i {~\trianglelefteq~}L\).

\end{exercise}

\begin{exercise}[?]

Show that \(L\) nilpotent is equivalent to there existing a finite \(n\)
such that for any set of elements \(\left\{{x_i}\right\}_{i=1}^n\),
\begin{align*}
( { \operatorname{ad}}_{x_1} \circ  { \operatorname{ad}}_{x_2} \circ \cdots \circ  { \operatorname{ad}}_{x_n})(y) = 0 \qquad \forall y\in L
.\end{align*}

\end{exercise}

\hypertarget{friday-august-26}{%
\section{Friday, August 26}\label{friday-august-26}}

\hypertarget{engels-theorem}{%
\subsection{3.2: Engel's theorem}\label{engels-theorem}}

\begin{proposition}[Nilpotent implies solvable]

Recall \(L\) is nilpotent if \(L^{n} = 0\) for some \(n\geq 0\) (the
descending central series) where \(L^{i+1} = [LL^{i}]\). Equivalently,
\(\prod_{i\leq n} { \operatorname{ad}}_{x_i} =0\) for any
\(\left\{{x_i}\right\}_{i\leq n} \subseteq L\). Note that
\(L^{i} \supseteq L^{(i)}\) by induction on \(i\) -- these coincide for
\(i=0,1\), and one can check
\begin{align*}
L^{(i+1)} = [L^{(i)} L^{(i)}] \subseteq [L L^{i}] = L^{i+1}
.\end{align*}

\end{proposition}

\begin{example}[Solvable does not imply nilpotent]

\({\mathfrak{b}}_n\) is solvable but not nilpotent, since
\({\mathfrak{b}}_n^1 = {\mathfrak{n}}_n\) but
\({\mathfrak{b}}_n^2 = {\mathfrak{n}}_n\) and the series never
terminates.

\end{example}

\begin{example}[?]

\({\mathfrak{n}}_n\) is nilpotent, since the number of diagonals with
zeros adds when taking brackets \([LL^{i}]\).

\end{example}

\begin{warnings}

\({\mathfrak{h}}\) is also nilpotent, since any abelian algebra is
nilpotent.

\end{warnings}

\begin{proposition}[?]

Let \(L\in\mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\), then

\begin{itemize}
\tightlist
\item
  If \(L\) is nilpotent then any subalgebra or homomorphic image of
  \(L\) is nilpotent.
\item
  If \(L/Z(L)\) is nilpotent then \(L\) is nilpotent.\footnote{Lift a
    series for the quotient, which is eventually in \(Z(L)\) since it
    was zero in the quotient, and then bracketing with \(Z(L)\)
    terminates.}
\item
  If \(L\neq 0\) is nilpotent, then \(Z(L) \neq 0\).\footnote{If
    \(L^n={n-1} \supseteq L^n=0\) then \([LL^{n-1}] = 0\) and thus
    \(L^{n-1} \subseteq Z(G)\).}
\end{itemize}

\end{proposition}

\begin{exercise}[?]

Show that if \(L/I\) and \(I{~\trianglelefteq~}L\) are nilpotent, then
\(L\) need not be nilpotent.

\end{exercise}

\begin{remark}

Distinguish \({ \operatorname{End} }(L)\) whose algebra structure is
given by associative multiplication and \({\mathfrak{gl}}(L)\) with the
bracket multiplication.

\end{remark}

\begin{definition}[ad-nilpotent]

An element \(x\in L\) is \textbf{ad-nilpotent} if
\({ \operatorname{ad}}_x \in { \operatorname{End} }(L)\) is a nilpotent
endomorphism.

\end{definition}

\begin{remark}

If \(L\) is nilpotent then \(x\in L\) is ad-nilpotent by taking
\(x_i = x\) for all \(i\). It turns out that the converse is true:

\end{remark}

\begin{theorem}[Engel's theorem]

If all \(x\in L\) are ad-nilpotent, then \(L\) is nilpotent.

\end{theorem}

\begin{proof}[?]

To be covered in an upcoming section.

\end{proof}

\begin{lemma}[?]

Let \(x\in {\mathfrak{gl}}(V)\) be a nilpotent linear transformation for
\(V\) finite-dimensional. Then
\begin{align*}
 { \operatorname{ad}}_x: {\mathfrak{gl}}(V)&\to {\mathfrak{gl}}(V) \\
y &\mapsto x\circ y - y\circ x
\end{align*}
is a nilpotent operator.

\end{lemma}

\begin{proof}[?]

Let \(\lambda_x, \rho_x \in { \operatorname{End} }({\mathfrak{gl}}(V))\)
be left and right multiplication by \(x\), which are commuting nilpotent
operators. The binomial theorem shows that if \(D_1, D_2\) are any two
commuting nilpotent endomorphisms of a vector space, then \(D_1\pm D_2\)
is again nilpotent. But then one can write
\({ \operatorname{ad}}_x = \lambda_x - \rho_x\).

\end{proof}

\begin{remark}

If \(x\in {\mathfrak{gl}}(V)\) is nilpotent then so is
\({ \operatorname{ad}}_x\). Conversely, if all
\({ \operatorname{ad}}_x\) for \(x\in L \leq {\mathfrak{gl}}(V)\) are
nilpotent operators then \(L\) is nilpotent by Engel's theorem.

\end{remark}

\begin{warnings}

The converse of the above lemma is not necessarily true: \(x\) being
ad-nilpotent does not imply that \(x\) is nilpotent. As a
counterexample, take \(x=I_n\in {\mathfrak{gl}}_n({\mathbf{C}})\), then
\({ \operatorname{ad}}_x = 0\) but \(x^k=x\) for any \(k\geq 1\).

\end{warnings}

\hypertarget{proof-of-engels-theorem}{%
\subsection{3.3: Proof of Engel's
theorem}\label{proof-of-engels-theorem}}

\begin{remark}

The following is related to the classical linear algebra theorem that
commuting operators admit a simultaneous eigenvector:

\end{remark}

\begin{theorem}[?]

Let \(L\) be a Lie subalgebra of \({\mathfrak{gl}}(V)\) for \(V\)
finite-dimensional. If \(L\) consists of nilpotent endomorphisms, then
there exists a nonzero \(v\in V\) such that \(Lv=0\).

\end{theorem}

\begin{proof}[?]

Proceed by induction on \(n = \dim L\) (assuming it holds for \emph{all}
vector spaces), where the \(n=1\) case is clear -- the characteristic
polynomial of such an operator is \(f(t) = t^n\), which has roots
\(t=0\) and every field contains zero. Once one has an eigenvalue, there
is at least one eigenvector.

For \(n > 1\), suppose \(K \leq L\) is a proper Lie subalgebra. By
hypothesis, \(K\) consists of nilpotent elements in
\({ \operatorname{End} }(V)\), so apply the previous lemma to see that
\({ \operatorname{ad}}(K) \subseteq { \operatorname{End} }(L)\) acts by
nilpotent endomorphisms of \(L\) since they are restrictions to \(L\) of
nilpotent endomorphisms of \({\mathfrak{gl}}(V)\). Since
\([KK] \subseteq K\), we can view
\({ \operatorname{ad}}(K) \subseteq { \operatorname{End} }(L/K)\) where
\(L/K\) is a vector space.

By the IH with \(V = L/K\), where \({ \operatorname{End} }(L/K)\) has
smaller dimension, one can find a nonzero \(x+K \in L/K\) such that
\({ \operatorname{ad}}(K)(x+K)=0\). Hence one can find an
\(x\in L\setminus K\) such that for all \(y\in K\) one has
\([yx] \in K\), so \(x\in N_L(K)\setminus K\). Thus
\(K \subsetneq N_L(K)\) is a proper containment.

\begin{quote}
To be continued.
\end{quote}

\end{proof}

\hypertarget{monday-august-29}{%
\section{Monday, August 29}\label{monday-august-29}}

\hypertarget{continuation-of-proof-and-corollaries}{%
\subsection{Continuation of proof and
corollaries}\label{continuation-of-proof-and-corollaries}}

\begin{remark}

Recall: we were proving that if \(L \leq {\mathfrak{gl}}(V)\) with \(V\)
finite dimensional and \(L\) consists of nilpotent endomorphisms, then
there exists a common eigenvector \(v\), so \(Lv = 0\).

\end{remark}

\begin{proof}[continued]

We're inducting on \(\dim L\) (over all \(V\)). Assuming \(\dim V > 1\),
we showed that proper subalgebras are strictly contained in their
normalizers:
\begin{align*}
K \lneq L \implies K \subsetneq N_L(K)
.\end{align*}
Let \(K\) be a maximal proper subalgebra of \(L\), then \(N_L(K) = L\)
by maximality and thus \(K\) is a proper ideal of \(L\). Then \(L/K\) is
a Lie algebra of some dimension, which must be 1 -- otherwise the
preimage in \(L\) under \(L\twoheadrightarrow L/K\) would be a
subalgebra of \(L\) properly between \(K\) and \(L\). Thus \(K\) is a
codimension 1 ideal in \(L\).Minimal model program Choosing any
\(z\in L\setminus K\) yields a decomposition
\(L = K \bigoplus { \mathbf{F} }z\) as vector spaces. Let
\(W \coloneqq\left\{{v\in V{~\mathrel{\Big\vert}~}Kv=0}\right\}\), then
\(W\neq 0\) by the IH.

\begin{claim}

\(W\) is an \(L{\hbox{-}}\)stable subspace.

To see this, let \(x\in L, y\in K, w\in W\). A useful trick:
\begin{align*}
y.(x.w) = w.(y.w) - [xy].w = 0
,\end{align*}
since the first term is zero and \([xy]\in K {~\trianglelefteq~}L\).

\end{claim}

Since \(z\curvearrowright W\) nilpotently, choose an eigenvector \(v\)
for \(z\) in \(W\) for the eigenvalue zero. Then \(z.v=0\), so \(Lv=0\).

\end{proof}

\begin{corollary}[Engel's theorem]

If all elements of a Lie algebra \(L\) are ad-nilpotent, then \(L\) is
nilpotent as an algebra.

\end{corollary}

\begin{proof}[?]

Induct on \(\dim L\). Note that
\({ \operatorname{ad}}(L) \leq {\mathfrak{gl}}(V)\) consists of
nilpotent endomorphisms. Use the theorem to pick \(x\in L\) such that
\({ \operatorname{ad}}(L).x = 0\), i.e.~\([L, x] = 0\),
i.e.~\(x\in Z(L)\) and thus \(Z(L)\) is nonzero. Now
\(\dim L/Z(L) < \dim L\), and a fortiori its elements are still
ad-nilpotent so \(L/Z(L)\) is nilpotent. By proposition 3.2b, \(L\) is
nilpotent.\footnote{Note that for arbitrary SESs, the 2-out-of-3
  property does not hold for nilpotency, but for the special cases of a
  quotient by the center it does.}

\end{proof}

\begin{corollary}[?]

Let \(o\neq L \leq {\mathfrak{gl}}(V)\) with \(\dim V < \infty\) be a
Lie algebra of nilpotent endomorphisms (as in the theorem).\footnote{Note
  that the assumption is not that \(L\) is a nilpotent algebra, but
  rather the stronger assumption on endomorphisms.} Then \(V\) has a
basis in which the matrices of \(L\) are all strictly upper triangular.

\end{corollary}

\begin{proof}[?]

Induct on \(\dim V\). Use the theorem to pick a nonzero \(v_1\) with
\(Lv_1=0\). Consider \(W\coloneqq V/{ \mathbf{F} }v_1\), and view
\(L \subseteq { \operatorname{End} }(V)\) as a subspace of
\({ \operatorname{End} }(W)\) -- these are still nilpotent
endomorphisms. By the IH, \(W\) has a basis
\(\left\{{\overline{v}_i}\right\}_{2\leq i \leq n}\) with respect to the
matrices in \(L\) (viewed as a subspace of
\({ \operatorname{End} }(W)\)) are strictly upper triangular. Let
\(\left\{{v_i}\right\} \subseteq V\) be their preimages in \(L\); this
basis has the desired properties. This results in a matrix of the
following form:

\includegraphics{figures/2022-08-29_09-38-58.png}

\end{proof}

\hypertarget{chapter-4-theorems-of-lie-and-cartan}{%
\subsection{Chapter 4: Theorems of Lie and
Cartan}\label{chapter-4-theorems-of-lie-and-cartan}}

\begin{remark}

From now on, assume \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\) is
algebraically closed and \(\operatorname{ch}(k) = 0\).

\end{remark}

\hypertarget{lies-theorem}{%
\subsection{4.1: Lie's Theorem}\label{lies-theorem}}

\begin{theorem}[Lie's Theorem]

Let \(L\neq 0\) be a solvable Lie subalgebra of \({\mathfrak{gl}}(V)\)
with \(\dim V < \infty\). Then \(V\) contains a common eigenvector for
all of \(L\).

\end{theorem}

\begin{proof}[?]

Induct on \(\dim L\). If \(\dim L = 1\), then \(L\) is spanned by 1
linearly operator \(x\) and over an algebraically closed field, \(x\)
has at least one eigenvector. For \(\dim L > 1\), take the following
strategy:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Identify \(K{~\trianglelefteq~}L\) proper of codimension 1,
\item
  By IH, find a common eigenvector for \(K\),
\item
  Verify that \(L\) stabilizes ``the'' subspace of all such common
  eigenvectors (much harder than before!)
\item
  In this subspace, find an eigenvector for some \(z\in L\setminus K\)
  with \(L = K \oplus { \mathbf{F} }z\).
\end{enumerate}

Off we go!

\textbf{Step 1}: Since \(L\) is solvable, we have \([LL]\) properly
contained in \(L\). In \(L/[LL]\) choose any codimension 1 subspace --
it is an ideal, which lifts to a codimension 1 ideal \(K \subset L\).

\textbf{Step 2}: Since subalgebras of solvable algebras are again
solvable, \(K\) is solvable. By the IH, pick a common nonzero
eigenvector \(v\) for \(K\). There exists a linear map
\(\lambda: K\to { \mathbf{F} }\) such that \(x.v = \lambda(x) v\) for
all \(x\in K\). Let
\(W \coloneqq\left\{{v\in V {~\mathrel{\Big\vert}~}y.v = \lambda(y) v\,\,\forall y\in K}\right\}\),
which is nonzero.

\textbf{Step 3}: Note \(L.W \subseteq W\). Let
\(w\in W, x\in L, y\in K\); we WTS \(y.(x.w) = \lambda(y)x.w\). Write
\begin{align*}
y.(x.w)
&= x.(y.w) - [xy].w \\
&= \lambda(y)(x.w) - \lambda([xy])w
,\end{align*}
where the second line follows since \([xy]\in K\). We then need
\(\lambda([xy]) = 0\) for all \(x\in L\) and \(y\in K\). Since
\(\dim V < \infty\), choose \(n\) minimal such that
\(\left\{{w, x.w, x^2.w,\cdots, x^n.w}\right\}\) is linearly dependent.
Set
\(W_i \coloneqq\mathop{\mathrm{span}}_{ \mathbf{F} }\left\{{w, x.w, \cdots, x^i.w}\right\}\),
so
\(W_0 = 0, W_1 = \mathop{\mathrm{span}}_{ \mathbf{F} }\left\{{w}\right\}\),
and so on, noting that

\begin{itemize}
\tightlist
\item
  \(\dim W_n = n\),
\item
  \(W_{n+k} = W_n\) for all \(k\geq 0\),
\item
  \(x.W_n \subseteq W_n\).
\end{itemize}

\begin{claim}

For all \(y\in K\),

\begin{align*}
y.x^i.w = \lambda(y) x^i.w \operatorname{mod}W_i
.\end{align*}

\end{claim}

To be continued!

\end{proof}

\hypertarget{wednesday-august-31}{%
\section{Wednesday, August 31}\label{wednesday-august-31}}

\hypertarget{section-4.1-continuing-the-proof}{%
\subsection{Section 4.1, continuing the
proof}\label{section-4.1-continuing-the-proof}}

\begin{remark}

Recall \(\dim L, \dim V < \infty\), \({ \mathbf{F} }\) is algebraically
closed, and \(\operatorname{ch}{ \mathbf{F} }= 0\). For
\(L \leq {\mathfrak{gl}}(V)\) solvable, we want a common eigenvector
\(v\in V\) for \(L\). Steps for the proof:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Find a \(K{~\trianglelefteq~}L\) proper of codimension 1.
\item
  Set
  \(W = \left\{{v\in V {~\mathrel{\Big\vert}~}x.v = \lambda(x) v \,\forall x\in K}\right\}\neq 0\)
  for some linear \(\lambda: K\to { \mathbf{F} }\).
\item
  Show \(L.W \subseteq W\); we needed to show \(\lambda([LK] ) = 0\).
\end{enumerate}

\end{remark}

\begin{proof}[Continued]

\textbf{Step 3}: Fix \(x\in L, w\in W\) and \(n\) minimal such that
\(\left\{{x^i w}\right\}_{i\leq n}\) is linearly dependent. For
\(i\geq 0\) set
\(W_i = { \mathbf{F} }\left\langle{w, xw, \cdots, x^{i-1}w}\right\rangle\).
Then \(\dim W_n = n, W_n = W_{n+i}\) for \(i\geq 0\), and
\(xW_n \subseteq W_n\).

\begin{claim}

For all \(y\in K\),
\begin{align*}
yx^i .w = \lambda(y) x^i w \operatorname{mod}W_i
.\end{align*}

\end{claim}

\begin{proof}[?]

This is proved by induction on \(i\), where \(i=0\) follows from how
\(W\) is defined. For \(i\geq 1\), use the commuting trick:
\begin{align*}
yx^i . w 
&= yxx^{i-1}w \\
&= (xy - [xy]) x^{i-1} w \\
&= x(y x^{i-1} w) - [xy]x^{i-1}w \\
&\equiv \lambda(y) x^i w - \lambda([xy])x^{i-1} w \operatorname{mod}W_{i-1} \\
&\equiv \lambda(y) x^i w - \lambda([xy])x^{i-1} w \operatorname{mod}W_{i} \qquad \text{since } W_{i-1} \leq W_i \\
&\equiv \lambda(y) x^i w \operatorname{mod}W_i
.\end{align*}

\end{proof}

Given this claim, for \(i=n\) this says that the matrices of any
\(y\in K\) with respect to the basis
\(\left\{{x^iw}\right\}_{0\leq i \leq n-1}\) is upper triangular with
diagonal entries all equal to \(\lambda(y)\). Thus
\({ \left.{{\operatorname{tr}(y)}} \right|_{{W_m}} } = n \lambda(y)\),
and so \([xy]\curvearrowright W_n\) with trace \(n \lambda([xy])\). On
the other hand, \(x,y\) both act on \(W_n\) (e.g.~by the formula in the
claim for \(yx^i.w\)) and so
\begin{align*}
{ \left.{{[xy]}} \right|_{{W_n}} } = { \left.{{xy}} \right|_{{W_n}} } - { \left.{{yx}} \right|_{{W_n}} }
,\end{align*}
thus \({ \left.{{ \operatorname{tr}([xy])}} \right|_{{W_n}} } = 0\).
Since \({ \mathbf{F} }\) is characteristic zero, we have
\(n \lambda([xy]) = 0 \implies \lambda([xy]) = 0\).

\textbf{Step 4}: By step 1, \(L = K \oplus { \mathbf{F} }z\) for some
\(z\in L\setminus K\). Viewing \(z: W\to W\) and using
\({ \mathbf{F} }= \overline{{ \mathbf{F} }}\), \(z\) has an eigenvector
\(v\in W\). Since \(v\in W\), it is also a common eigenvector for \(K\)
and thus an eigenvector for \(L\) by additivity.

\end{proof}

\begin{corollary}[A: Lie's theorem]

Let \(L\leq {\mathfrak{gl}}(V)\) be a solvable subalgebra, then \(L\)
stabilizes some flag in \(V\). In particular, there exists a basis for
\(V\) with respect to which the matrices in \(L\) are all upper
triangular.

\end{corollary}

\begin{remark}

Recall that for \(V \in{ \mathsf{Vect}}_{/ {{ \mathbf{F} }}}\), a
complete flag is an element of
\begin{align*}
\operatorname{Fl}(V) \coloneqq\left\{{ 0 = V^0 \subsetneq V^1 \subsetneq \cdots \subsetneq V^n = V {~\mathrel{\Big\vert}~}\dim V^i = i}\right\}
.\end{align*}
A subalgebra \(L\) \textbf{stabilizes} a flag if \(LV^i \subseteq V^i\)
for all \(i\), which implies there is a compatible basis (got by
extending one vector at a time from a basis for \(V^1\)) for which \(L\)
acts by upper triangular matrices.

\end{remark}

\begin{proof}[?]

Use the theorem and induct on \(n=\dim V\) as in Engel's theorem -- find
a common eigenvector for \(V^1\), since \(L\) stabilizes one gets an
action \(L\curvearrowright V^i/V^{i-1}\) which is smaller dimension.
Then just lift through the quotient.

\end{proof}

\begin{corollary}[B]

Let \(L\) be a solvable Lie algebra, then there exists a chain of ideals
\begin{align*}
0 = L_0 \subsetneq L_1 \subsetneq \cdots \subsetneq L_n = L
\end{align*}
such that \(\dim L_i = i\). \label{thm:corB}

\end{corollary}

\begin{proof}[?]

Consider \({ \operatorname{ad}}L \leq {\mathfrak{gl}}(L)\). Apply Lie's
theorem:
\(( { \operatorname{ad}}L)L_i \subseteq L_i \iff [LL_i] \subseteq L_i\),
making \(L_i{~\trianglelefteq~}L\) an ideal.

\end{proof}

\begin{corollary}[C]

Let \(L\) be solvable, then
\(x\in [LL]\implies { \operatorname{ad}}_L x\) is nilpotent. Hence
\([LL]\) is nilpotent by Lie's theorem.

\end{corollary}

\begin{proof}[?]

Find a flag of ideals by \autoref{thm:corB} and let
\(\left\{{x_1,\cdots, x_n}\right\}\) be a compatible basis. Then the
matrices
\(\left\{{ { \operatorname{ad}}_x{~\mathrel{\Big\vert}~}x\in L}\right\}\)
are all upper triangular. If \(x\in [LL]\), without loss of generality
\(x = [yz]\) for some \(y,z\in L\). Then
\begin{align*}
 { \operatorname{ad}}_x = [ { \operatorname{ad}}_y  { \operatorname{ad}}_z] =  { \operatorname{ad}}_y { \operatorname{ad}}_z -  { \operatorname{ad}}_z  { \operatorname{ad}}_y
\end{align*}
will be strictly upper triangular (since these are upper triangular and
the commutator cancels diagonals) and hence nilpotent.

\end{proof}

\hypertarget{section-4.3}{%
\subsection{Section 4.3}\label{section-4.3}}

\begin{remark}

We'll come back to 4.2 next time. For this section, assume
\({ \mathbf{F} }= \overline{{ \mathbf{F} }}\) and
\(\operatorname{ch}{ \mathbf{F} }= 0\). \textbf{Cartan's criterion} for
a semisimple \(L\) (i.e.~\(\mathop{\mathrm{Rad}}(L) = 0\)) involves the
\textbf{Killing form}, a certain nondegenerate bilinear form on \(L\).
Recall that if \(L\) is solvable then \([LL]\) is nilpotent, or
equivalently every \(x\in [LL]\) is ad-nilpotent.

\end{remark}

\begin{lemma}[Checking nilpotency with a trace condition (technical)]

Let \(A \subseteq B\) be subspaces of \({\mathfrak{gl}}(V)\) (really
\({ \operatorname{End} }(V)\) as a vector space) with \(V\)
finite-dimensional. Let
\begin{align*}
M\coloneqq\left\{{w\in {\mathfrak{gl}}(V) {~\mathrel{\Big\vert}~}[wB] \subseteq A}\right\}
\end{align*}
and suppose some \(w\in M\) satisfies \(\operatorname{tr}(wz) = 0\) for
all \(z\in M\). Then \(w\) is nilpotent.

\end{lemma}

\begin{proof}[?]

Later!

\end{proof}

\begin{definition}[Bilinear form terminology]

A \textbf{bilinear form} is a map
\begin{align*}
\beta({-}, {-}): L\times L\to { \mathbf{F} }
,\end{align*}
which is \textbf{symmetric} if \(\beta(x,y) = \beta(y,x)\) and
\textbf{associative} if \(\beta([xy], z) = \beta(x, [yz])\) for all
\(x,y,z\in L\). The \textbf{radical} of \(\beta\) is
\begin{align*}
\mathop{\mathrm{Rad}}(\beta) \coloneqq\left\{{w\in V{~\mathrel{\Big\vert}~}\beta(w, V) = 0}\right\}
,\end{align*}
and \(\beta\) is \textbf{nondegenerate} if
\(\mathop{\mathrm{Rad}}(\beta) = 0\).

\end{definition}

\begin{example}[?]

For \(L = {\mathfrak{gl}}(V)\), take
\(\beta(x,y)\coloneqq\operatorname{tr}(xy)\). One can check this is
symmetric, bilinear, and associative -- associativity follows from the
following:
\begin{align*}
[xy]z &= xyz-yxz\\
x[yz] &= xyz - xzy
.\end{align*}
Then note that \(y(xz)\) and \((xz)y\) have the same trace, since
\(\operatorname{tr}(AB) = \operatorname{tr}(BA)\).

\end{example}

\begin{proposition}[?]

If \(\beta\) is associative, then
\(\mathop{\mathrm{Rad}}(\beta) {~\trianglelefteq~}L\).

\end{proposition}

\begin{proof}[?]

Let \(z\in \mathop{\mathrm{Rad}}(\beta)\) and \(x,y\in L\). To see if
\([zx]\in \mathop{\mathrm{Rad}}(\beta)\), check
\begin{align*}
\beta([zx], y) = \beta(z, [xy]) = 0
\end{align*}
since \(z\in \mathop{\mathrm{Rad}}(\beta)\). Thus
\([zx] \in \mathop{\mathrm{Rad}}(\beta)\).

\end{proof}

\hypertarget{friday-september-02}{%
\section{Friday, September 02}\label{friday-september-02}}

\hypertarget{jordan-chevalley-decomposition}{%
\subsection{4.2: Jordan-Chevalley
decomposition}\label{jordan-chevalley-decomposition}}

\begin{remark}

Let \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\) of arbitrary
characteristic and
\(V\in{ \mathsf{Vect}}_{/ {{ \mathbf{F} }}}^{\mathrm{fd}}\) with
\(x\in { \operatorname{End} }_{ \mathbf{F} }(V)\). The JCF of \(x\) is
of the form \(D+N\) where \(D\) is diagonal and \(N\) is nilpotent where
\(D, N\) commute. Recall \(x\) is semisimple (diagonalizable) iff the
minimal polynomial of \(x\) has distinct roots.

\end{remark}

\begin{proposition}[?]

If \(x\in { \operatorname{End} }(V)\),

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  There is a decomposition \(x = x_s + x_n\) where \(x_s\) is semisimple
  and \(x_n\) is nilpotent. This is unique subject to the condition that
  \(x_s, x_n\) commute.
\item
  There are polynomials \(p(T), q(T)\) without constant terms with
  \(x_s = p(x), x_n = q(x)\). In particular, \(x_s, x_n\) commute with
  any endomorphism which commutes with \(x\).
\end{enumerate}

\end{proposition}

\begin{lemma}[?]

Let \(x\in {\mathfrak{gl}}(V)\) with Jordan decomposition
\(x = x_s + x_n\). Then
\({ \operatorname{ad}}_x = { \operatorname{ad}}_{x_s} + { \operatorname{ad}}_{x_n}\)
is the Jordan decomposition of \({ \operatorname{ad}}_x\) in
\({ \operatorname{End} }({ \operatorname{End} }(V))\).

\end{lemma}

\begin{proof}[?]

If \(x\in {\mathfrak{gl}}(V)\) is semisimple then so is
\({ \operatorname{ad}}_x\) since the eigenvalues for
\({ \operatorname{ad}}_x\) are differences of eigenvalues of \(x\). I.e.
if \(\left\{{v_1,\cdots, v_n}\right\}\) is an eigenbasis for \(V\) and
\(x.v_i = a_i v_i\) in this bases, we have
\([x e_{ij}] = (a_i - a_j) = e_{ij}\), so \(\left\{{e_{ij}}\right\}\) is
an eigenbasis for \({ \operatorname{ad}}_x\). If \(x\) is nilpotent then
\({ \operatorname{ad}}_x\) is nilpotent, since
\({ \operatorname{ad}}_x(y) = \lambda_x(y) - \rho_x(y)\) where
\(\lambda, \rho\) are left/right multiplication, and sums of nilpotents
are nilpotent. One can check
\([ { \operatorname{ad}}_{x_s} { \operatorname{ad}}_{x_n}] = { \operatorname{ad}}_{[x_s x_n} = 0\)
since they commute.

\end{proof}

\begin{remark}

One can show that if \(L\) is semisimple then
\({ \operatorname{ad}}(L) = \mathop{\mathrm{Der}}(L)\), which is used to
show that if \(L\) is an arbitrary Lie algebra then one has

\begin{itemize}
\tightlist
\item
  \(x = x_s + x_n\),
\item
  \([x_s x_n] = 0\),
\item
  \({ \operatorname{ad}}_{x_s}\) is semisimple and
  \({ \operatorname{ad}}_{x_n}\) is nilpotent.
\end{itemize}

This gives a notion of semisimplicity and nilpotency for Lie algebras
not of the form \({\mathfrak{gl}}(V)\).

\end{remark}

\begin{lemma}[?]

Let \(U\in \mathsf{Alg}_{/ {{ \mathbf{F} }}}^{\mathrm{fd}}\), then
\(\mathop{\mathrm{Der}}(U)\) is closed under taking semisimple and
nilpotent parts.

\end{lemma}

\begin{proof}[?]

Let \(\delta\in \mathop{\mathrm{Der}}(U)\) and write
\(\delta = \sigma + v\) be the Jordan decomposition of \(\delta\) in
\({ \operatorname{End} }(U)\). It STS \(\sigma\) is a derivation, so for
\(a\in { \mathbf{F} }\) define
\begin{align*}
U_a \coloneqq\left\{{x\in U {~\mathrel{\Big\vert}~}(\delta - a)^k x = 0 \,\,\text{for some } k}\right\}
.\end{align*}
Note \(U = \bigoplus _{a\in \Lambda} U_a\) where \(\Lambda\) is the set
of eigenvalues of \(\delta\), which are also the eigenvalues of
\(\sigma\) -- this is because \(\sigma, v\) are commuting operators, so
eigenvalues of \(\delta\) are sums of eigenvalues of \(s\) and \(v\).

\begin{claim}

For any \(a,b\in { \mathbf{F} }\), \(U_a U_b \subseteq U_{a+b}\).

\end{claim}

Assuming this, it STS \(\sigma(xy) = \sigma(x)y + x \sigma(y)\) when
\(x\in U_a, y\in U_b\) where \(a,b\) are eigenvalues. Using that
eigenvalues of \(\delta\) are also eigenvalues of \(\sigma\), since
\(xy\in U_{a+b}\) by the claim, \(\sigma(xy) = (a+b)xy\) and thus
\begin{align*}
\sigma(x)y + x \sigma(y) = axy + xby = (a+b)xy
.\end{align*}
So \(\sigma\in \mathop{\mathrm{Der}}(U)\).

\end{proof}

\begin{proof}[of claim]

A sub-claim:
\begin{align*}
(\delta - (a+b) 1) (xy) = \sum_{0\leq i\leq n} {n\choose i} (\delta - aI)^{n-i}x (\delta- b 1)^i y
.\end{align*}

\end{proof}

\hypertarget{wednesday-september-07}{%
\section{Wednesday, September 07}\label{wednesday-september-07}}

\hypertarget{cartans-criterion-for-semisimplicity}{%
\subsection{4.3: Cartan's criterion for
semisimplicity}\label{cartans-criterion-for-semisimplicity}}

\begin{remark}

For the rest of the course, \(V\) is a vector space of finite dimension.
Goal: get a criterion for semisimplicity.

\end{remark}

\begin{theorem}[Cartan's criterion for linear Lie algebras]

Let \(L\leq {\mathfrak{gl}}(V)\) be a linear Lie algebra and suppose
\(\operatorname{tr}(xz)=0\) for all \(x\in [LL]\) and \(z\in L\). Then
\(L\) is solvable.

\end{theorem}

\begin{lemma}[?]

Let \(A \subseteq B\) be subspaces of
\({ \operatorname{End} }(V) = {\mathfrak{gl}}(V)\) and define
\begin{align*}
M = \left\{{w\in {\mathfrak{gl}}(V) {~\mathrel{\Big\vert}~}[w, B] \subseteq A}\right\}
.\end{align*}
Suppose that \(w\in M\) satisfies \(\operatorname{tr}(wz) = 0\) for all
\(z\in M\). Then \(w\) is nilpotent.

\end{lemma}

\begin{proof}[of Cartan, assuming the lemma]

To show \(L\) is solvable, it STS that \([LL]\) is nilpotent since the
ideals used to check nilpotency are bigger than those to check
solvability. By Engel's theorem, it STS to show each \(w\in [LL]\) is
ad-nilpotent. Since \(L \leq {\mathfrak{gl}}(V)\), it STS to show each
\(w\in [LL]\) is a nilpotent endomorphism. As in the setup of the lemma,
set \(B = L, A = [LL]\), then
\begin{align*}
M \coloneqq\left\{{z\in {\mathfrak{gl}}(V) {~\mathrel{\Big\vert}~}[zL] \subseteq [LL] }\right\} \supseteq L \supseteq[LL]
.\end{align*}
Let \(w\in [LL] \subseteq M\), then note that
\(\operatorname{tr}(wz) = 0\) for all \(z\in L\), but we need to know
this for all \(z\in M\). Letting \(z\in M\) be arbitrary; by linearity
of the trace it STS \(\operatorname{tr}(wz) = 0\) on generators
\(w = [xy]\) on \([LL]\) for \(x,y\in L\). We thus WTS
\(\operatorname{tr}([xy]z) = 0\):
\begin{align*}
\operatorname{tr}([xy]z) &= \operatorname{tr}(x [yz] ) \\
&=\operatorname{tr}([yz] x) \qquad \in \operatorname{tr}(LMx) \subseteq \operatorname{tr}([LL]L) \\
&= 0
\end{align*}
by assumption. By the lemma, \(w\) is nilpotent.

\end{proof}

\begin{corollary}[Cartan's criterion for general Lie algebras]

Let \(L\in \mathsf{Lie} \mathsf{Alg}\) with
\(\operatorname{tr}( { \operatorname{ad}}_x { \operatorname{ad}}_y) = 0\)
for all \(x \in [LL]\) and \(y\in L\). Then \(L\) is solvable.

\end{corollary}

\begin{proof}[of corollary]

Use \({ \operatorname{ad}}: L\to {\mathfrak{gl}}(V)\), a morphism of Lie
algebras. Its image is solvable by Cartan's criterion above, and
\(\ker { \operatorname{ad}}= Z(L)\) which is abelian and hence a
solvable ideal.\footnote{The derived series terminates immediately for
  an abelian algebra.} Therefore \(L\) is solvable.

\end{proof}

\begin{proof}[of lemma]

Let \(w = s + n\) be the Jordan-Chevalley decomposition of \(w\). Choose
a basis for \(V\) such that this is the JCF of \(w\),
i.e.~\(s = \operatorname{diag}(a_1,\cdots, a_n)\) and \(n\) is strictly
upper triangular. Idea: show \(s=0\) by showing
\(A\coloneqq{\mathbf{Q}}\left\langle{a_1,\cdots, a_n}\right\rangle = 0\)
by showing \(A {}^{ \vee }= 0\), i.e.~any
\({\mathbf{Q}}{\hbox{-}}\)linear functional \(f: A\to {\mathbf{Q}}\) is
zero. If \(\sum a_i f(a_i) = 0\) then
\begin{align*}
0 = f(\sum a_i f(a_i)) = \sum f(a_i)^2 \implies f(a_i) = 0 \,\,\forall i
,\end{align*}
so we'll show this. Let
\(y = \operatorname{diag}( f(a_1), \cdots, f(a_n) )\), then
\({ \operatorname{ad}}_y\) is a polynomial (explicitly constructed using
Lagrange interpolation) in \({ \operatorname{ad}}_s\) without a constant
term. So do this for \({ \operatorname{ad}}_y\) and
\({ \operatorname{ad}}_s\) (see exercise). Since
\({ \operatorname{ad}}_s\) is a polynomial in \({ \operatorname{ad}}_w\)
with zero constant term, and since \({ \operatorname{ad}}_w: B\to A\),
we have \({ \operatorname{ad}}_s(B) \subseteq A\) and the same is thus
true for \({ \operatorname{ad}}_y\). So \(y\in M\) and \(w\in M\), and
applying the trace condition in the lemma with \(z\coloneqq y\) we get
\begin{align*}
0 = \operatorname{tr}(wy) = \sum a_i f(a_i)
,\end{align*}
noting that \(w\) is upper triangular and \(y\) is diagonal. So \(s=0\)
and \(w=n\) is nilpotent.

\end{proof}

\begin{exercise}[?]

Show \({ \operatorname{ad}}_y\) is a polynomial in
\({ \operatorname{ad}}_s\).

\end{exercise}

\begin{remark}

Recall that \(\mathop{\mathrm{Rad}}L\) is the unique maximal (not
necessarily proper) solvable ideal of \(L\). This exists, e.g.~because
sums of solvable ideals are solvable. Note that \(L\) is semisimple iff
\(\mathop{\mathrm{Rad}}L = 0\).

\end{remark}

\begin{definition}[Killing form]

Let \(L\in \mathsf{Lie} \mathsf{Alg}^{\mathrm{fd}}\) and define the
\textbf{Killing form}
\begin{align*}
\kappa: L\times L &\to { \mathbf{F} }\\
\kappa(x, y) &= \operatorname{tr}( { \operatorname{ad}}_x \circ  { \operatorname{ad}}_y)
.\end{align*}
This is an associative\footnote{Associative is \(f([xy]z) = f(x [yz])\),
  sometimes called \emph{invariant}.} bilinear form on \(L\).

\end{definition}

\begin{example}[?]

Let \(L = {\mathbf{C}}\left\langle{x, y}\right\rangle\) with
\([xy] = x\). In this ordered basis,
\begin{align*}
 { \operatorname{ad}}_x = {
\begin{bmatrix}
  {0} & {1} 
\\
  {0} & {0}
\end{bmatrix}
} \qquad  { \operatorname{ad}}_y = {
\begin{bmatrix}
  {-1} & {0} 
\\
  {0} & {0}
\end{bmatrix}
} 
,\end{align*}
and one can check \(\kappa(x,x) = \kappa(x, y) = \kappa(y, x) = 0\) and
\(\kappa(y,y) = 1\). Moreover
\(\mathop{\mathrm{Rad}}\kappa = {\mathbf{C}}\left\langle{x}\right\rangle\).

\begin{quote}
See the text for \(\kappa\) defined on \({\mathfrak{sl}}_2\).
\end{quote}

\end{example}

\begin{lemma}[?]

Let \(I {~\trianglelefteq~}L\). If \(\kappa\) is the Killing form of
\(L\) ad \(\kappa_I\) that of \(I\), then
\begin{align*}
\kappa_I = { \left.{{\kappa}} \right|_{{I\times I}} }
.\end{align*}

\end{lemma}

\begin{proof}[?]

Let \(x\in I\), then \({ \operatorname{ad}}_x(L) \subseteq I\) since
\(I\) is an ideal. Choosing a basis for \(I\) yields a matrix:

\includegraphics{figures/2022-09-07_09-56-24.png}

So if \(x,y\in I\), we have
\begin{align*}
\kappa(x,y) 
&= \operatorname{tr}( { \operatorname{ad}}_x \circ  { \operatorname{ad}}_y) \\
&= \operatorname{tr}( { \operatorname{ad}}_{I, x} \circ  { \operatorname{ad}}_{I, y}) \\
&= \kappa_I(x, y)
.\end{align*}

\end{proof}

\hypertarget{friday-september-09}{%
\section{Friday, September 09}\label{friday-september-09}}

\hypertarget{the-killing-form}{%
\subsection{5.1: The Killing form}\label{the-killing-form}}

\begin{remark}

For the rest of the course: \(k = { \overline{k} }\) and
\(\operatorname{ch}k = 0\). Theorem from last time: \(L\) is semisimple
iff its Killing form
\(\kappa(x, y) \coloneqq\operatorname{tr}( { \operatorname{ad}}_x { \operatorname{ad}}_y)\)
is nondegenerate.

\end{remark}

\begin{proof}[?]

Let \(S = \mathop{\mathrm{Rad}}(\kappa) {~\trianglelefteq~}L\), which is
easy to check using ``invariance'' (associativity) of the form. Given
\(s,s'\in S\), the restricted form
\(\kappa_S(x, y) = \operatorname{tr}( { \operatorname{ad}}_{S, s} { \operatorname{ad}}_{S, s'}) = \operatorname{tr}( { \operatorname{ad}}_{L, s} { \operatorname{ad}}_{L, s'})\),
which was proved in a previous lemma. But this is equal to
\(\kappa(s, s') = 0\). In particular, we can take \(s\in [SS]\), so by
(the corollary of) Cartan's criterion for solvable Lie algebras, \(S\)
is solvable as a Lie algebra and thus solvable as an ideal in \(L\).

\(\implies\): Since \(\mathop{\mathrm{Rad}}(L)\) is the sum of all
solvable ideals, we have \(S \subseteq \mathop{\mathrm{Rad}}(L)\), but
since \(L\) is semisimple \(\mathop{\mathrm{Rad}}(L) = 0\) and thus
\(S=0\).

\(\impliedby\): Assume \(S=0\). If \(I {~\trianglelefteq~}L\) is a
solvable ideal so \(I^{(n)} = 0\) for some \(n\geq 0\). If
\(I^{(n-1)} \neq 0\), it is a nonzero abelian ideal -- since we want to
show \(\mathop{\mathrm{Rad}}(L) = 0\), we don't want this to happen!
Thus it STS every abelian ideal is contained in \(S\).

So let \(I {~\trianglelefteq~}L\) be an abelian ideal, \(x\in I\),
\(y\in L\). Define an operator
\begin{align*}
A_{xy}^2 \coloneqq( { \operatorname{ad}}_x  { \operatorname{ad}}_y)^2: L \xrightarrow{ { \operatorname{ad}}_y} L \xrightarrow{ { \operatorname{ad}}_x} I \xrightarrow{ { \operatorname{ad}}_y} I \xrightarrow{ { \operatorname{ad}}_x} 0
,\end{align*}
which is zero since \([II] =0\). Thus \(A_{xy}\) is a nilpotent
endomorphism, which are always traceless, so
\(0 = \operatorname{tr}( { \operatorname{ad}}_x { \operatorname{ad}}_y) = \kappa(x, y)\)
for all \(y\in L\), and so \(x\in S\). Thus \(I \subseteq S\).

\end{proof}

\begin{warnings}

\(\mathop{\mathrm{Rad}}(\kappa) \subseteq \mathop{\mathrm{Rad}}(L)\)
always, but the reverse containment is not always true -- see exercise
5.4.

\end{warnings}

\hypertarget{simple-ideals-of-l}{%
\subsection{\texorpdfstring{5.2: Simple Ideals of
\(L\)}{5.2: Simple Ideals of L}}\label{simple-ideals-of-l}}

\begin{definition}[Direct sums of Lie algebras]

Let \(L_i\in \mathsf{Lie} \mathsf{Alg}_{/ {k}}\), then their
\textbf{direct sum} is the product \(L_1 \times L_2\) with bracket
\begin{align*}
[x_1 \oplus x_2, y_1 \oplus y_2] \coloneqq[x_1 y_1] \oplus [x_2 y_2]
.\end{align*}

\end{definition}

\begin{remark}

In particular, \([L_1, L_2] = 0\), and thus any ideal
\(I_1 {~\trianglelefteq~}L_1\) yields an ideal
\(I_1 \oplus 0 {~\trianglelefteq~}L_1 \oplus L_2\). Moreover, if
\(L = \bigoplus I_i\) is a vector space direct sum of ideals of \(L\),
this is automatically a Lie algebra direct sum since
\([I_i, I_j] = I_i \cap I_j = 0\) for all \(i\neq j\).

\end{remark}

\begin{warnings}

This is \emph{not} true for subalgebras! Also, in this theory, one
should be careful about whether direct sums are as vector spaces or (in
the stronger sense) as Lie algebras.

\end{warnings}

\begin{theorem}[?]

Let \(L\) be a finite-dimensional semisimple Lie algebra. Then there
exist ideals \(I_n\) of \(L\) such that \(L = \bigoplus I_i\) with each
\(I_j\) simple as a Lie algebra. Moreover every simple ideal if one of
the \(I_j\).

\end{theorem}

\begin{proof}[?]

Let \(I {~\trianglelefteq~}L\) and define
\begin{align*}
I^\perp \coloneqq\left\{{x\in L {~\mathrel{\Big\vert}~}\kappa(x, I) = 0 }\right\}
,\end{align*}
the orthogonal complement of \(I\) with respect to \(\kappa\). This is
an ideal by the associativity of \(\kappa\). Set
\(J\coloneqq I \cap I^\perp {~\trianglelefteq~}L\), then
\(\kappa([JJ], J) = 0\) and by Cartan's criterion \(J\) is a solvable
ideal and thus \(J = 0\), making \(L\) semisimple.

From the Endman-Wildon lemma in the appendix (posted on ELC, lemma
16.11), \(\dim L = \dim I + \dim I^\perp\) and \(L = I \oplus I^\perp\),
so now induct on \(\dim L\) to get the decomposition when \(L\) is not
simple. These are semisimple ideals since solvable ideals in the
\(I, I^\perp\) remain solvable in \(L\). Finally let
\(I{~\trianglelefteq~}L\) be simple, then \([I, L] \subseteq L\) is a
ideal (in both \(L\) and \(I\)), which is nonzero since \(Z(L) = 0\).
Since \(I\) is simple, this forces \([I, L] = I\). Writing
\(L = \bigoplus I_i\) as a sum of simple ideals, we have
\begin{align*}
I = [I, L] = [I, \bigoplus I_i] = \bigoplus [I, I_i] 
,\end{align*}
and by simplicity only one term can be nonzero, so \(I = [I, I_j]\) for
some \(j\). Since \(I_j\) is an ideal, \([I, I_j] \subseteq I_j\), and
by simplicity of \(I_j\) we have \(I = I_j\).

\end{proof}

\begin{corollary}[?]

Let \(L\) be semisimple, then \(L = [LL]\) and all ideals and
homomorphic images (but not subalgebras) are again semisimple. Moreover,
every ideal of \(L\) is a sum of simple ideals \(I_j\) of \(L\).

\end{corollary}

\begin{proof}[?]

Take the canonical decomposition \(L = \bigoplus I_i\) and check
\begin{align*}
[L, L] = [\bigoplus I_i, \bigoplus I_j] = \bigoplus [I_i, I_i] = \bigoplus I_i
,\end{align*}
where in the last step we've used that the \(I_i\) are (not?) abelian.
Let \(J {~\trianglelefteq~}L\) to write \(L = J \bigoplus J^\perp\),
both of which are semisimple as Lie algebras. In particular, if
\(\phi: L\to L'\), set \(J \coloneqq\ker \phi {~\trianglelefteq~}L\).
Then \(\operatorname{im}\phi = L/J \cong J^\perp\) as Lie algebras,
using the orthogonal decomposition, so \(\operatorname{im}\phi\) is
semisimple. Finally if \(J {~\trianglelefteq~}L\) then
\(L = J \oplus J^\perp\) with \(J\) semisimple, so by the previous
theorem \(J\) decomposes as \(J = \oplus K_i\) with \(K_i\) (simple)
ideals in \(J\) -- but these are (simple) ideals in \(L\) as well since
it's a direct sum. Thus the \(K_i\) are a subset of the \(I_j\), since
these are the only simple ideals of \(L\).

\end{proof}

\hypertarget{monday-september-12}{%
\section{Monday, September 12}\label{monday-september-12}}

\begin{remark}

Question from last time: does \(L\) always factor as
\(\mathop{\mathrm{Rad}}(L) \oplus L_{{\mathrm{ss}}}\) with
\(L_{{\mathrm{ss}}}\) semisimple? Not always, instead there is a
semidirect product decomposition
\(L = \mathop{\mathrm{Rad}}(L) \rtimes{\mathfrak{s}}\) where
\({\mathfrak{s}}\) is the \emph{Levi subalgebra}. Consider
\(L = {\mathfrak{gl}}_n\), then
\(\mathop{\mathrm{Rad}}(L) \neq {\mathfrak{h}}\) since
\([h, e_{ij}] = (h_i - h_j)e_{ij}\), so in fact this forces
\(\mathop{\mathrm{Rad}}(L) = {\mathbf{C}}I_n = Z(L)\) with complementary
subalgebra \({\mathfrak{s}}= {\mathfrak{sl}}_n\). Note that
\({\mathfrak{gl}}_n = {\mathbf{C}}I_n \oplus {\mathfrak{sl}}_n\) where
\({\mathfrak{sl}}_n = [L L]\) is a direct sum, and \({\mathfrak{gl}}_n\)
is \textbf{reductive}.

\end{remark}

\hypertarget{inner-derivations}{%
\subsection{5.3: Inner Derivations}\label{inner-derivations}}

\begin{theorem}[In semisimples, every derivation is inner]

If \(L\) is semisimple then
\({ \operatorname{ad}}(L) = \mathop{\mathrm{Der}}L\).

\end{theorem}

\begin{proof}[?]

We know \({ \operatorname{ad}}(L) \leq \mathop{\mathrm{Der}}L\) is a
subalgebra, and \(L\) semisimple implies
\(0 = Z(L) = \ker { \operatorname{ad}}\), so
\({ \operatorname{ad}}: L { \, \xrightarrow{\sim}\, } { \operatorname{ad}}(L)\)
is an isomorphism and \({ \operatorname{ad}}(L)\) is semisimple. The
Killing form of a semisimple Lie algebra is always nondegenerate, so
\(\kappa_{ { \operatorname{ad}}(L)}\) is nondegenerate, while
\(\kappa_{\mathop{\mathrm{Der}}L}\) may be degenerate. Recall that
\({ \operatorname{ad}}(L) {~\trianglelefteq~}\mathop{\mathrm{Der}}L\),
so
\([\mathop{\mathrm{Der}}L, { \operatorname{ad}}(L)] \subseteq { \operatorname{ad}}(L)\).
Define
\({ \operatorname{ad}}(L)^\perp {~\trianglelefteq~}\mathop{\mathrm{Der}}L\)
to be the orthogonal complement in \(\mathop{\mathrm{Der}}(L)\) with
respect to \(\kappa_{\mathop{\mathrm{Der}}L}\), which is an ideal by the
associative property.

Claim:
\({ \operatorname{ad}}(L) \cap { \operatorname{ad}}(L)^\perp = 0\), This
follows readily from the fact that \(\kappa_{ { \operatorname{ad}}(L)}\)
is nondegenerate and so
\(\mathop{\mathrm{Rad}}(\kappa_{ { \operatorname{ad}}(L)}) = 0\).

Note that \({ \operatorname{ad}}(L), { \operatorname{ad}}(L)^\perp\) are
both ideals, so
\([ { \operatorname{ad}}(L), { \operatorname{ad}}(L)^\perp] \subseteq { \operatorname{ad}}(L) \cap { \operatorname{ad}}(L)^\perp = 0\).
Let \(\delta \in { \operatorname{ad}}(L)^\perp\) and \(x\in L\), then
\(0 = [\delta, { \operatorname{ad}}_x] = { \operatorname{ad}}_{ \delta(x) }\)
where the last equality follows from an earlier exercise. Since
\({ \operatorname{ad}}\) is injective, \(\delta(x) = 0\) and so
\(\delta = 0\), thus \({ \operatorname{ad}}(L)^\perp = 0\). So we have
\(\mathop{\mathrm{Rad}}\kappa_{\mathop{\mathrm{Der}}L} \subseteq { \operatorname{ad}}(L)^\perp = 0\)
since any derivation orthogonal to all derivations is in particular
orthogonal to inner derivations, and thus
\(\kappa_{\mathop{\mathrm{Der}}L}\) is nondegenerate. Finally, we can
write
\(\mathop{\mathrm{Der}}L = { \operatorname{ad}}(L) \oplus { \operatorname{ad}}(L)^\perp = { \operatorname{ad}}(L) \oplus 0 = { \operatorname{ad}}(L)\).

\end{proof}

\hypertarget{abstract-jordan-decompositions}{%
\subsection{5.4: Abstract Jordan
Decompositions}\label{abstract-jordan-decompositions}}

\begin{remark}

Earlier: if \(A\in \mathsf{Alg}_{/ {{ \mathbf{F} }}}^{\mathrm{fd}}\),
not necessarily associative, \(\mathop{\mathrm{Der}}A\) contains the
semisimple and nilpotent parts of all of its elements. Applying this to
\(A = L\in \mathsf{Lie} \mathsf{Alg}\) yields
\(L \cong { \operatorname{ad}}(L) = \mathop{\mathrm{Der}}L\) ad
\({ \operatorname{ad}}_x = s + n \in { \operatorname{ad}}(L) + { \operatorname{ad}}(L)\),
so write \(s = { \operatorname{ad}}_{x_s}\) and
\(n = { \operatorname{ad}}_{x_n}\), then
\({ \operatorname{ad}}_x = { \operatorname{ad}}_{x_s} + { \operatorname{ad}}_{x_n} = { \operatorname{ad}}_{x_s + x_n}\)
so \(x = x_s + x_n\) by injectivity of \({ \operatorname{ad}}\),
yielding a definition for the semisimple and nilpotent parts of \(x\).
If \(L \leq {\mathfrak{gl}}(V)\), it turns out that these coincide with
the usual decomposition -- this is proved in section 6.4.

\end{remark}

\hypertarget{modules-chapter-6-complete-reducibility-of-representations}{%
\subsection{6.1: Modules (Chapter 6: Complete Reducibility of
Representations)}\label{modules-chapter-6-complete-reducibility-of-representations}}

\begin{definition}[$L\dash$modules and representations]

Let
\(L \in \mathsf{Lie} \mathsf{Alg}_{/ {{\mathbf{C}}}}^{\mathrm{fd}}\),
then a \textbf{representation} of \(L\) on \(V\) is a homomorphism of
Lie algebras \(\phi: L \to {\mathfrak{gl}}(V)\). For
\(V\in{ \mathsf{Vect}}_{/ {{\mathbf{C}}}}\) with an \emph{action} of
\(L\), i.e.~an operation
\begin{align*}
L\times V &\to V \\
(x, v) &\mapsto x.v
,\end{align*}
is an \textbf{\(L{\hbox{-}}\)module} iff for all
\(a,b\in {\mathbf{C}}, x,y\in L, v,w\in V\),

\begin{itemize}
\tightlist
\item
  (M1) \((ax+by).v = a(x.v) + b(y.v)\).
\item
  (M2) \(x.(av+bw) = a(x.v) + b(x.w)\).
\item
  (M3) \([xy].v = x.(y.v) - y.(x.v)\).
\end{itemize}

\end{definition}

\begin{remark}

An \(L{\hbox{-}}\)module \(V\) is equivalent to a representation of
\(L\) on \(V\). If \(\phi: L \to {\mathfrak{gl}}(V)\) is a
representation, define \(x.v \coloneqq\phi(x)v \coloneqq\phi(x)(v)\).
Conversely, for \(V\in {}_{L}{\mathsf{Mod}}\) define
\(\phi: L\to {\mathfrak{gl}}(V)\) by \(\phi(x)(v) \coloneqq x.v\).

\end{remark}

\begin{example}[?]

\(L\in {}_{L}{\mathsf{Mod}}\) using \({ \operatorname{ad}}\), this
yields the \textbf{adjoint representation}.

\end{example}

\begin{definition}[Morphism of $L\dash$modules]

A \textbf{morphism} of \(L{\hbox{-}}\)modules is a linear map
\(\psi: V\to W\) such that \(\psi(x.v) = x.\psi(v)\) for all
\(x\in L, v\in V\). It is an isomorphism as an \(L{\hbox{-}}\)module iff
it is an isomorphism of the underlying vector spaces.\footnote{It turns
  out that the inverse map of vector spaces \(\psi^{-1}: W\to V\) is
  again a morphism of \(L{\hbox{-}}\)modules.} In this case we say
\(V, W\) are \emph{equivalent} representations.

\end{definition}

\begin{example}[?]

Let \(L = {\mathbf{C}}x\) for \(x\neq 0\), then

\begin{itemize}
\item
  What is a representation of \(L\) on \(V\)? This amounts to picking an
  element of \({ \operatorname{End} }(V)\).
\item
  When are 2 \(L{\hbox{-}}\)modules on \(V\) equivalent? This happens
  iff the two linear transformations are conjugate in
  \({ \operatorname{End} }(V)\).
\end{itemize}

Thus representations of \(L\) on \(V\) are classified by Jordan
canonical forms when \(V\) is finite dimensional.

\end{example}

\begin{definition}[Submodules, irreducible/simple modules]

For \(V\in {}_{L}{\mathsf{Mod}}\), a subspace \(W \subseteq V\) is a
\textbf{submodule} iff it is in invariant subspace, so \(x.w\in W\) for
all \(x\in L, w\in W\). \(V\) is \textbf{irreducible} or \textbf{simple}
if \(V\) has exactly two invariant subspaces \(V\) and \(0\).

\end{definition}

\begin{warnings}

Note that this rules out \(0\) as being a simple Lie algebra.

\end{warnings}

\begin{definition}[Quotient modules]

For \(W\leq V \in {}_{L}{\mathsf{Mod}}\) a submodule, the
\textbf{quotient module} is \(V/W\) has underlying vector space \(V/W\)
with action \(x.(v+W) \coloneqq(x.v) + W\). This is well-defined
precisely when \(W\) is a submodule.

\end{definition}

\begin{warnings}

\(I{~\trianglelefteq~}L \iff { \operatorname{ad}}(I) \leq { \operatorname{ad}}(L)\),
i.e.~ideals corresponds to submodules under the adjoint representation.
However, irreducible ideals need not correspond to simple submodules.

\end{warnings}

\hypertarget{wednesday-september-14}{%
\section{Wednesday, September 14}\label{wednesday-september-14}}

\hypertarget{structure-theory}{%
\subsection{6.1: Structure theory}\label{structure-theory}}

\begin{definition}[Natural representation]

Note that all of the algebras \({\mathfrak{g}}\) we've considered
naturally act on column vectors in some \({ \mathbf{F} }^n\) -- this is
the \textbf{natural representation} of \({\mathfrak{g}}\).

\end{definition}

\begin{example}[?]

Letting \({\mathfrak{b}}_n\) be the upper triangular matrices in
\({\mathfrak{gl}}_n\), this acts on \({ \mathbf{F} }^n\). Taking a
standard basis
\({ \mathbf{F} }^n = V \coloneqq\left\langle{e_1,\cdots, e_n}\right\rangle_{ \mathbf{F} }\),
one gets submodules
\(V_i = \left\langle{e_1,\cdots, e_i}\right\rangle_{ \mathbf{F} }\)
which correspond to upper triangular blocks got by truncating the first
\(i\) columns of the matrix. This yields a submodule precisely because
the lower-left block is zero.

\end{example}

\begin{remark}

Let \(\phi: L\to {\mathfrak{gl}}(V)\) be a representation, noting that
\({ \operatorname{End} }(V)\) is an associative algebra. We can consider
the associative unital algebra \(A\) generated by the image \(\phi(L)\).
Note that the structure of \(V\) as an \(L{\hbox{-}}\)module is the same
as its \(A{\hbox{-}}\)module structure, so we can apply theorems/results
from the representation of rings and algebras to study Lie algebra
representations, e.g.~the Jordan-Holder theorem and Schur's lemma.

\end{remark}

\begin{definition}[Direct sums of $L\dash$modules]

Given \(V, W\in {}_{L}{\mathsf{Mod}}\), their vector space direct sum
admits an \(L{\hbox{-}}\)module structure using
\(x.(v, w) \coloneqq(x.v, x.w)\), which we'll write as
\(x.(v+w) \coloneqq xv + xw\).

\end{definition}

\begin{definition}[Completely reducible modules]

\(V\in {}_{L}{\mathsf{Mod}}\) is \textbf{completely reducible} iff \(V\)
is a direct sum of irreducible \(L{\hbox{-}}\)modules. Equivalently, for
each \(W\leq V\) there is a complementary submodule \(W'\) such that
\(V = W \oplus W'\).

\end{definition}

\begin{warnings}

``Not irreducible'' is strictly weaker than ``completely reducible'',
since a submodule may not admit an invariant complement -- for example,
the flag in the first example above.

\end{warnings}

\begin{example}[?]

The natural representation of \({\mathfrak{h}}_n\) is completely
reducible, decomposing as \(V_1 \oplus V_2 \oplus \cdots V_n\) where
\(V_i = { \mathbf{F} }e_i\).

\end{example}

\begin{definition}[Indecomposable modules]

A module \(V\) is \textbf{indecomposable} iff \(V\neq W \oplus W'\) for
proper submodules \(W, W' \leq V\). This is weaker than irreducibility.

\end{definition}

\begin{example}[?]

Consider the natural representation \(V\) for
\(L \coloneqq{\mathfrak{b}}_n\). Every nonzero submodule of \(V\) must
contain \(e_1\), so \(V\) is indecomposable if \(n\geq 1\).

\end{example}

\begin{remark}

Recall that the \textbf{socle} of \(V\) is the (direct) sum of all of
its irreducible submodules. If \(\mathop{\mathrm{Soc}}(V)\) is simple
(so one irreducible) then \(V\) is indecomposable, since every summand
must contain this simple submodule ``at the bottom''. For
\(L = {\mathfrak{b}}_n\), note that
\(\mathop{\mathrm{Soc}}(V) = { \mathbf{F} }e_1\).

\end{remark}

\begin{remark}

For the reminder of chapters 6 and 7, we assume all modules are
finite-dimensional over \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\).

\end{remark}

\begin{theorem}[Jordan-Holder]

Let
\(L\in\mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}^{\mathrm{fd}}\),
then there exists a \textbf{composition series}, a sequence of
submodules \(0 = V_0 \subseteq V_1 \subseteq \cdots \subseteq V_n = V\)
such that each composition factor (sometimes called \emph{sections})
\(V_i/V_{i-1}\) is irreducible/simple. Moreover, any two composition
series admit the same composition factors with the same multiplicities,
up to rearrangement and isomorphism.

\end{theorem}

\begin{example}[?]

If \(V = W \oplus W'\) with \(W, W'\) simple, there are two composition
series:

\begin{itemize}
\tightlist
\item
  \(0 \subseteq W \subseteq V\) with factors \(W, V/W \cong W'\),
\item
  \(0 \subseteq W' \subseteq V\) with factors \(W', V/W' \cong W\).
\end{itemize}

These aren't equal, since they're representations on different coset
spaces, but are isomorphic.

\end{example}

\begin{theorem}[Schur's lemma]

If \(\phi: L\to {\mathfrak{gl}}(V)\) is an irreducible representation,
then \({ \operatorname{End} }_L(V)\cong { \mathbf{F} }\).

\end{theorem}

\begin{proof}[?]

If \(V\) is irreducible then every \(f\in {}_{L}{\mathsf{Mod}}(V, V)\)
is either zero or an isomorphism since \(f(V) \leq V\) is a submodule.
Thus \({ \operatorname{End} }_L(V)\) is a division algebra over
\({ \mathbf{F} }\), but the only such algebra is \({ \mathbf{F} }\)
since \({ \mathbf{F} }= \overline{{ \mathbf{F} }}\).

Letting \(\phi\) be as above, it has an eigenvalue
\(\lambda \in { \mathbf{F} }\), again since
\({ \mathbf{F} }= \overline{{ \mathbf{F} }}\). Then
\(\phi - \lambda I \in { \operatorname{End} }_L(V)\) has a nontrivial
kernel, the \(\lambda{\hbox{-}}\)eigenspace. So
\(\phi - \lambda I = 0 \implies \varphi = \lambda I\).

\end{proof}

\begin{warnings}

Schur's lemma is not always true for Lie \emph{superalgebras}.

\end{warnings}

\begin{definition}[Trivial modules]

The \textbf{trivial \(L{\hbox{-}}\)module} is
\({ \mathbf{F} }\in {}_{L}{\mathsf{Mod}}\) equipped with the zero map
\(\varphi: L\to {\mathfrak{gl}}({ \mathbf{F} })\) where
\(x.1 \coloneqq 0\) for all \(x\in L\). Note that this is irreducible,
and any two such 1-dimensional trivial modules are isomorphic by sending
a basis \(\left\{{e_1}\right\}\) to \(1\in { \mathbf{F} }\).

More generally, an \(V \in {}_{L}{\mathsf{Mod}}\) is trivial iff
\(x.v = 0\) for all \(x\in L, v\in V\), and \(V\) is completely
reducible as a direct sum of copies of the above trivial module.

\end{definition}

\begin{definition}[Homs, Tensors, and duals]

Let \(V, W\in {}_{L}{\mathsf{Mod}}\), then the following are all
\(L{\hbox{-}}\)modules:

\begin{itemize}
\tightlist
\item
  \(V\otimes_{ \mathbf{F} }W\): the action is
  \(x.(v \otimes w) = (x.v)\otimes w + v\otimes(x.w)\).\footnote{Note
    that groups would act on each factor separately, and this is more
    like a derivative.}
\item
  \(\mathop{\mathrm{Hom}}_{ \mathbf{F} }(V, W)\): the action is
  \((x.f)(v) = x.(f(v)) - f(x.v) \in W\).
\item
  \(V {}^{ \vee }\coloneqq\mathop{\mathrm{Hom}}_{ \mathbf{F} }(V, { \mathbf{F} })\):
  the action is a special case of the above since \(x.w = 0\),
  so\footnote{One might expect an inverse from group theory, which
    differentiates to a minus sign.}
  \begin{align*}
  (x.f)(v) = -f(x.v)
  .\end{align*}
\end{itemize}

\end{definition}

\begin{remark}

These structures come from the Hopf algebra structure on the universal
associative algebra \(U({\mathfrak{g}})\), called the \textbf{universal
enveloping algebra}. Note that we also have
\begin{align*}
\mathop{\mathrm{Hom}}_{\mathbf{C}}(V, W)\underset{  {}_{L}{\mathsf{Mod}}}{\cong} V {}^{ \vee }\otimes_{ \mathbf{F} }W
.\end{align*}

\end{remark}

\hypertarget{friday-september-16}{%
\section{Friday, September 16}\label{friday-september-16}}

\hypertarget{casimir-element-of-a-representation}{%
\subsection{6.2: Casimir element of a
representation}\label{casimir-element-of-a-representation}}

\begin{remark}

Last time: \(L\) semisimple over \({\mathbf{C}}\) implies
\(\kappa(x,y)\coloneqq\operatorname{tr}( { \operatorname{ad}}_x { \operatorname{ad}}_y)\)
is nondegenerate. Using Cartan's criterion, we can show that for any
faithful representation \(\phi: L\to {\mathfrak{gl}}(V)\) we can define
a symmetric bilinear form \(\beta_\phi\) on \(L\) defined by
\(\beta_\phi(x, y) = \operatorname{tr}(\phi(x) \phi(y))\). Note that
\(\beta_{ { \operatorname{ad}}} = \kappa\). Since
\(\mathop{\mathrm{Rad}}(\beta_\phi) = 0\), it is nondegenerate. This
defines an isomorphism \(L { \, \xrightarrow{\sim}\, }L {}^{ \vee }\) by
\(x\mapsto \beta(x, {-})\), so given a basis
\({\mathcal{B}}\coloneqq\left\{{x_i}\right\}_{i\leq n}\) for \(L\) there
is a unique dual basis
\({\mathcal{B}}' = \left\{{y_i}\right\}_{i\leq n}\) for \(L\) such that
\(\beta(x_i, y_j) = \delta_{ij}\). Note that the \(y_i \in L\) are dual
to the basis \(\beta(x_i, {-}) \in L {}^{ \vee }\).

\end{remark}

\begin{example}[?]

For \(L\coloneqq{\mathfrak{sl}}_2({\mathbf{C}})\), the matrix of
\(\kappa\) is given by
\({ \begin{bmatrix}  {0} & {0} & {4} \\  {0} & {8} & {0} \\  {4} & {0} & {0} \end{bmatrix} }\)
with respect to the ordered basis
\({\mathcal{B}}=\left\{{x,h,y}\right\}\).\footnote{See Humphreys p.22.}
Thus
\({\mathcal{B}}' = \left\{{{1\over 4}y, {1\over 8}h, {1\over 4}x}\right\}\).

\end{example}

\begin{definition}[Casimir element]

Now let \(\phi: L\to {\mathfrak{gl}}(V)\) be a faithful irreducible
representation. Fix a basis \({\mathcal{B}}\) of \(L\) and define the
\textbf{Casimir element}
\begin{align*}
c_\phi = c_\phi(\beta) \coloneqq\sum_{i\leq n} \phi(x_i) \circ \phi(y_i) \in { \operatorname{End} }_{\mathbf{C}}(V)
.\end{align*}

\end{definition}

\begin{remark}

One can show that \(c_\phi\) commutes with \(\phi(L)\). Supposing
\(\phi\) is irreducible, \({ \operatorname{End} }_L(V) = {\mathbf{C}}\)
by Schur's lemma, so \(c_\phi\) is acts on \(V\) as a scalar. This
follows from
\begin{align*}
\operatorname{tr}(c_ \varphi) = \sum_{i\leq n} \operatorname{tr}( \varphi(x_i) \varphi(y_i) ) = \sum_{i\leq n} \beta(x_i, y_i) = n = \dim L
\implies
c_\phi = {\dim L \over \dim V} \operatorname{id}_V
,\end{align*}
since there are \(\dim V\) entries. In particular, \(c_\phi\) is
independent of the choice of \({\mathcal{B}}\). This will be used to
prove Weyl's theorem, one of the main theorems of semisimple Lie theory
over \({\mathbf{C}}\). If \(L\) is semisimple and \(\phi:\) is not
faithful, replace \(L\) by \(L/\ker \phi\). Since
\(\ker \phi {~\trianglelefteq~}L\) and \(L\) is semisimple,
\(\ker \phi\) is a direct sum of certain simple ideals of \(L\) and the
quotient is isomorphic to the sum of the remaining ideals. This yields a
representation
\(\overline{\phi}: L/\ker \varphi \to {\mathfrak{gl}}(V)\) which is
faithful and can be used to define a Casimir operator.

\end{remark}

\begin{example}[?]

Let \(L = {\mathfrak{sl}}_2({\mathbf{C}})\) and let
\(V = {\mathbf{C}}^2\) be the natural representation, so
\(\phi: L\to {\mathfrak{gl}}(V)\) is the identity. Fix
\({\mathcal{H}}= \left\{{x,h,y}\right\}\), then
\(\beta(u, v) = \operatorname{tr}( u v )\) since \(\phi(u) = u\) and
\(\phi(v) = v\). We get the following products:

\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.23}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.23}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.23}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.27}}@{}}
\toprule
& \({ \begin{bmatrix} {0} & {1} \\ {0} & {0} \end{bmatrix} }\) &
\({ \begin{bmatrix} {1} & {0} \\ {0} & {-1} \end{bmatrix} }\) &
\({ \begin{bmatrix} {0} & {0} \\ {1} & {0} \end{bmatrix} }\) \\
\midrule
\endhead
\({ \begin{bmatrix} {0} & {1} \\ {0} & {0} \end{bmatrix} }\) & 0 &
\({ \begin{bmatrix} {0} & {-1} \\ {0} & {0} \end{bmatrix} }\) &
\({ \begin{bmatrix} {1} & {0} \\ {0} & {0} \end{bmatrix} }\) \\
\({ \begin{bmatrix} {1} & {0} \\ {0} & {-1} \end{bmatrix} }\) & & I &
\({ \begin{bmatrix} {0} & {0} \\ {-1} & {0} \end{bmatrix} }\) \\
\({ \begin{bmatrix} {0} & {0} \\ {1} & {0} \end{bmatrix} }\) & & & 0 \\
\bottomrule
\end{longtable}

Thus
\(\beta = { \begin{bmatrix}  {0} & {0} & {1} \\  {0} & {2} & {0} \\  {1} & {0} & {0} \end{bmatrix} }\),
and \({\mathcal{B}}' = \left\{{y, {1\over 2}h, x}\right\}\), so
\begin{align*}
c_\phi = xy + {1\over 2}h^2 + yx = {
\begin{bmatrix}
  {1} & {0} 
\\
  {0} & {0}
\end{bmatrix}
} + {1\over 2}I + {
\begin{bmatrix}
  {0} & {0} 
\\
  {0} & {1}
\end{bmatrix}
} = {3\over 2}I = {\dim {\mathbf{C}}^2\over \dim {\mathfrak{sl}}_2({\mathbf{C}})} I
.\end{align*}

\end{example}

\hypertarget{complete-reducibility}{%
\subsection{6.3: Complete reducibility}\label{complete-reducibility}}

\begin{lemma}[?]

Let \(\phi: L\to {\mathfrak{gl}}(V)\) be a representation of a
semisimple Lie algebra, then \(\phi(L) \subseteq {\mathfrak{sl}}(V)\).
In particular, \(L\) acts trivially on any 1-dimensional
\(L{\hbox{-}}\)module since a \(1\times 1\) traceless matrix is zero.
The proof follows from \(L = [LL]\).

\end{lemma}

\begin{warnings}

Arbitrary reductive Lie algebras can have nontrivial 1-dimensional
representations.

\end{warnings}

\begin{theorem}[Weyl's theorem]

Let \(\phi: L\to {\mathfrak{gl}}(V)\) be a finite-dimensional
representation of a finite-dimensional semisimple Lie algebra over
\({\mathbf{C}}\). This \(\phi\) is completely reducible.

\end{theorem}

\begin{warnings}

This is not true for characteristic \(p\), or infinite dimensional
representations in characteristic 0.

\end{warnings}

\hypertarget{proof-of-weyls-theorem}{%
\subsubsection{Proof of Weyl's theorem}\label{proof-of-weyls-theorem}}

\begin{remark}

Replace \(L\) by \(L/\ker \phi\) if necessary to assume \(\phi\) is
faithful, since these yield the same module structures. Define a Casimir
operator \(c_\phi\) as before, and recall that complete reducibility of
\(V\) is equivalent to every \(L{\hbox{-}}\)submodule \(W\leq V\)
admitting a complementary submodule \(W''\) such that
\(V = W \oplus W''\). We proceed by induction on \(\dim V\), where the
dimension 1 case is clear.

\textbf{Case I}: \(\operatorname{codim}_V W = 1\),
i.e.~\(\dim (V/W) = 1\). Take the SES
\(W \hookrightarrow V \twoheadrightarrow V/W\).

\textbf{Case 1}: Suppose \(W' \leq W\) is a proper nonzero
\(L{\hbox{-}}\)submodule. Schematic:

\includegraphics{figures/2022-09-16_10-02-33.png}

Using the 2nd isomorphism theorem, there is a SES
\(W/W' \hookrightarrow V/W' \twoheadrightarrow V/W\). Since
\(\dim W' > 0\), \(\dim V/W' \leq \dim V\), so by the IH there is a
1-dimensional complement to \(W/W'\) in \(V/W'\). This lifts to
\(\tilde W' \leq V\) with \(W' \leq \tilde W\) with
\(\dim \tilde W/W' = 1\), and moreover
\(V/W' = W/W' \oplus \tilde W/W'\). Take the SES
\(W' \hookrightarrow\tilde W \twoheadrightarrow\tilde W/W'\) with
\(\dim \tilde W < \dim V\). Apply the IH again to get a subspaces
\(X \leq \tilde W \leq V\) with \(\tilde W = W' \oplus X\). We'll
continue by showing \(V = W \bigoplus X\).

\end{remark}

\hypertarget{monday-september-19}{%
\section{Monday, September 19}\label{monday-september-19}}

\hypertarget{proof-of-weyls-theorem-continued}{%
\subsection{Proof of Weyl's theorem
(continued)}\label{proof-of-weyls-theorem-continued}}

\begin{remark}

Recall: we're proving Weyl's theorem, i.e.~every finite-dimensional
representation of semisimple Lie algebra over \({\mathbf{C}}\) is
completely reducible. Strategy: show every \(W \leq_L V\) has a
complement \(W'' \leq_L V\) such that \(V = W \oplus W''\); induct on
\(\dim V\).

\end{remark}

\begin{proof}[of Weyl's theorem, continued]

\textbf{Case I}: \(\dim V/W = 1\).

\textbf{Case 1}: \(W\) is reducible. We got \(0 < W' < W < V\) (proper
submodules), represented schematically by a triangle. We showed
\(V/W' \cong W/W' \oplus \tilde W/W'\), since

\begin{itemize}
\tightlist
\item
  \(\tilde W \cap W \subseteq W'\),
\item
  \(V= W + \tilde W + W' = W + \tilde W\) since \(W' \subseteq W\).
\item
  \(\tilde W = W' \oplus X\) for some submodule
  \(X \leq_L \tilde W \leq_L V\).
\end{itemize}

Thus replacing \(\tilde W\) in the second point yields
\(V = W + \tilde W = W + W' +X = W + X\); we want to prove this sum is
direct. Since \(X\) is contained in \(\tilde W\), we can write
\begin{align*}
X \cap W 
&= (X \cap\tilde W) \cap W \\
&= X \cap(\tilde W \cap W) \qquad\text{by 1}\\
&\subseteq X \cap W' = 0 \qquad \text{by 2}
,\end{align*}
so \(V = W \oplus X\).

\textbf{Case 2}: Let \(c_\phi\) be the Casimir element of \(\phi\), and
note that \(c_\phi(W) \subseteq W\) since \(c_\phi\) is built out of
endomorphisms in \(\phi(L)\) sending \(W\) to \(W\) (since \(W\) is a
submodule). In fact, \(\phi(L)(V) = W\) since \(V/W\) is a 1-dimensional
representation of the semisimple Lie algebra \(L\), hence trivial. Thus
\(c_\phi(V) \subseteq W\) and thus \(\ker c_\phi \neq 0\) since
\(W < V\) is proper. Note also that \(c_\phi\) commutes with anything in
\(c_\phi(L)\) on \(V\), and so defines a morphism
\(c_\phi \in {}_{L}{\mathsf{Mod}}(V, V)\) and \(\ker c_\phi \leq_L V\).
On the other hand, \(c_\phi\) induces an element of
\({ \operatorname{End} }_{L}(W)\), and since \(W\) is irreducible,
\({ \left.{{c_\phi}} \right|_{{W}} } = \lambda \operatorname{id}_W\) for
some scalar \(\lambda\). This can't be zero, since
\(\operatorname{tr}({ \left.{{c_\phi}} \right|_{{W}} }) = {\dim L \over \dim W} > 0\),
so \(\ker c_\phi \cap W = 0\). Since \(\operatorname{codim}_V W = 1\),
i.e.~\(\dim W = \dim V - 1\), we must have \(\dim \ker c_\phi = 1\) and
we have a direct sum decomposition \(V = W \oplus \ker c_\phi\).

\begin{quote}
Use of the Casimir element in basic Lie theory: producing a complement
to an irreducible submodule.
\end{quote}

\textbf{Case 2}: Suppose \(0 < W < V\) with \(W\) any nontrivial
\(L{\hbox{-}}\)submodule; there is a SES
\(W \hookrightarrow V \twoheadrightarrow V/W\) Consider
\(H \coloneqq\hom_{\mathbf{C}}(V, W)\), then
\(H \in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) by
\((x.f)(v) \coloneqq x.(f(v)) - f(x.v)\) for \(f\in H, x\in L, v\in V\).
Let
\({\mathcal{V}}\coloneqq\left\{{f {~\mathrel{\Big\vert}~}H {~\mathrel{\Big\vert}~}{ \left.{{f}} \right|_{{W}} } = \alpha \operatorname{id}_W \,\text{ for some } \alpha \in {\mathbf{C}}}\right\} \subseteq H\).
For \(f\in V\) and \(w\in W\), we have
\begin{align*}
(x.f)(w) = x.f(w) - f(x.w) = \alpha x.w - \alpha x.w = 0
.\end{align*}
So let
\({\mathcal{W}}\coloneqq\left\{{f\in {\mathcal{V}}{~\mathrel{\Big\vert}~}f(W) = 0}\right\} \subseteq {\mathcal{V}}\),
then we've shown that \(L.{\mathcal{V}}\subseteq {\mathcal{W}}\). Now
roughly, the complement is completely determined by the scalar.
Rigorously, since \(\dim {\mathcal{V}}/{\mathcal{W}}= 1\), any
\(f\in {\mathcal{V}}\) is determined by the scalar
\({ \left.{{f}} \right|_{{W}} } = \alpha \operatorname{id}_W\): we have
\(f-\alpha \chi_W \in {\mathcal{W}}\) where \(\chi_W\) is any extension
of \(\operatorname{id}_W\) to \(V\) which is zero on \(V/W\), e.g.~by
extending a basis and having it act by zero on the new basis elements.

Now
\({\mathcal{W}}\hookrightarrow{\mathcal{V}}\twoheadrightarrow{\mathcal{V}}/{\mathcal{W}}\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\)
with \(\operatorname{codim}_{\mathcal{V}}{\mathcal{W}}= 1\). By Case I,
\({\mathcal{V}}= {\mathcal{W}}\oplus {\mathcal{W}}''\) for some
complement \(L{\hbox{-}}\)submodule \({\mathcal{W}}''\). Let
\(f: V\to W\) span \({\mathcal{W}}''\), then
\({ \left.{{f}} \right|_{{W}} }\) is a nonzero scalar -- a scalar since
it's in \({\mathcal{V}}\), and nonzero since it's in the complement of
\({\mathcal{W}}\). By rescaling, we can assume the scalar is 1, so
\(\operatorname{im}f = W\) and by rank-nullity
\(\dim \ker f = \dim V - \dim W\). Thus \(\ker f\) has the right
dimension to be the desired complement. It is an
\(L{\hbox{-}}\)submodule, since
\(L.f \subseteq {\mathcal{W}}'' \cap{\mathcal{W}}= 0\) since
\({\mathcal{W}}''\) is an \(L{\hbox{-}}\)submodule and
\(f\in {\mathcal{W}}\) since \(L.{\mathcal{V}}\subseteq {\mathcal{W}}\).
Noting that if \((x.f) = 0\) then \(x.(f(v)) = f(x.v)\), making \(f\) an
\({\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) morphism. Thus
\(W'' \coloneqq\ker f \leq_L V\), and \(W \cap W'' = 0\) so
\({ \left.{{f}} \right|_{{W}} } = \operatorname{id}_W\). Since the
dimensions add up correctly, we get \(V = W \oplus W''\).

\end{proof}

\hypertarget{preservation-of-jordan-decomposition}{%
\subsection{6.4: Preservation of Jordan
decomposition}\label{preservation-of-jordan-decomposition}}

\begin{theorem}[?]

Let \(L \leq {\mathfrak{gl}}(V)\) be a subalgebra with \(L\) semisimple
and finite-dimensional. Given \(x\in L\), there are two decompositions:
the usual JCF \(x = s + n\), and the abstract decomposition
\({ \operatorname{ad}}_x = { \operatorname{ad}}_{x_s} + { \operatorname{ad}}_{x_n}\).
\(L\) contains the semisimple and nilpotent parts of all of its
elements, and in particular the two above decompositions coincide.

\end{theorem}

\begin{proof}[Idea]

The proof is technical, but here's a sketch:

\begin{itemize}
\tightlist
\item
  Construct a subspace \(L \leq L' \leq_L {\mathfrak{gl}}(V)\) such that
  \(l'\) contains the semisimple and nilpotent parts of all elements
  where \(L\curvearrowright{\mathfrak{gl}}(V)\) by
  \({ \operatorname{ad}}: L\to {\mathfrak{gl}}({\mathfrak{gl}}(V))\).
\item
  Check \(L' {\mathfrak{q}}N_{{\mathfrak{gl}}(V)}(L)\) (the normalizer),
  so \([LL'] \subseteq L\).
\item
  Show \(L' = L\):

  \begin{itemize}
  \tightlist
  \item
    If \(L'\neq L\), use Weyl's theorem to get a complement with
    \(L' = L \oplus M\).
  \item
    Check \([LM] \subseteq [LL'] \subseteq L\) and \([LM] \subseteq M\)
    since \(M\leq_L M\), forcing \([LM] \subseteq L \cap M = 0\).
  \item
    Use Weyl's theorem on all of \(V\) splits it into sums of
    irreducibles, bracket against the irreducibles, and use specific
    properties of this \(L'\) to show \(M = 0\).
  \end{itemize}
\item
  Since \(s, n\in L\) when \(x=s+n\) and
  \({ \operatorname{ad}}_x = { \operatorname{ad}}_s + { \operatorname{ad}}_n = { \operatorname{ad}}_{x_s} + { \operatorname{ad}}_{x_n}\),
  the result follows from uniqueness of the abstract JCF that
  \(s=x_s, n=x_n\) (using that \({ \operatorname{ad}}\) is injective
  when \(L\) is semisimple since \(Z(L) = 0\)).
\end{itemize}

\end{proof}

\begin{corollary}[?]

If
\(L \in \mathsf{Lie} \mathsf{Alg}_{/ {{\mathbf{C}}}}^{{\mathrm{ss}}}\)
(not necessarily linear) and
\(\phi\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\), writing \(x=s+n\) the
abstract Jordan decomposition, \(\phi(x) = \phi(s) + \phi(n)\) is the
usual JCF of \(\phi(x)\in {\mathfrak{gl}}(V)\).

\end{corollary}

\begin{proof}[Sketch]

Consider \({ \operatorname{ad}}_{\phi(L)}\phi(s)\) and
\({ \operatorname{ad}}_{\phi(L)}\phi(n)\), which are semisimple (acts on
a vector space and decomposes into a direct sum of eigenspaces) and
nilpotent respectively and commute, yielding the abstract Jordan
decomposition of \({ \operatorname{ad}}_{\phi(x)}\). Now apply the
theorem.

\end{proof}

\hypertarget{ch.-7-representations-of-mathfraksl_2mathbfc-wednesday-september-21}{%
\section{\texorpdfstring{Ch. 7: Representations of
\({\mathfrak{sl}}_2({\mathbf{C}})\) (Wednesday, September
21)}{Ch. 7: Representations of \{\textbackslash mathfrak\{sl\}\}\_2(\{\textbackslash mathbf\{C\}\}) (Wednesday, September 21)}}\label{ch.-7-representations-of-mathfraksl_2mathbfc-wednesday-september-21}}

\begin{remark}

If \(L\in \mathsf{Lie} \mathsf{Alg}\) and \(s\) is a semisimple element,
then \(\phi(s)\) is semisimple in any finite-dimensional representation
\(\phi\) of \(L\). In particular, taking the natural representation,
this yields a semisimple operator. For the same reason,
\({ \operatorname{ad}}_s\) is semisimple. Similar statements hold for
nilpotent elements.

\end{remark}

\hypertarget{weights-and-maximal-vectors}{%
\subsection{7.1: Weights and Maximal
Vectors}\label{weights-and-maximal-vectors}}

\begin{remark}

Let \(L \coloneqq{\mathfrak{sl}}_2({\mathbf{C}})\), then recall

\begin{itemize}
\tightlist
\item
  \(x = { \begin{bmatrix}  {0} & {1} \\  {0} & {0} \end{bmatrix} }\)
\item
  \(h = { \begin{bmatrix}  {1} & {0} \\  {0} & {-1} \end{bmatrix} }\)
\item
  \(y= { \begin{bmatrix}  {0} & {0} \\  {1} & {0} \end{bmatrix} }\)
\item
  \([xy] = h\)
\item
  \([hx] = 2x\)
\item
  \([hy] = -2y\)
\end{itemize}

Goal: classify \({\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fd}}\). By
Weyl's theorem, they're all completely reducible, so it's enough to
describe the simple objects.

\end{remark}

\begin{definition}[Weight decomposition and weight spaces]

Note that
\(L \leq {\mathfrak{gl}}_2({\mathbf{C}}) = {\mathfrak{gl}}({\mathbf{C}}^2)\),
and since \(h\) is semisimple, \(\phi(h)\) acts semisimply on any
finite-dimensional representation \(V\) with
\(\phi: L\to {\mathfrak{gl}}(V)\). I.e. \(\phi(h)\) acts diagonally on
\(V\). Thus \(V = \bigoplus _{ \lambda} V_{ \lambda}\) which are
eigenspaces for the \(\phi(h)\) action, where
\begin{align*}
V_\lambda \coloneqq\left\{{v\in V {~\mathrel{\Big\vert}~}h.v = \lambda v, \lambda\in {\mathbf{C}}}\right\}
.\end{align*}
If \(V_\lambda \neq 0\) we say \(\lambda\) is a \textbf{weight} of \(h\)
in \(V\) and \(V_\lambda\) is the corresponding \textbf{weight space}.

\end{definition}

\begin{lemma}[?]

If \(v\in V_ \lambda\), then \(x.v \in V_ {\lambda+2}\) and
\(y.v \in V_{ \lambda- 2}\).

\end{lemma}

\begin{exercise}[?]

Prove this using the commutation relations.

\end{exercise}

\begin{definition}[Highest weights]

Note that if \(V\) is finite-dimensional then there can not be
infinitely many nonzero \(V_\lambda\), so there exists a
\(\lambda\in {\mathbf{C}}\) such that \(V_{ \lambda} \neq 0\) but
\(V_{ \lambda+ 2} = 0\). We call \(\lambda\) a \textbf{highest weight}
(h.w.) of \(V\) (which will turn out to be unique) and any nonzero
\(v\in V\) a \textbf{highest weight vector}.

\end{definition}

\hypertarget{classification-of-irreducible-mathfraksl_2mathbfchbox-modules}{%
\subsection{\texorpdfstring{7.2: Classification of Irreducible
\({\mathfrak{sl}}_2({\mathbf{C}}){\hbox{-}}\)Modules}{7.2: Classification of Irreducible \{\textbackslash mathfrak\{sl\}\}\_2(\{\textbackslash mathbf\{C\}\})\{\textbackslash hbox\{-\}\}Modules}}\label{classification-of-irreducible-mathfraksl_2mathbfchbox-modules}}

\begin{lemma}[?]

Let
\(V \in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fd}}, {\mathrm{irr}}}\)
and let \(v_0\in V_{ \lambda}\) be a h.w. vector. Set \(v_{-1} = 0\) for
\(i\geq 0\) and \(v_i \coloneqq{1\over i!}y^i v_0\), then for
\(i\geq 0\),

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(h.v_i = (\lambda- 2i)v_i\)
\item
  \(y.v_i = (i+1)v_{i+1}\)
\item
  \(x.v_i = ( \lambda- i + 1) v_{i-1}\).
\end{enumerate}

\end{lemma}

\begin{proof}[?]

In parts:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  By the lemma and induction on \(i\).
\item
  Clear!
\item
  Follows from
  \(i x.v_i = x.(y.v_{i-1}) = y.(x.v_{i-1}) + [xy].v_{i-1}\) and
  induction on \(i\).
\end{enumerate}

\end{proof}

\begin{remark}

Some useful facts:

\begin{itemize}
\item
  The nonzero \(v_i\) are linearly independent since they are
  eigenvectors of \(h\) with different eigenvalues -- this is a linear
  algebra fact.
\item
  The subspace of \(V\) spanned by the nonzero \(v_i\) is an
  \(L{\hbox{-}}\)submodule, but since \(V\) is irreducible the \(v_i\)
  must form a basis for \(V\).
\item
  Since \(V\) is finite-dimensional, there must be a smallest
  \(m\geq 0\) such that \(v_m \neq 0\) but \(v_{m+1} = 0\), and thus
  \(v_{m+k} = 0\) for all \(k\). Thus \(\dim_{\mathbf{C}}V = m+1\) with
  basis \(\left\{{v_0, v_1, \cdots, v_m}\right\}\).
\item
  Since \(v_{m+1} = 0\), we have \(0 = x.v_{m+1} = ( \lambda- m) v_m\)
  where \(v_m\neq 0\), so \(\lambda = m \in {\mathbf{Z}}_{\geq 0}\).
  Thus its highest weight is a non-negative integer, equal to 1 less
  than the dimension. We'll reserve \(\lambda\) to denote a highest
  weight and \(\mu\) an arbitrary weight.
\item
  Thus the weights of \(V\) are
  \(\left\{{m, m-2, \cdots, \star, \cdots, -m+2, -m}\right\}\) where
  \(\star = 0\) or \(1\) depending on if \(m\) is even or odd
  respectively, each occurring with multiplicity one (using that
  \(\dim V_{\mu} = 1\) if \(\mu\) is a weight of \(V\)).
\end{itemize}

\end{remark}

\begin{theorem}[?]

Let
\(V \in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fd}}, {\mathrm{irr}}}\)
for \(L\coloneqq{\mathfrak{sl}}_2({\mathbf{C}})\), then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Relative to \(h\), \(V\) is a direct sum of weight spaces \(V_\mu\)
  for \(\mu \in \left\{{m, m-2,\cdots, -m+2, -m}\right\}\) where
  \(m+1=\dim V\) and each weight space is 1-dimensional.
\item
  \(V\) has a unique (up to nonzero scalar multiples) highest weight
  vector whose weight (the highest weight of \(V\)) is
  \(m\in {\mathbf{Z}}_{\geq 0}\)
\item
  The action \(L\curvearrowright V\) is given explicitly as in the lemma
  if the basis is chosen in a prescribed fashion. In particular, there
  exists a unique finite-dimensional irreducible
  \({\mathfrak{sl}}_2{\hbox{-}}\)module of dimension \(m+1\) up to
  isomorphism.
\end{enumerate}

\end{theorem}

\begin{corollary}[?]

Let
\(V \in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fd}}, {\mathrm{irr}}}\),
then the eigenvalues of \(h\curvearrowright V\) are all integers, and
each occurs along with its negative with the same multiplicity.
Moreover, in a decomposition of \(V\) in a direct sum of irreducibles,
the number of simple summands is \(\dim V_0 + \dim V_1\).

\end{corollary}

\begin{remark}

Existence of irreducible highest weight modules of highest weight
\(m \geq 0\):

\begin{itemize}
\tightlist
\item
  \(m=0\): take the trivial representation \(V={\mathbf{C}}\).
\item
  \(m=1\): take \(V = {\mathbf{C}}^2\) with the natural representation.
\item
  \(m=2\): take \(V = L\) with the adjoint representation.
\end{itemize}

\end{remark}

\begin{remark}

The formula in the lemma can be used to construct an irreducible
representation of \(L\) having highest weight \(\lambda= m\) for any
\(m\in {\mathbf{Z}}_{\geq 0}\), which is unique up to isomorphism and
denoted \(L(m)\) (or \(V(m)\) in Humphreys) which has dimension \(m+1\).
In fact, the formulas can be used to define an infinite-dimensional
representation of \(L\) with highest weight \(\lambda\) for any
\(\lambda\in {\mathbf{C}}\), which is denoted \(M( \lambda)\) -- we just
don't decree that \(v_{m+1} = 0\), yielding a basis
\(\left\{{v_0, v_1,\cdots}\right\}\). This yields a decomposition into
1-dimensional weight spaces
\(M( \lambda) = \oplus _{i=1}^\infty M_{ \lambda-2i}\) (\textbf{Verma
modules}) where
\(M_{ \lambda-2i} = \left\langle{v_i}\right\rangle_{\mathbf{C}}\).

\end{remark}

\hypertarget{ch.-8-root-space-decompositions-friday-september-23}{%
\section{Ch. 8: Root space decompositions (Friday, September
23)}\label{ch.-8-root-space-decompositions-friday-september-23}}

\begin{remark}

Recall that the relations from last time can produce an
infinite-dimensional module with basis
\(\left\{{v_0,v_1,\cdots}\right\}\). Note that if
\(m\in {\mathbf{Z}}_{\geq 0}\), then
\(x.v_{m+1} = ( \lambda- m ) v_m = 0\). This says that one can't raise
\(v_{m+1}\) back to \(v_m\), so
\(\left\{{v_{m+1},v_{m+2} \cdots}\right\}\) spans a submodule isomorphic
to \(M(-m-2)\). Quotienting yields \(L(m) \coloneqq M(m) / M(-m-2)\),
also called \(V(m)\), spanned by \(\left\{{v_0, \cdots, v_m}\right\}\).
Note that \(M(-m-2)\) and \(L(m)\) are irreducible.

\end{remark}

\begin{remark}

Let
\(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{\mathbf{C}}}}^{{\mathrm{fd}}, {\mathrm{ss}}}\)
for this chapter.

\end{remark}

\hypertarget{maximal-toral-subalgebras-and-roots}{%
\subsection{8.1: Maximal toral subalgebras and
roots}\label{maximal-toral-subalgebras-and-roots}}

\begin{remark}

Let \(L\ni x = x_s + x_n \in L + L\) be the abstract Jordan
decomposition. then if \(x=x_n\) for every \(x\in L\) then \(L\) is
nilpotent which contradicts Engel's theorem. Thus there exists some
\(x=x_s\neq 0\).

\end{remark}

\begin{definition}[Toral subalgebras]

A \textbf{toral subalgebra} is any nonzero subalgebra spanned by
semisimple elements.

\end{definition}

\begin{example}[?]

The algebraic torus \(({\mathbf{C}}^{\times})^n\) which has Lie algebra
\({\mathbf{C}}^n\), thought of as diagonal matrices.

\end{example}

\begin{lemma}[?]

Let \(H\) be a maximal toral subalgebra of \(L\). Any toral subalgebra
\(H \subseteq L\) is abelian.

\end{lemma}

\begin{proof}[?]

Let \(T \leq L\) be toral and let \(x\in T\) be a basis element. Since
\(x\) is semisimple, it STS \({ \operatorname{ad}}_{T, x} = 0\).
Semisimplicity of \(x\) implies \({ \operatorname{ad}}_{L, x}\)
diagonalizable, so we want to show \({ \operatorname{ad}}_{T, x}\) has
no non-zero eigenvalues. Suppose that there exists a nonzero \(y\in T\)
such that \({ \operatorname{ad}}_{T, x}(y) = \lambda y\) for
\(\lambda \neq 0\). Then
\({ \operatorname{ad}}_{T, y}(x) = [yx] = -[xy] = - { \operatorname{ad}}_{T, x} = - \lambda y\neq 0\),
and since \({ \operatorname{ad}}_{T, y}(y) = -[yy] = 0\), \(y\) is an
eigenvector with eigenvalue zero. Since \({ \operatorname{ad}}_{T, y}\)
is also diagonalizable and \(x\in T\), write \(x\) as a linear
combination of eigenvectors for it, say \(x = \sum a_i v_i\). Then
\({ \operatorname{ad}}_{T, y}(x) = \sum \lambda_i a_i v_i\) and the
terms with \(\lambda_i = 0\) vanish, and the remaining element is a sum
of eigenvectors for \({ \operatorname{ad}}_{T, y}\) with nonzero
eigenvalues. \(\contradiction\)

\end{proof}

\begin{example}[?]

If \(L = {\mathfrak{sl}}_n({\mathbf{C}})\) then define \(H\) to be the
set of diagonal matrices. Then \(H\) is toral and in fact maximal: if
\(L = H \oplus {\mathbf{C}}z\) for some \(z\in L\setminus H\) then one
can find an \(h\in H\) such that \([hz] \neq 0\), making it nonabelian,
but toral subalgebras must be abelian.

\end{example}

\begin{definition}[Roots]

Recall that a commuting family of diagonalizable operators on a
finite-dimensional vector space can be simultaneously diagonalized.
Letting \(H \leq L\) be maximal toral, this applies to
\({ \operatorname{ad}}_L(H)\), and thus there is a basis in which all
operators in \({ \operatorname{ad}}_L(H)\) are diagonal. Set
\(L_{ \alpha} \coloneqq\left\{{x\in L {~\mathrel{\Big\vert}~}[hx] = \alpha(h) x\,\,\forall h\in H}\right\}\)
where \(\alpha: H\to {\mathbf{C}}\) is linear and thus an element of
\(H {}^{ \vee }\). Note that \(L_0 = C_L(H)\), and the set
\(\Phi\coloneqq\left\{{\alpha\in H {}^{ \vee }{~\mathrel{\Big\vert}~}\alpha\neq 0, L_\alpha\neq 0}\right\}\)
is called the \textbf{roots} of \(H\) in \(L\), and \(L_\alpha\) is
called a \textbf{root space}. Note that \(L_0\) is not considered a root
space. This induces a \textbf{root space decomposition}
\begin{align*}
L = C_L(H) \oplus _{ \alpha\in \Phi } L_\alpha
.\end{align*}

\end{definition}

\begin{remark}

Note that for classical algebras, we'll show \(C_L(H) = H\) and
corresponds to the standard bases given early in the book.

\end{remark}

\begin{example}[?]

Type \(A_n\) yields \({\mathfrak{sl}}_{n+1}({\mathbf{C}})\) and
\(\dim H = n\) for \(H\) defined to be the diagonal traceless matrices.
Define \({\varepsilon}_i\in H {}^{ \vee }\) as
\({\varepsilon}_i \operatorname{diag}(a_1,\cdots, a_{n+1}) \coloneqq a_i\),
the
\(\Phi \coloneqq\left\{{{\varepsilon}_i - {\varepsilon}_j {~\mathrel{\Big\vert}~}1\leq i\neq j\leq ,n+1}\right\}\)
and \(L_{{\varepsilon}_i - {\varepsilon}_j} = {\mathbf{C}}e_{ij}\). Why:
\begin{align*}
[h, e_{ij}] =
\left[ \sum a_k e_{kk}, e_{ij}\right] = a_i e_{ii} e_{ij} - a_j e_{ij} e_{jj} = (a_i - a_j) e_{ij} \coloneqq({\varepsilon}_i -{\varepsilon}_j)(h) e_{ij}
.\end{align*}

\end{example}

\hypertarget{monday-september-26}{%
\section{Monday, September 26}\label{monday-september-26}}

\hypertarget{continued}{%
\subsection{8.1 Continued}\label{continued}}

\begin{remark}

A Lie algebra is \textbf{toral} iff every element is semisimple -- this
exists because any semisimple Lie algebra contains at least one
semisimple element and you can take its span. Let \(H\) be a fixed
maximal toral subalgebra, then we have a root space decomposition
\begin{align*}
L = C_L(H) \oplus \bigoplus _{\alpha\in \Phi \subseteq H {}^{ \vee }} L_\alpha, \qquad L_\alpha \coloneqq\left\{{ x\in L {~\mathrel{\Big\vert}~}[hx] = \alpha(h) x \,\forall h\in H}\right\}
.\end{align*}
Let \(L\) be semisimple and finite dimensional over \({\mathbf{C}}\)
from now on.

\end{remark}

\begin{proposition}[?]

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item

  \begin{align*}
  [L_\alpha, L_\beta] \subseteq L_{ \alpha + \beta} \qquad \forall \alpha, \beta\in H {}^{ \vee }
  .\end{align*}
\item

  \begin{align*}
  x\in L_{\alpha}, \alpha\neq 0 \implies  { \operatorname{ad}}_x \text{ is nilpotent}
  .\end{align*}
\item
  If \(\alpha, \beta\in H {}^{ \vee }\) and \(\alpha + \beta\neq 0\)
  then \(L_ \alpha \perp L_ \beta\) relative to \(\kappa_L\), the
  Killing form for \(L\).
\end{enumerate}

\end{proposition}

\begin{proof}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Follows from the Jacobi identity.
\item
  Follows from (1), that \(\dim L < \infty\), and the root space
  decomposition. This is because
  \(L_{\alpha}\neq L_{\alpha + \beta}\neq L_{\alpha + 2\beta} \neq \cdots\)
  and there are only finitely many \(\beta\) to consider.
\item
  If \(\alpha + \beta\neq 0\) then \(\exists h\in H\) such that
  \((\alpha + \beta)(h) \neq 0\). For \(x\in L_\alpha, y\in L_ \beta\),
  \begin{align*}
    \alpha(h) \kappa(x,y) 
    &= \kappa([hx], y) \\
    &= -\kappa([xh], y) \\
    &= -\kappa(x, [hy]) \\
    &= -\beta(h)\kappa(x,y) \\
    &\implies ( \alpha + \beta)(h) \kappa(x,y) = 0 \\
    &\implies \kappa(x,y)=0 \text{ since } (\alpha + \beta)(h) \neq 0
    .\end{align*}
\end{enumerate}

\end{proof}

\begin{corollary}[?]

\({ \left.{{\kappa}} \right|_{{L_0}} }\) is nondegenerate, since
\(L_0 \perp L_{\alpha}\) for all \(\alpha\in \Phi\), but \(\kappa\) is
nondegenerate. Moreover, if \(L_{ \alpha} \neq 0\) then
\(L_{- \alpha}\neq 0\) by (3) and nondegeneracy.

\end{corollary}

\hypertarget{c_lh}{%
\subsection{\texorpdfstring{8.2: \(C_L(H)\)}{8.2: C\_L(H)}}\label{c_lh}}

\begin{proposition}[?]

Let \(H \leq L\) be maximal toral, then \(H = C_L(H)\).

\end{proposition}

\begin{proof}[?]

Skipped, about 1 page of dense text broken into 7 steps. Uses the last
corollary along with Engel's theorem.

\end{proof}

\begin{observation}

If \(L\) is a classical Lie algebra over \({\mathbf{C}}\), we choose
\(H\) to be diagonal matrices in \(L\), and \(x\in L\setminus H\) is
non-diagonal, then there exists an \(h\in H\) such that \([hx]\neq 0\).
Note that toral implies abelian and nonabelian implies nontoral, thus
there is no abelian subalgebra of \(L\) properly containing \(H\) --
adding any nontoral element at all to \(H\) makes it nonabelian. This
same argument shows \(C_L(H) = H\) since nothing else commutes with
\(H\). This implies that \(L = H \oplus_{ \alpha \in \Phi} L_\alpha\).

\end{observation}

\begin{corollary}[?]

\({ \left.{{\kappa}} \right|_{{H}} }\) is nondegenerate.

\end{corollary}

\begin{remark}

As a result, \(\kappa\) induces an isomorphism
\(H { \, \xrightarrow{\sim}\, }H {}^{ \vee }\) by
\(h\mapsto \kappa(h, {-})\) and
\(H {}^{ \vee } { \, \xrightarrow{\sim}\, }H\) by
\(\phi\mapsto t_\phi\), the unique element such that
\(\kappa(t_\phi, {-}) = \phi({-})\). In particular, given
\(\alpha\in \Phi \subset H {}^{ \vee }\) there is some
\(t_\alpha\in H\). The next 3 sections are about properties of \(\Phi\):

\begin{itemize}
\tightlist
\item
  Orthogonality,
\item
  Integrality,
\item
  Rationality.
\end{itemize}

\end{remark}

\hypertarget{orthogonality-properties-of-phi.}{%
\subsection{\texorpdfstring{8.3: Orthogonality properties of
\(\Phi\).}{8.3: Orthogonality properties of \textbackslash Phi.}}\label{orthogonality-properties-of-phi.}}

\begin{proposition}[Big!]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\Phi\) spans \(H {}^{ \vee }\).
\item
  If \(\alpha\in \Phi\) is a root then \(-\alpha\in \Phi\) is also a
  root.
\item
  Let \(\alpha\in \Phi, x\in L_{ \alpha}, y\in L_{- \alpha}\), then
  \([xy] = \kappa(x, y) t_\alpha\).
\item
  If \(\alpha\in \Phi\) then
  \([L_{ \alpha}, L_{- \alpha}] = {\mathbf{C}}t_\alpha\) is
  1-dimensional with basis \(t_\alpha\).
\item
  For any \(\alpha \in \Phi\), we have
  \(\alpha(t_\alpha) = \kappa(t_ \alpha, t_ \alpha) \neq 0\).
\item
  (Important) If
  \(\alpha\in \Phi, x_ \alpha\in L_{ \alpha}\setminus\left\{{0}\right\}\)
  then there exists some \(y_ \alpha L_{- \alpha}\) in the opposite root
  space such that
  \(x_ \alpha, y_ \alpha, h_ \alpha \coloneqq[x_ \alpha, y_ \alpha]\)
  span a 3-dimensional subalgebra \({\mathfrak{sl}}(\alpha) \leq L\)
  isomorphic to \({\mathfrak{sl}}_2({\mathbf{C}})\).
\item
  \(h_\alpha = {2 t_\alpha \over \kappa(t_ \alpha, t_ \alpha)}\),
  \(\alpha(h_ \alpha) = 2, h_{ \alpha} = h_{- \alpha}\).
\end{enumerate}

\end{proposition}

\begin{proof}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  If it does not span, choose \(h \in H\setminus\left\{{0}\right\}\)
  with \(\alpha(h) = 0\) for all \(\alpha\in \Phi\). Then
  \([h, L_ \alpha] = 0\) for all \(\alpha\), but \([h H] = 0\) since
  \(H\) is abelian. Using the root space decomposition, \([h L] =0\) and
  so \(h\in Z(L) = 0\) since \(L\) is semisimple. \(\contradiction\)
\item
  Follows from proposition 8.2 and \(\kappa(L_ \alpha, L_ \beta) = 0\)
  when \(\beta\neq \alpha\).
\item
  Let \(h\in H\), then
  \begin{align*}
  \kappa(h, [xy]) 
  &= \kappa([hx], y) \\
  &= \alpha(h) \kappa(x, y)\\
  &= \kappa(t_\alpha, h) \kappa(x, y) \\
  &= \kappa( \kappa(x, y) t_ \alpha, h) \\
  &= \kappa( h, \kappa(x, y) t_ \alpha) \\
  &\implies \kappa(h, [xy] - \kappa(x,y)t_\alpha) = 0 \\
  &\implies [xy] = \kappa (x,y) t_ \alpha
  ,\end{align*}
  where we've used that \([xy]\in H\) and \(\kappa\) is nondegenerate on
  \(H\) and \([L_{ \alpha}, L_{ - \alpha}] \subseteq L_0 = H\).
\item
  \begin{enumerate}
  \def\labelenumii{(\alph{enumii})}
  \setcounter{enumii}{2}
  \tightlist
  \item
    shows \([L_{ \alpha}, L_{ - \alpha}]\) is spanned by \(t_ \alpha\)
    if it is nonzero. Let \(x\in L_ \alpha\setminus\left\{{0}\right\}\),
    then if \(\kappa(x, L_{ - \alpha}) = 0\) then \(\kappa\) would have
    to be degenerate, a contradiction. So there is some
    \(y\in L_{ - \alpha}\) with \(\kappa(x, y) \neq 0\). Moreover
    \(t_\alpha\neq 0\) since \(\alpha\neq 0\) and
    \(\alpha\mapsto t_\alpha\) is an isomorphism. Thus
    \([xy] = \kappa(x,y) t_ \alpha\).
  \end{enumerate}
\item
  Suppose \(\alpha(t_\alpha) = \kappa(t_{ \alpha}, t_{ \alpha}) = 0\),
  then for \(x\in L_{\alpha}, y\in L_{ - \alpha}\), we have
  \([t_ \alpha, x] = \alpha(t_ \alpha)x = 0\) and similarly
  \([t_ \alpha, y] = 0\). As before, find
  \(x\in L_{ \alpha}, y\in L_{ - \alpha}\) with \(\kappa(x,y)\neq 0\)
  and scale one so that \(\kappa(x, y) = 1\). Then by (c),
  \([x, y] = t_ \alpha\), so combining this with the previous formula
  yields that \(S \coloneqq\left\langle{x, y, t_ \alpha}\right\rangle\)
  is a 3-dimensional solvable subalgebra.\footnote{Note that this don't
    actually exist! We're in the middle of a contradiction.} Taking
  \({ \operatorname{ad}}: L\hookrightarrow{\mathfrak{gl}}(L)\), which is
  injective by semisimplicity, and similarly
  \({ \left.{{ { \operatorname{ad}}}} \right|_{{S}} }: S\hookrightarrow { \operatorname{ad}}(S) \leq {\mathfrak{gl}}(L)\).
  We'll use Lie's theorem to show everything here is a commutator of
  upper triangular, thus strictly upper triangular, thus nilpotent and
  reach a contradiction.
\end{enumerate}

\end{proof}

\hypertarget{wednesday-september-28}{%
\section{Wednesday, September 28}\label{wednesday-september-28}}

\hypertarget{continued-proof}{%
\subsection{Continued proof}\label{continued-proof}}

Recall the proposition from last time:

\begin{proposition}[Big!]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\Phi\) spans \(H {}^{ \vee }\).
\item
  If \(\alpha\in \Phi\) is a root then \(-\alpha\in \Phi\) is also a
  root.
\item
  Let \(\alpha\in \Phi, x\in L_{ \alpha}, y\in L_{- \alpha}\), then
  \([xy] = \kappa(x, y) t_\alpha\).
\item
  If \(\alpha\in \Phi\) then
  \([L_{ \alpha}, L_{- \alpha}] = {\mathbf{C}}t_\alpha\) is
  1-dimensional with basis \(t_\alpha\).
\item
  For any \(\alpha \in \Phi\), we have
  \(\alpha(t_\alpha) = \kappa(t_ \alpha, t_ \alpha) \neq 0\).
\item
  (Important) If
  \(\alpha\in \Phi, x_ \alpha\in L_{ \alpha}\setminus\left\{{0}\right\}\)
  then there exists some \(y_ \alpha L_{- \alpha}\) in the opposite root
  space such that
  \(x_ \alpha, y_ \alpha, h_ \alpha \coloneqq[x_ \alpha, y_ \alpha]\)
  span a 3-dimensional subalgebra \({\mathfrak{sl}}(\alpha) \leq L\)
  isomorphic to \({\mathfrak{sl}}_2({\mathbf{C}})\).
\item
  \(h_\alpha = {2 t_\alpha \over \kappa(t_ \alpha, t_ \alpha)}\),
  \(\alpha(h_ \alpha) = 2, h_{ \alpha} = h_{- \alpha}\).
\end{enumerate}

\end{proposition}

\begin{proof}[continued]

\textbf{Part e}: We have
\(\alpha(t_ \alpha ) = \kappa(t_ \alpha, t_ \alpha)\), so suppose this
is zero. Pick \(x\in L_{ \alpha}, y\in L_{ - \alpha}\) such that
\(\kappa(x, y) = 1\), then

\begin{itemize}
\tightlist
\item
  \([t_ \alpha, x] = 0\),
\item
  \([t_ \alpha, y] = 0\),
\item
  \([x, y] = t_ \alpha\).
\end{itemize}

Set
\(S \coloneqq{\mathfrak{sl}}( \alpha) \coloneqq{\mathbf{C}}\left\langle{x,y,t_ \alpha}\right\rangle\)
and restrict
\({ \operatorname{ad}}: L\hookrightarrow{\mathfrak{gl}}(L)\) to \(S\).
Then \({ \operatorname{ad}}(S) \cong S\) by injectivity, and this is a
solvable linear subalgebra of \({\mathfrak{gl}}(L)\). Apply Lie's
theorem to choose a basis for \(L\) such that the matrices for
\({ \operatorname{ad}}(S)\) are upper triangular. Then use that
\({ \operatorname{ad}}_L([SS]) = [ { \operatorname{ad}}_L(S) { \operatorname{ad}}_L(S)]\),
which is strictly upper triangular and thus nilpotent. In particular,
\({ \operatorname{ad}}_L (t_ \alpha)\) is nilpotent, but since
\(t_\alpha\in H\) which is semisimple, so
\({ \operatorname{ad}}_L( t_ \alpha)\) is semisimple. The only things
that are semisimple \emph{and} nilpotent are zero, so
\({ \operatorname{ad}}_L( t_ \alpha) = 0 \implies t_\alpha = 0\). This
contradicts that \(\alpha\in H {}^{ \vee }\setminus\left\{{0}\right\}\).
\(\contradiction\)

\textbf{Part f}: Given
\(x_ \alpha\in L_ \alpha\setminus\left\{{0}\right\}\), choose
\(y_ \alpha \in L_{ - \alpha}\) and rescale it so that
\begin{align*}
\kappa(x_ \alpha, y _{\alpha} ) = {2\over \kappa(t_ \alpha, t_ \alpha)}
.\end{align*}
Set
\(h_ \alpha \coloneqq{2t_ \alpha\over \kappa(t_ \alpha, t_ \alpha) }\),
then by (c),
\([x_ \alpha, y_ \alpha] = \kappa( x_ \alpha, y_ \alpha) t_ \alpha = h_ \alpha\).
So
\begin{align*}
[ h_ \alpha, x_ \alpha] = {2\over \alpha(t_ \alpha) }[ t_ \alpha, x_ \alpha] = {2\over \alpha(t_ \alpha)} \alpha(t_ \alpha) x_ \alpha = 2x_ \alpha
,\end{align*}
and similarly \([h_ \alpha, y_ \alpha] = -2 y_ \alpha\). Now the span
\(\left\langle{x_ \alpha, h_ \alpha, y_ \alpha}\right\rangle \leq L\) is
a subalgebra with the same multiplication table as
\({\mathfrak{sl}}_2({\mathbf{C}})\), so
\(S \cong {\mathfrak{sl}}_2({\mathbf{C}})\).

\textbf{Part g}: Note that we would have
\(h_{ - \alpha} = {2 t_{ - \alpha} \over \kappa( t_ { - \alpha}, t_{- \alpha} ) } = - h_ \alpha\)
if \(t_{ \alpha} = t_{ - \alpha}\). This follows from the fact that
\(H {}^{ \vee } { \, \xrightarrow{\sim}\, }H\) sends
\(\alpha\mapsto t_\alpha, -\alpha\mapsto t_{- \alpha}\), but by
linearity \(-\alpha\mapsto -t_{ \alpha}\).

\end{proof}

\begin{corollary}[?]

\(L\) is generated as a Lie algebra by the root spaces
\(\left\{{L_ \alpha{~\mathrel{\Big\vert}~}\alpha\in \Phi}\right\}\).

\end{corollary}

\begin{proof}[?]

It STS
\(H \subseteq \left\langle{\left\{{L_\alpha}\right\}_{\alpha\in \Phi}}\right\rangle\).
Given \(\alpha\in \Phi\),
\begin{align*}
\exists x_\alpha\in L_ \alpha, y_ \alpha\in L_{- \alpha} \quad\text{such that}\quad
\left\langle{x_ \alpha, y_ \alpha, h_ \alpha\coloneqq[x_ \alpha, y_ \alpha] }\right\rangle \cong {\mathfrak{sl}}_2({\mathbf{C}})
.\end{align*}
Note any \(h_\alpha\in {\mathbf{C}}^{\times}t_ \alpha\) corresponds via
\(\kappa\) to some \(\alpha\in H {}^{ \vee }\). By (a), \(\Phi\) spans
\(H {}^{ \vee }\), so \(\left\{{t_\alpha}\right\}_{\alpha\in \Phi}\)
spans \(H\).

\end{proof}

\hypertarget{integrality-properties-of-phi}{%
\subsection{\texorpdfstring{8.4: Integrality properties of
\(\Phi\)}{8.4: Integrality properties of \textbackslash Phi}}\label{integrality-properties-of-phi}}

\begin{remark}

Any \(\alpha\in \Phi\) yields
\({\mathfrak{sl}}(\alpha) \cong {\mathfrak{sl}}(\alpha)\), and in fact
that the generators entirely determined by the choice of \(x_\alpha\).
View \(L\in {}_{{\mathfrak{sl}}(\alpha)}{\mathsf{Mod}}\) via
\({ \operatorname{ad}}\).

\end{remark}

\begin{lemma}[?]

If \(M \leq L \in {}_{{\mathfrak{sl}}(\alpha)}{\mathsf{Mod}}\) then all
eigenvalues of \(h_\alpha\curvearrowright M\) are integers.

\end{lemma}

\begin{proof}[?]

Apply Weyl's theorem to decompose \(M\) into a finite direct sum of
irreducibles in
\({}_{ {\mathfrak{sl}}_2({\mathbf{C}}) }{\mathsf{Mod}}\). The weights of
\(h_\alpha\) are of the form
\(m, m-2,\cdots, -m+2, -m\in {\mathbf{Z}}\).\footnote{This fails for
  infinite dimensional modules, e.g.~Verma modules. The highest weight
  can be any complex number.}

\end{proof}

\begin{example}[?]

Let
\(M = H + {\mathfrak{sl}}(\alpha) \leq L \in {}_{{\mathfrak{sl}}( \alpha)}{\mathsf{Mod}}\),
which one can check is actually a submodule since bracketing either
lands in \({\mathfrak{sl}}(\alpha)\) or kills elements. What does Weyl's
theorem say about this submodule? There is some intersection. Set
\(K \coloneqq\ker\alpha \subseteq H\), so
\(\operatorname{codim}_H K = 1\) by rank-nullity. Note that
\(h_\alpha \not\in K\), so \(M = K \oplus {\mathfrak{sl}}(\alpha)\).
Moreover \({\mathfrak{sl}}(\alpha)\curvearrowright K\) by zero, since
bracketing acts by \(\alpha\) which vanishes on \(K\). So
\(K\cong {\mathbf{C}}{ {}^{ \scriptscriptstyle\oplus^{n+1} } }\)
decomposes into trivial modules.

\end{example}

\begin{example}[?]

Let \(\beta\in \Phi\cup\left\{{0}\right\}\) and define
\(M \coloneqq\bigoplus _{c\in {\mathbf{C}}} L_{\beta + c\alpha}\), then
\(L \leq L \in {}_{{\mathfrak{sl}}( \alpha)}{\mathsf{Mod}}\). It will
turn out that
\(L_{\beta+ c \alpha} \neq 0 \iff c\in [-r, q] \subseteq {\mathbf{Z}}\)
with \(r, q\in {\mathbf{Z}}_{\geq 0}\).

\end{example}

\begin{proposition}[A]

Let \(\alpha\in \Phi\), then the root spaces
\(\dim L_{\pm \alpha} = 1\), and the only multiples of \(\alpha\) which
are in \(\Phi\) are \(\pm \alpha\).

\end{proposition}

\begin{proof}[?]

Note \(L_\alpha\) can only pair with \(L_{- \alpha}\) to give a
nondegenerate Killing form. Set
\begin{align*}
M \coloneqq\bigoplus _{c\in {\mathbf{C}}} L_{c \alpha} = H \oplus \bigoplus _{c \alpha\in \Phi} L_{c \alpha}
.\end{align*}
By Weyl's theorem, this decomposes into irreducibles. This allows us to
take a complement of the decomposition from before to write
\(M = H \bigoplus {\mathfrak{sl}}(\alpha) \oplus W\), and we WTS
\(W = 0\) since this contains all \(L_{c \alpha}\) where
\(c\neq \pm 1\). Since
\(H \subseteq K \oplus {\mathfrak{sl}}( \alpha)\), we have
\(W \cap H = 0\). If \(c_ \alpha\) is a root of \(L\), then
\(h_ \alpha\) has \((c \alpha)(h_ \alpha) = 2c\) as an eigenvalue, which
must be an integer by a previous lemma. So \(c\in {\mathbf{Z}}\) or
\(c\in {\mathbf{Z}}+ {1\over 2}\).

Suppose \(W\neq 0\), and let \(V(s)\) (or \(L(s)\) in modern notation)
be an irreducible \({\mathfrak{sl}}( \alpha){\hbox{-}}\)submodule of
\(W\) for \(s\in {\mathbf{Z}}_{\geq 0}\). If \(s\) is even, \(V(s)\)
contains an eigenvector \(w\) for \(h_ \alpha\) of eigenvalue zero by
applying \(y_\alpha\) \(s/2\) times. We can then write
\(w = \sum_{c\in {\mathbf{C}}} v_{c \alpha}\) with
\(v_{ ca } \in L_{ c \alpha}\), and by finiteness of direct sums we have
\(v_{c \alpha} = 0\) for almost every \(c\in {\mathbf{C}}\). Then
\begin{align*}
0
&= [h_ \alpha, w] \\
&= \sum_{c\in {\mathbf{C}}} [h_ \alpha, v_{ c \alpha} ] \\
&= \sum_{c\in {\mathbf{C}}} (c\alpha)( h _{\alpha} )v_{c \alpha} \\
&= \sum 2c v_{c \alpha} \\
&\implies v_{c \alpha} = 0 \text{ when } c\neq 0
,\end{align*}
forcing \(w\in H\), the zero eigenspace. But \(w\in W\), so
\(w\in W \cap H = 0\). \(\contradiction\)

\end{proof}

\hypertarget{friday-september-30}{%
\section{Friday, September 30}\label{friday-september-30}}

\hypertarget{section}{%
\subsection{8.4}\label{section}}

\begin{proposition}[A]

\(\alpha\in \Phi\implies \dim L_{\pm \alpha} = 1\), and
\(\alpha, \lambda \alpha\in \Phi\implies \lambda = \pm 1\).

\end{proposition}

\begin{proof}[of proposition A]

Consider
\(M \coloneqq\bigoplus \bigoplus _{ c\in {\mathbf{C}}} L_{c \alpha} \leq L\in {}_{{\mathfrak{sl}}( \alpha)}{\mathsf{Mod}}\).
Write
\({\mathfrak{sl}}(\alpha) = \left\langle{ x_ \alpha, h_ \alpha, y_ \alpha}\right\rangle\),
we decomposed \(M = K \oplus {\mathfrak{sl}}(\alpha) \oplus W\) where
\(\ker \alpha \leq H\) and \(W \cap H = 0\). WTS: \(W = 0\). So far,
we've shown that if \(L(s) \subseteq W\) for
\(s\in {\mathbf{Z}}_{\geq 0}\) (which guarantees finite dimensionality),
then \(s\) can't be even -- otherwise it has a weight zero eigenvector,
forcing it to be in \(H\), but \(W \cap H = 0\).

Aside: \(\alpha\in \Phi \implies 2\alpha\not\in \Phi\), since it would
have weight \((2\alpha)(h_ \alpha) = 2\alpha h_ \alpha = 4\), but
weights in irreducible modules have the same parity as the highest
weight and no such weights exist in \(M\) (only \(0, \pm 2\) in
\(K \oplus {\mathfrak{sl}}_2(\alpha)\) and only odd in \(W\)). Suppose
\(L(s) \subseteq W\) and \(s\geq 1\) is odd. Then \(L(s)\) has a weight
vector for \(h_ \alpha\) of weight 1. This must come from \(c=1/2\),
since \((1/2) \alpha (h_ \alpha) = (1/2) 2 = 1\), so this is in
\(L_{\alpha/2}\). However, by the aside, if \(\alpha\in \Phi\) then
\(\alpha/2\not\in\Phi\).

Thus it \(W\) can't contain any odd roots or even roots, so \(W = 0\).
Note also that \(L_{\pm\alpha}\not\subset K \oplus W\), forcing it to be
in \({\mathfrak{sl}}( \alpha)\), so
\(L_{ \alpha} = \left\langle{x_ \alpha}\right\rangle\) and
\(L_{- \alpha} = \left\langle{y _{\alpha} }\right\rangle\).

\end{proof}

\begin{proposition}[B]

Let \(\alpha, \beta\in \Phi\) with \(\beta\neq \pm \alpha\) and consider
\(\beta + k \alpha\) for \(n\in {\mathbf{Z}}\).

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\beta(h_ \alpha) \in {\mathbf{Z}}\).
\item
  \(\exists r,q\in {\mathbf{Z}}_{\geq 0}\) such that for
  \(k\in {\mathbf{Z}}\), the combination
  \(\beta + k \alpha\in \Phi \iff k \in [-r, q] \ni 0\). The set
  \(\left\{{ \beta + k \alpha {~\mathrel{\Big\vert}~}k\in [-r, q]}\right\} \subseteq \Phi\)
  is the \textbf{\(\alpha{\hbox{-}}\)root string through \(\beta\)}.
\item
  If \(\alpha + \beta\in \Phi\) then
  \([L_ \alpha L_ \beta ] = L_{ \alpha + \beta}\).
\item
  \(\beta- \beta(h_ \alpha) \alpha\in \Phi\).
\end{enumerate}

\end{proposition}

\begin{proof}[?]

Consider
\begin{align*}
M \coloneqq\bigoplus _{k\in {\mathbf{Z}}} L_{ \beta + k \alpha} \leq L \quad\in  {}_{{\mathfrak{sl}}( \alpha)}{\mathsf{Mod}}
.\end{align*}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(\beta(h _{\alpha} )\) is the eigenvalues of \(h_ \alpha\) acting on
  \(L_ \beta\). But by the lemma, \(\beta(h_ \alpha)\in {\mathbf{Z}}\).
\item
  By the previous proposition, \(\dim L_{ \beta+ k \alpha} = 1\) if
  nonzero, and the weight of \(h_\alpha\) acting on it is
  \(\beta( h _{\alpha} ) + 2k\) all different for distinct \(k\). By
  \({\mathfrak{sl}}_2{\hbox{-}}\)representation theory, we know the
  multiplicities of various weight spaces as the sum of dimensions of
  the zero and one weight spaces, and thus \(M\) is a single irreducible
  \({\mathfrak{sl}}(\alpha){\hbox{-}}\)module. So write \(M - L(d)\) for
  some \(d\in {\mathbf{Z}}_{\geq 0}\), then
  \(h_ \alpha\curvearrowright M\) with eigenvalues
  \(\left\{{d,d-2,\cdots, -d+2, -d}\right\}\). But
  \(h_ \alpha\curvearrowright M\) with eigenvalues
  \(\beta( h_ \alpha) + 2k\) for those \(k\in {\mathbf{Z}}\) with
  \(L_{\beta + k \alpha}\neq 0\). Since the first list is an unbroken
  string of integers of the same parity, thus the \(k\) that appear must
  also be an unbroken string. Define \(r\) and \(q\) by setting
  \(d = \beta(h_\alpha) + 2q\) and \(-d =\beta( h_ \alpha ) - 2r\) to
  obtain \([-r, q]\). Adding these yields
  \(0 = 2\beta( h_ \alpha) + 2q-2r\) and \(r-q = \beta(h_ \alpha)\).
\item
  Let \(M\cong L(d) \in {}_{{\mathfrak{sl}}(\alpha)}{\mathsf{Mod}}\) and
  \(x_\beta \in L_ \beta\setminus\left\{{0}\right\}\subseteq M\) with
  \(x_ \alpha\in L_{ \alpha}\). If \([x_ \alpha x_ \beta] = 0\) then
  \(x_ \beta\) is a maximal \({\mathfrak{sl}}(\alpha){\hbox{-}}\)vector
  in \(L(d)\) and thus \(d = \beta(h_ \alpha)\). But
  \(\alpha + \beta\in \Phi \implies \beta)(h_ \alpha) + 2\) is a weight
  in \(M\) bigger than \(d\), a contradiction. Thus
  \(\alpha + \beta\in \Phi \implies [x_ \alpha x_ \beta] \neq 0\). Since
  this bracket spans and \(\dim L_{ \alpha + \beta} = 1\), so
  \([L_ \alpha L_ \beta] = L_{ \alpha + \beta}\).
\item
  Use that \(q\geq 0, r\geq 0\) to write \(-r \leq -r+q \leq q\). Then
  \begin{align*}
    \beta - \beta(h_ \alpha) \alpha = \beta - (r-q) \alpha = \beta + (-r+q \alpha)\coloneqq\beta + \ell\alpha
    \end{align*}
  where \(\ell\in [-r, q]\). Thus \(\beta + \ell\alpha\in \Phi\) is an
  unbroken string by (b).
\end{enumerate}

\end{proof}

\begin{question}

Is it true that
\(\bigoplus_{k\in {\mathbf{Z}}} L_{\beta+ k \alpha} = \bigoplus _{c\in {\mathbf{C}}} L_{\beta + c\alpha}\)?
The issue is that \(c\in {\mathbf{Z}}+ {1\over 2}\) is still possible.

\end{question}

\hypertarget{rationality-properties-of-phi}{%
\subsection{\texorpdfstring{8.5: Rationality properties of
\(\Phi\)}{8.5: Rationality properties of \textbackslash Phi}}\label{rationality-properties-of-phi}}

\begin{remark}

Recall that \(\kappa\) restrict to a nondegenerate bilinear form on
\(H\) inducing \(H {}^{ \vee } { \, \xrightarrow{\sim}\, }H\) via
\(\phi\mapsto t_\phi\) where \(\kappa(t_\phi, {-}) = \phi({-})\).
Transfer to a nondegenerate symmetric bilinear form on \(H {}^{ \vee }\)
by \((\lambda, \mu) \coloneqq\kappa(t_\lambda, t_\mu)\). By prop 8.3 we
know \(H {}^{ \vee }\) is spanned by \(\Phi\), so choose a
\({\mathbf{C}}{\hbox{-}}\)basis
\(\left\{{ \alpha_1,\cdots, \alpha_n}\right\} \subseteq \Phi\). Given
\(\beta\in\Phi\), write \(\beta = \sum c_i \alpha_i\) with
\(c_i\in {\mathbf{C}}\).

\end{remark}

\begin{claim}

\(c_i \in {\mathbf{Q}}\) for all \(i\)!

\end{claim}

\hypertarget{monday-october-03}{%
\section{Monday, October 03}\label{monday-october-03}}

\hypertarget{integrality-and-rationality-properties}{%
\subsection{Integrality and Rationality
Properties}\label{integrality-and-rationality-properties}}

\begin{claim}

Setup:

\begin{itemize}
\item
  Decompose \(L = H \oplus \bigoplus _{ \alpha\in \Phi} L_{ \alpha}\)
\item
  Use the isomorphism
  \begin{align*}
  H & { \, \xrightarrow{\sim}\, }H {}^{ \vee }\\
  \varphi &\mapsfrom t_{ \varphi}
  \end{align*}
  to define \((\lambda, \mu) \coloneqq\kappa(t_ \lambda, t_ \mu)\) on
  \(H\).
\item
  Choose a basis
  \(\left\{{ \alpha_i}\right\} \subseteq \Phi \subseteq H {}^{ \vee }\)
\item
  For any \(\beta \in \Phi\), write \(\beta= \sum c_i \alpha_i\) with
  \(c_i\in {\mathbf{C}}\). Then
\end{itemize}

\begin{align*}
c_i\in {\mathbf{Q}}
.\end{align*}

\end{claim}

\begin{proof}[?]

Write \(( \beta, \alpha_j) = \sum c_i (\alpha_i, \alpha_j)\) and m
\begin{align*}
{2 (\beta, \alpha_j) \over (\alpha_j, \alpha_j) }
= \sum c_i {2 (\alpha_i, \alpha_j) \over (\alpha_j, \alpha_j) }
,\end{align*}
where the LHS is in \({\mathbf{Z}}\), as is
\(2( \alpha_i, \alpha_j) \over (\alpha_j, \alpha_j)\). On the other hand
\begin{align*}
{2 (\beta, \alpha_j) \over (\alpha_j, \alpha_j) }
= {2 (t_ \beta, t_{\alpha_j} ) \over \kappa(t_{ \alpha_j}, \kappa_{ \alpha_j} ) }
= \kappa(t_ \beta, h_{\alpha_j} )
= \beta(h_{ \alpha_j})
\end{align*}
using that
\(( \alpha_j, \alpha_j) = \kappa( t_{ \alpha_j}, t_{ \alpha_j} )\neq 0\)
from before.\footnote{More generally,
  \begin{align*}
  {2 (\lambda, \alpha)\over (\alpha, \alpha) } = \lambda(h_ \alpha) \qquad\forall \alpha\in \Phi
  .\end{align*}
} Since \(\left\{{ \alpha_i}\right\}\) is a basis for \(H {}^{ \vee }\)
and \(({-}, {-})\) is nondegenerate, the matrix
\([ ( \alpha_i, \alpha _j) ]_{1\leq i, j\leq n}\) is invertible. Thus so
is
\(\left[ 2 ( \alpha_i, \alpha_j) \over (\alpha_j, \alpha_j ) \right]_{1\leq i,j \leq n}\),
since it's given by multiplying each column as a nonzero scalar, and one
can solve for the \(c_i\) by inverting it. This involves denominators
coming from the determinant, which is an integer, yielding entries in
\({\mathbf{Q}}\).

\end{proof}

\begin{remark}

Given \(\lambda, \mu \in H {}^{ \vee }\) then
\begin{align*}
(\lambda, \mu) = \kappa(t_ \lambda, t_\mu) = \operatorname{Trace}( { \operatorname{ad}}_{t_ \lambda} \circ  { \operatorname{ad}}_{t_\mu} ) = \sum_{ \alpha\in \Phi} \alpha(t_ \lambda) \cdot \alpha(t_\mu)
,\end{align*}
using that both ads are diagonal in this basis, so their product is
given by the products of their diagonal entries. One can write this as
\(\sum_{ \alpha\in \Phi} \kappa(t_ \alpha, t_ \lambda) \kappa(t_ \alpha, t_\mu)\),
so we get a formula
\begin{align*}
( \lambda, \mu ) = \sum_{ \alpha\in \Phi} ( \alpha, \lambda) (\alpha, \mu), \qquad (\lambda, \lambda) = \sum_{ \alpha\in \Phi} (\alpha, \lambda)^2
.\end{align*}
Setting \(\lambda = \beta\) and dividing by \((\beta, \beta)^2\) yields
\begin{align*}
{1\over (\beta, \beta)} = \sum_{ \alpha\in \Phi} {(\alpha, \beta)^2 \over (\beta, \beta)^2} \in {1\over 4}{\mathbf{Z}}
,\end{align*}
since
\((\alpha, \beta)\over (\beta, \beta)\in {1\over 2} {\mathbf{Z}}\). So
\((\beta, \beta)\in {\mathbf{Q}}\) and thus
\((\alpha, \beta)\in {\mathbf{Q}}\) for all \(\alpha, \beta\in \Phi\).
It follows that the pairings \((\lambda, \mu)\) on the
\({\mathbf{Q}}{\hbox{-}}\)subspace \({\mathbb{E}}_{\mathbf{Q}}\) of
\(H {}^{ \vee }\) spanned by \(\left\{{ \alpha_i}\right\}\) are all
rational.

\end{remark}

\begin{claim}

\(({-}, {-})\) on \({\mathbb{E}}_{\mathbf{Q}}\) is still nodegenerate

\end{claim}

\begin{proof}[?]

If
\(\lambda\in {\mathbb{E}}_{\mathbf{Q}}, ( \lambda, \mu) =0 \forall \mu\in {\mathbb{E}}_{\mathbf{Q}}\),
then
\(( \lambda, \alpha_i) = 0 \forall i \implies (\lambda, \nu) = 0 \forall \nu\in H {}^{ \vee }\implies \lambda= 0\).

\end{proof}

\begin{remark}

Similarly,
\((\lambda, \lambda) = \sum_{ \alpha\in \Phi \subseteq {\mathbb{E}}_{\mathbf{Q}}} ( \alpha, \lambda)^2\)
is a sum of squares of rational numbers, and is thus non-negative. Since
\(( \lambda, \lambda) = 0 \iff \lambda= 0\), the form on
\({\mathbb{E}}_{\mathbf{Q}}\) is positive definite. Write
\({\mathbb{E}}\coloneqq{\mathbb{E}}_{\mathbf{Q}}\otimes_{\mathbf{Q}}{\mathbf{R}}= {\mathbf{R}}\left\{{\alpha_i}\right\}\),
then \(({-}, {-})\) extends in the obvious way to an
\({\mathbf{R}}{\hbox{-}}\)values positive definite bilinear form on
\({\mathbb{E}}\), making it a real inner product space.

\end{remark}

\begin{theorem}[?]

Let \(L, H, \Phi, {\mathbb{E}}_{/ {{\mathbf{R}}}}\) be as above, then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\Phi\) is a finite set which spans \({\mathbb{E}}\) and does not
  contain zero.
\item
  If \(\alpha\in \Phi\) then \(-\alpha\in \Phi\) and thus is the only
  other scalar multiple in \(\Phi\).
\item
  If \(\alpha, \beta \in \Phi\) then
  \begin{align*}
  \beta - \beta(h_ \alpha) \alpha = \beta - {2 (\beta, \alpha) \over ( \alpha, \alpha) } \alpha\in \Phi
  ,\end{align*}
  which only depends on \({\mathbb{E}}\). Note that this swaps
  \(\pm \alpha\).
\item
  If \(\alpha, \beta \in \Phi\) then
  \(\beta(h_\alpha) = {2(\beta, \alpha) \over (\alpha, \alpha)}\in {\mathbf{Z}}\).
\end{enumerate}

Thus \(\Phi\) satisfies the axioms of a \textbf{root system} in
\({\mathbb{E}}\).

\end{theorem}

\begin{example}[?]

Recall that for \({\mathfrak{sl}}_3({\mathbf{C}})\),
\(\kappa(x,y) = 6 \operatorname{Trace}(xy)\). Taking the standard basis
\(\left\{{v_i}\right\} \coloneqq\left\{{x_i, h_i, y_i \coloneqq x_i^t}\right\}\),
the matrix \(\operatorname{Trace}(v_i v_j)\) is of the form
\begin{align*}
{
\begin{bmatrix}
  {0} & {0} & {I}
\\
  {0} & {A} & {0}
\\
  {I} & {0} & {0}
\end{bmatrix}
}\qquad A \coloneqq{
\begin{bmatrix}
  {2} & {-1} 
\\
  {-1} & {2}
\end{bmatrix}
}
.\end{align*}
This is far from the matrix of an inner product, but the middle block
corresponds to the form restricted to \(H\), which is positive definite.
One can quickly check this is positive definite by checking positivity
of the upper-left \(k\times k\) minors, which here yields
\(\operatorname{det}(2) = 2, \operatorname{det}A = 4-1 = 3\).

\end{example}

\hypertarget{part-iii-root-systems}{%
\subsection{Part III: Root Systems}\label{part-iii-root-systems}}

\hypertarget{ch.-9-axiomatics.-9.1-reflections-in-a-euclidean-space}{%
\subsection{Ch. 9, Axiomatics. 9.1: Reflections in a Euclidean
Space}\label{ch.-9-axiomatics.-9.1-reflections-in-a-euclidean-space}}

\begin{remark}

Let \({\mathbb{E}}\) be a fixed real finite-dimensional Euclidean space
with inner product \((\alpha, \beta)\), we consider property (c) from
the previous theorem:
\begin{align*}
\beta - {2( \beta, \alpha) \over (\alpha, \alpha)} \in \Phi \qquad\forall \alpha, \beta\in \Phi
.\end{align*}

\end{remark}

\begin{definition}[Reflections]

A \textbf{reflection} in \({\mathbb{E}}\) is an invertible linear map on
an \(n{\hbox{-}}\)dimensional Euclidean space that fixes pointwise a
hyperplane \(P\) (of dimension \(n-1\)) and sending any vector
\(v\perp P\) to \(-v\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-03_09-52.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\begin{remark}

If \(\sigma\) is a reflection sending \(\alpha\mapsto - \alpha\), then
\begin{align*}
\sigma_\alpha(\beta) = \beta - {2( \beta, \alpha) \over (\alpha, \alpha)} \alpha \qquad \forall \beta\in {\mathbb{E}}
.\end{align*}
One can check that \(\sigma_\alpha^2 = \operatorname{id}\). Some notes
on notation:

\begin{itemize}
\tightlist
\item
  Humphreys writes
  \({\left\langle { \beta},~{ \alpha} \right\rangle} \coloneqq{2 ( \beta, \alpha) \over (\alpha, \alpha)}\)
  This is linear in \(\beta\) but not in \(\alpha\)!
\item
  More modern:
  \((\beta, \alpha {}^{ \vee }) \coloneqq{\left\langle { \beta},~{\alpha} \right\rangle}\)
  where \(\alpha {}^{ \vee }\coloneqq{2\alpha\over (\alpha, \alpha)}\)
  corresponds to \(h_\alpha\).
\item
  Modern notation for the map: \(s_\alpha\) instead of
  \(\sigma_\alpha\).
\end{itemize}

\end{remark}

\hypertarget{wednesday-october-05}{%
\section{Wednesday, October 05}\label{wednesday-october-05}}

\hypertarget{reflections-in-mathbben}{%
\subsection{\texorpdfstring{Reflections in
\({\mathbb{E}}^n\)}{Reflections in \{\textbackslash mathbb\{E\}\}\^{}n}}\label{reflections-in-mathbben}}

\begin{remark}

Recall the formula
\begin{align*}
s_\alpha( \lambda) = \lambda- (\lambda, \alpha {}^{ \vee })\alpha, \qquad \alpha {}^{ \vee }\coloneqq{2\alpha\over (\alpha, \alpha)}, \alpha\neq 0
,\end{align*}
which is a reflection through the hyperplane
\(P_\alpha\coloneqq\alpha\perp\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-05_09-13.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{lemma}[?]

Let \(\Phi \subseteq {\mathbb{E}}\) be a set that spans
\({\mathbb{E}}\), and suppose all of the reflections \(s_\alpha\) for
\(\alpha \in \Phi\) leave \(\Phi\) invariant. If
\(\sigma\in \operatorname{GL}({\mathbb{E}})\) leaves \(\Phi\) invariant,
fixes a hyperplane \(P\) pointwise, and sends some
\(\alpha\in \Phi\setminus\left\{{0}\right\}\) to \(-\alpha\), then
\(\sigma = s_\alpha\) and \(P = P_\alpha\).

\end{lemma}

\begin{proof}[?]

Let
\(\tau = \sigma s_ \alpha =\sigma s_{ \alpha}^{-1}\in \operatorname{GL}({\mathbb{E}})\),
noting that every \(s_\alpha\) is order 2. Then \(\tau( \Phi) = \Phi\)
and \(\tau( \alpha) = \alpha\), so \(\tau\) acts as the identity on the
subspace \({\mathbf{R}}\alpha\) and the quotient space
\({\mathbb{E}}/{\mathbf{R}}\alpha\) since there are two decompositions
\({\mathbb{E}}= P_ \alpha \oplus {\mathbf{R}}\alpha = P \oplus {\mathbf{R}}\alpha\)
using \(s_\alpha\) and \(\sigma\) respectively. So
\(\tau - \operatorname{id}\) acts as zero on
\({\mathbb{E}}/{\mathbf{R}}\alpha\), and so maps \({\mathbb{E}}\) into
\({\mathbf{R}}\alpha\) and \({\mathbf{R}}\alpha\) to zero, s
\((\tau - \operatorname{id})^2 = 0\) on \({\mathbb{E}}\) and its minimal
polynomial \(m_\tau(t)\) divides \(f(t) \coloneqq(t-1)^2\).

Note that \(\Phi\) is finite, so the vectors
\(\beta, \tau \beta, \tau^2 \beta, \tau^3 \beta,\cdots\) can not all be
distinct. Since \(\tau\) is invertible we can assume
\(\tau^k \beta = \beta\) for some particular \(k\). Taking the least
common multiple of all such \(k\) yields a uniform \(k\) that works for
all \(\beta\) simultaneously, so \(\tau^k \beta = \beta\) for all
\(\beta \in \Phi\). Since \({\mathbf{R}}\Phi = {\mathbb{E}}, \tau^k\)
acts as \(\operatorname{id}\) on all of \({\mathbb{E}}\), so
\(\tau^k - 1 = 0\) and so \(m_\tau(t) \divides t^k - 1\) for some \(k\).
Therefore \(m_\tau(t) \divides \gcd( (t-1)^2, t^k-1 ) = t-1\), forcing
\(\tau = \operatorname{id}\) and \(\sigma = s_ \alpha\) and
\(P = P_\alpha\).

\end{proof}

\hypertarget{abstract-root-systems}{%
\subsection{Abstract root systems}\label{abstract-root-systems}}

\begin{definition}[Root systems]

A subset \(\Phi \subseteq {\mathbb{E}}\) of a real Euclidean space is a
\textbf{root system} iff

\begin{itemize}
\tightlist
\item
  R1: \({\sharp}\Phi < \infty\), \({\mathbf{R}}\Phi = {\mathbb{E}}\),
  and \(0\not\in \Phi\),
\item
  R2: \(\alpha\in \Phi \implies -\alpha\in \Phi\) and no other scalar
  multiples of \(\alpha\) are in \(\Phi\),
\item
  R3: If \(\alpha\in \Phi\) then \(s_\alpha( \Phi) = \Phi\),
\item
  R4: If \(\alpha, \beta\in \Phi\) then
  \((\beta, \alpha {}^{ \vee }) = {2(\beta, \alpha) \over ( \alpha, \alpha) } \in {\mathbf{Z}}\).
\end{itemize}

Notably,
\(\ell\coloneqq{\left\lVert {\beta - s_\alpha\beta} \right\rVert}\) is
an integer multiple of \(\alpha\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-05_09-33.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{definition}

\begin{definition}[Weyl groups]

The \textbf{Weyl group} \(W\) associated to a root system \(\Phi\) is
the subgroup
\(\left\langle{s_\alpha, \alpha\in \Phi}\right\rangle \leq \operatorname{GL}({\mathbb{E}})\).

\end{definition}

\begin{remark}

Note that \({\sharp}W < \infty\): \(W\) permutes \(\Phi\) by (R3), so
there is an injective group morphism
\(W \hookrightarrow\mathrm{Perm}(\Phi)\), which is a finite group --
this is injective because if \(w\curvearrowright\Phi\) as
\(\operatorname{id}\), since \({\mathbf{R}}\Phi = {\mathbb{E}}\), by
linearity \(w\curvearrowright{\mathbb{E}}\) by \(\operatorname{id}\) and
\(w=\operatorname{id}\). Recalling that
\(s_ \alpha( \lambda) = \lambda- (\lambda, \alpha {}^{ \vee }) \alpha\),
we have \((s_ \alpha(\lambda), s_ \alpha(\mu)) = ( \lambda, \mu)\) for
all \(\lambda, \mu \in {\mathbb{E}}\). So in fact
\(W \leq {\operatorname{O}}({\mathbb{E}}) \leq \operatorname{GL}({\mathbb{E}})\),
which have determinant \(\pm 1\) -- in particular,
\(\operatorname{det}s_\alpha = -1\) since it can be written as a block
matrix \(\operatorname{diag}(1, 1, \cdots, 1, -1)\) by choosing a basis
for \(P_\alpha\) and extending it by \(\alpha\).

\begin{quote}
Note that one can classify finite subgroups of
\({\operatorname{SO}}_n\).
\end{quote}

\end{remark}

\begin{example}[?]

Let
\(\Phi = \left\{{ {\varepsilon}_i - {\varepsilon}_j {~\mathrel{\Big\vert}~}1\leq i,j \leq n+1, i\neq j}\right\}\)
be a root system of type \(A_n\) where
\(\left\{{{\varepsilon}_i}\right\}\) form the standard basis of
\({\mathbf{R}}^{n+1}\) with the standard inner product, so
\(({\varepsilon}_i, {\varepsilon}_j) = \delta_{ij}\). One can compute
\begin{align*}
s_{{\varepsilon}_i - {\varepsilon}_j}({\varepsilon}_k) = {\varepsilon}_k {2 ({\varepsilon}_k, {\varepsilon}_i - {\varepsilon}_j) \over ({\varepsilon}_i - {\varepsilon}_j, {\varepsilon}_i - {\varepsilon}_j)}({\varepsilon}_i - {\varepsilon}_j)
= {\varepsilon}_k - ({\varepsilon}_k, {\varepsilon}_i - {\varepsilon}_j)({\varepsilon}_i - {\varepsilon}_j) 
=
\begin{cases}
{\varepsilon}_j & k=i 
\\
{\varepsilon}_i & k=j
\\
{\varepsilon}_k & \text{otherwise}.
\end{cases}
=
{\varepsilon}_{(ij).k}
\end{align*}
where \((ij) \in S_{n+1}\) is a transposition, acting as a function on
the index \(k\). Thus there is a well-defined group morphism
\begin{align*}
W &\to S_{n+1} \\
s_{{\varepsilon}_i - {\varepsilon}_j} &\mapsto (ij)
.\end{align*}
This is injective since \(w\) acting by the identity on every
\({\varepsilon}_k\) implies acting by the identity on all of
\({\mathbb{E}}\) by linearity, and surjective since transpositions
generate \(S_{n+1}\). So \(W\cong S_{n+1}\), and \(A_n\) corresponds to
\({\mathfrak{sl}}_{n+1}({\mathbf{C}})\) using that
\begin{align*}
[h, e_{ij}] = (h_i - h_j) e_{ij} = ({\varepsilon}_i - {\varepsilon}_j)(h) e_{ij}
.\end{align*}
In \(G = {\operatorname{SL}}_n({\mathbf{C}})\) one can define
\(N_G(T)/C_G(T)\) for \(T\) a maximal torus.

\end{example}

\begin{exercise}[?]

What are the Weyl groups of other classical types?

\end{exercise}

\begin{lemma}[?]

Let \(\Phi \subseteq {\mathbb{E}}\) be a root system. If
\(\sigma\in \operatorname{GL}({\mathbb{E}})\) leaves \(\Phi\) invariant,
then for all \(\alpha\in \Phi\),
\begin{align*}
\sigma s_{ \alpha} \sigma = s_{ \sigma( \alpha)}, \qquad (\beta, \alpha {}^{ \vee }) = (\sigma( \beta), \sigma(\alpha) {}^{ \vee }) \,\,\forall \alpha, \beta\in \Phi
.\end{align*}
Thus conjugating a reflection yields another reflection.

\end{lemma}

\begin{proof}[?]

Note that \(\sigma s_ \alpha \sigma^{-1}\) sends \(\sigma( \alpha)\) to
its negative and fixes pointwise the hyperplane \(\sigma(P_\alpha)\). If
\(\beta \in \Phi\) then \(s_{ \alpha}( \beta) \in \Phi\), so
\(\sigma s_ \alpha ( \beta) \in \Phi\) and
\begin{align*}
(\sigma s_ \alpha \sigma^{-1}) ( \sigma( \beta)) = \sigma s_ \alpha(\beta) \in \sigma\Phi
,\end{align*}
so \(\sigma s_ \alpha \sigma^{-1}\) leaves invariant the set
\(\left\{{ \sigma( \beta) {~\mathrel{\Big\vert}~}\beta\in \Phi}\right\} = \Phi\).
By the previous lemma, it must equal \(s_{ \sigma( \alpha)}\), and so
\begin{align*}
( \sigma( \beta), \sigma( \alpha) {}^{ \vee }) = (\beta, \alpha {}^{ \vee })
\end{align*}
by applying both sides to \(\sigma(\beta)\).

\end{proof}

\begin{warnings}

This does not imply that
\((\sigma( \beta), \sigma( \alpha) ) = (\beta, \alpha)\)! With the
duals/checks, this bracket involves a ratio, which is preserved, but the
individual round brackets are not.

\end{warnings}

\hypertarget{friday-october-07}{%
\section{Friday, October 07}\label{friday-october-07}}

\begin{lemma}[?]

Let \(\Phi \subseteq {\mathbb{E}}\) be a root system with Weyl group
\(W\). If \(\sigma\in \operatorname{GL}({\mathbb{E}})\) leaves \(\Phi\)
invariant then
\begin{align*}
\sigma s_{\alpha} \sigma^{-1}= s_{ \sigma( \alpha)} \qquad\forall \alpha\in \Phi
\end{align*}
and
\begin{align*}
( \beta, \alpha {}^{ \vee }) = ( \sigma(\beta), \sigma(\alpha) {}^{ \vee }) \qquad \forall \alpha, \beta \in \Phi
.\end{align*}

\end{lemma}

\begin{warnings}

\begin{align*}
( \sigma( \beta), \sigma( \alpha) ) \neq (\beta, \alpha)
,\end{align*}
i.e.~the \(({-}) {}^{ \vee }\) is important here since it involves a
ratio. Without the ratio, one can easily scale to make these not equal.

\end{warnings}

\begin{definition}[?]

Two root systems
\(\Phi \subseteq {\mathbb{E}}, \Phi' \subseteq {\mathbb{E}}'\) are
\textbf{isomorphic} iff there exists
\(\phi: {\mathbb{E}}\to {\mathbb{E}}'\) of vector spaces such that
\(\phi(\Phi) = \Phi'\) such that
\begin{align*}
(\varphi( \beta), \varphi(\alpha) {}^{ \vee }) = (\beta, \alpha) \coloneqq{2 (\beta, \alpha) \over (\alpha, \alpha)} \qquad\forall \alpha, \beta \in \Phi
.\end{align*}

\end{definition}

\begin{example}[?]

One can scale a root system to get an isomorphism:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-19.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{remark}

Note that if \(\phi: \Phi { \, \xrightarrow{\sim}\, }\Phi'\) is an
isomorphism, then
\begin{align*}
\varphi(s_{ \alpha}( \beta)) = s_{ \varphi( \alpha)}( \varphi(\beta)) \qquad \forall \alpha, \beta\in \Phi \implies \varphi \circ s_{ \alpha} \varphi^{-1}= s_{ \varphi( \alpha)}
.\end{align*}
So \(\phi\) induces an isomorphism of Weyl groups
\begin{align*}
W & { \, \xrightarrow{\sim}\, }W' \\
s_{\alpha} &\mapsto s_{ \varphi( \alpha)}
.\end{align*}

By the lemma, an automorphism of \(\Phi\) is the same as an automorphism
of \({\mathbb{E}}\) leaving \(\Phi\) invariant. In particular,
\(W\hookrightarrow\mathop{\mathrm{Aut}}( \Phi)\).

\end{remark}

\begin{definition}[Dual root systems]

If \(\Phi \subseteq {\mathbb{E}}\) is a root system then the
\textbf{dual root system} is
\begin{align*}
\Phi {}^{ \vee }\coloneqq\left\{{ \alpha {}^{ \vee }{~\mathrel{\Big\vert}~}\alpha\in \Phi}\right\}, \qquad \alpha {}^{ \vee }\coloneqq{2\alpha\over (\alpha, \alpha)}
.\end{align*}

\end{definition}

\begin{exercise}[?]

Show that \(\Phi {}^{ \vee }\) is again a root system in
\({\mathbb{E}}\).

\end{exercise}

\begin{remark}

One can show \(W( \Phi) = W( \Phi {}^{ \vee })\) and
\({\left\langle {\lambda},~{ \alpha {}^{ \vee }} \right\rangle} \alpha {}^{ \vee }= {\left\langle { \lambda},~{ \alpha} \right\rangle} \alpha = (\lambda, \alpha {}^{ \vee })\)
for all \(\alpha\in \Phi, \lambda\in {\mathbb{E}}\), so
\(s_{\alpha {}^{ \vee }} = s_{\alpha}\) as linear maps on
\({\mathbb{E}}\).

\end{remark}

\hypertarget{examples}{%
\subsection{9.3: Example(s)}\label{examples}}

\begin{definition}[Ranks of root systems]

Let \(\Phi \subseteq {\mathbb{E}}\) be a root system, then
\(\ell \coloneqq\dim_{\mathbf{R}}{\mathbb{E}}\) is the \textbf{rank} of
\(\Phi\).

\end{definition}

\begin{remark}

Rank 1 root systems are given by choice of \(\alpha\in {\mathbf{R}}\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-31.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{remark}

Recall \({2( \beta, \alpha) \over (\alpha, \alpha)} \in {\mathbf{Z}}\),
and from linear algebra,
\({\left\langle {v},~{w} \right\rangle} = {\left\lVert {v} \right\rVert} \cdot {\left\lVert {w} \right\rVert} \cos( \theta)\)
and \({\left\lVert {\alpha} \right\rVert}^2 = ( \alpha, \alpha)\). We
can thus write
\begin{align*}
{\left\langle { \beta},~{ \alpha} \right\rangle} = {2( \beta, \alpha) \over (\alpha, \alpha)} = 2{{\left\lVert {\beta} \right\rVert}\over {\left\lVert {\alpha} \right\rVert}} \cos (\theta), \qquad {\left\langle {\alpha },~{\beta} \right\rangle}= 2{{\left\lVert {\alpha} \right\rVert}\over {\left\lVert {\beta} \right\rVert}} \cos( \theta)
,\end{align*}
and so
\begin{align*}
{\left\langle {\alpha },~{\beta} \right\rangle}{\left\langle {\beta },~{\alpha } \right\rangle}= 4\cos^2( \theta)
,\end{align*}
noting that
\(L_{ \alpha, \beta} \coloneqq{\left\langle {\alpha },~{\beta} \right\rangle}, {\left\langle {\beta },~{\alpha} \right\rangle}\)
are integers of the same sign. If positive, this is in QI, and if
negative QII. This massively restricts what the angles can be, since
\(0 \leq \cos^2( \theta) \leq 1\).

First, an easy case: suppose \(L_{ \alpha, \beta} = 4\), so
\(\cos^2( \theta) = 1\implies \cos( \theta) = \pm 1\implies \theta= 0, \pi\).

\begin{itemize}
\tightlist
\item
  If \(0\), then \(\alpha,\beta\) are in the same 1-dimensional subspace
  and thus \(\beta = \alpha\). In this case,
  \({\left\langle {\beta },~{\alpha } \right\rangle}= 2 = {\left\langle {\alpha },~{\beta} \right\rangle}\).
\item
  If \(\pi\), then \(\alpha = - \beta\). Here
  \({\left\langle {\beta },~{\alpha } \right\rangle}= -2\).
\end{itemize}

So assume \(\beta\neq \pm \alpha\), and without loss of generality
\({\left\lVert {\beta} \right\rVert}\geq {\left\lVert {\alpha} \right\rVert}\),
or equivalently
\({\left\langle {\alpha },~{\beta } \right\rangle}\leq {\left\langle {\beta },~{\alpha} \right\rangle}\).
Note that if \({\left\langle {\alpha },~{\beta} \right\rangle}\neq 0\)
then
\begin{align*}
{ {\left\langle {\beta },~{\alpha} \right\rangle}\over {\left\langle {\alpha },~{\beta} \right\rangle}} = {{\left\lVert { \beta} \right\rVert}^2 \over {\left\lVert { \alpha} \right\rVert}^2}
.\end{align*}

The other possibilities are as follows:

\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.26}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.26}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.07}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 6\tabcolsep) * \real{0.40}}@{}}
\toprule
\({\left\langle {\alpha},~{\beta} \right\rangle}\) &
\({\left\langle {\beta},~{\alpha} \right\rangle}\) & \(\theta\) &
\({\left\lVert {\beta} \right\rVert}^2/{\left\lVert {\alpha} \right\rVert}^2\) \\
\midrule
\endhead
0 & 0 & \(\pi/2\) & Undetermined \\
1 & 1 & \(\pi/3\) & 1 \\
-1 & -1 & \(2\pi/3\) & 1 \\
1 & 2 & \(\pi/4\) & 2 \\
-1 & -2 & \(3\pi/4\) & 2 \\
1 & 3 & \(\pi/6\) & 3 \\
-1 & -3 & \(5\pi/6\) & 3 \\
\bottomrule
\end{longtable}

Cases for the norm ratios:

\begin{itemize}
\tightlist
\item
  \(1: A_2\)
\item
  \(2: B_2 = C_2\)
\item
  \(3: G_2\)
\end{itemize}

These are the only three irreducible rank 2 root systems.

\end{remark}

\begin{lemma}[?]

Let \(\alpha, \beta\in\Phi\) lie in distinct linear subspaces of
\({\mathbb{E}}\). Then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  If \((\alpha, \beta) > 0\), i.e.~their angle is strictly acute, then
  \(\alpha - \beta\) is a root
\item
  If \((\alpha, \beta) < 0\) then \(\alpha + \beta\) is a root.
\end{enumerate}

\end{lemma}

\begin{proof}[?]

Note that (2) follows from (1) by replacing \(\beta\) with \(-\beta\).
Assume \((\alpha, \beta) > 0\), then by the chart
\({\left\langle {\alpha },~{\beta } \right\rangle}=1\) or
\({\left\langle {\beta },~{\alpha } \right\rangle}= 1\). In the former
case,
\begin{align*}
\Phi\ni s_{ \beta}( \alpha) = \alpha - {\left\langle {\alpha },~{\beta } \right\rangle}\beta = \alpha- \beta
.\end{align*}
In the latter,
\begin{align*}
s_{ \alpha}(\beta) = \beta- \alpha \in \Phi\implies - (\beta- \alpha) = \alpha- \beta\in \Phi
.\end{align*}

\end{proof}

\begin{remark}

Suppose \(\operatorname{rank}( \Phi) = 2\). Letting \(\alpha\in \Phi\)
be a root of shortest length, since \({\mathbf{R}}\Phi = {\mathbb{E}}\)
there is some \(\beta \in {\mathbb{E}}\) not equal to \(\pm \alpha\).
Without loss of generality assume \(\angle_{\alpha, \beta}\) is obtuse
by replacing \(\beta\) with \(-\beta\) if necessary:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-57.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Also choose \(\beta\) such that \(\angle_{ \alpha, \beta}\) is maximal.

\textbf{Case 0}: If \(\theta = \pi/2\), one gets
\({\mathbf{A}}_1\times {\mathbf{A}}_1\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/Moduli/sections/figures}{2022-10-07_09-59.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

We'll continue this next time.

\end{remark}

\hypertarget{monday-october-10}{%
\section{Monday, October 10}\label{monday-october-10}}

\hypertarget{classification-of-rank-2-root-systems}{%
\subsection{Classification of Rank 2 Root
Systems}\label{classification-of-rank-2-root-systems}}

\begin{remark}

If \(\beta\neq \pm \alpha\),

\begin{itemize}
\tightlist
\item
  \((\alpha, \beta) > 0 \implies \alpha - \beta\in \Phi\)
\item
  \((\alpha, \beta) < 0 \implies \alpha + \beta\in \Phi\)
\end{itemize}

\end{remark}

\begin{remark}

Rank 2 root systems: let \(\alpha\) be a root of shortest length, and
\(\beta\) a root with angle \(\theta\) between \(\alpha,\beta\) with
\(\theta \geq \pi/2\) as large as possible.

\begin{itemize}
\item
  If \(\theta = \pi/2\): \(A_1 \times A_1\).

  \includegraphics{figures/2022-10-10_09-16-04.png}
\item
  If \(\theta = 2\pi/3\): \(A_2\)
\end{itemize}

\includegraphics{figures/2022-10-10_09-20-14.png}

One can check
\({\left\langle {\alpha },~{\beta} \right\rangle}= 2(-1/2) = -1\) and
\({\left\langle {\alpha + \beta},~{ \beta} \right\rangle} = {\left\langle {\alpha },~{\beta } \right\rangle}+ {\left\langle {\beta },~{\beta } \right\rangle}= -1 + 2 = 1\).

\begin{itemize}
\tightlist
\item
  If \(\theta = 3\pi/4\): \(B_2\)
\end{itemize}

\begin{itemize}
\tightlist
\item
  If \(\theta = 5\pi/6\): \(G_2\)
\end{itemize}

One can check that using linearity of
\({\left\langle {{-}},~{{-}} \right\rangle}\) in the first variable that

\begin{itemize}
\tightlist
\item
  \(s_\alpha \beta = \beta + 3 \alpha\),
\item
  \(s_\alpha(\beta+ \alpha) = \beta+ 2 \alpha\),
\item
  \(s_ \beta(\beta+ 3 \alpha) = (\beta+ 3 \alpha) - {\left\langle { \beta+ 3 \alpha},~{ \beta} \right\rangle}= 2 \beta+ 3 \alpha \in \Phi\).
\end{itemize}

\end{remark}

\begin{remark}

Note that in each case one can see the root strings, defined as
\begin{align*}
R_\beta \coloneqq\left\{{\beta+ k \alpha {~\mathrel{\Big\vert}~}k\in {\mathbf{Z}}}\right\} \cap\Phi
.\end{align*}
Let \(r,q\in {\mathbf{Z}}_{\geq 0}\) be maximal such that
\(\beta-r \alpha, \beta + q \alpha\in \Phi\). The claim is that every
such root string is unbroken.

Suppose not, then there is some \(k\) with \(-r < k < q\) with
\(\beta + k \alpha \not\in \Phi\). One can then find a maximal \(p\) and
minimal \(s\) with \(p < s\) and \(\beta+p \alpha \in \Phi\) but
\(\beta + (p+1) \alpha \not\in \Phi\), and similarly
\(\beta + (s-1)\alpha\not\Phi\) but \(\beta + s \alpha\in \Phi\). By a
previous lemma, \((\beta+ p \alpha, \alpha) \geq 0\) and similarly
\((\beta+ s \alpha, \alpha) \leq 0\). Combining these,
\begin{align*}
p( \alpha, \alpha) \geq s (\alpha, \alpha) \implies p \geq s \text{ since } (\alpha, \alpha) > 0 \qquad\contradiction
.\end{align*}

The picture:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-10_09-44.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

So \(s_\alpha\) reverses the root string, since it sends the line
containing the root string to itself but reflects through \(P_\alpha\).
One can compute
\begin{align*}
\beta - r \alpha 
&= s_ \alpha(\beta + q \alpha) \\
&= (\beta+ q \alpha) - {\left\langle {\beta+ q \alpha},~{ \alpha} \right\rangle} \alpha \\
&= (\beta+ q \alpha) - {\left\langle { \beta},~{\alpha} \right\rangle}ga - 2q \alpha \\
&= \beta - \qty{{\left\langle {\beta },~{\alpha } \right\rangle}+ q} \alpha
,\end{align*}
so \(r = {\left\langle {\beta },~{\alpha} \right\rangle}\) and
\(r-q = {\left\langle {\beta },~{\alpha } \right\rangle}= \beta(h_ \alpha)\)
for a semisimple Lie algebra.

Supposing
\({\left\lvert {{\left\langle {\beta },~{\alpha} \right\rangle}} \right\rvert} \leq 3\).
Choose \(\beta\) in \(R_\alpha\) such that \(\beta-\alpha\) is not a
root and \(\beta\) is at the left end of the string and \(r=0\):

\includegraphics{figures/2022-10-10_09-51-39.png}

Then \(q = -{\left\langle {\beta },~{\alpha} \right\rangle}\), so the
root string contains at most 4 roots (for \(\Phi\) of any rank).

\end{remark}

\hypertarget{ch.-10-simple-roots-10.1-bases-and-weyl-chambers}{%
\subsection{Ch. 10 Simple Roots, 10.1 Bases and Weyl
Chambers}\label{ch.-10-simple-roots-10.1-bases-and-weyl-chambers}}

\begin{definition}[Base (simple roots)]

A subset \(\Delta \subseteq \Phi\) is a \textbf{base} (or more modernly
a \textbf{set of simple roots}) if

\begin{itemize}
\tightlist
\item
  B1: \(\Delta\) is a basis for \({\mathbb{E}}\),
\item
  B2: Each \(\beta\in \Phi\) can be written as
  \(\beta = \sum_{\alpha\in \Delta} k_\alpha \alpha\) with
  \(k_\alpha\in {\mathbf{Z}}\) with either all
  \(k_\alpha\in {\mathbf{Z}}_{\geq 0}\) or all
  \(k_\alpha \in {\mathbf{Z}}_{\leq 0}\).
\end{itemize}

\end{definition}

\begin{example}[?]

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  The roots labeled \(\alpha,\beta\) in the rank 2 cases were all simple
  systems.
\item
  For \(A_n\), a base is
  \(\left\{{{\varepsilon}_1 - {\varepsilon}_2, {\varepsilon}_2 - {\varepsilon}_3, \cdots, {\varepsilon}_{n} - {\varepsilon}_{n+1} }\right\}\),
  where
  \(\Phi = \left\{{{\varepsilon}_i - {\varepsilon}_j {~\mathrel{\Big\vert}~}i\neq j}\right\}\).
\end{enumerate}

\end{example}

\hypertarget{wednesday-october-12}{%
\section{Wednesday, October 12}\label{wednesday-october-12}}

\begin{remark}

Today: finding bases for root systems. It's not obvious they always
exist, but e.g.~in the previous rank 2 examples, \(\alpha,\beta\) formed
a base.

\end{remark}

\begin{definition}[Height, positive/negative roots]

Given a base \(\Delta \subseteq \Phi\), the \textbf{height} of a root
\(\beta = \sum_{\alpha\in \Delta} k_ \alpha \alpha\) is
\begin{align*}
\operatorname{ht}(\beta) \coloneqq\sum k_ \alpha
.\end{align*}
If all \(k_\alpha \geq 0\), we say \(\beta\) is \textbf{positive} and
write \(\beta\in \Phi^+\). Similarly, \(\beta\) is \textbf{negative} iff
\(k_\alpha \leq 0\) for all \(\alpha\), and we write
\(\beta\in \Phi^-\). This decomposes a root system into
\(\Phi = \Phi^+ {\textstyle\coprod}\Phi^-\), and moreover
\(-\Phi^+ = \Phi^-\).

\end{definition}

\begin{remark}

A choice of \(\Delta\) determines a partial order on \(\Phi\) which
extends to \({\mathbb{E}}\), where \(\lambda\geq \mu \iff \lambda- \mu\)
is a non-negative integer linear combination of elements of \(\Delta\).

\end{remark}

\begin{lemma}[?]

If \(\Delta \subseteq \Phi\) is a base and \(\alpha, \beta\in \Delta\),
then
\begin{align*}
\alpha\neq \beta\implies ( \alpha, \beta) \leq 0 \text{ and } \alpha- \beta\not\in \Phi
.\end{align*}

\end{lemma}

\begin{proof}[?]

We have \(\alpha\neq \pm \beta\) since \(\Delta\) is a linearly
independent set. By a previous lemma, if \(( \alpha, \beta) > 0\) then
\(\beta- \alpha \in \Phi\) by a previous lemma. \(\contradiction\)

\end{proof}

\begin{definition}[Regular]

An element \(\gamma\in {\mathbb{E}}\) is \textbf{regular} iff
\(\gamma \in {\mathbb{E}}\setminus\bigcup_{\alpha\in \Phi} P_ \alpha\)
where \(P_ \alpha= \alpha^\perp\), otherwise \(\gamma\) is
\textbf{singular}.

\end{definition}

\begin{lemma}[?]

Regular vectors exist.

\end{lemma}

\begin{remark}

The basic fact used is that over an infinite field, no vector space is
the \emph{union} of a finite number of proper subspaces. Note that this
is a union, not a sum!

Given a regular vector \(\gamma\in {\mathbb{E}}\), define
\begin{align*}
\Phi^+(\gamma) = \left\{{ \alpha\in \Phi {~\mathrel{\Big\vert}~}(\alpha, \gamma) > 0 }\right\}
,\end{align*}
the roots on the positive side of the hyperplane \(\alpha^\perp\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-12_09-31.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

This decomposes
\(\Phi = \Phi^+(\gamma) {\textstyle\coprod}\Phi^-(\gamma)\) where
\(\Phi^-(\gamma) \coloneqq- \Phi^+(\gamma)\). Note that \(\gamma\) lies
on the positive side of \(\alpha^\perp\) for every
\(\alpha\in \Phi^+(\gamma)\).

\end{remark}

\begin{definition}[Decomposable roots]

A positive root \(\beta\in \Phi^+\) is \textbf{decomposable} iff
\(\beta = \beta_1 + \beta_2\) for \(\beta_i \in \Phi^+( \gamma)\).
Otherwise \(\beta\) is \textbf{indecomposable}.

\end{definition}

\begin{theorem}[?]

There exists a base for \(\Phi\).

\end{theorem}

\begin{theorem}[?]

Let \(\gamma\in {\mathbb{E}}\) be regular. Then the set
\(\Delta(\gamma)\) of all indecomposable roots in \(\Phi^+( \gamma)\) is
a base for \(\Phi\). Moreover, any base for \(\Phi\) arises in this way.

\end{theorem}

\begin{proof}[in 5 easy steps]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Claim: each root in \(\Phi^+( \gamma)\) is in
  \({\mathbf{Z}}_{\geq 0} \Delta( \gamma)\).
\end{enumerate}

The proof: if not, pick \(\beta\in \Phi^+(\gamma)\) which cannot be
written this way and choose it such that \((\beta, \gamma)\) is maximal
(by finiteness). Since \(\beta\not\in \Delta( \gamma)\), it is
decomposable as \(\beta = \beta_1 + \beta_2\) with
\(\beta_i \in \Phi^+\). Now \((\beta, \gamma) = \sum (\beta_i, \gamma)\)
is a sum of nonnegative numbers, so
\((\beta_i, \gamma) < (\beta, \gamma)\) for \(i=1,2\). By minimality,
\(\beta_i\in {\mathbf{Z}}_{\geq 0 } \Delta(\gamma)\), but then by adding
them we get \(\beta\in {\mathbf{Z}}_{\geq 0 } \Delta(\gamma)\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\item
  Claim: if \(\alpha, \beta\in \Delta( \gamma)\) with
  \(\alpha\neq \beta\) then \((\alpha, \beta) \leq 0\). Note
  \(\alpha\neq - \beta\) since
  \((\alpha, \gamma), (\beta, \gamma) > 0\). By lemma 9.4, if
  \(( \alpha, \beta) > 0\) then \(\alpha - \beta\in \Phi\) is a root.
  Then one of \(\alpha- \beta, \beta- \alpha\in \Phi^+( \gamma)\). In
  the first case, \(\beta + (\alpha- \beta ) = \alpha\), decomposing
  \(\alpha\). In the second, \(\alpha + (\beta- \alpha) = \beta\), again
  a contradiction.
\item
  Claim: \(\Delta\coloneqq\Delta( \gamma)\) is linearly independent.
  Suppose \(\sum_{ \alpha\in \Delta} r_ \alpha \alpha = 0\) for some
  \(r_ \alpha \in {\mathbf{R}}\). Separate the positive terms
  \(\alpha\in \Delta'\) and the remaining \(\alpha\in \Delta''\) to
  write
  \({\varepsilon}\coloneqq\sum_{ \alpha\in \Delta'}r_ \alpha \alpha = \sum_{ \beta\in \Delta''} t_\beta \beta\)
  where now \(r_ \alpha, t_\beta > 0\). Use the two expressions for
  \({\varepsilon}\) to write
  \begin{align*}
    ({\varepsilon}, {\varepsilon}) = \sum _{ \alpha\in \Delta', \beta\in \Delta''} r_ \alpha t_ \beta (\alpha, \beta) \leq 0
    ,\end{align*}
  since \(r_ \alpha t_ \beta >0\) and \((\alpha, \beta) \leq 0\). So
  \({\varepsilon}= 0\), since \(({-}, {-})\) is an inner product. Write
  \(0 = (\gamma, {\varepsilon}) = \sum_{ \alpha\in \Delta'} r_\alpha (\alpha, \delta)\)
  where \(r_ \alpha > 0\) and \((\alpha, \gamma) > 0\), so it must be
  the case that \(\Delta' = \emptyset\). Similarly
  \(\Delta'' = \emptyset\), so \(r_\alpha = 0\) for all
  \(\alpha\in \Delta\).
\item
  Claim: \(\Delta( \gamma)\) is a base for \(\Phi\). Since
  \(\Phi = \Phi^+(\gamma){\textstyle\coprod}\Phi^-( \gamma)\), we have
  B2 by step 1. This also implies \(\Delta( \gamma)\) is a basis for
  \({\mathbb{E}}\), since we have linear independent by step 3. Thus
  \({\mathbf{Z}}\Delta ( \gamma) \supseteq\Phi\) and
  \({\mathbf{R}}\Phi = {\mathbb{E}}\).
\item
  Claim: every base of \(\Phi\) is \(\Delta( \gamma)\) for some regular
  \(\gamma\). Given \(\Delta\), choose \(\gamma\in {\mathbb{E}}\) such
  that \((\gamma, \alpha) > 0\) for all \(\alpha\in \Delta\). Then
  \(\gamma\) is regular by B2. Moreover
  \(\Phi^+ \subseteq \Phi^+( \gamma)\) and similarly
  \(\Phi^- \subseteq \Phi^-( \gamma)\), and taking disjoint unions
  yields \(\Phi\) for both the inner and outer sets, forcing them to be
  equal, i.e.~\(\Phi^{\pm} = \Phi^{\pm}( \gamma)\). One can check that
  \(\Delta \subseteq \Delta( \gamma)\) using
  \(\Phi^+ = \Phi^+( \gamma)\) and linear independence of \(\Delta\) --
  but both sets are bases for \({\mathbb{E}}\) and thus have the same
  cardinality \(\ell = \dim {\mathbb{E}}\), making them equal.
\end{enumerate}

\end{proof}

\hypertarget{bases-delta-for-phi-friday-october-14}{%
\section{\texorpdfstring{Bases \(\Delta\) for \(\Phi\) (Friday, October
14)}{Bases \textbackslash Delta for \textbackslash Phi (Friday, October 14)}}\label{bases-delta-for-phi-friday-october-14}}

\begin{remark}

From a previous discussion: given a rank \(n\) root system \(\Phi\) with
\(n\geq 2\), is
\({\mathbf{R}}\left\langle{ \alpha, \beta}\right\rangle \cap\Phi\)
always a rank 2 root system? The answer is yes! This follows readily
from just checking the axioms directly.

\end{remark}

\begin{remark}

For a regular
\(\gamma \in {\mathbb{E}}\setminus\bigcup_{\alpha\in \Phi} P_ \alpha\),
define
\(\Phi^+(\gamma) \coloneqq\left\{{ \beta\in \Phi {~\mathrel{\Big\vert}~}(\beta, \gamma) > 0}\right\}\)
and let \(\Delta( \gamma)\) be the indecomposable elements of
\(\Phi^+( \gamma)\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-14_09-16.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{theorem}[?]

\(\Delta( \gamma)\) is a base for \(\Phi\), and every base is of this
form.

\end{theorem}

\begin{definition}[?]

The connected components of
\({\mathbb{E}}\setminus\bigcup_{ \alpha\in \Phi} P_ \alpha\) are called
the \textbf{(open) Weyl chambers} of \({\mathbb{E}}\). Each regular
\(\gamma\) belongs to some Weyl chamber, which we'll denote
\(C(\gamma)\).

\end{definition}

\begin{remark}

Note that \(C(\gamma) = C(\gamma') \iff \gamma, \gamma'\) are on the
same side of every root hyperplane \(P_\alpha\) for \(\alpha\in \Phi\),
which happens iff
\(\Phi^+( \gamma) = \Phi^+(\gamma;) \iff \Delta( \gamma) = \Delta(\gamma')\),
so there is a bijection
\begin{align*}
\left\{{\substack{
  \text{Weyl chambers}
}}\right\}
&\rightleftharpoons
\left\{{\substack{
  \text{Bases for $\Phi$}
}}\right\}
.\end{align*}
Note also that \(W\) sends one Weyl chamber to another: any
\(s_ \alpha\) preserves the connected components
\({\mathbb{E}}\setminus\bigcup_{ \beta\in \Phi} P_{\beta}\), so if
\(\gamma\) is regular and \(\sigma\in W\) and
\(\sigma( \gamma) = \gamma'\) for some regular \(\gamma'\), then
\(\sigma(C(\gamma)) = C(\gamma')\). \(W\) also acts on bases for
\(\Phi\): if \(\Delta\) is a base for \(\Phi\), then \(\sigma( \Delta)\)
is still a basis for \({\mathbb{E}}\) since \(\sigma\) is an invertible
linear transformation. Since \(\sigma( \Phi) = \Phi\) by the axioms, any
root \(\alpha\in \Phi\) is of the form \(\sigma(\beta)\) for some
\(\beta\in \Phi\), but writing
\(\beta = \sum _{\alpha\in\Delta} k_ \alpha \alpha\) with all
\(k_\alpha\) the same sign,
\(\sigma( \beta) = \sum_{ \alpha\in \Delta} k_ \alpha \sigma( \alpha)\)
is a linear combination of elements in \(\sigma(\Delta)\) with
coefficients of the same sign.

The actions of \(W\) on chambers and bases are compatible: if
\(\Delta = \Delta( \gamma)\) then
\(\sigma( \Delta) = \Delta( \sigma( \gamma))\), since
\(\sigma(\Phi^+( \gamma)) = \Phi^+( \sigma( \gamma))\) since
\(W \leq {\operatorname{O}}({\mathbb{E}})\) and thus
\(( \sigma \gamma, \sigma \alpha) = (\gamma, \alpha)\).

\end{remark}

\begin{lemma}[A]

Fix a base \(\Delta \subset \Phi\), which decomposes
\(\Phi = \Phi^+ {\textstyle\coprod}\Phi^-\). If
\(\beta\in \Phi^+ \setminus\Delta\), then \(\beta- \alpha\in \Phi^+\)
for some \(\alpha\in\Delta\).

\end{lemma}

\begin{proof}[?]

If \(( \beta, \alpha)\leq 0\) for all \(\alpha\in \Delta\), then the
proof of theorem 10.1 would show \(\beta = 0\) by taking
\({\varepsilon}\coloneqq\beta\) in step 3. One can then find an
\(\alpha\in \Delta\) with \(( \beta, \alpha) > 0\), where clearly
\(\beta \neq \pm \alpha\). By lemma 9.4, \(\beta- \alpha\in \Phi\). Why
is this positive? Note that \(\beta\) has at least one coefficient for a
simple root (not the coefficient for \(\alpha\)) which is strictly
positive, and thus all coefficients are \(\geq 0\). This coefficient
stays the same in \(\beta- \alpha\), so its coefficients are all
non-negative by axiom B2 and \(\beta-\alpha\in \Phi^+\).

\end{proof}

\begin{corollary}[?]

Each \(\beta\in \Phi^+\) can be written as
\(\alpha_1 + \cdots + \alpha_k\) where \(\alpha_i\in \Delta\) not
necessarily distinct, such that each truncated sum
\(\alpha_1 + \cdots + \alpha_i\) for \(1\leq i \leq k\) is a positive
root. One proves this by induction on the height of \(\beta\).

\end{corollary}

\begin{lemma}[B]

Let \(\alpha \in \Delta\), then \(s_ \alpha\) permutes
\(\Phi^+ \setminus\left\{{ \alpha }\right\}\).

\end{lemma}

\begin{proof}[?]

Let \(\beta \in \Phi^+\setminus\left\{{ \alpha }\right\}\); if
\(\beta = \sum _{\gamma\in \Delta} k_ \gamma \gamma\) with
\(k_ \gamma \in {\mathbf{Z}}_{\geq 0}\), since \(\beta\neq \alpha\),
some \(k_ \gamma > 0\) for some \(\gamma \neq \alpha\). Using the
formula
\(s_ \alpha( \beta) = \beta- {\left\langle {\beta },~{\alpha } \right\rangle}\alpha\)
still has coefficient \(k_ \gamma\) for \(\gamma\). Thus
\(s_ \alpha( \beta) \in \Phi^+\) and \(s_ \alpha( \beta)\neq \alpha\)
since \(s_ \alpha(- \alpha) = \alpha\) and \(s_\alpha\) is bijective,
and so \(s_\alpha(\beta)\in \Phi^+\setminus\left\{{ \alpha }\right\}\).
As a result, \(s_\alpha\) permutes this set since it is invertible.

\end{proof}

\begin{corollary}[?]

Let
\begin{align*}
\rho \coloneqq{1\over 2}\sum_{ \beta\in \Phi^+} \beta \quad\in {\mathbb{E}}
.\end{align*}
Then \(s_\alpha( \rho) = \rho- \alpha\) for all \(\alpha\in \Delta\),
and \(s_ \alpha( \rho) = \rho\).

\begin{quote}
Note that Humphreys uses \(\delta\), but nobody uses this notation.
\end{quote}

\end{corollary}

\begin{lemma}[C, The Deletion Condition]

Let \(\alpha_1,\cdots \alpha_t \in \Delta\) be not necessarily distinct
simple roots, and write \(s_i \coloneqq s_{\alpha_i}\). If
\(s_1 \cdots s_{t-1}(\alpha_t) < 0\), then for some \(1\leq u \leq t\)
one has
\begin{align*}
s_1\cdots s_t = s_1\cdots s_{u-1} s_{u+1} s_{t-1}
,\end{align*}
so one can delete \(s_u\) and \(s_t\) to get a shorter product of
reflections.

\end{lemma}

\begin{proof}[?]

For \(0\leq i\leq t-1\), let
\(\beta_i \coloneqq s_{i+1} \cdots s_{t-1}( \alpha_t)\) and
\(\beta_{t-1} \coloneqq\alpha_t\). Since
\(\beta_0 \coloneqq s_1\cdots s_{t-1}( \alpha_t ) < 0\) and
\(\beta_{t-1} = \alpha_t > 0\), there must be a smallest index \(u\)
such that \(\beta_u > 0\). Note that \(u\geq 1\) since \(\beta_0\) is
negative. Then
\begin{align*}
s_u( \beta_u) 
&= s_u s_{u+1} \cdots s_{t-1}( \alpha_t) \\
&= \beta_{u-1} \\
& < 0
\end{align*}
by choice of \(u\). Noting that
\(\beta_u = s_{u+1}\cdots s_{t-1} (\alpha_t)\), by lemma B,
\(s_u = s_{\alpha_u}\) permutes roots other than \(\alpha_u\) since
\(\beta_u > 0\) and \(s_{\alpha_u}( \beta_u) < 0\). By lemma 9.2, write
\begin{align*}
s_{\alpha_u} = s_{\beta_u} = s_{ s_{u+1}\cdots s_{t-1}( \alpha_t ) } = (s_{u+1} \cdots s_{t-1}) s_{\alpha_u} (s_{u+1}\cdots s_{t-1})^{-1}
.\end{align*}
Multiply both sides on the left by \((s_1\cdots s_u)\) and on the right
by \((s_{u+1}\cdots s_{t-1})\) to obtain
\begin{align*}
(s_1 \cdots s_{u-1})(s_{u+1}\cdots s_{t-1}) = (s_1\cdots s_u)(s_{u+1}\cdots s_t), \qquad s_t \coloneqq s_{\alpha_t}
.\end{align*}

\end{proof}

\begin{corollary}[?]

If \(\sigma = s_1\cdots s_t\) is an expression for \(w\in W\) in terms
of simple reflections (which we don't yet know exists, but it does) with
\(t\) minimal, then \(\sigma( \alpha_t) < 0\).

\end{corollary}

\hypertarget{monday-october-17}{%
\section{Monday, October 17}\label{monday-october-17}}

\hypertarget{the-weyl-group}{%
\subsection{10.3: The Weyl group}\label{the-weyl-group}}

\begin{theorem}[?]

Fix a base for \(\Phi\).

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  (\(W\) acts transitively on the set of Weyl chambers) If
  \(\gamma\in {\mathbb{E}}\) is regular (not on a root hyperplane),
  there exists \(\sigma\in W\) such that
  \(( \sigma(\delta), \alpha)> 0\) for all \(\alpha\in \Delta\),
  i.e.~\(\sigma( \gamma) \in {\mathcal{C}}(\Delta)\), the dominant Weyl
  chamber relative to \(\Delta\).
\item
  (\(W\) acts transitively on bases) If \(\Delta'\) is another base for
  \(\Phi\), then there exists \(\sigma\in W\) such that
  \(\sigma( \Delta') = \Delta\), so \(W\) acts transitively on bases.
\item
  (Every orbit of \(W\Phi\) contains a simple root) If \(\beta\in \Phi\)
  then there exists a \(\sigma\in W\) such that
  \(\sigma( \beta)\in \Delta\).
\item
  (\(W\) is generated by simple roots)
  \(W = \left\langle{s_ \alpha {~\mathrel{\Big\vert}~}\alpha\in \Delta}\right\rangle\)
  is generated by \emph{simple} roots.
\item
  (Stabilizers are trivial) If \(\sigma( \Delta) = \Delta\) for some
  \(\sigma\in W\), then \(\sigma = 1\).
\end{enumerate}

\end{theorem}

\begin{proof}[?]

\textbf{Part c}: Set
\(W' \coloneqq\left\langle{s_ \alpha{~\mathrel{\Big\vert}~}\alpha\in \Delta}\right\rangle\),
we'll prove (c) with \(W\) replaced \(W'\), which is larger. First
suppose \(\beta\in \Phi^+\) and consider \(W' \beta \cap\Phi^+\). This
is nonempty since it includes \(\beta\) and is a finite set, so choose
\(\gamma\) in it of minimal height. Claim:
\(\operatorname{ht}(\gamma) = 1\), making \(\gamma\) simple. If not,
supposing \(\operatorname{ht}( \gamma) > 1\), write
\(\gamma = \sum_{ \alpha\in \Delta} k_ \alpha \alpha\) with
\(k_ \alpha > 0\). Since \(\gamma\neq 0\), we have
\((\gamma, \gamma) > 0\), so substitute to yield
\begin{align*}
0 < (\gamma, \gamma) = (\gamma, \sum_{\alpha \in \Delta} k_ \alpha \alpha) = \sum_{\alpha\in \Delta} k_ \alpha (\gamma, \alpha)
,\end{align*}
so \((\gamma, \alpha)>0\) for some \(\alpha \in \Delta\), and
\(s_{ \alpha} \gamma = \gamma - {\left\langle {\gamma },~{\alpha } \right\rangle}\alpha\in \Phi^+\)
is positive where
\({\left\langle {\gamma },~{\alpha } \right\rangle}> 0\). This is a
contradiction, since it has a smaller height. Note that if
\(\beta\in \Phi^-\) then \(-\beta\in \Phi^+\) and there exists a
\(\sigma\in W'\) such that \(\sigma( - \beta) = \alpha\in \Delta\). So
\(\sigma( \beta) = - \alpha\), and
\(s_ \alpha \sigma( \beta) = s_ \alpha( - \alpha) = \alpha \in \Delta\).

\textbf{Part d}: Given \(\beta\), pick \(\sigma\in W'\) such that
\(\sigma^{-1}( \beta) = \alpha\in \Delta\). Then
\begin{align*}
s_ \beta = s_{ \sigma( \alpha)} = \sigma s_{ \alpha} \sigma^{-1}\in W'
,\end{align*}
so \(W \leq W' \leq W\), making \(W = W'\).

\textbf{Parts a and b}: Recall
\(\rho = {1\over 2}\sum _{ \beta\in \Phi^+}\) and choose \(\sigma\in W\)
such that \(( \sigma(\delta), \rho)\) is maximal (picking from a finite
set). Given \(\alpha\in \Delta\), we have \(s_\alpha \sigma\in W\), and
so
\begin{align*}
( \sigma(\delta), \rho) 
&\geq ( \sigma(\delta), \rho) \\
&= (\sigma(\delta), s_ \alpha \rho) \\
&= (\sigma(\delta), \rho - \alpha) \\
&= (\sigma(\delta), \rho ) - ( \sigma( \delta), \alpha)
,\end{align*}
and so \(( \sigma( \delta), \alpha)\geq 0\) for all
\(\alpha\in \Delta\). Importantly, \(\gamma\) is regular, so this
inequality is structure for all \(\alpha\in \Delta\). So \(W\) acts
transitively on the Weyl chambers, and consequently on simple systems
(i.e.~bases for \(\Phi\)) by the discussion at the end of \(\S 10.1\).2

\textbf{Part e}: Suppose \(\sigma( \Delta) = \Delta\) and
\(\sigma \neq 1\), and write \(\sigma = \prod_{1\leq i \leq t} s_i\)
with \(s_i \coloneqq s_{\alpha_i}\) for \(\alpha_i \in \Delta\) with
\(t \geq 1\) minimal. Note \(\sigma( \Delta) = \Delta\) and
\(\alpha_t \in \Delta\), we have \(\sigma( \alpha_t) > 0\) and
\(\prod_{1\leq i\leq t}(\alpha_t) = \prod_{1\leq i \leq t-1}s_i (-\alpha_t)\)
so \(\prod_{1\leq i\leq t-1} s_i(\alpha_t) < 0\). This fulfills the
deletion condition, so
\(\prod_{1\leq i \leq t} = s_1\cdots \widehat{s_u}\cdots \widehat{s_t}\)
which is of smaller length.

\end{proof}

\begin{remark}

In type \(A_n\), \({\sharp}W(A_n) \approx n!\), and since bases biject
with \(W\) there are many choices of bases.

\end{remark}

\begin{definition}[?]

Let \(\Delta \subseteq \Phi\) be a base and write \(\sigma\in W\) as
\(\sigma = \prod_{1\leq i \leq t} s_{\alpha_i}\) with
\(\alpha_i\in \Delta\) and \(t\) minimal. We say this is a
\textbf{reduced expression} for \(\sigma\) and say \(t\) is the
\textbf{length} of \(\sigma\), denoted \(\ell( \sigma)\). By definition,
\(\ell(1) = 0\).

\end{definition}

\begin{remark}

Since \(W \leq \operatorname{GL}({\mathbb{E}})\), there is a map
\(\operatorname{det}: W\to \operatorname{GL}_1({\mathbf{R}}) = {\mathbf{R}}^{\times}\).
The determinant of a reflection is \(-1\) by writing it in a basis about
the fixed hyperplane, and so
\(\operatorname{det}\sigma = (-1)^{\ell( \sigma)}\) and in fact
\(\operatorname{det}: W\to \left\{{\pm 1}\right\}\). Thus
\(\ell( \sigma \sigma') \equiv \ell( \sigma) + \ell( \sigma)\operatorname{mod}2\).

Note also that if \(\sigma' = s_ \alpha\) for \(\alpha\) simple, then
\(\ell( \sigma s_{ \alpha}) = \ell( \sigma) \pm 1\). The proof:
\(\ell( \sigma s_ \alpha)\leq \ell( \sigma) + 1\), similarly for
\(\sigma s_ \alpha\), and use
\(\operatorname{det}( \sigma s_ \alpha) = - \operatorname{det}\sigma\).

\end{remark}

\begin{warnings}

Reduced expressions are not unique: for \(A_2\), one has
\(s_ \alpha s_ \beta s_ \alpha = s_ \beta s_ \alpha s_ \beta\), and
these two reflections do not commute.

\end{warnings}

\begin{remark}

Some temporary notation for this section: for \(\sigma\in W\), set
\begin{align*}
n( \sigma) \coloneqq{\sharp}( \Phi^- \cap\sigma(\Phi^+))
,\end{align*}
the number of positive roots that \(\sigma\) sends to negative roots.

\end{remark}

\begin{lemma}[A]

For all \(\sigma\in W\),
\begin{align*}
n( \sigma) = \ell( \sigma)
.\end{align*}

\end{lemma}

\begin{proof}[?]

Induct on \(\ell(\sigma)\): if zero, then \(\sigma = 1\) and
\(n(1) = 0\) since it fixes all positive roots. If
\(\ell( \sigma ) = 1\) then \(\sigma = s_{ \alpha}\) for some simple
\(\alpha\), and we know from the last section that \(\sigma\) permutes
\(\Phi^+\setminus\left\{{ \alpha }\right\}\) and
\(\sigma( \alpha) = - \alpha\), so \(n( \sigma) = 1\).

\end{proof}

\hypertarget{wednesday-october-19}{%
\section{Wednesday, October 19}\label{wednesday-october-19}}

\begin{proof}[of lemma A, continued]

We're proving
\(\ell( \sigma) = n(\sigma) \coloneqq{\sharp}( \Phi^- \cap\sigma(\Phi^-))\)
by induction on \(\ell(\sigma)\), where we already checked the zero
case. Assume the result for all \(\tau\) with
\(\ell( \tau) \leq \ell( \sigma)\) for \(\tau \in W\). Write
\(\sigma = s_1\cdots s_t\) with
\(s_i \coloneqq s_{ \alpha_i}, \alpha_i\in \Delta\) reduced. Set
\(\tau \coloneqq\sigma s_t = s_1\cdots s_{t-1}\) which is again reduced
with \(\ell(\tau) = \ell( \sigma) - 1\). By the deletion condition,
\(s_1 \cdots s_{t-1}( \alpha_t) > 0\), so
\(s_1\cdots s_{t-1}s_t (\alpha_t) = s_1 \cdots s_{t-1}(- \alpha_t) < 0\).
Thus \(n(\tau) = n( \sigma) - 1\), since \(s_t\) permutes
\(\Phi^+\setminus\left\{{ \alpha_t }\right\}\), so
\begin{align*}
\ell( \sigma) - 1 = \ell( \tau) = n( \tau) = n( \sigma) -1 \implies \ell( \sigma) = n( \sigma)
.\end{align*}

\end{proof}

\begin{remark}

This is useful for finding reduced expressions, or at least their
length: just compute how many positive roots change sign under
\(\sigma\). Using the deletion condition and lemma A, it's clear that
any expression for \(\sigma\) as a product of simple reflections can be
converted into a \emph{reduced} expression by deleting pairs of simple
reflections, and this terminates after finitely many steps.

\end{remark}

\begin{lemma}[B]

Recall that the open Weyl chambers are the complements of hyperplanes.
The closure of any Weyl chamber is a fundamental domain for the action
\(W\curvearrowright{\mathbb{E}}\).

\end{lemma}

\hypertarget{irreducible-root-systems}{%
\subsection{10.4: Irreducible root
systems}\label{irreducible-root-systems}}

\begin{definition}[Irreducible root systems]

A root system \(\Phi \subseteq {\mathbb{E}}\) is \textbf{irreducible} if
it cannot be partitioned into mutually orthogonal nonempty subsets.
Otherwise, \(\Phi\) is \textbf{reducible}.

\end{definition}

\begin{proposition}[?]

Let \(\Delta \subseteq \Phi\) be a simple system. Then \(\Phi\) is
irreducible iff \(\Delta\) is irreducible, i.e.~\(\Delta\) cannot be
partitioned into nonempty orthogonal subsets.

\end{proposition}

\begin{proof}[?]

\(\Phi\) reducible implies \(\Delta\) reducible: write
\(\Phi = \Phi_1 {\textstyle\coprod}\Phi_2\) where
\(( \Phi_1, \Phi_2) = 0\); this induces a similar partition of
\(\Delta\). Then
\((\Delta, \Phi_2) = 0 \implies ({\mathbb{E}}, \Phi_2) = 0 \implies {\mathbb{E}}= \emptyset\)
using nondegeneracy of the bilinear form. \(\contradiction\)

Now \(\Delta\) reducible implies \(\Phi\) reducible: write
\(\Delta =\Delta_1 {\textstyle\coprod}\Delta_2\) with
\((\Delta_1, \Delta_2) = 0\). Let \(\Phi_i\) be the roots which are
\(W{\hbox{-}}\)conjugate to an element of \(\Delta_i\). Then elements in
\(\Phi_i\)are obtained from \(\Delta_i\) by adding and subtracting only
elements of \(\Delta_i\), so \((\Phi_1, \Phi_2) = 0\) and
\(\Phi = \Phi_1 \cup\Phi_2\) by a previous lemma that every
\(\beta\in \Phi\) is conjugate to some \(\alpha\in \Delta\).

\end{proof}

\begin{lemma}[A]

Let \(\Phi \supseteq\Delta\) be irreducible. Relative to the partial
order \(\leq\) on roots, there is a unique maximal root
\(\tilde \alpha\). In particular, if \(\beta\in \Phi\) and
\(\beta\neq \tilde \alpha\), then
\(\operatorname{ht}( \beta) < \operatorname{ht}( \tilde \alpha)\) and
\((\tilde \alpha, \alpha) \geq 0\) for all \(\alpha\in \Delta\).
Moreover, one can write \(\tilde \alpha = \sum _{\alpha\in \Delta}\)
with \(k_\alpha > 0\), i.e.~it is a sum where every simple root appears.

\end{lemma}

\begin{proof}[?]

\textbf{Existence}: Let \(\tilde \alpha\) be any maximal root in the
ordering. Given \(\alpha \in \Delta, (\tilde \alpha, \alpha) \geq 0\) --
otherwise
\(s_ \alpha(\tilde \alpha)= \tilde \alpha-{\left\langle { \tilde \alpha},~{\alpha} \right\rangle} \alpha > \alpha\),
a contradiction. \(\contradiction\) Write
\(\tilde \alpha = \sum_{\alpha\in \Delta} k_ \alpha \alpha\) with
\(k_ \alpha \in {\mathbf{Z}}_{\geq 0}\), where it's easy to see these
are all non-negative. Suppose some \(k_\gamma = 0\), then
\((\tilde \alpha, \gamma)\leq 0\) -- otherwise
\(s_\gamma( \tilde \alpha) = \tilde \alpha - {\left\langle {\tilde \alpha},~{\gamma} \right\rangle} \gamma\)
has both positive and negative coefficients, which is not possible.
Since \((\tilde \alpha, \alpha) \geq 0\), we must have
\(( \tilde \alpha, \gamma) = 0\). So write
\begin{align*}
0 = (\tilde \alpha, \gamma) = \sum_{ \alpha\in \Delta} k_ \alpha( \alpha, \gamma) \leq 0
,\end{align*}
so \(( \alpha, \gamma)= 0\) whenever \(k_\alpha \neq 0\), otherwise this
expression would be strictly \(< 0\). Thus
\(\Delta = \Delta_1 {\textstyle\coprod}\Delta_2\) where
\(\Delta_1 = \left\{{\alpha\in \Delta {~\mathrel{\Big\vert}~}K_ \alpha\neq 0}\right\}\)
and
\(\Delta_2 = \left\{{ \alpha\in \Delta{~\mathrel{\Big\vert}~}k_ \alpha = 0 }\right\}\).
This is an orthogonal decomposition of \(\Delta\), since any
\(\gamma \in \Delta_2\) is orthogonal to any \(\alpha\in \Delta_1\).
Note that \(\Delta_1\neq \empty\) since \(\tilde\alpha \neq 0\), and if
\(\Delta_2\neq \empty\) then this is a contradiction, so \(\Delta_2\)
must be empty. So no such \(\gamma\) exists.

\textbf{Uniqueness}: let \(\tilde \alpha\) be any maximal root in the
ordering and let \(\tilde \alpha'\) be another such root. Then
\((\tilde \alpha, \tilde \alpha') = \sum_{\alpha\in \Delta} k_ \alpha (\alpha, \tilde \alpha')\)
with \(k_ \alpha > 0\) and \((\alpha, \tilde \alpha') \geq 0\). So
\((\tilde \alpha, \tilde \alpha ') > 0\) since \(\Delta\) is a basis for
\({\mathbb{E}}\) and anything orthogonal to a basis is zero by
nondegeneracy of the form. Since \(\tilde \alpha \neq 0\), it is not
orthogonal to everything. By Lemma 9.4, either
\(\tilde \alpha, \tilde \alpha '\) are proportional (which was excluded
in the lemma), in which case they are equal since they're both positive,
or otherwise \(a \coloneqq\tilde \alpha - \tilde \alpha' \in \Phi\) is a
root. In the latter case,
\(a > 0 \implies \tilde \alpha > \tilde \alpha'\) or
\(a< 0 \implies \tilde \alpha < \tilde \alpha'\), both contradicting
maximality.

\end{proof}

\begin{remark}

If \(\beta = \sum_{\alpha\in \Delta} m_ \alpha \alpha \in \Phi^+\), then
\(m_ \alpha \leq k_ \alpha\) for all \(\alpha\) since
\(\beta \leq \alpha\).

\end{remark}

\begin{lemma}[B]

If \(\Phi\) is irreducible then \(W\) acts irreducibly on
\({\mathbb{E}}\) (so there are no \(W{\hbox{-}}\)invariant subspaces).
In particular, the \(W{\hbox{-}}\)orbit of a root spans
\({\mathbb{E}}\).

\end{lemma}

\begin{proof}[?]

Omitted.

\end{proof}

\begin{lemma}[C]

If \(\Phi\) is irreducible, then at most two root lengths occur, denoted
\textbf{long} and \textbf{short} roots.

\end{lemma}

\begin{proof}[?]

Omitted.

\end{proof}

\begin{example}[?]

\(B_2\) has 4 long roots and 4 short roots, since they fit in a square:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-19_09-57.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

Similarly \(G_2\) has long and short roots, fitting into a star of
David.

\end{example}

\begin{lemma}[D]

If \(\Phi\) is irreducible then the maximal root \(\tilde \alpha\) is a
long root.

\end{lemma}

\begin{remark}

There is also a unique maximal short root.

\end{remark}

\begin{proof}[?]

Omitted.

\end{proof}

\hypertarget{friday-october-21}{%
\section{Friday, October 21}\label{friday-october-21}}

\hypertarget{the-cartan-matrix}{%
\subsection{11.1: The Cartan Matrix}\label{the-cartan-matrix}}

\begin{definition}[Cartan matrix]

Fix \(\Delta \subseteq \Phi\) a rank \(\ell\) root system with Weyl
group \(W\). Let
\(\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{\ell} }\right\}\)
and then the matrix \(A\) where
\(A_{ij} = {\left\langle {\alpha_i},~{\alpha_j} \right\rangle} = 2{{\left\langle {\alpha_i},~{\alpha_j} \right\rangle} \over {\left\langle {\alpha_j},~{\alpha_j} \right\rangle}}\)
is the \textbf{Cartan matrix of \(A\).} Note that changing the ordering
of \(\Delta\) permutes the rows and columns of \(A\), but beyond this,
\(A\) does not depend on the choice of \(\Delta\) since they are
permuted by \(W\) and \(W\) preserves the inner products and thus the
ratios defining the \emph{Cartan numbers} \(A_{ij}\). More
\(A\in \operatorname{GL}_\ell({\mathbf{Z}})\) since the inner product is
nondegenerate and \(\Delta\) is a basis for \({\mathbb{E}}\).

\end{definition}

\begin{example}[?]

Note that the diagonals are always 2. Some classical types:

\begin{itemize}
\tightlist
\item
  \(A_1 \times A_1: { \begin{bmatrix}  {2} & {0} \\  {0} & {2} \end{bmatrix} }\)
\item
  \(A_2: { \begin{bmatrix}  {2} & {-1} \\  {-1} & {2} \end{bmatrix} }\)
\item
  \(B_2: { \begin{bmatrix}  {2} & {-2} \\  {-1} & {2} \end{bmatrix} }\)
\item
  \(G_2: { \begin{bmatrix}  {2} & {-1} \\  {-3} & {2} \end{bmatrix} }\).
\end{itemize}

\end{example}

\begin{remark}

The Cartan matrix \(A\) determines the root system \(\Phi\) up to
isomorphism: if \(\Phi' \subseteq {\mathbb{E}}'\) is another root system
with base
\(\Delta' = \left\{{ { \alpha'}_{1}, \cdots, { \alpha'}_{\ell} }\right\}\)
with \(A'_{ij} = A_{ij}\) for all \(i, j\) then the bijection
\(\alpha_i \mapsto \alpha_i'\) extends to a bijection
\(\phi: {\mathbb{E}} { \, \xrightarrow{\sim}\, }{\mathbb{E}}'\) sending
\(\Phi\) to \(\Phi'\) which is an isometry,
i.e.~\({\left\langle {\varphi(\alpha)},~{\varphi( \beta)} \right\rangle} = {\left\langle {\alpha },~{\beta} \right\rangle}\)
for all \(\alpha, \beta \in \Phi\). Since \(\Delta, \Delta'\) are bases
of \({\mathbb{E}}\), this gives a vector space isomorphism
\(\phi(\alpha_i) \coloneqq\alpha_i'\). If \(\alpha, \beta\in \Delta\)
are simple, then
\begin{align*}
s_{\varphi( \alpha)}( \varphi( \beta)) 
&= \varphi( \beta)- {\left\langle {\beta'},~{\alpha'} \right\rangle}\phi( \alpha) \\
&= \varphi( \beta)-{\left\langle { \beta},~{ \alpha} \right\rangle} \phi( \alpha) \\
&= \phi(\beta- {\left\langle {\beta },~{\alpha } \right\rangle}\alpha) \\
&= \phi( s_ \alpha( \beta))
,\end{align*}
so this diagram commutes since these maps agree on the simple roots,
which form a basis:

\begin{center}
\begin{tikzcd}
    {\mathbb{E}}&& {\mathbb{E}}\\
    \\
    {\mathbb{E}}&& {\mathbb{E}}
    \arrow["\phi", from=1-1, to=1-3]
    \arrow["{s_\alpha}", from=1-3, to=3-3]
    \arrow["\phi"', from=3-1, to=3-3]
    \arrow["{s_\alpha}"', from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXEVFIl0sWzIsMCwiXFxFRSJdLFswLDIsIlxcRUUiXSxbMiwyLCJcXEVFIl0sWzAsMSwiXFxwaGkiXSxbMSwzLCJzX1xcYWxwaGEiXSxbMiwzLCJcXHBoaSIsMl0sWzAsMiwic19cXGFscGhhIiwyXV0=}{Link
to Diagram}
\end{quote}

Since \(W, W'\) are generated by reflections and
\(s_{ \varphi( \alpha)} = \varphi\circ s_ \alpha \circ \varphi^{-1}\)
for \(\alpha\in \Delta\), there is an isomorphism
\begin{align*}
W & { \, \xrightarrow{\sim}\, }W \\
s_ \alpha &\mapsto s_{ \varphi( \alpha)} = \varphi s_ \alpha \varphi \quad \forall \alpha \in \Delta
.\end{align*}
If \(\beta \in \Phi\), then \(\beta = w( \alpha)\) for some
\(\alpha\in \Delta\) and \(w\in W\) by theorem 10.3C. Thus
\(\phi( \beta) = ( \varphi \circ w \circ \varphi^{-1})( \varphi( \alpha))\in \Phi'\)
since \(\varphi\circ w \circ \varphi^{-1}\in W'\). Thus
\(\phi( \Phi) = \Phi'\). Using lemma 9.2,
\(s_{\varphi(\beta)} = \varphi s_ \beta \varphi^{-1}\), so \(\phi\)
preserves all of the Cartan integers
\({\left\langle {\beta },~{\gamma} \right\rangle}\) for all
\(\gamma, \beta\in \Phi\).

\end{remark}

\begin{remark}

Read the last paragraph of \(\S 11.1\) which gives an algorithm for
constructing \(\Phi^+\) from \(\Delta\) and \(A\).

\end{remark}

\hypertarget{coxeter-graphs-and-dynkin-diagrams}{%
\subsection{11.2: Coxeter graphs and Dynkin
diagrams}\label{coxeter-graphs-and-dynkin-diagrams}}

\begin{definition}[Coxeter graph]

If \(\alpha\neq \beta\in \Phi^+\) then
\({\left\langle {\beta },~{\alpha} \right\rangle}{\left\langle {\alpha },~{\beta } \right\rangle}= 0,1,2,3\)
from the table several sections ago. Fix
\(\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{\ell} }\right\}\),
then the \textbf{Coxeter graph} \(\Gamma\) of \(\Phi\) is the graph with
\(\ell\) vertices \(1,\cdots, \ell\) with vertices \(i, j\) connected by
\({\left\langle {\alpha_i},~{ \alpha_j} \right\rangle} {\left\langle {\alpha_j},~{\alpha_i} \right\rangle}\)
edges.

\end{definition}

\begin{example}[?]

Recall that the table was

\begin{longtable}[]{@{}
  >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.49}}
  >{\raggedright\arraybackslash}p{(\columnwidth - 2\tabcolsep) * \real{0.49}}@{}}
\toprule
\({\left\langle {\alpha },~{\beta} \right\rangle}\) &
\({\left\langle {\beta },~{\alpha} \right\rangle}\) \\
\midrule
\endhead
0 & 0 \\
-1 & -1 \\
-1 & -2 \\
-1 & -3 \\
\bottomrule
\end{longtable}

Here \(\alpha\) is the shorter root., although without loss of
generality in the first two rows we can rescale so that
\({\left\lVert {\alpha} \right\rVert}= {\left\lVert {\beta} \right\rVert}\).
The graphs for some classical types:

\includegraphics{figures/2022-10-21_09-42-15.png}

\end{example}

\begin{remark}

If \(\Phi\) has roots all of the same length, the Coxeter graph
determines the Cartan integers since \(A_{ij} = 0, 1\) for \(i\neq j\).
If \(i \to j\) is a subgraph of \(\Gamma\) then
\({\left\langle { \alpha_i},~{ \alpha_j} \right\rangle} = {\left\langle { \alpha_j},~{\alpha_i} \right\rangle} = -1\),
so \(\alpha_i, \alpha_j\) have the same length. However, if there are
roots of multiple lengths, taking the product to determine the number of
edges loses information about which root is longer.

\end{remark}

\begin{definition}[Dynkin diagram]

The \textbf{Dynkin diagram} of \(\Phi\) is the Coxeter graph \(\Gamma\)
where for each multiple edge, there is an arrow pointing from the longer
root to the shorter root.

\end{definition}

\begin{example}[?]

In rank 2:

\includegraphics{figures/2022-10-21_09-50-47.png}

We also have the following diagram for \(F_4\):

\includegraphics{figures/2022-10-21_09-51-32.png}

\end{example}

\begin{remark}

Note that \(\Phi\) is irreducible iff \(\Delta\) can not be partitioned
into two proper nonempty orthogonal subsets iff the Coxeter graph is
connected. In general, if \(\Gamma\) has \(t\) connected components, let
\(\Delta = \coprod_{1\leq i\leq t} \Delta_i\) be the corresponding
orthogonal partition of simple roots. Let
\({\mathbb{E}}_i = \mathop{\mathrm{span}}_{\mathbf{R}}\Delta_i\), then
\({\mathbb{E}}= \bigoplus_{1\leq i\leq t}{\mathbb{E}}_i\) is an
orthogonal direct sum decomposition into \(W{\hbox{-}}\)invariant
subspaces, which follows from the reflection formula. Writing
\(\Phi_i = ({\mathbf{Z}}\Delta_I) \cap\Phi\), one has
\(\Phi = \coprod_{1\leq i\leq t} \Phi_i\) since each root is
\(W{\hbox{-}}\)conjugate to a simple root and \({\mathbf{Z}}\Delta_i\)
is \(W{\hbox{-}}\)invariant and each \(\Phi_i \subseteq {\mathbb{E}}_i\)
is itself a root system. Thus it's enough to classify irreducible root
systems.

\end{remark}

\hypertarget{monday-october-24}{%
\section{Monday, October 24}\label{monday-october-24}}

\begin{remark}

Classifying root systems:
\(\Delta \subseteq \Phi \subseteq {\mathbb{E}}\) a base yields a
decomposition

\begin{itemize}
\tightlist
\item
  \({\mathbb{E}}= \bigoplus_{i=1}^t {\mathbb{E}}_i\),
\item
  \(\Phi= \bigoplus_{i=1}^t \Phi_i\),
\item
  \(\Delta = \bigoplus_{i=1}^t \Delta_i\), where are orthogonal direct
  sums with respect to \(({-}, {-})\). Note that the sub-bases
  \(\Delta_i\) biject with connected components of the Coxeter graph
  \(\Gamma\) of \(\Delta\). We saw
  \({\left\langle {\alpha_i},~{\alpha_j} \right\rangle} {\left\langle {\alpha_j},~{\alpha_i} \right\rangle} \in \left\{{0,1,2,3}\right\}\)
  is the number of edges between nodes \(i\) and \(j\) in \(\Gamma\),
  using that the first term is
  \(4\cos^2(\theta)\in [0, 3] \cap{\mathbf{Z}}\). It suffices to
  classify irreducible root systems, corresponding to connected Coxeter
  graphs. Recall arrows point from long to short roots.
\end{itemize}

\end{remark}

\begin{theorem}[?]

If \(\Phi\) is an irreducible root system of rank \(\ell\), then its
Dynkin diagram is one of the following:

\begin{itemize}
\tightlist
\item
  The four infinite families, corresponding to classical types:
\end{itemize}

\includegraphics{figures/2022-10-24_09-19-12.png}

\begin{itemize}
\tightlist
\item
  Exceptional classes
\end{itemize}

\includegraphics{figures/2022-10-24_09-19-36.png}

Types ADE are called \textbf{simply laced} since they have no multiple
edges.

\end{theorem}

\begin{remark}

Idea: classify possible connected Coxeter graphs, ignoring relative root
lengths. If \(\alpha, \beta\) are simple roots, note that for any \(c\),
\begin{align*}
{\left\langle { \alpha },~{ \beta } \right\rangle} {\left\langle { \beta },~{ \alpha } \right\rangle} = {2(c \alpha, \beta)\over (\beta, \beta)} {2( \beta, c \alpha) \over (c \alpha, c \alpha)} \in \left\{{0,1,2,3}\right\}
,\end{align*}
so \(\alpha \mapsto c\alpha\) leaves this number invariant and we can
assume all simple roots are unit vectors.

\end{remark}

\begin{definition}[?]

Let \({\mathbb{E}}\) be a finite dimensional Euclidean space, then a
subset
\(A =\left\{{{ {\varepsilon}_1, \cdots, {\varepsilon}_{n} } }\right\}\subseteq {\mathbb{E}}\)
of linearly independent unit vectors satisfying

\begin{itemize}
\tightlist
\item
  \(({\varepsilon}_i, {\varepsilon}_j) \leq 0\) for all \(i,j\),
\item
  \(4({\varepsilon}_i, {\varepsilon}_j)^2 = 4\cos^2(\theta) \in \left\{{0,1,2,3}\right\}\)
  for all \(i\neq j\) where \(\theta\) is the angle between
  \({\varepsilon}_i\) and \({\varepsilon}_j\)
\end{itemize}

is called \textbf{admissible}.

\end{definition}

\begin{example}[?]

Any base for a root system where each vector is normalized is
admissible.

\end{example}

\begin{remark}

To such an \(A\) we associate a graph \(\Gamma\) as before with vertices
\(1,\cdots, n\) where \(i,j\) are joined by
\(4({\varepsilon}_i, {\varepsilon}_j)^2\) edges. We'll determine all
connected graphs \(\Gamma\) that can occur, since these include all
connected Coxeter graphs.

\end{remark}

\hypertarget{proof-of-classification}{%
\subsection{Proof of classification}\label{proof-of-classification}}

\begin{proof}[Sketch]

An easy 10 steps:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  If some \({\varepsilon}_i\) are discarded, the remaining ones still
  form an admissible set in \({\mathbb{E}}\) whose graph is obtained
  from \(\Gamma\) by omitting the corresponding discarded vertices.
\item
  The number of pairs of vertices in \(\Gamma\) connected by at least
  one edge is strictly less than \(n\).
\end{enumerate}

Proof: Set \({\varepsilon}\coloneqq\sum_{i=1}^n {\varepsilon}_i\), which
is nonzero by linear independence. Then
\(0 < ({\varepsilon}, {\varepsilon}) = n + \sum_{i<j} 2 ({\varepsilon}_i, {\varepsilon}_j)\),
and if \(i< j\) are joined, so
\(({\varepsilon}_i, {\varepsilon}_j)\neq 0\), then
\(4({\varepsilon}_i, {\varepsilon}_j)^2 = 1,2,3\) and so
\(2({\varepsilon}_i, {\varepsilon}_j)\leq -1\). However, since the sum
is positive, there are \(\leq n-1\) such pairs.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  \(\Gamma\) contains no cycles.
\end{enumerate}

Proof: A cycle is subgraph that corresponds to an admissible subset
\(A' \subseteq A\) with \(m\) vertices and \(m\) edges corresponding to
\(m\) connected pairs. This contradicts (2).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{3}
\tightlist
\item
  No more than 3 edges can be incident to any vertex of \(\Gamma\).
\end{enumerate}

Let \({\varepsilon}\in A\) be a vertex, and suppose
\({ \eta _1, \cdots, \eta _{k} }\) are the vertices connected to
\({\varepsilon}\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-24_09-40.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

By (3), no two \(\eta_i, \eta_j\) are connected, so
\(( \eta_i, \eta_j) = 0\) for all \(i,j\). Apply Gram-Schmidt to
\(\left\{{{ \eta _1, \cdots, \eta _{k} } , {\varepsilon}}\right\}\) only
involves modifying \({\varepsilon}\) since
\(\left\{{{ \eta _1, \cdots, \eta _{k} }}\right\}\) are already
orthonormal. Call the new vector \(\eta_0\) and let
\(\left\{{{ \eta _0, \cdots, \eta _{k} }}\right\}\) be the resulting
orthonormal set. One can then write
\({\varepsilon}= \sum_{i=0}^k ({\varepsilon}, \eta_i)\eta_i\), which
implies \(({\varepsilon}, \eta_0)\neq 0\) by linear independence. Then
\(1 = ({\varepsilon}, {\varepsilon}) = \sum_{i=0}^k ({\varepsilon}, \eta_i)^2\),
but \(({\varepsilon}, \eta_0)^2 > 0\) so
\(\sum_{i=1}^k ({\varepsilon}, \eta_i)^2 < 1\) and thus
\(\sum_{i=1}^k 4 ({\varepsilon}, \eta_i)^2 < 4\). But this sum is the
number of edges incident to \({\varepsilon}\) in \(\Gamma\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{4}
\item
  The only connected graph which contains a triple edge is the Coxeter
  graph of \(G_2\) by (4), since the triple edge forces each vertex to
  already have 3 incident edges.
\item
  Let
  \(\left\{{{\varepsilon}_1,\cdots, {\varepsilon}_k}\right\} \subseteq A\)
  have a simple chain \(\cdot \to \cdot \to \cdots \to \cdot\) as a
  subgraph. If
  \(A' \coloneqq\left\{{A\setminus\left\{{ { {\varepsilon}_1, \cdots, {\varepsilon}_{k} } }\right\}}\right\} \cup\left\{{{\varepsilon}}\right\}\)
  where \({\varepsilon}\coloneqq\sum_{i=1}^k {\varepsilon}_i\), then
  \(A'\) is admissible. The corresponding graph \(\Gamma'\) is obtained
  by shrinking the chain to a point, where any edge that was incident to
  any vertex in the chain is now incident to \({\varepsilon}\), with the
  same multiplicity.
\end{enumerate}

Proof: number the vertices in the chain \(1,\cdots, k\). Linear
independence of \(A'\) is clear. Note
\(4({\varepsilon}_i, {\varepsilon}_{i+1})^2 = 1\implies 2({\varepsilon}_i, {\varepsilon}_{i+1}) \implies ({\varepsilon}, {\varepsilon}) = k + 2 \sum_{i< j} ({\varepsilon}_i, {\varepsilon}_j) = k + (-1)(k-1) = 1\).
Any
\(\eta\in A\setminus\left\{{ { {\varepsilon}_1, \cdots, {\varepsilon}_{k} } }\right\}\)
is connected to at most one of
\({ {\varepsilon}_1, \cdots, {\varepsilon}_{k} }\) since this would
otherwise form a cycle, so
\((\eta, {\varepsilon}) = (\eta, {\varepsilon}_i)\) for a single \(i\).
So
\(4(\eta, {\varepsilon})^2 = 4(\eta, {\varepsilon}_i)^2 \in \left\{{0,1,2,3}\right\}\)
and \((\eta, {\varepsilon}) = (\eta, {\varepsilon}_i) \leq 0\), which
verifies all of the admissibility criteria.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{6}
\tightlist
\item
  \(\Gamma\) contains no graphs of the following forms:
\end{enumerate}

\includegraphics{figures/2022-10-24_10-01-11.png}

Proof: collapsing the chain in the middle produces a vertex with 4
incident edges.

\end{proof}

\hypertarget{wednesday-october-26}{%
\section{Wednesday, October 26}\label{wednesday-october-26}}

\begin{quote}
Missed first 15m!
\end{quote}

\begin{proposition}[Step 8]

\includegraphics{figures/2022-10-26_09-28-53.png}

\end{proposition}

\begin{proposition}[Step 9]

The only connected \(\Gamma\) graphs of the second type in Step 8 are
either \(F_4\) or \(B_\ell = C_\ell\).

Compute
\begin{align*}
({\varepsilon}, {\varepsilon}) = \sum_{i=1}^k i^2 - \sum_{i=1}^{p-1} i(i+1) \\ 
= p^2 - \sum_{i=1}^{p-1} i \\
= p^2 - {p(p-1)\over 2} \\
= {p(p+1)\over 2}
,\end{align*}
and similarly \((\eta, \eta) = {q(q+1)\over 2}\). Note
\(4({\varepsilon}_p, \eta_q)^2 = 2\), so
\(({\varepsilon},\eta)^2 = p^2 q^2 ({\varepsilon}_p, \eta_q)^2 = {p^2q^2\over 2}\).
By Cauchy-Schwarz,
\(({\varepsilon}, \eta)^2< ({\varepsilon}, {\varepsilon}) (\eta, \eta)\),
where the inequality is strict since \(\eta, {\varepsilon}\) are
linearly independent. Then check
\begin{align*}
{p^2 q^2\over 2} &< {p(p+1)\over 2} \cdot {q(q+1)\over 2} \\
{p1\over 2} &< {p+1\over 2}\cdot {q+1\over 2} \\
2pq &< pq + p + q + 1
,\end{align*}
and so combining these yields \(pq-p-q+1 < 2\) and thus
\begin{align*}
(p-1)(q-1) < 2
.\end{align*}
Since \(p\geq q\geq 1\), this yields two possible cases:

\begin{itemize}
\tightlist
\item
  \(p=q=2 \leadsto F_4\)
\item
  \(q=1, p\in {\mathbf{Z}}_{\geq 0} \leadsto B_\ell = C_\ell\).
\end{itemize}

\end{proposition}

\begin{proposition}[Step 10]

The only connected \(\Gamma\) of type (d) are \(D_\ell, E_6, E_7, E_8\).

Set \({\varepsilon}\coloneqq\sum i{\varepsilon}_i\),
\(\eta \coloneqq\sum i\eta_i\), and \(\zeta = \sum i \zeta_i\). Note
that \({\varepsilon}, \eta, \zeta\) mutually orthogonal by inspecting
the graph, and \(\psi\) is not in their span. Let \(\theta_1\) (resp.
\(\theta_2, \theta_3\)) be the angles between \({\varepsilon}\) (resp.
\(\eta, \zeta\)) and \(\psi\). Since \({\varepsilon},\eta,\zeta\) are
linearly independent, the idea is to apply Gram-Schmidt to
\(\left\{{{\varepsilon},\eta,\zeta,\psi}\right\}\) without normalizing.
The first 3 are already orthogonal, so we get a new orthogonal basis
\(\left\{{\psi_1 \coloneqq{\varepsilon}, \psi_2\coloneqq\eta, \psi_3\coloneqq\zeta, \psi_0}\right\}\)
where \((\psi_0, \psi) \neq 0\). We can expand \(\psi\) in this basis to
write
\(\psi = \sum_{i=0}^3 \qty{\psi, {\psi_i\over {\left\lVert {\psi_i} \right\rVert}}} {\psi_i \over {\left\lVert {\psi_i} \right\rVert}}\).
Note that \((\psi, \psi) = 1\), and consequently
\(\sum_{i=1}^3 \qty{\psi, {\psi_i \over{\left\lVert {\psi_i} \right\rVert}}}^2 < 1 \implies \sum_{i=1}^3 \cos^2(\theta_i) < 1\).
So
\begin{align*}
\cos^2(\theta_1) + \cos^2( \theta_2) + \cos^2( \theta_3) < 1
.\end{align*}

As in Step (9), \(({\varepsilon}, {\varepsilon}) = {p(p-1)\over 2}\) and
similarly for \(\eta,\zeta\), and so
\begin{align*}
\cos^1( \theta_1) = {({\varepsilon}, \psi)^2 \over ({\varepsilon}, {\varepsilon}) (\psi, \psi)} 
= {(p-1)^2 ({\varepsilon}_{p-1}, \psi )^2 \over {p(p-1)\over 2} \cdot 1 } \\
= {p-1\over p} {1/4\over 1/2} \\
= {1\over 2}\qty{1-{1\over p}}
,\end{align*}
where we've used that \(4({\varepsilon}_{p-1},\psi ) = 1\). Similarly
(and summarizing),
\begin{align*}
\cos^2(\theta_1) &= {1\over 2}\qty{1-{1\over p}} \\
\cos^2(\theta_2) &= {1\over 2}\qty{1-{1\over q}} \\
\cos^2(\theta_3) &= {1\over 2}\qty{1-{1\over r}} \\ \\
&\implies {1\over 2}\qty{ 1 - {1\over p} + 1 - {1\over q} + 1 - {1\over r}} < 1 \\
&\implies p^{-1}+ q^{-1}+r^{-1}> 1
.\end{align*}
and since
\(p\geq q\geq r\geq 2 \implies p^{-1}\leq q^{-1}\leq r^{-1}\leq 2^{-1}\),
we have \({3\over r} > 1\) by replacing \(p,q\) with \(r\) above. So
\(r < 3\), forcing \(r=2\), and there is only one ``top leg'' in the
graph for (d) above.

We also have
\begin{align*}
{2\over q} \geq {1\over p} + {1\over q} > {1\over 2},
\qquad (\star)
.\end{align*}
so \(q<4\) forces \(q=2,3\).

\begin{itemize}
\tightlist
\item
  If \(q=2\), then \((\star)\) is true for any \(p\geq 2\), and the
  bottom leg has two vertices and this yields type \(D_\ell\).
\item
  If \(q=3\) then \({1\over p} > {1\over 2}-{1\over 3} = {1\over 6}\)
  implies \(p < 6\), forcing \(p=3,4,5\) corresponding to
  \(E_6, E_7, E_8\).
\end{itemize}

\end{proposition}

\begin{remark}

Note that the diagrams we've constructed are the only possible Coxeter
graphs of a root system, since normalizing any set of simple roots
yields an admissible set. This proves one direction of a correspondence,
but what are all possible Dynkin diagrams? Note that types
\(B_\ell, C_\ell\) have the same underlying Coxeter graph, and only
differ by directions on the multi-edges.

\end{remark}

\begin{question}

Does every connected Dynkin diagram correspond to an irreducible root
system.

Yes: types \(A,B,C,D\) can be constructed from root systems in classical
Lie algebras, and the corresponding Dynkin diagrams can be constructed
directly. The 5 exceptional types must be constructed directly.

\end{question}

\begin{question}

Does each irreducible root system occur as the root system of some
semisimple Lie algebra over \({\mathbf{C}}\)?

The answer is of course: yes!

\end{question}

\begin{quote}
Next time: starting Ch. V.
\end{quote}

\hypertarget{part-v-existence-theorem.-ch.-17-the-universal-enveloping-algebra-monday-october-31}{%
\section{Part V: Existence Theorem. Ch. 17: The Universal Enveloping
Algebra (Monday, October
31)}\label{part-v-existence-theorem.-ch.-17-the-universal-enveloping-algebra-monday-october-31}}

\hypertarget{the-tensor-algebra-and-symmetric-algebra}{%
\subsection{17.1: The tensor algebra and symmetric
algebra}\label{the-tensor-algebra-and-symmetric-algebra}}

\begin{remark}

Let \({ \mathbf{F} }\) be an arbitrary field, not necessarily
characteristic zero, and let
\(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\) be an arbitrary
Lie algebra, not necessarily finite-dimensional. Recall that the tensor
algebra \(T(V)\) is the \({\mathbf{Z}}_{\geq 0}\) graded unital algebra
where
\({\mathsf{gr}\,}_n T(V) = T^n(V) \coloneqq V{ {}^{ \scriptstyle\otimes_{{ \mathbf{F} }}^{n} } }\)
where \(T^0(V) \coloneqq{ \mathbf{F} }\). Note
\(T(V) = \bigoplus _{n\geq 0} T^n(V)\). If \(V\) has a basis
\(\left\{{x_k}\right\}_{k\in K}\) then
\(T(V) \cong { \mathbf{F} }\left\langle{x_k {~\mathrel{\Big\vert}~}k\in K}\right\rangle\),
a polynomial ring in the noncommuting variables \(x_k\). Degree \(n\)
monomials in this correspond to pure tensors with \(n\) components in
\(T(V)\).

There is an \({ \mathbf{F} }{\hbox{-}}\)linear map
\(V \xrightarrow{i} T(V)\), and \(T(V)\) satisfies a universal property:
given any linear map \(\phi\in {}_{{ \mathbf{F} }}{\mathsf{Mod}}(V, A)\)
where \(A\) has the structure of an associative algebra, there exists a
unique
\(\psi\in \mathsf{Assoc} \mathsf{Alg}_{/ {{ \mathbf{F} }}}(T(V), A)\)
making the diagram commute:

\begin{center}
\begin{tikzcd}
    V && {T(V)} \\
    \\
    && A
    \arrow["i", from=1-1, to=1-3]
    \arrow["{\exists !}", dashed, from=1-3, to=3-3]
    \arrow["\phi"', from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJWIl0sWzIsMCwiVChWKSJdLFsyLDIsIkEiXSxbMCwxLCJpIl0sWzEsMiwiXFxleGlzdHMgISIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDIsIlxccGhpIiwyXV0=}{Link
to Diagram}
\end{quote}

In fact, one can explicitly write \(\psi\) as
\(\psi(x_{k_1}\otimes\cdots x_{k_n}) = \phi(x_k)\cdots \phi(x_{k_n})\)
using the multiplication in \(A\).

The \textbf{symmetric algebra} and \textbf{exterior algebra} are defined
as
\begin{align*}
S(V) \coloneqq T(V)/\left\langle{x\otimes y -y\otimes x {~\mathrel{\Big\vert}~}x,y\in V}\right\rangle, 
 { {\bigwedge}^{\scriptscriptstyle \bullet}} (V) \coloneqq T(V)/\left\langle{x\otimes y + y\otimes x {~\mathrel{\Big\vert}~}x,y\in V}\right\rangle
.\end{align*}

\end{remark}

\begin{definition}[The Universal Enveloping Algebra]

Let \(L\in \mathsf{Lie} \mathsf{Alg}_{/ {{ \mathbf{F} }}}\) with basis
\(\left\{{x_k}\right\}_{k\in K}\). A \textbf{universal enveloping
algebra} for \(L\) is a pair \((U, i)\) where \(U\) is a unital
associative \({ \mathbf{F} }{\hbox{-}}\)algebra and \(i: L \to U_L\)
(where \(U_L\) is \(U\) equipped with the commutator bracket
multiplication) is a morphism of Lie algebras, i.e.
\begin{align*}
i([xy]) = i(x) i(y) - i(y) i(x) = [i(x) i(y) ] \quad \forall x,y\in L
.\end{align*}
It satisfies a universal property: for any unital associative algebra
\(A\) receiving a Lie algebra morphism \(j: L\to A_L\), there is a
unique \(\phi\) in the following:

\begin{center}
\begin{tikzcd}
    \mathsf{Lie} \mathsf{Alg}&& \mathsf{Assoc} \mathsf{Alg}\\
    L && U \\
    \\
    && A
    \arrow["i", from=2-1, to=2-3]
    \arrow["{\exists ! \phi}", dashed, from=2-3, to=4-3]
    \arrow["j"', from=2-1, to=4-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMiwxLCJVIl0sWzIsMywiQSJdLFsyLDAsIlxcQXNzb2NcXEFsZyJdLFswLDAsIlxcTGllXFxBbGciXSxbMCwxLCJMIl0sWzQsMCwiaSJdLFswLDEsIlxcZXhpc3RzICEgXFxwaGkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbNCwxLCJqIiwyXV0=}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

Uniqueness follows from the usual proof for universal objects.
Existence: let
\begin{align*}
U(L) \coloneqq T(L) / J, \qquad J \coloneqq\left\langle{x\otimes y - y\otimes x - [xy] {~\mathrel{\Big\vert}~}x,y\in L}\right\rangle
.\end{align*}
Warning: \(J\) is a two-sided ideal, but is not homogeneous!

One can form the required map:

\begin{center}
\begin{tikzcd}
    && {T(L)} \\
    L \\
    && {U(L)} \\
    \\
    && A
    \arrow["{i_*}"', from=2-1, to=1-3]
    \arrow["\pi", from=1-3, to=3-3]
    \arrow["{\exists ! \psi}", dashed, from=3-3, to=5-3]
    \arrow["{j\in \mathsf{Lie} \mathsf{Alg}(L, A)}"', from=2-1, to=5-3]
    \arrow["i"', from=2-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwxLCJMIl0sWzIsMCwiVChMKSJdLFsyLDIsIlUoTCkiXSxbMiw0LCJBIl0sWzAsMSwiaV8qIiwyXSxbMSwyLCJcXHBpIl0sWzIsMywiXFxleGlzdHMgISBcXHBzaSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFswLDMsImpcXGluIFxcTGllXFxBbGcoTCwgQSkiLDJdLFswLDIsImkiLDJdXQ==}{Link
to Diagram}
\end{quote}

This satisfies
\(\psi(x\otimes y - y\otimes x - [xy]) = j(x) j(y) - j(y)j(x) - j([xy]) = 0\)
using the properties of \(j\). \(\phi\) is unique because \(U(L)\) is
generated by 1 and \(\operatorname{im}i\), since \(T(L)\) is generated
by 1 and the image of \(L = T^1(L)\).

\end{remark}

\begin{remark}

If \(L\) is abelian, \(U(L) = S(L)\) is the symmetric algebra.

Note that \(J \subseteq \bigoplus _{n\geq 1} T^n(L)\) so
\({ \mathbf{F} }= T^0(L)\) maps isomorphically into \(U(L)\) under
\(\pi\). So \({ \mathbf{F} }\hookrightarrow U(L)\), meaning
\(U(L) \neq 0\), although we don't yet know if \(L\) injects into
\(U(L)\).

\end{remark}

\begin{theorem}[Poincaré-Birkhoff-Witt (PBW) Theorem]

Let \(L\) be a Lie algebra with basis \(\left\{{x_k}\right\}_{k\in K}\),
and filter \(T(L)\) by \(T_m \coloneqq\bigoplus _{i\leq m} T^i (L)\).
Then \(T_m\) is the span of words of length at most \(m\) in the basis
elements \(x_k\). Note \(T_m \cdot T_n \subseteq T_{m+n}\), and the
projection \(\pi: T(L) \twoheadrightarrow U(L)\) induces an increasing
filtration \(U_0 \subseteq U_1 \subseteq \cdots\) of \(U(L)\). Let
\(G^m \coloneqq U_m / U_{m-1}\) be the \(m\)th graded piece. The product
on \(U(L)\) induces a well-defined product
\(G^m \times G^n \to G^{m+n}\) since
\(U_{m-1}\times U_{n-1} \subseteq U_{m+n-2} \subseteq U_{m+n-1}\).
Extending this bilinearly to \(\bigoplus _{m\geq 0} G^m\) to form the
associated graded algebra of \(U(L)\).

\begin{quote}
Note that this construction generally works for any filtered algebra
where the multiplication is compatible with the filtration.
\end{quote}

\end{theorem}

\begin{example}[?]

Let \(L \coloneqq{\mathfrak{sl}}_2({\mathbf{C}})\) with ordered basis
\(\left\{{x,h,y}\right\}\). Then \(y\otimes h\otimes x\in T^3(L)\) --
denote the image of this monomial in \(U(L)\) by \(yhx \in U_3\). We can
reorder this:
\begin{align*}
yhx 
&= hyx + [yh]x
&= hyx + 2yx \qquad \in U_3 + U_2 \\
,\end{align*}
so in \(G^3\) we have \(yhx=hyx\). This is a general feature: reordering
introduces error terms of lower degree, which are quotiented out.
Continuing,
\begin{align*}
hyx + 2yx
&= hxy + h[yx] + 2xy + 2[yx] \\
&= hxy - h^2  + 2xy - 2h \\
&= xhy + [hx]y - h^2  + 2xy - 2h \\
&= xhy + 2xy - h^2  + 2xy - 2h \\
&= xhy + 4xy - h^2 - 2h \qquad \in U_3 + U_2 + U_2 + U_2
.\end{align*}

\end{example}

\hypertarget{wednesday-november-02}{%
\section{Wednesday, November 02}\label{wednesday-november-02}}

\begin{remark}

Clarification from last time: for
\(L\in \mathsf{Lie} {}_{{ \mathbf{F} }} \mathsf{Alg}\) over
\({ \mathbf{F} }\) an arbitrary field:

\begin{center}
\begin{tikzcd}
    && {T(L)} & {\trianglerighteq J \coloneqq\left\langle{x\otimes y-y\otimes x -[xy] {~\mathrel{\Big\vert}~}x,y\in L = T^1(L)}\right\rangle} \\
    L \\
    && {U(L)} \\
    && { A}
    \arrow["{i_0}", from=2-1, to=1-3]
    \arrow["i"', from=2-1, to=3-3]
    \arrow["\pi", from=1-3, to=3-3]
    \arrow["{\exists \psi}"', dashed, from=3-3, to=4-3]
    \arrow["{\exists ! \phi}"', curve={height=-30pt}, dashed, from=1-3, to=4-3]
    \arrow["j"', from=2-1, to=4-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwxLCJMIl0sWzIsMiwiVShMKSJdLFsyLDAsIlQoTCkiXSxbMywwLCJcXHRyaWFuZ2xlcmlnaHRlcSBKIFxcZGEgXFxnZW5ze3hcXHRlbnNvciB5LXlcXHRlbnNvciB4IC1beHldIFxcc3QgeCx5XFxpbiBMID0gVF4xKEwpfSJdLFsyLDMsIiBBIl0sWzAsMiwiaV8wIl0sWzAsMSwiaSIsMl0sWzIsMSwiXFxwaSJdLFsxLDQsIlxcZXhpc3RzIFxccHNpIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzIsNCwiXFxleGlzdHMgISBcXHBoaSIsMix7ImN1cnZlIjotNSwic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzAsNCwiaiIsMl1d}{Link
to Diagram}
\end{quote}

Then \(i\) is a Lie algebra morphism since
\(i([xy]) = i(x)i(y) - i(x)i(y) = [i(x) i(y)]\). We know
\begin{align*}
0 
&= \pi(x\otimes y - y\otimes x - [xy]) \\
&= \pi(i_0(x) i_0(y) - i_0(y) i_0(x) - i_0([xy])) \\
&= i(x) i(y) - i(y) i(x) - i([xy])
.\end{align*}

\end{remark}

\begin{remark}

Recall that we filtered
\(0 \subseteq U_1 \subseteq \cdots \subseteq U(L)\) and defined the
associated graded \(G^m = U_m / U_{m-1}\) and
\(G(L) \coloneqq\bigoplus _{m\geq 0} G^m\), and we saw by example that
\(yhx = hyx = xhy\) in \(G^3({\mathfrak{sl}}_2)\). There is a projection
map \(T^m(L) \to U(L)\) whose image is contained in \(U_m\), so there is
a composite map
\begin{align*}
T^m(L) \to U_m \to U_m/U_{m-1} = G^m
.\end{align*}
Since \(T(L) = \bigoplus _{m\geq 0} T^m(L)\), these can be combined into
an algebra morphism \(T(L) \to G(L)\). It's not hard to check that this
factors through
\(S(L) \coloneqq T(L)/\left\langle{x\otimes y - y\otimes x{~\mathrel{\Big\vert}~}x,y\in L}\right\rangle\)
since \(x\otimes y = y\otimes x + [xy]\) and the \([xy]\) term is in
lower degree. So this induces \(w: S(L) \to G(L)\), and the PBW theorem
states that this is an isomorphism of graded algebras.

\end{remark}

\begin{corollary}[C]

Let \(\left\{{x_k}\right\}_{k\in K}\) be an ordered basis for \(L\),
then the collection of monomials \(x_{k_1}\cdots x_{k_m}\) for
\(m\geq 0\) where \(k_1\leq \cdots k_m\) is a basis for \(U(L)\).

\end{corollary}

\begin{proof}[?]

The collection of such monomials of length exactly \(n\) forms a basis
for \(S^m(L)\), and via \(w\), a basis for \(G^m(L)\). In particular
these monomials form a linearly independent in \(U_m/U_{m-1}\), since
taking quotients can only increase linear dependencies, and hence these
are linearly independent in \(U_m\) and \(U(L)\). By induction on \(m\),
\(U_{m-1}\) has a basis consisting of all monomials of length
\(\leq m-1\). We can then get a basis of \(U_m\) by adjoining to this
basis of \(U_{m-1}\) any preimages in \(U_m\) of basis elements for the
quotient \(U_m/U_{m-1}\). So a basis for \(U_m\) is all ordered
monomials of length \(\leq m\). Since \(U(L) = \cup_{m\geq 0} U_m\),
taking the union of bases over all \(m\) yields the result.

\end{proof}

\begin{corollary}[B]

The canonical map \(i: L\to U(L)\) is injective.

This follows from taking \(m=1\) in the previous corollary.

\end{corollary}

\begin{corollary}[D]

Let \(H\leq L\) be a Lie subalgebra and extend an ordered basis for
\(H\), say \(\left\{{h_1, h_2,\cdots}\right\}\), and extend it to an
ordered basis \(\left\{{h_1,h_2,\cdots, x_1,x_2,\cdots}\right\}\). Then
the injection \(H\hookrightarrow L\) induces an injective morphism
\(U(H) \hookrightarrow U(L)\). Moreover,
\(U(L)\in {}_{U(H)}{\mathsf{Mod}}^{\mathrm{free}}\) with a basis of
monomials \(x_{k_1}x_{k_2}\cdots x_{k_m}\) for \(m\geq 0\).

This follows directly from corollary C.

\end{corollary}

\begin{remark}

We'll skip 17.4, which proves the PBW theorem. The hard part: linear
independence, which is done by constructing a representation of \(U(L)\)
in another algebra.

\end{remark}

\hypertarget{free-lie-algebras}{%
\subsection{17.5: Free Lie algebras}\label{free-lie-algebras}}

\begin{definition}[Free Lie algebras]

Let \(L \in \mathsf{Lie} {}_{{ \mathbf{F} }} \mathsf{Alg}\) which is
generated as a Lie algebras (so allowing commutators) by a subset
\(X \subseteq L\).\footnote{Note that \({\mathfrak{sl}}_2\) has a basis
  \(\left\{{x,h,y}\right\}\) but is freely generated by \(x,y\) since
  \(h=[xy]\).} We say \(L\) is \textbf{free on \(X\)} and write
\(L = L(X)\) if for any set map \(\phi: X\to M\) with
\(M\in \mathsf{Lie} {}_{{ \mathbf{F} }} \mathsf{Alg}\) there exists an
extension:

\begin{center}
\begin{tikzcd}
    & {} && \mathsf{Lie} {}_{{ \mathbf{F} }}  \mathsf{Alg}\\
    {\mathsf{Set}}& X && L \\
    \\
    &&& M
    \arrow[hook, from=2-2, to=2-4]
    \arrow["{\exists !\psi}", dashed, from=2-4, to=4-4]
    \arrow["\phi"', from=2-2, to=4-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwxLCJYIl0sWzMsMSwiTCJdLFszLDMsIk0iXSxbMywwLCJcXExpZVxcZmFsZyJdLFsxLDBdLFswLDEsIlxcU2V0Il0sWzAsMSwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwyLCJcXGV4aXN0cyAhXFxwc2kiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMCwyLCJcXHBoaSIsMl1d}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

Existence:

\begin{itemize}
\tightlist
\item
  Let \(V = V(X)\) be the free \({}_{ { \mathbf{F} }}{\mathsf{Mod}}\) on
  \(X\).
\item
  Let \(L(X) \leq T(V)_L\)\footnote{\(W_L\) is \(W\) made into a Lie
    algebra via \([xy] = xy-yx\).} be the Lie subalgebra generated by
  \(X\) (or equivalently by \(V\)), which has elements like
  \begin{align*}
  x,y,z\in X,\qquad x\otimes y-y\otimes x, \qquad z(x\otimes y-y\otimes x) - (x\otimes y - y\otimes x)z,\,\cdots
  .\end{align*}
\item
  Check that \(L(X)\) is free on \(X\) by letting \(\phi: X\to M\) for
  \(M\) a Lie algebra, then by the universal property of \(V(X)\) we get
  a unique linear map \(\tilde \phi: V(X) \to M\) extending \(\phi\):
\end{itemize}

\begin{center}
\begin{tikzcd}
    {V(X)} && {T(V)} \\
    \\
    M \\
    \\
    {U(M)}
    \arrow["\phi", from=1-1, to=3-1]
    \arrow[hook, from=3-1, to=5-1]
    \arrow[hook, from=1-1, to=1-3]
    \arrow["{\exists ! \tilde \phi \text{ (an algebra morphism)}}", dashed, from=1-3, to=5-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJWKFgpIl0sWzIsMCwiVChWKSJdLFswLDIsIk0iXSxbMCw0LCJVKE0pIl0sWzAsMiwiXFxwaGkiXSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDEsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMywiXFxleGlzdHMgISBcXHRpbGRlIFxccGhpIFxcdGV4dHsgKGFuIGFsZ2VicmEgbW9ycGhpc20pfSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

One checks that \(\tilde \phi\) restricts to a Lie algebra morphisms
\(\tilde \phi: L(X) \to U(M)\) whose image is the Lie subalgebra of
\(U(M)\) generated by \(M\) -- but this subalgebra is precisely \(M\),
since e.g.~\(U(M)\ni x\otimes y - y\otimes x = [xy]\in M\). Thus we can
view \(\tilde \phi\) as a map \(\tilde \phi: L(X)\to M\).

\end{remark}

\begin{remark}

One can check that \(U(L(X)) = T(V(X))\).

\end{remark}

\hypertarget{s-18-generators-and-relations-for-simple-lie-algebras-friday-november-04}{%
\section{\texorpdfstring{\(\S 18\): Generators and Relations for Simple
Lie Algebras (Friday, November
04)}{\textbackslash S 18: Generators and Relations for Simple Lie Algebras (Friday, November 04)}}\label{s-18-generators-and-relations-for-simple-lie-algebras-friday-november-04}}

\begin{remark}

Recall that the free Lie algebra of a set \(X\), \(L(X)\) satisfies a
universal property:

\begin{center}
\begin{tikzcd}
    X && {L(X)} \\
    \\
    && M
    \arrow["\iota", hook, from=1-1, to=1-3]
    \arrow["{\forall \phi}"', from=1-1, to=3-3]
    \arrow["{\exists ! \tilde \phi}", dashed, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJYIl0sWzIsMCwiTChYKSJdLFsyLDIsIk0iXSxbMCwxLCJcXGlvdGEiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDIsIlxcZm9yYWxsIFxccGhpIiwyXSxbMSwyLCJcXGV4aXN0cyAhIFxcdGlsZGUgXFxwaGkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=}{Link
to Diagram}
\end{quote}

\end{remark}

\hypertarget{s-18.1-relations-satisfied-by-l}{%
\subsection{\texorpdfstring{\(\S 18.1\): Relations satisfied by
\(L\)}{\textbackslash S 18.1: Relations satisfied by L}}\label{s-18.1-relations-satisfied-by-l}}

\begin{definition}[Relations]

Given an arbitrary \(L\in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}\) and
fix a set \(X\) of generators for \(L\) and form \(LX()\), then there is
a Lie algebra morphism \(\pi: L(X) \twoheadrightarrow L\) which is
surjective since \(X\) generates \(L\). Defining \(R\coloneqq\ker \pi\),
one has \(L \cong L(X)/R\), so \(R\) is called the \textbf{ideal of
relations}.

\end{definition}

\begin{remark}

Let
\(L \in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}^{{\mathrm{fd}}, {\mathrm{ss}}}_{{\mathbf{C}}}\),
let \(H \subseteq L\) be a maximal toral subalgebra, and \(\Phi\) its
root system. Fix a base
\(\Delta = \left\{{ \alpha_1, \cdots, \alpha_\ell}\right\} \subseteq \Phi\).
Recall
\begin{align*}
{\left\langle { \alpha_j},~{ \alpha_i} \right\rangle} \coloneqq{2 (\alpha_j, \alpha_i) \over (\alpha_i, \alpha_i)} = \alpha_j(h_i),\qquad g_i \coloneqq h_{\alpha_i} = {2 \alpha_i \over (\alpha_i, \alpha_i)}
.\end{align*}
The root strings are of the form
\(\beta - r \alpha, \cdots, \beta, \cdots, \beta+ q \alpha\) where
\(r-q = \beta(h_ \alpha)\). For any \(i\) we can fix a standard
\({\mathfrak{sl}}_2\) triple \(\left\{{x_i, h_i, y_i}\right\}\) such
that \(x_i\in L_{ \alpha_i}, y_i \in L_{- \alpha_i}, h_i\in [x_i y_i]\).

\end{remark}

\begin{proposition}[Serre relations]

\(L\) is generated as a Lie algebra by the \(3\ell\) generators
\(X\coloneqq\left\{{x_i, h_i, y_i {~\mathrel{\Big\vert}~}1\leq i\leq \ell}\right\}\)
subject to at least the following relations:

\begin{itemize}
\tightlist
\item
  S1: \([h_i h_j] = 0\),
\item
  S2: \([x_i y_j] = \delta_{ij} h_i\),
\item
  S3:
  \([h_i x_j] = {\left\langle { \alpha_j},~{\alpha_i} \right\rangle} x_j\)
  and
  \([h_i y_j] = - {\left\langle { \alpha_j},~{ \alpha_i} \right\rangle} y_j\).
\item
  \(\text{S}_{ij}^+\):
  \({ \operatorname{ad}}_{x_i}^{-{\left\langle { \alpha_j},~{\alpha_i} \right\rangle} + 1}(x_j) = 0\)
  for \(i\neq j\)
\item
  \(\text{S}_{ij}^-\):
  \({ \operatorname{ad}}_{y_i}^{-{\left\langle { \alpha_j},~{\alpha_i} \right\rangle} + 1}(x_j) = 0\)
  for \(i\neq j\)
\end{itemize}

\end{proposition}

\begin{proof}[?]

Recall that differences of simple roots are never roots, since the
coefficients have mixed signs. Since
\(\alpha_i - \alpha_j \not\in \Phi\), we have \([x_i y_j] = 0\) for
\(i\neq j\) since it would have to be in \(L_{\alpha_i - \alpha_j}\).
Consider the \(\alpha_i\) root string through \(\alpha_j\): we have
\(r=0\) from above, and the string is
\begin{align*}
\alpha, \alpha+ \alpha_i, \cdots, \alpha_j - {\left\langle {\alpha_j},~{\alpha_i} \right\rangle} \alpha_i
\end{align*}
since \(L_\beta = 0\) for
\(\beta \coloneqq\alpha_j - \qty{ {\left\langle {\alpha_j},~{ \alpha_i} \right\rangle} + 1} \alpha_i\).
The relations for \(S_{ij}^\pm\) follow similarly.

\end{proof}

\begin{remark}

Note that these relations are all described in a way that only involves
the Cartan matrix of \(\Phi\), noting that changing bases only permutes
its rows and columns.

\end{remark}

\begin{theorem}[Serre]

These five relations form a complete set of defining relations for
\(L\), i.e.~\(L \cong L(X)/R\) where \(R\) is the ideal generated by the
Serre relations above. Moreover, given a root system \(\Phi\) and a
Cartan matrix, one can define a Lie algebra using these generators and
relations that is finite-dimensional, simple, and has root system
\(\Phi\).

\end{theorem}

\hypertarget{s-18.2-consequences-of-serre-relations-s1-s2-s3}{%
\subsection{\texorpdfstring{\(\S 18.2\): Consequences of Serre relations
S1, S2,
S3}{\textbackslash S 18.2: Consequences of Serre relations S1, S2, S3}}\label{s-18.2-consequences-of-serre-relations-s1-s2-s3}}

\begin{remark}

Fix an irreducible root system \(\Phi\) of rank \(\ell\) with Cartan
matrix \(A\). Let \(\widehat{L} \coloneqq L(\widehat{X})\) where
\(\widehat{X} \coloneqq\left\{{\widehat{x}_i, \widehat{h}_i, \widehat{y}_i {~\mathrel{\Big\vert}~}1\leq i\leq \ell}\right\}\).
Let \(\widehat{K} {~\trianglelefteq~}\widehat{L}\) be the 2-sided ideal
generated by the relations S1, S2, S3. Let
\(L_0 \coloneqq\widehat{L}/\widehat{K}\) and write \(\pi\) for the
quotient map \(\widehat{L}\to L_0\) -- note that \(L_0\) is
infinite-dimensional, although it's not yet clear that \(L_0\neq 0\).
We'll study \(L_0\) by defining a representation of it, which is
essentially the adjoint representation of \(L_0\) acting on
\(\left\{{y_i}\right\}\).

\end{remark}

\begin{remark}

Recall that a representation of
\(M\in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}\) is a morphism
\(\phi\in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}(M, {\mathfrak{gl}}(V))\)
for \(V\in {}_{ { \mathbf{F} }}{\mathsf{Mod}}\). This yields a diagram

\begin{center}
\begin{tikzcd}
    {} && \mathsf{Assoc} \mathsf{Alg}\\
    M && {U(M)} \\
    \\
    && {{ \operatorname{End} }(V)}
    \arrow["\iota", hook, from=2-1, to=2-3]
    \arrow["\phi"', from=2-1, to=4-3]
    \arrow["{\exists !\phi}", dashed, from=2-3, to=4-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwxLCJNIl0sWzIsMSwiVShNKSJdLFsyLDMsIlxcZW5kbyhWKSJdLFsyLDAsIlxcQXNzb2NcXEFsZyJdLFswLDBdLFswLDEsIlxcaW90YSIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMiwiXFxwaGkiLDJdLFsxLDIsIlxcZXhpc3RzICFcXHBoaSIsMCx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to diagram}
\end{quote}

Conversely given an algebra morphism
\(\tilde \phi: U(M) \to { \operatorname{End} }(V)\), and restricting
\(\tilde \phi\) to \(M \subseteq U(M)\) gives a Lie algebra morphism
\(\phi: M\to { \operatorname{End} }(V) = {\mathfrak{gl}}(V)\). This
representations of \(M\) (using \({\mathfrak{gl}}(V)\)) correspond to
associative algebra representations of \(U(M)\) (using
\({ \operatorname{End} }(V)\)). Since \(U(M) = T(V(\widehat{X}))\),
using the various universal properties, having a representation \(V\) of
\(\widehat{L}\) is equivalent to having a set map
\(\widehat{X}\to { \operatorname{End} }(V)\), i.e.~elements of
\(\widehat{X}\) should act linearly on \(V\).

\end{remark}

\begin{remark}

Let \(V\) be the tensor algebra on a vector space with basis
\(\left\{{v_1,\cdots, v_\ell}\right\}\), thinking of each \(v_i\) being
associated to \(\widehat{y}_i\). Write
\(v_1 v_2\cdots v_t \coloneqq v_1 \otimes v_2\otimes\cdots\otimes v_t\),
and define elements of \({ \operatorname{End} }(V)\) by

\begin{itemize}
\tightlist
\item
  \(\widehat{h}_j \cdot 1 \coloneqq 0\),
\item
  \(\widehat{h}_j \cdot v_{i_1}\cdots v_{i_t} \coloneqq-\qty{ {\left\langle { \alpha_{i_1}},~{\alpha_j} \right\rangle} + \cdots {\left\langle {\alpha_{i_t}},~{\alpha_j} \right\rangle} }v_{i_1}\cdots v_{i_t}\),
\item
  \(\widehat{y}_j\cdot v_{i_1}\cdots v_{i_t} \coloneqq v_j v_{i_1} \cdots v_{i_t}\)
  for \(t\geq 0\),
\item
  \(\widehat{x}_j \cdot 1 \coloneqq 0\),
\item
  \(\widehat{x}_j \cdot v_i \coloneqq 0\) for all \(i\),
\item
  \(\widehat{x}_j \cdot v_{i_1}\cdots v_{i_t} \coloneqq v_{i_1} \qty{ \widehat{x}_j v_{i_2} \cdots v_{i_t}} - \delta_{i_1, j}\qty{{\left\langle {\alpha_{i_2} },~{\alpha_j} \right\rangle} + \cdots + {\left\langle {\alpha_{i_2} },~{\alpha_j} \right\rangle} }v_{i_2}\cdots v_{i_t}\)
  for \(t\geq 2\).
\end{itemize}

\end{remark}

\hypertarget{monday-november-07}{%
\section{Monday, November 07}\label{monday-november-07}}

\begin{remark}

Last time: constructing a semisimple Lie algebra that has a given root
system. Setup:

\begin{itemize}
\tightlist
\item
  \(\Delta = \left\{{\alpha_1,\cdots, \alpha_l}\right\}\).
\item
  \(\widehat{L}\) the free Lie algebra on
  \(\left\{{\widehat{x}_i, \widehat{h}_i, \widehat{y}_i}\right\}_{1\leq i\leq l}\).
\item
  \(\widehat{K}\) the ideal generated by the Serre relations.
\item
  \(L_0 \coloneqq\widehat{L}/\widehat{K}\) with quotient map \(\pi\).
\item
  \(\widehat{\phi}: \widehat{L}\to {\mathfrak{gl}}(V)\) a representation
  we constructed.
\item
  \(\widehat{H}\) the free Lie algebra on \(\left\{{h_i}\right\}\).
\end{itemize}

\end{remark}

\begin{theorem}[?]

\(\widehat{K} \subseteq \ker \widehat{\phi}\), so \(\widehat{\phi}\)
induces a representation \(\phi\) of \(L_0\) on \({\mathfrak{gl}}(V)\)

\begin{center}
\begin{tikzcd}
    {\widehat{L}} && {{\mathfrak{gl}}(V)} \\
    \\
    & {L_0}
    \arrow["\widehat{\phi}", from=1-1, to=1-3]
    \arrow["\pi"', from=1-1, to=3-2]
    \arrow["{\exists \phi}"', dashed, from=3-2, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXGhhdCBMIl0sWzIsMCwiXFxsaWVnbChWKSJdLFsxLDIsIkxfMCJdLFswLDEsIlxcaGF0XFxwaGkiXSxbMCwyLCJcXHBpIiwyXSxbMiwxLCJcXGV4aXN0cyBcXHBoaSIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{theorem}

\begin{proof}[?]

Straightforward but tedious checking of all relations, e.g.~
\begin{align*}
\widehat{\phi}(\widehat{h}_i) \circ \widehat{\phi}(\widehat{x}_j) - \widehat{\phi}(\widehat{x}_j )\widehat{\phi}(\widehat{h}_i) = {\left\langle { \alpha_j},~{\alpha_i} \right\rangle} \widehat{\phi}(\widehat{x}_j)
.\end{align*}

\end{proof}

\begin{theorem}[?]

In \(L_0\), the \(h_i\) form a basis for an
\(\ell{\hbox{-}}\)dimensional abelian subalgebra \(H\) of \(L_0\), and
moreover \(L_0 = Y \oplus H \oplus X\) where \(Y, X\) are the
subalgebras generated by the \(x_i\) and \(y_i\) respectively.

\end{theorem}

\begin{proof}[?]

\textbf{Steps 1 and 2}: :::\{.claim\} \(\pi(\widehat{H}) = H\) is
\(\ell{\hbox{-}}\)dimensional.

\end{proof}

Clearly the \(\widehat{h}_i\) span an \(\ell{\hbox{-}}\)dimensional
subspace of \(\widehat{L}\), so we need to show that \(\pi\) restricts
to an isomorphism \(\pi: \widehat{H} { \, \xrightarrow{\sim}\, }H\).
Suppose
\(\widehat{h} \coloneqq\sum_{j=1}^\ell c_j \widehat{h}_j \in \ker \pi\),
so \(\widehat{\phi }(\widehat{h}) = 0\). Thus
\begin{align*}
0 = \widehat{h} \cdot v_i = \sum_{j} c_j \widehat{h}_j \cdot v_i = - \sum_j c_j {\left\langle { \alpha_i},~{\alpha_j} \right\rangle}
= - \sum_j a_{ij} c_j \qquad \forall i
,\end{align*}
so \(A\mathbf{c} = 0\) where \(A\) is the Cartan matrix, and so
\(\mathbf{c} = \mathbf{0}\) since \(A\) is invertible (since it was
essentially a Gram matrix).

\textbf{Step 3}: Now
\(\sum { \mathbf{F} }x_i + \sum { \mathbf{F} }h_i + \sum { \mathbf{F} }y_i \xrightarrow{\pi} L_0\)
maps isomorphically to \(L_0\), and S2, S3 show that for each \(i\).
Then \({ \mathbf{F} }x_i + { \mathbf{F} }h_i + { \mathbf{F} }y_i\) is a
homomorphic image of \({\mathfrak{sl}}_2\), which is simple if
\(\operatorname{ch}{ \mathbf{F} }\neq 2\). Note
\(\pi( \widehat{h}_i) = h_i \neq 0\) in \(L_0\) by (1), so this subspace
of \(L_0\) is isomorphic to \({\mathfrak{sl}}_2({ \mathbf{F} })\). In
particular \(\left\{{x_i, h_i, y_i}\right\}\) is linearly independent in
\(L_0\) for each fixed \(i\). Supposing
\(0 = \sum_{j=1}^\ell (a_j x_j + b_j h_j + c_j y_j)\), applying
\({ \operatorname{ad}}_{L_0, h_i}\) for each \(i\) to obtain
\begin{align*}
0 = \sum_{j=1}^\ell \qty{ a_j{\left\langle {\alpha_j},~{\alpha_i} \right\rangle} x_j + b_j 0 - c_j {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} y_j  }
= \sum_{j=1}^\ell {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} (a_j x_j - x_y y_j) 
,\end{align*}
and by invertibility of \(A\) we have \(a_j x_j - c_j y_j = 0\) for each
\(j\). So \(a_j = c_j = 0\) for all \(j\), and \(\sum b_j h_j = 0\)
implies \(b_j = 0\) for all \(j\) from (1).

\textbf{Step 4}: \(H = \sum_{j=1}^\ell { \mathbf{F} }h_j\) is an
\(\ell{\hbox{-}}\)dimensional abelian subalgebra of \(L_0\) by (1) and
S1.

\textbf{Step 5}: Write
\([x_{i_1}\cdots x_{i_t}] \coloneqq[x_{i_1} [ x_{i_2} [ \cdots [x_{i_{t-1}} x_{i_t}] \cdots ]] \in X\)
for an iterated bracket, taken by convention to be bracketing from the
right. We have
\begin{align*}
 { \operatorname{ad}}_{L_0, h_j}([x_{i_1} \cdots x_{i_t}] ) = \qty{ {\left\langle { \alpha_{i_1} },~{ \alpha_j } \right\rangle} + \cdots + {\left\langle { \alpha_{i_t} },~{ \alpha_j } \right\rangle}    } [x_{i_1} \cdots x_{i_t}] \qquad t\geq 1
,\end{align*}
and similarly for \([y_{i_1}\cdots y_{i_t}]\).

\textbf{Step 6}: For \(t\geq 2\),
\([y_j [ x_{i_1} \cdots x_{i_t} ] ] \in X\), and similarly with the
roles of \(x_i, y_i\) reversed. This follows from the fact that
\({ \operatorname{ad}}_{L_0, y_j}\) acts by derivations, and using S2
and S3.

\textbf{Step 7}: It follows from steps 4 through 6 that \(Y+H+X\) is a
subalgebra of \(L_0\). One shows that
\([[ x_{i_1} \cdots x_{i_t}], [y_{i_1} \cdots y_{i_t} ]]\), which comes
down to the Jacobi identity and induction on \(s+t\). E.g.
\begin{align*}
[ [x_1 x_2], [y_3 y_4] ] = [x_1[ x_2 [y_3 y_4 ] ] ] - [x_2 [x_1 [y_3 y_4]] \in [x_1, { \mathbf{F} }y_3 + { \mathbf{F} }y_4] + \cdots \in H + \cdots
,\end{align*}
which lands in \(H\) since there are as many \(x_i\) as \(y_i\), whereas
if there are more \(x_i\) than \(y_i\) this lands in \(X\), and so on.
Since \(Y+H+X\) is a subalgebra that contains the generators
\(x_i, h_i, y_i\) of \(L_0\), it must be equal to \(L_0\).

\textbf{Step 8}: The decomposition \(L_0 = X + H + Y\) is a direct sum
decomposition of \(L_0\) into submodule for the adjoint action of \(H\).
Use the computation in the previous step to see that every element of
\(X\) is a linear combination of elements \([x_{i_1}\cdots x_{i_t}]\)
and similarly for \(Y\). These are eigenvectors for the action of
\({ \operatorname{ad}}_H \curvearrowright L_0\) by (5), and
eigenfunctions for \(X\) have the form
\(\lambda = \sum_{i=1}^\ell c_i \alpha_i\) with
\(c_i \in {\mathbf{Z}}_{\geq 0}\). The \(\lambda_i\) is referred to as a
\textbf{weight}, and \(c_i\) is the number of times \(i\) appears is an
index in \(i_1,\cdots, i_t\). So every weight space \(X_\lambda\) is
finite-dimensional, and the weights of \(Y\) are \(-\lambda\). Since the
weights in \(X, H, Y\) are all different, their intersections must be
trivial and the sum is direct. :::

\begin{remark}

\(L_0 = Y \bigoplus H \bigoplus X\) is known as the \textbf{triangular
decomposition}, where the \(x_i\) are on the super diagonal and bracket
to upper-triangular elements, and the \(y_i\) are their transposes.

\end{remark}

\hypertarget{wednesday-november-09}{%
\section{Wednesday, November 09}\label{wednesday-november-09}}

\begin{remark}

Progress so far: we start with the data of an irreducible root system
\(\Phi \supseteq\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{\ell} }\right\}\)
and Cartan matrix
\(A = ({\left\langle { \alpha_i},~{\alpha_j} \right\rangle})\) and Weyl
group \(W\). We set
\(L_0 \coloneqq{\left\langle{x_i, y_i, h_i {~\mathrel{\Big\vert}~}1\leq i\leq \ell}\right\rangle \over \left\langle{\text{S1, S2, S3}}\right\rangle} = Y \oplus H \oplus X\).
Letting \(h\in H\) act by \({ \operatorname{ad}}_h\), we get weight
spaces
\((L_0)_{ \lambda} \coloneqq\left\{{v\in L_0 {~\mathrel{\Big\vert}~}[hv] = \lambda(h) v \, \forall h\in H}\right\}\).

\end{remark}

\hypertarget{s-18.3-serres-theorem}{%
\subsection{\texorpdfstring{\(\S\) 18.3: Serre's
Theorem}{\textbackslash S 18.3: Serre's Theorem}}\label{s-18.3-serres-theorem}}

\begin{remark}

For \(i\neq j\), set
\begin{align*}
y_{ij} \coloneqq( { \operatorname{ad}}_{y_i})^{- {\left\langle {\alpha_j},~{ \alpha_i} \right\rangle} + 1}(y_j)
,\end{align*}
and similarly for \(x_{ij}\). Recall
\(\alpha_i(h_j) \coloneqq{\left\langle { \alpha_i },~{ \alpha_j } \right\rangle}\).

\end{remark}

\begin{lemma}[A]

\begin{align*}
 { \operatorname{ad}}_{x_k}(y_{ij}) = 0 \qquad \forall i\neq j, \forall k
.\end{align*}

\end{lemma}

\begin{proof}[?]

\textbf{Case 1}: \(k\neq i\).

In this case, \([x_k y_i] = 0\) and thus
\begin{align*}
( { \operatorname{ad}}_{x_k}) ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ ga_i } \right\rangle} +1 }(y_j) = ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} +1 } ( { \operatorname{ad}}_{x_k})(y_j)
.\end{align*}

\begin{itemize}
\tightlist
\item
  Case i, \(k\neq j\): then \(( { \operatorname{ad}}_{x_k})(y_j) = 0\).
\item
  Case ii, \(k=j\): then \(( { \operatorname{ad}}_{x_j})(y_j) = h_j\)
  and
  \(( { \operatorname{ad}}_{y_i})(h_j) = {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} y_i\).

  \begin{itemize}
  \tightlist
  \item
    Case a,
    \({\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} \neq 0\),
    then
    \begin{align*}
    ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} +1 } ( { \operatorname{ad}}_{x_j}) (y_j) = {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle}  }(y_i) = 0 
    .\end{align*}
  \item
    Case b,
    \({\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} = 0\),
    then \({\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} =0\).
    In this case we have
    \(( { \operatorname{ad}}_{y_i})^1 (h_j) = {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} y_i = 0\).
  \end{itemize}
\end{itemize}

\textbf{Case 2}: \(k=i\).

In this case, we saw that for any fixed \(i\),
\(\left\{{x_i, h_i, y_i}\right\}\) spans a standard
\({\mathfrak{sl}}_2\) triple in \(L_0\), so consider the
\({\mathfrak{sl}}_2{\hbox{-}}\)submodule of \(Y_J \leq L_0\) generated
by \(y_j\). Since \(i\neq j\), we know \([x_i y_j] = 0\), so \(y_j\) is
a maximal vector for \(Y_j\) with weight
\(m \coloneqq- {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle}\).

One can show by induction on \(t\) that the following formula holds:
\begin{align*}
( { \operatorname{ad}}_{x_i}) ( { \operatorname{ad}}_{y_i})^t (y_j) = t (m-t+1) ( { \operatorname{ad}}_{y_i})^{t-1} (y_j) \qquad t\geq 1
.\end{align*}
So in particular
\(( { \operatorname{ad}}_{x_i}) ( { \operatorname{ad}}_{y_i})^{m+1}(y_j) = 0\),
and the LHS is \(y_{ij}\).

\end{proof}

\begin{definition}[Locally nilpotent and the exponential]

An endomorphism \(x\in { \operatorname{End} }(V)\) is \textbf{locally
nilpotent} if for all \(v\in V\) there exists some \(n\) depending on
\(v\) such that \(x^n \cdot v = 0\). If \(x\) is locally nilpotent, then
define the \textbf{exponential} as
\begin{align*}
\exp(x) = \sum_{k\geq 0} {1\over k!} x^k = 1 + x + {1\over 2}x^2 + \cdots \qquad \in { \operatorname{End} }(V)
,\end{align*}
which is in fact an automorphism of \(V\) since its inverse is
\(\exp(-x)\).

\end{definition}

\begin{lemma}[B]

Suppose \({ \operatorname{ad}}_{x_i}, { \operatorname{ad}}_{y_i}\) are
locally nilpotent on \(L_0\) and define
\begin{align*}
\tau_i \coloneqq\exp( { \operatorname{ad}}_{x_i})\circ \exp( { \operatorname{ad}}_{-y_i}) \circ \exp( { \operatorname{ad}}_{x_i})
.\end{align*}
Then \(\tau_i((L_0)_\lambda) = (L_0)_{s_i (\lambda)}\) where
\(s_i \coloneqq s_{ \alpha_i} \in W\) for \(\alpha_i\in \Phi\). Here
\(\lambda\in H {}^{ \vee }\cong {\mathbf{C}}\left\langle{ \alpha_1, \cdots, \alpha_\ell}\right\rangle\)
since
\(H = {\mathbf{C}}\left\langle{h_1,\cdots, h_{\ell}}\right\rangle\),
using that \(A\) is invertible. We use the formula
\(s_{\alpha_i}(\alpha_j) = \alpha_j - {\left\langle { \alpha_j },~{ \alpha_i } \right\rangle} \alpha_i\)
and extending linearly to \(H {}^{ \vee }\) as done previously.

\end{lemma}

\begin{proof}[?]

Omitted. See \(\S 14.3\) and \(\S 2.3\) for a very similar calculation.

\end{proof}

\begin{theorem}[Serre]

The Lie algebra \(L\) generated by the \(3\ell\) elements
\(\left\{{x_i, h_i, y_i}\right\}_{1\leq i\leq \ell}\) subject to
relations S1-S3 and the remaining two relations \(S_{ij}^{\pm}\) (which
hold in any finite dimensional semisimple Lie algebra) is a finite
dimensional semisimple Lie algebra with maximal torus spanned by
\(\left\{{h_i}\right\}_{1\leq i\leq \ell}\) and with corresponding root
system \(\Phi\).

\end{theorem}

\hypertarget{proof-of-serres-theorem}{%
\subsection{Proof of Serre's Theorem}\label{proof-of-serres-theorem}}

\begin{proof}[?]

By definition, \(L \coloneqq L_0/K\) where \(K{~\trianglelefteq~}L_0\)
is generated by the elements \(x_{ij}, y_{ij}\) where \(i\neq j\).
Recall that \(X, Y\leq L_0\) are the subalgebras generated by the
\(x_i\) and \(y_i\) respectively, so let \(I\) (resp. \(J\)) be the
ideal in \(X\) (resp. \(Y\)) generated by the \(x_{ij}\) (resp.
\(y_{ij}\)) for \(i\neq j\). Clearly \(I, J \subseteq K\).

\begin{claim}[Step 1]

\begin{align*}
I, J {~\trianglelefteq~}L_0
.\end{align*}

\end{claim}

\begin{proof}[of Step 1]

We'll prove this for \(J\), and \(I\) is similar. Note
\(J{~\trianglelefteq~}Y\) and write
\(J = \left\langle{y_{ij} {~\mathrel{\Big\vert}~}i\neq j}\right\rangle\).
Fix \(1\leq k\leq \ell\), then
\(( { \operatorname{ad}}_{y_k}) (y_{ij}) \in J\) by definition. Recall
\(y_{ij} = ( { \operatorname{ad}}_{y_i})^{- {\left\langle { \alpha_i },~{ \alpha_j } \right\rangle} +1 }(y_{ij})\).
Note \(( { \operatorname{ad}}_{h_k})(y_{ij}) = c_{ijk} y_{ij}\) for some
constant \(c_{ijk} \in {\mathbf{Z}}\), and
\(( { \operatorname{ad}}_{x_k})(y_{ij}) = 0\) by lemma A above. Since
\(x_k, h_k, y_k\) generate \(L_0\), we have
\([L_0, y_{ij}] \subseteq J\). Using the Jacobi identity and that
\({ \operatorname{ad}}_z\) is a Lie algebra derivation for \(z\in L_0\),
it follows that \([L_0, J] \subseteq J\).

\begin{quote}
This essentially follows from \([h_\ell, Y] \subseteq Y\) and
\([x_\ell, Y] \subseteq H + Y\), and bracketing these against \(y_{ij}\)
lands in \(J\).
\end{quote}

\end{proof}

\begin{claim}[Step 2]

\begin{align*}
K = I + J
.\end{align*}

\end{claim}

\begin{proof}[of Step 2]

We have \(I+J \subseteq K\), but \(I+J {~\trianglelefteq~}L_0\) by claim
1 and it contains the generators of \(K\) -- since \(K\) is the smallest
such ideal, \(K \subseteq I+J\).

\end{proof}

\begin{observation}[Step 3]

We have a decomposition \(L_0 = Y \oplus H \oplus X\) as modules under
\({ \operatorname{ad}}_H\), and \(K = J \oplus 0 \oplus I\). Taking the
quotient yields
\(L \coloneqq L_0/K = Y/J \oplus H \oplus X/I \coloneqq N^- \oplus H \oplus N^+\).

\end{observation}

\begin{observation}[Step 4]

As in the proof last time,
\(\left\{{x_i, h_i, y_i}\right\} \subseteq L\) spans a copy of
\({\mathfrak{sl}}_2\). We deduce that
\(\sum_{1\leq i \leq \ell} { \mathbf{F} }x_i + { \mathbf{F} }h_i + { \mathbf{F} }y_i \subseteq L_0\)
maps isomorphically into \(L\), so we can identify \(x_i, h_i, y_i\)
with their images in \(L\), which are still linearly independent and
still generate \(L\) as a Lie algebra.

\end{observation}

\begin{observation}[Step 5]

For \(\lambda\in H {}^{ \vee }\), set
\(L_ \lambda\coloneqq\left\{{z\in L {~\mathrel{\Big\vert}~}[hz] = \lambda(h) z\, \forall h\in H }\right\}\)
and write \(\lambda > 0 \iff \lambda \in {\mathbf{Z}}_{\geq 0} \Delta\)
and similarly define \(\lambda < 0\). View
\(\alpha_i \in H {}^{ \vee }\), extended linearly as before. Note
\(H = L_0, N^+ = \sum_{\lambda> 0} L_{\lambda}, N^- = \sum_{\lambda<0} L_ \lambda\),
and thus
\begin{align*}
L = N^- \oplus H \oplus N^+
,\end{align*}
which is a direct sum since the eigenvalues in different parts are
distinct.

\end{observation}

\end{proof}

\hypertarget{serres-theorem-continued-friday-november-11}{%
\section{Serre's Theorem, Continued (Friday, November
11)}\label{serres-theorem-continued-friday-november-11}}

\begin{proof}[of Serre's theorem, continued]

Recall that we have
\begin{align*}
L = N^- \oplus H \oplus N^+ \coloneqq Y/\left\langle{(S_{ij}^-)}\right\rangle \oplus {\mathbf{C}}\left\langle{h_1,\cdots, h_\ell}\right\rangle \oplus X/\left\langle{ (S_{ij}^+) }\right\rangle
.\end{align*}

\begin{remark}[Step 6]

For \(1\leq i\leq \ell\), note \({ \operatorname{ad}}_{L, x_i}\) (and
similarly \({ \operatorname{ad}}_{L, y_i}\)) is locally nilpotent on
\(L\). Let \(M \subseteq L\) be the subspace of elements on which
\({ \operatorname{ad}}_{x_i}\) acts nilpotently. By the Leibniz rule,
\(( { \operatorname{ad}}_{x_i})^{m+n}([uv])\) when
\(( { \operatorname{ad}}_{x_i})^m(v) = 0\) and
\(( { \operatorname{ad}}_{x_y})^n(u) = 0\), so \(M \leq L\) is a Lie
subalgebra. By the Serre relations,
\(( { \operatorname{ad}}_{x_i})^2(h_j) = 0\) and
\(( { \operatorname{ad}}_{x_i})^3(y_j) = 0\), so the generators of \(L\)
are in \(M\) and thus \(L = M\).

\end{remark}

\begin{remark}[Step 7]

Defining
\(\tau_i \coloneqq\exp( { \operatorname{ad}}_{x_i}) \circ \exp( { \operatorname{ad}}_{-y_i}) \circ \exp( { \operatorname{ad}}_{x_i}) \in \mathop{\mathrm{Aut}}(L)\),
by lemma (B) we have \(\tau_i(L_ \lambda) = L_{s_i \lambda}\) where
\(s_i \coloneqq s_{\alpha_i}\) and \(\lambda\in H {}^{ \vee }\).

\end{remark}

\begin{remark}[Step 8]

Let \(\lambda, \mu \in H {}^{ \vee }\) and suppose \(w \lambda = \mu\);
we want to show \(\dim L_ \lambda = \dim L_{ \mu}\). Note \(W\) is
generated by simple reflections \(s_i\), so it STS this when \(w=s_i\),
whence it follows from lemma (B).

\end{remark}

\begin{remark}[Step 9]

Clearly \(\dim (L_0)_{\alpha_i} = 1\) since it's spanned by \(x_i\).
Then \(\dim (L_0)_{k \alpha_i} = 0\) for \(k\neq 0, \pm 1\), so
\(\dim L_{\alpha_i} \leq 1\) and \(\dim L_{k \alpha} = 0\) for
\(k\neq 0, \pm 1\). Since \(x_i\in L_{\alpha_i}\) has a nonzero image in
\(L\), \(\dim L_{\alpha_i} = 1\).

\end{remark}

\begin{remark}[Step 10]

If \(\beta\in \Phi\), conjugate it to a simple root using
\(\beta = w \alpha_i\) with \(w\in W, \alpha_i \in \Delta\). By step 8,
\(\dim L_{ \beta} = 1\) and \(L_{k \beta} = 0\) for \(k\neq 0, \pm 1\).

\end{remark}

\begin{remark}[Step 11]

Suppose \(L_ \lambda \neq 0\) where \(\lambda\neq 0\). Then
\(\lambda\in {\mathbf{Z}}_{\geq 0} \Delta\) or
\({\mathbf{Z}}_{\leq 0} \Delta\), i.e.~all coefficients are the same
sign. Suppose \(\lambda\not\in \Phi\), then
\(\lambda\in {\mathbf{Z}}\Phi\) by (10). Exercise 10.10 yields
\(\exists w\in W\) such that \(w \lambda{\mathbf{Z}}\Delta\) with both
positive and negative coefficients. Thus \(w \lambda\) can not be a
weight, and by step 8, \(0 = \dim L_{w \lambda} = \dim L_ \lambda\).

\end{remark}

\begin{remark}[Step 12]

Writing \(L = N^- \oplus H \oplus N^+\) with
\(H = L_0, N^+ = \sum_{ \lambda > 0} L_{ \lambda} = \sum_{\beta\in \Phi^+} L_{ \lambda}\)
and
\(N^- = \sum_{ \lambda < 0} L_{ \lambda} = \sum_{ \beta\in \Phi^-} L_ \beta\),
by step 10 we can conclude \(\dim L = \ell + {\sharp}\phi < \infty\).
This shows that \(H\) is toral, i.e.~its elements are ad-semisimple.

\end{remark}

\begin{remark}[Step 13]

We have that \(L\) is a finite-dimensional Lie algebra. To show
semisimplicity, we need to now \(L\) has no nonzero solvable ideals, and
as before it's ETS \(L\) has no nonzero \emph{abelian} ideals. Suppose
\(A \subseteq L\) is an abelian ideal; we WTS \(A = 0\). Since
\([H,A] \subseteq A\) and \(H\curvearrowright L\) diagonally,
\(H\curvearrowright A\) diagonally as well and thus
\begin{align*}
A = (A \cap H) \oplus \bigoplus _{\alpha\in \Phi} (A \cap L_{ \alpha})
.\end{align*}
If \(A \cap L_{\alpha} \neq 0\) then
\(A \cap L_{ \alpha} = L_{\alpha}\), which is 1-dimensional.

\begin{quote}
Note: the argument in Humphreys here may not be quite right, so we have
to do something different.
\end{quote}

Now \(\exists w\in W\) such that \(w \alpha = \alpha_i \in \Delta\) as
in step 8, so write \(w = s_{i_1}\cdots s_{i_t}\). Then
\(\tau_{i_1}\cdots \tau_{i_t} (L_ \alpha) = L_ \alpha\). Set
\(A' \coloneqq\tau(A)\), then \(A'{~\trianglelefteq~}L\) is necessarily
an abelian ideal and \(A' \cap L_{\alpha_i} = L_{\alpha_i}\).

So we can replace \(\alpha\) by a simple root and replace \(A\) by
\(A'\). Then \(x_i\in L_{\alpha_i} \subseteq A' {~\trianglelefteq~}L\),
but \(A'\ni -[y_i, x_i] = h_i\), but \([h_i, x_i] \neq 0\),
contradicting that \(A'\) is abelian. \(\contradiction\).

Note \([A', L_{\alpha_j}] \subseteq A' \subseteq H\) and
\([A', L_{\alpha_j}] \subseteq L_{\alpha_j}\), but
\(H\curvearrowright L_{\alpha_j}\) with eigenvalues \(\alpha_j\) and
thus \(A' \subseteq \cap_{j=1}^\ell \ker \alpha_j = 0\) since the
\(\alpha_j\) span \(H {}^{ \vee }\). So \(A' = 0\) and \(L\) is
semisimple.

\end{remark}

\begin{remark}[Step 14]

Since \(L = H \oplus \bigoplus _{\alpha\in \Phi} L_{ \alpha}\), it's
easy to check that \(C_L(H) = H\) by considering what happens when
bracketing against any nonzero element in \(\bigoplus L_ \alpha\). Thus
\(H\) is a maximal toral subalgebra with corresponding root system
\(\Phi\).

\end{remark}

\end{proof}

\begin{remark}

Next: part VI on representation theory, although we'll first cover
\(\S 13\) on weights, especially \(\S 13.1, \S 13.2\). Goal: Weyl's
character formula.

\end{remark}

\hypertarget{part-5-representation-theory-monday-november-14}{%
\section{Part 5: Representation Theory (Monday, November
14)}\label{part-5-representation-theory-monday-november-14}}

\hypertarget{s-13-abstract-theory-of-integral-weights}{%
\subsection{\texorpdfstring{\(\S 13\) Abstract theory of integral
weights}{\textbackslash S 13 Abstract theory of integral weights}}\label{s-13-abstract-theory-of-integral-weights}}

\begin{definition}[Integral weights and the root lattice]

Let \({\mathbb{E}}\supseteq\Phi \supseteq\Delta\) with Weyl group \(W\).
An element \(\lambda\in {\mathbb{E}}\) is an \textbf{integral weight} if
\({\left\langle {\lambda },~{\beta } \right\rangle}= (\lambda, \beta {}^{ \vee })\in {\mathbf{Z}}\)
for all \(\beta \in \Phi\), where
\(\beta {}^{ \vee }\coloneqq{2\beta \over (\beta, \beta)}\). We write
the set of all weights as \(\Lambda\), and write
\(\Lambda_r \coloneqq{\mathbf{Z}}\Phi\) for the \textbf{root lattice}.

\end{definition}

\begin{remark}

Recall
\(\Delta {}^{ \vee }\coloneqq\left\{{\alpha {}^{ \vee }{~\mathrel{\Big\vert}~}\alpha\in \Delta}\right\}\)
is a base for
\(\Phi {}^{ \vee }= \left\{{ \beta {}^{ \vee }{~\mathrel{\Big\vert}~}\beta \in \Phi }\right\}\),
and so
\begin{align*}
\lambda \in \Lambda\iff (\lambda, \alpha {}^{ \vee }) = {\left\langle {\lambda },~{\alpha } \right\rangle}\in {\mathbf{Z}}\forall \alpha \in \Delta
.\end{align*}

\end{remark}

\begin{definition}[Dominant weights]

A weight \(\lambda\in \Lambda\) is \textbf{dominant} iff
\({\left\langle {\lambda },~{\alpha } \right\rangle}\geq 0\) for all
\(\alpha\in \Delta\), and we denote the set of all such dominant weights
\(\Lambda^+\). The weight \(\lambda\) is \textbf{strongly dominant} if
\({\left\langle {\lambda },~{\alpha } \right\rangle}> 0\) for all
\(\alpha \in \Delta\). Writing
\(\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{\ell} }\right\}\),
let \(\left\{{ \lambda_i }\right\}_{1\leq i \leq \ell}\) be the dual
basis for \({\mathbb{E}}\) relative to
\({\left\langle {{-}},~{{-}} \right\rangle}\), so
\({\left\langle { \lambda_i},~{\alpha_i} \right\rangle} = \delta_{ij}\).
The \(\lambda_i\) are referred to as the \textbf{fundamental dominant
weights}, written \(\lambda_i = \omega_i = \varpi_i\).

\end{definition}

\begin{remark}

If \(\lambda \in \Lambda\) then one can write
\(\lambda = \sum_{i=1}^\ell m_i \lambda_i\) where
\(m_i \coloneqq{\left\langle { \lambda},~{\alpha_i} \right\rangle}\), so
\(\Lambda\) is a \({\mathbf{Z}}{\hbox{-}}\)lattice with lattice basis
\(\left\{{\lambda_i}\right\}_{1\leq i\leq \ell}\) containing the root
lattice as a sublattice, so in fact \(\Lambda_r = {\mathbf{Z}}\Delta\).
Writing the Cartan matrix as
\(A = ({\left\langle { \alpha_i},~{\alpha_j} \right\rangle})\) we have
\(\alpha_i = \sum_{j=1}^\ell {\left\langle {\alpha_i},~{\alpha_j} \right\rangle} \lambda_j\)
coming from the \(i\)th row of \(A\). So this matrix expresses how to
write simple roots in terms of fundamental dominant roots, and inverting
it allows writing the fundamental roots in terms of simple roots.

\end{remark}

\begin{fact}

The entires of \(A^{-1}\) are all nonnegative rational numbers, so each
fundamental dominant root is a nonnegative rational linear combination
of simple roots.

\end{fact}

\begin{example}[?]

For \(A_3\) one has
\(A = { \begin{bmatrix}  {2} & {-1} & {0} \\  {-1} & {2} & {-1} \\  {0} & {-1} & {2} \end{bmatrix} }\),
so
\begin{align*}
\alpha_1 &= 2 \lambda_1 - \lambda_2 \\
\alpha_2 &= - \lambda_1 + 2 \lambda_2 - \lambda_3 \\
\alpha_3 &= - \lambda_2 + 2 \lambda_3
.\end{align*}

\end{example}

\begin{definition}[Fundamental group]

The quotient \(\Lambda/ \Lambda_r\) is called the \textbf{fundamental
group} of \(\Phi\), and the index \(f\coloneqq[\Lambda: \Lambda_r]\) is
called its \textbf{index of connection}.

\end{definition}

\begin{remark}

The index is generally small:

\begin{itemize}
\tightlist
\item
  \(A_\ell\) has \(f=\ell + 1\)
\item
  \(f=1\) is obtained from \(F_4, G_2, E_8\),
\item
  \(f=2\) is obtained from types \(B, C, E_7\),
\item
  \(f=3\): \(E_6\),
\item
  \(f=4\): type \(D\).
\end{itemize}

\end{remark}

\hypertarget{s-13.2-dominant-weights}{%
\subsection{\texorpdfstring{\(\S 13.2\) Dominant
weights}{\textbackslash S 13.2 Dominant weights}}\label{s-13.2-dominant-weights}}

\begin{remark}

Note that
\begin{align*}
s_i \lambda_j = \lambda_j - {\left\langle { \lambda_j},~{\alpha_i} \right\rangle} \alpha_i = \lambda_j - \delta_{ij} \alpha_i
,\end{align*}
so \(\Lambda\) is invariant under \(W\). In fact, any sublattice of
\(\Lambda\) containing \(\Lambda_r\) is \(W{\hbox{-}}\)invariant.

\end{remark}

\begin{lemma}[A]

Each integral weight is \(W{\hbox{-}}\)conjugate to exactly one dominant
weight. If \(\lambda\) is dominant, then \(w \lambda\leq \lambda\) for
all \(w\in W\), and if \(\lambda\) is strongly dominant then
\(w \lambda= \lambda\iff w=1\).

\end{lemma}

\begin{proof}[Sketch]

Most of this follows from Theorem 10.3, exercise 10.14, lemma 10.3B, and
corollary 10.2C, along with induction on \(\ell(w)\). We'll omit the
details.

\end{proof}

\begin{remark}

The ordering \(\leq\) on \(\Lambda\) is not well-behaved with respect to
dominant weights, i.e.~one can have \(\mu \Lambda\) with
\(\mu\in \Lambda^+\) dominant but \(\lambda\not\in \Lambda^+\) not
dominant.

\end{remark}

\begin{example}[?]

Let \(\Phi\) be indecomposable of type \(A_1\) with two roots
\(\alpha, \beta\), then \(0\in \Lambda^+\) is dominant, but
\(0 < \alpha \in \Delta\) is not dominant:
\((\alpha,\beta) < 0 \implies {\left\langle {\alpha },~{\beta} \right\rangle}< 0\).

\end{example}

\begin{lemma}[?]

Let \(\lambda \in \Lambda^+\) be dominant, then the number of dominant
\(\mu \in \Lambda^+\) with \(\mu\leq \lambda\) is finite.

\end{lemma}

\begin{proof}[?]

Let \(\lambda, \mu\in \Lambda^+\) and write \(\lambda - \mu\) as a
nonnegative integer linear combination of simple roots. Note
\begin{align*}
0 \leq (\lambda+ \mu, \lambda- \mu ) = (\lambda, \lambda ) - (\mu, \mu) = {\left\lVert { \lambda} \right\rVert}^2 - {\left\lVert {\mu} \right\rVert}^2
,\end{align*}
so \(\mu\) lies in the compact set of vectors whose length is
\({\left\lVert {\lambda} \right\rVert}\) and also in the discrete set
\(\Lambda^+\). The intersection of a compact set and a discrete set is
always finite.

\end{proof}

\hypertarget{s-13.3-the-weight-rho}{%
\subsection{\texorpdfstring{\(\S 13.3\) The weight
\(\rho\)}{\textbackslash S 13.3 The weight \textbackslash rho}}\label{s-13.3-the-weight-rho}}

\begin{definition}[$\rho$]

\begin{align*}
\rho \coloneqq{1\over 2} \sum_{\alpha\in \Phi^+} \alpha
.\end{align*}

\end{definition}

\begin{remark}

This section shows \(\rho = \sum_{i = l}^\ell \lambda_i\) and
\({\left\lVert {\lambda+ \rho} \right\rVert}^2 \geq{\left\lVert {w \lambda+ \rho} \right\rVert}^2\)
when \(\lambda\) is the unique dominant weight in the orbit
\(W \lambda\).

\end{remark}

\hypertarget{s-13.4-saturated-sets-of-weights}{%
\subsection{\texorpdfstring{\(\S 13.4\): Saturated sets of
weights}{\textbackslash S 13.4: Saturated sets of weights}}\label{s-13.4-saturated-sets-of-weights}}

\begin{remark}

This section will be used later to analyze the set of weights in a
finite-dimensional module for semisimple Lie algebra over
\({\mathbf{C}}\).

\end{remark}

\hypertarget{s-20-weights-and-maximal-vectors.}{%
\subsection{\texorpdfstring{\(\S 20\): Weights and maximal
vectors.}{\textbackslash S 20: Weights and maximal vectors.}}\label{s-20-weights-and-maximal-vectors.}}

\begin{remark}

Let \(L\) be finite-dimensional semisimple over \({\mathbf{C}}\)
containing \(H\) its toral subalgebra. This corresponds to
\(\Phi \supseteq\Delta\) with Weyl group \(W\) and
\(\Phi \subseteq {\mathbb{E}}= {\mathbf{R}}\Phi\).

\end{remark}

\hypertarget{s-20.1}{%
\subsection{\texorpdfstring{\(\S 20.1\)}{\textbackslash S 20.1}}\label{s-20.1}}

\begin{definition}[Weight spaces and weights for $L\dash$modules]

Let \(V\) be a finite-dimensional \(L{\hbox{-}}\)module. By corollary
6.4, \(H\curvearrowright V\) semisimply (diagonally) and we can
simultaneously diagonalize to get a decomposition
\begin{align*}
V= \bigoplus _{\lambda\in H {}^{ \vee }} V_{\lambda}, \qquad V_{ \lambda} \coloneqq\left\{{v\in V {~\mathrel{\Big\vert}~}h.v = \lambda(h) v\,\,\forall h\in H}\right\}
.\end{align*}
If \(V_{\lambda}\neq 0\) then \(\lambda\) is a \textbf{weight}.

\end{definition}

\begin{example}[?]

If \(\phi = { \operatorname{ad}}\) and \(V=L\), then
\(L = H \oplus \oplus _{\alpha\in \Phi} L_{\alpha}\) where \(H = L_0\).

\end{example}

\begin{warnings}

If \(\dim V = \infty\), \(V_{ \lambda}\) still makes sense but \(V\) may
no longer decompose as a direct sum of its weight spaces. E.g. take
\(V = {\mathcal{U}}(L)\) and the left regular representation given by
left-multiplication in the algebra
\({\mathcal{U}}(L) \curvearrowright{\mathcal{U}}(L)\). This restricts to
\(L = L_0 \curvearrowright{\mathcal{U}}(L)\), the \emph{regular action}
of \(L\) on \({\mathcal{U}}(L)\). Note that there are no eigenvectors,
since taking a PBW basis one can write
\(\prod h_i^{n_i} \cdot \prod_{\alpha\in \Phi} x_{\alpha}^{n_ \alpha}\),
which strictly increases monomial degrees and thus there are no
eigenspaces. So \(V_ \lambda= 0\) for all \(\lambda\), i.e.~there are no
weight spaces at all.

\end{warnings}

\hypertarget{wednesday-november-16}{%
\section{Wednesday, November 16}\label{wednesday-november-16}}

\begin{remark}

Let \(L\in \mathsf{Lie} \mathsf{Alg}^{{\mathrm{fd}}, {\mathrm{ss}}}\)
containing \(H\) with \(\Phi, \Delta, W\) as usual. Recall that
\(V\in {}_{L}{\mathsf{Mod}}\implies V = \bigoplus _{\lambda\in H {}^{ \vee }} V_{ \lambda}\)
where
\(V_{ \lambda} \coloneqq\left\{{v\in V{~\mathrel{\Big\vert}~}h.v = \lambda(h)v\, \forall h\in H}\right\}\),
which we call a \textbf{weight space} when \(\lambda\neq 0\). Note that
if if \(V\) is \emph{any} representation of \(V\), even
finite-dimensional,
\(V' \coloneqq\bigoplus _{ \lambda\in H {}^{ \vee }} V_{ \lambda} \leq V\)
is always an \(L{\hbox{-}}\)submodule. The sum is still direct since the
terms correspond to eigenspaces with distinct eigenvalues. Note that if
\(h\in H, x\in L_{\alpha}, v\in V_{ \lambda}\), then
\begin{align*}
h.(x.v) 
&= x.(h.v) + [hx].v \\
&= \lambda(h) x.v + \alpha(h) x.v \\
&= (\lambda+ \alpha)(h) x.v
,\end{align*}
so \(L_{ \alpha} V_{ \lambda} \subseteq V_{ \lambda+ \alpha}\).

\end{remark}

\begin{lemma}[?]

Let \(V\in {}_{L}{\mathsf{Mod}}\), then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(L_{\alpha}\) maps \(V_{ \lambda}\) into \(V_{ \lambda + \alpha}\),
\item
  The sum \(V' \coloneqq\sum _{ \lambda\in H {}^{ \vee }} V_{ \lambda}\)
  is direct and \(V' \leq V\) is an \(L{\hbox{-}}\)submodule,
\item
  If \(\dim V < \infty\) then \(V = V'\).
\end{enumerate}

\end{lemma}

\hypertarget{s-20.2-highest-weight-modules}{%
\subsection{\texorpdfstring{\(\S 20.2\): Highest weight
modules}{\textbackslash S 20.2: Highest weight modules}}\label{s-20.2-highest-weight-modules}}

\begin{definition}[Maximal vectors]

A \textbf{maximal vector} in an \(L{\hbox{-}}\)module \(V\) is a nonzero
weight vector \(v\in V_{ \lambda}\) such that \(L_ \alpha.v = 0\) for
all positive roots \(\alpha \in \Phi^+\). Equivalently,
\(L_ \alpha .v = 0\) for all \(\alpha\in \Delta\).

\end{definition}

\begin{definition}[Highest weight vectors]

A \textbf{highest weight vector} is a nonzero \(v\in V_{ \lambda}\)
where \(\lambda\) is maximal among all weights of \(V\) with respect to
the ordering \(\leq\) corresponding to the choice of \(\Delta\).

\end{definition}

\begin{observation}

If \(v\) is a highest weight vector then \(v\) is necessarily a maximal
vector, since \(\lambda+ \alpha > \lambda\), but the converse is not
necessarily true.

\end{observation}

\begin{warnings}

I.e., the weight of a highest weight vector need not be maximal.

\end{warnings}

\begin{example}[?]

In \(\S 18\), \(L\) is constructed using the Serre relations to get
\(L_0 \twoheadrightarrow L\) where \(L\) involved \((S_{ij}^{\pm})\) and
\(L_0\) involved S1-S3. Recalling \(y_i^{m+1} y_j = y_{ij}\), since
\(x_k . y_{ij} = 0\), \(y_{ij}\) is a maximal vector in \(L_0\) as an
\(L{\hbox{-}}\)module but is \emph{not} a highest weight vector since
\(\mathrm{wt} y_{ij} = (-m+1) \alpha_i - \alpha_j < - \alpha_j = \mathrm{wt}(y_j)\)
and the weight is not maximal.

\end{example}

\begin{example}[?]

View \(L\in {}_{L}{\mathsf{Mod}}\) via \({ \operatorname{ad}}_L\), then
\(\S 10.4\) shows that there is a unique highest root \(\tilde \alpha\)
satisfying \(\tilde \alpha \geq \alpha\) for all \(\alpha\in \Phi\). Any
nonzero \(v\in L_{\tilde \alpha}\) is a highest weight vector for the
adjoint representation.

\end{example}

\begin{definition}[Borel subalgebras]

A \textbf{Borel subalgebra} of \(L\) is a maximal solvable subalgebra
\(B\leq L\).

\end{definition}

\begin{proposition}[?]

\(B \coloneqq H \oplus \bigoplus _{\alpha\in \Phi^+} L_ \alpha\) is a
Borel subalgebra of \(L\).

\end{proposition}

\begin{proof}[?]

If \(\alpha, \beta\in \Phi^+\) then
\([L_{ \alpha}, L_{ \beta}] = L_{\alpha + \beta}\) where
\(\alpha + \beta\in \Phi^+\) (if this is still a root), so \(H\leq L\).
One has
\(B^{(i)} \coloneqq[B^{(i-1)}, B^{(i-1)}] \subseteq \sum_{\operatorname{ht}( \beta) \geq 2^{i-1} } L_{\beta}\),
since bracketing elements of \(H\) together will vanish (since \(H\) is
abelian) and bracketing height 1 roots yields height 2, bracketing
height 2 yields height 4, and so on. Thus \(B\) is a solvable
subalgebra, since the height is uniformly bounded above by a finite
number. To see that its maximal, note that any subalgebra \(B' \leq L\)
containing \(B\), it must also contain some \(L_{ - \alpha}\) for some
\(\alpha\in \Phi^+\). But then
\(B' \supseteq{\mathfrak{sl}}_2({\mathbf{C}}) = L_{ - \alpha} \oplus [ L_{ - \alpha }, L_{ \alpha}] \oplus L_{ \alpha}\)
which is not solvable, so \(B'\) can not be solvable.

\end{proof}

\begin{remark}

Let \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fd}}\), then
\(V\in {}_{B}{\mathsf{Mod}}\) by restriction and by Lie's theorem \(V\)
must have a common eigenvector \(v\) for the action of \(B\). Since
\(B\supseteq H\), \(v\) is a weight vector,
\([B, B] = \bigoplus _{ \alpha\in \Phi^+} L_ \alpha\) acts by
commutators of operators acting by scalars, which commute, and thus this
acts by zero on \(v\) and makes \(v\) a maximal vector in \(V\). So any
finite dimensional \(L{\hbox{-}}\)module as a maximal vector.

\end{remark}

\begin{definition}[Highest weight modules]

A module \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\), possibly infinite
dimensional, is a \textbf{highest weight module} if there exists a
\(\lambda \in H {}^{ \vee }\) and a nonzero vector
\(v^+ \in V_{ \lambda}\) such that \(V\) is generated as an
\(L{\hbox{-}}\)module by \(v^+\), i.e.~\(U(L).v^+ = V\).

\end{definition}

\begin{remark}

Let
\(x_ \alpha \in L_ \alpha, y_ \alpha\in L_{- \alpha}, h_{\alpha} = [x_ \alpha, y_ \alpha]\)
be a fixed standard \({\mathfrak{sl}}_2\) triple in \(L\).

\end{remark}

\begin{theorem}[?]

Let \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) be a highest weight
module with maximal vector \(v^+ \in V_{ \lambda}\). Write
\(\Phi^+ = \left\{{ {\beta }_{1}, \cdots, {\beta }_{m} }\right\}\),
\(\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{\ell} }\right\}\),
then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(V\) is spanned by the vectors
  \(y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m} .v^+\) for
  \(i_j\in {\mathbf{Z}}_{\geq 0}\). In particular,
  \(V = \bigoplus _{ \mu\in H {}^{ \vee }} V_ \mu\).
\item
  The weights of \(V\) are of the form
  \(\mu = \lambda- \sum_{i=1}^\ell k_i \alpha_i\) with
  \(k_i\in {\mathbf{Z}}_{ \geq 0}\), and all weights \(\mu\) satisfy
  \(\mu \leq \lambda\).
\item
  For each \(\mu\in H {}^{ \vee }\), \(\dim V_{ \mu} < \infty\), and for
  the highest weight \(\lambda\), one has \(\dim V_{ \lambda} = 1\)
  spanned by \(v^+\).
\item
  Each \(L{\hbox{-}}\)submodule \(W\) of \(V\) is a direct sum of its
  weight spaces.
\item
  \(V\) is indecomposable in \({\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) with
  a unique maximal proper submodule and a corresponding unique
  irreducible quotient.
\item
  Every nonzero homomorphic image of \(V\) is also a highest weight
  module of the same highest weight.
\end{enumerate}

\end{theorem}

\begin{proof}[Sketch]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Use the PBW theorem and extend a basis for any \(B\leq L\) to a basis
  where the \(B\) basis elements come second. Writing
  \(L = N^ \oplus B\), one can decompose
  \(U(L) = U(N^-) \otimes_{\mathbf{C}}U(B)\) and get
  \(U(L) v^+ = U(N^-)U(B)v^+ = U(N^-)U(H \oplus N^+)v^+\)
\item
  Writing the \(\beta\) in terms of \(\alpha\) yields this expression.
\item
  Clear.
\end{enumerate}

We'll finish the rest next time.

\end{proof}

\hypertarget{s21.2-a-sufficient-condition-for-finite-dimensionality-monday-november-21}{%
\section{\texorpdfstring{\(\S21.2\): A sufficient condition for
finite-dimensionality (Monday, November
21)}{\textbackslash S21.2: A sufficient condition for finite-dimensionality (Monday, November 21)}}\label{s21.2-a-sufficient-condition-for-finite-dimensionality-monday-november-21}}

\begin{remark}

Last time: if \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fd}}\)
then \(V = L( \lambda)\) for some dominant weight
\(\lambda\in \Lambda^+\), yielding a necessary condition for
finite-dimensionality. Today: a sufficient condition.

\end{remark}

\begin{lemma}[?]

Write
\(\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{n} }\right\}\) and
set \(x_i \coloneqq x_{ \alpha_i}, y_i \coloneqq y_{\alpha_i}\). For
\(k\geq 0\) and \(1\leq i,j\leq \ell\), the following relations hold in
\(U(L)\):

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \([x_j, y_i^{k+1}] = 0\) for \(i\neq j\)
\item
  \([h_j, y_i^{k+1}] = - (k+1) \alpha_i(h_j) y_i^{k+1}\)
\item
  \([x_i, y_i^{k+1}] = (k+1) y_i^k(h_i - k\cdot 1)\)
\end{enumerate}

\end{lemma}

\begin{proof}[of (c)]

Use that \({ \operatorname{ad}}\) acts by derivations:
\begin{align*}
[x_i, y_i^{k+1}] 
&= x_i y_i^{k+1} - y_i^{k+1} x_i \\
&= x_i y_i y_i^k - y_i x_i y_i^k + y_i x_i y_i^k - y_i y_i^k x_i \\
&= [x_i y_i] y_i^k + y_i[x_i y_i^k] \\
&= h_i y_i^k + y_i [x_i y_i^k] \\
&= \qty{ y_i^k h_i - k \alpha_i (h_i) y_i^k } + y_i \qty{k y_i^{k-1} (h_i - (k-1)\cdot 1)  } \qquad\text{by (b) and induction} \\
&= y_i^k h_i - 2k y_i^k + ky_i^k (h_i - (k-1)\cdot 1) \\ 
&= (k+1)y_i^k h_i - (2k + k(k-1)) y_i^k \\
&= (k+1) y_i^k h_i - k(k+1) y_i^k \\
&= (k+1)y_i^k (h_i - k\cdot 1)
.\end{align*}

\end{proof}

\begin{theorem}[?]

Given \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}\), let
\(\Pi(V) \coloneqq\left\{{ \lambda\in H {}^{ \vee }{~\mathrel{\Big\vert}~}V_ \lambda\neq 0}\right\}\)
be the set of weights. If \(\lambda\in \Lambda^+\) is a dominant weight,
then
\(V \coloneqq L(\lambda)\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{irr}}\)
is finite-dimensional and \(\Pi(V)\) is permuted by \(W\) with
\(\dim V_ \mu = \dim V_ { \sigma \mu}\) for all \(\sigma\in W\).

\end{theorem}

\begin{proof}[?]

The main work is showing the last part involving equality of dimensions.
It STS this for a simple reflection \(s_i \coloneqq s_{\alpha_i}\) since
\(\sigma\) is a product of such reflections. Write
\(\phi: L\to {\mathfrak{gl}}(V)\) be the representation associated to
\(V\) -- the strategy is to show that \(\phi(x_i)\) and \(\phi(y_i)\)
are locally nilpotent endomorphisms of \(V\). Let
\(v^+ \in V_ \lambda\setminus\left\{{0}\right\}\) be a fixed maximal
vector and set \(m_i \coloneqq\lambda(h_i)\) so \(h_i.v^+ = m_i v^+\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Set \(w \coloneqq y_i^{m_i+1}. v^+\), then the claim is that \(w=0\).
  Supposing not, we'll show \(w\) is a maximal vector of weight not
  equal to \(\lambda\), and thus not a scalar multiple of \(v^+\). We
  have
  \({\operatorname{wt}}( w) = \lambda- (m_i +1) \alpha_i < \lambda\) (a
  strict inequality). If \(j\neq i\) then
  \(x_j.w = x_j y_i^{m_i + 1} v^+ = y_i^{m_i + 1} x_j v\) by part (a) of
  the lemma above, and this is zero since \(v^+\) is highest weight and
  thus maximal (recalling that these are distinct notions). Otherwise
  \begin{align*}
    x_i w 
    &= x_i y_i^{m_i + 1} v^+ \\
    &= y_i^{m_i+1} x_i v^+ + (m_i + 1) y_i^{m_i} (h_i - m_i\cdot 1)v^+ \\
    &= 0 + (m_i+1)y_i^{m_i}(m_i - m_i) v^+ = 0
    .\end{align*}
  So \(w\) is a maximal vector of weight distinct from \(\lambda\),
  contradicting corollary 20.2 since this would generate a proper
  submodule. \(\contradiction\)
\item
  Let
  \(S_i = \left\langle{x_i, y_i, h_i}\right\rangle \cong {\mathfrak{sl}}_2\),
  then the claim is that \(v^+, y_i v^+, \cdots, y_i^{m_i} v^+\) span a
  nonzero finite-dimensional \(S_i{\hbox{-}}\)submodule of \(V\). The
  span is closed under (the action of) \(h_i\) since all of these are
  eigenvectors for \(h_i\), and is closed under \(y_i\) since \(y_i\)
  raises generators and annihilates \(y_i^{m_i} v^+\), and is closed
  under \(x_i\) by part (c) of the lemma (since it lowers generators).
\item
  The sum of two finite-dimensional \(S_i{\hbox{-}}\)submodules of \(V\)
  is again a finite-dimensional \(S_i{\hbox{-}}\)submodule, so let
  \(V_i\) be the sum of all finite-dimensional
  \(S_i{\hbox{-}}\)submodules of \(V\) (which is not obviously
  finite-dimensional, since we don't yet know if \(V\) is
  finite-dimensional). The claim is \(V'\) is a nonzero
  \(L{\hbox{-}}\)submodule of \(V\). Let \(w\in V'\), then \(w\) is a
  finite sum and there exists a finite-dimensional
  \(S_i{\hbox{-}}\)submodule \(W\) of \(V\) with \(w\in W\). Construct
  \(U \coloneqq\sum_{ \alpha \in \Phi} x_ \alpha . W\) where
  \(x_ \alpha \coloneqq y_{- \alpha}\) if \(\alpha\in \Phi^-\), which is
  a finite-dimensional vector subspace of \(V\). Check that
  \begin{align*}
    h_i (x_ \beta . W) &= x_ \beta(h_i.W) + [h_i x_ \beta].W \subseteq x_ \beta.W \subseteq U \\
    x_i (x_ \beta.W) &= x_ \beta(x_i.W) + [x_i x_ \beta] .W \subseteq x_ \beta .W + x _{ \beta + \alpha_i}.W \subseteq U \\
    y_i(x_ \beta. W) &= x_{ \beta}(y_i.W) + [y_i x_ \beta].W \subseteq x_ \beta W + x_{ \beta - \alpha_i}.W \subseteq U
    ,\end{align*}
  and so \(U\) is a finite-dimensional \(S_i{\hbox{-}}\)submodule of
  \(V\) and thus \(U \subseteq V'\). So if \(w\in V'\) then
  \(x_ \alpha .w \in V'\) for all \(\alpha\in \Phi\) and \(V'\) is
  stable under a set of generators for \(L\), making \(V' \leq V\) an
  \(L{\hbox{-}}\)submodule. Since \(V'\neq 0\) (since it contains at
  least the highest weight space), it must be all of \(V\) since \(V\)
  is irreducible.
\item
  Given an arbitrary \(v\in V\), apply the argument for \(w\) in step 3
  to show that there exists a finite-dimensional
  \(S_i{\hbox{-}}\)submodule \(W \subseteq V\) with \(v\in W\). The
  elements \(x_i, y_i\) act nilpotently on any finite-dimensional
  \(S_i{\hbox{-}}\)module, and so in particular they act nilpotently on
  \(v\) and we get local nilpotence.
\item
  Now
  \(\tau_i \coloneqq e^{\phi(x_i)} \circ e^{\phi(-y_i)} \circ e^{\phi(x_i)}\)
  is well-defined. As seen before, \(\tau_i(V_ \mu) = V_{s_i \mu}\),
  i.e.~\(\tau_i\) is an automorphism that behaves like \(s_i\), and so
  \(\dim V_ \mu = \dim V_{ \sigma \mu}\) for all \(\mu\in \Pi(V)\) and
  \(\sigma\in W\). Now any \(\mu\in \Pi(V)\) is conjugate under \(W\) to
  a unique dominant weight \(\mu^+\), and by (4) \(\mu^+\in \Pi(V)\) and
  since the weights in \(V = L(\lambda)\) has only weights smaller than
  \(\lambda\), we have \(\mu^+ \leq \lambda\). Note
  \(\lambda \in \Lambda^+\) is dominant, and so by 13.2B there are only
  finitely many such weights \(\mu^+\). Now there are only finitely many
  conjugates of the finitely many possibilities for \(\mu^+\), so
  \({\sharp}\Pi(V) < \infty\). By the general theory of highest weight
  modules, all weight spaces \(V_{\mu} \leq L( \lambda)\) are
  finite-dimensional. Since \(\dim V_\mu < \infty\) for all
  \(\mu\in\Pi(V)\), we have \(\dim V < \infty\).
\end{enumerate}

\end{proof}

\begin{remark}

Skipping the next two sections \(\S 21.3\) on weight strings and weight
diagrams, and \(\S 21.4\) on generators and relations for
\(L( \lambda)\) for \(\lambda\in \Lambda^+\) dominant.

\end{remark}

\hypertarget{monday-november-28}{%
\section{Monday, November 28}\label{monday-november-28}}

\begin{remark}

Setup:
\(L\in {\mathsf{Lie}{\hbox{-}} \mathsf{Alg}}^{{\mathrm{fd}}, {\mathrm{ss}}}_{/ {{\mathbf{C}}}}\)
containing \(H\) a maximal toral subalgebra,
\(\Phi \supseteq\Phi^+ = \left\{{ {\beta }_{1}, \cdots, {\beta }_{m} }\right\} \supseteq\Delta = \left\{{ {\alpha }_{1}, \cdots, {\alpha }_{\ell} }\right\}\)
with Weyl group \(W\), and we have
\(x_i \in L_{ \alpha_i}, y_i\in L_{- \alpha_i}, h_i = [ x_i y_i]\). For
\(\beta\in \Phi^+\), we also wrote
\(x_ \beta\in L_ \beta, y_ \beta\in L_{- \beta}\). There is also a Borel
\(B = H \oplus \bigoplus _{\beta > 0} L_ \beta\) with
\(H \subseteq B \subseteq L\).

We saw that if \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fd}}\)
then \(V = \bigoplus _{\mu\in H {}^{ \vee }} V_ \mu\) and \(V\) is a
highest weight module of highest weight \(\lambda\) where
\(\lambda \in \Lambda^+ = \left\{{\nu \in H {}^{ \vee }{~\mathrel{\Big\vert}~}\nu(h_i) \in {\mathbf{Z}}_{\geq 0} \, 1\leq i\leq \ell}\right\}\).
Writing \(M( \lambda)\) for the Verma module
\({\mathcal{U}}(L) \otimes_{{\mathcal{U}}(B)} C_{ \lambda}\), there is a
unique irreducible quotient
\(M( \lambda) \twoheadrightarrow L( \lambda)\) with highest weight
\(\lambda\). It turns out that \(L ( \lambda)\) is finite-dimensional if
\(\lambda\in \Lambda^+\).

\end{remark}

\begin{question}

How can we understand \(L( \lambda)\) for \(\lambda\in \Lambda^+\)
better? What is \(\dim L( \lambda)\)? What is \(\dim L( \lambda)_\mu\)
(i.e.~the dimensions of weight spaces)?

\end{question}

\hypertarget{s-22-multiplicity-formulas-s-22.5-formal-characters}{%
\subsection{\texorpdfstring{\(\S 22\) Multiplicity Formulas; \(\S 22.5\)
Formal
characters}{\textbackslash S 22 Multiplicity Formulas; \textbackslash S 22.5 Formal characters}}\label{s-22-multiplicity-formulas-s-22.5-formal-characters}}

\begin{remark}

Let \(\Lambda \subseteq H {}^{ \vee }\) be the lattice of integral
weights. Note that
\(\lambda(h_\beta) \in {\mathbf{Z}}\,\, \forall \beta\in \Phi \iff \lambda(h_i) \in {\mathbf{Z}}\)
for all \(1\leq i\leq \ell\). Since
\(\Lambda\in {}_{{\mathbf{Z}}}{\mathsf{Mod}}\) there is a group algebra
\({\mathbf{Z}}[ \Lambda]\), the free \({\mathbf{Z}}{\hbox{-}}\)module
with basis \(e( \lambda)\) for \(\lambda\in \Lambda\), also written
\(e_ \lambda\) or \(e^{ \lambda}\). This has a ring structure given by
linearly extending
\(e( \lambda)\cdot e( \mu) \coloneqq e( \lambda+ \mu)\), so
\begin{align*}
\qty{ \sum a_ \lambda e( \lambda) } \cdot \qty{\sum b_ \mu e( \mu) } = \sum c_ \sigma e( \sigma), \qquad c_ \sigma\coloneqq\sum_{ \lambda+ \mu = \sigma} a_ \lambda b _{\mu}
.\end{align*}
Note that if \(V\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fd}}\)
then \(V = \bigoplus _{ \mu\in H {}^{ \vee }} V_ \mu\), where
\(V_{ \mu}\neq 0\implies \mu \in \Lambda\). In this case we can define
the \textbf{formal character}
\begin{align*}
\mathrm{ch}_V \coloneqq\sum_{ \mu\in \Lambda} (\dim V_ \mu) e( \mu) \in {\mathbf{Z}}[ \Lambda ]
.\end{align*}

\end{remark}

\begin{proposition}[?]

Let \(V, W\in {\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fd}}\), then
\begin{align*}
\mathrm{ch}_{V\otimes W} = \operatorname{ch}_V \cdot \operatorname{ch}_W
.\end{align*}

\end{proposition}

\begin{proof}[?]

Take dimensions in the formula
\((V\otimes W)_ \sigma = \sum_{ \lambda+ \mu = \sigma} V_ \lambda\otimes W_ \mu\).

\end{proof}

\begin{remark}

For \(\lambda\in \Lambda^+\), we have \(L( \lambda)\) (noting Humphreys
uses \(V( \lambda)\)), we write
\begin{align*}
\operatorname{ch}_ \lambda\coloneqq\operatorname{ch}_{L( \lambda)} = \sum m_ \lambda(\mu) e( \mu)
\end{align*}
where
\(m_{\lambda}( \mu) \coloneqq\dim L( \lambda)_ \mu\in {\mathbf{Z}}_{\geq 0}\).

\end{remark}

\begin{remark}

We now involve the Weyl group to make more progress: let \(W\) be the
Weyl group of \((L, H)\), then
\(W\curvearrowright{\mathbf{Z}}[ \Lambda]\) by
\(w . e( \mu) \coloneqq e(w \mu)\) for \(w\in W, \mu\in \Lambda\). So
\(W\to \mathop{\mathrm{Aut}}_{\mathsf{Ring}}({\mathbf{Z}}[ \Lambda])\),
and recalling that \(\dim L( \lambda)_ \mu = \dim L(\lambda)_{ w \mu}\),
we have
\begin{align*}
w. \operatorname{ch}_{\lambda} = \sum_\mu m_ \lambda( \mu) e(w \mu) = \sum_\mu m_ \lambda(w \mu) e(w \mu) = \operatorname{ch}_{\lambda}
,\end{align*}
so \(\operatorname{ch}_ \lambda\) are \(W{\hbox{-}}\)invariant elements
of the group algebra.

\end{remark}

\begin{proposition}[?]

Let \(f\in {\mathbf{Z}}[ \Lambda]^W\) be a \(W{\hbox{-}}\)invariant
element, then \(f\) can be written uniquely as a
\({\mathbf{Z}}{\hbox{-}}\)linear combination of the
\(\operatorname{ch}_ \lambda\) for \(\lambda \in \Lambda^+\), i.e.~these
form a \({\mathbf{Z}}{\hbox{-}}\)basis:
\begin{align*}
{\mathbf{Z}}[ \Lambda]^W = \left\langle{\operatorname{ch}_ \lambda{~\mathrel{\Big\vert}~}\lambda\in \Lambda}\right\rangle_{\mathbf{Z}}
.\end{align*}

\end{proposition}

\begin{proof}[?]

Recall every \(\lambda\in \Lambda\) is conjugate to a unique
\(\lambda\in \Lambda^+\). Since \(f\) is \(W{\hbox{-}}\)invariant, we
can write it as
\begin{align*}
f = \sum_{ \lambda\in \Lambda^+} c( \lambda) \qty{\sum_{w\in W} e( w \lambda)}
\end{align*}
where \(c( \lambda)\in {\mathbf{Z}}\) and almost all are zero. Given
\(\lambda\in \Lambda^+\) with \(c( \lambda)\neq 0\), a previous lemma
(13.2B) shows that
\({\sharp}\left\{{ \mu\in \Lambda^+ {~\mathrel{\Big\vert}~}\mu\leq \lambda}\right\} < \infty\)
by a compactness argument. Set
\begin{align*}
M_f \coloneqq\bigcup_{c( \lambda) \neq 0, \lambda\in \Lambda^+} \left\{{ \mu\in \Lambda^+ {~\mathrel{\Big\vert}~}\mu\leq \lambda}\right\}
,\end{align*}
all of the possible weights that could appear, then
\({\sharp}M_f < \infty\) since this is a finite union of finite sets.
Choose \(\lambda\in \Lambda^+\) maximal with respect to the property
that \(c( \lambda)\neq 0\), and set
\(f' \coloneqq f- c( \lambda) \operatorname{ch}_ \lambda\). Note that
\(f'\) is again \(W{\hbox{-}}\)invariant, since \(f\) and
\(\operatorname{ch}_ \lambda\) are both \(W{\hbox{-}}\)invariant, and
\(M_{ \operatorname{ch}_ \lambda} \subseteq M_f\). However
\(\lambda \not\in M_{f'}\) by maximality, since we've subtracted
\(c( \lambda) e(\lambda)\) off, so \({\sharp}M_{f'} < {\sharp}M_f\).
Inducting on \({\sharp}M_f\), \(f'\) is a
\({\mathbf{Z}}{\hbox{-}}\)linear combination of
\(\operatorname{ch}_{\lambda'}\) for \(\lambda' \in \Lambda\), and thus
so is \(f\). One checks the base case \(L(0) \cong {\mathbf{C}}\) where
everything acts with weight zero. Uniqueness is relegated to exercise
22.8.

\end{proof}

\begin{question}

Is there an explicit formula for \(\operatorname{ch}_ \lambda\) for
\(\lambda \in \Lambda^+\)? An intermediate goal will be to understand
characters of Verma modules \(\operatorname{ch}M( \lambda)\) -- note
that this isn't quite well-defined yet, since this is an
infinite-dimensional module and thus the character has infinitely many
terms and is not an element in \({\mathbf{Z}}[ \Lambda]\).

\end{question}

\hypertarget{s-23.2-characters-and-verma-modules}{%
\subsection{\texorpdfstring{\(\S 23.2\) Characters and Verma
Modules}{\textbackslash S 23.2 Characters and Verma Modules}}\label{s-23.2-characters-and-verma-modules}}

\begin{question}

Let \({\mathcal{Z}}(L) \subseteq {\mathcal{U}}(L)\) be the center of
\({\mathcal{U}}(L)\), not to be confused with \(Z(L) \subseteq L\) which
is zero since \(L\) is semisimple (since \(Z(L)\) is an abelian ideal).
How does \({\mathcal{Z}}(L) \curvearrowright M( \lambda)\)?

\end{question}

\begin{remark}

Note
\(M( \lambda) = {\mathcal{U}}(L) \otimes_{{\mathcal{U}}(B)} {\mathbf{C}}_{ \lambda} \cong {\mathcal{U}}(N^-) \otimes_{\mathbf{C}}{\mathbf{C}}_ \lambda\)
and write \(v^+\) for a nonzero highest weight vector of
\(M( \lambda)\). Let \(z\in {\mathcal{Z}}(L)\) and \(h\in H\), then
\begin{align*}
h.(z.v^+) = z.(h.v^+) = z.( \lambda(h) ) v^+ = \lambda(h) v^+
,\end{align*}
and \(x_ \alpha(z.v^+) = z(x_ \alpha.v^+) = 0\) for all
\(\alpha\in \Phi^+\), so \(z.v^+\) is a maximal vector in
\(M( \lambda)\) of weight \(\lambda\), i.e.~there exists
\(\chi_ \lambda(z)\in {\mathbf{C}}\) such that
\begin{align*}
z.v^+ = \chi_ \lambda( z) v^+
\end{align*}
since \(\dim M( \lambda)_ \lambda = 1\). Thus there is an algebra
morphism
\begin{align*}
\chi_ \lambda: {\mathcal{Z}}(L) \to {\mathbf{C}}
.\end{align*}

Pick a PBW basis for the Verma module \(M(\lambda)\), then
\begin{align*}
z. y_{\beta_1}^{i_1}\cdots y_{\beta_m }^{i_m} v^+ =
y_{\beta_1}^{i_1}\cdots y_{\beta_m }^{i_m} z. v^+ =
\chi_ \lambda(z) y_{\beta_1}^{i_1}\cdots y_{\beta_m }^{i_m} v^+ 
,\end{align*}
so \(z.m = \chi_ \lambda(z)m\) for all \(m\in M( \lambda)\), and thus
\({\mathcal{Z}}(L)\curvearrowright M(\lambda)\) by the character
\(\chi_ \lambda\). Consequently, \({\mathcal{Z}}(L)\) acts on any
subquotient of \(M( \lambda)\).

\end{remark}

\begin{question}

When is \(\chi_ \lambda= \chi_ \mu\) for two integral weights
\(\lambda, \mu\in \Lambda\)?

\end{question}

\hypertarget{wednesday-november-30}{%
\section{Wednesday, November 30}\label{wednesday-november-30}}

\begin{remark}

Recall: \({\mathcal{Z}}(L) \coloneqq Z(\mu(L))\) acts by a character
\(\chi_ \lambda: {\mathcal{Z}}(L) \to {\mathbf{C}}\) on \(M( \lambda)\)
and thus any subquotients. For
\(\lambda, \mu\in \Lambda \subseteq H {}^{ \vee }\), when is
\(\chi_ \lambda = \chi_ \mu\)?

\end{remark}

\begin{definition}[Linkage]

Two weights \(\lambda, \mu\in H {}^{ \vee }\) are \textbf{linked} iff
\(\exists w\in W\) such that \(\mu + \rho = w ( \lambda+ \rho)\), where
\(\rho \coloneqq{1\over 2}\sum_{ \beta\in \Phi^+} \beta\). In this case,
write \(\mu \sim \lambda\) for this equivalence relation,
i.e.~\(\mu = w( \lambda+ \rho) - \rho\). We'll write
\(w\cdot \lambda \coloneqq w( \lambda+ \rho)\) and call this the
\textbf{dot action} of \(W\) on \(H {}^{ \vee }\).

\end{definition}

\begin{warnings}

This defines an action of a group on a set, but it is \textbf{not} a
linear action.

\end{warnings}

\begin{proposition}[?]

Let \(\lambda\in \Lambda, \alpha\in \Delta\) and suppose
\(m\coloneqq\lambda(h_ \alpha) = {\left\langle {\lambda },~{\alpha} \right\rangle}\in {\mathbf{Z}}_{\geq 0}\)
and let \(v^+\) be a highest weight vector of \(M( \lambda)\). Then
\(w\coloneqq y_ \alpha^{m+1}\cdot v^+\) is maximal vector in
\(M( \lambda)\) of weight \(\lambda - (m+1 ) \alpha\).

\end{proposition}

\begin{proof}[?]

The proof that \(w\) is a maximal vector is step 1 in theorem 21.2,
which showed \(\dim L( \lambda) < \infty\) (using lemma 21.2 and
commutator relations in \({\mathcal{U}}(L)\)). Then check that
\begin{align*}
{ \operatorname{weight} }(w) = \lambda - (m+1) \alpha = \lambda- \qty{ {\left\langle {\lambda },~{\alpha } \right\rangle}+ 1} \alpha
.\end{align*}
In fact, for any \(\lambda\in H {}^{ \vee }, \alpha\in \Delta\) we can
define
\begin{align*}
\mu \coloneqq\lambda- \qty{ {\left\langle {\lambda },~{\alpha } \right\rangle}+ 1} \alpha
,\end{align*}
so in our case \({ \operatorname{weight} }(w) = \mu\). Note that
\begin{align*}
\mu 
&= \lambda- {\left\langle {\lambda},~{\alpha } \right\rangle}\alpha- \alpha \\
&= s_ \alpha( \lambda) + (s_ \alpha( \rho) - \rho) \\
&= s_ \alpha(\lambda+ \rho) - \rho\\
&= s_ \alpha \cdot \lambda
.\end{align*}
Now \(w\) generates a highest weight module \(W\leq M( \lambda)\) of
highest weight \(\mu = s_ \alpha \cdot \lambda\). Note that \(M( \mu)\)
is the universal highest weight module with highest weight \(\mu\),
i.e.~\(\exists! M(\mu) \twoheadrightarrow W\). This yields a
\(B{\hbox{-}}\)module morphism
\begin{align*}
{\mathbf{C}}_ \mu &\to W \\
1 &\mapsto w
,\end{align*}
which yields an \(L{\hbox{-}}\)module morphism
\({\mathcal{U}}(L) \otimes_{{\mathcal{U}}(B)} {\mathbf{C}}_ \mu \twoheadrightarrow W\).
So \(W\) is a nonzero quotient of \(M( \mu)\) and
\({\mathcal{Z}}(L)\curvearrowright W\) by \(\chi_ \mu\). On the other
hand \(W\leq M( \lambda)\) and so \({\mathcal{Z}}(L)\curvearrowright W\)
by \(\chi_ \lambda\), yielding \(\chi_ \lambda= \chi_ \mu\).

So we conclude that if \(\mu = s_ \alpha \cdot \lambda\) with
\({\left\langle {\lambda },~{\alpha} \right\rangle}\in {\mathbf{Z}}_{\geq 0}\),
then \(\chi_ \mu = \chi_ \lambda\).

\end{proof}

\begin{corollary}[?]

Let
\(\lambda\in \Lambda, \alpha \in \Delta, \mu = s_ \alpha \cdot \lambda\).
Then \(\chi_ \mu = \chi_ \lambda\).

\end{corollary}

\begin{proof}[?]

\(\mu = s_ \alpha \cdot \lambda = \lambda- (m+1) \alpha\) where
\(m \coloneqq{\left\langle {\lambda },~{\alpha} \right\rangle}\).

\begin{itemize}
\tightlist
\item
  Case 1: \(m=-1\), then \(\mu = \lambda\) and we're done.
\item
  Case 2: \(m\geq 0\), then \(\chi_ \lambda= \chi_ \mu\) by the
  proposition.
\item
  Case 3: \(m\leq -2\), then
  \begin{align*}
  {\left\langle {\mu },~{\alpha } \right\rangle}
  &= {\left\langle { \lambda- (m+1) \alpha},~{\alpha} \right\rangle} \\
  &= m-2(m+1) \\
  &= -m-2 \geq 0
  .\end{align*}
  We also have
  \(\mu = s_ \alpha\dot \lambda\implies s_ \alpha\cdot \mu= s_ \alpha^2 \cdot \lambda= \lambda\)
  by applying the action on both sides. By the proposition, swapping
  \(\mu, \lambda\) to get submodules of \(M( \mu)\) of highest weight
  \(\lambda\) and conclude \(\chi_ \lambda = \chi_ \mu\).
\end{itemize}

\end{proof}

\begin{corollary}[?]

Let \(\lambda, \mu\in \Lambda\), then if \(\mu\sim \lambda\) then
\(\chi_ \lambda=\chi_ \mu\).

\end{corollary}

\begin{proof}[?]

Say \(\mu = w\cdot \lambda\) for \(w\in W\), then write
\(w = s_{i_1}\cdots s_{i_t}\) and use induction on \(t\), where the base
case is the previous corollary.

\end{proof}

\begin{theorem}[Harish-Chandra ($\S 23.2$)]

If \(\lambda, \mu\) satisfy \(\chi_ \lambda= \chi_ \mu\), then
\(\lambda\sim \mu\).

\end{theorem}

\begin{remark}

Goal: find
\(\operatorname{ch}_ \lambda\coloneqq\operatorname{ch}_{L( \lambda)} \coloneqq\sum_{ \mu\in \Lambda} (\dim L( \lambda)_\mu ) e( \mu) \in {\mathbf{Z}}[ \Lambda]\)
for \(\lambda\in \Lambda^+\).

\end{remark}

\hypertarget{s-24-formulas-of-weyl-kostant-steinberg-s-24.1-functions-on-h-vee}{%
\subsection{\texorpdfstring{\(\S 24\): Formulas of Weyl, Kostant,
Steinberg; \(\S 24.1\) functions on
\(H {}^{ \vee }\)}{\textbackslash S 24: Formulas of Weyl, Kostant, Steinberg; \textbackslash S 24.1 functions on H \{\}\^{}\{ \textbackslash vee \}}}\label{s-24-formulas-of-weyl-kostant-steinberg-s-24.1-functions-on-h-vee}}

\begin{remark}

View \({\mathbf{Z}}[ \Lambda]\) as finitely-supported
\({\mathbf{Z}}{\hbox{-}}\)valued functions on \(\Lambda\) with elements
\(f = \sum_{ \lambda\in \Lambda} a_ \lambda e( \lambda)\) regarded as
functions \(f(\mu) \coloneqq a_\mu\). Thus
\(e( \lambda)( \mu) = \delta_{\lambda= \mu }\). The point of this
maneuver: Verma modules will be infinite dimensional, but
\({\mathbf{Z}}[ \Lambda]\) only handles finite sums. For
\(f,g\in \mathop{\mathrm{Hom}}_{\mathbf{Z}}( \Lambda, {\mathbf{Z}})\),
define
\((f\ast g) ( \sigma) = \sum_{ \lambda+ \mu} f( \lambda) g( \mu)\).
Consider the set \({\mathcal{X}}\) of functions
\(\mathop{\mathrm{Hom}}_{\mathbf{C}}(H {}^{ \vee }, {\mathbf{C}})\)
whose support is contained in a finite union of sets of the form
\(\lambda_{ \leq} \coloneqq\left\{{ \lambda - \sum_{ \mu\in \Phi^+} k_ \beta \beta{~\mathrel{\Big\vert}~}k_ \beta\in {\mathbf{Z}}_{\geq 0}}\right\}\).
One can show \({\mathcal{X}}\) is closed under convolution and
\({\mathcal{X}}\) becomes a commutative associative algebra containing
\({\mathbf{Z}}[ \Lambda]\) as a subring. Note that
\(\mathop{\mathrm{supp}}(f\ast g) \subseteq ( \lambda+ \mu)_{\leq }\).

\includegraphics{figures/2022-11-30_09-56-16.png}

Write \(e( \lambda)\) as \(e_{ \lambda}\) for
\(\lambda \in H {}^{ \vee }\), regarded as a function
\(e_ \lambda:H {}^{ \vee }\to {\mathbf{C}}\) where
\(e_ \lambda( \mu) = \delta_{ \lambda = \mu}\) is the characteristic
function of \(\lambda\), and note that
\(e_ \lambda \ast e_ \mu = e_ { \lambda+ \mu}\). Let
\(p = \operatorname{ch}_{M(0)}: H {}^{ \vee }\to {\mathbf{C}}\), then
\begin{align*}
M(0) = {\mathcal{U}}(L) \otimes_{{\mathcal{U}}(B)} {\mathbf{C}}_0 \cong {\mathcal{U}}(N^-) \otimes_{\mathbf{C}}{\mathbf{C}}_0 \quad\in   {}_{H}{\mathsf{Mod}},   {}_{N^-}{\mathsf{Mod}}
.\end{align*}
By PBW, \({\mathcal{U}}(N^-)\) has a basis
\(y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m}\) for
\(i_j\in {\mathbf{Z}}_{\geq 0}\) where
\(\Phi^+ = \left\{{ {\beta }_{1}, \cdots, {\beta }_{m} }\right\}\). The
weights of \(M(0)\) are \(\mu = - \sum_{j=1}^m i_j \beta_j\) and
\(\mu\in 0_{\leq}\). Note
\(\operatorname{ch}_{M(0)}( \mu) = \dim M(0)_ \mu\), and so
\(\operatorname{ch}_{M(0)}\in {\mathcal{X}}\) and thus
\(p\in {\mathcal{X}}\).

\end{remark}

\hypertarget{friday-december-02}{%
\section{Friday, December 02}\label{friday-december-02}}

\hypertarget{convolution-formulas}{%
\subsection{Convolution Formulas}\label{convolution-formulas}}

\begin{remark}

Last time:
\begin{align*}
{\mathcal{X}}\coloneqq\left\{{f\in \mathop{\mathrm{Hom}}_{\mathbf{C}}(H {}^{ \vee }, {\mathbf{C}}) {~\mathrel{\Big\vert}~}\mathop{\mathrm{supp}}f \subseteq \bigcup_{i=1}^n (\lambda_i - {\mathbf{Z}}_{\geq 0} \Phi^+) }\right\}
,\end{align*}
which is a commutative associative unital algebra under convolution,
where \(e_{ \lambda}(\mu) = \delta_{ \lambda \mu}\) for
\(\mu\in H {}^{ \vee }\) and
\(e_ \lambda\ast e_ \mu = e_{ \lambda+ \mu}\) with \(e_0 = 1\). We have
\(\operatorname{ch}_{M(0)}\) which records weights
\(\mu = \sum_{i=1}^m -i_j \beta_j\) with
\(i_j\in {\mathbf{Z}}_{\geq 0}\), and
\begin{align*}
\dim M(0)_\mu \coloneqq\operatorname{ch}_{M(0)}(\mu) = {\sharp}\left\{{\mathbf{i} \in {\mathbf{Z}}_{\geq 0}^m {~\mathrel{\Big\vert}~}\sum i_j \beta_j = -\mu }\right\} \coloneqq p(\mu)
\end{align*}
which is the \textbf{Kostant function} -- its negative is the Kostant
partition function, which records the number of ways of writing a weight
as a sum like this. We'll regard finding such a count as a known or easy
problem, since this can be done by enumeration.

Define the \textbf{Weyl function}
\(q\coloneqq\prod_{\alpha \in \Phi^+} (e_{\alpha\over 2} - e_{-{\alpha \over 2}})\)
where the product is convolution. For
\(\alpha\in \Phi+, f_ \alpha:H {}^{ \vee }\to {\mathbf{Z}}\), define
\begin{align*}
f_ \alpha( \lambda) \coloneqq
\begin{cases}
1 & \lambda = -k \alpha \text{ for some } k\in {\mathbf{Z}}_{\geq 0} 
\\
0 & \text{otherwise}.
\end{cases}
.\end{align*}
We can regard this as an infinite sum
\(f_ \alpha = e_0 + e_{ - \alpha} + e_{ -2 \alpha} + \cdots = \sum_{k\geq 0} e_{ -k \alpha}\).

\end{remark}

\begin{lemma}[A]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(p = \prod_{ \alpha\in \Phi^+} f_ \alpha\),
\item
  \((e_0 - e_{- \alpha}) \ast f_ \alpha = e_0\),
\item
  \(q = e_\rho \ast\prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha})\).
\end{enumerate}

\end{lemma}

\begin{proof}[of lemma A]

\textbf{Part a}: The coefficient of \(e_\mu\) in
\(\prod_{i=1}^m f_{ \beta_j}\) is the convolution
\begin{align*}
\sum_{i_1,\cdots, i_m \in {\mathbf{Z}}_{\geq 0}, -\sum i_j \beta_j = \mu } f_{\beta_1}(-i_1 \beta_1) \cdots f_{ \beta_m }(-i_m \beta_m) = p(\mu)
.\end{align*}

\textbf{Part b}:
\begin{align*}
(e_0 - e_{- \alpha}) \ast(e_0 + e_{- \alpha} + e_{-2 \alpha} + \cdots)
= e_0 + e_{ - \alpha} - e_{ - \alpha}+ e_{ -2 \alpha}- e_{ -2 \alpha} +\cdots = e_0
,\end{align*}
noting the telescoping. This can be checked rigorously by regarding
these as functions instead of series.

\textbf{Part c}: Recall
\(\rho = \sum_{ \alpha\in \Phi^+} {1\over 2}\alpha\), so
\(e_\rho = \prod_{ \alpha\in \Phi^+} e_{\alpha\over 2}\). Thus the RHS
is
\begin{align*}
\prod_{ \alpha\in \Phi^+} \qty{e_{\alpha\over 2} \ast(e_0 - e_{- \alpha}) } = \prod_{\alpha\in \Phi^+}(e_{\alpha\over 2} - e_{- \alpha\over 2} ) \coloneqq q
.\end{align*}
Note that \(q\neq 0\) since \(q(\rho) = 1\).

\end{proof}

\begin{lemma}[B]

Let \(w\in W\), recalling that \(w.e_{\alpha} = e_{w \alpha}\),
\begin{align*}
wq = (-1)^{\ell(w)} q
.\end{align*}

\end{lemma}

\begin{proof}[of lemma B]

ETS for \(\alpha \in \Delta\). Recall \(s_ \alpha\) permutes
\(\Phi^+ \setminus\left\{{ \alpha }\right\}\) and
\(s_ \alpha(\alpha) = - \alpha\), so \(s_ \alpha q\) permutes the
factors \((e_{ \beta\over 2} - e_{-{\beta\over 2}})\) for
\(\beta\neq \alpha\) and negates
\(e_{\alpha\over 2} - e_{- \alpha\over 2}\).

\end{proof}

\begin{lemma}[C]

\(q\) is invertible:
\begin{align*}
q\ast p \ast e_{\rho} = e_0 \quad \implies q^{-1}= \rho \ast e_\rho
.\end{align*}

\end{lemma}

\begin{proof}[of lemma C]

Use lemma A:
\begin{align*}
q\ast\rho \ast e_{-\rho} 
&= e_\rho \ast\qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha} ) } \ast p \ast e_{ - \rho} \qquad \text{by C} \\
&= \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha} ) } \ast p \\
&= \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha}) \ast f_{ \alpha} } \qquad \text{by A}\\
&= \prod_{ \alpha\in \Phi^+} e_0 \qquad \text{by B}\\
&= e_0
.\end{align*}

\end{proof}

\begin{lemma}[D]

For \(\lambda, \mu\in H {}^{ \vee }\),
\begin{align*}
\operatorname{ch}_{M( \lambda)}( \mu) = p( \mu - \lambda) = (p \ast e_ \lambda)( \mu) \implies \operatorname{ch}_{M( \lambda)} = p \ast e_{ \lambda}
.\end{align*}

\end{lemma}

\begin{proof}[of lemma D]

\(M( \lambda)\) has basis
\(y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m} \cdot v^+\) where \(v^+\)
is a highest weight vector of weight \(\lambda\). Note that
\begin{align*}
\mu = \lambda- \sum_{j=1}^M i_j \beta_j \iff \mu- \lambda= - \sum_{i=1}^m i_j \beta_j
,\end{align*}
and \(\dim M(\lambda)_ \mu = p( \mu- \lambda)\). Now check
\((p\ast e_ \lambda)( \mu) \coloneqq p( \mu - \lambda) e_ \lambda(\lambda) = p( \mu- \lambda)\).

\end{proof}

\begin{lemma}[E]

\begin{align*}
q \ast\operatorname{ch}_{M( \lambda)} = e_{ \lambda+ \rho}
.\end{align*}

\end{lemma}

\begin{proof}[of lemma E]

\begin{align*}
\text{LHS} 
\overset{D}{=} q \ast p \ast e_ \lambda 
\overset{C}{=} e_ \rho \ast e_ \lambda = e_{ \lambda+ \rho}
.\end{align*}

\end{proof}

\hypertarget{s-24.2-kostants-multiplicity-formula}{%
\subsection{\texorpdfstring{\(\S 24.2\) Kostant's Multiplicity
Formula}{\textbackslash S 24.2 Kostant's Multiplicity Formula}}\label{s-24.2-kostants-multiplicity-formula}}

\begin{remark}

Recall that characters of Verma modules are essentially known. For
\(\lambda\in \Lambda^+\), we have
\(\operatorname{ch}_{ \lambda} \coloneqq\operatorname{ch}_{L( \lambda)}\),
recalling that \(L( \lambda)\) is a finite-dimensional irreducible
representation. Goal: express this as a finite
\({\mathbf{Z}}{\hbox{-}}\)linear combination of certain
\(\operatorname{ch}_{M( \lambda)}\).

Fix \(\lambda\in H {}^{ \vee }\) and let \({\mathcal{M}}_ \lambda\) be
the collection of all \(L{\hbox{-}}\)modules satisfying the following:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(V = \bigoplus _{ \mu\in H {}^{ \vee }} V_ \mu\),
\item
  \({\mathcal{Z}}(L)\curvearrowright V\) by \(\chi_ \lambda\),
\item
  \(\operatorname{ch}_V \in {\mathcal{X}}\)
\end{enumerate}

Note that any highest weight module of highest weight \(\lambda\) is in
\({\mathcal{M}}_ \lambda\), and this is closed under submodules,
quotients, and finite direct sums. The Harish-Chandra theorem implies
that
\begin{align*}
{\mathcal{M}}_ \lambda = {\mathcal{M}}_{\mu} \iff \lambda\sim \mu\iff \mu = w. \lambda
.\end{align*}

\end{remark}

\begin{lemma}[?]

If \(0\neq V \in {\mathcal{M}}_ \lambda\) then \(V\) has a maximal
vector.

\end{lemma}

\begin{proof}[?]

By (3), the weights of \(V\) lie in a finite number of cones
\(\lambda_i - {\mathbf{Z}}_{\geq 0} \Phi^+\). So if \(\mu\) is a weight
of \(V\) and \(\alpha\in \Phi^+\), then \(\mu + k \alpha\) is not a
weight of \(V\) for \(k\gg 0\). Iterating this for the finitely many
positive weights, there exists a weight \(\mu\) such that
\(\mu + \alpha\) is not a weight for any \(\alpha\in \Phi^+\). Then any
\(0\neq v\in V_ \mu\) is a maximal vector.

\end{proof}

\begin{definition}[?]

For \(\lambda \in H {}^{ \vee }\), set
\begin{align*}
\Theta( \lambda) \coloneqq\left\{{ \mu\in H {}^{ \vee }{~\mathrel{\Big\vert}~}\mu\sim \lambda\, \text{ and } \, \mu\leq \lambda}\right\}
,\end{align*}
which by the Harish-Chandra theorem is a subset of \(W\cdot \lambda\)
which is a finite set.

\end{definition}

\begin{remark}

The following tends to hold in any setting with ``standard'' modules,
e.g.~quantum groups or superalgebras:

\end{remark}

\begin{proposition}[?]

Let \(\lambda\in H {}^{ \vee }\), then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(M( \lambda)\) has a composition series,
\item
  Each composition factor of \(M( \lambda)\) is of the form \(L( \mu)\)
  for some \(\mu\in \Theta( \lambda)\). Define
  \([M (\lambda): L( \mu)]\) to be the multiplicity of \(L(\mu)\) as a
  composition factor of \(M( \lambda)\).
\item
  \([M( \lambda): L( \lambda)] = 1\).
\end{enumerate}

\end{proposition}

\begin{proof}[?]

By induction on the number of maximal vectors (up to scalar multiples).
If \(M( \lambda)\) is irreducible then it's an irreducible highest
weight module, and these are unique up to isomorphism and so
\(M( \lambda) = L( \lambda)\) and we're done. Otherwise \(M( \lambda)\)
has a proper irreducible submodule \(V\), and
\(V\in {\mathcal{M}}_ \lambda\) by closure under submodules. By the
lemma, \(V\) has a maximal vector of some weight \(\mu\) which must be
strictly less than \(\lambda\), i.e.~\(\mu < \lambda\). As before,
\(\chi_{ \mu} = \chi_ \lambda\) and thus \(\mu\in \Theta( \lambda)\).
Consider \(V\) and \(M( \lambda)/ V\) -- each lies in
\({\mathcal{M}}_ \lambda\) and either has fewer weights linked to
\(\lambda\) than \(M( \lambda)\) has, or else it must have exactly the
same set of weights linked to \(\lambda\), just with smaller
multiplicities. By induction each of these has a composition series, and
these can be pieced together into a series for \(M\) since they fit into
a SES.

\end{proof}

\hypertarget{monday-december-05}{%
\section{Monday, December 05}\label{monday-december-05}}

\hypertarget{kostants-character-formula}{%
\subsection{Kostant's Character
Formula}\label{kostants-character-formula}}

\begin{remark}

Last time: \({\mathcal{M}}_ \lambda\) defined as a certain category of
\(L{\hbox{-}}\)modules for \(\lambda\in H {}^{ \vee }\), and we defined
\(\theta( \lambda) \coloneqq\left\{{ \mu\in H {}^{ \vee }{~\mathrel{\Big\vert}~}\mu \sim \lambda,\,\, \mu\leq \lambda}\right\} \subseteq W. \lambda\).
Proposition from last time:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(M(\lambda)\) has a composition series,
\item
  Each composition factor of \(M(\lambda)\) is of the form \(L( \mu)\)
  for some \(\mu \in \theta( \lambda)\),
\item
  \([M (\lambda): L( \mu) ] = 1\).
\end{enumerate}

Note that any character of \(M\) is the sum of the characters of its
composition factors.

\end{remark}

\begin{proof}[of proposition]

\textbf{Part b}: Each composition factor of \(M( \lambda)\) is in
\({\mathcal{M}}_ \lambda\), hence by the lemma has a maximal vector.
Since it's irreducible, it is a highest weight module \(L( \mu)\) for
some \(\mu\in \theta( \lambda)\).

\textbf{Part c}: \([M( \lambda) : L( \lambda)] = 1\) since
\(\dim M( \lambda)_ \lambda = 1\) and all other weights are strictly
less than \(\lambda\).

\end{proof}

\begin{remark}

Order \(\theta( \lambda) = \left\{{ \mu_1, \cdots, \mu_t}\right\}\) such
that \(\mu_\leq \mu_j \implies i\leq j\). In particular,
\(\mu_t = \lambda\). By the proposition,
\(\operatorname{ch}_{M( \mu_j)}\) is a
\({\mathbf{Z}}_{\geq 0}{\hbox{-}}\)linear combination of
\(\operatorname{ch}_{L(\mu_i)}\) where \(i\leq j\), and the coefficient
of \(\operatorname{ch}_{L(\mu_j)}\) is 1. Recording multiplicities in a
matrix, we get the following:

\begin{center}
\begin{tikzcd}
    & {\operatorname{ch}_{M(\mu_j)}} &&&&& {\,} \\
    {\operatorname{ch}_{L(\mu_i)}} && 1 & \bullet & \bullet & \bullet \\
    && 0 & 1 & \bullet & \bullet \\
    && 0 & 0 & \ddots & \bullet \\
    && 0 & 0 & 0 & 1 \\
    & {\,}
    \arrow[no head, from=1-2, to=6-2]
    \arrow[no head, from=1-2, to=1-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

This is an upper triangular unipotent matrix, and thus invertible.

\end{remark}

\begin{corollary}[?]

Let \(\lambda \in H {}^{ \vee }\), then
\begin{align*}
\operatorname{ch}_{L( \lambda)} = \sum_{ \mu \in \theta( \lambda)} c( \mu) \operatorname{ch}_{M( \mu)}, \qquad c( \mu)\in {\mathbf{Z}}, \, c( \lambda) = 1
.\end{align*}

\end{corollary}

\begin{remark}

Assume \(\lambda\in \Lambda^+\), and recall:

\begin{itemize}
\tightlist
\item
  \({\mathcal{X}}= \left\{{f: H {}^{ \vee }\to {\mathbf{C}}{~\mathrel{\Big\vert}~}\text{ the "forest" support condition }}\right\}\),
\item
  \(p=\operatorname{ch}_{M(0)}\),
\item
  \(p(\mu) = {\sharp}\left\{{ {\left[ {i_1:\cdots:i_m} \right]} \in {\mathbf{Z}}_{\geq 0}^m }\right\}\),
\item
  \(q\coloneqq\prod_{ \alpha\in \Phi^+} (e_{\alpha\over 2} - e_{-{\alpha\over 2}})\),
\item
  \(\mu = -\sum_{1\leq j\leq m} i_j \beta_j\),
\item
  \(wq = (-1)^{\ell(w)} q\) for \(w\in W\),
\item
  \(q \ast p \ast e_{-\rho} = e_0\),
\item
  \(\operatorname{ch}_{M(\lambda)} = p \ast e_ \lambda\),
\item
  \(q \ast\operatorname{ch}_{M( \lambda) } = e_{ \lambda+ \rho}\),
\item
  \(\operatorname{ch}_ \lambda\coloneqq\operatorname{ch}_{L( \lambda)} = \sum_{ \mu\in \theta( \lambda)} c( \mu) \operatorname{ch}_{M( \mu)}\),
\item
  \(q\ast\operatorname{ch}_{L( \lambda)} = \sum c( \mu) q \ast\operatorname{ch}_{M( \mu)} = \sum c( \mu) e_{ \mu + \rho}\).
  \(\qquad\star_1\)
\end{itemize}

Fixing \(w\in W\), we have
\begin{align*}
\sum_{ \mu\in \theta( \lambda) } c( \mu) e_{w (\lambda+ \rho)} 
&= w( q \ast\operatorname{ch}_ \lambda) \\
&= wq \ast w \operatorname{ch}_{ \lambda} \\
&= (-1)^{\ell(w)} q \operatorname{ch}_ \lambda \qquad \text{since $\operatorname{ch}_ \lambda$ is $W{\hbox{-}}$invariant} \\
&= (-1)^{\ell(w)} \sum_ \mu c( \mu) e_{ \mu+ \rho}
.\end{align*}
Since \(\lambda\in \Lambda^+\), \(W\) acts simply transitively on
\(\theta( \lambda) + \rho \coloneqq\left\{{v+\rho {~\mathrel{\Big\vert}~}v\in \theta( \lambda) }\right\}\).
Note \(\mu \sim \lambda\iff \mu+ \rho = w( \lambda+ \rho)\) for some
\(w\in W\), which is unique since \(\lambda+ \rho\) is strongly/strictly
dominant, and lemma 13.2A shows its stabilizer is the identity. So
\({\operatorname{Stab}}_W( \lambda+ \rho) = \left\{{1}\right\}\). The
equation \(\mu + \rho = w( \lambda+ \rho)\) implies
\(\mu + \rho \leq \lambda+ \rho\), since apply \(W\) to dominant
elements goes down in the partial order. Thus
\(\mu\in \theta(\lambda)\), and \(\theta( \lambda)\) consists of
precisely those \(\mu\) satisfying this equation, and
\begin{align*}
\theta( \lambda) = W\cdot \lambda
.\end{align*}

Continuing the computation, take \(\mu = \lambda\) on the LHS, so
\(w( \lambda+ \rho) = \mu+ \rho\) and
\begin{align*}
c( \lambda) e_{w( \lambda+ \rho)} = (-1)^{\ell(w)} c( \mu) e_{ \mu+ \rho}\implies c(\mu) = (-1)^{\ell(w)} c( \lambda)
.\end{align*}

Substituting this into \(\star_1\) yields
\begin{align*}
q \ast\operatorname{ch}_ \lambda = \sum_{w\in W} (-1)^{\ell(w)} e_{w ( \lambda+ \rho)} \qquad \star_2
,\end{align*}
so
\begin{align*}
\operatorname{ch}_ \lambda 
&= q \ast p \ast e_{ - \rho} \ast\operatorname{ch}_ \lambda\\
&= p \ast e_{- \rho} \ast\sum_{w\in W} (-1)^{\ell(w)} e_{w (\lambda+ \rho)} \\
&= \sum_{w\in W} (-1)^{\ell(w)} p \ast e_{w( \lambda+ \rho) - \rho}
&= \sum_{w\in W} (-1)^{\ell(w)} p \ast e_{w\cdot \lambda}
.\end{align*}

This yields the following:

\end{remark}

\begin{theorem}[Kostant]

For \(\lambda\in \Lambda^+\) dominant, the weight multiplicities in
\(L( \lambda)\) are given by
\begin{align*}
\dim L( \lambda)_\mu \coloneqq m_ \lambda( \mu) 
= \sum_{w\in W} (-1)^{\ell(w)} p( \mu+ \rho - w( \lambda+ \rho))
= \sum_{w\in W} (-1)^{\ell(w)} p( \mu - w\cdot \lambda)
.\end{align*}

\end{theorem}

\hypertarget{s-24.3-weyls-character-formula}{%
\subsection{\texorpdfstring{\(\S 24.3\) Weyl's Character
Formula}{\textbackslash S 24.3 Weyl's Character Formula}}\label{s-24.3-weyls-character-formula}}

\begin{lemma}[?]

\begin{align*}
q = \sum_{w\in W} (-1)^{\ell(w)} e_{w \rho}
.\end{align*}

\end{lemma}

\begin{proof}[?]

Take \(\lambda = 0\) in \(\star_2\), and use that
\(\operatorname{ch}_0 = e_0\) where \(L(0) \cong {\mathbf{C}}\).

\end{proof}

\begin{theorem}[Weyl's Character Formula]

Let \(\lambda \in \Lambda^+\), then
\begin{align*}
\qty{ \sum_{w\in W} (-1)^{ \ell(w)} e_{w\rho} } \ast\operatorname{ch}_{L( \lambda)} = \sum_{w\in W} (-1)^{\ell(w)} e_{w( \lambda+ \rho)}
.\end{align*}

\end{theorem}

\begin{proof}[?]

Apply \(\star_2\) and the lemma.

\end{proof}

\begin{corollary}[Weyl's Dimension Formula]

\begin{align*}
\dim L( \lambda) = { \prod_{ \alpha\in \Phi^+} {\left\langle { \lambda+ \rho},~{ \alpha} \right\rangle} \over\prod_{\alpha\in \Phi^+} {\left\langle {\rho },~{\alpha } \right\rangle}} = \sum_{ \mu \in \Pi( \lambda)} m_ \lambda( \mu)
.\end{align*}

\end{corollary}

\hypertarget{tuesday-december-06}{%
\section{Tuesday, December 06}\label{tuesday-december-06}}

\hypertarget{weyl-dimension-formula}{%
\subsection{Weyl Dimension Formula}\label{weyl-dimension-formula}}

\begin{remark}

Last time:

\begin{itemize}
\tightlist
\item
  \(q \coloneqq\sum_{ \alpha\in \Phi^+} (e_{\alpha\over 2} - e_{-{\alpha\over 2}}) = \sum_{w\in W} (-1)^{\ell(w)} e_{w \rho}\).
\item
  The WCF:
  \(q\ast\operatorname{ch}_{ \lambda} = \sum_{w\in W} (-1)^{\ell(w)} e_{w ( \lambda+ \rho)}\)
\item
  An alternative writing of the WCF:
  \begin{align*}
  \operatorname{ch}_ \lambda= { \sum_{w\in W} (-1)^{\ell(w)} e_{w (\lambda+ \rho)} \over \sum_{w\in W} (-1)^{\ell(w)} e_{w \rho}}
  ,\end{align*}
  where the denominator is denoted the \emph{Weyl denominator}.
\end{itemize}

\end{remark}

\begin{corollary}[Weyl dimension formula]

\begin{align*}
\dim L( \lambda) = { \prod_{ \sigma\in \Phi^+} {\left\langle { \lambda+ \rho},~{ \alpha} \right\rangle} \over \prod_{ \alpha\in \Phi^+} {\left\langle {\rho },~{\alpha } \right\rangle}} 
,\end{align*}
which is a quotient of two integers.

\end{corollary}

\begin{exercise}[?]

Show that \(W\) always has an equal number of even and odd elements, so
\(\sum_{w\in W} (-1)^{\ell(w)} = 0\).

\end{exercise}

\begin{proof}[?]

Note
\(\operatorname{ch}_{ \lambda} = \sum _{\mu} \dim L( \lambda)_ \mu e_ \mu\in {\mathbf{Z}}[ \Lambda]\),
and \(\dim L( \lambda) = \sum_{ \mu\in \Lambda} m_ \lambda( \mu)\).
Viewing \(\operatorname{ch}_ \lambda: \Lambda\to {\mathbf{Z}}\) as a
restriction of a function \(H {}^{ \vee }\to {\mathbf{C}}\),
\(\dim L( \lambda)\) is the sum of all values of
\(\operatorname{ch}_ \lambda\). Work in the
\({\mathbf{C}}{\hbox{-}}\)subalgebra \({\mathcal{X}}_0\) of
\({\mathcal{X}}\) generated by the characteristic functions
\(S \coloneqq\left\{{e_ \mu{~\mathrel{\Big\vert}~}\mu\in \Lambda}\right\}\);
this equals the span of \(S\) since
\(e_{ \mu} \ast e_{\nu} = e_{ \mu+ \nu}\). We have a map
\begin{align*}
v: {\mathcal{X}}_0 &\to {\mathbf{C}}\\
f &\mapsto \sum_{ \mu\in \Lambda} f( \mu)
,\end{align*}
which makes sense since \({\sharp}\mathop{\mathrm{supp}}f < \infty\).
This function sums the values we're after, so the goal is to compute
\(v( \operatorname{ch}_ \lambda)\). By the exercise, attempting to apply
this directly to the numerator and denominator yields \(0/0\), and we
get around this by using a variant of L'Hopital's rule. Define
\({\partial}_ \alpha (e_ \mu) = ( \mu, \alpha)e_ \mu\), extended
linearly to \({\mathcal{X}}_0\). In the basis \(S\) this operator is
diagonal, and this is a derivation relative to convolution:
\begin{align*}
{\partial}_ \alpha \qty{ e_ \mu \ast e_{ \nu} } 
&= {\partial}_ \alpha( e_{ \mu+ \nu} ) \\
&= ( \mu+ \nu, \alpha)e_{ \mu+ \nu} \\
&= \qty{ (\mu, \alpha) e_ \mu} \ast e_ \nu + e_ \mu \ast\qty{ (\nu, \alpha) e_ \nu} \\
&= ({\partial}_ \alpha e_ \mu) \ast e_ \nu + e_ \mu \ast({\partial}_{\alpha} e _{\nu})
.\end{align*}
Moreover they commute,
i.e.~\({\partial}_ \alpha {\partial}_ \beta = {\partial}_{\beta} {\partial}_{\beta}\).
Set \({\partial}\coloneqq\prod_{ \alpha\in \Phi^+} {\partial}_ \alpha\)
where the product here is composition, and view \({\partial}\) as an
\(m\)th order differential operator. Write
\(\omega( \lambda+ \rho)\coloneqq\sum_{w\in W} (-1)^{\ell(w)} e_{w (\lambda+ \rho)}\)
for \(\lambda\in \Lambda^+\), so \(q = \omega ( \rho)\). Rewriting the
WCF we have
\begin{align*}
\omega( \rho) \ast\operatorname{ch}_ \lambda = \omega( \lambda+ \rho) \qquad \star_1
,\end{align*}
and
\begin{align*}
\prod_{ \alpha\in \Phi^+}\qty{e_{ \alpha\over 2} - e_{- {\alpha\over 2}}} \ast\operatorname{ch}_ \lambda = \omega( \lambda+ \rho)
.\end{align*}
We now try to apply \({\partial}\) to both sides, followed by \(v\).
Note that if any two factors of \({\partial}\) hit the same factor on
the LHS, then noting that
\(v(e_{\alpha\over 2} - e_{-{ \alpha\over 2}}) = 0\), such terms will
vanish. So the total result will be zero unless all of the factors of
\({\partial}\) are applied to the \(q\) factor in the LHS.

So apply \(v\circ {\partial}\) to \(\star_1\) to get
\begin{align*}
v( {\partial}\omega( \rho)) v( \operatorname{ch}_ \lambda) = v( {\partial}\omega( \lambda+ \rho))
.\end{align*}
We can compute
\begin{align*}
v( {\partial}\omega ( \rho)) 
= v\qty{ {\partial}\sum_{w\in W} (-1)^{\ell(w)} e_{ w \rho} }
= \sum_{w\in W} (-1)^{\ell(w)} v({\partial}( e_{ w \rho} ) )
.\end{align*}
We have
\begin{align*}
v ( {\partial}( e_{ w \rho})) 
&= v\qty{ \qty{ \prod_{\alpha\in \Phi^+} {\partial}_\alpha } e_{ w \rho} } \\
&= v\qty{\prod_{ \alpha\in \Phi^+} (w \rho, \alpha) e_{ w \rho} } \\
&= \prod_{ \alpha\in \Phi^+} ( w \rho, \alpha) \\
&= \prod_{ \alpha\in \Phi^+} ( \rho, w^{-1}\alpha) \qquad \star_2
.\end{align*}
Note that \(w^{-1}\cdot \Phi^+\) is a permutation of \(\Phi^+\), just
potentially with some signs changed -- in fact, exactly \(n(w^{-1})\),
the number of positive roots sent to negative roots, and
\(n(w^{-1}) = \ell(w^{-1})\). Thus the above is equal to
\begin{align*}
(-1)^{\ell(w)} \prod_{ \alpha\in \Phi^+}( \rho, \alpha)
.\end{align*}
Continuing \(\star_2\), we have
\begin{align*}
v( {\partial}( e_{ w \rho})) 
&= \sum_{w\in W} (-1)^{\ell(w)} (-1)^{\ell(w)} \prod_{ \alpha\in \Phi^+}( \rho, \alpha) \\
&= {\sharp}W \prod_{ \alpha\in \Phi^+}( \rho, \alpha)
,\end{align*}
which is the LHS. Similarly for the RHS,
\begin{align*}
v({\partial}( \lambda+ \rho)) = {\sharp}W \prod_{ \alpha\in \Phi^+} (\lambda+ \rho, \alpha)
.\end{align*}

Taking the quotient yields
\begin{align*}
\dim L(\lambda) = 
{{\sharp}W \prod_{ \alpha\in \Phi^+}( \rho, \alpha)
\over 
{\sharp}W \prod_{ \alpha\in \Phi^+} (\lambda+ \rho, \alpha) } = 
{\prod_{ \alpha\in \Phi^+}( \rho, \alpha)
\over 
\prod_{ \alpha\in \Phi^+} (\lambda+ \rho, \alpha) }
.\end{align*}
Multiplying the numerator and denominator by
\(\prod_{ \alpha\in \Phi^+} {2\over (\alpha, \alpha)}\) yields
\begin{align*}
\prod_{ \alpha\in \Phi^+} {\left\langle {\rho},~{\alpha} \right\rangle}
\over 
\prod_{ \alpha\in \Phi^+} {\left\langle { \lambda+ \rho},~{\alpha} \right\rangle} 
.\end{align*}

\end{proof}

\begin{remark}

If \(\alpha\in \Phi^+\), using that \(\alpha {}^{ \vee }\) is a basis of
\(\Phi {}^{ \vee }\), one can write
\(\alpha {}^{ \vee }= \sum_{i=1}^\ell c_i^{ \alpha} \alpha_i {}^{ \vee }\)
for some \(c_{i}^{\alpha} \in {\mathbf{Z}}_{\geq 0}\) and
\(\lambda = \sum_{i=1}^\ell m_i \lambda_i\) for
\(m_i \in {\mathbf{Z}}_{ \geq 0}\), using that
\(( \rho, \alpha_i {}^{ \vee }) = {\left\langle { \rho},~{ \alpha_i} \right\rangle} = 1\)
one can rewrite the dimension formula in terms of the integers
\(c_i^{\alpha}\) and \(m_i\).

\end{remark}

\hypertarget{new-directions}{%
\subsection{New Directions}\label{new-directions}}

\begin{remark}

Where you could go after studying semisimple finite-dimensional Lie
algebras over \({\mathbf{C}}\):

\begin{itemize}
\item
  Infinite-dimensional representations of such algebras, e.g.~the Verma
  modules \(M( \lambda)\). One has a SES
  \(K( \lambda) \hookrightarrow M( \lambda) \twoheadrightarrow L( \lambda)\),
  which doesn't split since \(M( \lambda)\) is indecomposable.
\item
  Category \({\mathcal{O}}\), expressing characters of simples in terms
  of characters of Vermas.
\item
  Parabolic versions of Verma modules: we've looked at modules induced
  from \(B = H + N\), but one could look at parabolics
  \(P = U + \sum L_i\).
\item
  Coxeter groups, i.e.~groups generated by reflections, including Weyl
  groups. These can be infinite, which ones are finite?
\item
  Quantize Coxeter groups to get Hecke algebras, which are algebras over
  \({\mathbf{C}}[q, q^{-1}]\). See Humphreys.
\item
  Representations of Lie groups over \({\mathbf{R}}\), semisimple
  algebraic groups, representations of finite groups of Lie type (see
  the classification of finite simple groups, e.g.~algebraic groups over
  finite fields).
\item
  Characteristic \(p\) representation theory, which is much more
  difficult.
\item
  Infinite-dimensional Lie algebras over \({\mathbf{C}}\),
  e.g.~affine/Kac-Moody algebras using the Serre relations on
  generalized Cartan matrices. See also current algebras, loop algebras.
\item
  Quantum groups (quantized enveloping algebras), closely tied to
  modular representation theory.
\end{itemize}

\end{remark}

\hypertarget{useful-tricks}{%
\section{Useful Tricks}\label{useful-tricks}}

\begin{itemize}
\tightlist
\item
  \([[xy]z] = [x[yz]] + [[xz]y] = [x[yz]] - [y[xz]]\).
\item
  \(xz.w = zx.w + [xz].w\).
\item
  If \(N, M\) are upper triangular, \([NM]\) has zeros along the
  diagonal.
\end{itemize}

\hypertarget{summary-of-main-results}{%
\section{Summary of main results}\label{summary-of-main-results}}

\hypertarget{computation}{%
\subsection{Computation}\label{computation}}

\begin{itemize}
\tightlist
\item
  Bracketing elementary matrices:
  \begin{align*}
  [e_{ij}, e_{kl}] = \delta_{jk} e_{il} - \delta_{li} e_{kj}
  .\end{align*}
\end{itemize}

\hypertarget{classical-algebras}{%
\subsection{Classical Algebras}\label{classical-algebras}}

\begin{itemize}
\tightlist
\item
  \({\mathfrak{sl}}_2({ \mathbf{F} })\) is dimension 3, corresponds to
  type \(A_2\), and generated by
\end{itemize}

\begin{align*}
x=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right)\quad
h=\left(\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right)
y=\left(\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right)\quad \\ \\
[x, y]=h, \quad[h, x]=2 x, \quad[y, h]=2 y
.\end{align*}
- \({\mathfrak{sl}}_n({ \mathbf{F} })\) is dimension \(n^2-1\) and
corresponds to type \(A_{n-1}\).

\hypertarget{definitions}{%
\subsection{Definitions}\label{definitions}}

\begin{itemize}
\tightlist
\item
  \(N_L(S) = \left\{{x\in L {~\mathrel{\Big\vert}~}[xS] \subseteq S}\right\}\).
\item
  \(C_L(S) = \left\{{x\in L {~\mathrel{\Big\vert}~}[x S] = 0}\right\}\).
\item
  \(L\) is simple iff \(L\neq 0\) and
  \(\operatorname{Id}(L) = \left\{{0, L}\right\}\).
\item
  \(L\) is solvable iff
  \(L^{(n+1)} \coloneqq[ L^{(n)} L^{(n)}] \overset{n\to\infty}\to 0\).
\item
  \(L\) is semisimple iff
  \(\mathop{\mathrm{Rad}}(L) \coloneqq\sum_{I {~\trianglelefteq~}L \text { solvable}} I = 0\).
\item
  \(L\) is nilpotent iff
  \(L^{n+1} \coloneqq[L L^{n-1}] \overset{n\to\infty}\to 0\).
\item
  The Killing form is
  \(\kappa(x,y)\coloneqq\operatorname{Trace}( { \operatorname{ad}}_x \circ { \operatorname{ad}}_y)\).
\item
  Checking \({\mathsf{L}{\hbox{-}}\mathsf{Mod}}\) actions:

  \begin{itemize}
  \tightlist
  \item
    \(( \lambda_1 x + \lambda_2 y).\mathbf{v} = \lambda_1 (x.\mathbf{v}) + \lambda_2(y.\mathbf{v})\)
  \item
    \(x.(\lambda_1 \mathbf{v}+ \lambda_2 \mathbf{w}) = \lambda_1 (x.\mathbf{v}) + \lambda_2(x.\mathbf{w})\)
  \item
    \([xy].\mathbf{v} = x.(y.\mathbf{v}) - y.(x.\mathbf{v})\). This is
    the axiom that introduces weird formulas for homs/tensors/duals.
  \end{itemize}
\item
  Duals: \((x.f)(\mathbf{v}) \coloneqq-f(x.\mathbf{v})\)
\item
  Tensors:
  \(x.(\mathbf{v}\otimes\mathbf{w}) \coloneqq\qty{ (x.\mathbf{v})\otimes\mathbf{w}} + \qty{\mathbf{v}\otimes(x.\mathbf{w})}\).
\item
  The Casimir element: for \(\phi: L\to {\mathfrak{gl}}(V)\) an
  irreducible representation of \(L\) semisimple, define
  \(\beta(x, y) \coloneqq\operatorname{Trace}(\phi(x)\circ \phi(y))\).
  Pick a basis and dual basis
  \(\left\{{e_i}\right\}, \left\{{e_i {}^{ \vee }}\right\}\) with
  respect to \(\beta\) and define
  \(c_\phi(\beta) \coloneqq\sum \phi(e_i)\phi(e_i {}^{ \vee }) \in { \operatorname{End} }(V)\).

  \begin{itemize}
  \tightlist
  \item
    This is an endomorphism of \(V\) commuting with the
    \(L{\hbox{-}}\)action which has nonzero trace.
  \end{itemize}
\end{itemize}

\hypertarget{results}{%
\subsection{Results}\label{results}}

\begin{itemize}
\tightlist
\item
  Engel's theorem: if every \(x\in L\) is ad-nilpotent then \(L\) is
  nilpotent.

  \begin{itemize}
  \tightlist
  \item
    Since conversely (for free) \(L\) nilpotent implies
    \({ \operatorname{ad}}_x\) is nilpotent for every \(x\), this
    becomes an iff: \(L\) is nilpotent iff every \(x\in L\) is
    ad-nilpotent.
  \end{itemize}
\item
  Lie's theorem: if \(L\leq {\mathfrak{gl}}(V)\) is solvable then \(L\)
  stabilizes a flag in \(V\) (i.e.~\(L\) has an upper triangular basis).
\item
  Cartan's criterion: if \(L\leq {\mathfrak{gl}}(V)\) and
  \(\operatorname{Trace}(xy) = 0\) for all \(x\in [LL]\) and \(y\in L\),
  then \(L\) is solvable.
\item
  \(L\) is semisimple iff \(\kappa_L\) is nodegenerate.
\item
  If \(L\) is semisimple, it decomposes as \(L = \bigoplus L_i\) where
  the \(L_i\) are uniquely determined simple ideals, and every simple
  ideal of \(L\) is one such \(L_i\).
\item
  \(\ker (L \xrightarrow{ { \operatorname{ad}}_L} {\mathfrak{gl}}(V)) = Z(L)\)
  and simple algebras are centerless, so any simple Lie algebra is
  isomorphic to a linear Lie algebra \({\mathfrak{gl}}(V)\) for some
  \(V\), namely \(\operatorname{im} { \operatorname{ad}}_L\).
\item
  Schur's lemma: if \(L \xrightarrow{\phi} {\mathfrak{gl}}(V)\) is an
  irreducible representation, then
  \(C_{{\mathfrak{gl}}(V)}(\phi(L)) = {\mathbf{C}}I\), i.e.~the only
  endomorphisms of \(V\) commuting with every \(\phi(x)\) are scalar
  operators. Equivalently,
  \({ \operatorname{End} }_L(V) \cong {\mathbf{C}}\).
\item
  Weyl's theorem: if \(L\) is semisimple and
  \(\phi: L\to {\mathfrak{gl}}(V)\) is a finite-dimensional
  representation then \(\phi\) is completely reducible.
\end{itemize}

\hypertarget{problem-set-1}{%
\section{Problem Set 1}\label{problem-set-1}}

\hypertarget{section-1}{%
\subsection{Section 1}\label{section-1}}

\begin{problem}[Humphreys 1.1]

Let \(L\) be the real vector space \(\mathbf{R}^{3}\). Define
\([x y]=x \times y\) (cross product of vectors) for \(x, y \in L\), and
verify that \(L\) is a Lie algebra. Write down the structure constants
relative to the usual basis of \(\mathbf{R}^{3}\).

\end{problem}

\begin{solution}

It suffices to check the 3 axioms given in class that define a Lie
algebra:

\begin{itemize}
\item
  \textbf{L1 (Bilinearity)}: This can be quickly seen from the formula
  \begin{align*}
  a\times b = {\left\lVert {a} \right\rVert}\cdot {\left\lVert {b} \right\rVert}\sin(\theta_{ab}) \widehat{n}_{ab}
  \end{align*}
  where \(\widehat{n}_{ab}\) is the vector orthogonal to both \(a\) and
  \(b\) given by the right-hand rule. The result follows readily from a
  direct computation:
  \begin{align*}
  (ra)\times(tb) 
  &= {\left\lVert {ra} \right\rVert} \cdot {\left\lVert {tb} \right\rVert} \sin(\theta_{ra, tb}) \widehat{n}_{ra, tb} \\
  &= (rt) {\left\lVert {a} \right\rVert} \cdot {\left\lVert {b} \right\rVert} \sin(\theta_{a, b}) \widehat{n}_{a, b} \\
  &= (rt)\qty{a\times b}
  ,\end{align*}
  where we've used the fact that the angle between \(a\) and \(b\) is
  the same as the angle between any of their scalar multiples, as is
  their normal.
\item
  \textbf{L2}: that \(a\times a = 0\) readily follows from the same
  formula, since \(\sin( \theta_{a, a}) = \sin(0) = 0\).
\item
  \textbf{L3 (The Jacobi identity)}: This is most easily seen from the
  ``BAC - CAB'' formula,
  \begin{align*}
  a\times (b\times c) = b{\left\langle {a},~{c} \right\rangle} - c{\left\langle {a},~{b} \right\rangle}
  .\end{align*}
  We proceed by expanding the Jacobi expression:
  \begin{align*}
  a\times(b\times c) + c\times (a\times b) + b\times (c\times a)
  &= {\color{blue} b{\left\langle {a},~{c} \right\rangle} } - {\color{red} c{\left\langle {a},~{b} \right\rangle} }\\
  &\quad + {\color{green} a {\left\langle { c },~{ b } \right\rangle} } - {\color{blue} b {\left\langle { c },~{ a } \right\rangle} } \\
  &\quad + {\color{red} c {\left\langle { a },~{ b } \right\rangle} } - {\color{green} a {\left\langle { b },~{ c } \right\rangle} } \\
  &= 0
  .\end{align*}
\end{itemize}

For the structure constants, let \(\left\{{e_1, e_2, e_3}\right\}\) be
the standard Euclidean basis for \({\mathbf{R}}^3\); we can then write
\(e_i\times e_j = \sum_{k=1}^3 c_{ij}^k e_k\) and we would like to
determine the \(c_{ij}^k\). One can compute the following multiplication
table:

\begin{longtable}[]{@{}llll@{}}
\toprule
\(e_i\times e_j\) & \(e_1\) & \(e_2\) & \(e_3\) \\
\midrule
\endhead
\(e_1\) & \(0\) & \(e_3\) & \(-e_2\) \\
\(e_2\) & \(-e_3\) & \(0\) & \(e_1\) \\
\(e_3\) & \(e_2\) & \(-e_1\) & \(0\) \\
\bottomrule
\end{longtable}

Thus the structure constants are given by the antisymmetric Levi-Cevita
symbol,
\begin{align*}
c_{ij}^k = {\varepsilon}^{ijk} \coloneqq
\begin{cases}
0 & \text{if any index $i,j,k$ is repeated} 
\\
\operatorname{sgn}\sigma_{ijk} & \text{otherwise},
\end{cases}
\end{align*}
where \(\sigma_{ijk} \in S_3\) is the permutation associated to
\((i, j, k)\) in cycle notation and \(\operatorname{sgn}\sigma\) is the
sign homomorphism.

\begin{remark}

An example to demonstrate how the Levi-Cevita symbol works:

\begin{align*}
e_1\times e_2 = c_{12}^1 e_1 + c_{12}^2 e_2 + c_{12}^3 e_3 = 0 e_1 + 0 e_2 + 1e_3
\end{align*}
since the first two terms have a repeated index and
\begin{align*}
c_{12}^3 = {\varepsilon}_{1,2,3} = \operatorname{sgn}(123) = \operatorname{sgn}(12)(23) = (-1)^2 = 1
\end{align*}
using that \(\operatorname{sgn}\sigma = (-1)^m\) where \(m\) is the
number of transpositions in \(\sigma\).

\end{remark}

\end{solution}

\begin{problem}[Humphreys 1.6]

Let \(x \in {\mathfrak{gl}}_n({ \mathbf{F} })\) have \(n\) distinct
eigenvalues \(a_{1}, \ldots, a_{n}\) in \({ \mathbf{F} }\). Prove that
the eigenvalues of \({ \operatorname{ad}}_x\) are precisely the
\(n^{2}\) scalars \(a_{i}-a_{j}\) for \(1 \leq i, j \leq n\), which of
course need not be distinct.

\end{problem}

\begin{solution}

For a fixed \(n\), let \(e_{ij} \in {\mathfrak{gl}}_n({ \mathbf{F} })\)
be the matrix with a 1 in the \((i, j)\) position and zeros elsewhere.
We will use the following fact:
\begin{align*}
e_{ij} e_{kl} = \delta_{jk} e_{il}
,\end{align*}
where \(\delta_{jk} = 1 \iff j=k\), which implies that
\begin{align*}
[e_{ij} e_{kl} ] = e_{ij}e_{kl} - e_{kl}e_{ij} = \delta_{jk} e_{il} - \delta_{li}e_{kj}
.\end{align*}
Suppose without loss of generality\footnote{If \(x\) is not diagonal,
  one can use that \(x\) is diagonalizable over \({ \mathbf{F} }\) since
  \(x\) has distinct eigenvalues in \({ \mathbf{F} }\). So one can
  reduce to the diagonal case by a change-of-basis of
  \({ \mathbf{F} }^n\) that diagonalizes \(x\).} that \(x\) is diagonal
and of the form \(x = \operatorname{diag}(a_1, a_2, \cdots, a_n)\). Then
the eigenvectors of \(x\) are precisely the \(e_{ij}\), since a direct
check via matrix multiplication shows \(xe_{ij} = a_i e_{ij}\).

We claim that every \(e_{ij}\) is again an eigenvector of
\({ \operatorname{ad}}_x\) with eigenvalue \(a_i - a_j\). Noting that
the \(e_{ij}\) are also left eigenvectors satisfying
\(e_{ij}x = a_j e_{ij}\), one readily computes
\begin{align*}
 { \operatorname{ad}}_x e_{ij} 
\coloneqq[x, e_ij] 
= xe_{ij} - e_{ij} x 
= a_i e_{ij} - a_j e_{ij} = (a_i - a_j)e_{ij}
,\end{align*}
yielding at least \(n^2\) eigenvalues. Since \({ \operatorname{ad}}_x\)
expanded in the basis \(\left\{{e_{ij}}\right\}_{1\leq i, j \leq n}\) is
an \(n\times n\) matrix, this exhausts all possible eigenvalues.

\end{solution}

\begin{problem}[Humphreys 1.9, one Lie type only]

When \(\operatorname{ch}{ \mathbf{F} }=0\), show that each classical
algebra \(L=\mathrm{A}_{\ell}, \mathrm{B}_{\ell}, \mathrm{C}_{\ell}\),
or \(\mathrm{D}_{\ell}\) is equal to \([L L]\). (This shows again that
each algebra consists of trace 0 matrices.)

\end{problem}

\begin{solution}

We will check for this type \(A_n\), corresponding to
\(L \coloneqq{\mathfrak{sl}}_{n+1}\). Since \([LL] \subseteq L\), it
suffices to show \(L \subseteq [LL]\), and we can further reduce to
writing every basis element of \(L\) as a commutator in \([LL]\). Note
that \(L\) has a standard basis given by the matrices

\begin{itemize}
\tightlist
\item
  \(\left\{{x_i \coloneqq e_{ij} {~\mathrel{\Big\vert}~}i > j}\right\}\)
  corresponding to \({\mathfrak{n}}^-\),
\item
  \(\left\{{h_i \coloneqq e_{ii} - e_{i+1, i+1} {~\mathrel{\Big\vert}~}1\leq i \leq n}\right\}\)
  corresponding to \({\mathfrak{h}}\), and
\item
  \(\left\{{y_i \coloneqq e_{ij} {~\mathrel{\Big\vert}~}i < j}\right\}\)
  corresponding to \({\mathfrak{n}}^+\).
\end{itemize}

Considering the equation
\([e_{ij} e_{kl} ] = \delta_{jk} e_{il} - \delta_{li}e_{kj}\), one can
choose \(j=k\) to preserve the first term and \(l\neq i\) to kill the
second term. So letting \(t, i, j\) be arbitrary with \(i\neq j\), we
have
\begin{align*}
[e_{it} e_{tj}] = \delta_{tt} e_{ij} - \delta_{ij}e_{tt} = e_{ij}
,\end{align*}
yielding all of the \(x_i\) and \(y_i\). But in fact we are done, using
the fact that \(h_i = [x_i y_i]\).

\end{solution}

\begin{problem}[Humphreys 1.11]

Verify that the commutator of two derivations of an
\({ \mathbf{F} }{\hbox{-}}\)algebra is again a derivation, whereas the
ordinary product need not be.

\end{problem}

\begin{solution}

We want to show that
\([\mathop{\mathrm{Der}}(L) \mathop{\mathrm{Der}}(L)] \subseteq \mathop{\mathrm{Der}}(L)\),
so let \(f,g\in \mathop{\mathrm{Der}}(L)\). The result follows from a
direct computation; letting \(D \coloneqq[fg]\), we have
\begin{align*}
D(ab) = [fg](ab) 
&= (fg-gf)(ab) \\
&= fg(ab) - gf(ab) \\
&= f\qty{g(a)b + ag(b) } - g\qty{ f(a)b + af(b)} \\
&= f\qty{g(a)b} + f\qty{ag(b)} - g\qty{f(a)b} - g\qty{af(b)} \\
&= { {\color{blue} (fg)(a)b } + {\color{red} g(a)f(b)} } \\
&\quad + { {\color{red} f(a)g(b) } + {\color{green} a (fg)(b)} } \\
&\quad - { {\color{blue} (gf)(a) b } + {\color{red} f(a)g(b)} } \\
&\quad - { {\color{red} g(a)f(b) } - {\color{green} a(gf)(b)} } \\
&= {\color{blue} [fg](a) b} - {\color{green} a [fg](b) } \\
&= D(a)b - aD(b)
.\end{align*}

To see that ordinary products of derivations need not be derivations,
consider the operators
\(D_x \coloneqq{\frac{\partial }{\partial x}\,}, D_y = {\frac{\partial }{\partial y}\,}\)
acting on a finite-dimensional vector space of multivariate polynomials
of some bounded degree, as a sub \({\mathbf{R}}{\hbox{-}}\)algebra of
\({\mathbf{R}}[x,y]\). Take \(f(x,y) = x+y\) and \(g(x,y) = xy\), so
that \(fg = g f = x^2 y+ xy^2\). Then \([D_x D_y] = 0\) since mixed
partial derivatives are equal, but
\begin{align*}
D_x D_y (fg) = D_x \qty{x^2 + 2xy} = 2x + 2y \neq 0
.\end{align*}

\end{solution}

\begin{problem}[Humphreys 1.12]

Let \(L\) be a Lie algebra over an algebraically closed field
\({ \mathbf{F} }\) and let \(x \in L\). Prove that the subspace of \(L\)
spanned by the eigenvectors of \({ \operatorname{ad}}_x\) is a
subalgebra.

\end{problem}

\begin{solution}

Let \(E_x \subseteq L\) be the subspace spanned by eigenvectors of
\({ \operatorname{ad}}_x\); it suffices to show
\([E_x E_x] \subseteq E_x\). Letting \(y_i \in E_x\), we have
\({ \operatorname{ad}}_x(y_i) = \lambda_i y_i\) for some scalars
\(\lambda_i \in { \mathbf{F} }\), and we want to show
\({ \operatorname{ad}}_x([y_1 y_2]) = \lambda_{12} [y_1 y_2]\) for some
scalar \(\lambda_{12}\). Note that the Jacobi identity is equivalent to
\({ \operatorname{ad}}\) acting as a derivation with respect to the
bracket, i.e.~
\begin{align*}
 { \operatorname{ad}}_x([yz]) = [ { \operatorname{ad}}_x(y) z] + [y  { \operatorname{ad}}_x(z)] \implies [x[yz]] = [[xy]z] + [y[xz]]
.\end{align*}
The result then follows from a direct computation:
\begin{align*}
 { \operatorname{ad}}_x([y_1y_2]) 
&= [[xy_1]y_2] + [y_1 [xy_2]] \\
&= [ \lambda_1 y_1 y_2] + [y_1 \lambda_2 y_2] \\
&= (\lambda_1 + \lambda_2)[y_1 y_2]
.\end{align*}

\end{solution}

\hypertarget{section-2}{%
\subsection{Section 2}\label{section-2}}

\begin{problem}[Humphreys 2.1]

Prove that the set of all inner derivations
\({ \operatorname{ad}}_x, x \in L\), is an ideal of
\(\mathop{\mathrm{Der}}L\).

\end{problem}

\begin{solution}

It suffices to show
\([\mathop{\mathrm{Der}}(L) \mathop{\mathrm{Inn}}(L)] \subseteq \mathop{\mathrm{Inn}}(L)\),
so let \(f\in \mathop{\mathrm{Der}}(L)\) and
\({ \operatorname{ad}}_x \in \mathop{\mathrm{Inn}}(L)\). The result
follows from the following check:
\begin{align*}
[f { \operatorname{ad}}_x](l) 
&= (f\circ  { \operatorname{ad}}_x)(l) - ( { \operatorname{ad}}_x \circ f)(l) \\
&= f([xl]) - [x f(l)] \\
&= [f(x) l] + [x f(l)] - [x f(l)] \\
&= [f(x) l] \\
&=  { \operatorname{ad}}_{f(x)}(l), \qquad \text{and }  { \operatorname{ad}}_{f(x)} \in \mathop{\mathrm{Inn}}(L)
.\end{align*}

\end{solution}

\begin{problem}[Humphreys 2.2]

Show that \(\mathfrak{s l}_n( { \mathbf{F} })\) is precisely the derived
algebra of \(\mathfrak{g l}_n( { \mathbf{F} })\) (cf.~Exercise 1.9).

\end{problem}

\begin{solution}

We want to show
\({\mathfrak{gl}}_n({ \mathbf{F} })^{(1)} \coloneqq[{\mathfrak{gl}}_n({ \mathbf{F} }) {\mathfrak{gl}}_n({ \mathbf{F} })] = {\mathfrak{sl}}_n({ \mathbf{F} })\).

\(\subseteq\): This immediate from the fact that for any matrices \(A\)
and \(B\),
\begin{align*}
{\mathrm{tr}}([AB]) = {\mathrm{tr}}(AB -BA) = {\mathrm{tr}}(AB) - {\mathrm{tr}}(BA) = {\mathrm{tr}}(AB) - {\mathrm{tr}}(AB) = 0
.\end{align*}

\(\supseteq\): From a previous exercise, we know that
\([{\mathfrak{sl}}_n({ \mathbf{F} }) {\mathfrak{sl}}_n({ \mathbf{F} })] = {\mathfrak{sl}}_n({ \mathbf{F} })\),
and since
\({\mathfrak{sl}}_n({ \mathbf{F} }) \subseteq {\mathfrak{gl}}_n({ \mathbf{F} })\)
we have
\begin{align*}
{\mathfrak{sl}}_n({ \mathbf{F} }) = {\mathfrak{sl}}_n({ \mathbf{F} })^{(1)} \subseteq {\mathfrak{gl}}_n({ \mathbf{F} })^{(1)}
.\end{align*}

\end{solution}

\begin{problem}[Humphreys 2.5]

Suppose \(\operatorname{dim} L=3\) and \(L=[L L]\). Prove that \(L\)
must be simple. Observe first that any homomorphic image of \(L\) also
equals its derived algebra. Recover the simplicity of
\(\mathfrak{s l}_2( { \mathbf{F} })\) when
\(\operatorname{ch}{ \mathbf{F} }\neq 2\).

\end{problem}

\begin{solution}

Let \(I{~\trianglelefteq~}L\) be a proper ideal, then
\(\dim L/I < \dim L\) forces \(\dim L/I = 1,2\). Since
\(L\twoheadrightarrow L/I\), the latter is the homomorphic image of a
Lie algebra and thus \((L/I)^{(1)} = L/I\) by the hint. Note that in
particular, \(L/I\) is not abelian. We proceed by cases:

\begin{itemize}
\tightlist
\item
  \(\dim L/I = 1\).

  \begin{itemize}
  \tightlist
  \item
    In this case, \(L/I = { \mathbf{F} }x\) is generated by a single
    element \(x\). Since \([xx] = 0\) in any Lie algebra, we have
    \(({ \mathbf{F} }x)^{(1)} = 0\), contradicting that \(L/I\) is not
    abelian. \(\contradiction\)
  \end{itemize}
\item
  \(\dim L/I = 2\): Write \(L/I = { \mathbf{F} }x + { \mathbf{F} }y\)
  for distinct generators \(x,y\), and consider the multiplication table
  for the bracket.

  \begin{itemize}
  \tightlist
  \item
    If \([xy] = 0\), then \(L/I\) is abelian, a contradiction.
    \(\contradiction\)
  \item
    Otherwise, without loss of generality \([xy] =x\) as described at
    the end of section 1.4. In this case,
    \((L/I)^{(1)} \subseteq { \mathbf{F} }_x \subsetneq L/I\), again a
    contradiction. \(\contradiction\)
  \end{itemize}
\end{itemize}

So no such proper ideals \(I\) can exist, forcing \(L\) to be simple.

Applying this to \(L \coloneqq{\mathfrak{sl}}_2({ \mathbf{F} })\), we
have
\(\dim_{ \mathbf{F} }{\mathfrak{sl}}_2({ \mathbf{F} }) = 2^2-1 = 3\),
and from a previous exercise we know
\({\mathfrak{sl}}_2({ \mathbf{F} })^{(1)} = {\mathfrak{sl}}_2({ \mathbf{F} })\),
so the above argument applies and shows simplicity.

\end{solution}

\begin{problem}[Humphreys 2.10]

Let \(\sigma\) be the automorphism of
\(\mathfrak{s l}_2({ \mathbf{F} })\) defined in (2.3). Verify that

\begin{itemize}
\tightlist
\item
  \(\sigma(x)=-y\),
\item
  \(\sigma(y)=-x\),
\item
  \(\sigma(h)=-h\).
\end{itemize}

Note that this automorphism is defined as
\begin{align*}
\sigma = \exp( { \operatorname{ad}}_x)\circ \exp( { \operatorname{ad}}_{-y}) \circ \exp( { \operatorname{ad}}_x)
.\end{align*}

\end{problem}

\begin{solution}

We recall that
\(\exp { \operatorname{ad}}_x(y) \coloneqq\sum_{n\geq 0}{1\over n!} { \operatorname{ad}}_x^n(y)\),
where the exponent denotes an \(n{\hbox{-}}\)fold composition of
operators. To compute these power series, first note that
\({ \operatorname{ad}}_t(t) = 0\) for \(t=x,y,h\) by axiom \textbf{L2},
so
\begin{align*}
(\exp  { \operatorname{ad}}_t)(t) = 1(t) +  { \operatorname{ad}}_t(t) + {1\over 2}  { \operatorname{ad}}_t^2(t) + \cdots = 1(t) = t
\end{align*}
where \(1\) denotes the identity operator. It is worth noting that if
\({ \operatorname{ad}}_t^n(t') = 0\) for some \(n\) and some fixed
\(t,t'\), then it is also zero for all higher \(n\) since each
successive term involves bracketing with the previous term:
\begin{align*}
 { \operatorname{ad}}^{n+1}_t(t') = [t\,  { \operatorname{ad}}_t^n(t')] = [t\, 0] = 0
.\end{align*}

We first compute some individual nontrivial terms that will appear in
\(\sigma\). The first order terms are given by standard formulas, which
we collect into a multiplication table for the bracket:

\begin{longtable}[]{@{}llll@{}}
\toprule
& \(x\) & \(h\) & \(y\) \\
\midrule
\endhead
\(x\) & \(0\) & \(-2x\) & \(h\) \\
\(h\) & \(2x\) & \(0\) & \(-2y\) \\
\(y\) & \(-h\) & \(2y\) & \(0\) \\
\bottomrule
\end{longtable}

We can thus read off the following:

\begin{itemize}
\tightlist
\item
  \({ \operatorname{ad}}_x(y) = h\)
\item
  \({ \operatorname{ad}}_x(h) = -2x\)
\item
  \({ \operatorname{ad}}_{-y}(x) = [-y x] = [xy] = h\)
\item
  \({ \operatorname{ad}}_{-y}(h) = [-yh] = [hy] = -2y\)
\end{itemize}

For reference, we compute and collect higher order terms:

\begin{itemize}
\tightlist
\item
  \({ \operatorname{ad}}_x^n(y)\):

  \begin{itemize}
  \tightlist
  \item
    \({ \operatorname{ad}}_x^1(y) = h\) from above,
  \item
    \({ \operatorname{ad}}_x^2(y) = { \operatorname{ad}}_x([xy]) = { \operatorname{ad}}_x(h) = [xh] = -[hx] = -2x\),
  \item
    \({ \operatorname{ad}}_x^3(y) = { \operatorname{ad}}_x(-2x) = 0\),
    so \({ \operatorname{ad}}_x^{\geq 3}(y) = 0\).
  \end{itemize}
\item
  \({ \operatorname{ad}}_x^n(h)\):

  \begin{itemize}
  \tightlist
  \item
    \({ \operatorname{ad}}_x^1(h) = -2x\) from above,
  \item
    \({ \operatorname{ad}}_x^2(h) = { \operatorname{ad}}_x(-2x) = 0\),
    so \({ \operatorname{ad}}_x^{\geq 2}(h) = 0\).
  \end{itemize}
\item
  \({ \operatorname{ad}}_{-y}^n(x)\):

  \begin{itemize}
  \tightlist
  \item
    \({ \operatorname{ad}}_{-y}^1(x) = h\) from above,
  \item
    \({ \operatorname{ad}}_{-y}^2(x) = { \operatorname{ad}}_{-y}(h) = [-yh] = [hy] = -2y\),
  \item
    \({ \operatorname{ad}}_{-y}^2(x) = { \operatorname{ad}}_{-y}(-2y) = 0\),
    so \({ \operatorname{ad}}_{-y}^{n\geq 2}(x) = 0\).
  \end{itemize}
\item
  \({ \operatorname{ad}}_{-y}^n(h)\):

  \begin{itemize}
  \tightlist
  \item
    \({ \operatorname{ad}}_{-y}^1(h) = -2y\) from above, and so
    \({ \operatorname{ad}}_{-y}^{\geq 2}(h) = 0\).
  \end{itemize}
\end{itemize}

Finally, we can compute the individual terms of \(\sigma\):

\begin{align*}
(\exp  { \operatorname{ad}}_x)(x) &= x \\ \\ 
(\exp  { \operatorname{ad}}_x)(h) &= 1(h) +  { \operatorname{ad}}_x(h) \\ &= h + (-2x) \\ &= h-2x \\ \\
(\exp  { \operatorname{ad}}_x)(y) &= 1(y) +  { \operatorname{ad}}_x(y) + {1\over 2} { \operatorname{ad}}_x^2(y) \\ &= y + h + {1\over 2}(-2x) \\ &= y+h-x \\ \\
(\exp  { \operatorname{ad}}_{-y})(x) &=  1(x) +  { \operatorname{ad}}_{-y}(x) x + {1\over 2}  { \operatorname{ad}}^2_{-y}(x) \\ &= x + h +{1\over 2}(-2y) \\ &= x+h-y \\ \\
(\exp  { \operatorname{ad}}_{-y})(h) &=  1(h) +  { \operatorname{ad}}_{-y}(h) \\ &= h - 2y \\ \\
(\exp  { \operatorname{ad}}_{-y})(y) &=  y 
,\end{align*}
and assembling everything together yields

\begin{align*}
\sigma(x) 
&= (\exp  { \operatorname{ad}}_x \circ \exp  { \operatorname{ad}}_{-y} \circ \exp  { \operatorname{ad}}_x)(x) \\
&= (\exp  { \operatorname{ad}}_x \circ \exp  { \operatorname{ad}}_{-y})(x) \\
&= (\exp  { \operatorname{ad}}_x)(x+h-y) \\
&= (x) + (h-2x) - (y+h-x) \\
&= -y \\ \\
\sigma(y)
&= (\exp  { \operatorname{ad}}_x \circ \exp  { \operatorname{ad}}_{-y} \circ \exp  { \operatorname{ad}}_x)(y) \\
&= (\exp  { \operatorname{ad}}_x \circ \exp  { \operatorname{ad}}_{-y} )(y+h-x) \\
&= \exp  { \operatorname{ad}}_x\qty{(y) + (h-2y) - (x+h-y) } \\
&= \exp  { \operatorname{ad}}_x\qty{-x} \\
&= -x \\ \\
\sigma(h) 
&= (\exp  { \operatorname{ad}}_x \circ \exp  { \operatorname{ad}}_{-y} \circ \exp  { \operatorname{ad}}_x)(h) \\
&= (\exp  { \operatorname{ad}}_x \circ \exp  { \operatorname{ad}}_{-y} )(h-2x) \\
&= (\exp  { \operatorname{ad}}_x )( (h-2y) - 2(x+h-y) ) \\
&= (\exp  { \operatorname{ad}}_x )(-2x -h ) \\
&= -2(x) - (h-2x) \\
&= -h
.\end{align*}

\end{solution}

\hypertarget{problem-set-2}{%
\section{Problem Set 2}\label{problem-set-2}}

\hypertarget{section-3}{%
\subsection{Section 3}\label{section-3}}

\begin{problem}[Humphreys 3.1]

Let \(I\) be an ideal of \(L\). Then each member of the derived series
or descending central series of \(I\) is also an ideal of \(L\).

\end{problem}

\begin{solution}

To recall definitions:

\begin{itemize}
\tightlist
\item
  The derived series of \(L\) is
  \(L \supseteq L^{(0)} \coloneqq[LL] \supseteq L^{(1)} \coloneqq[[LL] [LL]] \supseteq\cdots\)
  and termination implies solvability.
\item
  The descending central series of \(L\) is
  \(L \supseteq L^1 \coloneqq[LL] \supseteq L^2 \coloneqq[L [LL]] \supseteq\cdots\),
  and termination implies nilpotency (and hence solvability since
  \([LL] \subseteq L \implies L^{(i)} \subseteq L^i\)).
\item
  \(I {~\trianglelefteq~}L \iff [L,I] \subseteq I\).
\end{itemize}

For the derived series, inductively suppose \(I \coloneqq I^{(i)}\) is
an ideal, so \([L I] \subseteq I\). We then want to show
\(I^{(i+1)} \coloneqq[I, I]\) is an ideal, so
\([L, [I, I] ] \subseteq [I, I]\). Letting \(l\in L\), and \(i,j\in I\),
one can use the Jacobi identity, antisymmetry of the bracket, and the
fact that \([I, I] \coloneqq L^{(i+1)} \subseteq I\) to write
\begin{align*}
[L, [I, I]] &\ni [l[ij]] \\
&= [[li]j] - [ [lj] i] \\
&\in [[L,I], I] - [[L,I], I] \\
&\subseteq [[L,I], I] \subseteq [I,I]
.\end{align*}

Similarly, for the lower central series, inductively suppose
\(I\coloneqq I^i\) is an ideal, so \([L, I] \subseteq I\); we want to
show \([L, [L, I]] \subseteq [L, I]\). Again using the Jacobi identity
and antisymmetry, we have
\begin{align*}
[L, [L, I]] 
&\ni [l_1, [l_2, i]] \\
&= [[i, l_1], l_2] + [[l_2, l_1], i] \\
&\subseteq [[I,L], L] + [ [L, L], I] \\
&\subseteq [I, L] + [L, I] \subseteq [L, I]
.\end{align*}

\end{solution}

\begin{problem}[Humphreys 3.4]

Prove that \(L\) is solvable (resp. nilpotent) if and only
\({ \operatorname{ad}}(L)\) is solvable (resp. nilpotent).

\end{problem}

\begin{solution}

\(\implies\): By the propositions in Section 3.1 (resp. 3.2), the
homomorphic image of any solvable (resp. nilpotent) Lie algebra is again
solvable (resp. nilpotent).

\(\impliedby\): There is an exact sequence
\begin{align*}
0 \to Z(L) \to L \xrightarrow{ { \operatorname{ad}}}   { \operatorname{ad}}(L) \to 0
,\end{align*}
exhibiting \({ \operatorname{ad}}(L)\cong L/Z(L)\). Thus if
\({ \operatorname{ad}}(L)\) is solvable, noting that centers are always
solvable, we can use the fact that the 2-out-of-3 property for short
exact sequences holds for solvability. Moreover, by the proposition in
Section 3.2, if \(L/Z(L)\) is nilpotent then \(L\) is nilpotent.

\end{solution}

\begin{problem}[Humphreys 3.6]

Prove that the sum of two nilpotent ideals of a Lie algebra \(L\) is
again a nilpotent ideal. Therefore, \(L\) possesses a unique maximal
nilpotent ideal. Determine this ideal for the nonabelian 2-dimensional
algebra \({ \mathbf{F} }x + { \mathbf{F} }y\) where \([xy]=x\), and the
3-dimensional algebra
\({ \mathbf{F} }x + { \mathbf{F} }y + { \mathbf{F} }z\) where

\begin{itemize}
\tightlist
\item
  \([xy] = z\)
\item
  \([xz]=y\)
\item
  \([yz] = 0\)
\end{itemize}

\end{problem}

\begin{solution}

To see that sums of nilpotent ideals are nilpotent, suppose
\(I^N = J^M = 0\) are nilpotent ideals. Then
\((I+J)^{M+N} \subseteq I^M + J^N\) by collecting terms and using the
absorbing property of ideals. One can now construct a maximal nilpotent
ideal in \(L\) by defining \(M\) as the sum of all nilpotent ideals in
\(L\). That this is unique is clear, since \(M\) is nilpotent, so if
\(M'\) is another maximal nilpotent ideal then \(M \subseteq M'\) and
\(M' \subseteq M\).

Consider the 2-dimensional algebra
\(L \coloneqq{ \mathbf{F} }x + { \mathbf{F} }y\) where \([xy] = x\) and
let \(I\) be the maximal nilpotent ideal. Note that \(L\) is not
nilpotent since \(L^k = { \mathbf{F} }x\) for all \(k\geq 0\), since
\(L^1 = { \mathbf{F} }x\) and \([L, { \mathbf{F} }x] = { \mathbf{F} }x\)
(since all brackets are either zero or \(\pm x\)). However, this also
shows that the subalgebra \({ \mathbf{F} }x\) is an ideal, and is in
fact a nilpotent ideal since \([{ \mathbf{F} }x, { \mathbf{F} }x] = 0\).
Although \({ \mathbf{F} }y\) is a nilpotent subalgebra, it is not an
ideal since \([L, { \mathbf{F} }y] = { \mathbf{F} }x\). So \(I\) is at
least 1-dimensional, since it contains \({ \mathbf{F} }x\), and at most
1-dimensional, since it is not all of \(L\), forcing
\(I = { \mathbf{F} }x\).

Consider now the 3-dimensional algebra
\(L \coloneqq{ \mathbf{F} }x + { \mathbf{F} }y + { \mathbf{F} }z\) with
the multiplication table given in the problem statement above. Note that
\(L\) is not nilpotent, since
\(L^1 = { \mathbf{F} }y + { \mathbf{F} }z = L^k\) for all \(k\geq 2\).
This follows from consider \([L, { \mathbf{F} }y + { \mathbf{F} }z]\),
where choosing \(x\in L\) is always a valid choice and choosing \(y\) or
\(z\) in the second slot hits all generators; however, no element
brackets to \(x\). So similar to the previous algebra, the ideal
\(J \coloneqq{ \mathbf{F} }x + { \mathbf{F} }y\) is an ideal, and it is
nilpotent since all brackets between \(y\) and \(z\) vanish. By similar
dimensional considerations, \(J\) must equal the maximal nilpotent
ideal.

\end{solution}

\begin{problem}[Humphreys 3.10]

Let \(L\) be a Lie algebra, \(K\) an ideal of \(L\) such that \(L / K\)
is nilpotent and such that
\({ \left.{{ { \operatorname{ad}}_x}} \right|_{{K}} }\) is nilpotent for
all \(x \in L\). Prove that \(L\) is nilpotent.

\end{problem}

\begin{solution}

Suppose that \(M \coloneqq L/K\) is nilpotent, so the lower central
series terminates and \(M^n = 0\) for some \(n\). Then
\(L^n \subseteq K\) for the same \(n\), and the claim is that \(L^n\) is
nilpotent. This follows from applying Engel's theorem: let
\(x\in L^n \subseteq K\), then
\({ \left.{{ { \operatorname{ad}}_x}} \right|_{{L^n}} } = 0\) by
assumption. So every element of \(L^n\) is ad-nilpotent, making it
nilpotent. Since \(0 = (L^n)^k = L^{n+k}\) for some \(k\), this forces
\(L\) to be nilpotent as well.

\end{solution}

\hypertarget{section-4}{%
\subsection{Section 4}\label{section-4}}

\begin{problem}[Humphreys 4.1]

Let \(L= {\mathfrak{sl}}(V)\). Use Lie's Theorem to prove that
\(\operatorname{Rad} L=Z(L)\); conclude that \(L\) is semisimple.

\begin{quote}
Hint: observe that \(\operatorname{Rad} L\) lies in each maximal
solvable subalgebra \(B\) of \(L\). Select a basis of \(V\) so that
\(B=L \cap \mathfrak{t}(n, \mathrm{~F})\), and notice that \(B^t\) is
also a maximal solvable subalgebra of \(L\). Conclude that
\(\operatorname{Rad} L \subset L \cap \mathfrak{d}(n, \mathrm{~F})\)
(diagonal matrices), then that \(\operatorname{Rad} L=Z(L) .]\)
\end{quote}

\end{problem}

\begin{solution}

Let \(R = \mathop{\mathrm{Rad}}(L)\) be the radical (maximal solvable
ideal) of \(L\). Using the hint, if \(S \leq L\) is a maximal solvable
subalgebra then it must contain \(R\). By (a corollary of) Lie's
theorem, \(S\) stabilizes a flag and thus there is a basis with respect
to which all elements of \(S\) (and thus \(R\)) are upper triangular.
Thus \(S \subseteq {\mathfrak{b}}\); however, taking the transpose of
every element in \(S\) again yields a maximal solvable ideal which is
lower triangular and thus contained in \({\mathfrak{b}}^-\). Thus
\(R \subseteq S \subseteq {\mathfrak{b}}\cap{\mathfrak{b}}^- = {\mathfrak{h}}\),
which consists of just diagonal matrices.

We have \(Z(L) \subseteq R\) since centers are solvable, and the claim
is that \(R \subseteq {\mathfrak{h}}\implies R \subseteq Z(L)\). It
suffices to show that \(R\) consists of scalar matrices, since it is
well-known that \(Z({\mathfrak{gl}}_n({ \mathbf{F} }))\) consists of
precisely scalar matrices, and this contains \(Z(L)\) since
\(L \leq {\mathfrak{gl}}_n({ \mathbf{F} })\) is a subalgebra. This
follows by letting \(\ell = \sum a_i e_{i,i}\) be an element of
\(\mathop{\mathrm{Rad}}(L)\) and considering bracketing elements of
\({\mathfrak{sl}}_n({ \mathbf{F} })\) against it. Bracketing elementary
matrices \(e_{i, j}\) with \(i\neq j\) yields
\begin{align*}
[e_{i,j}, \ell] = a_j e_{i, j} - a_i e_{i, j}
,\end{align*}
which must be an element of \(\mathop{\mathrm{Rad}}(L)\) and thus
diagonal, which forces \(a_j = a_i\) for all \(i, j\).

To conclude that \(L\) is semisimple, note that a scalar traceless
matrix is necessarily zero, and so \(Z({\mathfrak{sl}}(V)) = 0\). This
suffices since \(\mathop{\mathrm{Rad}}(L) = 0 \iff L\) is semisimple.

\end{solution}

\begin{problem}[Humphreys 4.3, Failure of Lie's theorem in positive characteristic]

Consider the \(p \times p\) matrices:
\begin{align*}
x=\left[\begin{array}{cccccc}
0 & 1 & 0 & . & . & 0 \\
0 & 0 & 1 & 0 & . & 0 \\
. & . & \cdot & \cdot & \cdot & \cdot \\
0 & \cdot & \cdot & \cdot & \cdot & 1 \\
1 & . & \cdot & \cdot & \cdots & 0
\end{array}\right]
,\qquad y = \operatorname{diag}(0,1,2,3,\cdots,p-1)
.\end{align*}

Check that \([x, y]=x\), hence that \(x\) and \(y\) span a two
dimensional solvable subalgebra \(L\) of \(\mathfrak{g l}(p, F)\).
Verify that \(x, y\) have no common eigenvector.

\end{problem}

\begin{solution}

Note that \(x\) acts on the left on matrices \(y\) by cycling all rows
of \(y\) up by one position, and similar yacts on the right by cycling
the columns to the right. Thus
\begin{align*}
xy - yx &= 
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & \vdots & \vdots & \ddots & 0 \\
0 & 0 & \cdots & 0 & p-1 \\
0 & 0 & \cdots & 0 & 0 
\end{bmatrix}
-
\begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
\vdots & \cdots & 0 & 2 & 0 \\
0 & 0 & \cdots & 0 & 3 \\
p-1 & 0 & \cdots & 0 & 0
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
-(p-1) & 0 & 0 & 0 & 0
\end{bmatrix}\\
&\equiv
\begin{bmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix} \qquad \in \operatorname{GL}_n({ \mathbf{F} }_p) \\
&= x
.\end{align*}
Thus \(L \coloneqq{ \mathbf{F} }x + { \mathbf{F} }y\) span a solvable
subalgebra, since \(L^{(1)} = { \mathbf{F} }x\) and so \(L^{(2)} = 0\).

Moreover, note that every basis vector \(e_i\) is an eigenvector for
\(y\) since \(y(e_i) = i e_i\), while no basis vector is an eigenvector
for \(x\) since \(x(e_i) = e_{i+1}\) for \(1\leq i \leq p-1\) and
\(x(e_p) = e_1\), so \(x\) cycles the various basis vectors.

\end{solution}

\begin{problem}[Humphreys 4.4]

For arbitrary \(p\), construct a counterexample to Corollary
\(\mathrm{C}\)\footnote{Corollary C states that if \(L\) is solvable
  then every \(x\in L^{(1)}\) is ad-nilpotent, and thus \(L^{(1)}\) is
  nilpotent.} as follows: Start with
\(L \subset {\mathfrak{gl}}_p({ \mathbf{F} })\) as in Exercise 3. Form
the vector space direct sum \(M=L \oplus { \mathbf{F} }^p\), and make
\(M\) a Lie algebra by decreeing that \({ \mathbf{F} }^p\) is abelian,
while \(L\) has its usual product and acts on \({ \mathbf{F} }^p\) in
the given way.

Verify that \(M\) is solvable, but that its derived algebra
\(\left(={ \mathbf{F} }x+ { \mathbf{F} }^p\right)\) fails to be
nilpotent.

\end{problem}

\begin{solution}

For pairs \(A_1 \oplus v_1\) and \(A_2 \oplus v_2\) in \(M\), we'll
interpret the given definition of the bracket as
\begin{align*}
[A_1 \oplus v_1, A_2 \oplus v_2] \coloneqq[A_1, A_2] \oplus (A_1(v_2) - A_2(v_1))
,\end{align*}
where \(A_i(v_j)\) denotes evaluating an endomorphism
\(A\in {\mathfrak{gl}}_p({ \mathbf{F} })\) on a vector
\(v\in { \mathbf{F} }^p\). We also define
\(L = { \mathbf{F} }x + { \mathbf{F} }y\) with \(x\) and \(y\) the given
matrices in the previous problem, and note that \(L\) is solvable with
derived series
\begin{align*}
L = { \mathbf{F} }x \oplus { \mathbf{F} }y \supseteq L^{(1)} = { \mathbf{F} }x \supseteq L^{(2)} = 0
.\end{align*}

Consider the derived series of \(M\) -- by inspecting the above
definition, we have
\begin{align*}
M^{(1)} \subseteq L^{(1)} \oplus { \mathbf{F} }^p = { \mathbf{F} }x \oplus { \mathbf{F} }^p
.\end{align*}
Moreover, we have
\begin{align*}
M^{(2)} \subseteq L^{(2)} \oplus { \mathbf{F} }^p = 0 \oplus { \mathbf{F} }^p
,\end{align*}
which follows from considering considering bracketing two elements in
\(M^{(1)}\): set \(w_{ij} \coloneqq A_i(v_j) - A_j(v_i)\), then
\begin{align*}
[ [A_1, A_2] \oplus w_{1,2}, \,\, [A_3, A_4] \oplus w_{3, 4} ] \hspace{16em}
\\
= [ [A_1, A_2], [A_3, A_4] ] \oplus [A_1, A_2](w_{3, 4}) - [A_3, A_4](w_{1, 2})
.\end{align*}
We can then see that \(M^{(3)} = 0\), since for any
\(w_i \in { \mathbf{F} }^p\),
\begin{align*}
[0 \oplus w_1, \, 0 \oplus w_2] = 0 \oplus 0(w_2)-0(w_1) = 0 \oplus 0
,\end{align*}
and so \(M\) is solvable.

Now consider its derived subalgebra
\(M^{(1)} = { \mathbf{F} }x \oplus { \mathbf{F} }^p\). If this were
nilpotent, every element would be ad-nilpotent, but let
\(v = {\left[ {1,1,\cdots, 1} \right]}\) and consider
\({ \operatorname{ad}}_{x \oplus 0}\). We have
\begin{align*}
 { \operatorname{ad}}_{x \oplus 0}(0 \oplus v) = [x \oplus 0, 0 \oplus v] = 0 \oplus xv = 0 \oplus v
,\end{align*}
where we've used that \(x\) acts on the left on vectors by cycling the
entries. Thus
\({ \operatorname{ad}}_{x \oplus 0}^n (0 \oplus v) = 0 \oplus v\) for
all \(n\geq 1\) and \(x \oplus 0 \in M^{(1)}\) is not ad-nilpotent.

\end{solution}

\hypertarget{problem-set-3}{%
\section{Problem Set 3}\label{problem-set-3}}

\hypertarget{section-5}{%
\subsection{Section 5}\label{section-5}}

\begin{problem}[5.1]

Prove that if \(L\) is nilpotent then the Killing form of \(L\) is
identically zero.

\end{problem}

\begin{solution}

Note that if \(L\) is nilpotent than every \(\ell \in L\) is
ad-nilpotent, so letting \(x,y\in L\) be arbitrary, their commutator
\(\ell \coloneqq[xy]\) is ad-nilpotent. Thus
\({ \operatorname{ad}}_{[xy]} \in { \operatorname{End} }(L)\) is a
nilpotent endomorphism of \(L\), which are always traceless.

The claim is the following: for any \(x,y\in L\),
\begin{align*}
\operatorname{Trace}( { \operatorname{ad}}_{[xy]}) = 0\implies 
\operatorname{Trace}( { \operatorname{ad}}_x \circ  { \operatorname{ad}}_y) = 0
,\end{align*}
from which it follows immediately that \(\beta\) is identically zero.

First we can use the fact that
\({ \operatorname{ad}}: L\to {\mathfrak{gl}}(L)\) preserves brackets,
and so
\begin{align*}
 { \operatorname{ad}}_{[xy]_L} = [ { \operatorname{ad}}_x  { \operatorname{ad}}_y]_{{\mathfrak{gl}}(L)} =  { \operatorname{ad}}_x \circ  { \operatorname{ad}}_y -  { \operatorname{ad}}_y \circ { \operatorname{ad}}_x
,\end{align*}
and so
\begin{align*}
0 = \operatorname{Trace}( { \operatorname{ad}}_{[xy]}) = \operatorname{Trace}( { \operatorname{ad}}_x  { \operatorname{ad}}_y -  { \operatorname{ad}}_y  { \operatorname{ad}}_x) = \operatorname{Trace}( { \operatorname{ad}}_x  { \operatorname{ad}}_y) - \operatorname{Trace}( { \operatorname{ad}}_y  { \operatorname{ad}}_x)
.\end{align*}
where we've used that the trace is an \({ \mathbf{F} }{\hbox{-}}\)linear
map \({\mathfrak{gl}}(L) \to { \mathbf{F} }\). This forces
\begin{align*}
\operatorname{Trace}( { \operatorname{ad}}_x  { \operatorname{ad}}_y) = - \operatorname{Trace}( { \operatorname{ad}}_y  { \operatorname{ad}}_x)
,\end{align*}
but by the cyclic property of traces, we always have
\begin{align*}
\operatorname{Trace}( { \operatorname{ad}}_x  { \operatorname{ad}}_y) = \operatorname{Trace}( { \operatorname{ad}}_y  { \operatorname{ad}}_x)
.\end{align*}
Combining these yields
\(\operatorname{Trace}( { \operatorname{ad}}_x { \operatorname{ad}}_y) = 0\).

\end{solution}

\begin{problem}[5.7]

Relative to the standard basis of \({\mathfrak{sl}}_3({ \mathbf{F} })\),
compute \(\operatorname{det}\kappa\). What primes divide it?

\begin{quote}
Hint: use 6.7, which says
\(\kappa_{{\mathfrak{gl}}_n}(x, y) = 2n \operatorname{Trace}(xy)\).
\end{quote}

\end{problem}

\begin{solution}

We have the following standard basis:
\begin{align*}
x_1=\left[\begin{array}{ccc}
\cdot & 1 & \cdot \\
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot
\end{array}\right] 
\quad &x_2=\left[\begin{array}{ccc}
\cdot & \cdot & 1 \\
\cdot & \cdot & \cdot \\
\cdot & \cdot & .
\end{array}\right] 
&x_3=\left[\begin{array}{ccc}
\cdot & \cdot & \cdot \\
\cdot & \cdot & 1 \\
\cdot & \cdot & \cdot
\end{array}\right]\\
h_1=\left[\begin{array}{ccc}
1 & \cdot & \cdot \\
\cdot & -1 & \cdot \\
\cdot & \cdot & \cdot
\end{array}\right] 
\quad &h_2=\left[\begin{array}{ccc}
\cdot & \cdot & \cdot \\
\cdot & 1 & \cdot \\
\cdot & \cdot & -1
\end{array}\right] &\\
y_1=\left[\begin{array}{ccc}
\cdot & \cdot & \cdot \\
1 & \cdot & \cdot \\
\cdot & \cdot & \cdot
\end{array}\right] 
\quad &y_2=\left[\begin{array}{lll}
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot \\
1 & \cdot & \cdot
\end{array}\right] 
&y_3=\left[\begin{array}{ccc}
\cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot \\
\cdot & 1 & \cdot
\end{array}\right]
.\end{align*}
For notational convenience, let \(\left\{{v_1,\cdots, v_8}\right\}\)
denote this ordered basis.

Direct computations show

\begin{itemize}
\tightlist
\item
  \([x_1 v_1] = [x_1 x_1] = 0\)
\item
  \([x_1 v_2] = [x_1 x_2] = 0\)
\item
  \([x_1 v_3] = [x_1 x_3] = e_{13} = x_2 = v_2\)
\item
  \([x_1 v_4] = [x_1 h_1] = -2e_{12} = -2x_2 = -2 v_2\)
\item
  \([x_1 v_5] = [x_1 h_2] = e_{12} = x_1 = v_1\)
\item
  \([x_1 v_6] = [x_1 y_1] = e_{11} - e_{22} = h_1 = v_4\)
\item
  \([x_1 v_7] = [x_1 y_2] = -e_{31} = -y_2 = v_6\)
\item
  \([x_1 v_8] = [x_1 y_3] = 0\)
\end{itemize}

Let \(E_{ij}\) denote the elementary \(8\times 8\) matrices with a 1 in
the \((i, j)\) position. We then have, for example,
\begin{align*}
 { \operatorname{ad}}_{x_1} 
&=  0 + 0 + E_{2,3} -2 E_{2, 4} + E_{1, 5} + E_{4, 6} + E_{6, 7} + 0 \\
&=
\left(\begin{array}{rrrrrrr}
\cdot & \cdot & \cdot & \cdot & 1 & \cdot & \cdot \\
\cdot & \cdot & 1 & -2 & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & 1 & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 1 \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot
\end{array}\right)
.\end{align*}

The remaining computations can be readily automated on a computer,
yielding the following matrices for the remaining
\({ \operatorname{ad}}_{v_i}\):

\begin{itemize}
\item
  \({ \operatorname{ad}}_{x_1} = 0 + 0 + E_{2,3} -2 E_{1, 4} + E_{1, 5} + E_{4, 6} + E_{8, 7} + 0\)
\item
  \({ \operatorname{ad}}_{x_2} = 0 + 0 + 0 -E_{2, 4} -E_{2, 5} - E_{3, 6} + (E_{4, 7} +E_{5, 7}) + E_{1, 8}\)
\item
  \({ \operatorname{ad}}_{x_3} = -E_{2,1} + 0 + 0 + E_{3, 4} -2 E_{3, 5} + 0 + E_{6, 7} + E_{5, 8}\)
\item
  \({ \operatorname{ad}}_{h_1} = 2E_{1,1} + E_{2, 2} - E_{3, 3} + 0 + 0 -2 E_{6, 6} - E_{7,7} + E_{8, 8}\)
\item
  \({ \operatorname{ad}}_{h_2} = -E_{1, 1} + E_{2,2} +2 E_{3,3} + 0 + 0 + E_{6,6} - E_{7,7} -2 E_{8,8}\)
\item
  \({ \operatorname{ad}}_{y_1} = -E_{4, 1} + E_{3, 2} + 0 + 2E_{6, 4} - E_{6, 5} + 0 + 0 - E_{7, 8}\)
\item
  \({ \operatorname{ad}}_{y_2} = E_{8,1} - (E_{4, 2} + E_{5, 2}) - E_{6, 3} + E_{7, 4} + E_{7, 5} +0 +0 + 0\)
\item
  \({ \operatorname{ad}}_{y_3} = 0 - E_{1, 2} - E_{5, 3} - E_{8, 4} +2 E_{8, 5} + E_{7, 6} + 0 + 0\)
\end{itemize}

Now forming the matrix
\((\beta)_{ij} \coloneqq\operatorname{Trace}( { \operatorname{ad}}_{v_i} { \operatorname{ad}}_{v_j})\)
yields
\begin{align*}
\beta = 
\left(\begin{array}{rrrrrrrr}
\cdot & \cdot & \cdot & \cdot & \cdot & 2 & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 6 & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 6 \\
\cdot & \cdot & \cdot & 12 & -6 & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & -6 & 12 & \cdot & \cdot & \cdot \\
2 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & 6 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & 6 & \cdot & \cdot & \cdot & \cdot & \cdot
\end{array}\right)
,\end{align*}
whence
\(\operatorname{det}(\beta) = (2\cdot 6\cdot 6)^2(12^2-36) = - 2^8 3^7\).

\end{solution}

\hypertarget{section-6}{%
\subsection{Section 6}\label{section-6}}

\begin{problem}[6.1]

Using the standard basis for \(L=\mathfrak{s l}_2({ \mathbf{F} })\),
write down the Casimir element of the adjoint representation of \(L\)
\emph{(cf.~Exercise 5.5)}. Do the same thing for the usual
(3-dimensional) representation of \(\mathfrak{s l}_3({ \mathbf{F} })\),
first computing dual bases relative to the trace form.

\end{problem}

\begin{solution}

A computation shows that in the basis
\(\left\{{e_i}\right\} \coloneqq\left\{{x,h,y}\right\}\), the Killing
form is represented by
\begin{align*}
\beta = {
\begin{bmatrix}
  {0} & {0} & {4}
\\
  {0} & {8} & {0}
\\
  {4} & {0} & {0}
\end{bmatrix}
} \implies \beta^{-T} = {
\begin{bmatrix}
  {0} & {0} & {1\over 4}
\\
  {0} & {1\over 8} & {0}
\\
  {1\over 4} & {0} & {0}
\end{bmatrix}
}
,\end{align*}
yielding the dual basis \(\left\{{e_i {}^{ \vee }}\right\}\) read from
the columns of \(\beta^{-T}\):

\begin{itemize}
\tightlist
\item
  \(x {}^{ \vee }= {1\over 4}y\),
\item
  \(h {}^{ \vee }= {1\over 8} h\),
\item
  \(y {}^{ \vee }= {1\over 4}x\).
\end{itemize}

Thus letting \(\phi = { \operatorname{ad}}\), we have
\begin{align*}
c_\phi 
&= \sum \phi(e_i)\phi(e_i {}^{ \vee }) \\
&=  { \operatorname{ad}}(x)  { \operatorname{ad}}(x {}^{ \vee }) +  { \operatorname{ad}}(h)  { \operatorname{ad}}(h {}^{ \vee }) +  { \operatorname{ad}}(y) { \operatorname{ad}}(y {}^{ \vee }) \\
&=  { \operatorname{ad}}(x)  { \operatorname{ad}}(y/4) +  { \operatorname{ad}}(h)  { \operatorname{ad}}(h/8) +  { \operatorname{ad}}(y) { \operatorname{ad}}(x/4) \\
&= {1\over 4}  { \operatorname{ad}}_x  { \operatorname{ad}}_y + {1\over 8}  { \operatorname{ad}}_h^2 + {1\over 4}  { \operatorname{ad}}_y  { \operatorname{ad}}_x
.\end{align*}

For \({\mathfrak{sl}}_3\), first take the ordered basis
\(\left\{{v_1,\cdots, v_8}\right\} = \left\{{x_1, x_2, x_3, h_1, h_2, y_1, y_2, y_3}\right\}\)
as in the previous problem. So we form the matrix
\((\beta)_{ij} \coloneqq\operatorname{Trace}(v_i v_j)\) by computing
various products and traces on a computer to obtain
\begin{align*}
\beta = 
\left(\begin{array}{rrrrrrrr}                                                                                                                              
\cdot & \cdot & \cdot & \cdot & \cdot & 1 & \cdot & \cdot \\                                                                                                                           
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 1 & \cdot \\                                                                                                                           
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 1 \\                                                                                                                           
\cdot & \cdot & \cdot & 2 & -1 & \cdot & \cdot & \cdot \\                                                                                                                          
\cdot & \cdot & \cdot & -1 & 2 & \cdot & \cdot & \cdot \\                                                                                                                          
1 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\                                                                                                                           
\cdot & 1 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\                                                                                                                           
\cdot & \cdot & 1 & \cdot & \cdot & \cdot & \cdot & \cdot                                                                                                                              
\end{array}\right) \implies 
\beta^{-T}
=
\left(\begin{array}{rrrrrrrr}
\cdot & \cdot & \cdot & \cdot & \cdot & 1 & \cdot & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 1 & \cdot \\
\cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & 1 \\
\cdot & \cdot & \cdot & \frac{2}{3} & \frac{1}{3} & \cdot & \cdot & \cdot \\
\cdot & \cdot & \cdot & \frac{1}{3} & \frac{2}{3} & \cdot & \cdot & \cdot \\
1 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & 1 & \cdot & \cdot & \cdot & \cdot & \cdot & \cdot \\
\cdot & \cdot & 1 & \cdot & \cdot & \cdot & \cdot & \cdot
\end{array}\right)
,\end{align*}
which yields the dual basis

\begin{itemize}
\tightlist
\item
  \(x_i {}^{ \vee }= y_i\)
\item
  \(h_1 {}^{ \vee }= {2\over 3}h_1 + {1\over 3}h_2\)
\item
  \(h_2 {}^{ \vee }= {1\over 3}h_1 + {2\over 3}h_2\)
\item
  \(y_i {}^{ \vee }= x_i\)
\end{itemize}

We can thus compute the Casimir element of the standard representation
\(\phi\) on a computer as
\begin{align*}
c_{\phi} 
&= \sum_i \phi(x_i)\phi(x_i {}^{ \vee }) + \phi(h_1)\phi(h_1 {}^{ \vee }) + \phi(h_2)\phi(h_2 {}^{ \vee }) + \sum_i \phi(y_i)\phi(y_i {}^{ \vee }) \\
&= \sum_i x_i y_i  + h_1 h_1 {}^{ \vee }+ h_2 h_2 {}^{ \vee }+ \sum_i y_i x_i \\
&= \sum_i \qty{x_i y_i + y_i x_i}  \\
&= {8\over 3}I
.\end{align*}

\end{solution}

\begin{problem}[6.3]

If \(L\) is solvable, every irreducible representation of \(L\) is one
dimensional.

\end{problem}

\begin{solution}

Let \(\phi: L\to V\) be an irreducible representation of \(L\). By Lie's
theorem. \(L\) stabilizes a flag in \(V\), say
\(F^\bullet = F^1 \subset \cdots F^n = V\) where
\(F^k = \left\langle{v_1,\cdots, v_k}\right\rangle\) for some basis
\(\left\{{v_i}\right\}_{i\leq n}\). Since \(\phi\) is irreducible, the
only \(L{\hbox{-}}\)invariant subspaces of \(V\) are \(0\) and \(V\)
itself. However, each \(F^k\) is an \(L{\hbox{-}}\)invariant subspace,
which forces \(n=1\) and \(F^1 = V\). Thus \(V\) is 1-dimensional.

\end{solution}

\begin{problem}[6.5]

A Lie algebra \(L\) for which \(\operatorname{Rad} L=Z(L)\) is called
reductive.\footnote{Examples: \(L\) abelian, \(L\) semisimple,
  \(L=\mathfrak{g l}_n({ \mathbf{F} })\).}

\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\tightlist
\item
  If \(L\) is reductive, then \(L\) is a completely reducible ad
  \(L\)-module. \footnote{If ad \(L \neq 0\), use Weyl's Theorem.} In
  particular, \(L\) is the direct sum of \(Z(L)\) and \([L L]\), with
  \([L L]\) semisimple.
\item
  If \(L\) is a classical linear Lie algebra (1.2), then \(L\) is
  semisimple. \emph{(Cf. Exercise 1.9.)}
\item
  If \(L\) is a completely reducible
  \({ \operatorname{ad}}(L){\hbox{-}}\)module, then \(L\) is reductive.
\item
  If \(L\) is reductive, then all finite dimensional representations of
  \(L\) in which \(Z(L)\) is represented by semisimple endomorphisms are
  completely reducible.
\end{enumerate}

\end{problem}

\begin{solution}

\textbf{Part 1}: If \({ \operatorname{ad}}(L) \neq 0\), as hinted, we
can attempt to apply Weyl's theorem to the representation
\(\phi: { \operatorname{ad}}(L)\to {\mathfrak{gl}}(L)\): if we can show
\({ \operatorname{ad}}(L)\) is semisimple, then \(\phi\) (and thus
\(L\)) will be a completely reducible
\({ \operatorname{ad}}(L){\hbox{-}}\)module. Assume \(L\) is reductive,
so \(\ker( { \operatorname{ad}}) = Z(L) = \mathop{\mathrm{Rad}}(L)\),
and by the first isomorphism theorem
\({ \operatorname{ad}}(L) \cong L/\mathop{\mathrm{Rad}}(L)\). We can now
use the fact stated in Humphreys on page 11 that for an arbitrary Lie
algebra \(L\), the quotient \(L/\mathop{\mathrm{Rad}}(L)\) is semisimple
-- this follows from the fact that
\(\mathop{\mathrm{Rad}}(L/\mathop{\mathrm{Rad}}(L)) = 0\), since the
maximal solvable ideal in the quotient would need to be a maximal proper
ideal in \(L\) containing \(\mathop{\mathrm{Rad}}(L)\), which won't
exist by maximality of \(\mathop{\mathrm{Rad}}(L)\). Thus
\({ \operatorname{ad}}(L)\) is semisimple, and Weyl's theorem implies it
is completely reducible.

To show that \(L = Z(L) \oplus [LL]\), we first show that it decomposes
as a sum \(L = Z(L) + [LL]\), and then that the intersection is empty so
the sum is direct. We recall that a Lie algebra is semisimple if and
only if it has no nonzero abelian ideals. Since \(L/Z(L)\) is
semisimple, we have \([L/Z(L), L/Z(L)] = L/Z(L)\) since it would
otherwise be a nonzero abelian ideal in \(L/Z(L)\). We can separately
identify \([L/Z(L), L/Z(L)] \cong [LL]/Z(L)\), since the latter is also
semisimple and the former is an abelian ideal in it. Combining these, we
have \([LL]/Z(L) \cong L/Z(L) \cong { \operatorname{ad}}(L)\), and so we
have an extension
\begin{align*}
0 \to Z(L) \to L \to [LL] \to 0
.\end{align*}
Since this sequence splits at the level of vector spaces,
\(L = Z(L) + [LL]\) as an \({ \operatorname{ad}}(L){\hbox{-}}\)module,
although the sum need not be direct. To show that it is, note that
\(Z(L) \leq L\) is an \({ \operatorname{ad}}(L){\hbox{-}}\)invariant
submodule, and by complete reducibility has an
\({ \operatorname{ad}}(L){\hbox{-}}\)invariant complement \(W\). We can
thus write \(L = W \oplus Z(L)\), and moreover \([LL] \subseteq W\), and
so we must have \(W = [LL]\) and \(L = [LL] \oplus Z(L)\).

Finally, to see that \([LL]\) is semisimple, note that the above
decomposition allows us to write \(L/Z(L) \cong [LL]\), and
\(\mathop{\mathrm{Rad}}(L/Z(L)) = \mathop{\mathrm{Rad}}(L/\mathop{\mathrm{Rad}}(L)) = 0\)
so \(\mathop{\mathrm{Rad}}([LL]) = 0\).

\textbf{Part 2}: Omitted for time.

\textbf{Part 3}: Omitted for time.

\textbf{Part 4}: Omitted for time.

\end{solution}

\begin{problem}[6.6]

Let \(L\) be a simple Lie algebra. Let \(\beta(x, y)\) and
\(\gamma(x, y)\) be two symmetric associative bilinear forms on \(L\).
If \(\beta, \gamma\) are nondegenerate, prove that \(\beta\) and
\(\gamma\) are proportional.

\begin{quote}
Hint: Use Schur's Lemma.
\end{quote}

\end{problem}

\begin{solution}

The strategy will be to define an irreducible \(L{\hbox{-}}\)module
\(V\) and use the two bilinear forms to produce an element of
\({ \operatorname{End} }_L(V)\), which will be 1-dimensional by Schur's
lemma.

The representation we'll take will be
\(\phi \coloneqq { \operatorname{ad}}: L\to {\mathfrak{gl}}(L)\), and
since \(L\) is simple, \(\ker { \operatorname{ad}}= 0\) since otherwise
it would yield a nontrivial ideal of \(L\). Since this is a faithful
representation, we will identify \(L\) with its image
\(V \coloneqq { \operatorname{ad}}(L) \subseteq {\mathfrak{gl}}(L)\) and
regard \(V\) as an \(L{\hbox{-}}\)module.

As a matter of notation, let \(\beta_x(y) \coloneqq\beta(x, y)\) and
similarly \(\gamma_x(y) \coloneqq\gamma(x, y)\), so that
\(\beta_x, \gamma_x\) can be regarded as linear functionals on \(V\) and
thus elements of \(V {}^{ \vee }\). This gives an
\({ \mathbf{F} }{\hbox{-}}\)linear map
\begin{align*}
\Phi_1: V &\to V {}^{ \vee }\\
x &\mapsto \beta_x
,\end{align*}
which we claim is an \(L{\hbox{-}}\)module morphism.

Assuming this for the moment, note that by the general theory of
bilinear forms on vector spaces, since \(\beta\) and \(\gamma\) are
nondegenerate, the assignments \(x\mapsto \beta_x\) and
\(x\mapsto \gamma_x\) induce vector space isomorphisms
\(V { \, \xrightarrow{\sim}\, }V {}^{ \vee }\). Accordingly, for any
linear functional \(f\in V {}^{ \vee }\), there is a unique element
\(z(f) \in V\) such that \(f(v) = \gamma(z(f), v)\). So define a map
using the representing element for \(\gamma\):
\begin{align*}
\Phi_2: V {}^{ \vee }&\to V \\
f &\mapsto z(f)
,\end{align*}
which we claim is also an \(L{\hbox{-}}\)module morphism.

We can now define their composite
\begin{align*}
\Phi \coloneqq\Phi_2 \circ \Phi_1: V &\to V \\
x &\mapsto z(\beta_x)
,\end{align*}
which sends an element \(x\in V\) to the element
\(z = z(\beta_x) \in V\) such that \(\beta_x({-}) = \gamma_z({-})\) as
functionals. An additional claim is that \(\Phi\) commutes with the
image
\(V \coloneqq { \operatorname{ad}}(L) \subseteq {\mathfrak{gl}}(L)\).
Given this, by Schur's lemma we have
\(\Phi\in { \operatorname{End} }_L(V) = { \mathbf{F} }\) (where we've
used that a compositions of morphisms is again a morphism) and so
\(\Phi = \lambda \operatorname{id}_L\) for some scalar
\(\lambda\in { \mathbf{F} }\).

To see why this implies the result, we have equalities of functionals
\begin{align*}
\beta(x, {-}) 
&= \beta_x({-}) \\
&= \gamma_{z(\beta_x) }({-}) \\
&= \gamma( z(\beta_x), {-})\\
&= \gamma( \Phi(x), {-}) \\
&= \gamma(\lambda x, {-}) \\
&= \lambda\gamma(x, {-})
,\end{align*}

and since this holds for all \(x\) we have
\(\beta({-}, {-}) = \lambda \gamma({-}, {-})\) as desired.

\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}

\begin{claim}

\(\Phi_1\) is an \(L{\hbox{-}}\)module morphism.

\end{claim}

\begin{proof}[?]

We recall that a morphism of \(L{\hbox{-}}\)modules \(\phi: V\to W\) is
an \({ \mathbf{F} }{\hbox{-}}\)linear map satisfying
\begin{align*}
\phi(\ell .\mathbf{x}) = \ell.\phi(\mathbf{x}) \qquad\forall \ell\in L,\,\forall \mathbf{x}\in V
.\end{align*}
In our case, the left-hand side is
\begin{align*}
\Phi_1(\ell . \mathbf{x}) \coloneqq\Phi_1(  { \operatorname{ad}}_\ell(\mathbf{x}) ) = \Phi_1([\ell,\mathbf{x}]) = \beta_{[\ell, \mathbf{x}]}
= \beta( [\ell, \mathbf{x}], {-})
.\end{align*}
and the right-hand side is
\begin{align*}
\ell.\Phi_1(\mathbf{x}) \coloneqq\ell.\beta_{\mathbf{x}} 
\coloneqq(y\mapsto -\beta_{\mathbf{x}}( \ell. y))
\coloneqq(\mathbf{y}\mapsto -\beta_{\mathbf{x}}( [\ell, \mathbf{y}] )) = -\beta(\mathbf{x}, [\ell, {-}])
.\end{align*}
By anticommutativity of the bracket, along with
\({ \mathbf{F} }{\hbox{-}}\)linearity and associativity of \(\beta\), we
have
\begin{align*}
\beta([\ell, \mathbf{x}], \mathbf{y}) = -\beta([\mathbf{x}, \ell], \mathbf{y}) = -\beta(\mathbf{x}, [\ell, \mathbf{y}]) \qquad \forall \mathbf{y}\in V
\end{align*}
and so the above two sides do indeed coincide.

\end{proof}

\begin{claim}

\(\Phi_2\) is an \(L{\hbox{-}}\)module morphism.

\end{claim}

\begin{proof}[?]

Omitted for time, proceeds similarly.

\end{proof}

\begin{claim}

\(\Phi\) commutes with \({ \operatorname{ad}}(L)\).

\end{claim}

\begin{proof}[?]

Letting \(x\in L\), we want to show that
\(\Phi \circ { \operatorname{ad}}_x = { \operatorname{ad}}_x \circ \Phi \in {\mathfrak{gl}}(L)\),
i.e.~that these two endomorphisms of \(L\) commute. Fixing
\(\ell\in L\), the LHS expands to
\begin{align*}
\Phi( { \operatorname{ad}}_x(\ell)) = z(\beta_{ { \operatorname{ad}}_x(\ell) }) = z(\beta_{[x\ell]})
,\end{align*}
while the RHS is
\begin{align*}
 { \operatorname{ad}}_x(\Phi(\ell)) =  { \operatorname{ad}}_x(z(\beta_\ell)) = [x, z(\beta_\ell)]
.\end{align*}
Recalling that \(\Phi(t) = z(\beta_t)\) is defined to be the unique
element \(t\in L\) such that
\(\beta(t, {-}) = \gamma(z(\beta_t), {-})\), for the above two to be
equal it suffices to show that
\begin{align*}
\beta([x, \ell], {-}) = \gamma( [x, z(\beta_\ell)], {-})
\end{align*}
as linear functionals. Starting with the RHS of this expression, we have
\begin{align*}
\gamma( [ x, z(\beta_\ell) ], {-})
&= -\gamma( [z(\beta_\ell), x], {-}) \quad\text{by antisymmetry}\\
&= -\gamma(z(\beta_\ell), [x, {-}]) \quad\text{by associativity of }\gamma \\
&= -\beta(\ell, [x, {-}]) \quad\text{by definition of } z(\beta_\ell) \\
&= -\beta([\ell, x], {-}) \\
&= \beta([x, \ell], {-})
.\end{align*}

\end{proof}

\end{solution}

\begin{problem}[6.7]

It will be seen later on that \(\mathfrak{sl}_n({ \mathbf{F} })\) is
actually simple. Assuming this and using Exercise 6, prove that the
Killing form \(\kappa\) on \(\mathfrak{s l}_n({ \mathbf{F} })\) is
related to the ordinary trace form by
\(\kappa(x, y)=2 n \operatorname{Tr}(x y)\).

\end{problem}

\begin{solution}

By the previous exercise, the trace pairing
\((x,y)\mapsto \operatorname{Trace}(xy)\) is related to the Killing form
by \(\kappa(x,y) = \lambda \operatorname{Trace}(x,y)\) for some
\(\lambda\) -- here we've used the fact that since
\({\mathfrak{sl}}_n({ \mathbf{F} })\) is simple,
\(\mathop{\mathrm{Rad}}(\operatorname{Trace}) = 0\) and thus the trace
pairing is nodegenerate. Since the scalar only depends on the bilinear
forms and not on any particular inputs, it suffices to compute it for
any pair \((x, y)\), and in fact we can take \(x=y\). For
\({\mathfrak{sl}}_n\), we can take advantage of the fact that in the
standard basis, \({ \operatorname{ad}}_{h_i}\) will be diagonal for any
standard generator \(h_i\in {\mathfrak{h}}\), making
\(\operatorname{Trace}( { \operatorname{ad}}_{h_i}^2)\) easier to
compute for general \(n\).

Take the standard \(h_{1} \coloneqq e_{11} - e_{22}\), and consider the
matrix of \({ \operatorname{ad}}_{h_1}\) in the ordered basis
\(\left\{{x_1,\cdots, x_k, h_1,\cdots, h_{n-1}, y_1,\cdots, y_k }\right\}\)
which has \(k + (n-1) + k = n^2-1\) elements where \(k= (n^2-n)/2\).
We'll first compute the Killing form with respect to this basis. In
order to compute the various \([h_1, v_i]\), we recall the formula
\([e_{ij}, e_{kl}] = \delta_{jk} e_{il} - \delta_{li} e_{kj}\). Applying
this to \(h_{1}\) yields
\begin{align*}
[h_{1}, e_{ij}] = [e_{11} - e_{22}, e_{ij}] = [e_{11}, e_{ij}] - [e_{22}, e_{ij}] = (\delta_{1i} e_{2j} - \delta_{1j} e_{i1}) - (\delta_{2i} e_{2j} - \delta_{2j} e_{i2})
.\end{align*}
We proceed to check all of the possibilities for the results as \(i, j\)
vary with \(i\neq j\) using the following schematic:
\begin{align*}
\left[\begin{array}{ c | c | c  }     \cdot & a & R_1 \, \cdots  \\    \hline  b & \cdot  & R_2\, \cdots \\  \hline \overset{C_1}{\vdots} & \overset{C_2}{\vdots} & M  \end{array}\right]
.\end{align*}

The possible cases are:

\begin{itemize}
\tightlist
\item
  \(a: i=1, j=2 \implies [h_1, e_{ij}] = 2e_{12}\), covering 1 case.
\item
  \(b: i=2, j=1 \implies [h_1, e_{ij}] = -2 e_{21}\) covering 1 case.
\item
  \(R_1: i=1, j>2 \implies [h_1, e_{ij}] = e_{1j}\) covering \(n-2\)
  cases.
\item
  \(R_2: i=2, j>2 \implies [h_1, e_{ij}] = e_{2j}\) covering \(n-2\)
  cases.
\item
  \(C_1: j=1, i>2 \implies [h_1, e_{ij}] = e_{i1}\) covering \(n-2\)
  cases.
\item
  \(C_2: j=2, i>2 \implies [h_1, e_{ij}] = e_{i2}\) covering \(n-2\)
  cases.
\item
  \(M: i, j > 2 \implies [h_1, e_{ij}] = 0\) covering the remaining
  cases.
\end{itemize}

Thus the matrix of \({ \operatorname{ad}}_{h_1}\) has \(4(n-2)\) ones
and \(2, -2\) on the diagonal, and \({ \operatorname{ad}}_{h_1}^2\) as
\(4(n-2)\) ones and \(4, 4\) on the diagonal, yielding
\begin{align*}
\operatorname{Trace}( { \operatorname{ad}}_{h_1}^2) = 4(n-2) + 2(4) = 4n
.\end{align*}
On the other hand, computing the standard trace form yields
\begin{align*}
\operatorname{Trace}(h_1^2) = \operatorname{Trace}(\operatorname{diag}(1,1,0,0,\cdots)) = 2
,\end{align*}
and so
\begin{align*}
\operatorname{Trace}( { \operatorname{ad}}_{h_1}^2) = 4n = 2n \cdot 2 = 2n\cdot \operatorname{Trace}(h_1^2) \implies \lambda = 2n
.\end{align*}

\end{solution}

\hypertarget{problem-set-3-1}{%
\section{Problem Set 3}\label{problem-set-3-1}}

\hypertarget{section-7}{%
\subsection{Section 7}\label{section-7}}

\begin{problem}[Humphreys 7.2]

\(M=\mathfrak{sl}(3, { \mathbf{F} })\) contains a copy of
\(L\coloneqq{\mathfrak{sl}}(2, { \mathbf{F} })\) in its upper left-hand
\(2 \times 2\) position. Write \(M\) as direct sum of irreducible
\(L\)-submodules (\(M\) viewed as \(L\) module via the adjoint
representation):
\begin{align*}
V(0) \oplus V(1) \oplus V(1) \oplus V(2)
.\end{align*}

\end{problem}

\begin{solution}

Noting that

\begin{itemize}
\tightlist
\item
  \(\dim V(m) = m +1\)
\item
  \(\dim {\mathfrak{sl}}_3({ \mathbf{F} }) = 8\)
\item
  \(\dim(V(0) \oplus V(1) \oplus V(1) \oplus V(2)) = 1+2+2+3 = 8\),
\end{itemize}

it suffices to find distinct highest weight elements of weights
\(0,1,1,2\) and take the irreducible submodules they generate. As long
as the spanning vectors coming from the various \(V(n)\) are all
distinct, they will span \(M\) as a vector space by the above dimension
count and individually span the desired submodules.

Taking the standard basis
\(\left\{{v_1,\cdots, v_8}\right\} \coloneqq\left\{{x_1, x_2, x_3, h_1, h_2, y_1, y_2, y_3}\right\}\)
for \({\mathfrak{sl}}_3({ \mathbf{F} })\) with \(y_i = x_i^t\), note
that the image of the inclusion
\({\mathfrak{sl}}_2({ \mathbf{F} }) \hookrightarrow{\mathfrak{sl}}_3({ \mathbf{F} })\)
can be identified with the span of
\(\left\{{w_1,w_2,w_3}\right\} \coloneqq\left\{{x_1, h_1, y_1}\right\}\)
and it suffices to consider how these \(3\times 3\) matrices act.

Since any highest weight vector must be annihilated by the
\(x_1{\hbox{-}}\)action, to find potential highest weight vectors one
can compute the matrix of \({ \operatorname{ad}}_{x_1}\) in the above
basis and look for zero columns:
\begin{align*}
 { \operatorname{ad}}_{x_1} = 
\left(\begin{array}{rrrrrrrr}
0 & 0 & 0 & -2 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0
\end{array}\right)
.\end{align*}
Thus \(\left\{{v_1 = x_1, v_2 = x_2, v_8 = y_3}\right\}\) are the only
options for highest weight vectors of nonzero weight, since
\({ \operatorname{ad}}_{x_1}\) acts nontrivially on the remaining basis
elements.

Computing the matrix of \({ \operatorname{ad}}{h_1}\), one can read off
the weights of each:
\begin{align*}
 { \operatorname{ad}}_{h_1} = 
\left(\begin{array}{rrrrrrrr}
2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & -2 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1
\end{array}\right)
.\end{align*}

Thus the candidates for highest-weight vectors are:

\begin{itemize}
\tightlist
\item
  \(x_1\) for \(V(2)\),
\item
  \(x_2\) for one copy of \(V(1)\),
\item
  \(y_3\) for the other copy of \(V(1)\),
\item
  \(h_1\) or \(h_2\) for \(V(0)\).
\end{itemize}

We can now repeatedly apply the \(y_1{\hbox{-}}\)action to obtain the
other vectors in each irreducible module.

For \(V(2)\):

\begin{itemize}
\tightlist
\item
  \(v_0 = x_1\) which has weight 2,
\item
  \(v_1 = y_1.v_0 = [y_1, x_1] = -h_1\) which has weight 0,
\item
  \(v_2 = {1\over 2} y_1^2.v_0 = {1\over 2} [y_1, [y_1, x_1]] = -y_1\)
  which has weight -2.
\end{itemize}

Since we see \(h_1\) appears in this submodule, we see that we should
later take \(h_2\) as the maximal vector for \(V(0)\). Continuing with
\(V(1)\):

\begin{itemize}
\tightlist
\item
  \(v_0 = x_2\) which has weight 1,
\item
  \(v_1 = y_1.v_0 = [y_1, x_2] = x_3\) which has weight -1.
\end{itemize}

For the other \(V(1)\):

\begin{itemize}
\tightlist
\item
  \(v_0 = y_3\) with weight 1,
\item
  \(v_1 = -y_2\) with weight -1.
\end{itemize}

For \(V(0)\):

\begin{itemize}
\tightlist
\item
  \(v_0 = h_2\).
\end{itemize}

We see that we get the entire basis of
\({\mathfrak{sl}}_3({ \mathbf{F} })\) this way with no redundancy,
yielding the desired direct product decomposition.

\end{solution}

\begin{problem}[Humphreys 7.5]

Suppose char
\({ \mathbf{F} }=p>0, L=\mathfrak{s l}(2, { \mathbf{F} })\). Prove that
the representation \(V(m)\) of \(L\) constructed as in Exercise 3 or 4
is irreducible so long as the highest weight \(m\) is strictly less than
\(p\), but reducible when \(m=p\).

\begin{quote}
Note: this corresponds to the formulas in lemma 7.2 parts (a) through
(c), or by letting \(L\curvearrowright{ \mathbf{F} }^2\) in the usual
way and extending \(L\curvearrowright{ \mathbf{F} }[x, y]\) by
derivations, so \(l.(fg) = (l.f)g + f(l.g)\) and taking the subspace of
homogeneous degree \(m\) polynomials
\(\left\langle{x^m, x^{m-1}y, \cdots, y^m}\right\rangle\) to get an
irreducible module of highest weight \(m\).
\end{quote}

\end{problem}

\begin{solution}

The representation \(V(m)\) in Lemma 7.2 is defined by the following
three equations, where \(v_0 \in V_{ m}\) is a highest weight vector and
\(v_k \coloneqq y^k v_0/k!\):

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(h\cdot v_i=(m-2 i) v_i\)
\item
  \(y \cdot v_i=(i+1) v_{i+1}\)
\item
  \(x \cdot v_i=(m-i+1) v_{i-1}\).
\end{enumerate}

Supposing \(m< p\), the vectors
\(\left\{{v_0, v_1,\cdots, v_m}\right\}\) still span an irreducible
\(L{\hbox{-}}\)module since it contains no nontrivial
\(L{\hbox{-}}\)submodules, just as in the characteristic zero case.

However, if \(m=p\), then note that
\(y.v_{m-1} = (m-1+1) v_m = 0 v_m = 0\) and consider the set
\(\left\{{v_0, \cdots, v_{m-1}}\right\}\). This spans an
\(m{\hbox{-}}\)dimensional subspace of \(V\), and the equations above
show it is invariant under the \(L{\hbox{-}}\)action, so it yields an
\(m{\hbox{-}}\)dimensional submodule of \(V(m)\). Since
\(\dim_{ \mathbf{F} }V(m) = m+1\), this is a nontrivial proper
submodule, so \(V(m)\) is reducible.

\end{solution}

\begin{problem}[Humphreys 7.6]

Decompose the tensor product of the two \(L\)-modules \(V(3), V(7)\)
into the sum of irreducible submodules:
\(V(4) \oplus V(6) \oplus V(8) \oplus V(10)\). Try to develop a general
formula for the decomposition of \(V(m) \otimes V(n)\).

\end{problem}

\begin{solution}

By a theorem from class, we know the weight space decomposition of any
\({\mathfrak{sl}}_2({\mathbf{C}}){\hbox{-}}\)module \(V\) takes the
following form:
\begin{align*}
V = V_{-m} \oplus V_{-m+2} \oplus \cdots \oplus V_{m-2} \oplus V_m
,\end{align*}
where \(m\) is a highest weight vector, and each weight space
\(V_{\mu}\) is 1-dimensional and occurs with multiplicity one. In
particular, since \(V(m)\) is a highest-weight module of highest weight
\(m\), we can write
\begin{align*}
V(3) &= \hspace{5.6em} V_{-3} \oplus V_{-1} \oplus V_1 \oplus V_3 \\
V(7) 
&= V_{-7} \oplus V_{-5} 
\oplus 
V_{-3} \oplus V_{-1} \oplus V_1 \oplus V_3 
\oplus 
V_{5} \oplus V_{7} 
,\end{align*}
and tensoring these together yields modules with weights between
\(-3 -7 = -10\) and \(3+7 = 10\):
\begin{align*}
V(3) \otimes V(7) 
&= 
V_{-10} 
\oplus V_{-8}{ {}^{ \scriptscriptstyle\oplus^{2} }  } \oplus 
V_{-6}{ {}^{ \scriptscriptstyle\oplus^{3} }  } \oplus V_{-4}{ {}^{ \scriptscriptstyle\oplus^{4} }  } 
\oplus V_{-2}{ {}^{ \scriptscriptstyle\oplus^{4} }  } \\
\qquad& \oplus V_0{ {}^{ \scriptscriptstyle\oplus^{4} }  } \\
\qquad& \oplus V_2{ {}^{ \scriptscriptstyle\oplus^{4} }  } \oplus V_4{ {}^{ \scriptscriptstyle\oplus^{4} }  } \oplus 
V_6{ {}^{ \scriptscriptstyle\oplus^{3} }  } \oplus V_8{ {}^{ \scriptscriptstyle\oplus^{2} }  } 
\oplus V_{10}
.\end{align*}
This can be more easily parsed by considering formal characters:
\begin{align*}
\operatorname{ch}(V(3)) &= e^{-3} + e^{-1} + e^{1} + e^3 =  \\ 
\operatorname{ch}(V(7)) &= e^{-7} + e^{-5} + e^{-3} + e^{-1} + e^1 + e^3 + e^5 + e^7 \\ 
\operatorname{ch}(V(3) \otimes V(7)) 
&= \operatorname{ch}(V(3))\cdot \operatorname{ch}(V(7)) \\ \\
&= (e^{-10} + e^{10}) 
+ 2(e^{-8} + e^{8}) 
+ 3( e^{-6} + e^6) \\
&\qquad + 4( e^{-4} + e^4) 
+ 4(e^{-2} + e^2) +4 
\\ \\
&= (e^{-10} + e^{10}) 
+ 2(e^{-8} + e^{8}) 
+ 3( e^{-6} + e^6) \\
&\qquad + 4\operatorname{ch}(V(4))
,\end{align*}
noting that
\(\operatorname{ch}(V(4)) = e^{-4} + e^{-2} + e^{2} + e^{4}\) and
collecting terms.

To see that \(V(3) \otimes V(7)\) decomposes as
\(V(4) \oplus V(6) \oplus V(8) \oplus V(10)\) one can check for equality
of characters to see that the various weight spaces and multiplicities
match up:
\begin{align*}
\operatorname{ch}(V(4) \oplus V(6) \oplus V(8) \oplus V(10))
&= \operatorname{ch}(V(4)) + \operatorname{ch}(V(6)) + \operatorname{ch}(V(8)) + \operatorname{ch}(V(10) \\ \\
&= 
\qty{e^{-4} + \cdots + e^4} +
\qty{e^{-6} + \cdots + e^6} \\
&\quad +\qty{e^{-8} + \cdots + e^8} +
\qty{e^{-10} + \cdots + e^{10}} \\ \\
&= 
2\operatorname{ch}(V(4))
+ (e^{-6} + e^6) \\
&\,\, + \operatorname{ch}(V(4)) + (e^{-6} + e^6) + (e^{-8} + e^8) \\
&\,\,+ \operatorname{ch}(V(4)) + (e^{-6} + e^6) + (e^{-8} + e^8) + (e^{-10} + e^{10})\\ \\
&= 4\operatorname{ch}(V(4)) + 3(e^{-6} + e^6) \\
&\,\, + 2(e^{-8} + e^8) + (e^{-10} + e^{10})
,\end{align*}
which is equal to \(\operatorname{ch}(V(3) \otimes V(7))\) from above.

More generally, for two such modules \(V, W\) we can write
\begin{align*}
V\otimes_{ \mathbf{F} }W = \bigoplus _{\lambda \in{\mathfrak{h}} {}^{ \vee }} \bigoplus _{\mu_1 + \mu_2 = \lambda} V_{\mu_1} \otimes_{ \mathbf{F} }W_{\mu_2}
,\end{align*}
where we've used the following observation about the weight of
\({\mathfrak{h}}\) acting on a tensor product of weight spaces:
supposing \(v\in V_{\mu_1}\) and \(w\in W_{\mu_2}\),
\begin{align*}
h.(v\otimes w) 
&= (hv)\otimes w + v\otimes(hw) \\
&= (\mu_1 v)\otimes w + v\otimes(\mu_2 w) \\
&= (\mu_1 v)\otimes w + (\mu_2 v)\otimes w \\
&= (\mu_1 + \mu_2)(v\otimes w)
,\end{align*}
and so \(v\otimes w \in V_{\mu_1 + \mu_2}\).

Taking \(V(m_1), V(m_2)\) with \(m_1 \geq m_2\) then yields a general
formula:
\begin{align*}
V(m_1) \otimes_{ \mathbf{F} }V(m_2) 
= \bigoplus _{n=-m_1-m_2}^{m_1+m_2} \bigoplus_{a + b = n} V_a \otimes_{ \mathbf{F} }V_b 
= \bigoplus_{n = m_1-m_2}^{m_2 + m_1} V(n)
.\end{align*}

\end{solution}

\hypertarget{section-8}{%
\subsection{Section 8}\label{section-8}}

\begin{problem}[Humphreys 8.9]

Prove that every three dimensional semisimple Lie algebra has the same
root system as \(\mathfrak{s l}(2, { \mathbf{F} })\), hence is
isomorphic to \(\mathfrak{s l}(2, { \mathbf{F} })\).

\end{problem}

\begin{solution}

There is a formula for the dimension of \(L\) in terms of the rank of
\(\Phi\) and its cardinality, which is more carefully explained in the
solution below for problem 8.10:
\begin{align*}
\dim {\mathfrak{g}}= \operatorname{rank}\Phi + {\sharp}\Phi
.\end{align*}
Thus if \(\dim L = 3\) then the only possibility is that
\(\operatorname{rank}\Phi = 1\) and \({\sharp}\Phi = 2\), using that
\(\operatorname{rank}\Phi \leq {\sharp}\Phi\) and that \({\sharp}\Phi\)
is always even since each \(\alpha\in \Phi\) can be paired with
\(-\alpha\in \Phi\). In particular, the root system \(\Phi\) of \(L\)
must have rank 1, and there is a unique root system of rank 1 (up to
equivalence) which corresponds to \(A_1\) and
\({\mathfrak{sl}}_2({ \mathbf{F} })\).

By the remark in Humphreys at the end of 8.5, there is a 1-to-1
correspondence between pairs \((L, H)\) with \(L\) a semisimple Lie
algebra and \(H\) a maximal toral subalgebra and pairs
\((\Phi, {\mathbb{E}})\) with \(\Phi\) a root system and
\({\mathbb{E}}\supseteq\Phi\) its associated Euclidean space. Using this
classification, we conclude that
\(L\cong {\mathfrak{sl}}_2({ \mathbf{F} })\).

\end{solution}

\begin{problem}[Humphreys 8.10]

Prove that no four, five or seven dimensional semisimple Lie algebras
exist.

\end{problem}

\begin{solution}

We can first write
\begin{align*}
{\mathfrak{g}}= {\mathfrak{n}}^- \oplus {\mathfrak{h}}\oplus {\mathfrak{n}}^+,\qquad
{\mathfrak{n}}^+ \coloneqq\bigoplus _{\alpha\in \Phi^+} {\mathfrak{g}}_{ \alpha},\quad 
{\mathfrak{n}}^- \coloneqq\bigoplus _{\alpha\in \Phi^+} {\mathfrak{g}}_{ -\alpha}
.\end{align*}
Writing
\(N\coloneqq{\mathfrak{n}}^+ \oplus {\mathfrak{n}}^- = \bigoplus _{\alpha\in \Phi} {\mathfrak{g}}_{\alpha}\),
we note that \(\dim_{ \mathbf{F} }{\mathfrak{g}}_{ \alpha} = 1\) for all
\(\alpha\in \Phi\). Thus \(\dim_{ \mathbf{F} }N = {\sharp}\Phi\) and
\begin{align*}
\dim_{ \mathbf{F} }{\mathfrak{g}}= \dim_{ \mathbf{F} }{\mathfrak{h}}+ {\sharp}\Phi
.\end{align*}
We can also use the fact that
\(\dim_{ \mathbf{F} }{\mathfrak{h}}= \operatorname{rank}\Phi \coloneqq\dim_{\mathbf{R}}{\mathbf{R}}\Phi\),
the dimension of the Euclidean space spanned by \(\Phi\), and so we have
a general formula
\begin{align*}
\dim_{ \mathbf{F} }{\mathfrak{g}}= \operatorname{rank}\Phi + {\sharp}\Phi
,\end{align*}
which we'll write as \(d=r+f\).

We can observe that \(f\geq 2r\) since if
\({\mathcal{B}}\coloneqq\left\{{ \alpha_1, \cdots, \alpha_r}\right\}\)
is a basis for \(\Phi\), no \(-\alpha_i\) is in \({\mathcal{B}}\) but
\(\left\{{\pm \alpha_1, \cdots, \pm \alpha_r}\right\} \subseteq \Phi\)
by the axiomatics of a root system. Thus
\begin{align*}
\dim_{ \mathbf{F} }{\mathfrak{g}}= r+f \geq r + 2r = 3r
.\end{align*}

We can now examine the cases for which \(d = r+f = 4,5,7\):

\begin{itemize}
\tightlist
\item
  \(r=1\): as shown in class, there is a unique root system \(A_1\) of
  rank 1 up to equivalence and satisfies \(f=2\) and thus \(d=3\), which
  is not a case we need to consider.
\item
  \(r=2\): this yields \(d \geq 3r = 6\), so this entirely rules out
  \(d=4,5\) as possibilities for a semisimple Lie algebra. Using that
  every \(\alpha\in \Phi\) is one of a pair
  \(+\alpha, -\alpha\in \Phi\), we in fact have that \(f\) is always
  even -- in other words, \(\Phi = \Phi^+ {\textstyle\coprod}\Phi^-\)
  with \({\sharp}\Phi^+ = {\sharp}\Phi^-\), so
  \(f \coloneqq{\sharp}\Phi = 2\cdot {\sharp}\Phi^+\). Thus
  \(d = r+f = 2 +f\) is even in this case, ruling out \(d=7\) when
  \(r=2\).
\item
  \(r\geq 3\): in this case we have \(d \geq 3r = 9\), ruling out
  \(d=7\) once and for all.
\end{itemize}

\end{solution}

\hypertarget{problem-set-5}{%
\section{Problem Set 5}\label{problem-set-5}}

\hypertarget{section-9}{%
\subsection{Section 9}\label{section-9}}

\begin{proposition}[Humphreys 9.2]

Prove that \(\Phi {}^{ \vee }\) is a root system in \(E\), whose Weyl
group is naturally isomorphic to \(\mathcal{W}\); show also that
\(\left\langle\alpha {}^{ \vee }, \beta {}^{ \vee }\right\rangle=\langle\beta, \alpha\rangle\),
and draw a picture of \(\Phi {}^{ \vee }\) in the cases
\(A_1, A_2, B_2, G_2\).

\end{proposition}

\begin{solution}

We recall and introduce some notation:
\begin{align*}
{\left\lVert { \alpha} \right\rVert}^2 &\coloneqq(\alpha, \alpha) \\ \\
{\left\langle { \beta},~{ \alpha} \right\rangle} &\coloneqq{2 (\beta, \alpha)\over {\left\lVert { \alpha} \right\rVert}^2} \\
&= {2 (\beta, \alpha) \over ( \alpha, \alpha)} \\ \\
s_\alpha( \beta) 
&= \beta - {2 (\beta, \alpha) \over {\left\lVert {\alpha} \right\rVert}^2 } \alpha \\
&= \beta - {2 (\beta, \alpha) \over (\alpha, \alpha) } \alpha \\ \\
\alpha {}^{ \vee }&\coloneqq{2 \over {\left\lVert {\alpha} \right\rVert}^2} \alpha = {2\over (\alpha, \alpha)} \alpha
.\end{align*}

\begin{claim}

\begin{align*}
{\left\langle {\alpha {}^{ \vee }},~{\beta {}^{ \vee }} \right\rangle} = {\left\langle {\beta },~{\alpha} \right\rangle}
.\end{align*}

\end{claim}

\begin{proof}[?]

This is a computation:
\begin{align*}
{\left\langle { \alpha {}^{ \vee }},~{ \beta {}^{ \vee }} \right\rangle} 
&= {2 (\alpha {}^{ \vee }, \beta {}^{ \vee }) \over {\left\lVert {\beta {}^{ \vee }} \right\rVert}^2 } \\
&= {2 (\alpha {}^{ \vee }, \beta {}^{ \vee }) \over (\beta {}^{ \vee }, \beta {}^{ \vee }) } \\
&= {2\qty{ {2\alpha\over {\left\lVert {\alpha} \right\rVert}^2}, {2 \beta\over {\left\lVert {\beta} \right\rVert}^2} } \over \qty{{2 \beta\over {\left\lVert {\beta} \right\rVert}^2}, {2 \beta\over {\left\lVert {\beta} \right\rVert}^2} } } \\
&= {2^3 {\left\lVert {\beta} \right\rVert}^4 (\alpha, \beta)\over 2^2 {\left\lVert {\alpha} \right\rVert}^2 {\left\lVert {\beta} \right\rVert}^2 (\beta, \beta)} \\
&= {2^3 {\left\lVert {\beta} \right\rVert}^4 (\alpha, \beta)\over 2^2 (\alpha, \alpha) {\left\lVert {\beta} \right\rVert}^2 {\left\lVert {\beta} \right\rVert}^2} \\
&= {2( \alpha, \beta) \over (\alpha, \alpha)} \\
&= {\left\langle {\beta},~{ \alpha} \right\rangle}
.\end{align*}

\end{proof}

\begin{claim}

\(\Phi {}^{ \vee }\) is a root system.

\end{claim}

\begin{proof}[?]

The axioms can be checked individually:

\begin{itemize}
\item
  \(R1\): there is a bijection of sets
  \begin{align*}
  \Phi & { \, \xrightarrow{\sim}\, }\Phi {}^{ \vee }\\
  \alpha &\mapsto \alpha {}^{ \vee }
  ,\end{align*}
  thus \({\sharp}\Phi {}^{ \vee }= {\sharp}\Phi < \infty\). To see that
  \({\mathbf{R}}\Phi {}^{ \vee }= {\mathbb{E}}\), for
  \(\mathbf{v}\in {\mathbb{E}}\), use the fact that
  \({\mathbf{R}}\Phi = {\mathbb{E}}\) to write
  \(\mathbf{v} = \sum_{ \alpha\in \Phi} c_\alpha \alpha\), then
  \begin{align*}
  \mathbf{v} 
  &= \sum_{ \alpha\in \Phi} c_ \alpha \alpha \\
  &= \sum_{ \alpha\in \Phi} c_{\alpha} {{\left\lVert { \alpha} \right\rVert}^2\over 2}\cdot {2\over {\left\lVert { \alpha} \right\rVert}^2} \alpha \\
  &\coloneqq\sum_{ \alpha\in \Phi} c_{\alpha} {{\left\lVert { \alpha} \right\rVert}^2\over 2} \alpha {}^{ \vee }\\
  &= \sum_{\alpha {}^{ \vee }\in \Phi {}^{ \vee }} d_{\alpha {}^{ \vee }} \alpha {}^{ \vee }, \qquad d_{ \alpha {}^{ \vee }} \coloneqq{1\over 2}c_ \alpha {\left\lVert { \alpha^2} \right\rVert}
  ,\end{align*}
  so \(\mathbf{v}\in {\mathbf{R}}\Phi {}^{ \vee }\). Finally,
  \(\mathbf{0}\not\in \Phi {}^{ \vee }\) since
  \({2\over (\alpha, \alpha)} \alpha\neq \mathbf{0}\) since
  \(\alpha \in \Phi\implies \alpha\neq \mathbf{0}\), and
  \(2/(\alpha, \alpha)\) is never zero.
\item
  \(R2\): It suffices to show that if
  \(\lambda \alpha {}^{ \vee }= \beta {}^{ \vee }\in \Phi {}^{ \vee }\)
  then \(\lambda = \pm 1\) and
  \(\beta {}^{ \vee }= \alpha {}^{ \vee }\). So suppose
  \(\lambda \alpha {}^{ \vee }= \beta {}^{ \vee }\), then
  \begin{align*}
  \lambda {2\over {\left\lVert { \alpha} \right\rVert}^2} \alpha = {2\over {\left\lVert { \beta} \right\rVert}^2} \beta \implies \beta = \lambda{{\left\lVert {\beta} \right\rVert}^2 \over {\left\lVert { \alpha} \right\rVert}^2} \alpha \coloneqq\lambda' \alpha
  ,\end{align*}
  and since \(\Phi\) satisfies \(R2\), we have \(\lambda' = \pm 1\) and
  \(\beta = \alpha\). But then
  \begin{align*}
  \pm 1 = \lambda {{\left\lVert { \beta} \right\rVert}^2 \over {\left\lVert { \alpha} \right\rVert}^2} = \lambda{{\left\lVert { \alpha} \right\rVert}^2 \over {\left\lVert { \alpha} \right\rVert}^2} = \lambda
  .\end{align*}
  Finally, if \(\alpha = \beta\) then
  \(\alpha {}^{ \vee }= \beta {}^{ \vee }\) since
  \(\Phi, \Phi {}^{ \vee }\) are in bijection.
\end{itemize}

\end{proof}

\begin{proof}[of R3 and R4]

Continuing:

\begin{itemize}
\item
  \(R3\): It suffices to show that if
  \(\alpha {}^{ \vee }, \beta {}^{ \vee }\in \Phi {}^{ \vee }\) then
  \(s_{\alpha {}^{ \vee }}(\beta {}^{ \vee }) = \gamma {}^{ \vee }\) for
  some \(\gamma {}^{ \vee }\in \Phi {}^{ \vee }\). This follows from a
  computation:
  \begin{align*}
  s_{ \alpha {}^{ \vee }}(\beta {}^{ \vee }) 
  &= \beta {}^{ \vee }- {\left\langle {\beta {}^{ \vee }},~{ \alpha {}^{ \vee }} \right\rangle} \alpha {}^{ \vee }\\
  &= \beta {}^{ \vee }- {\left\langle { \alpha},~{ \beta} \right\rangle} \alpha {}^{ \vee }\\
  &= {2\beta\over {\left\lVert { \beta} \right\rVert}^2 }- {\left\langle { \alpha},~{ \beta} \right\rangle} {2 \alpha\over {\left\lVert { \alpha} \right\rVert}^2 } \\
  &= {2\beta\over {\left\lVert { \beta} \right\rVert}^2 }- {2 (\alpha, \beta) \over (\beta, \beta) } {2 \alpha\over {\left\lVert { \alpha} \right\rVert}^2 } \\
  &= {2\beta\over {\left\lVert { \beta} \right\rVert}^2 }- {2 (\alpha, \beta) \over {\left\lVert {\beta} \right\rVert}^2 } {2 \alpha\over {\left\lVert { \alpha} \right\rVert}^2 } \\
  &= {2\over {\left\lVert {\beta} \right\rVert}^2} \qty{ \beta -  {2 (\alpha, \beta) \over {\left\lVert {\alpha} \right\rVert}^2 } \alpha } \\
  &= {2\over ( \beta, \beta)} \qty{ \beta -  {2 (\beta, \alpha) \over {\left\lVert {\alpha} \right\rVert}^2 } \alpha } \\
  &= {2\over ( \beta, \beta)} \sigma_{ \alpha}(\beta) \\
  &= {2\over ( \sigma_{ \alpha}(\beta), \sigma_{ \alpha}(\beta) )} \sigma_{ \alpha}(\beta) \\
  &\coloneqq(\sigma_ \alpha( \beta)) {}^{ \vee }
  ,\end{align*}
  where we've used that \(\sigma_{\alpha}\) is an isometry with respect
  to the symmetric bilinear form \(({-}, {-})\).
\item
  \(R4\): This follows directly from the formula proved in the claim at
  the beginning:
  \begin{align*}
  {\left\langle { \alpha {}^{ \vee }},~{\beta {}^{ \vee }} \right\rangle} = {\left\langle { \beta},~{ \alpha} \right\rangle}\in {\mathbf{Z}}
  ,\end{align*}
  since \(\alpha, \beta\in \Phi\) and \(\Phi\) satisfies \(R4\).
\end{itemize}

\end{proof}

\begin{claim}

There is an isomorphism of groups
\({\mathcal{W}}(\Phi) { \, \xrightarrow{\sim}\, }{\mathcal{W}}(\Phi {}^{ \vee })\).

\end{claim}

\begin{proof}[?]

There is a map of Weyl groups
\begin{align*}
\tilde \psi: {\mathcal{W}}(\Phi) & { \, \xrightarrow{\sim}\, }{\mathcal{W}}(\Phi {}^{ \vee }) \\
s_ \alpha &\mapsto s_{\alpha {}^{ \vee }}
,\end{align*}
which is clearly a bijection of sets with inverse
\(s_{\alpha {}^{ \vee }} \mapsto s_{ \alpha}\). Since it is also a group
morphism, this yields an isomorphism of groups.

\end{proof}

\begin{remark}

The following are pictures of \(\Phi {}^{ \vee }\) in the stated special
cases:

\begin{itemize}
\tightlist
\item
  \(A_1\): Writing
  \(\Phi(A_1) = \left\{{\alpha = e_1 - e_2, - \alpha = e_2 -e_1}\right\} \subseteq {\mathbf{R}}^2\),
  we have \((\alpha, \alpha) = \sqrt 2\) and thus
  \(\Phi(A_1) {}^{ \vee }= \left\{{\sqrt{2} \alpha, - \sqrt{2} \alpha}\right\}\):
\end{itemize}

\includegraphics{figures/2022-10-31_22-18-46.png}

\begin{itemize}
\tightlist
\item
  \(A_2\): Writing
  \(\Phi(A_2) = \left\{{e_1 - e_2, e_1 - e_3, e_2 - e_1, e_2 - e_3, e_3 - e_1, e_3 - e_2}\right\} \subseteq {\mathbf{R}}^3\),
  noting that every root has length \(\sqrt{2}\), the dual results in a
  scaled version of \(A_2\):
\end{itemize}

\includegraphics{figures/2022-10-31_22-58-28.png}

\begin{itemize}
\tightlist
\item
  \(B_2\): Let
  \(\Phi(B_2) = \left\{{e_1, -e_1, e_2, -e_2, e_1 + e_2, e_1 - e_2, -e_1 + e_2, -e_1 -e_2}\right\}\)
  with \(\alpha = e_1\) the short root at \(\beta = -e_1 + e_2\) the
  long root, taking the dual fixes the short roots \(\pm e_1\) and
  \(\pm e_2\), and normalizes the lengths of the long roots
  \(\pm e_1 \pm e_2\) to 1:
\end{itemize}

\includegraphics{figures/2022-10-31_23-18-28.png}

\begin{itemize}
\tightlist
\item
  \(G_2\): \(\Phi(G_2)\) is shown here in gray, with
  \(\Phi(G_2) {}^{ \vee }\) in green:
\end{itemize}

\includegraphics{figures/2022-10-31_23-45-39.png}

\end{remark}

\end{solution}

\begin{proposition}[Humphreys 9.3]

In Table 1, show that the order of \(\sigma_\alpha \sigma_\beta\) in
\(\mathcal{W}\) is (respectively) \(2,3,4,6\) when
\(\theta=\pi / 2, \pi / 3\) (or \(2 \pi / 3\) ), \(\pi / 4\) (or
\(3 \pi / 4\) ), \(\pi / 6\) (or \(5 \pi / 6\) ).

\begin{quote}
Note that \(\sigma_\alpha \sigma_\beta=\) rotation through \(2 \theta\).
\end{quote}

\end{proposition}

\begin{solution}

Given the hint, this is immediate: if \(s_\alpha s_\beta = R_{2\theta}\)
is a rigid rotation through an angle of \(2\theta\), then it's clear
that
\begin{align*}
R_{2 \cdot {\pi \over 2}}^2 = R_{2 \cdot {\pi \over 3}}^3 = R_{2\cdot {\pi \over 4}}^4 = R_{2\cdot {\pi \over 6}}^6 = \operatorname{id}
,\end{align*}
since these are all rotations through an angle of \(2\pi\).

To prove the hint, note that in any basis, a reflection has determinant
\(-1\) since it fixes an \(n-1{\hbox{-}}\)dimensional subspace (the
hyperplane \(H_\alpha\) of reflection) and negates its 1-dimensional
complement (generated by the normal to \(H_\alpha\)). On the other hand,
\(\operatorname{det}(s_\alpha s_\beta) = (-1)^2 = 1\) and is an isometry
that only fixes the intersection
\(H_\alpha = H_\beta = \left\{{\mathbf{0}}\right\}\), so it must be a
rotation.

To see that this is a rotation through an angle of exactly \(2\theta\),
consider applying \(s_\beta\circ s_\alpha\) to a point \(P\). Letting
\(H_\alpha, H_\beta\) by the corresponding hyperplanes. We then have the
following geometric situation:

\includegraphics{figures/2022-11-01_00-54-53.png}

We then have \(\theta_1 + \theta_2 = \theta\), noting that the angle
between \(\alpha\) and \(\beta\) is equal to the angle between the
hyperplanes \(H_\alpha, H_\beta\). The total angle measure between \(P\)
and \(s_\beta(s_\alpha(P))\) is then
\(2\theta_1 + 2\theta_2 = 2\theta\).

\end{solution}

\begin{proposition}[Humphreys 9.4]

Prove that the respective Weyl groups of
\(A_1 \times A_1, A_2, B_2, G_2\) are dihedral of order \(4,6,8,12\). If
\(\Phi\) is any root system of rank 2 , prove that its Weyl group must
be one of these.

\end{proposition}

\begin{solution}

In light of the fact that
\begin{align*}
D_{2n} = \left\langle{s, r {~\mathrel{\Big\vert}~}r^n = s^2 = 1, srs^{-1}= r^{-1}}\right\rangle
\end{align*}
where \(r\) is a rotation and \(s\) is a reflection, for the remainder
of this problem, let \(s \coloneqq s_\alpha\) and
\(r \coloneqq s_\alpha s_\beta\) after choosing roots \(\alpha\) and
\(\beta\).

\begin{itemize}
\item
  \(A_1\times A_1\): we have \(\Phi(A_2) = \pm e_1, \pm e_2\), and
  setting \(\alpha = e_1, \beta = e_2\) yields \(\theta = \pi/2\). We
  have
  \begin{align*}
  {\mathcal{W}}(A_2) = \left\{{\operatorname{id}, s_{\alpha}, s_{\beta}, s_\alpha s_\beta}\right\}
  \end{align*}
  where \(s_{\alpha}^2 = s_{\beta}^2 = 1\), \(s_\alpha s_\beta\) is
  rotation through \(2\theta = \pi\) radians, and
  \((s_\alpha s_\beta)^2 = \operatorname{id}\). Setting
  \(r= s_\alpha, s= s_\alpha s_\beta\) yields
  \(r^2 = s^2 = \operatorname{id}\) and \(srs = s\), which are the
  defining relations for \(D_4\).
\item
  \(A_2\): there is an inscribed triangle in the regular hexagon formed
  by the convex hull of the roots (see the dotted triangle below), and
  the reflections \(s_\alpha\) about the hyperplanes \(H_\alpha\)
  restrict to precisely the symmetries of this triangle, yielding
  \(D_{6}\): \includegraphics{figures/2022-11-01_01-35-46.png}
  Alternatively, choose a simple system
  \(\Delta = \left\{{\alpha= e_1, \beta= -e_1 - e_2}\right\}\), then
  \({\mathcal{W}}(A_2) = \left\langle{s_\alpha, s_\alpha s_\beta}\right\rangle\)
  is enough to generate the Weyl group. Since we have
  \(s \coloneqq s_\alpha \implies s^2 = 1\) and
  \(r\coloneqq s_\alpha s_\beta \implies r_3 = 1\) (since
  \(\theta = \pi/3\)), these satisfy the relations of \(D_6\).
\item
  \(B_2\): there is similarly a square on which the hyperplane
  reflections act on, highlighted with dotted lines here:
  \includegraphics{figures/2022-11-01_01-53-11.png} Since the
  \(s_\alpha\) act faithfully as the symmetries of a square, we have
  \({\mathcal{W}}(B_2)\cong D_{8}\). Alternatively, take
  \(\alpha = e_1\) and \(\beta = -e_1 + e_2\) and set
  \(s = s_\alpha, r = s_\alpha s_\beta\). Then
  \({\mathcal{W}}(B_2) = \left\langle{s, r}\right\rangle\) and since
  \(s^2 = r^4 = e\) (since here \(\theta = \pi/4\)) and they satisfy the
  proper commutation relation, this yields precisely the relations for
  \(D_{2n}, n=4\).
\item
  \(G_2\): In this case, the convex hull of the short roots form a
  hexagon, on which the hyperplane reflections precisely restrict to
  symmetries: \includegraphics{figures/2022-11-01_02-01-32.png}

  This yields \({\mathcal{W}}(G_2)\cong D_{12}\). Alternatively, take
  \(\alpha = e_1\) and \(\beta\) the long root in quadrant II, set
  \(s = s_\alpha, r= s_\alpha s_\beta\), then \(s^2 = r^6 = 1\) since
  \(\theta = \pi/6\) and again the commutation relations for
  \(D_{2n}, n=6\) are satisfied.
\end{itemize}

Finally, for any root system \(\Phi\) of rank 2, we will have
\({\mathcal{E}}(\Phi) = \left\langle{ s \coloneqq s_ \alpha, r\coloneqq s_ \alpha s _{\beta} }\right\rangle\).
Because \(\theta\) is restricted to one of the angles in Table 1 in
Humphreys \(\S 9.4\), i.e.~the angles discussed in problem 9.3 above,
the order of \(s\) is always 2 and the order of \(r\) is one of
\(4,6,8,12\). Since \(srs^{-1}=srs = r^{-1}\) in all cases, this always
yields a dihedral group.

\end{solution}

\hypertarget{section-10}{%
\subsection{Section 10}\label{section-10}}

\begin{proposition}[Humphreys 10.1]

Let \(\Phi {}^{ \vee }\) be the dual system of
\(\Phi, \Delta {}^{ \vee }=\left\{\alpha {}^{ \vee }\mathrel{\Big|}\alpha \in \Delta\right\}\).
Prove that \(\Delta {}^{ \vee }\) is a base of \(\Phi {}^{ \vee }\).

\begin{quote}
Compare Weyl chambers of \(\Phi\) and \(\Phi {}^{ \vee }\).
\end{quote}

\end{proposition}

\begin{solution}

Suppose that \(\Delta\) is a base of \(\Phi\). We can use the fact that
bases are in bijective correspondence with Weyl chambers via the
correspondence
\begin{align*}
\Delta \to \mathrm{WC}(\Delta) \coloneqq\left\{{v\in {\mathbb{E}}{~\mathrel{\Big\vert}~}(v, \delta) > 0 \,\, \forall \delta\in \Delta}\right\}
,\end{align*}
sending \(\Delta\) to all of the vectors making an acute angle with all
simple vectors \(\delta \in \Delta\), or equivalently the intersection
of the positive half-spaces formed by the hyperplanes \(H_{\delta}\) for
\(\delta\in \Delta\).

The claim is that
\(\mathrm{WC}(\Delta {}^{ \vee }) = \mathrm{WC}(\Delta)\), i.e.~the Weyl
chamber is preserved under taking duals. This follows the fact that if
\(v\in \mathrm{WC}(\Delta)\), then \((v,\delta)> 0\) for all
\(\delta\in \Delta\). Letting
\(\delta {}^{ \vee }\in \Delta {}^{ \vee }\), we have
\begin{align*}
(v,\delta {}^{ \vee }) > \qty{v, {2 \over (\delta, \delta)}\delta } = {2 (v, \delta) \over (\delta,\delta)} > 0
\end{align*}
using that every term in the last step is non-negative. Since this works
for every \(\delta {}^{ \vee }\in \Delta {}^{ \vee }\), this yields
\(v\in \mathrm{WC}(\Delta {}^{ \vee })\), and a similar argument shows
the reverse containment. So \(\Delta {}^{ \vee }\) corresponds to a
fundamental Weyl chamber and thus a base.

\end{solution}

\begin{proposition}[Humphreys 10.9]

Prove that there is a unique element \(\sigma\) in \(\mathcal{W}\)
sending \(\Phi^{+}\)to \(\Phi^{-}\)(relative to \(\Delta\) ). Prove that
any reduced expression for \(\sigma\) must involve all
\(\sigma_\alpha(\alpha \in \Delta)\). Discuss \(\ell(\sigma)\).

\end{proposition}

\begin{solution}

The existence and uniqueness of such an element follows directly from
the fact that \(W\) acts simply transitively on the set of bases, and
since \(\Delta\) and \(-\Delta\) are both bases, there is some
\(w_0\in W\) such that \(w_0(\Delta) = - \Delta\) and consequently
\(w_0(\Phi^+) = \Phi^-\). Since
\(\ell(\alpha) = n(\alpha) \leq {\sharp}\Phi^+\) for any root \(\alpha\)
and \(n(w_0) = {\sharp}\Phi^+\) by definition, \(w_0\) must be the
longest element in \(W\), i.e.~\(\ell(w_0)\) is maximal.

Any reduced expression for \(w_0\) must involve all \(s_\alpha\) -- if
not, and say \(s_\alpha\) doesn't occur in any reduced expression for
\(w_0\), then \(w_0\) does not change the sign of \(\alpha\) since every
\(s_\beta\) for \(\beta\neq \alpha \in \Delta\) changes the sign of
\(\beta\) and acts by permutations on
\(\Phi^+\setminus\left\{{\beta}\right\}\). However, in this case,
\(w_0' \coloneqq w_0s_\alpha\) satisfies \(n(w_0') = n(w_0) + 1\) since
\(w_0'\) necessarily changes the sign of \(\alpha\), contradicting
maximality of \(w_0\).

Finally, we have \(\ell(w_0) = n(w_0) = {\sharp}\Phi^+\).

\end{solution}

\hypertarget{section-11}{%
\subsection{Section 11}\label{section-11}}

\begin{proposition}[Humphreys 11.3]

Use the algorithm of (11.1) to write down all roots for \(G_2\).

Do the same for \(C_3\):
\begin{align*}
\left(\begin{array}{rrr}2 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -2 & 2\end{array}\right)
\end{align*}

\end{proposition}

\begin{solution}

Note that it suffices to find all positive roots, since
\(\Phi = \Phi^+ {\textstyle\coprod}\Phi^-\) once a simple system
\(\Delta\) is chosen. Since \({\sharp}\Phi(G_2) = 12\), it thus suffices
to find 6 positive roots. For \(G_2\), the Dynkin diagram indicates one
long and one short root, so let \(\alpha\) be short and \(\beta\) be
long. In this system we have
\begin{align*}
{\left\langle {\alpha },~{\alpha } \right\rangle}= {\left\langle {\beta },~{\beta } \right\rangle}&= 2 \\
{\left\langle {\alpha },~{\beta } \right\rangle}&= -1 \\
{\left\langle {\beta },~{\alpha } \right\rangle}&= -3
.\end{align*}

\begin{itemize}
\item
  The \(\beta\) root string through \(\alpha\): since
  \(\operatorname{ht}(\alpha) = 1\) and \(\beta-\alpha\not\in \Phi\), we
  have \(r=0\). Since
  \(q = -{\left\langle {\alpha },~{\beta } \right\rangle}= -(-1) = 1\),
  we obtain the string \(\alpha, \alpha + \beta\).
\item
  The \(\alpha\) root string through \(\beta\): since
  \(\operatorname{ht}( \beta) = 1\) and \(\alpha -\beta \not\in \Phi\)
  we have \(r=0\) again. Here
  \(q = - {\left\langle {\alpha },~{\beta } \right\rangle}= - (-3) = 3\),
  we obtain \(\beta, \beta+ \alpha, \beta + 2 \alpha, \beta+ 3 \alpha\)
\item
  We know that the \(\alpha\) root strings through any of the above
  roots will yield nothing new.
\item
  The \(\beta\) root strings through \(\alpha + \beta, \beta + 2\alpha\)
  turn out to yield no new roots.
\item
  The \(\beta\) root string through \(\beta + 3 \alpha\): since
  \((\beta + 3\alpha) - \beta = 3 \alpha\not\in\Phi\), using that only
  \(\pm \alpha\in \Phi\), we have \(r=0\). We also have
  \begin{align*}
  r-q = {\left\langle {\beta + 3 \alpha},~{ \beta} \right\rangle} = {\left\langle {\beta },~{\beta } \right\rangle}+ 3 {\left\langle {\alpha },~{\beta } \right\rangle}= 2 + 3(-1) =-1
  ,\end{align*}
  we have \(q=1\) and obtain \(\beta+ 3 \alpha, 2\beta + 3 \alpha\).
\end{itemize}

Combining these yields 6 positive roots:
\begin{align*}
\Phi^+(G_2) = \left\{{ \alpha, \alpha+ \beta, \beta, \beta+ 2 \alpha, \beta+ 3 \alpha, 2 \beta +3 \alpha}\right\} 
.\end{align*}

\begin{center}\rule{0.5\linewidth}{0.5pt}\end{center}

For \(C_3\), there are \(2\cdot 3^2 = 18\) total roots and thus 9
positive roots to find. Let \(\alpha, \beta, \gamma\) be the three
ordered simple roots, then the Cartan matrix specifies
\begin{align*}
{\left\langle {\alpha },~{\alpha } \right\rangle}= {\left\langle {\beta },~{\beta } \right\rangle}= {\left\langle {\gamma },~{\gamma } \right\rangle}&= 2 \\
{\left\langle {\beta },~{\alpha } \right\rangle}= {\left\langle {\alpha },~{\beta } \right\rangle}&= -1 \\
{\left\langle {\alpha },~{\gamma } \right\rangle}= {\left\langle {\gamma },~{\alpha } \right\rangle}&= 0 \\
{\left\langle {\beta },~{\gamma } \right\rangle}&= -1 \\
{\left\langle {\gamma },~{\beta } \right\rangle}&= -2
.\end{align*}

\begin{itemize}
\tightlist
\item
  The \(\alpha\) root string through \(\beta\):
  \(r=0,\, q = - {\left\langle {\alpha },~{\beta } \right\rangle}= 1\leadsto \alpha, \alpha + \beta\).
\item
  The \(\beta\) root string through \(\gamma\):
  \(r=0,\, q = - {\left\langle {\gamma },~{\beta } \right\rangle}= 2 \leadsto \gamma, \gamma+ \beta, \gamma + 2 \beta\).
\item
  The \(\gamma\) root string through \(\alpha\): here \(r=q=0\) since
  \({\left\langle {\gamma },~{\alpha } \right\rangle}= 0\).
\item
  The \(\alpha\) root string through \(\alpha + \beta\): \(r=0\) since
  \(\alpha + \beta - \gamma\not\in \Phi\), and
  \(r-1 = {\left\langle {\alpha+ \beta},~{ \gamma} \right\rangle}= -1\implies q=1 \leadsto \alpha + \beta+ \gamma\).
\item
  The \(\beta\) root string through \(\alpha + \beta + \gamma\): here
  \(r=1\) since \((\alpha + \beta + \gamma) - 2 \gamma \not\in \Phi\),
  and
  \(r-q = {\left\langle {\alpha+ \beta + \gamma},~{ \beta} \right\rangle} = -1 + 2 + 2 = -1 \implies q=2 \leadsto \alpha + 2 \beta + \gamma, \alpha + 3 \beta+ \gamma\).
\end{itemize}

This yields 9 positive roots:
\begin{align*}
\Phi^+(C_3) = \left\{{\alpha, \beta, \gamma, \alpha + \beta, \gamma + \beta, \gamma+ 2 \beta, \alpha+ \beta+ \gamma, \alpha+ 2 \beta + \gamma, \alpha + 3 \alpha + \gamma}\right\}
.\end{align*}

\end{solution}

\hypertarget{problem-set-6}{%
\section{Problem Set 6}\label{problem-set-6}}

\hypertarget{section-17}{%
\subsection{Section 17}\label{section-17}}

\begin{proposition}[17.1]

\end{proposition}

\begin{proposition}[17.3]

\end{proposition}

\hypertarget{section-18}{%
\subsection{Section 18}\label{section-18}}

\begin{problem}[18.1]

\end{problem}

\begin{problem}[18.4]

\end{problem}

\hypertarget{section-20}{%
\subsection{Section 20}\label{section-20}}

\begin{problem}[20.3]

\begin{quote}
Do one type that is not \(A_n\).
\end{quote}

\end{problem}

\begin{problem}[20.5]

\end{problem}

\begin{problem}[20.8]

\end{problem}

\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage

% Hook into amsthm environments to list them.
\addsec{Definitions}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
\cleardoublepage

\addsec{Theorems}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
\cleardoublepage

\addsec{Exercises}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
\cleardoublepage

\addsec{Figures}
\listoffigures
\cleardoublepage



\newpage
\printbibliography[title=Bibliography]


\end{document}
