# Friday, August 19

## Humphreys 1.3

:::{.definition title="Derivations"}
Let $A\in \Alg\slice\FF$, not necessarily associative (e.g. a Lie algebra).
An **$\FF\dash$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 $\Der_\FF(A) \leq \liegl_\FF(A)$.[^derivs]

[^derivs]: 
The usual product somehow involves "second-order terms", while the commutator product cancels higher order terms to give something first-order.

:::

:::{.warnings}
If $D, D'$ are derivations, then the composition $D\circ D'$ is *not* generally a derivation.
:::

:::{.definition title="Adjoint"}
If $L\in \Lie\Alg\slice\FF$, for $x\in L$ fixed define the **adjoint operator**
\[
\ad_x: L &\to L \\
y &\mapsto [x, y]
.\]
Note that $\ad_x\in \Der_\FF(A)$ by the Jacobi identity.
Any derivation of this form is an **inner derivation**, and all other derivations are **outer derivations**.
:::

:::{.remark}
Given $x\in K\leq L$, note that $K$ is a Lie subalgebra, and we'll want to distinguish $\ad_L x$ and $\ad_K x$ (which may differ).
Note that $\liegl_n(\FF) \geq \lieb = \lieh \oplus \lien$, where $\lieb$ are upper triangular, $\lieh$ diagonal, and $\lien$ strictly upper triangular matrices.
If $x\in \lieh$ then note that $\ad_\lieh x = 0$, but $\ad_\lieg x \lieh \neq 0$.
:::

## Humphreys 2.1

:::{.remark}
Some notes:

- $L\geq I$ is an **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.
- If $A, B \subseteq L$ define $[A, B] \da \spanof_\FF\ts{[a,b] \st a\in A,b\in B}$.
  *Not* taking the span generally won't even yield a subalgebra.
- For ideals $I,J\normal L$, $I+J, [I,J] \normal L$.
- $L/I \da \ts{x+I \st x\in L}$ with $[x+I, y+I] \da [x,y] + I$.[^well_def]
- Ideals are subalgebras (since this only requires closure under bracketing), but subalgebras are not necessarily ideals.
- Centers: $Z(L) = \ts{x\in L \st [xy] = 0 \forall y\in L} \leq L$.
- Derived ideals $L' \da L^1 \da [L, L] \normal L$.
- If $\lieg \geq \lieb = \lieh \oplus \lien$, and $\lien \normal \lieb$ using the fact that products of upper-triangular matrices involve multiplying diagonals, and bracketing/subtracting cancels the diagonal off. Moreover $\lieb/\lien \cong \lieb$.
- For $K \subseteq L$ a subspace, the normalizer is $N_L(K) \da \ts{x\in L\st [x, K] \subseteq K}$.
  If $K = N_L$ then $K$ is self-normalizing.
- The centralizer of $K$ in $L$ is $C_L(K) \da \ts{x\in L\st [x, K] = 0} \leq L$, which is a subalgebra by the Jacobi identity.

[^well_def]: 
One should check that this is well-defined.

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:::{.exercise title="?"}
Is $\lieh\normal \lieb$?
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:::{.definition title="Simple Lie algebras"}
A Lie algebra $L$ is **simple** if $L\neq 0$ and $\Id(L) = \ts{0, L}$ 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 $K\normal L$ since $[L,K] = 0$.
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:::{.example title="?"}
Let $L = \liesl_2(\CC)$, so $\trace(x) = 0$.
This has standard basis
\[
x = \matt 0100, \qquad
y = \matt 0010,\qquad
h = \matt 100{-1}. \\
[xy]=h,\quad [hx] = 2x,\quad [hy] = -2y
.\]
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:::{.exercise title="?"}
Prove that $\liesl_2(\CC)$ is simple.
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:::{.exercise title="?"}
Show that for $K\leq L$, the normalizer $N_L(K)$ is the largest subalgebra of $L$ in which $K$ is an ideal.
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:::{.exercise title="?"}
Show that $\lieh \subseteq \lieg\da \liesl_n(\CC)$ is self-normalizing subalgebra of $\lieg$.

> Hint: use $[h, e_{ij}] = (h_i - h_j) e_{ij}$ where $h = \diag(h_1,\cdots, h_n)$.
> The standard basis is $\lieh = \gens{e_{11} - e_{22}, e_{22} - e_{33}, \cdots, e_{n-1, n-1} - e_{n,n} }$.

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:::{.exercise title="?"}
What is $\dim \liesl_3(\CC)$? 
What is the basis for $\lieg$ and $\lieh$?
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## Humphreys 2.2

:::{.remark}
Notes: 

- Let $L, L'\in \cat C \da \Lie\Alg\slice \FF, \phi \in \cat C(L, L'), \ker \phi = \ts{x\in L\st \phi(x) = 0} \normal L$.
  - Note that if $x\in \ker \phi, y\in L$ then $\phi([xy]) = [\phi(x)\phi(y)] = [0\phi(y)] = 0$.
- A **representation** of $L$ is some $\phi\in \cat C(L, \liegl(V))$ for some $V\in \Vect\slice \FF$.
- The usual 3 isomorphism theorems for groups hold for Lie algebras.
- $\ad: L\to \liegl(L)$ where $x\mapsto \ad x$ is a representation.
- $\ker \ad = Z(L)$, so if $L$ is simple then $Z(L) = 0$ and $\ad$ is injective.
  Thus any simple Lie algebra is linear.
  - Compare to: any finite dimensional Lie algebra is linear.
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