# Wednesday August 14

Recall from last time that a Lie Algebra is a vector space with a bilinear bracket, which importantly satisfies the Jacobi identity:

\begin{align*}
[x, [y, z]] + [y, [x,z]] + [z, [x, y]] = \vector 0
.\end{align*}

Also recall the examples from last time:

- $A_\ell \iff \liesl(\ell + 1, F)$
- $B_\ell \iff \lieso(2\ell + 1, F)$
- $C_\ell \iff \liesp(2\ell, F)$
- $D_\ell \iff \lieso(2\ell, F)$

*Exercise:*
Characterize these matrix subalgebras in terms of basis elements, and compute their dimensions.

## Lie Algebras of Derivations

**Definition:**
An **$F\dash$algebra** $A$ is an $F\dash$vector space endowed with a bilinear map
$A^2 \to A,~ (x,y) \mapsto xy$.

**Definition:**
An algebra is **associative** if $x(yz) = (xy)z$.

> Modern interest: simple Lie algebras, which have a good representation theory. Take a look a Erdmann-Wildon (Springer) for an introductory look at 3-dimensional algebras.

**Definition:**
Any map $\delta: A^2 \to A$ that satisfies the Leibniz rule is called a **derivation** of $A$, where the rule is given by $\delta(xy) = \delta(x)y + x\delta(y)$.

**Definition:**
We define $\mathrm{Der}(A) = \theset{\delta \suchthat \delta\text{ is a derivation }}$.

Any Lie algebra $\lieg$ is an $F\dash$algebra, since $[\wait, \wait]$ is bilinear. Moreover, $\lieg$ is associative iff $[x, [y,z]] = 0$.

*Exercise:*
Show that $\mathrm{Der} \lieg \leq \liegl(\lieg)$ is a Lie subalgebra. One needs to check that $\delta_1, \delta_2 \in \lieg \implies [\delta_1, \delta_2] \in \lieg$.

*Exercise:*
Define the adjoint by $\ad_x: \lieg\selfmap,~ y \mapsto [x, y]$. Show that $\ad_x \in \mathrm{Der}(\lieg)$.

## Abstract Lie Algebras

**Fact:** 
Every finite-dimensional Lie algebra is isomorphic to a linear Lie algebra, i.e. a subalgebra of $\liegl(V)$. Each isomorphism type can be specified by certain *structure constants* for the Lie bracket.

*Example:*
Any $F\dash$vector space can be made into a Lie algebra by setting $[x,y] = 0$; such algebras are referred to as *abelian*.

Attempting to classify Lie algebras of dimension at most 2.

- 1 dimensional: We can write $\lieg = Fx$, and so $[x, x] = 0 \implies [\wait, \wait] = 0$. So every bracket must be zero, and thus every Lie algebra is abelian.
- 2 dimensional: Write $\lieg = Fx \oplus Fy$, the only nontrivial bracket here is $[x, y]$. Some cases:
  - $[x, y] = 0 \implies \lieg$ is abelian.
  - $[x, y] = ax + by \neq 0$. Assume $a\neq 0$ and set $x' = ax+by, y' = \frac y a$. Now compute $[x', y'] = [ax+by, \frac y a] = [x,y] = ax+by = x'$. Punchline: $\lieg \cong Fx' \oplus Fy', [x', y'] = x'$.

We can fill in a table with all of the various combinations of brackets:

\begin{center}
\begin{tabular}{l|ll}
$[\wait, \wait]$ & $x'$  & $y'$ \\ \hline
$x'$                                               & $0$   & $x'$ \\
$y'$                                               & $-x'$ & $0$
\end{tabular}
\end{center}

*Example:*
Let $V = \RR^3$, and define $[a,b] = a\cross b$ to be the usual cross product.

*Exercise:*
Look at notes for basis elements of $\liesl(2, F)$,
\begin{align*}
e=\left[\begin{array}{ll}{0} & {1} \\ {0} & {0}\end{array}\right],\\
h=\left[\begin{array}{cc}{1} & {0} \\ {0} & {-1}\end{array}\right],\\
f=\left[\begin{array}{ll}{0} & {0} \\ {1} & {0}\end{array}\right]
.\end{align*}

Compute the matrices of $\ad(e), \ad(h), \ad(g)$ with respect to this basis.

## Ideals

**Definition:**
A subspace $I \subseteq \lieg$ is called an **ideal**, and we write $I \normal \lieg$, if $x,y \in I \implies [x,y]\in I$.

Note that there is no need to distinguish right, left, or two-sided ideals. This can be shown using $[x,y] = [-y, x]$.

*Exercise:*
Check that the following are all ideals of $\lieg$:

- $\theset 0, \lieg$.
- $\mathfrak z (\lieg) = \theset{z\in \lieg \suchthat [x, z] = 0\quad \forall x\in \lieg}$
- The commutator (or derived) algebra $[\lieg, \lieg] = \theset{\sum_i [x_i, y_i] \suchthat x_i, y_i \in \lieg}$.
  - Moreover, $[\liegl(n, F),\liegl(n, F) ] = \liesl(n, F)$.


**Fact:** 
If $I, J \normal \lieg$, then

- $I+J = \theset{x+y\suchthat x\in I, y\in J} \normal \lieg$
- $I \intersect J \normal \lieg$
- $[I, J] = \theset{\sum_i [x_i, y_i] \suchthat x_i \in I, y_i \in J} \normal \lieg$

**Definition:**
A Lie algebra is **simple** if $[\lieg, \lieg] \neq 0$ (i.e. when $\lieg$ is not abelian) and has no non-trivial ideals. Note that this implies that $[\lieg, \lieg] = \lieg$.

**Theorem:**
Suppose that $\ch F \neq 2$, then $\liesl(2, F)$ is not simple.

*Proof:*

Recall that we have a basis of $\liesl(2, F)$ given by $B = \theset{e, h, f}$ where

- $[e, f] = h$,
- $[h, e] = 2e$,
- $[h, f] = -2f$.

So think of $[h, e] = \ad_h$, so $h$ is an eigenvector of this map with eigenvalues $\theset{0, \pm 2}$. Since $\ch F \neq 2$, these are all distinct. Suppose $\liesl(2, F)$ has a nontrivial ideal $I$; then pick $x = ae + bh + cf \in I$. Then $[e, x] = 0 - 2be + ch$, and $[e, [e,x]] = 0 - 0 + 2ce$. Again since $\ch F \neq 2$, then if $c\neq 0$ then $e\in I$. Now you can show that $h\in I$ and $f\in I$, but then $I = \liesl(2, F)$, a contradiction. So $c=0$.

Then $x = bh \neq 0$, so $h\in I$, and we can compute

\begin{align*}
2e = [h, e] \in I \implies e \in I, \\
2f = [h, -f] \in I \implies f \in I
.\end{align*}

which implies that $I = \liesl(2, F)$ and thus it is simple.

$\qed$