# Friday August 16

Last time, we looked at ideals such as $0, \lieg, Z(\lieg),$ and $[\lieg, \lieg]$.

**Definition:**
If $I \normal \lieg$ is an ideal, then the quotient $\lieg/I$ also yields a Lie algebra with the bracket given by $[x+I, y+I] = [x,y] + I$.

*Exercise:*
Check that this is well-defined, so that if $x + I = x' + I$ and $y+I = y' + I$ then $[x,y] + I = [x', y'] + I$.

## Homomorphisms and Representations

**Definition:**
A linear map $\phi: \lieg_{1} \to \lieg_{2}$ is a *Lie homomorphism* if $\phi[x,y] = [\phi(x), \phi(y)]$.

> Remark: $\ker \phi \normal \lieg_{1}$ and $\im\phi \leq \lieg_{2}$ are subalgebras.

**Fact:** 
There is a canonical way to set up a 1-to-1 correspondence $\theset{I \normal \lieg} \iff \theset{\hom \phi: \lieg \to \lieg'}$ where $I \mapsto (x \mapsto x + I)$ and the inverse is given by $\phi \mapsto \ker \phi$.

**Theorem (Isomorphism theorem for Lie algebras):**

- If $\phi: \lieg_{1} \to \lieg_{2}$ is a Lie algebra homomorphism, then $\lieg/\ker\phi \cong \im \phi$
- If $I,J \normal \lieg$ are ideals and $I \subset J$ then $J/I \normal \lieg g/I$ and $(\lieg/I)/(J/I) \cong \lieg/J$.
- If $I, J \normal \lieg$ then $(I+J)/J \cong I/(I\intersect J)$.

**Definition:**
A *representation* of a Lie algebra $\lieg$ is a Lie algebra homomorphism $\phi:\lieg \to \liegl(V)$ into a linear Lie algebra for some vector space $V$.

We call $V$ a $\lieg\dash$module with action $g\cdot v = \phi(g)(v)$.

*Example:* 
The *adjoint representation*:

\begin{align*}
\ad: \lieg \to \liegl(\lieg) \\
x \mapsto [x, \wait]
.\end{align*}

**Corollary:**
Any simple Lie algebra is isomorphic to a linear Lie algebra.

*Proof:* 

Since $\lieg$ is simple, the center $Z(\lieg) = 0$. We can rewrite the center as
\begin{align*}
Z(\lieg) = \theset{x\in\lieg \suchthat \ad_{x(y)} = 0 \quad \forall y\in\lieg} \\
= \ker \ad_{x}
.\end{align*}

Using the first isomorphism theorem, we have $\lieg/Z(\lieg) \cong \im \ad \subseteq \liegl (\lieg)$. But $\lieg/Z(\lieg) = \lieg$ here, so we are done.

## Automorphisms

**Definition:** 
An automorphism of $\lieg$ is an isomorphism $\lieg\selfmap$, and we define

\begin{align*}
\Aut(\lieg) = \theset{\phi:\lieg\selfmap \suchthat \phi \text{ is an isomorphism }}
.\end{align*}

**Proposition:**
If $\delta \in \mathrm{Der}(\lieg)$ is nilpotent, then
$$
\exp(\delta)\coloneqq\sum \frac{\delta^{n}} {n!} \in \Aut(\lieg).
$$

This is well-defined because $\delta$ is nilpotent, and a binomial formula holds:

\begin{align*}
\frac{\delta^{n([x,y])}}{n!} = \sum_{i=0}^{n} [\frac{\delta^{i}(x)}{i!}, \frac{\delta^{n-i}(y)}{(n-i)!}]
.\end{align*}

and for $n=1, \delta([x,y]) = [x, \delta(y)] + [\delta(x), y]$.

*Exercise:* 
Show that
\begin{align*}
[(\exp \delta)(x), (\exp\delta)(y)] = \sum_{n=0}^{k-1} \frac {\delta^{n}([x, y])} {n!}
.\end{align*}

*Example:* 
Let $\lieg = \liesl(2, \FF)$ and define
\begin{align*}
s = \exp(\ad_{e}) \exp(\ad_{-f}) \exp(\ad_{e}) \in \Aut \lieg
.\end{align*}

where $e,f$ are defined as (todo, see written notes).

Then define the Weyl group $W = \generators{s}$.

*Exercise:*
Check that $s(e) = -f, s(f) = -e, s(h) = -h$, and so the order of $s$ is 2 and $W = \theset{1, s}$.