# Monday September 16

Let $S = \exp(\ad e) \circ \exp(\ad -f) \circ \exp(\ad ei)$, which has the following matrix:

![Image](figures/2019-09-16-09:13.png)\

Where $\exp(\ad e) = 1 + \ad e + \frac 1 2 (\ad e)^2$, which would have the form

![Image](figures/2019-09-16-09:15.png)\

**Theorem:** 
If $\lieg$ is semisimple, then any finite dimensional $\lieg\dash$module $V$ is completely reducible, i.e. it splits into a direct sum of simple modules.

## Proof of Weyl's(?) Theorem

If $V$ itself is simple, then we're done, so suppose it is not.

Assume there exists a nonzero submodule $U \subsetneq V$.
It suffices to show that $V = U \oplus U'$ for some $U'$.

### Step 1: 

If $\dim V = 2$ and $\dim U = 1$.

Then $U, V / U$ are both trivial modules.
So $g\actson u = 0$ for all $u\in U$.
But then $g \actson (v + U) = U$ for all $v\in V$, since $g\actson v \in U$.

So for all $x,y \in lieg$ and all $v\in V$, we have $[x,y]\actson v = x\actson(y\actson v) - y\actson(x\actson v)$. But both of the terms in parenthesis are in $U$, and all elements in $\lieg$ kill elements in $U$, so this is zero. So $[\lieg, \lieg] \actson V$ trivially.

> Exercise: If $\lieg$ is semisimple, then $[\lieg, \lieg] = \lieg$.

So $\lieg \actson V$ trivially. Thus any $U'$ that is a complementary subspace of $U$ will be a submodule of $V$.

### Step 2:

Suppose $U$ is simple and $\dim U > 1$, so $\dim V / U = 1$.

Let $\Omega$ be the Casimir element on $U$ (faithful representation?).
Then $\Omega u = c u$ for some $c \in \FF$, and so $\Omega(U) \subseteq U$.

Since $\Omega: V \selfmap$ is a homomorphism, $\ker \Omega \subseteq V$ is a $\lieg\dash$submodule.
Then $\dim V / U = 1 \implies V/U$ is a trivial module.
So $\lieg \actson V/U = 0$, i.e. $\lieg \actson V \subseteq U$.

Then $\Omega(v) = \sum_i x_i \actson (y_i \actson v) \in U$ for all $v\in V$. 
What is the matrix of $\Omega$?

![Image](figures/2019-09-16-09:31.png)\

In particular, $\Tr(\Omega \mid_{V/U}) = 0$. So $\Tr(\Omega) = \Tr(\Omega\mid_U)$.
From 6.2, we know that $\Tr(\Omega) \neq 0 \implies c\neq 0$, where $c$ is the scalar appearing above.
So $\ker \Omega$ is 1-dimensional, and $\ker \Omega \intersect U = \theset 0$.

So take $U' = \ker \Omega$.

### Step 3:

Suppose $U$ is *not* simple, but $\dim V/U = 1$.

We will induct on the dimension of $U$. 
Pick a proper nonzero submodule $\overline U \subsetneq U$, so that $\dim U / \overline U < \dim U$. 
Now $V/U \cong (V / \overline U) / (U / \overline U)$ by an isomorphism theorem.
So $U / \overline U$ is a submodule of $V/\overline U$ of codimension 1.
Applying the inductive hypothesis, we obtain $V / \overline U = U / \overline U \oplus \overline V / \overline U$ for some $\overline V$ such that $U \subseteq \overline V \subseteq V$.

In particular, since $U \subseteq \overline V$ has codimension 1, $\dim \overline U < \dim U$.
So apply the inductive hypothesis again: $\overline V = \overline U \oplus U'$ for some $U'$, and $V = U \oplus U'$.

### Step 4: The general case

Recall that $\hom(V, U)$ is a $\lieg\dash$module where 
\begin{align*}
(g \actson \phi)(v) = g\actson\phi(v) - \phi(g\actson v)
.\end{align*}

Define
$$
S = \theset{\phi \in \hom(V, U) \suchthat \phi\mid_U \in F 1_U}.
$$

Then $S \leq \hom(V, U)$ as a submodule.
Define $T = \theset{\phi \in S \suchthat \phi\mid_U = 0}$.
Then $T \leq S$ as a submodule, and $\lieg(S) \subseteq T$.

Now each $\phi\in S$is determined ($\mod T$) by the scalar $\phi\mid_U$.
Note that $\dim(S/T) = 1$.
By steps 1-3, we know that $S = T \oplus T'$ for some $T' \subseteq S$ of dimension 1.
Then $T' = \mathrm{span}_\FF(f)$ for some nonzero map $f: V \to U$ such that $f(u) = cu$ for some $c \neq 0$.

Then $\lieg(T \oplus T') = \lieg (S) \subseteq T \implies \lieg(T') = 0$.
So for all $g\in \lieg$, we have $0 = (g\actson f)(v) = f \actson f(v) - f(g\actson v)$.
Then $f: V \to U$ is a lie algebra homomorphism, $\ker f = U'$, and thus $V = U \oplus U'$. 
$\qed$


Some consequences of Weyl's theorem:

## Preservation of Jordan Decomposition

Recall that when $\lieg \in \liegl(V)$ is a linear lie algebra, then for $x\in \lieg$ we have:

Jordan Decomposition: $x = x_s + x_n$ where $x_s, x_n \in \mathrm{End}(V)$.

**Abstract Jordan Decomposition:** 
\begin{align*}
\lieg \mapsvia{\ad} \ad(\lieg) \\
x \mapsto \ad x \\
x_s \leftarrow(\ad x)_s \\
x_n \leftarrow  (\ad x)_n
.\end{align*}

and so $x = x'_s + x'_n$ for some $x'$. The theorem will be that these recover the usual Jordan decomposition.

**Theorem:**
If $\lieg \in \liegl(V)$ is semisimple and $V$ is finite dimensional, then $x_s, x_n \in \lieg$, and $x_s = x'_s, x'_n$.

**Corollary:**
If $\lieg$ is semisimple and finite dimensional and $\phi: \lieg \to \liegl(V)$ is a finite dimensional representation, then if $x = x_s + x_n$ is the abstract Jordan decomposition, then $\phi(x) = \phi(x_s) + \phi(x_n)$ is the Jordan decomposition in $\liegl(V)$.

*Example:*
If $\lieg = \liesl(2, \CC)$ is semisimple and finite dimensional, and $h$ is diagonal, then by JD $h = h + 0, \phi(h) = \phi(h) + 0$.
Then $h \actson V$ semisimply, or $V = \bigoplus_{\lambda \in \CC} V_\lambda$, where $V_\lambda = \theset{v\in V \suchthat h\actson v = \lambda v}$ are the eigenspaces.

![Image](figures/2019-09-16-10:05.png)\