# Monday, September 19

## Proof of Weyl's theorem (continued)

:::{.remark}
Recall: we're proving Weyl's theorem, i.e. every finite-dimensional representation of semisimple Lie algebra over $\CC$ 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 $\dim V$.
:::

:::{.proof title="of Weyl's theorem, continued"}
**Case I**: 
$\dim V/W = 1$.

**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

- $\tilde W \intersect W \subseteq W'$,
- $V= W + \tilde W + W' = W + \tilde W$ since $W' \subseteq W$.
- $\tilde W = W' \oplus X$ for some submodule $X \leq_L \tilde W \leq_L V$.

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
\[
X \intersect W 
&= (X \intersect \tilde W) \intersect W \\
&= X \intersect (\tilde W \intersect W) \qquad\text{by 1}\\
&\subseteq X \intersect W' = 0 \qquad \text{by 2}
,\]
so $V = W \oplus X$.

**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 $c_\phi \in\mods{L}(V, V)$ and $\ker c_\phi \leq_L V$.
On the other hand, $c_\phi$ induces an element of $\Endo_{L}(W)$, and since $W$ is irreducible, $\ro{c_\phi}{W} = \lambda \id_W$ for some scalar $\lambda$.
This can't be zero, since $\trace(\ro{c_\phi}{W}) = {\dim L \over \dim W} > 0$, so $\ker c_\phi \intersect W = 0$.
Since $\codim_V W = 1$, i.e. $\dim W = \dim V - 1$, we must have $\dim \ker c_\phi = 1$ and we have a direct sum decomposition $V = W \oplus \ker c_\phi$.

> Use of the Casimir element in basic Lie theory: producing a complement to an irreducible submodule.

**Case 2**:
Suppose $0 < W < V$ with $W$ any nontrivial $L\dash$submodule; there is a SES
\(
W \injects V \surjects V/W
\)
Consider $H \da \hom_\CC(V, W)$, then $H \in \lmod$ by $(x.f)(v) \da x.(f(v)) - f(x.v)$ for $f\in H, x\in L, v\in V$.
Let $\mcv \da \ts{f \st H \st \ro f W = \alpha \id_W \,\text{ for some } \alpha \in \CC} \subseteq H$.
For $f\in V$ and $w\in W$, we have
\[
(x.f)(w) = x.f(w) - f(x.w) = \alpha x.w - \alpha x.w = 0
.\]
So let $\mcw \da \ts{f\in \mcv \st f(W) = 0} \subseteq \mcv$, then we've shown that $L.\mcv \subseteq \mcw$.
Now roughly, the complement is completely determined by the scalar.
Rigorously, since $\dim \mcv/\mcw = 1$, any $f\in \mcv$ is determined by the scalar $\ro{f}{W} = \alpha \id_W$: we have $f-\alpha \chi_W \in \mcw$ where $\chi_W$ is any extension of $\id_W$ 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 $\mcw \injects \mcv \surjects \mcv/\mcw \in \lmod$ with $\codim_\mcv \mcw = 1$.
By Case I, $\mcv = \mcw \oplus \mcw''$ for some complement $L\dash$submodule $\mcw''$.
Let $f: V\to W$ span $\mcw''$, then $\ro f W$ is a nonzero scalar -- a scalar since it's in $\mcv$, and nonzero since it's in the complement of $\mcw$.
By rescaling, we can assume the scalar is 1, so $\im f = W$ 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 $L\dash$submodule, since $L.f \subseteq \mcw'' \intersect \mcw = 0$ since $\mcw''$ is an $L\dash$submodule and $f\in \mcw$ since $L.\mcv \subseteq \mcw$.
Noting that if $(x.f) = 0$ then $x.(f(v)) = f(x.v)$, making $f$ an $\lmod$ morphism.
Thus $W'' \da \ker f \leq_L V$, and $W \intersect W'' = 0$ so $\ro f W = \id_W$.
Since the dimensions add up correctly, we get $V = W \oplus W''$.



:::

## 6.4: Preservation of Jordan decomposition

:::{.theorem title="?"}
Let $L \leq \liegl(V)$ 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 $\ad_x = \ad_{x_s} + \ad_{x_n}$.
$L$ contains the semisimple and nilpotent parts of all of its elements, and in particular the two above decompositions coincide.
:::

:::{.proof title="Idea"}
The proof is technical, but here's a sketch:

- Construct a subspace $L \leq L' \leq_L \liegl(V)$ such that $l'$ contains the semisimple and nilpotent parts of all elements where $L\actson \liegl(V)$ by $\ad: L\to \liegl(\liegl(V))$.
- Check $L' \lieq N_{\liegl(V)}(L)$ (the normalizer), so $[LL'] \subseteq L$.
- Show $L' = L$:
  - If $L'\neq L$, use Weyl's theorem to get a complement with $L' = L \oplus M$.
  - Check $[LM] \subseteq [LL'] \subseteq L$ and $[LM] \subseteq M$ since $M\leq_L M$, forcing $[LM] \subseteq L \intersect M = 0$.
  - 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 $M = 0$.
- Since $s, n\in L$ when $x=s+n$ and $\ad_x = \ad_s + \ad_n = \ad_{x_s} + \ad_{x_n}$, the result follows from uniqueness of the abstract JCF that $s=x_s, n=x_n$ (using that $\ad$ is injective when $L$ is semisimple since $Z(L) = 0$).
:::

:::{.corollary title="?"}
If $L \in \Lie\Alg\slice \CC^{\ss}$ (not necessarily linear) and $\phi\in \lmod$, writing $x=s+n$ the abstract Jordan decomposition, $\phi(x) = \phi(s) + \phi(n)$ is the usual JCF of $\phi(x)\in \liegl(V)$.
:::

:::{.proof title="Sketch"}
Consider $\ad_{\phi(L)}\phi(s)$ and $\ad_{\phi(L)}\phi(n)$, 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 $\ad_{\phi(x)}$.
Now apply the theorem.
:::