# Friday, September 16

## 6.2: Casimir element of a representation

:::{.remark}
Last time: $L$ semisimple over $\CC$ implies $\kappa(x,y)\da \trace(\ad_x \ad_y)$ is nondegenerate.
Using Cartan's criterion, we can show that for any faithful representation $\phi: L\to \liegl(V)$ we can define a symmetric bilinear form $\beta_\phi$ on $L$ defined by $\beta_\phi(x, y) = \trace(\phi(x) \phi(y))$.
Note that $\beta_{\ad} = \kappa$.
Since $\Rad(\beta_\phi) = 0$, it is nondegenerate.
This defines an isomorphism $L \iso L\dual$ by $x\mapsto \beta(x, \wait)$, so given a basis $\mcb \da \ts{x_i}_{i\leq n}$ for $L$ there is a unique dual basis $\mcb' = \ts{y_i}_{i\leq n}$ for $L$ such that $\beta(x_i, y_j) = \delta_{ij}$.
Note that the $y_i \in L$ are dual to the basis $\beta(x_i, \wait) \in L\dual$.
:::

:::{.example title="?"}
For $L\da\liesl_2(\CC)$, the matrix of $\kappa$ is given by $\mattt 0 0 4 0 8 0 4 0 0$ with respect to the ordered basis $\mcb=\ts{x,h,y}$.[^humphreys_p22]
Thus $\mcb' = \ts{{1\over 4}y, {1\over 8}h, {1\over 4}x}$.

[^humphreys_p22]: 
See Humphreys p.22.

:::

:::{.definition title="Casimir element"}
Now let $\phi: L\to \liegl(V)$ be a faithful irreducible representation.
Fix a basis $\mcb$ of $L$ and define the **Casimir element**
\[
c_\phi = c_\phi(\beta) \da \sum_{i\leq n} \phi(x_i) \circ \phi(y_i) \in \Endo_\CC(V)
.\]
:::

:::{.remark}
One can show that $c_\phi$ commutes with $\phi(L)$.
Supposing $\phi$ is irreducible, $\Endo_L(V) = \CC$ by Schur's lemma, so $c_\phi$ is acts on $V$ as a scalar.
This follows from
\[
\trace(c_ \varphi) = \sum_{i\leq n} \trace( \varphi(x_i) \varphi(y_i) ) = \sum_{i\leq n} \beta(x_i, y_i) = n = \dim L
\implies
c_\phi = {\dim L \over \dim V} \id_V
,\]
since there are $\dim V$ entries.
In particular, $c_\phi$ is independent of the choice of $\mcb$.
This will be used to prove Weyl's theorem, one of the main theorems of semisimple Lie theory over $\CC$.
If $L$ is semisimple and $\phi:$ is not faithful, replace $L$ by $L/\ker \phi$.
Since $\ker \phi \normal L$ and $L$ is semisimple, $\ker \phi$ is a direct sum of certain simple ideals of $L$ and the quotient is isomorphic to the sum of the remaining ideals.
This yields a representation $\bar\phi: L/\ker \varphi \to \liegl(V)$ which is faithful and can be used to define a Casimir operator.
:::

:::{.example title="?"}
Let $L = \liesl_2(\CC)$ and let $V = \CC^2$ be the natural representation, so $\phi: L\to \liegl(V)$ is the identity.
Fix $\mch = \ts{x,h,y}$, then $\beta(u, v) = \trace( u v )$ since $\phi(u) = u$ and $\phi(v) = v$.
We get the following products:

|                 	| $\matt 0100$ 	| $\matt 100{-1}$ 	| $\matt 0010$       	|
|-----------------	|--------------	|-----------------	|--------------------	|
| $\matt 0100$    	| 0            	| $\matt 0{-1}00$ 	| $\matt 1000$       	|
| $\matt 100{-1}$ 	|              	| I               	| $\matt 0 0 {-1} 0$ 	|
| $\matt 0010$    	|              	|                 	| 0                  	|

Thus $\beta = \mattt 001 020 100$, and $\mcb' = \ts{y, {1\over 2}h, x}$, so
\[
c_\phi = xy + {1\over 2}h^2 + yx = \matt 1000 + {1\over 2}I + \matt 0001 = {3\over 2}I = {\dim \CC^2\over \dim \liesl_2(\CC)} I
.\]

:::

## 6.3: Complete reducibility

:::{.lemma title="?"}
Let $\phi: L\to \liegl(V)$ be a representation of a semisimple Lie algebra, then $\phi(L) \subseteq \liesl(V)$.
In particular, $L$ acts trivially on any 1-dimensional $L\dash$module since a $1\times 1$ traceless matrix is zero.
The proof follows from $L = [LL]$.
:::

:::{.warnings}
Arbitrary reductive Lie algebras can have nontrivial 1-dimensional representations.
:::

:::{.theorem title="Weyl's theorem"}
Let $\phi: L\to \liegl(V)$ be a finite-dimensional representation of a finite-dimensional semisimple Lie algebra over $\CC$.
This $\phi$ is completely reducible.
:::

:::{.warnings}
This is not true for characteristic $p$, or infinite dimensional representations in characteristic 0.
:::

### Proof of Weyl's theorem

:::{.remark}
Replace $L$ by $L/\ker \phi$ if necessary to assume $\phi$ is faithful, since these yield the same module structures.
Define a Casimir operator $c_\phi$ as before, and recall that complete reducibility of $V$ is equivalent to every $L\dash$submodule $W\leq V$ admitting a complementary submodule $W''$ such that $V = W \oplus W''$.
We proceed by induction on $\dim V$, where the dimension 1 case is clear.

**Case I**: $\codim_V W = 1$, i.e. $\dim (V/W) = 1$.
Take the SES $W \injects V \surjects V/W$.

**Case 1**:
Suppose $W' \leq W$ is a proper nonzero $L\dash$submodule.
Schematic:

<!-- Xournal file: /home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures/2022-09-16_09-54.xoj -->

![](figures/2022-09-16_10-02-33.png)

Using the 2nd isomorphism theorem, there is a SES $W/W' \injects V/W' \surjects V/W$.
Since $\dim W' > 0$, $\dim V/W' \leq \dim V$, so by the IH there is a 1-dimensional complement to $W/W'$ in $V/W'$.
This lifts to $\tilde W' \leq V$ with $W' \leq \tilde W$ with $\dim \tilde W/W' = 1$, and moreover $V/W' = W/W' \oplus \tilde W/W'$.
Take the SES $W' \injects \tilde W \surjects \tilde W/W'$ with $\dim \tilde W < \dim V$.
Apply the IH again to get a subspaces $X \leq \tilde W \leq V$ with $\tilde W = W' \oplus X$.
We'll continue by showing $V = W \bigoplus X$.
:::