# Friday, April 23

## Applications Chevalley-Eilenberg Complex

:::{.remark}
Last time: $V_n(\lieg) \da \Ug\tensor_k \Extalg^n \lieg \surjectsvia{\eps} k$ is a projective resolution in $\liegmod$.
Note that we can introduce negative signs to easily interchange $\modsleft{\lieg}$ and $\modsright{\lieg}$.
:::

:::{.corollary title="Chevalley-Eilenberg"}
Let $M\in \modsright{\lieg}$, then 
\[
H_*(\lieg; M) \cong \Tor_*^{\Ug}(M, k)
\]
is the homology of the following complex:
\[
M\tensor_\Ug V_*(\lieg) 
\da M \tensor_\Ug \Ug \tensor_k \Extalg^* \lieg 
\cong M\tensor_k \Extalg^* \lieg
,\]
where we have a concrete differential $d$ on $\Extalg^* \lieg$ and we can define $\partial \da \one \tensor d$.
If $M\in \modsleft{\lieg})$ (which is more convenient for cohomology), then
\[
H^*(\lieg; M) \cong \Ext_\Ug(k, M)
\]
is the cohomology of the cochain complex
\[
\Hom_\Ug(V_*(\lieg), M) 
\da \Hom_{\Ug}(\Ug \tensor_k \Extalg^* \lieg, M)
\cong \Hom_k(\Extalg^*\lieg, M)
.\]
:::

:::{.remark}
This is very concrete!
Standard trick for exterior algebras: any $n\dash$cochain $f\in \Hom_k(\Extalg^n \lieg, M)$ can be viewed as an alternating $k\dash$multilinear function $f(x_1, \cdots, x_n): \lieg\to M$.
The cochain differential should increase degree, so we define
\[
\Theta_1 f(\elts{x}{n}) = \sum_{i=1}^{n+1} (-1)^{i+1} x_i \cdot f(x_1, \cdots, \hat{x_i}, \cdots, x_n)
+ \sum_{i < j} (-1)^{i+1} f( [x_i x_j], x_1, \cdots, \hat{x_i}, \cdots, \hat{x_j}, \cdots, x_n)
.\]
Note that the tor definition has the arguments switched compared to the original definition.
This is to set up the tensor cancellation of $\cdots \tensor_\Ug \Ug \cdots$.
Swapping factors and introducing signs makes this work for left $\lieg\dash$modules.
:::

:::{.corollary title="?"}
If $k$ is a field and $\dim_k \lieg = n$, then for any $M \in \liegmod$, 
\[
H^i(\lieg; M) = 0 = H_i(\lieg; M) && \forall i\geq n+1
.\]
:::

:::{.proof title="?"}
This follows from the fact that \( \Extalg^{\geq n+1} \lieg =0 \).
:::

:::{.example title="?"}
Take $\lieg = \liesl_2(\CC)$, then $\dim_\CC \lieg = 3$ (4 dimensions and one linear condition).
Then $H^i(\lieg; m) = 0$ for all $i > 3$.
:::

## Brief Intro to Semisimple Lie Algebras (Weibel 7.8)

:::{.remark}
Public service section since we won't have a Lie algebras course next Fall.
Semisimples: the most important and interesting classes of Lie algebras!
These occur frequently and we can prove a lot about them.
We'll assume $\lieg$ is a finite dimensional Lie algebra over a field $k$, where we'll soon assume $\ch(k) = 0$.
:::

:::{.definition title="Simple Lie Algebras"}
A Lie algebra $\lieg$ is **simple** if it has no ideals other than $0$ and $\lieg$ and $[\lieg\lieg] \neq 0$ (i.e. $\lieg$ is not abelian).
:::

:::{.remark}
Recall that $\liealg^\Ab \approx \kmod$ are vector spaces, and so if $\lieg$ is abelian it automatically has a chain of ideals by just taking vector subspaces.
These are closed under brackets since bracketing is zero.
So if $\dim_k \lieg \geq 2$, there are nontrivial ideals, so the abelian condition rules out all 1-dimensional Lie algebras -- they're all abelian by taking a generator, bracketing it with itself, and noting you get zero.
So there's only one 1-dimensional Lie algebra over any field $k$: the abelian one.
:::

:::{.remark}
The derived algebra $[\lieg\lieg] \normal \lieg$ is a subalgebra and always an ideal, so if $\lieg$ is simple then $[\lieg \lieg] = \lieg$.
So $\liegl_n(\CC)$ is not simple, since $[\liegl_n \liegl_n] = \liesl_n$ by taking traces.
:::

:::{.remark}
The vector space sum of any two solvable ideals is again a solvable ideal.
Note that this works for products of solvable subgroups $N, H\leq G$ with $N$ normal. 
Use 2-out-of-3 property for solvable groups and quotient by $N$.
By finite-dimensionality, we can find a maximal solvable ideal:
:::

:::{.definition title="?"}
For $\dim_k \lieg < \infty$, define the **radical** to be $\rad \lieg \da \sum I_j$ be the sum of all solvable ideals $I_j\normal \lieg$.
We say $\lieg$ is **semisimple** if $\rad \lieg = 0$.
:::

:::{.lemma title="?"}
Simple implies semisimple.
:::

:::{.lemma title="?"}
$\lieg / \rad\lieg$ is always semisimple.
:::

:::{.remark}
There shouldn't be any solvable ideals in this quotient, otherwise you could lift.
Next up, our most powerful tool for semisimple Lie algebras:
:::

:::{.definition title="?"}
Recall that for $x\in\lieg$ we can define $\ad_x \in \Endo_k(\lieg)$ where $\ad_x(y) \da [x, y]$.
It has a well-defined (and basis-independent) trace, so define the **Killing form**[^killing_form_name]:

\[
\kappa(x, y) \da \tr(\ad_x \circ \ad_y) \in k && x,y\in\lieg
.\]

[^killing_form_name]: 
Named for a mathematician *named* Killing.

:::

:::{.remark}
This is a symmetric bilinear form since traces don't depend on the order of products.
It has another nice property, $\lieg\dash$invariance: 
\[
\kappa([xy], z] = \kappa(x, [yz])
.\]
:::

:::{.proposition title="Cartan's Criterion"}
Let $\ch(k) = 0$ and $\dim_k \lieg < \infty$.
Then $\lieg$ is semisimple $\iff \kappa$ is nondegenerate.
:::

:::{.proof title="?"}
Omitted, see Humphreys.
:::

:::{.theorem title="?"}
Let $\ch(k) = 0$, then $\lieg$ is semisimple $\iff \lieg \cong \bigoplus_{i=1}^r \lieg_i$ as a direct sum/product of simple ideals, so $[\lieg_i \lieg_j] = 0$ for $i\neq j$ and $[\lieg_i \lieg_i ] = \lieg_i$.
In particular, every ideal of $\lieg$ is a sum of sum of certain $\lieg_i$'s, and $\lieg = [\lieg \lieg]$.
:::

:::{.remark}
These are like "orthogonal" ideals.
So we can study semisimple Lie algebras by just studying simple Lie algebras.
:::

:::{.observation}
Reminder: if $M\in \liegmod(\Triv)$, then any derivation $D \in \Der(\lieg, M)$ satisfies $D([xy]) = 0$ for all $x,y\in \lieg$.
This follows from expanding the Leibniz rule and using trivial modules act by zero.
There is an isomorphism
\[
\Der(\lieg, M) \cong \Hom_\kmod(\lieg^\ab, M) && \lieg^\ab \da \lieg / [\lieg \lieg]
.\]
Recall that $H^1$ is related to derivations.
:::

:::{.corollary title="?"}
Let $\lieg \in \liegmod(\semisimple)$ with $\dim_k \lieg < \infty$, then
\[
H^1(\lieg;k ) = 0 = H_1(\lieg; k) 
.\]
:::

:::{.proof title="?"}
Since $[\lieg \lieg] = \lieg$, we have $\lieg^\ab = 0$.
By Weibel theorem 7.4.1, one can check that $H_1(\lieg; k) \cong \lieg^\ab = 0$.
We also had $\Der(\lieg, k) \surjects H^1(\lieg; k)$ (it was outer derivations), the left-hand side is isomorphic to $\Hom_k(\lieg^\ab; k)$.
:::

:::{.theorem title="?"}
Let $\lieg \in \liealg(\semisimple)$ with $\dim_k \lieg < \infty$ and $\ch(k) = 0$.
Then if $k\neq M$ is a simple $\lieg\dash$module (where simple means no proper nontrivial $\lieg\dash$invariant submodules), then
\[
H^i(\lieg; M) = 0 = H_i(\lieg; M)
.\]
:::

:::{.proof title="?"}
Omitted.
This uses the Casimir operator for $M$, which is in the center $Z(\Ug)$.
:::