# Monday November 25

Today:
The Conjugacy theorem

December 2nd: 
Kac-Moody Algebras (i.e. infinite-dimensional lie algebras)

December 4th:
Summary of semisimple lie algebras over $\CC$

Last time:
We had the following goal: If $L(\mu)$ is a composition factor of $M(\lambda)$, then $\mu \in W\cdot \lambda$.
We then get a central character $\chi: Z(U(\lieg)) \to \CC$, e.g. $\chi_\lambda: Z\to \CC$.

Any $z\in Z$ acts on the highest weight module $M$ of highest weight $\lambda$ by $\chi_\lambda(z)$.

*Example:*
$\lieg = \liesl(2, \CC)$ and $\lambda \in \ZZ_{\geq 0}$.

![Image](figures/2019-11-25-09:18.png)\

We now want to prove this using the twisted Harish-Chandra homomorphism, where here we have
$$
\psi: Z \to U(\lieh) = S(\lieg) \\
z \mapsto \iota \circ \xi(z)
$$

where 

\begin{align*}
\iota: S(\lieh) &\to S(\lieh) \\
\rho(h_1, \cdots, h_\ell) &\mapsto \rho(h_1-1, \cdots, h_\ell - 1)
.\end{align*}

For example, $h^2 + 2h \mapsto (h-1)^2 + 2(h-1)$.

**Theorem (Harish-Chandra):**
For all $\lambda, \mu \in \lieh^*$, we have $\chi_\mu = \chi_\lambda \iff \mu \in W \cdot \lambda$.

*Proof (sketch):*

1. Assuming *Chevalley's restriction theorem*, we have

\begin{align*}
P(\lieg)^G \cong P(\lieh)^W
\end{align*}

where $G \definedas \generators{\exp(\ad x) \mid x\in \lieg} \subseteq \aut(\lieg)$.

Then the map $Z \to S(\lieh)^W$ given by $z \mapsto \psi(z)$ is an isomorphism.

> Note: the RHS denote a subset invariant under the Weyl group action.

*Example:*
$\lieg = \liesl(2, \CC)$ and $W = \theset{e, s}$ where $s(h) = -h$.
Then $\Omega = h^2 + 2h + fe$, and 

$$
\psi(\Omega) = (h-1)^2 + 2(h-1) = h^2-1 \in S(\lieh)^W
$$ 

because 
$$
S \cdot (h^2-1) = (-h)^2 - 1 = h^2 - 1
.$$

2. 

$\impliedby$ 
It suffices to prove the case $\lambda, \mu \in \Lambda$ since $\Lambda$ is dense in $\lieh^*$ in the Zariski topology.

We can check that 


\begin{align*}
\chi_\lambda &= \chi_{W\cdot \lambda} \\
\iff \lambda(\xi(z)) &= (W\cdot \lambda) (\xi(z)) \forall z\in Z \\
\iff (\lambda + \rho)(\psi(z)) &= ((\lambda + \rho))(\psi(z)) \forall z\in Z \\
&= (\lambda + \rho)(W\inv \cdot \psi(z)) \\
&= (\lambda + \rho)(\psi(z)) \quad \text{by (1)}
.\end{align*}

$\implies$
Suppose that $\chi_\lambda = \chi_\mu$ but $\mu \not\in W\cdot \lambda$.

Construct $g \in S(\lieh)^W$ such that $g(W\cdot \lambda) = 1$ and $g(W\cdot \mu) = 0$.

By 1, there exists a $z = \psi\inv(g) \in Z$ such that 
$$
\chi_\lambda(z) = (\lambda + \rho) (g) = g(\lambda) + g(\rho) \neq g(\mu) + g(\rho) = \cdots = \chi_\mu(z).
$$
$\qed$

## Cartan Subalgebra (Chapter 15)

Recall that the Cartan subalgebra (CSA) is equal to the maximal toral subalgebra, which is nilpotent and self-normalizing.
Then $\rank \lieg = \dim \lieh$ is well-defined by any CSA, since they are all conjugate under $G$.

### Engel Subalgebras

**Definition:**
The *Engel subalgebra* $x\in \lieg$ of $\lieg$ is the generalized eigenspace of $\ad x$ with eigenvalue 0.
We can then define $\lieg_{0, x} \definedas \theset{y\in\lieg \suchthat (\ad x)^n (y) = 0 \text{ for } n \gg 0}$.

An element $x\in\lieg$ is *regular* if $\dim \lieg_{0, x}$ is minimal.

Some facts:

a. $\lieg_{0, x} = N_\lieg(\lieg_{0, x})$
b. If $x$ is regular then $\lieg_{0, x}$ is nilpotent
c. Combining (a) + (b), if $x$ is regular then $\lieg_{0, x}$ is a Cartan subalgebra.

*Example:*
$\lieg = \liesl(2, \CC)$, then 

- $\lieg_{0, e} = \lieg$ since $\ad e$ kills everything eventually.
- $\lieg_{0, h} = \CC h$, a 1-dimensional algebra spanned by $h$, and $h$ is regular, and $\CC h = \lieh$ is a CSA, which is nilpotent.


### CSAs

**Theorem:**
Let $\lieh \leq \lieg$, then

a. $\lieh$ is a CSA $\iff$ $\lieh = \lieg_{0, x}$ for some regular $x\in \lieg$
b. If $\lieg$ is semisimple and $\ch \FF \neq 0$, then $\lieh$ is maximal toral $\iff \lieh$ is a CSA.


*Proof (sketch):*

*Proof of (a):*

$\impliedby$: 
Easy to check.

$\implies$:
Suppose $\lieh$ is nilpotent and $\lieh \subseteq  \lieg_{0, x}$ for all $x\in\lieh$.
Then suppose that $\lieh \not\in \lieg_{0, x}$ for all $x\in\lieh$.

So pick $z\in \lieh$ such that $\dim \lieg_{0, z} \leq \dim \lieg_{0, x}$ for all $x\in \lieh$.
Then $\lieg_{0, z} \subseteq \lieg_{0, x}$ for all $x\in \lieh$.
This implies that $x\in \lieh \actson  \lieg{0, z} / \lieh$ is a nilpotent action (where this quotient is nonzero).
By Theorem 3.3, there exists a $y+\lieh \neq \lieg$ such that $\ad \lieh(y+\lieh) = \lieh$, or there exists a $y\not\in\lieh$ such that $[\lieh, y] \subseteq \lieh$
with $\lieh = N_\lieg(\lieh)$.
$\qed$

*Proof of (b):*

$\implies$:
$\lieh$ is abelian and thus nilpotent, so $\lieg = \lieh + \sum_\alpha \lieg_\alpha$ with $[\lieh, \lieh] = \theset{0} \subseteq \lieh$, and $[\lieh, \lieg_\alpha] \subseteq \lieg_\alpha$
Thus $N_\lieg(\lieh)$.

$\impliedby$:
*Next time.*