# Friday November 22

*Remark:*
For $\lieg$ semisimple, studying $\mathrm{Rep}(\lieg)$ is too hard. 
So we study category $\mathcal O$, which contains simple modules $L(\lambda)$ for $\lambda \in \lieg^*$.

Case 1, $\lambda \in \Lambda^+$:
In the finite-dimensional setting, we use Weyl's character formula.

Case 2, $\lambda \not\in\Lambda^+$:
If suffices to consider $\lambda \in \Lambda$, then we apply Soergel's translation functor $\Bbb V$.
Then $\L(\lambda)$ for $\lieg$ corresponds to $L(\lambda^\sharp)$ for $\lieg^\sharp$ such that $\lambda^\sharp \in \Lambda(\lieg^\sharp)$.

For $\lambda \in \Lambda$, it suffices to consider $\lambda \in W\cdot 0$ using Jantzen's translation functor.

Then $\mathrm{ch} L(w\cdot 0) = \sum_{x\leq W} (-1)^{\ell(w) - \ell(x)} P_{w_0w, w_0x}(1) \mathrm{ch}M(x\cdot 0)$.

The $x\leq w$ index indicates the Bruhat order on $W$, and $P$ is the Kazhdan-Lusztig polynomial and $w_0$ is the longest element in $W$.

*Example:*
Type $A_2$, the $W = \Sigma_3, w_0 = s_{\alpha_1} s_{\alpha_2} s_{\alpha_1} = s_{\alpha_2} s_{\alpha_1} s_{\alpha_2} = \abs{3~2~1}$.

Last time:
If $L(\mu)$ is a composition factor of $M(\lambda)$, then $\mu \in W\cdot \lambda$.

## Central Characters (Ch. 23)

### Action of the Center (23.2)

Let $Z \definedas Z(U(\lieg))$ be the center of the universal enveloping algebra.
Then there is a Casimir element $\Omega \in Z$, and $\Omega \actson L(\lambda)$ by scalar multiplication.

**Definition/Proposition:**
For $\lambda \in \lieh^*$, its *central character* is $\chi_\lambda: Z \to \CC$ such that $z\cdot m = \chi_\lambda(z)m$ for all $z\in Z, m\in M$, where $M$ is a highest weight module with highest weight $\lambda$ and $v^+$ is a highest weight vector.

*Proof:*
For all $h\in \lieh$, we have
$h.(z.v^+) = z.(h.v^+)$ since $z\in Z$, but this equals $\lambda(h) z.v^+$.
Then $z.v^+ \in M_\lambda = \CC v^+$, so $z.v^+ = \chi_\lambda(z) v^+$ for some $\chi_\lambda(z)\in \CC$.

An arbitrary element in $M = U(\lieg).v^+$ is $m=x.v^+$ for $x\in U(\lieg)$.
Then

\begin{align*}
z.m &= z.(x.v^+) \\
&= x.(z.v^+) \\
&= \chi_\lambda(z) x.v^+ \\
&= \chi_\lambda (z) m
.\end{align*}

$\qed$

*Remark:*
We also have $Z\actson M(\lambda)$ and any submodule or composition factor by the same scalar.

*Example:*
Let $\lieg = \liesl(2, \CC)$ and $\Omega = h^2 + 2h + fe \in Z$.

Take $\lambda \in \Lambda^+ \equiv \ZZ_{\geq 0}$ and $v^+ \in M(\lambda)_\lambda$.
Then $\Omega.v^+ = (h^2 + 2h + fe)v^+ = (\lambda^2 + 2\lambda)v^+$, which means that $\chi_\lambda(\Omega) = \lambda(\lambda+2) \in \CC$.

### Harish-Chandra Theorem

**Definition:**
The *Harish-Chandra* homomorphism is the algebra homomorphism 

\begin{align*}
\xi: Z &\to U(\lieh) \\
\vector{f}^{\vector a} \vector h^{\vector b} \vector e^{\vector c} &\mapsto 
\begin{cases}
\vector h^{\vector b} & if \vector a = \vector 0 = \vector c \\
0 & else
\end{cases}
.\end{align*}


*Example:*
$\xi(\Omega) = h^2 + 2h$.

*Lemma:*
$\chi_\lambda(z) = \lambda(\xi(z))$ implies that $\Omega \actson L(\lambda), M(\lambda)$ by $(\lambda + \rho, \lambda)$.

*Proof*:
If $z = \vector f^{\vector a}\vector h^{\vector b}\vector e^{\vector c}$ with $\vector c \neq \vector 0$, the $z.v^+ = 0$ which implies that both sides are zero.
If $\vector c = \vector 0$, then $\vector a = \vector 0$.
Otherwise $z\in U(\lieg)_\beta$ for some $\beta\neq 0$, so there exists an $h\in\lieh$ such that $[h, z] = \beta(h) z \neq 0$, while $[h, z] = 0$ and $z\in Z$.

Thus $\chi_\lambda(z) = \lambda(\vector h^{\vector b}) = \lambda(\xi(z))$ if $z = \vector h^{\vector b}$. 
$\qed$

Recall that $\Omega = \sum_{j=1}^\ell h_j h'_j + \sum_{i=1}^m (e_i t_i + f_i e_i)$ for $h_i = [e_i, f_i] = e_if_i - f_i e_i$.

Then if we have a basis $\theset{h_i, e_i, f_i}$, we can produce a dual basis $\theset{h'_i, e'_i, f'_i}$ with respect to the killing form.
Thus $\Omega = \sum_j h_j h_j' + \sum_{i=1}^m h_i + 2f_i e_i$ and $\xi(\Omega) = \sum_j h_j h'_j + \sum{i=1}^m h_i$.

Now by writing $t_\lambda = \sum_i a_i h_i = \sum_i b_i h_j'$, where $\kappa(t_\lambda, h_j) = b_j, \kappa(t_\lambda, h'_j)$, and $\kappa(t_\lambda, t_\lambda) = \sum a_i b_i$, we can write

\begin{align*}
\lambda(\xi(\Omega)) = \sum_j \lambda(h_j) \lambda(h'_j) + \sum_i \lambda(h_i) 
&= \sum_j \kappa(t_\lambda, h_j)\kappa(t_\lambda, h]_j) + \sum_i \lambda(h_i) \\
&= \kappa(t_\lambda, t_\lambda) + \sum_i \lambda(h_i) \\
&= \kappa(t_\lambda, t_\lambda) + \sum_{i=1}^m (\lambda, \alpha_i) \\
&= \kappa(t_\lambda, t_\lambda) + (\lambda, \sum_{\alpha\in\Phi^+} \alpha) \\
&= \kappa(t_\lambda, t_\lambda) + (\lambda, 2\rho\\
&= (\lambda, \lambda) + (\lambda, 2\rho)  \\
&= (\lambda + 2\rho, \lambda)
.\end{align*}

$\qed$

*Example:*
$\lambda(\xi(\Omega)) = \lambda(h^2 + h) = \lambda^2 + 2\lambda = (\lambda+2, \lambda)$ under $\lieh^* \equiv \ZZ$ where $\alpha \mapsto 2, \rho \mapsto 1$.

**Definition:**
The *twisted Harish-Chandra homomorphism* is the algebra homomorphism $\psi = \zeta \circ \xi: Z \to S(\lieh)$ where 

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

*Example:*
$\xi(\Omega) = h^2 + 2h$, so 
$$
\psi(\Omega) = \zeta(h^2 + 2h) = (h-1)^2 + 2(h-1) = h^2 - 1
.$$

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

**Corollary:**
If $L(\mu)$ is a composition factor of $M(\lambda)$, then $Z \actson M(\lambda)$ by the same scalar $\chi_\lambda(z) = \chi_\mu(z)$ for all $z$.

Then $\chi_\lambda = \chi_\mu \implies \mu = W \cdot \lambda$.

*Remark:*
Assuming this theorem, this completes the proof of the Weyl Character Formula.