# Friday January 31st

Theorem (Duals of Simple Quotients of Vermas)
: 	A useful formula: $L(\lambda)\dual \cong L(-w_0)$.

Proof
: $L(\lambda)\dual$ is a finite dimensional module, and $(x\cdot f)(v) = -f(x\cdot v)$, so $L(\lambda)\dual \cong L(\nu)$ for some $\nu \in \Lambda^+$.
	The weights of $L(\lambda)\dual$ are the negatives of the weight of $L(\lambda)$.
	Thus the lowest weight of $L(\lambda)$ is $w_0\lambda$, since $w_o$ reverses the partial order on $\lieh\dual$, i.e $w_0 \Phi^+ = \Phi^-$

	Then 
	$$
	\mu \in \Pi(L(\mu)) \implies w_0 \mu \in \Pi(L(\lambda)) \implies w_0\mu \leq \lambda
	.$$
	This shows that the lowest weight of $L(\lambda)$ is $w_0 \lambda$, thus the highest weight $L(\lambda)\dual$ is $-w_0 \lambda$ by reversing this inequality.

	The inner product is $W$ invariant and is its own inverse, so we can move it to the other side.

## 1.7: Action of $Z(\lieg)$

Next big goal:
Every module in $\OO$ has a *finite* composition series (Jordan-Holder series, the quotients are simple).
Leads to Kazhdan-Lustzig conjectures from 1979/1980, which were solved, but are still open in characteristic $p$ case.

The technique we'll use the Harish-Chandra homomorphism, which identifies $\mcz(\lieg)$ explicitly.

It's commutative, a subalgebra of a Noetherian algebra, no zero divisors -- could be a quotient, but then it'd have zero divisors, so this seems to force it to be a polynomial algebra on some unknown. 

> Also note that $\mcz(\lieg) \definedas Z(U(\lieg))$.

Recall:
$\mcz(\lieg)$ acts locally finitely on any $M\in \OO$ -- this is by theorem 1.1e, i.e. $v\in M_\mu$ and $z\in \mcz(\lieg)$ implies that $zv\in M_\mu$.
(The calculation just follows by computing the weight and commuting things through.)

Let $\lambda \in \lieh\dual$ and $M = U(\lieg)v^+$ a highest weight module of highest weight $\lambda$.
Then for $z\in \mcz(\lieg)$, $z\cdot v^+ \in M_\lambda$ which is 1-dimensional.
Thus $z$ acts by scalar multiplication here, and $z\cdot v^+ = \chi_\lambda(z) v^+$.
Now if $u\in U(\lieu^-)$, we have $$z\cdot(u\cdot v^+) = u\cdot(z\cdot v^+) = u(\chi_\lambda(z)v^+) = \chi_\lambda(z) u\cdot v^+.$$
Thus $z$ acts on *all* of $M$ by the scalar $\chi_\lambda(z)$.

Exercise
: Show that $\chi_\lambda$ is a nonzero additive and multiplicative function, so $\chi_\lambda: \mcz(\lieg) \to \CC$ is linear and thus a morphism of algebras.
	Conclude that $\ker \chi_\lambda$ is a maximal ideal of $\mcz(\lieg)$.

Note: this is called the *infinitesimal character*.

Note that $\chi_\lambda$ doesn't depend on which highest weight module $M_\lambda$ was chosen, since they're all quotients of $M(\lambda)$.
In fact, every submodule and subquotient of $M(\lambda)$ has the same infinitesimal character.

Definition (Central/Infinitesimal Character)
: 	$\chi_\lambda$ is called the *central (or infinitesimal) character*, and $\hat\mcz(\lieg)$ denotes the set of all central characters.
		More generally, any algebra morphism $\chi: \mcz(\lieg) \to \CC$ is referred to as a central character.
    Central characters are in one-to-one correspondence with maximal ideals of $\mcz(\lieg)$, where 
    
    \begin{align*}
    \chi & \iff \ker \chi \\
    \CC[x_1, \cdots, x_n] &\iff \generators{x_1 - a_1, \cdots, x_n - a_n}
    \end{align*}
    
    where $[a_1, \cdots, a_n] \in \CC^n$.

Next goal:
Describe $\chi_\lambda(z)$ more explicitly.

Using PBW, we can write $z\in \mcz(\lieg) \subset U(\lieg) = U(\lien^-) U(\lieh) U(\lien)$.
Some observations:

1. Any PBW monomial in $z$ ending with a factor in $\lien$ will kill $v^+$, and hence can not contribute to $\chi_\lambda(z)$.
2. Any PBW monomial in $z$ beginning with a factor in $\lien^-$ will send $v^+$ to a lower weight space, so it also can't contribute.

So we only need to see what happens in the $\lieh$ part.
A relevant decomposition here is 
$$
U(\lieg) = U(\lieh) \oplus \qty{ \lien^- U(\lieg) + U(\lieg)\lien^+  }
.$$

Exercise
: Why is this sum direct?

Let $\mathrm{pr}: U(\lieg) \to U(\lieh)$ be the projection onto the first factor.
Then $\chi_\lambda(z) = \lambda(\mathrm{pr} z)$ for all $z\in \mcz(\lieg)$.
Then if $\mathrm{pr}(z) = h_1^{m_1} \cdots h_\ell^{m_\ell}$, we can extend the action on $\lieh$ to all polynomials in elements of $\lieh$ (which is in fact evaluation on these monomials), and thus $\chi_\lambda(z) = \lambda(h_1)^{m_1} \cdots \lambda(h_\ell)^{m_\ell}$.

Note that for $\lambda \in \lieh\dual$, we've extended this to the "evaluation map" $\lambda: U(\lieg) \cong S(\lieh)$, the symmetric algebra on $\lieh$.

Why is this the correct identification? 
The RHS is $T(\lieh) / \generators{x\tensor y - y\tensor x - [xy]}$, but notice that the bracket vanishes since $\lieh$ is abelian, and this is the exact definition of the symmetric algebra.

Thus $\chi_\lambda = \lambda \circ \mathrm{pr}$.

Observation:

\begin{align*}
\lambda(\mathrm{pr}(z_1 z_2))
&= \chi_\lambda(z_1 z_1)\\
&= \chi_\lambda(z_1) \chi_\lambda(z_2) \\
&= \cdots \\
&= \lambda( \mathrm{pr}(z_1) \mathrm{pr}(z_2) )
.\end{align*}

Exercise
: 	Show $\intersect_{\lambda \in \lieh\dual} \ker \lambda = \theset{0}$.

Definition (Harish-Chandra Morphism)
: 	Let $\xi = \restrictionof{\mathrm{pr}}{\mcz(\lieg)}: \mcz(\lieg) \to U(\lieh)$.

Definition (Twisted Harish-Chandra Morphism)
: 	$\xi$ is an algebra morphism, and is referred to as the *Harish-Chandra homomorphism*.

See page 23 for interpretation of $\xi$ without reference to representations.

Questions:

1. Is $\xi$ injective?
2. What is $\im \xi \subset U(\lieh)$?

When does $\chi_\lambda = \chi_\mu$?
Proved last time: we introduced the $\cdot$ action and proved that $M(s_\alpha \cdot \lambda) \subset M(\lambda)$ where $\alpha \in \Delta$.
It'll turn out that the statement holds for all $\lambda \in W$.

Wednesday:
Section 1.8.