# Monday, November 28


:::{.remark}
Setup: $L\in \liealg^{\fd, \semisimple}\slice \CC$ containing $H$ a maximal toral subalgebra, $\Phi \contains \Phi^+ = \tlset \beta m \contains \Delta = \tlset \alpha \ell$ with Weyl group $W$, and we have $x_i \in L_{ \alpha_i}, y_i\in L_{- \alpha_i}, h_i = [ x_i y_i]$.
For $\beta\in \Phi^+$, we also wrote $x_ \beta\in L_ \beta, y_ \beta\in L_{- \beta}$.
There is also a Borel $B = H \oplus \bigoplus _{\beta > 0} L_ \beta$ with $H \subseteq B \subseteq L$.

We saw that if $V\in \lmod^\fd$ then $V = \bigoplus _{\mu\in H\dual} V_ \mu$ and $V$ is a highest weight module of highest weight $\lambda$ where $\lambda \in \Lambda^+ = \ts{\nu \in H\dual \st \nu(h_i) \in \ZZ_{\geq 0} \, 1\leq i\leq \ell}$.
Writing $M( \lambda)$ for the Verma module $\mcu(L) \tensor_{\mcu(B)} C_{ \lambda}$, there is a unique irreducible quotient $M( \lambda) \surjects L( \lambda)$ with highest weight $\lambda$.
It turns out that $L ( \lambda)$ is finite-dimensional if \( \lambda\in \Lambda^+ \).
:::

:::{.question}
How can we understand $L( \lambda)$ for \( \lambda\in \Lambda^+ \) better?
What is $\dim L( \lambda)$? What is $\dim L( \lambda)_\mu$ (i.e. the dimensions of weight spaces)?
:::

## $\S 22$ Multiplicity Formulas; $\S 22.5$ Formal characters

:::{.remark}
Let \( \Lambda \subseteq H\dual \) be the lattice of integral weights.
Note that \( \lambda(h_\beta) \in \ZZ \,\, \forall \beta\in \Phi \iff \lambda(h_i) \in \ZZ \) for all $1\leq i\leq \ell$.
Since \( \Lambda\in \zmod \) there is a group algebra $\ZZ[ \Lambda]$, the free $\ZZ\dash$module with basis $e( \lambda)$ for \( \lambda\in \Lambda \), also written $e_ \lambda$ or $e^{ \lambda}$.
This has a ring structure given by linearly extending $e( \lambda)\cdot e( \mu) \da e( \lambda+ \mu)$, so
\[
\qty{ \sum a_ \lambda e( \lambda) } \cdot \qty{\sum b_ \mu e( \mu) } = \sum c_ \sigma e( \sigma), \qquad c_ \sigma\da \sum_{ \lambda+ \mu = \sigma} a_ \lambda b _{\mu}
.\]
Note that if $V\in \lmod^\fd$ then $V = \bigoplus _{ \mu\in H\dual} V_ \mu$, where $V_{ \mu}\neq 0\implies \mu \in \Lambda$.
In this case we can define the **formal character** 
\[
\mathrm{ch}_V \da \sum_{ \mu\in \Lambda} (\dim V_ \mu) e( \mu) \in \ZZ[ \Lambda ]
.\]
:::

:::{.proposition title="?"}
Let $V, W\in \lmod^\fd$, then
\[
\mathrm{ch}_{V\tensor W} = \character_V \cdot \character_W
.\]
:::

:::{.proof title="?"}
Take dimensions in the formula $(V\tensor W)_ \sigma = \sum_{ \lambda+ \mu = \sigma} V_ \lambda\tensor W_ \mu$.
:::

:::{.remark}
For \( \lambda\in \Lambda^+ \), we have $L( \lambda)$ (noting Humphreys uses $V( \lambda)$), we write 
\[
\character_ \lambda\da\character_{L( \lambda)} = \sum m_ \lambda(\mu) e( \mu)
\]
where $m_{\lambda}( \mu) \da \dim L( \lambda)_ \mu\in \ZZ_{\geq 0}$.
:::

:::{.remark}
We now involve the Weyl group to make more progress: let $W$ be the Weyl group of $(L, H)$, then $W\actson \ZZ[ \Lambda]$ by $w . e( \mu) \da e(w \mu)$ for $w\in W, \mu\in \Lambda$.
So $W\to \Aut_{\Ring}(\ZZ[ \Lambda])$, and recalling that $\dim L( \lambda)_ \mu = \dim L(\lambda)_{ w \mu}$, we have
\[
w. \character_{\lambda} = \sum_\mu m_ \lambda( \mu) e(w \mu) = \sum_\mu m_ \lambda(w \mu) e(w \mu) = \character_{\lambda}
,\]
so $\character_ \lambda$ are $W\dash$invariant elements of the group algebra.
:::

:::{.proposition title="?"}
Let $f\in \ZZ[ \Lambda]^W$ be a $W\dash$invariant element, then $f$ can be written uniquely as a $\ZZ\dash$linear combination of the $\character_ \lambda$ for \( \lambda \in \Lambda^+ \), i.e. these form a $\ZZ\dash$basis:
\[
\ZZ[ \Lambda]^W = \gens{\character_ \lambda\st \lambda\in \Lambda}_\ZZ
.\]

:::

:::{.proof title="?"}
Recall every \( \lambda\in \Lambda \) is conjugate to a unique \( \lambda\in \Lambda^+ \).
Since $f$ is $W\dash$invariant, we can write it as 
\[
f = \sum_{ \lambda\in \Lambda^+} c( \lambda) \qty{\sum_{w\in W} e( w \lambda)}
\]
where $c( \lambda)\in \ZZ$ and almost all are zero.
Given \( \lambda\in \Lambda^+ \) with $c( \lambda)\neq 0$, a previous lemma (13.2B) shows that $\size\ts{ \mu\in \Lambda^+ \st \mu\leq \lambda} < \infty$ by a compactness argument.
Set 
\[
M_f \da \Union_{c( \lambda) \neq 0, \lambda\in \Lambda^+} \ts{ \mu\in \Lambda^+ \st \mu\leq \lambda}
,\]
all of the possible weights that could appear, then $\size M_f < \infty$ since this is a finite union of finite sets.
Choose \( \lambda\in \Lambda^+ \) maximal with respect to the property that $c( \lambda)\neq 0$, and set $f' \da f- c( \lambda) \character_ \lambda$.
Note that $f'$ is again $W\dash$invariant, since $f$ and $\character_ \lambda$ are both $W\dash$invariant, and $M_{ \character_ \lambda} \subseteq M_f$.
However \( \lambda \not\in M_{f'} \) by maximality, since we've subtracted $c( \lambda) e(\lambda)$ off, so $\size M_{f'} < \size M_f$.
Inducting on $\size M_f$, $f'$ is a $\ZZ\dash$linear combination of $\ch_{\lambda'}$ for $\lambda' \in \Lambda$, and thus so is $f$.
One checks the base case $L(0) \cong \CC$ where everything acts with weight zero.
Uniqueness is relegated to exercise 22.8.
:::

:::{.question}
Is there an explicit formula for $\character_ \lambda$ for $\lambda \in \Lambda^+$?
An intermediate goal will be to understand characters of Verma modules $\character M( \lambda)$ -- note that this isn't quite well-defined yet, since this is an infinite-dimensional module and thus the character has infinitely many terms and is not an element in $\ZZ[ \Lambda]$.
:::

## $\S 23.2$ Characters and Verma Modules

:::{.question}
Let $\mcz(L) \subseteq \mcu(L)$ be the center of $\mcu(L)$, not to be confused with $Z(L) \subseteq L$ which is zero since $L$ is semisimple (since $Z(L)$ is an abelian ideal).
How does $\mcz(L) \actson M( \lambda)$?
:::

:::{.remark}
Note $M( \lambda) = \mcu(L) \tensor_{\mcu(B)} \CC_{ \lambda} \cong \mcu(N^-) \tensor_\CC \CC_ \lambda$ and write $v^+$ for a nonzero highest weight vector of $M( \lambda)$.
Let $z\in \mcz(L)$ and $h\in H$, then
\[
h.(z.v^+) = z.(h.v^+) = z.( \lambda(h) ) v^+ = \lambda(h) v^+
,\]
and $x_ \alpha(z.v^+) = z(x_ \alpha.v^+) = 0$ for all \( \alpha\in \Phi^+ \), so $z.v^+$ is a maximal vector in $M( \lambda)$ of weight \( \lambda \), i.e. there exists $\chi_ \lambda(z)\in \CC$ such that 
\[
z.v^+ = \chi_ \lambda( z) v^+
\]
since $\dim M( \lambda)_ \lambda = 1$.
Thus there is an algebra morphism
\[
\chi_ \lambda: \mcz(L) \to \CC
.\]

Pick a PBW basis for the Verma module $M(\lambda)$, then
\[
z. y_{\beta_1}^{i_1}\cdots y_{\beta_m }^{i_m} v^+ =
y_{\beta_1}^{i_1}\cdots y_{\beta_m }^{i_m} z. v^+ =
\chi_ \lambda(z) y_{\beta_1}^{i_1}\cdots y_{\beta_m }^{i_m} v^+ 
,\]
so $z.m = \chi_ \lambda(z)m$ for all $m\in M( \lambda)$, and thus $\mcz(L)\actson M(\lambda)$ by the character $\chi_ \lambda$.
Consequently, $\mcz(L)$ acts on any subquotient of $M( \lambda)$.
:::

:::{.question}
When is $\chi_ \lambda= \chi_ \mu$ for two integral weights \( \lambda, \mu\in \Lambda \)?
:::