# Friday, December 02

## Convolution Formulas

:::{.remark}
Last time: 
\[
\mcx \da \ts{f\in \Hom_\CC(H\dual, \CC) \st \supp f \subseteq \Union_{i=1}^n (\lambda_i - \ZZ_{\geq 0} \Phi^+) }
,\]
which is a commutative associative unital algebra under convolution, where $e_{ \lambda}(\mu) = \delta_{ \lambda \mu}$ for $\mu\in H\dual$ and $e_ \lambda\convolve e_ \mu = e_{ \lambda+ \mu}$ with $e_0 = 1$.
We have $\ch_{M(0)}$ which records weights $\mu = \sum_{i=1}^m -i_j \beta_j$ with $i_j\in \ZZ_{\geq 0}$, and
\[
\dim M(0)_\mu \da \ch_{M(0)}(\mu) = \size \ts{\vector i \in \ZZ_{\geq 0}^m \st \sum i_j \beta_j = -\mu } \da p(\mu)
\]
which is the **Kostant function** -- its negative is the Kostant partition function, which records the number of ways of writing a weight as a sum like this.
We'll regard finding such a count as a known or easy problem, since this can be done by enumeration.

Define the **Weyl function** $q\da \prod_{\alpha \in \Phi^+} (e_{\alpha\over 2} - e_{-{\alpha \over 2}})$ where the product is convolution.
For \( \alpha\in \Phi+, f_ \alpha:H\dual \to \ZZ \), define
\[
f_ \alpha( \lambda) \da 
\begin{cases}
1 & \lambda = -k \alpha \text{ for some } k\in \ZZ_{\geq 0} 
\\
0 & \text{otherwise}.
\end{cases}
.\]
We can regard this as an infinite sum $f_ \alpha = e_0 + e_{ - \alpha} + e_{ -2 \alpha} + \cdots = \sum_{k\geq 0} e_{ -k \alpha}$.
:::

:::{.lemma title="A"}
\envlist

a. $p = \prod_{ \alpha\in \Phi^+} f_ \alpha$,
b. $(e_0 - e_{- \alpha}) \convolve f_ \alpha = e_0$,
c. $q = e_\rho \convolve \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha})$.

:::

:::{.proof title="of lemma A"}
**Part a**:
The coefficient of $e_\mu$ in $\prod_{i=1}^m f_{ \beta_j}$ is the convolution
\[
\sum_{i_1,\cdots, i_m \in \ZZ_{\geq 0}, -\sum i_j \beta_j = \mu } f_{\beta_1}(-i_1 \beta_1) \cdots f_{ \beta_m }(-i_m \beta_m) = p(\mu)
.\]

**Part b**:
\[
(e_0 - e_{- \alpha}) \convolve (e_0 + e_{- \alpha} + e_{-2 \alpha} + \cdots)
= e_0 + e_{ - \alpha} - e_{ - \alpha}+ e_{ -2 \alpha}- e_{ -2 \alpha} +\cdots = e_0
,\]
noting the telescoping. 
This can be checked rigorously by regarding these as functions instead of series.

**Part c**:
Recall $\rho = \sum_{ \alpha\in \Phi^+} {1\over 2}\alpha$, so $e_\rho = \prod_{ \alpha\in \Phi^+} e_{\alpha\over 2}$.
Thus the RHS is 
\[
\prod_{ \alpha\in \Phi^+} \qty{e_{\alpha\over 2} \convolve (e_0 - e_{- \alpha}) } = \prod_{\alpha\in \Phi^+}(e_{\alpha\over 2} - e_{- \alpha\over 2} ) \da q
.\]
Note that $q\neq 0$ since $q(\rho) = 1$.

:::

:::{.lemma title="B"}
Let $w\in W$, recalling that $w.e_{\alpha} = e_{w \alpha}$,
\[
wq = (-1)^{\ell(w)} q
.\]
:::

:::{.proof title="of lemma B"}
ETS for $\alpha \in \Delta$.
Recall $s_ \alpha$ permutes \( \Phi^+ \smts{\alpha} \) and $s_ \alpha(\alpha) = - \alpha$, so $s_ \alpha q$ permutes the factors $(e_{ \beta\over 2} - e_{-{\beta\over 2}})$ for $\beta\neq \alpha$ and negates $e_{\alpha\over 2} - e_{- \alpha\over 2}$.
:::

:::{.lemma title="C"}
$q$ is invertible: 
\[
q\convolve p \convolve e_{\rho} = e_0 \quad \implies q\inv = \rho \convolve e_\rho
.\]
:::

:::{.proof title="of lemma C"}
Use lemma A:
\[
q\convolve \rho \convolve e_{-\rho} 
&= e_\rho \convolve \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha} ) } \convolve p \convolve e_{ - \rho} \qquad \text{by C} \\
&= \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha} ) } \convolve p \\
&= \qty{ \prod_{ \alpha\in \Phi^+} (e_0 - e_{ - \alpha}) \convolve f_{ \alpha} } \qquad \text{by A}\\
&= \prod_{ \alpha\in \Phi^+} e_0 \qquad \text{by B}\\
&= e_0
.\]
:::

:::{.lemma title="D"}
For \( \lambda, \mu\in H\dual \), 
\[
\ch_{M( \lambda)}( \mu) = p( \mu - \lambda) = (p \convolve e_ \lambda)( \mu) \implies \ch_{M( \lambda)} = p \convolve e_{ \lambda}
.\]
:::

:::{.proof title="of lemma D"}
$M( \lambda)$ has basis $y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m} \cdot v^+$ where $v^+$ is a highest weight vector of weight \( \lambda \).
Note that
\[
\mu = \lambda- \sum_{j=1}^M i_j \beta_j \iff \mu- \lambda= - \sum_{i=1}^m i_j \beta_j
,\]
and $\dim M(\lambda)_ \mu = p( \mu- \lambda)$.
Now check $(p\convolve e_ \lambda)( \mu) \da p( \mu - \lambda) e_ \lambda(\lambda) = p( \mu- \lambda)$.
:::

:::{.lemma title="E"}
\[
q \convolve \ch_{M( \lambda)} = e_{ \lambda+ \rho}
.\]
:::

:::{.proof title="of lemma E"}
\[
\text{LHS} 
\equalsbecause{D} q \convolve p \convolve e_ \lambda 
\equalsbecause{C} e_ \rho \convolve e_ \lambda = e_{ \lambda+ \rho}
.\]
:::

## $\S 24.2$ Kostant's Multiplicity Formula

:::{.remark}
Recall that characters of Verma modules are essentially known.
For \( \lambda\in \Lambda^+ \), we have $\ch_{ \lambda} \da \ch_{L( \lambda)}$, recalling that $L( \lambda)$ is a finite-dimensional irreducible representation.
Goal: express this as a finite $\ZZ\dash$linear combination of certain $\ch_{M( \lambda)}$.

Fix \( \lambda\in H\dual \) and let $\mcm_ \lambda$ be the collection of all $L\dash$modules satisfying the following:

1. $V = \bigoplus _{ \mu\in H\dual} V_ \mu$,
2. $\mcz(L)\actson V$ by $\chi_ \lambda$,
3. $\ch_V \in \mcx$

Note that any highest weight module of highest weight \( \lambda \) is in $\mcm_ \lambda$, and this is closed under submodules, quotients, and finite direct sums.
The Harish-Chandra theorem implies that 
\[
\mcm_ \lambda = \mcm_{\mu} \iff \lambda\sim \mu\iff \mu = w. \lambda
.\]
:::

:::{.lemma title="?"}
If $0\neq V \in \mcm_ \lambda$ then $V$ has a maximal vector.
:::

:::{.proof title="?"}
By (3), the weights of $V$ lie in a finite number of cones $\lambda_i - \ZZ_{\geq 0} \Phi^+$.
So if $\mu$ is a weight of $V$ and $\alpha\in \Phi^+$, then $\mu + k \alpha$ is not a weight of $V$ for $k\gg 0$.
Iterating this for the finitely many positive weights, there exists a weight $\mu$ such that $\mu + \alpha$ is not a weight for any $\alpha\in \Phi^+$.
Then any $0\neq v\in V_ \mu$ is a maximal vector.
:::

:::{.definition title="?"}
For $\lambda \in H\dual$, set
\[
\Theta( \lambda) \da \ts{ \mu\in H\dual \st \mu\sim \lambda\, \text{ and } \, \mu\leq \lambda}
,\]
which by the Harish-Chandra theorem is a subset of $W\cdot \lambda$ which is a finite set.
:::

:::{.remark}
The following tends to hold in any setting with "standard" modules, e.g. quantum groups or superalgebras:
:::

:::{.proposition title="?"}
Let $\lambda\in H\dual$, then

a. $M( \lambda)$ has a composition series,
b. Each composition factor of $M( \lambda)$ is of the form $L( \mu)$ for some $\mu\in \Theta( \lambda)$.
Define $[M (\lambda): L( \mu)]$ to be the multiplicity of $L(\mu)$ as a composition factor of $M( \lambda)$.
c. $[M( \lambda): L( \lambda)] = 1$.

:::

:::{.proof title="?"}
By induction on the number of maximal vectors (up to scalar multiples).
If $M( \lambda)$ is irreducible then it's an irreducible highest weight module, and these are unique up to isomorphism and so $M( \lambda) = L( \lambda)$ and we're done.
Otherwise $M( \lambda)$ has a proper irreducible submodule $V$, and $V\in \mcm_ \lambda$ by closure under submodules.
By the lemma, $V$ has a maximal vector of some weight $\mu$ which must be strictly less than $\lambda$, i.e. $\mu < \lambda$.
As before, $\chi_{ \mu} = \chi_ \lambda$ and thus $\mu\in \Theta( \lambda)$.
Consider $V$ and $M( \lambda)/ V$ -- each lies in $\mcm_ \lambda$ and either has fewer weights linked to \( \lambda \) than $M( \lambda)$ has, or else it must have exactly the same set of weights linked to $\lambda$, just with smaller multiplicities.
By induction each of these has a composition series, and these can be pieced together into a series for $M$ since they fit into a SES.
:::