# Monday, December 05

## Kostant's Character Formula

:::{.remark}
Last time: $\mcm_ \lambda$ defined as a certain category of $L\dash$modules for \( \lambda\in H\dual \), and we defined $\theta( \lambda) \da \ts{ \mu\in H\dual \st \mu \sim \lambda,\,\, \mu\leq \lambda} \subseteq W. \lambda$.
Proposition from last time: 

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

Note that any character of $M$ is the sum of the characters of its composition factors.
:::

:::{.proof title="of proposition"}
**Part b**: Each composition factor of $M( \lambda)$ is in $\mcm_ \lambda$, hence by the lemma has a maximal vector.
Since it's irreducible, it is a highest weight module $L( \mu)$ for some \( \mu\in \theta( \lambda) \).

**Part c**:
$[M( \lambda) : L( \lambda)] = 1$ since $\dim M( \lambda)_ \lambda = 1$ and all other weights are strictly less than $\lambda$.
:::

:::{.remark}
Order $\theta( \lambda) = \ts{ \mu_1, \cdots, \mu_t}$ such that $\mu_\leq \mu_j \implies i\leq j$.
In particular, $\mu_t = \lambda$.
By the proposition, $\ch_{M( \mu_j)}$ is a $\ZZ_{\geq 0}\dash$linear combination of $\ch_{L(\mu_i)}$ where $i\leq j$, and the coefficient of $\ch_{L(\mu_j)}$ is 1.
Recording multiplicities in a matrix, we get the following:

\begin{tikzcd}
	& {\ch_{M(\mu_j)}} &&&&& {\,} \\
	{\ch_{L(\mu_i)}} && 1 & \bullet & \bullet & \bullet \\
	&& 0 & 1 & \bullet & \bullet \\
	&& 0 & 0 & \ddots & \bullet \\
	&& 0 & 0 & 0 & 1 \\
	& {\,}
	\arrow[no head, from=1-2, to=6-2]
	\arrow[no head, from=1-2, to=1-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=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)

This is an upper triangular unipotent matrix, and thus invertible.
:::

:::{.corollary title="?"}
Let $\lambda \in H\dual$, then
\[
\ch_{L( \lambda)} = \sum_{ \mu \in \theta( \lambda)} c( \mu) \ch_{M( \mu)}, \qquad c( \mu)\in \ZZ, \, c( \lambda) = 1
.\]
:::

:::{.remark}
Assume \( \lambda\in \Lambda^+ \), and recall:

- $\mcx = \ts{f: H\dual \to \CC \st \text{ the "forest" support condition }}$, 
- $p=\ch_{M(0)}$, 
- $p(\mu) = \size\ts{ \tv{i_1:\cdots:i_m} \in \ZZ_{\geq 0}^m }$,
- $q\da \prod_{ \alpha\in \Phi^+} (e_{\alpha\over 2} - e_{-{\alpha\over 2}})$,
- $\mu = -\sum_{1\leq j\leq m} i_j \beta_j$,
- $wq = (-1)^{\ell(w)} q$ for $w\in W$,
- $q \convolve p \convolve e_{-\rho} = e_0$,
- $\ch_{M(\lambda)} = p \convolve e_ \lambda$,
- $q \convolve \ch_{M( \lambda) } = e_{ \lambda+ \rho}$,
- $\ch_ \lambda\da \ch_{L( \lambda)} = \sum_{ \mu\in \theta( \lambda)} c( \mu) \ch_{M( \mu)}$,
- $q\convolve \ch_{L( \lambda)} = \sum c( \mu) q \convolve \ch_{M( \mu)} = \sum c( \mu) e_{ \mu + \rho}$. $\qquad\star_1$

Fixing $w\in W$, we have
\[
\sum_{ \mu\in \theta( \lambda) } c( \mu) e_{w (\lambda+ \rho)} 
&= w( q \convolve \ch_ \lambda) \\
&= wq \convolve w \ch_{ \lambda} \\
&= (-1)^{\ell(w)} q \ch_ \lambda \qquad \text{since $\ch_ \lambda$ is $W\dash$invariant} \\
&= (-1)^{\ell(w)} \sum_ \mu c( \mu) e_{ \mu+ \rho}
.\]
Since \( \lambda\in \Lambda^+ \), $W$ acts simply transitively on $\theta( \lambda) + \rho \da \ts{v+\rho \st v\in \theta( \lambda) }$.
Note \( \mu \sim \lambda\iff \mu+ \rho = w( \lambda+ \rho) \) for some $w\in W$, which is unique since \( \lambda+ \rho \) is strongly/strictly dominant, and lemma 13.2A shows its stabilizer is the identity.
So $\stab_W( \lambda+ \rho) = \ts{1}$.
The equation $\mu + \rho = w( \lambda+ \rho)$ implies $\mu + \rho \leq \lambda+ \rho$, since apply $W$ to dominant elements goes down in the partial order.
Thus $\mu\in \theta(\lambda)$, and $\theta( \lambda)$ consists of precisely those $\mu$ satisfying this equation, and
\[
\theta( \lambda) = W\cdot \lambda
.\]

Continuing the computation, take $\mu = \lambda$ on the LHS, so $w( \lambda+ \rho) = \mu+ \rho$ and 
\[
c( \lambda) e_{w( \lambda+ \rho)} = (-1)^{\ell(w)} c( \mu) e_{ \mu+ \rho}\implies c(\mu) = (-1)^{\ell(w)} c( \lambda)
.\]

Substituting this into $\star_1$ yields
\[
q \convolve \ch_ \lambda = \sum_{w\in W} (-1)^{\ell(w)} e_{w ( \lambda+ \rho)} \qquad \star_2
,\]
so
\[
\ch_ \lambda 
&= q \convolve p \convolve e_{ - \rho} \convolve \ch_ \lambda\\
&= p \convolve e_{- \rho} \convolve \sum_{w\in W} (-1)^{\ell(w)} e_{w (\lambda+ \rho)} \\
&= \sum_{w\in W} (-1)^{\ell(w)} p \convolve e_{w( \lambda+ \rho) - \rho}
&= \sum_{w\in W} (-1)^{\ell(w)} p \convolve e_{w\cdot \lambda}
.\]

This yields the following:
:::

:::{.theorem title="Kostant"}
For \( \lambda\in \Lambda^+ \) dominant, the weight multiplicities in $L( \lambda)$ are given by 
\[
\dim L( \lambda)_\mu \da m_ \lambda( \mu) 
= \sum_{w\in W} (-1)^{\ell(w)} p( \mu+ \rho - w( \lambda+ \rho))
= \sum_{w\in W} (-1)^{\ell(w)} p( \mu - w\cdot \lambda)
.\]
:::

## $\S 24.3$ Weyl's Character Formula

:::{.lemma title="?"}
\[
q = \sum_{w\in W} (-1)^{\ell(w)} e_{w \rho}
.\]
:::

:::{.proof title="?"}
Take $\lambda = 0$ in $\star_2$, and use that $\ch_0 = e_0$ where $L(0) \cong \CC$.
:::

:::{.theorem title="Weyl's Character Formula"}
Let $\lambda \in \Lambda^+$, then
\[
\qty{ \sum_{w\in W} (-1)^{ \ell(w)} e_{w\rho} } \convolve \ch_{L( \lambda)} = \sum_{w\in W} (-1)^{\ell(w)} e_{w( \lambda+ \rho)}
.\]
:::

:::{.proof title="?"}
Apply $\star_2$ and the lemma.
:::

:::{.corollary title="Weyl's Dimension Formula"}
\[
\dim L( \lambda) = { \prod_{ \alpha\in \Phi^+} \inp { \lambda+ \rho}{ \alpha} \over\prod_{\alpha\in \Phi^+} \inp \rho \alpha } = \sum_{ \mu \in \Pi( \lambda)} m_ \lambda( \mu)
.\]
:::