# Wednesday, November 30 :::{.remark} Recall: $\mcz(L) \da Z(\mu(L))$ acts by a character $\chi_ \lambda: \mcz(L) \to \CC$ on $M( \lambda)$ and thus any subquotients. For \( \lambda, \mu\in \Lambda \subseteq H\dual \), when is $\chi_ \lambda = \chi_ \mu$? ::: :::{.definition title="Linkage"} Two weights \( \lambda, \mu\in H\dual \) are **linked** iff $\exists w\in W$ such that $\mu + \rho = w ( \lambda+ \rho)$, where $\rho \da {1\over 2}\sum_{ \beta\in \Phi^+} \beta$. In this case, write $\mu \sim \lambda$ for this equivalence relation, i.e. $\mu = w( \lambda+ \rho) - \rho$. We'll write $w\cdot \lambda \da w( \lambda+ \rho)$ and call this the **dot action** of $W$ on $H\dual$. ::: :::{.warnings} This defines an action of a group on a set, but it is **not** a linear action. ::: :::{.proposition title="?"} Let \( \lambda\in \Lambda, \alpha\in \Delta \) and suppose $m\da \lambda(h_ \alpha) = \inp \lambda \alpha\in \ZZ_{\geq 0}$ and let $v^+$ be a highest weight vector of $M( \lambda)$. Then $w\da y_ \alpha^{m+1}\cdot v^+$ is maximal vector in $M( \lambda)$ of weight \( \lambda - (m+1 ) \alpha \). ::: :::{.proof title="?"} The proof that $w$ is a maximal vector is step 1 in theorem 21.2, which showed $\dim L( \lambda) < \infty$ (using lemma 21.2 and commutator relations in $\mcu(L)$). Then check that \[ \weight(w) = \lambda - (m+1) \alpha = \lambda- \qty{ \inp \lambda \alpha + 1} \alpha .\] In fact, for any \( \lambda\in H\dual, \alpha\in \Delta \) we can define \[ \mu \da \lambda- \qty{ \inp \lambda \alpha + 1} \alpha ,\] so in our case $\weight(w) = \mu$. Note that \[ \mu &= \lambda- \inp{\lambda} \alpha \alpha- \alpha \\ &= s_ \alpha( \lambda) + (s_ \alpha( \rho) - \rho) \\ &= s_ \alpha(\lambda+ \rho) - \rho\\ &= s_ \alpha \cdot \lambda .\] Now $w$ generates a highest weight module $W\leq M( \lambda)$ of highest weight $\mu = s_ \alpha \cdot \lambda$. Note that $M( \mu)$ is the universal highest weight module with highest weight $\mu$, i.e. $\exists! M(\mu) \surjects W$. This yields a $B\dash$module morphism \[ \CC_ \mu &\to W \\ 1 &\mapsto w ,\] which yields an $L\dash$module morphism $\mcu(L) \tensor_{\mcu(B)} \CC_ \mu \surjects W$. So $W$ is a nonzero quotient of $M( \mu)$ and $\mcz(L)\actson W$ by $\chi_ \mu$. On the other hand $W\leq M( \lambda)$ and so $\mcz(L)\actson W$ by $\chi_ \lambda$, yielding $\chi_ \lambda= \chi_ \mu$. So we conclude that if $\mu = s_ \alpha \cdot \lambda$ with $\inp \lambda \alpha\in \ZZ_{\geq 0}$, then $\chi_ \mu = \chi_ \lambda$. ::: :::{.corollary title="?"} Let \( \lambda\in \Lambda, \alpha \in \Delta, \mu = s_ \alpha \cdot \lambda \). Then $\chi_ \mu = \chi_ \lambda$. ::: :::{.proof title="?"} $\mu = s_ \alpha \cdot \lambda = \lambda- (m+1) \alpha$ where $m \da \inp \lambda \alpha$. - Case 1: $m=-1$, then $\mu = \lambda$ and we're done. - Case 2: $m\geq 0$, then $\chi_ \lambda= \chi_ \mu$ by the proposition. - Case 3: $m\leq -2$, then \[ \inp \mu \alpha &= \inp{ \lambda- (m+1) \alpha}{\alpha} \\ &= m-2(m+1) \\ &= -m-2 \geq 0 .\] We also have $\mu = s_ \alpha\dot \lambda\implies s_ \alpha\cdot \mu= s_ \alpha^2 \cdot \lambda= \lambda$ by applying the action on both sides. By the proposition, swapping \( \mu, \lambda \) to get submodules of $M( \mu)$ of highest weight $\lambda$ and conclude $\chi_ \lambda = \chi_ \mu$. ::: :::{.corollary title="?"} Let \( \lambda, \mu\in \Lambda \), then if $\mu\sim \lambda$ then $\chi_ \lambda=\chi_ \mu$. ::: :::{.proof title="?"} Say $\mu = w\cdot \lambda$ for $w\in W$, then write $w = s_{i_1}\cdots s_{i_t}$ and use induction on $t$, where the base case is the previous corollary. ::: :::{.theorem title="Harish-Chandra ($\S 23.2$)"} If \( \lambda, \mu \) satisfy $\chi_ \lambda= \chi_ \mu$, then \( \lambda\sim \mu \). ::: :::{.remark} Goal: find $\character_ \lambda\da \character_{L( \lambda)} \da \sum_{ \mu\in \Lambda} (\dim L( \lambda)_\mu ) e( \mu) \in \ZZ[ \Lambda]$ for \( \lambda\in \Lambda^+ \). ::: ## $\S 24$: Formulas of Weyl, Kostant, Steinberg; $\S 24.1$ functions on $H\dual$ :::{.remark} View $\ZZ[ \Lambda]$ as finitely-supported $\ZZ\dash$valued functions on \( \Lambda \) with elements $f = \sum_{ \lambda\in \Lambda} a_ \lambda e( \lambda)$ regarded as functions $f(\mu) \da a_\mu$. Thus $e( \lambda)( \mu) = \delta_{\lambda= \mu }$. The point of this maneuver: Verma modules will be infinite dimensional, but $\ZZ[ \Lambda]$ only handles finite sums. For $f,g\in \Hom_\ZZ( \Lambda, \ZZ)$, define $(f\convolve g) ( \sigma) = \sum_{ \lambda+ \mu} f( \lambda) g( \mu)$. Consider the set $\mcx$ of functions $\Hom_\CC(H\dual, \CC)$ whose support is contained in a finite union of sets of the form $\lambda_{ \leq} \da \ts{ \lambda - \sum_{ \mu\in \Phi^+} k_ \beta \beta\st k_ \beta\in \ZZ_{\geq 0}}$. One can show $\mcx$ is closed under convolution and $\mcx$ becomes a commutative associative algebra containing $\ZZ[ \Lambda]$ as a subring. Note that $\supp(f\convolve g) \subseteq ( \lambda+ \mu)_{\leq }$. ![](figures/2022-11-30_09-56-16.png) Write $e( \lambda)$ as $e_{ \lambda}$ for $\lambda \in H\dual$, regarded as a function $e_ \lambda:H\dual \to \CC$ where $e_ \lambda( \mu) = \delta_{ \lambda = \mu}$ is the characteristic function of $\lambda$, and note that $e_ \lambda \convolve e_ \mu = e_ { \lambda+ \mu}$. Let $p = \character_{M(0)}: H\dual \to \CC$, then \[ M(0) = \mcu(L) \tensor_{\mcu(B)} \CC_0 \cong \mcu(N^-) \tensor_\CC \CC_0 \quad\in \mods{H}, \mods{N^-} .\] By PBW, $\mcu(N^-)$ has a basis $y_{\beta_1}^{i_1}\cdots y_{ \beta_m}^{i_m}$ for $i_j\in \ZZ_{\geq 0}$ where $\Phi^+ = \tsl \beta 1 m$. The weights of $M(0)$ are $\mu = - \sum_{j=1}^m i_j \beta_j$ and $\mu\in 0_{\leq}$. Note $\character_{M(0)}( \mu) = \dim M(0)_ \mu$, and so $\character_{M(0)}\in \mcx$ and thus $p\in \mcx$. :::