# Wednesday, October 28 ## Review of Last Time Suppose we have two weights in the same facet, i.e. they're in the same stabilizer under the action of the affine Weyl group: ![Weights in the same facet](figures/image_2020-10-28-13-56-17.png) We had a theorem: if $\lambda, \mu$ are in the same facet, then $\mathcal{B}_\lambda \cong \mathcal{B}_\mu$ is an equivalence of categories, where the map is via the translation functors. ## Description of $T_\lambda^\mu {H^i(w\cdot \lambda) }$ We can write \[ T_\lambda^\mu \qty{H^i(w\cdot \lambda)} &= \pr_\mu \qty{L(\nu_1) \tensor \pr_\lambda\qty{H^i(w\cdot \lambda)} } \\ &= \pr_\mu \qty{L(\nu_1) \tensor {H^i(w\cdot \lambda)} } \\ &= \pr_\mu \qty{L(\nu_1) \tensor {R^i \ind_B^G w\cdot \lambda} } \\ &= \pr_\mu\qty{R^i \ind_B^G \qty{L(\nu_1) \tensor w\cdot \lambda } } .\] Take a composition series by $B\dash$modules of $L(\nu_1) \tensor w\cdot \lambda$, say \[ 0 = M_0 \subseteq M_1 \cdots \subseteq M_r = L(\nu_1) \tensor w\cdot \lambda .\] where $M_j / M_{j-1} \cong \lambda+j + w\cdot \lambda$ and $\lambda_j < \lambda_{j'} \implies j < j'$, i.e. we can order them in a decreasing way. Consider the SES \begin{tikzcd} 0 \ar[r] & M_{j-1} \ar[r] & M_j \ar[r] & M_{j} / M_{j-1} \ar[r] & 0 \end{tikzcd} where applying $\pr_\mu(\wait)$ induces the LES \begin{tikzcd} \cdots \ar[r] & \pr_\mu M_{j-1} \ar[r] & \pr_\mu M_j \ar[r] & \pr_\mu \left( M_{j} / M_{j-1} \right) \ar[r] & \cdots \end{tikzcd} We know that \[ \pr_\mu H^i\qty{\lambda_j + w\cdot \lambda} = \begin{cases} H^i(\lambda_j + w\cdot \lambda ) & \lambda+j + w\cdot \lambda \in W_p\cdot \mu \\ 0 & \text{else} \end{cases} ,\] i.e. this projection is the identity for weights linked to $\mu$ and zero otherwise. We also have \[ \pr_\mu H^i(M_r) = T_\lambda^\mu H^i(w\cdot \lambda) .\] :::{.theorem title="?"} Let $\lambda, \mu \in \bar{C}_\ZZ$ and $F$ be a facet with $\lambda \in F$. If $\mu \in \bar{F}$, then we have \[ T_\lambda^\mu\qty{H^i(w\cdot \lambda)} = H^i(w\cdot \mu) \qquad \forall w\in W_p .\] ::: :::{.example title="?"} ![Image](figures/image_2020-10-28-14-12-31.png) Here consider $H_0(\lambda) \mapsvia{T_\lambda^\mu} H_0(\mu) = 0$, since $\mu$ is outside of the dominant region (in orange.) We also have $H^0(w\cdot \lambda) \to H^0(w\cdot \mu) \neq 0$, since this falls *into* the dominant region. ::: :::{.proof title="?"} Let $\lambda \in F$ and $\mu\in\bar{F}$. Then $\stab_{W_p}(\lambda) \subseteq \stab_{W_p}(\mu)$. By a previous technical lemma, we had a formula for computing $\ch T_\lambda^\mu V$, which involved considering \[ w_1 \in {\stab_{W_p}(\lambda) \over \stab_{W_p}(\lambda) \intersect \stab_{W_p}(\mu)} .\] In this case, we get $w_1 = \id$, since the top and bottom are equal. By that lemma, there exists a unique $\ell$ such that $w\cdot \lambda + \lambda_\ell \in W_p\cdot \mu$, where $\lambda_\ell$ is a weight of $L(\nu_1)$. From the LES, we have \begin{tikzcd} \cdots \ar[r] & \pr_\mu M_{j-1} \ar[r] & \pr_\mu M_j \ar[r] & \pr_\mu \left( M_{j} / M_{j-1} \right) = \lambda_j + w\cdot \lambda \ar[r] & \cdots \end{tikzcd} where the last term will only be nonzero in restricted cases. We can thus conclude that \[ \pr_\mu(H^i(M_j)) = \begin{cases} 0 & j< \ell \\ H^i(w\cdot \mu) & j\geq \ell. \end{cases} \] Setting $j=r$, we have \[ T_\lambda^\mu \qty{H^i(w\cdot\lambda)} = \pr_\mu H^j(M_r) = H^i(w\cdot \mu) .\] ::: Suppose $\lambda \in \bar{C}_\ZZ$ and $\mu \in C_\ZZ$. What happens when you translate $\lambda$ (blue) off of a wall? $T_\lambda^\mu\qty{H^0(w\cdot \lambda)}$ has a filtration with factors $H^0(w_1\cdot \mu)$ and $H^0(w_2\cdot \mu)$ (shown in green). ![Filtration coming from translating off of a wall](figures/image_2020-10-28-14-25-12.png){width=350px} If $w\lambda$ is a vertex with $\mu \in C_\ZZ$, then $T_\lambda^\mu(H^0(w\cdot \lambda))$ has six factors: ![Weight where the translation has six factors](figures/image_2020-10-28-14-27-05.png){width=350px} :::{.proposition title="?"} Suppose $\lambda \in \bar{C}_\ZZ$ and $\mu \in C_\ZZ$, and let $w\in W_p$ where $w\cdot \lambda \in X(T)_+$. Then $T_\lambda^\mu (H^0(w\cdot \lambda))$ has a filtration such that all of the composition factors are of the form $H^0(ww_1 \cdot \mu)$ where $w_1\in \stab_{W_p}(\lambda)$ and each of the factors occurs at most once. ::: Recall that $\hat F$ denotes the *upper closure*. :::{.proposition title="?"} Let $\lambda, \mu \in \bar{C}_\ZZ$ be in the bottom alcove, where $\mu \in \bar{F}_1$ but $\lambda\in F_1$. Let $F$ be the facet containing $w\cdot\lambda$, then \[ T_\lambda^\mu(L(w\cdot \lambda)) = \begin{cases} L(w\cdot \mu) & w\cdot \mu \in \hat{F} \\ 0 & \text{else}. \end{cases} \] ::: :::{.example title="?"} In this situation, we have $T_\lambda^\mu(L(\lambda)) = 0$: ![Image](figures/image_2020-10-28-14-35-11.png){width=350px} If instead $\mu \in \hat{C}_\ZZ$, we have $T_\lambda^\mu( L(\lambda)) = L(\mu)$: ![Image](figures/image_2020-10-28-14-36-02.png){width=350px} ::: **Big Question**: What happens to $L(w\cdot \lambda)$ when translating away from a wall? :::{.definition title="Walls"} A facet $F$ is a **wall** $\iff \abs{ \Phi_0^+(F) } = 1$. In this case, there exists a unique $\alpha$ such that $\inner{\lambda + \rho}{\alpha\dual} = n_\alpha p$. ![Example of a wall](figures/image_2020-10-28-14-39-23.png) ::: :::{.remark} Note that $s_F = s_{\beta, n_p}$ where $n_p = \inner{\lambda + \rho}{\beta\dual}$ acts on the wall as the identity and reflects across it: ![Image](figures/image_2020-10-28-14-41-36.png) Here $\stab_{W_p}(\lambda) = \ts{1, s_F}$. ::: :::{.proposition title="?"} Consider the following situation: ![Image](figures/image_2020-10-28-14-43-39.png) 1. $[T_\mu^\lambda (L(w\cdot \mu)) : L(w\cdot \lambda)] = 2$, appearing once in the socle and once in the head. 2. $L(w\cdot \lambda) = \soc_G T_\mu^\lambda(L(w\cdot \mu)) = T_\mu^\lambda(L(w\cdot \mu)) / \Rad T_\mu^\lambda (L(w\cdot \mu))$. ![Heart of the module](figures/image_2020-10-28-14-46-57.png) ::: **Big Problems**: 1. When is the heart semisimple? 2. Determine the composition factors in the heart? Given these, you could compute dimensions of irreducible representations.