# Friday, October 09 Last time: Bott-Borel-Weil. Stated for characteristic zero, working toward a generalization. Let $\Delta$ be the set of simple roots, and $\alpha\in \Delta$. We can form a Levi decomposition $P_\alpha \da L_\alpha \semidirect U_\alpha$: ![Image](figures/image_2020-10-09-13-58-02.png) We have $B \subseteq P_\alpha \subseteq G$. The dot action is given by the following: Let $W$ be the Weyl group, then $W$ acts on $X(T)$ by $w\cdot \lambda = w(\lambda + \rho) - \rho$, where \[ \rho = {1\over 2} \sum_{\alpha\in \Phi^+} \alpha = \sum_{i=1}^n w_n .\] We obtained a formula \[ S_\alpha \cdot \lambda = \lambda - \inner{\lambda + \rho}{\alpha\dual} \alpha .\] ## Bott-Borel-Weil Theory :::{.proposition title="?"} Let $\alpha\in\Delta$ be simple and $\lambda \in X(T)$ be an arbitrary weight. Then - $U_\alpha$ acts trivially on $\ind_B^{P_\alpha} \lambda$. - (Kempf's Vanishing for $P_\alpha$) If $\inner{\lambda}{\alpha\dual} = r \geq 0$, then \[ R^i \ind_B^{P_\alpha} \lambda = 0 \qquad \text{for } i \geq 0 ,\] and $\dim \ind_B^{P_\alpha}\lambda = r + 1$. - If $\inner{\lambda}{\alpha\dual} = -1$, then $R^i \ind_B^{P_\alpha} \lambda = 0$ for all $i$. - If $\inner{\lambda}{\alpha\dual} \leq -2$, then - $R^i \ind_B^{P_\alpha} \lambda = 0$ for $i \neq 1$, and - $\dim R^1 \ind_B^{P_\alpha} \lambda = r+1$ Note: we have \[ \ind_B^{P_\alpha} \lambda = S^r(V) \qquad &\text{when } \inner{\lambda}{\alpha\dual} = r \geq 0 \\ R^1 \ind_B^{P_\alpha} = S^r(V)\dual \qquad&\text{where $V$ is a 2-dim representation and } \inner{\lambda}{\alpha\dual} \leq -2 \\ &\text{and } r = \abs{\inner{\lambda}{\alpha\dual}} - 1 .\] ::: This gives us an analog of $A_1$ or $\SL_2$ theory. Also note that we have Serre duality: \[ H^1(\lambda) = H^0( - (\lambda + 2\rho) )\dual .\] :::{.corollary title="?"} Let $\alpha\in \Delta$ and $\lambda\in X(T)$, and suppose $\lambda$ is dominant with respect to $\alpha$, i.e. $\inner{\lambda}{\alpha\dual} \geq 0$. - If $\ch(k) = 0$ then $\ind_B^{P_\alpha}\lambda = R^1 \ind_B^{P_\alpha} s_\alpha \cdot \lambda$ - If $\ch(k) = p$ and if there exists an $s, m$ with $0 \dim P_\alpha/B = 1 .\] So the resulting spectral sequence will only be supported on the first two lines, and $E_3 = E_\infty$. Note the differential will be of bidegree $\del_r \leadsto (r, 1-r)$, and $E_2$ will look like the following, ![Image](figures/image_2020-10-09-14-30-47.png) Recall that $R^i \ind_B^G \lambda \da H^i(\lambda)$ :::{.proposition title="?"} Let $\alpha\in\Delta$ and $\lambda \in X(T)$. 1. If $\inner{\lambda}{\alpha\dual} = -1$, then $H^\wait(\lambda) = 0$. 2. If $\inner{\lambda}{ \alpha\dual} \geq 0$, then $H^i(\lambda) = R^i \ind_B^{P_\alpha} \lambda$ for all $i\geq 0$. 3. If $\inner{\lambda}{\alpha\dual} \leq -2$, then \[ H^i(\lambda) = R^{i-1} \ind_{P_\alpha}^G \qty{ R^1 \ind_B^{P_\alpha} \lambda } \qquad \forall i .\] 4. Suppose $\inner{\lambda}{\alpha\dual} \geq 0$. If $\ch(k) = 0$, or $\ch(k) = p> 0$ and $\inner{\lambda}{\alpha\dual} = sp^n - 1$, then \[ H^i(\lambda) = H^{i+1}(s_\alpha\cdot \lambda) .\] ::: :::{.proof title="of a"} If $\inner{\lambda}{\alpha\dual} = -1$, then $R^\wait \ind_B^{P_\alpha} \lambda = 0$. But this is what appears as the "coefficients" in the spectral sequence, so $E_2^{\wait, \wait} = 0$ and this $R^\wait \ind_B^{P_\alpha} = 0$. ::: :::{.proof title="of b"} If $\inner{\lambda}{\alpha\dual} = 0$, then $R^j \ind_B^{P_\alpha} \lambda = 0$ for all $j>0$. Thus only the bottom line survives, and the spectral sequence degenerates on page 2. Thus $E_2^{1, 0} = R^i \ind_B^G \lambda$, where the LHS is equal to $R^i \ind_{P_\alpha}^G \qty{\ind_B^{P_\alpha} \lambda }$. ::: :::{.proof title="of c"} If $\inner{\lambda}{\alpha\dual} = -2$, then $R^i \ind_B^{P_\alpha} \lambda = 0$ for $i\neq 1$, so only $i=1$ survives Then \[ R^{i-1} \ind_{P_\alpha}^G \qty{ \ind_B^{PP_\alpha} \alpha} = R^i \ind_B^G \lambda ,\] so there is some dimension shifting. ::: :::{.proof title="of d"} If $\inner{\lambda}{\alpha\dual} \geq 0$, then by (b), \[ H^i(\lambda) &= R^i \ind_{P_\alpha}^G \qty{ \ind_B^{P_\alpha} \lambda } && \text{by c}\\ &= R^i \ind_{P_\alpha}^G \qty{ R^1 \ind_B^{P_\alpha} s_\alpha\cdot \lambda } && \text{by corollary}\\ &= H^{i+1}(s_\alpha\cdot \lambda) .\] We can then check that \[ s_\alpha \cdot \lambda &= \lambda - \inner{\lambda + \rho}{\alpha\dual}\alpha \\ &= \lambda - \qty{ \inner{\lambda}{\alpha\dual} + 1 }\alpha && \text{using } \inner{\rho}{\alpha\dual} = 1 \\ \\ \implies \inner{s_\alpha \cdot \lambda}{\alpha\dual} &= \inner{\lambda}{\alpha\dual} - \qty{ \inner{\lambda}{\alpha\dual}+1 }\inner{\alpha}{\alpha\dual} \\ &= \inner{\lambda}{\alpha\dual} - \qty{ \inner{\lambda}{\alpha\dual}+1 }2 \\ &= -\inner{\lambda}{\alpha\dual} - 2 \\ &\leq -2 .\] ::: Now define \[ \bar{C}_{\ZZ} &\da \ts{ \lambda \in X(T) \st 0 \leq \inner{\lambda+\rho}{\beta\dual} \,\forall \beta \in \Phi^+ } \qquad\text{ if } \ch(k) = 0 \\ &\da \ts{ \lambda \in X(T) \st 0 \leq \inner{\lambda+\rho}{\beta\dual} \leq \ch(k) \,\forall \beta \in \Phi^+ } \qquad\text{if } \ch(k) = p .\] Idea: ![Image](figures/image_2020-10-09-14-45-08.png) ![Image](figures/image_2020-10-09-14-45-20.png) :::{.theorem title="Bott-Borel-Weil Generalization, due to Andersen"} a. If $\lambda \in \bar{C}_\ZZ$ and $\lambda \not\in X(T)_+$, then $H^0(w\cdot \lambda) = 0$. b. If $\lambda \in \bar{C}_\ZZ \intersect X(T)_+$, then for all $w\in W$, \[ H^i(w\cdot \lambda) = \begin{cases} H^0(\lambda) & i= \ell(w) \\ 0 & \text{otherwise} \end{cases} .\] ::: Note that this covers everything in the $\ch(k) = 0$ case, but only gives the following hexagon in the $\ch(k) = p$ case: ![Image](figures/image_2020-10-09-14-48-41.png) :::{.remark} **Open Problem**: Determine $\ch H^i(\lambda)$ for $\lambda\in X(T)$ in characteristic $p>0$. Andersen provided necessary an sufficient conditions for $H^1(\lambda) \neq 0$ and computed $\soc_G H^1(\lambda)$. :::