# Verma module and vanishing of extension groups

Let $$\mathfrak{g}$$ be a finite dimensional complex semisimple Lie algebra with Cartan subalgebra $$\mathfrak{h}$$. Let $$W$$ be the associated Weyl group and let $$\Phi$$ be its root system. We write $$\Phi^+$$ for the set of positive roots in $$\Phi$$.

Fix a subset of simple roots $$I$$ and let $$W_I$$ be the corresponding standard parabolic subgroup of $$W$$, with longest element $$w_I$$ and root system $$\Phi_I\subseteq \Phi$$. Let $$\Phi_I^+:=\Phi_I\cap\Phi^+$$.

Define $$\Lambda^+_I := \{\nu \in \mathfrak{h}^* : \langle\nu,\alpha^\lor\rangle \in \mathbb{Z}^{\ge 0} \ \text{for all }\alpha \in \Phi^+_I\}.$$

Consider $$\lambda \in \Lambda^+_I$$ and assume $$\lambda$$ is integral. We define $${}^IW := \{w\in W: w where $$<$$ is the Bruhat ordering on $$W$$.

Denote by $$\Delta$$ the simple system corresponding to the positive system $$\Phi^+$$ in $$\Phi$$. The orbit $$W\cdot\lambda$$ contains a unique $$\mu\in\mathfrak{h}^*$$ that is antidominant in the sense that $$\langle \mu+\rho,\alpha^{\lor}\rangle\not\in\mathbb{Z}^{>0}$$ for all $$\alpha\in \Phi^+$$, where $$\rho = \frac{1}{2} \sum_{\alpha \in \Phi^+} \alpha$$.

The set of singular simple roots associated to $$\mu$$ in $$\Delta$$ is defined by $${\Sigma_\mu} : = \{\alpha\in \Delta: \langle\mu+\rho,\alpha^\lor\rangle=0\}.$$

The subgroup $$W_{\Sigma_\mu} := \{w\in W: w(\mu+\rho)=\mu+\rho\}\subseteq W$$ is then the isotropy group of $$\mu$$.

Let $${}^IW^{\Sigma_\mu} : = \{w\in {}^IW: w where $$<$$ is the Bruhat ordering on $$W$$.

Let $$M(\eta)$$ be the Verma module with highest weight $$\eta$$ and $$L(\eta)$$ be its unique simple quotient.

Let $$w\in {}^IW^{\Sigma_\mu}$$, suppose $$\mathrm{Ext}^i_{\mathcal{O}}\left(M(w_Ix\cdot\mu),L(w_Iw\cdot\mu)\right)=\{0\}$$ for all $$i\ge 0$$ and for all $$x\in {}^IW^{\Sigma_\mu}-\{w\}$$. Does this imply $$w=e$$?

• Note that Venkataramana's "correction" of the header is wrong in this context where non-Verma modules occur. Commented Aug 1, 2019 at 1:52
• More substantively, I'd start with the case $I=\Delta$. where more is known. The general case is probably too difficult right now. Commented Aug 1, 2019 at 1:55
• @JIm Humphreys: excuse me: I did not correct the header at all! I corrected some grammar at the end. Commented Aug 1, 2019 at 2:36
• If we consider $I=\Delta$, we get $W_I=W$ and hence ${}^I W=\{e\}$. This implies that ${}^IW^{\Sigma_\mu}=\emptyset$ for nonempty ${\Sigma_\mu}$. Commented Aug 1, 2019 at 7:33
• @Venkataramana: Sorry to have attributed this change to you. The string of edits here is hard to follow, and at first I had the impression that you were responsible. At any rate, the current header is more accurate. (But motivation for the question is lacking, and I suspect the general case is very difficult.) Commented Aug 1, 2019 at 14:05

The answer is No. We need the following lemma:

Lemma: $$W_I\cap W_J=W_{I\cap J}$$.

to prove the following proposition:

Proposition: $$e\in {}^IW^{\Sigma_\mu}\iff I\cap \Sigma_\mu=\emptyset$$.

Proof: Recall that $${}^IW^{\Sigma_\mu} : = \{w\in {}^IW: w. Suppose $$e\in {}^IW_{[\lambda]}^{\Sigma_\mu}$$, we have $$s_\alpha\in {}^IW$$ for all $$\alpha\in \Sigma_\mu$$. Then $$W_{\Sigma_\mu}\subseteq {}^IW$$. Since $$W_I\cap {}^IW=\{e\}$$ and , we get $$W_{\Sigma_\mu}\cap W_I=\{e\}$$. Note that $$W_{I\cap\Sigma_\mu}=W_{\Sigma_\mu}\cap W_I$$ and hence $$W_{I\cap\Sigma_\mu}=\{e\}$$. This implies that $$I\cap\Sigma_\mu=\emptyset$$.

Suppose $$I\cap\Sigma_\mu=\emptyset$$. We have $$W_{\Sigma_\mu}\cap W_I=W_{I\cap\Sigma_\mu}=\{e\}$$. Suppose $$W_{\Sigma_\mu}\not\subseteq {}^IW$$ on contrary, there is $$w\in W_{\Sigma_\mu}, w\not\in {}^IW$$. Let $$w=w_1w^{1}$$ where $$w_1\in W_I$$ and $$w^1\in {}^IW$$. Since $$w\not\in {}^IW$$, we have $$w_1\neq e$$. Note that $$w_1\le w$$ implies $$w_1\in W_{\Sigma_\mu}$$ since $$W_{\Sigma_\mu}$$ is a parabolic subgroup of $$W$$. This contradicts to the fact that $$W_{\Sigma_\mu}\cap W_I=\{e\}$$. This implies $$W_{\Sigma_\mu}\subseteq {}^IW$$. In particular, $$e for all $$\alpha\in\Sigma_\mu$$. Therefore, $$e\in {}^IW^{\Sigma_\mu}$$.

It is easy to construct example with $$I\cap \Sigma_\mu\neq\emptyset$$, which does not allow us to have $$e\in {}^IW^{\Sigma_\mu}$$.