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<s_\alpha w \ \text{for all }\alpha\in I\}, $ 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<ws_\alpha\in {}^IW\ \text{for all }\alpha\in {\Sigma_\mu}\}, $ 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 $w=e$?

  • $\begingroup$ Note that Venkataramana's "correction" of the header is wrong in this context where non-Verma modules occur. $\endgroup$ – Jim Humphreys Aug 1 '19 at 1:52
  • $\begingroup$ More substantively, I'd start with the case $I=\Delta$. where more is known. The general case is probably too difficult right now. $\endgroup$ – Jim Humphreys Aug 1 '19 at 1:55
  • $\begingroup$ @JIm Humphreys: excuse me: I did not correct the header at all! I corrected some grammar at the end. $\endgroup$ – Venkataramana Aug 1 '19 at 2:36
  • $\begingroup$ If we consider $I=\Delta$, we get $W_I=W$ and hence ${}^I W=\{e\}$. This implies that ${}^IW^{\Sigma_\mu}=\emptyset$ for nonempty ${\Sigma_\mu}$. $\endgroup$ – James Cheung Aug 1 '19 at 7:33
  • $\begingroup$ @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.) $\endgroup$ – Jim Humphreys Aug 1 '19 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 ${}^IW^{\Sigma_\mu} : = \{w\in {}^IW: w<ws_\alpha\in {}^IW\ \text{for all }\alpha\in {\Sigma_\mu}\}$. 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, $e<es_\alpha\in {}^IW$ 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 $e\in {}^IW^{\Sigma_\mu}$.


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