4
$\begingroup$

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 $w=e$?

$\endgroup$
9
  • $\begingroup$ Note that Venkataramana's "correction" of the header is wrong in this context where non-Verma modules occur. $\endgroup$ Aug 1, 2019 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. $\endgroup$ Aug 1, 2019 at 1:55
  • $\begingroup$ @JIm Humphreys: excuse me: I did not correct the header at all! I corrected some grammar at the end. $\endgroup$ Aug 1, 2019 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 ${\Sigma_\mu}$. $\endgroup$ Aug 1, 2019 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.) $\endgroup$ Aug 1, 2019 at 14:05

1 Answer 1

0
$\begingroup$

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 $e\in {}^IW^{\Sigma_\mu}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.