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 $w=e$?
> 
> 
> 



Verma module and vanishing of extension groups
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 $e\in {}^IW^{\Sigma\_\mu}$.


