3
$\begingroup$

Let $P_{x,w}(q)$ be the Kazhdan Lusztig polynomial. It is well-known that $P_{x,w}(q)\neq 0\iff x\le w$.

By the interpretation of the Kazhdan Lusztig polynomial in terms of extension group, it holds that $x\le w \iff \mathrm{Ext}_{\mathcal{O}}^i(M(x\cdot(-2\rho)),L(w\cdot(-2\rho)))\neq 0$ for some $i\ge 0$, where $M(\eta)$ is the Verma module, $L(\eta)$ is its unique simple quotient and $\rho$ is the half sum of positive roots.

Let $\mu$ be an integral, antidominant weight, $\Delta$ be the set of simple roots, $\Sigma=\{\alpha\in\Delta:\langle\mu+\rho,\alpha^\lor\rangle=0\}$, $W^{\Sigma}=\{w\in W:w<ws_\alpha,\forall \alpha\in\Sigma_\mu\}$ and $x,w\in W^{\Sigma}$.

It is not hard to show $\mathrm{Ext}_{\mathcal{O}}^i(M(x\cdot\mu),L(w\cdot\mu))\neq 0$ for some $i\ge 0\implies x\le w$.

I would like to know whether the converse:

$x\le w\implies \mathrm{Ext}_{\mathcal{O}}^i(M(x\cdot\mu),L(w\cdot\mu))\neq 0$ for some $i\ge 0$

is true or not. If it is true, I would like to know why. If not, any counter-example?

$\endgroup$
  • $\begingroup$ Maybe you wanted to write $Ext(M(x \cdot \mu), L(w \cdot \mu))$? $\endgroup$ – Rafael Mrđen Sep 2 at 9:47
  • $\begingroup$ Thank you for reminding me. It is a typo and I have corrected them accordingly. $\endgroup$ – James Cheung Sep 2 at 10:01
2
$\begingroup$

It is not true in general. Take $\mathfrak{sl}_3$, with simple reflections $s,t$, such that $s$ is singular. Put $x=e$, $w= st$; both are in $W^\Sigma$.

I claim that $Ext_\mathcal{O}^i(M(\mu),L(st \cdot \mu)) = 0$ for all $i \geq 0$.

Denote by $\lambda$ a regular, integral antidominant weight. By [2, Corollary 1.3.3.] we have

$$ \dim Ext_\mathcal{O}^i(M(\mu),L(st \cdot \mu)) = \begin{cases} \dim Hom(M(\lambda),L(st \cdot \lambda))=0 & \colon i=0 \\ \dim Ext_\mathcal{O}^i(M(\lambda),L(st \cdot \lambda)) - \dim Ext_\mathcal{O}^{i-1}(M(s \cdot \lambda),L(st \cdot \lambda)) &:i>0 \end{cases}$$.

We know that $P_{e,st} = P_{s,st} = 1$. From KL polynomial in terms of extension groups in the regular case [1, Theorem 8.11.], it follows that the above always cancels. (For $i=2$ we have $1-1=0$.)


[1] Humphreys: Representations of semisimple Lie algebras in the BGG category O. Graduate Studies in Mathematics, 94. American Mathematical Society, Providence, RI, 2008.

[2] Irving: Singular blocks of the category O, Math Z (1990) 204: 209.


EDIT: General explanation: A singular block of $\mathcal{O}$ is Koszul-dual to the regular block of the parabolic category $\mathcal{O}^{\mathfrak{p}}$, where $\mathfrak{p}$ is given by the same singularity set. By this duality, the dimensions of Ext-group correspond to the multiplicities of simples inside generalized Verma modules. But strange things may happen with simples inside generalized Vermas, in particular, some simples that are "expected" to appear by the Bruhat order, may not appear. This is because a generalized Verma is a quotient of the usual Verma, where you kill all the submodules that are not in $\mathcal{O}^{\mathfrak{p}}$. But a lot of composition factors that actually exist in $\mathcal{O}^{\mathfrak{p}}$ are killed in the quotient.

$\endgroup$
  • $\begingroup$ I want to know more about Koszul-dual and the fact that the dimensions of Ext-group correspond to the multiplicities of simples inside generalized Verma modules. Do you have a reference for that? Please help me. $\endgroup$ – James Cheung Sep 3 at 6:22
  • $\begingroup$ You can read a bit in Humphreys Category O book, the last section. The canonical reference is Beilinson-Ginzburg-Soergel: Koszul Duality Patterns in Representation Theory, J. Amer. Math. Soc. 9 (1996), 473-527. $\endgroup$ – Rafael Mrđen Sep 3 at 6:55
  • $\begingroup$ More precisely, Proposition 1.3.1. in the paper I mentioned. $\endgroup$ – Rafael Mrđen Sep 3 at 6:56
  • $\begingroup$ Thank you very much. $\endgroup$ – James Cheung Sep 3 at 7:02
  • $\begingroup$ @Rafael: What do you mean by "$s$ is singular" in your second sentence? $\endgroup$ – Jim Humphreys Sep 7 at 21:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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