According to p.8 of the note KOSZUL DUALITY AND APPLICATIONS IN REPRESENTATION THEORY by Geordie Williamson. enter image description here

Let $M(\eta)$ be the Verma module of weight $\eta$, $L(\eta)$ be its unique simple quotient and $w_0$ be the longest element in $W$.

The strange formula in our notation is $$[M(x\cdot 0):L(y\cdot 0)]=\sum_{i\ge 0}\dim\mathrm{Ext}_{\mathcal{O}}^i(M(w_0x\cdot 0):L(w_0y\cdot 0))$$ for $x,y\in W$.

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

Do we still have $$[M(x\cdot \mu):L(y\cdot \mu)]=\sum_{i\ge 0}\dim\mathrm{Ext}_{\mathcal{O}}^i(M(w_0x\cdot \mu):L(w_0y\cdot \mu))$$ for $x,y\in W^\Sigma$?


2 Answers 2


No. If the left-hand side is in the singular block, the right hand-side should be in the regular parabolic block, i.e. its Koszul dual.

(The regular block for the Borel subalgebra is Koszul-self-dual.)

See Beilinson-Ginzburg-Soergel: Koszul Duality Patterns in Representation Theory, J. Amer. Math. Soc. 9 (1996), 473-527.

EDIT: A counterexample to your question is in my answer to your previous similar question: https://mathoverflow.net/a/339679/15292


This is a more concrete explanation for what Rafael Mrđen had said.

Let $x,w\in W^\Sigma$ and $\ell(x,w)=\ell(w)-\ell(x)$.

It is well-known that $\mathrm{ch}L(w\cdot\mu)=\sum_{x\in W^\Sigma}(-1)^{\ell(x,w)}P^{\Sigma}_{x,w}(1)\mathrm{ch}M(x\cdot\mu)$, where $P^\Sigma_{x,w}(q)=\sum_{i\ge 0}q^{\frac{\ell(x,w)-i}{2}}\dim\mathrm{Ext}_{\mathcal{O}}^i(M(x\cdot\mu),L(w\cdot\mu))$.

Let $P_{u,v}(q)$ be the Kazhdan Lusztig polynomial of $W$.

Let $P_{u,v}^{\Sigma,q}(q)$ be the parabolic Kazhdan Lusztig polynomial of $W^\Sigma$ of type $q$.

Let $P_{u,v}^{\Sigma,-1}(q)$ be the parabolic Kazhdan Lusztig polynomial of $W^\Sigma$ of type $-1$.

By Section 8.2 of KOSTANT MODULES IN BLOCKS OF CATEGORY $\mathcal{O}_S$ (taking $S=\emptyset$, $J=\Sigma$), it holds that $P^\Sigma_{x,w}(q)=\sum_{t\in W_\Sigma}(-1)^{\ell(t)}P_{xt,w}(q)$.

Note that $W_\Sigma=\{w\in W: w\cdot\mu=\mu\}=\langle s_\alpha\in W: \alpha\in\Sigma\rangle$. It is well-known that $P_{x,w}^{\Sigma,q}(q)=\sum_{t\in W_\Sigma}(-1)^{\ell(t)}P_{xt,w}(q)$. Hence, $P^\Sigma_{x,w}(q)=P_{x,w}^{\Sigma,q}(q)$. Let $w_\Sigma$ be the longest element in $W_\Sigma$. It is well-known that $P_{x,w}^{\Sigma,-1}(q)=P_{xw_\Sigma,ww_\Sigma}(q)$.

Now we try to compute $[M(x\cdot\mu):L(y\cdot\mu)]$ by using inverse parabolic Kazhdan Lusztig polynomials.

By Corollary 3.7 (iii) of Hiroyuki Tagawa---Some Properties of Inverse Weighted Parabolic Kazhdan Lusztig Polynomials. It holds that $\sum_{x\le w\le y, w\in W^\Sigma}(-1)^{\ell(x)+\ell(w)}P_{x,w}^{\Sigma,q}(q)P_{w_0yw_\Sigma,w_0ww_\Sigma}^{\Sigma,-1}(q)=\delta_{x,y}=\begin{cases} 1, x=y\\ 0, x\neq y\\ \end{cases}$.

Since $x\not\le w\implies P_{x,w}^{\Sigma,q}(q)=0$, $w\not\le y\iff w_0yw_\Sigma\not\le w_0ww_\Sigma\implies P_{w_0yw_\Sigma,w_0ww_\Sigma}^{\Sigma,-1}(q)=0$ and $(-1)^{\ell(x)+\ell(w)}=(-1)^{\ell(x,w)}$. It holds that $\sum_{w\in W^\Sigma}(-1)^{\ell(x,w)}P_{x,w}^{\Sigma}(q)P_{w_0y,w_0w}(q)=\delta_{x,y}$.

Then $$\sum_{w\in W^\Sigma}P_{w_0y,w_0w}(1)\mathrm{ch}L(w\cdot\mu)$$ $$=\sum_{w\in W^\Sigma}\sum_{x\in W^\Sigma}(-1)^{\ell(x,w)}P^{\Sigma}_{x,w}(1)P_{w_0y,w_0w}(1)\mathrm{ch}M(x\cdot\mu)$$ $$=\sum_{x\in W^\Sigma}\sum_{w\in W^\Sigma}(-1)^{\ell(x,w)}P^{\Sigma}_{x,w}(1)P_{w_0y,w_0w}(1)\mathrm{ch}M(x\cdot\mu)$$ $$=\sum_{x\in W^\Sigma}\delta_{x,y}\mathrm{ch}M(x\cdot\mu)$$ $$=\mathrm{ch}M(y\cdot\mu).$$

Hence $$[M(y\cdot\mu):L(w\cdot\mu)]=P_{w_0y,w_0w}(1)$$ and then $$[M(x\cdot\mu):L(y\cdot\mu)]$$ $$=P_{w_0x,w_0y}(1)$$ $$=\sum_{i\ge 0}\dim\mathrm{Ext}_\mathcal{O}^i(M(w_0x\cdot(-2\rho)),L(w_0y\cdot(-2\rho))).$$.


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