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I am currently reading a paper from Sankaran and Vanchinathan where they compute certain Kazhdan-Lusztig polynomials.

Sankaran, P.; Vanchinathan, P.: Small resolutions of Schubert varieties and Kazhdan-Lusztig polynomials. Publ. Res. Inst. Math. Sci. 31 (1995), no. 3, 465-480.

Let $G$ be a complex semisimple algebraic group with Weyl group $W$ and $\tau \leq \lambda$ elements in $W$. For some $\lambda$, they can take resolutions called small and use the fact that the Kazhdan-Lusztig polynomial $P_{\tau, \lambda}(q)$ is equal to the Poincaré polynomial in $q^{1/2}$ of the fibre of $\tau P/P$. The resolutions are constructed as a tower of locally trivial fibrations with fibers being Schubert varieties.

My question concerns the theory behind their computation of the Poincaré polynomial. Let $F_q$ be a finite field with $q$ elements. Then they claim that the value of the polynomial at $q$ is given by the number of rational points of the fibre if the varieties are considered over $F_q$ (all varieties are well defined over any field in this case). Why is this true? I guess this derives from a more general and well known theorem about the Poincaré polynomial and counting points over finite fields.

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I don't think the relationsip is so simple in general, so it probably depends on some special properties of the varieties under consideration. One condition that might be useful is if the varieties have a cellular decomposition; of course Schubert varieties do, but from your question it is not clear exactly which varieties they are computing the Poincare polynomials of. –  ulrich Mar 26 '12 at 13:34
    
They are constructed as a tower of locally trivial fibrations with fibers being Schubert varieties. I will add that to the question. –  Benjamin Schmidt Mar 26 '12 at 13:57

2 Answers 2

up vote 5 down vote accepted

Disclaimer: I know almost nothing about Schubert varieties and K-L polynomals.

The general relationship between $\mathbf F_q$-points and cohomology is the Grothendieck-Lefshetz trace formula $$ \# X(\mathbf F_q) = \sum (-1)^i \mathrm{tr} (\mathrm{Frob}_q | H^i_c(X \otimes \mathbf{\overline F}_q,\mathbf Q_\ell)).$$ If $X$ is proper, so $H^i = H^i_c$, then this formula shows that the Poincaré polynomial gives the number of $\mathbf F_q$-points if all the eigenvalues of $\mathrm{Frob}_q$ on $H^i$ are equal to $q^{i/2}$. This is usually stated by saying that $H^i$ is "pure Tate of weight $i$".

A notable example of when you know that the cohomology is pure and Tate type is when the variety $X$ has an algebraic cell decomposition, i.e. a stratification where each stratum is isomorphic to an affine $n$-space $\mathbf A^n$. To prove this, one needs to use the long exact sequence of Galois representations $$ \cdots \to H^i_c(U) \to H^i(X) \to H^i(Z) \to H^{i+1}_c(U) \to \cdots, $$ where $U \subset X$ is open and $Z$ its closed complement. One takes $U \cong \mathbf A^n$ an open stratum of $X$. $Z$ has cohomology of pure Tate type (and in particular no odd cohomology!) by noetherian induction. This proves the claim for Grassmannians (the Bruhat decomposition) and also general Schubert varieties.

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I added the fact that the resolution is constructed as a tower of locally trivial fibrations with fibers being Schubert varieties. Therefore, I am really in the situation you describe. Where do you actually use the smoothness of the Schubert variety here? –  Benjamin Schmidt Mar 26 '12 at 14:01
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Smoothness is not needed since for projective varieties $H^i = H^i_c$ for all $i$. –  ulrich Mar 26 '12 at 14:34
    
Thanks. This seems to be exactly what I was searching for. Is there any nice reference for these statements? –  Benjamin Schmidt Mar 26 '12 at 14:39
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Thanks ulrich! I must've been thinking it would be needed at some point since smooth + proper implies purity, but you are right that the cell decomposition implies the purity even if you are in the non-smooth case. I edited to reflect this. –  Dan Petersen Mar 26 '12 at 15:06
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@Benjamin: everything I wrote in my answer is standard étale cohomology. The original reference is SGA4-5 (which I have not read), but you could look at Freitag & Kiehl's book. –  Dan Petersen Mar 26 '12 at 15:15

This is worked out in detail in the special case of type A in:

MR0646823 (83i:14045) Lascoux, Alain ; Schützenberger, Marcel-Paul Polynômes de Kazhdan & Lusztig pour les grassmanniennes. Young tableaux and Schur functors in algebra and geometry (Toruń, 1980), pp. 249--266, Astérisque, 87–88, Soc. Math. France, Paris, 1981.

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