Poincaré Polynomial and Counting Rational Points

Question

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.

Answer 1

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.

Answer 2

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.