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.



Poincaré Polynomial and Counting Rational Points
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.


