# Richardson varieties over finite fields

Let G be a semisimple algebraic group over some algebraically closed field K, and suppose we have fixed a Borel subgroup B and an opposite Borel subgroup B'. Let P be a parabolic subgroup containing B and consider the partial flag variety X = G/P. The closures of the B-orbits on X are the Schubert varieties, and the closures of the B'-orbits on X are the opposite Schubert varieties. Their intersections are Richardson varieties.

As far as I am aware, all of this stuff can be defined over the integers (anyway I'm only interested in the case when G is a symplectic or orthogonal group), so we can ask about how many F_q-rational points these various varieties have. For a Schubert varieties Z I know how to do this: consider the poset of Schubert varieties contained in Z. It's graded, and specializing q into the rank generating function of this poset gives me the number of points. For opposite Schubert varieties it's similar.

My question is what can one say about Richardson varieties? What I want to say is to just do the same thing: the Schubert varieties correspond to lower intervals in the Bruhat order, and the Richardson varieties correspond to arbitrary intervals. So it's tempting to say just take the rank generating function of this interval and specialize at q. Does this work? What if I assume something like G/P is minuscule? If that's not enough, I really only care about the case when the opposite Schubert variety is replaced by the open B' orbit on X.

A related question: do the Richardson varieties have nice cell decompositions like the Schubert varieties?

My feeling is that something like this is relatively easy and known (I just can't find it), or hopelessly complicated. Please inform me that I'm in the first case.

The intersections of opposite Schubert cells have a very nice decomposition into products of tori and affine spaces due to Deodhar which, of course, induces such a decomposition of the Richardson. This decomposition is defined over $\mathbb{Z}$ (actually it works in any building), so it lets you count points, and the strata are combinatorially described by special subwords of a reduced decomposition of one of the words. I recommend reading the paper of Marsh and Rietsch.
• I second Ben's suggestion. Specifically, you want to look at section 4.4 and section 3. They explain exactly the Coxeter combinatorics which governs the stratification. (They're working over $G/B$ not $G/P$, but I doubt that makes a big difference.) The restrictions that you suggest don't seem to simplify the question enough that it has an obviously easily-statable answer. – Hugh Thomas supports Monica Nov 12 '09 at 22:31
• Indeed it doesn't make a big difference: for each open Richardson in $X^\circ_w \cap X^v_\circ \subseteq G/P$, the open Richardson $X^\circ_w \cap X^v_\circ \subseteq G/B$ maps to it isomorphically, if $w,v \in W^P$ i.e. are minimal coset representatives in $W/W_P$. – Allen Knutson Mar 19 '15 at 14:22
This is both well known and complicated. The number of points on a Richardson variety over $\mathbb{F}\_q$ is given by the R-polynomials of Kazhdan and Lusztig. These are not the more famous Kazhdan-Lusztig polynomials, but they are related and are introduced in the same paper. There is a simple recursion for R-polynomials (see the Wikipedia link). My impression (but I am not an expert) is that experts do not think there will ever be a simple description like the rank generating function of some simple poset. It is true that, in many small cases, the description you give as the rank generating function of an interval is correct. I believe that there should be a geometric description of when this happens (maybe if the Richardson is smooth?) and also a pattern avoidance description, but I don't know the details.
Regarding the questions about cell decompositions: The obvious decomposition of a Richardson, that is to say, as intersections of Schubert cells with opposite Schubert cells, is a stratification. This means that the closure of each piece is a union of such pieces. The pieces are not $\mathbb{A}^n$'s. I don't believe Richardson's can be stratified by $\mathbb{A}^n$'s, but I'm not sure.