Posted by: Steven Sam | April 28, 2009

## q-analogues and homogeneous spaces

While I was working on my final paper for the course on quivers that I’m taking this semester, I came across the following result (the notation $\mathbf{F}_{p^r}$ means the finite field with $p^r$ elements, and $\mathbf{C}$ is the field of complex numbers):

Theorem. Let X be a variety defined over Z and assume for some prime p, and all r>0, that the function $\#X({\bf F}_{p^r})$ is obtained by plugging in $p^r$ into some polynomial P(t). Then P(t) has integral coefficients, and P(1) is the Euler characteristic of $X({\bf C})$.

If you’re not comfortable with varieties, just think of the solution set of some collection of polynomials in multiple variables, and if you don’t know what the rest of the words mean, that won’t matter much for the stuff starting with the next paragraph. Here I’m using the notation X(F) to denote the solutions for X over F, and I’m thinking of $X({\bf C})$ as a complex analytic space. The proof uses $\ell$-adic cohomology (with compact support) and the Grothendieck–Lefschetz trace formula. I won’t get into that, but I’d like to use this as an excuse to talk about the q-analogues of the natural numbers.

Let’s fix a field with q elements, and work over this field. First let’s use the theorem on projective space $\mathbf{P}^{n-1}$, whose points are the one-dimensional subspaces (lines) in an n-dimensional vector space V. The number of lines of an n-dimensional vector space is $(q^n-1)/(q-1) = q^{n-1} + q^{n-2} + \cdots + q + 1$ because any nonzero vector defines a line, and each line is spanned by q-1 different vectors. Let’s denote this number [n] because substituting q=1 gives n, which is the Euler characteristic of n-1 dimensional complex projective space. Now let’s move onto complete flag varieties Flag(V) for V n-dimensional: the elements are just increasing chains $V_1 \subset V_2 \subset \cdots \subset V_n$ where $\dim V_i = i$. The number of flags is $[ n ] [ n - 1 ] \cdots [ 2 ] [1]$ because to define a flag, we first pick $v_1$ nonzero to get $V_1 = \langle v_1 \rangle$, and in general, after we pick $v_1, \dots, v_{i-1}$, we can pick any lift $v_i$ of a nonzero vector $\overline{v}_i \in V / \langle v_1, \dots, v_{i-1} \rangle$, and the space $V_i = \langle v_1, \dots, v_i \rangle$ is independent of the choice of lift. So let’s call this number $[ n ] !$ because specializing q=1 gives n!, which we now know is the Euler characteristic of the complex complete flag variety.

Let’s take a look at the Grassmannian Gr(k,V), whose points are the k-dimensional subspaces of an n-dimensional vector space V. To count its number of points, notice that we have a map Flag(n) to Gr(k,n) which just sends a flag $V_1 \subset \cdots \subset V_n$ to the subspace $V_k$. This map is obviously surjective, but what are the fibers? It follows from the definitions that the preimage of a k-dimensional subspace W is Flag(W) x Flag(V/W), so the number of points is just $\displaystyle \frac{[ n ] ! }{ [ k ] ! [ n - k ] !}$. And we’ll call this number $\left[ \begin{matrix} n \\ k \end{matrix} \right]$ because it specializes to $\binom{n}{k}$ upon setting q=1, and as before, this is the Euler characteristic of the complex Grassmannian Gr(k,V) where now V is a complex n-dimensional vector space.

Now we can even do partial flag varieties for tuples $(d_1, \dots, d_r)$, whose points consists of partial flags $V_1 \subset \cdots \subset V_r$ where $\dim V_i = d_i$. The number of points is going to be $\left[ \begin{matrix} n \\ d_1 \end{matrix} \right] \left[ \begin{matrix} n-d_1 \\ d_2 - d_1 \end{matrix} \right] \cdots \left[ \begin{matrix} n-d_1 - \cdots - d_{r-1} \\ d_r - d_{r-1} - \cdots - d_1 \end{matrix} \right]$, and setting q=1 gives the analogous product of binomial coefficients for the Euler characteristic.

In all of the cases, it turned out that each of the polynomials in q had positive coefficients. So we might ask what these numbers are counting, if anything. It turns out that setting $q=t^2$ gives the generating function $\sum_{i \ge 0} \dim_{\mathbf{C}} \mathrm{H}^i(X; \mathbf{Z}) t^i$ where X is whichever variety we are talking about, and H refers to plain old singular cohomology. I can give a better answer though: in each of the cases of interest, there is an explicit cellular decomposition of variety in question, and the number of i-dimensional cells is exactly the coefficient of $q^i$ in the corresponding polynomial. We could discuss this at length, but for now, I’ll just refer the interested reader to William Fulton’s book Young Tableaux.

And for those who don’t know what a homogeneous space is: a homogeneous space is a manifold (resp. algebraic variety) with a transitive smooth (resp. algebraic) action of a Lie (resp. algebraic) group. And one can verify that all of the examples I’ve discussed are homogeneous spaces for the group $\mathbf{GL}(n)$ of $n \times n$ invertible matrices. For fun, the reader can work out the counts for the symplectic and orthogonal groups and/or figure out what the right analogues of Grassmannians and flag varieties are.

## Responses

1. “Theorem. Let X be a variety defined over Z and assume for some prime p, and all r>0, that the function is obtained by plugging in into some polynomial P(t). Then P(t) has integral coefficients, and P(1) is the Euler characteristic of (computed using cohomology with compact support).”

Under what conditions is this true? It seems like an interesting theorem. (Is this related to the Weil conjectures? I remeber something like this was true for elliptic curves.)

2. I don’t think we have to assume anything other than what I said. I have it written up here:
http://math.mit.edu/~ssam/writings/calderochapoton.pdf (Lemma 2.2)

I’m not very familiar with the Weil conjectures, but I think the technique used to prove this theorem is utilized.