Posted by: Steven Sam | November 23, 2009

Finite field counts and the Grothendieck ring of varieties

Lately some of us at MIT have been thinking about counting \mathbf{F}_q-rational points on some classes of varieties related to linear algebra that provide natural q-analogues for various classes of permutations. One thing we came across was some classes that have the same counts over every finite field. Yan wanted me to post about the following, so I’ll delay my post on the K-theory of the Grassmannian until next time.

We’ll consider varieties defined over a fixed field K. Form the free Abelian group on the isomorphism classes of such varieties. If Z is a closed subvariety of X, then we impose the relation

[X] = [Z] + [X \setminus Z].

We can put a product structure on this group via

[X] \cdot [Y] = [X \times Y]

though it will not be relevant for this post. Related to this product structure is a paper by Bjorn Poonen which shows that if the characteristic is 0, then this ring is not an integral domain. And presumably the result is true over positive characteristic also, but the paper uses the existence of resolution of singularities. This is the Grothendieck ring of varieties. This is at least one way to make sense of statements of the form: \mathbf{P}^2 = \mathbf{A}^2 + \mathbf{P}^1 = \mathbf{A}^2 + \mathbf{A}^1 + \mathbf{A}^0.

In particular this works over a finite field, and then the equations can be turned into equations of numbers by replacing [X] with \#X(\mathbf{F}_q).

Here’s one such example. Let G be a semisimple group defined over an algebraically closed field K of characteristic p > 0, let B be a Borel subgroup, and let X = G/B be its full flag variety. Then B acts on X by left multiplication, and the orbits are indexed by elements of the Weyl group W. In particular, each orbit is an affine space, and its dimension is the length of the corresponding element. So we get a decomposition

\displaystyle [X] = \sum_{w \in W} [\mathbf{A}^{\ell(w)}].

All of this stuff is defined over the integers, so we can actually work in a finite field. Then we get the equation

\displaystyle \#X(\mathbf{F}_q) = \sum_{w \in W} q^{\ell(w)}.

The odd spin group \mathbf{Spin}_{2n+1}(K) (the double cover of the special orthogonal group \mathbf{SO}_{2n+1}(K)) and the symplectic groups \mathbf{Sp}_{2n}(K) have the same Weyl groups, so in particular, their flag varieties are equal in the Grothendieck ring of varieties, even though the varieties themselves are not isomorphic for n \ge 3. One way of seeing this is via the Borel–Weil construction: the global sections of a line bundle on G/B is either 0 or an indecomposable module called a Weyl module. The dimension of the Weyl module (after we pick a way to index the line bundles) is independent of characteristic since it has a \mathbf{Z}-form which is a free Abelian group. In general the sections contain a unique irreducible representation as a submodule, and all irreducible representations arise in this way.

In characteristic 0, indecomposable is the same thing as irreducible, so we can calculate the dimensions using the Weyl character formula. Since these multisets of dimensions are different, we can’t have an isomorphism.

Does anyone know of a better reason why they are not isomorphic varieties?

-Steven

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Categories

%d bloggers like this: