5
$\begingroup$

I am looking for an example of a smooth irreducible quasiprojective variety $X$ over ${\mathbb C}$, such that when reduced over finite fields ${\mathbb F_q}$, the number of its points is a polynomial $P(q)$ of $q$ with nonnegative (integer) coefficients, but $X$ has some odd cohomology.

Background: as discussed in an answer to this question, if a variety $X$ is paved by affine spaces, then it only has $(p,p)$ cohomology, and the number of its ${\mathbb F_q}$-points equals $P(q)$, where $P(t)$ is the Poincare polynomial of (compactly supported cohomology of) $X$. Note that the coefficients of $P$ are necessarily non-negative, given by the number of affine cells in the paving of a fixed dimension. In the appendix to this paper, N.Katz proves a kind of converse to this statement: if the number of points of $X$ over a finite field is given by a polynomial $P(q)$ of $q$, then this polynomial determines the so-called $E$-polynomial $E(x,y)$ of $X$ by the formula $E(x,y)=P(xy)$. The $E$-polynomial is a partial Euler characteristic, where we remember the weights of (compactly supported) cohomology but not the degrees.

Of course varieties of the latter type can have odd cohomology; the typical example is $X={\mathbb C}^*$, with point count polynomial $P(q)=q-1$. $X$ of course has odd cohomology. A slighly more complicated example, due to N.Katz, shows that $X$ can also have non-$(p,p)$ cohomology. But in these examples, the polynomial $P$ has some negative coefficients.

Hence the question: can $X$ have positive polynomial count, but still some odd cohomology (which cancels in the $E$-polynomial)? Note that $X$ can't be smooth projective, since then its cohomology would be pure so any odd cohomology would have to show up in the $E$-polynomial.

There may of course be a trivial example which I am missing.

$\endgroup$
2
  • $\begingroup$ I am confused about the example of Katz that you mention -- if the $E$-polynomial of $X$ is a function of $xy$, then how can $X$ have cohomology not of $(p,p)$ type? $\endgroup$ Mar 30, 2012 at 12:22
  • $\begingroup$ The $E$-poly involves a signed summation in the cohomology degree - so there can be cancellation. There is some non-$(p,p)$ stuff that cancels out. $\endgroup$
    – Balazs
    Mar 30, 2012 at 20:05

1 Answer 1

7
$\begingroup$

The answer to your question is yes:

Let $X$ be the blowup of $\mathbb{A}^1 \times (\mathbb{A}^1 - \{0\})$ in a point. The number of points over a field of $q$ elements is $q(q-1) + q = q^2$ and $X$ has non-trivial $H^1$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.