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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.

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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? –  Dan Petersen Mar 30 '12 at 12:22
    
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. –  Balazs Mar 30 '12 at 20:05
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1 Answer

up vote 5 down vote accepted

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 $H^1$.

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Thanks, that was fast... –  Balazs Mar 30 '12 at 9:47
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