Weil Conjectures for nonprojective algebraic varieties

If we replace projective variety with algebraic variety in the statement of the Weil conjectures what happens? To me it seems the statement still makes sense. But is it still true?

-
Downvoted for having a profoundly uninformative title. – JSE Nov 25 '09 at 2:37
Title edited. Better now? – David Speyer Nov 25 '09 at 4:40

2 Answers

Correctly restated, the conjectures hold for any variety $V$ (not necessarily complete or nonsingular) over a finite field $k$.

Dwork proved that the zeta function $Z(V,t)$ of $V$ is a rational function of $t$.

Grothendieck (et al.) expressed $Z(V,t)$ as the alternating product of the characteristic polynomials of the Frobenius map $F$ acting on the etale cohomology groups of $V$ with compact support.

Deligne showed (Weil II) that for each positive integer $i$ and each eigenvalue $a$ of $F$ acting on the $i$th etale cohomology group of $V$ with compact support, there exists an integer $j\leq i$ such that all the complex conjugates of $a$ have absolute value $q^{j/2}$ where $q=|k|$.

When $V$ is nonsingular and complete, these statements together with Poincare duality, give Weil's original conjectures.

-
What does "correctly restated" mean? – John McCarthy Nov 25 '09 at 13:21
He "correctly restated" them in his answer. – David Zureick-Brown Nov 25 '09 at 17:30
Weil conjectures can even be extended to Artin stacks of finite type over a finite field, if correctly restated. – shenghao Dec 19 '09 at 6:30
Would the coefficients still have integer coefficients (rather than just rational)? – Makhalan Duff Oct 24 '11 at 18:49

As usually stated, most of the statements break. Try looking at $\mathbb{A}^1$ minus a point.

There are still a lot of interesting things to say about the relationship between cohomology and point counting (the supertrace of the Frobenius on the compactly supported cohomology still gives you the number of points), but things like the functional equation and Riemann hypothesis depend heavily on Poincare duality, which, of course, fails pretty badly.

-
A comment and a question: (1) It is still true that the zeta function is rational. (2) I always get confused about what can be said about the eigenvalues of Frobenius acting on compactly supported H^i(X) for X smooth but not projective. What's the right statement? – David Speyer Nov 24 '09 at 15:53
I believe rationality still holds. This should just be looking at action of the Frobenius on c.s. Cohomology. The statement on eigenvalues is that they're always bigger than the Riemann hypothesis would lead you to guess (but I think are still algebraic integers). – Ben Webster Nov 25 '09 at 8:30
two comments: (1) for compact supp cohomology, the absolute values of eigenvalues are LESS than.... (2) for proper varieties with quotient singularities, we still have functional equation and RH, because they are coarse moduli spaces of proper smooth DM stacks, for which H^i is pure of weight i. – shenghao Dec 23 '09 at 5:45