[Edit] Let me make question more focused. It is about details of Weil conjectures.

Rationality of zeta function does NOT require the manifold to be smooth & projective, so zeta function is a factor of some polynomials:

$ \frac{P_1(T)\dotsb P_{2n-1}(T)}{P_0(T)\dotsb P_{2n}(T)}, $

Question What happens if manifold is not required to be smooth and projective, is there some cohomology theory which gives / bounds DEGREES of those polynomials in zeta function ?

Can some version of intersection cohomology appear in singular case ? At least for small resolutions ?

If manifold is affine and smooth - is it enough to consider l-adic cohomologies with compact support ? (Seems - YES by Ben's answer: Grothendieck trace formula )


If manifold is SMOOTH and PROJECTIVE, then as far as I understand degrees of those polynomials equals to dimensions of l-adic cohomologies (which in the case of GOOD REDUCTION coincide with dim $H^i$ over complex numbers (Betti numbers)).

OLD version of question:

Weil conjectures are "highly influential proposals by André Weil (1949) on the generating functions (known as local zeta-functions) derived from counting the number of points on algebraic varieties over finite fields". See nice brief surveys on Wikipedia or Tao2014.

Conjectures consist of several statements:

  • Rationality of zeta-function

  • Functional equation and Poincaré duality for the zeta function

  • Riemann hypothesis

  • (Betti numbers) If X is a (good) "reduction mod p" of a non-singular projective variety Y defined over a number field embedded in the field of complex numbers, then the degree of Pi is the ith Betti number of the space of complex points of Y.

Item 2 clearly needs that variety to be smooth and projective. While items 1,3 holds for any quasi-projective (e.g. affine) manifold (see Tao2014).

Question: What about the last item of the conjectures for affine (or quasi-projective) manifolds (not necessarily smooth) - is there any cohomological interpretation of degrees of polynomials P_i (which represent the zeta function) ?

For example. Consider the commuting variety: pairs of matrices which commute or pairs/triples/n-tuples of non-degenerate matrices which commute. Number of points over finite field is known (at least for pairs) see e.g. MO271752.

Question 2: What about particular cases of such manifolds ?

  • $\begingroup$ it is true for all varieties $V$, with no assumption on smoothness or properness. $P_i(T) = {\rm det}(1-FT| H^i_c(V, \mathbb Q_{\ell}))$ is the characteristic polynomial of Frobenius acting on the cohomology with compact support of the variety $V$. Here $\ell$ is any prime different from the characteristic of the finite field. So degree of $P_i$ is always the dimension of $H^i_c(V)$. $\endgroup$ – guest Aug 7 '17 at 0:14
  • $\begingroup$ @guest thank you ! What is the reference ? $\endgroup$ – Alexander Chervov Aug 7 '17 at 4:46

The number of points always has an interpretation as a supertrace of Frobenius on the compactly supported cohomology, by the Grothendieck trace formula. The issue is that it can be quite difficult to understand how Frobenius acts in this case. On a compact smooth variety, it's constrained by Poincare duality (which is basically the Riemann hypothesis); otherwise, it's quite difficult to control, though it can be in special situations. The most important is that cohomology classes of degree $2d$ that are dual to rational subvarieties still have the expected eigenvalue $p^d$; unfortunately, most varieties will have cohomology classes that don't arise this way.

EDIT: To address the degrees of the polynomials: the degrees of the polynomials are given by the Betti numbers of the etale cohomology of the variety in characteristic $p$. The roots of the polynomials are the inverses of eigenvalues of Frobenius acting there. For all but finitely $p$, this will be the same as the cohomology of complex points of the variety, but there's no especially obvious condition for which finitely many p you have to avoid. As Will points out, in the smooth case, good reduction is enough; I'm not sure there's an analogous notion of good reduction in the non-smooth case.

  • 2
    $\begingroup$ The other issue is that it is harder to give a good definition of good reduction. $\endgroup$ – Will Sawin Aug 4 '17 at 1:50
  • $\begingroup$ Thank you for your answer! To clarify: does degrees of polynoms equal to dimensions of those cohomologies ? $\endgroup$ – Alexander Chervov Aug 4 '17 at 5:26
  • $\begingroup$ If I understand correctly Wikipedia link requires the scheme to be smooth... $\endgroup$ – Alexander Chervov Aug 5 '17 at 16:20
  • $\begingroup$ @AlexanderChervov You're right, though I'm not sure where they're using that assumption. There's a version without smoothness assumptions in Theorem 1.2 of math.stanford.edu/~conrad/Weil2seminar/Notes/L20.pdf $\endgroup$ – Ben Webster Aug 5 '17 at 18:13
  • $\begingroup$ Thank you for the reference, but I need time to think ... I think looking on examples like cusp curve will help me... $\endgroup$ – Alexander Chervov Aug 6 '17 at 12:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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