MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is exacly the statement of Poincaré duality for smooth projective varieties over finite fields and twisted constant $\mathbf{Z}_\ell$ sheaves? Where can I find a proof?

By twisted constant $\mathbf{Z}_\ell$ sheaf, I mean a system of $\mathbf{Z}/\ell^n$-sheaves that are constructible and étale locally constant, e.g. the system $(\mu_{\ell^n}) = \mathbf{Z}_\ell(1)$.

I'm interested in the finite field case of Poincaré duality. Presumably, the formulation is something like $H^i(X, F) \times H^{2d+1-i}(X, F') \to H^{2d+1}(X, ?) = \mathbf{Z}_\ell$. Now, I want to know what $F'$ and $?$ is.

share|cite|improve this question
The phrase "twisted constant" sounds funny. Since you make simplifying assumptions ("projective" rather than "quasi-projective"), do you also want $\mathbb{Q}_ {\ell}$-sheaves instead? The answer is simpler in that case since then both sides of the duality use cohomology, without Ext's (but to prove the result one uses torsion sheaves, and hence Ext's). Or is the point of the question precisely to not invert $\ell$, and/or to encode a Galois-equivariance condition (since you mention non-sep. closed base field)? Please clarify your motivation so it is clearer what properties matter to you. – Boyarsky Jul 8 '10 at 12:59
I have edited the question accordingly. – Timo Keller Jul 8 '10 at 17:03
Let $X_s = X_ {k_s}$, $F_s = F_ {k_s}$ on $X_s$. Since you avoid Ext's, you must require the stalks of $F$ to be free, so I will assume this. By Leray the natural map $H^i(X,F) \rightarrow H^0(k,H^i(X_s,F_s))$ is surjective with kernel $H^1(k,H^{i-1}(X_s,F_s))$. In particular, $H^{2d+1}(X,\mathbf{Z}_ {\ell}(d)) = H^1(k,H^{2d}(X_s,\mathbf{Z}_ {\ell}(d)))$. Thus, using trace map and $G_k \simeq \widehat{\mathbf{Z}}$, this is $\mathbf{Z}_ {\ell}$. So try $? = \mathbf{Z}_ {\ell}(d)$, $F' = F^{\vee}(d)$, and look for orthogonality in cup products. Try Artin-Mazur, or Milne ADT, for dimension 1? – Boyarsky Jul 8 '10 at 18:23
What's the title of Artin-Mazur? What do you mean by $F^\vee$ and what by "look for orthogonality in cup products"? – Timo Keller Jul 8 '10 at 18:58
By $F^{\vee}$ I mean linear dual, as in usual Poincare duality (you don't want Ext's, so you need freeness on $F$-stalks). Contemplate the cup product and the Leray filtrations: maybe sub's annihilate each other and pair perfectly against the cokernels? That would do it...hmm, cohomology could have torsion, so you ought to invert $\ell$; otherwise the Ext's come up. See Zink's Appendix 2 of Haberland's Galois cohomology book for $S$-integers of number fields (with ref. to Ann. ENS notes of Mazur), and Google "Artin-Verdier duality"; adapt it to curves over finite fields. Check Milne's ADT. – Boyarsky Jul 8 '10 at 19:13

The main case can be found in Milne's article specifically Theorems 1.13, 1.14 on page 310. The idea, briefly, is as follows: Given a sheaf $F$ on a variety $X$ over a finite field $k$, then over an algebraic closure $\bar{k}$ of $k$, the group $H^i_{et}(\bar{X}, F)$ becomes a $Gal(\bar{k}/k)$-module. There is a spectral sequence involving the $H^j(Gal(\bar{k}/k), H^i_{et}(\bar{X}, F))$ which converges to $H^n_{et}(X,F)$. This is true over any perfect field.

When you have duality over $\bar{k}$ (e.g. $X$ smooth proper and $F$ nice), combine it with duality in Galois cohomology (in our case, the group is very simple: $\hat{Z}$) to get duality over $k$. The duality theorems now reflect the $k$: if Poincare duality for $X$ of dimension $d$ over $\bar{k}$ pairs $H^i$ with $H^{2d-i}$, over $k$ the pairing will be between $H^i$ and $H^{2d +m -i}$ where $m$ is the cohomological dimension (assumed finite) of the Galois group ($m=1$ in the case of a finite field).

Hope this helps.

share|cite|improve this answer

I guess you know about Theorem 11.1 in Milne's book Étale cohomology. It is over a separably closed field though (i.e. not finite).

share|cite|improve this answer
Yes, I'm interested in the finite field case. – Timo Keller Jul 8 '10 at 16:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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