Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a nonsingular projective variety of dimension $n$ over a field $k$, and $\omega_X$ be its canonical sheaf. Let $G$ be a finite subgroup of the automorphism group $Aut_k(X)$, and $\mathcal{F}$ a locally free $G$-equivariant sheaf on $X$. Then $G$ acts on all the cohomology groups $H^i(X, \mathcal{F})$. Is the Serre duality $$ H^i(X, \mathcal{F})\times H^{n-i}(X, \mathcal{F}^\vee\otimes \omega_X)\to H^n(X, \omega_X)=k$$ a $G$-equivariant perfect pairing? Where can I find a reference to this result?

Thank you.

share|improve this question
4  
The isomorphism $\mathrm H^n(X, \omega_X) \simeq k$ is universal, that is, invariant under isomorphisms. The universal pairing is functorial under pullbacks. The result follows from this. –  Angelo Mar 20 '13 at 11:14
    
@Angelo: Perhaps the OP seeks a reference to justify the invariance you mention? Many references on Serre duality make the construction of the trace in a manner that is not sufficiently intrinsic to render the triviality apparent. It is equivalent to show that the natural composite map $H^n(X,\Omega^n_{X/k}) \rightarrow H^n(X,g^{\ast}(\Omega^n_{X/k})) \rightarrow H^n(X,\Omega^n_{X/k})$ is the identity (1st step pullback, 2nd step canonical at sheaf level); settling projective spaces "by bare hands" is a bit unpleasant (though easy by using the structure of the automorphism group). –  user28172 Mar 20 '13 at 14:41
    
@nosr, I tried to prove what you suggested and indeed it is not too difficult. However, I was trying to find a reference to include in a paper. –  Jiangwei Xue Mar 26 '13 at 17:24
add comment

1 Answer 1

In a restricted situation you may consult

Peskin, Barbara R. On the dualizing sheaf of a quotient scheme. Comm. Algebra 12 (1984), no. 15-16, pp. 1855–1869.

Of course, the general treatment is

Hashimoto, Mitsuyasu Equivariant twisted inverses. in Foundations of Grothendieck duality for diagrams of schemes, pp. 261–478, Lecture Notes in Math., 1960, Springer, Berlin, 2009.

but it requires a great deal of machinery, derived categories, etc.

Hope this is of some help

share|improve this answer
add comment

Your Answer

 
discard

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.