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

Hi,

I have a question related to Serre Duality: if I have a smooth projective variety $X$ with dualizing sheaf $\omega$ and two coherent sheaves $F$ and $G$ on $X$, then how can I get a canonical map

$Ext^{i}(F,G)\rightarrow Ext^{n-i}(G, F \otimes \omega)^{\star}$ ?

I know that in the case that $F$ and $G$ are locally free, one gets it and then it is an isomorphism. But I don't see how you get it for coherent ones.

Thank you

share|improve this question
3  
Crossposted from Math.SE: math.stackexchange.com/q/56877 –  Theo Buehler Aug 11 '11 at 13:08

1 Answer 1

This works directly when one of $F, G$ is locally free. I am not sure whether it is true when both are merely assumed to be coherent (e.g. I don't see how to get the map). (In general, even the generalization of Serre duality -- Grothendieck duality -- tells you how to hom out of $\mathbf{R}\Gamma \mathcal{F}$ (or more generally derived push-forward) of a sheaf $\mathcal{F}$ into some complex of abelian groups, and this doesn't seem to tell you about $\mathrm{Ext}$ functors up top, in $X$, though perhaps I'm missing something). See below for the extension without local freeness hypotheses.

Namely, there is a map $H^n(X, \omega) \to k$ (the "integration" map*). To get the map $$\mathrm{Ext}^i(F, G) \to \mathrm{Ext}^{n-i}(G, F \otimes \omega)^*$$ (which is natural), we need a pairing $$\mathrm{Ext}^i(F, G) \times \mathrm{Ext}^{n-i}(G, F \otimes \omega) \to k.$$ To do this, we can use the Yoneda product to pair these to $\mathrm{Ext}^n(F, F \otimes \omega)$. If $F$ is locally free, then this naturally maps to $\mathrm{Ext}^n(O_X, F \otimes F^{\vee} \otimes \omega)$, which in turns maps to $H^n(X, \omega)$ (by coevaluation) and thus to $k$. If $G$ is locally free, we can similarly write both sides as $\mathrm{Ext}^i(F \otimes G^{\vee}, \omega)$ and $\mathrm{Ext}^{n-i}(O_X, G^{\vee} \otimes F \otimes \omega)^*$, and we get the pairing and isomorphism just as in Hartshorne.

Now if we fix one of $F, G$, we get a $\delta$-functor in the other. So if the natural transformation is an isomorphism when both are locally free, it is an isomorphism when one is locally free and the other merely coherent (since on a projective scheme, every coherent sheaf has a locally free presentation, and we can use the "finite presentation trick").

*Here the comparison is as follows: on a compact complex manifold $X$ of dimension $n$, if $\omega$ denotes the sheaf of holomorphic $(n,0)$-forms, we have (Dolbeaut isomorphism) $$H^n(X, \omega) = \frac{(n,n)\mathrm{-forms}}{\overline{\partial}\mathrm{-exact\ top forms}}$$ and so we can define the map as integration, legitimately (because a $\overline{\partial}$-exact top form is exact in the usual sense, this is well-defined).

Edit: As above, the obstacle to defining the map was that there was no natural trace morphism $$\mathrm{Ext}^n(F, F \otimes \omega) \to H^n(X, \omega)$; given one, the same arguments would answer your question for the case of $F, G$ both only assumed coherent. As Donu Arapura observes below, there *is* a natural way to define the trace. Reason: $F$ can be replaced by a bounded complex of locally frees in the derived category (it is a "perfect" complex) since we are working over a *smooth* variety (in particular, this means that any locally free resolution can be truncated at a finite stage to still yield a locally free one, by Serre's theorem on the finiteness of global dimension). For a bounded complex of locally frees $K^\bullet$, we can define a map $$\mathrm{Ext}^n(K^\bullet, K^\bullet \otimes \omega) \to H^n(X, \omega)$$ by taking the "partial trace." One can think of the former as consisting of maps $K^\bullet \to K^\bullet \otimes \omega[n]$, or $\mathbf{R}\underline{Hom}(K^\bullet, K^\bullet) \to \omega[n]$. (By the conditions on $K^\bullet$, the derived internal hom is the same as the usual sheaf hom.) Since there is a natural map from $\mathcal{O}_X$ to the derived internal hom (given by the identity), we can define the trace. To show that map you are interested in becomes an isomorphism, we use the same "finite presentation" trick.

share|improve this answer
2  
I was going to refrain from answering the original, but... since $X$ is smooth you can resolve $F,G$ by bounded complexes of locally free sheaves. In other words, they are perfect, there is a trace (Illusie) $$Ext^n(F,F\otimes \omega)\to H^n(X,\omega)\cong k$$ Combine this with Yoneda as you did, and it should work in general. –  Donu Arapura Aug 11 '11 at 14:42
    
Dear Donu: Thanks. I'll add that to the answer. –  Akhil Mathew Aug 11 '11 at 17:02
4  
Dear Akhil, you're welcome. And if I may say so, I'm impressed. –  Donu Arapura Aug 11 '11 at 17:41

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.