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

This is a cross-post of the following math.stackexchange question: http://math.stackexchange.com/questions/188760/cup-product-and-hypercohomology

I always found the cup product slightly mysterious. Recently I discovered the following interesting theorem (in Voisin's book Hodge theory and complex algebraic geometry I, chapter 4.3):

For the setup, let $(X, \mathcal{O})$ be a ringed space, $\mathcal{F}$, $\mathcal{G}$ sheaves of $\mathcal{O}$-modules, $\mathcal{F}^\bullet, \mathcal{G}^\bullet$ acyclic resolutions of $\mathcal{F}, \mathcal{G}$, and $\mathcal{H}^\bullet$ an acyclic resolution of $\mathcal{F} \otimes \mathcal{G}$. Suppose given a morphism of complexes $$\phi^\bullet: Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet) \to \mathcal{H}^\bullet,$$ (where $Tot$ denotes the total (simple) complex associated to a double complex). This data naturally yields homomorphisms $$H^p(X, \mathcal{F}) \otimes H^q(X, \mathcal{G}) \to H^{p+q}(X, \mathcal{F} \otimes \mathcal{G}) \quad(*).$$

The theorem is this: if $\phi^\bullet$ is compatible with the resolutions (that is, the evident triangle involving $\mathcal{F}\otimes\mathcal{G}$, $Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet)$ and $\mathcal{H}^\bullet$ is commutative), then the induced morphism $(*)$ on cohomology is the cup product pairing.

The proof says, somewhat mysteriously to me, that the result follows by defining cup products on hypercohomology, and then using commutativity. While I know about hypercohomology, it is unclear what cup products should even mean in this situation. Can you explain what Voisin means, or provide a reference?

Note: the theorem essentially says that all such $\phi^\bullet$ induce the same morphism on cohomology (independent of the resolutions even), so we need not acutally know here what the cup product pairing is.

Thanks in advance.

EDIT: Since this may not have been clear, my question is this: How do you prove the above theorem, potentially by defining a cup product on hypercohomology and exploiting its properties?

It is quite easy to construct, for arbitray left-bounded complexes $\mathcal{F}^\bullet$ and $\mathcal{G}^\bullet$, a canonical product $\mathbb{H}^p(\mathcal{F}^\bullet) \otimes \mathbb{H}^q(\mathcal{G}^\bullet) \to \mathbb{H}^{p+q}(Tot(\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet))$ (by using "Godement double-resolutions" and the standard fact that godement resolutions remain resolutions after tensoring). It is not clear to me if this is a sensible construction, since $\mathcal{F}^\bullet \otimes \mathcal{G}^\bullet$ is not stable under replacing $\mathcal{F}^\bullet$, $\mathcal{G}^\bullet$ by quasi-isomorphis complexes.

EDIT 2: the above claim is false; I made a mistake in my computation. As Donu points out below, the natural target for cup product in hypercohomology is the "derived tensor" $\mathcal{F}^\bullet \otimes^L \mathcal{G}\bullet$. This is indeed what the construction I had in mind yields.

share|improve this question
2  
It would probably help if you told us what does the cup product means in the situations you do understand :-) –  Mariano Suárez-Alvarez Aug 31 '12 at 18:26
1  
In any case, already in the book by Cartan and Eilenberg they construct cup products just like Voisin does in terms of arbitrary resolutions. Maybe reading that and mentally adding hyper all over the place (although I was told here a while ago that the cool guys have dropped the hyper later and just say cohomology! :-) ) can answer your question. –  Mariano Suárez-Alvarez Aug 31 '12 at 18:28
    
My question is not really "what is the cup product in hypercohomology" but "how do I prove this theorem". I'll check out cartan-eilenberg. –  Tom Bachmann Aug 31 '12 at 18:44
add comment

1 Answer

up vote 2 down vote accepted

Any two natural approaches will likely yield the same product up to sign, but this is not much of an answer. If you want a precise reference comparing the Cech-style cup product with the product using resolutions, take a look at Godement's Théorie des Faisceaux.


After rereading the post-edited question, I realize you were partly wondering about a good definition of cup product for hypercohomology. Unfortunately, due to lack of time, this will be a bit sketchy, but here goes. You can identify $$\mathbb{H}^i(X,\mathcal{F}^\bullet)\cong Ext^i(\mathcal{O}, \mathcal{F}^\bullet)\cong Hom_D(\mathcal{O}[-i], \mathcal{F}^\bullet)$$ where $D$ is the derived category of (bounded if you like) sheaves of $\mathcal{O}$-modules. The natural target for the cup product is the hypercohomology of the derived tensor product $\mathcal{F}^\bullet\otimes^L \mathcal{G}^\bullet\in D$ obtained by tensoring flat acyclic resolutions (although there are some technical issues that I'm suppressing if want boundedness). From the above formulas, one gets $$\mathbb{H}^i(X, \mathcal{F}^\bullet)\otimes \mathbb{H}^j(X, \mathcal{G}^\bullet)\to \mathbb{H}^{i+j}(X, \mathcal{F}^\bullet\otimes^L \mathcal{G}^\bullet)$$ To prove that this coincides with some other notion of cup product is going to be a slog.

share|improve this answer
    
More precisely, Thm. 6.2.1 of Ch. II of Godement (after reading the Alexander-Spanier method of construction in section 6.1 of Ch. II), as well as various other worked examples in that part of Chapter II, for cohomology of ordinary sheaves. –  user22479 Aug 31 '12 at 20:43
    
Thanks for the answer. I'll check out the reference as soon as I have time. –  Tom Bachmann Sep 1 '12 at 12:02
    
6.2.1 is indeed precisely what I was looking for. Thanks very much. –  Tom Bachmann Sep 2 '12 at 8:11
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.