2
$\begingroup$

I have a finite category $\mathcal{C}$, along with a functor $F \colon \mathcal{C}^{\mathrm{op}} \to \mathsf{GradedCommRings}$. If $F_j$ is $j$-th graded piece of $F$, then I write $H^i(\mathcal{C},F_j)$ for the $i$-th derived inverse limit of the diagram $\mathcal{C}^{\mathrm{op}} \to \mathsf{Ab}$ of abelian groups. Equivalently, it's the $i$-th sheaf cohomology of the sheaf $F_j$, where I regard $\mathcal{C}$ as the site with trivial Grothendieck topology.

I have computed the various $H^i(\mathcal{C},F_j)$. Assembling them, there should be a cup product structure $H^i(\mathcal{C},F_j) \otimes H^{i'}(\mathcal{C},F_{j'}) \to H^{i+i'}(\mathcal{C},F_{j + j'})$. I would like to compute this product structure.

The only method I'm aware of is through sheaf cohomology, involving explicit resolutions, tensor products, and total complexes (see [1]). Unfortunately, I do not have an explicit resolution of $F$ or $F \otimes F$: it seems too complicated to do by hand, especially because my $F(c)$ are typically infinitely generated. (In my computation of $H^i(\mathcal{C},F_j)$ I circumvented this by using spectral sequences but these obscure the product structure.)

I'm led to the following questions:

  • Does anyone know of a more efficient method for computing cup products of presheaf cohomology / derived limits?
  • If not, is there computer software that might be capable of taking over some of the tasks outline above?

[1] : R.D. Swan. Cup products in sheaf cohomology, pure injectives, and a substitute for projective resolutions.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Cup products in sheaf (and presheaf) cohomology are often easy to compute by resolving the source (in the projective model structure, say), not the target. For an example of resolving the source in this manner, see The Yoneda pairing, hypercohomology, and cup product

In the case under consideration, one can equip the category of presheaves of chain complexes on C with a projective model structure. The latter has an explicit cofibrant replacement functor, which can be used to write down an explicit projective resolution. The cofibrant replacement functor is precisely the classical bar construction applied to the adjunction between presheaves of chain complexes on C and Ob(C)-indexed chain complexes.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Is the link really talking about the same thing? There is always a Yoneda pairing $\operatorname{Ext}^p(\mathcal{F},\mathcal{G}) \otimes \operatorname{Ext}^q(\mathcal{G},\mathcal{H}) \to \operatorname{Ext}^{p+q}(\mathcal{F},\mathcal{H})$ but as far as I understand that's a different product. Also, what would this 'explicit cofibrant replacement functor' be? Do you perhaps have an example? $\endgroup$
    – Mr. Palomar
    Commented Jan 12, 2021 at 14:10
  • $\begingroup$ Perhaps I should add that the sheaves I'm interested in take values not merely in rings but in $k$-algebras, for $k$ a field, which means that the underlying $j$-th graded pieces produce vector spaces. This gives it a more 'combinatorial' flavour, and for this reason I'm hoping that something like GAP or Sage has the capacity to take over the work of finding the resolutions. $\endgroup$
    – Mr. Palomar
    Commented Jan 12, 2021 at 14:21
  • $\begingroup$ @Mr.Palomar: The Yoneda pairing is not the cup product, the link was giving an example how to resolve the source instead of the target. An explicit description of the cofibrant replacement functor is given in the last sentence: it is the bar construction applied to the free-forgetful adjunction between Ob(C)-indexed chain complexes over k and presheaves of chain complex over k on C. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 12, 2021 at 14:49
  • $\begingroup$ @Mr.Palomar: A more explicit description of the source resolution can only be given for a more explicit description of the category C. $\endgroup$
    – Dmitri Pavlov
    Commented Jan 12, 2021 at 15:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .