9
$\begingroup$

I'm trying to understand how the nlab's definition of cohomology works concretely. There, it is (very convincingly) claimed that every incarnation of "cohomology" in mathematics is a special case of the following:

Given $X,A$ in an $\infty$-category $\mathcal C$, and given a $n$-fold delooping $B^nA$ of $A$, define $H^n(X,A)$ as the set of connected components of the mapping space $\mathcal C(X,B^n A)$. This also implies $H^{n-i}(X,A)\cong \pi_i\mathcal C(X,B^n A)$.

To get the cohomology of a space or site $X$, one takes $\mathcal C$ to be the $\infty$-category presented by Jardine's model structure on simplicial presheaves on $X$. Now I want to see how Cech cohomology actually computes this. A reasonable thing to do is the following: By results of Dugger, Hollander and Isaksen we can replace (the constant presheaf) $X$ by a hypercover $H_\bullet$, such that there is a weak equivalence $X\simeq\mathrm{hocolim}H_\bullet$. Then we have that $\mathcal C(X,B^nA)$ is weakly equivalent to $\mathrm{holim}_i\mathcal C(H_i,B^nA)$, which is easy to calculate if the hypercover has the property that $B^nA$ is fibrant on its simplices. Typically (always?) this will be the case for sufficiently fine hypercovers.

This is all very good, and it's how Cech cohomology is usually described in this setting, except for the fact that it's not how we actually calculate Cech cohomology in practice. Rather, instead of the simplicial mapping space $\mathrm{holim}_i \mathcal C(H_i,B^nA)$, in practice we take the cosimplicial simplicial set $\mathcal C(H_\bullet,A)$ (which is usually just a cosimplicial set if $A$ is for example a discrete abelian group), and then we take its --for lack of a better word-- cohomotopy groups (at least in the abelian case, where by co-Dold-Kan, a cosimplicial abelian group is just a cochain complex), which turn out to be the cohomology groups. Somehow in doing so, we got around the need to write down a delooping $B^nA$ (although the information is still there in the abelian structure).

  • Are there any mistakes in the above?
  • Where should I conceptually place the cosimplicial set $\mathcal C(H_\bullet,A)$?
  • In derived functor cohomology one takes cochain resolutions of $A$, which again by co-Dold-Kan is the same thing as cosimplicial resolutions. Can one also do this the nonabelian setting?
  • All of the above suggests defining $$H^n(X,A)=\pi_{-n}\mathcal C(X,A).$$Is there some way to make sense of this, i.e. perhaps by turning $\mathcal C(X,A)$ into some kind of truncated spectrum, such that it is somehow conceptually obvious that the cosimplicial objects calculate the negative-degree homotopy groups whereas the simplicial objects calculate the positive-degree ones?
$\endgroup$
4
  • 3
    $\begingroup$ The correct homotopical understanding of sheaf cohomology is that $H^n(X;A)$ is just the $\pi_{-n}$ of the global sections of the sheaf of spectra associated to the presheaf $H\circ A$, where $H$ is the Eilenberg-MacLane functor. Your $\mathcal{C}(H_\bullet;A)$ is just a cosimplicial abelian group that you use to compute these derived global sections (that is, its totalization in spectra computes the correct global sections). $\endgroup$ Commented Jan 18, 2017 at 15:15
  • $\begingroup$ For example, when $X$ is a space and $A$ is an abelian group, the sheaf of spectra is simply sending an open subset $U$ to $F(\Sigma^\infty_+U,HA)$, thus recovering your simpler description. $\endgroup$ Commented Jan 18, 2017 at 15:27
  • $\begingroup$ @DenisNardin OK, so do cosimplicial simplicial sets present spectra? I guess the problem is that I've gained some intuition for simplicial sets and objects, but I don't really know what sort of animal a cosimplicial object is. Simplicial sets always have homotopy groups, but cosimplicial sets only have cohomotopy groups if we have an abelian group structure on top? $\endgroup$
    – user00000
    Commented Jan 18, 2017 at 15:29
  • $\begingroup$ I think you are trying to do too many things at once. I'll see if I can write a proper answer this evening, or maybe someone else will beat me to the punch. $\endgroup$ Commented Jan 18, 2017 at 15:35

1 Answer 1

2
$\begingroup$

Your target BnA is an HZ-module. HZ-modules are Quillen equivalent to chain complexes, and the delooping functor on HZ-modules corresponds to the shift functor on chain complexes.

The space C(Hi,BnA) can be computed as BnC(Hi,A) (this is how I interpret the rather vague statement “BnA is fibrant on its simplices”), and by the above correspondence this is simply a shift of C(Hi,A), i.e., a shift of your cosimplicial object C(H,A).

Cosimplicial homotopy limits of chain complexes can be computed as the totalization (see Reference for homotopy (co)limits of (co)chain complexes via totalization of double complexes) of the corresponding bicomplex obtained via Dold-Kan, which shows that your construction is equivalent to nLab's.

In derived functor cohomology one takes cochain resolutions of A, which again by co-Dold-Kan is the same thing as cosimplicial resolutions. Can one also do this the nonabelian setting?

Yes, injective resolutions of sheaves can be expressed in the nonabelian setting as fibrant replacements in the injective model structure on simplicial presheaves. This would give a different but equivalent computation to the above one (which uses the projective model structure).

Is there some way to make sense of this, i.e. perhaps by turning C(X,A)C(X,A) into some kind of truncated spectrum?

Yes, start with the original sheaf of abelian groups, apply the Eilenberg-MacLane spectrum functor (respectively simply place it in chain degree 0), and then ∞-sheafify the resulting presheaf of spectra (respectively chain complexes). Evaluating the resulting ∞-sheaf on some X and taking π−n will give you Hn(X,A).

This point of view is explained in Ken Brown's famous paper.

$\endgroup$

You must log in to answer this question.

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