Do we have non-abelian sheaf cohomology?

Lets $X$ be a complex manifold (algebraic variety), $N$ an integer, and consider the sheaf $F$ defined by:

$F(U)$ ={ holomorphic maps $f: U\rightarrow GL(N,\mathbb{C})$ } with multiplicative structure.

Can we define $H^i(X,F)$ ? Note that for $N=1$, this would be just $H^i(X,O_X^*)$.

• Dear Mohammad: Is there an application in mind? For $i = 1$ the definition of the non-abelian formalism is just a convenient way to do the bookkeeping when working with torsors (e.g., for $i=1$ what you ask is (almost) by def'n the set of isom. classes of rank-$n$ holomorphic vector bundles, as you may know) but it has no subtlety (unlike derived functors). E.g., gives a slicker version of the bare-hands way of getting the ${\rm{H}}^2$-obstruction to lifting projective bundle to vector bundle, but is really the same argument. Do you want $i=2$ to analyze something of interest for $i = 1$? – BCnrd Aug 23 '10 at 17:26
• I don't have an application in mind. I started by H^1 which as you said gives the rank-n vector bundles. I was wondering if there is a definition for i>2. That's why I asked the question. – Mohammad F. Tehrani Aug 24 '10 at 21:46

The quick reply is: not really for $i \gt 2$, and not in the way you perhaps expect for $i=2$, see below.

EDIT (Feb 2017): Debremaeker's PhD thesis [0] has now been translated into English and placed on the arXiv: Cohomology with values in a sheaf of crossed groups over a site, arXiv:1702.02128

The comment on Charles' answer about 'teaching you never to ask that question again' is partly true, partly not. The lesson to learn from Giraud is that really one does not use groups for coefficients of higher cohomology. For a start, Giraud's $H^2(X,G)$ is not functorial with respect to group homomorphisms $G\to H$! One also does not get the exact sequences that one expects (this is due to the lack of functoriality). But this is not a problem with his definition of the cohomology set, but a problem with what category you believe the coefficients lie in. This is because the coefficient object of Giraud's cohomology is actually the crossed module $AUT(G) = (G \to Aut(G))$, and the assignment $G \mapsto AUT(G)$ is not functorial. (Aside: Giraud contains lots of other important things on stacks and gerbes and sites and so on, so the book is not a waste of time by any means)

But little-known work by Debremaeker[1-3] from the 1970s fixed this up and showed that really the Giraud cohomology was functorial with respect to morphisms of crossed modules. This has been recently extended by Aldrovandi and Noohi [4] by showing that it is functorial with respect to weak maps of crossed modules aka butterflies/papillion.

It was realised by John E. Roberts (no relation) and Ross Street that the most general nonabelian cohomology has as coefficient objects higher categories. In fact, we now know that the coefficients of $n^{th}$ degree cohomology is an $n$-category (usually an $n$-groupoid, though), even when we are talking about usual abelian cohomology.

Everything I've talked about is just for groups etc in Set, but it can all be done internal to a topos, i.e. for sheaves of groups, and more generally a Barr-exact category (and probably weaker, but Barr-exact means that the monadic description of cohomology therein due to Duskin (probably going back to Beck) works fine).

[0] R. Debremaeker, Cohomologie met waarden in een gekruiste groepenschoof op een situs, PhD thesis, 1976 (Katholieke Universiteit te Leuven). English translation: Cohomology with values in a sheaf of crossed groups over a site, arXiv:1702.02128

[1] R. Debremaeker, Cohomologie a valeurs dans un faisceau de groupes croises sur un site. I, Acad. Roy. Belg. Bull. Cl. Sci. (5), 63, (1977), 758 -- 764.

[2] R. Debremaeker, Cohomologie a valeurs dans un faisceau de groupes croises sur un site. II, Acad. Roy. Belg. Bull. Cl. Sci. (5), 63, (1977), 765 -- 772.

[3] R. Debremaeker, Non abelian cohomology, Bull. Soc. Math. Belg., 29, (1977), 57 -- 72.

[4] E. Aldrovandi and B. Noohi, Butterflies I: Morphisms of 2-group stacks, Advances in Mathematics, 221, (2009), 687 -- 773.

• Just to expand on that a little: what Giraud calls $H^2(X,G)$ is perhaps better called $H^1(X,AUT(G))$. You really can't define $H^i$ for i>1 with nonabelian coefficients, because you can't "deloop" a nonabelian group more than once, but you can associate to a nonabelian group G another group (or 2-group) AUT(G) that looks a little like a delooping of it. In particular, one should beware that when G is abelian, $H^1(X,AUT(G))$ (Giraud's $H^2(X,G)$) is bigger than the classical abelian $H^2(X,G)$, since AUT(G) is bigger than the classical delooping BG. – Mike Shulman Aug 24 '10 at 4:16
• Expanding a bit more, try looking at Larry Breen's notes for his talks at the Minnesota conference. arxiv.org/abs/math.CT/0611317 as a way into this. The paper by Aldrovandi and Noohi that David mentions is very nice and readable if you have some background in the area. – Tim Porter Aug 24 '10 at 6:31
• There is also nLab: ncatlab.org/nlab/show/nonabelian+cohomology . – Urs Schreiber Aug 24 '10 at 8:15
• I'll probably live to regret the joke. Thanks for the explanation. – Donu Arapura Aug 24 '10 at 17:13
• Could not find this mentioned anywhere, so decided to do it here: in "Combinatorial Differential Forms" (AiM, 2001) Breen and Messing give explicit description of three first differentials of a nonabelian version of the de Rham complex. – მამუკა ჯიბლაძე Feb 8 '17 at 6:41

Look up Giraud's "Cohomologie Non-abelienne", it should answer most of your questions. And, for the record, for $i=1$, by an analagous argument, you get a cohomology set (only a group if abelian) which is the moduli of vector bundles of rank $n$, or more generally principal $G$-bundles, if you replace $GL(N,\mathbb{C})$ with another group $G$.

Edit: "answer most of your questions" is somewhat stronger than what I actually meant. It's a place to start looking, and to see why the answer is far from trivial. I've heard vaguely that Lurie has some answers to this question (perhaps in Higher Topos Theory?) but I haven't had a chance to go through in detail and look at it.

• I don't quite understand what Giraud's book has to do with this specific question. Does it really define cohomology sets for i>1 for a sheaf of non-abelian groups? – Lennart Meier Aug 23 '10 at 17:18
• Charles, Giraud may not "answer most of your questions", but it will certainly teach you to never ask that question again. More seriously, I don't think he goes beyond $i=2$. – Donu Arapura Aug 23 '10 at 17:29
• Ok, yeah, I have to agree with both comments. I didn't think too hard about what exactly is in Giraud, and it is more somewhere to start looking. – Charles Siegel Aug 23 '10 at 20:33

Chapter 4 of Jean-Luc Brylinski's Loop spaces, characteristic classes and geometric quantization talks about non-abelian sheaf cohomology in degree 1 (i.e., $i=1$ in the notation in the question). I heard that there was a generalisation for $i>1$ due to Giraud, but I never studied it and cannot give you a reference.