Why does non-abelian group cohomology exist?

Question

If $K$ is a non-abelian group on which a group $G$ acts via automorphisms, we can define 1-cocycles and 1-coboundaries by mimicking the explicit formulas coming from the bar resolution in ordinary group cohomology, and thus we have a reasonable notion of $H^1(G, K)$.

It turns out we have a part of the expected long exact sequence, until this construction breaks down for building $H^i$ when $i > 1$, where the long exact sequence stops. There are other analogues to ordinary group cohomology as well. The only proof I've ever seen of any of this is by hand. Is there some deeper explanation of why non-abelian group cohomology exists (and then ceases to exist)?

Answer 1

Topologically, you could say that this is true because $K(A,1)$ exists for nonabelian groups $A$. When the action of $G$ on $A$ is trivial, at least, $H^1(G,A)$ should be homotopy classes of maps from $K(G,1)$ to $K(A,1)$ (the same way $H^n(G,A) = H^n(K(G,1);A) =$ homotopy classes of maps $K(G,1) \to K(A,n)$ for $A$ abelian). In a similar way, $H^0$ is defined with coefficients in any pointed set.

Answer 2

To elaborate on Eric's answer, I believe that $H^{1-n}(G, A)$ is $\pi_n$ of the homotopy fixed point space $K(A, 1)^{hG}$. That exact sequence which ends at $H^1$ - which is only a set, while $H^0$ is a group - is actually the long exact sequence of the fibration $K(A, 1)^{hG} \to K(B, 1)^{hG} \to K(C, 1)^{hG}$ in disguise.

That reindexing $1-n$ is related to the $1$ in $K(A, 1)$; if $A$ is abelian, then we can replace all the occurrences of $1$ by $r$ for any $r \geq 0$, giving arbitrarily long segments of the long exact sequence. (Or you can use the language of spectra: $H^{-n}(G, A) = \pi_n((HA)^{hG})$. Note that $(HA)^{hG}$ has nonzero homotopy groups only in non-positive dimension.)

$K(A, 1)^{hG}$ is a groupoid (it only has homotopy in dimensions 0 and 1) and I expect it should be the groupoid of some kind of extensions of $G$ by $A$ (where morphisms are isomorphisms of extensions which are the identity on $G$ and $A$). $H^1(G, A)$ would then classify those extensions, and $H^0(G, A) = A^G$ would be the automorphism group of the basepoint, which I guess is the semidirect product (this is easy to check).

Answer 3

I don't know if this is a "deep" or "shallow" explanation, but if anyone is still reading this thread, here is a different explanation. I'll start with the preliminary comment that the cohomology of a group $K$ is a special case of cohomology of topological spaces. In topology in general, you get the same phenomenon that $H^k(X,G)$ is well-defined either when $G$ is abelian or $k=1$.

Consider the definition of simplicial cohomology for locally finite simplicial complexes. Or, more generally, CW cohomology for locally finite, regular complexes — regular means mainly that each attaching map is embedded. You can define a $k$-cochain with coefficients in a group $G$ (or even in any set) as a function from the $k$-cells to $G$. In attempting to define the coboundary of a cochain $c$ on a $k+1$-cell $e$, you should multiply together the values of $c$ on the facets of $e$. The obvious problem is that if $G$ is non-abelian, the product is order-dependent. However, if $k=1$, geometry gives you a gift: The facets are cyclically ordered, and what you mainly wanted to know is whether the product is trivial. The criterion of whether a cyclic word is trivial is well-defined in any group, not just abelian groups. A similar but simpler phenomenon occurs for the notion of a coboundary: If $e$ is an oriented edge and $c$ is a 0-cochain, there is a non-abelian version of modifying a 1-cochain by $c$ because the vertices of $e$ are an ordered pair.

So far this is just a more geometric version of Eric Wofsey's answer. It is very close to the fact that $\pi_1$ is non-abelian while higher homotopy groups are abelian — and therefore non-commutative classifying spaces exist only for $K(G,1)$. However, in this version of the explanation, something extra appears when $X=M$ is a 3-manifold.

If $M$ is a 3-manifold, then not only are the edges of a face cyclically ordered, the faces incident to an edge are also cyclically ordered. It turns out that, at least at the level of computing the cardinality of $H^1(M,G)$, you can let $G$ be both non-commutative and non-cocommutative. In other words, $G$ can be replaced by a finite-dimensional Hopf algebra $H$ which does not need to be commutative or cocommutative. Finiteness is necessary because it is a counting invariant. The resulting invariant $\#(M,H)$ was a topic of my PhD thesis and is explained in Involutory Hopf algebras and 3-manifold invariants and Non-involutory Hopf algebras and 3-manifold invariants. Although the motivation is original, the invariant is a special case of more standard quantum invariants defined by other people. (The same construction was also later found by three physicists, but I can't remember their names at all right now.)

Many 3-manifolds are also classifying spaces of groups, so for these groups there is the same notion of noncommutative, non-cocommutative group cohomology.

Answer 4

A concrete and arithmetically useful way to interpret it without appeal to explicit cocycle formulas is to express everything in the language of torsors. More specifically, for arithmetic purposes if the group G is Gal(F'/F) for a Galois extension F'/F and if the group K is H(F') for an F-group scheme H of finite type and K is equipped with the evident left G-action then ${\rm{H}}^1(G,K)$ is the set of isomorphism classes of H-torsors over F which split over F' (i.e.,, admit an F'-rational point). The low-degree exact sequence can then be expressed entirely in such terms, using pushouts and pullbacks with torsors. (Implicit in the argument is effectivity of Galois descent for H-torsors, which uses that H is quasi-projective over F.)

This is useful in settings as varied as H an abelian variety and H a linear algebraic group, and even the non-smooth case. In fact, when using non-smooth H it is rather restrictive to use Galois cohomology (but not unnatural if studying Tate-Shafarevich sets with coefficients in an Aut-functor, such as for a projective variety), and in such cases the "right" variant that is often more useful is to work with torsors for the fppf topology over F. The torsor viewpoint also gives a useful perspective when working over richer base rings than fields, such as rings of S-integers in a global field, even in the case of a smooth coefficient group (over the ring of S-integers), for which the etale topology is "enough". See section 5.3 in Chapter I of Serre's book on Galois cohomology for the Galois case, Milne's "Etale cohomology" book for generalization with flat and 'etale topologies, and Appendix B in my paper on "Finiteness theorems for algebraic groups over function fields" for a concrete fleshing out of the dictionary between the torsor and Galois languages (where I work with affine group schemes, due to the context of that paper).

Some papers of Mazur and Grothendieck (not as co-authors...) on abelian varieties make creative use of the torsor viewpoint when working with Tate-Shafarevich groups. The exact sequence for Brauer groups in global class field theory also has a useful interpretation via torsors; see Grothendieck's papers on Brauer groups for more in that direction (and somewhere in there he also discusses Tate-Shafarevich). Beware that when the base is not a field (or even when it is a field but we relax "quasi-projective" to "locally finite type" on $H$, such as for Aut-schemes of projective or proper varieties) then effectivity of descent for torsors is not at all clear, even with quasi-projective hypotheses, and so the torsors often need to be understood to be taken in the category of algebraic spaces (for which fppf descent is always effective). In the Néron Models book they have a discussion (somewhere in Chapter 6, I think) on effectivity of descent for torsors if one wishes to avoid algebraic spaces (under suitable hypotheses on the "coefficient group"), but working with algebraic spaces isn't so bad once one gets used to them and it is a more natural setting due to their better general behavior with respect to descent.

Answer 5

The topological viewpoint of Eric's answer applies to cohomology with nontrivial actions, too. If the action of G on A is not trivial, then the group cohomology $H(G;A)$ can be identified with the topological cohomology of $K(G,1)$ with local coefficients in the coefficient system (= locally constant sheaf) which is $A$ with its action of $\pi_1(K(G,1)) = G$. So the real question is then, why is the sheaf cohomology $H^1(X;A)$ defined with coefficients in a sheaf of nonabelian groups, but not $H^n$ for $n > 1$? This then essentially follows from the same argument ($K(A,1)$ exists for any group $A$, but other $K(A,n)$'s only for $A$ abelian) but applied in the category (or $(\infty, 1)$-category, or model category, or whatever) of sheaves of spaces over $X$.

There is also a sort of "higher nonabelian cohomology". For a nonabelian group $A$, you can't make $K(A,2)$ but you can make $B(\operatorname{hAut}(K(A,1)))$ where $\operatorname{hAut}$ denotes the topological monoid of self-homotopy-equivalences, and you can think about homotopy classes of maps from a space $X$ (such as $K(G,1)$) into $B(\operatorname{hAut}(K(A,1)))$ as a sort of "nonabelian $H^2$". If $A$ happens to be abelian, this construction contains the usual abelian $H^2$ via the map $K(A,2) = B(K(A,1)) \to B(\operatorname{hAut}(K(A,1)))$ given by letting $K(A,1)$ act on itself via left translation (since it is a topological group whenever $A$ is abelian). But even in this case, the "nonabelian $H^2$" contains much more than the usual abelian $H^2$, so it's a little misleading to call it "nonabelian $H^2$".

Answer 6

Extensions exist for non-abelian groups too.