# Non-abelian Ext functor and non-abelian $H^2$

Let $$G$$ be a group and $$0\rightarrow K\rightarrow M\rightarrow N\rightarrow 0$$ a short exact sequence of groups. Now these are abelian groups, if I want to show that $$\text{Hom}(G,M)\rightarrow \text{Hom}(G,N)$$ is surjective, I would show that $$\text{Ext}^1(G,K)=0$$. However, if I'm studying the same question for non-abelian groups, then I do not have the tool of derived categories at my disposal. Can this be overcome with (non-abelian) cohomology?

• Yes (in some sense). Aug 1 '20 at 19:30
• @MikhailBorovoi How? Aug 1 '20 at 19:38
• See the reference in my comment to this question. Aug 1 '20 at 20:45
• @MikhailBorovoi, you can link to specific comments, not just questions. I think you mean this comment. Aug 1 '20 at 22:07
• I took the liberty to edit the title and to add the tag galois-cohomology. This is relevant to my answer. Roll back if you don't like this. Aug 3 '20 at 20:55

EDITED, taking into account the comments of Donu Arapura.

As JLA wrote, a homomorphism $$f\colon G\to N$$ gives an extension $$$$\label{e:E} 1\to K\to E\to G\to 1.\tag{E}$$$$ This extension defines a homomorphism $$b\colon G\to \operatorname{Out} K$$ called the band (lien, kernel) of \eqref{e:E}. By definition, $$H^2(G,K,b)$$ is the set of isomorphism classes of extensions \eqref{e:E} bound by $$b$$.

A cohomology class $$\eta(E)\in H^2(G,K,b)$$ is called neutral if the extension \eqref{e:E} splits, that is, there exists a homomorphism $$G\to E$$ such that the composite homomorphism $$G\to E\to G$$ is the identity automorphism of $$G$$. In this case we obtain an action $$\varphi$$ of $$G$$ on the normal subgroup $$K$$ of $$E$$, and we obtain an isomorphism $$E\overset{\sim}{\to}K\rtimes_\varphi G$$ with the semidirect product.

There may be more that one neutral class in $$H^2(G,K,b)$$: they correspond to semidirect products with different actions $$\varphi$$ of $$G$$ on $$K$$. I have read that there may be no neutral elements, but I don't know examples. (In the Galois cohomology setting, for a connected reductive group, by Douai's theorem there always exists a neutral element in nonabelian $$H^2$$; see [2], Proposition 3.1).

If $$K$$ is abelian, then $$\operatorname{Out} K = \operatorname{Aut} K$$, so $$b$$ is just an action of $$G$$ on $$K$$, and $$H^2(G,K,b)$$ is the usual abelian group cohomology $$H^2(G,K)$$, where $$G$$ acts on $$K$$ via $$b$$.

The set $$H^2(G,K,b)$$ can be described in terms of cocycles. See Section 1.14 in Springer [1].

The band $$b$$ defines an action of $$G$$ on the center $$Z=Z(K)$$, and we may consider the usual (abelian) group cohomology $$H^2(G,Z)$$. From the cocyclic description of $$H^2(G,K,b)$$ it is clear that $$H^2(G,Z)$$ naturally acts on $$H^2(G,K,b)$$.

Moreover, if the set $$H^2(G,K,b)$$ is nonempty, then $$H^2(G,Z)$$ acts on it simply transitively; see Mac Lane, Homology, Theorem IV.8.8. The set $$H^2(G,K,b)$$ is nonempty if and only if a certain obstruction $$\operatorname{Obs}(G,K,b)\in H^3(G,Z)$$ vanishes; see Mac Lane, Theorem IV.8.7.

Note that we should not think that $$H^2(G,K,b)$$ "equals" $$H^2(G,Z)$$. First, $$H^2(G,K,b)$$ does not have a distinguished unit element. Secondly, $$H^2(G,K,b)$$ has a distinguished subset $$N^2(G,K,b)$$ of neutral elements. This is important because in many applications one uses nonabelian $$H^2$$ in order to determine whether a given extension \eqref{e:E} is split or not.

As far as I know, nonabelian $$H^2$$ is mostly used in the Galois cohomology setting. Namely, if $$k$$ is an algebraic closure of a field $$k_0$$ of characteristic 0, $$G=\operatorname{Gal}(k/k_0)$$, and $$Y$$ is a quasi-projective $$k$$-variety with additional structure (say, an algebraic group or a homogeneous space) such that for any $$\sigma\in G=\operatorname{Gal}(k/k_0)$$ there exists an isomorphism $$\alpha\colon\sigma Y\overset{\sim}{\to}Y$$, then it defines an extension $$1\to \operatorname{Aut} Y\to E\to G\to 1,$$ where $$E$$ is the set of such pairs $$(\alpha,\sigma)$$ with a suitably defined composition law. We obtain the cohomology class $$\eta(Y)\in H^2(k_0,\operatorname{Aut} Y,b)$$ of this extension for a suitable band $$b$$. The variety $$Y$$ (with additional structure) admits a $$k_0$$-model if and only if $$\eta(Y)$$ is neutral, that is, the extension splits; see this question.

For nonabelian $$H^2$$ in Galois cohomology see:

[1] T. A. Springer, Non-abelian $$H^2$$ in Galois cohomology, in: Algebraic Groups and Discontinuous Subgroups, Proc. Sympos. Pure Math. 9, Amer. Math. Soc., Providence, 1966, 164-182.

[2] M. Borovoi, Abelianization of the second nonabelian Galois cohomology. Duke Math. J. 72 (1993), 217-239.

[3] Flicker, Scheiderer, Sujatha, Grothendieck's theorem on non-abelian $$H^2$$ and local–global principles. J. Amer. Math. Soc. 11 (1998), no. 3, 731–750.

• If you are interested in nonabelian $H^2$ in Galois cohomology, please ask a separate question, and I will answer it when I have time Aug 3 '20 at 11:11
• This is a very nice answer. I do have a couple of questions. (1) When you say $\eta(E)$ is neutral, does that mean there exists a lift of $b$ to $G\to Aut(K)$, such that $E$ is the semi direct product? So then it is clear, there may be more than one, or perhaps none. (2) What is the relationship to the old results of Eilenberg-Maclane "Cohomology theory in abstract groups. II." Annals (1947)? As I recall, they show that the set of extension classes is either empty (with obstructions in $H^3$) or parameterized by $H^2(G, Z)$, where $Z$ is the centre of $K$. Aug 3 '20 at 16:54
• Of course it's your right not to name your paper if you don't want, and I apologise for an unwanted edit; but too many old MO comments become less useful when "here"-type links fade, so I'll put the name of "this preprint" here if it's all right: Borovoi - Extending the exact sequence of nonabelian $H^1$, using nonabelian $H^2$ with coefficients in crossed modules. Aug 3 '20 at 21:07
• Thanks again for the interesting answer. Concerning an example. Let $F$ be a nonabelian free group. Choose a surjection $f:F\to G$, where $G$ is finite and nontrivial. Let $K$ be the kernel. $K$ is again nonabelian and free, so it must have trivial centre. By your answer, the nonabelian $H^2$ consists of a single element. This cannot be neutral, because the sequence can't split, as $F$ is torsion free. Aug 4 '20 at 15:11
If you have a morphism $$f:G\to N\,,$$ then you get a $$K$$-extension of $$G$$ by pulling back the $$K$$-extension of $$N\,.$$ The morphism $$f$$ lifts to a morphism into $$M$$ if and only if this extension is trivial. So you could show the map you want is surjective by showing that all $$K$$-extensions of $$G$$ are trivial.
If $$K$$ is abelian, then isomorphism classes of $$K$$-extensions correspond to classes in (abelian) group cohomology.