1) Consider a group $G$ and a $G$-module $A$. Then it is well-known that there is a $1-1$ correspondence between

  • elements of $H^2(G,A),$ and
  • group extensions $1\rightarrow A \rightarrow H\rightarrow G \rightarrow 1,$ inducing the given action of $G$ on $A$, modulo the equivalence of extensions.

So my question is:

Is there a similar interpretation of $H^n(G,A)$ for $n >2$? In particular, do the elements of $H^n(G,A)$ correspond to the $(n-1)$-extensions $$1 \rightarrow A \rightarrow H_1 \rightarrow \dots \rightarrow H_{n-1} \rightarrow G \rightarrow 1,$$ again modulo the obvious equivalence? If not, is there another natural interpretation?

I am aware that the fact that $H^1$ does not have an interpretation of this type probably suggests the negative answer (however, in the "analogous" module case, $\mathrm{Ext}^0$ does not correspond to isomorphisms, but to homomorphisms in general). But I would be interested in other interpretations as well (in particular, if this interpretation generalizes the one for $H^1$ and $H^2$). Also I do not see whether and how could the grop $G$ induce an action on $A$ in the case of $n$-extensions.

2) Also, since $H^2(G, A) \simeq \mathrm{Ext}_{\mathbb{Z}[G]}^{2}(\mathbb{Z}, A)$ and there is a correspondence between elements of $\mathrm{Ext}_{\mathbb{Z}[G]}^{n}(\mathbb{Z}, G)$ and $2$-extensions $$0 \rightarrow A \rightarrow H_1 \rightarrow H_{2} \rightarrow \mathbb{Z} \rightarrow 0,$$ so there is a correspondence between extensions of $G$ by $A$ (with the given action) ad $2$-extensions of this form.

How does this correspondence look like? And can it be generalized to get the correspondence as in 1)?

Thanks in advance for any help and/or references.

  • 2
    $\begingroup$ Read Ken Brown's book Cohomology of Groups, because it explains $H^3$ as crossed-module extensions, and there is an extension-interpretation of $H^1$. In general, the higher cohomologies were described by MacLane (the appropriate references are given in this book) but are essentially intractable. $\endgroup$ – Chris Gerig Nov 11 '14 at 17:43
  • 2
    $\begingroup$ You could also look at the paper D.F. Holt, An interpretation of the cohomology groups $H^n(G, M)$, Journal of Algebra Volume 60, Issue 2, October 1979, Pages 307–318. The historical note at the end by Saunders MacLane provides a lot of information on the literature on this topic. $\endgroup$ – Derek Holt Nov 11 '14 at 21:16
  • $\begingroup$ I'll put in another vote for Brown's book, which gives a good group-theoretic treatment of group cohomology. For a speciifc example, following Chris Gerig's comment above, Brown constructs a particular obstruction in $H^3$ (with coefficients in a particular module) that vanishes iff an extension of $G$ by some fixed (nonabelian) $N$ exists. $\endgroup$ – anomaly Nov 12 '14 at 5:41

Here is an interpretation of $H^{n+1}(G, A)$ for every $n$: it is precisely (derived) global sections of the (derived) local system on the classifying space / Eilenberg-MacLane space $BG \cong K(G, 1)$ given by the induced action of $G$ on $B^{n+1} A \cong K(A, n+1)$.

The reason you might care about this is that for $n \ge 0$ this is the same as classifying (weak, pointed, connected) homotopy types $X$ fitting into a fiber sequence

$$B^n A \to X \to BG$$

with fixed action of $\pi_1$ of the base on the fiber, and for $n \ge 2$ this is the same thing as classifying (weak, pointed, connected) homotopy types $X$ with $\pi_1(X) \cong G, \pi_n(X) \cong A$, and with the action of $\pi_1(X)$ on $\pi_n(X)$ fixed. For $n = 1$ you recover group extensions, but this case is somewhat anomalous as it's the only case where the fiber and the base have homotopy groups in the same dimension, so while it's concrete it's also misleading.

In a slogan,

higher group cohomology solves an extension problem for higher groups, thought of as (weak, pointed, connected) homotopy types via the homotopy hypothesis.

The topological picture is that nice homotopy types can be described as "iterated extensions" of Eilenberg-MacLane spaces, with the extension data you need to build a nice space $X$ out of the Eilenberg-MacLane spaces $B^n \pi_n(X)$ being given by a sequence of cohomology classes called its Postnikov invariants. In the nicest case (e.g. when $X$ is simply connected), if these all vanish, then $X$ is homotopy equivalent to $\prod_n B^n \pi_n(X)$, so the Postnikov invariants can be thought of as measuring the extent to which $X$ fails to be a product of Eilenberg-MacLane spaces. In the above case, the class in $H^{n+1}(G, A)$ corresponding to $X$ measures the extent to which $X$ fails to be a semidirect product $B^n A \rtimes BG$.

The categorical picture is that in the same way that groups act on sets, higher groups act on objects in higher categories: for example, the natural things that act on categories themselves are not groups but 2-groups, which from the perspective of the homotopy hypothesis can be thought of as (weak, pointed, connected) homotopy types with vanishing $\pi_n, n \ge 3$. Higher groups can be built out of groups using extra data related to coherence isomorphisms, and that extra data is what Postnikov invariants are encoding.

Example. 2-groups can also be thought of as particularly nice monoidal categories, and if $X$ is a 2-group with $\pi_1(X) \cong G, \pi_2(X) \cong A$ (here $\pi_1(X)$ corresponds to the group of invertible objects in $X$ thought of as a monoidal category and $\pi_2(X)$ corresponds to the group of automorphisms of the unit for the monoidal product), and fixed action of $G$ on $A$, then the only additional data needed to specify $X$ is a Postnikov invariant in $H^3(G, A)$ which turns out to come, in the categorical picture, from the associator of the monoidal product. The pentagon condition here becomes the cocycle condition.

Subexample. Where do 2-groups come from? Any category has associated to it an automorphism 2-group, but this is hard to compute in general. Groups are in particular one-object categories, so any group $G$ has an automorphism 2-group. Its $\pi_1$ is the outer automorphism group $\text{Out}(G)$ and its $\pi_2$ is the center $Z(G)$. The action of $\pi_1$ on $\pi_2$ is the obvious one. This 2-group is relevant to the complete description of the extension problem for groups where the normal subgroup is not required to be abelian, and in particular the fact that its Postnikov invariant does not vanish in general is responsible for the obstruction living in $H^3$ mentioned by anomaly in the comments.

  • 1
    $\begingroup$ Thank you for this answer. So the natural interpretation is in fact topological one and it has (more or less) algebraic analogue provided that the homotopy hypothesis holds. Do I understand at least this much correctly? Also, is it possible to bypass the hypothesis, formulating the connection only by means of the higher groups? $\endgroup$ – Pavel Čoupek Nov 13 '14 at 13:18
  • $\begingroup$ @PavelC: the lesson of the homotopy hypothesis is that homotopy theory is higher group theory; calling higher groups "topological" doesn't recognize the many ways higher groups can arise that have nothing to do with topological spaces (e.g. as higher automorphism groups). You can bypass the homotopy hypothesis but then you need to define what a higher group is. The homotopy hypothesis provides one of the easiest definitions (it's a weak, pointed, connected homotopy type); without that you need to work harder, although for 2-groups you can still be relatively explicit. $\endgroup$ – Qiaochu Yuan Nov 13 '14 at 23:40

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