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I've been revisiting group cohomology, and I realized that there is something I never quite understood. Let $G$ be a finite group, and let $A$ be a $G$-module (i.e. $\mathbb{Z}[G]$-module). Then the second cohomology $H^2(G,A)$ classifies group extensions $1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$ such that the action of $G$ on $A$ by left conjugation jibes with the given action of $G$ on $A$.

The interpretation of the first cohomology group, however, I still find somewhat elusive. On the one hand, there is the following interpretation: $H^1(G,A)$ is isomorphic to $\operatorname{Aut} ( 1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1) / \sim$ (where $\sim$ is conjugation by an element of $A$, and where $\operatorname{Aut}$ is taken in the category of group extensions of $G$ by $A$) by $\langle \rangle \mapsto (e \mapsto \langle f(e)\rangle e)$, where $f$ is the map $E\rightarrow G$, and where $\langle \rangle$ is a $1$-cocycle. This is indeed an isomorphism.

However, there is another interpretation! Some places say that there is a bijection between $H^1(G,A)$ and sections of $f:E\rightarrow G$ up to conjugation by an element of $A$! Here the problem is that I can't quite decipher if this should mean set-theoretic sections or group-theoretic sections.

Here's how I'm thinking about this: you can let $\operatorname{Aut}(E)/\sim$ act on the set of set-theoretic sections of $f$ by $\phi$ acts on $\lambda$ taking it to the section $\phi\circ \lambda$. It seems like this action is free. Furthermore, it seems like it is transitive. Indeed, if you want to take $\lambda$ to $\lambda'$ you can use the automorphism $e\mapsto \lambda'(f(e))\lambda^{-1}(f(e))e$ ($e\mapsto \lambda'(f(e))\lambda^{-1}(f(e))$ is a $1$-cocycle). So it seems like this gives a (non-canonical) bijection between $H^1(G,A)$ and the set-theoretic sections. But something is wrong -- this action takes the group-theoretic sections to group-theoretic sections! So it can't act freely and transitively on the set of set-theoretic sections...

I'm not quite sure what the resolution of this is since most sources don't distinguish between group-theoretic sections and set-theoretic sections.

Also, if you can inform me a way to think about this that relates these two interpretations with the interpretation that the first cohomology classifies torsors, I would be very happy indeed!

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What is $E$ supposed to be in both of your definitions of $H^1$? Also, if $A$ is really a $\mathbb{C}G$-module and $G$ is finite then first and second cohomology are zero. –  mt_ Dec 27 '11 at 18:08
    
$E$ is some (any) specific group extension of $G$ by $A$. As for your second comment, I'm not at all sure what you mean. The first and second group cohomologies are sometimes trivial, but usually not. –  Nicole Dec 27 '11 at 18:14
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Nicole, he means $A$ should probably be a $\mathbb{Z}[G]$ module. –  user641 Dec 27 '11 at 20:45
    
Ah! I will correct this. –  Nicole Dec 27 '11 at 20:55
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1 Answer

up vote 4 down vote accepted

My standard way to think about $H^1(G,A)$ is as functions $G \to A$ such that $f(gh)= g \cdot f(h) + f(g)$ ("cocycles"), modulo functions of the form $g \mapsto (g-1)\cdot a$ for some $a \in A$ ("coboundaries"). This is what you get by working with the bar resolution.

In answer to your question about sections, it's group homomorphisms that you need. You can choose the semidirect product $A \rtimes G$ as $E$. Then $H^1(G,A)$ as defined above is in bijection with the group homomorphisms $G \to A \rtimes G $ that split the projection $A\rtimes G\to G$, up to the action by conjugation of $A \subset A \rtimes G$. Any such map is of the form $g \mapsto (\alpha(g), g)$ where $\alpha: G \to A$; you can see that $\alpha$ must be a cocycle if the map is to be a group homomorphism, and that conjugation by $A$ affects $\alpha$ by a coboundary $g \mapsto (g-1)\cdot a$.

Returning to the sequence definition, an automorphism of your sequence (again with $E = A \rtimes G$) is really a group homomorphism $E \to E$ that induces the identity on $A$ and on $A \rtimes G / A \cong G$. Thus you can write it as $(a,g) \mapsto (a \alpha(g), g)$, where $\alpha : G \to A$. This map being a group homomorphism forces $\alpha$ to be a cocycle (compute the image of $(1,gh)$, compare to the product of the images of $(1,g)$ and $(1,h)$), and that conjugation by elements of $A$ changes $\alpha$ by a coboundary. So the sequence definition agrees with the `standard' one. This shows that your two definitions are equivalent.

I don't know the definition of a torsor I'm afraid so I can't help with your last question. Could you say what it means?

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Aha... So what you're saying is that I can't just choose any group extension in this interpretation - I have to work with the trivial extension (the semi-direct product). I see... Can you tell me where I was wrong with my argument in the question? It seemed like I defined an action of $H^1(G,A)$ on the set-theoretic sections of any given extension (not nec. the semi-direct product extension) which is free and transitive. Where did I go wrong? –  Nicole Dec 28 '11 at 0:01
    
P.S. You can find "torsor" on wiki. The idea is that $H^1(X,G)$, where $X$ is a "space" (e.g. scheme, manifold, topological space, ...) and $G$ is a "group object" (e.g. group scheme, group variety, finite group, ...), classifies $G$-torsors over $X$. Think of this naively as principal $G$-bundles over a topological space $X$. –  Nicole Dec 28 '11 at 0:04
    
@Nicole: consider $G=\mathbb{Z}/2=A=$ with trivial action. You could choose $\mathbb{Z}/2 \times \mathbb{Z}/2$ or $\mathbb{Z}/4$ in the middle, and for the latter choice there won't be any splitting maps. So you really must choose (something isomorphic to) the split extension if you want to get the first cohomology back. What you say about torsors seems to go in a different direction: $G$ is appearing as the coefficients, not the group so this will be some sort of non-abelian cohomology in general. –  mt_ Dec 28 '11 at 0:18
    
Having thought about it a bit more, I think that torsors should be classified by $H^1(G,G)$ in some appropriate theory. The boundaries should be maps like $g \mapsto [g,a]$ and the fact that boundaries are cycles should come from the commutator identity $[xz,y]=[x,y]^z[z,y]$. –  mt_ Dec 28 '11 at 0:53
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I didn't read your earlier comment about what was wrong with your argument properly. You say you defined an action of $H^1$-via-sequences on $H^$-via-splittings. (note first if any group-theoretic splitting exists then $E$ is isomorphic to the semidirect product). Second, it's not really Aut(E) you want (the autos must preserve A pointwise and induce the identity on E/A). You don't really mean $\lambda^{-1}$. What you construct may not be an automorphism of $E$ if $\lambda,\lambda'$ aren't homomorphisms (it may not even send $1\to 1$, try the $\mathbb{Z}/2$ example) –  mt_ Dec 28 '11 at 12:14
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