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I have a discrete group $G$ and classes $x,y\in H^1(G;\mathbb{Q})$ (group cohomology with coefficients in the rationals viewed as a trivial $G$-module) such that the Massey product $$\alpha:=\langle x, x, y\rangle\in H^2(G;\mathbb{Q})$$ is defined and has zero indeterminacy.

I would like to find a short exact sequence of $G$-modules $0\to \mathbb{Q}\to M\to N\to 0$ and a class $a\in H^1(G;N)$ which maps to $\alpha$ under the associated Bockstein operator $\beta\colon\thinspace H^1(G;N)\to H^2(G;\mathbb{Q})$.

Is this always possible? If so, is there a procedure for doing this? If not, are there known necessary or sufficient conditions (on the group $G$, or the classes $x$ and $y$)?

What I know

I'm attempting to use the interpretations of low-dimensional group cohomology described in Chapter IV of the book of Brown. Using the standard resolution, the classes $x$, $y$ are represented by derivations $d_x,d_y\colon\thinspace G\to \mathbb{Q}$, which I can write down explicitly. I am also able to write down cochains $\lambda,\mu\in C^1(G;\mathbb{Q})$ (essentially just functions $G\to \mathbb{Q}$) satisfying $\delta\lambda = d_x\cup d_x$ and $\delta\mu=d_x\cup d_y$. This enables me to write down explicitly the central extension $$0\to\mathbb{Q}\to E\to G\to 1$$ which represents the class $\alpha$. But I'm not sure where to go from here.

One approach would to look for a $G$-module $M$ which has $\mathbb{Q}\hookrightarrow M$ as a trivial submodule, and such that $\alpha$ is in the kernel of the induced map $H^2(G;\mathbb{Q})\to H^2(G; M)$. But again I'm not sure how to go about doing this.

Any general suggestions would be appreciated.

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If $Q$ has trivial $G$-action, isn't $H^1(G,Q) = Hom(G,Q)$? –  SGP Jun 17 '11 at 22:13
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Also, if $Q \hookrightarrow M$ is an embedding into an injective $G$-module, then $H^i(G,N) = H^{i+1}(G,Q)$ for all $i > 0$; here $N$ is the quotient $M/Q$; this is called dimension-shifting..see Serre's Local Fields book or any other standard reference. –  SGP Jun 17 '11 at 22:18
    
@SGP: Yes it is. –  Mark Grant Jun 18 '11 at 8:05
    
@SGP: Thanks, I read about dimension shifting soon after posting the question! So the question trivially has a positive answer. If you want to post as an answer I'd be happy to accept it (although Fernando's answer is also an interesting take). –  Mark Grant Jun 18 '11 at 8:12
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1 Answer

up vote 2 down vote accepted

If $\mathbb{Q}G$ is the group algebra, group cohomology is just $H^{\ast}(G,\mathbb{Q})=\operatorname{Ext}^{\ast}_{\mathbb{Q}G}(\mathbb{Q},\mathbb{Q})$. Now take any ring $R$, any pair of $R$-modules $A$ and $B$, and any class $\alpha\in \operatorname{Ext}^{n}(A,B)$. Take a Yoneda $n$-extension representing $\alpha$:

$$B\hookrightarrow M_n\rightarrow M_{n-1}\rightarrow\cdots\rightarrow M_2\rightarrow M_1 \twoheadrightarrow A$$

Factor the homomorphism $M_n\rightarrow M_{n-1}$ as

$$M_n\twoheadrightarrow N \hookrightarrow M_{n-1}$$

You can split the Yoneda $n$-extension representing $\alpha$ into a Yoneda $1$-extension and a Yoneda $(n-1)$-extension representing $\beta\in \operatorname{Ext}^{n-1}(A,N)$

$$B\hookrightarrow M_n\twoheadrightarrow N$$

$$N\hookrightarrow M_{n-1}\rightarrow\cdots\rightarrow M_2\rightarrow M_1 \twoheadrightarrow A$$

This $\beta\in \operatorname{Ext}^{n-1}(A,N)$ is kicked by $\alpha\in \operatorname{Ext}^{n}(A,B)$ in the long exact sequence associated to $B\hookrightarrow M_n\twoheadrightarrow N$

$$\cdots\rightarrow \operatorname{Ext}^{n-1}(A,N) \rightarrow \operatorname{Ext}^{n}(A,B) \rightarrow\cdots$$

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