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This may seem like a pie-in-the-sky speculation question, but I have good reasons for asking this.

Is there any sense in which $H^{-1}(G;M)$ is defined for a group $G$ and a $G$-module $M$?

The reason I suspect this may make sense comes from some of my work with Kassabov and Vogtmann, where we consider cohomology spaces $H^{2n-3}(\mathrm{Out}(F_n);S(V^{\otimes n}))$. Here $V=H_1(\Sigma_{g,1};\mathbb Q)$ is the first homology of a surface of genus $g$ with one boundary component, $S(V^{\otimes n})$ is the symmetric algebra on $V^{\otimes n}$ and $\mathrm{Out}(F_n)$ acts on $S(V^{\otimes n})\cong S(V\otimes \mathbb Q^n)$ via the $\mathrm{GL}_n(\mathbb Z)$-action on $\mathbb Q^n$. (Taking the base field to be $\mathbb Q$.)

The reason we were interested in these spaces is that we could embed the abelianization of the Lie algebra $\mathfrak h^+_V$ which is the target of the Johnson homomorphism on $\mathrm{Mod}(g,1)$ into the direct sum of these cohomology groups. Well, almost. We had a separate discussion for the $n=1$ case, where you would be looking at $H^{-1}(\mathrm{Out}(F_1);S(V))=H^{-1}(\mathbb Z_2;S(V))$. Here $\mathbb Z_2$ acts on the symmetric algebra $S(V)$ by $(-1)^{k}$ on $S^k(V)$, and the answer "should be" $\oplus_{k=1}^\infty S^{2k+1}(V)$, which in some sense is the anti-invariant part of $S(V)$. So one hypothesis might be that $H^{-1}(G;M)= M/M^G$, but I don't see how to justify that.

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Well, there is Tate cohomology $\widehat{H}^\ast(G)$ defined for $\ast\in\mathbb{Z}$, which packages group homology and cohomology together (positive integers is cohomology, negative integers is homology). – Chris Gerig Jun 24 '13 at 1:33
In particular, $\widehat{H}^{-1}(G,M)=Ker(N)\subset H_0(G,M)$ where $N:M_G\to M^G$ is the norm map. – Chris Gerig Jun 24 '13 at 1:47
And Tate-Farrell cohomology, which works for some infinite groups. The don't quite give the $H^{-1}$ you want, but close. – anon Jun 24 '13 at 1:51
Yes, Tate cohomology isn't giving me the "right" answer. For rational vector spaces like $S^k(V)$ the norm map is an iso, so $\hat{H}^{-1}=0$. – Jim Conant Jun 24 '13 at 4:07

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