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Given a group $G$ and $G$-modules $M,N$ with $M$ $\mathbb{Z}$-free then it's well known that $$Ext_{\mathbb{Z}G}^i(M,N) \cong H^i(G,Hom(M,N))$$ for all $i \ge 0$ (a reference is Brown, Cohomology of Groups, Proposition 2.2).

But what happens if $M$ is not $\mathbb{Z}$-free ? Is it still possible to express $Ext_{\mathbb{Z}G}^i(M,N)$ by cohomology groups of $G$ (maybe in form of a spectral sequence) ?

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    $\begingroup$ I think this is either eq. (5) in Chap. XVI, Sect. 4 of Cartan-Eilenberg or you can get what you want by the same method. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Dec 20, 2012 at 22:10
  • $\begingroup$ Thanks for the reference. (+1) for fishing out the appropriate formula from the mass of formulas in CE. $\endgroup$
    – Mark Opitz
    Commented Dec 21, 2012 at 0:26

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There is a long exact sequence

$$0 \to H^1(G,Hom(M,N)) \to Ext_{\mathbb{Z}G}^1(M,N) \to \cdots $$ $$\begin{array}{lll} \cdots & \to & H^i(G,Hom(M,N)) \to Ext_{\mathbb{Z}G}^i(M,N) \newline & \to & H^{i-1}(G,Ext_{\mathbb{Z}}^1(M,N))\to H^{i+1}(G,Hom(M,N)) \to \cdots \end{array}$$

For, as pointed out by Will and Mariano, there is a spectral sequence $$H^i(G, Ext^j_\mathbb Z (M,N)) \Rightarrow Ext_{\mathbb ZG}^{i+j}(M,N)$$ and since the projective dimension of $\mathbb{Z}$ is one, the spectral sequence takes the form $$E_2 = \quad \begin{array}{ccccc} \vdots & \vdots & \vdots & \vdots & \newline 0 & 0 & 0 & 0 & \cdots \newline \bullet & \bullet & \bullet & \bullet & \cdots \newline \bullet & \bullet & \bullet & \bullet & \cdots \end{array}$$ Now the relation $E_\infty =E_3 = \ker(d_2)/\text{im}(d_2)$ yields the exact sequence $$0 \to \ker(d_2^{i-2,1}) \to E_2^{i-2,1} \to E_2^{i,0} \to H^i \to \ker(d_2^{i-1,1}) \to 0$$ where $H^i=Ext_{\mathbb ZG}^i(M,N)$ is the abutment.

Remark: The statement remains true if we replace $\mathbb{Z}$ by an hereditary commutative ring $R$, $\mathbb{Z}G$ by an augmented $R$-projective $R$-algebra $A$ and $H^\ast(G,-)$ by $Ext_A^\ast(R,-)$.

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  • $\begingroup$ augmented R-projective R-algebra A = group ring RG? $\endgroup$
    – Marc Palm
    Commented Dec 21, 2012 at 11:46
  • $\begingroup$ Marc, I'm not sure what you mean by this equality. $\endgroup$
    – Ralph
    Commented Dec 21, 2012 at 12:02
  • $\begingroup$ I meant to say, eg RG is such an algebra? mathoverflow.net/questions/674/… $\endgroup$
    – Marc Palm
    Commented Dec 21, 2012 at 13:25
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    $\begingroup$ Yes. The augmentation is given by $RG \to R,\;g \mapsto 1$ and $RG=\bigoplus_g Rg$ is $R$-free. $\endgroup$
    – Ralph
    Commented Dec 21, 2012 at 13:43
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As you've probalby noticed, we can't just take group cohoology. If $G$ is trivial, we are just taking exts in the category of abelian groups, which are nontrivial, but the group cohomology is trivial.

The spectral sequence idea is completely correct. The functor $Hom_{\mathbb ZG}(M,-)$ is the composition of the functor $Hom_{\mathbb Z}(M,-)$ with the functor $H^0(G,-)$. Here we need to see $Hom$ as a $G$-module. $G$ acts by taking $f$ to $g^{-1} \circ f \circ g$.

Now we use the spectral sequence for a composition of derived functors. $Hom_{\mathbb Z}$ preserves projectives so that makes sense.

So we get a spectral sequence sending $H^i(G, Ext^j_\mathbb Z (M,N))$ to $Ext^i_{\mathbb ZG}(M,N)$.

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