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I want to consider relative group cohomology: the construction is as follows. I have a subgroup $H\subseteq G$ (and note that I don't want to assume that $H$ is normal in $G$), and a $\mathbb Z[G]$-module $M$. Then we have the standard chain complxes $C^\ast(G;M)$ and $C^\ast(H,M)$, and there is a natural morphism $C^\ast(G,M)\to C^\ast(H,M)$, which induces the "restriction homomorphism" on group cohomology $\operatorname{res}:H^\ast(G,M)\to H^\ast(H,M)$. Let us define the "relative group cohomology" as the cohomology of the chain complex which fits into the exact sequence: $$ 0\to C^\ast(G,H;M)\to C^\ast(G;M)\to C^\ast(H;M)\to 0 $$ (i.e. $C^\ast(G,H;M)$ is defined to be the kernel of the second map). I haven't ever heard of these "relative group cohomology" groups, but it seems like a very natural idea to me, and what I'm trying to do is define algebraically the cellular cohomology groups $H^\ast(K(G,1),K(H,1);M)$ (in case we have a $K(H,1)$ which is naturally a subcomplex of a $K(G,1)$). If anyone has a good reference for these I'd like to know! Note that by definition, the relative group cohomology groups $H^\ast(G,H;M)$ fit into a natural long exact sequence: $$ \cdots\to H^\ast(G,H;M)\to H^\ast(G;M)\to H^\ast(H;M)\to\cdots $$ and this is what one would expect the cellular cohomology groups $H^\ast(K(G,1),K(H,1);M)$ to satisfy. If I've messed this construction up, please tell me.


How do I understand the relative group cohomology in terms of derived functors? We know that $H^\ast(G;M)=\operatorname{Ext}^\ast_{\mathbb Z[G]}(\mathbb Z,M)$ and $H^\ast(H;M)=\operatorname{Ext}^\ast_{\mathbb Z[H]}(\mathbb Z,M)$. But since these are $\operatorname{Ext}$'s in different categories, it doesn't seem clear how to fit a third into the exact sequence. What I'd like is some $\operatorname{Ext}$ definition of the relative group cohomology groups I've defined above.

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I've tried the following, but it seems to give the "wrong" answer. We can get everything into the same category by observing that $M^H=\operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z[G/H],M)$, and thus the cohomology is given by $H^\ast(H;M)=\operatorname{Ext}^\ast(\mathbb Z[G/H],M)$ (from now on, all $\operatorname{Ext}$'s are in the category of $\mathbb Z[G]$-modules). Furthermore (and correct me if I am wrong), the restriction homomorphism $H^\ast(G,M)\to H^\ast(H,M)$ is induced by the "sum coefficients" morphism $\mathbb Z[G/H]\to\mathbb Z$ (giving the map $\operatorname{Ext}^\ast(\mathbb Z,M)\to\operatorname{Ext}^\ast(\mathbb Z[G/H],M)$). So, now it looks like we get what we want, but now comes a surprise. The "first argument"s of the $\operatorname{Ext}$'s fit into a short exact sequence: $$ 0\to\ker\to\mathbb Z[G/H]\to\mathbb Z\to 0 $$ and thus we have a long exact sequence of $\operatorname{Ext}$: $$ \cdots\to\operatorname{Ext}^\ast(\mathbb Z,M)\to\operatorname{Ext}^\ast(\mathbb Z[G/H],M)\to\operatorname{Ext}^\ast(\ker,M)\to\cdots $$ But now it looks like $\operatorname{Ext}^\ast(\ker,M)$ is not giving the relative group cohomology groups we want: the long exact sequence isn't the same as the one above, it's gotten flipped around. I guess this doesn't entirely disqualify the construction, since perhaps we have $H^\ast(G,H;M)=\operatorname{Ext}^{\ast-1}(\ker,M)$, but in this case I'd still like an explanation for why this dimension shifting happens.

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There is a $K(H,1)$ which is a subcomplex of $K(G,1)$: the usual one given by the bar construction. – Mariano Suárez-Alvarez May 29 '11 at 15:53
Have you read the construction of relative cohomology in Benson's representations and cohomoogy vol 1 section 3.9? The dimension shifting behaviour is because your short exact sequence is the "relative projective cover" of $\mathbb{Z}$. – M T May 29 '11 at 18:01
I think your construction is already known in the literature. See: Satoru Takasu: Relative homology and relative cohomology theory of groups. J. Fac. Sci. Univ. Tokyo. Sect. I 8(1959), 75-110. I haven't looked into the paper, but from Math Sci Net review: "For a group $\Pi$ and a subgroup $\pi$, define $I = Ker(\mathbb{Z}\Pi \otimes_{\mathbb{Z}\pi} \mathbb{Z} \to \mathbb{Z})$ and set $H^q(\Pi,\pi:A) = \operatorname{Ext}^{q-1}_{\mathbb{Z}\pi}(I,A)$ for $q\ge 1$." There is also an interpretation via standard chain complexes. – Ralph May 30 '11 at 16:32
The original reference for this construction seems to be: W.S. Massey: Some problems in algebraic topology and the theory of fibre bundles, Ann. of Math. 62 (1955) 327-359, Problem 22. In a footnote, he says "relative homology groups of groups have been considered by K. de Leeuw in his PhD Thesis "The relative cohomology theory of finite groups and algebraic number theory" (Princeton University, 1954)". I wonder whether it is the same construction. – Daniel Moskovich Jul 25 '12 at 3:36
up vote 9 down vote accepted

Your argument is correct, and you do get that the relative group cohomology in your terms is a shift of these Ext-groups.

One reason why the dimension-shifting behavior occurs is that, with original gradings, you can't possibly have that $H^i(G,H;M)$ are the derived functors of $H^0(G,H;M)$. In fact, $C^0(G,H;M)$ is always the zero group, and so $H^0(G,H;M)$ is always zero; its derived functors are zero.

In this case, what your argument shows is that $H^{i+1}(G,H;M)$ is the i'th derived functor of $H^1(G,H;M)$.

If I might be permitted to wax philosophical, once one has moved to chain complexes the general yoga of triangulated categories says that one shouldn't really worry about the distinction between a kernel and the shift of a cokernel. Up to weak equivalence of chain complexes, one can always move between these.

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Here are some remarks: the situation we would like to model algebraically is this: suppose $H$ is a subgroup of a group $G$ and $M$ is a $G$-module. Let $L_M$ be the corresponding local system on $BG$. Then we have a map of classifying spaces $f:BH\to BG$ obtained as follows: take $EG$, a contractible space on which $G$ acts freely and quotient it by $H$; the result will be $BH$ and it maps to $BG=EG/G$. The local system $L_M$ pulls back to $BH$, which gives a map $f^*:H^*(BG,L_M)\to H^*(BH,f^{-1}L_M)$. This is, of course, the map $H^*(G,M)\to H^*(H,M)$.

A side remark: if $H$ is normal, then $f$ is the projection of a principal $G/H$-bundle, which gives the classifying map $g:BG\to B(G/H)$. Now, if one replaces $g$ with a Serre fibration, the fiber will be $BH$ [this is not completely obvious; perhaps I'll add a reference later] and the local system restricted to the fiber will be isomorphic to $f^{-1}L_M$. Taking the Leray spectral sequence of this fibration one gets the Serre-Hochschild spectral sequence.

Now, coming to the derived functors picture: the cohomology $H^i(G,M)$ is the $i$-th derived functor of the invariants functor from the category of $G$-modules to the category of abelian groups. In other words, we have to derive the functor $M\mapsto Hom(\mathbb{Z},M)$. There are two ways to do that. First, we can replace $\mathbb{Z}$ with a projective resulotion; this is what people usually do since this resolution can be written down explicitly and the resulting $Hom$ complex is the standard cochain complex $C^*(G,M)$. But we can take an injective resolution of $M$; it would work just as well. So $H^i(G,M)$ is nothing but $Hom_{D G-mod}(\mathbb{Z},M[i])$ (unless I've messed it up and there should be minuses somewhere).

Every $G$-module is an $H$-module, which gives a functor $f^{-1}:D G-mod\to D H-mod$. This functor corresponds to pulling sheaves from $BG$ back to $BH$ and it has a right adjoint $f_*$, which corresponds to pushing sheaves forward from $BH$ to $BG$. The construction of $f_*$ is not too difficult, but requires some work, see Bernstein-Lunts, Equivariant sheaves and functors, part I. We have a natural adjunction morphism $M\to f_* f^{-1}M$; if we apply $Hom_{DG-mod}(\mathbb{Z},-)$ to the shifts of this morphism, we get the maps $H^*(G,M)\to H^*(H,M)$. Now, the relative cochain complex of the posting computes $Hom_{DG-mod}(\mathbb{Z},-)$ of an object $A$ of $DG-mod$ such that $A\to M\to f_* f^{-1}M\to A[1]$ is a distinguished triangle.

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