Let $G$ and $\Gamma$ be discrete groups, and let $\phi\colon\thinspace G\to \Gamma$ be a homomorphism.

Define its cohomological dimension $\operatorname{cd}\phi$ to be the least integer $d$ such that $\phi^\ast\colon H^i(\Gamma;M)\to H^i(G;M)$ is the zero homomorphism for all $i>d$ and all $\Gamma$-modules $M$ (where $M$ is regarded as a $G$-module via $\phi$).

Given that cohomological dimension of groups is such a well-studied invariant, I would have expected to find references to this relative notion in the literature. Alas, I cannot.

Are there any references considering cohomological dimension of homomorphisms?

and more specifically

Does anyone know an example of a surjective homomorphism $\phi$ as above for which $$\operatorname{cd} \phi < \min\lbrace \operatorname{cd}G, \operatorname{cd} \Gamma \rbrace?$$

EDIT: Thanks to Tom and Ralph's answers, I have been able to prove the following precise statement:

Let $$ 1\to A \to G \stackrel{\phi}{\to} \Gamma \to 1$$ be a central extension, where $H_\ast(A)$ is free and of finite type, and $\Gamma$ is a duality group with $\operatorname{cd}\Gamma = n$. Then $\operatorname{cd}\phi = n$.

Proof. We will show that $0\neq \phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma)\to H^n(G;\mathbb{Z}\Gamma)$. This follows from the Lyndon-Hochschild-Serre spectral sequence. Since the action of $\Gamma$ on $A$ is trivial, and $\mathbb{Z}\Gamma$ is a trivial $A$-module, the $E_2$ term has $$H^p(\Gamma;H^q(A;\mathbb{Z}\Gamma))\cong H^p(\Gamma;H^q(A)\otimes\mathbb{Z}\Gamma)$$ in the $(p,q)$-position. Since $\Gamma$ is a duality group, this is zero for $p\neq n$. Hence there are no non-trivial differentials, and the edge homomorphism $$\phi^\ast\colon\thinspace H^n(\Gamma;\mathbb{Z}\Gamma) \to H^n(G;\mathbb{Z}\Gamma)$$ is an isomorphism. $\Box$

Tom's answer shows that either centrality or finite type is necessary in the above statement. I haven't accepted it yet because I'm hoping someone will give an example with $\operatorname{cd} G <\infty$.

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    $\begingroup$ If asked to make up a def for "cd of a homo", I might have looked at vanishing relative cohomology in high dimensions (i.e. "induced map is an iso in high dimensions" rather than "induced map is zero in high dimensions". $\endgroup$ – Tom Goodwillie Feb 22 '12 at 12:39
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    $\begingroup$ @Tom: That would also be an interesting definition, which seems to also be absent from the literature! I have good reasons for asking about this one though, which I won't go into here. $\endgroup$ – Mark Grant Feb 22 '12 at 14:59
  • $\begingroup$ I would just like to remark on what info I immediately can say on this: 1) If $\phi$ is an inclusion then $\phi^*$ (being the restriction map) is nonzero in infinitely many degrees, so $cd(\phi)=\infty$. And 2) If we instead use Tom's suggestion of the definition, then $cd(\phi)=\infty$ for the case where $\phi$ is an inclusion and $G$ controls p-fusion in $\Gamma$ (Mislin's Theorem). $\endgroup$ – Chris Gerig Feb 22 '12 at 20:00
  • $\begingroup$ @Chris: thanks for your comments. I guess you are assuming that $G$ and $\Gamma$ are finite? Or does your 1) work for any inclusion of groups of infinite cohomological dimension? $\endgroup$ – Mark Grant Feb 22 '12 at 20:07
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    $\begingroup$ This is Richard Swan's result from The Nontriviality of the Restriction Map in the Cohomology of Groups, and it holds for $G$ a compact Lie group and closed subgroup $\Gamma$. $\endgroup$ – Chris Gerig Feb 23 '12 at 6:53

How about this? Let $\Gamma$ be free abelian of rank $2$. Poincare duality in the torus identifies $H^2(\Gamma;M)$ with $H_0(\Gamma;M)$, so that in particular there is a natural surjection $M\to H^2(\Gamma;M)$.

Let $F$ be the particular $\Gamma$-module $\mathbb Z\Gamma$, free module of rank one for the group ring. A generator of $H^2(\Gamma;F)=H_0(\Gamma;F)=\mathbb Z$ determines an extension of $\Gamma$ by $F$. Call it $G$.

$\Gamma$ has cohomological dimension $2$. $G$ has cohomological dimension at least $2$, doesn't it? (EDIT: Yes, just because it has a subgroup $F$ with infinite cd.) But for every $\Gamma$-module the map $H^2(\Gamma;M)\to H^2(G;M)$ is zero, because using that natural surjection $M\to H^2(\Gamma;M)$ it's enough to prove this in the case $M=F$, where it is clear.

EDIT: To spell out this last step, there is a surjective map $M\to H_0(\Gamma;M)$ (for every group $\Gamma$), natural in $M$. For this particular group, there is also a natural isomorphism $H_0(\Gamma;M)\to H^2(\Gamma;M)$. Following this by your natural map $H^2(\Gamma;M)\to H^2(G;M)$, we get a map $M\to H^2(G;M)$, natural in $M$, and we just have to show that it takes every element $x\in M$ to $0$. But there is a map $F\to M$ taking a generator to $x$, so by naturality it is enough to see that you get the zero map when $M$ is $F$.

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  • $\begingroup$ That's a nice example. $\endgroup$ – Ralph Feb 23 '12 at 3:39
  • $\begingroup$ Wonderful! I'm not sure I follow your last step though. Is it something like: choose a surjection of modules $F'\to M$ with $F'$ free. Then the long exact coefficient sequence shows that $H^2(\Gamma;F')\to H^2(\Gamma;M)$ is surjective. So its enough to show the map vanishes with $F'$ coeffs. But this doesn't use duality...? (I get that its clear for $M=F$.) $\endgroup$ – Mark Grant Feb 23 '12 at 8:11

One more example refering to the second question: Let $$\Gamma = \left\lbrace \left. \begin{pmatrix} 1 & \ast & \ast \newline & 1 & \ast \newline & & 1 \end{pmatrix} \right\vert\ \ast \in \mathbb{Z} \right\rbrace$$ be the group of integral upper triangular matrices with unit diagonal. $\Gamma$ fits into the non-split central extension $$0 \to \mathbb{Z} \to \Gamma \to \mathbb{Z}^2 \to 0$$ that corresponds to a generator $\epsilon \in H^2(\mathbb{Z}^2;\mathbb{Z}) = \mathbb{Z}$.

Claim 1: $cd(\Gamma) = 3$

Since $\mathbb{Z}$ resp. $\mathbb{Z}^2$ has cd $1$ resp. $2$, the LHS spectral sequence $E_2^{ij} = H^i(\mathbb{Z}^2;H^j(\mathbb{Z};M))$ shows $cd(\Gamma) \le 3$. Moreover, by positional reasons $E_\infty^{2,1}=E_2^{2,1}$ and in particular $E_\infty^{2,1} = \mathbb{Z}$ for $M = \mathbb{Z}$. Hence $cd(\Gamma) = 3$.

Claim 2: The inflation map $\text{inf}: H^2(\mathbb{Z}^2;M) \to H^2(\Gamma;M)$ is zero.

Since the image of inflation is just $E_\infty^{2,0} = \text{coker}(d_2^{0,1})$, it is sufficient to show that $d_2^{0,1}$ is surjective. Since the action of $\Gamma$ on $M$ is induced by the action of $\mathbb{Z}^2$, it follows that $\mathbb{Z}$ acts trivially on $M$. Futhermore, $\mathbb{Z}^2$ acts trivially on $H^\ast(\mathbb{Z};M)$ because $\mathbb{Z}$ is central. Therefore $E_2^{0,1} = Hom(\mathbb{Z},M)$.

Let $\alpha \in Hom(\mathbb{Z},M)$. Then $$d_2^{0,1}: Hom(\mathbb{Z},M) \to H^2(\mathbb{Z}^2;M)$$ is given by $d_2(\alpha)= - \alpha^\ast(\epsilon)$ (well-known formula) where $$\alpha^\ast: H^2(\mathbb{Z}^2;\mathbb{Z}) \to H^2(\mathbb{Z}^2;M)$$ is induced by $\alpha$ on the coefficients. By using a projective resolution or by Poincare duality one easily sees that $$\alpha^\ast : \mathbb{Z} \to M/\lbrace gm-m \mid g \in \mathbb{Z}^2 \rbrace =: \bar{M}$$ is just $\alpha$ composed with the natural projection. Identifying $Hom(\mathbb{Z},M) = M$ now shows that $d_2^{0,1}: M \to \bar{M}$ is the natural projection and hence surjective.

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    $\begingroup$ I don't think so. When you say $Hom(\mathbb Z,M)$, do you mean Hom of abelian groups or Hom of $\Gamma$-modules? The latter is correct. So in fact this $d_2$ is mapping the subgroup of invariants in $M$ to the quotient group of co-invariants, and might not be surjective. $\endgroup$ – Tom Goodwillie Feb 23 '12 at 4:37
  • $\begingroup$ Tom, you are right. Thanks for pointing out. The problem in my argument is in detail: Let $C \to G \to Q$ be a central extension and $M$ a $G$-module. Then the action of $G$ (and hence $Q$) on $H^\ast(C;M)$ is induced by the pair $(c \mapsto gcg^{-1}; m \mapsto gm)$ (see Brown's book before Prop. 8.1). Since $C$ is central, the first map is the identity. But the second usually isn't. Thus my statement that $Q$ acts trivally on $H^\ast(C;M)$ is wrong. I let stand the "answer" for a while so that people who are intersted can see the comment and will delete it in some hours. $\endgroup$ – Ralph Feb 23 '12 at 11:50
  • $\begingroup$ @Ralph: I like this example very much, and hope we can save it. Please don't delete it yet! $\endgroup$ – Mark Grant Feb 23 '12 at 12:51
  • $\begingroup$ OK, I see now that the conclusion is false. For if we take $M=\mathbb{Z}[\mathbb{Z}^2]$ (the free module of rank one) then the corresponding LHSSS has only one nonzero column (since $\mathbb{Z}^2$ is a Poincaré duality group) and so the map $H^2(\mathbb{Z}^2;M)\to H^2(\Gamma;M)$ is an iso in this case. Both your's and Tom's answers have been very helpful though. I will post again if I have any more thoughts on this. $\endgroup$ – Mark Grant Feb 24 '12 at 8:44

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