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Hello, I want to show that the cohomological dimension (say over Z or R) of some group $K$ is 1. $K$ occurs in an exact sequence $1 \to K \to \pi_1(X) \to \pi_1(C) \to 1$, where $\pi_1(X)$ has cohomological dimension 3 (in the same coefficients) and $C$ is a curve of genus greater than 2.

So I want a kind of additivity but this is not true in general. If I look at the associated fibration $BK \to B\pi_1(X) \to C$ and use Leray-Serre spectral sequence, I have some information on the cohomology of $BK$ and in fact can solve the problem if I assume that the action of the fundamental group of $B$ on the cohomology of the fiber is trivial. But I'm not familiar with cohomology with local coefficients and don't manage to show the general case.

Someone can help me ? (or solve this problem more directly ?) (or this is false in general ?)

mister_jones

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An obvious restatement of your question would be: does every epimorphism $G\to S$ from a group of cohomological dimension 3 to a (non-abelian) surface group have free kernel? –  Mark Grant Jan 26 '11 at 16:21
    
Yes and in fact this is my original problem, where G is a Kähler group and the epimorphism is induced by the Albanese map (G has 1-dimensional Albanese image). –  mister_jones Jan 26 '11 at 16:35
    
Is arxiv.org/abs/0709.4350 relevant? –  Mark Grant Jan 26 '11 at 16:56
    
I don't think so because in a way I try to prove something stronger. In fact we can adapt the proof in this article to show that if the cohomology of G satisfies 3-dimensional Poincaré duality, then we have a contradiction. What I want to prove is that there is no Kähler group of cohomological dimension one, without assumptions of Poincaré duality. –  mister_jones Jan 26 '11 at 18:02
    
I won't claim it's false, but it not obvious that it should be true Mr J. If the (outer) action of $\pi_1(C)$ on $K$ is sufficiently complicated, then it's conceivable that $H^j(K,M)\not= 0$ for $j>1$ but that $H^i(\pi_1(C), H^j(K,M))=0$ (so that it dies in Hochsild-Serre). –  Donu Arapura Jan 26 '11 at 19:05

1 Answer 1

up vote 6 down vote accepted

It is false. The spectral sequence shows that cohomological dimension of group extensions is subadditive. It is not additive in general as every group is resolved by free groups, eg, $F_\infty\to F_3\to \mathbb Z^3$.

For your hypotheses, let $G=A*B$ be the free product of a three dimensional group $A$, say, $\mathbb Z^3$, and a surface group $B$. The dimension of the free product is the maximum of the dimensions of the factors, so 3. There is a natural map $G\to B$ that is the identity on $B$ and trivial on $A$. The kernel is 3-dimensional because it contains $A$, which is 3-dimensional.

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In the OP's case of interest, the group has one end. Are there counterexamples there? –  Richard Kent Jan 26 '11 at 20:18
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A one-ended example: $\mathbb Z\times(B*B)\to B$. –  Ben Wieland Jan 26 '11 at 20:56
    
Oh, right. Thanks. –  Richard Kent Jan 26 '11 at 20:58

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