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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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  • $\begingroup$ 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? $\endgroup$
    – Mark Grant
    Commented Jan 26, 2011 at 16:21
  • $\begingroup$ 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). $\endgroup$
    – mister_jones
    Commented Jan 26, 2011 at 16:35
  • $\begingroup$ Is arxiv.org/abs/0709.4350 relevant? $\endgroup$
    – Mark Grant
    Commented Jan 26, 2011 at 16:56
  • $\begingroup$ 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. $\endgroup$
    – mister_jones
    Commented Jan 26, 2011 at 18:02
  • $\begingroup$ 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). $\endgroup$
    – Donu Arapura
    Commented Jan 26, 2011 at 19:05

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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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  • $\begingroup$ In the OP's case of interest, the group has one end. Are there counterexamples there? $\endgroup$
    – Autumn Kent
    Commented Jan 26, 2011 at 20:18
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    $\begingroup$ A one-ended example: $\mathbb Z\times(B*B)\to B$. $\endgroup$
    – Ben Wieland
    Commented Jan 26, 2011 at 20:56

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