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



Cohomological dimension of a group, fibration and local coefficients
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.


