second betti number of surface groups

Question

Let $\Gamma$ be a surface group. Denote $b^i(\Gamma):=\operatorname{dim}_{\Bbb Z}H^i(\Gamma, \Bbb Z) $.

Using group cohomology for the trivial action of $\Gamma$ on $\Bbb Z$, we can see that $H^0(\Gamma, \Bbb Z)\cong \Bbb Z $ by taking the invariants of the trivial action.

We can see also that $H^1(\Gamma, \Bbb Z)\cong \operatorname{Hom}_{Grp}(\Gamma,\Bbb Z)\cong \Bbb Z^{\oplus 2g} $ since $\Bbb Z$ is abelian.

How about $b^2(\Gamma)$? It is known that the Euler characteristic of a Riemann surface is $2g-2$. Does it mean that $b^2(\Gamma)=1$? can we deduce it from some kind of "integral duality" theorem that gives $b^0(\Gamma)=b^2(\Gamma)$? Thank you.

Answer 1

The 2nd cohomology group of the fundamental group of a surface $X$ of genus $\ge 1$ is isomorphic to $H^2(X)$, hence ${\mathbb Z}$.