# Cohomology of Complements by an analytic subset?

Good moring,

Let $\Omega$ be a domain in $\mathbb{C}^n$ and $S\subset\Omega$ an analytic subset of codimension 1. What can we say about the cohomology group $H^1(\Omega\backslash S, \mathbb{Z})$? E.g, when $\Omega$ is a ball?

Any help is appreciated. Thanks in advance.

Duc Anh

I am not sure about cohomology with coefficients in $\mathbb Z$, but if we replace $\mathbb Z$ by $\mathbb Q$ and if $\Omega$ is a ball, then Poincar\'e duality implies that $H^1(\Omega\setminus S,\mathbb Q)$ is dual to $H^{2n-1}_c(\Omega\setminus S,\mathbb Q)$ (cohomology with compact support). Now the exact sequence $$H^{2n-2}_c(\Omega)\to H^{2n-2}_c(S)\to H^{2n-1}_c(\Omega\setminus S)\to H^{2n-1}_c(\Omega)$$ shows that $H^{2n-1}_c(\Omega\setminus S)\cong H^{2n-2}_c(S)$; the latter is isomorphic to $H^{2n-2}_c(S_{\mathrm{smooth}})$, that is, to $\mathbb Q^N$, where $N$ is the number of components of $S$.
When $\Omega$ is a ball $H^1(\Omega\setminus S)\cong H_{2n-2}(S^c)$ by Alexander duality. Here $S^c$ is the one-point compactification of $S$.
• thank you very much. So what can we say about the homology of $S$? Do you know any reference? – Đức Anh Jan 22 '13 at 6:40
• @Gregory: are you sure? Suppose that $n=2$, $\Omega=\{(z,w)\colon |z|^2+|w|^2<1\}$, $S=\{(z,w)\in\Omega\colon w=0\}$. Then $H^1(\Omega\setminus S,\mathbb Z)=\mathbb Z$ but $H_2(S)=0$. – Serge Lvovski Jan 22 '13 at 7:16