How to caculate $H^2(X,\mathcal{O^*})$

Question

Let $X$ be a complex torus of dimension 2. I wonder that if there is a torsion free normal subgroup of $H^2(X,\mathcal{O^*})$. I intended to use Selberg's Lemma to deduce, but it requires that $H^2(X,\mathcal{O^*})$ is a linear group. I have no idea how to describe $H^2(X,\mathcal{O^*})$. Is there some methods to calculate it or just to show it is virtually torsion free.

Answer 1

We have an exact sequence $H^1(X,\mathcal{O}^{\times}) \rightarrow H^2(X,\mathbb{Z}) \rightarrow H^2(X,\mathcal{O}) \rightarrow H^2(X,\mathcal{O}^{\times}) \rightarrow H^3(X,\mathbb{Z}) \rightarrow H^3(X,\mathcal{O})$.

Now, $H^2(X,\mathbb{Z}) \cong \mathbb{Z}^6$ by properties of tori, $H^2(X,\mathcal{O}) \cong H^0(X,\Omega^2)^*=\mathbb{C}$ by Serre duality, $H^3(X,\mathbb{Z})=\mathbb{Z}^4$ by properties of tori, $H^3(X,\mathcal{O})=0$ because $X$ is smooth of dimension $2$ and $\mathcal{O}$ is coherent.

Thus we have an exact sequence $\mathrm{Pic}(X) \rightarrow \mathbb{Z}^6 \rightarrow \mathbb{C}\rightarrow H^2(X,\mathcal{O}^{\times}) \rightarrow \mathbb{Z}^4 \rightarrow 0$.

In particular, we have an exact sequence $0 \rightarrow Q \rightarrow H^2(X,\mathcal{O}^{\times}) \rightarrow \mathbb{Z}^4 \rightarrow 0$, which then splits, so that $H^2(X,\mathcal{O}^{\times})$ is the product of $\mathbb{Z}^4$ and a divisible group (which is the quotient of $\mathbb{C}$ by a free abelian subgroup of rank between $6-r$ and $6$, where $r$ is the Néron-Severi rank of $X$). In particular, it has a torsion-free subgroup.