I want to know where the mistake is in the following argument.
Let $X$ be an affine elliptic curve over $\mathbb{C}$ (for example, let $X=\operatorname{Spec}\mathbb{C}[x,y]/(y^2-x^3-x)$).
The exponential sheaf sequence induces the following exact sequense:
$$
H^1(X,\mathcal{O}_X)\to \operatorname{Pic}X\to H^2(X,\mathbb{Z})\to H^2(X,\mathcal{O}_X).
$$
Since $X$ if affine, $H^1(X,\mathcal{O}_X)=H^2(X,\mathcal{O}_X)=0$ due to the Serre's theorem.
Thus we have $\operatorname{Pic}X\cong H^2(X,\mathbb{Z})$.
However, I don't think this is true because:
- $\operatorname{Pic}X$ is isomorphic to the projective elliptic curve $\bar{X}$ (the compactification of $X$) as a group. So, $\operatorname{Pic}X\ne 0$.
- $X$ is homeomorphic to a punctured torus and it is homotopy-equivalent to the wedge sum $S^1 \vee S^1$. So, $H^2(X,\mathbb{Z})=H^2(S^1 \vee S^1,\mathbb{Z})=H^2(S^1, \mathbb{Z})\oplus H^2(S^1, \mathbb{Z})=0$.
What is my misunderstanding?
Thank you in advance.
Best Answer
There is nothing wrong with everything that you have written. Your source of confusion is between the analytic Picard group, $ H^1(X, \mathcal{O}_X^*) $ which is indeed zero as you compute in the second bullet point, versus the algebraic Picard group which you compute as $ \bar{X} $ in the first bullet point and I am guessing you do this from the (injective, in this case) short exact sequence $$ 0 \rightarrow \mathbb{Z} \rightarrow Cl(\bar{X}) \rightarrow Cl(X) \rightarrow 0 $$where the first map sends an integer $ n $ to the class of $ n $ times the punctured point.
There is obviously a map from the algebraic Picard group to the analytic Picard group (If a line bundle is algebraically trivializable, it is obviously holomorphically trivializable). Your example just shows that the map is zero in this case. In general the map is neither injective, nor surjective.