Zariski vs analytic cohomology of $\mathcal O_X^\times$

Question

Let $X/\mathbf C$ be a smooth proper variety. Is it true that $H^1(X, \mathcal O_X^\times) = H^1(X^{an}, \mathcal O_{X^{an}}^\times)?$ GAGA doesn't apply, because $\mathcal O_X^\times$ is not coherent (let alone an $\mathcal O_X$-module)...

Answer 1

It is well known that $H^1 (X, \mathscr{O}_X^\times)$ is (isomorphic to) the algebraic Picard group (= group of algebraic line bundles) and that $H^1 (X^\mathrm{an}, \mathscr{ O}_{X^\mathrm{an}}^\times)$ is (isomorphic to) the analytic Picard group (mutatis mutandis). So it suffices to show that there is an isomorphism between the two Picard groups.

We have the following result of Grothendieck, extending the result of Serre:

Theorem. Let $X$ be a proper $\mathbb{C}$-scheme. Then the functor that sends a coherent sheaf $\mathscr{F}$ on $X$ to the coherent sheaf $\mathscr{F}^\mathrm{an}$ on $X^\mathrm{an}$ is an equivalence of categories.

This is Théorème 4.4 in [SGA 1, Exposé XII]. Since the functor $\mathscr{F} \mapsto \mathscr{F}^\mathrm{an}$ preserves $\otimes$, it follows that we have the required isomorphism of Picard groups.