# $\mathcal{O}_X\cong \mathcal{O}_{X_{h}}$?

If $$X$$ is $$\mathbb{C}P^n$$ as a projective variety, and $$X_h$$ is the corresponding analytic structure. Then do we have an isomorphism $$\mathcal{O}_X\cong \mathcal{O}_{X_{h}}$$ for the structure sheaves?

More generally, do we have $$\mathcal{O}_X\cong \mathcal{O}_{X_{h}}$$ if $$X$$ is a nonsingular projective variety?

• What do you mean by $\mathcal{O}_X$? If you mean its algebraic structure sheaf, that's a sheaf on the Zariski topology, whereas $\mathcal{O}_{X_h}$ is a sheaf on the Euclidean topology... – Eric Wofsey Feb 20 at 23:07
• Even on a Zariski open subset these two sheaves does not coincide. For example $\mathbb{C}\subset\mathbb{P}^1$ is an open subset. We have $\mathcal{O}_X(\mathbb{C})=\mathbb{C}[t]$ whereas $\mathcal{O}_{X_h}(\mathbb{C})$ is the set of all holomorphic functions on $\mathbb{C}$. – Roland Feb 20 at 23:15
• @EricWofsey : Let $f : X_h\to X$ be the identity map, which is continuous because analytic topology is finer than Zariski topology, and we have $f^*\mathcal{O}_U = \mathcal{O}_{U_h}$. I am thinking if $f^*$ can get an isomorphism. – Katherine Feb 20 at 23:55