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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?

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    $\begingroup$ 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... $\endgroup$ – Eric Wofsey Feb 20 at 23:07
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    $\begingroup$ 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 $\mathbb{C}$. $\endgroup$ – Roland Feb 20 at 23:15
  • $\begingroup$ @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. $\endgroup$ – Katherine Feb 20 at 23:55

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