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I am currently studying complex geometry following the book by Huybrechts. I wonder if two surfaces(smooth) in $\mathbb{CP}^n$ with the same hodge number are isomorphic. If yes, I would like some hints for the proof. If not, I would appreciate a counterexample. Recall hodge numbers are the dimensions of the Dolbeault cohomology. Thank you.

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Here are a class of counterexamples.

One can just consider the ruled surfaces $F_n$. These surfaces are not isomorphic for different $n$. However they have the same Hodge diamonds.

One reference is Arapura's notes https://www.math.purdue.edu/~arapura/preprints/partIV.pdf see Corollary 17.3.4 (Durfee).

And https://www.math.purdue.edu/~arapura/preprints/partII.pdf see Example 11.1.2

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    $\begingroup$ Not only is it the case that $F_n$ and $F_m$ are not biholomorphic for $n \neq m$, depending on $n$ and $m$, they may not even be homotopy equivalent: $F_{2k}$ is diffeomorphic to $S^2\times S^2$ while $F_{2k+1}$ is diffeomorphic to $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$. $\endgroup$
    – Michael Albanese
    Jun 9 at 22:26

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