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.
Hodge number of complex surfaces
algebraic-geometrycomplex-geometryhomology-cohomologysurfaces
Related Solutions
Books
1.You can refer to Complex Manifolds written by James Morrow and Kunihiko Kodaira.
It is an excellent primer including a lot of calculations and details.
2.You can refer to Principal of Algebraic Geometry written by Griffiths and Harris and Complex Analytic and Differential Geometry written by Jean-Pierre Demailly.
They have more differential-geometric points of view.
3.You can refer to Hodge Theory and Complex Algebraic Geometry 1 written by Claire Voisin.
It may be more difficult and advanced and have more algebraic points of view.
Notes and Lectures
1.Notes about complex manifolds which is a wonderful supplement of the Huybrechts' book.
:https://www.math.stonybrook.edu/~cschnell/pdf/notes/complex-manifolds.pdf
2.New lectures written by Hossein Movasati about the Hodge theory.
:http://w3.impa.br/~hossein/myarticles/hodgetheory.pdf
3.If you interested in Kahler Manifolds,you can see the lectures written by Werner Ballmann.
:http://people.mpim-bonn.mpg.de/hwbllmnn/archiv/kaehler0609.pdf
Videos
1.Hans-Joachim Hein's Complex Geometry that consisting of ten classes is nice. See here ：https://m.youtube.com/results?search_query=complex+geometry
2.Complex Analytic and Algebraic Geometry by Qaisar Latif.
See here： https://m.youtube.com/playlist?list=PLWy-AYZriyMGZjZQ7I8kYhZhtlUzFcWcz
It depends on numbers of irreducible components of $D$ as well as the relations of their fundamental classes in $H^2(X)$.
Assume that $X$ is smooth projective, and $D$ is a normal crossing divisor. There is an exact sequence $$H^1(X,X-D,\mathbb Q)\to H^1(X,\mathbb Q)\xrightarrow{f} H^1(X-D,\mathbb Q)\to H^2(X,X-D,\mathbb Q)\xrightarrow{g} H^2(X,\mathbb Q)$$
So to determine if $f$ is isomorphism, we need to compute $H^i(X,X-D,\mathbb Q)$, $i=1,2$ first.
When $D$ is smooth, or it has only one component, Thom isomorphism implies that $H^i(X,X-D)$$\cong H^{i-2}(D)$, which is $\mathbb Q$ when $i=2$ and $0$ when $i=1$. The Gysin homomorphism $g$ sends a generator to the fundamental class of $D$, which is not trivial, so $f$ is an isomorphism.
In general, by Fujiki's paper Duality of Mixed Hodge Structures of Algebraic Varieties, there is a duality between mixed Hodge structures
$$H^{2n-i}(D,\mathbb Q)\times H^i(X,X-D,\mathbb Q)\to \mathbb Q.$$
Set $i=1,2$, we find $h^1(X,X-D,\mathbb Q)=0$ and $h^2(X,X-D,\mathbb Q)=h^2(D,\mathbb Q)=k$, where $k$ is the number of irreducible components of $D$. The exact sequence becomes
$$0\to H^1(X,\mathbb Q)\xrightarrow{f} H^1(X-D,\mathbb Q)\to \mathbb Q^k\xrightarrow{g} H^2(X,\mathbb Q).$$
$g$ sends each component $D_i$ of $D$ to its fundamental class. So the conclusion is
Claim: $H^1(X,\mathbb Q)\to H^1(X-D,\mathbb Q)$ is an isomorphism if and only if the mixed Hodge structure $H^1(X-D,\mathbb Q)$ is pure of weight one if and only if the fundamental classes $[D_i]$ of each component of $D$ are linearly independent in $H^2(X,\mathbb Q)$.
For example, when $X=\mathbb P^2$, $D=L_1\cup L_2$ is union of two lines, the relation $[L_1]=[L_2]$ tells us that the mixed Hodge structure on $H^1(\mathbb P^2\setminus (L_1\cap L_2))$ is concentrated on weight two (and pure).
When $X=Bl_p\mathbb P^2$ is blowup of one point on $\mathbb P^2$ and $D=L\cup E$, where $D$ is the strict transform of a line through $p$, and $E$ is the exceptional divisor, the MHS on $H^1(X\setminus D)$ is pure.
To give an example where $H^1(X\setminus D)$ is not pure, just take any $X=\mathbb P^1\times C$ where $C$ is a curve with genus $g\ge 1$. Take $D$ to be the union of two disjoint fibers of $X\to \mathbb P^1$, then $W_1H^1(X\setminus D)=H^1(X)=\mathbb Q^{2g}$ and $W_2/W_1\cong \mathbb Q$, $H^1(X\setminus D)$ is not pure.
Best Answer
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