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This is related to the following statement. Let $S\subset P^3$ be degree $h$ smooth hypersurface. Then use degree $h$ embedding of $i:P^3\to P^N$ with $N$ appropriately choosen according to $h$. One obtains $S\to P^N$ embedding and this makes $S$ being a submanifold of $P^N$ cut out by $i(P^3)\cap H$ where $H$ is a hyperplane in $P^N$. Now it follows from Leftschetz theorem that $H^1(P^3,C)\cong H^1(S,C)$ and $H^1(P^3,C)=0\implies H^1(S,C)=0\implies H^0(S,\Omega^1)=0$.(The last step follows from Kahlerian condition and application of Hodge decomposition Theorem.)

The book claims that if $S$ is a 2-dimensional compact complex manifold(non-Kahler) and $H^1(S,C)=0$, then $H^0(S,\Omega^1)=0$.

$\textbf{Q:}$ The example of non-Kahler compact complex surface given in the book is Hopf surface like $S^1\times S^{2n-1}$.(In this case, I could choose $n=2$ which yields a complex 2 dimensional manifold.) However, from Kunneth formula, I see $H^1(S^1\times S^3)=H^1(S^1)\otimes H^0(S^3)\oplus H^0(S^1)\otimes H^1(S^3)=Z$ with $Z$ coefficients.) So I do not know whether there is such an example. What is the example of 2 complex dimensional compact non-Kahler manifold with $H^1(S,C)=0$?

Ref. Kodaira, Complex Manifolds and Deformation of Complex Structures Chpt 3, Sec 6 (d) Surfaces in $P^3$.

user45765
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  • See https://mathoverflow.net/questions/57535/examples-of-non-kahler-surfaces-with-explicit-non-kahler-metric – KReiser Jul 30 '19 at 22:10
  • @KReiser My bad, I knew that example already and I forgot to put down the further constraint $H^1(S,C)=0$ which rules out that example. – user45765 Jul 30 '19 at 23:19
  • @KReiser I found this post https://math.stackexchange.com/q/1668216. I guess the book did not say $b_1$ being even is sufficient to determine kahler. And in this case, it is already kahler. Hence, the result follows as well. Is this correct? – user45765 Jul 30 '19 at 23:43
  • By my reading, it would seem so. I think it might be best to answer your own question and cite the linked post you found. – KReiser Jul 30 '19 at 23:49
  • @KReiser Ah, Thanks a lot. – user45765 Jul 30 '19 at 23:50

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According to the post Compact complex surfaces with $h^{1,0} < h^{0,1}$, for complex compact surfaces, $X$ is Kahler iff $b_1(X)$ is even.

In the setting of the book, $b_1(S)=0$. Since it is already compact complex surface, one deduces that $S$ is Kahler. Hence, there is no such example of non-Kahler compact complex surfaces with $b_1=0$.

user45765
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