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Jun 22, 2018 at 11:32 history edited Francesco Polizzi CC BY-SA 4.0
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Jun 22, 2018 at 11:27 comment added Francesco Polizzi Deear Georges, you are right and right :-) Thank you very much for pointing out this, I will correct the answer.
Jun 22, 2018 at 9:29 comment added Georges Elencwajg In their book Barth, Hulek, Peters and Van de Ven claim that indeed the image of $H^1(M,\mathbb Z)$ in $H^1(M,\mathcal O)$ is not a full lattice in the general case of a non-Kähler manifold. Here is the relevant page: books.google.fr/…
Jun 22, 2018 at 9:13 comment added Georges Elencwajg Dear Francesco, here are the questions I ask myself: 1) We have $\dim H^1(M,\mathcal O)=h^{0,1.}$ (by Dolbeault) but you write $h^{1,0}$ for that number. The equality $h^{0,1.}=h^{1,0}$ requires a Kähler structure, I think. 2) Is the fact that $H^1(M, \, \mathbb{Z})$ is a full lattice in $H^1(M, , \mathcal{O}_M)$ not dependant on $M$ being Kähler?
Jun 22, 2018 at 8:26 comment added Francesco Polizzi I added the assumption compact in the answer.
Jun 22, 2018 at 8:25 history edited Francesco Polizzi CC BY-SA 4.0
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Jun 21, 2018 at 20:49 comment added Francesco Polizzi Dear Georges, I think that compact, complex manifold is enough. If not, where am I using the Kähler assumption?
Jun 21, 2018 at 20:43 comment added Georges Elencwajg Dear Francesco, aren't you using some hypothesis on $M$ in your answer like, say, compact kähler ?
Jun 21, 2018 at 9:54 history edited Francesco Polizzi CC BY-SA 4.0
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Jun 21, 2018 at 9:36 history edited Francesco Polizzi CC BY-SA 4.0
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Jun 21, 2018 at 9:30 history edited Francesco Polizzi CC BY-SA 4.0
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Jun 21, 2018 at 9:24 history answered Francesco Polizzi CC BY-SA 4.0