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Given a Kähler manifold $(X,\omega)$ we know that its Kähler class lies in an open cone of $H^{1,1}(X) \cap H^2 (X,\mathbb{R})$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ we should be able to find a Kähler class belonging to $H^{1,1}(X) \cap H^2 (X,\mathbb{Q})$. Such a class would be represented by a closed positive form of type (1,1), therefore X would be projective thanks to Kodaira theorem.

I know this reasoning has to be wrong since not every Kähler manifold is projective. Can you tell me where?

Thanks.

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    $\begingroup$ Take a line in $\mathbf R^2$ with irrational slope. Can you find a nonzero point on it with rational coordinates? $\endgroup$
    – user5117
    Feb 27, 2015 at 15:52
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    $\begingroup$ @Artie Prendergast-Smith. I think you should write a (short) answer instead of a comment, because this comment is indeed the answer! The OP just want to know where he is wrong in his reasoning, and not really examples of Kähler non projective compact complex manifolds (it seems to me that he is aware of this)! $\endgroup$
    – diverietti
    Feb 27, 2015 at 17:09

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Since Artie Prendergast-Smith is not expanding his comment in an answer, let me do it. As I said in the comments, his comment is essentially THE answer to the OP question. But let me give some more details.

So, let $V_\mathbb Q$ be a finite dimensional $\mathbb Q$-vector space, and let $V_\mathbb R:=V_\mathbb Q\otimes_\mathbb Q\mathbb R$ and $V_\mathbb C:=V_\mathbb Q\otimes_\mathbb Q\mathbb C$ (at the end of the day, thanks to the universal coefficient theorem, you have to think at $V_\mathbb K$, for $\mathbb K=\mathbb Q, \mathbb R, \mathbb C$, as $H^2(X,\mathbb K)$).

Then, on $V_\mathbb C=V_\mathbb R\oplus i V_\mathbb R$, you have a natural conjugation and $V_\mathbb R=\{v\in V_\mathbb C\mid v=\bar v\}$.

Now, suppose you have a decomposition (a so-called pure Hodge structure of weight $2$) on $V_\mathbb C$: $$ V_\mathbb C=V^{2,0}\oplus V^{1,1}\oplus V^{0,2}, $$ where $V^{p,q}=\overline{V^{q,p}}$. Finally, call $$ V^{1,1}_\mathbb R=V_\mathbb R\cap V^{1,1}=\{v\in V^{1,1}\mid v=\bar v\}. $$ Observe, first of all, that, as in the last sentence of semyon alesker's answer, if $V^{2,0}=\{0\}$, then $V_\mathbb C=V^{1,1}$, and thus $$ V^{1,1}_\mathbb R=V_\mathbb R. $$ In this case, your reasoning works, since $V_\mathbb Q$ is dense in $V_\mathbb R=V^{1,1}_\mathbb R$ and therefore any open set in $V^{1,1}_\mathbb R$ (in particular any open cone...) contains some ''rational'' point.

In the general case, everything depends on the relative positions of $V_\mathbb R$ and $V^{2,0}$, $V^{0,2}$, and $V^{1,1}$. Now, the hypothesis of $X$ being Kähler can be translated in $V^{1,1}_\mathbb R\ne\{0\}$, so that in fact $V_\mathbb R$ always intersects in a non trivial way $V^{1,1}$, but it principle it may very well happen (and it happens, indeed, cf. the first part of the answer of semyon alesker), that $V_\mathbb R\not\subset V^{1,1}$ so that $\dim V^{1,1}_\mathbb R<\dim V_\mathbb R$.

In this case an open set in $V^{1,1}_\mathbb R$, is just an open set (for the induced topology) in a (real) vector subspace of $V_\mathbb R$, and not in the whole $V_\mathbb R$! Now, of course a dense subset of $V_\mathbb R$ (like $V_\mathbb Q$ is, but now I am speaking in general) needs not to be dense in all subsets of $V_\mathbb R$; even worst it may very well have empty intersection with another subset of $V_\mathbb R$. In the case of $V_\mathbb Q$, it is dense and always contains the zero vector, so the worst case is that it intersects $V^{1,1}_\mathbb R$ only in zero (and this is the case for a generic torus!).

And here comes the comment of Artie Prendergast-Smith. Just think at $V_\mathbb Q=\mathbb Q^3$ endowed with the following decomposition: $$ V_\mathbb C=V^{2,0}\oplus V^{1,1}\oplus V^{0,2}, $$ where $V^{2,0}=\operatorname{Span}_\mathbb C(e_1+ie_3)$, $V^{1,1}=\operatorname{Span}_\mathbb C(e_1+\sqrt 2 e_2)$ and $V^{0,2}=\operatorname{Span}_\mathbb C(e_1-ie_3)$ (here $\{e_1,e_2,e_3\}$ is the canonical basis of $V_\mathbb K$). Then, $V^{p,q}=\overline{V^{q,p}}$ and $V^{1,1}_\mathbb R=\operatorname{Span}_\mathbb R(e_1+\sqrt 2 e_2)$. Then, no non-zero vector in $V^{1,1}_\mathbb R$ lies in $\mathbb Q^3$.

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  • $\begingroup$ "may very well have empty intersection" is not correct since the $0$ vector is always in this intersection $\endgroup$
    – YangMills
    Mar 2, 2015 at 4:12
  • $\begingroup$ Dear YangMills, I think this is a misunderstanding: I was speaking about a general dense subset, saying in parenthesis to keep in mind the case of $V_\mathbb Q$. But I admit that the way I wrote it can be misleading. I try to fix it. $\endgroup$
    – diverietti
    Mar 2, 2015 at 7:40
  • $\begingroup$ It is funny that this example be essentially the solution, but never appears for Hodge structures (what would $V^{0,1}$ be?). $\endgroup$
    – ACL
    Mar 2, 2015 at 8:45
  • $\begingroup$ @ACL. Sorry but I am not sure I understand well your comment... The solution to what? And what do you mean about the $V^{01}$? Cheers, Simone. $\endgroup$
    – diverietti
    Mar 2, 2015 at 9:50
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    $\begingroup$ @ArtiePrendergast-Smith: it was my pleasure! I didn't want to "steal" your answer, but at some point I saw that you weren't answering and so... Anyway, maybe I shall modify my $\mathbb Q^2$ in $\mathbb Q^3$, to avoid the problem pointed out by ACL. $\endgroup$
    – diverietti
    Mar 3, 2015 at 9:25
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It might happen that $H^{1,1}(X)\cap H^2(X,\mathbb{Q})=0$ while $H^{1,1}(X)\cap H^2(X,\mathbb{R})\ne 0$. This is the case e.g. when $X=\mathbb{C}^n/\Lambda$, where $\Lambda$ is a generic lattice.

Nevertheless your argument can be easily applied to prove that a compact Kahler manifold is algebraic provided $H^{2,0}(X)=0$.

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