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I would like to find a proof for Remark 0.5 in the following article of Claire Voisin:

https://webusers.imj-prg.fr/~claire.voisin/Articlesweb/fanosymp.pdf

She writes in this remark the following:

Remark 0.5 A compact Kähler manifold $X$ which is rationally connected satisfies $H^2(X, {\cal O}_X) = 0$, hence is projective.

I understand that a Kähler manifold with $H^2(X, {\cal O}_X) = 0$ is projective. However, I don't understand why a Kähler manifold that is rationally connected has $H^2(X, {\cal O}_X) = 0$. Indeed, the definition for rational connectedness that Voisin is using is the following:

Definition 0.3 A compact Kähler manifold $X$ is rationally connected if for any two points $x, y\in X$, there exists a (maybe singular) rational curve $C\subset X$ with the property that $x\in C$, $y\in C$.

So my question is the following: How to prove that $H^2(X, {\cal O}_X)$ for a compact Kähler manifold $X$ that satisfies Definition 0.3? Is this easy/hard/well-known?

PS. As Donu Arapura correctly says below the vanishing of $H^2(X, {\cal O}_X)$ for rationally connected projective manifolds is a classical fact. However I want a proof of such a vanishing for Kähler manifolds (to show that they are projective). So I want to know if this vanishing is a well known fact or a couple of pages are needed to prove it?

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    $\begingroup$ By Hodge duality, it's enough to show $H^0(X,\Omega^2)=0$. For this, see p 202 of Kollár's Rational Curves on Algebraic Varieties. (I assume the argument still works in the Kahler case, but haven't checked.) $\endgroup$
    – Donu Arapura
    Commented May 29, 2019 at 15:23
  • $\begingroup$ Dear Donu, thanks. If you look in this book on page 202 Corollary 3.8, you'll see that Kollar proves the vanishing for separably rationally connected manifolds, not for the above definition that Voisin gives. How do you prove that Voisin's definition implies separably rationally connected for Kahler manifolds? $\endgroup$
    – aglearner
    Commented May 29, 2019 at 17:53
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    $\begingroup$ The "usual argument" that rationally connected manifolds (with definition above) have vanishing global sections of every tensor power $\Omega_{X/\mathbb{C}}^{\otimes n}$, $n>0$ generalizes from projective manifolds to closed Kaehler manifolds. For closed Kaehler manifolds, use Douady spaces to parameterize rational curves (rather than Hilbert schemes). $\endgroup$
    – Jason Starr
    Commented May 29, 2019 at 19:00
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    $\begingroup$ @aglearner "... or maybe you think this might be written down somewhere?" It is probably written down every time a mathematician needs to use it. In some sense, this is Lemma 4.5 of one of my own preprints: arxiv.org/pdf/1803.06412v1.pdf That lemma shows that the definition used by Voisin implies the existence of "free" rational curves. Combined with the "rational quotient" (which Campana defined in the Kaehler category), this quickly implies vanishing of all sections of $\Omega_{X/\mathbb{C}}^{\otimes n}$, $n>0$. There may be a more direct reference . . . $\endgroup$
    – Jason Starr
    Commented May 29, 2019 at 20:24
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    $\begingroup$ Thanks for the reference Jason! I believe that there are not so many 6 word sentences that make sense in mathematics, the correct ones (not completely trivial ones) are worth to be proven. For this reason I would like to ask you if you could transform your last comment into a sketch proof of the fact that "Rationally connected Kahler manifolds are projective." $\endgroup$
    – aglearner
    Commented May 29, 2019 at 22:24

1 Answer 1

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This result follows from Corollaire (p.212) of this paper by F. Campana, Coréduction algébrique d'un espace analytique faiblement kählérien compact, Invent. Math. 63 (1981), no. 2, 187–223.

I had the same problem of finiding a citable reference for this result some time ago, and this one did the job.

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  • $\begingroup$ That's great, it is a nice result. Thank you! (in fact I had a feeling that some result of Campana might be relevant :) ) $\endgroup$
    – aglearner
    Commented Jun 6, 2019 at 19:16
  • $\begingroup$ In fact, would you mind to quote Campana's result in your answer, since he proves a statement stronger than what I was asking for - curves don't need to be rational. $\endgroup$
    – aglearner
    Commented Jun 6, 2019 at 19:45

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