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Jun 6, 2019 at 19:18 vote accept aglearner
Jun 6, 2019 at 18:29 answer added YangMills timeline score: 4
May 31, 2019 at 11:06 comment added Jason Starr @aglearner If you want more details, you are welcome to e-mail me.
May 30, 2019 at 2:42 history edited Michael Albanese CC BY-SA 4.0
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May 30, 2019 at 0:57 history edited Michael Albanese CC BY-SA 4.0
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May 29, 2019 at 22:46 history edited aglearner CC BY-SA 4.0
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May 29, 2019 at 22:24 comment added aglearner 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."
May 29, 2019 at 20:24 comment added Jason Starr @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 . . .
May 29, 2019 at 19:23 comment added aglearner Dear Jason, thanks for these comments! Is this some kind of folklore, or maybe you think this might be written down somewhere?
May 29, 2019 at 19:00 comment added Jason Starr 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).
May 29, 2019 at 17:57 history edited aglearner CC BY-SA 4.0
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May 29, 2019 at 17:53 comment added aglearner 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?
S May 29, 2019 at 16:08 history suggested Kevin Casto CC BY-SA 4.0
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May 29, 2019 at 15:48 review Suggested edits
S May 29, 2019 at 16:08
May 29, 2019 at 15:23 comment added Donu Arapura 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.)
May 29, 2019 at 14:56 history asked aglearner CC BY-SA 4.0