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How we can prove the short time existence for Kahler Ricci flow on compact manifolds. I know that by linearization and Detork's trick we can prove the short time existence for Ricci flow, but for proving on Kahler Ricci flow we have to use of second order parabolic equations. Of course on noncompact manifolds I have seen a nice proof but In this question I mean on compact manifolds.

edited Jul 12 at 0:06

asked Jun 13 at 13:48

Hassan Jolany
1,046●1●2●12

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arxiv.org/pdf/1205.1039.pdf See chapter 8. In particular, you don't need any tricks, because KRF is a parabolic pde for the potential! – Otis Chodosh Jun 13 at 14:20

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Detork's trick? – Igor Rivin Jun 13 at 14:35

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Also, comments like "I am just a master student, when I finished my PhD thesis after I will think for being Mathematician, So if you think you are mathematician solve my question!" are, in my view, extremely rude, and do not do your cause any favours – Yemon Choi Jun 18 at 7:33

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My email to the address you provided bounced back, so I will have to say this in public: Please be nicer. Further infractions will lead to longer suspensions. – S. Carnahan♦ Jun 18 at 7:58

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The short time existence follows for example from the book of G.Lieberman "Second Order Parabolic Differential Equations", Lemma 14.8 or, if you prefer, also Theorem 14.10. There the equation is considered on a domain in Euclidean space, but similar arguments apply to the case of closed manifolds. Alternatively, you can easily modify the proof of Theorem 5.1 in Hamilton's original 1982 paper (p.263), by using Holder spaces and the usual IFT for Banach spaces instead of Frechet spaces and the Nash-Moser IFT (because the scalar equation of the KRF is STRICTLY parabolic). – YangMills Jun 23 at 19:16

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Deane Yang · Accepted Answer · 2012-06-13 16:19:09Z UTC

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First, the Kahler-Ricci flow is the Ricci flow applied to a Kahler metric (which is a Riemannian metric), so any theorem and proof for local time existence of the Ricci flow applies to the Kahler-Ricci flow.

Second, as far as I know, any proof for short time existence that works for noncompact manifolds will work without any change for a compact manifold (without boundary).

Third, as Otis has pointed out, the Kahler-Ricci flow can be written as an honest nonlinear parabolic PDE for the potential. If you believe that a linear parabolic PDE can be solved, then the nonlinear theorem is easily proved using energy integral estimates and the implicit function theorem.

answered Jun 13 at 16:19

Deane Yang
12.6k●1●33●82

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You have it backwards. The Kahler-Ricci flow is a special case of the Ricci flow. So any theorem about the Ricci flow applies to the Kahler-Ricci flow. As for the second point, you have to work through the proof step by step and verify that it can be adapted to the compact case. As for the third point, I unfortunately don't have time to write out the proof and I also don't know the right reference. I'm sorry about this. – Deane Yang Jun 13 at 20:19

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I don't understand your objection to the first point. The Kahler-Ricci flow is the exact same equation as the Ricci flow. The only difference is that the initial data is a Riemannian metric that happens to be a Kahler metric. You can check that the Ricci flow preserves the Kahler structure (including the complex structure). So the resulting Ricci flow of Kahler metrics is known as the Kahler-Ricci flow. – Deane Yang Jun 14 at 12:18

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Haskell, your last comments are completely irrelevant to the question about Kahler-Ricci flow. My answer and comments already provide a complete and correct explanation of the situation. You just choose not to believe me, which I find baffling. You should try to check the details and either verify that I'm correct or provide a precise reason why I'm not. Citing a theorem about isometric embeddings certainly proves absolutely nothing. – Deane Yang Jun 16 at 14:36

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Dear Haskell, The real part of a Kaehler metric is a Riemannian metric. The Kaehler-Ricci flow of a Kaehler metric is precisely the Ricci flow of that associated Riemannian metric. Moreover it is well known that if a Riemannian metric is the real part of a Kaehler metric, then it remains so under Ricci flow. So Deane's explanation is correct: the short time existence for the Ricci flow implies the short time existence for the Kaehler-Ricci flow. – Ben McKay Jun 17 at 18:10

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Haskell, you're being quite dismissive and disrespectful of people (like YangMills) who know this stuff quite a bit better than you do. At this point, we're all repeating ourselves, because you continue to disbelieve what we say. I will edit my answer to summarize the situation. – Deane Yang Jun 18 at 13:21

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Short time existence on normalized Kahler Ricci flow

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