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Oct 23, 2012 at 14:12 answer added Mina timeline score: 0
Oct 21, 2012 at 17:08 vote accept Pavel
Oct 21, 2012 at 16:39 answer added Peter Dalakov timeline score: 2
Oct 21, 2012 at 16:31 comment added Pavel yes. This is actually yes!
Oct 21, 2012 at 15:46 comment added Peter Dalakov Or maybe you are asking how the existence of the Kaledin-Feix metric gives you a special lagrangian embedding?
Oct 21, 2012 at 14:30 comment added Peter Dalakov I am not exactly sure what you mean by "explanation", but since you're asking for references, have a look at Birte Feix's thesis "Hyperkaehler metrics on cotangent bundles". There she constructs the HK metric in a different way. See also mathoverflow.net/questions/46752/…
Oct 21, 2012 at 6:03 history edited Pavel CC BY-SA 3.0
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Oct 21, 2012 at 6:01 comment added Pavel ok, I see. lets assume that the given Kaehler manifold is also real analytic. Is it then possible? how can one explain this? what are the ingredients in showing this?
Oct 20, 2012 at 22:19 comment added Robert Bryant Your question suggests that you are looking for an isometric embedding of the given Kähler manifold as a special Lagrangian in a Calabi-Yau manifold, but you don't mention this requirement in the text. I'll just point out that the induced metric on any special Lagrangian submanifold of a Calabi-Yau manifold is necessarily real-analytic, so it follows that it is not possible, in general to isometrically embed a given Kähler manifold as a special Lagrangian in some Calabi-Yau manifold.
Oct 20, 2012 at 12:49 comment added Spiro Karigiannis or by Calabi: archive.numdam.org/ARCHIVE/ASENS/ASENS_1979_4_12_2/…
Oct 20, 2012 at 12:48 comment added Spiro Karigiannis try looking at papers of Stenzel: math.osu.edu/~stenzel.3/research/publications/ricci-flat.pdf
Oct 20, 2012 at 10:24 history asked Pavel CC BY-SA 3.0