Timeline for Can you make the cotangent bundle to a complex manifold?
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 6, 2012 at 5:14 | comment | added | Eric Zaslow | The Nijenhuis tensor lives upstairs and doesn't have so many symmetries, while the curvature tensor lives downstairs (with half the number of vectors) with more symmetries -- so the relation is not direct, and involves the splitting of T(TM) into TM+TM. Maybe the exact relation is in Kobayashi-Nomizu? | |
| Jan 5, 2012 at 21:20 | comment | added | Matthias Ludewig | What do you mean with "essentially the curvature"? Since the Nijnhuis Tensor N is a (1,2) tensor, whereas the Riemann-Tensor R is a (1,3) tensor, I would think that N has to contain less information than R. Also, if you calculate N in canonical coordinates with respect to the "canonical J", you get an expression in $g_{ij}$ and $g^{ij}$ and derivative of both, whereas R consists of the first and second derivatives of the metric. | |
| Jan 5, 2012 at 15:06 | comment | added | Eric Zaslow | The metric-induced almost complex structure is not integrable unless M is flat. (The Nijenhuis tensor is essentially the curvature.) Now interpreting Tim Perutz's excellent response to the question you cite: if M is compact, then there is an integrable complex structure on TM by Eliashberg's result. It is not Kahler, in general. A Kahler metric exists on TM = TM in a neighborhood of the zero section ("Grauert tube"). | |
| Jan 5, 2012 at 13:11 | history | edited | Matthias Ludewig | CC BY-SA 3.0 |
edited title
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| Jan 5, 2012 at 12:33 | history | asked | Matthias Ludewig | CC BY-SA 3.0 |