# Can you make the cotangent bundle to a complex manifold?

The cotangent bundle of a manifold has a canonical symplectic form and if we choose a riemannian metric on $M$, we can give it an almost complex structure.

Is this structure integrable, and if it isn't in general, what are the conditions on the manifold for it to be integrable? Can we give $T^*M$ a complex structure in some other way?

Furthermore, what are the conditions on $M$ to ensure that the symplectic form is a Kaehler form?

I read this thread: Kähler structure on cotangent bundle?, but I honestly didn't understand, how far it answers these questions.

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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"). – Eric Zaslow Jan 5 '12 at 15:06
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. – Matthias Ludewig Jan 5 '12 at 21:20
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? – Eric Zaslow Jan 6 '12 at 5:14