The total space of cotangent bundle of any manifold M is a symplectic manifold.
Is it true\false\unknown that for any M, $T^*M$ has Kähler structure?
Please support your claim with reference or counterexample.
Kähler Structure on Cotangent Bundle
complex-geometrydg.differential-geometry
Best Answer
This is true! I assume $M$ compact.
Method 1. Real algebraic geometry. Cf. this article. By a version of the Nash-Tognoli embedding theorem, one can realise $M$ as an real affine algebraic variety $V_\mathbb{R}$, cut out by polynomials $f_i \in \mathbb{R}[x_1,\dots,x_N]$. The complex variety $V_\mathbb{C}$ will then be smooth in a small neighbourhood $U$ of $V_\mathbb{R}$, hence Kaehler in that region, with $V_{\mathbb{R}}$ as a Lagrangian submanifold. But $U$ is diffeomorphic to $T^\ast M$. The resulting symplectic structure on $T^\ast M$ may be non-standard; via the Lagrangian neighbourhood theorem, you can take the symplectic form to be the canonical one if you'll settle for a Kaehler structure only near the zero-section.
Method 2. Eliashberg's existence theorem for Stein structures. See Cieliebak-Eliashberg's unfinished book, Symplectic geometry of Stein manifolds, Theorem 9.5. We observe that $T^\ast M$ has an almost complex structure $J$ (one compatible with the canonical 2-form, for instance) and a bounded-below, proper Morse function $\phi$ whose critical points have at most the middle index (namely, the norm-squared plus a small multiple of a Morse function pulled back from $M$). In this situation Eliashberg, via an amazing chain of deformations, finds an integrable complex structure $I$ homotopic to $J$ such that $dd^c \phi$ is non-degenerate. This makes $T^\ast M$ Stein! His theorem only applies in dimensions $\geq 6$ (this paper of Gompf explains what you have to check in dimension 4), so without doing those checks or appealing to other methods, the case of $M$ a surface is left out.
I think that the more precise version of Eliashberg's theorem, which may not yet be in the book, would tell us that the Stein structure is homotopic to an easy-to-write-down Weinstein structure on $T^\ast M$ involving its canonical symplectic structure $\omega_{can}$, hence that $dd^c\phi$ is symplectomorphic to $\omega_{can}$.