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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.

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Do you mean a Kahler structure where the Kahler form is the canonical symplectic form? Or just any Kahler form at all? (Not that I know the answer in either case...) –  Jason DeVito Jun 2 '10 at 2:50

7 Answers 7

up vote 29 down vote accepted

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 $\omega_{can}$.

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does not your first argument need M to be an analytic manifold? I found the papers mentioned by Zehmisch and the one in the comment below that interesting as well. –  Mohammad F. Tehrani Mar 7 '13 at 19:04
Right, but Whitney proved that every smooth manifold is diffeomorphic to a real analytic manifold. Kai's answer is indeed a good one (I didn't know of the work he cited). –  Tim Perutz Mar 7 '13 at 22:27

MR1131444 (93e:32018) Guillemin, Victor(1-MIT); Stenzel, Matthew(1-MIT) Grauert tubes and the homogeneous Monge-Ampère equation. J. Differential Geom. 34 (1991), no. 2, 561–570. 32F07 (32E10)

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Also this one: Complex structures on tangent bundles of Riemannian manifolds, Szoke –  Mohammad F. Tehrani Mar 7 '13 at 19:02

In a paper by Goldman, Kapovich, and Leeb, it is pointed out that a fuchsian (surface) group embedded into the isometries of complex hyperbolic space has quotient the tangent bundle to the surface. Since the tangent and cotangent bundles are diffeomorphic (e.g., they may be identified using a Riemannian metric), the cotangent bundle will admit a Kahler structure. However, I'm not sure if this is compatible with the natural symplectic form on the cotangent bundle (which I'm guessing is implicitly required in your question).

There are restrictions on the fundamental groups of closed Kahler manifolds, but I do not know of restrictions on open Kahler manifolds (however I am far from being an expert).

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I think it is false, in general. I have heard in a talk that $T*M$ of Riemannian manifolds with non-constant curvature are "standard" examples of strictly almost Kahler manifolds. Quick google search gives me arxiv.org/pdf/math/0308227 whose theorem 3 seems to give an answer.

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this paper discusses one particular choice of almost complex structure. and my calculations shows that he has not considered a good choice. –  Mohammad F. Tehrani Jun 2 '10 at 13:11

The tangent bundle $TM$ of a Riemannian manifold M has a natural Kähler structure with the Kähler form agreeing with the canonical symplectic form of $TM$ coming from the cotangent bundle.

To see this, pick local coordinates $\mathbf{x}=(x_1,\ldots,x_n)$ on M and let the metric be given by a positive definite matrix A $$g = d\mathbf{x}^TAd\mathbf{x}$$ Introduce complex coordinates $\mathbf{z}=\mathbf{x}+i\ d{\mathbf{x}}$ and lift the the metric to a Hermitian metric $h$ on $TM$ $$h = d\mathbf{z}^*Ad\mathbf{z} = (d\mathbf{x}-i\ d^2\mathbf{x})^T\ A\ (d\mathbf{x}+i\ d^2\mathbf{x}) $$ (Here $d^2\mathbf{x}=(d^2x_1,\ldots,d^2x_n)$ are coordinates on the second order tangent space.)

The Kähler form is $$ \Omega = -\text{Im}\ h(d\mathbf{z}_1, d\mathbf{z}_2) = d^2\mathbf{x}_1^T\ A\ d\mathbf{x}_2 - d\mathbf{x}_1^T\ A\ d^2\mathbf{x}_2 $$

and since the momentums (cotangent coordinates) are $\mathbf{p}=d\mathbf{x}^TA$, the Kähler form becomes $$ \Omega = d\mathbf{p}_1\ d\mathbf{x}_2 - d\mathbf{p}_2\ d\mathbf{x}_1$$ which is the canonical symplectic form of $T^*M$.

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Your calculation is local, and it seems to me that the transition functions of $TM$ are not holomorphic. If $f:\mathbb{R}\to \mathbb{R}$ is a diffeomorphism, does $F(x+iy):=f(x)+if'(x)y$ define a holomorphic function? –  Tim Perutz Jun 3 '10 at 14:09
As far as I computed, this is true iff curvature is totally zero which means manifold is affine. So as Tim said your calculation is local and curvature enters into the story when you want to construct some thing global. –  Mohammad F. Tehrani Jun 5 '10 at 1:03

In the refernce mentioned by Zemisch, Guillemin and Stenzel prove:

Theorem. For an analytic manifold L and analytic metric g on L, there is a $\sigma$-invariant neighborhood ($\sigma(x,v)=(x,-v)$) of $L\subset T^*L$ with a unique complex structure on that such that

i- $\sigma$ is an anti-holomorphic involution

ii- The one form $Im \bar\partial h$, where $h=|v|^2$ is the square of length of $v$ with respect to $g$, is the standard one-form $\sum y_i dx^i$. (This would imply $\sqrt{-1}\partial \bar\partial h$ is the standard Kahler form).

This is indeed an impressive result.

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False. You need a Riemannian metric on $T^*M$ to construct a Kähler structure. But $T^*M$ does not come with a natural metric. (Since that would imply that M itself would have an intrinsic metric.)

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OP didn't ask about a "natural" Kahler structure. For example, every smooth manifold $M$ admits a Riemannian structure, although not a canonical one. –  Victor Protsak Jun 2 '10 at 8:09

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