Kähler structure on cotangent bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:20:57Z http://mathoverflow.net/feeds/question/26776 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle Kähler structure on cotangent bundle? Mohammad F.Tehrani 2010-06-02T02:15:27Z 2013-03-08T01:32:45Z <p>The total space of cotangent bundle of any manifold M is a symplectic manifold.</p> <p>Is it true\false\unknown that for any M, $T^*M$ has Kähler structure?</p> <p>Please support your claim with reference or counterexample.</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/26779#26779 Answer by Ian Agol for Kähler structure on cotangent bundle? Ian Agol 2010-06-02T03:07:48Z 2010-06-02T03:07:48Z <p>In a paper by <a href="http://www.ams.org/mathscinet-getitem?mr=1807952" rel="nofollow">Goldman, Kapovich, and Leeb</a>, 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). </p> <p>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). </p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/26781#26781 Answer by Tim Perutz for Kähler structure on cotangent bundle? Tim Perutz 2010-06-02T04:01:44Z 2010-06-02T04:18:53Z <p>This is true! I assume $M$ compact.</p> <p><b>Method 1.</b> Real algebraic geometry. Cf. <a href="http://arxiv.org/abs/math/0701783" rel="nofollow">this article</a>. 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.</p> <p><b>Method 2.</b> Eliashberg's existence theorem for Stein structures. See Cieliebak-Eliashberg's unfinished book, <a href="http://math.stanford.edu/~eliash/Public/Site/Eilenberg_Lectures_files/stein7.pdf" rel="nofollow">Symplectic geometry of Stein manifolds</a>, 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 <a href="http://arxiv.org/abs/0810.4511" rel="nofollow">Gompf</a> 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.</p> <p>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 $\omega_{can}$.</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/26795#26795 Answer by Kant was right for Kähler structure on cotangent bundle? Kant was right 2010-06-02T07:10:05Z 2010-06-02T07:10:05Z <p>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.)</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/26804#26804 Answer by Bugs Bunny for Kähler structure on cotangent bundle? Bugs Bunny 2010-06-02T09:46:35Z 2010-06-02T09:46:35Z <p>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.</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/26913#26913 Answer by I rather hang out at sci-math for Kähler structure on cotangent bundle? I rather hang out at sci-math 2010-06-03T11:51:50Z 2010-06-03T12:10:41Z <p>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.</p> <p>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.)</p> <p>The Kähler form is $$\Omega = -\text{Im}\ h(d\mathbf{z}_1, d\mathbf{z}_2) = d^2\mathbf{x}_1^T\ A\ d\mathbf{x}_2 - d\mathbf{x}_1^T\ A\ d^2\mathbf{x}_2$$</p> <p>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 $T^*M$.</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/69739#69739 Answer by Kai Zehmisch for Kähler structure on cotangent bundle? Kai Zehmisch 2011-07-07T18:25:28Z 2011-07-07T18:25:28Z <p>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)</p> http://mathoverflow.net/questions/26776/kahler-structure-on-cotangent-bundle/123931#123931 Answer by Mohammad F.Tehrani for Kähler structure on cotangent bundle? Mohammad F.Tehrani 2013-03-08T01:32:45Z 2013-03-08T01:32:45Z <p>In the refernce mentioned by Zemisch, Guillemin and Stenzel prove:</p> <p>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</p> <p>i- $\sigma$ is an anti-holomorphic involution </p> <p>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).</p> <p>This is indeed an impressive result.</p>