Hamiltonian polar action with Lagrangian section - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:38:15Z http://mathoverflow.net/feeds/question/116294 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116294/hamiltonian-polar-action-with-lagrangian-section Hamiltonian polar action with Lagrangian section Renato G Bettiol 2012-12-13T16:11:23Z 2012-12-17T01:53:13Z <blockquote> <p>I am looking for examples of Hamiltonian polar isometric actions of a compact Lie group on a Kahler-Einstein (or perhaps just Kahler) manifold, that admits a Lagrangian section.</p> </blockquote> <p>Recall that an isometric action on $M$ is <em>polar</em> if there exists a submanifold $\Sigma\subset M$, called <em>section</em>, that meets all orbits orthogonally. Any section is automatically a totally geodesic submanifold, and there may be many different sections on $M$. The examples I am looking for should have a section $\Sigma\subset M$ that is also a Lagrangian submanifold, i.e., $\omega|_\Sigma=0$ and $\dim M=2\dim \Sigma$.</p> <blockquote> <p>So far, the only example I can see is the standard torus action of $T^n$ on $\mathbb C P^n$, with the Fubini metric.</p> </blockquote> <p>This action is Hamiltonian and isometric. It is also polar (in fact, by a Theorem of Podesta-Thorbergsson, if a torus $T^n$ acts on a compact Kahler manifold $M^{2n}$ of complex dimension $n$ and positive Euler characteristic, then this action is polar). Finally, the usual totally real embedding of $\mathbb R P^n$ into $\mathbb C P^n$ gives a Lagrangian section for the action.</p> <hr> <p>With such strong hypothesis I would imagine that such actions are perhaps even classified, but I could not find anything in the literature about Lagrangian sections. Most classification results have to do with <em>coisotropic</em> actions, where it is required that the <em>orbits</em> satisfy $\omega|_{G(x)}=0$; while I am interested in an "orthogonal" version of that, i.e., I want the <em>section</em> to be <em>coisotropic</em> (even more, <em>Lagrangian</em>).</p> http://mathoverflow.net/questions/116294/hamiltonian-polar-action-with-lagrangian-section/116496#116496 Answer by Robert Bryant for Hamiltonian polar action with Lagrangian section Robert Bryant 2012-12-16T00:06:17Z 2012-12-17T01:53:13Z <p>I don't know what the global classification might be, but examining the structure equations for such a structure shows that the 'local' classification is reasonable.</p> <p>Let $(M,g,\omega)$ be a Kähler manifold (of real dimension $2n$) with a symplectic and isometric polar action $G\times M\to M$ and let $\Sigma^n\subset M$ be a Lagrangian section. (Since the consideration will be local, this can be just a local section for now.) Because the section has dimension $n$, the generic orbits (which meet $\Sigma$ orthogonally) must be of (real) codimension $n$ and must, themselves be Lagrangian. Let us restrict our attention to the open subset of $M$ consisting of $n$-dimensional $G$-orbits. </p> <p>As the OP observes, $\Sigma$ must be totally geodesic. If $m\in\Sigma$ is fixed and $G_m\subset G$ is the stabilizer of $m$, then $G_m$, which preserves the $n$-dimensional orbit $G\cdot m$ and hence its tangent space, must preserve the orthogonal to this tangent space, i.e., $T_m\Sigma$. Hence, $G_m$ must preserve $\Sigma$ since the image of $\Sigma$ under an element of $G_m$ must be a totally geodesic Lagrangian submanifold that is tangent to $\Sigma$ at $m$. Thus the images of $\Sigma$ under $G$ define a Lagrangian foliation of $M$ that is transverse to the Lagrangian foliation defined by the orbits of $G$.</p> <p>Following along this kind of argument, one sees that, locally, one can choose local holomorphic coordinates $z = (z^i) = x + i\ y$ so that a Kähler potential $\phi$ for $\omega$ can be found that is a function of $x$ alone and the action of $G$ is just translation in the $y$-directions. In particular, $G$ is abelian, and, in order to make the group compact, one has to take $y\in \mathbb{R}^n$ to be defined modulo a lattice $\Lambda\subset\mathbb{R}^n$. Locally, the section $\Sigma$ can be taken to be defined by $y=0$.</p> <p>In order for the potential $\phi = \phi(x)$ to define a Kähler metric, $\phi$ has to be convex relative to the affine structure on $\mathbb{R}^n$. Then, in order for this to define a Kähler-Einstein metric, $\phi$ has to satisfy a Monge-Ampère equation. </p> <p>I believe that there is a section in a Chapter of Besse's <em>Einstein manifolds</em> that describes these metrics (which, I think, were first seriously investigated by Calabi). However, I don't have a copy of the book here at home with me this weekend.</p> <p>Note that this is just a local classification on a dense open set. Of course, to deal with the case of $M$ compact, one is dealing with the case of toric Kähler manifolds, but I'm not sure which toric Kähler manifolds admit a polar section. However, I would imagine that this must be well-studied; one should look for more information in the toric literature. In particular, I recommend two papers (out of many possibilities):</p> <p>V. Guillemin, <em>Kähler metrics on toric varieties</em>, J. Diff. Geom. <strong>40</strong> (1994), 285–309. MR 95h:32029</p> <p>and </p> <p>M. Abreu, <em>Kähler geometry of toric varieties and extremal metrics</em>, Internat. J. Math. <strong>9</strong> (1998), 641–651. MR 99j:58047</p>