When is a symplectic manifold equivalent to a cotangent bundle? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:46:25Z http://mathoverflow.net/feeds/question/35900 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35900/when-is-a-symplectic-manifold-equivalent-to-a-cotangent-bundle When is a symplectic manifold equivalent to a cotangent bundle? Per Vognsen 2010-08-17T19:44:44Z 2010-08-20T22:16:31Z <p>Let $X$ be a differentiable manifold. Its cotangent bundle $T^*X$ carries a canonical 1-form $\alpha$ whose exterior differential $\omega = d\alpha$ endows $T^*X$ with the structure of a symplectic manifold.</p> <p>But what about the converse question? Which symplectic manifolds are cotangent bundles? </p> <p>Clearly a necessary condition is that $\omega$ must be exact, so cohomological obstructions are relevant. Is that all? Compact symplectic manifolds have non-trivial de Rham cohomology in grade two, so the cohomological test passes muster for that important class of examples.</p> <p>I'm also interested in examples of manifolds where different symplectic forms (modulo exact 2-forms) give qualitatively different dynamics with the same Hamiltonian. As symplectic structures on the same manifold are all locally equivalent by Darboux's theorem, one expects such phenomena would occur only on a global scale.</p> http://mathoverflow.net/questions/35900/when-is-a-symplectic-manifold-equivalent-to-a-cotangent-bundle/35901#35901 Answer by Mohammed Abouzaid for When is a symplectic manifold equivalent to a cotangent bundle? Mohammed Abouzaid 2010-08-17T20:12:33Z 2010-08-17T21:07:58Z <p>Using the h-principle, Gromov showed that there is a symplectic form on $\mathbb{R}^6$ which admits $S^3$ as a Lagrangian submanifold. Using holomorphic curves, he showed that the standard symplectic form on $T^* \mathbb{R}^3$ does not admit any such Lagrangian. There is now a whole industry of building exotic symplectic forms on non-compact manifolds (see papers of Seidel-Smith, Mark McLean, ...).</p> <p>Probably the only reasonable answer to characterising cotangent bundles uses the existence of a Lagrangian foliation by planes. If you have a foliation parametrised by a manifold which admits a Lagrangian section, then you have yourself an open subset of a cotangent bundle (this is just Weinstein's theorem). You can't drop the condition of the existence of a section precisely because you can add the pull back of a $2$-form on the base. If your symplectic form is "complete" then the existence of a Lagrangian section is a cohomological condition. Pick any section: If the pullback of $\omega$ doesn't vanish, then you don't have a cotangent bundle. If it vanishes in cohomology, you can use a primitive $1$-form to flow your section to a Lagrangian.</p> <p>Added Remark:</p> <p>I want to point out that the methods we have for producing different symplectic forms do not proceed by writing down different $2$-forms on the same space. Rather, you find some construction of symplectic manifolds (using some general notion of symplectic surgery) which produces a large class of symplectic manifolds, then you prove that some of these result in the same smooth manifold. The existence of a diffeomorphism is obtained abstractly, so I do not know of examples where we can write down a Hamiltonian whose dynamics for two different symplectic forms can be compared.</p> http://mathoverflow.net/questions/35900/when-is-a-symplectic-manifold-equivalent-to-a-cotangent-bundle/36068#36068 Answer by Will Merry for When is a symplectic manifold equivalent to a cotangent bundle? Will Merry 2010-08-19T09:05:00Z 2010-08-20T22:16:31Z <p>In response to your last paragraph, the so-called "twisted" cotangent bundles provide examples where different symplectic forms exhibit very different dynamics with the same Hamiltonian.</p> <p>Suppose $\omega=d\alpha$ is the standard symplectic form on a cotangent bundle $\pi:T^{*}X\to X$, where $X$ is a closed manifold. Let $\sigma$ denote a closed non-exact two-form on $X$, and consider a new family of two-forms $\omega_{s}$ for $s\in [0,\infty)$ defined by $\omega_{s}:=\omega-s\pi^{*}\sigma$. It's easily checked that $\omega_{s}$ is again a symplectic form on $T^{*}X$ for each $s\in [0,\infty)$ (it's closed as $\sigma$ is closed and non-degenerate as $d\pi$ vanishes on "vertical" tangent vectors). </p> <p>Fix a Riemannian metric $g$ on $X$, and let $H:T^{*}X\to X$ denote the standard "kinetic energy" Hamiltonian defined by $H(x,p):=\frac{1}{2}|p|^{2}$, and let $\xi_{s}$ denote the symplectic gradient of $H$ with respect to $\omega_{s}$ (i.e. $i_{\xi_{s}}\omega_{s}=-dH$). Let $\phi_{s}$ denote the flow of $\xi_{s}$.</p> <p>Let $S^{*}X$ denote the unit cosphere bundle of $X$. Since $H$ is autonomous, the flow $\phi_{s}$ preserves $S^*{X}$ for each $s\in[0,\infty)$. The point is that the dynamics of $\phi_{s}$ on $S^{*}X$ can vary dramatically depending on $s$.</p> <p>As a concrete example of this, consider a closed hyperbolic surface $X=\mathbb{H}^{2}/\Gamma$, where $\Gamma$ is a cocompact lattice of $\mathrm{PSL}(2,\mathbb{R})$. Let $\sigma$ denote the area form on $X$. Note that for $s=0$, $\phi_{0}$ is just the cogeodesic flow. For $0\le s&lt;1$, the dynamics of $\phi_{s}$ is Anosov and conjugate (after rescaling) to the cogeodesic flow. All closed orbits are non-contractible. In this case the unit cosphere bundle is a contact type hypersurface in the symplectic manifold $(T^{*}X,\omega_{s})$. For $s=1$ we get the horocycle flow. There are no closed orbits at all, and the unit cosphere bundle is not of contact type (in fact, it's not even stable). For $s>1$ all the orbits are closed and contractible. The unit cosphere bundle is again of contact type, but with the opposite orientation.</p> <p>Perhaps the best place to read about this is Ginzburg's survey article "On closed trajectories of a charge in a magnetic field: An application of symplectic geometry", which is in the book "Contact and symplectic geometry" (CUP,1994). The recent paper "Symplectic topology of Mane's critical values" by Cieliebak, Frauenfelder and Paternain contains lots of examples of this sort of behaviour.</p>