Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:51:04Z http://mathoverflow.net/feeds/question/9335 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9335/why-is-every-symplectomorphism-of-the-unit-disk-hamiltonian-isotopic-to-the-ident Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity? Ilya Grigoriev 2009-12-19T01:54:23Z 2010-02-16T14:53:06Z <p>That is, for any symplectomorphism $\psi: D^2 \to D^2$, there should be a time-dependent Hamiltonian <em>H<sub>t</sub></em> on <em>D<sup>2</sup></em> such that the corresponding flow at time 1 is equal to $\psi$.</p> <p>I found this in claim a paper, and I think it should be easy, but nothing comes to mind. I'd be happy with a reference to a page in McDuff-Salomon, but I couldn't find this there immediately.</p> <p>Thanks!</p> http://mathoverflow.net/questions/9335/why-is-every-symplectomorphism-of-the-unit-disk-hamiltonian-isotopic-to-the-ident/9339#9339 Answer by Greg Kuperberg for Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity? Greg Kuperberg 2009-12-19T02:40:30Z 2009-12-19T20:54:43Z <p>It is a theorem of Smale that the group of orientation-preserving diffeomorphisms of $D^2$, rel boundary, is contractible. If the diffeomorphisms can move the boundary, you can establish a homotopy equivalence between that and the circle. The diffeomorphisms do not have to preserve area. Then, a <a href="http://mathoverflow.net/questions/7817/normal-coordinates-for-a-manifold-with-volume-form" rel="nofollow">theorem of Moser</a> establishes a deformation retract from diffeomorphisms to volume-preserving diffeomorphisms. Moser's result is easier to see if you have a closed manifold, but it extends to manifolds with boundary with the doubling trick. Together, this indirectly gives you a curve of symplectomorphisms connecting the identity to $\psi$, since in two dimensions the symplectic structure is just a volume structure. Finally if you have a smooth curve of area-preserving diffeomorphisms of a disk, I think there is a time-dependent Hamiltonian obtained by integrating the corresponding vector field.</p> <p><hr /></p> <p>I shared the same concern that Ilya expresses in the comments, but after considering it, here is why I think that it works. To have a clean view of the boundary conditions, let's double the disk to the sphere and let everything be equivariant with respect to reflection across the equator.</p> <p>Moser's theorem truly is a deformation retraction. Let $M$ be a Riemannian manifold, let $\mu$ be a volume form on $M$ (not necessarily Riemannian volume), and let $\phi_\alpha:M \to M$ be a family of diffeomorphisms of $M$ that may or may not preserve $\mu$. Then <code>$\mu_\alpha = (\phi_\alpha)_*(\mu)$</code> is "wrong". Let $\mu_{\alpha,t}$ be a family of volume forms defined as the weighted geometric mean of $\mu_\alpha$ and $\mu$: $$\mu_{\alpha,t} = \mu^t \mu_\alpha^{1-t}.$$ Then there is a corresponding Moser flow $\phi_{\alpha,t}$ such that $\phi_{\alpha,0} = \phi_\alpha$ and $\phi_{\alpha,1}$ is volume-preserving. Moreover, $\phi_{\alpha,t} = \phi_\alpha$ for all $t$ if $\phi_\alpha$ is already volume-preserving for some fixed $\alpha$.</p> <p>In particular, if $\phi_t$ is a curve of diffeomorphisms as produced by Smale's theorem with $\phi_0$ the identity, then Moser gives you an improvement $\phi_{t,s}$ such that $\phi_{t,1}$ is then what you want. What worried us is whether $\phi_{1,1} = \phi_1$; if $\phi_1$ is area-preserving, then it is true.</p> http://mathoverflow.net/questions/9335/why-is-every-symplectomorphism-of-the-unit-disk-hamiltonian-isotopic-to-the-ident/9392#9392 Answer by Ilya Grigoriev for Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity? Ilya Grigoriev 2009-12-19T23:55:08Z 2009-12-20T00:00:52Z <p>I just wanted to expand two points of Greg's answer. Both are rather trivial additions, but took me a short while to understand, so I'm putting them here for completeness's sake and my own future reference.</p> <p>First, here's a picture of Greg's idea of fixing the problem with Moser's trick. In his notation, we have volume forms $\mu_{\alpha,t} = \mu^t \mu_\alpha^{1-t}$. They are defined because the two volume form must have the same sign everywhere (since everything is orientation-preserving; note that because of this I doubt that this argument can be generalized to symplectic forms in higher dimensions).</p> <p>The obvious, but wrong, solution (described in my comments to his answer) would be to apply Moser's theorem to $\mu_{\alpha,0}$ in this notation. This would correspond to flowing along the horizontal axes of the picture below. However, Greg's idea is to flow in a different direction: we fix $\alpha$ and vary $t$. Strictly speaking, we should also show that the resulting flow $\phi_{\alpha,1}$ will be smooth, but this should follow from the proof of Moser's theorem quite easily.</p> <pre><code> Moser flow, t defines phi_alpha,t (mu_{alpha,1} = mu) 1 ^ ^ | . | . | . | . (mu_{alpha,0} = mu_alpha) 0 ----------------&gt; alpha (phi_{alpha,0} = phi_alpha) </code></pre> <p>(Note that, for all <em>t</em>, $\mu_{0,t}=\mu_{1,t}=\mu$) </p> <p><hr /></p> <p>Secondly, Greg's answer implies that there is an isotopy $\psi_t: D^2 \to D^2$ such that each $\psi_t$ is volume preserving (or, equivalently, a symplectomorphism), $\psi_0$ is the identity, and $\psi_1=\psi$. Here's how we find the time-dependent Hamiltonian such that this is the Hamiltonian flow. Let <em>X<sub>t</sub></em> be the time-dependent vector field that is the derivative of the flow $\psi_t$. The fact our flow is volume-preserving is equivalent to the fact that the 1-form $\iota_{X_t} \omega$ is be closed for all $t$. Since we are on a disk, this form will also be exact. So, let $H_t$ be the function such that $dH_t = \iota_{X_t} \omega$. The Hamiltonian flow of the function $H_t$ is precisely $\psi_t$; this is immediate from the definitions.</p>