Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity? - MathOverflow

Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?

2009-12-19T01:54:23Z

That is, for any symplectomorphism $\psi: D^2 \to D^2$, there should be a time-dependent Hamiltonian H_t on D² such that the corresponding flow at time 1 is equal to $\psi$.

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.

Thanks!

Answer by Greg Kuperberg for Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?

2009-12-19T02:40:30Z

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 theorem of Moser 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.

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.

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 $\mu_\alpha = (\phi_\alpha)_*(\mu)$ 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 $\alpha$.

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.

Answer by Ilya Grigoriev for Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?

2009-12-19T23:55:08Z

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.

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).

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.

                                         Moser flow,
                                t        defines phi_alpha,t
        (mu_{alpha,1} = mu)   1 ^          ^
                                |          .
                                |          .
                                |          .
                                |          .
   (mu_{alpha,0} = mu_alpha)  0 ----------------> alpha
 (phi_{alpha,0} = phi_alpha)

(Note that, for all t, $\mu_{0,t}=\mu_{1,t}=\mu$)

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 X_t 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.