Square root for Hamiltonian diffeomorphisms - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:51:45Z http://mathoverflow.net/feeds/question/101858 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101858/square-root-for-hamiltonian-diffeomorphisms Square root for Hamiltonian diffeomorphisms Mohammad F.Tehrani 2012-07-10T15:08:02Z 2013-04-20T20:12:17Z <p>Let $\psi_t: X\to X$, $t \in [0,1]$, be a path Hamiltonian diffeomorphism on a symplectic manifold $X$, given by functions $H_t$. If $H_t \equiv H$ is independent of $t$ then </p> <p>$$\psi_1 = \psi_{\frac{1}{2}}^2$$</p> <p>and therefore the Hamiltonian diffeomorphism $\psi_1$ has a Hamiltonian square root.</p> <p>Is the same thing true for any arbitrary Hamiltonian $\psi_1$, i.e. is there another Hamiltonian $\phi$ such that $\phi^2 = \psi_1$ ?</p> http://mathoverflow.net/questions/101858/square-root-for-hamiltonian-diffeomorphisms/101878#101878 Answer by Mohammad F.Tehrani for Square root for Hamiltonian diffeomorphisms Mohammad F.Tehrani 2012-07-10T19:32:35Z 2012-07-10T19:32:35Z <p>I got this answer from Dusa McDuff (and she got it from some body else):</p> <p>Suppose given $f:[0,1]\to [0,1]$ such thqt 0 is repelling fixed point and 1 is attracting fixed point and there are no others.</p> <p>So $f'(0) = \lambda >1$, and $f'(1)=\mu &lt; 1$.</p> <p>A thm says that in suitable local coords near $0$ $f$ is simply mult by $\lambda$ (this is a linearization them). Therefore f has a unique square root on [0,1). Similarly, it has a unique square root on (0,1].</p> <p>But in general the coords at the two ends will NOT be compatible so there is no square root on [0,1].</p> <p>Now consider a smooth $f: S^2\to S^2$ with two non-deg fixed points $p_0,p_1$ with a homoclinic orbit $A$ between them. i.e. there is an arc $A$ which at one end is the unstable manifold of $p_0$ and at the other is the stable manifold of $p_1$. Now restrict f to A.</p> <p>(There is a stable manifold thm that says that locally these invariant submanifodls exist etc.)</p> http://mathoverflow.net/questions/101858/square-root-for-hamiltonian-diffeomorphisms/128192#128192 Answer by Nate Bottman for Square root for Hamiltonian diffeomorphisms Nate Bottman 2013-04-20T20:12:17Z 2013-04-20T20:12:17Z <p>In a <a href="http://arxiv.org/abs/1304.4067" rel="nofollow">short paper</a> posted last week, Peter Albers and Urs Frauenfelder prove that if $(M,\omega)$ is any closed symplectic manifold, then in any $\mathcal{C}^\infty$-neighborhood of the identity in $\text{Ham}(M,\omega)$ there is a Hamiltonian diffeomorphism that does not have a square root in $\text{Ham}(M,\omega)$ (the square root is not required to lie in this neighborhood).</p> <p>The key is an observation of Milnor's that for any $k > 0$, an obstruction to a self-diffeomorphism of a manifold having a square root is that it has an odd number of $2k$-cycles.</p>