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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

$$ \psi_1 = \psi_{\frac{1}{2}}^2 $$

and therefore the Hamiltonian diffeomorphism $\psi_1$ has a Hamiltonian square root.

Is the same thing true for any arbitrary Hamiltonian $\psi_1$, i.e. is there another Hamiltonian $\phi$ such that $\phi^2 = \psi_1$ ?

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2 Answers

In a short paper 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).

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.

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good to know, thanks –  Mohammad F. Tehrani May 14 '13 at 15:12
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I got this answer from Dusa McDuff (and she got it from some body else):

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.

So $f'(0) = \lambda >1$, and $f'(1)=\mu < 1$.

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].

But in general the coords at the two ends will NOT be compatible so there is no square root on [0,1].

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.

(There is a stable manifold thm that says that locally these invariant submanifodls exist etc.)

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