MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\phi^2 = \psi_1$ ?

share|cite|improve this question

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.)

share|cite|improve this answer

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.

share|cite|improve this answer
good to know, thanks – Mohammad F. Tehrani May 14 '13 at 15:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.