## Periodic orbits of Hamiltonian flows

I have a stupid question :

Lets $H_t$ be a 1-periodic family of Hamiltonian functions on symplectic manifold $(M,w)$. corresponding to that, we considers the 1-periodic Hamiltonian vector fields $X_t$, $\iota_{X_t}w=-dH_t$, and the Hamiltonian flow, $\Psi_t$, given by the equation $$\dot{x}=X_t(x(t))$$.

The goal of Floer homology is to count the 1-periodic orbits of this flow. lets $x(t)$ be one of them, so one can consider the the linearization of $\Psi_t$ along this orbit to get the map $D\Psi_1(x(0)) : T_{x(0)}M \rightarrow T_{x(0)}M$.

According to definition, a periodic orbit is called non-degenerate if $det(I-D\Psi_1(x(0)))\neq 0$. But how could it be possible because $D\Psi_1(x(0)) (X_0(x(0)))=X_0(x(0))$.

Reference : page 14 of http://students.washington.edu/rlmill/superseminar/salamon.pdf

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The equality you wrote will hold only in the case of an autonomous (time-independent) flow. In the "forced" case that you consider, it will not hold generically. – Guy Katriel May 27 2010 at 17:24
@Guy: True, and even more: there are examples where it does hold and the orbit is still non-degenerate (i.e. X_0(x(0))=0). take e.g. $H=x^2+y^2$ in $\mathbb{R}^2$ – Thomas Kragh May 27 2010 at 17:44
@Thomas: I believe in the example you gave the orbit (any of them) would be degenerate according to the definition given by Mohammad. For autonomus system one needs a different definition of "non-degenerate". – Guy Katriel May 27 2010 at 18:00
@Guy: only when the orbits are not constant, here the orbit is the constant orbit equal to $0$, and the differential of the time 1 flow at 0 is the same as the time 1 flow - i.e. a rotation of angle 2 - so almost a third of the way round the point. So it is non-degenerate. It works in this case because the obvious action of $S^1$ which for non constant orbits creates an $S^1$ family of orbits proving degeneracy does not work here. – Thomas Kragh May 27 2010 at 18:52
One can prove that Hamiltonian Floer homology is isomorphic to Morse homology by using an autonomous Hamiltonian $\epsilon H$ where $\epsilon$ is chosen small enough that the only 1-periodic Hamiltonian orbits are constant orbits at the critical points. Such systems are non-degenerate if $H$ is Morse. It is, I think, correct that big orbits of an autonomous Hamiltonian are degenerate, for the reason that Mohammad points out and Guy clarifies. – Tim Perutz May 27 2010 at 19:07