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Edit: The previous version of this question contained 2 part. In this new version, I deleted the first part and move it to a new question.

Is There a polynomial Hamiltonian $H(x,y,z,w)=zP(x,y)+wQ(x,y)$, such that the corresponding Hamiltonian vector field in $\mathbb{R}^{4}$ has at least one periodic orbit and the total number of periodic orbits is finite?

The motivation for this question is that, for such particular Hamiltonian, the first two components of $X_{H}$ is a polynomial planar vector field on the $x-y$ plane. On the other hand, a generic algebraic vector field on $\mathbb{R}^{2}$ has only a finite number of periodic orbits.

By non trivial periodic orbit, I mean, a periodic solution which minimum period is non zero.(That is a periodic orbit not a singular point).

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I am assuming by you formulation you allow ANY orbit time and that you are not counting the multiple covers. If this is so, then the last question is equivalent to asking whether such a vector field on $\mathbb{R}^2$ can have all the periodic orbits non-degenerate (also its multiple covers). –  Thomas Kragh Jan 28 '14 at 11:54
@ThomasKragh thanks for comment. In my question(part 1 and part 2), I count the total number of periodic orbit in $\mathbb{R}^{4}$, Including the multiple cover. –  Ali Taghavi Jan 28 '14 at 20:14
But then it is only the time 1 periodic ones? If not you always get infinitely many? –  Thomas Kragh Jan 29 '14 at 14:37
@ThomasKragh my question has a geometric nature.(no matter of parametrization).To remove the missunderestanding, I restate my question:Let $H:\mathbb{R}^{4} \to \mathbb{R}$ be a Hamiltonian. The corresponding hamiltonian vector field $X_{H}$, defines a one dimensional foliation on $\mathbb{R}^{4}-S$ where $S$ is the set of critical points of $H$. My question :Is there a polynomial Hamiltonian $H$ which corresponding foliation has a finite number of compact leaves(diffeomorphic to $\mathbb{S}^{1}$), and the number of compact leaves is not zero? –  Ali Taghavi Jan 30 '14 at 8:22
Well - then you are not counting the multiple covers, as each of these leaves corresponds to a set of periodic orbits of time $T,2T,3T,\dots$, for some $T>0$, and only the one with time $T$ is not a multiple cover. –  Thomas Kragh Jan 31 '14 at 6:22

1 Answer 1

up vote 4 down vote accepted

Answer to part two: No - this cannot be - even if you relax the conditions and only assume the vector field to be smooth.

Indeed, the question can be generalized somewhat: let $X : N \to TN$ be any smooth vector field on a manifold $N$. Then we may define a Hamiltonian

$H : T^*N \to \mathbb{R} \qquad$ by $\qquad H(q,p)=p(X_q)$.

Here $T^*N$ denotes the cotangent bundle (which has a canonical symplectic structure), and hence it makes sense to evaluate a cotangent vector $p \in T^*_qN$ on a vector in $T_qN$. This generalizes your second question with $X(x,y)=(Q(x,y),P(x,y))$ and $N=\mathbb{R}^2$.

The flow of the associated Hamiltonian vector field $X_H$ has flow $\Phi_t$ the same as the flow $\varphi_t$ of $X$ but lifted to the unique symplectomorphism, which is linear in $T_q^*N \to T^*_{\varphi(q)}N$. This is also given as:

$\Phi_t(q,p)=(\varphi_t(q),p\circ (D_q\varphi_t)^{-1})$

Indeed, if you (in local coordinates on $N$ - so we are now back in $\mathbb{R}^{2n}=T^*\mathbb{R}^n$) differentiate this w.r. to $t$ at $t=0$ you get the vector field:

$(X_q,-p \circ (\nabla X)_q)$

This is (up to a sign) precisely equal to $J_0\nabla H$.

Consequences of this are: Any periodic orbit (i.e. a fixed point $(q,p)=\Phi_T(q,p)$ must project to a similar orbit for $X$, and have: $p=p \circ (D_q\varphi_T)^{-1}$. This means that any eigenvector with eigenvalue 1 for $D_q \varphi_T$ defines such an orbit - so if 1 is an eigenvalue there are infinitely many. Since this return map takes any vector tangent to the orbit to itself we have such an eigenvector.

About the first question: I doubt it: indeed, no orbit can be non-degenerate since these are stable (for fixed orbit time $T_0$) under pertubations - hence there must be one with period $T$ for each $T\in[T_0-\epsilon,T_0+\epsilon]$ close by (for small enough $\epsilon>0$).

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Thank you very much for your beautiful answer.Is the flow on cotangent bundle equivalent to the flow on tangent bundle generated by "Variational equation"? –  Ali Taghavi Jan 31 '14 at 20:10
is it reasonable that I delete the first part of the question and i post the first part as a new independent question?I can not understand your statement on the first part my question. –  Ali Taghavi Jan 31 '14 at 20:13
I have looked up "variational equation" and I don't think so. Indeed, that seems to depend on a variation of $X$. I don't know about rewriting, I would think not? –  Thomas Kragh Feb 2 '14 at 7:11
By variational vector field I mean the following:I describe it in euclidean coordinate: Let F be a vector field on $U\subset \mathbb{R}^{n}$, the variational vector field on $TU=\{(x,v)\in U\times \mathbb{R}^{n}\}$ is $(F(x), DF(x).v)$ –  Ali Taghavi Feb 3 '14 at 13:52
That does look equivalent (maybe you need a reversal of time or simply a sign somewhere - to account for the inverse in my formula). In any case, in general if $N$ has Riemannian structure you can define your variational vector field, and this is probably the same using the canonical identification of cotangent vectors with tangent vectors defined by the Riemannian structure. –  Thomas Kragh Feb 3 '14 at 15:05

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