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Is there a Hamiltonian $H:\mathbb{R}^{2n} \to \mathbb{R}$ with the following property:

For two regular values $a<b$ for which $[a,b]$ consists of regular values, the dynamics of $X_H$ on $H^{-1}(a)$ is not topological equivalent to the dynamic of $X_H$ on $H^{-1}(b)$?

Here $X_H$ is the hamiltonian vector field associated to the Hamiltonian $H$.

The motivation for this question is the following post. The answer to the following post would be negative if the answer to the above question is negative:

An algebraic Hamiltonian vector field with a finite number of periodic orbits (2)

Ali Taghavi
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    Do you assume the Palais-Smale condition on the function $H$? (e.g., $H$ is coercive). Otherwise in your hypotheses already the sub-manifolds $H^{-1}(a)$ and $H^{-1}(b)$ may well differ topologically (for instance: one is empty, the other is not, $H$ is bounded below but does not have a minimum) – Pietro Majer Jul 23 '23 at 16:53
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    In other words maybe you want “$a$ is a regular value” to be understood in a strong sense (“$|\nabla H|$ is bounded away from $0$ on $H^{-1}(a)$“). – Pietro Majer Jul 23 '23 at 16:59
  • @PietroMajer Thank you very much for your comment and the interesting point of Palais Smale condition. No I did not assume the Palais Smale condition. But I remember a theorem in Hirsch book differential topology that every two non empty level set are diffeomorphic provided there is no critical value between a and b. He use normalized gradient vector field to prove this theorem. – Ali Taghavi Jul 24 '23 at 06:41
  • So I wonder if every two regular level set as above have similar dynamics? If there would be a counter example then it would be interesting to prove it with extra assumption Palais Smale condition. – Ali Taghavi Jul 24 '23 at 06:46
  • @PietroMajer Am I mistaken on diffeomorphicity of every two regular level set as above without PS condition? May be in Hirsch book we had the compactness of M which automatically implies the PS condition? – Ali Taghavi Jul 24 '23 at 06:52
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    Yes, that is true for compact manifolds, or more generally, assuming PS condition, orherwise it may be false. Consider e.g. f(x):=exp(x) on R. No critical points, the level { f=-1} is empty, tge level {f=1} is not. – Pietro Majer Jul 24 '23 at 06:55
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    There is a "critical point at infinity" at the level c=0, that is a sequence with $f(x_n)\to c$, $df(x_n)\to0$, $x_n\to\infty$ (in the 1 pt compactification). It can be an obstruction to deformation just as a true critical point – Pietro Majer Jul 24 '23 at 07:03
  • @PietroMajer according to definition PS condition you mentioned above I think $exp$ satisfies this condition. On the other hand I think "exp" is not a counter example to the question in my last comment. Because in my last comment i search for an example of two non empty non diffeomorphic regular level set with values a and b such that [a b] consist of regular values. You said that in PS condition it is impossible but what about non PS condition – Ali Taghavi Jul 25 '23 at 15:13
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    Actually $f(x)= \exp(x)$ on $\mathbb R$ does not satisfy PS, precisely at the level $c=0$, because there is e.g. the non compact sequence $x_n=-n$ such that $f'(x_n)\to0$ and $f(x_n)\to c$. Empty vs singleton is just the simplest pair of non diffeomorphic objects, but if we want two non-empty level sets, take $f(x)=xe^x$. The only critical level is the minimum $-1/e$ so $[-1/3, 1]$ consists of regular values, {f=1} is a singleton, {f=-1/3} is a doubleton. – Pietro Majer Jul 25 '23 at 17:17
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    For another example: take any smooth $f:M\to\mathbb R$ with two non empty non diffeomorphic regular level sets with values a and b. Then remove the critical set $\text{crit}(f):={x\in M: df(x)=0}$ from $M$: now $[a,b]$ consists of regular values of $g:=f$ restricted to $N:=M\setminus\text{crit}(f)$, since $g$ has no critical points. (And of course, if $\text{crit}(f)$ was non-empty, $g$ does not satisfy PS). – Pietro Majer Jul 25 '23 at 17:27
  • @PietroMajer yes I see. So it seems that a precise connected compact counter example is not straightforward. Yes? – Ali Taghavi Jul 26 '23 at 04:47
  • What counterexample would you like exactly? – Pietro Majer Jul 26 '23 at 19:46
  • @PietroMajer Befor I talk about the exact counterexample, I would like to say that the mapping $exp$ satisfies an equivalent statment you wrote in one of your previous comment: "the norm of gradient is far from zero on every level set" I think it is not equivalent to PS condition. – Ali Taghavi Jul 27 '23 at 08:48
  • By counter example I mean a connected manifold M with an interval of regular values [a,b] (no level set is empty) but $H^{-1}(a)$ is not diffeomorphic to $H^{-1} (b)$. I mean we emphasis on connected ness of M. (We forget compactness condition). A modified question: a connected manifold and a regular map without critical point which has two non diffeomorphic non empty level set. Thanks again for your attention to my question. – Ali Taghavi Jul 27 '23 at 08:53
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    Yes you are right of course, that was incorrect, sorry, one should say "$|\nabla H|$ is bounded away from $0$ on $H^{-1}([a-\epsilon,a+\epsilon])$ for some $\epsilon>0$" (this is "$a$ regular value and PS holds at level $a$") – Pietro Majer Jul 27 '23 at 09:43
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    Also, this way it is easy to build examples of smooth functions on $\mathbb R^n$ with no critical points and non homeomorphic level sets. Start from an $f:\mathbb R^n\to\mathbb R$ that has e.g. $\text{crit}(f)\subset \mathbb R^n\setminus\Omega$, $f^{-1}(a)\subset \Omega$, $f^{-1}(b)\subset \Omega$, and $\Omega$ is an open set diffeomorphic to $\mathbb R^n$ via a differ $\phi:\mathbb R^n\to\Omega$. Then $g:=f\circ\phi:\mathbb R^n\to\mathbb R$ has no critical points, $g^{-1}(a)\sim f^{-1}(a)$ and $g^{-1}(b)\sim f^{-1}(b)$. – Pietro Majer Jul 27 '23 at 10:23
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    I did not recalculate this, but I think the following is true: even if you do not have compactness issues the dynamics can change drastically. You can have a hamiltonian on $R^4$ whose regular level set at $1$ is a sphere, where all orbits are closed, but at $2$ it is an ellipse with very few closed orbits. – Thomas Rot Jul 27 '23 at 12:03
  • @PietroMajer yes that is very good. Thank you – Ali Taghavi Jul 27 '23 at 18:58
  • @ThomasRot I think you mean ellipsoid not ellipse. Moreover I think on sphere you have two singularities yes?(not possible qll solutions are closed orbit). I would appreciate if you wrote such a Hamiltonian – Ali Taghavi Jul 27 '23 at 19:00
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    Here the sphere is $\mathbb S^3$, all solutions are great circles. His Hamiltonian could even be a strictly convex coercive function I think. Very nice – Pietro Majer Jul 27 '23 at 20:01
  • @PietroMajer Yes I was mistaken since the energy level must have codimension 1. So 3 dimension. Could you please write the Hamiltonian you are talking about – Ali Taghavi Jul 28 '23 at 08:04
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    Btw, suppose H satisfy $[J\nabla H,\nabla H]$ on $\mathbb R^{2n}$ w.r.to the standard symokectic form. Then the Hamiltonian flow and the gradient flow of $H$ commute, and the your thesis holds. I – Pietro Majer Jul 28 '23 at 11:57

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This is again with the caveat that I have not done the calculations. Let $a>1$ be an irrational number. Let $H:\mathbb{R}^4\rightarrow \mathbb R$ be a function such that

$ H(x,y,z,w)=(x^2+y^2+z^2+w^2) $

when $x^2+y^2+z^2+w^2=1$ and

$ H(x,y,z,w)=x^2+y^2+z^2+aw^2 $

when $x^2+y^2+z^2+aw^2=2$. You can convince yourself that it is possible to find such a function without critical points in the region

$$ 1\leq x^2+y^2+z^2+w^2\qquad \text{and}\qquad x^2+y^2+z^2+aw^2\leq 2 $$

(i.e. by interpolating the values in the region in between radially)

At the level set $H^{-1}(1)$ all orbits are closed because these are two uncoupled harmonic oscillators with the same period. At the level $H^{-1}(2)$ the motion is also described by two harmonic oscillators, but now with periods that are irrational with respect to each other. The only periodic orbits are those where one of the oscillators is not in motion. So there are two of them.

Thomas Rot
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  • My +1 and thanks for your attention to my question and your answer. I think some thing is missing. It seems that you actually use the following statement but I think it is not true: "If $N$ is codimension one submanifold and $H_1$ and $H_2$ are constant on $N$ then the two Hamiltonian flow are the same on $N$" . This is not true for example put $N=$ unit circle in the plane, $H_1\equiv 0$ and $H_2=x^2+y^2-1$ The first Hamiltonian consist of singularities and the second one consist of a periodic orbit, two different dynamic. Am I missing some thing? – Ali Taghavi Jul 28 '23 at 10:34
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    I use the following fact: If $N$ is a regular level set of two different Hamiltonians $H_1$ and $H_2$, then the dynamics on $N_1$ with respect to $H_1$ is a rescaling of the dynamics of $H_2$ on $N$. – Thomas Rot Jul 28 '23 at 10:59
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    See for example the discussion in Weinstein's On the hypotheses of Rabinowitz' periodic orbit theorems – Thomas Rot Jul 28 '23 at 11:02