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Disclaimer: I don't know if this question is well suited for this site, but I have posted this The original question on Math.StackExchange with no answer, so I have thought to post it even hereconsisted of two parts. The first one
has been answered negatively (see
below the answers of Sam Lisi and
Alejandro). It remains the second one.
Background
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, X_H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$
If $N$ is a smooth submanifold contained in $\text{per},$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
Question
In Guillemin, Stenberg, Geometric Asymptotics, on page 170between pages 170-171, I have additionally found that, when all integral curves of $X_H$ are periodic, we can take $N=\text{per}$ in $(\ast),$ which should mean that in such a case $\text{per}$ is a smooth submanifold of $\mathbb R\times M.$
In order to understand justify this last point I was wondering myself:
- If $X$ is a non singular vector field on $M,$ all of whose integral curves are periodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth?
- otherwise, how to prove that in such a case $\text{per}$ is a submanifold?
What I have tried about point 2
Probably I am missing something because my guess is that if there were a principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$-orbits are the trajectories of $X$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
But I don't know how to proceed without this additional hypothesis.
Edit
Edit1 (After Sebastian's answer)answer about point 1): As illustration of my difficulties with point 1, I imagine that $M$ is the Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }y=0\\2&\text{if }y\neq 0\end{cases}$$
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Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, X_H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$
If $N$ is a smooth submanifold contained in $\text{per},$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are periodic, we can take $N=\text{per}$ in $(\ast),$ which should mean that in such a case $\text{per}$ is a smooth submanifold of $\mathbb R\times M.$
In order to understand this last point I was wondering myself:
If $X$ is a non singular vector field on $M,$ all of whose integral curves are periodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}$ is a submanifold?
Probably I am missing something because my guess is that if there were a principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$-orbits are the trajectories of $X$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
But I don't know how to proceed without this additional hypothesis.
Edit (After Sebastian's answer): As illustration of my difficulties I imagine that $M$ is the Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }x=0\\2y=0\\2&\text{if }x\neq y\neq 0\end{cases}$$
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Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, H)$X_H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}_H:=\{(t,x)\in\mathbb \text{per}:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$
If $N$ is a smooth submanifold contained in $\text{per}_H,$ \text{per},$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are periodic, we can take $N=\text{per}_H$ N=\text{per}$ in $(\ast),$ which should mean that in such a case $\text{per}_H$ \text{per}$ is a smooth submanifold of $\mathbb R\times M.$
In order to understand this last point I was wondering myself:
If $X$ is a non singular vector field on $M,$ all of whose integral curves of $X_H$ are periodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}_H$ \text{per}$ is a submanifold?
Probably I am missing something because my guess is that if there were a principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$-orbits are the trajectories of $X_H$ X$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
But I don't know how to proceed without this additional hypothesis.
Edit (Added After Thomas Rot's comment)Sebastian's answer): As illustration of my difficulties I shold make explicit the imagine that $M$ is the constant integral curves shouldn't be considered as periodic solutions, i.e Moebius band $[0,1]\times\mathbb R/\sim$ and $X=\frac{\partial}{\partial x}$ then the period should be positive.is $$\tau([(x,y)]_{\sim})=\begin{cases}1&\text{if }x=0\\2&\text{if }x\neq 0\end{cases}$$
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Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}_H:={(t,x)\mid\Phi(t,x)=x}.$\text{per}_H:=\{(t,x)\in\mathbb R\times M\mid\Phi(t,x)=x\}.$
If $N$ is a smooth submanifold contained in $\text{per}_H,$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are closedperiodic, we can take $N=\text{per}_H$ in $(\ast),$ which should mean that in such a case $\text{per}_H$ is a smooth submanifold of $\mathbb R\times M.$
Starting from
In order to understand this last point I was wondering myself:
If all integral curves of $X_H$ are closedperiodic, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}_H$ is a submanifold?
My
Probably I am missing something because my guess is that :if there were principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$ orbits S^1$-orbits are the trajectories of $X_H$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $\zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action. But I don't know how to proceed without this additional hypothesis.
Edit (Added After Thomas Rot's comment): I shold make explicit the the constant integral curves shouldn't be considered as periodic solutions, i.e the period should be positive.
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Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}_H:={(t,x)\mid\Phi(t,x)=x}.$
If $N$ is a smooth submanifold contained in $\text{per}_H,$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are closed, we can take $N=\text{per}_H$ in $(\ast),$ which should mean that in such a case $\text{per}_H$ is a smooth submanifold of $\mathbb R\times M.$
Starting from this I was wondering myself:
If all integral curves of $X_H$ are closed, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}_H$ is a submanifold?
My guess is that:
if there were principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$ orbits are the trajectories of $X_H$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $X_H=\tau\zeta,$ \zeta=\tau X_H,$ where $\zeta$ is the infinitesimal generator of the action.
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Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}_H:={(t,x)\mid\Phi(t,x)=x}.$
If $N$ is a smooth submanifold contained in $\text{per}_H,$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are closed, we can take $N=\text{per}_H$ in $(\ast),$ which should mean that in such a case $\text{per}_H$ is a smooth submanifold of $\mathbb R\times M.$
Starting from this I was wondering myself:
If all integral curves of $X_H$ are closed, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}_H$ is a submanifold?
My guess is that:
if there were principal bundle structure $(M,p,X,\mathbb S^1)$ such that the $\mathbb S^1$ orbits are the trajectories of $X_H$ then the period $\tau:M\to\mathbb R$ should be smooth because of the relation $X_H=\tau\zeta,$ where $\zeta$ is the infinitesimal generator of the action.
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Given a vector field all of whose integral curves are closed, is the period a smooth function?
Disclaimer: I don't know if this question is well suited for this site, but I have posted this question on Math.StackExchange with no answer, so I have thought to post it even here.
I am reading about the energy-period relation for Hamiltonian Systems.
In Weinstein's formulation (cf. Abraham, Marsden, Foundations of Mechanics 2nd Ed, page 198) this relation amounts to:
$(\ast)$ Given an Hamiltonian system $(M,\omega, H)$, let be $\Phi$ the flow of $X_H$ and $\text{per}_H:={(t,x)\mid\Phi(t,x)=x}.$
If $N$ is a smooth submanifold contained in $\text{per}_H,$ then $\left.dt\wedge dH\right|_N=0,$ i.e. $t=t(H)$ on $N,$ (the period depends only on the energy.)
In Guillemin, Stenberg, Geometric Asymptotics, on page 170, I have additionally found that, when all integral curves of $X_H$ are closed, we can take $N=\text{per}_H$ in $(\ast),$ which should mean that in such a case $\text{per}_H$ is a smooth submanifold of $\mathbb R\times M.$
Starting from this I was wondering myself:
If all integral curves of $X_H$ are closed, and $\tau(p)$ denotes the period of the orbit through $p,$ then $\tau:M\to\mathbb R$ is smooth? otherwise how to prove that in such a case $\text{per}_H$ is a submanifold?
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