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The hamiltonian flow box theorem, as stated in Abraham and Marsden's Foundations of Mechanics, says that:

Given an hamiltonian system $(M,\omega,h)$ with $dh(x_0)\neq 0$ for some $x_0$ in $M$, there is a symplectic chart $(U,\phi)$ on $M$ centered at $x_0$ such that $\phi_{\ast}h(x)=h(x_0)+\omega_0(\phi_{\ast}X_h(x_0),x)$, where $\omega_0$ is the canonical symplectic form.

*Question:*I know some different proofs of this theorem, but I would know if, at your knowledge, in the literature there is a proof which uses the Moser's trick as in the proof of the Darboux' theorem.


In Abraham and Marsden there is a proof using the contact structure associated to the symplectic one and its canonical transormations. I know even that it has an extension in a theorem of Cartan which says: Given a $2n$-dimensional symplectic manifold $(M,\omega)$, it is possible to extend to a system of symplectic coordinates on $(M,\omega)$ any set of local functions $f_1,\ldots,f_k,g_1,\ldots,g_l$ on $M$ such that $f_1,\ldots,f_k$ are independent and in involution, $g_1,\ldots,g_l$ are independent and in involution, and $\{f_i,g_j\}=\delta_{ij}$ for any $i,j$.

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2 Answers 2

up vote 6 down vote accepted

My guess would be that you can find such a proof in the literature, since the Moser trick is such a powerful tool, though I don't know where.

Instead let me sketch a proof of the fact that any two Hamiltonian systems $(M_i,\omega_i,h_i)$ are locally isomorphic around non deg. points $x_i\in M_i, i=0,1$ using the Moser trick. That's the statement I see behind the flow box theorem.

First as usal one proves the linearized fact (i.e. any two $2n$ dim vector spaces $V_i$ supplied with non-degenerate two-forms $\omega_i$ and non degenerate one forms $v_i, i=0,1$ are isomorphic). Using this one constructs a local diffeomorphism between $(M_0,\omega_0,h_0)$ and $(M_1,\omega_1,h_1)$ satisfying:

  1. point $x_0$ goes to $x_1$
  2. the symplectic forms coincide on the above points and
  3. the Hamiltonian function $h_0$ goes to $h_1$.

Now we have a manifold $M$ with two symplectic forms $\omega_0,\omega_1$ coinciding at $x\in M$ and one function $h$ which is nondegenerate at $x$. Next you try the usual Moser trick to morph $\omega_1$ to $\omega_2$ with the flow of a time dependent vector field $X_t$, imposing the additional requirement that $X_t$ preserve the function $h$. I.e. $L_{X_t}h=0$. Hence $X_t$ should lie in the $2n-1$ dimensional distribution $\ker dh$.

At some point in the Moser trick one chooses a one-form $\alpha$ such that $d\alpha=\omega_1-\omega_0$, and here we have the freedom to fullfill the additional restriction since we can add any closed one form $df$ to $\alpha$. We want the result $\alpha+df$ to lie in the $2n-1$ dimensional subspace of one-forms satisfyinig $i_{Y_t}\alpha'=0$, where $Y_t$ denotes the Hamiltonian vector field associated to $h$ w.r.t. the symplectic structure $\omega_t$. This can always be achieved since $Y_t$ is non deg and you are solving the equation $Y_t(f)=g_t$ where $g_t=-i_{Y_t}\alpha$.

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Dear Michael, Thank you very much. Your answer is very enlightening and properly what I was searching for. I have tried to read it careful. In order to express you my appreciation I have posted an answer in which I write what I have understood with just a slight modification about your condition 3. –  Giuseppe Tortorella Apr 14 '11 at 20:59
    
You're welcome! I'll try to explain how i though of point 3 as soon as I have a moment. –  Michael Bächtold Apr 15 '11 at 6:19

Warning: I have posted this as an answer and not as comment, not to gain in reputation, but just in order to have enough space to write to Michael what I had understood of his answer that has been very much useful to me.

Let $(M_0,\omega_1)$ and $(M_1,\omega_1)$ be symplectic manifolds of the same dimension. If $h_i$ is a smooth function on $M_i$ with $dh_i(x_i)\neq 0$ for some $x_i\in M_i$, and $i=0,1$, then there exists a local diffeomorphism $\phi$ from $M_0$ to $M_1$ such that $\phi(x_0)=x_1, \phi_{\ast}\omega_0=\omega_1$, and $d\phi_{\ast}h_0=dh_1$.

For the result from linear algebra reported in Michael's answer, there a local diffeomorphism $\psi$ from $M_0$ to $M_1$ such that $\psi(x_0)=x_1, \psi_{\ast}\omega_0(x_1)=\omega_1(x_1)$, and $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.

So we have now a manifold $M$ with symplectic forms $\Omega_0$ and $\Omega_1$ coinciding at a point $x_0$, and smooth regular functions $H_0$ and $H_1$ with $dH_0(x_0)=dH_1(x_1)$. With no loss of generality we can assume also $H_0(x_0)=H_1(x_0)$.

Let us introduce the following time dependent forms on $M$:
$H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.

In order to construct the required local diffeomorphism using the Moser's trick, we need a time dependent local vector field $X_t$ around $x_0$ satisfying: $di_{X_t}\Omega_t+\tilde{\Omega}=0$, $i_{X_t}dH_t+\tilde{H}=0$, $X_t(x_0)=0$, for $t\in\[0,1\]$. Really the third condition follows from the second one because $\tilde{H}(x_0)=0$ and $H_t$ is regular.

Let $\alpha$ be a local primitive of $\tilde{\Omega}$ vanishing at $x_0$. The first condition becomes $i_{X_t}\Omega_t=-\alpha+df_t$, and determines a unique $X_t$ for each smooth function $f=\{f_t\}_t$ on a neighborhood of $M\times\[0,1\]$.

Finally the second condition becomes the following one only on $f=\{f_t\}_t$: $\mathcal{L}(Y_t).(f_t)=g_t\equiv\tilde{H}-i_{Y_t}\alpha$. Where $Y_t$ is the Hamiltonian vector field corresponding to $H_t$ w.r.t. $\Omega_t$.

A solution for this equation $\mathcal{L}(Y_t+0\frac{\partial}{\partial t}).f=g$ could be constructed using the method of characterics, considering that $Y\equiv Y_t+0\frac{\partial}{\partial t}$ is non singular because such is $dH_t$.

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That seems absolutely right. –  Michael Bächtold Apr 15 '11 at 6:14
    
@Michael: If the final step invoking the method of characteristic is correct, then you could also manage the case when $H$ is replaced by a set of functions that are independent and in involution. So you take in account an hamiltonian systems with a set of its first integral. –  Giuseppe Tortorella Apr 15 '11 at 6:25

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