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 $i,j$.


2 Answers 2


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 $g_t=-i_{Y_t}\alpha$.

  • $\begingroup$ 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. $\endgroup$
    – agt
    Apr 14, 2011 at 20:59
  • $\begingroup$ You're welcome! I'll try to explain how i though of point 3 as soon as I have a moment. $\endgroup$ Apr 15, 2011 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 $dH_t$.

  • $\begingroup$ That seems absolutely right. $\endgroup$ Apr 15, 2011 at 6:14
  • $\begingroup$ @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. $\endgroup$
    – agt
    Apr 15, 2011 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.