On the proof of the hamiltonian flow box theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T09:03:03Z http://mathoverflow.net/feeds/question/61697 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61697/on-the-proof-of-the-hamiltonian-flow-box-theorem On the proof of the hamiltonian flow box theorem Giuseppe 2011-04-14T13:16:16Z 2011-04-16T15:19:09Z <p>The hamiltonian flow box theorem, as stated in Abraham and Marsden's Foundations of Mechanics, says that:</p> <p>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.</p> <p>*<em>Question:</em>*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.</p> <hr> <p>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 $i,j$.</p> http://mathoverflow.net/questions/61697/on-the-proof-of-the-hamiltonian-flow-box-theorem/61719#61719 Answer by Michael for On the proof of the hamiltonian flow box theorem Michael 2011-04-14T16:07:14Z 2011-04-14T18:19:27Z <p>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. </p> <p>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.</p> <p>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:</p> <ol> <li>point $x_0$ goes to $x_1$</li> <li>the symplectic forms coincide on the above points and</li> <li>the Hamiltonian function $h_0$ goes to $h_1$.</li> </ol> <p>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 $\ker dh$.</p> <p>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 $g_t=-i_{Y_t}\alpha$.</p> http://mathoverflow.net/questions/61697/on-the-proof-of-the-hamiltonian-flow-box-theorem/61749#61749 Answer by Giuseppe for On the proof of the hamiltonian flow box theorem Giuseppe 2011-04-14T21:01:23Z 2011-04-16T15:19:09Z <p>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.</p> <p>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 $d\phi_{\ast}h_0=dh_1$.</p> <p>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 $d\psi_{\ast}h_0(x_1)=dh_1(x_1)$.</p> <p>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 $H_0(x_0)=H_1(x_0)$.</p> <p>Let us introduce the following time dependent forms on $M$:<br> $H_t=H_0+t\tilde{H}=H_0+t(H_1-H_0)$ and $\Omega_t=\Omega_0+t\tilde{\Omega}=\Omega_0+t(\Omega_1-\Omega_0)$.</p> <p>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.</p> <p>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 $M\times[0,1]$.</p> <p>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. $\Omega_t$.</p> <p>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 $dH_t$.</p>