Not sure if this is too much physics to be here...

Consider $$H:\mathbb{R}^{2N+1}\rightarrow\mathbb{R}$$ of class $C^2$, let $H(x,y,z)$ such that $x\in\mathbb{R}^N$, $y\in\mathbb{R}^N$ and $z\in\mathbb{R}$. Let $\varphi$ be the flow associate with the Hamiltonian system $$\dot{x}_i=-\frac{\partial H}{\partial y_i}$$ $$\dot{y}_i=\frac{\partial H}{\partial x_i}$$ $$\dot{z}=1$$ I have to prove that if $\eta$ is a 1-form given by $\eta=\sum_{i=1}^Nx_i \ dy_i-H \ dz$ and $c$ is a closed curve in $\mathbb{R}^{2N+1}$, then for all $s$ we have $$\int_{\varphi(s,c)}\eta=\int_c\eta.$$

Thank you in advance!

  • $\begingroup$ How is this a Hamiltonian system? As presented down here the phase space is not even-dimensional. What is the symplectic form associated to it? $\endgroup$ – Novo Jul 29 '13 at 14:54
  • $\begingroup$ I agree with @Novo that this system is not Hamiltonian in usual sense. As an advice: since $\varphi(s, \dot)$ is a diffeomorphism, you can do the change of variables in l.h.s. integral, returning back to the parameterisation of curve $c$. $\endgroup$ – Evgeny Jul 29 '13 at 15:00
  • $\begingroup$ Ok, but if it isn't a Hamiltonian system, then what is? And how can I solve it? And thank you for the observations. $\endgroup$ – diff_math Jul 29 '13 at 15:39
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    $\begingroup$ The system is a time-dependent Hamiltonian system. The $x-y$ subspace has a symplectic form. You can see this in the form of the equations for the $\dot{x}$ and $\dot{y}$. $\endgroup$ – Robert Lewis Jul 29 '13 at 18:12
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    $\begingroup$ What is $\varphi(s,c)$? $\endgroup$ – user7530 Jul 29 '13 at 18:14

By Stokes $\int_c\eta = \int_D d \eta$ for any disc $D$ with $\partial D = c$ (which exists because all curves are contractible in $\mathbb{R}^{2n+1}$), and then also $\int_{\phi(s,c)} \eta = \int_{\phi(s,D)} d \eta$

Thus it suffices to show that Lie drivative of $d \eta$ is $0$.

By Cartan's magic formula

$$L_X d \eta = d (i_X (d\eta)) = d (i_X \omega -i_X (dH \wedge dz))$$

We have $i_X dz =1$.

We then compute $i_X \omega = dH - \frac{\partial H}{\partial z} dz$

and thus $d H = i_X \omega + \frac{\partial H}{\partial z} dz$ which contracted with $X$ again gives

$i_X dH = i_X (\frac{\partial H}{\partial z} dz)= \frac{\partial H}{\partial z}.$

Plugging this in we get

$$L_X d \eta=d (i_X \omega -i_X (dH \wedge dz)) = d( dH - \frac{\partial H}{\partial z} dz + (i_X dH) \wedge dz - d H \wedge (i_X dz))= d (0)=0.$$


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