Lagrangian flow preserves symplectic form

Question

Let $X$ be a configuration space and $L: TX \rightarrow \mathbb{R}$ a Lagrangian. Then I want to show that the Lagrangian flow $F^t(x(0),x'(0)) = (x(t),x'(t))$ preserves the symplectic form just like the Hamiltonian flow does.

My idea was to use the Euler Lagrange equations.

So let $F^t(x(0) + \varepsilon \xi(0),x'(0)+ \varepsilon \xi'(0)) = (x_{\varepsilon}(t),x'_{\varepsilon}(t))$ be the flow of the Euler Lagrange equations.

Then we have after the usual integration by parts trick that $$dS_t(x(0),x'(0))(\xi(0),\xi'(0)) := d_{\varepsilon}|_{\varepsilon=0} \int_0^t L(x_{\varepsilon}(\tau),x'_{\varepsilon}(\tau)) d\tau = \sum_{k=1}^{n} \partial_2L(F^\tau((x(0),x'(0))) \xi_k(\tau)|_0^t.$$

Note, that here the Euler Lagrange equation vanishes due to the property of the flow being a solution to the equations for all times.

This last equation almost looks like $dS_t = (F^t)^* \theta - \theta,$ where $\theta = \sum_{k=1}^{n} (\partial_2 L)_k dx^k$ which would give us of course the result, because then $0=d(dS_t) = (F^t)^* \omega - \omega$ by the usual commutation of the pullback with the exterior derivative.

For this we would need $(F^t)^*\omega(x(0),x'(0))(\xi(0),\xi'(0)) = \sum_{k=1}^{n} \partial_{x_k'(0)} L(F^t(x(0),x'(0)))\xi_k(t)$ holds which is only the case if this $\xi(t)= dF^t_{x,x'}(\xi(0),\xi'(0))$ according to my notation above, but I don't see that this equality holds.

Answer 1

You should indeed come to the conclusion that $dS_t=F_t^*\theta-\theta$.

Note that your $1$-form $\theta$ can be written in terms of the summation convention as $\theta=\frac{\partial L}{\partial \dot{x}^k}\,dx^k$. Now fix $(x,\dot{x})\in TX$ and let $(x(t),\dot{x}(t))=F^t(x,\dot{x})$. Also fix a tangent vector $(\delta x,\delta\dot{x})\in T_{(x,\dot{x})}X$ and set $$(\delta x(t),\delta\dot{x}(t))=T_{(x,\dot{x})}F^t(\delta x,\delta\dot{x}).$$ The pullback of $\theta$ along $F^t$ can be written

$$[(F^t)^*\theta](\delta x,\delta\dot{x})=\frac{\partial L}{\partial\dot{x}^k}(x(t),\dot{x}(t))\delta x^k(t).$$

Next notice that your function $S_t:TX\rightarrow\mathbb{R}$ can be written

$$S_t(x,\dot{x})=\int_0^tL(x(\tau),\dot{x}(\tau))\,d\tau.$$ As you noticed, only the boundary terms contribute to $dS_t$, in particular, $$dS_t(\delta x,\delta\dot{x})=\frac{\partial L}{\partial \dot{x}^k}(x(t),\dot{x}(t))\delta x^k(t)-\frac{\partial L}{\partial x^k}(x,\dot{x})\delta x^k=(F^{t*}\theta-\theta)(\delta x,\delta\dot{x}).$$