1
$\begingroup$

I have a question on the proof of Lemma 1.1.10 of the second edition of McDuff and Salamon's book Introduction to Symplectic Topology:

Lemma 1.1.10: Given a smooth time-dependent Hamiltonian function

\begin{equation} \mathbb{R}\times \mathbb{R}^{2n}\to \mathbb{R}: (t,x,y)\mapsto H_t(x,y) \end{equation}

define $\phi_H^{t,t_0}(z_0) := z(t)$ where $z(t)$ is the solution to the ODE (the Hamilton equation)

\begin{equation} \dot{z} = - J_0 \nabla H(z) \end{equation}

with initial condition $z(t_0) = z_0$. Here,

\begin{equation} J_0 := \begin{pmatrix} 0 & - \mathbf{1} \\ \mathbf{1} & 0 \\ \end{pmatrix} \in M_{2n}(\mathbb{R}) \end{equation}

Then $\phi_H^{t,t_0}$ is a symplectomorphism wherever it is defined.

The proof starts out as follows: Let $z_0\in \mathbb{R}^{2n}$ and define

\begin{equation} \Phi(t) := d \phi_H^{t,t_0}(z_0)\in \mathbb{R}^{2n\times 2n} \end{equation}

Then for every $\zeta_0\in \mathbb{R}^{2n}$, the function $\zeta(t) := \Phi(t) \zeta_0$ is the solution to the linearized differential equation

\begin{equation} \dot{\zeta} = d X_{H_t}(z) \zeta \end{equation}

where

\begin{equation} X_{H_t}(z) := - J_0 \nabla H_t(z) \end{equation}

My question: I am not sure how to deduce this last equation from the Hamilton equation. I have tried expanding $X_{H_t}(z)$ in the variable $z$, but I have not made progress.

$\endgroup$
0
$\begingroup$

You have $\zeta(t)= \frac{d}{ds}|_{s=0}\phi^{t,t_0}_H (z_0 +s\zeta_0)$. Therefore we can compute $$ \dot{\zeta} =\frac{d}{du}|_{u=t}\frac{d}{ds}|_{s=0}\phi^{u,t_0}_H (z_0 +s\zeta_0)=\frac{d}{ds}|_{s=0}X_{H_t}(\phi^{u,t_0}_H (z_0 +s\zeta_0))=dX_{H_t}(z(t))d\phi^{t,t_0}_{H}(z_0)\zeta_0$$

$\endgroup$

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.