# Linearized ODE Step in Lemma 1.1.10 of McDuff & Salamon

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.

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