# Prove that the isotopy generated by a time-dependent symplectic vector field is a symplectomorphism

Let $$M$$ a compact and connected smooth manifold. Suppose $$X_t$$ is a time-dependent symplectic vector field and let $$\phi_t$$ be the isotopy generated by $$X_t$$. Prove that $$\phi_t ∈ Symp(M, \omega)$$ for all $$t$$.

I know that:

Given a symplectic form $$\omega$$ on $$M$$, we denote by $$Symp(M, \omega)$$ the space of symplectomorphisms of $$(M, \omega)$$, that is, diﬀeomorphisms $$\Phi : M \to M$$ such that $$\Phi^∗\omega = \omega$$

A vector field $$X ∈ \scr X(M)$$, is called symplectic if $$i_X\omega$$ is closed as a 1-form on M. A time-dependent vector field $$X_t$$ is called symplectic if $$X_t$$ is a symplectic vector field for all $$t$$

Also:

My try:

I have to prove that $$\phi_t$$ is smooth and $$\phi_t^∗\omega = \omega$$. Since it is an isotopy it is already a smooth map $$M \to M$$ by definition so I guess I just have to prove that $$\phi_t^∗\omega = \omega$$.

So I tried doing this $$\phi_t^*\omega(Y,Z)=\omega (\phi_t(X),\phi_t(Y))$$

but it does not seem to go anywhere. Moreover it feels wrong that $$\phi_t$$ acts on a vector field X, when it should be acting on points of the manifold by definition, but that is what I get by definition of pullback. What is going on here?

Since $$X_t$$ is symplectic, $$i_{X_t}\omega$$ is closed, that means $$d(i_{X_t}\omega)=0$$. So I tried using this formula

to use the given condition and I got:

$$0=d(i_{X_t}\omega)(X_t,Y)={X_t}(i_{X_t}\omega(Y))-Y(i_{X_t}\omega({X_t}))-i_{X_t}\omega([{X_t},Y])$$ $$={X_t}(\omega(X_t,Y))-Y(\omega(X_t,X_t))-i_{X_t}\omega({X_t},[{X_t},Y])= {X_t}(\omega(X_t,Y))-i_{X_t}\omega({X_t},[{X_t},Y])$$

but I am not sure this helps

How should I proceed?

Since $$\phi_0=Id$$, to show $$\phi_t^*\omega = \omega$$, we only need $$\frac{d}{dt}\phi_t^*\omega = 0.$$ But this by definition is Lie derivative. So by the Cartan formula, \begin{align*} \frac{d}{dt}\phi_t^*\omega &= L_{X_t}\omega = (di_{X_t} + i_{X_t}d)\omega = di_{X_t}\omega + i_{X_t} d\omega = 0 + 0 = 0, \end{align*} since $$X_t$$ is a symplectic vector field and $$\omega$$ is a symplectic form.