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: enter image description here

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

enter image description here

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?


1 Answer 1


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.


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