symplectic vector field

Symplectic vector fields

For a symplectic manifold $(M,\omega )$, a symplectic vector field is a smooth vector field $X\in \mathcal{X}(M)$ with flow $\Phi$ such that for each $t\in \mathbb{R}$ that ${\Phi }_t$ is defined, then ${\Phi }_t$ is a symplectomorphism. In other words, ${\Phi }_t^{\ast }\omega =\omega$

Closed vector fields

For a vector field $X\in \mathfrak{X}(M)$, the following are equivalent:

• $X$ is symplectic
• ${\mathcal{L}}_X\omega =0$
• $d({\iota }_X\omega )=0$ (i.e. ${\iota }_X\omega$ is closed)

Proof

($1⟹2$) Assume $X$ is symplectic, thus ${\Phi }_t^{\ast }\omega =\omega$. Then for every $p\in M$, $v,w\in T_{p}M$, we have

$$\begin{array}{rlr}({\mathcal{L}}_X\omega {)}_p(v,w)& =\underset{t\to 0}{\lim }\frac{(d{\Phi }_t{)}_p^{\ast }({\omega }_{{\Phi }_t(p)})(v,w)-{\omega }_p(v,w)}{t}& \text{Flow definition of Lie derivative}\\ & =\underset{t\to 0}{\lim }\frac{{\omega }_{{\Phi }_t(p)}((d{\Phi }_t{)}_p(v),(d{\Phi }_t{)}_p(w))-{\omega }_p(v,w)}{t}& \\ & =\underset{t\to 0}{\lim }\frac{({\Phi }_t^{\ast }\omega {)}_p(v,w)-{\omega }_p(v,w)}{t}& \\ & =\underset{t\to 0}{\lim }\frac{{\omega }_p(v,w)-{\omega }_p(v,w)}{t}& \text{Φt is symplectomorphism.}\\ & =0& \end{array}$$

($2⟹3$) For this part we can use Cartan’s magic formula which says:

$${\mathcal{L}}_X\omega =d{\iota }_X\omega +{\iota }_{X}d\omega$$

$\omega$ is a symplectic form and is thus closed, so $d\omega =0$. Thus, we see

$$\begin{array}{rl}{\mathcal{L}}_X\omega & =d{\iota }_X\omega +{\iota }_{X}d\omega \\ 0& =d{\iota }_X\omega +0\end{array}$$

Therefore, $d({\iota }_X\omega )=0$.

It is pretty simple to follow the logic back up to show that both conditions 2 and 3 imply 1.