Sufficient condition for a flow to be symplectic

Question

Suppose $(M,\omega)$ is a symplectic manifold and $X$ a smooth vector field defined on $M$ such that its corresponding flow $\{g_s\}_{s\in\mathbb{R}}$ is defined for all $s$ in sone some $U_s\subset M$. Is it true that if the Lie derivative $\mathcal{L}_X \omega=0$, then the flow is symplectic; i.e. $g_s^*\omega=\omega$?

Answer 1

For simplicity, I'll just assume that the flow of $X$ is defined everywhere; i.e $g:\Bbb{R} \times M \to M$. I think a stronger statement is true as well. If $T$ is any tensor field on $M$, then we have \begin{align} \mathcal{L}_XT = 0 \quad \iff \text{for all $s\in \Bbb{R}$, } g_s^*T = T \end{align} The proof of this follows from the "flow definition" of Lie-derivatives, that $\mathcal{L}_XT := \dfrac{d}{ds}\bigg|_{s=0}g_s^*T$ (everything is of course interpreted pointwise).

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From this definition, the "if" part ($\impliedby$) of the statement is clear. For the "only if" ($\implies$) part, note that since flows have the group property that $g_{s_1 + s_2} = g_{s_1}\circ g_{s_2}$, it follows that for every $\lambda\in \Bbb{R}$, \begin{align} \dfrac{d}{ds}\bigg|_{s=\lambda} g_s^*T &= g_{\lambda}^*(\mathcal{L}_XT) \end{align} (this is just a 1-2 line calculation). So, based on the assumption $\mathcal{L}_XT = 0$, the above identity shows that for every $\lambda\in \Bbb{R}$, $\dfrac{d}{ds}\bigg|_{s=\lambda} g_s^*T = 0$. This means $\lambda \mapsto g_{\lambda}^*T$ is a constant function (notice that we're implicitly using the fact $\Bbb{R}$ is connected here). By evaluating at $\lambda = 0$, we see the "constant" (in this case a constant tensor field with respect to $\lambda$) is $T$.

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From here, you can of course specialize to the case where $T = \omega$ is the symplectic form. But nothing in the proof is simplified by assuming this from the beginning (side note: applying to the case where $T=g$ is the metric tensor from Riemannian geometry shows that the flow of a vector field consists of isometries if and only if $\mathcal{L}_Xg = 0$; i.e Lie derivative of the metric vanishes).