Prove that $\max _{M} f-\min _{M} f$ is invariant with the choice of a $K$-invariant real symplectic form $\omega$ on manifold $M$ Let $X \in \mathfrak{k}:=\operatorname{Lie}(K)$ be a real vector field on a compact connected smooth manifold $M$ with an effective action of a compact real Lie group $K$. By choosing a $K$-invariant real symplectic form $\omega$ on $M$, assume $f \in C^{\infty}(M)_{\mathbb{R}}$ is such that
$$
d f=i_{X} \omega
$$
Show that the value $\max _{M} f-\min _{M} f$ is independent of the choice of $\omega$ as far as $\omega$ defines the same de Rham cohomology class $[\omega]$.
This is a question from Yau's contest in 2019, I was confused by how relate the conditions with the conclusion. From the conditions, we know that $M$ is $2n$ dimensional for some $n $ and $M $ is a symplectic manifold, so it is orientable, and the n fold product of the form $\omega$ must be a multiple of its volume form. I have considered the case $n=1$ and assume $\omega$ just be its volume   form, I conjectured that the conclusion may be deduced from integration along some specific curve, but I am really not sure and do not know how to deal with higher dimension problem?
For reference, I found some theorem about symplectic manifold in GTM218, they may be helpful for solving the problem.
 A: I am not sure this is a complete solution; it might still be interesting and/or useful.
Let $\omega$ and $\tilde{\omega}$ be $K$-invariant symplectic forms in the same cohomology class. I think that, as $K$ is compact, we can assume that there is a $K$-invariant $1$-form $\eta$ such that $$\tilde{\omega}=\omega+\mathrm{d}\eta.$$ Contracting with $X$, we get $\iota_X\tilde{\omega}=\mathrm{d}f+\iota_X\mathrm{d}\eta$. By Cartan's formula $$\iota_X\mathrm{d}\eta=\mathcal{L}_X\eta-\mathrm{d}(\iota_X\eta)$$ but as $\eta$ is $K$-invariant, $\mathcal{L}_X\eta=0$. So, we see that $$\iota_X\tilde{\omega}=\mathrm{d}\left(f-\eta(X)\right).$$ In other words, if $\tilde{f}$ is defined by $\mathrm{d}\tilde{f}=\iota_X\tilde{\omega}$, we have found that $\tilde{f}=f-\eta(X)$ up to a global constant.
Now, what are the critical points of $f$? These are $p\in M$ identified by the condition $\mathrm{d}f(p)=0$. But as $\omega$ is symplectic, $\mathrm{d}f$ vanishes only at the points where $X$ itself is $0$. As $\tilde{\omega}$ is also symplectic, the same can be said for the critical points of $\tilde{f}$. So

*

*$f$ and $\tilde{f}$ have the same critical points;

*the critical points are the zeroes of $X$.

If $p$ is one of these points, we find $f(p)=\tilde{f}(p)+\eta(X)(p)=\tilde{f}(p)$ up to a constant on $M$. In particular, the difference of maximum and minimum of $f$ or $\tilde{f}$ will be the same.
