Two equivalent definitions of (strongly) Hamiltonian actions?

Question

Let $(M, \omega)$ be symplectic acted on by a Lie group $G$ and suppose $i_{\underline{X}} \omega$ is exact.

Notation:

• For $X \in \frak{g}$, $\underline{X}$ is the fundamental vector field on $M$ associated to $X$.
• $H_X \in \mathcal{C}^\infty(M)$ is the hamiltonian of $X \in \mathfrak{g}$ i.e. $i_{\underline{X}} \omega = d H_X$.
• The moment map $\mu:M \rightarrow \frak{g}^*$ is defined by $\mu(x)(X):= H_X(x)$.

I am used to defining (strongly) Hamiltonian actions with the following conditions, which I have seen in multiple texts:

$$ \omega(\underline{X}, \underline{Y}) = H_{[X,Y]} $$ for any $X,Y \in \frak{g}$, and the association $X \in \mathfrak{g} \mapsto H_X$ is $\mathbb{R}$-linear.

In the 1982 paper of Duistermaat and Heckman, "On the variation..." where they work with an abelian $G$, the condition reduces to: $$ \omega(\underline{X}, \underline{Y}) = 0 $$

What they say is that this is equivalent to the fact that $G$ acts along the fibers of the moment map, which is another way to say: $$ H_X(gx) = H_X(x)$$ for any $x \in M, g \in G, X \in \frak{g}$.

How would one show this?

Answer 1

As commented, when $\mathfrak g$ is abelian and the action is strongly Hamiltonian in the general sense, we have $$ \omega(\underline X,\underline Y)=H_{[X,Y]}=H_0=0 $$ thus the general notion of strongly Hamiltonian action reduces in this case to the condition $\omega(\underline X,\underline Y)=0$.

Next the derivative along $\underline Y$ of the function on $M$ given by $x\mapsto H_{X}(x)$ is $$ \underline Y(H_{X})=\{H_{Y},H_{X}\}=0 $$ because all Hamiltonians are in involution. Hence $H_{X}(x)$ in invariant under the infinitesimal action of $G$ and so $$ H_{X}(gx) $$ does not depend on $g\in G$.

EDIT: to address the comment, more details on the last statement. Any $g\in G$ can be written as $\exp(Y)$ for some $Y\in\mathfrak g$ (here we use that $G$ is connected and abelian). Therefore the function of $t\in[0,1]$ given by $$ f(t):=H_X(\exp(t Y)x) $$ (for any fixed $x\in M$) satisfies $$ f(0)=H_X(x),\qquad f(1)=H_X(gx). $$ However, $f(t)$ is constant because $$ \frac{\mathrm d}{\mathrm d t}f(t) = \frac{\mathrm d }{\mathrm d t} H_X(\exp(t Y)x)=\{H_{Y},H_{X}\}(\exp(t Y)x)=0 $$ and $[0,1]$ is connected.