Suppose $(M, \omega)$ be a symplectic manifold equipped with $G$-action $\Phi : G \times M \to M$. Assume $\Phi^*_g \omega = \omega$.
Let $J$ be an equivariant momentum map of $M$, $J : M \to \mathrm{Lie}(G)^*$. In here, equivariance of $J$ means
\begin{align}
\mathrm{Ad}_g^* J = J \circ \Phi_{g}
\end{align}
Then, does it always hold,
\begin{align}
J_{\mathrm{d}_{J(x)}f}(x) = f \circ J(x)
\end{align} for any $x \in M$ and $f : C^\infty(\mathrm{Lie}(G)^*, \mathbb{R})$? In here, I use the identification,
\begin{align}
\mathrm{d}_{J(x)}f : T\mathrm{Lie}(g)^* \cong \mathrm{Lie}(g)^* \overset{linear}{\to} \mathbb{R}
\end{align}
If $f$ is linear functional, i.e. $f \in \mathrm{Lie}(G)$, then I can show this. But, does it holds for general case?
If it's not true, how can I approach to prove
$$
\{ f \circ J, g \circ J \}_M = \{ f, g \}_{\mathrm{Lie}(G)^*} \circ J
$$ where
$$
\{ f, g\}_M = \omega(X_f, X_g)\\
\{ f, g\}_{\mathrm{Lie}(G)^*} = \mu \mapsto \mu([\mathrm{d}_\mu f , \mathrm{d}_\mu g])
$$?
The reason why I tried to prove the above lemma is,
I get the result that It's enough to prove
$$
\{f \circ J, g \circ J\}_M = \{J_{\mathrm{d}_{J(x)}f}(x), J_{\mathrm{d}_{J(x)}g}(x) \}_M
$$
Best Answer
Your claim that for all $f\in C^{\infty}(\mathfrak{g}^{*})$ and $x\in M$, $$ J_{d_{J(x)}f}(x)=f(J(x)) $$ is not true. For instance, if $f\equiv 1$ then the LHS is zero, while the RHS is equal to $1$.
Note that $J:(M,\{\cdot,\cdot\}_{M})\rightarrow (\mathfrak{g}^{*},\{\cdot,\cdot\}_{\mathfrak{g}^{*}})$ matches the Poisson brackets exactly when for all $f\in C^{\infty}(\mathfrak{g}^{*})$, the Hamiltonian vector fields $X_{f}$ and $X_{J^{*}f}$ are $J$-related. In fact, it suffices to check this condition for functions $f\in C^{\infty}(\mathfrak{g}^{*})$ whose differentials span the cotangent spaces of $\mathfrak{g}^{*}$. Such a collection is given by $\{h_{v}:v\in\mathfrak{g}\}$, where $h_{v}\in C^{\infty}(\mathfrak{g}^{*})$ denotes the linear function on $\mathfrak{g}^{*}$ corresponding with $v\in\mathfrak{g}$.
For all $v\in\mathfrak{g}$, we have $J_{v}=J^{*}(h_{v})$. Taking Hamiltonian vector fields, we get $$ v^{\sharp}=X_{J^{*}(h_{v})}, $$ where $v^{\sharp}$ denotes the fundamental vector field of the action corresponding to $v$.
Show that $v^{\sharp}$ and $X_{h_{v}}$ are $J$-related. This is where equivariance of $J$ comes into play.
We now know that $X_{J^{*}(h_{v})}$ and $X_{h_{v}}$ are $J$-related, so the conclusion follows.