[proofplan] We prove conservation componentwise. Fixing $\xi \in \mathfrak g$, the scalar function $J_\xi(x)=J(x)(\xi)$ differentiates along a Hamiltonian trajectory by the chain rule. The momentum map identity converts this derivative into a symplectic pairing, and the Hamiltonian vector field identity converts that pairing into $-dH(\xi_M)$. Since $H$ is invariant under the $G$-action, its derivative in every infinitesimal action direction vanishes, so every component $J_\xi$ is constant; separation by all $\xi \in \mathfrak g$ then implies $J$ itself is constant. [/proofplan] [step:Differentiate the Hamiltonian invariance in infinitesimal action directions] Fix $\xi \in \mathfrak g$. Define the infinitesimal generator $\xi_M: M \to TM$ by \begin{align*} \xi_M(x) = \left.\frac{d}{ds}\right|_{s=0} \exp(s\xi)\cdot x. \end{align*} Since $H$ is $G$-invariant, for every $x \in M$ and every $s$ for which the expression is defined, \begin{align*} H(\exp(s\xi)\cdot x) = H(x). \end{align*} Differentiating this identity at $s=0$ gives \begin{align*} dH_x(\xi_M(x)) = 0. \end{align*} Thus $dH(\xi_M)=0$ as a smooth real-valued function on $M$. [guided] Fix an element $\xi \in \mathfrak g$. The Lie algebra element $\xi$ determines a one-parameter subgroup $s \mapsto \exp(s\xi)$ of $G$, and the corresponding infinitesimal generator is the vector field \begin{align*} \xi_M: M \to TM, \quad x \mapsto \left.\frac{d}{ds}\right|_{s=0} \exp(s\xi)\cdot x. \end{align*} This vector field records the velocity at $x$ of the orbit curve obtained by flowing in the group direction $\xi$. The hypothesis that $H$ is $G$-invariant means that $H$ is constant on every $G$-orbit. In particular, for each fixed $x \in M$, the real-valued curve \begin{align*} s \mapsto H(\exp(s\xi)\cdot x) \end{align*} is constant near $s=0$. Differentiating this constant curve at $s=0$ gives \begin{align*} 0 = \left.\frac{d}{ds}\right|_{s=0} H(\exp(s\xi)\cdot x). \end{align*} By the chain rule, the derivative of $H$ at $x$ applied to the velocity of the orbit curve is exactly \begin{align*} \left.\frac{d}{ds}\right|_{s=0} H(\exp(s\xi)\cdot x) = dH_x(\xi_M(x)). \end{align*} Therefore \begin{align*} dH_x(\xi_M(x)) = 0 \end{align*} for every $x \in M$. Since $x$ was arbitrary, this proves $dH(\xi_M)=0$ on all of $M$. [/guided] [/step] [step:Compute the derivative of each momentum component along a Hamiltonian trajectory] Let $I \subset \mathbb R$ be an interval, and let $\gamma: I \to M$ be an integral curve of $X_H$, so \begin{align*} \dot{\gamma}(t) = X_H(\gamma(t)) \end{align*} for every $t \in I$. Define \begin{align*} J_\xi: M \to \mathbb R, \quad x \mapsto J(x)(\xi). \end{align*} For each $t \in I$, the chain rule gives \begin{align*} \frac{d}{dt}(J_\xi \circ \gamma)(t) = d(J_\xi)_{\gamma(t)}(X_H(\gamma(t))). \end{align*} Using the momentum map identity $dJ_\xi=\iota_{\xi_M}\omega$, this becomes \begin{align*} \frac{d}{dt}(J_\xi \circ \gamma)(t) = \omega_{\gamma(t)}(\xi_M(\gamma(t)),X_H(\gamma(t))). \end{align*} By skew-symmetry of the two-form $\omega$, \begin{align*} \omega_{\gamma(t)}(\xi_M(\gamma(t)),X_H(\gamma(t))) = -\omega_{\gamma(t)}(X_H(\gamma(t)),\xi_M(\gamma(t))). \end{align*} Using $\iota_{X_H}\omega=dH$, we obtain \begin{align*} \omega_{\gamma(t)}(X_H(\gamma(t)),\xi_M(\gamma(t))) = dH_{\gamma(t)}(\xi_M(\gamma(t))). \end{align*} The previous step gives $dH_{\gamma(t)}(\xi_M(\gamma(t)))=0$, hence \begin{align*} \frac{d}{dt}(J_\xi \circ \gamma)(t) = 0. \end{align*} Thus $J_\xi \circ \gamma$ is constant on $I$. [/step] [step:Recover conservation of the full momentum map from its scalar components] We have proved that for every $\xi \in \mathfrak g$, the real-valued function \begin{align*} J_\xi \circ \gamma: I \to \mathbb R \end{align*} is constant. Fix $t_0 \in I$. For every $t \in I$ and every $\xi \in \mathfrak g$, \begin{align*} J(\gamma(t))(\xi) = J_\xi(\gamma(t)) = J_\xi(\gamma(t_0)) = J(\gamma(t_0))(\xi). \end{align*} Two elements of $\mathfrak g^*$ are equal exactly when they agree on every $\xi \in \mathfrak g$. Therefore \begin{align*} J(\gamma(t)) = J(\gamma(t_0)) \end{align*} for every $t \in I$. Hence $J \circ \gamma$ is constant along every Hamiltonian trajectory, which proves the theorem. [/step]