[proofplan] Fix an element $\xi\in\mathfrak g$ and compute the derivative of the function $\mu^\xi$ along an arbitrary trajectory of $X_H$. The moment map identity rewrites this derivative as the symplectic pairing $\theta(\xi_M,X_H)$, and skew-symmetry of $\theta$ converts it to $-dH(\xi_M)$ using the Hamiltonian vector field convention. Finally, $G$-invariance of $H$ implies that $H$ has zero derivative in every infinitesimal action direction, so the derivative of $\mu^\xi$ along the Hamiltonian flow vanishes wherever the flow is defined. [/proofplan]

[step:Compute the derivative of a moment map component along a Hamiltonian trajectory] Fix $\xi\in\mathfrak g$ and $p\in M$. Define the open interval \begin{align*} I_p=\{t\in\mathbb R:(t,p)\in D\} \end{align*} and define the trajectory $\gamma_p:I_p\to M$ by \begin{align*} \gamma_p(t)=\varphi_t(p). \end{align*} By the defining property of the flow of $X_H$, \begin{align*} \frac{d}{dt}\gamma_p(t)=X_H(\gamma_p(t)) \end{align*} for every $t\in I_p$. Since $\mu^\xi:M\to\mathbb R$ is smooth, the chain rule gives \begin{align*} \frac{d}{dt}\mu^\xi(\gamma_p(t))=d\mu^\xi_{\gamma_p(t)}\bigl(X_H(\gamma_p(t))\bigr). \end{align*} [/step]

[step:Use the moment map identity and the Hamiltonian convention to identify the derivative]Let $x\in M$ be arbitrary. The moment map convention gives \begin{align*} d\mu^\xi_x\bigl(X_H(x)\bigr)=\theta_x\bigl(\xi_M(x),X_H(x)\bigr). \end{align*} Since $\theta_x$ is skew-symmetric, \begin{align*} \theta_x\bigl(\xi_M(x),X_H(x)\bigr)=-\theta_x\bigl(X_H(x),\xi_M(x)\bigr). \end{align*} The Hamiltonian vector field convention $dH=\iota_{X_H}\theta$ gives \begin{align*} \theta_x\bigl(X_H(x),\xi_M(x)\bigr)=dH_x\bigl(\xi_M(x)\bigr). \end{align*} Therefore \begin{align*} d\mu^\xi_x\bigl(X_H(x)\bigr)=-dH_x\bigl(\xi_M(x)\bigr). \end{align*}[/step]

[guided]We now translate the derivative of $\mu^\xi$ into an expression involving the derivative of $H$. Fix an arbitrary point $x\in M$. The moment map convention in the statement says that the one-form $d\mu^\xi$ is the contraction of $\theta$ with the fundamental vector field $\xi_M$. Concretely, this means that for every vector $v\in T_xM$, \begin{align*} d\mu^\xi_x(v)=\theta_x\bigl(\xi_M(x),v\bigr). \end{align*} Apply this equality with $v=X_H(x)$ to obtain \begin{align*} d\mu^\xi_x\bigl(X_H(x)\bigr)=\theta_x\bigl(\xi_M(x),X_H(x)\bigr). \end{align*} The next move is only a sign change, but it is exactly where the convention matters. Since $\theta_x$ is an alternating [bilinear form](/page/Bilinear%20Form) on $T_xM$, it is skew-symmetric, so \begin{align*} \theta_x\bigl(\xi_M(x),X_H(x)\bigr)=-\theta_x\bigl(X_H(x),\xi_M(x)\bigr). \end{align*} The Hamiltonian vector field $X_H$ is defined in the statement by $dH=\iota_{X_H}\theta$. This means that for every $v\in T_xM$, \begin{align*} dH_x(v)=\theta_x\bigl(X_H(x),v\bigr). \end{align*} Using this with $v=\xi_M(x)$ gives \begin{align*} \theta_x\bigl(X_H(x),\xi_M(x)\bigr)=dH_x\bigl(\xi_M(x)\bigr). \end{align*} Combining these three equalities yields \begin{align*} d\mu^\xi_x\bigl(X_H(x)\bigr)=-dH_x\bigl(\xi_M(x)\bigr). \end{align*} This identity is the algebraic heart of the theorem: conservation of $\mu^\xi$ will follow once we know that $H$ has zero derivative in the infinitesimal symmetry direction $\xi_M$.[/guided]

[step:Differentiate the invariance of $H$ in the infinitesimal action direction] Let $x\in M$ be arbitrary. Define the smooth curve $c_x:\mathbb R\to M$ by \begin{align*} c_x(s)=\exp(s\xi)\cdot x. \end{align*} By definition of the fundamental vector field, \begin{align*} c_x'(0)=\xi_M(x). \end{align*} Since $H$ is $G$-invariant, the function $H\circ c_x:\mathbb R\to\mathbb R$ is constant with value $H(x)$. Differentiating at $s=0$ gives \begin{align*} 0=\left.\frac{d}{ds}\right|_{s=0}H(c_x(s))=dH_x\bigl(c_x'(0)\bigr)=dH_x\bigl(\xi_M(x)\bigr). \end{align*} Thus \begin{align*} dH_x\bigl(\xi_M(x)\bigr)=0 \end{align*} for every $x\in M$. [/step]

[step:Conclude that the moment map component is constant along the flow] For every $t\in I_p$, apply the identity from the preceding steps at the point $x=\gamma_p(t)$. We obtain \begin{align*} \frac{d}{dt}\mu^\xi(\varphi_t(p))=d\mu^\xi_{\gamma_p(t)}\bigl(X_H(\gamma_p(t))\bigr)=-dH_{\gamma_p(t)}\bigl(\xi_M(\gamma_p(t))\bigr)=0. \end{align*} Since $\xi\in\mathfrak g$, $p\in M$, and $t\in I_p$ were arbitrary, every component $\mu^\xi$ is conserved along every trajectory of the Hamiltonian vector field $X_H$ on its flow domain. [/step]