[proofplan] The proof is the infinitesimal content of symplectic reduction. First, the reduced Hamiltonian vector field is the projection of the original Hamiltonian vector field along the quotient map $\pi_\mu$. Second, the kernel of $d(\pi_\mu)_m$ is exactly the tangent space to the $G_\mu$-orbit through $m$. Therefore vanishing of the reduced vector field is equivalent to $X_H(m)$ being vertical for the quotient, which is exactly the infinitesimal condition for a relative equilibrium at fixed momentum. [/proofplan] [step:Identify the reduced Hamiltonian vector field as the projection of $X_H$] Since $X_H$ is tangent to $J^{-1}(\mu)$, define the restricted vector field \begin{align*} X_H^\mu:J^{-1}(\mu)&\to T(J^{-1}(\mu)) \end{align*} by \begin{align*} X_H^\mu(m)&=X_H(m) \end{align*} for each $m\in J^{-1}(\mu)$. We claim that $X_H^\mu$ is $\pi_\mu$-related to the reduced Hamiltonian vector field $X_{H_\mu}:M_\mu\to TM_\mu$, meaning \begin{align*} d(\pi_\mu)_m(X_H(m))=X_{H_\mu}(\pi_\mu(m)) \end{align*} for every $m\in J^{-1}(\mu)$. Let $m\in J^{-1}(\mu)$ and let $v\in T_mJ^{-1}(\mu)$. Since $\pi_\mu:J^{-1}(\mu)\to M_\mu$ is a submersion, every tangent vector at $\pi_\mu(m)$ is of the form $d(\pi_\mu)_m(v)$ for some $v\in T_mJ^{-1}(\mu)$. Using $\pi_\mu^*\omega_\mu=i_\mu^*\omega$ and $H_\mu\circ\pi_\mu=H\circ i_\mu$, we compute \begin{align*} \omega_\mu(\pi_\mu(m))(d(\pi_\mu)_mX_H(m),d(\pi_\mu)_mv) = \omega(m)(X_H(m),v). \end{align*} By the defining equation for the Hamiltonian vector field $X_H$, \begin{align*} \omega(m)(X_H(m),v)=dH_m(v). \end{align*} By $H_\mu\circ\pi_\mu=H\circ i_\mu$, \begin{align*} dH_m(v)=d(H_\mu)_{\pi_\mu(m)}(d(\pi_\mu)_mv). \end{align*} Thus \begin{align*} \omega_\mu(\pi_\mu(m))(d(\pi_\mu)_mX_H(m),d(\pi_\mu)_mv) = d(H_\mu)_{\pi_\mu(m)}(d(\pi_\mu)_mv). \end{align*} Since $d(\pi_\mu)_m(v)$ ranges over all of $T_{\pi_\mu(m)}M_\mu$ and $\omega_\mu$ is nondegenerate, this proves \begin{align*} d(\pi_\mu)_m(X_H(m))=X_{H_\mu}(\pi_\mu(m)). \end{align*} [guided] The point of reduction is that the dynamics on the quotient are obtained by pushing forward the original dynamics. We now verify that statement from the definitions. Because $X_H$ is tangent to $J^{-1}(\mu)$, the value $X_H(m)$ lies in $T_mJ^{-1}(\mu)$ for every $m\in J^{-1}(\mu)$. Hence it makes sense to apply $d(\pi_\mu)_m$ to $X_H(m)$. Define \begin{align*} X_H^\mu:J^{-1}(\mu)&\to T(J^{-1}(\mu)) \end{align*} by \begin{align*} X_H^\mu(m)&=X_H(m). \end{align*} We want to show that this restricted vector field projects to $X_{H_\mu}$. Fix $m\in J^{-1}(\mu)$ and $v\in T_mJ^{-1}(\mu)$. The reduced symplectic form is defined by \begin{align*} \pi_\mu^*\omega_\mu=i_\mu^*\omega. \end{align*} Evaluating this identity on the pair $(X_H(m),v)$ gives \begin{align*} \omega_\mu(\pi_\mu(m))(d(\pi_\mu)_mX_H(m),d(\pi_\mu)_mv) = \omega(m)(X_H(m),v). \end{align*} Now use the defining equation for $X_H$: for every tangent vector $w\in T_mM$, \begin{align*} \omega(m)(X_H(m),w)=dH_m(w). \end{align*} Applying this with $w=v$ gives \begin{align*} \omega(m)(X_H(m),v)=dH_m(v). \end{align*} Finally, the reduced Hamiltonian is defined by \begin{align*} H_\mu\circ\pi_\mu=H\circ i_\mu. \end{align*} Differentiating at $m$ in the direction $v\in T_mJ^{-1}(\mu)$ gives \begin{align*} d(H_\mu)_{\pi_\mu(m)}(d(\pi_\mu)_mv)=dH_m(v). \end{align*} Combining the preceding identities yields \begin{align*} \omega_\mu(\pi_\mu(m))(d(\pi_\mu)_mX_H(m),d(\pi_\mu)_mv) = d(H_\mu)_{\pi_\mu(m)}(d(\pi_\mu)_mv). \end{align*} Since $\pi_\mu$ is a submersion, $d(\pi_\mu)_m(v)$ ranges over all vectors in $T_{\pi_\mu(m)}M_\mu$. Therefore $d(\pi_\mu)_mX_H(m)$ satisfies the defining equation for the Hamiltonian vector field of $H_\mu$ at $\pi_\mu(m)$. By nondegeneracy of $\omega_\mu$, \begin{align*} d(\pi_\mu)_m(X_H(m))=X_{H_\mu}(\pi_\mu(m)). \end{align*} [/guided] [/step] [step:Identify the vertical tangent space with the infinitesimal $G_\mu$-orbit] For each $m\in J^{-1}(\mu)$, the quotient projection $\pi_\mu:J^{-1}(\mu)\to M_\mu$ has fibers equal to $G_\mu$-orbits: \begin{align*} \pi_\mu^{-1}(\pi_\mu(m))=G_\mu\cdot m. \end{align*} Since the $G_\mu$-action is free and proper, the quotient theorem for free proper Lie group actions gives that $\pi_\mu$ is a principal $G_\mu$-bundle and a surjective submersion. Hence the vertical tangent space of this principal bundle is \begin{align*} \ker d(\pi_\mu)_m=T_m(G_\mu\cdot m). \end{align*} The orbit map through $m$ is the smooth map \begin{align*} \Phi_m:G_\mu&\to J^{-1}(\mu) \end{align*} defined by \begin{align*} \Phi_m(g)&=g\cdot m. \end{align*} Its differential at the identity sends $\xi\in\mathfrak g_\mu$ to the infinitesimal generator value $\xi_M(m)$. Therefore \begin{align*} T_m(G_\mu\cdot m)=\{\xi_M(m):\xi\in\mathfrak g_\mu\}. \end{align*} Consequently, \begin{align*} \ker d(\pi_\mu)_m=\{\xi_M(m):\xi\in\mathfrak g_\mu\}. \end{align*} [/step] [step:Show that a fixed momentum relative equilibrium gives a reduced equilibrium] Assume that $m$ is a relative equilibrium at momentum $\mu$. Then there exists $\xi\in\mathfrak g_\mu$ such that \begin{align*} X_H(m)=\xi_M(m). \end{align*} By the vertical tangent space computation, \begin{align*} \xi_M(m)\in\ker d(\pi_\mu)_m. \end{align*} Therefore \begin{align*} d(\pi_\mu)_m(X_H(m))=0. \end{align*} Using the projection identity from the first step, \begin{align*} X_{H_\mu}(\pi_\mu(m))=d(\pi_\mu)_m(X_H(m))=0. \end{align*} Thus $\pi_\mu(m)$ is an equilibrium of the reduced Hamiltonian system. [/step] [step:Show that a reduced equilibrium gives a fixed momentum relative equilibrium] Assume that $\pi_\mu(m)$ is an equilibrium of the reduced Hamiltonian system. Then \begin{align*} X_{H_\mu}(\pi_\mu(m))=0. \end{align*} By the projection identity, \begin{align*} d(\pi_\mu)_m(X_H(m))=0. \end{align*} Hence \begin{align*} X_H(m)\in\ker d(\pi_\mu)_m. \end{align*} Using the vertical tangent space computation, there exists $\xi\in\mathfrak g_\mu$ such that \begin{align*} X_H(m)=\xi_M(m). \end{align*} This is precisely the infinitesimal relative equilibrium condition at fixed momentum $\mu$. It remains only to connect the infinitesimal condition with the trajectory statement. Let \begin{align*} \gamma:\mathbb R&\to M \end{align*} be the Hamiltonian trajectory through $m$, so $\gamma(0)=m$ and $\dot{\gamma}(t)=X_H(\gamma(t))$ wherever the trajectory is defined. Let \begin{align*} \sigma:\mathbb R&\to M \end{align*} be the group-orbit curve defined by \begin{align*} \sigma(t)&=\exp(t\xi)\cdot m. \end{align*} Because $H$ is $G$-invariant, the Hamiltonian vector field is $G$-equivariant. Hence, for all $t$ in the common domain of definition, \begin{align*} X_H(\sigma(t))=d(\exp(t\xi))_m(X_H(m)). \end{align*} Using $X_H(m)=\xi_M(m)$ and the fact that the one-parameter subgroup $\exp(t\xi)$ commutes with its generator, this becomes \begin{align*} X_H(\sigma(t))=\xi_M(\sigma(t)). \end{align*} Also, \begin{align*} \dot{\sigma}(t)=\xi_M(\sigma(t)). \end{align*} Thus $\sigma$ solves the Hamiltonian equation with the same initial condition as $\gamma$. By the local existence and uniqueness theorem for smooth ordinary differential equations, applied to the smooth vector field $X_H:M\to TM$, two integral curves with the same initial value agree on every connected component of their common interval of definition containing $0$. Hence \begin{align*} \gamma(t)=\exp(t\xi)\cdot m \end{align*} for every $t$ in the common interval of definition of the Hamiltonian trajectory and the group-orbit curve. Since $\xi\in\mathfrak g_\mu$, the curve $\exp(t\xi)$ lies in $G_\mu$, and therefore \begin{align*} \gamma(t)\in G_\mu\cdot m\subset J^{-1}(\mu). \end{align*} So $m$ is a relative equilibrium whose trajectory stays in the fixed momentum level. [/step] [step:Conclude the equivalence] The preceding two steps prove both implications: \begin{align*} m \text{ is a relative equilibrium at momentum } \mu \iff X_{H_\mu}(\pi_\mu(m))=0. \end{align*} Equivalently, $m$ is a relative equilibrium whose trajectory remains in $J^{-1}(\mu)$ if and only if $\pi_\mu(m)$ is an equilibrium of the reduced Hamiltonian $H_\mu$ on $M_\mu$. [/step]