[proofplan] We first use $G$-invariance of $H$ to show that $dH$ annihilates every infinitesimal generator $\xi_M$. The momentum map identity and the Hamiltonian vector field identity then imply that every component $J_\xi$ has zero derivative along $X_H$, so $J$ is conserved and each level set $J^{-1}(\mu)$ is invariant under the Hamiltonian flow. On the reduced level set, $H$ is constant on $G_\mu$-orbits, hence descends uniquely to a smooth function $H_\mu$. Finally, we compare the defining identities for the Hamiltonian vector fields after pulling back by $\pi_\mu$; the nondegeneracy of the reduced symplectic form identifies the projected vector field with $X_{H_\mu}$. [/proofplan]

[step:Use invariance of $H$ to annihilate infinitesimal generators]For $\xi\in\mathfrak g$, let $\xi_M:M\to TM$ be the infinitesimal generator of the action, so $\xi_M(x)\in T_xM$ is the tangent vector at $t=0$ to the curve $t\mapsto \exp(t\xi)\cdot x$. Since $H$ is $G$-invariant, for every $x\in M$ and every $\xi\in\mathfrak g$ the smooth map $h_{x,\xi}:\mathbb R\to\mathbb R$ defined by \begin{align*} h_{x,\xi}(t)=H(\exp(t\xi)\cdot x) \end{align*} is constant. Differentiating at $t=0$ gives \begin{align*} dH_x(\xi_M(x))=0. \end{align*}[/step]

[guided]Fix $\xi\in\mathfrak g$ and $x\in M$. The infinitesimal generator $\xi_M$ is defined by differentiating the one-parameter subgroup action: \begin{align*} \xi_M(x)=\frac{d}{dt}\Big|_{t=0}\exp(t\xi)\cdot x. \end{align*} To use the $G$-invariance of $H$, define the curve-level function $h_{x,\xi}:\mathbb R\to\mathbb R$ by \begin{align*} h_{x,\xi}(t)=H(\exp(t\xi)\cdot x). \end{align*} Because $H(g\cdot y)=H(y)$ for all $g\in G$ and $y\in M$, this function is constant and equals $H(x)$ for every $t$. Therefore its derivative at $t=0$ is zero. By the chain rule, \begin{align*} 0=h_{x,\xi}'(0)=dH_x(\xi_M(x)). \end{align*} Thus $dH$ vanishes on every tangent direction generated by the [group action](/page/Group%20Action). This is the infinitesimal form of $G$-invariance.[/guided]

[step:Show that every momentum component is constant along the Hamiltonian flow]Let $X_H:M\to TM$ denote the Hamiltonian vector field of $H$, defined by the identity $i_{X_H}\omega=dH$. Let $\gamma:I\to M$ be an integral curve of $X_H$ on an interval $I\subset\mathbb R$, so $\gamma'(t)=X_H(\gamma(t))$ for every $t\in I$. Fix $\xi\in\mathfrak g$. Using the chain rule, the momentum map identity $i_{\xi_M}\omega=dJ_\xi$, the Hamiltonian vector field identity $i_{X_H}\omega=dH$, and skew-symmetry of $\omega$, we obtain \begin{align*} \frac{d}{dt}J_\xi(\gamma(t))=dJ_\xi|_{\gamma(t)}(X_H(\gamma(t))). \end{align*} Also, \begin{align*} dJ_\xi|_{\gamma(t)}(X_H(\gamma(t)))=\omega_{\gamma(t)}(\xi_M(\gamma(t)),X_H(\gamma(t))). \end{align*} By skew-symmetry and the defining identity for $X_H$, \begin{align*} \omega_{\gamma(t)}(\xi_M(\gamma(t)),X_H(\gamma(t)))=-\omega_{\gamma(t)}(X_H(\gamma(t)),\xi_M(\gamma(t)))=-dH_{\gamma(t)}(\xi_M(\gamma(t))). \end{align*} The final expression is zero by the preceding step. Hence \begin{align*} \frac{d}{dt}J_\xi(\gamma(t))=0 \end{align*} for every $\xi\in\mathfrak g$ and every $t\in I$. Since the components $J_\xi$ determine $J$ as an element of $\mathfrak g^*$, the map $t\mapsto J(\gamma(t))$ is constant. In particular, if $\gamma(t_0)\in J^{-1}(\mu)$ for some $t_0\in I$, then $\gamma(t)\in J^{-1}(\mu)$ for every $t\in I$.[/step]

[guided]Let $X_H:M\to TM$ denote the Hamiltonian vector field of $H$, defined by $i_{X_H}\omega=dH$. Let $\gamma:I\to M$ be an integral curve of $X_H$, meaning that $\gamma'(t)=X_H(\gamma(t))$ for all $t\in I$. We want to prove that $J(\gamma(t))$ does not change with $t$. Since $J(\gamma(t))$ takes values in $\mathfrak g^*$, it is enough to test it against an arbitrary $\xi\in\mathfrak g$. For fixed $\xi$, the component function is the smooth map $J_\xi:M\to\mathbb R$ defined by \begin{align*} J_\xi(x)=J(x)(\xi). \end{align*} The chain rule gives \begin{align*} \frac{d}{dt}J_\xi(\gamma(t))=dJ_\xi|_{\gamma(t)}(\gamma'(t))=dJ_\xi|_{\gamma(t)}(X_H(\gamma(t))). \end{align*} Now the momentum map convention says $i_{\xi_M}\omega=dJ_\xi$. Evaluated on the vector $X_H(\gamma(t))$, this gives \begin{align*} dJ_\xi|_{\gamma(t)}(X_H(\gamma(t)))=\omega_{\gamma(t)}(\xi_M(\gamma(t)),X_H(\gamma(t))). \end{align*} The symplectic form is skew-symmetric, so \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*} The Hamiltonian vector field is defined by $i_{X_H}\omega=dH$, so evaluating this identity on $\xi_M(\gamma(t))$ gives \begin{align*} \omega_{\gamma(t)}(X_H(\gamma(t)),\xi_M(\gamma(t)))=dH_{\gamma(t)}(\xi_M(\gamma(t))). \end{align*} Combining the last three displayed identities, \begin{align*} \frac{d}{dt}J_\xi(\gamma(t))=-dH_{\gamma(t)}(\xi_M(\gamma(t))). \end{align*} The previous step proved that $dH(\xi_M)=0$ everywhere, hence \begin{align*} \frac{d}{dt}J_\xi(\gamma(t))=0. \end{align*} Because this holds for every $\xi\in\mathfrak g$, the functional $J(\gamma(t))\in\mathfrak g^*$ has constant value. Therefore the entire momentum map $J$ is conserved along the Hamiltonian flow, not merely one selected component.[/guided]

[step:Descend the invariant Hamiltonian to the reduced space]Let $i_\mu:J^{-1}(\mu)\to M$ denote the inclusion map, let $M_\mu:=J^{-1}(\mu)/G_\mu$ denote the reduced orbit space, and let $\pi_\mu:J^{-1}(\mu)\to M_\mu$ denote the quotient projection. By the [Marsden-Weinstein reduction theorem](/theorems/6847), the free and proper action of $G_\mu$ on the regular level set gives a unique symplectic form $\omega_\mu$ on $M_\mu$ satisfying $\pi_\mu^*\omega_\mu=i_\mu^*\omega$. Because $J$ is $\operatorname{Ad}^*$-equivariant, for $g\in G_\mu$ and $x\in J^{-1}(\mu)$ we have \begin{align*} J(g\cdot x)=\operatorname{Ad}_g^*J(x)=\operatorname{Ad}_g^*\mu=\mu. \end{align*} Thus $G_\mu$ acts on $J^{-1}(\mu)$. Since $H$ is $G$-invariant, for every $g\in G_\mu$ and $x\in J^{-1}(\mu)$, \begin{align*} (H\circ i_\mu)(g\cdot x)=H(g\cdot x)=H(x)=(H\circ i_\mu)(x). \end{align*} Therefore $H\circ i_\mu$ is constant on the fibers of the quotient projection $\pi_\mu$. Since the action of $G_\mu$ on $J^{-1}(\mu)$ is free and proper, $M_\mu=J^{-1}(\mu)/G_\mu$ is equipped with the [quotient topology](/page/Quotient%20Topology) and the unique smooth structure for which the quotient projection \begin{align*} \pi_\mu:J^{-1}(\mu)\to M_\mu \end{align*} is a surjective submersion. We use the smooth quotient descent property for a surjective submersion: a smooth function on the total space that is constant on the fibers descends uniquely to a smooth function on the base. Applying this property to the smooth function $H\circ i_\mu:J^{-1}(\mu)\to\mathbb R$, there is a unique smooth map $H_\mu:M_\mu\to\mathbb R$ such that \begin{align*} H_\mu\circ\pi_\mu=H\circ i_\mu. \end{align*} Uniqueness follows from surjectivity of $\pi_\mu$: if $K_\mu:M_\mu\to\mathbb R$ also satisfies $K_\mu\circ\pi_\mu=H\circ i_\mu$, then for every $y\in M_\mu$ choose $x\in J^{-1}(\mu)$ with $\pi_\mu(x)=y$, and obtain \begin{align*} K_\mu(y)=K_\mu(\pi_\mu(x))=H(i_\mu(x))=H_\mu(\pi_\mu(x))=H_\mu(y). \end{align*}[/step]

[guided]The purpose of this step is to turn the invariant function on the constraint surface into a function on the orbit space. Let $i_\mu:J^{-1}(\mu)\to M$ denote the inclusion map, let $M_\mu:=J^{-1}(\mu)/G_\mu$ denote the reduced orbit space, and let $\pi_\mu:J^{-1}(\mu)\to M_\mu$ denote the quotient projection. By the Marsden-Weinstein reduction theorem, the free and proper action of $G_\mu$ on the regular level set gives a unique symplectic form $\omega_\mu$ on $M_\mu$ satisfying $\pi_\mu^*\omega_\mu=i_\mu^*\omega$. First we check that the constraint surface is preserved by the subgroup that fixes $\mu$. If $g\in G_\mu$ and $x\in J^{-1}(\mu)$, equivariance of $J$ gives \begin{align*} J(g\cdot x)=\operatorname{Ad}_g^*J(x)=\operatorname{Ad}_g^*\mu=\mu. \end{align*} Thus $g\cdot x\in J^{-1}(\mu)$, so $G_\mu$ acts on $J^{-1}(\mu)$. Now restrict $H$ to the level set through the inclusion $i_\mu:J^{-1}(\mu)\to M$. For $g\in G_\mu$ and $x\in J^{-1}(\mu)$, the $G$-invariance of $H$ gives \begin{align*} (H\circ i_\mu)(g\cdot x)=H(g\cdot x)=H(x)=(H\circ i_\mu)(x). \end{align*} Therefore $H\circ i_\mu$ is constant on each $G_\mu$-orbit, and the fibers of $\pi_\mu:J^{-1}(\mu)\to M_\mu$ are exactly those orbits. The hypotheses that the $G_\mu$-action is free and proper are used here: they imply that $M_\mu=J^{-1}(\mu)/G_\mu$ carries the quotient topology and the unique smooth structure for which $\pi_\mu$ is a surjective submersion. For a surjective submersion, a smooth function on the total space descends to a smooth function on the base exactly when it is constant on fibers. Applying this descent property to $H\circ i_\mu$ produces a unique smooth map $H_\mu:M_\mu\to\mathbb R$ satisfying \begin{align*} H_\mu\circ\pi_\mu=H\circ i_\mu. \end{align*} The uniqueness is forced by surjectivity. If $K_\mu:M_\mu\to\mathbb R$ also satisfies $K_\mu\circ\pi_\mu=H\circ i_\mu$, then for any $y\in M_\mu$ we choose $x\in J^{-1}(\mu)$ with $\pi_\mu(x)=y$ and compute \begin{align*} K_\mu(y)=K_\mu(\pi_\mu(x))=H(i_\mu(x))=H_\mu(\pi_\mu(x))=H_\mu(y). \end{align*} Thus $K_\mu=H_\mu$.[/guided]

[step:Show that $X_H$ projects to a vector field on the quotient]Since $\mu$ is a regular value of $J$, the [regular value theorem](/theorems/3903) implies that $J^{-1}(\mu)$ is an embedded submanifold of $M$, with inclusion map $i_\mu:J^{-1}(\mu)\to M$. The conservation of $J$ implies that $X_H(x)\in T_xJ^{-1}(\mu)$ for every $x\in J^{-1}(\mu)$, so the restriction $Y:J^{-1}(\mu)\to TJ^{-1}(\mu)$ defined by $Y(x)=X_H(x)$ is a smooth vector field on $J^{-1}(\mu)$. We next show that $Y$ is $G_\mu$-invariant. For $g\in G_\mu$, let $L_g:M\to M$ be the action diffeomorphism defined by \begin{align*} L_g(x)=g\cdot x. \end{align*} The action is symplectic because it is Hamiltonian, so $L_g^*\omega=\omega$, and $H\circ L_g=H$ by $G$-invariance. For $x\in M$ and $v\in T_{g\cdot x}M$, \begin{align*} \omega_{g\cdot x}\bigl((dL_g)_xX_H(x),v\bigr)=\omega_x\bigl(X_H(x),(dL_{g^{-1}})_{g\cdot x}v\bigr). \end{align*} Using $i_{X_H}\omega=dH$, \begin{align*} \omega_x\bigl(X_H(x),(dL_{g^{-1}})_{g\cdot x}v\bigr)=dH_x\bigl((dL_{g^{-1}})_{g\cdot x}v\bigr). \end{align*} Since $H\circ L_{g^{-1}}=H$, the chain rule gives \begin{align*} dH_x\bigl((dL_{g^{-1}})_{g\cdot x}v\bigr)=dH_{g\cdot x}(v). \end{align*} Thus $(dL_g)_xX_H(x)$ satisfies the defining equation for $X_H(g\cdot x)$. By nondegeneracy of $\omega$, \begin{align*} (dL_g)_xX_H(x)=X_H(g\cdot x). \end{align*} Restricting to $g\in G_\mu$ and $x\in J^{-1}(\mu)$ shows that $Y$ is $G_\mu$-invariant. Since the $G_\mu$-action on $J^{-1}(\mu)$ is free and proper, $\pi_\mu:J^{-1}(\mu)\to M_\mu$ is a principal $G_\mu$-bundle. The quotient projectability theorem for principal bundles says that a smooth vector field on the total space is $\pi_\mu$-projectable exactly when its flow preserves fibers, equivalently when it is invariant under the principal action in the sense that $(dL_g)_xY(x)=Y(g\cdot x)$ for every $g\in G_\mu$ and $x\in J^{-1}(\mu)$. Hence $Y$ is $\pi_\mu$-projectable. Let $\overline{Y}:M_\mu\to TM_\mu$ be the projected vector field, characterized by \begin{align*} (d\pi_\mu)_xY(x)=\overline{Y}(\pi_\mu(x)). \end{align*}[/step]

[guided]We first need a vector field on the constraint surface. Since $\mu$ is a regular value of $J$, the regular value theorem gives that $J^{-1}(\mu)$ is an embedded submanifold of $M$ with inclusion $i_\mu:J^{-1}(\mu)\to M$. Conservation of $J$ says that an integral curve of $X_H$ starting in $J^{-1}(\mu)$ stays in $J^{-1}(\mu)$, which is equivalent to \begin{align*} X_H(x)\in T_xJ^{-1}(\mu) \end{align*} for every $x\in J^{-1}(\mu)$. Therefore the restriction $Y:J^{-1}(\mu)\to TJ^{-1}(\mu)$ defined by $Y(x)=X_H(x)$ is a smooth vector field on the level set. Next we verify the invariance condition required for projectability. For $g\in G_\mu$, let $L_g:M\to M$ be the action diffeomorphism $L_g(x)=g\cdot x$. Because the action is Hamiltonian, it is symplectic, so $L_g^*\omega=\omega$. Because $H$ is $G$-invariant, $H\circ L_g=H$ and also $H\circ L_{g^{-1}}=H$. For $x\in M$ and $v\in T_{g\cdot x}M$, symplecticity gives \begin{align*} \omega_{g\cdot x}((dL_g)_xX_H(x),v)=\omega_x(X_H(x),(dL_{g^{-1}})_{g\cdot x}v). \end{align*} The defining equation $i_{X_H}\omega=dH$ gives \begin{align*} \omega_x(X_H(x),(dL_{g^{-1}})_{g\cdot x}v)=dH_x((dL_{g^{-1}})_{g\cdot x}v). \end{align*} Differentiating $H\circ L_{g^{-1}}=H$ at $g\cdot x$ gives \begin{align*} dH_x((dL_{g^{-1}})_{g\cdot x}v)=dH_{g\cdot x}(v). \end{align*} Thus $(dL_g)_xX_H(x)$ satisfies the same Hamiltonian vector field identity at $g\cdot x$ as $X_H(g\cdot x)$. Since $\omega$ is nondegenerate, the solution to that identity is unique, so \begin{align*} (dL_g)_xX_H(x)=X_H(g\cdot x). \end{align*} Restricting to $g\in G_\mu$ and $x\in J^{-1}(\mu)$ proves that $Y$ is $G_\mu$-invariant. Now we use the precise quotient result. The free and proper action of $G_\mu$ on $J^{-1}(\mu)$ makes $\pi_\mu:J^{-1}(\mu)\to M_\mu$ a principal $G_\mu$-bundle. The quotient projectability theorem for principal bundles says that a smooth vector field on the total space is projectable to the base exactly when it is invariant under the principal action; here invariance means $(dL_g)_xY(x)=Y(g\cdot x)$ for every $g\in G_\mu$ and $x\in J^{-1}(\mu)$. This condition is precisely what makes the value of $(d\pi_\mu)_xY(x)$ independent of the representative $x$ chosen in a fiber. Since $Y$ is invariant, there is a unique smooth vector field $\overline{Y}:M_\mu\to TM_\mu$ such that \begin{align*} (d\pi_\mu)_xY(x)=\overline{Y}(\pi_\mu(x)) \end{align*} for every $x\in J^{-1}(\mu)$.[/guided]

[step:Identify the projected vector field with the reduced Hamiltonian vector field]Let $X_{H_\mu}:M_\mu\to TM_\mu$ denote the Hamiltonian vector field of $H_\mu$ on the symplectic manifold $(M_\mu,\omega_\mu)$, defined by $i_{X_{H_\mu}}\omega_\mu=dH_\mu$. We prove that $\overline{Y}=X_{H_\mu}$. Fix $x\in J^{-1}(\mu)$ and let $z=\pi_\mu(x)\in M_\mu$. Let $w\in T_zM_\mu$. Since $\pi_\mu$ is a submersion, choose $v\in T_xJ^{-1}(\mu)$ such that \begin{align*} (d\pi_\mu)_xv=w. \end{align*} Using the defining relation $\pi_\mu^*\omega_\mu=i_\mu^*\omega$, we get \begin{align*} \omega_\mu|_z(\overline{Y}(z),w)=\omega_x((di_\mu)_xY(x),(di_\mu)_xv). \end{align*} Since $(di_\mu)_xY(x)=X_H(x)$ and $i_{X_H}\omega=dH$, \begin{align*} \omega_x((di_\mu)_xY(x),(di_\mu)_xv)=dH_x((di_\mu)_xv). \end{align*} Because $H_\mu\circ\pi_\mu=H\circ i_\mu$, the chain rule gives \begin{align*} dH_x((di_\mu)_xv)=dH_\mu|_z((d\pi_\mu)_xv)=dH_\mu|_z(w). \end{align*} Therefore \begin{align*} \omega_\mu|_z(\overline{Y}(z),w)=dH_\mu|_z(w) \end{align*} for every $w\in T_zM_\mu$. By the defining equation and uniqueness of the Hamiltonian vector field on the symplectic manifold $(M_\mu,\omega_\mu)$, \begin{align*} \overline{Y}=X_{H_\mu}. \end{align*}[/step]

[guided]The goal is to prove that the vector field obtained by projecting $X_H$ is exactly the Hamiltonian vector field of the descended Hamiltonian $H_\mu$. Let $X_{H_\mu}:M_\mu\to TM_\mu$ denote the Hamiltonian vector field of $H_\mu$ on $(M_\mu,\omega_\mu)$, defined by $i_{X_{H_\mu}}\omega_\mu=dH_\mu$. Let \begin{align*} \overline{Y}:M_\mu&\to TM_\mu \end{align*} be the vector field defined by \begin{align*} (d\pi_\mu)_xY(x)=\overline{Y}(\pi_\mu(x)), \end{align*} where $Y=X_H|_{J^{-1}(\mu)}$. Fix a point $x\in J^{-1}(\mu)$ and write $z=\pi_\mu(x)$. To verify the Hamiltonian vector field identity at $z$, take an arbitrary tangent vector $w\in T_zM_\mu$. Since $\pi_\mu$ is a submersion, its differential \begin{align*} (d\pi_\mu)_x:T_xJ^{-1}(\mu)\to T_zM_\mu \end{align*} is surjective, so there exists $v\in T_xJ^{-1}(\mu)$ with $(d\pi_\mu)_xv=w$. Now use the definition of the reduced symplectic form. The identity \begin{align*} \pi_\mu^*\omega_\mu=i_\mu^*\omega \end{align*} means that for tangent vectors $a,b\in T_xJ^{-1}(\mu)$, \begin{align*} \omega_\mu|_{\pi_\mu(x)}((d\pi_\mu)_xa,(d\pi_\mu)_xb)=\omega_x((di_\mu)_xa,(di_\mu)_xb). \end{align*} Apply this with $a=Y(x)$ and $b=v$. Since $Y(x)=X_H(x)$ as a vector tangent to the level set, this yields \begin{align*} \omega_\mu|_z(\overline{Y}(z),w)=\omega_x(X_H(x),(di_\mu)_xv). \end{align*} The Hamiltonian vector field identity for $X_H$ gives \begin{align*} \omega_x(X_H(x),(di_\mu)_xv)=dH_x((di_\mu)_xv). \end{align*} Finally, differentiate the defining relation for the reduced Hamiltonian, \begin{align*} H_\mu\circ\pi_\mu=H\circ i_\mu. \end{align*} The chain rule gives \begin{align*} dH_\mu|_z((d\pi_\mu)_xv)=dH_x((di_\mu)_xv). \end{align*} Since $(d\pi_\mu)_xv=w$, we have \begin{align*} dH_x((di_\mu)_xv)=dH_\mu|_z(w). \end{align*} Combining the identities, \begin{align*} \omega_\mu|_z(\overline{Y}(z),w)=dH_\mu|_z(w). \end{align*} This holds for every $w\in T_zM_\mu$. The two-form $\omega_\mu$ is symplectic, hence nondegenerate, so there is only one vector at $z$ satisfying this identity. Therefore $\overline{Y}(z)=X_{H_\mu}(z)$. Since $z$ was arbitrary, $\overline{Y}=X_{H_\mu}$ on $M_\mu$.[/guided]

[step:Conclude that the Hamiltonian flow descends to the reduced flow]Let $\Phi_t^H:M\to M$ denote the local Hamiltonian flow of $X_H$, and let $\Phi_t^{H_\mu}:M_\mu\to M_\mu$ denote the local Hamiltonian flow of $X_{H_\mu}$. The conservation of $J$ shows that $\Phi_t^H$ preserves $J^{-1}(\mu)$ whenever the flow is defined. Since $X_H|_{J^{-1}(\mu)}$ is $\pi_\mu$-related to $X_{H_\mu}$, uniqueness of integral curves gives \begin{align*} \pi_\mu(\Phi_t^H(x))=\Phi_t^{H_\mu}(\pi_\mu(x)) \end{align*} for every $x\in J^{-1}(\mu)$ and every time $t$ for which both sides are defined. Thus the Hamiltonian flow of $H$ on $J^{-1}(\mu)$ projects to the Hamiltonian flow of $H_\mu$ on $(M_\mu,\omega_\mu)$, completing the proof.[/step]

[guided]Let $\Phi_t^H:M\to M$ be the local flow of $X_H$, and let $\Phi_t^{H_\mu}:M_\mu\to M_\mu$ be the local flow of $X_{H_\mu}$. Conservation of $J$ proves that if $x\in J^{-1}(\mu)$ and $\Phi_t^H(x)$ is defined, then \begin{align*} \Phi_t^H(x)\in J^{-1}(\mu). \end{align*} Thus the original Hamiltonian flow restricts to a local flow on the constraint surface. The previous step proved that the restricted vector field $Y=X_H|_{J^{-1}(\mu)}$ is $\pi_\mu$-related to $X_{H_\mu}$, meaning \begin{align*} (d\pi_\mu)_xY(x)=X_{H_\mu}(\pi_\mu(x)) \end{align*} for every $x\in J^{-1}(\mu)$. Fix $x\in J^{-1}(\mu)$. Let $I_x\subset\mathbb R$ denote the interval of definition of the curve $t\mapsto \Phi_t^H(x)$, and define the curve $c:I_x\to M_\mu$ by \begin{align*} c(t)=\pi_\mu(\Phi_t^H(x)) \end{align*} for every $t\in I_x$. Differentiating with the chain rule and using the projectability relation gives \begin{align*} c'(t)=(d\pi_\mu)_{\Phi_t^H(x)}X_H(\Phi_t^H(x))=X_{H_\mu}(\pi_\mu(\Phi_t^H(x)))=X_{H_\mu}(c(t)). \end{align*} Also $c(0)=\pi_\mu(x)$. Therefore $c$ is an integral curve of $X_{H_\mu}$ with initial value $\pi_\mu(x)$. By uniqueness of integral curves for smooth vector fields, \begin{align*} \pi_\mu(\Phi_t^H(x))=\Phi_t^{H_\mu}(\pi_\mu(x)) \end{align*} for every time $t$ for which both sides are defined. This is exactly the statement that the Hamiltonian flow of $H$ on $J^{-1}(\mu)$ projects to the Hamiltonian flow of $H_\mu$ on the reduced symplectic manifold $(M_\mu,\omega_\mu)$.[/guided]