[proofplan] The regular value hypothesis identifies the tangent space of the level set with $\ker dJ_x$. The moment map identity then identifies this kernel with the symplectic orthogonal of the tangent space to the full $G$-orbit. One inclusion follows by checking that infinitesimal generators from the coadjoint stabilizer are tangent to the level set and annihilate its tangent space. For the reverse inclusion, a vector in the restricted kernel lies in the double symplectic orthogonal of the full orbit tangent space, hence in the full orbit tangent space, and tangency to the level set forces its infinitesimal generator to lie in the [Lie algebra](/page/Lie%20Algebra) of $G_\mu$. [/proofplan]

[step:Identify the level set tangent space with the symplectic orthogonal of the orbit tangent space]Since $\mu$ is a regular value of the smooth map $J:M\to\mathfrak g^*$, the regular level set theorem gives that $J^{-1}(\mu)$ is an embedded submanifold of $M$ and, for every $x\in J^{-1}(\mu)$, \begin{align*} T_xJ^{-1}(\mu)=\ker dJ_x. \end{align*} Define the linear subspace \begin{align*} E_x:=T_x(G\cdot x)\subset T_xM. \end{align*} By definition of the orbit map, this is \begin{align*} E_x=\{\xi_M(x):\xi\in\mathfrak g\}. \end{align*} For $v\in T_xM$ and $\xi\in\mathfrak g$, the moment map identity gives \begin{align*} dJ_x(v)(\xi)=d\langle J,\xi\rangle_x(v)=\omega_x(\xi_M(x),v). \end{align*} Since $\omega_x$ is skew-symmetric, the condition $dJ_x(v)=0$ is equivalent to \begin{align*} \omega_x(v,\xi_M(x))=0 \end{align*} for every $\xi\in\mathfrak g$. Therefore \begin{align*} T_xJ^{-1}(\mu)=E_x^\omega, \end{align*} where \begin{align*} E_x^\omega:=\{v\in T_xM:\omega_x(v,e)=0\text{ for every }e\in E_x\}. \end{align*}[/step]

[guided]The first task is to turn the geometric level-set condition $J(y)=\mu$ into a linear condition at $x$. Because $\mu$ is a regular value of $J:M\to\mathfrak g^*$, the regular level set theorem applies to this smooth finite-dimensional map. Its conclusion is that $J^{-1}(\mu)$ is an embedded submanifold and that its tangent space at $x$ is the kernel of the differential: \begin{align*} T_xJ^{-1}(\mu)=\ker dJ_x. \end{align*} Now define \begin{align*} E_x:=T_x(G\cdot x)\subset T_xM. \end{align*} The orbit map is the smooth map $G\to M$ sending $g$ to $g\cdot x$, and its differential at the identity sends $\xi\in\mathfrak g$ to the fundamental vector $\xi_M(x)$. Hence \begin{align*} E_x=\{\xi_M(x):\xi\in\mathfrak g\}. \end{align*} We next compare $\ker dJ_x$ with the symplectic orthogonal of $E_x$. For $v\in T_xM$ and $\xi\in\mathfrak g$, evaluating the covector $dJ_x(v)\in\mathfrak g^*$ on $\xi$ gives \begin{align*} dJ_x(v)(\xi)=d\langle J,\xi\rangle_x(v). \end{align*} The moment map convention is \begin{align*} d\langle J,\xi\rangle=\iota_{\xi_M}\omega. \end{align*} Therefore \begin{align*} dJ_x(v)(\xi)=\omega_x(\xi_M(x),v). \end{align*} Because $\omega_x$ is skew-symmetric, this vanishes for every $\xi\in\mathfrak g$ exactly when \begin{align*} \omega_x(v,\xi_M(x))=0 \end{align*} for every $\xi\in\mathfrak g$. Since the vectors $\xi_M(x)$ are precisely the elements of $E_x$, this condition is exactly $v\in E_x^\omega$, where \begin{align*} E_x^\omega:=\{v\in T_xM:\omega_x(v,e)=0\text{ for every }e\in E_x\}. \end{align*} Thus \begin{align*} T_xJ^{-1}(\mu)=\ker dJ_x=E_x^\omega. \end{align*}[/guided]

[step:Show stabilizer orbit directions lie in the restricted kernel] Let \begin{align*} \operatorname{ad}_\xi^*\mu:=\left.\frac{d}{dt}\right|_{t=0}\operatorname{Ad}_{\exp(t\xi)}^*\mu\in\mathfrak g^* \end{align*} denote the infinitesimal coadjoint action of $\xi\in\mathfrak g$ on $\mu$. Let \begin{align*} \mathfrak g_\mu:=\{\xi\in\mathfrak g:\operatorname{ad}_\xi^*\mu=0\} \end{align*} be the Lie algebra of $G_\mu$. The tangent space to the $G_\mu$-orbit at $x$ is \begin{align*} T_x(G_\mu\cdot x)=\{\xi_M(x):\xi\in\mathfrak g_\mu\}. \end{align*} Let $\xi\in\mathfrak g_\mu$ and set \begin{align*} u:=\xi_M(x)\in T_xM. \end{align*} Since $J$ is equivariant and $J(x)=\mu$, differentiating the curve \begin{align*} t\mapsto J(\exp(t\xi)\cdot x) \end{align*} at $t=0$ gives \begin{align*} dJ_x(u)=\operatorname{ad}_\xi^*\mu=0. \end{align*} Thus $u\in\ker dJ_x=T_xJ^{-1}(\mu)$. Also $u\in E_x$ by definition of $E_x$. Since $T_xJ^{-1}(\mu)=E_x^\omega$, every $w\in T_xJ^{-1}(\mu)$ satisfies \begin{align*} \omega_x(u,w)=0. \end{align*} Therefore \begin{align*} T_x(G_\mu\cdot x)\subseteq\ker\left(\omega_x|_{T_xJ^{-1}(\mu)}\right). \end{align*} [/step]

[step:Use the double symplectic orthogonal to force a kernel vector into the full orbit direction] Let \begin{align*} v\in\ker\left(\omega_x|_{T_xJ^{-1}(\mu)}\right). \end{align*} Then $v\in T_xJ^{-1}(\mu)$ and \begin{align*} \omega_x(v,w)=0 \end{align*} for every $w\in T_xJ^{-1}(\mu)$. Since $T_xJ^{-1}(\mu)=E_x^\omega$, this says precisely that \begin{align*} v\in (E_x^\omega)^\omega. \end{align*} The finite-dimensional double-orthogonal identity for symplectic complements, [citetheorem:10039], applied to the symplectic [vector space](/page/Vector%20Space) $(T_xM,\omega_x)$ and the subspace $E_x\subset T_xM$, gives \begin{align*} (E_x^\omega)^\omega=E_x. \end{align*} Hence $v\in E_x=T_x(G\cdot x)$. [/step]

[step:Use tangency to the level set to reduce the full orbit direction to a stabilizer orbit direction] Since $v\in E_x$, there exists $\xi\in\mathfrak g$ such that \begin{align*} v=\xi_M(x). \end{align*} Since also $v\in T_xJ^{-1}(\mu)=\ker dJ_x$, we have \begin{align*} 0=dJ_x(v)=dJ_x(\xi_M(x)). \end{align*} Equivariance of $J$ gives, as above, \begin{align*} dJ_x(\xi_M(x))=\operatorname{ad}_\xi^*\mu. \end{align*} Therefore $\operatorname{ad}_\xi^*\mu=0$, so $\xi\in\mathfrak g_\mu$. Consequently \begin{align*} v=\xi_M(x)\in T_x(G_\mu\cdot x). \end{align*} This proves \begin{align*} \ker\left(\omega_x|_{T_xJ^{-1}(\mu)}\right)\subseteq T_x(G_\mu\cdot x). \end{align*} Together with the opposite inclusion, the desired equality follows. [/step]