Let $G$ be a connected Lie group acting on a smooth manifold $X$. Let $\mathfrak{g}$ be the Lie algebra of $G$. Consider the induced action on the cotangent space $T^*X$ which is always Hamiltonian with (G-equivariant) moment map denoted by $\mu: T^*X \rightarrow \mathfrak{g}^*$.
Question: Given $\lambda \in \mathfrak{g}^*$ a $G$-invariant point. We have: $\alpha \in \mu^{-1}(\lambda)$ is a smooth point if and only if the differential at $\alpha$, $d_\alpha\mu$ is surjective.
What I have I know, $$Ker(d_\alpha\mu) = T_\alpha ( G\dot\alpha)^{\perp \omega}$$ where $\perp \omega$ means the orthogonal complement with respect to the canonical symplectic form on $T^*M$ and $$Im(d_\alpha\mu)=Lie(G_{\alpha})^\perp$$, where $G_\alpha$ is the stablizier of $\alpha$ in $G$ and $Lie(G_{\alpha})^\perp=\{f \in Lie(G)^* : f(Lie(G_\alpha))=0\}$.
I think (for what I read) that this should be a quite standard result. Could anyone give me a hint or a reference for understanding how to prove it?
The following is an argument which holds tells us that for $f:M\to N$ a smooth map,if $\alpha\in f^{-1}(\lambda)$ where $d_\alpha f$ is surjective, $\alpha$ is a smooth point of $f^{-1}(\lambda)$.
If $d_{\alpha}\mu$ is surjective then there is a neighborhood $U$ of $\alpha$ where $d\mu$ is surjective (by lower semicontinuity of the rank), then $\mu\vert_{U}$ is a submersion. The implicit function theorem says that $\mu\vert_{U}^{-1}(\lambda)=U\cap \mu^{-1}(\lambda)$ is a submanifold of $T^*X$ and hence $\alpha$ is a smooth point of $\mu^{-1}(\lambda)$ by the following argument. Because $\mathcal{O}(\mu^{-1}(\lambda))_{\alpha}\cong\mathcal{O}(\mu\vert_{U}^{-1}(\lambda))_{\alpha}$ we see that the maximal ideals of these rings are isomorphic and hence if $\mathfrak{m}_{\alpha}$ and $\tilde{\mathfrak{m}}_{\alpha}$ are the maximal ideals of these stalks, then $\mathfrak{m}_{\alpha}/\mathfrak{m}_{\alpha}^2\cong \tilde{\mathfrak{m}}_{\alpha}/\tilde{\mathfrak{m}}_{\alpha}^2\cong \mathbb{R}^{\dim M-\dim N}$ as desired.
Here's some intuition behind the dimension given to the submanifold. Let $f: M\to N$ be a submersion ($d_xf$ is surjective for each $x\in M$). Let $V$ be a coordinate neighborhood of $f(x)=\lambda$ in $N$ and $U$ a coordinate neighborhood of $x$ which maps into $V$. We can represent $f$ as a function now $\mathbb{R}^{m}\to\mathbb{R}^n$ and elements of $f^{-1}(\lambda)$ will be solutions of the set of equations \begin{align} f^1(x^1,\cdots, x^m)&=\lambda^1\\ f^2(x^1,\cdots, x^m)&=\lambda^2\\ \vdots\quad&\\ f^n(x^1,\cdots,x^m)&=\lambda^n \end{align} So we can think of $f^{-1}(\lambda)$ as the intersection of $n$ hypersurfaces of codimension $1$, and in the tangent space of the intersection will be given by the intersection of $m$ codimension $1$ hyperplanes, yielding a dimension of $m-n$.
The reverse direction requires more time than I have right now, but I may be able to expand upon this later.