Let $(M,\omega)$ be a symplectic compact connexe manifold endowed with a hamiltonian action of a torus $T$. Let $\mu : M \longrightarrow {Lie(T)}^*,$ be a moment map associated to this action.
We say that a point $m \in M $ is critical point of $\mu$, if $T_m \mu : T_m M \longrightarrow {Lie(T)}^*$, the tangent map of $\mu$ is not surjective.
How can one prove that the above definition of critical points of the moment map $\mu$ is equivalent to the following statements:
(1). $m \in M $ is critical point of $\mu$ if and only if the stabilizer of m, stab(m), contains a sub-torus of T of dimension 1.
(2). The tangent map $T_m \mu$ is not surjective if there exists an element $X \in Lie(T)-\lbrace 0 \rbrace$, such that $\langle T_m \mu, X \rangle = 0$.
Any feedback would be appreciated.
We have $(ker T_m\mu)^\perp =T_mO$, if $O$ is the orbit of the action of the torus, and $\perp$ denote the symplectic orthogonal.
let us accept this basic fact. then the equivalence is obvious : $T_m\mu$ is not surjective iff the co-dimension of its kernel is smalle than the dimension of the image, namely the dimension of the torus, and the dimension of the orbit is smaller than the dimension of the torus iff the stabilizer contains a sub-torus.
The proof of the "basic fact" is well known : $T_mO$ is the set of fondamental vectors, or the image by the transpose of the derivative of the moment moment : $(T \mu )^t= T_mO$. the result follows as the image of the transpose of a linear map is just the orthogonal of the kernel of this map.