Momentum Map-Submersion Let $(M,\omega)$ be a symplectic manifold and $G$ a Lie group acting hamiltonian on $M$, such that the momentum map $\Phi \colon M \to \mathfrak{g}^*$ is $G$-equivariant w.r.t. the coadjoint-action on $\mathfrak{g}^*$.
Assuming that $W := \Phi(M)$ is a manifold, is it true that then $\Phi$ is a submersion on $W$? If not, could someone maybe give an example to see why it's not always the case?
 A: It is not always the case. Notice that it is a question about the (non-)surjectivity of the derivative $\mathrm{d}\Phi$ at some point $m \in M$.

Write $\langle -, - \rangle : \mathfrak{g}^{\ast} \times \mathfrak{g} \to \mathbb{R}$ for the duality pairing. For $m \in M$ and $\xi \in \mathfrak{g}$, we have by definition of a moment map that
$$ \langle \Phi(m), \xi \rangle = H_{\xi}(m)  $$
for some hamiltonian $H_{\xi} \in C^{\infty}(M)$ (note that $\Phi^{\ast} : \mathfrak{g} \to C^{\infty}(M) : \xi \mapsto H_{\xi}$ is the so-called comoment map). Taking derivatives of this relation with respect to $m$, we get the relation between 1-forms on $M$
$$ \langle \mathrm{d}\Phi , \xi \rangle = \mathrm{d}H_{\xi} = X_{\xi} \lrcorner \, \omega $$
where $X_{\xi}$ is the (hamiltonian) vector field associated to the infinitesimal action of $\mathfrak{g}$ on $M$.
The above equality is to be understood as follows : for $m \in M$ and $v \in T_mM$,
$$ \langle \mathrm{d}\Phi_m(v) , \xi \rangle = \omega_m(X_{\xi}(m), v), $$
where we think of $\mathrm{d}\Phi_m(v) \in T_{\Phi(m)}\mathfrak{g}^{\ast} \cong \mathfrak{g}^{\ast}$.
Since $\omega$ is nondegenerate, this relation allows for an implicit description of the image of $\mathrm{d}\Phi_m$. Indeed, consider the annihilator of this image,
$$ N_m := (\mathrm{Im}(\mathrm{d}\Phi_m))^0 := \{ \,  \xi \in \mathfrak{g} \; | \; \langle \mathrm{d}\Phi_m(v), \xi \rangle = 0 \quad \forall v \in T_mM \} . $$
Hence, $\xi \in N_m$ if and only if $X_{\xi} \lrcorner \, \omega = 0$ at $m$, that is (by nondegeneracy of the symplectic form) if and only if $X_{\xi}(m) = 0$. This is precisely the Lie algebra to the stabilizer $G_m$ of $m$. To summarize, $N_m = Lie(G_m)$.

These considerations show that $\Phi : M \to \Phi(M)$ might not be a submersion at those points $m \in M$ where the dimension of the stabilizer increases. Typically, this is the case of those points $m$ whose image by $\Phi$ are on the boundary of $W = \Phi(M)$. You might think this is not a satisfactory counterexample, since the tangent plane to $W$ at a point on its boundary is arguably ill-defined.
Here is a better example, for which I let you work out the details. Consider the diagonal action of $SO(3)$ on $(S^2 \times S^2, \omega_0 \oplus \omega_0)$ that is $M \cdot (x,y) = (Mx, My)$, where $SO(3)$ acts on $S^2$ in the usual way. One can show that this is a hamiltonian action, and the image of the associated moment map is a closed 3-dimensional ball in $\mathfrak{so}(3)^{\ast}$ centered at the origin. The origin is the image by $\Phi$ on those points on the anti-diagonal $\bar{\Delta} = \{(x, -x) \, | \, x \in S^2 \}$. At those points, $\Phi$ is not a submersion.
