Are the orbits of a symplectic toric manifold the fibers of its moment map? A symplectic toric manifold, by definition, carries an effective torus action generated by a moment map. The orbits of the torus action are of course contained in the fibers of the moment map, but does each orbit have to be a whole fiber?
Looking at Delzant's construction of a toric manifold from its moment polytope, I think the answer is yes, but I'd like to make sure there are no obvious counterexamples before I try to come up with a proof.
An answer either way with a citable reference would be much appreciated.
 A: If a $2n$-dimensional symplectic manifold $(M, \omega)$ carries an effective Hamiltonian action by a $n$-dimensional torus, then it follows from the work of Delzant that, indeed, any fiber of the moment map is an orbit of the action. This is not true however if the torus has dimension smaller than $n$, as is shown by the $S^1$-action on $(M, \omega) = (S^2 \times S^2, \omega_0 \times \omega_0)$ given by $t \cdot (x,y) = (t \cdot x, y)$, where $t \cdot x$ is the usual action by rotation and $\omega_0$ is the standard symplectic form on $S^2$.
On the one hand, consider any Hamiltorian action of a torus $T$ on a symplectic manifold $(M, \omega)$ with associated moment map $\mu$. Given any point $m \in M$, let $T_m \subset T$ denote the stabilizer of $m$. The element $\mu(m)$ belongs to a face of $\mu(M)$ of dimension the codimension of $T_m$ in $T$ ; This follows from Theorem 32.3 in Guillemin-Sternberg's book Symplectic Techniques in Physics. Note that this codimension equals the dimension of the orbit $m^T$. 
On the other hand, for an effective Hamiltonian action by a torus of dimension half the dimension of $M$, Lemma 2.2 in Delzant's article Hamiltoniens périodiques et images convexes de l'application moment states (among other things) that (in the notation above) each fiber $\mu^{-1}(\mu(m))$ of the moment map $\mu$ is a torus (in particular, a connected manifold) of dimension the dimension of the face of $\mu(M)$ to which $\mu(m)$ belongs. From the preceding paragraph, we conclude that the (connected) fiber $\mu^{-1}(\mu(m))$ has the same dimension as the orbit $m^T$ which it contains. This forces equality of the two sets.
N.B. - In fact, Lemma 2.2 in Delzant's article states this result explicitly : $\mu$ is the quotient map of the torus action. 
