Suppose $M$ is a symplectic toric manifold. This means there is a compact torus
$T$ that has a Hamiltonian action on $M$, with moment map $\mu:M \to \mathfrak t^\*$, and $\dim(M)=2\dim(T)$. Can one tell from the moment polytope $\mu(M)$ whether $M$ is Fano?



Condition on moment polytope for a toric manifold to be Fano
The basic answer is "yes, *of course,* because the toric variety is uniquely determined by the polytope. But no, because it's the wrong polytope for the question of Fanoness."


The question is whether the anticanonical class, not the given class $[\omega]$, is ample. Translate the polytope to contain $0$ in the interior. Now translate the facets in/out from the origin, until they're at lattice distance $1$ from the origin. (I.e. a facet is $\{\vec v \in \mathfrak t^\*\ :\ \langle \vec v, X \rangle = c > 0\}$ for some unique shortest $X \in \ker(\exp: \mathfrak t {\longrightarrow} T)$; move it to $c=1$). In this way you've computed the moment polytope w.r.t. the anticanonical class = the sum of the toric divisors (each with coefficient $c=1$). The question is whether each old facet still defines a facet of the new polytope.


(It's possible that one of these hyperplanes might completely miss the new polytope. That corresponds to the anticanonical bundle having basepoints along that divisor.)


Non-example: the trapezoid giving $\widetilde{\mathbb P^2}$. Then the intersection of the exceptional curve's facet with the resulting unit triangle is just a corner, not a facet. So no basepoints, but the exceptional curve blows down in the anticanonical non-embedding.


