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?
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$\begingroup$ what does it mean for a symplectic manifold to be Fano? or do you want tell if it is Fano for some kahler metric compatible with the symplectic structure? $\endgroup$– Dima SustretovCommented Oct 1, 2015 at 9:32
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1$\begingroup$ @dima-sustretov A symplectic toric manifold can alternately be viewed as a symplectic quotient of a torus acting on $\mathbb C^n$, so it comes with a K\"ahler structure. $\endgroup$– AnonCommented Oct 1, 2015 at 9:50
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$\begingroup$ If you can access the book McDuff, Salamon: J-holomorphic curves and symplectic topology, read Section 11.3.1 on page 409. I think the (positive) answer is there. $\endgroup$– Oldřich SpáčilCommented Oct 1, 2015 at 11:25
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$\begingroup$ It's better to add the condition that $M$ is compact, otherwise you will allow examples like $Bl_0(\mathbb{C}^n)$, which is "Fano" in the sense that $c_1(M)>0$. Also "monotone" is a better word than "Fano" in the symplectic category. $\endgroup$– YHBKJCommented Oct 1, 2015 at 18:02
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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.
answered Oct 1, 2015 at 12:04