From Delzant polytope to lattice polytope

Question

By definition, an $n$-dimensional Delzant polytope $P$ is not necessarily a lattice polytope. But is there a natural way (or operations) to turn $P$ into a lattice polytope using the fact that the edge vectors incident to any vertex of $P$ form an integral basis of $\mathbb{Z}^n$? And what do these operations mean in symplectic geometry/topology?

The background of this question is: when people explain the equivalence between symplectic toric manifold and smooth projective toric variety, they often assume the moment polytope to be both lattice and Delzant. So I feel there must be a reason for this assumption.

Answer 1

There is a way to turn a Delzant polytope into a lattice polytope, but there is no natural or canonical way. Note that the Delzant condition implies that the normal vector to each facet (codimension 1 face) is integral. By letting the defining equation for the hyperplane containing each facet rational (by slightly translating hyperplanes), you can make the coordinates of all vertices rational without changing the combinatorial structure. Now the polytope is integral after scaling. As you see, this is far from being canonical.

The above operation changes the cohomology class represented by the symplectic form. So you can think of it as choosing a different symplectic form on the same smooth manifold. If you view a symplectic toric manifold as a symplectic reduction, you are taking a different regular value of the moment map.

Some people regard symplectic toric manifolds as smooth projective varieties, but strictly speaking, that is wrong. Any symplectic toric manifold is diffeomorphic to a smooth projective variety, but it may not be symplecally embedded into a projective space equipped with the Fubini-Study form $\omega_{FS}$. In fact, by Kodaira embedding theorem, a symplectic toric manifold $(M, \omega)$ can be (symplectically) embedded into $\left(\mathbb{P}^N, \omega_{FS}\right)$ for some $N$ if and only if $[\omega] \in H^2(M, \mathbb{Z})$. Even if we allow scalar multiples of $\omega_{FS}$, the class $[\omega]$ can only be a multiple of an integral one.

For an easy example, consider $(S^2 \times S^2, \sigma \oplus \lambda\sigma)$ where $\sigma$ is a volume form on $S^2$ and $\lambda>0$ is irrational. It can never be embedded into a projective space preserving the symplectic structure. You can take rational $\lambda$ to find an embedding, but then you are dealing with a different symplectic manifold.

Answer 2

If you have a lattice polytope then you get a projective toric variety. If your polytope is not a lattice polytope then you can still construct a symplectic toric manifold (e.g. by performing symplectic cuts) but it won't be the underlying symplectic manifold of a projective variety.

To see this, note that the cohomology class of the symplectic form on a projective toric variety is integral because its ultimately pulls back from the Fubini-Study form on projective space. If your polytope is not lattice then one of the edges has non-integer affine length, so the symplectic area of the corresponding sphere is not an integer, so there cohomology class of the symplectic form is not integral.

Let $a_1,\ldots,a_N$ be the affine lengths of the edges of your polytope. If the ratios $a_i/a_j$ are all rational then you can rescale to make them integers and this rescaling of your polytope will be a lattice polytope.

Otherwise, I guess (though I didn't check) you should be able to change the constants in the inequalities that define your polytope by an arbitrarily small amount to get these ratios to be rational, and hence find a small deformation of your symplectic toric manifold which is (a rescaling of) a projective toric variety.