Question

A Delzant polytope in R^n by definition is a simple, rational, and smooth convex polytope in R^n (Ana Cannas da Silva's book for notions.) Do you guys have any insight of the definition, for example, anything we can say about the shape? They satisfy some rigidity conditions? All related comments are welcome!

Answer 1

The standard model of a vertex which satisfies the Delzant condition is the positive "quadrant" $x_i \geq 0$ of $\mathbb{R}^n$ near the origin. In general a polytope is Delzant if and only if every vertex can be taken to this standard model by some element of $\mathrm{GL}(n, \mathbb Z)$.

The motivation for this definition, coming from symplectic geometry, is the following fact, due to Archimedes. Consider the standard radius-one sphere in 3-space enclosed in a cylinder of radius one. Project the sphere outwards onto the cylinder, orthogonal to the cylinder's axis. Archimedes' theorem says that this map preserves areas. In terms of Delzant polytopes, you should think of the sphere as 2-dimensional toric manifold. The corresponding "polytope" is the interval $[-1,1]$. Archimedes' theorem says that if you take the cylinder $S^1\times [-1,1]$ and collapse the circles $S^1\times{\pm 1}$ you obtain the sphere as a symplectic manifold. To see how to generalise this to higher dimensions, take the n-fold product of the cylinder $S^1\times [0,\infty)$ and consider the map to $\mathbb{R}^n$ given by forgetting the $S^1$-factors. The image is the positive "quadrant" mentioned in the first paragraph above. Archimedes' theorem tells us that if we collapse the preimage of the boundary of this quadrant in a controlled way we obtain something which admits a smooth symplectic structure. I.e., over the orgin we collapse the whole $T^n$-fibre, more generally, over each coordinate $r$-plane we contract the corresponding orthogonal copy of $T^{n-r}$ inside $T^n$. (To do this recall each $S^1$-factor corresponds to a direction in $\mathbb{R}^n$).

Now the meaning of Delzant's condition should be clear: given a Delzant polytope P we built a symplectic manifold by taking $P \times T^n$ and then collapsing parts of the boundary. To decide how, we map each vertex to the standard model described above and collapse as in that case. Archimedes theorem tells us the resulting object is a smooth symplectic manifold with a torus action.

Answer 2

As "all related comments are welcome" I should perhaps say that if you add a "reflexive" condition (the dual polytope is also lattice polytope) we get smooth Fano ones of which there are finitely many in each dimension, but the number grows pretty quickly. The attempts to classify these have been moderately successful (see www.imf.au.dk/publications/phd/2008/imf-phd-2008-moe.pdf for example), but they are still somewhat wild.

Answer 3

Well, they are fairly rigid in that the normals to their faces have to be lattice vectors. So the only deformations are to slide the faces perpendicular to themselves. (On the other hand, you may always do this, so they are not completely rigid.) And "simple" is a restriction on their shape. The image of a Delzant polytope under $GL_n(\mathbb{Z})$ is another Delzant polytope, so it only makes sense to ask about properties which are $GL_n(\mathbb{Z})$ invariant. That tends to rule out obvious geometric properties (e.g. curvature) so I'm not sure what else to say.

UPDATE: In the theme of "all related comments are welcome", here is a speculation that I haven't thought about seriously. I do know one geometric property which is preserved by $GL_n(\mathbb{R})$. For $K$ a compact convex set, let $A$ be the volume of the smallest ellipsoid containing $K$, and $a$ the volume of the largest ellipsoid contained in $K$. Let Define $s:=a/A$ to be the sphericity of $K$. (Not sure if this is the standard name.) So $s=1$ for ellipsoids, and smaller for everything else. Delzant polytopes tend to be kind of fat, so I wonder if one could prove a lower bound for their sphericity which was better than for ordinary polytopes.