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!



look into Delzant Polytope
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. 


