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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!

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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.

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Thank you! Your first paragraph is brief and sharp! In your second paragraph you said "more generally, over each coordinate r-plane we contract the corresponding orthogonal copy of T^(n−r) inside T^n"; that is the same as to contract the corresponding copy of T^r inside T^n, right? For example, take a 3-fold product (labeled x,y,z in order) of S^1 \cross [0, \infinity), then over the xy-coordinate plane, we contract the last copy, i.e., the z copy of S^1 inside T^3 = S^1 \cross S^1 \cross S^1, right? –  Wayne Nov 13 '09 at 4:54
    
Firstly, credit where credit's due; this cute modern interpretation of Archimedes' theorem was explained to me by Richard Thomas. Secondly, your 3-dim example is correct, but the preceding statement - that contracting the orthogonal T^{n-r} in T^n it is the same as contracting T^r - is not. The first gives an r-torus, the second an (n-r)-torus. This is evident in your example: we must contract the third S^1 - leaving a 2-torus - and not the product of the first and second S^1 which would leave a single S^1. Hope that helps! –  Joel Fine Nov 13 '09 at 12:11
    
Thank you! It definitely helped me. It is clear to me now and sorry I misunderstood the language at first. Also sorry for the delay as I was offline over the weekend. –  Wayne Nov 16 '09 at 2:14
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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.

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Great! Thank you! Lattice polytopes are not necessarily Delzant, but reflexive lattice polytopes must be! –  Wayne Nov 13 '09 at 4:22
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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.

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It is not true that any GL_n(Z) element takes a Delzant polytope to another, e.g. the triangle in the plane with vertexes (0,0) (0,1) (2,0) is not but the triangle with vertexes (0,0) (0,1) (1,0) is. But for those with determinate 1 or -1 it is true. I think the smoothness condition is subtle. That the normals to the faces are lattice vectors follows basically from the rational condition. I do not quite get why the ONLY deformations are to slide the faces perpendicular to themselves but I agree with you we can always do that. –  Wayne Nov 11 '09 at 16:14
    
The definition of GL_n(Z) that I am familiar with is that a matrix M is in GL_n(Z) if M and M^{-1} have integer entries. This forces determinant \pm 1. –  David Speyer Nov 11 '09 at 16:47
    
How would you smoothly change a lattice vector while keeping it a lattice vector? If you say that you can't, then the normals are fixed and the only deformations are to slide the facets while keeping the normals the same. By the way, this isn't a purely rhetorical question. I would be interested in knowing of some other notion of deformation here. –  David Speyer Nov 11 '09 at 17:06
    
@David: Make sense. Thank you! –  Wayne Nov 11 '09 at 17:57
    
David: There are a zillion different ways to gauge the roundness of a convex body; no one of them will be called "the" sphericity. I'm not sure if yours has a name, but are you partial to that choice? For instance you could look at the Banach-Mazur distance, which is the ratio of the circumradius to the inradius (concentrically), minimized over the action of GL(n,R). –  Greg Kuperberg Nov 12 '09 at 15:55
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