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This is probably a silly question, but I figured that if there is a good answer, this would be a good place to ask.

Ever since I got my hands on the book "Toric Varieties" by Cox, Little and Schenck, I've been excited to learn about the different combinatorial properties of polytopes that one can deduce from the corresponding toric varieties. In fact, toric varieties can prove combinatorial theorems not only about polytopes but also about many other objects living in $\mathbb Z^n$. One such thing would be tilings of $\mathbb R^d$ by integral polytopes.

I believe the following comes as a natural question:

Can one tell if a convex polytope $P$ tiles Euclidean space by looking at its corresponding projective toric variety? Can one deduce properties of the tiling this way?

In case there is no simple answer, do tilings by polytopes correspond to algebraic gadgets in the same spirit that polytopes are in bijection with projective toric varieties with a specified ample line bundle?

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2 Answers 2

up vote 11 down vote accepted

A related question (but not exactly the one you asked) is:

Can one tell if a convex polytope $P$ and its translations by $\mathbb Z^n$ tile $\mathbb R^n$? Which polytopes $P$ have this property?

Fix some positive quadratic form $q$ on $\mathbb R^n$ and the corresponding distance function. Let $P^0$ be the set of points in $\mathbb R^n$ which are closer to $0$ than to any other integral (i.e. in $\mathbb Z^n$) point. The closure $P$ of $P^0$ is called the Voronoi polytope w.r.t. $q$. Then $P$ obviously has the above property.

Voronoi's conjectured circa 1907 that the opposite is true, i.e. any such $P$ is a Voronoi polytope w.r.t. some $q$.

This conjecture is known for $n\le 4$ due to Delaunay and for zonotopes by Erdahl "Zonotopes, Dicings, and Voronoi Conjecture on Parallelohedra". It is still open in general, I believe.

So what is special about the toric variety $X_P$ corresponding to $P$? I am not sure. If you look at the Delaunay tiling which is dual to the Voronoi tiling $P+\mathbb Z^n$, then the polytopes in that tiling and the corresponding toric varieties have a clear geometric meaning: they describe degenerations of principally polarized abelian varieties. But this is a dual picture.

Note by the way that Delaunay polytopes have vertices in $\mathbb Z^n$, so they indeed correspond to projective polarized toric varieties. In contrast, the Voronoi polytope for a generic $q$ will have irrational vertices. Also, when you vary $q$ continuously, the Voronoi polytope will vary continuously. But the Delaunay polytopes will jump discretely, and there are only finitely many Delaunay polytopes modulo $GL(n,\mathbb Z)$.

One place where the Voronoi tilings appear is tropical geometry. Indeed, a principally polarized tropical abelian variety $A$ is just the real torus $\mathbb R^n / \mathbb Z^n$ together with the positive definite form $q$. Then the $(n-1)$-skeleton of the Voronoi tiling modulo $\mathbb Z^n$ is the theta divisor on $A$. See Mikhalkin-Zharkov http://arxiv.org/abs/math/0612267 for more details.

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Voronoi's conjecture is still open, see e.g. "Affine Equivalent Classes of Parallelohedra" by Dolbilin, Itoh and Nara, 2011. The conjecture was proven by Voronoi himself for "primitive" tilings, i.e., where the dual polyhedral complex is simplicial. Conjecturally, "primitive" tilings are dense. –  Misha Apr 11 '12 at 0:44

Apparently, yes, see this beautiful talk of Valery Alexeev's.

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Wow, thanks! Embarrassingly, this material goes a bit over my head. I think I can see how he associates a tiling to certain varieties but it's not clear to me if there is an inverse to this process. Maybe there is a lowbrow explanation of this somewhere? –  Gjergji Zaimi Apr 10 '12 at 0:28
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I certainly cannot claim to have any sort of deep understanding of this. Perhaps one can pester Alexeev directly?! –  Igor Rivin Apr 10 '12 at 0:52
    
I will try to do that, but first I must make an attempt to understand this talk and the references. –  Gjergji Zaimi Apr 10 '12 at 1:20
    
You will be a sadder and wiser man when you do (and perhaps you can keep the sadness and pass on the wisdom to the rest of us). –  Igor Rivin Apr 10 '12 at 3:28
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@Igor: As far as I can tell, VA is Valery Alexeev. –  Mark Sapir Apr 10 '12 at 12:39

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