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Let $\Lambda$ denote a lattice of $\mathbb{R}^n$, i.e. $$\Lambda = \left\{\sum_{k=1}^n n_i\mathbf{a}_i\ \bigg|\ n_i\in\mathbb{Z}\right\},$$ for $n$ linearly independent vectors $\{\mathbf{a}_i\}$ in $\mathbb{R}^n$. The Voronoi cell of the lattice (attached to point $\mathbf{p}\in\Lambda$) $V(\mathbf{p})$ is the set of points closer to $\mathbf{p}$ than to any other lattice point, i.e. $$V(\mathbf{p}) = \left\{\mathbf{x}\in\mathbb{R}^n\mid \forall \mathbf{q}\in\Lambda\ \ \ \|\mathbf{x}-\mathbf{p}\|\le\|\mathbf{x}-\mathbf{q}\|\right\}.$$ Since the lattice has translational symmetry, it follows that the Voronoi cell attached to any point is the same, so we can drop the attachment to $\mathbf{p}$ and denote the Voronoi cell of a lattice as $V(\Lambda)$.

Let $\mathrm{Aut} (\Lambda)$ denote the automorphism group of the lattice, the subgroup of the isometries of $\mathbb{R}^n$ which leaves the lattice invariant and fixes the origin. It is clear that each automorphism of the lattice induces an automorphism of $V(\Lambda)$. It follows that we must have $\mathrm{Aut}(\Lambda) \le \mathrm{Aut}(V(\Lambda))$.

My question is whether the converse holds. Do we necessarily have $\mathrm{Aut}(V(\Lambda))\le \mathrm{Aut}(\Lambda)$? In other words, do the symmetries of a lattice's Voronoi cell determine the lattice's symmetries? Intuition seems to say yes, but I can't seem to find a simple proof (or counter-example) of this fact.

EuYu
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  • You will need to give some more detail. What are your sources? Also, I would have said the automorphism group fixes the origin. – Will Jagy Jan 18 '15 at 02:24
  • @WillJagy I am not sure what you mean by sources. I feel the question is pretty clear and self-contained, what details should I add? The automorphism group of the lattice doesn't fix the origin (since they include translations for example). – EuYu Jan 18 '15 at 02:58
  • *Sphere Packings, Lattices and Groups* by Conway and Sloane, chapter 3, section 4.1, .....``that fix the origin.'' In the first edition page 90. – Will Jagy Jan 18 '15 at 03:07
  • @WillJagy Thanks for the reference. If it's conventional to take the automorphism group as origin preserving then I will edit that in. – EuYu Jan 18 '15 at 03:16
  • http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries – cactus314 Jan 21 '15 at 20:39
  • It's hard to say since finite subgroups of $O(n)$ can be quite varied. https://en.wikipedia.org/wiki/Crystallographic_restriction_theorem – cactus314 Jan 21 '15 at 21:02

1 Answers1

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The Voronoi cell $V$ determines the lattice $\Lambda$.

Consider an ($n-1$)-face $F$ of $V$. The hyperplane of the face $F$ is the perpendicular bisector of the segment between $O$ and a lattice point $p$. Then $V+p$ is the neighboring Voronoi cell that shares face $F$. Marching through the Voronoi cells we can find all cells and thus all lattice points at the centers of the cells.

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Since every automorphism of $V$ maps $V$ to itself, they also map $\Lambda$ into itself. $Aut(V(\Lambda))\subset Aut(\Lambda)$.

G.Kós
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    This proof looks good to me. I got to the same point except for "marching through" the cells. Two comments. (1) An initial fact that needs to be observed is that there is a finite set $\Lambda_0$ of lattice points such that the Voronoi cell of $O$ in $\Lambda$ is the Voronoi cell of $O$ in $\Lambda_0$. This guarantees that the cell is a convex polytope. (2) For the marching part, take any broken-line path from $O$ to a chosen lattice point $A$ that avoids all the faces of dimension $\leq n-2$. This guarantees that when you cross from one cell to another, you do it via an $n-1$-face. – user208259 Jan 22 '15 at 18:39
  • Thank you for the proof. There are a few things I would formalize a bit, but the overall direction seems great. In particular your proof is just the fact that the facet vectors of a lattice generate it, something which I should've thought of :). – EuYu Jan 23 '15 at 08:27