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So let $L$ be a full dimensional lattice in $\mathbb{R}^{n}$. Then the Voronoi-cell of the lattice are precisely the points in $\mathbb{R}^{n}$ that are at least as close to the origin, as to any other lattice points. We denote the Voronoi cell as $$ \operatorname{vor}L := \{\mathbf{v}\in\mathbb{R}^{n}\vert\forall \mathbf{l}\in L\colon d(\mathbf{v},\mathbf{0})\le d(\mathbf{v},\mathbf{l})\}. $$ The translates of $\operatorname{vor}{L}$ by lattice vectors give a partition of $\mathbb{R}^{n}$. Also $\operatorname{vor}L$ consists of the intersections of the following half spaces corresponding to lattice vectors: $$ H_{\mathbf{l}}:=\{\mathbf{v}\in\mathbb{R}^{n}\vert \mathbf{v}\cdot\mathbf{l}\le\mathbf{l}\cdot\mathbf{l}/2\} $$ where $\mathbf{l}\in L$. This also gives that $\operatorname{vor}L$ is centerally symmetric since it is intersection of centerally symetric sets, namely the sets $H_{\mathbf{l}}\cap H_{-\mathbf{l}}$. The facets of the voronoi cell are of the form: $$ F_{\mathbf{l}}:=\operatorname{vor}L\cap \{\mathbf{v}\in\mathbb{R}^{n}\vert \mathbf{v}\cdot\mathbf{l}=\mathbf{l}\cdot\mathbf{l}/2\}. $$

Now i think that $F_{\mathbf{l}}$ should allways be symmetric by $\mathbf{l}/2$ meaning that if $\mathbf{l}/2+\mathbf{v}\in F_{\mathbf{l}}$ then $\mathbf{l}/2-\mathbf{v}\in F_{\mathbf{l}}$. My argument is that $\mathbf{l}/2+\mathbf{v}\in F_{\mathbf{l}}$ is true then $-\mathbf{l}/2-\mathbf{v}\in F_{-\mathbf{l}}$, and i think $F_{-\mathbf{l}}+\mathbf{l}=F_{\mathbf{l}}$ should hold which would implie my claim. Im having trouble with actually proving $F_{-\mathbf{l}}+\mathbf{l}=F_{\mathbf{l}}$. I think it should hold but im not sure.

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    $\begingroup$ Seems to be a typo in your first display: the condition does not mention $\mathbf I$. $\endgroup$
    – Lubin
    Commented Nov 7 at 16:00
  • $\begingroup$ $\mathbf{l}$ is any lattice point. Or we could assume that it is a lattice point such that $F_{\mathbf{l}}\neq \emptyset$ $\endgroup$
    – Péter Fazekas
    Commented Nov 7 at 18:07
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    $\begingroup$ Isn't it clear that translating by l the Voronoi facet dual to edge (0,-l) gives the Voronoi facet dual to (0,l)? It is a lattice symmetry $\endgroup$
    – alesia
    Commented Nov 7 at 19:31
  • $\begingroup$ Yeah but this is basically my question: whats the proof of this? I mean it seems obvious but how does one actually prove it? $\endgroup$
    – Péter Fazekas
    Commented Nov 7 at 21:32
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    $\begingroup$ For a reference, see the paragraph after Theorem 1.1 of R. M. Erdahl, Zonotopes, Dicings, and Voronoi’s Conjecture on Parallelohedra (sciencedirect.com/science/article/pii/…). $\endgroup$
    – Richard Stanley
    Commented Nov 8 at 0:13

1 Answer 1

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This is a community wiki answer based on the comments by alesia and Richard Stanley.

You already gave the hardest part of the argument. To finish it off, note that translating by $\mathbf{l}$ the Voronoi facet dual to edge $(\mathbf{0},-\mathbf{l})$ gives the Voronoi facet dual to $(\mathbf{0},\mathbf{l})$ because it is an isometry and a symmetry of the lattice.

That every facet of a Voronoi polytope is centrally symmetric is mentioned in the paragraph after Theorem 1.1 of R. M. Erdahl, Zonotopes, Dicings, and Voronoi’s Conjecture on Parallelohedra.

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