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If $\mathscr P\subset \mathbb R^d$ is a discrete point configuration, take the Voronoi diagram of $\mathscr P$ and call $\mathscr P'$ the vertices of that diagram.

I would like to know if configurations such that $\mathscr P = \mathscr P''$ have an established name, and if people studied them.

Some first facts and questions that may make the above property interesting:

  1. If $\mathscr P$ has the property, then $\mathscr P'$ has it.
  2. If $\mathscr P$ has the property, then if we join the vertices of $\mathscr P$ to their nearest neighbors in $\mathscr P'$, we form a bipartite graph with edges of equal length. Not all such graphs give $\mathscr P$ satisfying the property, but does there exist a simple extra convexity condition that can be added to get it?
  3. It seems that $\mathscr P$ that have the property are unbounded, is that true? I could not prove or find a counterexample so far.
  4. Do there exist $\mathscr P$ that have the property and that have accumulation points?
  5. Examples: (1) All simple lattices I tried as $\mathscr P$ have the property. Are there lattices that do not have the property? (2) Vertices of rhombic tilings of $\mathbb R^2$ have the property. In particular this includes Penrose tilings.
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    $\begingroup$ For $d=1$, you can describe $\mathcal{P}$ by an increasing sequence $(x_n)$, and you have explicit linear equations giving you a sequence $(x_n'')$ defining $\mathcal{P}''$. Then the condition is that there is a $k$ such that $(x_n'')=(x_{n+k})$. For each $k$, you get a linear induction formula, so that $(x_n)$ should be a linear combination of products of polynomials and exponentials. They thus cannot be bounded. This leaves open the possibility of a accumulation point, but I have not worked out the computations ($k=0,1,2$ do not work if I did not messed up). $\endgroup$
    – Benoît Kloeckner
    Jul 9, 2020 at 15:38
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    $\begingroup$ @BenoîtKloeckner thanks! in one dimension are there examples not congruent to the integers? (cf. point 2 in the list of facts I mention) $\endgroup$
    – Mircea
    Jul 9, 2020 at 18:05
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    $\begingroup$ Ah, you're right pointing this out! Point 2 shows that when $d=1$ all example must be congruent to lattices: consecutive points must be at distance twice the constant length of edges of the bipartite graph. This make my comment moot. $\endgroup$
    – Benoît Kloeckner
    Jul 10, 2020 at 16:27

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