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I've been working on a problem I'm working on in ergodic theory (finding Alpern lemmas for measure-preserving $\mathbb R^d$ actions) and have found some neat tilings, that I presume were previously known.

They are periodic tilings of $\mathbb R^d$ by a single prototile consisting of any box with a smaller box removed from a corner. The two-dimensional case is illustrated in the figure below:

There exist versions of this in arbitrary dimensions, irrespective of the size of the boxes. Furthermore, there appear to be many essentially different tilings using the same prototile.

Does anybody recognize the $d$-dimensional version, or have a name for it?

Thanks

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  • $\begingroup$ This is essentially hexagonal tiling, isn't it? $\endgroup$
    – Yuri Bakhtin
    Commented Oct 12, 2011 at 20:49
  • $\begingroup$ I guess so in 2 dimensions. My question is really about the higher-dimensional version. $\endgroup$
    – Anthony Quas
    Commented Oct 12, 2011 at 20:58
  • $\begingroup$ If you take the centers of the boxes, do you get the plain old cartesian lattice up to an affine-linear transformation? $\mathbb Z^d \subset \mathbb R^d$ ? $\endgroup$
    – Ryan Budney
    Commented Oct 12, 2011 at 21:46
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    $\begingroup$ Yes. The prototile is a fundamental domain for $\mathbb R^d/M\mathbb Z^d$ for some matrix $M$. $\endgroup$
    – Anthony Quas
    Commented Oct 12, 2011 at 22:44
  • $\begingroup$ Take any nice region in $\mathbb{R}^d$. Pick a lattice s.t. the translates of the region by lattice vectors cover $\mathbb{R}^d$. Pick any self-consistent scheme to remove overlaps (this smells nontrivial in general). Then you have a tessellation. $\endgroup$
    – Steve Huntsman
    Commented Oct 13, 2011 at 0:06

2 Answers 2

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Kolountzakis worked with some tilings of this sort in his paper "Lattice tilings by cubes: whole, notched and extended", Electr. J. Combinatorics 5 (1998), 1, R14.

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  • $\begingroup$ Thanks very much for the reference! Apparently Kolountzakis re-proved using harmonic analysis a result of Sherman Stein that gave a direct proof that the "notched cubes" tile $\mathbb R^d$ ("notched cubes" being exactly what is being asked about in the question) $\endgroup$
    – Anthony Quas
    Commented Oct 13, 2011 at 8:47
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Hi Anthony,

I maybe should walk down the hall... but this is easier.

Dual to the $d$-cubical tiling of $\mathbb R^d$ would be the "cross-polytope" tiling. This is the tiling made up of the duals -- the vertices are the centres of the $d$-cubes, the edges of the cross-polytope are the faces of the $d$-cubes, and so on.

To me it looks like you get your tiling from the cross-polytope tiling, by simply scaling up each tile appropriately -- scaling each tile at one of its vertices, and doing the scaling symmetrically with respect to the translation symmetry of the tiling. So as you scale, part of the tile vanishes (from a growing tile eating it up) and part gets created (via scaling).

Your picture appears to be consistent with something like that. Or rather than the cross-polytope tiling, it could the the same idea but with the cubical tiling.

edit:

Take this procedure for generating a tiling of $\mathbb R^n$. Let $M$ be an $n \times n$ invertible matrix with real entries. Let $\vec v \in \mathbb Z^n$. In the lexicographical order on $\mathbb Z^n$ we can lay down a "tile" being $[0,1]^n + M\vec v$, where whenever we place a tile, it overwrites any old tile that it may be placed on top of. Provided the norm of the matrix is small enough, this procedure writes over the entire plane. It produces a tiling a fair bit more general than what you're talking about. I'd call it perhaps a linear overlapping translation of the cubical tiling. But I don't know if there's standard names for such thing. Perhaps "lizard scales" ?

The tiling in your picture looks like something like this, with a 2x2 matrix with entries (left to right then top to bottom) $2/3, -1/3, 1/3, 3/4$. I seem to have forgotten how to typeset a $2\times 2$ matrix in mathjax or whatever is powering this website nowadays...

Does that make sense?

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  • $\begingroup$ Wouldn't this give you something where translates in the original lattice directions didn't have a dense orbit? This, in fact, was the main reason for me to look at this: say the big boxes are unit cubes. If you take a point and repeatedly move by 1 in the `vertical' direction, then the images become dense in the prototile (for some choices of the sub-box). $\endgroup$
    – Anthony Quas
    Commented Oct 12, 2011 at 22:58
  • $\begingroup$ I'm not following your comment. In my edited response, would this condition be satisfied if the matrix $M$ had irrational entries? $\endgroup$
    – Ryan Budney
    Commented Oct 13, 2011 at 0:03
  • $\begingroup$ My condition was that multiples of $e_d$, the $d$th coordinate vector, are dense in $\mathbb R^d/M\mathbb Z^d$. Having irrational entries is the right kind of condition to get this, but is not sufficient: basically you have to ensure that $e_d$ doesn't belong to any proper closed subgroup of $\mathbb R^d/M\mathbb Z^d$. $\endgroup$
    – Anthony Quas
    Commented Oct 13, 2011 at 0:25
  • $\begingroup$ Is it obvious that for any BOX$\setminus$box, it arises as a "linear overlap tiling of a cubic tiling"? $\endgroup$
    – Anthony Quas
    Commented Oct 13, 2011 at 1:06
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    $\begingroup$ If you take vertices at the centres of the d-cubes and edges through the faces of the d-cubes, you just get the cubical tiling back again; it's self-dual. $\endgroup$
    – Nick Matteo
    Commented Oct 1, 2013 at 15:51

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