Suppose we have compact space, like sphere or torus in particular dimension $d$. Is it possible to construct aperiodic tiling in such setting? It seems obvious, answer is yes, because we may just divide its surface on finite number of basic shapes ( "curved polygons") but every one of those may be different than another. It is relatively easy I suppose.

Some of curved polygons above may be isomorphic tiles according to Euclidean group, translation and rotation, action in $d+1$ dimensional space, in which case we may say they have the same shape.

What is the smallest number of basic tiles of the same shape for which we may get aperiodic tilling? In particular: is it possible to construct aperiodic Penrose tiling on sphere? Torus? On cylinder?

If answers are known, could you take some references?

Edit: of course you may always use two different, "curved polygons" so probably question need some refactoring. Tiling I am asking should be defined as some kind of limit, where tiles are small in comparison to surface of the compact space of sphere or torus. So to make it nontrivial and mathematically interesting, more details are needed. From the other side, intuitive meaning here's clear. We want to exclude trivial cases of very few, very deformed polygons, and focus rather on very small tiles, building something like mesh rather than just dividing surface in finite number if shapes.

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    $\begingroup$ Take one tile to be some small subset of the compact space and a second tile to be its complement. Does that tiling count as aperiodic to you? Why/why not? $\endgroup$
    – Wojowu
    Dec 25, 2017 at 11:27
  • $\begingroup$ Of course it is aperiodic, so that is basically my answer in second sentence of the question. But they are not used"the same" shapes of tilling. Because tiles are not isomorphic due to Euclidean group ( translation+rotation in d+1 dimensional space). So it basically could be answer, and you have let's say n tiles. Is this minimal number of tiles required? $\endgroup$
    – kakaz
    Dec 25, 2017 at 11:58
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    $\begingroup$ I see, I've missed the fact that you require all tiles to have the same shape. $\endgroup$
    – Wojowu
    Dec 25, 2017 at 12:01
  • $\begingroup$ I rather require it is minimal number of different shapes. Of course they may be every different than others, which is clearly not minimal. $\endgroup$
    – kakaz
    Dec 25, 2017 at 12:02
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    $\begingroup$ I have no intuition for what it means for a tiling on a torus or a sphere to be periodic.... $\endgroup$
    – Dan Rust
    Dec 27, 2017 at 1:26

1 Answer 1


Not sure if this is what you are looking for, but some of the compact spaces you mentioned have a natural group operation. And you can talk abou aperiodic tilings in any LCA group.

For example the unit circle $U(1)\simeq \mathbb R/ \mathbb Z $ is a group. A tiling of this group is simply a partition in arcs.

Some of those partitions are periodic, for example pick the vertices of a regular $n$-gon, but if for example the arcs have pairwise distinct lenghts, the tiling is automatically periodic.

Same way, the torrus is a group, you can think about it as $U(1) \times U(1)$.

Note As you mention, you can also pick a group which acts on your space, and talk about periodicity/aperiodicity with respect to this action. This actually makes a lot of sense, since the set of periods of any tiling in $\mathbb R^d$ is just the stabilizer of the translation action of $\mathbb R^d$ on $\mathbb R^d$.

Thus, an element is periodic if and only if its stabilizer is non-trivial.

The advantage of the group action approach over the compact group is that in this case you can speak about "fully periodic elements".

But you have to also be careful. If your space is for example $U(1)$, the "natural" action to consider is the rotation by elements in $\mathbb R$, but then, since this action is not faithful, every element is automatically periodic. So you need make sure that your action is faithful.


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