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Does exist a Einstein tile analogous for covering the surface of a 3D sphere?

I Google for it and found a guy making a football (soccer) ball using a Einstein tile but he must added some other extra shape in order to make the ball, showing that a tessellation of the sphere with that shape was not possible.

example

But I didn't find some analogous shape to cover a sphere with only one aperiodic figure (here a short YT video explanation).

But neither I understand if it makes any sense to talk about aperiodic shape over a sphere since you are going to come back where you started after one turn.

So I would like to separate the question in sections:

  1. It is possible to cover a 3D sphere/ball using tiles of only one shape? (already answered affirmatively in the comments)
  2. It is possible for this unique shape to be aperiodic? (maybe in a $2\pi R$-mod sense)

Hope you could share examples in the affirmative scenario, or instead give reference where it is shown is impossible, or maybe it is still an open problem?

answer to comment

@MoisheKohan the best I can do to define anything related to my questions is looking for a spherical analogue to the tiles found by David Smith.

Update 1

User @JaapScherphuis noted that in the case of a sphere talking about an aperiodic tiling make little sense since it is not infinite, and also many paths will comes back to himself. So to make sense of the question, think about an sphere of radius $0<R<\infty$, for any point on the surface of the sphere, the individual tile should be aperiodic on any geodesic that crosses this point on it is $2\pi R$ extension in all directions, an the tiles of this unique kind should cover the whole sphere surface.

User @JaapScherphuis also shared a paper in his comments where are shown some tilings with unique triangular shapes that cover the whole sphere surface, so that answers sub-question 1.

An I hope this update introduce a restriction senseful enough to make an analogous version of David Smith's tiles for the surface of the sphere (I think this also answers @MoisheKohan comment).

I think from the video you could get the intuition behind the question: any recommendations for improving the restrictions in order to make this question rigorous in the math sense are very welcomed - feel free to edit the question.

Joako
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  • Please, specify what you mean by a "tile" when working with the sphere. For instance, does the entire sphere qualify as a tile? – Moishe Kohan Oct 29 '24 at 03:31
  • @MoisheKohan If you honestly believe your example is a valid solution, I encourage you to give it as an answer explainning your reasoning - Dealing with answers in comments – Joako Oct 29 '24 at 09:50
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    Of course, not: I was trying to give an absurd example. I was asking you to give your definition. So far, I do not see it. – Moishe Kohan Oct 29 '24 at 13:56
  • @MoisheKohan part of the question is focus in trying to understand what a definition of this could be... I am not familiarized enough with these topics for giving you any, so choose the one you prefer for an answer – Joako Oct 31 '24 at 00:34
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    There are several spherical tilings of the surface of a sphere, for example you can project any of the platonic solids to the surface of the sphere to get a tiling with 4, 6, 8, 12, or 20 copies of a single tile. As for non-periodicity, that makes no sense to me here since periodicity implicitly implies infinite repetitions. You could however look at the number of symmetries a spherical tiling has, and consider one that has no symmetries (other than the trivial identity) to be the equivalent of a non-periodic plane tiling. [...] – Jaap Scherphuis Oct 31 '24 at 10:17
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    .... You can look at this paper to see some interesting ones. – Jaap Scherphuis Oct 31 '24 at 10:17
  • @GEdgar my apologies but I don't how to answer your comment, I don't know what $S^3$ means... I made the question in an intuitive way knowing very little about geometry on the surface of the sphere. – Joako Nov 02 '24 at 20:22
  • @MoisheKohan ‘s example doesn’t feel exactly wrong. I would consider it a base-case of sorts for a good definition of tiling the sphere. – Sidharth Ghoshal Nov 05 '24 at 23:52
  • @SidharthGhoshal My answer to him was not fake: I don't know how to properly define what would be a right intuition to define the analogous tile, I was expecting to gain insight with his answer... any recommendation about how to rigurously define the problem are very welcomed – Joako Nov 06 '24 at 00:56
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    Here's a start, its probably wrong, but could help synthesize a right answer: "a tiling of the sphere is a collection of congruent simply connected geometric sets in R^3 of hausdorff dimension 2 such that the completion of their union is exactly equal to the surface of the sphere". And here we say 2 sets are congruent if there is a distance and angle preserving map between them. – Sidharth Ghoshal Nov 06 '24 at 02:55
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    The elements of collection are referred to as "tiles". In this case @MoisheKohan gives us the base case of a single tile covering the sphere. The two hemispheres also is a tiling. And then the more standard stuff corresponding to the platonic solids as well. But you are free to consider more irregular tilings with this definition too. – Sidharth Ghoshal Nov 06 '24 at 02:57
  • @SidharthGhoshal but how to define it such us your mentioned tillings got excluded and only irregular kind are kept? irregular such they are aperiodic in any directions on a $2\pi R$ sense? – Joako Nov 20 '24 at 17:31
  • If I understand correctly irregular tilings in the euclidean plane lack translational symmetry (although they may have rotational/reflectional symmetry). Regarding the surface of the sphere symmetries are always a combination of rotations or reflections so you want a sphere tiling in the sense of the comments I gave above, upon which no finite subgroup of the rotation group of the sphere acts. – Sidharth Ghoshal Nov 20 '24 at 21:23
  • @SidharthGhoshal thanks for the definition, but I am not familiar with topology, so I barely understand the idea of what you shared... but feel free to edit the question if that makes more sense than what I have updated last time – Joako Dec 18 '24 at 23:47

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