Polyhedra that cover the sphere more than once

Question

Spherical polyhedra can be thought of as tilings of the sphere. I am interested in the possibility of double covering or multiple covering tilings of the sphere, but I can't find much information about them and would like to know where I can learn more. (It might be that they are known by a different terminology.)

A "double covering tiling" would mean that as you add tiles to the sphere there will be a place where they don't line up, but if you keep adding tiles on top of the existing ones and keep going, they will line up once every part of the sphere is covered by exactly two tiles. This idea can be extended to triple cover tilings and so on, assuming those exist.

This idea has to be made precise in the right way - that's also part of the question. In the comments, Ivan Neretin gives an example that would make the question trivial, but it can be excluded by stipulating that the angles have to add up to $2\pi$ at each vertex, so let's add that as a requirement. (edit: or in fact let's not, since that would exclude multiple coverings entirely.)

It seems that double cover tilings exist and are known. Wikipedia's page on uniform polyhedra says "There are some non-orientable polyhedra that have double covers satisfying the definition of a uniform polyhedron," but doesn't give further details except to say that they aren't usually counted as uniform polyhedra. A web page on tiling the sphere with triangles states that the spherical triangle with angles 90$^\circ$, 75$^\circ$ and 45$^\circ$ will give a double covering of the sphere and that a 75$^\circ$-60$^\circ$-60$^\circ$ triangle gives a five-fold tiling.

In general I'm looking for more information about multiple covering sphere tilings, but here are my specific questions about them:

• Are there any/many other known examples of double coverings, aside from the 90$^\circ$-75$^\circ$-45$^\circ$ triangle?

• Are there any examples of double covering tilings that are made of regular polygons and are vertex transitive, as hinted at on the Wikipedia page? I would like to see a specific example.

• What about triple and higher order coverings - are there known examples beside the 75$^\circ$-60$^\circ$-60$^\circ$ triangle, and are there any composed of regular polygons? Do there exist $n$-fold coverings for every $n$?

• Are there "infinite covering tilings" in the sense that you can keep adding tiles in the same repeated pattern but the edges will never quite line up, so each part of the sphere will be covered by an infinite number of tiles? If so, are there such tilings where all the tiles are regular polygons?

• Is there a set of tiles such that you can always add more tiles, but not in a repeating pattern? This would be a spherical analog of aperiodic tilings of the plane, such as Penrose tiles.

Answer 1

What about projecting the great stellated dodecahedron = {5/2, 3}, the small stellated dodecahedron = {5/2, 5}, the great icosahedron = {3, 5/2}, or the great dodecahedron = {5, 5/2} onto the sphere? These 4 are the stary (non-convex) $regular$ polyhedra. - The set of the $uniform$ polyhedra contains even more stary figures. Many of them are orientable, so that those too can be projected onto the sphere to provide spherical tilings, which casually provide multicovors.

--- rk

Answer 2

You even could consider {5/2, 4}. Again 4 is integral and the vertex configuration of 4 regular planar pentagrams is still less than the full circuit, so has an angular deficit, i.e. provides a spherical arrangement which furthermore has a local spherical vertex angle of $2\pi$ too. - On the other hand the numerators (4 and 5) are incommensurate (being ruled out as $finite$ Coxeter group, so the produced tiling does never end, i.e. it covers the sphere infinitely often.

--- rk

Answer 3

We can ask how many times a set of identical regular pentagons can tile a sphere without simply duplicating a tiling with a smaller covering. It turns out that we can tile a sphere one or three times in this way, but not twice! (We can have two coverings, but only by duplicating the single dodechedral covering.)

The single tiling is, of course, the regular dodecahedron. For the triple tiling, start with a regular icosahedron. With each of the twelve vertices we associate a pentagonal loop formed from the five vertices adjacent to the selected one. We can then define the regions bounded by the pentagonal loops as faces of a stellated polyhedron, which is the $\{5,5/2\}$ great dodecahedron described by Dr. Richatd Klitzing.

We can see that the great dodecahedron tiles the sphere thrice in either of two ways. First, we note that each triangular face of the original icosahedron is counted three times within the pentagonal loops, and the sphere would be fully tiled with just one count of each triangle. Second, and more formally, we note that the pentagons projected onto the sphere have vertex angles of $144°$ and thus an angular sum of $720°$ versus $540°$ for a planar pentagon. Taking the difference of $180°$ per pentagon times twelvebpentagons gives $2160°$ of total spherical excess, which is three times the $720°$ that would cover the sphere once.

So what goes wrong if we try for exactly two coverings with regular pentagons? We identify two necessary conditions for a double cover:

1. The total excess angular sum above the planar value is $1440°$.

2. The vertex angle at each pentagon divides $360n°$, where $n\le 2$ with only two coverings of the sphere available.

Now suppose that we have $m$ regular pentagons covering the sphere twice. Then by condition (1) each vertex angle must measure

$[108+1440/(5m)]°=[108+288/m]°$

Condition (2) then implies that $3m+8$ must divide $20m$, which is possible only if $m=8$ or $m=24$. But $m=8$ corresponds to a $144°$ vertex angle, and we have just seen that this is actually the great dodecahedral tiling covering the sphere three times with twelve pentagons. With five pentagons meeting at each vertex of the great dodecahedron, we cannot remove just one-third of the faces in regular fashion. Meanwhile the $m=24$ case, with vertex angles of $120°$, is just a doubling of the regular dodecahedral single covering.