Spaces of Tilings

How many ways can you place a regular tetrahedron inside of a sphere if it is completely unmarked? Infinitely many, sure; and in fact the space of ways forms a spherical 3-manifold. This is far from an isolated occurence - the space of ways to inscribe a dodecahedron in a sphere is the Poincare homology sphere!

More generally, spaces of tilings of spheres, euclidean and hyperbolic spaces lead to an interesting class of three manifolds - unit tangent bundles of geometric 2-orbifolds - and we will look at a couple of pretty examples here.

Getting Comfortable

Let's get started with an easy example. Consider the tiling of the real line by unit intervals:

Our goal is to understand the space of ways to do this: the points of this space are the tilings of $\mathbb{R}$ by unit intervals. Hopefully its intuitively clear that this space is a circle, and in fact we can illustrate this by drawing the single closed curve of tilings below:

Our goal here is to look at the more interesting examples that arise from tilings of two dimensional spaces, but before doing so let's back up a bit and define things in a little greater generality.

Spacings of Tilings

Consider some geometry $(X,\textrm{Isom}(X))$ and some partition $P$ of $X$ into a bunch of subsets (which, when we get more specific, will be the tiles of a tesselation). A different way of breaking $X$ up into the same sort of pieces would be to look at all the translates of the pieces in $P$ by some isometry of $X$: the "space of ways to break $X$ up like $P$" can be parameterized by the isometries of $X$. ( In the one-dimensional example above, if we consider only orientation preserving isometries $\textrm{Isom}(\mathbb{R})$ is itself a copy of $\mathbb{R}$, acting by translation.) This parameterization often overcounts - particularly in the case that $P$ is a tiling of $X$ by congruent shapes - in this case there are nontrivial isometries of $X$ which bring the tiling back to itself - not tile-for-tile, as a whole collection. (Think for instance of unit translation of the tiling of the real line discussed above, or a 60-degree rotation about a vertex of the equilateral tiling below).

Denote by $\textrm{Sym}(P)$ the sugbgroup of $\textrm{Isom}(X)$ which permutes the tiles of $P$, then the space we are interested in is the coset space
$$\textrm{Tilings}_X(P):=\frac{\textrm{Isom}(X)}{\textrm{Sym}(P)}$$

Euclidean Tilings

The orientation preserving geometry of the Euclidean plane has as its isometries $\textrm{Isom}^+(\mathbb{E}^2)\simeq \textrm{SO}(2)\rtimes\mathbb{R}^2$ which is diffeomorphic to $\mathbb{S}^1\times\mathbb{R}^2$ - and so topoloically, our spaces of tilings will be quotients of this.

Let's look at an easy example; the following is a tiling of the plane with only nontrivial translational symmetry

Here, we almost don't need the general stuff to figure out the space of tilings - its the three torus! There is a circles worth of translations in the $x$ direction leading to distinct tilings, and a circles worth in the $y$ direction. Rotating by any angle about any point of the tiling also gives a distinct tiling (it doesn't land back on itself until a full $2\pi$ rotation) - and so that gives a third circle's worth.

(Three videos of this happening below)

More precisely, we can see this by noting that the group of isometries which take this tiling back to itself is just $G\cong \mathbb{Z}\oplus\mathbb{Z}$ generated by unit $x$ and $y$ translations. The coset space of this in the translation subgroup of $\textrm{Isom}(\mathbb{E}^2)$ forms a torus, and the rotations "come along for the ride" - as each rotation about each point gives a distinct way to place the tiling in the plane. Thus, our coset space looks like
$$\frac{\textrm{Isom}(\mathbb{E}^2)}{G}\cong \mathbb{T}^2\times\mathbb{S}^1\cong\mathbb{T}^3$$

What about a slightly more complicated example - where the rotations do come into play? Consider the following tiling:

This can be brought back to itself by a translation in the $x$ direction of length $2$, or a unit translation in the $y$ direction - but now rotation about each point is note equal. For instance, rotating about the center of a rectangle brings this tiling back to itself after only $\pi$:

As does rotation about the midpoint of any edge

And similarly rotation about a vertex.

Rotating about any other point leads to no special symmetries, but this information already gives us some useful facts about our manifold of tilings. In particular, given any tiling, rotating this (about, say $0\in\mathbb{R}^2$) gives a circles worth of tilings - and so our manifold admits a circle action - its a Seifert fibered space! The fact that there are four exceptional points about which rotation gives extra symmetry (the center of a long edge, short edge, center, and vertex) means that four of these circles correspond to exceptional orbits, half as long as the others. That is, there are four exceptional fibers in the Seifert fibration about which generic fibers wrap twice.

A little more thought allows us to even specify more precisely what manifold this is: it turns out to be the Seifert fiber space over the orbifold $\mathbb{S}^2(2,2,2,2)$: the pillowcase with four cone points each of angle $\pi$.

Let's look at one more tiling here: what happens if we just have unmarked squares?

Just like above, rotating the possible tilings decomposes the space into a bunch of circular orbits, and if we hope to understand its structure as a Seifert Fiber space, we should look for rotations which have extra symmetry. Here, rotating about the center of a square or a vertex by $\pi/2$ returns the tiling back to itself

And rotation about any edge by $\pi$ does as well - its easy to convince yourself there are no nontrivial symmetries given by rotation about other points. As up to translation all centers and vertices are equivalent, then up to rotation about a vertex all edges are equivalent, there really are only three exceptional fibers here: two are of length $\pi/2$ and the other is of length $\pi$, so our manifold of tilings is a Seifert fiber space over some base orbifold with three cone points: two of order $4$ and one of order $2$. Again, a little more thought shows that this base is the two-sphere, and so the space of tilings can be described as the SFS over $\mathbb{S}^2(2,4,4)$.

Something we have yet to mention (but has been true all along) is that we may profitably think about these spaces as the unit tangent bundles to their base orbifolds - as they are quotients of $\textrm{Isom}(\mathbb{E}^2)$ by discrete subgroups acting by translation, but this space itself can be identified with the unit tangent bundle of $\mathbb{E}^2$.

In fact, this gives us a way to classify all the manifolds of Euclidean tilings! Every tiling of the Euclidean plane is preserved by some discrete group, and the quotient of $\mathbb{E}^2$ by this gives rise to a Euclidean orbifold. The rotations about points in the plane are encoded by circles above each point on this orbifold - its unit tangent bundle. So, how many Euclidean orbifolds are there? (Here we are considering orientation preserving groups so there will be no mirror edges)

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