# If you know that a shape tiles the plane, does it also tile other surfaces?

For instance there is a hexagonal tiling of the plane. There is also one using quadrilaterals. It seems intuitive that both of these tilings also apply on a torus. Is it the case that anything that maps 1-to-1 to a plane surface has a "complete" tiling with regular polygons? It seems like I should be able to tile hexes over a torus, and even over the sphere; although this seems like an edge case, since it's not 'on' the sphere except with a certain granularity and so on. I saw a Buckminster Fuller house the other days and have been wondering about this.

I will try to spell out the sense of the one-to-one mapping requirement a bit more clearly. Can I tile any surface homeomorphic to the plane? To what degree are constraints around gluing edges together the same as those for performing a tiling? It seems plausible to me that tiling is at least somewhat independent of homeomorphicity -- for instance I can imagine that there could be certain constructions permitting smooth tiling even over particularly awkward edges. (That is to say it seems possible that tiling properties don't need to have quite the same "algebra" as proximity relations; that possibly a slightly different homeomorphic relation is required?)

I note that on further reflection it seems like I'm conflating two tiling strategies in the first graph. The first being an 'internal' tiling, i.e., drawing on the actual surface. The second style being an 'approximation' tiling (like the way the Bucky house approximates a sphere with hexagonal tiles.) I hope this clarifies somewhat: that is, the question is whether the first kind of tiling is 'limited' to manifolds homeomorphic to the plane.

Some assumptions that might help clarify my intent:

1. By tiling I mean the isometric "filling" of a surface by compact tiles; and in particular I am concerned with the case of simple polygonal tilings, e.g., quadrilaterals and hexagons
2. By surface I think I mean a 2-d Riemannian manifold (at any rate: I am particularly curious about the cases of the torus, cylinder and 2-sphere)
• I am not sure about this, but I feel like any surface which can be assembled out of a planar diagram homeomorphic to the plane should also be able to be tiled. A torus is formed by taking a square, and gluing the opposite edges together into a cylinder, and then a torus. This makes me think that surfaces formed in a similar way should also work out. – Alfred Yerger Dec 29 '14 at 20:57
• You have to be a little more specific about what you mean by 'maps 1-to-1 to a plane surface'. For instance, does (e.g.) a two-holed donut count? A cross-cap? These map onto planar surfaces with particularly awkward edge-gluing rules... – Steven Stadnicki Dec 29 '14 at 20:59
• @AlfredYerger I think you need to be careful with that line of reasoning. A Mobius strip can't be tiled with a shape that is asymmetrical in the "short" direction because the tiles would be superimposed with their reflections. – Reinstate Monica Dec 29 '14 at 22:06
• Note also that any size square will tile a plane, but to tile a torus, the sides of the torus must be integral multiples of the edge of the square. – MJD Dec 29 '14 at 22:35
• @MJD I'm not sure I understand your statement correctly, but it looks wrong to me: the torus obtained by gluing the opposite sides of the square with side length 2 can be tiled by two squares of side length $\sqrt 2$ (rotated by $\pi/4$). – Start wearing purple Dec 31 '14 at 15:51

You are asking several questions, I understand only the first one, the rest will require some major clarification before they become answerable:

Question 1. Let $M$ is a Riemannian surface homeomorphic to the plane. Does $M$ admit a tiling? Here a tiling means a partition of $M$ into pairwise isometric relatively compact regions with piecewise-smooth boundary, such that two distinct tiles intersect along at most one boundary curve.

This question has a very easy an negative answer. For instance, start with the Euclidean plane $E^2$ and modify its flat metric on an open ball $B$, so that the new metric has nonzero (at some point) curvature in $B$ and remains flat (i.e., of zero curvature) outside of $B$. (This modification can be even made so that the surface $M$ is isometrically embedded in the Euclidean 3-space $E^3$: start with the flat plane in $E^3$ and make a little bump on it.) The resulting manifold admits no tiling, since all but finitely many tiles would be disjoint from $B$ and, hence, have zero curvature metric. At the same time, at least one tile $T_1$ will contain points where the curvature is not zero. Therefore, $T_1$ cannot be isometric to the tiles which are disjoint from $B$.

The same (or similar) construction can be used for any (connected, noncompact) smooth 2-dimensional manifold $S$, no matter what its topology is. This manifold always admits a metric $g_0$ of constant curvature $-1$. Now, modify the metric $g_0$ on a small ball $B\subset S$ making it into a metric of nonconstant curvature in $B$. Since $S$ is non-compact, the same argument as above shows that the new Riemannian manifold admits no tiling. If $S$ is compact, then, of course, each metric on $S$ admits a tiling, consisting of a single tile. If you require at least two tiles, one can always construct metrics which do not admit tilings, but this, requires more work.

However, if your manifold $M$ is, say, a flat 2-torus, i.e., genus 1 closed surface equipped with a zero curvature metric, then such $M$ always admits a tiling consisting of more than one tile. Let me know if you want to see a proof (it is quite easy).

I believe you mean 2-sphere, which is a 2-d surface that lives in 3-d space. None of the tilings of the plane extend to the 2-sphere because of the Euler characteristic. You can tile the 2-sphere with pentagons, but not with squares or hexagons (except for the trivial case of dividing a great circle into 4 or 6 segments and using two squares or hexagons.)

• Doesn't this depend on your definition of 'square'? The cube is a tiling of a 2-sphere by squares, meeting 3 at each vertex... – Steven Stadnicki Jan 2 '15 at 18:24
• @StevenStadnicki: that is true. It isn't the same tiling, but I allowed that with the trivial one I mentioned. You also get triangles 3,4,5 at a vertex. – Ross Millikan Jan 2 '15 at 18:37
• Naturally. And your tiling with pentagons also doesn't correspond to any tiling of the plane, so I presumed you were already working under different rules. (And of course, it's still the case that you can't tile the sphere with hexagons regardless unless you use the trivial tiling, because they can't meet fewer than 3 at a vertex, and that tiling is planar rather than spherical.) – Steven Stadnicki Jan 2 '15 at 18:52