# Tilings of the plane and meromorphic functions on the plane

This question has three up-votes on m.s.e. but isn't getting any answers.

Every textbook says every doubly-periodic meromorphic function on $\mathbb C$ has a fundamental domain that is a parallelogram. So someone asked whether one could do with a tiling of the plane by regular hexagons what is thus done with parallegorams: Is there some meromorphic function whose restriction to each hexagon in the tiling is a shift of its restriction to any of the other hexagons? And indeed there is: Dixon's elliptic functions do just that. (That in no way conflicts with the existence of the aforementioned parallelogram: just find one whose four vertices are centers of four suitably located hexagons.)

This inspires two other questions another question:

• What about other periodic tilings? For example, there is a periodic tiling by hexagons, squares, and triangles. Might the restriction of some doubly periodic meromorphic function to any of those be a shift or maybe a shift followed by a rotation, of the restriction to any of the others? For that tiling, rotation as well as translation becomes relevant. For which tilings does such a thing exist?

• What about aperiodic tilings with a finite set of sizes and shapes of tiles? We'd want a function meromorpic on the whole plane whose restriction to any tile is a shift-plus-rotation of its restriction to any of the infinitely many tiles of the same shape. [The answer to this one is obvious.]

• Periodic tilings: I think you are being misled by the pretty patterns. If the automorphism group acts transitively on the polygons you want the function to repeat on then you're just looking for functions which are invariant under the automorphism group, so this is why the hexagon question becomes the parallelogram question. Aperiodic tilings -- this isn't going to happen. If a holo fn is the same on two shapes one of which is a rotation of the other then by the identity principle it's invariant by that rotation and now you're in trouble -- this will surely turn into a proof that allfnsarecst – znt Aug 28 '16 at 17:29
• @znt : Maybe you should make this an answer. It doesn't fully answer the question of which periodic tilings this can be done with, but maybe just a bit more does it. $\qquad$ – Michael Hardy Aug 28 '16 at 18:05
• @znt : That I'm extraordinarily rusty at this is demonstrated by the point about aperiodic tilings, since that's obvious. Moving on to periodic tilings, the one with a square on each side of a hexagon and then triangular interstices could follow the pattern I describe with the squares only if there's a lot of rotational symmetry in the hexagons, and then you might as well just make each hexagon consist of six triangular tiles. But then not all trianglular tiles would behave the same way: it would depend on whether the triangle is surrounded by three squares. $\qquad$ – Michael Hardy Aug 28 '16 at 18:23
• No, this: projecteuclid.org/download/pdf_1/euclid.mmj/1028998184 This should be thought of as conformal, rather than euclidean tiling. The finite version is currently fashionable under the name of "dessins d'enfants". – Adam Epstein Aug 29 '16 at 17:15
• Not unrelated is recent work of Chris Bishop on "quasiconformal folding". – Adam Epstein Aug 29 '16 at 18:15