# Reference request: embedding the hyperbolic triangulation in $\mathbb{R}^3$

Let $T_d$ be the infinite valence $d$ triangulation of the hyperbolic plane, where each triangle is equilateral and $d \ge 7$. Question: Is there an isometric embedding from $T_d \to \mathbb{R}^3$? Here an isometric embedding means that each triangle is flat, with Gaussian curvature only at the vertices.

I've been told that Thurston and others have tackled this problem and failed, so apparently it is fairly hard. Actual question: Are there references discussing existing attempts?

The problem is likely very different for even vs. odd $d$, since for even $d$ we can almost fold along geodesics down to a single triangle. That is, given any $\epsilon > 0$, there is an almost isometric embedding of $T_{2d}$ into an $\epsilon$ neighborhood of a single triangle, where the isometry is off by a factor of $1 \pm \epsilon$.

For odd $d$, at least, my intuition is that it is completely unreasonable that such an embedding exists: the number of triangles in a ball grows exponentially with the radius due to the exponential growth of hyperbolic space, which results in an exponential number of arbitrarily closely layered sheets.

Here is the radius 5 portion of $T_7$ in the Poincare disk model, together with an embedding into $\mathbb{R}^3$:

• Maybe you could precise a few points: how exactly is defined $T_d$? I guess the triangle are regular, and that the valence is the valence of vertices. More important, I guess that the "isometric" embedding you seek is combinatorially isometric but realized with flat triangle, is that right? And of course, you do not mean isometric for the extrinsic distance, but that might be better to say so explicitly. – Benoît Kloeckner Feb 24 '14 at 18:59
• Comments hopefully addressed, let me know if you have other suggestions. I'm not sure why the images are so poorly arranged. – Geoffrey Irving Feb 24 '14 at 19:06
• There's an isometric immersion of $T_7$ into $R^3$, which may be obtained by taking a dodecahedron, and attaching 4 octahedra to non-adjacent faces, and repeating (one obtains an infinite diamond lattice, with dodecahedra representing atoms, and octahedra representing bonds). Here, I'm assuming you want the triangles to be Euclidean equilateral triangles. – Ian Agol Feb 24 '14 at 20:40
• @IanAgol: Thanks! I'm having trouble visualizing that, though. How do you attach an octahedron to a pentagonal face of a dodecahedron? Do you mean icosahedron? – Geoffrey Irving Feb 24 '14 at 21:04
• yes, icosahedron, not dodecahedron! – Ian Agol Feb 24 '14 at 23:29

Ian Agol's example in the comments generalizes -- you can probably tile any triply periodic surface with equilateral triangles (after some fiddling you might even make them flat triangles) and the tiling will admit a covering map from a tiling of the hyperbolic plane because of the negative Gaussian curvature. I am not sure if you can get all $T_d$ this way (the tilings you get don't necessarily have constant valence).
Here's a picture I just found of $T_8$ on Schwarz's D-surface by Doug Dunham.