6
$\begingroup$

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 $\mathbb{R}^3$:

$T_7$ out to radius 5, Poincare disk model $T_7$ out to radius 5, embedding

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$ – Benoît Kloeckner Feb 24 '14 at 18:59
  • $\begingroup$ Comments hopefully addressed, let me know if you have other suggestions. I'm not sure why the images are so poorly arranged. $\endgroup$ – Geoffrey Irving Feb 24 '14 at 19:06
  • 2
    $\begingroup$ 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. $\endgroup$ – Ian Agol Feb 24 '14 at 20:40
  • $\begingroup$ @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? $\endgroup$ – Geoffrey Irving Feb 24 '14 at 21:04
  • $\begingroup$ yes, icosahedron, not dodecahedron! $\endgroup$ – Ian Agol Feb 24 '14 at 23:29
2
$\begingroup$

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).

I learned about this connection between triply periodic surfaces and hyperbolic tilings from this paper by Ramsden, Robins and Hyde. They also have a nice website with more information. They consider more general tilings, not just of equilateral triangles. As I said, to make these into the flat equilateral triangular tilings you are looking for you may need to do some fiddling around.

Here's a picture I just found of $T_8$ on Schwarz's D-surface by Doug Dunham.

Doug Dunham's tiling

$\endgroup$
2
  • $\begingroup$ Like Ian Agol's comment, this addresses isometric immersions, not embeddings. I'm a bit too tired to think through more details at the moment, so apologies for the sloppy description above. $\endgroup$ – j.c. Feb 24 '14 at 21:44
  • $\begingroup$ Thank you for the wonderful picture! Embeddings are certainly much harder, but having pictures for the immersion case is great. $\endgroup$ – Geoffrey Irving Feb 24 '14 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.