Hilbert proved that there's no complete regular ($C^k$ for sufficiently large $k$) isometric embedding of the hyperbolic plane into $\mathbb{R}^3$. On the other hand, the pseudosphere is locally isometric to the hyperbolic plane up to its cusps (though it has the topology of a cylinder). What's the largest hyperbolic disk (with Gaussian curvature -1) that can be smoothly (or $C^2$, say) isometrically embedded in $\mathbb{R}^3$?

edit: This doesn't seem to be getting many views, so I'll bump this by adding in a rather easy lower bound from the pseudosphere. First, the pseudosphere is parametrized by the region $PS=\{z | Im z ≥ 1, -π < Re z ≤ π\}$ on the upper half plane model of $H^2$. Let $z=x+iy$, so that ordered pairs $(x,y)∈ H^2$ when $y>0$.

Next, Euclidean circles drawn on in the upper-half plane model with center $(x,y\cosh r)$ and radius $y\sinh r$ correspond to hyperbolic circles with center $(x,y)$ and radius $r$. I can fit a Euclidean circle of radius $π$ centered at $(0,1+π)$ into the region $PS$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

edit2: fixed mistakes in formulas above (didn't affect the bound). Here're some pictures:

disk in the pseudosphere in the upper half plane model disk on the embedded pseudosphere


I didn't see the exact answer to your question in the Borisenko paper, since section 2.4 only seems to address immersions of subsets of ℍ2 into ℝ3. However, a perturbation of the pseudosphere, Dini's surface, which is an isometrically embedded horodisk, seems to do the trick since it contains arbitrarily large disks in the hyperbolic plane. See Dini's Surface at the Geometry Center.

alt text

  • $\begingroup$ Thank you. I believe you mean that Dini's surface is an isometrically embedded horodisk, though? (The horocycle ought to be the cusp of Dini's surface.) I had been meaning to come back to update this question after following up on some of the references in Borisenko's paper, where I found that the immersions weren't particularly close to what I had in mind. $\endgroup$ – j.c. Nov 2 '09 at 3:12
  • $\begingroup$ yes, thanks, I changed it to horodisk. $\endgroup$ – Ian Agol Nov 2 '09 at 3:32
  • 2
    $\begingroup$ Robert Bryant shows that Dini's surface is not a horodisk in his answer here mathoverflow.net/questions/149842/… . Instead it is the region between a geodesic and a curve of constant geodesic curvature. $\endgroup$ – j.c. Nov 25 '13 at 15:29
  • 1
    $\begingroup$ @j.c.: thanks for the correction - I didn't realize this, but now that I read Robert's answer, I should have realized it wasn't a horodisk. I still think the answer works, by varying the parameter (in the limit, it should approach the immersed horodisk of the pseudosphere in the appropriate sense). $\endgroup$ – Ian Agol Nov 25 '13 at 23:51
  • $\begingroup$ @IanAgol it looks like the parameter is bounded by the curvature though. What about a fixed curvature? $\endgroup$ – PyRulez Oct 19 '19 at 15:22

Several russian geometers have addressed this question. I suggest you a survey on Isometric immersions by A. Borisenko (2001, I think) in Russian Mathematical Surveys (it is in English)

  • $\begingroup$ Thanks! If you mean iop.org/EJ/abstract/0036-0279/56/3/R01, then it looks like section 2.4 has exactly what I've been looking for! $\endgroup$ – j.c. Oct 21 '09 at 21:43
  • $\begingroup$ Could you also post an answer here, please? $\endgroup$ – Ilya Nikokoshev Nov 2 '09 at 0:06
  • $\begingroup$ I will edit the question above with an answer (and hopefully some graphics) at some point in the future. $\endgroup$ – j.c. Nov 2 '09 at 3:13

I can't comment on Noah's answer, so:

The reason for the $C^1$ condition is that Nash's $C^1$ embedding theorem says that any Riemannian $k$-manifold with a short embedding into $\mathbb{R}^n$ has an isometric $C^1$ embedding into $\mathbb{R}^n$ for any $n > k$. In particular, there is an isometric $C^1$ embedding of the hyperbolic plane into $\mathbb{R}^3$. The very beautiful proof proceeds by "pleating" the embedding to be closer and closer to isometric, while keeping it contained in a not-much-larger region.

  • $\begingroup$ Thanks, I didn't know that about the pleating. Like I said, I'm more interested in the smooth cases, but I guess I'll look into the proof of Nash-Kuiper at some point as well. $\endgroup$ – j.c. Oct 17 '09 at 21:24
  • $\begingroup$ for n chosen sufficiently larger than k. $\endgroup$ – Narasimham Oct 26 '15 at 21:19

I think that you can get an arbitrarily large disk. The proof is by crochet. Since there's a pattern for crocheting constant negative curvature disks where you increase the radius as you go and since we live in 3-space, it follows that you can get arbitrarily large disks.

See this TED talk for some cool applications of hyperbolic crochet to biology, or this article for a more rigorous explanation.

  • $\begingroup$ It's my reading of the second link that these crotcheted surfaces are not even C^1 though. Thanks for the links though. $\endgroup$ – j.c. Oct 15 '09 at 0:47
  • 2
    $\begingroup$ I should point out that the crochet article linked above has references to research literature. In particular, it mentions that there are no C^2 embeddings of the whole plane, but there is a C^1 embedding. $\endgroup$ – S. Carnahan Oct 15 '09 at 0:50
  • $\begingroup$ That's actually the reason I put in C^2 in my question. There's a recent book on isometric embeddings, by Han and Hong, which tackles related problems from an analysis viewpoint. There seem to be plenty of results on local isometric embedding without any explicit bounds or constructions, unless there are general ways to back out bounds from such proofs that I'm not aware of. $\endgroup$ – j.c. Oct 15 '09 at 0:56
  • $\begingroup$ I'm confused, my reading of that link is that they don't give a C^1 embedding of the whole hyperbolic plane. But I can't see how they wouldn't be C^1 for arbitrarily large (but finite) radius. $\endgroup$ – Noah Snyder Oct 15 '09 at 1:14
  • $\begingroup$ Yeah, I was just looking at the journal version of that article to see if there were more details, and there the sentence reads "The finite surfaces described here can apparently be extended indefinitely, but they appear always not to be differentiably embedded (see Figure 12)." Figure 12 is a picture of one of their crotcheted surfaces with the center pulled up so that it kind of looks like a pseudosphere, with the caption "Crotcheted pseudosphere". $\endgroup$ – j.c. Oct 15 '09 at 1:20

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.