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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 $$\mathrm{PS}=\{z \mid \mathrm{Im} z \ge 1,\; -\pi < Re z \le π\}$$ 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 $\mathrm{PS}$. This corresponds to a hyperbolic disk of radius $\operatorname{arctanh}(π/(1+π)) \sim 0.993$.

Surely one can do better?

Edit 2: 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

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4 Answers 4

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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 (or rather family of surfaces), which are isometrically embedded one-sided tubular neighborhoods of a geodesic in the hyperbolic plane (see https://mathoverflow.net/a/149884/1345), seems to do the trick since they contain arbitrarily large disks in the hyperbolic plane by varying the parameter so that the radius of the tubular neighborhood grows. See Dini's Surface at the Geometry Center.

alt text

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  • $\begingroup$ 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.
    Commented Nov 2, 2009 at 3:12
  • $\begingroup$ yes, thanks, I changed it to horodisk. $\endgroup$
    – Ian Agol
    Commented Nov 2, 2009 at 3:32
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    $\begingroup$ 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.
    Commented Nov 25, 2013 at 15:29
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    $\begingroup$ @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
    Commented Nov 25, 2013 at 23:51
  • $\begingroup$ @IanAgol it looks like the parameter is bounded by the curvature though. What about a fixed curvature? $\endgroup$
    – Christopher King
    Commented Oct 19, 2019 at 15:22
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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)

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  • $\begingroup$ 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.
    Commented Oct 21, 2009 at 21:43
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    $\begingroup$ Could you also post an answer here, please? $\endgroup$
    – Ilya Nikokoshev
    Commented Nov 2, 2009 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.
    Commented Nov 2, 2009 at 3:13
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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.

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  • $\begingroup$ 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.
    Commented Oct 17, 2009 at 21:24
  • $\begingroup$ for n chosen sufficiently larger than k. $\endgroup$
    – Narasimham
    Commented Oct 26, 2015 at 21:19
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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.

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  • $\begingroup$ 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.
    Commented Oct 15, 2009 at 0:47
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    $\begingroup$ 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 ♦
    Commented Oct 15, 2009 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.
    Commented Oct 15, 2009 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
    Commented Oct 15, 2009 at 1:14
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    $\begingroup$ @j.c. Link is broken $\endgroup$
    – Akiva Weinberger
    Commented Jan 4, 2023 at 21:03

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