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Let $S=[(x,y)\in\mathbb{H}^{2}:0< x< 2\pi]$ where $\mathbb{H}^{2}$ is a hyperplane with standard metric. I.e., a strip whose boundary components are geodesics, both approaching a common infinite point.

My simple question is about whether it can be isometrically embedded in $\mathbb{R}^{3}$.

I don't remember exactly but I read some articles about an isometric embedding from a portion of $\mathbb{H}^{2}$ into $\mathbb{R}^{3}$. For example, it might have been proved that infinite polygons of some type can be isometrically embedded but I could not catch the meaning of the type the author said about. Or, I read in another article that every(?) equidistant strip in $\mathbb{H}^{2}$ can be embedded in $\mathbb{R}^{3}$. If this is true, the answer to my question would be positive.

Is there anyone who know about this content precisely?

How about $T=[(x,y)\in\mathbb{H}^{2}:0<\sqrt{{x}^{2}+{y}^{2}}<{e}^{2\pi}]$? This strip also has geodesic boundary components but both components are approaching different infinite point.

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The earlier MO question, "Largest hyperbolic disk embeddable in Euclidean 3-space?" (with beautiful pictures!) is possibly relevant: mathoverflow.net/questions/533 –  Joseph O'Rourke Jun 1 '12 at 11:42
@O'Rourke: Yeah, pablo seems to give a nice answer. But I think the pictures there are not that helpful to me actually... Thanks. –  KENSO Jun 1 '12 at 15:35
"Hyperbolic plane" is much more common and less confusing than "hyperplane"! –  Benoît Kloeckner Jun 1 '12 at 19:34
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1 Answer

up vote 7 down vote accepted

The comment given by Joseph gives indeed the answer... In Borisenko's survey (iop.org/EJ/abstract/0036-0279/56/3/R01), one can find in section 2.4 the following result due to Poznyak: "Any polygon belonging to one of the classes $M_1$ and $M_2$ can be regularly and isometrically embedded in $E^3$".

The polygons considered are intersections of a finite or denumerable set of closed halfplanes whose boundaries have no points in common. The class $M_1$ (which is the only one needed here) is defined as the class of all polygons for which there is a horocycle $O$ in the plane such that the greatest lower bound of the lengths of the orthogonal projections of the sides on this horocycle is positive.

As observed by Poznyak in his original article, any polygon with finitely many sides which does not contain a halfplane belongs to the class $M_1$. The band $S$ you described in your question is of this type, hence it can be isometrically immersed in the euclidean space $E^3$.

Here is the reference to the english translation of Poznyak's article:

“Isometric embedding in $E^3$ of certain non-compact domains of the Lobachevsky plane”, Mat. Sb. 102 (1977), 3–12; English transl., Math. USSR-Sb. 31 (1977), 1–8.

Concerning your surface $T$, it is a half-plane and thus cannot be isometrically immersed in $E^3$ (as explained in section 2.4 of the survey).

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@Bonnot: I see what you mean, especially the type ${M}_{1}$ of polygon. And sorry that there is a typo in my question: $T=[(x,y)\in\mathbb{H}^{2}:1<\sqrt{{x}^{2}+{y}^{2}}<{e}^{2\pi}]$. If I'm not mistaken, $T$ is also able to be embedded in ${E}^{3}$, right? Then one more question. What is the definition of polygons in $\mathbb{H}^{2}$ in general? Do they allow the ideal parts to be an edge?? –  KENSO Jun 1 '12 at 15:31
@Kenso: this new definition of $T$ changes things, now your new $T$ can be embedded (being also of type $M_1$). Usually hyperbolic polygons are closed convex set in the hyperbolic plane that can be expressed as the intersection of a locally finite collection of closed half-planes. Here Poznyak allows some ideal parts as you said, and calls those polygons (like S) "ideal polygons". One last comment: in the def. of $M_1$ one needs to add that every side has another one parallel to it (like in the surface S). –  Sylvain Bonnot Jun 1 '12 at 16:12
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