I have a difficulty with hyperbolic geometry.

Let $\mathbb{H}^{2}$ be a 2-dimensional hyperbolic plane.

(i.e., upper half plane in $\mathbb{R}^{2}$ with a metric $\frac{ds}{y}$)

(or, upper half plane in $\mathbb{C}$ with a metric $\frac{|dz|}{\textrm{Im}(z)}$ )

You may have heard about pseudosphere in $\mathbb{R}^{3}$. Let's denote the half-pseudosphere by $P$

This can be obtained by glueing both side($x=0$ and $x=2\pi$ parts) of

$\left[(x,y)\in\mathbb{H}^{2} : 0\leq x\leq 2\pi, y>1 \right]$

Denoting this quotient space by $A$(Note that $A$ is homeomorphic to a cylinder), we can now get a "globally isometrically embedding" map $\rho:A \rightarrow \mathbb{R}^{3}$ with $\rho(A)=P$

To be specific, $\rho(x,y) = (t-\tanh(t), \textrm{sech}(t)\ \cos(x), \textrm{sech}(t)\sin(x))$ where $t=\textrm{arccosh}(y)$

Now the question I have is following

: Is there another possible (global and isometric)embedding $\rho$ from $A$ into $\mathbb{R}^{3}$ ?

Actually I'm interested in $\rho(A)$ and by calculating, one can easily find that if $\rho(A)$ is a "surface of revolution", then it should be $P$(up to isometry of $\mathbb{R}^3$"

Thus I'm looking for $\rho(A)$ which is different from a surface of revolution.

Any idea?

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    $\begingroup$ Yes, check out Dini's surface: mathoverflow.net/questions/533/… $\endgroup$ – Ian Agol May 28 '12 at 20:10
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    $\begingroup$ @Ian: Kenso is apparently interested in embeddings/immersions of the cyclic quotient $A$ of the horodisk rather than of the horodisk itself, see his conversation with Robert below. $\endgroup$ – Misha May 30 '12 at 4:59
  • $\begingroup$ @Agol: I think Dini's surface is even not homeomorphic to a cylinder. $\endgroup$ – KENSO May 30 '12 at 5:37

You should be looking at the theory of Bäcklund transformations for surfaces of Gaussian curvature $K=-1$. There is a large literature on this, and there are many examples of pseudospherical immersions that are not surfaces of revolution. You should especially look at the work of Chuu-Lian Terng in this area. She, together with her husband, Richard Palais, have developed this theory quite a bit in the past 30 years, and they have some excellent graphics for these surfaces. Whether they can answer your specific question about isometrically embedding your particular domain $A$ into $\mathbb{R}^3$, I don't know, but that would be a good place to start.

  • $\begingroup$ Thank you so much, but one more question to verify? Do those immersions you mentioned have images each of which is "homeomorphic to a cylinder", in spite that they are not kinds of surfaces of revolution? $\endgroup$ – KENSO May 28 '12 at 11:48
  • $\begingroup$ @KENSO: No, typically, the images are not homeomorphic to a cylinder. One place to start would be at virtualmathmuseum.org/Surface/gallery_o.html, and to look at the file Pseudosphericalsurfaces.pdf that can be found there. $\endgroup$ – Robert Bryant May 28 '12 at 12:13
  • $\begingroup$ @Bryant: Then I'm still confused. The (half-)pseudosphere, or equivalently the set $A$, is homeomorphic to a (half-)cylinder so the isometrical image I want should be homeomorphic as well. Then what do you mean? Sorry for my poor understanding... $\endgroup$ – KENSO May 28 '12 at 12:27
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    $\begingroup$ @KENSO: I see what you mean. I wasn't taking the identification of the two 'sides' of $A$ as part of the problem. (By the way, in the definition of $A$, you have a typo, a '$\pi$' where there should be a '$2\pi$'.) I don't know about this full cylinder, but I believe there are embedded pseudospherical cylinders in $\mathbb{R}^3$ with one end asymptotic to a cusp that are not surfaces of revolution. Terng should be able to answer this. $\endgroup$ – Robert Bryant May 28 '12 at 15:05
  • $\begingroup$ @Bryant: Yes, my mistake. I editted the question as a whole. Anyway, do you mean only curvature $K=-1$ by a pseudospherical surface? I don't know exactly the definition of "pseudospherical", but if it means $K=-1$, I agree that there are many pseudospherical cylinders that are not surfaces of revolution in $\mathbb{R}^{3}$. But I think given $A$ has much more "rigidity"... $\endgroup$ – KENSO May 30 '12 at 4:24

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