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I found in several books by Aminov, Gromov and other current authors that deal with isometric dives, that the horocircle can be immersed isometrically in the pseudosphere (or failing that, on the Dini' surface),

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although visually I found several references on this platform or in books, but I have not been able to find what would be the explicit immersion, any suggestions?

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We know from Hilbert's theorem that the whole plane cannot be isometrically immersed in $\mathbb{R}^3$ but with this at least part of it is, hence my interest in getting explicitly which immersion could be. Knowing that there is at least a part of the hyperbolic plane that can be isometrically immersed in $\mathbb{R}^3$, the next question I will work on will be: What is the largest region of the hyperbolic plane with this characteristic?

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    $\begingroup$ What exactly do you mean by an explicit immersion? Do you want a closed form map from some model of a horodisk into R^3? $\endgroup$
    – Magma
    Jul 26, 2021 at 23:23

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Consider the upper half-plane model of $H^2$, it has a horodisk $D = \{(x, y) \in H^2 : y > 1\}$.

Let $t \mapsto (f(t), g(t))$ be a parametrization of a tractrix starting at $(0, 1)$ with $\lim_{t \to \infty}g(t) = 0$.

Now the map

$$\varphi: \mathbb D \to \mathbb R^3, (x, y) \mapsto (g(\ln(y)) \cos x,g(\ln(y)) \sin x, f(\ln(y)))$$

covers a pseudosphere isometrically.

Unrolling the pseudosphere yields a horodisk in $H^2$, and unrolling Dini's surface yields not a horodisk, but a different shape: the convex hull of a hypercycle (in other words, the area between the hypercycle and its axis). The horodisk has infinite radius, and the other shape may have arbitrary finite width depending on how tightly the outer helix of the corresponding Dini's surface is wound. Both shapes have infinite area, but neither contains the other.

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  • $\begingroup$ Thank you very much for your interesting answer, I am left with a question look: if $(u,v,w)=(g(\ln(y)) \cos x,g(\ln(y)) \sin x, f(\ln(y)))$ then $\varphi^*g=du^2+dv^2+dw^2=g^2(\ln(y)) dx^2+\frac{(g'(\ln(y)))^2+(f'(\ln(y)))^2}{y^2}dy^2$, so that means that it doesn't work for all parameterizations of the pseudosphere, not even for one parameterized by arc length. Were you thinking of some $(f(t), g(t))$ in particular? But it is possible that I have confused something. $\endgroup$
    – Zaragosa
    Jul 27, 2021 at 18:15
  • $\begingroup$ Yes, I was thinking of a parametrization by arc length. Note that $(f')² + (g')² = 1$ and $g(t) = e^{-t}$. $\endgroup$
    – Magma
    Jul 27, 2021 at 18:54
  • $\begingroup$ Great thank you so much. One last question, how did you come up with the $\ln(y)$ out there in an article I read that $\arccosh(y)$ can also be applied, but how specifically do these work make everything work well? what motivates you? $\endgroup$
    – Zaragosa
    Jul 27, 2021 at 23:24
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    $\begingroup$ In the half-plane model, the length of a line segment from $(x,1)$ to $(x,y)$ is $\ln(y)$, so that's how far I need to go along the tractrix to ensure the correct geodesic distance between $\varphi(x, 1)$ and $\varphi(x, y)$. $\endgroup$
    – Magma
    Jul 28, 2021 at 12:25

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