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I know that the pseudosphere is a regular surface with Gaussian curvature $-1$ that is not complete, also this surface is not complete. Hilbert's theorem ensures that there is no isometric immersion of the hyperbolic plane $\mathbb{H}^2$ into $\mathbb{R}^3$, but if we remove a point from the hyperbolic plane, can it be immersed isometrically in $\mathbb{R}^3$? What could it be?

I read this: Dini's surface provides a analytic isometric embedding of the hyperbolic plane. That region includes hyperbolic disks of any finite radius, though big disks will be wrapped many times, very tightly around the screw axis of Dini's surface. But how could I prove it? I have already tried a few days without success.

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    $\begingroup$ The Dini surface, I guess $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 12, 2021 at 4:41
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    $\begingroup$ Does this answer your question? Largest hyperbolic disk embeddable in Euclidean 3-space? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 12, 2021 at 4:43
  • $\begingroup$ Thanks for the comments and I have indeed read this: Dini's surface provides a analytic isometric embedding of the hyperbolic plane. That region includes hyperbolic disks of any finite radius, though big disks will be wrapped many times, very tightly around the screw axis of Dini's surface. But how could I prove it? I have already tried a few days without success. $\endgroup$
    – Zaragosa
    Commented Jul 12, 2021 at 7:05
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    $\begingroup$ Concerning the geometry of Dini's surface and the portion of the hyperbolic plane that it represents, you might find useful the discussion at the MO question mathoverflow.net/questions/149842/…. $\endgroup$
    – Robert Bryant
    Commented Jul 16, 2021 at 10:53
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    $\begingroup$ Also, you have to be careful, because there is more than one Dini surface with Gauss curvature -1. In fact, if you look at the formulae, you'll see the two positive parameters $(a,b) = (\cos\tau,\sin\tau)$ (with $0<\tau<\pi/2$), and for the surface with those given parameters, there is a maximal radius $r(\tau)$ for a hyperbolic disc in that surface. In fact, $r(\tau) = \tfrac12\log\bigl((1+b)/(2-2a)\bigr)$. Thus, to get a hyperbolic disk of a given radius embedded in a Dini surface, you have to choose $\tau$ sufficiently small. $\endgroup$
    – Robert Bryant
    Commented Jul 19, 2021 at 19:34

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