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I am interested in prescribing the boundary for $\mathbb{R}^3$-embeddings of simply connected closed subsets of the hyperbolic plane. In particular, can a hyperbolic disk isometrically embed into $\mathbb{R}^3$, and have a circle for its boundary?

This is doable for spherical disks, so maybe it could be possible for hyperbolic disks.

If it is not possible then perhaps a weaker statement holds. Can a circle in $\mathbb{R}^3$ be the boundary of a simply connected surface of constant negative Gaussian curvature?

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  • $\begingroup$ Can you ask your question more precisely? What is given and what are you trying to find? What are the quantifiers in your question? $\endgroup$
    – Lee Mosher
    Commented May 15, 2021 at 3:54
  • $\begingroup$ Hello, I will attempt that. Perhaps you can specify what you find imprecise though? Equip the upper half plane with the standard hyperbolic metric. Equip it with the distance function obtained by taking infima of paths. A disk of radius $r$ is defined in the usual way and will all be isometric. Can such a disk embed into $\mathbb{R}^3$ so that the boundary gets mapped to a circle? In other words, can a surface in $\mathbb{R}^3$ equipped with the pullback metric under inclusion be isometric to a disk in $\mathbb{H}^2$, and have its boundary component be a circle in $\mathbb{R}^3$? $\endgroup$ Commented May 15, 2021 at 4:09

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No. A hyperbolic disc or radius $r>0$ has circumference $2\pi\sinh(r)>2\pi r$. For an isometric embedding $\iota:S\to M$, the diameter of $\iota(S)$ in $M$ is no more than the diameter of $S$. Since a circle in $\mathbb{E}^3$ of circumference $2\pi\sinh(r)$ has diameter greater than $2r$, this leads to a contradiction.

The weaker case can be addressed using two isoperimetric-type inequalities:

  • Among simply connected regions in the hyperbolic plane $\mathbb{H}^2$ with fixed perimiter $P$, geodesic discs have the maximal area.
  • Among simply connected surfaces in Euclidean space $\mathbb{E}^3$ whose boundary is a circle, the flat disc minimizes the area.

The desired embedding cannot satisfy both of these inequalities simultaneously.

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  • $\begingroup$ You wouldn’t happen to know about the weaker proposition, would you? $\endgroup$ Commented May 15, 2021 at 6:29
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    $\begingroup$ @GeoffreySangston That one is also false, but, as far as I can tell, it requires some isoperimetric inequalities to prove. $\endgroup$
    – Kajelad
    Commented May 15, 2021 at 6:57
  • $\begingroup$ Thank you for this answer by the way. I believe it has taught me something. Now I’ll leave for another day what the requirements are on a smooth curve in $\mathbb{R}^3$, other than length and diameter, for it to be the boundary of an embedded hyperbolic disk. I saw your second respond after submitting this. I would triple upvote you if I could. Thanks! $\endgroup$ Commented May 15, 2021 at 7:03

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