Can a hyperbolic disk isometrically embed into $\mathbb{R}^3$, with a circle for its boundary?

Question

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?

Answer 1

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.