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I've recently learned about Amsler's surface, a surface of constant negative Gaussian curvature. If I understand things correctly, there is a whole family of such surfaces, differing in the angle of intersection for the two lines that generate it. But I guess I'm only interested in the most symmetric version, where these lines are orthogonal. I'd like to know how large a portion of the hyperbolic plane I can embed on this surface.

So if the unit of length is chosen such that the Gaussian curvature of the surface becomes $-1$, what is the radius of a circle which is centered at the intersection of the asymptotic lines and just touches the cuspidal edges of the surface? A “circle” here would be the set of points on the surface with fixed geodesic distance to a central point. That intrinsic circle would not be a planar circle in the 3D embedding of the surface.

The following illustration of Amsler's surface was taken from the Gallery of pseudospherical surfaces by A. Ovchinninkov. I'm not sure how accurately it matches what I ask for, since there are other figures on the web which look somewhat different from this.

Fig. 14 of the referenced document

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up vote 2 down vote accepted

this question was studied in http://arxiv.org/abs/1005.4442 . The figure above was taken from Amsler's original paper and the dashed curve closely matches the largest geodesic circle cut from the Amsler surface.

Interestingly, the authors show that arbitrarily large isometric embeddings of the hyperbolic plane exist. They show this by taking an Amsler surface with an angle of $\phi_n=\frac{\pi}{n}$ between the two generating lines and periodically extending the surface bounded between these two lines. However, it is still true that there does not exist an isometric embedding of the entire hyperbolic plane.

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