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Let us consider the tractroid (pseudosphere) obtained by rotation from the tractrix curve. The surface is not defined on the "big rim", so it is not a complete set. Hilbert's theorem states that there exists no complete regular surface of constant negative Gaussian curvature immersed in $\mathbb{R}^{3}$.

Then, if we truncate the tractroid surface with two planes that are orthogonal to the rotation axis, apparently we obtain a compact surface with negative curvature... but that it not possible by Hilbert's theorem. Is there a loss of regularity along the borders?

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It's compact, but obviously not complete.

(Wait, are you including the truncating surfaces in the result? Then it fails to be differentiable at the edges, so it doesn't have a Gaussian curvature there, much less a constant negative one).

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To what surface do you refer when you say "compact"? And what do you mean with "compact but not complete" when we are in metric spaces? –  user14174 Sep 8 '11 at 21:13
    
"Compact" is from your question: "we obtain a compact surface ..." I hope you know which one you meant there; I am only guessing. Perhaps you meant that the truncated pseudosphere is a topologically compact set? In that case it is not a manifold at the boundary. A complete surface is one where every geodesic can be extended arbitrarily far in each direction. It's one of the assumptions in Hilbert's theorem. –  Henning Makholm Sep 8 '11 at 21:26

The " big rim" is a cuspidal edge or equator; it has infinity and zero principal curvatures. In between them are zero normal curvature asymptotic lines having constant geodesic torsion across cusp, even if curvatures vary continuously from - infinity to + infinity, across the rim.

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