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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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2 Answers 2

up vote 2 down vote accepted

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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