Hilbert theorem and constant negative curvature surfaces

Question

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?

Henning Makholm · Accepted Answer · 2011-09-07 22:05:48Z

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).

answered Sep 7 '11 at 22:05

Henning Makholm
53.9k363135

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

asked	1 year ago
viewed	197 times
active	1 year ago

Hilbert theorem and constant negative curvature surfaces

1 Answer

Your Answer

Hilbert theorem and constant negative curvature surfaces

1 Answer

Your Answer

Sign up or log in

Post as a guest

Related