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Hilbert's theorem tells us that there is no immersion in $\mathbb{R}^3$ with negative Gauß curvature that is complete. Despite, there are some models of surfaces with negative Gauß-curvature like the Pseudo-sphere. So what is the problem with these models?

Question 1

How do I see that for example the pseudosphere is not complete?

Then, it is possible to construct models of hyperbolic space by defining a metric tensor on objects like the Poincaré half-plane or Poincaré disc.

Question 2

Are these objects proper in the sense that they are complete models of hyperbolic space? (although they are not immersions in any $\mathbb{R}^n$)?

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I understand that your question is about the hyperbolic plane, not hyperbolic space of arbitrary dimension.

The Pseudosphere is usually drawn as a symmetric object with a sharp edge; this object is not a smooth surface because of that edge. One half of that object is a smooth surface of constant negative curvature. It has the boundary (the edge), unlike the hyperbolic space. If you remove the boundary, it is not complete because it is not a closed subset of $\mathbb R^3$. A sequence of points converging to a point on the edge is Cauchy but does not converge within the surface. So, whichever way we cut it, we either don't have a smooth surface-without-boundary, or we don't have completeness.

Poincaré disk is an isometric model of the hyperbolic plane, and so is the halfplane with the hyperbolic metric. Yes, they share all properties that the hyperbolic plane has. They serve as concrete representatives of the isometry class of complete surfaces of constant curvature $-1$.

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  • $\begingroup$ NO the Poincaré disk and the Poincaré halfplane models of hyperbolic geometry are not isometric, they are conformal $\endgroup$ – Willemien Dec 28 '14 at 8:24
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    $\begingroup$ When equipped with their hyperbolic metric, they are isometric. For each $n$ there is a unique, up to isometry, simply-connected complete $n$-manifold of constant negative curvature $-1$. See Wikipedia. $\endgroup$ – user147263 Dec 28 '14 at 8:30
  • $\begingroup$ @Willemien -- Last time I checked, they are isometric. We define the metric on the disk so that a certain diffeomorphism from the half-plane to the disk is an isometry. $\endgroup$ – Robin Goodfellow Dec 28 '14 at 18:51
  • $\begingroup$ @RobinGoodfellow I don't understand what "isometric invariant" means, I will do some study and maybe make a question of my own about it. but it looks behavior is right (but then what about hilberts Theorem en.wikipedia.org/wiki/… ? ) $\endgroup$ – Willemien Dec 28 '14 at 20:47
  • $\begingroup$ @RobinGoodfellow see math.stackexchange.com/questions/1083930/… $\endgroup$ – Willemien Dec 28 '14 at 22:08

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