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The Hilbert theorem states that there exists no complete regular surface S of constant negative gaussian curvature $K$ immersed in $R^3$.

Ok.. so I'm guessing that the surface of revolution of the Tractrix, which has constant negative curvature, is not complete or regular. Furthermore, in Minkowski 3d space we can define a surface with the following equation:

$-y_{0}^{2}+y_{1}^{2}+y_{2}^{2}=R^2$

i.e. the locus of points equidistant of the origin. This surface has constant negative curvature. In fact, a simple projection of this surface on the $y_1y_2$ plane gives us the Poincaré disc, a commonly used hyperbolic model, and other models. Is this surface also not complete or regular? If so, what about the Poincaré disc then?

I'm having a hard time understanding Hilbert's theorem. Also, I understand that the surface of revolution of the Tractrix is "smaller" than the surface defined in the Minkowski 3d space (and consequently the Poincaré disc and the other hyperbolic models), in the sense that you can fit one inside the other. Is this relevant? (sorry for the lack of technical terms ;)

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Hilbert's theorem refers to surfaces that are immersed in $R^3$ with its usual inner product. With an immersed manifold, it is implicit that the inner product on the tangent space has to be the one inherited from the ambient space.

In the case of the hyperboloid model of the hyperbolic plane, for the curvature to be negative, the inner product on the tangent space at each point has to be the one inherited by Minkowski space, that is, $<x,y> = x_1y_1 + x_2y_2 - x_3y_3$.

Another way to see that the hyperboloid model is not of constant negative curvature as a submanifold of $R^3$ is to use the intuitive idea that a surface with negative curvature is saddle-shaped at every point (whereas a surface with positive curvature is dome-shaped at every point). The hyperboloid is not saddle-shaped, but the surface of revolution of the tractrix is. Actually the hyperboloid immersed inside $R^3$ with its usual inner product would have positive curvature everywhere if I'm not mistaken.

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This was fast! thanks :) –  Diego May 31 '11 at 21:17
    
(Hit enter for new line and posted a incomplete comment) So, you mean the theorem is specifically for $R^3$ with the diag(1,1,1) metric? Where can I find this theorem in its more precise form? I can only find simplified versions of it. –  Diego May 31 '11 at 21:20

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