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I'm trying to get an intuitive grasp of the relations between the one- and two-sheet hyperboloids and the two-dimensional hyperbolic sphere.

Anthony Zee uses the two-sheet hyperboloid $x^2+y^2-z^2=-1$ to derive the two-dimensional hyperbolic sphere $H^2$ on pages 92-93 of his Einstein Gravity in a Nutshell.

The two-sheet hyperboloid seems to have a positive scalar curvature, like the normal sphere $S^2$, whereas the one-sheet hyperboloid has a negative scalar curvature. Why do we need the two-sheet hyperboloid to derive $H^2$? Wouldn't the one-sheet hyperboloid make more sense, given that its scalar curvature is negative.

Andy Miles
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Embedded in standard Minkowski space $\mathbf{R}^{2, 1}$, each sheet of the two-sheeted hyperboloid acquires a Riemannian metric of constant negative curvature.

Qualitatively, the induced metric on the upper sheet $H$ is Riemannian because each tangent plane is spacelike. To show the curvature of $H$ is constant and negative, it may be easiest to show that the identity component of the orthogonal group $O(2, 1)$ acts transitively (and isometrically) on $H$, and to show the curvature of $H$ is negative at $(0, 0, 1)$. For the latter, stereographic projection from $(0, 0, -1)$ to the open unit disk $\{(x, y, z) : x^{2} + y^{2} < 1, z = 0\}$ defines an isometry with the Poincaré metric, see for example Problem in understanding models of hyperbolic geometry.

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Andrew D. Hwang
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  • So to obtain a Riemannian manifold with negative curvature, the two-sheeted hyperboloid needs to be embedded into Minkowski space, but the one-sheeted one into Euclidean $\mathbb R^3$. Is that correct? – Andy Miles Dec 02 '16 at 19:37
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    The metric induced by the Euclidean metric on the one-sheeted hyperboloid has negative Gaussian curvature, yes, though of course the curvature isn't constant. (In fact, the integral of the Gaussian curvature is finite.) – Andrew D. Hwang Dec 02 '16 at 22:01