# Why is the hyperboloid model a hyperbolic model?

Everyone presents the hyperboloid as a hyperbolic model, but I don't understand what this means. I thought that a hyperbolic model was a Riemannian variety isometric to the Poincaré disk. Eventough we have a Riemannian structure on the hyperboloid, it doesn't have constant -1 curvature, so it's not isometric to the Poincaré disk.

• Lee’s Riemannian geometry text goes through these examples very nicely, I think that will help greatly. Commented Sep 3, 2023 at 20:54
• But it does have constant sectional curvature $-1$. Why do you think not? Commented Sep 3, 2023 at 22:35

## 2 Answers

The underlying set of the hyperboloid model is $$H^2 = \{(x,y,z)\in\Bbb R^3 \mid x^2+y^2-z^2=-1\mbox{ and }z>0\},$$however, the metric it inherits from $$\Bbb R^3$$ is not the standard Riemannian metric $$\mathtt{g}_E = {\rm d}x^2+{\rm d}y^2+{\rm d}z^2$$, but instead the Lorentzian metric $$\mathtt{g}_M = {\rm d}x^2+{\rm d}y^2-{\rm d}z^2$$.

Both $$\mathtt{g}_E$$ and $$\mathtt{g}_M$$ are flat and restrict to Riemannian metrics on $$H^2$$. The Gaussian curvature of the restriction of $$\mathtt{g}_E$$ to $$H^2$$ is not constant (this is what probably prompted you to ask this question), but it turns out that the restriction of $$\mathtt{g}_M$$ to $$H^2$$ does have constant Gaussian curvature equal to $$-1$$.

The manifold $$H^2$$ equipped with the metric induced from $$\mathtt{g}_M$$ is indeed isometric to the Poincare disk.

$$\newcommand{\ft}{\mathfrak{t}}$$ $$\newcommand{\Two}{\mathbb{II}}$$ $$\newcommand{\rR}{\mathrm{R}}$$ $$\newcommand{\R}{\mathbb{R}}$$ Here is a treatment covering a slightly larger class of conic hypersurfaces, covering both the sphere and the hyperboloid.

Let $$\ft$$ be a symmetric nondegenerate matrix in $$\R^{N\times N}$$. We write $$x^{\ft}y$$ for the (semi-Riemannian) product $$x^T\ft y$$, where $$x^T$$ is the usual transpose of a matrix, where $$x, y\in \R^N$$ are considered as column vectors. For $$\epsilon = \pm 1$$, consider the manifold $$M$$ defined by $$x^{\ft}x = \epsilon.$$ We assume $$M$$ is nonempty, or $$\epsilon\ft$$ is not negative definite. Then $$M$$ is smooth as the Jacobian of the constraint is $$2x\ft$$ is of full rank $$1$$. An any $$x\in M$$, the tangent space is defined by the equation $$x^{\ft}\eta = 0$$, the projection to the tangent space $$T_xM$$ is given by $$\Pi(x)\omega = \omega - \epsilon xx^{\ft}\omega$$ which is compatible with the inner product on $$\R^N$$ defined by $$\ft$$, we check easily $$x^{\ft} \Pi(x)\omega=0$$ and $$\omega - \Pi(x)\omega$$ is proportional to $$x$$, thus normal to tangent vectors. From here, the second fundamental form for two tangent vectors $$\xi, \eta$$ is $$\Two(\xi, \eta) = (D_{\xi} \Pi)\eta = -\epsilon x \xi^{\ft}\eta -\epsilon \xi x^{\ft}\eta = -\epsilon x \xi^{\ft}\eta$$ and by the Gauss-Codazzi theorem, the sectional curvature numerator is $$(\rR_{\xi\eta}\eta)^{\ft}\xi = \Two(\xi, \xi)^{\ft}\Two(\eta, \eta) - \Two(\xi, \eta)^{\ft}\Two(\xi, \eta) =\epsilon( \xi^{\ft}\xi\eta^{\ft}\eta- (\xi^{\ft}\eta)^2),$$ where we use the relation $$x^{\ft}x =\epsilon$$.

With $$N = n+1$$, for the sphere, $$\ft=I_{n+1}, \epsilon =1$$, for the hyperboloid model $$\ft = diag(1,\cdots,1, -1)$$, $$\epsilon=-1$$ and the sectional curvature is $$\epsilon$$ in both cases.

For other choices of $$\ft$$, the metric on $$TM$$ may be only semi-Riemannian, $$\xi^{\ft}\xi\eta^{\ft}\eta- (\xi^{\ft}\eta)^2$$ may be zero for linearly independent $$\xi, \eta$$.

For three tangent vectors, we can compute $$\rR_{\xi\eta}\phi$$ directly to get the same result. In fact, the Christoffel function, the operator version of the Christoffel symbol is $$\Gamma(\xi, \eta) = -(D_{\xi} \Pi)\eta = \epsilon x \xi^{\ft}\eta$$ and the curvature is $$\rR_{\xi\eta}\phi = (D_{\xi}\Gamma)(\eta, \phi) - (D_{\xi}\Gamma)(\eta, \phi) + \Gamma(\xi, \Gamma(\eta, \phi)) - \Gamma(\eta, \Gamma(\xi, \phi))\\ = \epsilon \xi \eta^{\ft}\phi - \epsilon \eta\xi^{\ft}\phi + x\xi^{\ft}x\eta^{\ft}\phi - x\eta^{\ft}x\xi^{\ft}\phi\\ =\epsilon (\xi \eta^{\ft}\phi - \eta\xi^{\ft}\phi).$$ Thus, the sectional curvature numerator is $$\xi^{\ft}(\rR_{\xi\eta}\eta) = \epsilon \xi^{\ft}(\xi \eta^{\ft}\eta - \eta\xi^{\ft}\eta)\\ =\epsilon( \xi^{\ft}\xi\eta^{\ft}\eta- (\xi^{\ft}\eta)^2).$$