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.

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

2 Answers 2


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 $$\xi^{\ft}(\rR_{\xi\eta}\eta) = \epsilon \xi^{\ft}(\xi \eta^{\ft}\eta - \eta\xi^{\ft}\eta)\\ =\epsilon( \xi^{\ft}\xi\eta^{\ft}\eta- (\xi^{\ft}\eta)^2). $$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .