Let $\mathbb{H}^n_K$ denote the set of points of the hyperboloid model, which models the hyperbolic space with sectional curvature $K<0$. So

$$ \mathbb{H}^n_K=\left\{x\in\mathbb{R}^{n+1} \,\middle|\, \langle x,x\rangle_*=\frac{1}{K} \,\land\,x_1>0\right\} $$

where $\langle x,y\rangle_*=-x_1y_1+\sum_{i=2}^{n+1}x_iy_i$ denotes the Lorentz inner product.

Question: How do you derive the metric tensor $g$ for any sectional curvature $K$?

My solution so far:

The first coefficient of a point $\mathbf{x}$ on the manifold is a function of the remaining coefficients:

$$ \mathbf{x}=(x_1,\underbrace{x_2,\ldots, x_{n+1}}_{=\mathbf{x'}})\in\mathbb{H}^n_K \Longleftrightarrow x_1=\sqrt{||\mathbf{x'}||_2^2-\frac{1}{K}}. $$

So a point $\mathbf{x}\in\mathbb{H}^n_K$ is given as a function of $n$ parameters: $$ \mathbf{x}(\underbrace{u_1,\ldots,u_n}_{=\mathbf{u}}) = \left( \underbrace{\sqrt{||\mathbf{u}||_2^2-\frac{1}{K}}}_{x_1}, \overbrace{\underbrace{u_1}_{x_2}, \ldots,\underbrace{u_i}_{x_{(i+1)}},\ldots, \underbrace{u_n}_{x_{n+1}}}^{=\mathbf{x'}}\right)^T \in\mathbb{H}^n_K. $$

Now, a set of tangent vectors $\{\mathbf{x}_1,\ldots,\mathbf{x}_n\}$ spanning the tangent space $\mathcal{T}_{\mathbf{x}}\mathbb{H}^n_K$ is given by $$ \mathbf{x}_i(u_1,\ldots,u_n) := \frac{\partial \mathbf{x}}{\partial u_i} = \left( \frac{u_i}{\sqrt{||\mathbf{u}||_2^2-\frac{1}{K}}}, 0,\ldots,0,\underbrace{1}_{(i+1)\text{th index}},0,\ldots,0 \right)^T\in\mathbb{R}^{n+1}. $$

Indeed, one can see that $\langle\mathbf{x},\mathbf{x}_i\rangle_{*}=-u_i+u_i=0$ as we require from a tangent vector $\mathbf{x}_i$.

Now, the coefficients of the first fundamental form for the basis $\{\mathbf{x}_1,...,\mathbf{x}_n\}$ of the tangent space are given by:

$$ g_{ij}(\mathbf{x}) = \langle\mathbf{x}_i,\mathbf{x}_j\rangle_{*} = \frac{-x_{i+1}x_{j+1}}{||\mathbf{x}'||_2^2-\frac{1}{K}} + \delta_{ij} $$

Note that use the Lorentz inner product here, because the intrinsic hyberbolic geometry is induced by the extrinsic Lorentzian geometry.

For convenience, we'll define: $$ \mathbf{g}(\mathbf{x}) = \begin{pmatrix} g_{11}(\mathbf{x}) & \cdots & g_{1n}(\mathbf{x})\\ \vdots & \ddots & \vdots\\ g_{n1}(\mathbf{x}) & \cdots & g_{nn}(\mathbf{x})\\ \end{pmatrix}\in\mathbb{R}^{n\times n}. $$

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    $\begingroup$ If you replace $p_\alpha$ by $q_\alpha=\frac{p_\alpha}{(-K)^{1/2}}$, then the equation for $H_K$ reduces to the equation for $H_1$. Thus, if you know the metric tensor for the latter, you can recover the metric tensor for the former. $\endgroup$ Commented Apr 2, 2019 at 18:24
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    $\begingroup$ @GiuseppeNegro This means that if $g_K$ denotes the metric tensor for curvature $K$, we have that $g_K=\frac{p}{\sqrt{|K|}}$? $\endgroup$
    – ndrizza
    Commented Apr 2, 2019 at 18:41
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    $\begingroup$ Terrible choice of notation; if $p_\alpha$ denotes points then you should use $g_{\alpha\beta}$ or $h_{\alpha\beta}$ to denote metric tensor. Anyway, I think there should be a $|K|$, not a $\sqrt{|K|}$, and I don't know if it is in the numerator or in the denominator. $\endgroup$ Commented Apr 2, 2019 at 18:42

2 Answers 2


You should stop thinking of the metric tensor as a matrix itself. It is not. Matrices are only a device for facilitating computations, and you don't have a natural choice of basis for the tangent spaces to form a matrix for a metric given in an arbitrary manifold. In other words, to form a matrix, one needs to choose a coordinate system for the manifold.

There is nothing wrong in saying that the metric tensor in $\Bbb H^d_K$ is just the pull-back of the standard Lorentz metric of $\Bbb R^{d+1}_1$ via the inclusion $\Bbb H^d_K\hookrightarrow \Bbb R^{d+1}_1$, and this just happens to be Riemannian metric. I don't see why one would insist on using a $d+1$ order matrix to represent a metric in a manifold of order $d$. The matrix would be the same, ${\rm diag}(-1,1,\ldots,1)$, the only difference being that now this matrix accepts fewer inputs. In other words, it's like using a matrix representation of a linear map to compute it's restriction to a subspace but promising you won't evaluate it in vectors outside your subspace.

As far as understanding the geometry of the submanifold, there are much more efficient ways, bypassing this awkward approach with matrices. To compute the sectional curvature of $\Bbb H^d_K$, though, is by using the following consequence of the Gauss formula (which you can see in any Riemannian geometry book, and remains valid in the pseudo-Riemannian case): $$K(X,Y) =\overline{K}(X,Y) +\frac{\langle \alpha(X,X),\alpha(Y,Y)\rangle -\langle\alpha(X,Y),\alpha(X,Y)\rangle}{\langle X,X\rangle\langle Y,Y\rangle},$$where $K$ and $\overline{K}$ denote the sectional curvatures of the submanifold and ambient manifold, and $\alpha$ denotes the second fundamental form, satisfying $\overline{\nabla}_XY=\nabla_XY+\alpha(X,Y)$ for all vector fields tangent to the submanifold. Here $\nabla$ and $\overline{\nabla}$ denote the Levi-Civita connections of the submanifold and of the ambient manifold.

Now, note that $\xi(p) = \sqrt{-K}p$ is a unit normal timelike vector to $\Bbb H^d_K$. So $$\alpha(X,Y)=-\langle \alpha(X,Y),\xi\rangle\ \xi = -\langle A_{\xi}(X),Y\rangle \xi = \langle \overline{\nabla}_X\xi,Y\rangle \xi = \sqrt{-K}\langle X,Y\rangle \xi.$$So if $(X,Y)$ is an orthonormal basis for any $2$-plane tangent to $\Bbb H^d_K$, we have $$K(X,Y) = \langle \sqrt{-K}\xi,\sqrt{-K}\xi\rangle = -K\langle \xi,\xi\rangle = K.$$

  • $\begingroup$ I'm a computer science student. So I didn't understand everything. Are you saying that the metric tensor is $g=diag(-1,1,...,1)$ for any curvature $K$? If not, how would you write it in the matrix or $g_{ij}$-form? $\endgroup$
    – ndrizza
    Commented May 7, 2019 at 18:00
  • $\begingroup$ Again: representing a metric tensor using a matrix requires the choice of coordinates in your manifold. You have natural coordinates for $\Bbb R^{d+1}$ but not for $\Bbb H^d_K$. To represent the metric restricted to $\Bbb H^d$ using an order $d$ matrix, one could try to mimic hyperspherical coordinates using hyperbolic trigonometric functions or trying to do some sort of stereographic projection, but as far as computing the curvature goes, this is way too much effort for nothing. My point is that [...] $\endgroup$
    – Ivo Terek
    Commented May 7, 2019 at 18:07
  • $\begingroup$ [...] the question "how to write $g_{ij}$'s for $\Bbb H^d$?", written just like that, doesn't quite compile (or better, it does with a lot of warnings). In general, if you have a coordinate system in a manifold, you can write the metric components with respect to these coordinates as $g_{ij} = g(\partial_i,\partial_j)$, where all the $\partial_i$ are coordinate fields tangent to the manifold. $\endgroup$
    – Ivo Terek
    Commented May 7, 2019 at 18:08
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    $\begingroup$ I know I am being repetitive, but you have to understand that "$g=(-1,1,\ldots,1)$ for $K=-1$" doesn't quite make sense, as again you're trying to describe a metric in a manifold of dimension $d$ using a matrix of order $d+1$. A metric tensor is NOT a matrix, it is a smooth choice of inner products in the tangent spaces of the manifold, which happens to be expressed by a matrix of SAME dimension of the manifold, once a coordinate system has been chosen. You are not choosing coordinates for $\Bbb H^d_K$ here. $\endgroup$
    – Ivo Terek
    Commented May 7, 2019 at 18:17
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    $\begingroup$ @ndrizza The matrix given in Appendix A of that paper represents the metric tensor of Minkowski space (with respect to the standard basis), not hyperbolic space. It is an order $n+1$ matrix representing the metric tensor of a dimension $n+1$ manifold. $\endgroup$
    – Ivo Terek
    Commented May 7, 2019 at 19:15

The metric tensor is simply the (Lorentzian) dot product. $g(a,b)=a\cdot b$. That is essentially the reason we use the hyperboloid model: the intrinsic hyberbolic geometry is induced by the extrinsic Lorentzian geometry.

To represent $g$ as a matrix, you need to have a basis for the tangent space (as you said). The usual way of getting such a basis is to use some coordinate system for the hyperboloid:


$$e_i=\frac{\partial x}{\partial u_i}$$

Now any tangent vector $a$ can be expressed as



$$a\cdot b=\left(\sum_ia_ie_i\right)\cdot\left(\sum_jb_je_j\right)$$

$$=\sum_{i,j}a_ib_j(e_i\cdot e_j)$$

and the metric tensor is

$$g_{ij}=e_i\cdot e_j=\frac{\partial x}{\partial u_i}\cdot\frac{\partial x}{\partial u_j}$$

See this answer of mine for some examples of coordinate systems. There I have $d=2,\,K=-1$, but some of them generalize to higher $d$. And if $x$ parametrizes the hyperboloid with curvature $-1$, then $x/\sqrt{-K}$ parametrizes the hyperboloid with curvature $K$. Compare this with spherical geometry: if $x$ is a point on the unit sphere, then $Rx$ is a point on the sphere of radius $R$ (and curvature $1/R^2$).

The notation there is a little different from here; the metric is expressed in terms of coordinate differentials:


  • $\begingroup$ I noticed that old answer doesn't mention the Beltrami-Klein or Poincare disc coordinates. Maybe I'll add it to this or that answer. $\endgroup$
    – mr_e_man
    Commented May 7, 2019 at 19:42
  • $\begingroup$ I've added a solution in your suggested style to my question. What do you think of it? $\endgroup$
    – ndrizza
    Commented May 7, 2019 at 22:44
  • $\begingroup$ It looks correct, though somewhat confusing with the different $\mathbf x$'s and $\mathbf u$'s and $i+1$'s. $\endgroup$
    – mr_e_man
    Commented May 7, 2019 at 22:55

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