Metric Tensor of Hyperboloid Model for Hyperbolic Space with Curvature $K$

Question

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.

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

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

Answer 1

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:

$$x=x(u_1,u_2,\cdots,u_d)$$

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

Now any tangent vector $a$ can be expressed as

$$a=a_1e_1+a_2e_2+\cdots+a_de_d$$

so

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

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

and the metric tensor is

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

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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:

$$ds^2=\sum_{i,j}g_{ij}\,du_i\,du_j$$

Answer 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 $$K(X,Y) = \langle \sqrt{-K}\xi,\sqrt{-K}\xi\rangle = -K\langle \xi,\xi\rangle = K.$$