Hyperbolic Geometry: Representation of the metric in a chart

Question

Let $\mathbb{R}^{1,n}$ denote Lorentzian $n$-space, i.e., $\mathbb{R}^n$ equipped with the Lorentzian inner product $$\langle x,y \rangle = - x_0 y_0 + x_1 y_1 + \cdots + x_n y_n.$$

Define $\mathbb{H}^n : = \left \{ \xi \in \mathbb{R}^{1,n} : \langle \xi, \xi \rangle = -1, \ \xi_0 > 0 \right \}$ to be hyperbolic $n$-space. I have been battling with the following seemingly standard/trivial exercises in Jöst's Riemannian Geometry and Geometric Analysis and have had little success:

5) Show that $\langle \cdot, \cdot \rangle$ induces a Riemannian metric on the tangent spaces $T_p \mathbb{H}^n \subset T_p \mathbb{R}^{n+1}$ for $p \in \mathbb{H}^n$.

6) Let $s = (-1, 0, ..., 0) \in \mathbb{R}^{n+1}$, and define $$f(x) = s - \frac{2(x-s)}{\langle x-s, x-s \rangle}.$$ Show that $f : \mathbb{H}^n \longrightarrow \{ \xi \in \mathbb{R}^n : | \xi | < 1 \}$ is a diffeomorphism, and that in this chart, the metric assumes the form $$\frac{4}{(1 - | \xi |^2)^2} d\xi_i \otimes d\xi_i.$$

Thoughts/Progress:

The fact that $\langle \cdot, \cdot \rangle$ defines an inner product is clear, but I cannot convince myself that $\langle \cdot, \cdot \rangle$ is positive definite. For Q6., I am aware that a similar question has been posted on here, but the solution is not clear to me.

Note that I do not want answers to these problems, just guidance.

Thanks in advance.

Answer 1

We will prove $5)$. As differential manifold, $\mathbb{H}^n$ is a regular submanifold of $\mathbb{R}^{n+1}$ so we can identify tangent vectors in $\mathbb{H}^n$ as velocities of differentiable curves in $\mathbb{H}^n$. Take a point $p=(p_0,\ldots,p_n)$ in $\mathbb{H}^n$ and a differentiable curve $\sigma : (-\varepsilon,\varepsilon)\to \mathbb{H}^n$ such that $\sigma(0)=p$. Write the coordinates of $\sigma$ as $\sigma_0,\ldots,\sigma_n$ and let $v_i=\frac{d\sigma_i}{dt}(0)$ for every $i$. So we want to prove that $\sum_{i=1}^nv_i^2>v_0^2$.

We have that $\langle \sigma(t),\sigma(t) \rangle =-1$ for every $t$. So in particular at $0$ we have that

$$ \sum_{i=1}^np_i^2 = p_0^2-1 $$

On the other hand, by the product rule for derivatives we have that

$$ \sum_{i=1}^nv_ip_i = v_0p_0 $$

So, by the Cauchy-Schwarz inequality and the first equation, it follows that

$$ v_0^2p_0^2=\bigg{(}\sum_{i=1}^nv_ip_i\bigg{)}^2\leq \bigg{(}\sum_{i=1}^nv_i^2\bigg{)} \bigg{(}\sum_{i=1}^np_i^2\bigg{)} = \bigg{(}\sum_{i=1}^nv_i^2\bigg{)} \bigg{(}p_0^2-1\bigg{)} < \bigg{(}\sum_{i=1}^nv_i^2\bigg{)} p_0^2 $$

We know that $p_0>0$ by definition of $\mathbb{H}^n$ so the proposition follows.

EDIT: Sketch of $6)$. Let $\mathbb{D}^n=\{\xi\in\mathbb{R}^n:|\xi|<1\}$. Note that $f:\mathbb{H}^n\to\mathbb{D}^n$ is defined by the formula $$ f(x_0,\ldots,x_n)=\bigg{(}\frac{x_1}{1+x_0},\ldots,\frac{x_n}{1+x_0}\bigg{)} $$

and it is differentiable. The inverse $f^{-1}:\mathbb{D}^n\to\mathbb{H}^n$, defined by the formula

$$ f^{-1}(\xi_1,\ldots,\xi_n)=\bigg{(}\frac{1+|\xi|^2}{1-|\xi|^2},\frac{2\xi_1}{1-|\xi|^2},\ldots,\frac{2\xi_n}{1-|\xi|^2}\bigg{)}, $$

is differentiable so $f$ is a diffeomorphism. Prove that the pullback by $f^{-1}$ of the metric defined in $\mathbb{H}^n$ is equal to the metric given on $\mathbb{D}^n$, that is

$$ \bigg{\langle} \frac{\partial f^{-1}}{\partial \xi_i}(\xi),\frac{\partial f^{-1}}{\partial\xi_j}(\xi) \bigg{\rangle} = \frac{4\delta_{ij}}{(1-|\xi|^2)^2} $$

for every $i,j\in \{1,\ldots,n\}$ and every $\xi\in\mathbb{D}^n$.

Note that $6)$ implies $5)$ because the pullback by a diffeomorphism of a riemannian metric is a riemannian metric.