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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 $$frac{4}{(1 - | xi |^2)^2} dxi_i otimes dxi_i.$$

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.

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 $v_i=frac{dsigma_i}{dt}(0)$ 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

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

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

EDIT: Sketch of $6)$. Let $mathbb{D}^n={xiinmathbb{R}^n:|xi|<1}$. Note that $f:mathbb{H}^ntomathbb{D}^n$ 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 $f^{-1}:mathbb{D}^ntomathbb{H}^n$, defined by the formula

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

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

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

for every $i,jin {1,ldots,n}$ and every $xiinmathbb{D}^n$.

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