Help with definition of hyperbolic angle

Question

I am currently reading about the Hyperbolic Space in the book Metric Spaces of Non-Positive Curvature of Martin R. Bridson and André Haefliger and they define the space $\mathbb{E}^{n,1}$ as $\mathbb{R}^{n+1}$ equipped with the following bilinear form $$ \left[u,v\right] = \left(\sum_{i=1}^{n}u_{i}v_{i}\right) - u_{n+1}v_{n+1} $$ and the Hyperbolic n-Space $\mathbb{H}^{n}$ as: $$ \mathbb{H}^{n} = \{x \in \mathbb{E}^{n,1} : \left[x,x\right] = -1, x_{n+1}>0\}. $$ Now given $A \in \mathbb{H}^{n}$ and a vector $u\in \mathbb{E}^{n,1}$ such that $\left[u,u\right] = 1$ and $\left[u,A\right] = 0$, then a hyperbolic segment is an image of an compact interval of the path $c : \mathbb{R} \to \mathbb{H}^{n}$, $c\left(t\right) = \left(\cosh{t}\right)A+\left(\sinh{t}\right)u$, we call $u$ the initial vector of the segment. They define the angle between two hyperbolic segments issuing from a point of $\mathbb{H}^{n}$, with initial vectors $u$ and $v$ is the unique number $\alpha \in \left[0,\pi\right]$ such that $$ \cos{\alpha} = \left[u,v\right]. $$

Ok, here is my problem, I don't see how the angle is well-defined, for this, I would need to have $\left|\left[u,v\right]\right|\leq 1$, I tried to use the definition of $\left[\cdot,\cdot\right]$ and using Cauchy-Schwarz inequality some times but it was useless, any help?

Answer 1

You cannot use Cauchy-Schwarz without further justification since it is only valid in general for semidefinite forms and this form is indefinite.

I hope you agree that if $u, v$ live in a definite subspace then $[u, v] = \cos\alpha$ for some $0 \leq \alpha \leq \pi$; they are unit vectors and this is just the usual expression for angle in a Euclidean space. Notice that $[\cdot,\cdot]$ is a form with signature $(n,1)$, meaning that any orthogonal basis for $\mathbb E^{n,1}$ must have $n$ vectors $[x,x] > 0$ and 1 vector with $[x,x] < 0$. In particular, for any $y$ with $[y,y] < 0$ we have the orthogonal decomposition $\mathbb E^{n,1} = \mathbb Ry\oplus y^\perp$ with $y^\perp$ the orthogonal complement of $y$, and this $y^\perp$ must be a positive-definite space.

By definition $u, v \in A^\perp$ with $[A,A] = -1 < 0$, so $u, v$ live in a definite space and the $\alpha$ in $\cos\alpha = [u,v]$ is well-defined.