Different representations of hyperbolic space

Question

Two equivalent representations (not quite sure if that's the right word) of hyperbolic space $\mathcal{H}^n$ are

1. the zero set of $- x_0^2 + x_1^2 + \ldots x_n^2 =-1 $
2. the quadratic space: vector space $\mathbb{R}^n$ with quadratic form $q= - x_1^2 +x_2^2 + \ldots x_n^2$

( For example, see https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.70.5236&rep=rep1&type=pdf and https://arxiv.org/abs/1910.07112)

However, with spheres $\mathcal{S}^n$ we take

1. the zero set of $ x_0^2 + x_1^2 + \ldots x_n^2 =1$
2. the quadratic space: the vector space $\mathbb{R}^n$ with quadratic form $q= x_1^2 + \ldots + x_n^2$

What is weird to me is that in the first representation of $\mathcal{H}^n$, the quadratic form is equal to a negative one. Although writing $ x_0^2 - x_1^2 - \ldots- x_n^2 =1 $, would have a lot of negative signs, it seems to be in better correspondence with the spherical case. However, if we do this, shouldn't the quadratic form in the second representation change and be $q = x_1^2 -x^2_2 - \ldots - x_n^2$? This would then change the signature, right?

So my questions are:

A) Why do we define hyperbolic space as the zero set of $- x_0^2 + x_1^2 + \ldots x_n^2 =-1 $ instead of the zero set of $ x_0^2 - x_1^2 - \ldots- x_n^2 =1 $?

B) If we define it as the zero set of $ x_0^2 - x_1^2 - \ldots- x_n^2 =1 $, shouldn't we say that the quadratic form (for the second representation) is then $q= x_1^2 - x_2^2 - \ldots- x_n^2$?

Answer 1

The first important feature in both cases --- $\mathcal H^n$ and $\mathcal S^n$ --- is that the equation defining that surface can also be used to define a quadratic form on $\mathbb R^{n+1}$. Namely, the surface is a level set of the quadratic form. (You mention quadratic forms on $\mathbb R^n$; see my last paragraph for remarks on that issue).

The second and key feature is that when that quadratic form on $\mathbb R^{n+1}$ is restricted to the tangent space at each point of the surface, the restricted form on that plane is positive definite (of rank $n$), and hence defines a Riemannian metric on the surface.

In particular, when you restrict $-x_1^2+x_2^2+...+x_n^2$ to each tangent space of the hyperplane $-x_1^2+x_2^2+...+x_n^2=-1$ then the result is positive definite of rank $n$. For example, at the point $x_1=1$, $x_2=...=x_n=0$, the tangent hyperplane is given by the equation $x_1=0$; and the restricted quadratic form on that plane is $x_2^2+...+x_n^2$ which is positive definite. With a bit of analytic geometry work you can verify that the restricted form is positive definition on the tangent hyperplane of every point.

But when you restrict the quadratic form $-x_1^2+x_2^2+...+x_n^2$ to the tangent spaces of the hyperplane $-x_1^2+x_2^2+...+x_n^2=1$, you will get a form of type $(-1,1,...,1)$ (with $n-1$ 1's), and hence is indefinite and does not define a Riemannian metric on the surface. For example, at the point $x_2=1$ and $x_1=x_3=...=x_n=0$, the tangent hyperplane is $x_2=0$, and the restricted quadratic form is $-x_1^2+x_3^2+...+x_n^2$ which has signature $(-1,1,..,1)$.

Remarks: Let me just add, it's not clear to me why you associate a quadratic form on $\mathbb R^n$ with the $\mathcal H^n \subset \mathbb R^{n+1}$, and it is similarly not clear to me why you associate a quadratic form on $\mathbb R^n$ with $\mathcal S^n \subset \mathbb R^{n+1}$. Instead, what you should be associating to each of $\mathcal H^n$ and $\mathcal S^n$ is a quadratic form on $\mathbb R^{n+1}$, which is how the earlier portions of my answer are written.