1
$\begingroup$

Previously, Wikipedia has this to say about the Hyperboloid Model:

If (x0, x1, ..., xn) is a vector in the (n + 1)-dimensional coordinate space Rn+1, the Minkowski quadratic form is defined to be

$Q(x_0, x_1, \ldots, x_n) = x_0^2 - x_1^2 - \ldots - x_n^2.$

The vectors vRn+1 such that Q(v) = 1 form an n-dimensional hyperboloid S (trimmed)...

The Minkowski bilinear form B is the polarization of the Minkowski quadratic form Q (trimmed), Explicitly,

$B((x_0, x_1, \ldots, x_n), (y_0, y_1, \ldots, y_n)) = x_0y_0 - x_1 y_1 - \ldots - x_n y_n .$

The hyperbolic distance between two points u and v of S+ is given by the formula

$d(\mathbf{u}, \mathbf{v}) = \operatorname{arcosh}(B(\mathbf{u}, \mathbf{v}))$

I feel this strongly implies that the metric tensor is $B$, with signature (+,-,-,...). However, if we consider the example point (1,0,0,0) in H3, its tangent space will spanned by {(0,1,0,0),(0,0,1,0),(0,0,0,1)}, and when we apply the quadratic form to these basis vectors the result for all of them is -1.

Indeed, rather than being positive-definite, the tangent space is negative-definite, and this extends to the entire manifold.

It seems like the proper metric is the one with signature (-,+,+,...), which will then be positive-definite and yield a Riemannian space. Am I missing something here?

Edit: I found a tacit admission that the signature is backwards in the section on reflections:

For two points $\mathbf p, \mathbf q \in \mathbb{H}^n, \mathbf p \neq \mathbf q$, there is a unique reflection exchanging them.

Let $\mathbf u = \frac {\mathbf p - \mathbf q}{\sqrt{-Q(\mathbf p - \mathbf q)}}$.

If you have to throw in a negative to make your norm work, you're doing it wrong.

$\endgroup$
3
  • 2
    $\begingroup$ I agree with you. $\endgroup$
    – Deane
    Aug 1, 2022 at 4:23
  • 2
    $\begingroup$ Remember: Anybody can edit Wikipedia articles, do not assume that everything there is written by competent people. $\endgroup$ Aug 1, 2022 at 4:25
  • $\begingroup$ I think the person who wrote it was more familiar with Minkowski space, where IIUC the choice of signature is a more arbitrary convention. The upside is that anyone (even less-than-competent people) can fix things on Wikipedia, too. :) But it'll take a careful and thorough edit to reverse the signature without breaking the article. $\endgroup$
    – D0SBoots
    Aug 1, 2022 at 9:45

0

You must log in to answer this question.