1
$\begingroup$

Unless I am very wrong, the following seems to be true:

If the angle between two vectors in $\mathbb{R}^{n}_{++}$ is small, then the value of the Hilbert projective metric between them is also small.

I am looking for a reference to a precise statement of this notion.

P.S. What I mean by the angle between $u$ and $v$ is $\arccos{\frac{\langle u,v \rangle}{||u|| \cdot ||v||}}$.

P.P.S. If $u,v \in \mathbb{R}^{n}_{++}$ then $d(u,v)=\log{\frac{\max{(u_{i}/v_{i})}}{\min{(u_{i}/v_{i})}}}$ is the Hilbert projective metric.

$\endgroup$
2
  • $\begingroup$ It would help matters if you were to define what you mean by "Hilbert projective metric". $\endgroup$
    – Igor Rivin
    Jul 11, 2013 at 20:21
  • $\begingroup$ @IgorRivin I added a definition. $\endgroup$ Jul 11, 2013 at 21:21

1 Answer 1

2
$\begingroup$

This is equivalent to the following: in the Klein model of hyperbolic space, if two points are close together, then their hyperbolic distane is small. This is true locally (e.g., fix one point), but not true globally (in other words, no bound on Euclidean distance implies a fixed bound on hyperbolic distance -- the distortion grows as you approach the boundary).

$\endgroup$
3
  • $\begingroup$ The local version may well do the job for my needs. Can you please direct me to a reference where it's spelt out? Thanks a mil! $\endgroup$ Jul 11, 2013 at 21:23
  • $\begingroup$ @FelixGoldberg: The "local statement" is just the restatement of the fact that Hilbert metric is continuous as a function of two variables. Given the explicit formula that you have, continuity is clear. More generally, if you have a bounded open convex domain in the affine space, continuity of its Hilbert metric is also clear, since it is given by a cross-ratio. $\endgroup$
    – Misha
    Jul 11, 2013 at 22:48
  • $\begingroup$ @Misha: my thoughts precisely :) $\endgroup$
    – Igor Rivin
    Jul 12, 2013 at 0:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.