Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For some computational project, I'm interested in the pairwise distance matrix between random points on a unit square of $\mathbb{R}^2$.

I now want to extend this case to non-zero curvature 2D spaces, but I don't see what is the proper way to spread random points on such spaces. Does one define the random distribution on $[0,1]^2$ and maps it to the space through an appropriate coordinate transform, or is there a way to do it directly ?

How would you expect the distance matrix to change with curvature ?

Thank you for your answers !

share|cite|improve this question
1  
Most likely you're going to want to have this depend on the area of your surface: if two regions have the same area, then they have equal probability of containing a random point. – Aaron Mazel-Gee Jul 11 '11 at 18:47
    
@Aaron Mazel-Gee : That's what I was more or less thinking also, but I'm lacking the mathematical link there. For example, on a Euclidean space, the infinitesimal area unit is dx x dy so a uniform distribution on both x and y is ok. On a sphere, this small area can be expressed as rdrd$\theta$, so a uniform distribution on $\theta$ and a uniform distribution over $r^2$ might be ok. And the same goes for a hyperbolic space ? – AlexPof Jul 11 '11 at 18:51
    
That's the idea, but I think you're using polar coordinates where you should be using spherical coordinates...? – Aaron Mazel-Gee Jul 11 '11 at 19:07
    
@Aaron Mazel-Gee : sorry, it got mixed up with something else. So for a unit sphere : $ sin \phi d\phi d\theta $, so I have to choose a uniform distribution over $\theta$ and $cos \phi$. If I'm guessing correctly, on the hyperbolic half-plane it would be over x and $1/y$, and other models of hyperbolic space give rather complicated formulas. – AlexPof Jul 11 '11 at 19:19
    
Yes, I think that sounds about right. I don't think the Poicare disk model would be so bad, either. – Aaron Mazel-Gee Jul 11 '11 at 19:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.