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Homework Help: A special cross-ratio

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Hi all,

    I wasn't sure whether to ask this here or not, but I've tried to make progress and so far had no such luck. I've not managed to make any headway on this so far, I don't think I can quite get my head around the geometry that's in play.

    If u,v[itex]\in \mathbb{C}[/itex] correspond to points P, Q on [itex]S^2[/itex], and d denotes the angular distance from P to Q on [itex]S^2[/itex], show that [itex]-\tan^2(\frac{d}{2})[/itex] is the cross ratio of the points [itex]u, v, \frac{-1}{u^*}, \frac{-1}{v^*}[/itex], taken in an appropriate order (which you should specify). (The star denotes complex conjugation - I'm not sure how to do the 'bar' in latex!)

    Now I'm useless at geometry, but if I recall correctly, [itex]\frac{-1}{u^*}[/itex] would correspond to the stereographic projection of the point (-P), right? And likewise with v - other than that however, I really can't see a smart way to do this. I certainly don't want to try all 6 permutations of the 4 points and see what pops up on the cross ratio, but at the same time I can't see intuitively where the [itex]-\tan^2(\frac{d}{2})[/itex] could have come from in order to try and work out how to take the cross ratio to get the desired result. I'm aware that the angle between the 2 points P and Q is equal to d, but I can't seem to work anything out in the complex plane rather than the sphere. Please help!

    Many thanks in advance.
  2. jcsd
  3. Jan 25, 2010 #2
    Anyone? I'll take any help you can offer! Thankyooou :)
  4. Jan 26, 2010 #3


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    Hi Mathmos6! :smile:

    If you choose your projection point at the midpoint of PQ, then the four projected points are in a line at distances ±tan(d/4) and ±1/tan(d/4) from the origin. :wink:
  5. Jan 26, 2010 #4
    Thanks so much, i didn't appreciate that the cross ratio is invariable under rotation so you can just place the points in a more convenient position and work it out that way. Have a good day!
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