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I was thinking about rewriting https://en.wikipedia.org/w/index.php?title=Hyperbolic_motion&oldid=738425823

(the article is a bit in chaos at the moment)

And that started me puzzeling with the three fundamental hyperbolic motions of the poincare half plane

in this version they are given as:

  • $ p \to q = (x + c, y ), c \in R $(left or right shift)
  • $ p \to q = (sx, sy ), s > 0 $ (dilation)
  • $ p \to q = ( r^{−1} \cos a, r^{−1} \sin a ) $ (inversion in unit semicircle).

And my puzzeling made me think that the third of them is wrong

Now I am looking for more advanced mathematicians that can tell me that i am right or wrong and especially why.

I think his motion is not orientation preserving.

To make the motion orientation preserving it should be: $$ (r \cos a \ , \ r \sin a) \to ( - r^{-1} \cos a \ , \ r^{-1} \sin a ) $$

Also when I start it to imagine it as an ordinary Mobius transformation:

For the halfplane $( y > 0 \ z=0 ) $ to be projected to itself the point of projection must be in the y= 0 plane

so one option is:

  • the Mobius sphere is a sphere of radius 1 above the origin centered around (0,0,1)
  • if P is the original point on the $ z = 0 $ and $ y > 0 $ half plane
  • A is the point where the segment P -(0,0,2) intersects the Mobius sphere (where the point P is projected on the mobius sphere)

  • $A'$ is the rotation of $A$ a half turn around the $ y = 0 $ and $z = 1 $ axis

  • $P'$ is the projection of $A'$ from (0,0,2) to the the $ z = 0 $ plane

And with this the point P' ends at at the other side of the x=0 plane as P.

So should the formula not be $$ (r \cos a\ , \ r \sin a) \to ( - r^{-1} \cos a \ , \ r^{-1} \sin a ) $$

(circle inversion followed by an reflection in the x=0 ray) or why am I wrong here?

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closed as unclear what you're asking by Lee Mosher, John B, C. Falcon, Daniel W. Farlow, Rohan Dec 23 '16 at 6:34

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ To my eyes you wrote -1/z, which maps the upperhalf plane to itself. $\endgroup$ – Charlie Frohman Dec 21 '16 at 21:57
  • $\begingroup$ Your question is unclear. You've given a formula for a function, but then you've asked if it should be a different formula instead. Well, those two formulas define different functions, and without more information there's not much more to be said. $\endgroup$ – Lee Mosher Dec 22 '16 at 13:36