I am puzzeling with the following:

Given a point $ A = ( a_x, a_y) : a_x^2+ a_y^2 \le 1 $ in the Poincare Disk model ( https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model )

to which point does this point map on the Poincare half plane model. (https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model )

Given that:

  • $(0.0 )$ maps to $(0,1)$ (or to $(0,2)$ or $(0,u)$ )
  • $(0,-1) $ maps to $y= \infty$ and
  • $(0,1) $ maps to $(0,0)$ ?

The Poincare disk and upper half-plane models are related by a Möbius transformation that maps the disk to the plane. Such a transformation has the form $$ f(z) \;=\; \frac{az+b}{cz+d} $$ for some constants $a,b,c,d\in\mathbb{C}$ such that $ad-bc\ne 0$.

In my mind, the "usual" Möbius transformation $f$ satisfies $f(-i) = 0$, $f(1) = 1$, $f(-1)=-1$, and $f(i)=\infty$, though there may be other conventions. Solving for the coefficients gives the formula for $f$: $$ f(z) \;=\; \frac{z + i}{iz+1}. $$ The inverse of this maps the upper half-plane to the disk, and is given by $$ f^{-1}(z) \;=\; \frac{z-i}{-iz+1} $$

Note For the specific requirements you gave for $f$ where $f(0) = i$, $f(-i)=\infty$, and $f(i) = 0$, you would want the $180^\circ$ rotation of the function $f$ above, i.e. $f(z) = \dfrac{z-i}{iz-1}$.

  • $\begingroup$ you are right with your remark "* The requirements on f that you asked for aren't achievable because 0 in the disk can't map to 1 on the half plane, since 0 lies in the interior of the disk and 1 lies on the boundary of the half plane." made a mistake with the coordinates, corrected my question , now it should be possible, can you help a bit more? $\endgroup$
    – Willemien
    Jun 21 '15 at 17:37
  • $\begingroup$ @Willemien I've changed my note to account for the new requirements. $\endgroup$
    – Jim Belk
    Jun 21 '15 at 17:44
  • $\begingroup$ thanks, so if i am correct the point $A = ( a_x, a_y)$ maps to $(2 a_x /e , ( (1- a_x^2 -a_y^2) /e) $ where $ e = a_x^2 + (a_y+1)^2 $ $\endgroup$
    – Willemien
    Jun 22 '15 at 21:34

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