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Is there a Möbius transformation mapping the upper half plane onto itself that interchange two preassigned points in the upper half plane? If so , how many such Möbius transformations are there?

Here are steps that I followed:

1.we know we can transform upper half plane onto unit circle.

  1. I want to find a Möbius transformation that maps unit circle to unit circle but interchanging the preassigned points.

3. Finally , if I find a transformation that maps the unit circle to the upper half plane.Then I think we are done.

I have problem in finding the Möbius transformation that maps unit circle to unit circle with interchanging property that already mention.

Does someone have any idea about this?

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    $\begingroup$ the transformations that take the upper half plane to itself are easier to describe, and work with, than what you suggest $\endgroup$ – Will Jagy Oct 1 '14 at 4:12
  • $\begingroup$ Question with a related answer $\endgroup$ – ccorn Oct 1 '14 at 5:56
  • $\begingroup$ What I think , we need three transformation to do this. But , I am having problem to see how to interchange the points $\endgroup$ – user178061 Oct 2 '14 at 16:36
  • $\begingroup$ You know that you can move the problem to the unit disk. I think that can indeed simplify matters somewhat. Can you show that there exists an automorphism of the unit disk that interchanges $0$ and any preassigned point $a$ in the unit disk? Can you then show that there is an automorphism of the unit disk interchanging two arbitrary preassigned $a,b$ in the unit disk? $\endgroup$ – Daniel Fischer Oct 2 '14 at 21:16
  • $\begingroup$ It makes sense. This is what I was trying to do but I have the problem to show there is an automorphism of the unit disk interchanging two arbitrary preassigned points. $\endgroup$ – user178061 Oct 3 '14 at 3:08
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A Möbius transformation that maps the upper half plane to itself must map the boundary, i.e. the real line, to itself as well. With the standard definition for the transformation, $$f(z) = \frac{az+b}{cz+d},$$ it is fairly straightforward to see that the parameters $a, b, c, d$ must be real to map the real line to itself (try plugging in $z=0$, $z=\infty$, and then one other real number).

Then since $f$ is supposed to interchange $z_1$ and $z_2$, that means applying $f$ twice should send $z_1$ back to itself, and the same for $z_2$. Note that $f(f(z))$ is again another Möbius transformation with real coefficients. You can now solve the equation for the fixed points of the transformation (or look up the formula). This should answer the question about whether the transformation is possible.

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  • $\begingroup$ I did not get how applying f twice takes the points back to itself and how does this interchanges the points? $\endgroup$ – user178061 Oct 14 '14 at 3:51
  • $\begingroup$ Note that $f^2$ will have two fixed points in the upper half plane, and has real coefficients, so $f^2$ is simply the identity map. $\endgroup$ – runway44 Mar 26 at 19:19
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Uniqueness of the Transformation

Some understanding of group actions can help us show that such a transformation (using real coefficients) must be unique (assume $w\ne z$). If $f$ and $g$ are two such transformations, then $f^{-1}\circ g$ is a transformation that fixes both $z$ and $w$. But writing out the equation for $z$ to be a fixed point of $\frac{az+b}{cz+d}$ reveals (again, for real values $a,b,c,d$) there can only be one fixed point $z$ in the upper half-plane, unless it is the identity. But if it is the identity, $f^{-1}\circ g=\mathrm{id}$ implies $f=g$ are the same transformation.

Geometry of the Transformation

The upper half-plane is a model for hyperbolic geometry, which has a metric. The geodesics are semicircles with diameters on the real axis. Around every point there is a group of "hyperbolic rotations" which fix it and whose derivative there rotates tangent vectors. In particular, around $i$, it is $\mathrm{SO}(2)$ (however the kernel of its action is $\{\pm I_2\}$ so maybe we want to think of it as $\mathrm{PSO}(2)$). Given two points $z$ and $w$ we may construct the unique semicircle arc between them. On that arc is a "hyperbolic midpoint" which is the same hyperbolic distance from both $z$ and $w$ (I expect this is not the midpoint according to Euclidean arclength, though). We can swap $z$ and $w$ by applying a $180^{\circ}$ hyperbolic rotation around this midpoint. (Note that if the midpoint was e.g. $i$, we'd need a $90^{\circ}$ rotation matrix in $\mathrm{SO}(2)$ to achieve this effect.)

Formula for the Transformation

The equations that say $f$ swaps $w$ and $z$ are given by

$$ \frac{aw+b}{cw+d}=z, \qquad \frac{az+b}{cz+d}=w. $$

Clearing denominators yields

$$ \begin{cases} aw+b = c(wz)+dz \\ az\,+\,b = c(zw)+dw \end{cases} $$

Subtracting gives $a(w-z)=d(z-w)$ which implies $d=-a$, and the equation becomes

$$ a(w+z)+b=c(wz). \tag{$\ast$} $$

Taking imaginary parts yields $a\mathrm{Im}(w+z)=c\,\mathrm{Im}(wz)$. This allows us to solve:

$$ \begin{cases} a = \lambda \mathrm{Im}(wz) \\ c = \lambda \mathrm{Im}(w+z) \end{cases} $$

where $\lambda\in\mathbb{R}^{\times}$. We may solve for $b$ by taking real parts of $(\ast)$ to finally get

$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \lambda \begin{pmatrix} \mathrm{Im}(wz) & \mathrm{Re}(wz)\mathrm{Im}(w+z)-\mathrm{Im}(wz)\mathrm{Re}(w+z) \\ \mathrm{Im}(w+z) & -\mathrm{Im}(wz) \end{pmatrix} $$

Special Example

Taking $z=w=i$ gives a $90^{\circ}$ rotation matrix, yielding a $180^{\circ}$ hyperbolic rotation around $i$. This is the limiting case of $z,w\to i$ with midpoint $\to i$.

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