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I was confronted with this exercise in the book Hyperbolic Geometry by Anderson which states:

In each case, find $m \in Möb(\mathbb{H})$ such that the property holds, or prove that no such $m$ exists.

The example in question is:

m takes $(-t,0,t)$ to $(-1,\infty,1)$ where $t \in ℝ$

So what it is asking for is a Möbius transformation from the first triple to the second. In case the notation isn't clear, $Möb(\mathbb{H})$ is the set of Möbius transformations which preserve $\mathbb{H}$, such that

$Möb(\mathbb{H})= {\{m \in Möb | m(\mathbb{H})=\mathbb{H}\}}$

Of course this is all in the upper-half plane model. So far I have been unable to come up with a Möbius transformation which comes anywhere close to these.

Attempt at a solution: The standard procedure for Möbius transformations between particular triples has not worked for me. This standard procedure goes as follows.

Define m(z) as the Möbius transformation taking $(z1,z2,z3)$ to $(0,1,\infty)$. Then

$m(z) = \dfrac{az+b}{cz+d}$ = $\dfrac{(z2-z3)z-z1(z2-z3)}{(z2-z1)z-z3(z2-z1)}$

This means $a=(z2-z3), b=-z1(z2-z3), c=(z2-z1), d= z3(z2-z1)$. If we want to take $(z1,z2,z3)$ to another triple $(w1,w2,w3)$, we find $m(w)$ and our final transformation is given by $m^{-1}(w)$ composed with $m(z)$.

But if we do this, we find that the Möbius transformation for the second pair of points, $(w1,w2,w3)=(-1,∞,1)$ contains many infinities.

$m(w) = \dfrac{((\infty-1)z+1(\infty-1))}{((\infty+1)z-1(\infty+1))} = \dfrac{(\infty z+\infty)}{(\infty z-\infty)}$

Which appears to be nonsense.

I am unsure if there is a Möbius transformation which satisfies these conditions because while $Möb$ acts sharply $3$-transitive over $\mathbb{C}$, $Möb(\mathbb{H})$ acts merely transitively over $\mathbb{H}$, not $2$- or $3$-transitively.

Thanks in advance for any insight into this problem.

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Since the composition of two Möbius transformations is again a Möbius transformation, we can attempt to build a transformation in multiple steps: We can map $0$ to $\infty$ with the inversion $I(z) : = -\frac{1}{z}$, in which case the desired map is $$\Phi \circ I,$$ where $\Phi$ maps $(I(-t), I(0), I(t)) = (\frac{1}{t}, \infty, -\frac{1}{t})$ to $(-1, \infty, 1)$. Since $\Phi$ maps $\infty$ to $\infty$, it must be linear. The only linear map that sends $(\frac{1}{t}, -\frac{1}{t})$ to $(-1, 1)$ is $z \mapsto - tz$, but this is only a Möbius transformation if $t < 0$.

In general, Möbius transformations do not act $3$-transitively on $\mathbb{R} \cup \{\infty\}$, but it does act transitively on the set of "oriented" triples of real numbers.

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  • $\begingroup$ Hi Travis, Does $Φ$ have an explicit formula? I remember using an argument mapping $0$ to $∞$ using an inversion, but that was with pairs of points on the imaginary positive axis and it is not immediately obvious to me how the map works here. I was also discussing with my professor yesterday whether Mobius transformations act 3-transitively on the extended real line but we did not go beyond a very rudimentary examination. Also, thank you for the clarifying edits, I am terrible with Tex. $\endgroup$ – Jake Weeks Jan 29 '15 at 6:37
  • $\begingroup$ Yes, we showed that if $t < 0$ then $\Phi$ is $-tz$; if $ \geq 0$, there is no element of $Mob(H)$ that satisfies the conditions. We can identify $Mob(H)$ with the set of transformations $z \mapsto \frac{az + b}{cz + d}$ where $a, b, c, d \in \mathbb{R}$ and $ad - bc = 1$. The larger group of transformations $\frac{az + b}{cd + d}$ where $a, b, c, d \in \mathbb{R}$ and $ad - bc \in \{\pm 1\}$ does act $3$-transitively on the extended real line, but such transformations preserve the upper half-plane if $ad - bc = 1$ and exchange the upper and lower half-planes if $ad - bc = -1$. $\endgroup$ – Travis Jan 29 '15 at 6:47
  • $\begingroup$ Ah, I see now. That clarifies everything. Thank you very much! $\endgroup$ – Jake Weeks Jan 29 '15 at 7:22
  • $\begingroup$ You're welcome, I'm glad you found it helpful. $\endgroup$ – Travis Jan 29 '15 at 7:31

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