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.
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.