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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.