Let $\mu(z) = \frac{az+b}{cz+d}$ be a Möbius transformation. I want to show that $$\mu(\overline{\mathbb{R}}) = \overline{\mathbb{R}} \iff a, b, c, d \in \mathbb{R}.$$ What would be an elegant, and hopefully short way to prove this statement?

I have tried to show one implication first but then I arrive at a lot of different cases and I don't think the way to show the statement is to make an awkward case-by-case analysis. Other exercises given by the author of the lecture notes are much easier - e.g. I showed that $\mu(\mathbb{D}) = \mathbb{D}$ and $\mu(0)=0$ if and only if $\mu(z) = \zeta z$ for $\zeta \in S^1$, where $\mathbb{D}$ denotes the open unit disk in $\mathbb{C}$ and $S^1$ denotes the unit circle in $\mathbb{C}$ - which is why I believe there must be a more elegant way to approach this problem.

Thanks for any answers in advance.

EDIT: As Chris and yoyo have pointed out, the statement is not correct. The correct statement would probably be (correct me if I'm wrong again)

$$\mu(\overline{\mathbb{R}}) = \overline{\mathbb{R}} \iff a, b, c, d \in \lambda \mathbb{R}$$ for some constant $\lambda \in \mathbb{C}$.

  • $\begingroup$ The claim is false. E.g. take $\mu(z)=\frac{iz+0}{0z+i}$. $\endgroup$ Dec 21 '11 at 18:55
  • $\begingroup$ what you state isn't exactly true, eg multiply $a,b,c,d$ all by $i$ to get the same function, but there will be a constant $\lambda$ st $a,b,c,d\in\lambda\mathbb{R}$. $\endgroup$
    – yoyo
    Dec 21 '11 at 19:00
  • $\begingroup$ I have edited my original question. Is the new statement correct? $\endgroup$
    – Huy
    Dec 21 '11 at 19:05
  • 5
    $\begingroup$ Note that the Moebious transform is uniquely determined by its value on three points. In particular, $$ \tilde{\mu} = \frac{(z - z_0)(z_1 - z_{\infty})}{(z - z_{\infty})(z_1 - z_0)}$$ is the Moebious transform which maps $z_0, z_1$ and $z_{\infty}$ to $0, 1$ and $\infty$, respectively. Now let $z_0 = \mu(0)$, $z_1 = \mu(1)$ and $z_{\infty} = \mu(\infty)$. These are elements of $\bar{\mathbb{R}}$, so $\tilde{\mu}$ defines a Moebious transform associated to a real matrix such that $\mu^{-1} = \tilde{\mu}$. Hence $\mu$ itself is also associated to a real matrix. $\endgroup$ Dec 21 '11 at 19:36
  • $\begingroup$ @sos440: Why not as an answer? :) $\endgroup$
    – Huy
    Dec 21 '11 at 22:13

One direction is clear: if there exists $\lambda\in\mathbb C$ such that $a,b,c,d\in\lambda\mathbb R$, then after canceling $\lambda$ we have all real coefficients. The converse was proved by sos440 in the comments: I copy the comment with improved formatting.

Note that the Möbius transform is uniquely determined by its value on three points. In particular, $$\tilde \mu(z)=\frac{(z−z_0)(z_1−z_\infty)}{(z−z_\infty)(z_1−z_0)} \tag{*}$$ is the Möbius transform that maps $z_0$, $z_1$ and $z_\infty$ to $0$, $1$ and $\infty$, respectively. Given a transform $\mu$ that fixes the extended real line, let $z_0=\mu (0)$, $z_1=\mu(1)$ and $z_\infty=\mu(\infty)$ in (*). These are elements of $\overline{\mathbb R}$, so $\tilde \mu$ defines a Möbius transform associated to a real matrix. Since $\mu^{-1}=\tilde \mu$, it follows that $\mu$ itself is associated to a real matrix.


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