Prove that TFAE for a Möbius transformation \(T\) given by \(T(z) = {az + b \over cz + d}\):
- \(T\) maps \(\RR\union \theset{\infty}\) to itself.
- It is possible to choose \(a,b,c,d\) to be real numbers.
- \(\bar{T(z)} = T(\bar z)\) for every \(z\in \CP^1\).
- There exist \(\alpha\in \RR, \beta \in \CC\setminus \RR\) such that \(T(\alpha) = \alpha\) and \(T(\bar \beta) = \bar{T(\beta)}\).
Solution
1. (a) \(\Leftrightarrow\) (b) \(\Leftrightarrow\) (c).
1.1. (a) \(\Rightarrow\) (b): \(T\) maps \(\RR \cup \{\infty\}\) to itself, so it sends three real points to three real points; a Möbius map is determined by three values, and a map sending three real points to three real points has real coefficients (up to a common real scalar).
the cross-ratio of real points is real.
1.2. (b) \(\Rightarrow\) (c): if \(a,b,c,d \in \RR\), then \(\overline{T(z)} = \frac{\bar a \bar z + \bar b}{\bar c \bar z + \bar d} = \frac{a\bar z + b}{c\bar z + d} = T(\bar z)\).
conjugation commutes with real coefficients.
1.3. (c) \(\Rightarrow\) (a): if \(z \in \RR\), then \(\bar z = z\), so \(\overline{T(z)} = T(\bar z) = T(z)\), hence \(T(z) \in \RR\); thus \(T\) maps \(\RR\) to \(\RR\) and \(\RR \cup \{\infty\}\) to itself.
2 applied to real \(z\).
2. (d) \(\Rightarrow\) (a).
2.1. \(T\) fixes \(\alpha \in \RR\) and satisfies \(T(\bar\beta) = \overline{T(\beta)}\) for \(\beta \notin \RR\).
hypothesis.
2.2. The map \(z \mapsto \overline{T(\bar z)}\) is a Möbius map agreeing with \(T\) at the three distinct points \(\alpha, \beta, \bar\beta\).
at \(\alpha\) (real), \(\overline{T(\bar\alpha)} = \overline{T(\alpha)} = \overline{\alpha} = \alpha = T(\alpha)\); at \(\beta\), \(\overline{T(\bar\beta)} = T(\beta)\) by hypothesis; at \(\bar\beta\), \(\overline{T(\beta)} = T(\bar\beta)\) by hypothesis.
2.3. Hence \(\overline{T(z)} = T(\bar z)\) for all \(z\), so \(T\) maps \(\RR \cup \{\infty\}\) to itself (by <1>2.3).
2 and 2.3.
3. (a) does NOT imply (d): the statement “TFAE” is false as written.
3.1. \(T(z) = -\frac{1}{z}\) has real coefficients, so it satisfies (a), (b), and (c).
\(a = 0\), \(b = -1\), \(c = 1\), \(d = 0\) are real.
3.2. But \(T\) has no real fixed point.
\(T(z) = z\) means \(-\frac{1}{z} = z\), i.e. \(z^2 = -1\), so the fixed points are \(z = \pm i \notin \RR\).
3.3. Hence (d) fails for \(T\), so (a) \(\not\Rightarrow\) (d).
1 and 2.
4. Q.E.D.
1 proves (a) \(\Leftrightarrow\) (b) \(\Leftrightarrow\) (c); 2 proves (d) \(\Rightarrow\) (a); 3 shows (a) \(\not\Rightarrow\) (d), so the four statements are not all equivalent.