(Can omit; related to the discussion of symmetry) Let \(T\) be the Möbius transformation \[ T(z)=\frac{az+b}{cz+d}. \] Consider the following conditions.
\(T\) maps \(\RR \cup \{ \infty \}\) onto itself.
It is possible to choose \(a , b , c , d \in \mathbb { R }\)
\({ \overline { { T z } } } = T ( { \overline { { z } } } )\) for every \(z \in \mathbb { C } \cup \infty\)
There exist \(\alpha \in \mathbb { R }\) and \(\beta \in \mathbb { C } \backslash \mathbb { R }\) satisfying \(T(\alpha) = \alpha\) and \(T ( \overline { { \beta } } ) = \overline { { T \beta } }\)
Prove that (a), (b), and (c) are equivalent. Prove that (d) implies these conditions, and show that the converse need not hold.
Erratum: the source asks to prove that all four printed conditions are equivalent. This is false. The Möbius transformation \[ T(z)=-\frac1z \] has real coefficients and therefore satisfies (a)–(c), but its fixed-point equation is \(z^2=-1\), so it has no real fixed point and cannot satisfy (d). The corrected statement keeps the four source conditions and asks for (d) \(\Rightarrow\) (a) \(\Leftrightarrow\) (b) \(\Leftrightarrow\) (c).
Solution
Let \[ \widehat{\RR}=\RR\cup\{\infty\} \] and let \[ J:\widehat{\CC}\longrightarrow\widehat{\CC}, \qquad J(z)=\bar z, \qquad J(\infty)=\infty. \] The fixed-point set of \(J\) is exactly \(\widehat{\RR}\).
1. Condition (b) implies condition (c).
Choose a representative \[ T(z)=\frac{az+b}{cz+d} \] with \[ a,b,c,d\in\RR, \qquad ad-bc\neq0. \] For every finite \(z\) at which the displayed fractions are finite, \[ \overline{T(z)} = \frac{a\bar z+b}{c\bar z+d} = T(\bar z). \] The same identity holds at the pole and at \(\infty\) in the Riemann sphere, because real coefficients make the pole and the corresponding extended values conjugation-invariant. Hence \[ J\circ T=T\circ J \] on all of \(\widehat{\CC}\), which is condition (c).
2. Condition (c) implies condition (a).
Assume \[ J\circ T=T\circ J. \] If \[ x\in\widehat{\RR}, \] then \(J(x)=x\), so \[ J(T(x)) = T(J(x)) = T(x). \] Thus \(T(x)\) is fixed by \(J\), and therefore \[ T(\widehat{\RR})\subseteq\widehat{\RR}. \]
Taking inverses in \[ J\circ T=T\circ J \] gives \[ T^{-1}\circ J = J\circ T^{-1}. \] The same argument applied to \(T^{-1}\) gives \[ T^{-1}(\widehat{\RR})\subseteq\widehat{\RR}. \] Equivalently, \[ \widehat{\RR}\subseteq T(\widehat{\RR}). \] Hence \[ T(\widehat{\RR})=\widehat{\RR}, \] which is condition (a).
3. Condition (a) implies condition (b).
Assume \[ T(\widehat{\RR})=\widehat{\RR}. \] Choose three distinct finite points \[ x_1,x_2,x_3\in\RR \] none of which is the pole of \(T\). Then \[ y_j=T(x_j)\in\RR \] for \(j=1,2,3\), and the \(y_j\) are distinct because \(T\) is injective.
Define \[ A(z) = \frac{(z-x_2)(x_3-x_1)} {(z-x_1)(x_3-x_2)} \] and \[ B(z) = \frac{(z-y_2)(y_3-y_1)} {(z-y_1)(y_3-y_2)}. \] Both are Möbius transformations with real coefficients, and \[ \begin{aligned} A(x_1)&=\infty,&A(x_2)&=0,&A(x_3)&=1,\\ B(y_1)&=\infty,&B(y_2)&=0,&B(y_3)&=1. \end{aligned} \] Therefore \[ S=B\circ T\circ A^{-1} \] fixes the three points \[ \infty,\quad0,\quad1. \] A Möbius transformation fixing \(\infty\) has zero lower-left coefficient; fixing \(0\) then forces its upper-right coefficient to vanish; fixing \(1\) forces the two remaining diagonal coefficients to agree up to the common projective scalar. Hence \(S\) is the identity.
Thus \[ T=B^{-1}\circ A. \] Both \(A\) and \(B^{-1}\) have real coefficients, so their composition admits real coefficients. This is condition (b).
4. Conditions (a), (b), and (c) are equivalent.
Steps <1>1–<1>3 give the cycle \[ \text{(b)} \Longrightarrow \text{(c)} \Longrightarrow \text{(a)} \Longrightarrow \text{(b)}. \]
5. Condition (d) implies condition (c).
Assume condition (d). Thus there are \[ \alpha\in\RR, \qquad \beta\in\CC\sm\RR \] such that \[ T(\alpha)=\alpha \] and \[ T(\bar\beta)=\overline{T(\beta)}. \] Define the Möbius transformation \[ S=J\circ T\circ J. \]
Since \(\alpha\) is real, \[ S(\alpha) = J(T(\alpha)) = \alpha = T(\alpha). \] Using the second hypothesis in (d), \[ \begin{aligned} S(\beta) &= J(T(\bar\beta))\\ &= J(\overline{T(\beta)})\\ &= T(\beta), \end{aligned} \] and also \[ S(\bar\beta) = J(T(\beta)) = \overline{T(\beta)} = T(\bar\beta). \] The three points \[ \alpha,\quad\beta,\quad\bar\beta \] are distinct. Two Möbius transformations agreeing at three distinct points are equal, so \[ S=T. \] Thus \[ J\circ T\circ J=T, \] and composing on the right with \(J\) gives \[ J\circ T=T\circ J. \] This is condition (c).
6. Condition (d) therefore implies each of (a), (b), and (c).
Step 5 gives (c), and step 4 gives the equivalence of (a)–(c).
7. The converse implication from (a)–(c) to (d) is false.
Consider \[ T(z)=-\frac1z. \] It is represented by the real matrix \[ \begin{pmatrix} 0&-1\\ 1&0 \end{pmatrix}, \] so it satisfies condition (b), and hence conditions (a) and (c) by step 4.
Its fixed points satisfy \[ z=-\frac1z, \] equivalently \[ z^2=-1. \] Thus its only fixed points are \[ z=\pm i. \] Neither is real, and \(\infty\) is not fixed because \(T(\infty)=0\). Consequently there is no \[ \alpha\in\RR \] with \(T(\alpha)=\alpha\), so condition (d) fails.
8. The true implication pattern for the four printed conditions is\[ \boxed{ \text{(d)} \Longrightarrow \text{(a)} \Longleftrightarrow \text{(b)} \Longleftrightarrow \text{(c)}, } \]and the implication from (a)–(c) back to (d) need not hold.
Step 4 proves the three-way equivalence, step 6 proves the implication from (d), and step 7 disproves the converse.
9. Q.E.D.
Step 8 proves the equivalence of (a)–(c), the implication from (d), and the failure of the converse.