Show any involutive Möbius transformation must have two distinct fixed points.

Question

Let $T$ be a Möbius transformation such that $T(T(z)) = z$ for all $z \in \mathbb{\hat{C}}$. That is, $T = T^{-1}$. I want to show that $T$ must have two distinct fixed points.

I have tried using a general $$ T(z) = \frac{az + b}{cz + d} \ , $$ attempting to solve for $z$, with no luck. I also know that $T(z)$ may only have two distinct fixed points if and only if $c \neq 0$ or $c = 0$ and $a \neq d$.

Are there any points that are guaranteed to be fixed under these conditions? How could I manage to find these points?

Answer 1

Every non-identity Möbius transformation has either one or two fixed points. A Möbius transformation with only one fixed point (the so-called “parabolic case”) is conjugate to a translation, i.e. $$ T(z) = z + a $$ if the fixed point is $\infty$, or $$ \frac{1}{T(z) - z_0} = \frac{1}{z-z_0} + a $$ if the fixed point is $z_0 \in \Bbb C$, for some $a \ne 0$. In either case $T\circ T$ is not the identity.

Answer 2

Every mobius transform can be represented as some $2 \times 2$ complex matrix $A$, in such a way that we consider $A$ and $\lambda A$ to represent the same transformation for $\lambda$ any non-zero scalar.

Let $A$ be a $2 \times 2$ matrix. We suppose that $A^2 = \lambda I$ for some $\lambda \neq 0$, so that projectively the matrix is an involution. The characteristic polynomial of the matrix is $x^2 - \lambda = (x+\sqrt \lambda)(x - \sqrt \lambda)$. Therefore the matrix has two distinct eigenvectors. When those eigenvectors are interpreted as homogeneous coordinates, they are fixed points.

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An alternative answer might be to notice that, without loss of generality, one of the fixed points can always be moved to zero via conjugation. So let $f(z)$ be a Mobius transformation: The general form of $f(z)$ when $f(0)=0$ is $f(z)=\frac{a}{\frac{c}{z} + d}$. Then taking into account the fact that $f(f(z)) = z$ reduces that to $f(z) = - \frac{1}{d + \frac{1}{z}}$, at which point the remaining fixed point is at $z=-2/d$. If $d = \infty$, which is the only way for the other fixed point to be at $0$, then $f(z)$ is no longer a Mobius transformation, let alone an involution.

The idea here was inspired by Martin R.