Classification of analytic involutions

Question

Does there exist a classification of all analytic involutions? I only know $x\mapsto x$, $x\mapsto -x$, $x\mapsto\frac{1}{x}$, and "shifts" thereof.

Answer 1

It highly depends on the domain you work on (you put $x \mapsto 1/x$ as an example so I assume you don't only want it for the whole plane).

I can give you an answer for $\mathbb{C}$ and $\mathbb{D}$ the unit disk. Since any open simply connected subset of $\mathbb{C}$ is biholomorphic to one of those, it already covers a lor of cases.

The unit disk :

Let us start with $\mathbb{D}$ which is the easiest since we know that all its biholomorphic transformations are the $$ f : x \longmapsto \lambda\frac{x + a}{\overline{a}x + 1}, $$ with $a \in \mathbb{D}$ and $\lambda$ such that $|\lambda| = 1$. Thus, for all $x$, \begin{align*} f(f(x)) & = \lambda\frac{f(x) + a}{\overline{a}f(x) + 1}\\ & = \lambda\frac{\lambda\frac{x + a}{\overline{a}x + 1} + a}{\overline{a}\lambda\frac{x + a}{\overline{a}x + 1} + 1}\\ & = \lambda\frac{\lambda(x + a) + a(\overline{a}x + 1)}{\overline{a}\lambda(x + a) + \overline{a}x + 1}\\ & = \lambda\frac{(\lambda + |a|^2)x + (\lambda + 1)a}{(\lambda + 1)\overline{a}x + \lambda|a|^2 + 1}\\ & = \lambda\frac{\lambda + |a|^2}{\lambda|a|^2 + 1}\frac{x + \frac{(\lambda + 1)a}{\lambda + |a|^2}}{\frac{(\lambda + 1)\overline{a}}{\lambda|a|^2 + 1}x + 1}\\ & = \mu\frac{x + b}{\overline{b}x + 1}. \end{align*} Here, we set $$ b = \frac{(\lambda + 1)a}{\lambda + |a|^2}, \qquad \mu = \lambda\frac{\lambda + |a|^2}{\lambda|a|^2 + 1} = \frac{\lambda + |a|^2}{\overline{\lambda} + |a|^2}, $$ which satisfy $b \in \mathbb{D}$ and $|\mu| = 1$. We know that this transformation is the identity if and only if $b = 0$ and $\mu = 1$.

$\mu = 1$ is equivalent to $\lambda = \overline{\lambda}$ i.e. $\lambda$ is real. But since $|\lambda| = 1$, it implies that $\lambda = \pm 1$. If $\lambda = -1$, then $b = 0$ automatically. If $\lambda = 1$, $b = 0$ if and only if $a = 0$. In this last case, $f = \mathrm{Id}$.

Therefore, outside of the identity, the involutions of the disk are given by the functions of the form $$ f : x \longmapsto -\frac{x + a}{\overline{a}x + 1}, $$ with $a \in \mathbb{D}$.

The plane :

For the plane, it is much easier since the biholomorphisms of the plane are exactly the affine maps $$ f : x \longmapsto ax + b, $$ with $a$ and $b$ complex numbers. In this case, $$ f(f(x)) = a(ax + b) + b = a^2x + (a + 1)b. $$ This function is the identity if and only if $a^2 = 1$ and $(a + 1)b = 0$. It means that $a = \pm 1$ and if $a = 1$, we must have $b = 0$. Once more, this last case corresponds to $f = \mathrm{Id}$ and the other ones are the $$ f : x \longmapsto -x + b. $$