Automorphisms of the Upper Half Plane

Welcome back to our little series on automorphisms of four (though, for all practical purposes, it's really three) different Riemann surfaces: the unit disc, the upper half plane, the complex plane, and the Riemann sphere. Last time, we proved that the automorphisms of the unit disc take on a certain form. Today, our goal is to prove a similar result about automorphisms of the upper half plane.

If you missed the introductory/motivational post for this series, be sure to check it out here!

Also in this series:

Automorphisms of the Upper Half Plane

Theorem: Every automorphism $f$ of the upper half plane $\mathcal{U}$ is of the form $f(z)=\frac{az+b}{c z+d}$ where $a,b,c,d\in\mathbb{R}$ and $ad-bc=1$.

Proof. First let's suppose $f\in\text{Aut}(\mathcal{U})$ and recall (or observe) that the function $g:\mathcal{U}\to\Delta$ given by $$g(z)=\frac{z-i}{z+i}$$ is a conformal map. (Here, $\Delta$ denotes the unit disc.) Define $F=g\circ f\circ g^{-1}:\Delta\to\Delta$ so that $F\in\text{Aut}(\Delta)$, that is, $F$ is an automorphism of $\Delta$. By our previous post, there exists $a,b\in\mathbb{C}$ with $|a|^2-|b|^2=1$ such that $$F(t)=\frac{at+b}{\bar bt + \bar a}, \qquad t\in\Delta.$$ Now let $z=g^{-1}(t)\in\mathcal{U}$ so that $g(z)=t$. Then since $f=g^{-1}\circ F \circ g$ and $g^{-1}(w)=\frac{i+wi}{1-w}$ for any $w\in\Delta$ we have $$f(z)=g^{-1}(F(g(z)))=g^{-1}\left( \frac{ a\left(\frac{z-i}{z+i} \right) + b }{\bar b\left(\frac{z-i}{z+i} \right) + \bar a }   \right)$$ which reduces to \begin{align} \frac{z(ai+bi+\bar a i +\bar bi) -\bar a +\bar b - b +a }{z(\bar a +\bar b - a - b) +\bar ai-\bar bi-bi+ai }. \end{align} Since  $a=x+iy$ and $b=p+iq$ for real numbers $x,y,p,q$, we see that \begin{align*} ai=-yix \qquad&\text{and}\qquad \bar ai=y+ix,\\ bi=-q+ip \qquad&\text{and}\qquad \bar bi=q+ip,\\ \bar a - a = -2iy \qquad&\text{and}\qquad \bar b - b=-2iq. \end{align*} Thus (1) becomes $$\frac{z(x+p) + (y-q)}{z(-y-q)+(x-p)}$$ which is of the form $\frac{Az+B}{Cz+D}$ where $A,B,C,D\in\mathbb{R}$ and $AD-BC=(x^2-p^2)+(y^2-q^2)=(x^2+y^2)-(p^2+q^2)=|a|^2-|b|^2=1$ as desired.

Conversely suppose $f(z)=\frac{az+b}{cz+d}$ for $a,b,c,d\in\mathbb{R}$ where $ad-bc=1$ for $z\in\mathcal{U}$. To show $f\in\text{Aut}(\mathcal{U})$, the only potentially tricky part is to show that $f(z)\in\mathcal{U}$, i.e. $\text{Im}f(z)>0$. (Note that $f$ is a linear fractional transformation, so it's holomorphic and invertible and it's inverse is also holomorphic.) But in fact it's not hard at all! Simply notice that if $z=x+iy$ then \begin{align*} f(z)&=\frac{(az+b)(c\bar z+d)}{|cz+d|^2}\\ &=\frac{ac(x^2+y^2) +bc(x-iy) + ad(x+iy) + bd  }{|cz+d|^2} \end{align*} and so Im$f(z)=\frac{ad-bc}{|cz+d|^2}y=\frac{\text{Im}(z)}{|cz+d|^2}$ and this is strictly greater than zero since $\text{Im}(z)>0$.    

$\square$    

Next time:  We'll prove a similar result about the automorphisms of the complex plane.

Related Posts

The Most Obvious Secret in Mathematics

Category Theory

Lebesgue Measurable But Not Borel

Analysis

One Unspoken Rule of Measure Theory

The Back Pocket

A Non-Measurable Set

Analysis
Leave a comment!