Let $T(z)=\frac{a_1z + b_1}{c_1z+d_1}$ $S(z)=\frac{a_2z+b_2}{c_2z+d_2}$ be two Mobius transformations. Show that if ...

Question

Let $T(z)=\frac{a_1z + b_1}{c_1z+d_1}$ $S(z)=\frac{a_2z+b_2}{c_2z+d_2}$ be two Mobius transformations. Show that if $T = S$, then there is $\lambda \in \mathbb C$ such that $a_1 = \lambda a_2, b_1= \lambda b_2, c_1 = \lambda c_2, d_1 = \lambda d_2$.

$\frac{a_1z + b_1}{c_1z+d_1} \leftrightarrow c_1z^2 +(d_1 - a_1)z + b_1$

$\frac{a_2z+b_2}{c_2z+d_2} \leftrightarrow c_2z^2 +(d_2 - a_2)z + b_2 = 0$

$c_1z^2 +(d_1 - a_1)z + b_1 = c_2z^2 +(d_2 - a_2)z + b_2 = 0$

$c_1 = 1 c_2 $

$b_1 = 1 b_2 $

$d_1 - a_1 = d_2 - a_2 \to a_1 = d_1 = 1 a_2 = 1 d_2$.

This must not be right because it's too simple, but it was the only idea I had.

Thank's for any help.

Answer 1

Your idea is nice, but you do not carry it out properly.

Each Möbius transformation $T$ has the form $T(z) = \dfrac{az + b}{cz + d}$ with a matrix $M = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$ such that $\det M = ad - bc \ne 0$. Let us write $T = \theta_M$. Note that $\det M \ne 0$ implies

1. If $a = 0$, then $b, c \ne 0$.

2. If $b = 0$, then $a, d \ne 0$.

Given $M_i = \left(\begin{matrix}a_i & b_i \\ c_i & d_i\end{matrix}\right)$, we are expected to prove that $\theta_{M_1} = \theta_{M_2}$ implies $M_1 = \lambda M_2$ for some $\lambda \in \mathbb C$. Note that in this case automatically $\lambda \ne 0$ (otherwise $M_1 = 0$ and $\det M_1 = 0$).

$\theta_{M_1} = \theta_{M_2}$ means that for all $z$ $$(a_1z + b_1)(c_2z +d_2) = (a_2z + b_2)(c_1z +d_1) ,$$ i.e. $$a_1c_2z^2 + (a_1d_2 +b_1c_2)z + b_1d_2 = a_2c_1z^2 + (a_2d_1 +b_2c_1)z + b_2d_1 .$$ This gives the three equations $$a_1c_2 = a_2c_1 \tag{1} $$ $$a_1d_2 +b_1c_2 = a_2d_1 +b_2c_1 \tag{2}$$ $$b_1d_2 = b_2d_1 \tag{3}$$

Case 1: $a_2 = 0$.
By 1. we see that $b_2, c_2 \ne 0$. Thus $ \lambda = b_1/b_2$ is a well-defined complex number and $(1)$ shows that $a_1 = 0$. By definition $b_1 = \lambda b_2$. From $(2)$ we get $c_1 = (b_1/b_2) \cdot c_2 = \lambda c_2$ and similarly from $(3)$ $d_1 = \lambda d_2$. Trivially $a_1 = \lambda a_2$.

Case 2: $b_2 = 0$.
Analogous as Case 1. Explicitly, from 2. we conclude that $a_2,d_2 \ne 0$. Thus $ \lambda = a_1/a_2$ is a well-defined complex number and $(3)$ shows that $b_1 = 0$. By definition $a_1 = \lambda a_2$. From $(2)$ we get $d_1 = (a_1/a_2) \cdot d_2 = \lambda d_2$ and similarly from $(1)$ $c_1 = \lambda c_2$. Trivially $b_1 = \lambda b_2$.

Case 3: $a_2, b_2 \ne 0$.
Then $\lambda = a_1/a_2$ and $\mu = b_1/b_2$ are well-defined complex numbers. This means that $a_1 = \lambda a_2$ and $b_1 = \mu b_2$ and from $(1), (3)$ we get $c_1 = \lambda c_2$ and $d_1 = \mu d_2$. Inserting in $(2)$ gives $$\lambda(a_2d_2 - c_2b_2) = \mu(a_2d_2 - c_2b_2) .$$ Since the second factor is $\ne 0$, we see that $\lambda = \mu$.

Answer 2

If you want to avoid calculation: You could use that to all Möbius transformations $T(z) = \frac{a_1z+b_1}{c_1z+d_1}$ and $S(z) = \frac{a_2z+b_2}{c_2z+d_2}$, there exist unique classes $[A_T],[A_S]\in GL(2)/\mathbb{C}^*$, with representatives given by \begin{align*} A_T &= \left(\begin{matrix}a_1 & b_1 \\ c_1 & d_1\end{matrix}\right),\\ A_S &=\left(\begin{matrix}a_2 & b_2 \\ c_2 & d_2\end{matrix}\right), \end{align*} so that that $$[A_T\cdot A_S] = [A_{T\circ S}].$$ Here, $\mathbb{C}^*:=\mathbb{C}\setminus\{0\}$ has to be factured out, because multiplying such a matrix by any non-zero $\lambda$ would yield another matrix with the same property (just as multiplying each coefficient $a,b,c,d$ in in the initial Möbius-function would do).

Then, assuming that $T=S$, you would get $E_2 = A_T\cdot A_T^{-1} \equiv A_T\cdot A_S^{-1} \mod \mathbb{C}^*$, which is equivalent to $A_T \equiv A_S \mod \mathbb{C}^*$, i.e. $A_T = \lambda\cdot A_S$ for some $\lambda\in\mathbb{C}^*$.

Answer 3

(Assuming $a_1, a_2, b_1, b_2, c_1, c_2, d_1, d_2$ all non-null to keep the answer short: these have to be dealt with in a few separate cases).

You can use specific values for $z$:

• $z=0$ gives $b_1/d_1=b_2/d_2$
• $z\rightarrow +\infty$ gives $a_1/c_1=a_2/c_2$
• $z=-b_1/a_1$ (zero) gives $a_2/a_1=b_2/b_1$
• (as an alternative to any of the 3 previous lines) $z\rightarrow -d_1/c_1$ (pole) gives $c_2/c_1=d_2/d_1$

So posing $a_1/a_2=\lambda$, we get
$\lambda=c_1/c_2=b_1/b_2=d_1/d_2$