Show that Möbius transformation $S$ commute with $T$ if $S$ and $T$ have the same fixed point.

Question

Let $T$ be a Möbius Transformation such that $T$ is not the identity. Show that Möbius transformation $S$ commute with $T$ if $S$ and $T$ have the same fixed point.

Here is what I know so far

1) if $T$ has fixed points says $z_1$ and $z_2$ then $S^{-1}TS$ has fixed point $S^{-1}(z_1)$ and $S^{-1}(z_2)$

2) if $T$ is dilation then $0$ and $\infty$ are its only fixed point, but if $T$ is translation then only $\infty$ is its fixed point.

Assume that $S$ and $T$ has the same fixed points $z_1$ and $z_2 $ then by 1)

$S^{-1}TS$ and $T^{-1}ST$ have the same fixed point $S^{-1}(z_1)=T^{-1}(z_1)$ and $S^{-1}(z_2)=T^{-1}(z_2)$

I know that $T$ is not the identity, but I can't assume it is dilation or translation to use 2), because it can also be inverse, right?

I wonder if anyone would please have me a hand from here.

Answer 1

Assume that $S$ and $T$ are two Moebius transformations of the extended $z$-plane $\overline{\Bbb C}$ having the same fixed points $z_1$, $z_2\in{\Bbb C}$, $\>z_1\ne z_2$. The we can introduce "temporarily" in $\overline{\Bbb C}$ a new complex coordinate $$w:={z-z_1\over z-z_2}\ .$$ The point $z=z_1$ gets the $w$-coordinate $w_1={z_1-z_1\over z-z_2}=0$, and the point $z=z_2$ gets the $w$-coordinate $w_2=\infty$. In terms of this new coordinate both $S$ and $T$ are again given by Moebius expressions, but now the fixed points have $w$-coordinate values $0$ and $\infty$; whence $S$ and $T$ appear as dilations: $$S(w)=\lambda w, \quad \lambda\ne 1;\qquad T(w)=\mu w,\quad \mu\ne1\ .$$ When expressed in terms of $w$ the two transformations obviously commute; therefore they have to commute as well when expressed in terms of the original coordinate $z$. – A similar argument takes care of the case $z_1\in{\Bbb C}$, $z_2=\infty$.

When $S$ and $T$ both have exactly one fixed point $z_0\in {\Bbb C}$ then we can replace $z_0$ by $w_0=\infty$ as before. Now $S$ appears as $$S(w)=\alpha w+\beta\ .$$ When $\alpha\ne1$ then $S$ would have a second fixed point $w_2={\beta\over 1-\alpha}$. It follows that $S$ and $T$ are of the form $$S(w)=w+\beta,\quad\beta\ne0;\qquad T(w)=w+\gamma,\quad \gamma\ne0\ ;$$ whence commute.

Answer 2

I thought about this the other day.

Suppose that

$M_1(z) = \frac{a_1z + b_1}{c_1z + d_1}$ and $M_2(z) = \frac{a_2z + b_2}{c_2z + d_2}$

are Möbius transforms with the same two fixed points $z_0$ and $z_1$.

Then $\;c_1 z^2 + (d_1 - a_1)z - b_1 = c_1(z-z_0)(z-z_1)\;$ and

$\;c_2 z^2 + (d_2 - a_2)z - b_2 = c_2(z-z_0)(z-z_1).$

With $c_1 = kc_2$ for some $k,$ we get $\;b_1 = -kc_2 z_0 z_1 = k b_2 ~~~~~(*),\;$

and $\;d_1 - a_1 = -kc_2 (z_0+z_1) = k(d_2 - a_2) ~~~~~ (**)$.

Now, $M_1(M_2(z)) = \frac{(a_1a_2 + b_1c_2)z + a_1b_2 + b_1d_2}{(c_1a_2 + d_1c_2)z + c_1b_2 + d_1d_2}$

[Using (*)]

$= \frac{(a_1a_2 + kb_2c_2)z + b_2(a_1+kd_2)}{c_2(ka_2+d_1)z + c_2b_1 + d_2d_1}$

[Using (**) and (*)]

$= \frac{(a_2a_1 + b_2c_1)z + b_1a_2+ b_2d_1}{(c_2 a_1 + c_1d_2)z + c_2b_1 + d_2d_1} = M_2(M_1(z)).$