Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is an exercise saying that locally (around $i$) $\tau\mapsto-1/\tau$ is a 180-degree rotation around $i$. I can prove it using some basic calculation. But there is hint saying that consider $$ \begin{bmatrix}1&-i\\1&i\end{bmatrix}\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}1&-i\\1&i\end{bmatrix}^{-1} $$

Can anyone explain to me how this hint works?

share|cite|improve this question
up vote 0 down vote accepted

Those are the matrix forms of Möbius transformations. The middle one is the transformation in question. The hint is to conjugate the transformation as indicated, and see what form the resulting transformation has. Following the hint, we see that $$\left[\begin{array}{cc}1 & -i\\1 & i\end{array}\right]\left[\begin{array}{cc}0 & -1\\1 & 0\end{array}\right]\left[\begin{array}{cc}1 & -i\\1 & i\end{array}\right]^{-1} = \left[\begin{array}{cc}-i & 0\\0 & i\end{array}\right],$$ so the transformation in question is conjugate to a rotation about $z=0$ by $180^\circ$. Undoing the conjugation of transformations gives us the equivalent representation $$\left[\begin{array}{cc}0 & -1\\1 & 0\end{array}\right]=\left[\begin{array}{cc}1 & -i\\1 & i\end{array}\right]^{-1}\left[\begin{array}{cc}-i & 0\\0 & i\end{array}\right]\left[\begin{array}{cc}1 & -i\\1 & i\end{array}\right].\tag{1}$$

I've never been much a fan of using the matrix forms, except to simplify the calculation, so I'll mostly be using an alternative notation from here on. In particular, if $\alpha,\beta,\gamma,\delta\in\Bbb C$ with $\alpha\delta-\beta\gamma\neq 0$, then $$\left[z\mapsto\frac{\alpha z+\beta}{\gamma z+\delta}\right]$$ will be my alternative way to represent the Möbius transformation representable by the matrix $$\left[\begin{array}{cc}\alpha & \beta\\\gamma & \delta\end{array}\right].$$ As with matrix representations, the operation being represented is composition, so $$\left[z\mapsto\frac{az+b}{cz+d}\right]\left[z\mapsto\frac{\alpha z+\beta}{\gamma z+\delta}\right]$$ represents the transformation $z\mapsto\frac{\alpha z+\beta}{\gamma z+\delta},$ followed by the transformation $z\mapsto\frac{az+b}{cz+d}.$

Rewriting $(1)$ with the alternative notation gives us $$\left[z\mapsto\frac{-1}{z}\right]=\left[z\mapsto\frac{z-i}{z+i}\right]^{-1}[z\mapsto-z]\left[z\mapsto\frac{z-i}{z+i}\right].\tag{$1'$}$$ Recall that if $\alpha,\beta,\gamma,\delta\in\Bbb C$ with $\alpha\delta-\beta\gamma\neq 0$, then $$\left[z\mapsto\frac{\alpha z+\beta}{\gamma z+\delta}\right]^{-1}=\left[z\mapsto\frac{\delta z-\beta}{-\gamma z+\alpha}\right].$$ This lets us rewrite $(1')$ as $$\left[z\mapsto\frac{-1}{z}\right]=\left[z\mapsto\frac{iz+i}{-z+1}\right][z\mapsto-z]\left[z\mapsto\frac{iz+i}{-z+1}\right]^{-1}.\tag{2}$$ Now, observe that $$\begin{align}\left[z\mapsto\frac{iz+i}{-z+1}\right] &= \left[z\mapsto\frac{iz+i}{-z+1}-i+i\right]\\ &= \left[z\mapsto\frac{2iz}{-z+1}+i\right]\\ &= [z\mapsto z+i]\left[z\mapsto\frac{2iz}{-z+1}\right],\end{align}$$ so it follows from $(2)$ that $$[z\mapsto z+i]^{-1}\left[z\mapsto\frac{-1}{z}\right][z\mapsto z+i]=\left[z\mapsto\frac{2iz}{-z+1}\right][z\mapsto-z]\left[z\mapsto\frac{2iz}{-z+1}\right]^{-1}.\tag{3}$$ Then $\left[z\mapsto\frac{-1}z\right]$ is locally (near $i$) a rotation about $i$ by $180^\circ$ if and only if the expression on the right-hand side of $(3)$ is locally (near $0$) a rotation about $0$ by $180^\circ$. (Why?)

A few observations about the transformation $\left[z\mapsto\frac{2iz}{-z+1}\right]$ will be enough to let us draw the desired conclusion. Said transformation fixes $0$, and is analytic with a non-zero derivative (so conformal) away from $z=1$. In particular, it is locally (near $0$) the transformation $[z\mapsto 2iz]$. (Why?) Likewise $\left[z\mapsto\frac{2iz}{-z+1}\right]^{-1}$ is locally (near $0$) the transformation $\left[z\mapsto\frac{z}{2i}\right]=[z\mapsto 2iz]^{-1}$. Since $[z\mapsto 2iz][z\mapsto -z]=[z\mapsto -z][z\mapsto 2iz]$, then the desired conclusion follows.

Let me know if you've any other questions.

share|cite|improve this answer
Yeah I know they are Mobius transformations and after conjugate, I get $z\mapsto -z$, which is a 180-degree rotation. But I wonder whether the transformation I'm conjugating by has a influence on the final consequence? – hxhxhx88 Jan 8 '13 at 9:48
@hxhxhx88: A good question. The answer is yes. I'll add the details to my answer. – Cameron Buie Jan 8 '13 at 20:16
It's in progress, but I need to run an errand. I'll finish it up, and post when I return. – Cameron Buie Jan 8 '13 at 20:54
beautiful explanation, thank you very much! – hxhxhx88 Jan 9 '13 at 1:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.