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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? |
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Those are the matrix forms of Mobius 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 Mobius 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. |
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