I am struggling to understand what prof. Penrose says in his "The Road to Reality" book (9.3):
There are many holomorphic mappings of the Riemann sphere to itself which send each hemisphere to itself, but which do not preserve the north or south poles (i.e. the points $z=0$ or $z=\infty$).These preserve the positive/negative frequency splitting but do not preserve the individual Fourier components $e^{-in\chi}$ or $e^{in\chi}$.
Then an exercise is proposed to show explicitely what these mappings are.
The first question I have in mind is whether that these holomorphic mappings of the sphere to itself necessarily correspond to Möbius transformations in the $\mathbb{C}$ plane.
Then, rather than the full solution, I am trying to find at least one example of a Möbius transformation that has the behaviour described; the only mapping that comes into my mind is a rotation through the "vertical" axis (hence a simple complex multiplication $az$ in terms of Möbius transformation) but this of course preserves the poles. I really cannot figure out how I can rotate or translate this sphere so that the north hemisphere still maps to the north, withouth keeping the north pole fixed. It really seems to me that prof. Penrose is meaning something else here, which I am not getting.
Last but not least, I am missing the connection between these holomorphic mappings and the fourier components of a function.
Any hint is welcome, thanks!
Best Answer
If $h$ is analytic and non-zero at $e^{it}$ then (since $h(z)=\sum_k c_k (z-e^{it})^k$) so is $g(z)=1/\overline{h(\overline{z}^{\ -1})}$
If $h(e^{it}) = e^{i\theta}$ then $g(e^{it}) = e^{i\theta}$.
Thus $h$ is meromorphic $\Bbb{C\to C}$ and $|h(z)|=1$ on $|z|=1$ means that $g-h$ is analytic around $e^{it}$ and vanishes on $e^{ix},x-t\in (-\epsilon,\epsilon)$ ie. $g=h$.
Next since $h$ is analytic $\Bbb{P^1(C)\to P^1(C)}$ then it is a rational function $ h(z)=C\frac{\prod_j(z-a_j)}{\prod_i(z-b_j)}$, that $h$ sends each hemisphere to itself gives $|h(z)|=1$ on $|z|=1$ so that $h=g$ and hence $$h(z)=e^{i\theta}\prod_{j=1}^J \frac{z-a_j}{1-\overline{a_j}z}$$ Finally each $|a_j|<1$ since otherwise $h$ would have some zero on the $\infty$-hemisphere.