I'm trying to understand a proof that the holomorphic maps of the Riemann sphere are the rational functions from this book:
All rational maps are analytic from the extended plane to itself. For the converse, suppose $f(z) \in \mathbb{C}$ for all $z \in \mathbb{C}_\infty$. Then $f$ is entire and bounded and thus constant. We can therefore assume that $f(z_0) = \infty$ for some $z_0 \in \mathbb{C}$ (consider $f(1/z)$ if necessary). By continuity of $f$ the point $z_0$ cannot be an essential singularity of $f$. In other words, $z_0$ is a removable singularity or a pole. By the uniqueness theorem, the poles cannot accumulate in $\mathbb{C}_\infty$. Since the latter is compact, there can thus only be finitely many poles. Hence, after subtracting the principal part of the Laurent series of $f$ around each pole in $\mathbb{C}$ from $f$, we obtain an entire function which grows at most like a polynomial. By Liouville's theorem, such a function must be a polynomial and we are done.
I have a few questions regarding this:
- Using the continuity argument can we also conclude that $z_0$ must be a pole?
- Why does the uniqueness theorem imply the poles cannot accumulate in $\mathbb{C}_\infty$?
- Perhaps slightly less relevant to this proof, does $f : \mathbb{C}_\infty \to \mathbb{C}_\infty$ holomorphic imply $f$ is meromorphic? I'm using Conway's definition that $f$ is meromorphic iff it is analytic except for poles.
Best Answer
Yes. It must be either a pole or an essential singularity, and essential singularities were ruled out. Though to be exact, we also have to show that it is isolated, first, which is done in the next step.
If they did accumulate, then their accumulation point would be an accumulation point of points on the domain where $f$ is $\infty$. The function which is $\infty$ everywhere is a holomorphic function with such an accumulation point, and since it is unique, $f$ must be this same constant function. This would make $z_0$ neither a pole, nor removable, nor essential, by the way. But it would make $f$ rational, depending on how you define rational. But I admit that this could have been communicated better by the author.
No, meromorphic functions make a distinction between $\infty$ and other points, while holomorphic ones do not. The function which is a constant $\infty$ is not meromorphic because the singularities are not isolated, for instance, but it is holomorphic as a function $\mathbb C^\infty\to\mathbb C^\infty$.