4

I'm working/familiar with the following defintion of a pole in $\mathbb{C}$:

Let $D\subset \mathbb{C}$ be an open domain and $f:D\to\mathbb{C}$ a function. $z_0\in\mathbb{C}$ is called a pole of $f$, if $\vert f(z)\vert\to\infty$ for $z\to z_0$.

Now I try to understand what it means for a point to be a pole on the Riemann sphere, but I have no rigorous definition for it (I do have a definition for being holomorphic at $\infty$). So let's expand the above definition to the Riemann sphere $\hat{\mathbb{C}}:=\mathbb{C}\cup\{\infty\}$, especially for the point $z_0=\infty$. We take the above definition, and apply it to $z_0=\infty$. When doing this, I always come to situations, where it seems to be problematic. Here is a simple example:

$f(z)=z$ on $\hat{\mathbb{C}}$, then $f(\infty)=\infty$. That means (by above definition) that $\infty$ is a pole of $f$. But I would expect $f$ to be entire in $\hat{\mathbb{C}}$.

Where is my mistake?

reference for the definition (page 4): https://analysis.math.uni-kiel.de/vorlesungen/meromorphic.17/Entire_Meromorphic.pdf

M.W.
  • 63

2 Answers2

2

We don't usually speak of entire functions on the sphere, because a holomorphic complex-valued function on the sphere is constant.

The identity function on the sphere, by contrast, is meromorphic, and has a simple pole at $\infty$ as you suspect.

Andrew D. Hwang
  • 82,053
  • After thinking about your answer, I came to the following question: Wikipedia article "Riemann sphere" says the following (§2, sentence 2): "[...] any rational function on the complex plane can be extended to a holomorphic function on the Riemann sphere, with the poles of the rational function mapping to infinity."

    So a rational function on $\mathbb{C}$ is always holomorphic on $\hat{\mathbb{C}}$, but it's obviously not constant.

     – M.W. Aug 14 '24 at 09:19
  • 1
    (i) You're perfectly correct, but (ii) that's why it's necessary to specify "a holomorphic complex-valued function." :) A non-constant holomorphic map from the Riemann sphere to itself, such as the extension to the sphere of a polynomial or rational function on $\mathbf{C}$, isn't complex-valued: It achieves the value $\infty$ somewhere on the sphere. – Andrew D. Hwang Aug 14 '24 at 13:03
  • 1
    that helps a lot and clarifies my problem. thank you! – M.W. Aug 14 '24 at 13:08
1

Your definition of a pole at infinity does not make sense because you are applying the familiar definition where it does not apply. There is also another issue, because your definition of pole is incomplete. The complete definition goes like this:

Let $D \subset \mathbb C$ be an open domain and $f : D \to \mathbb C$ a function. $z_0 \in \mathbb C$ is called a pole of $f$ if the following two conditions hold:

  1. there exists $r > 0$ such that $\{z \in \mathbb C \mid 0 < |z-z_0| < r\} \subset D$;
  2. $|f(z)| \to \infty$ as $z \to z_0$.

With this corrected definition, your mistake becomes clear: $z=\infty$ does not satisfy requirement 1.

But you can extend the familiar definition to allow $\infty$, by adding this clause:

In addition, $\infty$ is called a pole of $f$ if $z=0$ is a pole of $f(1/z)$. Equivalently, there exists $R > 0$ such that $\{z \in \mathbb C \mid |z|>R\} \subset D$ and such that $|f(1/z)| \to \infty$ as $z \to 0$.

Lee Mosher
  • 135,265