Extending a holomorphic function $\mathbb{C} \rightarrow \mathbb{C}$ to be holomorphic on the Riemann sphere $\mathbb{C}^{*}$

Question

How can I quickly see if a function $f:\mathbb{C} \rightarrow \mathbb{C}$ can be extended to $f:\mathbb{C}^* \rightarrow \mathbb{C}^*$? (where $\mathbb{C}^*$ denotes the Riemann surface $\mathbb{C} \cup \{{\infty}\} $)?

From the definition we can quickly see that a function $f:X_1 \rightarrow X_2$ between two riemann surfaces is holomorphic if and only if $f \circ \phi^{-1}: \mathbb{C} \rightarrow X_2$ is holomorphic for all charts $\phi$. (where $\mathbb{C}$ is equipped with the chart $id:\mathbb{C} \rightarrow \mathbb{C}$).

And a criterion tells us that such a mapping $f \circ \phi^{-1}: \mathbb{C} \rightarrow \mathbb{C}^{*}$ is holomorphic exactly when $f$ is meromorphic. So together we know that: $f:\mathbb{C}^* \rightarrow \mathbb{C}^*$ is holomorphic, if and only if,

$$f \circ \phi_0^{-1}: z \mapsto f(z)$$ and $$f \circ \phi_{\infty}^{-1}= \begin{cases} f(\frac{1}{z}), & z \neq 0,\\ f(\infty), & z = 0. \end{cases}$$ are meromorphic, where $\phi_0, \phi_\infty$ denotes the charts of $\mathbb{C}^*$.

This criterion is useful for proofs, and it is quick to check the first requirement if we know that $f$ is meromorphic. How can we quickly check that $f \circ \phi_\infty^{-1}$ is meromorphic? Or are there any other more useful criteria? Most of the time I need to quickly decide whether a function is extensible or not, so any feedback would be greatly appreciated, thanks!

Edit: After reading some of the comments, I think we only have to check that $f(1/z)$ is not an essential singularity at $z = 0$ and $f$ has finitely many poles. Since this would imply that $f \circ \phi_\infty^{-1}$ is meromorphic. Is this correct?

Answer 1

To restate the question more precisely: when can a meromorphic function $f:\mathbb{C}\to\mathbb{C}$ be extended to a function $\mathbb{C}^*\to\mathbb{C}^*$ that is holomorphic as a map between Riemann surfaces?

This is equivalent to asking if $f(\infty)$ can be defined so that this extended $f$ is meromorphic on the whole Riemann sphere.

This is equivalent to:

• The original $f$ is "bounded" in a neighborhood of $\infty$. That is, either (i) for some $r,b\in\mathbb{R}^+$, we have $|f(z)|<b$ whenever $|z|>r$, or else (ii) for some $r,b\in\mathbb{R}^+$, we have $|f(z)|>b$ whenever $|z|>r$.

(i) is bounded in the usual sense: the values lie in some disk around 0, whenever $z$ is large enough. (ii) is "bounded" in the sense that the values lie in some "disk around $\infty$", for large enough $z$. (People sometimes say "bounded away from 0" for this.)

Proof: We can move the domain to a neighborhood of 0 by looking at $g(z)=f(1/z)$. If this is bounded in that neighborhood, then the Riemann removable singularities theorem says that $g(z)$ can be extended to a function holomorphic at 0, by setting $g(0)=\lim_{z\to 0}g(z)$ ($=a$, say). (The theorem asserts that the limit exists.) Then set $f(\infty)=a$ and $f$ is holomorphic at $\infty$.

If $f(z)$ is "bounded" in sense (ii) above, then look at $g(z)=1/f(1/z)$: this moves both the domain and range to a neighborhood of $0$. Now we have $g(z)$ bounded near 0, so we set $g(0)=\lim_{z\to 0}g(z)=a$ as before. If $a\neq0$, then $\lim_{z\to 0}f(1/z)=1/a$ and we have the previous case. Otherwise $\lim_{z\to 0}g(z)=0$ and so $\lim_{z\to 0}f(1/z)=\infty$, so set $f(\infty)=\infty$ and this extended $f$ is meromorphic at $\infty$.

On the other hand, if we can extend $f$ meromorphically to include $\infty$ in its domain, then the original $f$ is obviously bounded near $\infty$ if $f(\infty)$ is finite, and "bounded" in sense (ii) near $\infty$ if the new $f$ has a pole at $\infty$.

It is well-known that a function meromorphic on the whole sphere must be rational. So that's another equivalent condition.

By one of Weierstrass's theorems, if $f(1/z)$ has an essential singularity at 0, then it cannot be bounded there; likewise for $1/f(1/z)$. So another equivalent condition is that $f$ does not have an essential singularity at $\infty$.

These all imply that the original $f$ has only finitely many poles. But by itself, that condition is not enough. For example, $e^z/z$ has only one pole plus an essential singularity at $\infty$. Or more simply still, $e^z$ has no poles and an essential singularity at $\infty$.

(Note that if $f$ with domain $\mathbb{C}^*$ has infinitely many poles, then these will have an accumulation point where $f$ will have an essential singularity. By initial assumption, $f$ has no essential singularities in the finite plane.)

Finally, all this "moving the domain (or range) from a neighborhood of $\infty$ to 0" is just an instance of the use of charts, as you noted: near $\infty$ we use the chart $z\mapsto 1/z$.