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Then Riemann sphere is defined by charts $(\mathbb C,Id_{\mathbb C})$ and $(\mathbb C-\{0\}\cup\{\infty\},\phi)$, $\phi(z) = \frac{1}{z},$ if $z \neq 0$, $\phi(z) = 0$ if $z = \infty$. I was told that if $f$ is an analytic function from the Riemann sphere to the Riemann sphere, then $f$ can only be a rational function.

However, I think about defining$\ $ $f(z)= e^z$, when $z \in \mathbb C$ and $f(z) = \infty$ when $z = \infty$. Isn't this a well-defined holomorphic function between Riemann spheres?

z.z
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2 Answers2

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$\mathbb{C}_\infty$ is the Riemann sphere. You can show that an analytic function $f : \mathbb{C}_\infty \to \mathbb{C}_\infty$ is a rational function by using the theory of Riemann surfaces : showing that the Riemann sphere is compact, so that $f(z)$ has finitely many zeros and poles and $f(z)\frac{\prod_{i=1}^I (z-a_i)}{\prod_{j=1}^J (z-b_j)}$ is bounded and hence constant (by Liouville's theorem).

Otherwise, there is a traditional complex analysis proof :

  • $f(z)$ is analytic $\mathbb{C}\to \mathbb{C}_\infty$ means that for every $z_0 \in \mathbb{C}$ : $f(z)$ or $1/f(z)$ is analytic around $z_0$. This is exactly the definition of $f(z)$ is meromorphic ($\mathbb{C} \to \mathbb{C}$).

  • By the isolated zeros theorem, an analytic function has finitely many zeros on any compact, and a meromorphic function has finitely many zeros and poles on any compact.

  • If $g(z)$ is meromorphic and $G(z) = g(1/z)$ is analytic at $z=0$, then there is a $R$ such that $G(z)$ doesn't have any pole on $|z| \le R$, and hence $g(z)$ doesn't have any pole on $|z| \ge 1/R$, but $|z| \le 1/R$ is compact, so that $g(z)$ has finitely many poles.

  • If $f(z)$ is analytic $\mathbb{C}_\infty \to \mathbb{C}_\infty$, the same argument shows that $f(z)$ has finitely many poles and finitely many zeros.

Finally, let $\tilde{f}(z) = f(z)\frac{\prod_{i=1}^I (z-a_i)}{\prod_{j=1}^J (z-b_j)}$ where $a_i,b_j$ are the poles and the zeros of $f(z)$, you get that $\tilde{f}(z) $ and $\tilde{f}(1/z)$ are entire $\implies$ it is constant (by Liouville's theorem) and $f(z)$ is a rational function.

reuns
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  • Why would you want to divide $f$ by the zeros of it? – z.z Nov 08 '16 at 01:31
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    @z.z for obtaining an entire function $\tilde{f}(z)$ such that $1/\tilde{f}(z)$ is entire too $\implies \tilde{f}(z)$ is constant by Liouville's theorem – reuns Nov 08 '16 at 01:34
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    @z.z Now if you don't get why $e^{z}$ isn't analytic $\mathbb{C}\infty \to \mathbb{C}\infty$, what you need to prove is that $f(z)$ is analytic $\mathbb{C}\to \mathbb{C}\infty$ means that for every $z_0 \in \mathbb{C}$ : $f(z)$ or $1/f(z)$ is analytic around $z_0$. And that $f(z)$ is analytic $\mathbb{C}\infty\to \mathbb{C}\infty$ means that $f(z)$ and $f(1/z)$ are analytic $\mathbb{C}\to \mathbb{C}\infty$ – reuns Nov 08 '16 at 02:05
  • I see but do you need to consider the order of zeros and poles in your construction of $\tilde f$? – z.z Nov 08 '16 at 02:10
  • @z.z Try and see – reuns Nov 08 '16 at 02:11
  • Why is $\tilde f$ constant? Both $e^{z}$ and $\frac{1}{e^{z}}$ are entire but $e^{z}$ is not entire. – z.z Dec 12 '16 at 03:45
  • @z.z You need to show $\tilde{f}$ is holomorphic $\mathbb{C}\infty \to \mathbb{C}$. Since $\mathbb{C}\infty $ is compact you get that $\tilde{f}$ is bounded, and you can apply Liouville's theorem – reuns Dec 12 '16 at 05:04
  • And I think I meant $\tilde{f}(z)$ and $\tilde{f}(1/z)$ are entire – reuns Dec 12 '16 at 05:07
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You were told correctly.

In your example, $e^z$ is not analytic at $z=\infty$, in fact not continuous. Note that e.g. $$\lim_{x \to -\infty} e^x = 0$$

More generally, in order for the limit (on the Riemann sphere) of $f(z)$ at complex $\infty$ to exist, the singularity of $f(z)$ at complex $\infty$ must be either removable (i.e. $\lim_{z \to \infty} f(z)$ exists as a complex number) or a pole (i.e. $\lim_{z \to \infty} |f(z)| = +\infty$). The singularity of $\exp(z)$ at $\infty$ is an essential singularity.

Robert Israel
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  • But if I strictly follows the definition to examine the coordinate representation of $f$, it should be $\phi f Id^{-1}{\mathbb C} = \frac{1}{e^z}$ and $Id{\mathbb C} f \phi^{-1} = {e^\frac{1}{z}}$, those two are holomorphic. – z.z Nov 07 '16 at 23:14
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    $e^{1/x}$ is not holomorphic at $0$. – Robert Israel Nov 07 '16 at 23:16
  • But the coordinate representation of that one is defined on $C - {0}$, since $\infty$ is not in the domain of $Id_{\mathbb C}$ – z.z Nov 07 '16 at 23:19