I asked a question a couple days ago so please let me know if I'm asking too frequently on here, I'm not sure what the etiquette generally is. In short, I'm looking to prove that for P, Q distinct points in the complex plane $\mathbb{C}$, $\mathbb{C}-\{P,Q\}$ is not conformally equivalent to either $\mathbb{C}$ or $\mathbb{C}^{*}$ (the punctured plane), and that the same is true for any domain whose complement has more than one point in $\mathbb{C}$, using typical techniques in complex analysis.

The only caveat is I hope to use this in order to show using the uniformization theorem that $\mathbb{C}-\{P,Q\}$ is not uniformed by the unit disk, so I would prefer to avoid proving this using the uniformization theorem to avoid a circular argument. Since this is homework, I wouldn't expect too much detail in the answer (unless you care to give it!), but I'm not sure what methods are conventionally used to prove 2 structures are not conformally equivalent. In general, is there any rule on a Riemann surface, compact or otherwise, which determines whether it can be uniformized by the unit disk or the complex plane? (For example, based on genus? This latter question is merely a matter of curiosity.) Could anyone help please? Many thanks in advance.

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    $\begingroup$ Conformally equivalent surfaces are a fortiori homeomorphic. Therefore it is enough to prove that ${\mathbb C}\setminus\lbrace P, Q\rbrace$ is not homeomorphic to either ${\mathbb C}$ or ${\mathbb C}^*$. $\endgroup$ Jan 10, 2011 at 12:31
  • $\begingroup$ With regard to your last question, there is a well-understood theory of uniformization of Riemann surfaces, the upshot of which is that "most" Riemann surfaces are uniformized by the unit disk. If the surface is compact, then it is a question of genus: $g =0$ is a Riemann sphere, and uniformizes itself; $g = 1$ is uniformzied by the complex plane; $g \geq 2$ is uniformized by the unit disk. $\endgroup$
    – Matt E
    Jan 10, 2011 at 16:38

1 Answer 1


$\mathbb C\setminus \{P,Q\}$ is not conformally equivalent to either $\mathbb C$ or $\mathbb C^∗$

As stated, this question admits a topological answer: the spaces are not even homeomorphic (e.g., their fundamental groups are different). This was a comment by Christian Blatter.

I hope to use this in order to show using the uniformization theorem that $\mathbb C\setminus \{P,Q\}$ is not uniformized by the unit disk

But the twice-punctured plane (thrice-punctured sphere) is uniformized by the disk. See Complex Analysis by Ahlfors, or MathOverflow questions


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