# Question about extended complex plane

I know that by Liouville's Theorem it can be shown that the open unit disk $\mathbb{D}$ is not conformally equivalent to the complex plane $\mathbb{C}$. Is it true then that the extended complex plane $\mathbb{C}^{*}$ is also not conformally equivalent to $\mathbb{D}$? How is that justified? If it is true, then we would have that since $\mathbb{C}^{*}$ is not conformally equivalent to $\mathbb{D}$ and $\mathbb{D}$ is not conformally equivalent to $\mathbb{C}$, it follows that $\mathbb{C}^{*}$ is not conformally equivalent to $\mathbb{C}$ by transitivity, correct? Any helpful input would be highly appreciated!

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## 2 Answers

Beware! “Not equivalent” is (almost) never a transitive relation: I'm not my sister, she's not me, but I am me nevertheless.

But, yes, $\mathbb C$, $\mathbb D$ and the extended complex plane $\overline {\mathbb C}$ are three different things. Actually, $\overline {\mathbb C}$ is compact, so it not even homeomorphic to $\mathbb C$ or $\mathbb D$ (and homemorphic is a weaker relation than conformally equivalent!)

And as you said, Liouville's theorem already proves that $\mathbb D$ and $\mathbb C$ are not conformally equivalent (even if they are homeomorphic), so we have proven that these three things are different.

By the way, $\mathbb C^*$ is a terrible notation for the extended complex plane (aka the Riemann sphere): this notation evokes $\mathbb C\setminus \{0\}$...

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They are all distinct. The extended plane is simply the ordinary sphere, with the North Pole being labelled $\infty.$ The bijection, conformal, between the sphere minus $\infty$ and the plane, or $\mathbb C,$ is given by stereographic projection. Note that stereographic projection identifies the equator of the sphere with the unit circle $|z| = 1, \; z \in \mathbb C.$

The easy part is that the sphere is compact as a topological space and the other two are not, so the sphere is not homeomorphic, or diffeomorphic, to either. More difficult is the lack of a conformal mapping between the unit disk and $\mathbb C,$ as there are obvious diffeomorphisms between them. Hardest of all, by far, is the Riemann Mapping Theorem.

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