Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

2 Answers 2

up vote 4 down vote accepted

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 $\mathbb C\setminus \{0\}$...

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.