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Let $H$ be a cyclic group of rotations of the unit disc $D=\{z\in \mathbb{C}:|z|<1\}$. For the sake of definiteness let's assume that $H$ is of order $3$. Consider the quotient map $\pi: D\rightarrow D/H\simeq D$ ($\simeq$ means 'homeomorphic').

There is a theorem saying that whenever a group $G$ acts freely and properly discontinuously (in the sense of the definition that I stated here) on a complex manifold $X$, the quotient space $X/G$ has a (unique) complex structure such that the quotient map is a regular holomorphic covering. In the above case, $H$ acts freely and properly discontinuously on $D^\ast :=D\setminus\{0\}$, so the quotient map $D^\ast\rightarrow D^\ast/H\simeq D^\ast$ is a regular covering which becomes holomorphic once one puts the Riemann surface structure described in the theorem on the base space $D^\ast$.

I was told about the following way of constructing a chart for the point $\pi(0)=0$. Consider the slit disc $D_s=D\setminus[0,1]$. Then $\pi^{-1}(D_s)$ is equal to one of the three (in our case) open sectors of the disc $D$ (which are permuted cyclically by elements of $H$). Further, if one applies the map $z\mapsto z^3$ to this sector, one gets $D_s$ again. So the map $(\pi^{-1})^3$ maps $D_s$ homeomorphically onto $D_s$, and this homeomorphism can be extended to the homeomorphism from $D$ to $D$.

But what I don't understand is the following: if $\pi(D)$ is homeomorphic to $D$ and say $\pi(0)=0$, then why can't one just take this $D$ along with the homeomorphism as a chart for the point $\pi(0)$? What is the sense in constructing the homeomorphism $(\pi^{-1})^3$ between two copies of the slit unit disc to get a homeomorphism from $D$ to $D$ which is God-given (viz. the identity map e.g.)? Also, is there any explanation on what makes the point $\pi(0)$ special (i.e., why it's not obvious how to construct a chart for this point) except for that saying that this is due to the fact that $H$ does not act nicely (=freely) at the origin so that one cannot apply an appropriate theorem?

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    $\begingroup$ Your map $\mathrm{id}$ is not a well-defined map $D/H \rightarrow \mathbb{C}$ so it isn't a chart on $D/H$. Also $\pi^{-1}(D_s)$ is not one of the three sectors but rather the union of all of them. $\endgroup$ – user399601 Jan 28 '17 at 20:27
  • $\begingroup$ Of course you are right about $\pi^{-1}(D_s)$. I believe instead I should have said that $D_s$ is homeomorphic to one of the three sectors via the (suitable restriction of the) map $\pi$ and its inverse - that would be correct, right? Also, I can't figure how to see that the map you are talking about is not well-defined... $\endgroup$ – Cary Jan 28 '17 at 20:48
  • $\begingroup$ Points in $D/H$ do not correspond to unique points in $\mathbb{C}$. Away from $0$ you can choose representatives in a reasonable (continuous) way but around $0$ you run into big problems. $\endgroup$ – user399601 Jan 28 '17 at 20:51

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