# Universal cover of n times punctured plane

• May 23rd 2010, 07:54 AM
Bruno J.
Universal cover of n times punctured plane
I'm looking for some insight into the relationship between the complex plane punctured $\displaystyle n$ times and its universal cover. I understand that the universal cover of the once-punctured plane is the plane itself, with the corresponding uniformizing function being the exponential (whose automorphism group $\displaystyle \cong \mathbb{Z}$ is isomorphic to the fundamental group of the base and to the group of deck transformations of the cover). I understand also that that the universal cover of the twice punctured plane is the unit disc (or upper half-plane), with the elliptic modular function $\displaystyle \lambda=k^2$ being the corresponding uniformizing function (whose automorphism group $\displaystyle \cong \mbox{free group on two generators} \cong \Gamma(2) \triangleleft \mbox{PSL}(2, \mathbb{Z})$) is once again isomorphic to the fundamental group of the base, and to the group of deck transformations of the cover).

In general, what is the universal cover of the $\displaystyle n$-times punctured plane, and what is the corresponding uniformizing function? I suppose that the universal cover is the upper-half plane for $\displaystyle n\geq 2$, with a modular function as the uniformizing function. However, this would imply that $\displaystyle \mbox{PSL}(2, \mathbb{Z})$ contains a copy of the free group on $\displaystyle n$ generators as a subgroup, which I doubt very much! It's impressive enough that it contains a copy of the free group on two generators...

Any pointers are greatly appreciated! (Nod)
• May 23rd 2010, 11:31 AM
Opalg
I don't know enough algebraic topology to answer this question, but I do know that $\displaystyle \mathbb{F}_2$, the free group on two generators, contains a copy of $\displaystyle \mathbb{F}_n$ as a subgroup, and therefore so does $\displaystyle \text{PSL}(2,\mathbb{Z})$. Here, n can be any positive integer or even infinity. If a, b are generators of $\displaystyle \mathbb{F}_2$ then (if I remember correctly) you can take $\displaystyle a^kb^ka^k\ (1\leqslant k\leqslant n)$ as generators for a copy of $\displaystyle \mathbb{F}_n$.
• May 23rd 2010, 12:28 PM
Bruno J.
Quote:

Originally Posted by Opalg
I don't know enough algebraic topology to answer this question, but I do know that $\displaystyle \mathbb{F}_2$, the free group on two generators, contains a copy of $\displaystyle \mathbb{F}_n$ as a subgroup, and therefore so does $\displaystyle \text{PSL}(2,\mathbb{Z})$. Here, n can be any positive integer or even infinity. If a, b are generators of $\displaystyle \mathbb{F}_2$ then (if I remember correctly) you can take $\displaystyle a^kb^ka^k\ (1\leqslant k\leqslant n)$ as generators for a copy of $\displaystyle \mathbb{F}_n$.

That's awesome! So I guess the possibility of the uniformizing function being a modular function is not ruled out.
• Apr 21st 2012, 09:23 PM
rfurman
Re: Universal cover of n times punctured plane
Hi Bruno,
Did you find a good resolution to this question? It's something that I am quite interested in as well. (I've heard the term Schottky space and Schottky group come up in this context)
Ralph