Universal cover of n times punctured plane

Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
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)
 

Opalg

MHF Hall of Honor
Aug 2007
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Leeds, UK
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 \(\displaystyle \mathbb{F}_n\).
 
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Bruno J.

MHF Hall of Honor
Jun 2009
1,266
498
Canada
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 2012
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0
Stanford, CA
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