# Universal cover of n times punctured plane

#### Bruno J.

MHF Hall of Honor
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
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 $$\displaystyle \mathbb{F}_n$$.

• Bruno J.

#### Bruno J.

MHF Hall of Honor
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.

#### rfurman

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