There is a beautiful way to see that the congruence subgroup $\Gamma(2)$ is free on two generators: the action of $\Gamma(2)$ on $\mathbb{H}$ is free and properly discontinuous, and there is a modular function $\lambda$ with respect to $\Gamma(2)$ coming from Legendre normal form such that $\mathbb{H}/\Gamma(2) \xrightarrow{\lambda} \mathbb{C} - \{ 0, 1 \}$ is an isomorphism. (Details.) It follows that $\mathbb{H}$ is the universal cover of $\mathbb{C} - \{0, 1 \}$, hence that $\Gamma(2)$ is isomorphic to the fundamental group of $\mathbb{C} - \{0, 1 \}$.

However, the action of $\Gamma(1) \simeq PSL_2(\mathbb{Z})$ on $\mathbb{H}$ is not properly discontinuous free; there are problems, which maybe I should call "ramification," at the points $i, e^{ \frac{\pi i}{3} }, e^{ \frac{2\pi i}{3} }$. This is supposed to be responsible for the fact that $\Gamma(1)$ is not free, but is instead the free product of a cyclic group of order $2$ and a cyclic group of order $3$, where the former somehow comes from the behavior at $i$ and the latter somehow comes from the behavior at the sixth roots of unity. That's what I've been told, anyway, but I don't know how the argument actually goes. What general context does it fit into? (Is "monodromy" a keyword here?)

  • $\begingroup$ Just a quick comment: the action is properly discontinuous; what it is not is free. $\endgroup$ – Pete L. Clark Mar 18 '10 at 20:53
  • $\begingroup$ If you run around in a circle, and some structure you're carrying with you undergoes an automorphism, then that automorphism is the monodromy of your structure attached to the circle. Here is an unhelpful link: monodromy.com $\endgroup$ – S. Carnahan Mar 19 '10 at 5:55

The key word here is "Bass-Serre Theory" -- using the action on the hyperbolic plane, you can easily cook up a nice action of $PSL_2(\mathbb{Z})$ on a tree. This is all described nicely in Serre's book "Trees".

EDIT: Let me give a few more details. It turns out that a group $G$ splits as a free produce of two subgroups $G_1$ and $G_2$ if and only if $G$ acts on a tree $T$ (nicely, meaning that it doesn't flip any edges) with quotient a single edge $e$ (not a loop) such that the following holds. Let $e'$ be a lift of $e$ to $T$ and let $x$ and $y$ be the vertices of $e'$. Then the stabilizers of $x$ and $y$ are $G_1$ and $G_2$ and the stabilizer of $e'$ is trivial.

If you stare at the fundamental domain for the action of $PSL_2(\mathbb{Z})$ on the upper half plane, then you will see an appropriate tree staring back at you. There is a picture of this in Serre's book.

EDIT 2: This point of view also explains why finite-index subgroups $\Gamma$ of $PSL_2(\mathbb{Z})$ tend to be free. If you restrict the action on the tree $T$ to $\Gamma$, then unless $\Gamma$ contains some conjugate of the order 2 or order 3 elements stabilizing the vertices, then $\Gamma$ will act freely. This means that the quotient $T/\Gamma$ will have fundamental group $\Gamma$. Since $T/\Gamma$ is a graph, this implies that $\Gamma$ is free.

  • $\begingroup$ There's also Hatcher's book: math.cornell.edu/~hatcher/TN/TNch1.pdf $\endgroup$ – Ryan Budney Mar 18 '10 at 20:48
  • $\begingroup$ Another nice sources is Scott and Wall's paper "Topological methods in group theory"; however, I don't think that Scott-Wall or Hatcher gives the example of PSL_2(Z) (though I might be wrong -- I don't have either with me right now). $\endgroup$ – Andy Putman Mar 18 '10 at 20:59

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