Commutator subgroup of $\Gamma(2).$

Question

As "everybody" knows the commutator subgroup of $SL(2, Z)$ is $\Gamma(2).$ But what about its commutator subgroup? Is there any nice description of its elements? Is there any way to estimate the Hausdorff dimension of its limit set?

Answer 1

To rephrase Alex Bartel's answer geometrically: the quotient of the hyperbolic plane by $\Gamma(2)$ is a 3-punctured sphere, whose homology is free abelian of rank 2.

Added comment: To clarify the relationship of the groups: neither $SL(2,\mathbb Z)$ nor $\Gamma(2)$ act effectively on $\mathbb H^2$: the center of $SL(2,\mathbb Z)$, generated by $-I$, acts trivially. So, $\pi_1(\mathbb H^2/\Gamma(2))$ is indeed the commutator subgoup of $SL(2,\mathbb Z)$, but as Alex Bartel says, it has index 12 in $SL(2,\mathbb Z)$. $\Gamma(2)$ itself (the subgroup of $SL(2,\mathbb Z)$ congruent to the identity mod 2) is isomorphic to the product of this fundamental group with the order 2 group $\left < -I \right > $.

Any space has a universal abelian cover, whose group of deck transformations is its homology; the homology of the universal abelian cover is what you're asking for. The universal abelian cover is easier to see for the punctured torus, which has the same fundamental group --- it's just the plane minus holes for the punctures. The homology of this cover is coordinatized by linking number with the holes.

To picture the universal abelian cover of the 3-punctured sphere: enlarge the punctures to make boundary components, forming a pair of pants. Cut the pair of pants along 3 seams into two hexagons. Use a template consisting of the plane divided into equilateral triangles with an alternating coloring, say red and blue; color the hexagons red and blue. Above each red triangle, put a red hexagon above it in $\mathbb R^3$ so that three of the hexagon's edges project to the midpoints of edges of the red triangles, and the other three connect the top of one vertical edge to the bottom of the counterclockwise vertical edge. Arrange the blue hexagons in the same way, over the blue triangles. Half of the edges of hexagons string together to form lines that weave in and out, in a hexagonal weave: these cover the three boundary components of the pair of pants.

Answer 2

I think that the commutator subgroup of $SL(2,\mathbb{Z})$ is an index 2 subgroup of $\Gamma(2)$, rather than $\Gamma(2)$ itself. Indeed, $\Gamma(2)$ has index 6 in $SL(2,\mathbb{Z})$, but you can write down explicitly an onto homomorphism from $SL(2,\mathbb{Z})$ to $\mathbb{Z}/12\mathbb{Z}$, so you immediately know that $\Gamma(2)$ is too big. Keith Conrad has a lovely set of notes on this.

This commutator subgroup is isomorphic to the free non-abelian group of rank 2. It is known that the commutator subgroup of a free group of rank >1 is free of infinite rank. In particular, for the free group $SL(2,\mathbb{Z})'\cong F(a,b)=G$ of rank 2, $G'$ is free on the set $[a^n,b^m]$. So I doubt that you will get a better explicit description than this.

Answer 3

The commutator subgroup of $SL(2,\mathbb{Z})$ has index 12 and $\Gamma(2)$ has index 6, but $\Gamma(2)$ does not contain the commutator subgroup since the quotient group $SL(2,\mathbb{Z})/\Gamma(2)$ is nonabelian. Geometrically, the quotient of the hyperbolic plane by $\Gamma(2)$ is a three-punctured sphere and the quotient by the commutator subgroup is a punctured torus. In both cases a fundamental domain for the action consists of two adjacent ideal triangles in the Farey tesselation by ideal triangles. For $\Gamma(2)$, two pairs of adjacent edges of the fundamental domain are identified to get the three-punctured sphere, while for the commutator subgroup it is opposite pairs of edges of the fundamental domain that are identified. The symmetry group of the three-punctured sphere is $SL(2,\mathbb{Z})/\Gamma(2)$, the nonabelian group of order 6, while for the punctured torus the symmetry group is cyclic of order 6, the quotient of $SL(2,\mathbb{Z})$ by the commutator subgroup enlarged by $-I$. (The subgroup $\Gamma(2)$ contains $-I$ but the commutator subgroup does not.)