# Abelianisation of certain congruence subgroups in GL_2(Z)

$$\DeclareMathOperator\SL{SL}\newcommand{\ab}{\mathrm{ab}}$$Denote by $$\Gamma(m) = \left\{ \left( \begin{array}{cc} 1 +ma_{11} & m a_{12} \\ ma_{21} & 1 +ma_{22} \end{array} \right) \in \SL_2(\mathbb{Z}) \mid a_{ij} \in \mathbb{Z} \right\}$$ the principal congruence subgroup of level $$m$$ of $$\SL_2(\mathbb{Z})$$. Also consider $$h = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right)$$ and $$g = \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$$. I was wondering about the following:

(1) Is the abelianisation of $$\langle \Gamma(m), h,g \rangle$$ finite?

(2) Suppose $$m= 2^{n}$$ is a $$2$$-power. Is then $$\langle \Gamma(m), h,g \rangle^{\ab}$$ finite and even an elementary abelian $$2$$-group? If yes, what about non $$2$$-powers?

In fact I am wondering about the above questions for the subgroup $$V_m =\left\{ \left( \begin{array}{cc} 1 +ma_{11} & m a_{12} \\ ma_{21} & 1 +ma_{22} \end{array} \right) \in \SL_2(\mathbb{Z}) \mid a_{11} \equiv a_{22}, a_{12}\equiv a_{21} \mod 2 \in \mathbb{Z} \right\}$$ of $$\Gamma(m)$$. Question (1) for $$\langle \Gamma(m), h,g \rangle$$ or $$\langle V_m, h,g \rangle$$ is however equivalent, but (2) a priori not.

I know that $$\Gamma(m)^{\ab}$$ is a free abelian group (because when $$m \geq 3$$ the group is free). However the abelianisation of $$\mathrm{GL}_2(\mathbb{Z}) \cong D_{12} \star_{C_2\times C_2} D_8$$ is finite. For $$m=2,4$$ one can show that (2) holds (via an explicit set of generators).

Thanks for any input and help!

• The set of matrices you define are not in $\mathrm{SL}_2$, you need to add "$\cap\mathrm{SL}_2(\mathbb{Z})$".
– YCor
Commented Feb 21, 2023 at 18:32
• You are right, I forgot to add that. I have edited it now. Commented Feb 21, 2023 at 19:07

$$\newcommand{\H}{\mathcal{H}}\DeclareMathOperator{\GL}{GL}\DeclareMathOperator{\SL}{SL}\DeclareMathOperator{\Hom}{Hom}\newcommand{\Z}{\mathbb{Z}}\newcommand{\Q}{\mathbb{Q}}$$No, the rank of the abelianisation goes to infinity with $$m$$. This can probably be obtained in many ways : Riemann-Hurwitz, trace formula, Shapiro's lemma and amalgamated product decompositions, etc.
Let me sketch one proof. Assume $$m\ge 3$$ and let $$\H$$ be the upper half-space with its usual action of $$\GL_2(\Z)$$. Let $$\Gamma'(m) = \langle \Gamma(m),g,h\rangle$$ and $$\Gamma''(m) = \Gamma'(m) \cap \SL_2(\Z) = \langle \Gamma(m),S\rangle$$ where $$S = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$$. The rank of the abelianisation of $$\Gamma'(m)$$ is the first Betti number of $$\H/\Gamma'(m)$$, which is half that of its orientation covering $$\H/\Gamma''(m)$$. We can compute the last Betti number as $$\dim H^1(\Gamma''(m),\Q) = \dim H^1(\Gamma(m),\Q)^{\langle S\rangle} = \dim H^1(\Gamma(m),\Q)^{\langle -1, S\rangle}$$ $$= \dim \Hom_{G_m}(\sigma,H^1(\Gamma(m),\Q))$$, where $$G_m = \SL_2(\Z/m\Z)$$ and $$\sigma$$ is the permutation representation $$\Q[G_m/\langle -1,S\rangle]$$ of $$G_m$$. You can get a formula for the last dimension by using for instance Proposition 4.18 (with $$N=m$$, $$k=2$$) in The cohomology of lattices in $$\SL_2(\mathbb{C})$$ by Finis, Grunewald and Tirao, and the order of magnitude is a constant times $$\# G_m$$, which tends to infinity with $$m$$.
• The action of $\mathrm{GL}_2(\mathbb{R})$ is the usual one by linear fractional transformations as for $\mathrm{SL}_2(\mathbb{R})$, except that if the determinant is negative you postcompose with the complex conjugation. It should be easy, yes: the formula only involves trace and dimension of fixed points on a permutation representation. $\dim \mathbb{C}[G/H]^U = \# U\backslash G / H$ and $\mathrm{trace}(g \mid \mathbb{C}[G/H]) =$ number of elements $x\in G/H$ fixed by $g$ (equivalently $g\in xHx^{-1}$). Commented Feb 23, 2023 at 21:44