On congruence subgroups of $GL_2^+$

Question

I am reading the book Holomorphic Hilbert Modular Forms by Paul Garrett. The author considers congruence subgroups of general linear groups of positive determinant. More precisely, let $F$ be a totally real number field with ring of integers $\mathfrak o$. For a non-zero ideal $\mathfrak n$ of $\mathfrak o$, the principal congruence subgroup $\Gamma (\mathfrak n)$ of level $\mathfrak n$ is defined as $\Gamma (\mathfrak n):=\{\gamma\in GL_2^+(\mathfrak o): \gamma\equiv E\mod\mathfrak n\}$. Let $Z(\mathfrak o)$ be the center of $GL^+_2(\mathfrak o)$. Then any subgroup $\Gamma$ of $GL_2^+(\mathfrak o)$ such that $Z(\mathfrak o)\Gamma$ contains some $\Gamma (\mathfrak n)$ with finite index is called a congruence subgroup of $GL_2^+(\mathfrak o)$.

The author claims that $SL_2(\mathfrak o)$ is a congruence subgroup without proof. My question is: why is the cliam right?

Since $E+E_{ij}$ belongs to $GL_2^+(\mathfrak o)$, we see that $Z(\mathfrak o)$ consists of diagonal matrices. So the determinat of any member of $Z(\mathfrak o)SL_2(\mathfrak o)$ must be a square. In general, by Dirichlet's unit theorem, $SL_2(\mathfrak o)$ cannot have level one. If the claim were right, then what is the level of $SL_2(\mathfrak o)$ ?

I am a beginner of Hilbert modular forms and I am quite confused. Thanks for any help.

Answer 1

$\mathcal{O}_F^{\times}/\mathcal{O}_F^{\times 2}$ is finite, and furthermore we know that some $x \in \mathcal{O}_F^{\times}$ is a square iff it is a square mod all but finitely many prime ideals (this is a toy version of the “Hasse principle”, I guess?).

In particular, by choosing one prime for each representative of nontrivial classes in $\mathcal{O}_F^{\times}/\mathcal{O}_F^{\times 2}$, you can construct a nonzero ideal $\mathfrak{n} \subset \mathcal{O}_F$ such that units that reduce to squares mod $\mathfrak{n}$ are squares.

Then you can show that $SL_2(\mathcal{O}_F)Z(\mathcal{O}_F) \supset \Gamma(\mathfrak{n})$, and the inclusion is necessarily with finite index (since everything is contained in $GL_2^+(\mathcal{O}_F)$, of which $\Gamma(\mathfrak{n})$ is a finite-index normal subgroup).

Answer 2

The case $F = \mathbb{Q}(\sqrt{2})$ is a good one to fix ideas.

In this situation, the unit group $\mathfrak{o}^\times$ is isomorphic to $C_2 \times \mathbb{Z}$, generated by $-1$ and $u = \sqrt{2} - 1$. Note that $u$ is not totally positive; so the group $\mathfrak{o}^{\times+}$ of totally-positive units is infinite cyclic generated by $u^2$. In particular, every totally-positive unit is the square of a unit.

Since any element of $GL_2^+(\mathfrak{o})$ has determinant in $\mathfrak{o}^{\times+}$, we can multiply it by a matrix $\begin{pmatrix} u^n \\ & u^n \end{pmatrix} \in Z(\mathfrak{o})$ for some $n \in \mathbb{Z}$ in order to get something in $SL_2(\mathfrak{o})$. Hence we have $GL_2^+(\mathfrak{o}) = SL_2(\mathfrak{o}) Z(\mathfrak{o})$; and this would work for any real quadratic field in which the fundamental unit has norm -1.

On the other hand, in $\mathbb{Q}(\sqrt{3})$ the fundamental unit is $2 + \sqrt{3}$ which has norm $+1$; so squares of units have index 2 in $\mathfrak{o}^{\times +}$, and the index of $ SL_2(\mathfrak{o}) Z(\mathfrak{o})$ in $GL_2^+(\mathfrak{o})$ is 2.

As for your question "what is the level of $SL_2(\mathfrak{o})$?": the answer is that it's not uniquely defined. For subgroups of $SL_2(\mathbb{Z})$ we have the rather nice property that if $\Gamma$ has level $N$ and also level $M$, then in fact it has level $gcd(M, N)$, so there is a uniquely defined "smallest" level for $\Gamma$. But subgroups of $GL_2^+$ (or of $GL_1$) don't work like that (because the Strong Approximation Theorem doesn't hold). As an example, for $\mathbb{Q}(\sqrt{3})$, note that if $\mathfrak{P} = (2 \sqrt{3} + 1)$ and $\mathfrak{Q} = 4 + \sqrt{3}$ (prime ideals of norm 11 and 13 respectively), then any unit congruent to 1 modulo either $\mathfrak{P}$ or $\mathfrak{Q}$ is a square of a unit; so $Z(\mathfrak{o}) SL_2(\mathfrak{o})$ has level $\mathfrak{P}$ and level $\mathfrak{Q}$ (but it doesn't have level $gcd(\mathfrak{P}, \mathfrak{Q}) = 1$).