SL$_n(\mathbb{Z})$-normalizer of a principal congruence subgroup

1.   Introduction

Denote by SL$_n(\mathbb{Q})$ the set

and SL$_n(\mathbb{Z})$ the set

This set SL$_n(\mathbb{Z})$ is usually referred to as "modular group". It is well known that SL$_2(\mathbb{Z})$ is closely related to modular forms, see ^[3]. Many important subgroups of SL$_n(\mathbb{Z})$ have been widely studied, such as

Definition 1.1 Let $n, q\geq 2$ be positive integers. The principal congruence subgroup of level $q$ is defined as

Remark 1.2 A different but equivalent definition can refer to ^[4]. We always denote by $I_n$ the $n\times n$ identity matrix throughout the paper, and use notation $\textbf{E}_{ij}$ to represent the matrix with all entries are 0 but the $(i, j)$ entry is 1. For a matrix $A$, denoted by $A_{ij}$ its $(i, j)$ entry.

Naturally, $\Gamma_q(n)$ could be addressed in the larger group SL$_n(\mathbb{Q})$. In this paper, we concentrate our attention on the the following subset $\Lambda$ of SL$_n(\mathbb{Z})$.

Definition 1.3 Let $n, q\geq 2$ be positive integers. The ${\rm SL}_n(\mathbb{Z})$-normalizer of $\Gamma_q(n)$ is defined as

and matrix in $\Lambda$ is called an ${\rm SL}_n(\mathbb{Z})$-normalization element of $\Gamma_q(n)$.

This notion, which ties in nicely with "The congruence subgroup problem" (see ^[7], or ^[4]), is inspired by observing subgroup topologies in SL$_n(\mathbb{Q)}$ (see ^[4]). As well known the normalizer of a congruence subgroup has acquired significance because it is related to some simple group in ^[2]. It has also played an important role in work on Weierstrass points on the Riemann surfaces in ^[5].

In order to figure out $\Lambda$, it is a natural approach to transform a matrix in SL$_n(\mathbb{Q})$ into the set of integral matrices by multiplying a suitable positive integer. The following concept plays a key role in our discussion on the structure of $\Lambda$.

Definition 1.4 Let $n, r$ be positive integers, and $B\in {\rm SL}_n(\mathbb{Q})$. We call $r$ the least common denominator ($\text{LCD}$ for convention) of $B$, denoted by $c_B$, if $r$ is the least positive integer such that $rB$ is an integral matrix.

With above preparation, the remainder of this paper is organized as follows. In Section 2, the fundamental relationship between SL$_n(\mathbb{Q})$ and $\Lambda$ is investigated. It is obvious that $\Lambda$ contains SL$_n(\mathbb{Z})$, and we will show that this inclusion is proper when $q$ is not square-free. With the aid of Smith normal form of an integral matrix, we first establish a necessary and sufficient condition for determining whether a matrix in SL$_n(\mathbb{Q})$ belongs to $\Lambda$. This statement does not give an explicit relationship between $c_B$ and $q$ in the case $n > 2$ yet. Fortunately one necessary condition is obtained, which enables us to bound the multiplicity of arbitrary prime factor of the LCD of a matrix in $\Lambda$ by factoring $q$. Other than that, we also show that $\Lambda$ is not a group, and not even a semigroup. In spite of this, $\Lambda$ admits many subgroups distinguished from SL$_n(\mathbb{Q})$. In section 3, two families of subgroups that are maximal in $\Lambda$ are presented in the case $n = 2$.

2.   LCD and the basic structure of $\Lambda$

In this section, we will show that it can be determined whether a matrix in SL$_n(\mathbb{Q})$ belongs to $\Lambda$ when its LCD is specified. In order to go on our following discussion, we need a widely understood theorem (see ^[1,6,8,9]), which arises from basic module theory over a principal ideal domain. The Smith normal form of integral matrices, which will be presented in the following paragraph, is our main tool in this section.

Let $n$ be a positive integer, and $A$ an integral matrix of order $n$. Then there exists unimodular matrices $P, Q$ such that $PAQ$ is a diagonal matrix

where $d_i$, unique up to the sign, are nonzero integers for all $1\leq i\leq l$, and $d_i\mid d_{i+1}$. Moreover, $d_i$ is called the $i$-th invariant factor of $A$, and (2.1) is the Smith normal form of $A$.

With the aid of this result, we give a characterization for the LCD of matrices in $\Lambda$ as following theorem states.

Theorem 2.1 Let $n, q\geq 2$ be positive integers, and $B\in {\rm SL}_n(\mathbb{Q})$. Then $B$ is an ${\rm SL}_n(\mathbb{Z})$-normalization element of $\Gamma_q(n)$ if and only if the $n$-th invariant factor $d_n$ of $c_{B}B$ divides $q$.

Proof. Suppose that $B$ is an SL$_n(\mathbb{Z})$-normalization element of $\Gamma_q(n)$ and $A = c_{B}B$. Let $P, Q$ be unimodular matrices such that $PAQ$ is the Smith normal form in (2.1). It follows from the fact $B$ is invertible that the number $l$ in (2.1) is equal to the order $n$ of $B$. It is obvious that $d_i\mid d_{i+1}$, and a positive integer $r$ is the LCD of $A$ if and only if the greatest common divisor of all entries of $rA$ is 1. Thus $d_1 = \pm 1$. Set $C = PBQ$, then $C$ is also an SL$_n(\mathbb{Z})$-normalization element of $\Gamma_q(n)$. Hence, for any $X = (x_{ij})_{n\times n} \in\Gamma_q(n)$, we have

Take $X = I_n+q\textbf{E}_{1n}$. It follows from above statements that $(1, n)$ entry of $CXC^{-1}$, which is equal to $\pm qd_n^{-1}$, is an integer, and thus this means that $d_n\mid q$. This completes the proof of necessity.

Conversely, if $d_n\mid q$, then $d_i\mid q$ for all $1\leq i\leq n$. We can assert from this fact that all entries in $CXC^{-1}$ are integers, and thus the sufficiency follows.

It is worth pointing out that in the case $n=2$, we have $d_1d_2=\pm d_2=c_B^{2}$, and thus the fact $d_2$ of $c_{B}B$ divides $q$ is equivalent to that $c^2_B$ divides $q$.

Unfortunately, above theorem does not give an explicit relationship between $c_B$ and $q$ in the case $n > 2$. However, the following proposition can help us to understand the problem stated to some extent.

Theorem 2.2 Let $n, q\geq 2$ be positive integers, $p_1^{e_1}\cdots p_k^{e_k}$ the prime decomposition of $q$, and $B = (b_{ij})_{n \times n}\in {\rm SL}_n(\mathbb{Q})$. If $B$ is an ${\rm SL}_n(\mathbb{Z})$-normalization element of $\Gamma_q(n)$, then $c_B = p_1^{f_1}\cdots p_k^{f_k}$, where $0\leq f_i\leq e_i-1$ for all $1\leq i\leq k$. In particular, if $q$ is square-free, then $\Lambda = {\rm SL}_n(\mathbb{Q})$.

Proof. Let $A = c_{B}B$, and $PAQ$ be the Smith normal form of $A$. We claim here that the following conclusion is true. Claim: Suppose that $p$ is a prime number and $f$ a positive integer. Then $p^f\mid c_B$ implies $p^{f+1}\mid d_n$.

If above assertion was established, then the result follows by Theorem 2.1.

Assume the claim is false, then $d_n$ is divisible by at most $f$-th power of $p$, so does $d_i$, $i = 1, \cdots, n$. Note here that $d_1 = \pm 1$, hence $d_1\cdots d_n$ is divisible by at most $(n-1)f$-th power of $p$. On the other hand, the determinants of $PBQ$ and $B$ are both equal to 1. Therefore, $d_1\cdots d_n = c_B^{n}$, and this means that $p^{nf}\mid d_1\cdots d_n = c_B^{n}$. So, above statements are contrary to the fact $(n-1)f < nf$, and thus the assumption is not correct.

Corollary 2.3 Let $q\geq 2$ be a square-free positive integer, then the normalizer of $\Gamma_q(n)$ in ${\rm SL}_n(\mathbb{Q})$ is ${\rm SL}_n(\mathbb{Z})$.

The previous conclusions present some description for the elements of $\Lambda$. It has to be pointed out that as the subset of the group SL$_n(\mathbb{Q})$, $\Lambda$ does not possess a fine structure. In fact, we can show that $\Lambda$ is not a semigroup of the group SL$_n(\mathbb{Q})$.

Proposition 2.4 Let $n\geq 2$ be a positive integer, $q = m^{2s}r$, where $m > 1, r\geq 1, s\geq 1$ are integers with $m^2\nmid r$. Then ${\rm SL}_n(\mathbb{Z})$ is properly contained in $\Lambda$, and $\Lambda$ is also a proper subset of ${\rm SL}_n(\mathbb{Q})$, but not a subsemigroup. In addition, any subgroup of ${\rm SL}_n(\mathbb{Q})$ which contains ${\rm SL}_n(\mathbb{Z})$ must have an element, which does not belong to $\Lambda$.

Proof. It is clear that $B = {\rm diag}(\frac{1}{m}, 1, \cdots, 1, m)\notin {\rm SL}_n(\mathbb{Z})$. Moreover, $m^2$ is the $n$-th invariant factor of $c_{B}B$ and divides $q$, hence $B\in \Lambda$, and thus we can get the first proper inclusion. On the other hand, we know that LCD of matrices in ${\rm SL}_n(\mathbb{Q})$ is unbounded. However, the fact $B\in \Lambda$ implies $c_{B} < q$ by Theorem 2.2, then the second inclusion follows.

In what follows, we show that $\Lambda$ is not a subsemigroup. As a matter of fact, it suffices to verify that there exist two matrices whose product does not belong to $\Lambda$. Let $K = {\rm diag}(\frac{1}{m^s}, 1, \cdots, 1, m^s)$. Analogously, $m^{2s}$ is the $n$-th invariant factor of $c_{K}K$ and divides $q$, hence $K\in\Lambda$. With $B$ defined as previous paragraph, set $J = BK$. Then the $n$-th invariant factor of $c_{J}J$ is $m^{2s+2}$ which does not divide $q$, the conclusion follows.

The last assertion can be restated as: Let $D\in\Lambda\setminus{\rm SL}_n(\mathbb{Z})$, then the subgroup $H$ generated by ${\rm SL}_n(\mathbb{Z})\bigcup\{D\}$ must have an element which does not belong to $\Lambda$.

Let $A = c_DD$, and $PAQ$ be the Smith normal form of $A$. We note that $PDQ\in H$, and thus $s$-th power of $PDQ$

is also in the group $H$. It is easily deduced from $d_1 = \pm 1$ that $c_{(PDQ)^s} = c_D^s$. If $(PDQ)^s\in\Lambda$, then $c_D^s$ divides $q$ by Theorem 2.2. However $c_D > 1$ for $D\notin{\rm SL}_n(\mathbb{Z})$, and $s$ can be chose arbitrarily, it is absurd.

3.   Maximal subgroups in $\Lambda$

In view of sparsity of set $\Lambda$ as Proposition 2.4 pointed out, we are led to focus our attention on subgroups of ${\rm SL}_n(\mathbb{Q})$ those are maximal in $\Lambda$, which is strictly defined as follows.

Definition 3.1 With $n, q$ defined as Proposition 2.4, we call a subgroup $M$ of ${\rm SL}_n(\mathbb{Q})$ contained in $\Lambda$ a maximal ${\rm SL}_n(\mathbb{Z})$-normalizer or simply maximal normalizer, if the following conditions hold

1. $M$ is a subgroup of ${\rm SL}_n(\mathbb{Q})$;

2. If $H$ is a subgroup of ${\rm SL}_n(\mathbb{Q})$ and properly contains $M$, then there is a matrix $A\in H\setminus\Lambda$.

It follows immediately from Proposition 2.4 that ${\rm SL}_n(\mathbb{Z})$ satisfies above two conditions, and we call it the trivial maximal normalizer which is not our focus. In fact, we are interested in the following problem.

Problem 3.2 Find all nontrivial maximal normalizers.

To our knowledge, the problem we mentioned above is difficult to solve completely. In the rest of this section, two families of nontrivial maximal normalizers will be presented in the case $n = 2$. In what follows, we focus attention on the case $n = 2$. In other words, we will address related problems in ${\rm SL}_2(\mathbb{Q})$.

Proposition 3.3 Let $q = \mu^2\nu > 1$ be a positive integer, where $\nu$ is a square-free integer. If $\tau_1$ and $\tau_2$ are two coprime positive integers with product $\tau_1\tau_2$ dividing $\mu$, then

is a maximal normalizer. And if either $\tau_1$ or $\tau_2$ is greater than $1$, then $H(\tau_1, \tau_2)$ is nontrivial.

Proof. By Theorem 2.1, we know that the ${\rm SL}_2(\mathbb{Z})$-normalizer of $\Gamma_q(2)$ is $\Lambda = \{B\in {\rm SL}_2(\mathbb{Q}): \ \ c_B \mid\mu\}$.

It is easy to verify that $H(\tau_1, \tau_2)$ is a subset of $\Lambda$, and also a subgroup of ${\rm SL}_2(\mathbb{Q})$. Therefore, we only need to show $H(\tau_1, \tau_2)$ is maximal in $\Lambda$. Namely, we have to prove that for any $K\in\Lambda\setminus H(\tau_1, \tau_2)$, the subgroup $L$ generated by $H(\tau_1, \tau_2)\bigcup\{K\}$ must have an element, which does not belong to $\Lambda$.

To this end, let $A = c_KK$. First, we show there exists $J\in H(\tau_1, \tau_2)$ such that $JA$ is a upper triangular matrix. If $A_{21} = 0$, it is trivial. And when $A_{11} = 0$, we need only take $J$ as $\frac{\tau_1}{\tau_2}\textbf{E}_{12}-\frac{\tau_1}{\tau_2}\textbf{E}_{21}$. As a consequence, we can assume that neither of $A_{11}, A_{21}$ is equal to 0. Let $\alpha = \gcd(A_{11}\tau_2, A_{21}\tau_1)$, and $c' = A_{21}\frac{\tau_1}{\alpha}$, $d' = -A_{11}\frac{\tau_2}{\alpha}$. As well known that there exists two integers $a', b'$ such that $a'd'-b'c' = 1$, so we can take

Set $D = JK$, then $D_{11}D_{22} = 1$ as the determinant of $JK$ is $1$. We show that the subgroup generated by $D$ is not contained in $\Lambda$. The scenarios will be considered for the following two cases:

Case 1. $D_{11}, D_{22}\notin\{\pm 1\}$. Let $D_{11} = \frac{\omega_1}{\theta_1}$ be the reduced fraction with $\theta_1 > 1$. Since $D\in L$ implies $D^m\in L$, we deduce $c_{D^m}\mid\mu$, i.e., $\theta_1^m\mid\mu$ for any positive integer $m$, but this is in contradiction with $\theta_1 > 1$.

Case 2. $D_{11}, D_{22}\in\{\pm 1\}$. We can assume $D_{11} = D_{22} = 1$ because of $-I_2\in H(\tau_1, \tau_2)$. Let $D_{12} = \frac{\beta}{\gamma}$ be the reduced fraction. Note that $D\notin H(\tau_1, \tau_2)$, then the ratio $\frac{\beta\tau_2}{\gamma\tau_1}$ of $\frac{\beta}{\gamma}$ and $\frac{\tau_1}{\tau_2}$ is not an integer. For any positive integer $m$, let

By this equality, we can obtain the following recurrence relation

Hence $a_m$ is a polynomial in $\frac{\beta\tau_2}{\gamma\tau_1}$ with integral coefficients. We note here that $\frac{\beta\tau_2}{\gamma\tau_1}$ is not an integer, so we can take the reduced fraction $\frac{\beta\tau_2}{\gamma\tau_1} = \frac{\omega}{\theta}$ with $\theta > 1$. Suppose now

where $s_i$ is an integer for $1\leq i\leq m$, and $m$ is any positive integer. The right hand side of (3.2) is obviously reduced also, and this implies $\theta^m\mid\mu$. Similar to case 1, it is absurd.

Remark 3.4 It is worth pointing out that the group $H(\tau_1, \tau_2)$ in above proposition is a conjugate subgroup of ${\rm SL}_2(\mathbb{Z})$ in ${\rm SL}_2(\mathbb{Q})$. It is natural to raise a problem here: which conjugate subgroups of ${\rm SL}_2(\mathbb{Z})$ in ${\rm SL}_2(\mathbb{Q})$ are maximal normalizers.

It is not hard to see from above discussion that if the subgroup $\left < A\right >$, generated by $A\in \Lambda$, is contained in $\Lambda$, then there must exist a maximal normalizer in $\Lambda$ which contains $A$. In order to go on our story, the following definition is needed.

Definition 3.5 Let $A\in {\rm SL}_n(\mathbb{Q})$. $A$ is called $\sigma-$stable if there exists a positive integer $\sigma$ such that $c_{A^m}\leq\sigma$ for any integer $m$. Otherwise, $A$ is called unbounded.

When a matrix $A\in {\rm SL}_n(\mathbb{Q})$ is $\sigma-$stable, we call $A$ stable instead of $\sigma-$stable for convenience if we do not need to know $\sigma$ exactly.

Example 3.6 Let $q = p^2$, where $p$ is a prime number. Then the matrix

belongs to the ${\rm SL}_n(\mathbb{Z})$-normalizer of $\Gamma_q(n)$, and is $p^{n-1}-$stable. It is not hard to check that $B$ is not $p-$stable when $n > 2$, and thus $\Lambda$ does not contain the cyclic group generated by $B$, i.e., no maximal normalizer contains $B$.

Having taken a short tour to the general case, we now concentrate our attention on the case $n = 2$ again. The following result gives some description for stable matrix in ${\rm SL}_2(\mathbb{Q})$.

Lemma 3.7 Let $A\in {\rm SL}_2(\mathbb{Q})$. Then $A$ is stable if and only if the trace $tr(A)$ is an integer. In particular, under assumptions of Proposition 3.3, there exists a maximal normalizer which contains $A$ whenever $tr(A)$ is an integer and $c_A\mid\mu$.

Proof. Set

It is easy to show the following recurrence relation holds

where $f_0 = 1, f_1 = tr(A)$ and $f_{m+1} = tr(A)f_m-f_{m-1}$ for any positive integer $m$. Hence $f_m$ is a polynomial in $tr(A)$ with integral coefficients, and $tr(A)$ is an integer implies in turn $f_m$ is also an integer. Then we can deduce $c_{A^m}\mid c_A$ for $m\geq1$ by (3.3). When $m\leq-1$, a recurrence relation analogous to (3.3) holds clearly, and thus the sufficiency follows.

Conversely, suppose $tr(A)$ is not an integer, we show that $c_{A^{m+1}}$ is unbounded for $m\geq 1$. Let $tr(A) = \frac{\omega}{\theta}$ be the reduced fraction with $\theta > 1$, we divide proof of the necessity into the following two cases as Proposition 3.3:

Case 1. $b = 0$ and $c = 0$. Proof is the same as Case 1 in Proposition 3.3.

Case 2. $b\neq 0$ or $c\neq 0$. It is sufficient to deal with $b\neq 0$. Let $b = \frac{\alpha}{\beta}$ be the reduced fraction. Set

and $m\geq f+1$. Let $\frac{\alpha_1}{\theta_m}$ be the reduced fraction of $\frac{\alpha}{\theta^m}$. Then $\alpha_1$ and $\theta$ are coprime numbers, and $p^{m-f}\mid\theta_m$ as long as $p$ is a prime divisor of $\theta$. Hence $b_{m+1}$ can be expressed as

where $s_i$ is an integer for $0\leq i\leq m-1$. It is easy to check that $\theta_m$ and the integer in bracket of (3.4) are coprime, and the proof is the analogy of Case 2 in Proposition 3.3. Then the denominator in the reduced fraction of $b_{m+1}$ is divisible by $\theta_m$, i.e., $\theta_m\mid c_{A^{m+1}}$. This completes the proof of necessity.

With the help of above lemma, we can give another maximal normalizer in $\Lambda$ except Proposition 3.3 presented.

Proposition 3.8 With $q, \mu$ defined as Proposition 3.3. Let

be a matrix in ${\rm SL}_2(\mathbb{Z})$-normalizer $\Lambda$ of $\Gamma_q(2)$, where $a, b, c, d$ are integers, $d, \mu$ are coprime and $a+d\equiv 0({\rm mod} \ \mu)$. Let $\lambda = d_{-1}b$ where $1\leq d_{-1}\leq\mu-1$ and $dd_{-1}\equiv 1({\rm mod} \ \mu)$. Then the unique maximal normalizer $H$ which contains $A$ is composed of

where $l, k, m, s$ are integers satisfy that $l(m\mu-\lambda s)-ks = \mu$ and $la+kc\equiv lb+kd\equiv 0 ({\rm mod} \ \ \mu)$.

Proof. It is not hard to verify that $c_A = \mu$ and $A\in H\subseteq\Lambda$. The product of any two matrices in $H$ can be expressed as

where $t_i = m_i\mu-\lambda s_i$ for $i = 1, 2$, and

Note here that $d(\lambda l_i+k_i) = (dd_{-1})l_ib+k_id\equiv 0({\rm mod}\ \mu)$ for $i = 1, 2$, so $\lambda l_i+k_i\equiv 0({\rm mod}\ \mu)$. Let

It is easy to see $l(m\mu-\lambda s)-ks = \mu$ for $\rm{det}P = 1$. Since

we deduce $la+kc\equiv 0({\rm mod}\ \mu)$ from the congruence $l_2a+k_2c\equiv 0({\rm mod}\ \mu)$ and $a-\lambda c = a(1-dd_1)+d_1\mu^2\equiv 0({\rm mod}\ \mu)$. Similarly, we can get $lb+kd\equiv 0({\rm mod}\ \mu)$, and this implies in turn $P\in H$. On the other hand, the inverse of arbitrary matrix in $H$ is

where

and

and thus $H$ is a subgroup of ${\rm SL}_2(\mathbb{Q})$.

In what follows, we show that $H$ is the maximal and the unique normalizer. In fact, we will prove that any $B\in\Lambda$ contained in the same maximal normalizer with $A$ must be in $H$.

Set

and compute $AB$ and $BA$ respectively. It follows from the fact $c_{AB}$ and $c_{BA}$ divide $\mu$ that $au+bs\equiv av+bt\equiv bs+dt\equiv 0({\rm mod}\ \mu)$. As $d$ and $\mu$ are coprime, we can find integers $l, k, m$ such that $u = l\mu+\lambda s$, $v = k\mu+\lambda t, t = m\mu-\lambda s$. On the other hand, we get by Lemma 3.7 that

Substitute $u, v, t$ in (3.5), we have

By Lemma 3.7 again, we obtain that $a+d\equiv 0({\rm mod}\ \mu)$, so $1+ad_1\equiv(a+d)d_1\equiv 0({\rm mod}\ \mu)$, and thus $la+kc\equiv 0({\rm mod}\ \mu)$.

Finally, let $D = \mu^3 ABA$, then

The fact $ABA\in\Lambda$ implies $D_{12}\equiv 0({\rm mod}\ \mu^2)$, and then $a(lb+kd)\equiv 0({\rm mod}\ \mu)$, that is $lb+kd\equiv 0({\rm mod}\ \mu)$. Our proof of necessity is completed.

Remark 3.9 For coprime numbers $d, \mu$, the maximal normalizer $H$ in Proposition 3.8 is clearly different from $H(\tau_1, \tau_2)$ in Proposition 3.3.

4.   Conclusions

We give some remarks about our results obtained above to conclude our paper. In this paper, we study a subset $\Lambda$ of ${\rm SL}_n(\mathbb{Q})$, where a matrix $B$ belongs to $\Lambda$ if and only if the conjugate subgroup $B\Gamma_q(n)B^{-1}$ of principal congruence subgroup $\Gamma_q(n)$ is contained in modular group ${\rm SL}_2(\mathbb{Z})$. We demonstrated that this subset $\Lambda$ is fairly loose, and even not a semigroup. However, we presented in previous two sections two families of nontrivial maximal normalizers, i.e., two families of subgroups that are maximal in $\Lambda$. It is worth being pointed out that the maximal normalizers we provided are only one part of solutions to Problem 3.2 when $n = 2$. In the light of Proposition 3.3 and 3.8, we need to consider matrices whose diagonals are not invertible or $0$ modulo $\mu$. As for the case $n > 2$, Example 3.6 indicates that it is not enough to consider $tr(A)$ only. However, it seems difficult to our knowledge to get a simple formula which is analogous to (3.3).

Acknowledgments

We thank the referees for careful reading of the manuscript and the suggestions on the writing of the paper. This work was supported by National Nature Science Foundation of China(No. 11971382).

Conflict of interest

All authors declare no conflicts of interest in this paper.