The cusp amplitudes and quasi-level of a congruence subgroup of SL2 over any Dedekind domain

TL;DR

This paper generalizes properties of classical modular groups to groups over Dedekind domains, focusing on cusp amplitudes, levels, and introducing quasi-levels to better understand congruence subgroups.

Contribution

It extends algebraic properties of SL_2(Z) to SL_2(D) over Dedekind domains, introduces quasi-amplitude and quasi-level concepts, and provides new conditions for congruence subgroups.

Findings

Extended properties of cusp amplitudes and levels to Dedekind domains

Introduced quasi-amplitude and quasi-level notions

Derived new necessary conditions for congruence subgroups

Abstract

We extend some algebraic properties of the classical modular group SL_2(Z) to equivalent groups in the theory of Drinfeld modules, in particular properties which are important in the theory of modular curves. We study cusp amplitudes and the level of a (congruence) subgroup of SL_2(D) for any Dedekind domain D, as ideals of D. In particular, we extend a remarkable result of Larcher. We introduce finer notions of quasi-amplitude and quasi-level, which are not required to be ideals and encode more information about the subgroup. Our results also provide several new necessary conditions for a subgroup of SL_2(D) to be a congruence subgroup.

Equations

$${\mathfrak{c}}_{\min }=\bigcap _{\mathfrak{q}\in \mathcal{A}(H)}\mathfrak{q}\text{mbox}and{\mathfrak{c}}_{\max }=\sum _{\mathfrak{q}\in \mathcal{A}(H)}\mathfrak{q}.$$

$$ql(H)=l(H).$$

$$|D/l(H)|dividesn!,$$

$$|D/l(H)|dividesn.$$

$$T(\alpha,r):=\left[\begin{array}[]{cc}\alpha&r\\ 0&\alpha^{-1}\end{array}\right]\;\mathrm{and}\;T(r):=T(1,r)=\left[\begin{array}[]{cc}1&r\\ 0&1\end{array}\right].$$

$$S(r):=\left[\begin{array}[]{cc}1&0\\ r&1\end{array}\right]\;\mathrm{and}\;R(r):=\left[\begin{array}[]{cc}1+r&r\\ -r&1-r\end{array}\right].$$

$$S{L}_2(R,\mathfrak{q})=\left\{X\in S{L}_2(R):X\equiv I_2(\mathrm{mod}\mathfrak{q})\right\}.$$

$$E_2(R,\mathfrak{q})\le N{E}_2(R,\mathfrak{q})⊴S{L}_2(R,\mathfrak{q}).$$

$$B_2(R)=\{T(\alpha ,r):\alpha \in R^{\ast },r\in R\}.$$

$$N_H=\bigcap _{g\in G}{H}^g.$$

$$S{L}_2(R,{\mathfrak{q}}^{\prime })\le C$$

$$X=\left[\begin{array}[]{cc}1+x&y\\ z&1+t\end{array}\right],$$

$$\theta :S{L}_2(R,\mathfrak{q})⟶{\mathfrak{r}}^3$$

$$\theta (X)=(\overline{x},\overline{y},\overline{z}).$$

$$S{L}_2(R,\mathfrak{q})/S{L}_2(R,{\mathfrak{q}}^{\prime })\cong ({\mathfrak{r}}^{+}{)}^3,$$

$$E_2(R,\mathfrak{q})\cdot S{L}_2(R,{\mathfrak{q}}^{\prime })=S{L}_2(R,\mathfrak{q}+{\mathfrak{q}}^{\prime }).$$

$$E_2(D,\mathfrak{q})\cdot S{L}_2(D,{\mathfrak{q}}^{\prime })=S{L}_2(D,\mathfrak{q}+{\mathfrak{q}}^{\prime }).$$

$$S{L}_2(D,\mathfrak{q})\cdot S{L}_2(D,{\mathfrak{q}}^{\prime })=S{L}_2(D,\mathfrak{q}+{\mathfrak{q}}^{\prime }).$$

$$H\ge S{L}_2(D,\mathfrak{q}).$$

$$S{L}_2(R)/S{L}_2(R,yR)\cong S{L}_2({\mathbb{F}}_q[x]).$$

$$G=S{L}_2(D)$$

$$G(\mathfrak{q})=S{L}_2(D,\mathfrak{q})$$

$$ql(N):=\{r\in D:T(r)\in N\}.$$

$$[D:ql(N)]\le [S{L}_2(D):N]<\infty ,$$

$$S{L}_2({\mathcal{O}}_S,\mathfrak{q})=N{E}_2({\mathcal{O}}_S,\mathfrak{q})$$

$$S{L}_2({\mathcal{O}}_{S^{\prime }},{\mathfrak{q}}^{\prime })=N{E}_2({\mathcal{O}}_{S^{\prime }},{\mathfrak{q}}^{\prime })$$

$$N{E}_2(D)=N{E}_2(D,\mathfrak{q}+{\mathfrak{q}}^{\prime })=N{E}_2(D,\mathfrak{q})\cdot N{E}_2(D,{\mathfrak{q}}^{\prime })$$

$$=S{L}_2(D,\mathfrak{q})\cdot N{E}_2(D,{\mathfrak{q}}^{\prime })=S{L}_2(D,\mathfrak{q}+{\mathfrak{q}}^{\prime })=S{L}_2(D),$$

$$PGL_{2}(D)\cong\mathop{\mbox{\LARGE$*$}}_{\ell\in{\mathbb{P}}^{1}(k)}\varDelta(\ell)$$

$$k=\bigcup _{n\in \mathbb{N}}{\mathbb{F}}_{2^{5^n}}\mathrm{and}D=k[x,y]\mathrm{with}{y}^2+y=x^3.$$

Full text

We study cusp amplitudes and the level of a (congruence) subgroup of $SL_{2}(D)$ for any Dedekind domain $D$, as ideals of $D$. In particular, we extend a remarkable result of Larcher.

We introduce finer notions of quasi-amplitude and quasi-level which are not required to be ideals and encode more information about the subgroup.

Our results also provide several new necessary conditions for a subgroup of $SL_{2}(D)$ to be a congruence subgroup.

Keywords: Dedekind domain, congruence subgroup, cusp amplitude, level, quasi-level, index.

Mathematics Subject Classification (2010): 11F06, 20G30, 20H05

Introduction

This paper is part of an ongoing project which aims to extend algebraic results from the classical modular group, $SL_{2}({\mathbb{Z}})$, to equivalent groups occurring in the context of Drinfeld modular curves [Ge]. We are especially concerned with those results which have proved applicable to the classical theory of modular forms. Our hope is that our results will prove useful to experts in the theory of Drinfeld modules. Here we are primarily concerned with the cusp amplitudes and level of a subgroup of such a group. One of our principal aims is to demonstrate that these concepts, as well as much of the classical theory of congruence subgroups, can be extended to linear groups defined over any Dedekind domain $D$.

Congruence subgroups, beginning in the $19$th century with $SL_{2}({\mathbb{Z}})$, are usually defined for a matrix group with entries in an arithmetic Dedekind domain $A$ (or, more generally, an order in such a domain). The definition involves a non-zero $A$-ideal. Since every proper quotient of $A$ is finite, every congruence subgroup is necessarily of finite index. This leads naturally to the converse question (for the particular matrix group), namely the celebrated Congruence Subgroup Problem, which has attracted considerable attention for many years. For $D$ which are not arithmetic, i.e. $D$ which might have proper infinite quotients, there is no longer a close connection between the congruence subgroups of $SL_{2}(D)$ and its finite index subgroups. For example, if $k$ is an infinite field, then $SL_{2}(k[t])$ has infinitely many congruence subgroups and no proper finite index subgroups.

The groups $SL_{2}(D)$ include a number of very important special cases, for example, the classical modular group $SL_{2}({\mathbb{Z}})$, the Bianchi groups, where $D$ is the ring of integers in an imaginary quadratic number field, and groups occurring in the context of Drinfeld modular curves [Ge]. Other examples are $SL_{2}$ over a ring of $S$-integers of a number field or of a function field in one variable over any constant field, $SL_{2}$ over any local ring and $SL_{2}$ over any principal ideal domain.

When generalizing concepts from the classical modular group $SL_{2}({\mathbb{Z}})$ to $SL_{2}(R)$ for more general rings $R$, one sometimes has to make a choice between loss of structure and loss of information. For example, let $H$ be a subgroup of $SL_{2}(R)$. Then the set of all $a\in R$ such the translation matrix ${1\ a\choose 0\ 1}$ is contained in $H$ is an additive subgroup of $R$. Hence in the special case $R={\mathbb{Z}}$ this is automatically an ideal. In the general case one can choose between considering this set and losing the ideal structure or considering the biggest ideal contained in this set and thus losing information about $H$.

This is one reason why it is sometimes difficult to extend results that hold for $SL_{2}({\mathbb{Z}})$ to more general groups $SL_{2}(R)$. Either the associated objects do not have enough structure and hence one cannot prove enough about them, or they have the necessary structure but they don’t encode all the desired information. This also indicates that one should look at both of the possible generalizations and how they are related.

Let us make this precise by introducing the most important instance of this dichotomy structure versus information.

Let $R$ be a commutative ring with identity and let $H$ be a subgroup of $SL_{2}(R)$. The classical definition of the level $l(H)$ of $H$ opts for structure over information and defines it as the largest $R$-ideal ${\mathfrak{q}}$ such that for every $a\in{\mathfrak{q}}$ the translation matrix ${1\ a\choose 0\ 1}$ is contained in every conjugate of $H$ in $SL_{2}(R)$. We emphasize information over structure and call the set of all $a\in R$ such that the translation matrix ${1\ a\choose 0\ 1}$ is contained in every conjugate of $H$ in $SL_{2}(R)$ the quasi-level of $H$, denoted by $ql(H)$. Implicitly the quasi-level already occurs for example in the proof of Proposition 1 in [Se1].

Obviously, the level of $H$ is the largest $R$-ideal contained in the quasi-level of $H$. Clearly for the special case $R={\mathbb{Z}}$ we always have $ql(H)=l(H)$. In general however, as we shall see, they can differ by “as much as possible”. By definition the level of a subgroup is not merely an additive subgroup of $R$ and it can be shown to have much stronger properties than the quasi-level. However as a parameter for studying $H$ it is less useful than $ql(H)$, mainly for the following reason. If $H$ is of finite index in $SL_{2}(R)$ then $ql(H)$ is non-zero. On the other hand for some $R$ (including $R=k[t]$, where $k$ is a finite field) it is known that $SL_{2}(R)$ has infinitely many finite index subgroups of level zero. Moreover for any finite index subgroup $H$ important information is provided by equations involving the size of the quasi-level of $H$ and its index in $SL_{2}(R)$.

Now let $D$ be any Dedekind domain with quotient field $F$. The group $SL_{2}(D)$ acts on the projective line ${\mathbb{P}}^{1}(F)=F\cup\{\infty\}$ as a set of linear fractional transformations. Let $H$ be a subgroup of $SL_{2}(D)$. Clearly $H$ acts on $O_{\infty}=\{g(\infty):g\in SL_{2}(D)\}$, the $SL_{2}(D)$-orbit containing $\infty$. We refer to the orbits of the $H$-action on $O_{\infty}$ as the $H$-cusps.

The classical definition of cusp amplitude again prefers structure over information and defines the cusp amplitude of the $H$-cusp containing $g(\infty)$ as a certain $D$-ideal ${\mathfrak{c}}(H,g)$ associated to the stabilizer of $g(\infty)$ in $H$. See Section 2 for details.

As the cusp amplitude often loses too much information about the stabilizer of the cusp, we also introduce what we call the quasi-amplitude. This represents the other possibility to generalize the notion of cusp-amplitude from $SL_{2}({\mathbb{Z}})$ to $SL_{2}(D)$, going for more information and less structure by dropping the requirement that it is an ideal and considering all translation matrices in $H^{g}=\{g^{-1}hg\ :\ h\in H\}$.

Again, in general the quasi-amplitude ${\mathfrak{b}}(H,g)$ is only an additive subgroup of $D$. Obviously, the cusp amplitude ${\mathfrak{c}}(H,g)$ is the largest ideal contained in ${\mathfrak{b}}(H,g)$. In situations where one can control how far the two can be apart, one then obtains results that generalize theorems for the classical modular group.

Let $\mathcal{A}(H)$ denote the set of all the cusp amplitudes of $H$. It is easy to see that the intersection of all these cusp amplitudes is nothing else than the level of $H$. In our first principal result we prove that for one important class of subgroups the set $\mathcal{A}(H)$ has the following surprising properties.

Theorem A. Let $H$ be a subgroup of $SL_{2}(D)$ and let

[TABLE]

If $H$ is a congruence subgroup, then ${\mathfrak{c}}_{\mathrm{min}},\;{\mathfrak{c}}_{\mathrm{max}}\in\mathcal{A}(H).$

*In particular, if $H$ is a congruence subgroup, there is an $H$-cusp in the $SL_{2}(D)$-orbit of $\infty$ whose cusp amplitude equals the level of $H$.

*This extends the results of a remarkable paper of Larcher [La] for the particular case $D={\mathbb{Z}}$. For this case Stothers [St] has also proved that $\mathcal{A}(H)$ has a minimum by an alternative method. In extending Larcher’s proofs we have simplified his approach in a number of ways. Most importantly we have avoided his use of the Dirichlet theorem on primes in an arithmetic progression (for ${\mathbb{Z}}$).

Strictly speaking, our definition of the cusp amplitude differs slightly from the one given in [La]. The results in [La] are formulated for $PSL_{2}({\mathbb{Z}})$, or equivalently, for congruence subgroups that contain $-I_{2}$, and for those both definitions coincide.

Cusp amplitudes were originally introduced for subgroups of the modular group $SL_{2}({\mathbb{Z}})$ and play an important role in the theory of modular forms and modular curves. Theorem A holds (trivially) for every normal subgroup of $SL_{2}(D)$ since then $\mathcal{A}(H)$ reduces to a single ideal. However it is known that it does not hold in general for non-normal non-congruence subgroups. Moreover we show that Theorem A does not hold in general for the quasi-amplitudes of a congruence subgroup.

The intersection of all the quasi-amplitudes (resp. cusp amplitudes) of $H$ is $ql(H)$, its quasi-level (resp. $l(H)$, its level). As previously stated, for the classical case $D={\mathbb{Z}}$ they clearly coincide. Our second principal result shows, rather surprisingly, that this is also true for “many” congruence subgroups of $SL_{2}(D)$.

We give a name to the precise condition as it will be used in several places in the paper.

Condition L. Let $H$ be a subgroup of $SL_{2}(D)$ of non-zero level $l(H)={\mathfrak{q}}$. We say that $l(H)$ (or $H$) satisfies Condition L if both the following two conditions hold:

• (i)

${\mathfrak{q}}+(2)=D$;

• (ii)

There are no prime ideals ${\mathfrak{p}}$ for which $|D/{\mathfrak{p}}|=3$ and $\mathrm{ord}_{{\mathfrak{p}}}({\mathfrak{q}})=1$.

Note that both conditions are satisfied if, for example, ${\mathfrak{q}}+(6)=D$, so in particular always for $char(D)=p\geq 5$. Condition L is also automatically satisfied if $D$ contains a field that is a nontrivial extension of ${\mathbb{F}}_{3}$.

Theorem B. Let $H$ be a congruence subgroup of $SL_{2}(D)$ whose level $l(H)={\mathfrak{q}}$ satisfies Condition L. Then

[TABLE]

*In other words, then the quasi-level of $H$ is automatically an ideal.

*There are examples of (normal) congruence subgroups which show that both restrictions on ${\mathfrak{q}}$ from Condition L are necessary in Theorem B. For non-congruence subgroups $H$ Theorem B breaks down completely, i.e. $ql(H)$ and $l(H)$ can “differ by as much as possible”. More precisely we prove that $SL_{2}(k[t])$, where $k$ is a field, contains normal, non-congruence subgroups whose quasi-level has $k$-codimension 1 in $k[t]$ and whose level can take any possible value (including zero). (There are no proper subgroups of $SL_{2}(k[t])$ whose quasi-level is $k[t]$.)

Presumably one of the most interesting and in practice one of the most useful results on congruence subgroups would be to bound the level in terms of the index of the subgroup, i.e. to generalize the classical result that the level of a congruence subgroup of $SL_{2}({\mathbb{Z}})$ can not be bigger than its index. In Section 5 we discuss different generalizations of this. The proofs use many of the previous results. We highlight a version that holds for general $D$.

Theorem C. Let $D$ be any Dedekind domain and let $H$ be a congruence subgroup of index $n$ in $SL_{2}(D)$ whose level $l(H)$ satisfies Condition L. Then

[TABLE]

and if $H$ is a normal subgroup then even

[TABLE]

The structure of the paper is as follows:

In Section 1 we clarify by means of theorems and counterexamples how the properties ‘Every subgroup of finite index has non-zero level’, ‘Every subgroup of non-zero level is a congruence subgroup’ and ‘Every subgroup of finite index is a congruence subgroup’ are related. In Section 2 we present the quite lengthy proof of Theorem A. In Section 3 we introduce quasi-amplitudes. In Section 4 we define the quasi-level and prove Theorem B. Using the previous results, we then can in Section 5 prove Theorem C and similar relations between the level and the index of a congruence subgroup.

On the way we also prove several other interesting results. Whenever appropriate we show by means of examples the necessity of conditions in theorems and the limitations of certain definitions.

Of course, every result that holds for congruence subgroups can also be reformulated as a necessary condition that a subgroup must satisfy in order to stand a chance of being a congruence subgroup. For the classical modular group $SL_{2}({\mathbb{Z}})$ many such criteria are known, and some of them are very useful in practice (compare Example 2.12). In order to get an overview, we now summarize the criteria that our results furnish. Several of them seem to be new or at least new in such a general setting.

Theorem D. Let $D$ be any Dedekind domain and ${\mathfrak{q}}$ a non-zero ideal of $D$. Let $H$ be a subgroup of $G=SL_{2}(D)$ with level $l(H)={\mathfrak{q}}$ and quasi-level $ql(H)$. In order for $H$ to be a congruence subgroup, the following conditions are necessary:

• a)

(Corollary 1.4)  $G({\mathfrak{q}})\subseteq H$.

• b)

(Corollary 2.8)  The ideal-theoretic sum of all cusp amplitudes of $H$ is again a cusp amplitude of $H$.

• c)

(Theorem 2.10)  The intersection of all cusp amplitudes of $H$ is again a cusp amplitude of $H$.

• d)

(Corollary 2.11)  The level ${\mathfrak{q}}$ of $H$ is a cusp amplitude of $H$.

• e)

(Lemma 4.2)  For any $\alpha\in D$ that is invertible modulo ${\mathfrak{q}}$ we have $\alpha^{2}ql(H)\subseteq ql(H)$.

Now suppose moreover that the level ${\mathfrak{q}}$ of $H$ satisfies Condition L. Then the following conditions are necessary for $H$ to be a congruence subgroup:

• f)

(Theorem 4.6)  $ql(H)=l(H)$.

• g)

(Corollary 5.7)  $|D/l(H)|\ \hbox{\it divides}\ (|G:H|)!$.

Finally suppose that in addition $H$ is a normal subgroup of $G$ (and that $l(H)$ satisfies Condition L). Then the following conditions are necessary for $H$ to be a congruence subgroup:

• h)

(Corollary 4.7)  Every quasi-amplitude of $H$ is actually a cusp amplitude of $H$.

• i)

(Theorem 5.6)  $|D/l(H)|\ \hbox{\it divides}\ |G:H|$.

With the exception of a) in general none of the conditions is sufficient for $H$ to be a congruence subgroup. Conditions e), f), and h) hold automatically if $D={\mathbb{Z}}$ and Remark 2.12 points out non-congruence subgroups of $SL_{2}({\mathbb{Z}})$ that satisfy conditions b), c), d), g) and i).

On the other hand, none of the conditions is trivial in the sense of automatically true for every $D$. Remark 2.12 provides examples of non-congruence subgroups of $SL_{2}({\mathbb{Z}})$ that violate conditions b), c) and d), and Theorem 4.12 allows the construction of normal non-congruence subgroups that satisfy Condition L, but violate the conditions e), f), g), h) and i).

If $H$ is a congruence subgroup such that $l(H)$ does not satisfy Condition L, then Examples 4.10 and 5.4 show that none of the conditions f), g), h) and i) need hold, even if $H$ is normal. Compare also Example 4.9 for conditions f) and h).

Finally, the condition that $H$ is normal cannot simply be dropped for parts h) and i). Examples 3.2 and 3.5 exhibit non-normal congruence subgroups $H$ that satisfy Condition L, but neither h) nor i).

1. Congruence subgroups

Let $R$ be a (commutative) ring and let $R^{*}$ be its group of units. For each $r\in R,\alpha\in R^{*}$, we put

[TABLE]

In addition we define

[TABLE]

Let $E_{2}(R)$ be the subgroup of $SL_{2}(R)$ generated by all $T(r)$ and $S(r)$ with $r\in R$. For each $R$-ideal ${\mathfrak{q}}$ we denote by $E_{2}(R,{\mathfrak{q}})$ the normal subgroup of $E_{2}(R)$ generated by all $T(q)$ with $q\in{\mathfrak{q}}$. By definition we note that $E_{2}(R,R)=E_{2}(R)$. Let $NE_{2}(R,{\mathfrak{q}})$ be the normal subgroup of $SL_{2}(R)$ generated by $E_{2}(R,{\mathfrak{q}})$. We put $NE_{2}(R,R)=NE_{2}(R)$. We also define

[TABLE]

where $I_{2}$ denotes the $2$-dimensional unit matrix. Obviously

[TABLE]

Occasionally we will also need the Borel group, consisting of the upper triangular matrices,

[TABLE]

If $H$ is a subgroup of $G$, we denote the core of $H$ in $G$, i.e. the biggest normal subgroup of $G$ contained in $H$ by

[TABLE]

Definition. The level of a subgroup $H$ of $SL_{2}(R)$, $l(H)$, is the largest ideal ${\mathfrak{q}}^{\prime}$, say, for which $NE_{2}(R,{\mathfrak{q}}^{\prime})\leq H$. The level is well-defined since $NE_{2}(R,{\mathfrak{q}}_{1})\centerdot NE_{2}(R,{\mathfrak{q}}_{2})=NE_{2}(R,{\mathfrak{q}}_{1}+{\mathfrak{q}}_{2})$.

Since $NE_{2}(R,{\mathfrak{q}}^{\prime})$ is the normal subgroup of $SL_{2}(R)$ generated by all $T(r)$ with $r\in{\mathfrak{q}}^{\prime}$, we can equivalently say that the level of $H$ is the largest ideal ${\mathfrak{q}}^{\prime}$ such that the core $N_{H}$ of $H$ contains all translation matrices $T(r)$ with $r\in{\mathfrak{q}}^{\prime}$.

Definition. A subgroup $C$ of $SL_{2}(R)$ is called a congruence subgroup if

[TABLE]

for some ${\mathfrak{q}}^{\prime}\neq\{0\}$.

It is clear that if $R/{\mathfrak{q}}^{\prime}$ is finite then $C$ is of finite index in $SL_{2}(R)$.

Let ${\mathfrak{q}},\;{\mathfrak{q}}^{\prime}$ be ideals for which ${\mathfrak{q}}\supseteq{\mathfrak{q}}^{\prime}\supseteq{\mathfrak{q}}^{2}$. We will require some properties of the quotient group $SL_{2}(R,{\mathfrak{q}})/SL_{2}(R,{\mathfrak{q}}^{\prime})$. We put ${\mathfrak{r}}={\mathfrak{q}}/{\mathfrak{q}}^{\prime}$ and we denote the image of any $q\in{\mathfrak{q}}$ in ${\mathfrak{r}}$ by $\overline{q}$. Let $X\in SL_{2}(R,{\mathfrak{q}})$. Then

[TABLE]

for some $x,y,z,t\in{\mathfrak{q}}$. We define a map

[TABLE]

by

[TABLE]

(Note that $x+t\equiv 0\;(\mathop{\mathrm{mod}}{\mathfrak{q}}^{\prime}).$)

Lemma 1.1. With the above notation, the map $\theta$ induces the following isomorphism

[TABLE]

where ${\mathfrak{r}}^{+}$ is the additive group of ${\mathfrak{r}}$.

*Moreover the quotient group $SL_{2}(R,{\mathfrak{q}})/SL_{2}(R,{\mathfrak{q}}^{\prime})$ is generated by the images of the elements $S(q),T(q),R(q)$, where $q\in{\mathfrak{q}}$.

Proof.* *See the proof of [MSt, Theorem 4.1]. $\Box$

To obtain more precise properties of congruence subgroups we require some restrictions on $R$. We recall that $R$ is said to be an $SR_{2}$-ring if, for all $a,b\in R$ such that $aR+bR=R$, there exists $c\in R$ for which $(a+cb)\in R^{*}$. Every semi-local ring, for example, is an $SR_{2}-$ring ([B, Theorem 3.5, p.239]).

Theorem 1.2. Let ${\mathfrak{q}},{\mathfrak{q}}^{\prime}$ be $R$-ideals such that $R/{\mathfrak{q}}^{\prime}$ is an $SR_{2}$-ring. Then

[TABLE]

Proof.* *See the proof of [B, (9.3) Corollary, p. 267]. $\Box$

Since every proper quotient of a Dedekind domain is semi-local the following is an immediate consequence of Theorem 1.2.

Corollary 1.3. Let $D$ be a Dedekind domain and let ${\mathfrak{q}},{\mathfrak{q}}^{\prime}$ be $D$-ideals.

• (i)

If ${\mathfrak{q}}^{\prime}\neq\{0\}$ then

[TABLE]

• (ii)

[TABLE]

The classical version of our next result (for the special case $D={\mathbb{Z}}$) is due to Fricke. (See [W].)

Corollary 1.4. Let $D$ be a Dedekind domain and let $H$ be a subgroup of $SL_{2}(D)$ with $l(H)={\mathfrak{q}}\neq(0)$. Then $H$ is a congruence subgroup if and only if

[TABLE]

Remarks 1.5.**

• a)

By virtue of Theorem 1.2, Corollaries 1.3 and 1.4 also apply to the case where $R$ is a Noetherian domain of Krull dimension one. For example all orders in algebraic number fields are of this type. One “standard” example of a Noetherian domain of Krull dimension one which is not Dedekind is the ring ${\mathbb{Z}}[\sqrt{-3}]$. A Dedekind domain $D$ is an integrally closed Noetherian domain of Krull dimension one.

• b)

But Corollaries 1.3 and 1.4 do in general not hold for an integrally closed Noetherian domain of Krull dimension bigger than one. Take for example $R={\mathbb{F}}_{q}[x,y]$, the polynomial ring in two variables over a finite field. Then

[TABLE]

Let $f(x)\in{\mathbb{F}}_{q}[x]$ be irreducible of degree at least $2$. By Theorem 4.12 below or by [MSch2, Theorem 5.5] then $SL_{2}({\mathbb{F}}_{q}[x])$ has a normal non-congruence subgroup $N$ of finite index and level $(f(x))$. The inverse image of $N$ under reduction modulo $y$ is a normal subgroup of finite index in $SL_{2}(R)$ of level $yR+f(x)R$ that contains $SL_{2}(R,yR)$ but not $SL_{2}(R,yR+f(x)R)$.

From now on $D$ will always be a Dedekind domain. We will sometimes use the abbreviations

[TABLE]

and

[TABLE]

for each $D$-ideal ${\mathfrak{q}}$.

Lemma 1.6. *If $D$ has characteristic $p>0$ and every unit of $D$ has finite order, then $D^{*}\cup\{0\}$ is a field $k$. Moreover, $k$ is the biggest algebraic extension of ${\mathbb{F}}_{p}$ contained in the field of fractions of $D$.

Proof. *Obviously $K$, the quotient field of $D$, contains the prime field ${\mathbb{F}}_{p}$. Let $k$ be the biggest algebraic extension of ${\mathbb{F}}_{p}$ contained in $K$.

Let $u$ be a unit of $D$ that has finite order. Then $u$ is a root of unity and hence contained in an algebraic extension of ${\mathbb{F}}_{p}$. Thus $u\in k$. Conversely, every element of $k^{*}$ is a root of unity, so it is in $D$ since $D$ is integrally closed, and obviously it is a unit of $D$. $\Box$

Lemma 1.7. *If $D$ contains an infinite field, then every subgroup of finite index in $SL_{2}(D)$ has level $D$.

Proof. *The following proof is inspired by the proof of [Se1, Proposition 1, p.491].

If $H$ is a subgroup of finite index in $SL_{2}(D)$, then its core is a normal subgroup of $SL_{2}(D)$ and still has finite index. So it suffices to prove the lemma for a normal subgroup $N$ of finite index in $SL_{2}(D)$. We define

[TABLE]

This set will be investigated in more detail in Section 4. Here we only need that $ql(N)$ is a subgroup of $(D,+)$ and that $u^{2}ql(N)\subseteq ql(N)$ for every $u\in D^{*}$. The last claim can be seen by conjugating with the diagonal matrix with entries $u$ and $u^{-1}$, as $N$ is normal.

Let $k$ be the infinite field in $D$. Then obviously $k^{*}\subseteq D^{*}$. So $ql(N)$ is stable under multiplication with $x^{2}$ and even with $x^{2}-y^{2}$ for all $x,y\in k$.

If the characteristic of $k$ is different from $2$, then it is easy to see that every element of $k$ is of the form $x^{2}-y^{2}$ with $x,y\in k$. Thus $ql(N)$ is a subspace of the $k$-vector space $D$. Since the quotient space $D/ql(N)$ is finite because

[TABLE]

but $k$ is infinite, we must have $ql(N)=D$.

If the characteristic is $2$, the squares in $k$ still form an infinite field $k_{2}$, and we can apply the same proof using $k_{2}$-vector spaces. $\Box$

Proposition 1.8. *If $D$ has characteristic [math] or if $D^{*}$ is infinite, then every subgroup of finite index in $SL_{2}(D)$ has non-zero level.

Proof. *As in the previous proof we can assume that the subgroup $N$ is normal. Again we show that $ql(N)$ contains a non-zero ideal. The first two cases were already treated in the proof [Se1, Proposition 1, p.491], namely:

If $char(D)=0$ and $[SL_{2}(D):N]=n$, then $T(nr)=(T(r))^{n}$ lies in $N$ for every $r\in D$, so $ql(N)$ contains the ideal $nD$.

If $char(D)=p>0$ and $D^{*}$ contains a unit $u$ of infinite order, then $D$ and $ql(N)$ are ${\mathbb{F}}_{p}[u^{2}]$-modules. So the quotient group $D/ql(N)$ is also an ${\mathbb{F}}_{p}[u^{2}]$-module, and since it is finite (whereas ${\mathbb{F}}_{p}[u^{2}]$ is isomorphic to a polynomial ring), it is annulated by some nonzero element $a$ of ${\mathbb{F}}_{p}[u^{2}]$. Thus $aD\subseteq ql(N)$.

Finally, if $char(D)=p>0$ and there are no units of infinite order, then Lemma 1.6 implies $D^{*}=k^{*}$ where $k$ is an infinite algebraic extension of ${\mathbb{F}}_{p}$. So in this case every subgroup of finite index has level $D$ by Lemma 1.7. $\Box$

We recall from the introduction that if $D$ is of arithmetic type then every congruence subgroup of $SL_{2}(D)$ is of finite index. It is known that for some such $D$ (including the classical case $R={\mathbb{Z}}$) the converse does not hold. At this point we record a class of arithmetic Dedekind domains for which the converse does hold.

Let $K$ be a global field. In other words, $K$ is either an algebraic number field, i.e. a finite extension of ${\mathbb{Q}}$, or $K$ is an algebraic function field of one variable with finite constant field, i.e., $K$ is a finite extension of a rational function field ${\mathbb{F}}_{q}(t)$.

Let $S$ be a proper subset of all the places of $K$ which contains all archimedean places. Let ${\mathcal{O}}_{S}$ be the set of all elements of $K$ that are integral outside $S$. Then ${\mathcal{O}}_{S}$ is an arithmetic Dedekind domain whose prime ideals correspond to the places of $K$ outside $S$.

Note that we do not assume that $S$ is finite (as in the standard definition of the ring of $S$-integers of $K$). If $S$ is infinite, then ${\mathcal{O}}_{S}^{*}$ is not finitely generated. If $S$ contains all places of $K$ except one, then ${\mathcal{O}}_{S}$ is a discrete valuation ring. The next theorem is a minor extension of a famous result of Serre [Se1].

Theorem 1.9. *Let $K$ be a global field and ${\mathcal{O}}_{S}$ the ring of $S$-integers where $S$ is an infinite set. Then every subgroup of finite index in $SL_{2}({\mathcal{O}}_{S})$ has non-zero level, and every subgroup of non-zero level is a congruence subgroup.

*Proof. The first claim follows from Proposition 1.8. The second claim is equivalent to showing

[TABLE]

for every non-zero ideal ${\mathfrak{q}}$ of ${\mathcal{O}}_{S}$. We use a “local” argument to reduce this to a suitable case where $S$ is finite. Take any $g\in SL_{2}({\mathcal{O}}_{S},{\mathfrak{q}})$. Then the entries of $g$, being elements of $K$, lie in ${\mathcal{O}}_{S^{\prime}}$ for some finite subset $S^{\prime}$ of $S$. We may assume that $|S^{\prime}|>1$ and that $S^{\prime}$ contains at least one non-archimedean place. Let ${\mathfrak{q}}^{\prime}$ be the ideal ${\mathfrak{q}}\cap{\mathcal{O}}_{S^{\prime}}$ of ${\mathcal{O}}_{S^{\prime}}$. Then

[TABLE]

by [Se1, Théorème 2 (b), p.498]. Since obviously $g\in SL_{2}({\mathcal{O}}_{S^{\prime}},{\mathfrak{q}}^{\prime})$ and $NE_{2}({\mathcal{O}}_{S^{\prime}},{\mathfrak{q}}^{\prime})\subseteq NE_{2}({\mathcal{O}}_{S},{\mathfrak{q}})$, we have shown $SL_{2}({\mathcal{O}}_{S},{\mathfrak{q}})\subseteq NE_{2}({\mathcal{O}}_{S},{\mathfrak{q}})$. The converse inclusion always holds. $\Box$

We present another class of Dedekind domains that have the same congruence subgroup property.

Theorem 1.10. *If $D$ has only finitely many maximal ideals, then every subgroup of finite index in $SL_{2}(D)$ has nonzero-level, and every subgroup (finite index or not) of non-zero level is a congruence subgroup.

*Proof. Let ${\mathfrak{a}}$ be the product of the maximal ideals of $D$. Since the ideals ${\mathfrak{a}}$, ${\mathfrak{a}}^{2}$, ${\mathfrak{a}}^{3},\ldots$ are all different, ${\mathfrak{a}}$ must be an infinite set. Now every element of the form $1+a$ with $a\in{\mathfrak{a}}$ obviously does not lie in any of the maximal ideals; so it must be a unit. Thus $D^{*}$ is infinite. So if $H$ is a subgroup of finite index in $SL_{2}(D)$, it has non-zero level by Proposition 1.8.

Now let $H$ be a subgroup, not necessarily of finite index, and assume that $H$ has non-zero level, say $l(H)={\mathfrak{q}}$. As $D$ is semi-local and hence an $SR_{2}$-ring by [B, Theorem 3.5, p.239], we can apply Theorem 1.2 with ${\mathfrak{q}}^{\prime}$ being the zero ideal, and obtain $SL_{2}(D,{\mathfrak{q}})\leq H$. $\Box$

In the previous two results the reason for the congruence subgroup property was that $NE_{2}(D,{\mathfrak{q}})=SL_{2}(D,{\mathfrak{q}})$ for every ideal ${\mathfrak{q}}$ of $D$. We now construct examples with a different reason. First a very general fact.

Lemma 1.11. *If $NE_{2}(D)\neq SL_{2}(D)$, then $NE_{2}(D,{\mathfrak{q}})\neq SL_{2}(D,{\mathfrak{q}})$ for every non-zero ideal ${\mathfrak{q}}$ of $D$.

*Proof. Let $NE_{2}(D)\neq SL_{2}(D)$. Then $D$ has infinitely many maximal ideals by Theorem 1.10. Asssume $NE_{2}(D,{\mathfrak{q}})=SL_{2}(D,{\mathfrak{q}})$. Choosing an ideal ${\mathfrak{q}}^{\prime}$ with ${\mathfrak{q}}+{\mathfrak{q}}^{\prime}=D$ and using Corollary 1.3 we would obtain

[TABLE]

[TABLE]

a contradiction. $\Box$

Example 1.12. Let $k$ be an algebraically closed field of characteristic [math]. Let $D=k[x,y]$ with $y^{2}=x^{3}+Ax+B$ such that the cubic polynomial $x^{3}+Ax+B\in k[x]$ has no multiple roots.

Then $NE_{2}(D,{\mathfrak{q}})\neq SL_{2}(D,{\mathfrak{q}})$ for every non-zero ideal ${\mathfrak{q}}$ of $D$. So $SL_{2}(D)$ has non-congruence subgroups of every level. But every subgroup of finite index is a congruence subgroup for the trivial reason that $SL_{2}(D)$ has no subgroups of finite index.

To see all this, we use Takahashi’s description [Ta] of the action of $GL_{2}(D)$ on the Bruhat-Tits tree of $GL_{2}(k((t)))$ where $t=\frac{x}{y}$. Note that since every element of $k^{*}$ is a square, there exists a natural isomorphism $PSL_{2}(D)\cong PGL_{2}(D)$.

Since the fundamental domain of this action is a connected graph without loops [Ta], by the theory of groups acting on trees [Se2], the group is generated by the stabilizers of the vertices. By [Ta, Theorem 5] these stabilizers are built from subgroups isomorphic to $k$, $k^{*}$ and $PSL_{2}(k)$. As the groups $k$ and $k^{*}$ are infinitely divisible, they have no subgroups of finite index; and since $k$ is infinite, $PSL_{2}(k)$ also has no finite index subgroups. Hence a subgroup of finite index in $PGL_{2}(D)$ would have to contain all stabilizers and so be equal to $PGL_{2}(D)$.

Moreover, the fact that the fundamental domain has a vertex with trivial stabilizer and the description of its neighbours [Ta] shows that $PGL_{2}(D)$ is a free product

[TABLE]

where $\varDelta(\infty)$ is GL_{2}(k)\mathop{\mbox{\Large}}_{B_{2}(k)}B_{2}(D) modulo the center. This shows that modulo scalar matrices $SL_{2}(D)/NE_{2}(D)$ is isomorphic to the infinite group \mathop{\mbox{\LARGE}}\limits_{\ell\in k}\varDelta(\ell).

To appreciate this example we point out that even if a Dedekind domain $D$ has no finite quotients, this does not imply that the group $SL_{2}(D)$ has no subgroups of finite index.

Example 1.13. Let

[TABLE]

Then $D^{*}=k^{*}$, and since every element of $k^{*}$ is a square we again have $PSL_{2}(D)\cong PGL_{2}(D)$. By Takahashi’s results [Ta] again we have a free product

[TABLE]

with $\varDelta(\infty)$ equal to SL_{2}(k)\mathop{\mbox{\Large*}}_{B_{2}(k)}B_{2}(D) modulo scalar matrices. However, since the equation $y^{2}+y=1$ has no solution in $k$, by [Ta, Theorem 5] this time

[TABLE]

So $L={\mathbb{F}}_{4}k$ is the unique quadratic extension of $k$. Every element of $k^{*}$ is a third power of an element of $k^{*}$. Moreover, ${\mathbb{F}}_{8}\nsubseteq L$, so ${\mathbb{F}}_{64}\nsubseteq L$. Thus $L$ contains the $3$-rd roots of unity, but not the $9$-th roots of unity. Hence the third powers in $L^{*}$ form a subgroup of index $3$ that contains $k^{*}$. So $L^{*}/k^{*}$ has a subgroup of index $3$. Because of the free product there exists a surjective homomorphism from $PSL_{2}(D)$ to $\varDelta(1)$ whose kernel is the normal subgroup generated by all other $\varDelta(\ell)$. Thus $SL_{2}(D)$ contains a normal non-congruence subgroup of index $3$ and level $D$.

We mention two types of Dedekind domains for which $SL_{2}(D)$ does not have the congruence subgroup property.

Every subgroup of finite index in $SL_{2}({\mathbb{Z}})$ has non-zero level, but $SL_{2}({\mathbb{Z}})$ has finite index non-congruence subgroups.

Let ${\mathcal{O}}_{S}$ be the ring of $S$-integers in a global function field $K$ where $|S|=1$. Then $SL_{2}({\mathcal{O}}_{S})$ has uncountably many finite index subgroups of level zero [MSch2, Corollary 3.6] and $SL_{2}({\mathcal{O}}_{S})$ has finite index non-congruence subgroups of almost every level [MSch2, Theorem 5.5].

Remark 1.14. This leaves open the question whether there exist examples $SL_{2}(D)$ that contain finite index subgroups of level zero but in which every finite index subgroup of non-zero level is a congruence subgroup.

By Lemmas 1.6 and 1.7 and Proposition 1.8 such a $D$ must be an ${\mathbb{F}}_{q}$-algebra with unit group ${\mathbb{F}}_{q}^{*}$. One might be tempted to think that the rings ${\mathcal{O}}_{S}$ in a global function field $K$ with $|S|=1$ are the only instances of such rings. This is true for finitely generated ${\mathbb{F}}_{q}$-algebras $D$, because then the Krull dimension of $D$ equals the transcendence degree of its quotient field.

However, O. Goldman [Go] has constructed Dedekind domains $D$ with $D^{*}={\mathbb{F}}_{q}^{*}$ such that $D/{\mathfrak{m}}$ is a finite field for every maximal ideal ${\mathfrak{m}}$ of $D$, and the quotient field $K$ of $D$ has transcendence degree $n$ over ${\mathbb{F}}_{q}$ where $n$ can be any given natural number. We thank Peter Vámos for pointing out this reference to us.

2. Cusp amplitudes

Throughout $D$ denotes a Dedekind domain with quotient field $F(\neq D)$. For each $x\in D$ we denote the principal $D$-ideal $xD$ by $(x)$. We write $(x,y)=z$, if $(x)+(y)=(z)$. All the results of Section $1$ apply to $SL_{2}(D)$ which from now on we denote by $G$.

We recall the action of $G$ on $\hat{F}={\mathbb{P}}^{1}(F)$. Let g=\left[\begin{array}[]{cc}a&b\\ c&d\end{array}\right] be an element of $G$. Then

[TABLE]

In particular $G$ acts transitively on $\hat{F}$ if and only if $D$ is a principal ideal domain. In general the $G$-orbit in $\hat{F}$ containing $\infty$ is

[TABLE]

Let $H$ be a subgroup of $G$. For each $g\in G$, let $H^{g}$ denote the conjugate subgroup $g^{-1}Hg$. For each $z\in\hat{F}$, we denote the stabilizer of $z$ in $H$ by

[TABLE]

It is clear that

[TABLE]

for all $g\in G$ and $z\in\hat{F}$, and that

[TABLE]

The following correspondence is obvious.

Definition. Let $z_{1},z_{2}\in\hat{F}$. We write

[TABLE]

for some $h\in H$.

Lemma 2.1. Let $r,s\in G$. Then

[TABLE]

Lemma 2.1 provides the following bijection:

[TABLE]

between the orbits of the $H$-action on $O_{\infty}$ and the $(H,G_{\infty})$ double cosets in $G$. We refer to the elements of $H\backslash O_{\infty}$ as the $H$-cusps.

It is clear that the double coset $HgG_{\infty}$ is a union of a set $\mathcal{S}$ of (right) $H$-cosets and that

[TABLE]

where $z=g(\infty)$. From now on let $\{g_{\lambda}:\lambda\in\Lambda\}$ be a complete set of representatives for the double coset space $H\backslash G/G_{\infty}$. Then $\{z_{\lambda}=g_{\lambda}(\infty):\lambda\in\Lambda\}$ is a complete set of representatives for the $H$-cusps in $O_{\infty}$, by Lemma 2.1. Our next lemma is an immediate consequence of the above.

Lemma 2.2. Let $H$ have finite index in $G$. Then, with the above notation,

[TABLE]

It follows from the above that, if $g\in G$ and $z=g(\infty)$, then every unipotent matrix in $G_{z}$ is of the form $gT(x)g^{-1}$, for some $x\in D$. It is obvious that $\{x\in D:gT(x)g^{-1}\in H\}$ is a subgroup of (the additive group of) $D$. Now let $g_{i}\in G,(i=1,2)$. Suppose now that $Hg_{1}G_{\infty}=Hg_{2}G_{\infty}$. Then $g_{1}=hg_{2}s$, for some $h\in H$ and $s\in G_{\infty}$. It is clear that

[TABLE]

where $\mu\in D^{*}$ is completely determined by $s$. We are now able to make the following definition.

Definition. For each $g\in G$, let ${\mathfrak{c}}(H,g)$ denote the largest $D$-ideal ${\mathfrak{q}}$ with the property that

[TABLE]

for all $q\in{\mathfrak{q}}$.

It is clear from the above that ${\mathfrak{c}}(H,g)={\mathfrak{c}}(H,g^{\prime})$, whenever $g(\infty)\equiv g^{\prime}(\infty)\;(\mathop{\mathrm{mod}}H)$. We call ${\mathfrak{c}}(H,g)$ the cusp amplitude of the $H$-cusp containing $g(\infty)$.

It is clear that, for each $g\in G$, there exists a unique $\lambda^{\prime}\in\Lambda$ such that

[TABLE]

Definition. We denote a complete set of cusp amplitudes for $H$ by

[TABLE]

The following is an immediate consequence of the above.

Lemma 2.3. For each subgroup $H$ of $G$,

[TABLE]

Remarks 2.4.

• (i)

If $N$ is a normal subgroup then ${\mathfrak{c}}(N,g_{1})={\mathfrak{c}}(N,g_{2})$, for all $g_{1},g_{2}\in G$. In this case $\mathcal{A}(N)$ involves a single $D$-ideal.

• (ii)

If $l(H)$ is non-zero (for example when $H$ is a congruence subgroup), then $\mathcal{A}(H)$ involves only finitely many ideals.

• (iii)

To avoid any possible confusion we emphasize that if $D$ is not a principal ideal domain the cusps (and the cusp amplitudes) of $H$ we consider are only a part of the $H$-orbits on ${\mathbb{P}}^{1}(F)$, namely those contained in the $G$-orbit of $\infty$.

It is easily verified that cusp amplitudes are invariant under conjugation as given below.

Lemma 2.5. Let $k\in G$. Then

[TABLE]

*for all $g\in G$.

We require a more detailed description of the unipotent matrices in each $G_{z}$, where $z\in O_{\infty}$. If $z=g(\infty)$ then $gT(-x)g^{-1}\in G_{z}$ is of the form

[TABLE]

where $(a,b)=1$. In this case $z=a/b$, when $b\neq 0$, and $z=\infty$, when $b=0$. Note that $U(1,0;x)=T(x)$ and $U(0,1;x)=S(x).$

Before our principal results we record this well-known useful property of Dedekind domains, which follows from the Chinese Remainder Theorem (CRT).

Lemma 2.6. Let ${\mathfrak{p}}_{1},\cdots,{\mathfrak{p}}_{t}$ be distinct prime $D$-ideals and let $\alpha_{1},\cdots,\alpha_{t}$ be non-negative integers, where $t\geq 1$. Then there exists $d\in D$ such that

[TABLE]

*where $1\leq i\leq t.$

For the first principal result our approach is more direct than that of Larcher [La]. In particular we avoid his use of the Dirichlet theorem on primes in an arithmetic progression (for ${\mathbb{Z}}$.)

Theorem 2.7. Let $H$ be a congruence subgroup of $G$ and let ${\mathfrak{q}}_{i}$ be a non-zero $D$-ideal contained in ${\mathfrak{c}}(H,g_{i})$, where $i=1,2$. Then there exists $g_{0}\in G$ such that

[TABLE]

Proof.* *Now $H\geq G({\mathfrak{q}})$, for some ${\mathfrak{q}}\neq\{0\}$. It is clear that, if $X,Y\in G$ and $X\equiv Y(\mathop{\mathrm{mod}}{\mathfrak{q}})$, then $X\in H$ if and only if $Y\in H$.

By Lemma 2.5 it is sufficient to prove the theorem for the case $g_{1}=I_{2}$. We may also assume that ${\mathfrak{q}}_{1}+{\mathfrak{q}}_{2}\neq{\mathfrak{q}}_{i}$ and (by Corollary 1.4 and Lemma 2.3) that ${\mathfrak{q}}_{i}\supseteq{\mathfrak{q}}$, where $i=1,2$.

Now $U(1,0;q)=T(q)\in H$, for all $q\in{\mathfrak{q}}_{1}$ and $U(c,d;q)\in H$, for all $q\in{\mathfrak{q}}_{2}$, for some $c,d\in D$, with $(c,d)=1$. Let ${\mathfrak{q}}_{1}+{\mathfrak{q}}_{2}={\mathfrak{q}}_{0}$. For our purposes it is sufficient to find $a,b$, with $(a,b)=1$ and ideals $\overline{{\mathfrak{q}}}_{i}$, contained in ${\mathfrak{q}}_{i}$, where $i=1,2$, such that $\overline{{\mathfrak{q}}}_{1}+\overline{{\mathfrak{q}}}_{2}={\mathfrak{q}}_{0}$, with the following properties. For all $\lambda\in\overline{{\mathfrak{q}}}_{1}$ and all $\mu\in\overline{{\mathfrak{q}}}_{2}$, there exist $\lambda_{0}\in{\mathfrak{q}}_{1}$ and $\mu_{0}\in{\mathfrak{q}}_{2}$ such that:

• (a)

$U(a,b;\lambda)\equiv T(\lambda_{0})\;(\mathop{\mathrm{mod}}{\mathfrak{q}})$;

• (b)

$U(a,b;\mu)\equiv U(c,d;\mu_{0})\;(\mathop{\mathrm{mod}}{\mathfrak{q}}).$

We may then take $g_{0}$ to be any matrix of the form

[TABLE]

By definition ${\mathfrak{q}}_{i}={\mathfrak{q}}_{i}^{\prime}{\mathfrak{q}}_{0}$, where $i=1,2$. Then ${\mathfrak{q}}_{1}^{\prime}+{\mathfrak{q}}_{2}^{\prime}=D$. Let ${\mathfrak{q}}_{i}^{\prime\prime}$ be the “smallest” divisor of ${\mathfrak{q}}_{0}^{-1}{\mathfrak{q}}$ all of whose prime divisors also divide ${\mathfrak{q}}_{i}^{\prime}$, where $i=1,2$. (Clearly ${\mathfrak{q}}_{i}^{\prime\prime}\subseteq{\mathfrak{q}}_{i}^{\prime}$.) Then ${\mathfrak{q}}_{1}^{\prime\prime}+{\mathfrak{q}}_{2}^{\prime\prime}=D$ and

[TABLE]

say, where ${\mathfrak{r}}+{\mathfrak{q}}_{1}^{\prime\prime}{\mathfrak{q}}_{2}^{\prime\prime}=D$.

Since $(a,b)=1$ the congruence (a) is equivalent to

[TABLE]

Since $(c,d)=1$ the congruence (b) is equivalent to:

[TABLE]

for some $x\in D$ such that $\mu_{0}\equiv\mu x\;(\mathop{\mathrm{mod}}{\mathfrak{q}})$. A consequence of $(**)$ is that, for all $\mu\in\overline{{\mathfrak{q}}}_{2}$,

[TABLE]

We begin by finding $a,b\in D$ for which

• (i)

$ad\equiv bc\;(\mathop{\mathrm{mod}}{\mathfrak{q}}_{1}^{\prime\prime})$,

• (ii)

$(a,b)=1$,

• (iii)

$b\in{\mathfrak{q}}_{2}^{\prime\prime}{\mathfrak{r}}$.

By Lemma 2.6 we may choose $d_{0}\in(c)+{\mathfrak{q}}_{1}^{\prime\prime}$ and $b\in{\mathfrak{q}}_{2}^{\prime\prime}{\mathfrak{r}}$ such that

• (i)

$(dd_{0})+{\mathfrak{q}}_{1}^{\prime\prime}=(bc)+{\mathfrak{q}}_{1}^{\prime\prime},$

• (ii)

$(d_{0})+(b)=D.$

By applying the CRT to the factors in the prime decomposition of ${\mathfrak{q}}_{1}^{\prime\prime}$ we can find $y\in D$ such that

• (i)

$ydd_{0}\equiv bc\;(\mathop{\mathrm{mod}}{\mathfrak{q}}_{1}^{\prime\prime}),$

• (ii)

$(y)+{\mathfrak{q}}_{1}^{\prime\prime}=D.$

(If ${\mathfrak{p}}^{\alpha}||{\mathfrak{q}}_{1}^{\prime\prime}$, where ${\mathfrak{p}}$ is prime with $\alpha>0$, and $dd_{0}\equiv bc\equiv 0\;(\mathop{\mathrm{mod}}{\mathfrak{q}}_{1}^{\prime\prime})$, take $y=1$.) Now for each prime divisor ${\mathfrak{p}}$ of $(b)$ we again use the CRT to find $z\in D$ for which

• (i)

$z\equiv y\;(\mathop{\mathrm{mod}}{\mathfrak{p}}^{\alpha})$, where ${\mathfrak{p}}^{\alpha}||{\mathfrak{q}}_{1}^{\prime\prime}$,

• (ii)

$z\equiv 1\;(\mathop{\mathrm{mod}}{\mathfrak{p}})$, when ${\mathfrak{p}}\nmid{\mathfrak{q}}_{1}^{\prime\prime}$.

The elements $a=zd_{0}$ and $b$ (as above) satisfy the requirements.

We can use the CRT to find $x\in D$ such that

[TABLE]

(Suppose that ${\mathfrak{p}}^{\alpha}||{\mathfrak{q}}_{1}^{\prime\prime}$. If $c\notin{\mathfrak{p}}$, take $x^{\prime}\equiv c^{-2}a^{2}\;(\mathop{\mathrm{mod}}{\mathfrak{p}}^{\alpha})$. If $d\notin{\mathfrak{p}}$, take $x^{\prime}\equiv d^{-2}b^{2}\;(\mathop{\mathrm{mod}}{\mathfrak{p}}^{\alpha})$.) Then, for all $\mu\in\overline{{\mathfrak{q}}}_{2}={\mathfrak{q}}_{2}^{\prime\prime}{\mathfrak{q}}_{0}{\mathfrak{r}}$, the congruences $(**)$ are satisfied.

In addition, for all $\lambda\in\overline{{\mathfrak{q}}}_{1}={\mathfrak{q}}_{1}^{\prime\prime}{\mathfrak{q}}_{0}$, the congruences $(*)$ are satisfied (with $\lambda_{0}=-a^{2}\lambda$). This completes the proof. $\Box$

Corollary 2.8. Let $H$ be a congruence subgroup of $G$ and let

[TABLE]

Then

[TABLE]

Proof.* *As previously noted $\mathcal{A}(H)$ involves only finitely many distinct ideals ${\mathfrak{q}}_{1}\cdots{\mathfrak{q}}_{t}$, say. By repeated applications of Theorem 2.7 it follows that there exists $g^{\prime}\in G$ such that

[TABLE]

But ${\mathfrak{c}}(H,g^{\prime})\in\mathcal{A}(H)$ and so ${\mathfrak{c}}(H,g^{\prime})={\mathfrak{q}}_{1}+\cdots+{\mathfrak{q}}_{t}={\mathfrak{c}}_{\mathrm{max}}$. $\Box$

We now come to our second principal result. Here our approach is similar to that of Larcher.

Lemma 2.9. Let $H$ be a congruence subgroup of $G$ of level ${\mathfrak{q}}$ and let ${\mathfrak{q}}_{\infty}={\mathfrak{c}}(H,I_{2})$ and ${\mathfrak{q}}_{0}={\mathfrak{c}}(H,g_{0})$, where $g_{0}(\infty)=0$. Then

[TABLE]

Proof.* *Suppose to the contrary that ${\mathfrak{q}}_{0}{\mathfrak{q}}_{\infty}+{\mathfrak{q}}\neq{\mathfrak{q}}$. We will prove that there exists an ideal ${\mathfrak{q}}^{\prime}$ such that

[TABLE]

which contradicts Corollary 1.4.

Recall that ${\mathfrak{q}}_{\infty}\supseteq{\mathfrak{q}}$ and ${\mathfrak{q}}_{0}\supseteq{\mathfrak{q}}$, by Lemma 2.3. We assume for now that

[TABLE]

where ${\mathfrak{p}}{\mathfrak{q}}^{\prime}={\mathfrak{q}}$, for some prime ideal ${\mathfrak{p}}$. Then $T(x),S(x)\in H$, for all $x\in{\mathfrak{q}}^{\prime}$. There are two possibilities.

Case 1: ${\mathfrak{p}}+{\mathfrak{q}}^{\prime}=D.$

By Corollary 1.3

[TABLE]

Now $D/{\mathfrak{p}}$ is a field and so $SL_{2}(D/{\mathfrak{p}})$ is generated by elementary matrices. It follows that

[TABLE]

and hence that $G({\mathfrak{q}}^{\prime})\leq H$.

Case 2: ${\mathfrak{p}}+{\mathfrak{q}}^{\prime}={\mathfrak{p}}.$

Note that here Lemma 1.1 applies to the quotient group $G({\mathfrak{q}}^{\prime})/G({\mathfrak{q}})$, since $({\mathfrak{q}}^{\prime})^{2}\subseteq{\mathfrak{q}}$. For all $x\in{\mathfrak{q}}_{\infty}$ and all $y\in{\mathfrak{q}}_{0}$, the element

[TABLE]

Now

[TABLE]

It follows that $H\cap G({\mathfrak{q}}^{\prime})$ contains elements

[TABLE]

for all $z\in{\mathfrak{q}}^{\prime}$ and also $S(z)$ and $T(z)$ (as mentioned before Case 1). This implies that $G({\mathfrak{q}}^{\prime})\leq H$, by Lemma 1.1.

If ${\mathfrak{q}}_{\infty}{\mathfrak{q}}_{0}+{\mathfrak{q}}\neq{\mathfrak{q}}^{\prime}$ we replace ${\mathfrak{q}}_{0}$, say, with ${\mathfrak{q}}^{*}$, divisible by ${\mathfrak{q}}_{0}$, where ${\mathfrak{q}}_{\infty}{\mathfrak{q}}^{*}+{\mathfrak{q}}={\mathfrak{q}}^{\prime}$, and repeat the above argument. $\Box$

An alternative proof of our second principal result (for the case $D={\mathbb{Z}}$) can be found in [St].

Theorem 2.10. Let $H$ be a congruence subgroup of $G$ and let

[TABLE]

Then

[TABLE]

Proof.* *Recall from Lemma 2.3 that ${\mathfrak{c}}(H,g)\supseteq{\mathfrak{q}}$, for all $g\in G$, where ${\mathfrak{q}}=l(H)$. We will assume that

[TABLE]

for all $g\in G$, and obtain a contradiction.

By Lemma 2.5 it is sufficient to prove the theorem for the case where

[TABLE]

(See Corollary 2.8.) Let ${\mathfrak{c}}(H,I_{2})={\mathfrak{q}}_{\infty}$ and let ${\mathfrak{c}}(H,g_{0})={\mathfrak{q}}_{0}$, where $g_{0}\in G$ is any element for which $g_{0}(\infty)=0$. ( For example, $g_{0}=T(-1)S(1)$.) By Corollary 2.8 it follows that ${\mathfrak{q}}_{\infty}\supseteq{\mathfrak{q}}_{0}$. Now ${\mathfrak{q}}={\mathfrak{q}}_{\infty}{\mathfrak{q}}_{\infty}^{\prime}={\mathfrak{q}}_{0}{\mathfrak{q}}_{0}^{\prime}$, say. Let ${\mathfrak{r}}={\mathfrak{q}}_{\infty}+{\mathfrak{q}}_{\infty}^{\prime}$. Then ${\mathfrak{q}}_{0}^{\prime}\supseteq{\mathfrak{r}}$, by Lemma 2.9.

By Lemma 2.6 we can choose $z\in D$ for prime ideals ${\mathfrak{p}}$ such that

• (i)

$z\in{\mathfrak{p}}$, when ${\mathfrak{p}}|{\mathfrak{r}},\;{\mathfrak{p}}\nmid{\mathfrak{q}}_{0}^{\prime}$,

• (ii)

$z\notin{\mathfrak{p}}$, when ${\mathfrak{p}}|{\mathfrak{q}}_{0}^{\prime}$.

(Possibly $z=1$.) Consider the $H$-cusp, ${\mathfrak{c}}(H,g_{z})$, where $g_{z}\in G$ is given by

[TABLE]

Then $g_{z}(\infty)=z$ and ${\mathfrak{c}}(H,g_{z})={\mathfrak{q}}_{z}$ is determined by the matrices

[TABLE]

where $x\in D$. Let ${\mathfrak{q}}={\mathfrak{q}}_{z}{\mathfrak{q}}_{z}^{\prime}$. If $t=T(z)$, then, by Lemma 2.5,

[TABLE]

Now $t^{-1}(\infty)=\infty$ and $t^{-1}g_{z}(\infty)=0$. Applying Lemma 2.9 therefore to $H^{t}$, it follows that ${\mathfrak{q}}_{z}^{\prime}\supseteq{\mathfrak{r}}$. (By definition $l(H)=l(H^{t})={\mathfrak{q}}$.) Note that by our initial hypothesis ${\mathfrak{q}}_{0}^{\prime},{\mathfrak{q}}_{z}^{\prime}\neq D$. There are two possibilities.

Case 1: ${\mathfrak{q}}_{0}^{\prime}+{\mathfrak{q}}_{z}^{\prime}\neq D.$

Choose a prime divisor ${\mathfrak{p}}$ of ${\mathfrak{q}}_{0}^{\prime}+{\mathfrak{q}}_{z}^{\prime}$. Let ${\mathfrak{q}}={\mathfrak{p}}{\mathfrak{q}}^{\prime}$. Then $({\mathfrak{q}}^{\prime})^{2}\subseteq{\mathfrak{q}}$, since ${\mathfrak{p}}$ divides ${\mathfrak{q}}_{\infty}$ and ${\mathfrak{q}}_{\infty}^{\prime}$. It is clear that ${\mathfrak{q}}^{\prime}$ is divisible by ${\mathfrak{q}}_{\infty},{\mathfrak{q}}_{0},{\mathfrak{q}}_{z}$. It follows that $T(x),S(x),Q(x)\in H$, for all $x\in{\mathfrak{q}}^{\prime}$.

Now by definition ${\mathfrak{p}}+(z)=D$ and so ${\mathfrak{q}}+(z){\mathfrak{q}}^{\prime}={\mathfrak{q}}^{\prime}$. Hence, for all $q\in{\mathfrak{q}}^{\prime}$, there exist $q^{\prime}\in{\mathfrak{q}}^{\prime}$ such that $zq^{\prime}\equiv q\;(\mathop{\mathrm{mod}}{\mathfrak{q}})$. As in the proof of Lemma 2.9 (Case 2) it follows from Lemma 1.3 that $G({\mathfrak{q}}^{\prime})\leq H$, which contradicts Corollary 1.4.

Case 2: ${\mathfrak{q}}_{0}^{\prime}+{\mathfrak{q}}_{z}^{\prime}=D.$

Choose a prime ${\mathfrak{p}}$ dividing ${\mathfrak{q}}_{z}^{\prime}$. Since ${\mathfrak{q}}_{z}^{\prime}\supseteq{\mathfrak{r}}$ and ${\mathfrak{p}}+{\mathfrak{q}}_{0}^{\prime}=D$, it follows that $z\in{\mathfrak{p}}$. As before let ${\mathfrak{q}}={\mathfrak{p}}{\mathfrak{q}}^{\prime}$. Then ${\mathfrak{q}}^{\prime}\subseteq{\mathfrak{q}}_{z}$, so $Q(x)\in H$ and

[TABLE]

for all $x\in{\mathfrak{q}}^{\prime}$. It follows that $S(x)\in H$, for all $x\in{\mathfrak{q}}^{\prime}+{\mathfrak{q}}_{0}$. But ${\mathfrak{q}}^{\prime}+{\mathfrak{q}}_{0}\neq{\mathfrak{q}}_{0}$, since ${\mathfrak{q}}_{0}\nsupseteq{\mathfrak{q}}^{\prime}$, which contradicts the maximality of ${\mathfrak{c}}(H,g_{0})$. This completes the proof. $\Box$

From Theorem 2.10 we obtain in particular the following necessary condition for a subgroup to be a congruence subgroup.

Corollary 2.11. *If $H$ is a congruence subgroup of $G$, then there exists some $g\in G$ with $l(H)={\mathfrak{c}}(H,g)$.

*All the results in this section are, of course, motivated by the classical modular group. It is appropriate therefore that we include a detailed account of how they apply to this important special case.

Example 2.12. Cusp amplitudes were first introduced for finite index subgroups of the modular group $SL_{2}({\mathbb{Z}})$ (motivated by the theory of modular forms). In this case the cusp $\infty$ is ${\mathbb{P}}^{1}({\mathbb{Q}})$, since ${\mathbb{Z}}$ is a PID. Let $H$ be a subgroup of finite index in $G$ containing $\pm I_{2}$. For each $g\in G$, the ${\mathbb{Z}}$-ideal ${\mathfrak{c}}(H,g)$ is non-zero, generated by a (unique) positive integer $n(H,g)$, say. Since ${\mathbb{Z}}^{*}=\{\pm 1\}$ it follows that in this case

[TABLE]

where $z=g(\infty)$. The double coset space $H\backslash G/G_{\infty}$ is finite, represented by elements $g_{1},\cdots,g_{t}$, say, of $G$. Let $n_{i}=n(H,g_{i})$, where $1\leq i\leq t$. We assume that $n_{1}\leq\cdots\leq n_{t}$. The sequence $(n_{1},\cdots,n_{t})$ is called the cusp-split of $H$ and satisfies the important cusp-split equation

[TABLE]

If $n_{0}(\geq 1)$ is a generator of the (${\mathbb{Z}}$-)ideal $l(H)$, then $n_{0}=\mathrm{lcm}\{n_{1},\cdots,n_{t}\}$, by Lemma 2.3.

Suppose now that $H$ is a congruence subgroup of index $\mu$ in $SL_{2}({\mathbb{Z}})$. Then, by Corollary 2.8 and Theorem 2.10,

[TABLE]

There are two immediate consequences, namely,

[TABLE]

and

[TABLE]

(See Corollary 1.4.) The inequality $(*)$ can be used [St] to obtain an upper bound for the number of congruence subgroups of $SL_{2}({\mathbb{Z}})$ of bounded index.

Among the index $7$ subgroups of $SL_{2}({\mathbb{Z}})$ are those with cusp-splits $(3,4)$ and $(2,5)$. (See [St].) They are non-congruence by Corollary 2.8 or Theorem 2.10 (or inequality $(*)$). On the other hand these results provide only necessary conditions. For example, it is known [AS] that there exist subgroups with cusp-splits $(1,6)$ and $(1,1,7)$ which, despite being consistent with Corollary 2.8 and Theorem 2.10, are non-congruence since neither satisfies $(**)$. However non-congruence subgroups do exist for which Corollary 2.8, Theorem 2.10 and $(**)$ all hold. It is known [AS] there exists a non-congruence subgroup with cusp-split $(8)$.

As previously emphasised we wish to demonstrate explicitly that our principal results, Corollary 2.8 and Theorem 2.10, do extend to $D$ which are not of arithmetic type.

Example 2.13. Let $D=k[t]$, the polynomial ring over a field $k$. Then $D$ is of arithmetic type if and only if $k$ is finite. Let ${\mathfrak{p}}$ be a prime $k[t]$-ideal.

We define

[TABLE]

Now $T(x)\in G_{0}({\mathfrak{p}})$ if and only if $x\in k+{\mathfrak{p}}$. Let $g\in G$. Then $z=g(\infty)=a/b$, where $a,b\in k[t]$, with $(a,b)=1$. If $U(a,b;x)\in G_{0}({\mathfrak{p}})$, then $xb^{2}\equiv 0\;(\mathop{\mathrm{mod}}{\mathfrak{p}})$. There are two possibilities.

**(i): $b\notin{\mathfrak{p}}$.

**In this case $x\in{\mathfrak{p}}$ and hence ${\mathfrak{c}}(G_{0}({\mathfrak{p}}),g)={\mathfrak{p}}$.

**(ii): $b\in{\mathfrak{p}}$.

**In this case $a\notin{\mathfrak{p}}$ and $U(a,b;x)\equiv T(-xa^{2})\;(\mathop{\mathrm{mod}}{\mathfrak{p}})$. Thus $U(a,b;x)\in G_{0}({\mathfrak{p}})$ if and only if $x\in c^{2}k+{\mathfrak{p}}$, where $ac\equiv 1\;(\mathop{\mathrm{mod}}{\mathfrak{p}})$. We are reduced to two possible outcomes.

If $\mathop{\mathrm{dim}}_{k}(k[t]/{\mathfrak{p}})>1$, then

[TABLE]

for all $g\in G$. We note that $G_{0}({\mathfrak{p}})$ is an example of a non-normal subgroup all of whose cusp-amplitudes are equal.

Suppose then that $\mathop{\mathrm{dim}}_{k}(k[t]/{\mathfrak{p}})=1$ (equivalently, $k[t]/{\mathfrak{p}}\cong k$). (This always happens when $k$ is algebraically closed.) It is easily verified that there are precisely two $(G_{0}({\mathfrak{p}}),S_{\infty})$ double cosets in $G=SL_{2}(k[t])$ and that $\{T(1),S(1)\}$ is a complete set of representatives for the $G_{0}({\mathfrak{p}})$-cusps, $G_{0}({\mathfrak{p}})\backslash\widehat{k(t)})$. It follows that

[TABLE]

3. Quasi-amplitudes

Definition. As usual let $H$ be a subgroup of $G$. We define the quasi-amplitude

[TABLE]

Clearly the cusp amplitude ${\mathfrak{c}}(H,g)$ is the biggest $D$-ideal contained in ${\mathfrak{b}}(H,g)$.

Note that, in contrast to ${\mathfrak{c}}(H,g)$, the quasi-amplitude ${\mathfrak{b}}(H,g)$ does not just depend on the $H$-cusp containing $g(\infty)$. More precisely it is easily verified that

[TABLE]

for some $u\in D^{*}$. It is clear that if $H$ is normal in $G$, then ${\mathfrak{b}}(H,g_{1})={\mathfrak{b}}(H,g_{2})$, for all $g_{1},g_{2}\in G$.

We put

[TABLE]

It is clear that, for all $g\in G$, the subgroup $(H^{g})_{\infty}$ normalizes $U$, so that $U\cdot(H^{g})_{\infty}$ is a subgroup of $G$. We then have the following one-one correspondences

[TABLE]

Definition. If $H$ is of finite index in $G$ we define

[TABLE]

Our next result is a generalization of the cusp split formula for subgroups of $SL_{2}({\mathbb{Z}})$.

Theorem 3.1. Let $H$ be a subgroup of finite index in $G$. If $\{g_{\lambda}(\infty):\lambda\in\Lambda\}$ is a complete set of representatives for the $H$-cusps in $O_{\infty}$, then

[TABLE]

Proof.* *We note that, for all $g\in G$,

[TABLE]

The proof follows from Lemma 2.2. $\Box$

Example 3.2. Let ${\mathcal{C}}={\mathbb{F}}_{9}[t]$. Then

[TABLE]

So there exists a (non-normal) congruence subgroup $H$ of $SL_{2}({\mathbb{F}}_{9}[t])$ of index $6$ and level $(t)$.

Obviously $(t)\subseteq{\mathfrak{b}}(H,I_{2})\subseteq{\mathbb{F}}_{9}[t]$. Actually, both inclusions are proper: ${\mathfrak{b}}(H,I_{2})$ cannot equal ${\mathbb{F}}_{9}[t]$ since $H/\pm SL_{2}({\mathcal{C}},(t))\cong A_{5}$ does not contain a subgroup of order $9$; on the other hand

[TABLE]

so ${\mathfrak{b}}(H,I_{2})$ cannot equal $(t)$. In particular, this shows that ${\mathfrak{b}}(H,I_{2})$ is not an ${\mathbb{F}}_{9}$-subspace of ${\mathbb{F}}_{9}[t]$, although, being an additive group, it is clearly an ${\mathbb{F}}_{3}$-subspace. All in all we see that

[TABLE]

where $\beta$ is an element from ${\mathbb{F}}_{9}^{*}$.

Let $\zeta\in{\mathbb{F}}_{9}$ be a primitive $8$-th root of unity. Then $H$ cannot contain any matrices of the form

[TABLE]

since $H/\pm SL_{2}({\mathcal{C}},(t))$ has no elements of order $4$. Hence $|G_{\infty}:H_{\infty}|$ is divisible by $2$. Together with $|{\mathcal{C}}:{\mathfrak{b}}(H,I_{2})|=3$ we see that $|G_{\infty}:H_{\infty}|=6$ and hence that $\infty$ is the only cusp of $H$.

Here we see the problem discussed earlier. The matrix

[TABLE]

represents the same cusp as $I_{2}$, namely $\infty$, but

[TABLE]

where $i$ is a primitive $4$-th root of unity in ${\mathbb{F}}_{9}$.

Proposition 3.3.

• a)

If $q>3$, then $SL_{2}({\mathbb{F}}_{q}[t])$ has no proper, normal subgroup of finite index with only one cusp.

• b)

If $q\leq 3$, then for every positive integer $e$ there are uncountably many normal subgroups of index $q^{e}$ in $SL_{2}({\mathbb{F}}_{q}[t])$ that have only one cusp.

Proof. a) If $N$ is a normal subgroup of finite index in $G$ with only one cusp, then Theorem 3.1 shows that the Borel group $G_{\infty}$ contains a system of coset representatives for $N$ in $G$. Thus $G/N$ is a semidirect product of an elementary abelian $p$-group where $p=char({\mathbb{F}}_{q})$ and a cyclic group of order $m$ where $m|(q-1)$. If $m>1$, then $G/N$ contains a normal subgroup of index $m$, in contradiction to the fact that the minimal index of a normal subgroup in $G$ is $|PSL_{2}({\mathbb{F}}_{q})|$. Compare [MSch1, Theorem 6.2]. If $m=1$, then $G/N$ contains a normal subgroup of index $p$, again contradicting the same minimal index.

b) Recall Nagao’s Theorem

[TABLE]

Compare for example [Se2, exercise 2, p.88]. We define

[TABLE]

Since for $q\in\{2,3\}$ all diagonal matrices are in the center of $SL_{2}({\mathbb{F}}_{q}[t])$, this amalgamated product shows that the identity on $V$ extends to a surjective group homomorphism $\phi:SL_{2}({\mathbb{F}}_{q}[t])\to V$ whose kernel is the normal subgroup generated by $SL_{2}({\mathbb{F}}_{q})$. Now for each of the uncountably many ${\mathbb{F}}_{q}$-subspaces $W$ of codimension $q^{e}$ in $V$ the inverse image $\phi^{-1}(W)$ is a normal subgroup of $SL_{2}({\mathbb{F}}_{q}[t])$ of index $q^{e}$, and Theorem 3.1 shows that it has only one cusp. $\Box$

The following lemma will be required later on.

Lemma 3.4. *Let $H$ be a congruence subgroup of $G$ of level ${\mathfrak{q}}$. Assume that $x\in{\mathfrak{b}}(H,I_{2})$. Then for every $\alpha\in D$ that is invertible modulo ${\mathfrak{q}}$ there exists an element $g\in G$ with $\alpha^{2}x\in{\mathfrak{b}}(H,g)$.

*Proof. Since $\alpha$ is invertible mod ${\mathfrak{q}}$, there exists $\delta\in D$ such that $\alpha\delta-1=:\gamma\in{\mathfrak{q}}$. Let $g^{-1}={\alpha\ 1\choose\gamma\ \delta}\in G$. Then $g^{-1}Hg$ contains

[TABLE]

Moreover,

[TABLE]

is an element of $G({\mathfrak{q}})$, which is a subgroup of $g^{-1}Hg$. Hence $T(\alpha^{2}x)\in g^{-1}Hg$.

$\Box$

We conclude this section with two examples which show that in general neither Corollary 2.8 nor Theorem 2.10 hold for the quasi-amplitudes of a congruence subgroup. If $H$ is any subgroup of $G$, we denote the subgroup $\langle H,-I_{2}\rangle$ by $\pm H$.

Example 3.5. Here we use $G$ to denote the Bianchi group $SL_{2}({\mathcal{O}}_{d})$, where ${\mathcal{O}}_{d}$ is the ring of integers in the imaginary quadratic number field ${\mathbb{Q}}(\sqrt{-d})$. We will assume that $d\equiv\ 1(\mathop{\mathrm{mod}}\ 3)$, which ensures that ${\mathfrak{q}}_{0}=(3)$ is a prime ideal in $D={\mathcal{O}}_{d}$. Then $D/{\mathfrak{q}}_{0}\cong{\mathbb{F}}_{9}$ and

[TABLE]

Hence there exists a (non-normal) subgroup $H$ of $G$ for which

[TABLE]

So $H$ is a congruence subgroup of level ${\mathfrak{q}}_{0}$ and index $6$.

Let $g\in G$. Then ${\mathfrak{q}}_{0}\subseteq{\mathfrak{b}}(H,g)\subseteq D$. Suppose that ${\mathfrak{b}}(H,g)=D$. Then $T(x)\in H^{g}$, for all $x\in D$, which implies that $A_{5}$ has a subgroup of order $9$. Hence ${\mathfrak{b}}(H,g)\neq D$. On the other hand, by Theorem 3.1,

[TABLE]

It follows that $|D:{\mathfrak{b}}(H,g)|=3$.

Now suppose moreover that $d\neq 1$; then $D^{*}=\{\pm 1\}$. With the above notation therefore, there exists $\lambda\in\Lambda$ such that ${\mathfrak{b}}(H,g)=\pm{\mathfrak{b}}(H,g_{\lambda})$. By Theorem 3.1 we deduce that

[TABLE]

Hence there are precisely two $(H,G_{\infty})$-double cosets in $G$. Then

[TABLE]

for some $b\in D\setminus{\mathfrak{q}}_{0}$. For a representative of the other double coset we fix $\alpha\in D$ which maps onto a primitive root of ${\mathbb{F}}_{9}\cong D/{\mathfrak{q}}_{0}$. By Lemma 3.4 there exists $g_{0}\in G$ for which

[TABLE]

Then $g_{0}$ can represent the other double coset. We conclude that

[TABLE]

Clearly this set has neither a minimum nor maximum member under set-theoretic containment.

Example 3.6. For our second example of this type we return to Example 2.13. Let ${\mathfrak{p}}$ be a prime ideal in $k[t]$, where $k$ is a field. We recall that

[TABLE]

From Example 2.13 it follows that, for all $g\in G_{0}$,

[TABLE]

for some $a\in k[t]$. It is clear that ${\mathfrak{b}}_{0}(a)={\mathfrak{b}}_{0}(b)$, if $a\equiv b\;(\mathop{\mathrm{mod}}{\mathfrak{p}})$, and that ${\mathfrak{b}}_{0}(a)={\mathfrak{p}}$ if and only if $a\in{\mathfrak{p}}$. We now restrict our attention to those ${\mathfrak{b}}_{0}(a)$ for which $a\notin{\mathfrak{p}}$.

Let $K=k[t]/{\mathfrak{p}}$. For each pair $x,y\in K^{*}$ we write

[TABLE]

Then $\rho$ is an equivalence relation on $K^{*}$. Let $\{a_{\omega}:\omega\in\Omega\}$ be a subset of $k[t]$ which maps bijectively onto a complete set of representatives for the $\rho$-classes. It follows that

[TABLE]

This set always has a minimal member, i.e. ${\mathfrak{p}}$. However it has a maximal member only when it reduces to $\{k+{\mathfrak{p}},{\mathfrak{p}}\}$. This happens only when $x^{2}\in k^{*}$, for all $x\in K^{*}$. It is easy to find examples of $K$ without this property when $k$ is not algebraically closed.

4. Quasi-level

Definition. Let $H$ be a subgroup of $G$. We define the quasi-level of $H$ as

[TABLE]

Since ${\mathfrak{b}}(H,g)={\mathfrak{b}}(H^{g},I_{2})$, we see that

[TABLE]

where $N_{H}$ is the core of $H$ in $G$.

However, Example 3.2 shows that in contrast to Lemma 2.3 the intersection over $g\in G$ can in general not be replaced by the intersection over a system of representatives of the cusps (even when $H$ is a congruence subgroup).

The second equality shows that our definition of quasi-level coincides with the one we gave in [MSch2]. Actually $ql(H)={\mathfrak{b}}(N_{H},g)$ for any $g\in G$. But Example 3.5 shows that in contrast to Corollary 2.11 there is in general no $g\in G$ for which $ql(H)={\mathfrak{b}}(H,g)$ (even when $H$ is a congruence subgroup).

We summarize the basic properties:

Lemma 4.1.

• (i)

$ql(H)=\{d\in D:T(d)\in N_{H}\}.$**

• (ii)

$ql(H)$* is an additive subgroup of $D$ with the property that*

[TABLE]

• (iii)

$ql(H)\supseteq l(H).$**

• (iv)

$ql(H)=ql(H^{g})=ql(N_{H})={\mathfrak{b}}(N_{H},g)$* for all $g\in G$.*

• (v)

$l(H)=l(H^{g})=l(N_{H})={\mathfrak{c}}(N_{H},g)$* for all $g\in G$.*

• (vi)

$l(H)$* is the largest $D$-ideal contained in $ql(H)$.*

• (vii)

$|D:ql(H)|\leq|SL_{2}(D):N_{H}|$.

For congruence subgroups we can combine Lemmas 3.4 and 4.1 to obtain the following extension of Lemma 4.1(ii).

Lemma 4.2. Let $H$ be a congruence subgroup and let $\alpha\in D$ be invertible modulo $l(H)$. Then

[TABLE]

We will also show that when $H$ is a congruence subgroup the inequality in Lemma 4.1(iii) becomes an equality (in “most” cases). For this purpose we require a number of preliminaries.

Lemma 4.3. Let $H$ be a congruence subgroup of $G$ and let $l(H)={\mathfrak{q}}={\mathfrak{q}}_{1}{\mathfrak{q}}_{2}$, where ${\mathfrak{q}}_{1}+{\mathfrak{q}}_{2}=D$. Then

[TABLE]

Proof.* *Let the required level be ${\mathfrak{q}}_{2}^{\prime}$. Then ${\mathfrak{q}}_{2}^{\prime}\supseteq{\mathfrak{q}}_{2}$. Now

[TABLE]

and so

[TABLE]

Hence ${\mathfrak{q}}\supseteq{\mathfrak{q}}_{1}{\mathfrak{q}}_{2}^{\prime}$ by Corollary 1.4. The result follows. $\Box$

Lemma 4.4. Let $N$ be a normal congruence subgroup of level ${\mathfrak{q}}={\mathfrak{q}}_{1}{\mathfrak{q}}_{2}$, where ${\mathfrak{q}}_{1}+{\mathfrak{q}}_{2}=D$. Let $N_{0}=(N\cap G({\mathfrak{q}}_{1}))\cdot G({\mathfrak{q}}_{2})$ and $\overline{N}=N\cdot G({\mathfrak{q}}_{2})$. Then

[TABLE]

Proof.* *Now $G=G({\mathfrak{q}}_{1})\cdot G({\mathfrak{q}}_{2})$, by Corollary 1.3. It follows that

[TABLE]

The result follows. $\Box$

Our next lemma is almost certainly well-known. In the absence of a reference we provide a proof.

Lemma 4.5. *Let $L$ be a local ring for which $2\in L^{*}$. Then $PSL_{2}(L)$ has trivial centre. *

Proof.* *Note that, since $2\in L^{*}$, the only involutions in $L^{*}$ are $\pm 1$. Let

[TABLE]

Then, if $g$ maps into the centre of $PSL_{2}(L)$, it follows that, for all $x\in L$,

[TABLE]

where $\lambda^{2}=\mu^{2}=1$. If $\gamma\in L^{*}$, then $\lambda=1$ and so $\gamma x=0$. Thus $\gamma\notin L^{*}$ and similarly $\beta\notin L^{*}$. We deduce that $\alpha,\delta\in L^{*}$. From the first of the above equations it follows that $\alpha=\lambda(\alpha+x\gamma)$ and hence that $2x\alpha\gamma+x^{2}\gamma^{2}=0$. The latter equation holds for $x=\pm 1$ and so $4\alpha\gamma=0$. From the above $\gamma=0$ and similarly $\beta=0$. It then follows from the above that $\alpha=\delta$, i.e. $g=\pm I_{2}$. $\Box$

Before coming to our next principal result we make another definition.

Definition. For a subgroup $H$ of $G=SL_{2}(D)$ we define $o(H)$ as the ideal of $D$ generated by all elements $a-d,\ b,\ c$ with ${a\ b\choose c\ d}\in H$. Somewhat unfortunately, $o(H)$ is sometimes called the order of $H$. Every matrix in $H$ is congruent modulo $o(H)$ to a scalar matrix $xI_{2}$ for some $x\in D$, and $o(H)$ is the smallest ideal of $D$ with this property. Obviously

[TABLE]

Conversely, for each $D$-ideal ${\mathfrak{q}}$ we define

[TABLE]

Then

[TABLE]

So it is clear that if

[TABLE]

then

[TABLE]

**Theorem 4.6. **Let $H$ be a congruence subgroup of $SL_{2}(D)$ such that $l(H)={\mathfrak{q}}$ satisfies Condition L (from the Introduction). Then

[TABLE]

*equivalently, the quasi-level is actually an ideal. *

Proof.* *By Lemma 4.1(iv), (v) we may assume that $H=N_{H}$ (i.e. $H\unlhd G$). From the above it is sufficient to prove that $N_{H}\leq Z({\mathfrak{q}})$, and hence that $N_{H}\leq Z({\mathfrak{p}}^{\alpha})$, where ${\mathfrak{p}}$ is any prime ideal for which $\alpha=\mathrm{ord}_{{\mathfrak{p}}}({\mathfrak{q}})>0$.

Let $L$ denote the local ring $D/{\mathfrak{p}}^{\alpha}$ and let

[TABLE]

denote the natural map. Now $SL_{2}(L)$ is generated by elementary matrices by [K, Theorem 1] and so (again by [K, Theorem 1])

[TABLE]

for all $D$-ideals ${\mathfrak{r}}$, where $\overline{{\mathfrak{r}}}$ is the image of ${\mathfrak{r}}$ in $L$. (In particular $\pi$ is an epimorphism.)

Let ${\mathfrak{q}}={\mathfrak{q}}^{\prime}{\mathfrak{p}}^{\alpha}$. Now suppose that $\pi(N_{H}\cap G({\mathfrak{q}}^{\prime}))$ is not central in $SL_{2}(L)$. Then by [K, Theorem 3] the hypotheses on ${\mathfrak{q}}$ and the above ensure that

[TABLE]

for some $\beta<\alpha$ which contradicts Lemma 4.3 (with ${\mathfrak{q}}_{2}={\mathfrak{p}}^{\alpha}$ and ${\mathfrak{q}}_{1}={\mathfrak{q}}^{\prime}$). It follows that

[TABLE]

The map $\pi$ extends to an epimorphism

[TABLE]

By Lemma 4.4 $\;\overline{\pi}(N_{H})$ is central in $PSL_{2}(L)$. We now apply Lemma 4.5 to conclude that

[TABLE]

$\Box$

For normal subgroups we can reformulate Theorem 4.6 as follows.

Corollary 4.7. Let $N$ be a normal congruence subgroup of $G$. If the level of $N$ satisfies Condition L, then

[TABLE]

*that is, the quasi-amplitudes of $N$ are actually the cusp amplitudes. *

**Proof. Since $N$ is normal any quasi-amplitude is equal to the quasi-level and any cusp amplitude is equal to the level. $\Box$

Remark 4.8. A word of warning is in order here. If $ql(H)=l(H)$ for a (non-normal) congruence subgroup $H$ and $H$ has only one cusp, this does **not imply ${\mathfrak{b}}(H,g)={\mathfrak{c}}(H,g)$, not even if $l(H)$ satisfies Condition L. See Example 3.2. Ultimately the problem is caused by diagonal matrices that are not central.

McQuillan [Mc, Theorem 1] has proved, for the special case $D={\mathbb{Z}}$, that, if $N$ is a normal congruence subgroup of $G$ of level ${\mathfrak{q}}$, then $N\leq Z({\mathfrak{q}})$, using a similar approach. We now provide a pair of examples to show that both restrictions in Theorem 4.6 are necessary.

Example 4.9. Our first example [M1, Example 2.3] shows that Theorem 4.6 can fail when ${\mathfrak{q}}$ is not prime to $2$. Let ${\mathfrak{p}}$ be a prime $D$-ideal for which $2\in{\mathfrak{p}}^{2}$. We recall from Lemma 1.1 that

[TABLE]

where the latter group is additive. Let $\Lambda=\{t^{2}+{\mathfrak{p}}^{4}:t\in{\mathfrak{p}}\}$. We define a subgroup $K$, where $G({\mathfrak{p}}^{4})\leq K\leq G({\mathfrak{p}}^{2})$, by

[TABLE]

Since $2{\mathfrak{p}}^{2}\subseteq{\mathfrak{p}}^{4}$, it is easily verified that $K$ is a subgroup of $G$, normalized by $S(x),T(x)$, for all $x\in D$. Since $K$ is also normalized by $G({\mathfrak{p}}^{4})$, it follows from Theorem 1.2 that $K$ is normal in $G$.

Clearly $ql(K)=\{t^{2}+q:t\in{\mathfrak{p}},q\in{\mathfrak{p}}^{4}\}$. Now suppose that $l(K)\neq{\mathfrak{p}}^{4}$. Then $G({\mathfrak{p}}^{3})\leq K$. Let $h$ be a generator of ${\mathfrak{p}}^{3}\;(\mathop{\mathrm{mod}}{\mathfrak{p}}^{4})$. It follows that there exists $k\in D$ such that $h\equiv k^{2}\;(\mathop{\mathrm{mod}}{\mathfrak{p}}^{4})$. We conclude that $l(K)={\mathfrak{p}}^{4}$.

Explicit examples that satisfy the requirement $2\in{\mathfrak{p}}^{2}$ are, among others, $D={\mathbb{Z}}[\sqrt{-2}]$ with ${\mathfrak{p}}=(\sqrt{-2})$, or to take a local example, $D={\mathbb{Z}}_{2}[\sqrt{2}]$ with ${\mathfrak{p}}=(\sqrt{2})$. More generally, this example actually works for any Dedekind domain of characteristic $2$ with any nonzero prime ideal ${\mathfrak{p}}$ because then trivially $2=0\in{\mathfrak{p}}$.

Example 4.10. Suppose that ${\mathfrak{q}}={\mathfrak{m}}_{1}{\mathfrak{m}}_{2}$, where ${\mathfrak{m}}_{1}+{\mathfrak{m}}_{2}=D$ and $|D/{\mathfrak{m}}_{i}|=3\;(i=1,2)$. (Consider, for example, $D={\mathbb{Z}}[\sqrt{-2}]$ with ${\mathfrak{m}}_{1}=(1+\sqrt{-2})$ and ${\mathfrak{m}}_{2}=(1-\sqrt{-2})$ or $D={\mathbb{F}}_{3}[t]$ with ${\mathfrak{m}}_{1}=(t)$ and ${\mathfrak{m}}_{2}=(t+1)$.) Then, by Corollary 1.3(ii),

[TABLE]

From the well-known structure of $SL_{2}({\mathbb{F}}_{3})$ it follows that there exists a normal subgroup, $N$, of $G$, containing $G({\mathfrak{q}})$, such that

[TABLE]

Now let

[TABLE]

Since $9=3^{2}$, $M\unlhd G$ and $|G:M|=3$. Obviously, $ql(M)$ contains $1$. If $1\in l(M)$, then $l(M)=D$, in which case $M=G(D)=G$. Thus $ql(M)\neq l(M)$.

In particular, there exists a normal congruence subgroup of index $3$ in $SL_{2}({\mathbb{F}}_{3}[t])$ that has level $t(t+1)$.

Remarks 4.11.

• a)

If the level of a congruence subgroup $H$ is a prime ideal ${\mathfrak{p}}$, then $ql(H)=l(H)$. This follows immediately from the simplicity of the group $PSL_{2}(D/{\mathfrak{p}})$, when $|D/{\mathfrak{p}}|>3$. The cases for which $|D/{\mathfrak{p}}|\leq 3$ can be checked directly.

• b)

If $D$ is any arithmetic Dedekind domain, the quasi-level of a congruence subgroup is not “too far from” its level. For a normal congruence subgroup $N$ the relation between $l(N)$ and $o(N)$ is described in [M2, Theorems 3.6, 3.10 and 3.14]. See also the end of Section 3 of [M2].

For example, for a finite index subgroup $H$ of $SL_{2}({\mathbb{Z}}[\sqrt{11}])$ we obtain

[TABLE]

from [M2, Theorem 3.6] since $2$ is ramified and $3$ is inert in ${\mathbb{Z}}[\sqrt{11}]$. Actually even $4ql(H)\subsetneq l(H)$ since $ql(N_{H})=o(N_{H})$ would mean that $ql(H)$ is an ideal and hence equal to $l(H)$.

Our final result demonstrates that for a non-congruence subgroup there is in general almost no connection between its quasi-level and level (in contrast with Theorem 4.6). We note that there is no proper normal subgroup of $SL_{2}(k[t])$ whose quasi-level is $k[t]$. (Since $k[t]$ is a Euclidean ring, $SL_{2}(k[t])$ is generated by $T(r),S(r)$, where $r\in k[t]$.)

Theorem 4.12. Let $k$ be any field and let $f\in k[t]$ with $\deg(f)\geq 2$. Suppose that $f(0)\neq 0$ and, further, that $f^{\prime}(0)\neq 0$, when $\deg(f)=2$. Then there exists a normal non-congruence subgroup $N$ of $SL_{2}(k[t])$ of level $(f)$ with the following properties.

• (i)

$N\cdot SL_{2}(k)=SL_{2}(k[t])$.

• (ii)

$l(N)=(f)$.

• (iii)

$ql(N)$* has $k$-codimension $1$ in $k[t]$.*

Proof.* *We note that, by hypothesis, $t\nmid f$. We define the $k$-subspace

[TABLE]

where $d=\deg(f)$. Let $N=\varDelta(Q)$ be the normal subgroup of $SL_{2}(k[t])$ generated by all $T(q)$, where $q\in Q$. Since $SL_{2}(k[t])$ is generated by all $T(r),S(r)$, where $r\in k[t]$, part (i) follows. In addition

[TABLE]

by [M3, Theorem 3.8]. Part (iii) follows. Suppose that $l(\varDelta(Q))\neq(f)$. Then

[TABLE]

for some polynomial divisor $h$ of $f$, with $\deg(h)<\deg(f)$. Then by the definition of $Q$, $h$, and hence $f$, must be divisible by $t$. Part (ii) follows.

Finally, suppose that $N$ is a congruence subgroup. Since, by hypothesis $t$ is prime to $l(N)=(f)$,

[TABLE]

by Lemma 4.2. If $\deg(f)>2$, then $t^{d}\in Q$. If $\deg(f)=2$, then $t^{3}\in Q$. Now for this case $tf\in Q$ and so $t^{2}\in Q$, by the extra hypothesis. In either case $(t)\subseteq Q$, which implies that $Q=k[t]$, a contradiction. The proof is complete. $\Box$

Remarks 4.13.

• a)

Obviously the group $N$ in Theorem 4.12 shows that Lemmas 3.4 and 4.2 do not hold in general for non-congruence subgroups, even if they are normal.

• b)

The restriction on the degree of $f$ in Theorem 4.12 is necessary. It is well-known that, if $\deg(f)\leq 1$, then every subgroup of $SL_{2}(k[t])$ of level $(f)$ is a congruence subgroup.

• c)

Several versions of Theorem 4.12 are already known for subgroups of level zero. (See Section 4 of [MSch2].)

5. Level and index

We now make use of quasi-amplitudes to extend the index/level inequality $(*)$ for finite index subgroups of $SL_{2}({\mathbb{Z}})$ (in Example 2.12) to other arithmetic Dedekind domains.

First, let $K$ be an algebraic number field of degree $d$ over ${\mathbb{Q}}$. We recall the definition of ${\mathcal{O}}_{S}$ from Section 1, where $S$ is a suitable, not necessarily finite, set of places of $K$.

Lemma 5.1. Let $L$ be a subgroup of ${\mathcal{O}}_{S}$ such that ${\mathcal{O}}_{S}/L$ has exponent $e$. Then $L$ contains the ${\mathcal{O}}_{S}$-ideal generated by $e$.

In particular, every finite index subgroup $L$ of ${\mathcal{O}}_{S}$ contains an ${\mathcal{O}}_{S}$-ideal ${\mathfrak{a}}$ such that

[TABLE]

Proof. The first claim is obvious since $ea\in L$ for every $a\in{\mathcal{O}}_{S}$.

For the second claim let $|{\mathcal{O}}_{S}:L|=n$. Then $L$ contains $n{\mathcal{O}}_{S}$. Denote the ring of integers of $K$ by ${\mathcal{O}}_{K}$. If $n{\mathcal{O}}_{K}=\prod{\mathfrak{p}}_{i}^{e_{i}}$ is the decomposition of $n{\mathcal{O}}_{K}$ into a product of prime ideals of ${\mathcal{O}}_{K}$, then

[TABLE]

So

[TABLE]

[TABLE]

$\Box$

Theorem 5.2. Let $K$ be an algebraic number field with $[K:{\mathbb{Q}}]=d$ and let ${\mathcal{O}}_{S}$ be the ring of $S$-integers of $K$ (with $|S|$ not necessarily finite). If $H$ is a congruence subgroup of $SL_{2}({\mathcal{O}}_{S})$, then

[TABLE]

Proof. By Corollary 2.11 there exists $g\in SL_{2}({\mathcal{O}}_{S})$ with $l(H)={\mathfrak{c}}(H,g)$. Let $L$ be the quasi-amplitude ${\mathfrak{b}}(H,g)$. Then $|{\mathcal{O}}_{S}:L|\leq|SL_{2}({\mathcal{O}}_{S}):H|$, and the Theorem follows from Lemma 5.1. $\Box$

Remarks 5.3.

• a)

Theorem 5.2 also holds for normal non-congruence subgroups of $SL_{2}({\mathcal{O}}_{S})$ since then every cusp amplitude is equal to the level.

• b)

If $|S|>1$ and $S$ contains at least one real or non-archimedean place, or if $|S|=\infty$, then by [Se1] resp. Theorem 1.9 the inequality in Theorem 5.2 holds for all subgroups of finite index in $SL_{2}({\mathcal{O}}_{S})$.

• c)

Lubotzky [Lu, (1.6) Lemma] has given a version of Theorem 5.2 for more general algebraic groups using different methods.

Next we exhibit some examples for which the inequality in Theorem 5.2 is sharp.

Example 5.4. Let $K$ be a number field of degree $d$ over ${\mathbb{Q}}$. Let $D={\mathcal{O}}_{K}$ denote the ring of integers of $K$ (i.e. the ring of $S$-integers of $K$, where $S$ consists precisely of the archimedean places of $K$). We suppose that the ideal ${\mathfrak{q}}_{0}=(2)=2D$ splits into the product of $d$ distinct prime ideals in $D$. Let $G=SL_{2}(D)$. Then, using Corollary 1.3,

[TABLE]

where for $1\leq i\leq d$,

[TABLE]

Let ${\mathfrak{p}}$ be any prime ideal dividing ${\mathfrak{q}}_{0}$ and let ${\mathfrak{q}}_{0}={\mathfrak{p}}{\mathfrak{p}}^{\prime}$. Then, again by Corollary 1.3, under the first isomorphism above

[TABLE]

for some $j$. For each $i$, let $N_{i}$ be the normal subgroup of $P_{i}$ of order $3$. There exists an epimorphism

[TABLE]

such that $N=\mathrm{ker}\;\theta$ contains $N_{1}\times\cdots\times N_{d}$ but not any $P_{i}$. Let $M/G({\mathfrak{q}}_{0})$ be the inverse image of $N$ in $G/G({\mathfrak{q}}_{0})$. If $l(M)\neq{\mathfrak{q}}_{0}$, then, by Corollary 1.4, $M$ contains $G({\mathfrak{p}}^{\prime})$, for some prime divisor ${\mathfrak{p}}$ of ${\mathfrak{q}}_{0}$, which contradicts the above. Hence $l(M)={\mathfrak{q}}_{0}$. We conclude that $M$ is a normal congruence subgroup with

[TABLE]

By the way, $|G:M|=2$ also shows

[TABLE]

Remark 5.5. Actually, for each $d$ there are infinitely many number fields $K$ of degree $d$ that satisfy the condition in Example 5.4. This can be seen as follows:

By Dirichlet’s Theorem on primes in arithmetic progressions there are infinitely many odd primes that are congruent to $1$ modulo $d$. Pick two of these, say $p_{1}$ and $p_{2}$ with $p_{i}=m_{i}d+1$. Then $2^{m_{i}d}\equiv 1\ \mathop{\mathrm{mod}}\ p_{i}$ and hence $2^{m_{1}m_{2}d}\equiv 1\ \mathop{\mathrm{mod}}\ p_{1}p_{2}$. So by the decomposition law in cyclotomic fields the inertia degree of $(2)$ in the $p_{1}p_{2}$-th cyclotomic field ${\mathbb{Q}}(\zeta_{p_{1}p_{2}})$ divides $m_{1}m_{2}d$. Since ${\mathbb{Q}}(\zeta_{p_{1}p_{2}})$ has degree $m_{1}m_{2}d^{2}$ over ${\mathbb{Q}}$, the decomposition field of $(2)$ in ${\mathbb{Q}}(\zeta_{p_{1}p_{2}})$ is an abelian extension of ${\mathbb{Q}}$ whose degree is divisible by $d$. Thus it contains a subfield $K$ of degree $d$ over ${\mathbb{Q}}$ with the desired property.

Alternatively we could argue as follows: $Gal({\mathbb{Q}}(\zeta_{p_{1}p_{2}})/{\mathbb{Q}})\cong{\mathbb{Z}}/m_{1}d{\mathbb{Z}}\oplus{\mathbb{Z}}/m_{2}d{\mathbb{Z}}$ and the Frobenius at $2$ generates a cyclic subgroup; so its fixed field is an abelian extension of ${\mathbb{Q}}$ whose degree is divisible by $d$.

Taking two other primes $p_{1}$, $p_{2}$ one obtains a different field $K$ since ${\mathbb{Q}}(\zeta_{p_{1}p_{2}})$ and hence $K$ is unramified outside $p_{1}p_{2}$.

By an analogous proof for each $d$ there are infinitely many number fields $K$ of degree $d$ such that $(3)$ splits completely in ${\mathcal{O}}_{K}$. For these one can similarly construct a normal congruence subgroup of index $3$ in $SL_{2}({\mathcal{O}}_{K})$ with level $3{\mathcal{O}}_{K}$. (Compare Example 4.10 for a special case.)

Thus it looks like Theorem 5.2 is optimal. Also, it is not possible to prove a function field analogue of Theorem 5.2 along the same lines, as there cannot be a function field analogue of the inequality in Lemma 5.1. Even in ${\mathbb{F}}_{q}[t]$ one can construct additive subgroups $L$ of index $q$ such that the biggest ideal ${\mathfrak{a}}$ contained in $L$ is the zero ideal or any prescribed nontrivial ideal. (This is actually the key point for many constructions in [MSch2]. See also Theorem 4.12.)

However, somewhat surprisingly, under the condition that the level is prime to certain ideals one can give a relation between the level and the index (if finite) of a congruence subgroup of $SL_{2}(D)$ that is valid for any Dedekind domain $D$.

Theorem 5.6. Let $D$ be any Dedekind domain and $N$ a normal congruence subgroup of $SL_{2}(D)$. If the level of $N$ satisfies Condition L, then

[TABLE]

Proof. If $N$ is normal, the level is equal to any cusp amplitude. Under Condition L this cusp amplitude is equal to the quasi-amplitude by Corollary 4.7. So the result follows from Theorem 3.1. $\Box$

Examples 3.2 and 3.5 show that even for “relatively simple” rings like ${\mathbb{F}}_{9}[t]$ or ${\mathbb{Z}}[\sqrt{-13}]$ we cannot expect

[TABLE]

for non-normal congruence subgroups $H$ that satisfy Condition L. But of course we have

Corollary 5.7. Let $D$ be any Dedekind domain and $H$ a congruence subgroup of $SL_{2}(D)$ of index $n$. If the level of $H$ satisfies Condition L, then

[TABLE]

Proof. $H$ has the same level as its core in $SL_{2}(D)$, and the index of this core divides $n!$. $\Box$

Corollary 5.7 is stronger than Theorem 5.2 if $n$ is sufficiently small compared to $d$.

Examples 4.10 and 5.4 show that Condition L cannot simply be dropped in Theorem 5.6 and Corollary 5.7. But for arithmetic Dedekind domains $D$ one can prove somewhat weaker results than Theorem 5.6 and Corollary 5.7 for congruence subgroups that do not satisfy Condition L. The key is that by results of [M2] one can still control the relation between $l(N)$ and $ql(N)$. See our Remark 4.11 b). We content ourselves with the function field case. Since the case of characteristic $p>3$ is already fully covered by Theorem 5.6, we only have to deal with characteristic $2$ and $3$.

Theorem 5.8. Let $K$ be a global function field with constant field ${\mathbb{F}}_{q}$ and let $D={\mathcal{O}}_{S}$ where $|S|$ is finite. Let $N$ be a normal congruence subgroup of $SL_{2}(D)$.

• a)

If $char({\mathbb{F}}_{q})=3$, then

[TABLE]

where $r$ is the number of prime ideals ${\mathfrak{p}}$ of $D$ with $|D/{\mathfrak{p}}|=3$.

• b)

If $char({\mathbb{F}}_{q})=2$, then

[TABLE]

where $s$ is the number of prime ideals ${\mathfrak{p}}$ of $D$ with $|D/{\mathfrak{p}}|=2$.

Proof. We only give the proof for characteristic $2$ as they are almost the same.

By [M2, Theorem 3.14] (compare also the end of Section 3 in [M2]) we have ${\mathfrak{s}}^{2}(o(N))^{2}\leq l(N)$ where ${\mathfrak{s}}$ is the product of all prime ideals ${\mathfrak{p}}$ in $D$ with $|D/{\mathfrak{p}}|=2$. Hence $|D/l(N)|$ divides $4^{s}|D/o(N)|^{2}$, which divides $4^{s}|D/ql(N)|^{2}$ since $ql(N)$ is an ${\mathbb{F}}_{q}$-subspace of $o(N)$. As $N$ is normal, we have $ql(N)={\mathfrak{b}}(N,g)$, and the claim follows from Theorem 3.1. $\Box$

Remarks 5.9.

• a)

If $K$ is a number field and $|S|$ is finite, then using [M2, Theorem 3.6] and the end of Section 3 of [M2], for a normal congruence subgroup $N$ of $SL_{2}({\mathcal{O}}_{S})$ we obtain in the worst case

[TABLE]

where $d=|K:{\mathbb{Q}}|$.

• b)

Theorems 5.6 and 5.8 can be used to improve [MSch2, Proposition 4.5 b)] in the sense that the condition that ${\mathfrak{m}}$ is a maximal ideal is not really needed there. This condition was used to show that the constructed groups of level ${\mathfrak{m}}$ are non-congruence subgroups. But if $\deg({\mathfrak{m}})$ is big enough, then Theorem 5.6 (resp. Theorem 5.8) guarantees the non-congruence property.

Acknowledgements. Parts of this paper were developed during research visits of the second author at Glasgow University. The hospitality of their Mathematics Department is gratefully acknowledged. The final version was written while the second author was holding a visiting position at the National Center for Theoretical Sciences (NCTS) in Hsinchu, Taiwan. We are also grateful to Peter Vámos for pointing out the reference [Go] to us.