Preamble

Functional Analysis and Its Applications, Vol. 2, No. 1, 1968, pp. 36–39.

Construction of $\Gamma$-Rational Groups for Certain Discrete Subgroups $\Gamma$ of the Group $SL(2, \mathbb{R})$

D. A. Kazhdan

We begin by recalling the definition of a $\Gamma$-rational group. Let $N$ be a group and $\Gamma$ be its subgroup. The $\Gamma$-rational group $\tilde{\Gamma}$ is defined as the subgroup of $N$ consisting of elements $n$ such that the subgroup $\Gamma \cap n\Gamma n^{-1}$ has finite index in both $\Gamma$ and $n\Gamma n^{-1}$. It is easily verified that such elements indeed form a group. Typically, the case of interest is when $N$ is a semisimple real group and $\Gamma$ is a discrete subgroup such that the quotient space $N/\Gamma$ has finite volume (see \cite{1}).

The $\Gamma$-rational group is easily identified when $\Gamma$ is an arithmetic group. Specifically, if $N = G_{\mathbb{R}}$ is the group of real points of an algebraic group $G$ defined over $\mathbb{Q}$, and $\Gamma$ is commensurable with $G_{\mathbb{Z}}$, then it is obvious that $\tilde{\Gamma} \supset G_{\mathbb{Q}}$. If the center of $G$ is trivial, then $\tilde{\Gamma} = G_{\mathbb{Q}}$ (see \cite{1}). Thus, for arithmetic groups $\Gamma$, their $\Gamma$-rational groups $\tilde{\Gamma}$ are dense in $N$. In \cite{1}, it was conjectured that the converse holds: if $\Gamma$ is a discrete subgroup of a semisimple group such that the volume of the quotient space is finite and the group $\tilde{\Gamma}$ is everywhere dense in $N$, then $\Gamma$ is an arithmetic subgroup.

This hypothesis will be tested for a certain class of discrete subgroups $\Gamma$ in $SL(2, \mathbb{R})$, which includes, for example, the Hecke groups. More precisely, we shall prove the following:

Theorem. Let $\Gamma$ be a discrete subgroup of $SL(2, \mathbb{R})$ such that the quotient space $SL(2, \mathbb{R})/\Gamma$ has finite volume but is not compact. Suppose the group $\tilde{\Gamma}$ consists of matrices whose elements are algebraic integers. Then the group $\tilde{\Gamma}$ is either discrete or conjugate to a subgroup of $SL(2, \mathbb{Z})$.

In the following, we shall always assume that all unipotent elements in $\Gamma$ are conjugate within $\Gamma$ to elements $\pm T_0^n$, where $n \in \mathbb{Z}$ and $T_0$ is a specific element in $\Gamma$. This assumption simplifies the proof slightly but is not essential. We recall several properties of the group $\Gamma$ following from conditions (a) and (b). Property (a) implies that $\Gamma$ contains unipotent elements, and there exist only finitely many distinct $\Gamma$-conjugacy classes of primitive elements (i.e., elements that are not powers of another element in $\Gamma$). Furthermore, property (a) implies that $\Gamma$ is finitely generated; consequently, property (b) implies that the elements of all matrices in $\Gamma$ lie in a number field $k$, which is a finite extension of $\mathbb{Q}$. From (a), it also follows that the field $k$ can be considered generated over $\mathbb{Q}$ by the traces of the matrices in $\Gamma$. The primary part of the proof involves constructing a canonical system of representatives for the elements of $\Gamma \backslash \tilde{\Gamma}$ and consists of several steps.

Construction of $\Gamma$-Rational Groups

Step I. The group $\tilde{\Gamma}$ consists of matrices with algebraic coefficients. More precisely, the image of any element $\gamma \in \tilde{\Gamma}$ in the group $PL(2, \mathbb{R})$ is a point defined over $\bar{\mathbb{Q}}$. (The group $PL(2, \mathbb{R})$ is viewed as the group of real points of an algebraic group defined over $\mathbb{Q}$, such that the projection $\pi: SL(2, \mathbb{R}) \to PL(2, \mathbb{R})$ is defined over $\mathbb{Q}$.) This assertion follows immediately from the fact that $\Gamma$ is a Zariski-dense subset of $SL(2, \mathbb{R})$ and that $PL(2, \mathbb{R})$ is the automorphism group of $SL(2, \mathbb{R})$.

Step II. For any element $\tau \in \tilde{\Gamma}$, there exists $\gamma \in \Gamma$ such that $\gamma \tau$ takes the form $\begin{pmatrix} 1 & \alpha \\ 0 & 1 \end{pmatrix}$, where $\alpha$ is a rational number. To prove this, consider the elements $\tau T_0 \tau^{-1}$. Since $\tau \in \tilde{\Gamma}$, there exists $n \in \mathbb{Z}, n \neq 0$, such that $\tau T_0^n \tau^{-1} \in \Gamma$. Therefore, there exists an element $\gamma \in \Gamma$ such that $\gamma \tau T_0^n \tau^{-1} \gamma^{-1} = T_0^m$. It follows directly that $\gamma \tau$ has the form $\alpha_0 = m/n$.

Step III. Consider the subgroup $\Gamma_A$ in $\tilde{\Gamma}$ consisting of matrices with algebraic integer coefficients. We will now prove that $\Gamma_A$, like $\Gamma$, is a discrete subgroup of $SL(2, \mathbb{R})$. We will show that $\Gamma_A$ is not everywhere dense in $SL(2, \mathbb{R})$, as it is obvious that any subgroup of $SL(2, \mathbb{R})$ containing $\Gamma$ is either discrete or everywhere dense. Consider an element $\gamma' \in \Gamma_A$. We have shown that there exists an element $\tau \in \Gamma$ such that $\tau \gamma'$ has the form $\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$. Since $\tau \gamma' \in \Gamma_A$, both $a$ and $a^{-1}$ are algebraic integers; thus $a = \pm 1$. By multiplying by $-I$, we can ensure $a = 1$. Thus, for any element $\gamma' \in \Gamma_A$, there exists $\tau \in \Gamma$ such that $\tau \gamma'$ has the form $\begin{pmatrix} 1 & \alpha \\ 0 & 1 \end{pmatrix}$. The discreteness of the group now follows from the following lemma.

Lemma. Let $\Gamma$ and $\Gamma'$ be two subgroups of $SL(2, \mathbb{R})$ such that:
a) $\Gamma$ is a discrete subgroup containing an element of the form $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$;
b) For any element $\tau' \in \Gamma'$, there exists an element $\tau \in \Gamma$ such that $\tau \tau'$ has the form $\begin{pmatrix} 1 & \alpha \\ 0 & 1 \end{pmatrix}$.
Then the subgroup $\Gamma'$ is not everywhere dense.

Proof. Consider the action of $SL(2, \mathbb{R})$ on the two-dimensional affine plane. From (a), it follows that the orbit of the point $(1,0)$ under $\Gamma$ is discrete. From (b), it follows that the orbit of $(1,0)$ under $\Gamma'$ coincides with its orbit under $\Gamma$ and is therefore also discrete. The lemma is proved.

Thus, we have shown that $\Gamma_A$ is discrete. Since $\Gamma \subset \Gamma_A$ and is a subgroup of finite index, all conditions imposed on $\Gamma$ are also satisfied for $\Gamma_A$. Therefore, we shall henceforth assume $\Gamma_A = \Gamma$.

Step IV. For each prime $p$, consider the subgroup $\Gamma_p$ in $\tilde{\Gamma}$ consisting of matrices whose coefficients have the form $a/p^k$, where $a$ is an algebraic integer and $k \in \mathbb{Z}$. We now prove that if $\tilde{\Gamma}$ is not a discrete subgroup of $SL(2, \mathbb{R})$, then there exists a prime $p$ such that $\Gamma_p$ is also not discrete.

We know that for any $\tau \in \tilde{\Gamma}$, there exists $\gamma \in \Gamma$ such that $\gamma \tau$ has the form $\begin{pmatrix} 1 & \alpha \\ 0 & 1 \end{pmatrix}$, where $\alpha = m/n$ is a rational number. It follows from the lemma that if $\tilde{\Gamma}$ is not discrete, there exists an element in $\tilde{\Gamma}$ of the form $\begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}$, where $c = m/n$, with $m$ and $n$ being coprime and $n > 1$.

1. Analysis of the Prime $p$

Let $p$ be a prime such that $n = p^k n'$, where $n' \not\equiv 0 \pmod{p}$ and $k > 0$. Then it is obvious that the elements $T_0 \tau^m T_0^{-1}$ lie in $\Gamma_p$ for all natural numbers $m$. Thus, it suffices to prove that if $\tilde{\Gamma}$ is not a subgroup of $SL(2, \mathbb{Z})$, then $\Gamma_p$ is discrete for all $p$. We now construct a system of representatives for $\Gamma \backslash \Gamma_p$. Every element can be brought to the form $\begin{pmatrix} a & b \\ 0 & a^{-1} \end{pmatrix}$ by left multiplication by an element $\gamma \in \Gamma$, where $a = p^k$. Let us fix an element $\gamma_p \in \Gamma_p$ of the form $\begin{pmatrix} p^k & b \\ 0 & p^{-k} \end{pmatrix}$ for which $k$ is minimal. Any element of $\Gamma_p$ can be represented as a product $\gamma_p^n \gamma \gamma_p^m$, where $n, m \in \mathbb{Z}$ and $\gamma$ has the form $\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$. We now prove that $x = j/p^k$ is such that the matrix $\begin{pmatrix} 1 & x \\ 0 & 1 \end{pmatrix}$ is a primitive unipotent element in $\Gamma$. Indeed, since $u \in \Gamma_p$, then $u = a/p^k$ for some algebraic integer $a$. But this implies $\Gamma_A = \Gamma$, meaning $k=0$. Thus, we have shown that in each left coset $\Gamma \backslash \Gamma_p$, there is an element of the form $\gamma_p^k u_l$, where $l$ is an integer such that $0 \le l < p^k$.

Step VI. We now introduce a norm $N$ on $\Gamma_p$, defined as the maximum of the $p$-adic norms of the matrix coefficients (the norm is chosen such that $\|p\|_p = 1/p$). Consider the subset $X_k$ in $\Gamma_p$ consisting of elements $\gamma_p$ such that $N(\gamma_p) = k$. The set $X_k$ is bi-invariant with respect to the subgroup $\Gamma$, and from the preceding arguments, it follows immediately that $X_k$ consists of no more than $2p^k$ left cosets.

We now proceed to the proof of the theorem. From the previous discussion, we only require the final result (Step VI). Since $\Gamma$ consists of matrices with algebraic integer coefficients from the field $k$, we can define congruence subgroups in $\Gamma$. We shall now study the completion $\hat{\Gamma}_p$ of $\Gamma$ with respect to these congruence subgroups. It is clear that $\hat{\Gamma}_p$ is a closed subgroup of $SL(2, \hat{O}_k)$, the completion of $SL(2, O_k)$ at $p$. It is also clear that $SL(2, \hat{O}_k)$ is isomorphic to the direct product of groups $SL(2, O_{\mathfrak{p}})$, where $\mathfrak{p}$ are the divisors of $p$ in $k$.

We now show that $\hat{\Gamma}_p$ is isogenous to a product of groups of type $SL(2, O_{\mathfrak{p}})$. Indeed, since $\hat{\Gamma}_p$ is a closed subgroup of the product $\prod SL(2, O_{\mathfrak{p}})$, it is isogenous to a direct product of a group $G_1$ of the same type, a solvable group $G_2$, and a $p$-adic group of rank zero $G_3$. Since the projection of $SL(2, \hat{O}_k)$ onto any factor is a monomorphism, the projection of $\hat{\Gamma}_p$ onto $G_i$ is a monomorphism for $i=1,2,3$. This implies $G_3 = \{e\}$, as a group of rank zero cannot contain unipotent elements, and $G_2 = \{e\}$, as $\Gamma$ is not solvable. Thus, $\hat{\Gamma}_p$ is a subgroup of $SL(2, \hat{O}_k)$ isomorphic to a direct product of groups of the form $SL(2, O_{\mathfrak{p}})$.

We shall prove that $\hat{\Gamma}_p$ is isomorphic to $SL(2, \mathbb{Z}_p)$ only if $\Gamma$ is a subgroup of $SL(2, \mathbb{Z})$. Suppose otherwise. Consider a Galois field $k_1$ containing $k$ and an automorphism $\sigma$ that is non-trivial on $k$. We have two representations of $\Gamma$ in $SL(2, \mathbb{C})$: the original $\phi$ and the representation $\phi^\sigma$ obtained via the automorphism. Since $k$ is generated over $\mathbb{Q}$ by the traces of $\Gamma$, these representations are inequivalent. Since the completion of $\Gamma_A$ coincides with the completion of $\Gamma$, we obtain two representations $\phi$ and $\phi_p$ of $\hat{\Gamma}_p$ such that the projections of $\phi(\hat{\Gamma}_p)$ and $\phi^\sigma(\hat{\Gamma}_p)$ onto any factor $SL(2, O_{\mathfrak{p}})$ are non-trivial. It follows that the restrictions of $\phi$ and $\phi_p$ to any subgroup of finite index in $\hat{\Gamma}_p$ are inequivalent, which is only possible if $\hat{\Gamma}_p \cong SL(2, \hat{O}_k)$.

Remark. One could also prove that $\hat{\Gamma}_p \cong SL(2, \hat{O}_k)$. Now consider the completion $\hat{\Gamma}$ of the group with respect to all congruence subgroups in $\Gamma$. It is easy to see that if $\hat{\Gamma}_p$ is isomorphic to a product of groups $SL(2, O_{\mathfrak{p}})$, then $\hat{\Gamma}$ is isomorphic to a product of groups $SL(2, k_{\mathfrak{p}})$, where $k_{\mathfrak{p}}$ are $p$-adic fields. If an element $\gamma_p \in \Gamma_p$ is such that $N(\gamma_p) = k$, then the norm of its projection onto any factor in $\hat{\Gamma}$ is also $k$. From this, and the fact that $\hat{\Gamma} \cong SL(2, \hat{O}_k)$ (in the case where $\Gamma \not\subset SL(2, \mathbb{Z})$), it follows that the closure $\bar{X}_k$ of the set $X_k$ in $\hat{\Gamma}$ consists of more than $2p^k$ left cosets relative to $\hat{\Gamma}_p$, as $X_k$ is bi-invariant under $\hat{\Gamma}_p$. This contradicts Step VI. The resulting contradiction proves the theorem.

Moscow State University
Received October 20, 1967

LITERATURE CITED

\cite{1} Piatetski-Shapiro, I. I., Shafarevich, I. R., "Galois theory of transcendental extensions and uniformization," Izv. Akad. Nauk SSSR Ser. Mat., 30 (1966).