Counting congruence subgroups

Abstract

Let Γ denote the modular group SL(2,Z) and Cn(Γ) the number of congruence subgproups of Γ of index at most n. We prove that lim n→∞ log Cn(Γ) (log n)2/ log log n = 3−2 √ 2 4 . Some extensions of this result for other arithmetic groups are presented as well as a general conjecture. §0. Introduction Let k be an algebraic number field, O its ring of integers, S a finite set of valuations of k (containing all the archimedean ones), and OS = { x ∈ k ∣∣ v(x) ≥ 0, ∀v ∈ S}. Let G be a semisimple, simply connected, connected algebraic group defined over k with a fixed embedding into GLd. Let Γ = G(OS) = G ∩ GLd(OS) be the corresponding S-arithmetic group. We assume that Γ is an infinite group. For every non-zero ideal I of OS let Γ(I) = Ker ( Γ → GLd(OS/I) ) . A subgroup of Γ is called a congruence subgroup if it contains Γ(I) for some I. For n > 0, define Cn(Γ) = # { congruence subgroups of Γ of index at most n } . Theorem 1. There exist two positive real numbers α− and α+ such that for all sufficiently large positive integers n n log n log log nα− ≤ Cn(Γ) ≤ n log n log log nα+ . This theorem is proved in [Lu], although the proof of the lower bound presented there requires the prime number theorem on arithmetic progressions in an interval where its validity depends on the GRH (generalized Riemann hypothesis for arithmetic progressions). The first two authors research is supported in part by the NSF. The third author’s Research is supported in part by OTKA T 034878. All three authors would like to thank Yale University for its hospitality. Typeset by AMS-TEX 1 2 DORIAN GOLDFELD ALEXANDER LUBOTZKY LASZLO PYBER In §2 below, we show that by appealing to a theorem of Linnik [Li1, Li2] on the least prime in an arithmetic progression, the proof can be made unconditional. Following [Lu] we define: α+(Γ) = lim logCn(Γ) λ(n) , α−(Γ) = lim logCn(Γ) λ(n) , where λ(n) = (log n) 2 log log n . It is not difficult to see that α+ and α− are independent of both the choice of the representation of G as a matrix group, as well as independent of the choice of S. Hence α± depend only on G and k. The question whether α+(Γ) = α−(Γ) and the challenge to evaluate them for Γ = SL2(Z) and other groups were presented in [Lu]. It was conjectured by Rademacher that there are only finitely many congruence subgroups of SL2(Z) of genus zero. This counting problem has a long history. Petersson [Pe, 1974] proved that the number of all subgroups of index n and fixed genus goes to infinity exponentially as n → ∞. Dennin [De, 1975] proved that there are only finitely many congruence subgroups of SL2(Z) of given fixed genus and solved Rademacher’s conjecture. It does not seem possible, however, to accurately count all congruence subgroups of index at most n in SL2(Z) by using the theory of Riemann surfaces of fixed genus. Here we prove: Theorem 2. α+(SL2(Z)) = α−(SL2(Z)) = 3−2 √ 2 4 = 0.0428932 . . . We believe that SL2(Z) represents the general case and we expect that α+ = α− for all groups. The proof of the lower bound in Theorem 2 is based on the Bombieri-Vinogradov Theorem [Bo], [Da], [Vi], i.e., the Riemann hypothesis on the average. The upper bound, on the other hand, is proved by first reducing the problem to a counting problem for subgroups of abelian groups and then solving that extremal counting problem. We will, in fact, show a more remarkable result: the answer is independent of O! Theorem 3. Let k be a number field with Galois group g = Gal(k/Q) and with ring of integers O. Let S be a finite set of primes, and OS as above. Assume GRH (generalized Riemann hypothesis) for k and all cyclotomic extensions k(ζ ) with a rational prime and ζ a primitive th root of unity. Then α+(SL2(OS)) = α−(SL2(OS)) = 3 − 2 √ 2 4 . The GRH is needed only for establishing the lower bound. It can be dropped in many cases by appealing to a theorem of Murty and Murty [MM] which generalizes the Bombieri– Vinogradov Theorem cited earlier. COUNTING CONGRUENCE SUBGROUPS 3 Theorem 4. Theorem 3 can be proved unconditionally for k if either (a) g = Gal(k/Q) has an abelian subgroup of index at most 4 (this is true, for example, if k is an abelian extension); (b) d = deg[k : Q] < 42. We conjecture that for every Chevalley group scheme G, the upper and lower limiting constants, α±(G(OS)), depend only on G and not on O. In fact, we have a precise conjecture, for which we need to introduce some additional notation. Let G be a Chevalley group scheme of dimension d = dim(G) and rank = rk(G). Let κ = |Φ+| denote the number of positive roots in the root system of G. Letting R = R(G) = d− 2 = κ , we see that R = +1 2 , (resp. , , −1, 3, 6, 6, 9, 15) if G is of type A (resp. B , C , D , G2, F4, E6, E7, E8). Conjecture. Let k,O, and S be as in Theorem 3, and suppose that G is a simple Chevalley group scheme. Then α+(G(OS)) = α−(G(OS)) = (√ R(R + 1) −R )2 4R2 . The conjecture reflects the belief that “most” subgroups of H = G(Z/mZ) lie between the Borel subgroup B of H and the unipotent radical of B. Our proof covers the case of SL2 and we are quite convinced that this will hold in general. For general G, we do not have such an in depth knowledge of the subgroups of G(Fq) as we do for G = SL2, yet we can still prove: Theorem 5. Let k,O, and S be as in Theorem 3. Let G be a simple Chevalley group scheme of dimension d and rank , and R = R(G) = d− 2 , then: (a) Assuming GRH or the assumptions of Theorem 4; α−(G(OS)) ≥ (√ R(R + 1) −R )2

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