On the Strong Density Conjecture for Integral Apollonian Circle Packings

ON THE STRONG DENSITY CONJECTURE FOR INTEGRAL APOLLONIAN CIRCLE PACKINGS JEAN BOURGAIN AND ALEX KONTOROVICH Abstract. We prove that a set of density one satisfies the Strong Density Conjecture for Apollonian Circle Packings. That is, for a fixed integral, primitive Apollonian gasket, almost every (in the sense of density) sufficiently large, admissible (passing local obstructions) integer is the curvature of some circle in the packing. Contents 1. Introduction2 2. Preliminaries I: The Apollonian Group and Its Subgroups5 3. Preliminaries II: Automorphic Forms and Representations 12 4. Setup and Outline of the Proof 16 5. Some Lemmata 22 6. Major Arcs 35 7. Minor Arcs I: Case q < Q0 38 8. Minor Arcs II: Case Q0 ≤ Q < X 41 9. Minor Arcs III: Case X ≤ Q < M 45 References 50 arXiv:1205.4416v1 [math.NT] 20 May 2012 Date: May 22, 2012. Bourgain is partially supported by NSF grant DMS-0808042. Kontorovich is partially supported by NSF grants DMS-1209373, DMS-1064214 and DMS-1001252. 1 2 JEAN BOURGAIN AND ALEX KONTOROVICH 1333 976 1584 1108 516 1440864 1077 1260 909 616 381 1621 436 1669 1581 772 1872 1261 204 1365 1212 1876 1741 669 156 253 1756 1624 376 1221 1384 1540 877 1317 525 876 1861 1861 700 1836 541 1357 901 589 85 1128 1144 1381 1660 1036 360 1629 1189 844 7961501 1216 309 468 189 877 1477 1624 661 1416 1732 621 1141 1285 1749 1821 1528 1261 1876 1020 1245 40 805 744 1509 781 1429 616 373 885 453 76 1861 1173 1492 912 1356 469 285 1597 1693 6161069 724 1804 1644 1308 1357 1341 316 1333 1384 861 996 460 1101 10001725 469 1284 181 1308 132 1168 1213 637 1204 205 1476 1536 412 1756 1564 709 1869 604 733 733 1501 1221 1792 1144 208 1645 981 1077 24 304 1069 120 1480 1405 1501 421 789 1180 420 892 1533 1573 1005 1084 556 316 685 1773 712 612 1884 376 1093 1165 888 1008 -11 1084 1084 1693 616 1504 1300 1128 1804 1528 1536 741 11891741 1389 901 517 1792 397 237 1540 157 1140 1824 1104 1725 628 1125 1197 61 388 1548 717 1044 1144 640 1669 1629 1869 952 1069 1264 336 453 1165 853 21 1612 1837 132 1284 1636 1392 829 1693 1168 805 1381 964 472 688 1384 780 1696 1405 616 237 1293 1504 472 1693 1176 544 213 1021 1560 1597 1644 1597 376 348 1237 1632 565 1800 1077 549 1141 861 1840 244 1293 1749 756 1396 748 565 1204 1645 880 28 997 1272 1836 349 765 1621 1581 1584 160 973 1893 1461 1461 853 1140 469 405 1048 948 96 1669 1120 1597 813 1540 1861 1524 733 1837 1221 1197 952 1876 52 1797 717 1165 624 516 253 1605 349 1636 925 216 1680 1080 1416 492 117 813 1216 1773 616 1525 1701 288 1228 1293 876 541 736 1396 1792 1584 Figure 1. The Apollonian gasket with root quadruple t v0 = (−11; 21; 24; 28) . 1. Introduction 1.1. The Strong Density Conjecture. Let P be an Apollonian circle packing, or gasket, see Fig.1. The number b(C) shown inside a circle C 2 P is its \bend" or curvature, that is, the reciprocal of its radius (the bounding circle has nega- tive orientation). Soddy [Sod37] first observed the existence of integral packings, meaning that there are gaskets P so that b(C) 2 Z for all C 2 P. Let B = BP := fb(C): C 2 Pg be the set of all bends in P. We call a packing primitive if gcd(B) = 1. From now on, we restrict our attention to a fixed primitive integral Apollonian gasket P. Lagarias, Graham et al [LMW02, GLM+03] initiated a detailed study of Diophantine properties of B, with two separate families of problems: studying B with multiplicity (that is, studying circles), or without multiplicity (studying the integers which arise). In the present paper, we are concerned with the latter (see e.g. [KO11, FS10] for the former). In particular, the following striking local-to-global conjecture for B + is given in [GLM 03]. Let A = AP denote the admissible integers, that is, those passing local (congruence) obstructions: A := fn 2 Z : n 2 B(mod q); for all q ≥ 1g: Conjecture 1.1 (Strong Density Conjecture [GLM+03], p. 37). If n 2 A and n 1, then n 2 B. That is, every sufficiently large admissible number is the bend of a curvature in the packing. INTEGRAL APOLLONIAN GASKETS 3 We restate the conjecture in the following way. For N ≥ 1, let B(N) := B \ [1;N]; and A (N) := A \ [1;N]: Then the Strong Density Conjecture is equivalent to the assertion that #B(N) = #A (N) + O(1); as N ! 1. (1.2) In her thesis, Fuchs [Fuc10] proved that A is a union of arithmetic progressions with modulus dividing 24, that is, n 2 A () n(mod 24) 2 A (24); (1.3) cf. Lemma 2.21. Hence obviously #A (24) #A (N) = · N + O(1): 24 In fact the local obstructions are easily determined from the so-called root quadruple v0 = v0(P); (1.4) that is, the column vector of the four smallest curvatures in B, as in Fig.1; see Remark 2.23. 1.2. Partial Progress and Statement of the Main Theorem. The history of this problem is as follows. The first progress towards (1.2) was already made in [GLM+03], who showed that #B(N) N 1=2: (1.5) Sarnak [Sar07] improved this to N #B(N) ; (1.6) (log N)1=2 and then Fuchs [Fuc10] showed N #B(N) : (log N)0:150::: Finally Bourgain and Fuchs [BF10] settled the so-called \Positive Den- sity Conjecture," that #B(N) N: The purpose of this paper is to prove that a set of density one (within the admissibles) satisfies the Strong Density Conjecture. 4 JEAN BOURGAIN AND ALEX KONTOROVICH Theorem 1.7. Let P be a primitive, integral Apollonian circle packing, and let A (N); B(N) be as above. Then there exists some absolute, effectively computable η > 0, so that as N ! 1, #B(N) = #A (N) + O(N 1−η): (1.8) 1.3. Methods. Our main approach is through the Hardy-Littlewood-Ramanujan circle method, combining two new ingredients. The first, applied to the major arcs, is effective bisector counting in infinite volume hyperbolic 3-folds, recently achieved by I. Vinogradov [Vin12], as well as the uniform spectral gap over congruence towers of such, established in [Var10, BV11, BGS09]. The second ingredient is the minor arcs analysis, inspired by that given recently by the first-named author in [Bou11], where it was proved that the prime curvatures in a packing constitute a positive proportion of the primes. (Obviously (1.8) implies that 100% of the admissible prime curvatures appear.) 1.4. Plan for the Paper. A more detailed outline of the proof, as well as the setup of some relevant exponential sums, is given in x4. Before we can do this, we need to recall the Apollonian group and some of its subgroups in x2, and also some technology from spectral and representation theory of infinite volume hyperbolic quotients, described in x3. Some lemmata are reserved for x5, the major arcs are estimated in x6, and the minor arcs are dealt with in xx7-9. 1.5. Notation. 2πix We use the following standard notation. Set e(x) = e and eq(x) = x e( q ). We use f g and f = O(g) interchangeably; moreover f g means f g f. Unless otherwise specified, the implied constants may depend at most on the packing P (or equivalently on the root quadruple v0), which is treated as fixed. The symbol 1{·} is the indi- cator function of the event {·}. The greatest common divisor of n and m is written (n; m), their least common multiple is [n; m], and !(n) denotes the number of distinct prime factors of n. The cardinality of a finite set S is denoted jSj or #S. The transpose of a matrix g is written gt. The prime symbol 0 in Σ 0 means the range of r(mod q) is r(q) restricted to (r; q) = 1. Finally, pjkq denotes pj j q and pj+1 - q. Acknowledgements. The authors are grateful to Peter Sarnak for illuminating discussions. INTEGRAL APOLLONIAN GASKETS 5 2. Preliminaries I: The Apollonian Group and Its Subgroups 2.1. Descartes Theorem and Consequences. Descartes' Circle Theorem states that a quadruple v of (oriented) curvatures of four mutually tangent circles lies on the cone F (v) = 0; (2.1) where F is the Descartes quadratic form: F (a; b; c; d) = 2(a2 + b2 + c2 + d2) − (a + b + c + d)2: (2.2) Note that F has signature (3; 1), and let 3 G := SOF (R) = fg 2 SL(3; R): F (gv) = F (v); for all v 2 R g be the real special orthogonal group preserving F . It follows immediately that for b; c and d fixed, there are two solutions a; a0 to (2.1), and a + a0 = 2(b + c + d): Hence we observe that a can be changed into a0 by a reflection, that is, t 0 t (a; b; c; d) = S1 · (a ; b; c; d) ; where the reflections 0−1 2 2 21 01 1 B 1 C B2 −1 2 2C S1 = B C ;S2 = B C ; @ 1 A @ 1 A 1 1 01 1 01 1 B 1 C B 1 C S3 = B C ;S4 = B C ; @2 2 −1 2A @ 1 A 1 2 2 2 −1 generate the so-called Apollonian group A = hS1;S2;S3;S4i : (2.3) 2 It is a Coxeter group, free save the relations Sj = I, 1 ≤ j ≤ 4.

Load more