Asia Pacific Mathematics Newsletter The Farey Structure of the Gaussian Integers Katherine Stange Arrange three circles so that every pair is mutu- ally tangent. Is it possible to add another tan- gent to all three? The answer, as described by Apollonius of Perga in Hellenistic Greece, is yes, and, indeed, there are exactly two solutions [12, Problem XIV, p.12]. The four resulting circles are called a Descartes quadruple, and it is impossible to add a fifth. There is a remarkable relationship between their four curvatures (inverse radii): 2(a2 + b2 + c2 + d2) = (a + b + c + d)2. Rene´ Descartes is first credited with this obser- vation, in correspondence with Princess Elizabeth of Bohemia in 1643 [4, p.49]. Nobel prize winning radiochemist Frederick Soddy published his own rediscovery in Nature, shortly before World War II, in the form of a poem [15], which begins: Fig. 1. In the top row, the first few stages of the iterative For pairs of lips to kiss maybe construction of an Apollonian circle packing. At bottom, an Involves no trigonometry. approximation of the finished packing, with curvatures shown. This not so when four circles kiss The outer circle has curvature −6, the sign indicating that its Each one the other three . interior is “outside.” If we choose three mutually tangent circles of in- which curvatures appear in an integral Apollo- teger curvatures a, b and c, then the two solutions nian circle packing. It is conjectured [5, 6] that of Apollonius correspond to the two solutions to the only obstructions are local: in other words, the resulting quadratic equation in d. If one of in a given packing, any sufficiently large integer these is integral, so is the other. Thus we discover not ruled out by a specific congruence restriction that, if we begin with a Descartes quadruple of modulo 24 will appear. This question has become integer curvatures, we can add a new circle of a demonstration piece for the newly emerging integer curvature. This quintuple contains several theory of thin groups, and sophisticated tools fresh quadruples of integer curvatures, and so have been brought to bear on its partial solution: follow other new integral circles, in the same namely, that a density one collection of such manner. Continuing in this way forever, we create integers appears [1, 2]. an integral Apollonian circle packing, an infinite Momentarily putting aside the geometry, the fractal arrangement of disjoint and tangent circles question is a recursive one: given a Descartes with integer curvatures, as in Fig. 1. quadruple of curvatures a, b, c, d, we obtain a new Related circle arrangements have appeared for integer d satisfying centuries, from Japanese sangaku temple art [10] d + d = 2(a + b + c). to Vi Hart’s popular YouTube videos [8]. But it is only in the past decade or two that number The much more classical question of the values theorists have begun to answer the question of represented by a quadratic form is very similar. If 10 December 2016, Volume 6 No 2 Asia Pacific Mathematics Newsletter −3 3 1 1 isometries are obtained by extending the action of PSL Z PSL Z 2 ( ) on the boundary. The action of 2 ( ) −2 1 2 1 0 1 on H has a fundamental region as delineated by −3 3 2 2 geodesics in Fig. 3. Also shown in that figure is 1 1 Conway’s topograph, now reincarnated as a sub- −1 1 set of the walls between images of this region: to −2 2 find it, choose the walls of finite length, i.e. those 3 0 3 R 1 1 1 not approaching the boundary . The regions of −2 2 the topograph now correspond to the cusps in −1 1 Q 3 3 the picture: one for each element of . (Each such region has a horocircle inscribed in it, a circle Fig. 2. A portion of Conway’s topograph. Each region is labelled by an element of Q. The parallelogram law, with respect to the tangent to all sides. The collection of such circles ∞ central wall, reads f ( − 1, 1) + f (1, 1) = 2(f (1, 0) + f (0, 1)). (discounting the one at ) is famously known as the Ford circles.) Also shown in Fig. 3, in green, is the orbit f is a quadratic form, then the parallelogram law of the geodesic line from 0 to ∞. The green states geodesics are in bijection with the unimodular + + − = + f (u v) f (u v) 2(f (u) f (v)). pairs of Q, by associating to each line its two In this way, we can generate the primitive values boundary points. This green structure could be of an integral binary quadratic form recursively termed a Farey fractal, as it illustrates the well- R from its values on any triple of primitive vectors known Farey subdivision of : beginning with the = / ∞ = / u, v, u + v ∈ Z2. two intervals created by 0 0 1 and 1 0, / / This observation led Conway to study these subdivide each interval (a b, c d) at its mediant + / + values in visual form on a topograph capturing (a c) (b d). Each green arc is the “top” of a the recursion [3]. Primitive vectors in Z2 (those hyperbolic triangle formed by this subdivision. with no common factor between their coordi- This is called the upper half plane Farey diagram in nates), considered up to sign, are in bijection with Hatcher’s rich treatment [9, Chapter 1]. Q := Q ∪ {∞}. The topograph is an infinite tree Now let us move up one dimension. Consider S H3 breaking the plane into regions corresponding to the upper half space , a model of sitting C = C ∪ {∞} the elements of Q, as in Fig. 2. In this picture, above its boundary, . The geodesic C two regions are adjacent (sharing a wall) if and planes are the hemispheres orthogonal to . The only if the corresponding elements a/b, c/d ∈ Q hyperbolic isometries of this model are exactly the C are unimodular, i.e. ad − bc = ±1. The recurrence extensions of the Mobius¨ transformations on , PSL C c c relation above relates the values of f on the four which can be expressed as 2 ( ) where regions surrounding any one wall (u and v on ei- is complex conjugation. PSL Z ther side, and u±v at either end). In Conway’s tree, The analogue of 2 ( ) of interest to us in PSL O O given two regions that share a boundary edge (are this setting is 2 ( K) where K is the ring of “tangent”), there are exactly two ways to choose integers of an imaginary quadratic field. This is a third so that all three are mutually “tangent”. a discrete subgroup of hyperbolic isometries, and Three such “mutually tangent” elements of Q are it has a fundamental domain, which is a volume called a superbasis. The values of an integral binary cut out by several geodesic planes. quadratic form on a superbasis determine all the There are many analogies to the upper half values through the recursion of the parallelogram plane. In H, the set Q was the orbit of the single law. cusp of the fundamental region. The number of PSL O These two questions — values of a quadratic cusps of the fundamental region of 2 ( K)is O form and curvatures in a packing — can be the class number of K. What is the analogue unified with a little hyperbolic geometry. Let of the Farey fractal? We take the orbit of one us consider H, the upper half plane. This is a geodesic plane. Restricting to the boundary C, model of H2, the hyperbolic plane, in which the this gives an orbit of circles under the collection PSL O geodesics run along Euclidean circles orthogonal of Mobius¨ transformations 2 ( K). A natural to its boundary, R := R ∪ {∞}. The hyperbolic choice is R = R ∪ {∞}: then we obtain a Schmidt December 2016, Volume 6 No 2 11 Asia Pacific Mathematics Newsletter H SL Z Fig. 3. A portion of the upper half plane . In shaded grey, the usual fundamental region for the action of 2 ( ). In black and red, the boundaries subdividing H into images of this fundamental region. In red, Conway’s topograph. In green, the orbit of the geodesic line from 0 to ∞ (which also breaks up H into fundamental triangles). [7, Theorem 6.1] and more generally [18, Theorem 1.3]). In other words, there is a subgroup of PSL O 2 ( K) which generates any integral Apollo- nian circle packing. The orbit of this subgroup is shown in Fig. 4. This subgroup is called the Apollonian group, and it is a so-called thin group, i.e. of infinite index in its Zariski closure. Similar thin groups appear in other Schmidt arrangements, giving rise to other Apollonian- like packings with integrality properties [18]. The Schmidt arrangements themselves reflect the arithmetic of their respective fields; for example, the Schmidt arrangement of K is connected if and O Fig. 4. The Schmidt arrangement of the Gaussian integers. The only if K is Euclidean [17, Theorem 1.5] (see + box between 0 and 1 i is shown, including only circles with Fig. 5). curvatures at most 20. The Schmidt arrangement is periodic under translation by Z[i]. The Apollonian strip packing (which is Now let us return to the question of the bounded by two horizontal lines through 0 and i) is highlighted integers represented by forms and curvatures. in black. First, consider H. The unimodular pair (a/b, c/d) has separation (distance between the elements) arrangement [17, 18], named for Asmus Schmidt’s 1/bd.