The Euclidean Algorithm in Circle/Sphere Packings Arseniy (Senia) Sheydvasser October 25, 2019 I Choose a circle C with center (x0; y0) and radius R. I To invert a point (x; y) through, measure the distance r between (x0; y0) and (x; y), and move (x; y) to distance R=r from (x0; y0) (along the same ray). Circle/Sphere Inversions 3 3 2 2 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 -1 -2 -2 I To invert a point (x; y) through, measure the distance r between (x0; y0) and (x; y), and move (x; y) to distance R=r from (x0; y0) (along the same ray). Circle/Sphere Inversions 3 3 2 2 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 -1 -2 -2 I Choose a circle C with center (x0; y0) and radius R. Circle/Sphere Inversions 3 3 2 2 1 1 -3 -2 -1 1 2 3 -3 -2 -1 1 2 3 -1 -1 -2 -2 I Choose a circle C with center (x0; y0) and radius R. I To invert a point (x; y) through, measure the distance r between (x0; y0) and (x; y), and move (x; y) to distance R=r from (x0; y0) (along the same ray). Circle/Sphere Inversions Circle/Sphere Inversions Question n Let Γ be a subgroup of M¨ob(R ), and S an n-sphere. What does the orbit Γ:S look like? Can we compute it effectively? Circle/Sphere Inversions Definition n M¨ob(R ) is the group generated by n-sphere reflections in n R [ f1g. Circle/Sphere Inversions Definition n M¨ob(R ) is the group generated by n-sphere reflections in n R [ f1g. Question n Let Γ be a subgroup of M¨ob(R ), and S an n-sphere. What does the orbit Γ:S look like? Can we compute it effectively? Motivation Question What analogs of the Apollonian circle packing are there? Motivation Question n What do hyperbolic quotient manifolds H =Γ look like? 3.0 2.5 2.0 1.5 1.0 0.5 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 SL(2; Z) SL(2; Z[i]) Motivation 0 M¨ob (R) SL(2; R)={±1g a b Let be a matrix in I c d M¨ob0( 2) SL(2; )={±1g R C SL(2; R) or SL(2; C). I z 7! (az + b)(cz + d)−1 is 0 3 M¨ob (R ) an orientation-preserving M¨obiustransformation. −1 0 4 z 7! (az + b)(cz + d) is M¨ob (R ) I an orientation-reversing . M¨obiustransformation. Accidental Isomorphisms Question n How do you even represent elements in M¨ob(R )? Accidental Isomorphisms Question n How do you even represent elements in M¨ob(R )? 0 M¨ob (R) SL(2; R)={±1g a b Let be a matrix in I c d M¨ob0( 2) SL(2; )={±1g R C SL(2; R) or SL(2; C). I z 7! (az + b)(cz + d)−1 is 0 3 M¨ob (R ) an orientation-preserving M¨obiustransformation. −1 0 4 z 7! (az + b)(cz + d) is M¨ob (R ) SL(2; H)={±1g I an orientation-reversing . M¨obiustransformation. Accidental Isomorphisms Question n How do you even represent elements in M¨ob(R )? 0 M¨ob (R) SL(2; R)={±1g a b Let be a matrix in I c d M¨ob0( 2) SL(2; )={±1g R C SL(2; R) or SL(2; C). I z 7! (az + b)(cz + d)−1 is 0 3 M¨ob (R ) ??? an orientation-preserving M¨obiustransformation. −1 0 4 z 7! (az + b)(cz + d) is M¨ob (R ) SL(2; H)={±1g I an orientation-reversing . M¨obiustransformation. ??? 3 I We'll consider the case n = 3, M¨ob(R ). I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. 3 I We'll consider the case n = 3, M¨ob(R ). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. 3 I We'll consider the case n = 3, M¨ob(R ). I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). Vahlen's Matrices I Vahlen, 1901: For any n, there is an isomorphism between n M¨ob(R ) and a group of 2 × 2 matrices with entries in a (subset of a) Clifford algebra, quotiented by {±1g. 3 I We'll consider the case n = 3, M¨ob(R ). I Define (w + xi + yj + zk)z = w + xi + yj − zk and H+ = quaternions fixed by z (i.e. with no k-component). z a b z z + z z SL (2; H) = 2 Mat(2; H) ab ; cd 2 H ; ad − bc = 1 c d Equivalently, z 0 k T 0 k SL (2; H) = γ 2 SL(2; H) γ γ = −k 0 −k 0 Inverses are given as follows: a b−1 dz −bz = c d −cz az What is SLz(2; H) as a Group? z a b z z + z z SL (2; H) = ab ; cd 2 H ; ad − bc = 1 c d Inverses are given as follows: a b−1 dz −bz = c d −cz az What is SLz(2; H) as a Group? z a b z z + z z SL (2; H) = ab ; cd 2 H ; ad − bc = 1 c d Equivalently, z 0 k T 0 k SL (2; H) = γ 2 SL(2; H) γ γ = −k 0 −k 0 What is SLz(2; H) as a Group? z a b z z + z z SL (2; H) = ab ; cd 2 H ; ad − bc = 1 c d Equivalently, z 0 k T 0 k SL (2; H) = γ 2 SL(2; H) γ γ = −k 0 −k 0 Inverses are given as follows: a b−1 dz −bz = c d −cz az 3 I Every orientation-preserving element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. 3 I Every orientation-reversing element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. M¨ob(R3) as SLz(2; H) 3 + I There is an action on R [ f1g = H [ f1g defined by a b :z = (az + b)(cz + d)−1 c d M¨ob(R3) as SLz(2; H) 3 + I There is an action on R [ f1g = H [ f1g defined by a b :z = (az + b)(cz + d)−1 c d 3 I Every orientation-preserving element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. 3 I Every orientation-reversing element of M¨ob(R ) can be written as z 7! (az + b)(cz + d)−1. I We will ask that Γ is arithmetic. I Note that SLz(2; H) can be seen as real solutions to a set of polynomial equations. I Roughly, an arithmetic group is the set of integer solutions to that set of polynomial equations. I Not quite true|can only define up to commensurability|but ignore that. Arithmetic Groups Definition What sort of subgroups Γ of SLz(2; H) should we consider? I We will ask that Γ is discrete. I We will ask that Γ carries some sort of algebraic structure. I Note that SLz(2; H) can be seen as real solutions to a set of polynomial equations. I Roughly, an arithmetic group is the set of integer solutions to that set of polynomial equations. I Not quite true|can only define up to commensurability|but ignore that. Arithmetic Groups Definition What sort of subgroups Γ of SLz(2; H) should we consider? I We will ask that Γ is discrete. I We will ask that Γ carries some sort of algebraic structure. I We will ask that Γ is arithmetic. I Not quite true|can only define up to commensurability|but ignore that. Arithmetic Groups Definition What sort of subgroups Γ of SLz(2; H) should we consider? I We will ask that Γ is discrete. I We will ask that Γ carries some sort of algebraic structure. I We will ask that Γ is arithmetic. I Note that SLz(2; H) can be seen as real solutions to a set of polynomial equations.