Generators and Relations for a discrete subgroup of SL 2 (C) SL 2 (C) joint work with A. del Río and E. Jespers Ann Kiefer Groups St Andrews 5th August 2013
Goal: Generators and Relations for a subgroup of the Hilbert Modular Group K = Q( d) with d a square-free positive integer { ω = 1+ d 1, if d 1 mod 4; d 0, with d 0 = 2, if d 1 mod 4. O = Z[ω] Goal of this work Finding generators and relations for SL 2 (O). 2 / 19
Motivation: Units in Group Rings 3 / 19
Motivation: Units in Group Rings G finite group. 3 / 19
Motivation: Units in Group Rings G finite group. Open Problem: Finding generators and relations for a subgroup of finite index of U(ZG). 3 / 19
Motivation: Units in Group Rings G finite group. Open Problem: Finding generators and relations for a subgroup of finite index of U(ZG). QG = n i=1 M n i (D i ), D i a division algebra, let O i be an order in D i for every 1 i n. 3 / 19
Motivation: Units in Group Rings G finite group. Open Problem: Finding generators and relations for a subgroup of finite index of U(ZG). QG = n i=1 M n i (D i ), D i a division algebra, let O i be an order in D i for every 1 i n. Finding generators and relations for U(ZG), up to commensurability, reduces to finding generators and relations for SL ni (O i ) for every 1 i n. 3 / 19
Exceptional Components An exceptional component is a component which is equal to a non-commutative division ring different from a totally definite quaternion algebra, M 2 (D) with D equal to Q or Q( d) with d > 0, M 2 (D) with D a totally definite rational quaternion algebra H(a, b, Q). 4 / 19
Exceptional Components An exceptional component is a component which is equal to a non-commutative division ring different from a totally definite quaternion algebra, M 2 (D) with D equal to Q or Q( d) with d > 0, M 2 (D) with D a totally definite rational quaternion algebra H(a, b, Q). Idea for some of these cases Discontinuous actions on hyperbolic spaces. 4 / 19
Discontinuous Actions and Fundamental Domains based on groups acting discontinuously on hyperbolic space H n 5 / 19
Discontinuous Actions and Fundamental Domains based on groups acting discontinuously on hyperbolic space H n Theorem Let X be a finitely compact metric space. A group Γ of isometries of X acts discontinuously on X if and only if it is discrete. 5 / 19
Discontinuous Actions and Fundamental Domains based on groups acting discontinuously on hyperbolic space H n Theorem Let X be a finitely compact metric space. A group Γ of isometries of X acts discontinuously on X if and only if it is discrete. Definition A fundamental domain of the discontinuous group Γ < Iso(X ) is a closed subset F X satisfying the following conditions: the boundary of F has Lebesgue measure 0, the members of {g(f ) g Γ} are mutually disjoint, and X = g Γ g(f). 5 / 19
Example: Action of SL 2 (Z) on H 2 SL 2 (Z) discrete in SL 2 (R) 6 / 19
Example: Action of SL 2 (Z) on H 2 SL 2 (Z) discrete in SL 2 (R) SL 2 (R) = group of isometries of H 2 6 / 19
Example: Action of SL 2 (Z) on H 2 SL 2 (Z) discrete in SL 2 (R) SL 2 (R) = group of isometries of H 2 H 2 = {Z = x + yi x, y R, y > 0} 6 / 19
Example: Action of SL 2 (Z) on H 2 SL 2 (Z) discrete in SL 2 (R) SL 2 (R) = group of isometries of H 2 H 2 = {Z = x + yi x, y R, y > 0} Action of SL 2 (Z) on H 2 ( ) a b Z = az + b c d cz + d H2 6 / 19
Fundamental Domain of SL 2 (Z) Fundamental Domain of SL 2 (Z) acting on H 2 7 / 19
The Poincaré Theorem Theorem (Poincaré) Let F be an exact, convex fundamental domain, which is a polyhedron, for a discrete group Γ of H n. Then Γ is generated by {g Γ F g(f) is a side of F}. Relations There are two kinds of relations pairing relations: established based on the sides of the polyhedron, cycle relations: established based on the edges of the polyhedron. 8 / 19
Example of SL 2 (Z) side parings: 9 / 19
Example of SL 2 (Z) side parings: ( ) 1 1, 0 1 9 / 19
Example of SL 2 (Z) side parings: ( ) 1 1, 0 1 ( ) 1 1, 0 1 9 / 19
Example of SL 2 (Z) side parings: ( ) 1 1, 0 1 ( ) 1 1, 0 1 ( ) 0 1. 1 0 9 / 19
Example of SL 2 (Z) side parings: ( ) 1 1, 0 1 ( ) 1 1, 0 1 ( ) 0 1. 1 0 SL 2 (Z) = ( ) ( ) 1 1 0 1, 0 1 1 0 9 / 19
Link to U(ZG) Exceptional Components where this may be applied: H ( a, b, Q( d) ) for d a square-free positive integer, M 2 (Q( d)) for d a square-free positive integer. 10 / 19
Link to U(ZG) Exceptional Components where this may be applied: Idea H ( a, b, Q( d) ) for d a square-free positive integer, M 2 (Q( d)) for d a square-free positive integer. these groups act as isometries via Möbius action on H 3 orders in these groups are discrete in SL 2 (C) discontinuous action on H 3 generators and relations via Poincaré theorem Has been done in joint work with E. Jespers, S. O. Juriaans, A. De A. E Silva, A. C. Souza Filho. 10 / 19
Fundamental Domain of H ( 1, 1, Z [ 1+ 23 2 ]) ( [ ]) Fundamental Domain of H 1, 1, Z 1+ 23 2 acting on H 3 11 / 19
More Difficult Context Other example of exceptional component D = H ( 1, 1, Z[ξ n ]) where ξ n is a n-th primitve root of unity. 12 / 19
More Difficult Context Other example of exceptional component D = H ( 1, 1, Z[ξ n ]) where ξ n is a n-th primitve root of unity. most easiest case U(Z(Q 8 C 7 )) U(H(Z[ξ 7 ])) 12 / 19
More Difficult Context Other example of exceptional component D = H ( 1, 1, Z[ξ n ]) where ξ n is a n-th primitve root of unity. most easiest case U(Z(Q 8 C 7 )) U(H(Z[ξ 7 ])) problem: U(H(Z[ξ 7 ])) SL 2 (Z[ξ 7 ]) 12 / 19
More Difficult Context Other example of exceptional component D = H ( 1, 1, Z[ξ n ]) where ξ n is a n-th primitve root of unity. most easiest case U(Z(Q 8 C 7 )) U(H(Z[ξ 7 ])) problem: U(H(Z[ξ 7 ])) SL 2 (Z[ξ 7 ]) SL 2 (Z[ξ 7 ]) not discrete in SL 2 (C), but 12 / 19
More Difficult Context Other example of exceptional component D = H ( 1, 1, Z[ξ n ]) where ξ n is a n-th primitve root of unity. most easiest case U(Z(Q 8 C 7 )) U(H(Z[ξ 7 ])) problem: U(H(Z[ξ 7 ])) SL 2 (Z[ξ 7 ]) SL 2 (Z[ξ 7 ]) not discrete in SL 2 (C), but we have discreteness in SL 2 (C) SL 2 (C) SL 2 (C) hence: discontinuous action on H 3 H 3 H 3 Question: Does the Poincaré Theory still work in this case? 12 / 19
Test Case K = Q( d) with d a square-free positive integer { ω = 1+ d 1, if d 1 mod 4; d 0, with d 0 = 2, if d 1 mod 4. O = Z[ω] If a O, a denotes its algebraic conjugate. We consider the group SL 2 (O). 13 / 19
Test Case K = Q( d) with d a square-free positive integer { ω = 1+ d 1, if d 1 mod 4; d 0, with d 0 = 2, if d 1 mod 4. O = Z[ω] If a O, a denotes its algebraic conjugate. We consider the group SL 2 (O). Embedding of SL 2 (O) in SL 2 (R) SL 2 (R) A SL 2 (O) ( A, A ) SL 2 (R) SL 2 (R) 13 / 19
Test Case K = Q( d) with d a square-free positive integer { ω = 1+ d 1, if d 1 mod 4; d 0, with d 0 = 2, if d 1 mod 4. O = Z[ω] If a O, a denotes its algebraic conjugate. We consider the group SL 2 (O). Embedding of SL 2 (O) in SL 2 (R) SL 2 (R) Lemma A SL 2 (O) ( A, A ) SL 2 (R) SL 2 (R) SL 2 (O) is discrete in SL 2 (R) SL 2 (R) and has thus a discontinuous action on H 2 H 2. 13 / 19
Action of SL 2 (O) on H 2 H 2 H 2 H 2 = {(Z 1, Z 2 ) Z i = x i + y i i, x i, y i R, y i > 0}. 14 / 19
Action of SL 2 (O) on H 2 H 2 H 2 H 2 = {(Z 1, Z 2 ) Z i = x i + y i i, x i, y i R, y i > 0}. ( ) (( ) ( a b a b a b A =, )) c d c d c d. 14 / 19
Action of SL 2 (O) on H 2 H 2 H 2 H 2 = {(Z 1, Z 2 ) Z i = x i + y i i, x i, y i R, y i > 0}. ( ) (( ) ( a b a b a b A =, )) c d c d c d. A(Z 1, Z 2 ) = ( az1 + b cz 1 + d, a Z 2 + b ) c Z 2 + d H 2 H 2 14 / 19
Problems 15 / 19
Problems Construction of a fundamental domain F? 15 / 19
Problems Construction of a fundamental domain F? possible constructions but no longer a polyhedron 15 / 19
Problems Construction of a fundamental domain F? possible constructions but no longer a polyhedron Concerning generators: find an adequate definition for a side of F. 15 / 19
Problems Construction of a fundamental domain F? possible constructions but no longer a polyhedron Concerning generators: find an adequate definition for a side of F. Concerning relations: find an adequate definition for an edge of F. 15 / 19
Problems Construction of a fundamental domain F? possible constructions but no longer a polyhedron Concerning generators: find an adequate definition for a side of F. Concerning relations: find an adequate definition for an edge of F. F has finitely many sides. 15 / 19
Problems Construction of a fundamental domain F? possible constructions but no longer a polyhedron Concerning generators: find an adequate definition for a side of F. Concerning relations: find an adequate definition for an edge of F. F has finitely many sides. Generalize the proof of the Poincaré theorem to this case. 15 / 19
Generators of SL 2 (O) ( ) 1 1 P 1 =, 0 1 ( ) 1 ω P 2 =, 0 1 ( ) ɛ0 0 P 3 = 0 ɛ 1, where ɛ 0 is the smallest unit of O greater 0 than 1. 16 / 19
Generators of SL 2 (O) ( ) 1 1 P 1 =, 0 1 ( ) 1 ω P 2 =, 0 1 ( ) ɛ0 0 P 3 = 0 ɛ 1, where ɛ 0 is the smallest unit of O greater 0 than 1. Theorem If O is a PID, there exists a finite set X 1, such that, SL 2 (O) = P 1, P 2, P 3, P c,d (c, d) X 1, where P c,d is constructed in the following ( way: ) for each a b (c, d) X 1 choose (a, b) so that P c,d = SL c d 2 (O). 16 / 19
Relations for SL 2 (O) Theorem Let F be the fundamental domain of SL 2 (O) as defined before. Then the following is a presentation of SL 2 (O): Generators: the side-pairing transformations of F, Relations: the pairing relations and the cycle relations. 17 / 19
Further Work Projects: 18 / 19
Further Work Projects: generalize the theory to K = Q( d) with O not PID, 18 / 19
Further Work Projects: generalize the theory to K = Q( d) with O not PID, no finite set X 1 any more 18 / 19
Further Work Projects: generalize the theory to K = Q( d) with O not PID, no finite set X 1 any more generalize the theory to more copies of hyperbolic (3 )space, 18 / 19
Further Work Projects: generalize the theory to K = Q( d) with O not PID, no finite set X 1 any more generalize the theory to more copies of hyperbolic (3 )space, more symmetries showing up 18 / 19
Further Work Projects: generalize the theory to K = Q( d) with O not PID, no finite set X 1 any more generalize the theory to more copies of hyperbolic (3 )space, more symmetries showing up attack the concrete problem of U(Z(Q 8 C 7 )). 18 / 19
Thank you for your attention. 19 / 19