Poincaré Bisectors and Application to Units

Transcription

1 Poincaré Bisectors and Application to Units joint work with E. Jespers, S. O. Juriaans, A. De A. E Silva, A. C. Souza Filho Ann Kiefer July 15, 2011

2 Motivation

3 Motivation Open Problem: Finding generators of a subgroup of finite index of U(ZG).

4 Motivation Open Problem: Finding generators of a subgroup of finite index of U(ZG). Definition A simple algebra A is said to be of exceptional type if it is either a non-commutative division algebra (which is not a totally definite quaternion algebra), or M(2, Q), or M(2, Q( d)), d > 0, or M(2, D), where D is a rational quaternion division algebra H(a, b, Q). G finite group and QG no exceptional Wedderburn components methods to solve the problem. Idea for other cases Hyperbolic Geometry.

5 Hyperbolic Geometry

6 Hyperbolic Geometry Let H 3 be the upperhalf space model of hyperbolic 3-space, i.e. H 3 = {z + rj C + Rj : z C, r R + }. Let B 3 be the ball model of the hyperbolic 3-space, i.e. B 3 = {z + rj C + Rj z 2 + r 2 < 1}. There exists a bijective isometry η 0 : H 3 B 3. Iso + (H 3 ) = Iso + (B 3 ) = PSL(2, C).

7 Hyperbolic Geometry Let H 3 be the upperhalf space model of hyperbolic 3-space, i.e. H 3 = {z + rj C + Rj : z C, r R + }. Let B 3 be the ball model of the hyperbolic 3-space, i.e. B 3 = {z + rj C + Rj z 2 + r 2 < 1}. There exists a bijective isometry η 0 : H 3 B 3. Iso + (H 3 ) = Iso + (B 3 ) = PSL(2, C). Definition The ( action ) of PSL(2, C) on H 3 is given by a b (P) = (ap + b)(cp + d) c d 1, where (ap + b)(cp + d) 1 is calculated in the classical quaternion algebra.

8 Fundamental Domain

9 Fundamental Domain Definition A closed subset F H 3 is called a fundamental domain of the discontinuous group Γ < Iso(H 3 ) if the following conditions are satisfied: F meets each Γ-orbit at least once, F meets each Γ-orbit at most once, the boundary of F has Lebesgue measure zero.

10 Dirichlet Fundamental domain

11 Dirichlet Fundamental domain Take γ PSL(2, C) and c H 3 a point that is not fixed by γ, D γ (c) = {z H 3 ρ(z, c) ρ(z, γ(c))}.

12 Dirichlet Fundamental domain Take γ PSL(2, C) and c H 3 a point that is not fixed by γ, D γ (c) = {z H 3 ρ(z, c) ρ(z, γ(c))}. Dirichlet Fundamental Domain Γ PSL(2, C) discrete group, Γ c = {γ Γ γ(c) = c}, F c the fundamental domain of Γ c, D Γ (c) = γ Γ\Γ c D γ (c). Then F = F c D Γ (c) is a Dirichlet or Poincaré fundamental domain of Γ with centre c. Proposition If Γ is geometrically finite, Γ = Γ c, γ γ(f) F.

13 Bisector = Isometric Sphere

14 Bisector = Isometric Sphere Definition Let γ be a Möbius transformation. There is a unique sphere Σ Ψ(γ) in B 3, on which γ acts as an euclidean isometry in B 3. This sphere is called the isometric sphere.

15 Bisector = Isometric Sphere Definition Let γ be a Möbius transformation. There is a unique sphere Σ Ψ(γ) in B 3, on which γ acts as an euclidean isometry in B 3. This sphere is called the isometric sphere. Theorem Let γ SL(2, C) with γ / SU(2, C) and let Σ Ψ(γ) be the isometric sphere of γ. Then Σ Ψ(γ) = {u B 3 ρ(0, u) = ρ(u, Ψ(γ 1 )(0))}, the bisector of the geodesic segment linking 0 and Ψ(γ 1 )(0).

16 Bisector = Isometric Sphere Definition Let γ be a Möbius transformation. There is a unique sphere Σ Ψ(γ) in B 3, on which γ acts as an euclidean isometry in B 3. This sphere is called the isometric sphere. Theorem Let γ SL(2, C) with γ / SU(2, C) and let Σ Ψ(γ) be the isometric sphere of γ. Then Σ Ψ(γ) = {u B 3 ρ(0, u) = ρ(u, Ψ(γ 1 )(0))}, the bisector of the geodesic segment linking 0 and Ψ(γ 1 )(0). In general in H 3, η 1 0 (Σ Ψ(γ)) is a bisector but not an isometric sphere.

17 Bisectors in H 3

18 Bisectors in H 3 ( ) a b Let γ = PSL(2, C) with c 0. c d Proposition In H 3, Σ γ = η 1 0 (Σ Ψ(γ)) is the bisector of the geodesic linking j and γ 1 (j). Moreover 1. Σ γ is an Euclidean sphere if and only if a 2 + c 2 1. In this case, its center and its radius are respectively given by P γ = (ab+cd) a 2 + c 2 1, R2 γ = 1+ Pγ 2. a 2 + c 2 2. Σ γ is a plane if and only if a 2 + c 2 = 1. In this case Re(vz) + v 2 2 = 0, z C is a defining equation of Σ γ, where v = ab + cd.

19 Dirichlet Algorithm of Finite Covolume (DAFC)

20 Dirichlet Algorithm of Finite Covolume (DAFC) We work in B 3.

21 Dirichlet Algorithm of Finite Covolume (DAFC) We work in B Embed Γ in PSL(2, C). 2. Find a system of algebraic equations (X) whose set of solutions is Γ. 3. Take f : Γ R, where f (γ) = γ Take an increasing sequence (r n ) such that (r n ) = Im(f ). 5. For each n N draw all the bisectors of the elements of the set f 1 (r n ). 6. Stop when B n f (γ) r n0 IntΣ Ψ(γ 1 ), for some n 0 N. 7. Use η 1 0 to obtain a fundamental domain in H n of a subgroup of finite index.

22 SL 1 (H( 1, 1, o K ))

23 SL 1 (H( 1, 1, o K )) Let K = Q( d) with d 7 mod 8, o K be a maximal order in K, i.e. o K = Z[ 1+ d 2 ].

24 SL 1 (H( 1, 1, o K )) Let K = Q( d) with d 7 mod 8, o K be a maximal order in K, i.e. o K = Z[ 1+ d 2 ]. Proposition (well known) Γ = SL 1 (H( 1, 1, o K )) acts discretely and cocompactly on H 3.

25 SL 1 (H( 1, 1, o K )) Let K = Q( d) with d 7 mod 8, o K be a maximal order in K, i.e. o K = Z[ 1+ d 2 ]. Proposition (well known) Γ = SL 1 (H( 1, 1, o K )) acts discretely and cocompactly on H 3. F 0 Γ 0 = i, j = Q 8 and thus η 1 0 (F 0) = {z C z < 1, Im(z) > 0}.

26 SL 1 (H( 1, 1, o K )) Let K = Q( d) with d 7 mod 8, o K be a maximal order in K, i.e. o K = Z[ 1+ d 2 ]. Proposition (well known) Γ = SL 1 (H( 1, 1, o K )) acts discretely and cocompactly on H 3. F 0 Γ 0 = i, j = Q 8 and thus η 1 0 (F 0) = {z C z < 1, Im(z) > 0}. DAFC D Γ (0).

27 d = 15

28 d = 15 Figure: Fundamental domain of SL 1 (H( 1, 1, Z[ ]))

29 d = 23

30 d = 23 Figure: Fundamental domain of S 23 SL 1 (H( 1, 1, Z[ ])

31 Cases we are able to treat:

32 Cases we are able to treat: Lemma Let G be a finite group and A = H( 1, 1, Q[ d]), a non-split division algebra. Then A is not a Wedderburn component of QG.

33 Cases we are able to treat: Lemma Let G be a finite group and A = H( 1, 1, Q[ d]), a non-split division algebra. Then A is not a Wedderburn component of QG. But: Q( d)q 8 = 4Q( d) H(Q( d)) U(H(Z[ d])) (1 + 2H(Z[ d])) U(Z[ d]q 8 ).

34 Cases we are able to treat: Lemma Let G be a finite group and A = H( 1, 1, Q[ d]), a non-split division algebra. Then A is not a Wedderburn component of QG. But: Q( d)q 8 = 4Q( d) H(Q( d)) U(H(Z[ d])) (1 + 2H(Z[ d])) U(Z[ d]q 8 ). Group rings with Wedderburn component of type H(a, b, Q( d), M(2, Q( d)), d > 0,

35 G nilpotent

36 G nilpotent G a nilpotent finite group of nilpotency class n. QG does not have simple components that are non-commutative division algebras (except totally definite quaternion algebras) or M(2, H(Q)).

37 G nilpotent G a nilpotent finite group of nilpotency class n. QG does not have simple components that are non-commutative division algebras (except totally definite quaternion algebras) or M(2, H(Q)). Some special units in U(ZG) in this case: the bicylic units in ZG, b (n), where b 1 = b, with b a Bass cyclic unit in ZG and, for 2 i n, b (i) = g Z i b g (i 1), 1 + E 11 ge 22 and 1 + E 22 ge 11, g G, with G/N = Q 8 C n so that 2 has even order in U(Z n ) (with E 11 and E 22 some special matrix units in a Wedderburn component of QG of the form M(2, Q(ξ n ))).

38 Theorem (based on Ritter, Sehgal; Jespers, Leal) The group generated by the units mentioned before and the congruence groups Γ 2,N (mo) with N a normal subgroup of G so that 1. O = Z, m = 8 N and G/N = D 8, 2. O = Z[ 2], m = 8 N and G/N = D 16, 3. O = Z[i], m = 8 N and G/N = D + 16, 4. O = Z[ 3], m = 24 N and G/N = D 8 C 3, 5. O = Z[ 3], m = 24 N and G/N = Q 8 C 3, 6. O = Z[i], m = 32 N and G/N = D +, is of finite index in U(ZG).

39 Γ 2 (2Z[i])

40 Γ 2 (2Z[i]) ( ) ( 1 2i, 0 1 ( 1 + 2i i is of finite index in Γ 2 (2Z[i]). ) ( 1 0, 2i 1 ), ) ( ) 1 0,, 2 1 ( ) 1 2i 2i 2i 1 + 2i

41 Γ 2 (2Z[i]) ( ) ( 1 2i, 0 1 ( 1 + 2i i is of finite index in Γ 2 (2Z[i]). ) ( 1 0, 2i 1 ), ) ( ) 1 0,, 2 1 ( ) 1 2i 2i 2i 1 + 2i

42 Γ 2 (2Z[ 2])

43 Γ 2 (2Z[ 2]) ( ) ( ) ( ) ,,..., , 2 is of finite index in Γ 2 (2Z[ 2]).

44 Γ 2 (2Z[ 2]) ( ) ( ) ( ) ,,..., , 2 is of finite index in Γ 2 (2Z[ 2]).

45 What is left? For G a finite nilpotent group, it remains to deal with simple components of type H(Q(ξ n )) (n a positive integer so that 2 has odd order modulo n), M(2, H(Q)).

46 Thank you for your attention.