Cosets of unimodular groups over Dedekind domains. Talk for the Research Seminar Computational Algebra and Number Theory in Düsseldorf

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1 Lehrstuhl A für Mathematik Marc Ensenbach Aachen, den 19. November 2008 Cosets of unimodular groups over Dedekind domains Talk for the Research Seminar Computational Algebra and Number Theory in Düsseldorf In this talk, a formula for counting right cosets of unimodular groups over Dedekind domains and some applications of this formula are presented. 1 D E T E R M I N A N TA L D I V I S O R T H E O RY O V E R D E D E K I N D D O M A I N S This first section presents the notions of norm-finite Dedekind domains (as the fundamental algebraic structure for this talk) and elementary divisors; furthermore, the notation used in this talk is fixed. 1.1 D e f i n i t i o n : A domain o is called Dedekind domain, if the following conditions are satisfied: (i) Every ideal a / {{0}, o} of o has a unique prime ideal decomposition a = p v 1 1 pv m m with pairwise distinct prime ideals p 1,..., p m of o and v 1,..., v m N 0. (ii) For ideals a, b of o there exists an ideal c of o with ac = b (in other words: a divides b, denoted by a b) if and only if a b. A Dedekind domain o is called norm-finite, if o/ao < holds for every a o. E x a m p l e s : The ring Z + Zω where ω = 5 is a norm-finite Dedekind domain. 1.2 In general, every full ring of integers of an algebraic number field is a norm-finite Dedekind domain. 1.3 N o tat i o n : In the following denote by o a norm-finite Dedekind domain and by K its field of fractions. Furthermore define G = GL 2 (K) and U = GL 2 (o) as well as I = o 2 2 G. 1

2 1.4 D e f i n i t i o n : Let A = ( a b c d ) I. Then one defines the first and second determinantal divisor of A by d 1 (A) = ao + bo + co + do and d 2 (A) = (det A)o, respectively. Furthermore, e 1 (A) := d 1 (A) and e 2 (A) := d 2 (A)d 1 (A) 1 are called the elementary divisors of A; and f 1 (A) := e 1 (A) = d 1 (A) and f 2 (A) := e 2 (A)e 2 (A) 1 = d 2 (A)d 1 (A) 2 are called the fundamental factors of A. The relation between determinantal divisors and (double) cosets of U is given in the following theorem. 1.5 Th e o r e m : Given A I and B I, then the following two assertions are equivalent: (i) UAU = UBU (where UAU := {PAQ P, Q U}). (ii) d 1 (A) = d 1 (B) and d 2 (A) = d 2 (B). 2 C O U N T I N G R I G H T C O S E T S In this section we are going to outline the derivation of a formula for the number of right cosets in a given double coset of U. The following lemma shows that this is a reasonable task. 2.1 L e m m a : Let A I. Then there exist m N and A 1,..., A m I satisfying UAU = UA 1 UA m (with denoting the pairwise disjoint union). Instead of developing the desired theorem in this talk lemma by lemma in general, it seems more sensible to present the whole approach first in the nicer classical case of o = Z in order to keep a clear view on the principal ideas. 2.2 E x a m p l e : Let o = Z. Since o is a principal ideal domain, every right coset UB for B I has a unique representative B = a b 0 d with a, d > 0 and 0 b < d, known as the Hermite normal form of B. Given this normal form, one can pose the following idea for calculating the number of right cosets in a given double coset UAU for A I: Generate all possible normal forms (in a sensible way) and decide whether they belong to U AU. The latter can be carried out using the above mentioned characterisation of UAU = UB U: Test whether d 1 (A) = d 1 (B ) and d 2 (A) = d 2 (B ) hold. As a concrete example construct every right coset representative B as above contained in UAU where A = ( ). Since d 2(A) = d 2 (B ) is a necessary condition for UAU = UB U, 2

3 the equation (det A)Z = (det B )Z and thus 4 = ad has to be satisfied. So we have three possible cases: (i) a = 4 and d = 1, (ii) a = 2 and d = 2, and (iii) a = 1 and d = 4. For these cases determine, for which values of b the equation UAU = UB U is fulfilled. To this end, it suffices to test whether d 1 (A) = d 1 (B ) holds, since a and d have already been constructed to satisfy d 2 (A) = d 2 (B ). Case (i): Since d = 1 and 0 b < d = 1, only the case b = 0 has to be analysed. We have d 1 (B ) = az + bz + dz = 4Z + 0Z + 1Z = 1Z = d 1 (A), so ( ) is a right coset representative in U AU. Case (ii): Since d = 2 and 0 b < d = 2, the cases b = 0 and b = 1 have to be analysed. For b = 0 we have d 1 (B ) = 2Z + 0Z + 2Z = 2Z = 1Z = d 1 (A), so ( ) is not an element of UAU. For b = 1 we have d 1 (B ) = 2Z + 1Z + 2Z = 1Z = d 1 (A), so ( ) belongs to UAU. Case (iii): Since d = 4 and 0 b < d = 4, the cases b {0, 1, 2, 3} have to be analysed. Due to a = 1, we have d 1 (B ) = 1Z = d 1 (A) in any of these cases, so ( b ) belongs to UAU for every b {0, 1, 2, 3}. Summarising, in U AU we have the following right coset representatives: ,,,,, Generalising these considerations, a formula for the number µ(a) of right cosets contained in UAU can be stated: µ(a) = d N d det A det A {b N 0 b < d, d Z + bz + dz = d 1 (A)}. The cardinality of {b N 0 b < d, az + bz + dz = d 1 (A)} can be calculated explicitly, which finally leads to a product formula for µ(a). The first main ingredient of the approach in the classical case was the Hermite normal form. In the general case, another normal form can be constructed (not as nice as in the classical case, but solving the issue of a uniquely determined representative). 2.3 L e m m a : Let B = ( r s ) I. Define b = det B and choose a o such that ao + bo = ro + so holds (always possible in the given setting). Furthermore, choose a transversal T of (o ab 1 o)/(ba 1 o ab 1 o). Then there exists a uniquely determined c T satisfying a c 1 U b ba 1 = UB. c With this normal form, a first elementary formula for the number of right cosets with a given gcd of the first column can be given. (All elements of a right coset of U have the same gcd of the first column, so it is possible to talk about the gcd of the first column of a right coset.). 3

4 2.4 C o ro l l a r y : The number of right cosets in UAU with a as gcd of the first column (denoted by µ a (A)) can be calculated by µ a (A) = {c T a + (c 1)o + ca 1 d 2 (A) = d 1 (A)}. The cardinality on the right-hand side of the formula in 2.4 can be calculated explicitly mostly by combinatorical arguments, which are omitted in this talk. 2.5 Th e o r e m : Let a be an ideal of o. If d 1(A) a d 2 (A)d 1 (A) 1, then µ a (A) can be calculated by µ a (A) = N(d 2(A)) N(a)N(d 1 (A)) (1 N(p) 1 ); p prime ideal p ad 1 (A) 1 +d 2 (A)d 1 (A) 1 a 1 otherwise, µ a (A) = 0 holds. Summing up µ a (A) for all ideals a of o (and doing more combinatorics), we obtain the desired formula for µ(a). 2.6 Th e o r e m : µ(a) = N(f 2 (A)) p prime ideal p f 2 (A) (1 + N(p) 1 ). 3 A P P L I C AT I O N S T O C O N G R U E N C E S U B G R O U P S In this section, an application of 2.6 to the calculation of indices of certain congruence subgroups is presented. 3.1 D e f i n i t i o n : For any ideal a of o define the congruence subgroup U 0 [a] = {( a b c d ) U b a}. In order to apply 2.6 in the context of subgroups of U, we need a transformation of cosets of U in UAU to cosets of other subgroups in U. This can be achieved be the following idea: If UAU = UA 1 A m, write UA i = UAQ i with Q i U (possible since A i UAU (hence A i = P i AQ i for some P i, Q i U and thus UA i = UP i AQ i = UAQ i )) and interpret the Q i as representatives of right cosets of a certain subgroup of U. This yield the following identity. 4

5 3.2 L e m m a : For any A I we have U \ UAU = (U A 1 UA) \ U. Specialising A = ( m ) with m o, one obtains U A 1 UA = U 0 [mo] and hence by the last lemma U 0 [mo] \ U = µ(a). Now 2.6 yields the following formula. 3.3 Th e o r e m : For any m o we have U 0 [mo] \ U = N(mo) p prime ideal p m (1 + N(p) 1 ). 4 A P P L I C AT I O N S T O H E C K E A L G E B R A S The formula 2.6 also has applications in the field of abstract Hecke algebras (this is where it originally comes from). This section begins with the introduction of these algebras. 4.1 D e f i n i t i o n : Denote by H the complex vector space spanned by {1 UAU A I}, where 1 M : G {0, 1} denotes the characteristic function of the set M. For A, A 1,..., A k, B, B 1,..., B m I with define UAU = UA 1 A k and UBU = UB 1 UB m 1 UAU 1 UBU = k m 1 UAi B j i=1 j=1 and extend this operation bilinearly to an operation on H. The obtained algebra is called Hecke algebra (precisely: abstract Hecke algebra related to the Hecke triple (G, I, U)). 4.2 R e m a r k : It can be proven that is well-defined in the sense that the result of 1 UAU 1 UBU is independent of the choice of representatives A 1,..., A k and B 1,..., B m. Furthermore, f g for arbitrary f, g H is again an element of H. A slight transformation of the definition of yields a formula for the coefficents of 1 UAU 1 UBU in the representation with respect to the canonical basis in L e m m a : Choose A, A 1,..., A k, B, B 1,..., B m I as in 4.1. Then for every C I we have (1 UAU 1 UBU )(C) = {(i, j) A i B j UC, 1 i k, 1 j m} 5

6 This lemma gives an idea for an algorithm to explicitly calculate those coefficients for given A and B. 4.4 holds: Th e o r e m : Let A, B I. If the following algorithm terminates, we obtain a set D I and coefficients c C N for every C D such that 1 UAU 1 UBU = c C 1 UCU C D (1) Decompose UAU and UBU into disjoint right cosets UA 1,..., UA k and UB 1,..., UB m, respectively. (2) Let D =. (3) For every pair (i, j) with 1 i k and 1 j m test whether there exists a C D with UA i B j U = UC U; if this is not the case, add the element A i B j to D and set c Ai B j = 1; otherwise, if additionally UC = UA i B j is fulfilled, increase c C by 1. For the execution of this algorithm the explicit construction of a right coset decomposition of UAU and UBU is important. With 2.6 we can give a probabilistic algorithm that carries out this task. 4.5 Th e o r e m : Let A I and S a finite system of generators of U. If the following algorithm terminates, we obtain a right transversal R of U \ UAU: (1) Calculate k = µ(a). (2) Set R = {A}. (3) Randomly choose a l N ( not too big ). (4) Set Q = E 2. Repeat l times the following step: Randomly choose an element C of S {P 1 P S} and use Q C instead of Q in the next iteration. (5) If there exists no B R with UAQ = UB, add the element AQ to R. (6) If R < k, go back to (3), otherwise stop. 4.6 E x a m p l e : Let ω = 5 and o = Z + Zω. Using the introduced algorithms, for A = ( ) (with µ(a) = N(2o)(1 + N(2o + (1 + ω)o) 1 ) = 4 ( ) = 6) we obtain the right coset decomposition 1 UAU = 1 UA UA6 with ( ω A 1 =, A 2 =, A 3 = ω 2 0 A 4 =, A 5 =, A = ), ( ω 1 ) 6

7 and the product with 1 UAU 1 UAU = 1 UA 1 U UA 2 U + 1 UA 3 U A =, A 2 = , A 3 = ω. Formulas for the numbers of right cosets can not only help in the explicit calculation of products, but can also yield theoretical results on the structure of Hecke algebras. The following reduction theorem is an application of Th e o r e m : Let A = ( 1 0 we have 0 a ) and B = ( b (1 UAU 1 UBU )(C) = ) as well as C = ( c ) for a, b, c o. Then { 1, if c abo, 0 otherwise. 7