Bounded generation for congruence subgroups of {\rm Sp}_4(R)
aa r X i v : . [ m a t h . G R ] J a n BOUNDED GENERATION FOR CONGRUENCE SUBGROUPS OF Sp ( R ) ALEXANDER A. TROST
Abstract.
This paper describes a bounded generation result concerning the minimal nat-ural number K such that for Q ( C , R ) := { Aε φ (2 x ) A − | x ∈ R, A ∈ Sp ( R ) , φ ∈ C } , onehas N C , R = { X · · · X K |∀ ≤ i ≤ K : X i ∈ Q ( C , R ) } for certain rings of algebraicintegers R and the principal congruence subgroup N C , R in Sp ( R ) . This gives an explicitversion of an abstract bounded generation result of a similar type as presented by Morris[15, Theorem 6.1(1)]. Furthermore, the result presented does not depend on several number-theoretic quantities unlike Morris’ result. Using this bounded generation result, we furthergive explicit bounds for the strong boundedness of Sp ( R ) for certain examples of rings R, thereby giving explicit versions of earlier strong boundedness results in [21]. We further givea classification of normally generating subsets of Sp ( R ) for R a ring of algebraic integers. Introduction
Bounded generation is a classic topic in the discussion of non-uniform lattices. For agroup G, it is usually about finding a collection Z , . . . , Z N of ’nice’ (often cyclic) subgroupsof G such that G = Z · · · Z N . There have been some fascinating papers about this topicin the last decade, for example Morris’ paper [15] about bounded generation for SL n ( R ) from a model theoretic viewpoint, Rapinchuk, Morgan and Sury’s paper [14] about boundedgeneration of SL ( R ) for R a ring of S-algebraic integers with infinitely many units andNica’s paper [17] about bounded generation of SL n ( F [ T ]) by root elements.A new variant of bounded generation that has emerged is the study of bounded generationnot by way of subgroups but by collections of conjugacy classes: That is for a group G anda subset T of G that generates G and is closed under conjugation, one asks if there is a(minimal) natural number L := L ( G, T ) such that G = { t · · · t L |∀ ≤ i ≤ L : t i ∈ T ∪ T − } and if such a L exists how it depends on T and on G. This type of problem has been studiedfor various different groups like diffeomorphism groups by Burago, Ivanov and Polterovich[4], finite simple groups by Liebeck, Lawther and Shalev [12],[11] and Lie Groups and linearalgebraic groups by Kedra, Libman and Martin [10]. Further, for Φ an irreducible root systemof rank at least and R a ring of algebraic integers, one can use classical bounded generationresults for G (Φ , R ) to see that L ( G (Φ , R ) , T ) always exists as observed by Burago, Ivanovand Polterovich [4, Example 1.6] and Kedra and Gal [7, Theorem 1.1]. But there are choicesfor T and G that give rise to new questions about bounded generation and in this paperwe are interested in two in particular: First, the case of G being the principal congruencesubgroup N I, Φ in G (Φ , R ) for I a proper ideal in R and T = Q (Φ , I ) := { Aε φ ( x ) A − | x ∈ I, A ∈ G (Φ , R ) , φ ∈ Φ } . Note that this uses that Q (Φ , I ) generates N I, Φ , which is trueaccording to Milnor’s, Serre’s and Bass’ solution for the Congruence subgroup problem [2,Theorem 3.6, Corollary 12.5], if R has infinitely many units or I = 2 R, which are the twocases we are interested in in this paper. A result by Morris [15, Theorem 6.1(1)] implies hat the quantity L ( N I, Φ , Q (Φ , I )) =: K ( I, Φ) exists and has an upper bound depending onthe number of generators of the ideal I , rank(Φ) , | R/I | as well as [ K : Q ] for K the numberfield containing R. Crucially, this upper bound does not depend on the algebraic structureof
R/I or R , but only on some numbers associated with the tuple (Φ , R, I, K ) . However, webelieve that the situation is better than described by Morris, if R has infinitely many units: Conjecture 1.
Let R be a ring of algebraic integers with infinitely many units, I a non-trivialideal in R , Φ an irreducible root system of rank at least . Then there is a minimal constant K := K (Φ) proportional to rank(Φ) and independent of R and I such that N I, Φ = Q (Φ , I ) K . We want to provide supporting evidence for this conjecture in the case that I = 2 R is aprime ideal, namely in the sub-case that each element of R is a sum of a unit in R and anelement of R. I call such rings R -pseudo-good. The motivating example is, of course, thering of integers Z , but we will show that some rings of quadratic and cubic integers R with R prime have this property as well. We will give bounds on K ( C , R ) for such rings: Theorem 1.
Let R be a R -pseudo-good ring of S-algebraic integers.(1) If R has infinitely many units, than K ( C , R ) ≤ .(2) If R is a principal ideal domain, then K ( C , R ) ≤ . This is essentially shown by determining how ’expensive’ a classical proof for the existenceof the Bruhat decomposition on Sp ( R/ R ) is in terms of Q ( C , R ) when forced on Sp ( R ) .Theorem 1 does not depend on the ring R and the corresponding field extension K | Q atleast if R has infinitely many units, which supports Conjecture 1.Second, we study the case that T is a union of finitely many conjugacy classes C , . . . , C k generating G (Φ , R ) . This problem has been studied in recent years, too. For example,we have shown in [21, Theorem 5.13] that independently of the specific conjugacy classes C , . . . , C k the quantity L ( G (Φ , R ) , T ) has an upper bound L k only depending on k, Φ and R. For a group G and k ∈ N , the minimal L k that works for all conjugacy classes C , . . . , C k generating G is denoted by ∆ k ( G ) , if it exists. In fact, our result [21, Theorem 5.13] showsthat there is a C (Φ , R ) ∈ N with k ≤ ∆ k ( G (Φ , R )) ≤ C (Φ , R ) k for k sufficiently big.Further explicit bounds in [10, Corollary 6.2] and by myself [20, Theorem 3] strongly suggestfor R a ring of integers with infinitely many units and Φ irreducible of rank at least that ∆ k ( G (Φ , R )) has an upper bound proportional to rank(Φ) · k and a lower bound proportionalto rank(Φ) · k with proportionality factors independent of R. I believe that ∆ k ( G (Φ , R )) hasan upper bound proportional to rank(Φ) · k , too, and a research strategy suggested to us by theanonymous referee of [21] would imply this. This strategy however depends on Conjecture 1.Our results in [21] showed further that the discussion of ∆ k ( G ) for G = Sp ( R ) or G ( R ) is more involved than for higher rank root systems Φ and involves certain number theoreticproblems related to what I call bad primes of R . Our result [21, Theorem 5.13] does still showthat ∆ k (Sp ( R )) has upper and lower bounds proportional to k for k sufficiently big, but asof now explicit bounds on ∆ k (Sp ( R )) have not appeared in the literature. This is partlydue to the fact giving explicit bounds for ∆ k (Sp ( R )) requires one to consider the value of K ( C , R ) , because contrary to higher rank groups G (Φ , R ) , it is hard to construct all rootelements of Sp ( R ) from a collection of generating conjugacy classes T . So Theorem 1 canbe used to derive explicit bounds on ∆ k (Sp ( R )) for R -pseudo-good rings R : Theorem 2.
Let k ∈ N be given, p a prime greater than . Further, let D be a positive,square-free number with D ≡ mod such that there are a, b positive odd numbers with D = a ± and such that R D the ring of algebraic integers in the number field Q [ √ D ] , isa principal ideal domain. Then(1) ∆ k (Sp ( R D ) ≤ k ,(2) ∆ k (Sp ( Z [ p − ])) ≤ k ,(3) ∆ k (Sp ( Z )) ≤ k and(4) ∆ k (cid:16) Sp ( Z [ √− ]) (cid:17) ≤ k hold. We also provide an improved version of [21, Theorem 6.3] to demonstrate that even for k ≥ r ( R ) , the number of bad primes r ( R ) of R can not be ignored when determining ∆ k (Sp ( R )) : Theorem 3.
Let R be a ring of S-algebraic integers in a number field. Further let r := r ( R ) := |{P| P divides 2R, is a prime ideal and R/ P = F }| be given. Then ∆ k (Sp ( R )) ≥ k + r ( R ) for all k ∈ N with k ≥ r ( R ) . Lastly, we prove the following theorem classifying normally generating subsets of Sp ( R ) , whose proof we promised in an earlier paper [21, Corollary 6.8]: Theorem 4.
Let R be a ring of S-algebraic integers and A : Sp ( R ) → Sp ( R ) / [Sp ( R ) , Sp ( R )] the abelianization homomorphism. Then S ⊂ Sp ( R ) normally generates Sp ( R ) precisely if Π( S ) = ∅ and A ( S ) generates Sp ( R ) / [Sp ( R ) , Sp ( R )] . This paper is divided into six sections: In the first section, we introduce necessary def-initions and notations. The second section explains how to provide values for K ( C , R ) for R -pseudo-good rings. The third section gives some examples of R -pseudo-good ringsof algebraic integers and proves Theorem 1. In the fourth section, we restate and slightlyrephrase [21, Theorem 3.1] in order to prove Theorem 2. The fifth and sixth section proveTheorem 4 and Theorem 3 respectively. Acknowledgments
I want to thank Benjamin Martin for his continued support and advice.1.
Definitions
Let G be a group and S a finite subset of G . In this paper k · k S : G → N ∪ { + ∞} denotesthe word norm given by the conjugacy classes C G ( S ) on G , that is k k S := 0 , k X k S := min { n ∈ N |∃ A , . . . , A n ∈ C G ( S ∪ S − ) : X = A · · · A n } for X ∈ hh S ii − { } and k X k S := + ∞ for X / ∈ hh S ii . We also set B S ( k ) := { A ∈ G |k A k S ≤ k } for k ∈ N and k G k S = diam( k · k S ) of G as the minimal N ∈ N , such that B S ( N ) = G oras + ∞ if there is no such N . If S = { A } , then we write k · k A instead of k · k { A } and B A ( k ) instead of B { A } ( k ) . Further define for k ∈ N the invariant ∆ k ( G ) := sup { diam( k · k S ) | S ⊂ G with | S | ≤ k and hh S ii = G } ∈ N ∪ { + ∞} with ∆ k ( G ) defined as −∞ , if there is no normally generating set S ⊂ G with | S | ≤ k. Thegroup G is called strongly bounded , if ∆ k ( G ) is finite or −∞ for all k ∈ N . We also define: ∆ ∞ ( G ) := sup { diam( k · k S ) | S ⊂ G with | S | < ∞ and hh S ii = G } ∈ N ∪ { + ∞} ith ∆ ∞ ( G ) defined as −∞ , if there is no finite, normally generating set S ⊂ G . Also note ∆ k ( G ) ≤ ∆ ∞ ( G ) for all k ∈ N .We will omit defining the simply-connected split Chevalley-Demazure groups G (Φ , R ) andthe corresponding root elements ε α ( x ) in this paper. Instead, we use a representation of thecomplex, simply-connected Lie group Sp ( C ) that gives the following, classical definition of G ( C , R ) = Sp ( R ) for R a commutative ring with Definition 1.1.
Let R be a commutative ring with and let Sp ( R ) := { A ∈ R × | A T J A = J } be given with J = I − I ! We can choose a system of positive simple roots { α, β } in C such that the Dynkin-diagramof this system of positive simple roots has the following form βα C : Then subject to the choice of the maximal torus in Sp ( C ) as diagonal matrices in Sp ( C ) ,the root elements for positive roots in G ( C , R ) = Sp ( R ) can be chosen as: ε α ( t ) = I + t ( e , − e , ) , ε β ( t ) = I + te , , ε α + β ( t ) = I + t ( e + e ) and ε α + β ( t ) = I + te for all t ∈ R. Root elements for negative roots φ ∈ C and x ∈ R are then ε φ ( x ) = ε − φ ( x ) T . Next, let φ ∈ C . Then the group elements ε φ ( t ) are additive in t ∈ R , that is ε φ ( t + s ) = ε φ ( t ) ε φ ( s ) holds for all t, s ∈ R . Further, a couple of commutator formulas, expressed inthe next lemma, hold. We will use the additivity and the commutator formulas implicitlythroughout the thesis usually without reference. Lemma 1.2. [8, Proposition 33.2-33.5]
Let R be a commutative ring with and let a, b ∈ R be given.(1) If φ, ψ ∈ C are given with = φ + ψ / ∈ C , then ( ε φ ( a ) , ε ψ ( b )) = 1 . (2) If α, β are positive, simple roots in C with α short and β long, then ( ε α + β ( b ) , ε α ( a )) = ε α + β ( ± ab ) and ( ε β ( b ) , ε α ( a )) = ε α + β ( ± ab ) ε α + β ( ± a b ) . We also define the Weyl group elements and diagonal elements in G (Φ , R ) : Definition 1.3.
Let R be a commutative ring with and let Φ be a root system. Define for t ∈ R ∗ and φ ∈ Φ the elements: w φ ( t ) := ε φ ( t ) ε − φ ( − t − ) ε φ ( t ) . We will often write w φ := w φ (1) . We also define h φ ( t ) := w φ ( t ) w φ (1) − for t ∈ R ∗ and φ ∈ Φ . Remark . Let
Π = { α , . . . , α u } be a system of simple, positive roots in the root system Φ . If w = w α i · · · w α ik is an element of the Weyl group W (Φ) , then there is an element e w ∈ G (Φ , R ) defined by e w := w α i (1) · · · w α ik (1) . We will often denote this element e w of G (Φ , R ) by w as well.Further, we use the following concept: efinition 1.5. Let R be a commutative ring with , Φ an irreducible root system, φ ∈ Φ and let S ⊂ G (Φ , R ) be given. Then for k ∈ N define the subset ε ( S, φ, k ) of R as { x ∈ R | ε φ ( x ) ∈ B S ( k ) } . Further for A ∈ G (Φ , R ) , set ε ( A, φ, k ) := ε ( { A } , φ, k ) . The subgroup U + (Φ , R ) , called the subgroup of upper unipotent elements of G (Φ , R ) , isthe subgroup of G (Φ , R ) generated by the root elements ε φ ( x ) for x ∈ R and φ ∈ Φ apositive root. Similarly, one can define U − (Φ , R ) , the subgroup of lower unipotent elements of G (Φ , R ) by root elements for negative roots. We also define B (Φ , R ) := B + (Φ , R ) := B ( R ) as the subgroup of G (Φ , R ) generated by the sets U + (Φ , R ) and { h φ ( t ) | t ∈ R ∗ , φ ∈ Φ } . Thissubgroup is called the upper Borel subgroup of G (Φ , R ) . Similarly, one defines the lowerBorel subgroup B − (Φ , R ) of G (Φ , R ) . Further for a non-trivial ideal I ⊂ R , we denote the group homomorphism G (Φ , R ) → G (Φ , R/I ) induced by the quotient map π I : R → R/I by π I as well. This group homomor-phism is commonly called the reduction homomorphism induced by I . Next, we define:
Definition 1.6.
Let R be a commutative ring with , I an ideal in R , Φ an irrducible rootsystem and S a subset of G (Φ , R ) . Then define the following two subsets of maximal idealsin R : (1) V ( I ) := { m maximal ideal in R | I ⊂ m } and(2) Π( S ) := { m maximal ideal of R | ∀ A ∈ S : π m ( A ) central in G (Φ , R/m ) } R -pseudo-good rings and possible values for K ( C , R ) First, we define R -pseudo-good rings: Definition 2.1.
Let R be a commutative ring with such that the set of coset representatives X of R in R can be chosen with the following properties:(1) each x ∈ X − R is a unit in R, (2) ∈ X and(3) if R = 2 R , then ∈ X. Then we call R a R -pseudo-good ring. Remark . If R is R -pseudo-good, then either R/ R is a field or is a unit in R. Thisis the case, because each element ¯ x in R/ R − { } can be written as ¯ x = x + 2 R for some x ∈ X a unit. But then ¯ x is itself a unit and hence each non-zero element of R/ R is a unitand so R is a field. On the other hand, R = 2 R implies that ∈ R is a unit. We shouldmention that R -pseudo-goodness is our own concept named so as an homage to good rings[23].We want to point out that being R -pseudo-good is equivalent to the map R ∗ → R/ R, u u + 2 R mapping onto R/ R − { } . Further, define for a R -pseudo-good ring R with thecorresponding set of coset representatives X , the set B R := { ε α + β ( x ) ε α + β ( x ) ε β ( x ) ε α ( x ) h α ( t ) h β ( s ) | t, s ∈ X ∩ R ∗ , x , x , x , x ∈ X } . The goal of this section is to prove:
Proposition 2.3.
Let R be a R -pseudo-good ring such that is not a unit and let J ∈ N be given such that Sp ( R ) = ( U + ( C , R ) U − ( C , R )) J or Sp ( R ) = ( U − ( C , R ) U + ( C , R )) J .Then K ( C , R ) ≤ J + 6 holds. emark . Technically, we should assume that also R is a ring of algebraic integers butthis does not play a role in the proof, so we will omit this assumption. Furthermore, wewill write N instead of N C , R and Q instead of Q ( C , R ) in this section to simplify thenotation.To show this, we define the usual Cayley word norms on finitely generated groups next: Definition 2.5.
Let G be a group and S ⊂ G with S = S − a generating set of G be given.Then define the function l S : G → N by l S (1) := 0 and by l S ( x ) := min { n ∈ N |∃ s , . . . , s n ∈ S : x = s · · · s n } for x = 1 . Remark . Note that in this definition S does not normally generate G but generate G. We also need the following definition:
Definition 2.7.
Let G be a group and S ⊂ G be given with S = S − a generating set of G .Further, let w = s · · · s n be given with all s i ∈ S .(1) The tuple (or string) ( s , . . . , s n ) ∈ S n is called an expression for w in terms of S oflength n . If n = l S ( w ) holds, then the tuple ( s , . . . , s n ) is called a minimal expressionfor w (with respect to S ). (2) An element w ′ ∈ G is called a subword of ( s , . . . s n ) if there is a sequence of integers ≤ i < i < · · · < i k ≤ n such that w ′ = s i · · · s i k and l S ( w ′ ) = k .If G is the Weyl group W (Φ) of an irreducible root system Φ , then the generating set S is usually chosen as the set F = { w α , . . . , w α u } of fundamental reflections associated to asystem of positive, simple roots Π = { α , . . . , α u } . The group W (Φ) further contains a unique element w with the property that l F ( w ) ismaximal, named the longest element of W (Φ) . According to [18, Appendix, p. 151, (24)Theo-rem], this element w can equivalently characterized by the property that w ( φ ) is a negativeroot in Φ for each positive root φ ∈ Φ . Next, note that the Weyl group W ( C ) is generatedby the set F := { w α , w β } for α and β chosen as in the beginning of Section 1. Then oneeasily checks that w = ( w α · w β ) and l F ( w ) = 4 holds.To prove Proposition 2.3, we show the following proposition now: Proposition 2.8.
Let R be a R -pseudo-good ring, w = s (1)1 · · · s (1) k and w = s (2)1 · · · s (2) k el-ements of W ( C ) with s (1)1 , . . . , s (1) k , s (2)1 , . . . , s (2) k elements of F and l F ( w ) = k and l F ( w ) = k . Then up to multiplication by l F ( w ) elements of Q , each element of ( B ( C , R ) w B ( C , R )) · ( B ( C , R ) w B ( C , R )) is an element of B ( C , R ) wB ( C , R ) for w some subword of the (pos-sibly non-minimal) expression ( s (1)1 , . . . , s (1) k , s (2)1 , . . . , s (2) k ) . We first show how to derive Proposition 2.3 from this proposition:
Proof.
To begin, note that we may assume wlog that each A ∈ N ⊂ Sp ( R ) can be writtenas A = J Y i =1 u + i u − i or all u + i elements of U + ( C , R ) and all u − i elements of U − ( C , R ) . But each element w − u − i w is a product of root elements of positive roots in C and hence an element of B ( C , R ) . But then w − = − w implies u + i u − i = ( u + i w )( w − u − i w ) w − ∈ ( B ( C , R ) w ) · ( B ( C , R ) w ) ⊂ ( B ( C , R ) w B ( C , R )) . holds for all i. This implies A ∈ ( B ( C , R ) w B ( C , R )) J .But l F ( w ) = 4 holds, so according to Proposition 2.8, the matrix A can be written as aproduct b ′ wb ′ for b ′ , b ′ ∈ B ( C , R ) and w ∈ W ( C ) after multiplication by l F ( w )(2 J − ≤ J − elements of Q. But each element of B ( C , R ) wB ( C , R ) is conjugate to an elementof B ( C , R ) w. Observe that each element of B ( C , R ) has the form ε α + β ( t α + β ) ε α + β ( t α + β ) ε β ( t β ) ε α ( t α ) h α ( s α ) h β ( s β ) for t α + β , t α + β , t β , t α ∈ R and s α , s β ∈ R ∗ . Hence after multiplication with elements of Q, we may assume that t α + β , t α + β , t β , t α are elements of the set X of coset representatives of R in R given by R -pseudo-goodness instead. Furthermore, h α ( s α ) = w α ( s α ) w − α holds and w α ( s α ) = ε α ( s α ) ε − α ( − s − α ) ε α ( s α ) . Note, that all elements of X − { } are units in R and so we can consider the set Y := {− x − | x ∈ X − { }} ∪ { } . One easily checks that this set Y is also a set of coset representatives of R in R. Thusafter multiplication with elements of Q , we may assume that s α is an element of X − { } .Similarly, we may assume after multiplication by elements of Q that s β is an element of X − { } . So each element of B ( C , R ) w agrees with an element of B R w after multiplicationby elements of Q. To summarize: Up to multiplication by at most J − J +6 elements of Q, eachelement A of N can be rewritten as an element of B R w for some w ∈ W ( C ) . Next, rememberthat N = ker( π R : Sp ( R ) → Sp ( R/ R )) . We are going to show that B R w ∩ N = ∅ implies w = I and B R w ∩ N = { I } . Together with Q ⊂ N, this implies K ( C , R ) ≤ J + 6 . To show that B R w ∩ N = ∅ implies w = I and B R w ∩ N = { I } , assume there is an A = bw ∈ B R w ∩ N for some w ∈ W ( C ) . Observe that π R ( A ) = I . But π R ( b ) is an elementof B ( R/ R, C ) of Sp ( R/ R ) . Further, slightly abusing notation, we obtain π R ( w ) = w andhence π R ( A ) is an element of B ( R/ R, C ) w. But R/ R is a field. Hence by the uniquenessof the Bruhat-decomposition for Sp ( R/ R ) [18, Chapter 3, p. 26, Theorem 4’], we obtain π R ( b ) = w = I . But according to the definition of B R and remembering that X is a set ofcoset-representatives of R in R , this implies b ∈ { h α ( t ) h β ( s ) | t, s ∈ X ∩ R ∗ } . So there are t, s ∈ X ∩ R ∗ with A = h α ( t ) h β ( s ) = t st − t − s − t But π R ( A ) = I and hence t ≡ mod R. But ∈ X and so t = 1 . Then s = 1 follows thesame way. Hence A = I . This finishes the proof of Proposition 2.3. (cid:3) emark . This proof can also be used to see that Q ( C , R ) does indeed generate N C , R for R a R -pseudo-good ring without using the complete congruence subgroup property.To prove Proposition 2.8, we need: Lemma 2.10.
Let R be a R -pseudo-good ring. Then up to multiplication by an element of Q , we have ( B ( C , R ) w α B ( C , R )) · ( B ( C , R ) w α B ( C , R )) ⊂ B ( C , R ) ∪ ( B ( C , R ) w α B ( C , R )) . The same holds for β instead of α. Proof.
Let b , b , b ′ , b ′ ∈ B ( C , R ) be given. Note that we may write b b ′ as b b ′ = ε α ( a ) u P − α h for a ∈ R , ε α + β ( b ) ε α + β ( c ) ε β ( d ) = u P −{ α } for b, c, d ∈ R and h ∈ { h α ( t ) h β ( s ) | t, s ∈ R ∗ } . Together with w − α = − w α , this implies: b w α b b ′ w α b ′ = b w α ε α ( a ) u P − α hw α b ′ = b ε − α ( ± a ) w α [ u P − α ( − h )] w − α b ′ . Next, w α [ u P − α ( − h )] w − α is an element of B ( C , R ) , because w α u P − α w − α is a product of rootelements associated to positive roots in C and w α ( − h ) w − α is an element of { h α ( t ) h β ( s ) | t, s ∈ R ∗ } as required. Thus b w α b b ′ w α b ′ ∈ B ( C , R ) ε − α ( ± a ) B ( C , R ) holds.There are two possible cases now. Either a is an element of R , then we are done aftermultiplying with one element of Q. On the other hand, if a / ∈ R holds, then as R is R -pseudo-good, there is a unit x ∈ R such that a ≡ − x − mod R. Hence after multiplyingwith one element of Q , we may assume a = − x − and so we obtain ε − α ( a ) = ε α ( − x )( ε α ( x ) ε − α ( − x − ) ε α ( x )) ε α ( − x )= ε α ( − x ) w α ( x ) ε α ( − x ) = ε α ( − x ) h α ( x ) w α ε α ( − x ) . But ε α ( − x ) h α ( x ) and ε α ( − x ) are elements of B ( C , R ) , so ε − α ( a ) is an element of B ( C , R ) w α B ( C , R ) .Hence b w α b b ′ w α b ′ ∈ B ( C , R ) ε − α ( ± a ) B ( C , R ) ⊂ B ( C , R ) · ( B ( C , R ) w α B ( C , R )) · B ( C , R )= B ( C , R ) w α B ( C , R ) holds after multiplication with up to one element of Q. (cid:3) Next, we are going to prove the Proposition 2.8:
Proof.
Slightly abusing notation, we set T ( w , w ) := ( s (1)1 , . . . , s (1) k , s (2)1 , . . . , s (2) k ) . We willfirst show by induction on l F ( w ) that ( B ( C , R ) w B ( C , R )) · ( B ( C , R ) w B ( C , R )) ⊂ [ w subword of T ( w ,w ) B ( C , R ) wB ( C , R ) holds up to multiplication by l F ( w ) elements of Q .For l F ( w ) = 0 , we obtain B ( C , R ) w B ( C , R ) = B ( C , R ) and hence the claim is obvious.So let w ∈ W ( C ) be given with l F ( w ) ≥ and assume without loss of generality that = w ′ w α and l F ( w ) = l F ( w ′ ) + 1 . Then by induction hypothesis ( B ( C , R ) w B ( C , R )) · ( B ( C , R ) w B ( C , R ))= ( B ( C , R ) w B ( C , R )) · ( B ( C , R ) w ′ ) · ( w α B ( C , R )) ⊂ [ w subword of T ( w ,w ′ ) B ( C , R ) wB ( C , R ) · w α B ( C , R )= [ w subword of T ( w ,w ′ ) ( B ( C , R ) wB ( C , R ) · w α B ( C , R )) holds up to multiplication by l F ( w ′ ) elements of Q. Hence it suffices to consider the specialcase w = w α . We distinguish two cases: First l F ( w w α ) > l F ( w ) and second l F ( w w α )
In this section, we will study certain examples of rings of algebraic integers R to proveTheorem 1 and Theorem 2. This will require to talk about bounded generation by rootelements and about R -pseudo-goodness to apply Proposition 2.3.3.1. R -pseudo-good rings of algebraic integers and bounded generation. First,the definition of S-algebraic integers:
Definition 3.1. [16, Chapter I, §11] Let K be a finite field extension of Q . Then let S be a finite subset of the set V of all valuations of K such that S contains all archimedeanvaluations. Then the ring O S is defined as O S := { a ∈ K | ∀ v ∈ V − S : v ( a ) ≥ } and O S is called the ring of S -algebraic integers in K. Rings of the form O S are called ringsof S-algebraic integers. Then for R a ring of S-algebraic integers, the group Sp ( R ) is boundedly generated byroot elements. The following theorem combines bounded generation results by several peopleamong them Tavgen [19], Rapinchuk, Morgan and Sury [14] as well as Carter and Keller [5]. Theorem 3.2.
Let K be a number field and R a ring of S-algebraic integers in K .(1) If R is a principal ideal domain, than Sp ( R ) = ( U + ( C , R ) · U − ( C , R )) or Sp ( R ) =( U − ( C , R ) · U + ( C , R )) .(2) If R has infinitely many units, than Sp ( R ) = ( U + ( C , R ) · U − ( C , R )) or Sp ( R ) =( U − ( C , R ) · U + ( C , R )) . Using Theorem 3.2 and Proposition 3.3, we can prove Theorem 1:
Proof.
According to Theorem 3.2, we can choose J in Proposition 2.3 as , if R is a principalideal domain and as , if R has infinitely many units. Thus using Proposition 2.3, we obtainthe claim of the theorem. (cid:3) We will provide some examples of rings of algebraic integers that are R -pseudo-good: Proposition 3.3.
Let D be a square-free, positive integer with D ≡ mod such that oneof the two following conditions holds:(1) There is a unit u ∈ R D with tr Q [ √ D ] | Q ( u ) odd.(2) There are odd numbers a, b ∈ N such that b D = a ± .Then R D is R D -pseudo-good and K ( C , R D ) ≤ .Proof. Due to [13, Theorem 25], we know that D ≡ mod implies that R D is a primeideal in R D . Thus we obtain that R D / R D is a field extension of F of degree two, that is R D / R D = F . Further we know for each x ∈ R D that it is a root of χ x ( T ) := T − tr Q [ √ D ] | Q ( x ) · T + N Q [ √ D ] | Q ( x ) . The polynomial χ x ( T ) is an element of Z [ T ] . So x + 2 R D is a root of the polynomial ¯ χ x ( T ) := T − (tr Q [ √ D ] | Q ( x ) + 2 Z ) · T + ( N Q [ √ D ] | Q ( x ) + 2 Z ) in F [ T ] . But if we assume that u ∈ R D is a unit with tr Q [ √ D ] | Q ( u ) odd, then this implies twothings: First, the norm N Q [ √ D ] | Q ( u ) is either or − and so N Q [ √ D ] | Q ( x ) + 2 Z = ¯1 . Second, he trace tr Q [ √ D ] | Q ( u ) also reduces to ¯1 in F . Thus, u + 2 R D is a root of the polynomial ¯ χ x ( T ) = T + T + ¯1 . But this implies that u + 2 R D can not be R D . Consequently, thegroup homomorphism R ∗ D → ( R D / R D ) ∗ can not be trivial. However, R D / R D is the field F and so ( R D / R D ) ∗ has three elements, which is a prime. Thus R ∗ D → R D / R D − { } must be surjective and hence R D is indeed R -pseudo-good. This finishes the proof of thefirst part of the proposition. For the second part, we will show that the fraction x := a + b √ D is an element of R D and a unit with the property that tr Q [ √ D ] | Q ( x ) is odd, which will finishthe proof by applying the first part. To this end, observe first that N Q [ √ D ] | Q ( x ) = a + b √ D · a − b √ D a − b D ± by assumption. Second, observe that tr Q [ √ D ] | Q ( x ) = a + b √ D a − b √ D a is odd. But now remember that x is a root of χ x ( T ) ∈ Z [ T ] and thus an element of R D . Lastly, R D has infinitely many units according to Dirichlet’s Unit Theorem [16, Corollary 11.7] andthus Theorem 1 implies K ( C , R D ) ≤ . (cid:3) Quite many squarefree, positive D with D ≡ mod satisfy the properties required inthe second part of Proposition 3.3. For example, all the ones below except for D = 37 satisfy them, as seen by the following equations: · + 4 , ·
13 = 3 + 4 , ·
21 = 5 − , ·
29 = 5 + 4 , ·
53 = 7 + 4 , ·
61 = 39 + 4 , ·
69 = 25 − , ·
77 = 9 − , ·
85 = 9 + 4 , ·
93 = 29 − However, the ring R has fundamental unit √ as can be seen by applying [16, Chap-ter 1§7,Excercise 1]. But then the image of the units R ∗ in ( R / R ) ∗ agrees with thecyclic subgroup (multiplicatively) generated by the image of √ in ( R / R ) ∗ . How-ever, one easily checks that √ maps to R . Thus R can not be R -pseudo-good.Furthermore, if D ≤ − is squarefree with D ≡ mod , then R D can not be R D -pseudo-good either, because rings of quadratic imaginary numbers have at most one unit if D = − . But there are other examples of rings of algebraic integers that are R -pseudo-good besidessome quadratic rings of integers: Proposition 3.4.
Let p be a prime number greater than and consider the polynomial Q p ( T ) := T + pT − ∈ Z [ T ] Then Q p ( T ) is irreducible with three distinct, real roots. Let x p the biggest root of Q p ( T ) andlet R be the ring of algebraic integers in the number field K := Q [ x p ] . Then R is R -pseudo-good and K ( C , R ) ≤ holds.Proof. Reducing Q p ( T ) modulo yields the irreducible polynomial T + T + 1 ∈ F [ T ] andhence Q p ( T ) itself is irreducible as well. We leave it as an excercise to the reader to showthat all roots of Q p ( T ) are real and different. Next, we are going to show that R is R -pseudo-good. To this end, we will first show that R is a prime ideal. Note that [ K : Q ] = 3 and hence there are the following possibilities for the prime factorization of R in R :(1)2 R = P · P · P , (2)2 R = P · P , (3)2 R = P , (4)2 R = P · Q , (5)2 R = S ith R/ P i = F for i = 1 , , and P , P and P being distinct prime ideals, R/ Q = F and R/S = F . Observe that the following equation holds: x p + px p = x p ( x p + p ) . Thus not only is x p a unit in R but also x p + p is a unit. Hence setting ¯ x p := x p +2 R , we obtainthat both ¯ x p and ¯ x p + 1 are units in the ring R/ R. But all the possible R/ R correspondingto the prime factorizations (1) through (4) have the quotient ring R/ P = F . Thus F wouldhave a unit u such that u + 1 is also a unit, but this is clearly impossible. Hence R mustbe a prime ideal itself. But then R/ R must be the field F . But the unit group F ∗ hasorder , which is a prime number. Thus to prove the R -pseudo-goodness of R, it suffices toshow that there is a unit in R which maps to a non-trivial unit of R/ R. However, observethat as x p (¯ x p + 1) holds in R/ R, it is impossible that ¯ x p is equal to . Thus R is a R -pseudo-good ring. The inequality K ( C , R ) ≤ follows from Theorem 1, because R has infinitely many units according to Dirichlet’s Unit Theorem [16, Corollary 11.7]. (cid:3) Proving strong boundedness for Sp ( R ) explicitly To prove Theorem 2, we will rephrase [21, Theorem 3.1]. To this end, we note twostatements:
Lemma 4.1. [21, Lemma 3.4]
Let R be a commutative ring with such that ( R : 2 R ) < ∞ and such that G := Sp ( R ) is boundedly generated by root elements. Further define Q ′ := { ε φ (2 x ) | x ∈ R, φ ∈ C } . and N ′ := hh Q ′ ii and let k · k Q ′ : N ′ → N be the conjugation invariant word norm on N ′ defined by Q ′ . (1) Then the group G/N ′ is finite.(2) Then there is a minimal K ′ ( C , R ) ∈ N such that k N ′ k Q ′ ≤ K ′ ( C , R ) .Remark . If R is a ring of algebraic integers, then the groups N C , R and N ′ as well as theconstants K ( C , R ) and K ′ ( C , R ) are the same due to the congruence subgroup property[2, Theorem 3.6, Corollary 12.5].We also need: Theorem 4.3. [21, Theorem 3.2]
Let R be a commutative ring with . Then there is aconstant L ( C , R ) ∈ N such that for A ∈ Sp ( R ) , there is an ideal I ( A ) ⊂ R with thefollowing two properties:(1) V ( I ( A )) ⊂ Π( { A } ) and(2) I ( A ) ⊂ ε ( A, φ, L ( C , R )) for all φ ∈ C . We can recast [21, Theorem 3.1] slightly more explicitly as follows:
Theorem 4.4.
Let R be a commutative ring with and ( R : 2 R ) < + ∞ such that Sp ( R ) isboundedly generated by root elements. Further, let L ( C , R ) ∈ N be as in Theorem 4.3 and K ′ ( C , R ) ∈ N as in Lemma 4.1. Then for all k ∈ N , one has ∆ k (Sp ( R )) ≤ L ( C , R ) · K ′ ( C , R ) · k + ∆ ∞ (Sp ( R ) /N ′ ) Remark . The group Sp ( R ) /N ′ is finite as observed in Lemma 4.1 and hence ∆ ∞ (Sp ( R ) /N ′ ) is a well-defined natural number. o to prove Theorem 2, we have to determine L ( C , R ) , K ′ ( C , R ) = K ( C , R ) and ∆ ∞ (Sp ( R ) /N ′ ) = ∆ ∞ (Sp ( R ) /N C , R ) = ∆ ∞ (Sp ( R/ R )) for the rings of S-algebraic inte-gers mentioned in Theorem 2. First, regarding K ( C , R ) : Note that Z and Z [ √− ] areprincipal ideal domains. This is obvious for Z and follows for Z [ √− ] , because Z [ √− ] isa euclidean domain by way of using the norm map N Q [ √− | Q : Z (cid:20) √− (cid:21) → Z . Further, R D and Z [ p − ] have infinitely many units. Lastly, all the rings Z , Z [ p − ] , R D and Z [ √− ] are R -pseudo-good: This is clear for Z , Z [ p − ] and follows for R D from Proposi-tion 3.3. For R := Z [ √− ] this follows from X := { , , √− , −√− } being a set of cosetrepresentatives of R in R satisfying the definition of R -pseudo-goodness. Thus applyingTheorem 1, we obtain K ( C , Z ) ≤ , K ( C , Z [ 1 + √−
32 ]) ≤ , K ( C , Z [ p − ]) ≤ and K ( C , R D ) ≤ . Next, we determine L ( C , R ) and ∆ ∞ (Sp ( R/ R )) . In regards to L ( C , R ) , we state thefollowing: Theorem 4.6.
Let R be a principal ideal domain. Then L ( C , R ) as in Theorem 4.3 can bechosen as . We will omit the proof as it is rather lengthy and similar to the proofs of [20, Theorem 2.3]and [10, Proposition 6.17]. The interested reader however can find the proof in the author’sPhD thesis [22, Theorem 4.2.1]. We note that all the rings R from Theorem 2 are principalideal domains. This is true by assumption for R D , obvious for Z and Z [ p − ] and we sawalready that Z [ √− ] is euclidean. Thus for all rings R as in Theorem 2 the constant L ( C , R ) can be chosen as . Next, we must determine ∆ ∞ (Sp ( R/ R )) for these rings R : Proposition 4.7.
Let K be either F or F . (1) If K = F , then ∆ ∞ (Sp ( K )) ≤ .(2) If K = F , then ∆ ∞ (Sp ( K )) ≤ .Proof. For the purposes of this proof, we define the following invariant for a group G : cn( G ) := min { n ∈ N |∀ C a conjugacy class in G with G = h C i : G = C n } with cn( G ) being defined as + ∞ , if the corresponding set is empty. Then one easily obtains ∆ ( G ) ≤ cn( G ) for all groups G. Next, let S be a normally generating subset of Sp ( K ) . Wedistinguish the two cases for K . First, assume K = F . Observe that Sp ( K ) = PSp ( K ) ,because each scalar matrix in Sp ( K ) must be I as char( K ) = 2 . But as K = F , the group Sp ( K ) = PSp ( K ) is simple by [18, Chapter 4, p. 33, Theorem 5]. Hence any non-trivialelement A S ∈ S also normally generates Sp ( K ) . But in the second case K = F , it iswell-known that the group Sp ( K ) is isomorphic to the permutation group S . This grouphowever only has three normal subgroups namely { } , S and the alternating subgroup A . Thus if we pick an A S ∈ S , that does not lie in A , then necessarily A S normally generates p ( K ) . So for each normally generating set S of Sp ( K ) there is an A S ∈ S that normallygenerates Sp ( K ) for both K = F and K = F . This implies: ∆ ∞ (Sp ( K )) ≥ ∆ (Sp ( K )) ≥ sup {k Sp ( K ) k A S | S normally generates Sp ( K ) }≥ sup {k Sp ( K ) k S | S normally generates Sp ( K ) } = ∆ ∞ (Sp ( K )) . Hence to give upper bounds on ∆ ∞ (Sp ( K )) , it suffices to give upper bounds on cn(Sp ( K )) . First, for Sp ( F ) = S , the invariant cn( S ) can be determined to be from the main resultin [3]. Second for K = F , we use that the paper [9] contains a list of the invariants cn( G ) for simple groups G with less than elements calculated using a computer algebrasystem and states on page that cn(Sp ( F )) = 4 . This yields ∆ ∞ (Sp ( F )) ≤ . (cid:3) Note next that Z / Z = F = Z [ p − ] / Z [ p − ] and R D / R D = F = Z [ √− ] / Z [ √− ] hold. Hence Proposition 4.7 implies ∆ ∞ (Sp ( Z ) /N ′ ) ≤ ≥ ∆ ∞ (Sp ( Z [ p − ]) /N ′ )∆ ∞ (Sp ( R D ) /N ′ ) ≤ ≥ ∆ ∞ (Sp ( Z [ 1 + √−
32 ]) /N ′ ) Combining these bounds on ∆ ∞ (Sp ( R ) /N ′ ) , the value of L ( C , R ) = 320 from Theorem 4.6and the bounds on K ( C , R ) determined before with Theorem 4.4, we obtain Theorem 2.5. Classifying normal generating subsets of Sp ( R ) In this section, we show Theorem 4. To prove this, we need the following:
Lemma 5.1.
Let R be a commutative ring with and of characteristic and let S ⊂ Sp ( R ) be given with Π( S ) = ∅ . Then ε α + β (1) ε α + β (1) ∈ hh S ii and ν ( R ) := ( x − x | x ∈ R ) ⊂ { y ∈ R |∀ φ ∈ C : ε φ ( x ) ∈ hh S ii} hold.Proof. We will first show that ε α + β (1) ε α + β (1) ∈ hh S ii . Second, we will show that anynormal subgroup of Sp ( R ) containing ε α + β (1) ε α + β (1) also contains ε φ ( y ) for any φ ∈ C and y ∈ ν ( R ) . For the first claim note that Π( S ) = ∅ implies R = X A ∈ S ( x ( A ) ij , x ( A ) ii − x ( A ) jj | ≤ i = j ≤ for A = ( x ( A ) ij ) . But using Freshmen’s dream, this implies that there are A , . . . , A k ∈ S aswell as u ( l ) ij , t ( l ) ij ∈ R with k X l =1 X ≤ i = j ≤ ( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) . However using [6, Theorem 2.6, 4.2, 5.1, 5.2], we know that the element ε α + β (( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) ) · ε α + β (( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) ) is contained in the normal subgroup generated by A l for all ≤ i = j ≤ and all ≤ l ≤ k. But the elements of the root subgroups ε α + β ( R ) and ε α + β ( R ) commute and this implies hat the normal subgroup generated by S contains the product Y ≤ l ≤ k Y ≤ i = j ≤ ε α + β (( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) ) · ε α + β (( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) )= ε α + β ( k X l =1 X ≤ i = j ≤ ( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) ) · ε α + β ( k X l =1 X ≤ i = j ≤ ( u ( l ) ij x ( A l ) ij + t ( l ) ij ( x ( A l ) ii − x ( A l ) jj )) )= ε α + β (1) ε α + β (1) This finishes the first step. For the second step, let M be the normal subgroup of Sp ( R ) generated by ε α + β (1) ε α + β (1) . Then we obtain for x ∈ R that(1) M ∋ ( ε α + β (1) ε α + β (1) , ε − β ( x )) = ε α ( x ) ε α + β ( x ) . This yields that M ∋ w β w α w β ε α ( x ) ε α + β ( x ) w − β w − α w − β = w β w α ε α + β ( x ) ε α + β ( x ) w − α w − β = w β ε α + β ( x ) ε β ( x ) w − β = ε α ( x ) ε − β ( x ) . On the other hand (1) implies for x, y ∈ R that M ∋ ( ε α ( y ) ε α + β ( y ) , ε − ( α + β ) ( x )) = ε α ( y ) ( ε α + β ( y ) , ε − ( α + β ) ( x )) ε − β (2 xy )= ε α ( xy ) ε − β ( x y ) . But this implies that ε − β ( x y − xy ) = ε − β ( x y ) ε − β ( xy ) = (cid:0) ε − β ( x y ) ε α ( xy ) (cid:1) · ( ε α ( xy ) ε − β ( xy )) ∈ M. This implies in particular that the ideal ν ( R ) := ( x − x | x ∈ R ) is contained in { z ∈ R |∀ φ ∈ C long: ε φ ( z ) ∈ M } . However, [21, Lemma 4.8] implies that if ε φ ( z ) ∈ M for all φ ∈ C long, then ε φ ( z ) ∈ M for all φ ∈ C . This finishes this proof. (cid:3) We can prove Theorem 4 now:
Proof.
According to [21, Corollary 3.11], the set S normally generates Sp ( R ) precisely if Π( S ) = ∅ and the image of S normally generates the quotient Sp ( R ) /N ′ for N ′ = hh ε φ (2 x ) | x ∈ R, φ ∈ C ii . As before, the congruence subgroup property [2, Theorem 3.6, Corollary 12.5] can be usedto identify the normal subgroup N ′ as the kernel of the reduction homomorphism π R :Sp ( R ) → Sp ( R/ R ) and hence Sp ( R ) /N ′ = Sp ( R/ R ) . So S normally generates Sp ( R ) precisely if Π( S ) = ∅ and S maps to a normal generating set of the finite group Sp ( R/ R ) . So to finish the proof of this corollary, it suffices to show that Π( S ) = ∅ and S mappingto a generating set of the abelianization of Sp ( R ) , implies that Sp ( R/ R ) is normallygenerated by the image ¯ S of S. But R/ R is a finite ring and so is semi-local. Thus [1,Corollary 2.4] implies that Sp ( R/ R ) is generated by root elements. Hence it suffices toshow that the normal subgroup ¯ W of Sp ( R/ R ) normally generated by ¯ S contains all rootelements. However, [21, Lemma 4.8] implies that if { ε α (¯ x ) | ¯ x ∈ R/ R } is a subset of ¯ W , than ¯ W contains all root elements. Hence it suffices to show that ¯ R := { ¯ x ∈ R/ R | ε α (¯ x ) ∈ ¯ W } = R/ R. o this end, observe that Π( S ) = ∅ implies Π( ¯ S ) = ∅ . But then Lemma 5.1 implies that ν ( R/ R ) := (¯ x − ¯ x | ¯ x ∈ R/ R ) is a subset of ¯ R and ε α + β (1) ε α + β (1) is an element of ¯ W .Next, let the ideal R in R split into primes as follows: R = r Y i =1 P l i i ! · s Y j =1 Q k j j ! with [ R/ P i : F ] = 1 for ≤ i ≤ r and [ R/ Q j : F ] > for ≤ j ≤ s. Using the ChineseRemainder Theorem, this implies that R/ R = ( r Y i =1 R/ P l i i ) × ( s Y j =1 R/ Q k j j ) . As the next step, we show that { (0 + P l , . . . , P l r r , x + Q k , . . . , x + Q k s s ) | x ∈ R } ⊂ ¯ R . To this end for each j = 1 , . . . , s , pick an element x j ∈ R such that neither x j nor x j − areelements of Q j . Such elements do exist, because otherwise [ R/ Q j : F ] = 1 would hold. Thisimplies that (0 + P l , . . . , P l r r , x ( x −
1) + Q k , . . . , x s ( x s −
1) + Q k s s ) is not only an element of ν ( R/ R ) ⊂ ¯ R , but also that ( x ( x − Q k , . . . , x s ( x s − Q k s s ) is a unit in the quotient ring Q sj =1 R/ Q k j j . But then [21, Proposition 6.4] implies { (0 + P l , . . . , P l r r , x + Q k · · · Q k s s ) | x ∈ R } ⊂ ¯ R . Next, we show that S mapping to a generating set of the abelianization of Sp ( R ) , impliesfurther that { ( x + P l · · · P l r r , Q k · · · Q k s s ) | x ∈ R } ⊂ ¯ R . First, observe that according to the proof of [21, Theorem 6.3], the abelianization homo-morphism of Sp ( R ) is uniquely defined by A : Sp ( R ) → F r , ε φ ( x ) ( x + P , . . . , x + P r ) for all φ ∈ C . According to the same proof the abelianization of Sp ( R ) factors through Sp ( R/ R ) . Also the element ε α + β (1) ε α + β (1) of Sp ( R/ R ) clearly vanishes under theabelianization map. Thus the abelianization map of Sp ( R ) factors through Sp ( R/ R ) / ¯ M for ¯ M the normal subgroup of Sp ( R/ R ) generated by ε α + β (1) ε α + β (1) . Second, we leave it as an exercise to the reader to show that ¯ M also contains the set { ε φ (¯ x ) ε φ (¯ x ) | ¯ x ∈ R/ R, φ , φ ∈ C } . But ¯ M is contained in ¯ W and so if φ , . . . , φ n ∈ C and x , . . . , x n ∈ R are given with ¯ X = n Y i =1 ε φ i ( x i + 2 R ) , an element of ¯ W , then ε α ( x + · · · + x n + 2 R ) is also an element of ¯ W and both ¯ X and ε α ( x + · · · + x n + 2 R ) map to ( x + · · · + x n + P , . . . , x + · · · + x n + P r ) under theabelianization map.Third, S maps to a generating set of the abelianization of Sp ( R ) and hence by the previousobservation, there must be a y ∈ R such that ( y + P , . . . , y + P r ) = (1 , , . . . , and ( y + P l , . . . , y + P l r r , y + Q k , . . . , y + Q k s s ) ∈ ¯ R . ut we already know that (0 + P l , . . . , P l r r , y + Q k , . . . , y + Q k s s ) ∈ ¯ R and so we obtainthat ( y + P l , y + P l · · · P l r r , Q k · · · Q l s s ) is an element of ¯ R . Next, observe that if z + 2 R is an element of ¯ R , then so is u · ( z + 2 R ) for any unit in R/ R. This can be seenby conjugating ε α ( z + 2 R ) with h β ( u − ) . But y is an element of P · · · P r and hence (1 + P l , y − P l · · · P l r r , Q k · · · Q k s s ) is a unit in R/ R and we already know by Lemma 5.1 that (0 + P l , (1 − y ) y + P l · · · P l r r , Q k · · · Q k s s ) is an element of ¯ R . So, we can conclude that (0 + P l , y + P l · · · P l r r , Q k · · · Q k s s ) is an element of ¯ R . Thus finally ( y + P l , P l · · · P l r r · Q k · · · Q l s s ) is an element of ¯ R . But y + P l is a unit in R/ P l by choice and hence for any unit u ∈ R/ P l the element ( u + P l , P l · · · P l r r · Q k · · · Q l s s ) is an element of ¯ R . But each element of R/ P l is a sumof at most two units according to [21, Proposition 6.4] and hence ( x + P l , P l · · · P l r r ·Q k · · · Q l s s ) is an element of ¯ R for all x ∈ R. But running through similar arguments forthe other prime ideals P , . . . , P r , then finally yields that { ( x + P l · · · P l r r , Q k · · · Q l s s ) | x ∈ R } ⊂ ¯ R . So summarizing, we finally obtain that indeed ¯ R = R/ R and this finishes the proof. (cid:3) Lower bounds for ∆ k (Sp ( R )) always depend on the number r ( R ) In this section, we prove Theorem 3. The proof is quite similar to the proof of [20,Theorem 2,3], so we will be rather brief about it. First, we need the following:
Proposition 6.1.
Let K be a field, t ∈ K − { } . Then the element E := ε β ( t ) normallygenerates Sp ( K ) and k Sp ( K ) k E ≥ . Further, k Sp ( K ) k E ≥ holds, if K = F . Proof.
The first claim is the content of [20, Proposition 5.1]. The second claim can be seenby noting that there is an isomorphism between Sp ( F ) and S that maps ε β (1) to thetransposition (4 , . Hence it suffices to show that S is normally generated by (4 , and k S k (4 , ≥ . The first claim is well-known and to see the the second one observe that for σ ∈ S the number of orbits of the induced group action of the cyclic subgroup h σ i on { , . . . , } only depends on the conjugacy class of σ in S and not on the permutation σ itself. However, for k ∈ { , . . . , } , a product of k transpositions in S has at least − k suchorbits in { , . . . , } . Thus the cyclce (1 , , , , , , which gives rise to just one such orbit,cannot be written as a product of at most transpositions and hence k S k (4 , ≥ . . (cid:3) This proposition can be used now to prove Theorem 3:
Proof.
We can assume without loss that R = 2 R. Let R split into distinct prime ideals asfollows: R = r Y i =1 P l i i ! · s Y j =1 Q k j j ! with [ R/ P i : F ] = 1 for ≤ i ≤ r and [ R/ Q j : F ] > for ≤ j ≤ s. Next, let c be the classnumber of R . Pick elements x , . . . , x r ∈ R such that P ci = ( x i ) for all i. Also choose r + 1 istinct prime ideals V r +1 , . . . , V k in R different from all the P , . . . , P r , Q , . . . , Q s . Passing tothe powers V cr +1 , . . . , V ck , we can find elements v r +1 , . . . , v k ∈ R with V cr +1 = ( v r +1 ) , . . . , V ck =( v k ) . Further, define the following elements for ≤ u ≤ r,r u := Y ≤ i = u ≤ r x i ! · v r +1 · · · v k . For k ≥ u ≥ r + 1 set r u := x · · · x r · Y r +1 ≤ u = q ≤ k v q ! . We consider the set S := { ε β ( r ) , . . . , ε β ( r k ) } in Sp ( R ) . Then clearly Π( S ) = ∅ holds.Further, according to the proof of [21, Theorem 6.3], the abelianization map A : Sp ( R ) → Sp ( R ) / [Sp ( R ) , Sp ( R )] = F r ( R )2 is uniquely defined through A ( ε φ ( x )) = ( x + P , . . . , x + P r ( R ) ) for all x ∈ R and Φ ∈ C . But then S clearly maps to a generating set of F r ( R )2 . ThusTheorem 4 implies that S is indeed a normally generating set of Sp ( R ) . Next, consider themap π : Sp ( R ) → r Y i =1 Sp ( R/ P i ) × k Y j = r +1 Sp ( R/V j ) , X ( π P ( X ) , . . . , π P r ( X ) , π V r +1 ( X ) , . . . , π V k ( X )) for the π P i : Sp ( R ) → Sp ( R/ P i ) and π V j : Sp ( R ) → Sp ( R/V j ) being the reductionhomomorphisms. Then note that k Sp ( R ) k S ≥ k r Y i =1 Sp ( R/ P i ) ! × k Y j = r +1 Sp ( R/V j ) ! k π ( S ) = r X i =1 k Sp ( R/ P i ) k ε β ( r i + P i ) ! + k X j = r +1 k Sp ( R/V j ) k ε β ( r j + V j ) ! . So to finish the proof, it suffices to apply Proposition 6.1 to obtain k Sp ( R/ P i ) k ε φ ( x i + P i ) ≥ and k Sp ( R/V j ) k ε β ( r j + V j ) ≥ for all i = 1 , . . . , r and j = r + 1 , . . . , k . (cid:3) Closing remarks
We should also mention that contrary to how similar the question of strong boundednessof G ( R ) appeared to Sp ( R ) in [21], it is actually easier to determine ∆ k ( G ( R )) than ∆ k (Sp ( R )) . This is mainly due to the fact that the ’intermediate’ normal subgroup weconstruct in the proof of [21, Theorem 5.13] is not the R -congruence subgroup N G , R butinstead the bigger group N G = hh{ ε φ (2 x ) | x ∈ R, φ ∈ G short } ∪ { ε φ ( x ) | x ∈ R, φ ∈ G long }ii . We might address this in a future paper. eferences [1] Eiichi Abe and Kazuo Suzuki. On normal subgroups of Chevalley groups over commutative rings. TohokuMath. J. (2) , 28(2):185–198, 1976.[2] H. Bass, J. Milnor, and J.-P. Serre. Solution of the congruence subgroup problem for SL n ( n ≥ and Sp n ( n ≥ . Inst. Hautes Études Sci. Publ. Math. , (33):59–137, 1967.[3] J. L. Brenner. Covering theorems for FINASIGs. VIII. Almost all conjugacy classes in A n have exponent ≤ . J. Austral. Math. Soc. Ser. A , 25(2):210–214, 1978.[4] Dmitri Burago, Sergei Ivanov, and Leonid Polterovich. Conjugation-invariant norms on groups of geo-metric origin. In
Groups of diffeomorphisms , volume 52 of
Adv. Stud. Pure Math. , pages 221–250. Math.Soc. Japan, Tokyo, 2008.[5] David Carter and Gordon Keller. Bounded elementary generation of SL n ( O ) . Amer. J. Math. ,105(3):673–687, 1983.[6] Douglas L. Costa and Gordon E. Keller. Radix redux: normal subgroups of symplectic groups.
J. ReineAngew. Math. , 427:51–105, 1992.[7] Światosław R. Gal and Jarek Kędra. On bi-invariant word metrics.
J. Topol. Anal. , 3(2):161–175, 2011.[8] James E. Humphreys.
Linear algebraic groups, corrected fifth printing . Springer-Verlag, New York-Heidelberg, 1975. Graduate Texts in Mathematics, No. 21.[9] S. Karni. Covering numbers of groups of small order and sporadic groups. In
Products of conjugacyclasses in groups , volume 1112 of
Lecture Notes in Math. , pages 52–196. Springer, Berlin, 1985.[10] Jarek Kędra, Assaf Libman, and Ben Martin. On boundedness properties of groups.
In preparation. ,https://arxiv.org/abs/1808.01815.[11] R. Lawther and Martin W. Liebeck. On the diameter of a Cayley graph of a simple group of Lie typebased on a conjugacy class.
J. Combin. Theory Ser. A , 83(1):118–137, 1998.[12] Martin W. Liebeck and Aner Shalev. Diameters of finite simple groups: sharp bounds and applications.
Ann. of Math. (2) , 154(2):383–406, 2001.[13] Daniel A. Marcus.
Number fields . Universitext. Springer, Cham, 2018. Second edition of [ MR0457396],With a foreword by Barry Mazur.[14] Aleksander V. Morgan, Andrei S. Rapinchuk, and Balasubramanian Sury. Bounded generation of SL over rings of S -integers with infinitely many units. Algebra Number Theory , 12(8):1949–1974, 2018.[15] Dave Witte Morris. Bounded generation of
SL( n, A ) (after D. Carter, G. Keller, and E. Paige). NewYork J. Math. , 13:383–421, 2007.[16] Jürgen Neukirch.
Algebraic number theory , volume 322 of
Grundlehren der Mathematischen Wis-senschaften [Fundamental Principles of Mathematical Sciences] . Springer-Verlag, Berlin, 1999. Trans-lated from the 1992 German original and with a note by Norbert Schappacher, With a foreword by G.Harder.[17] B. Nica. On bounded elementary generation for sl n over polynomial rings.
Isr. J. Math. , pages 403–410,2018.[18] Robert Steinberg.
Lectures on Chevalley groups , volume 66 of
University Lecture Series . AmericanMathematical Society, Providence, RI, 2016.[19] O. I. Tavgen. Bounded generability of Chevalley groups over rings of S -integer algebraic numbers. Izv.Akad. Nauk SSSR Ser. Mat. , 54(1):97–122, 221–222, 1990.[20] Alexander Trost. Explicit strong boundedness for higher rank symplectic groups. https://arxiv.org/pdf/2012.12328.pdf , 2020.[21] Alexander Trost. Strong boundedness of simply connected Split Chevalley groups defined over rings. https://arxiv.org/pdf/2004.05039.pdf , 2020, submitted.[22] Alexander Trost. Quantitative aspects of normal generation of split Chevalley groups.
PhD thesis , Uni-versity of Aberdeen 2020, supervised by B. Martin.[23] Peter Vámos. 2-good rings.
Q. J. Math. , 56(3):417–430, 2005.
University of Aberdeen
Email address : [email protected]@abdn.ac.uk