Non-virtually abelian anisotropic linear groups are not boundedly generated
Pietro Corvaja, Andrei Rapinchuk, Jinbo Ren, Umberto Zannier
aa r X i v : . [ m a t h . G R ] J a n NON-VIRTUALLY ABELIAN ANISOTROPIC LINEAR GROUPS ARE NOTBOUNDEDLY GENERATED
PIETRO CORVAJA, ANDREI S. RAPINCHUK, JINBO REN, AND UMBERTO M. ZANNIER
Abstract.
We prove that if a linear group Γ ⊂ GL n ( K ) over a field K of characteristic zero isboundedly generated by semi-simple (diagonalizable) elements then it is virtually solvable. As a con-sequence, one obtains that infinite S -arithmetic subgroups of absolutely almost simple anisotropicalgebraic groups over number fields are never boundedly generated. Our proof relies on Laurent’stheorem from Diophantine geometry and properties of generic elements. Introduction
An abstract group Γ has the property of bounded generation (BG) if there exist elements γ , . . . , γ r ∈ Γ (not necessarily distinct) such that(BG) Γ = h γ i · · · h γ r i , where h γ i denotes the cyclic subgroup of Γ generated by an element γ ∈ Γ. (In this case, we also saythat Γ is boundedly generated by γ , . . . , γ r .) Obviously, every group satisfying (BG) is finitely gen-erated. Conversely, every finitely generated (virtually) abelian, or more generally (virtually) nilpotent,group has bounded generation - note that such a group is also linear. On the other hand, the fact thatthere exist non-virtually solvable linear groups having (BG) is rather nontrivial. Historically, thefirst examples emerged as a consequence of the result of D. Carter and G. Keller [5] that for any n ≥ O of algebraic integers, every unimodular ( n × n )-matrix with entries in O is a product ofa bounded number (with a bound depending on n and the discriminant of O ) of elementary matricesover O . Indeed, this fact immediately implies the existence for Γ = SL n ( O ) of a presentation (BG) inwhich the γ i ’s are suitable elementary (in particular, unipotent) matrices. Although SL ( O ) does nothave (BG) when O is either Z or the ring of integers of an imaginary quadratic field, it does have (BG)in all other arithmetic situations, i.e. when O is the ring of S -integers of a number field with infiniteunit group O × . Again, this is derived from the fact that in the situation at hand every matrix inSL ( O ) is a product of at most 9 elementary matrices [20], and then the resulting presentation (BG)involves a mix of unipotent and semi-simple (diagonalizable) matrices, with the unipotent ones defi-nitely present. The result of [5] was extended by O.I. Tavgen [33] to all Chevalley groups of rank > S -arithmetic subgroups in isotropic, butnot necessarily quasi-split, orthogonal groups of quadratic forms over number fields was established(under some natural assumptions) in [8]. These (and some other) results generated the expectationthat higher rank S -arithmetic subgroups of absolutely almost simple algebraic groups over numberfields should have (BG).To put this problem into perspective, we recall that while being a purely combinatorial propertyof groups, bounded generation has a number of interesting consequences and applications in differentareas. First, a group having (BG) in which every finite index subgroup has finite abelianization is SS -rigid , i.e. has only finitely many inequivalent completely reducible complex representations ineach dimension [22, Appendix A]. Second, it was shown in [17] and [21] that S -arithmetic subgroupsof absolutely almost simple algebraic groups that have (BG) under some natural assumptions possessthe congruence subgroup property in the sense that the corresponding congruence kernel is finite.Third, property (BG) played a crucial role in the proof of the Margulis-Zimmer conjecture on com-mensurated subgroups for higher rank S -arithmetic subgroups of Chevalley groups [32], estimation ofKazhdan constants [15], [31] and other situations. For the sake of completeness, we also mention that Mathematics Subject Classification.
Primary 11F06, Secondary 11D72. the natural analog of (BG) is of significance in the theory of profinite groups; in particular, the pro- p groups with bounded generation are precisely the p -adic analytic groups [7]. Thus, bounded generationhas long been regarded as an abstract property that can potentially provide a uniform approach tosome important problems for S -arithmetic subgroups including Serre’s congruence subgroup and theMargulis-Zimmer conjectures, and also explain some rigidity phenomena. This would be particularlyvaluable in the anisotropic case where many results involving arithmetic groups often rely on ad hoc techniques. We recall that a semi-simple algebraic group G over a field K of characteristic zero is K -anisotropic (i.e., has K -rank zero) if and only if the group G ( K ) contains no nontrivial unipotentelements [3], hence consists entirely of semi-simple elements. Building on this characterization, we willcall anisotropic any subgroup Γ ⊂ GL n ( K ) (in other words, any linear group) that contains onlysemi-simple elements. It should be noted that while quite a few examples of S -arithmetic groups with(BG) have been found over the years, none of these was anisotropic, which brings us to the following. Question A.
Can (BG) possibly hold for an infinite S -arithmetic subgroup of an anisotropic absolutelyalmost simple algebraic group? To approach this question, one may first try to re-examine the nature of presentations (BG) thatarise in the known examples of boundedly generated groups. As we pointed out above, in the presen-tations (BG) for SL n , n ≥
2, over the rings of algebraic S -integers in appropriate situations that arederived from bounded generation of these groups by elementary matrices, some or even all elements γ i are unipotent. So, one may wonder if in these (or some other) examples one can produce a presenta-tion (BG) with all the γ i ’s being semi-simple – in which we would say that Γ is boundedly generatedby semi-simple elements. Along these lines, one can ask the following general question which in a waysubsumes Question A. Question B.
Which linear groups are boundedly generated by semi-simple elements?
The goal of this paper is to give, in the case of arbitrary linear groups over a field of characteristiczero, a powerful necessary condition for bounded generation by semi-simple elements. This condition,in particular, leads to a negative answer to Question A.
Theorem 1.1.
Let Γ ⊂ GL n ( K ) be a linear group over a field K of characteristic zero, which is notvirtually solvable. Then in any possible presentation (BG) for Γ at least two of the elements γ i must benon-semi-simple. In particular, a linear group over a field of characteristic zero boundedly generatedby semi-simple elements is virtually solvable. There exist virtually solvable finitely generated linear groups that do not admit bounded generationby semi-simple elements (see Example 7.1), so Theorem 1.1 is not a criterion. However, it yields thefollowing criterion in the case of anisotropic groups.
Corollary 1.2.
An anisotropic linear group Γ ⊂ GL n ( K ) over a field of characteristic zero has (BG)if and only if it is finitely generated and virtually abelian. To formulate our result for S -arithmetic groups, we need to introduce one additional notation. Let G be a linear algebraic group defined over a number field K , and let S be a finite set of valuations of K containing all archimedean ones. We set G S := Y v ∈ S G ( K v ) , where K v denotes the completion of K with respect to v , and recall that for G semi-simple, the non-compactness of G S is equivalent to the fact that the S -arithmetic subgroups of G are infinite (cf. [22, § Theorem 1.3.
Let G be an algebraic group over a number field K , and let S be a finite set ofvaluations of K containing all archimedean ones. Assume that G possesses a K -defined semi-simple K -anisotropic normal subgroup H such that the group H S is non-compact. Then the S -arithmeticsubgroups of G are not boundedly generated. In particular, infinite S -arithmetic subgroups of absolutelyalmost simple K -anisotropic groups are not boundedly generated. OUNDED GENERATION 3
Finally, we would like to make a remark about the methods used to prove these results. One generaltechnique that has been used to show that various groups, including free amalgamated products andHNN-extensions subject to certain conditions , lattices in rank one groups etc., do not have boundedgeneration involves bounded cohomology, cf. [10]-[12], [13]-[14]. More precisely, one observes that if agroup is boundedly generated then its second bounded cohomology is a finite-dimensional real vectorspace, and then disproves bounded generation by showing that in the cases of interest this spaceis actually infinite dimensional. Unfortunately, this approach cannot be used to show the absenceof bounded generation in the higher rank S -arithmetic subgroups of anisotropic absolutely almostsimple algebraic groups as for these groups the second bounded cohomology vanishes [4]. Instead, ourmethod hinges on the results from Diophantine geometry (more specifically, Laurent’s theorem – seeTheorem 4.2 below) and uses the existence of generic elements in Zariski-dense subgroups [23]-[25].The application of these techniques, particularly in such combination, in the context of group theoryappears to be novel, so it would be interesting to see whether these can be employed to tackle someother group-theoretic problems.The structure of the paper is the following. In §
2, we show that it is enough to prove Theorem 1.1for a subgroup Γ ⊂ GL n ( K ) where K is a number field. In § §§ n ( K ) (where K is a number field) generated by elements which are semi-simple with one possibleexception – see Theorem 4.1. We then prove both Theorems 1.1 and 1.3 and Corollary 1.2 in § § A reduction to linear groups over number fields
The goal of this section is to reduce the proof of Theorem 1.1 to the case of linear groups overnumber fields. The argument is based on the following proposition that enables us to construct asuitable specialization.
Proposition 2.1.
Let R be a finitely generated Q -algebra without zero divisors. Given a non-virtuallysolvable subgroup Γ ⊂ GL n ( R ) and semi-simple elements γ , . . . , γ r ∈ Γ , there exists a Q -algebrahomomorphism θ : R → F to a number field F such that for the corresponding group homomorphism Θ : GL n ( R ) → GL n ( F ) , the image Θ(Γ) is not a virtually solvable group and each of the elements Θ( γ ) , . . . , Θ( γ r ) is semi-simple. We note that similar statements but without the assertion of the semi-simplicity of the images ofthe given semi-simple elements can be found in [9, §
3] and [18, Proposition 16.4.13]. For the proof weneed the following lemma. Given a group ∆, we let D m (∆) denote the m -th term of the derived seriesof ∆, and for ℓ ∈ N let ∆ ( ℓ ) denote the (normal) subgroup of ∆ generated by the ℓ -th powers of itselements. Lemma 2.2.
There exists ℓ = ℓ ( n ) ∈ N such that for any virtually solvable subgroup ∆ ⊂ GL n ( E ) ,where E is a field of characteristic zero, we have D n +1 (∆ ( ℓ ) ) = { I n } . Proof.
We can assume without loss of generality that E is algebraically closed. Let G be the Zariski-closure of ∆. Since ∆ is virtually solvable, the connected component G ◦ is solvable, hence triangular-izable by the Lie-Kolchin theorem (cf. [2, Corollary 10.5]). It follows that(1) D n ( G ◦ ) = { I n } . On the other hand, according to Bass’s generalization of Jordan’s theorem [1, Theorem 1], there exists j ∈ N depending only on n such that G/G ◦ has an abelian normal subgroup H of index at most j .Set ℓ = j !. Then for the canonical morphism f : G → G/G ◦ we have f (∆ ( ℓ ) ) = f (∆) ( ℓ ) ⊂ H, See § P. CORVAJA, A. RAPINCHUK, J. REN, AND U. ZANNIER and therefore f ( D (∆ ( ℓ ) )) = { e } , i.e. D (∆ ( ℓ ) ) ⊂ G ◦ . Combining this with (1), we obtain ourclaim. (cid:3) Beginning the proof of Proposition 2.1, let us first show that(2) D n +1 (Γ ( ℓ ) ) = { I n } , where ℓ = ℓ ( n ) is the constant from Lemma 2.2. Assume the contrary, and let G and H denote theZariski-closures of Γ and Γ ( ℓ ) , respectively; clearly, H is a normal subgroup of G . For the power map µ : G → G, g g ℓ , we have µ (Γ) ⊂ H , and therefore µ ( G ) ⊂ H . This means that the quotient G/H is an algebraic groupof the finite exponent ℓ , hence finite since char E = 0. On the other hand, since D n +1 (Γ ( ℓ ) ) = { I n } ,the group H is solvable. This means that the finite index subgroup Γ ∩ H ⊂ Γ is solvable, making Γvirtually solvable. This contradicts our assumption proving (2).According to (2), we can pick an element g ∈ D n +1 (Γ ( ℓ ) ) \ { I n } , and let a be any nonzero entry ofthe matrix g − I n . Next, let L be the field of fractions of R , and let f i ( t ) ∈ L [ t ] will be the minimalpolynomial of the matrix γ i ∈ GL n ( R ). Replacing R with a larger finitely generated Q -subalgebra R ′ ⊂ L , we can assume that f i ∈ R [ t ]. Since γ i is semi-simple, the polynomial f i does not havemultiple roots, hence its discriminant d i is = 0. Set d = d · · · d r . Replacing R by the localization R (cid:2) ad (cid:3) , we can assume that both a and d are invertible in R . Let m be a maximal ideal of R . Since R is a finitely generated Q -algebra, it follows from a version of Nullstellensatz (cf. [16, Ch. IX, Corollary1.2]) that F := R/ m is a finite extension of Q . We will now show that the canonical homomorphism θ : R → F is as required.First, since a ∈ R × , we have θ ( a ) = 0, and therefore Θ( g ) = I n . Thus, for ∆ = Θ(Γ) we have D n +1 (∆ ( ℓ ) ) = Θ( D n +1 (Γ ( ℓ ) )) = { I n } , which by Lemma 2.2 implies that ∆ is not virtually solvable. Second, since d ∈ R × , for each i = 1 , . . . , r we have θ ( d i ) = 0, which means that the (monic) polynomial ¯ f i ( t ) ∈ F [ t ], obtained by applying θ to the coefficients of f i , does not have multiple roots. Since ¯ f i (Θ( γ i )) = 0, the minimal polynomialof the matrix Θ( γ i ) does not have multiple roots either, implying that this matrix is semi-simple, asrequired. Reduction 2.3.
If the first assertion of Theorem 1.1 is valid for all subgroups ∆ ⊂ GL n ( F ) where F is a number field (i.e., every such subgroup that has a presentation (BG) in which all elements γ i ,with one possible exception, are semi-simple is necessarily virtually solvable) then it is also valid forall subgroups Γ ⊂ GL n ( K ) where K is any field of characteristic zero. Indeed, assume that there is a non-virtually solvable subgroup Γ ⊂ GL n ( K ) that admits a pre-sentation (BG) in which all matrices γ i , with one possible exception, are semi-simple. Let R be the Q -subalgebra generated by the entries of all the γ i ’s and their inverses; clearly Γ ⊂ GL n ( R ). ThenProposition 2.1 yields a homomorphism θ : R → F to a number field F such that for the correspondinghomomorphism Θ : GL n ( R ) → GL n ( F ), the group ∆ = Θ(Γ) is not virtually solvable and all matricesΘ( γ ) , . . . , Θ( γ r ), with one possible exception, are semi-simple. Applying Θ to (BG), we obtain∆ = h Θ( γ ) i · · · h Θ( γ r ) i . Then by our assumption ∆ must be virtually solvable, which is not the case by our construction. Acontradiction, justifying the reduction.3.
On generic elements and their eigenvalues
The proof of Theorem 1.1 requires a result (see Proposition 3.5 below) stating that given a finiteset of elements in a linear group over a field of characteristic zero whose Zariski-closure has semi-simple connected component, the group always contains a semi-simple element with the eigenvaluesmultiplicatively independent from those of the given elements. This fact is interesting in its own right,and its proof relies on the existence and properties of generic elements , which we will now recall.
OUNDED GENERATION 5
Let G be a semi-simple algebraic group over a field K , and let T be a maximal K -torus of G .The absolute Galois group G = Gal( K sep /K ) naturally acts on the character group X ( T ), and thisaction leaves the corresponding root system Φ := Φ( G, T ) invariant, yielding a (continuous) grouphomomorphism ρ T : G −→
Aut(Φ)to the automorphism group of Φ.
Definition 3.1.
A maximal K -torus T is said to be generic over K if the image im ρ T contains theWeyl group W (Φ) ⊂ Aut(Φ) . Furthermore, a regular semi-simple element γ ∈ G ( K ) is K - generic ifthe K -torus T := Z G ( g ) ◦ is generic over K . We refer the reader to [24, §
9] for a discussion of these notions. The following existence theoremis proved in [23] (see also [24, Theorem 9.6]); its extension to Zariski-dense subgroups of absolutelyalmost simple algebraic groups over fields of positive characteristic is given in [25].
Theorem 3.2.
Let G be a semi-simple algebraic group over a finitely generated field K of characteristiczero , and let Γ ⊂ G ( K ) be a finitely generated Zariski-dense subgroup. Then Γ contains a K -genericsemi-simple element without components of finite order. Here the components of an element are understood in terms of the decomposition G = G · · · G r asan almost direct product of absolutely almost simple groups. We note that a K -generic regular semi-simple element without components of finite order generates a Zariski-dense subgroup of a maximaltorus that contains it, making this torus unique, cf. [25, p. 22]. We also note that Lemma 3.3.
Let G be a semi-simple algebraic group over a finitely generated field K of characteristiczero, and let Γ ⊂ G ( K ) be a finitely generated Zariski dense subgroup. Then for any finitely generatedextension L of K , there exists a regular semi-simple element γ ∈ Γ without components of finite orderthat satisfies the following condition: for the torus T := Z G ( γ ) ◦ and any character χ ∈ X ( T ) , the factthat χ ( γ ) ∈ L × implies χ ( γ ) = 1 .Proof. Since L is itself a finitely generated field of characteristic zero, we can use Theorem 3.2 to finda regular semi-simple element γ ∈ Γ without components of finite order that is generic over L . Let T = Z G ( γ ) ◦ , and let χ ∈ X ( T ) be a character such that χ ( γ ) ∈ L × . Then for any σ ∈ Gal(
L/L ) wehave ( σ ( χ ))( γ ) = σ ( χ ( σ − ( γ ))) = χ ( γ ) . As we observed earlier, the cyclic group h γ i is Zariski-dense in T , so the above equation yields σ ( χ ) = χ .On the other hand, the fact that T is L -generic implies that X ( T ) does not contain any nontrivialGal( L/L )-fixed elements. Thus, χ = 0 and χ ( γ ) = 1. (cid:3) For γ ∈ GL n ( K ), we let Λ( γ ) denote the subgroup of K × generated by all eigenvalues of γ in K × .We observe that if γ lies in T ( K ) for some K -torus T ⊂ GL n , then Λ( γ ) coincides with the set of allvalues χ ( γ ) of the characters χ ∈ X ( T ). Definition 3.4.
Let γ, γ , . . . , γ r ∈ GL n ( K ) . We say that the eigenvalues of γ are multiplicativelyindependent from those of γ , . . . , γ r if (3) Λ( γ ) ∩ [Λ( γ ) · · · Λ( γ r )] = { } . Proposition 3.5.
Let Γ ⊂ GL n ( K ) be a finitely generated linear group over a field K of characteristiczero, and let G be its Zariski-closure. Assume that the connected component G ◦ is a nontrivial semi-simple group. Then for any γ , . . . , γ r ∈ Γ there exists a semi-simple element γ ∈ Γ ∩ G ◦ whoseeigenvalues are multiplicatively independent from those of γ , . . . , γ r and for which the subgroup Λ( γ ) is nontrivial and torsion-free. I.e., a finitely generated extension of Q . P. CORVAJA, A. RAPINCHUK, J. REN, AND U. ZANNIER
Proof.
Since Γ is finitely generated, we can assume that the field K is also finitely generated. BySelberg’s Lemma (cf. [26, 6.11]), we can choose a finite index subgroup Γ ′ ⊂ Γ ∩ G ◦ which is neat ,i.e. satisfies the property that for any δ ∈ Γ ′ the subgroup Λ( δ ) is torsion-free; obviously, Γ ′ is Zariski-dense in G ◦ . Let L be the extension of K generated by the eigenvalues of γ , . . . , γ r , and let γ ∈ Γ ′ bea regular semi-simple element provided by Lemma 3.3 for the group G ◦ . To show that this element isas required, we only need to verify condition (3). However, any element x ∈ Λ( γ ) ∩ [Λ( γ ) · · · Λ( γ r )]can be written in the form χ ( γ ) for some character χ of the maximal torus Z G ◦ ( γ ) ◦ and, on the otherhand, belongs to Λ( γ ) · · · Λ( γ r ) ⊂ L × . So, x = 1 by Lemma 3.3 completing the argument. (cid:3) Remark 3.6.
1. The assertion of Proposition 3.5 is false for virtually solvable subgroups Γ ⊂ GL n ( K ).2. Using the adjoint representation, it is not difficult to show that any linear group Γ ⊂ GL n ( K )over a field of characteristic zero with semi-simple connected component of the Zariski-closure containsa finitely generated subgroup having the same Zariski-closure as Γ. This observation enables one todrop the assumption of finite generation of Γ in all statements of this section.4. The key matrix statement: Proof in Case 1
The main results of the paper will be derived in § Theorem 4.1.
Assume that the matrices γ , . . . , γ r ∈ GL n ( Q ) , with one possible exception, are semi-simple. Then for any semi-simple matrix γ ∈ GL n ( Q ) that has an eigenvalue λ ∈ Q × which isnot a root of unity and for which h λ i ∩ [Λ( γ ) . . . Λ( γ r )] = { } (cf. Definition 3.4), the intersection h γ i ∩ h γ i · · · h γ r i is finite. In particular, h γ i 6⊂ h γ i · · · h γ r i . The proof of this theorem critically depends on the following result from Diophantine geometry –see [6, Theorem 2.7].
Theorem 4.2 (Laurent’s theorem) . Let Ω be a finitely generated subgroup of ( Q × ) N for some N > ,and let Σ ⊂ Ω be a subset. Then the Zariski closure of Σ in the torus T = ( G m ) N is a finite union oftranslates of algebraic subgroups of T . The proof of Theorem 4.1 will be divided into two cases.
Case 1.
All elements γ , . . . , γ r are semi-simple. This case will be treated below. Assuming that Theorem 4.1 has been established in this case,it remains to consider the case where among the elements γ , . . . , γ r exactly one element, say γ s (1 ≤ s ≤ r ), is not semi-simple. The Jordan decomposition γ s = συ , where σ and υ are commutingmatrices with σ semi-simple and υ unipotent, leads to the inclusion h γ i · · · h γ r i ⊂ h γ i · · · h γ s − ih σ ih υ ih γ s +1 i · · · h γ r i . Thus, it is enough to prove that under the assumptions made in Theorem 4.1, the intersection h γ i ∩ [ h γ i · · · h γ s − ih σ ih υ ih γ s +1 i · · · h γ r i ]is finite. On the other hand, we have Λ( σ ) = Λ( γ s ) and Λ( υ ) = { } , so the assumption on λ in thetheorem translates into h λ i ∩ [Λ( γ ) · · · Λ( γ s − )Λ( σ )Λ( υ )Λ( γ s +1 ) · · · Λ( γ r )] = { } . So, re-labeling the elements, we see that to complete the proof of Theorem 4.1, it is enough to consider
Case 2.
For some s ∈ { , . . . , r } , the elements γ , . . . , γ s − , γ s +1 , . . . γ r are semi-simple and theelement γ s = 1 is unipotent. In this section we will treat Case 1, postponing Case 2 until the next section. We begin by estab-lishing the following application of Theorem 4.2 that will be used in both cases.
OUNDED GENERATION 7
Proposition 4.3.
Suppose we are given a polynomial f ( x, x , . . . , x d ) ∈ Q [ x, x , . . . , x d ] and algebraicnumbers µ, µ , . . . , µ d ∈ Q × , with µ not a root of unity. Assume that h µ i ∩ h µ , . . . , µ d i = { } . Then the set M of integers m ∈ Z such that there exist a ( m ) , . . . , a d ( m ) ∈ Z satisfying f (cid:16) µ m , µ a ( m )1 , . . . , µ a d ( m ) d (cid:17) = 0 and f (cid:16) x, µ a ( m ) , . . . , µ a d ( m ) d (cid:17) is non-constant , is finite.Proof. For each m ∈ M we fix a d -tuple ( a ( m ) , . . . , a d ( m )) ∈ Z d as in the above description of M .Let N = 1 + d and T = ( G m ) N be an N -dimensional Q -split torus, the coordinate functions on whichwill be denoted x, x , . . . , x d . Furthermore, we let Ω denote the subgroup h µ i × h µ i × · · · × h µ d i of T ( Q ) = ( Q × ) N ; clearly, Ω is finitely generated. Consider the subset Σ ⊂ Ω consisting of the elements σ ( m ) := (cid:16) µ m , µ a ( m )1 , . . . , µ a d ( m ) d (cid:17) for m ∈ M . By Laurent’s Theorem 4.2, the Zariski-closure Σ of Σin T is of the form(4) Σ = b [ ℓ =1 s ℓ T ℓ for some s ℓ ∈ T ( Q ) and some algebraic subgroups T ℓ of T for ℓ = 1 , . . . , b . We may assume that thisdecomposition of Σ is minimal, i.e. none of the cosets can be dropped. Lemma 4.4.
None of the subgroups T ℓ is of the form G m × T ′ ℓ for some algebraic subgroup T ′ ℓ of ( G m ) N − (last ( N − components).Proof. By construction, the polynomial f ( x, x , . . . , x d ) vanishes on Σ, hence also vanishes on Σ. Inparticular, it vanishes on each coset s ℓ T ℓ . Suppose that for some ℓ ∈ { , . . . , b } we have T ℓ = G m × T ′ ℓ .Then(5) s ℓ T ℓ = G m × ( s ′ ℓ T ′ ℓ ) , where s ′ ℓ is the projection of s ℓ to ( G m ) N − . The minimality of (4) implies that s ℓ T ℓ contains an element (cid:16) µ m , µ a ( m )1 , . . . , µ a d ( m ) d (cid:17) ∈ Σ. Due to (5), this means that f vanishes on G m × n(cid:16) µ a ( m )1 , . . . , µ a d ( m ) d (cid:17)o .But this is impossible since by our assumption the polynomial f (cid:16) x, µ m , µ a ( m )1 , . . . , µ a d ( m ) d (cid:17) is non-constant. (cid:3) Now, if M is infinite, then by the pigeonhole principle, one can find m , m ∈ M , m = m , suchthat the elements σ ( m ) and σ ( m ) belong to the same coset s ℓ T ℓ . Then(6) σ ( m ) σ ( m ) − = (cid:16) µ m − m , µ a ( m ) − a ( m )1 , . . . , µ a d ( m ) − a d ( m ) d (cid:17) belongs to T ℓ . According to [2, Ch. III, 8.2], the subgroup T ℓ is the intersection of the kernels of allcharacters of T that vanish on it. Writing a character χ ∈ X ( T ) as(7) χ ( x, x , . . . , x d ) = x k x k · · · x k d d , we observe that if k = 0 then ker χ is of the form G m × T ′ . So, it follows from Lemma 4.4 that thereexists a character χ ∈ X ( T ) that vanishes on T ℓ and for which in the corresponding presentation (7)we have k = 0. Furthermore, since σ ( m ) σ ( m ) − ∈ T ℓ , it follows from (6) that µ k ( m − m ) ∈ h µ , . . . , µ d i . We obtain a contradiction completing thereby the proof of Proposition 4.3. (cid:3)
Proof of Theorem 4.1 in Case 1.
As usual, we let diag( u , . . . , u n ) denote the diagonal matrix withdiagonal entries u , . . . , u n . Since γ, γ , . . . , γ r are semi-simple, there exist g, g , . . . , g r ∈ GL n ( Q ) suchthat g − γg = diag( λ , . . . , λ n ) and g − i γ i g i = diag( λ i , . . . , λ in ) for i = 1 , . . . , r. P. CORVAJA, A. RAPINCHUK, J. REN, AND U. ZANNIER
Besides, we may assume that λ = λ . We introduce d = rn indeterminates x , . . . , x n , . . . , x r , . . . , x rn ,and let p ( x , . . . , x rn ) ∈ Q [ x , . . . , x rn ] denote the polynomial representing the (1 , g − · " r Y i =1 (cid:0) g i · diag( x i , . . . , x in ) · g − i (cid:1) · g. We set f ( x, x , . . . , x rn ) = x − p ( x , . . . , x rn ) observing that the polynomial f ( x, x , . . . , x rn ) isnon-constant for any x , . . . , x rn ∈ Q .Now, let J = { m ∈ Z | γ m ∈ h γ i · · · h γ r i } . Then for each m ∈ J we can make a choice of integers a ( m ) , . . . , a r ( m ) so that γ m = γ a ( m )1 · · · γ a r ( m ) r , and consequently λ m = p (cid:16) λ a ( m )11 , . . . , λ a ( m )1 n , . . . , λ a r ( m ) r , . . . , λ a r ( m ) rn (cid:17) . Then f (cid:16) λ m , λ a ( m )11 , . . . , λ a r ( m ) rm (cid:17) = 0 . This means that J is contained in the set M constructed in Proposition 4.3 for the polynomial f andtaking µ to be λ and µ , . . . , µ d to be λ , . . . , λ rn . We note that then h µ , . . . , µ d i = Λ( γ ) · · · Λ( γ r ) . It follows that the assumptions of Proposition 4.3 do hold in our situation, so M is finite. So, J isalso finite, and our claim follows.5. The key matrix statement: Proof in Case 2
In this section we will prove Theorem 4.1 in Case 2 (cf. § γ , . . . , γ s − , γ s +1 , . . . , γ r are semi-simple and the element γ s = 1 is unipotent. Again, we need toshow that the set J = { m ∈ Z | γ m ∈ h γ i · · · h γ r i} is finite. Emulating the construction in §
4, we find g, g , . . . , g s − , g s +1 , . . . , g r ∈ GL n ( Q ) so that g − γg = diag( λ , . . . , λ n ) and g − i γ i g i = diag( λ i , . . . , λ in ) for i = 1 , . . . , s − , s + 1 , . . . , r, where we may assume that λ = λ . On the other hand, γ s = I n + ν where ν = 0 and ν n = 0. Thenby the binomial expansion, for any m ∈ Z we have γ ms = n − X k =0 (cid:18) mk (cid:19) ν k where as usual (cid:18) mk (cid:19) = m ( m − · · · ( m − k + 1) k ! for k ≥ (cid:18) m (cid:19) = 1 for any m ∈ Z . It follows that there exists an ( n × n )-matrix A ( z ) with entries in Q [ z ] such that γ ms = A ( m ) for all m ∈ Z . We now introduce the indeterminates x ij for i ∈ { , . . . , s − , s + 1 , . . . , r } and j ∈ { , . . . , n } .Then for any α, β ∈ { , . . . , n } we let p αβ ( x ij , z ) denote the polynomial in Q [ x , . . . , x rn , z ] thatrepresent the ( α, β )-entry of the matrix P ( x ij , z ) := g − · " s − Y i =1 (cid:0) g i · diag( x i , . . . , x in ) · g − i (cid:1) · A ( z ) · r Y i = s +1 (cid:0) g i · diag( x i , . . . , x in ) · g − i (cid:1) g. Set q ( x, x ij , z ) := x − p ( x ij , z ). For each m ∈ J we can fix a ( m ) , . . . , a r ( m ) ∈ Z such that γ m = γ a ( m )1 · · · γ a r ( m ) r . Then for all m ∈ J we have(8) q (cid:16) λ m , λ a ( m )11 , . . . , λ a r ( m ) rn , a s ( m ) (cid:17) = 0Furthermore, since the matrix g − γg is diagonal, we in addition have(9) p αβ (cid:16) λ a ( m )11 , . . . , λ a r ( m ) rn , a s ( m ) (cid:17) = 0 whenever α = β, OUNDED GENERATION 9 for all m ∈ J . Assume that J is infinite. To obtain a contradiction, we will eliminate z from the pairof polynomials q ( x, x ij , z ) and ˜ p αβ ( x ij , z ) for some α = β , where ˜ p αβ is a suitable truncation of p αβ , togenerate a polynomial f ( x, x ij ) having the following property: the set M constructed in Proposition4.3 for this f by taking µ = λ and µ , . . . , µ d to be λ , . . . , λ rn contains an infinite subset J ′ ⊂ J . Itwill be obvious that the assumptions of the proposition hold in our situation which will imply M isactually finite and provide the required contradiction.We will now pick a suitable pair of indices α, β ∈ { , . . . , n } , α = β , and construct a requiredtruncation ˜ p αβ ( x ij , z ). For any such pair we write p αβ ( x ij , z ) = ψ αβt αβ ( x ij ) z t αβ + · · · + ψ αβ ( x ij ) with ψ αβk ( x ij ) ∈ Q [ x ij ] , and then for k = 1 , . . . , t αβ setΨ αβk = n m ∈ J | ψ αβk (cid:16) λ a ( m )11 , . . . , λ a r ( m ) rn (cid:17) = 0 o . Lemma 5.1. \ α = β t αβ \ k =1 Ψ αβk = ∅ .Proof. Assume the contrary, and let m be an element of this intersection. Then all off-diagonal entriesof the matrix P ( z ) := P (cid:16) λ a ( m )11 , . . . , λ a r ( m ) rn , z (cid:17) are independent of z . Thus, since P ( a s ( m )) = g − · (cid:16) γ a ( m )1 · · · γ a s ( m ) s · · · γ a r ( m ) r (cid:17) · g = g − γ m g is diagonal, the matrix P ( z ) is diagonal for any z . In particular, P ( a s ( m ) + 1) = g − · (cid:16) γ a ( m )1 · · · γ a s ( m )+1 s · · · γ a r ( m ) r (cid:17) · g is diagonal. Then P ( a s ( m )) − P ( a s ( m ) + 1) = (cid:16) γ a s +1 ( m ) s +1 · · · γ a r ( m ) r g (cid:17) − · γ s · (cid:16) γ a s +1 ( m ) s +1 · · · γ a r ( m ) r g (cid:17) is also diagonal, contradicting the fact that γ s is a nontrivial unipotent matrix. (cid:3) Thus, J = [ α = β t αβ [ k =1 J \ Ψ αβk , so there exist α = β and k ∈ { , . . . , t αβ } such that J \ Ψ αβk is infinite. We fix one such pair ( α, β )and let t ≥ largest integer ≤ t αβ for which J \ Ψ αβt is infinite. Since J \ Ψ αβk is finite for t < k ≤ t αβ , the set J ′ := (cid:16) J \ Ψ αβt (cid:17) \ t αβ \ k = t +1 Ψ αβk is still infinite. Let ˜ p αβ ( x ij , z ) = ψ αβt ( x ij ) z t + · · · + ψ αβ ( x ij ) . It follows from (9) and our construction that for all m ∈ J ′ we have(10) ˜ p αβ (cid:16) λ a ( m )11 , . . . , λ a r ( m ) rn , a s ( m ) (cid:17) = 0 and ψ αβt (cid:16) λ a ( m )11 , . . . , λ a r ( m ) rn (cid:17) = 0 . Let p ( x ij , z ) = φ e ( x ij ) z e + · · · + φ ( x ij ) (this polynomial was denoted p in the proof of Case 1) sothat q ( x, x ij , z ) = − φ e ( x ij ) z e − · · · − φ ( x ij ) z + ( x − φ ( x ij )) . Referring to [16, Ch. IV, §
8] for the basic facts about resultants, we consider the resultant of thepolynomials q and ˜ p αβ with respect to the variable z : R z ( q, ˜ p αβ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − φ e ( x ij ) . . . − φ ( x ij ) x − φ ( x ij ). . . . . . . . . . . . − φ e ( x ij ) . . . − φ ( x ij ) x − φ ( x ij ) ψ αβt ( x ij ) ψ αβt − ( x ij ) . . . ψ αβ ( x ij ). . . . . . . . . . . . ψ αβt ( x ij ) ψ αβt − ( x ij ) . . . ψ αβ ( x ij ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t rows e rowsWe will view R z ( q, ˜ p αβ ) as a polynomial f ( x, x ij ) ∈ Q [ x, x ij ]. It is easy to see that deg x f = t and thecoefficient of x t is ± [ ψ αβt ( x ij )] e . It follows from (8) and (10) that for any m ∈ J ′ the polynomials q (cid:16) λ m , λ a ( m )11 , . . . , λ a r ( m ) rn , z (cid:17) and ˜ p αβ (cid:16) λ a ( m )11 , . . . , λ a r ( m ) rn , z (cid:17) have a common root z = a s ( m ), implying that f (cid:16) λ m , λ a ( m )11 , . . . , λ a r ( m ) rn (cid:17) = 0(cf. [16, Ch. IV, Prop. 8.1]). It also follows from (10) that f (cid:16) x, λ a ( m )11 , . . . , λ a r ( m ) rn (cid:17) is a non-constant polynomial. This means that J ′ is contained in the set M defined in Proposition 4.3 for ourpolynomial f and by taking µ = λ and µ , . . . , µ d to be λ , . . . , λ rn . Since h λ i ∩ [Λ( γ ) · · · Λ( γ r )] = { } by assumption, the condition h µ i ∩ h µ , . . . , µ d i = { } of Proposition 4.3 holds, allowing us to concludethat the set M is finite. This contradicts the fact that J ′ is infinite and completes the proof ofTheorem 4.1. 6. Proof of the main results
Proof of Theorem 1.1.
We need to show that if a linear group Γ ⊂ GL n ( K ) admits a presentation(BG) where at most one of the γ i ’s fails to be semi-simple, then Γ is virtually solvable. Accordingto Reduction 2.3, it is enough to consider the case where K is a number field. Letting G denote theZariski-closure of Γ, we consider the radical R of the connected component G ◦ (which is a normalsubgroup of G ) and the corresponding canonical morphism ϕ : G → G/R = G ′ . Clearly, G and R , hence also G ′ and ϕ , are defined over K , and in particular, we can choose a faithful K -definedrepresentation G ′ ֒ → GL n ′ . Since ϕ takes semi-simple elements to semi-simple elements (cf. [2, Ch. I,4.4]), among the elements γ ′ i = ϕ ( γ i ) there is at most one non-semi-simple. Furthermore, the subgroupΓ ′ = ϕ (Γ) of G ′ ( K ) ⊂ GL n ′ ( K ) has a presentationΓ ′ = h γ ′ i · · · h γ ′ r i . If we assume that Γ is not virtually solvable, then the connected component of the Zariski-closure G ′ of Γ ′ will be a nontrivial semi-simple group. Thus, what we need to show is that if a linear groupΓ ⊂ GL n ( K ) over a number field K is such that the connected component G ◦ of its Zariski-closure G is a nontrivial semi-simple group, then Γ cannot have a presentation (BG) where all γ i , with onepossible exception, are semi-simple.Assume the contrary. Using Proposition 3.5, we can find a semi-simple element γ ∈ Γ ∩ G ◦ whoseeigenvalues are multiplicatively independent from those of the elements γ , . . . , γ r in (BG) and forwhich the subgroup Λ( γ ) is nontrivial and torsion-free. In particular, γ has an eigenvalue λ which isnot a root of unity and for which h λ i ∩ [Λ( γ ) · · · Λ( γ r )] = { } . Then according to Theorem 4.1 wehave h γ i 6⊂ h γ i · · · h γ r i , which obviously contradicts (BG). (cid:3) OUNDED GENERATION 11
We would like to point out that as proved in [34], every matrix in Γ p = SL ( Z [1 /p ]) (where p isa prime) is a product of ≤ §
5] demonstrates that for p > p that are not products of 4 elementaries. These facts suggest that itmay be possible to upgrade Theorem 1.1 to a statement that for a non-virtually solvable linear groupΓ over a field of characteristic zero, any presentation (BG) must involve at least 5 non-semi-simpleelements, with 5 being the best possible bound. Proof of Corollary 1.2.
Let Γ ⊂ GL n ( K ) be an anisotropic linear group over a field K of characteristiczero that has bounded generation. Then Γ is boundedly generated by semi-simple elements, hencevirtually solvable by Theorem 1.1. So, if we let G denote the Zariski-closure of Γ, then the connectedcomponent G ◦ is solvable. Let U be the unipotent radical of G ◦ ; then the quotient G ◦ /U is a torus (cf.[2, Ch. III, Theorem 10.6]). On the other hand, since Γ is anisotropic, the restriction of the quotientmap G ◦ → G ◦ /U to Γ ∩ G ◦ is injective, so the latter is isomorphic to a subgroup of the abelian group( G ◦ /U )( K ). Thus, Γ ∩ G ◦ is abelian, making Γ virtually abelian. The finite generation of Γ is obvious.Conversely, if Γ is finitely generated and has an abelian subgroup ∆ of finite index, then ∆ itself isfinitely generated, hence has bounded generation, implying that Γ has bounded generation as well. (cid:3) Proof of Theorem 1.3.
It is well-known and easy to show that a finite index subgroup of a group hasbounded generation if and only if the group does. (Incidentally, this implies that if one S -arithmeticsubgroup of an algebraic group has bounded generation then all S -arithmetic subgroups do.) So,passing to the connected component, we may assume G to be connected. Let R be the radical of G ,and let ϕ : G → D be the quotient map to the semi-simple group D = G/R (everything is definedover K ). Furthermore, let π : D → D be the isogeny to the corresponding adjoint group. Then for ψ = π ◦ ϕ , the restriction ψ | H is an isogeny (because H is semi-simple) and the image ψ ( H ) is anormal K -subgroup of D . Since D is adjoint, ψ ( H ) is a direct factor of D , and therefore coincideswith the adjoint group H . Composing ψ with the projection D → H , we obtain a surjective morphism η : G → H . Now, if Γ ⊂ G ( K ) is an S -arithmetic subgroup having (BG), then ∆ = η (Γ) is an S -arithmetic subgroup of H ( K ) (cf. [22, Theorem 5.9]) that also has (BG). Since the group H S isnoncompact, H has a normal K -defined almost K -simple subgroup H ′ with noncompact H ′ S . Then H ′ is a direct factor of H and for the corresponding projection ρ : H → H ′ , the image Φ = ρ (∆) is an S -arithmetic subgroup of H ′ ( K ) having (BG). But since H is K -anisotropic, so is H ′ , and thereforeΦ ⊂ H ′ ( K ) is an anisotropic linear group. So, applying Corollary 1.2, we obtain that Φ is virtuallyabelian. On the other hand, by Borel’s Density Theorem for S -arithmetic groups (cf. [22, Theorem4.10] for usual arithmetic subgroups), Φ is Zariski-dense in H ′ . However, being a connected groupthat coincides with its derived subgroup, H ′ cannot have a Zariski-dense virtually abelian subgroup.A contradiction, proving the theorem. (cid:3) Final remarks
First, here is an example of a solvable finitely generated linear group without bounded generationwhich shows that Theorem 1.1 is not a criterion.
Example 7.1
Let x be a variable. Consider the groupΓ = (cid:26) (cid:18) x i a x − i (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) i ∈ Z , a ∈ A := Z [ x, x − ] (cid:27) . It is solvable and finitely generated. In fact, it is generated by the following three matrices t = (cid:18) x x − (cid:19) , u = (cid:18) (cid:19) , and v = (cid:18) x (cid:19) . To see that these matrices indeed generate Γ, one observes that Γ =
T U (semi-direct product) where T = h t i and U = (cid:26) (cid:18) a (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a ∈ A (cid:27) . Then our claim follows from the relations t i ut − i = (cid:18) x i (cid:19) and t i vt − i = (cid:18) x i +1 (cid:19) . Let us show that Γ does not have bounded generation. For a = P a n x n ∈ A , we define the width w ( a ) to be the number of terms with nonzero coefficients, and for M, N > A N = { a ∈ A | w ( a ) ≤ N } and B M,N = A N x M − . We note the following simple properties:(a) A N + A N ⊂ A N + N ;(b) A N A N ⊂ A N N , and consequently for any N > a ∈ A , we have aA N ⊂ A w ( a ) N (in particular, xA N = A N );(c) for any M i , N i > i = 1 ,
2) there exist
M, N > B M ,N + B M ,N ⊂ B M,N .Properties (a) and (b) are obvious, and to prove (c) one observes that(11) A N x M − A N x M − f ( x ) A N + f ( x ) A N x M M − f ( x ) = x M M − x M − f ( x ) = x M M − x M − . Since w ( f ) = M + 1 and w ( f ) = M + 1, using (a) and (b) we immediately see that the right-handside of (11) is contained in B M,N with M = M M and N = ( M + 1) N + ( M + 1) N .We also set U M,N = (cid:26) (cid:18) a (cid:19) (cid:12)(cid:12)(cid:12)(cid:12) a ∈ A ∩ B M,N (cid:27) , and note that since xA N = A N , hence xB M,N = B M,N , for all
M, N , the set U M,N is normalized by T . Lemma 7.2.
Given any γ , . . . , γ n ∈ Γ (not necessarily distinct) there exist M, N > such that h γ i · · · h γ n i ⊂ T U
M,N . Proof.
We will induct on n . First, let n = 1 and let γ = γ be of the form γ = t i · (cid:18) a (cid:19) . Since our claim in this case is obvious if i = 0, we can assume that i = 0. Writing γ in the form γ = β − t i β , where β = (cid:18) b (cid:19) with b = a − x − i , one easily finds that for any d ∈ Z we have γ d = β − t id β = t id · (cid:18) c d (cid:19) where c d = a ( x − id − x − i − . It follows that h γ i ⊂ T U
M,N with M = 2 | i | and N = 2 w ( a ).Let us assume that n > M , N are chosen so that h γ i · · · h γ n − i ⊂ T U M ,N . Furthermore, using the case n = 1 we choose M , N so that h γ n i ⊂ T U M ,N . Applying property (c) above we can find
M, N such that B M ,N + B M ,N ⊂ B M,N . Since U M ,N isnormalized by T , we obtain h γ i · · · h γ n − ih γ n i ⊂ ( T U M ,N )( T U M ,N ) = T ( U M ,N U M ,N ) ⊂ T U
M,N , as required. (cid:3) OUNDED GENERATION 13
Now, to complete the proof of the fact that Γ is not boundedly generated, it remains to show thatfor any
M, N we have A B M,N . For this, let us consider the polynomial f ( x ) = 1 + x ( M +1) + · · · + x N ( M +1) and assume that f ( x ) ∈ B M,N . Then ( x M − f ( x ) ∈ A N . However, the polynomials f ( x ) and x M f ( x )do not contain the same monomials, so w (( x M − f ( x )) = 2 w ( f ( x )) = 2( N + 1) > N, a contradiction. (cid:3) For comparison, we recall that a solvable group of integral matrices is polycyclic (see [28, p. 26]),hence has bounded generation.Now, there are series of S -arithmetic subgroups of anisotropic absolutely almost simple simplyconnected algebraic groups over number fields for which the congruence kernel is known to be finite (cf.[27, Theorems 3 and 4]). Then the profinite completions of these groups have bounded generation asprofinite groups [21, Theorem 2]. At the same time, according to Theorem 1.3 the groups themselves donot have bounded generation. Thus, we obtain examples of finitely generated residually finite groupswithout bounded generation whose profinite completions do have bounded generation as profinitegroups. To the best of our knowledge, no such examples demonstrating the difference between thenotions of bounded generation in the discrete and profinite sense were previously available.Next, we would like to discuss bounded generation of amalgamated products. Let Γ = Γ ∗ Γ Γ be a free amalgamated product of two groups Γ and Γ along a common subgroup Γ . It was shownin [11] and [14] that if the number of double cosets Γ \ Γ i / Γ is > i ∈ { , } then Γdoes not have bounded generation. On the other hand, the group Γ = SL (cid:16) Z h p i(cid:17) ( p a prime), whichis an amalgamated product [30, Ch. 2, § ∗ Γ Γ where both Γ and Γ are isomorphicto SL ( Z ), and Γ is identified with the subgroup Γ ( p ) consisting of matrices (cid:18) a bc d (cid:19) satisfying c ≡ p ), so the fact that | Γ \ Γ i / Γ | = 2 for i = 1 , over F p = Z /p Z .) Nevertheless, if H is the division algebra of usual quaternions (correspondingto the pairs ( − , − Q and G = SL , H is the associated norm 1 group, then by Theorem 1.3the group ∆ = G ( R ) of points over the ring R = Z h p , q i , where p and q are distinct odd primes,does not have bounded generation. Surprisingly, this happens despite the fact that ∆ shares manygroup-theoretic properties with Γ, viz. ∆ has a presentation ∆ ∗ ∆ ∆ where ∆ , ∆ are virtuallyfree groups with ∆ having a description very similar to that of Γ ; both Γ and ∆ do not havenoncentral normal subgroups of infinite index; both Γ and ∆ are SS -rigid (in fact, super-rigid). Thetrue reason behind the fundamental distinction between Γ and ∆ as far as bounded generation isconcerned remains elusive at this point. So, we would like to propose the following. Problem I.
Give a criterion or at least a verifiable and general enough sufficient condition for a freeamalgamated product
Γ = Γ ∗ Γ Γ to have bounded generation. Of course, one is particularly interested in a sufficient condition that would explain bounded gener-ation of SL (cid:16) Z h p i(cid:17) from the group-theoretic perspective, so the case where Γ and Γ are (virtually)free groups is of special significance. On the other hand, our Theorem 1.3 seems to suggest that whenΓ and Γ are both surface groups, and a common subgroup Γ = Γ , Γ has index ≥ ∗ Γ Γ never has bounded generation.Speaking about linear groups, one can ask if all linear groups over a field of characteristic zero thathave bounded generation can be obtained by some natural operations from S -arithmetic groups. Amore specific question is whether a linear group which is a nontrivial amalgamated product and hasbounded generation at the same time must be S -arithmetic? As we have already mentioned in the introduction, bounded generation of S -arithmetic subgroupsof absolutely almost simple algebraic groups, under some natural assumptions, implies the congruencesubgroup property. So, the study of bounded generation in this context was seen as a new approach toSerre’s Congruence Subgroup Conjecture [29], particularly for anisotropic groups where one cannot useunipotent elements. Since now we know that infinite S -arithmetic subgroups of anisotropic absolutelyalmost simple algebraic groups are not boundedly generated, one may wonder about the possible waysto modify this strategy. In this regard, we would like to point out that it was shown in [21] that underthe same natural assumptions, the congruence subgroup property follows from a weaker property ofpolynomial index growth for an S -arithmetic subgroup Γ:(PIG) there exists positive constants c and k such that for any integer n > ( n ) generated by the n -th powers of elements of Γ has finite index in Γ bounded by cn k ,or even its weaker version(PIG) ′ for any n >
0, the subgroup Γ ( n ) is of finite index in Γ and for fixed n and a prime p thereexist c, k > ( np α ) ] ≤ cp kα for all α > §
7] that condition (PIG) ′ for SL n ( Z ), n ≥
3, can be verified by quitestraightforward computations that rely only on the well-known commutator identities for elementarymatrices. So, it may be realistic to verify (PIG) ′ in other situations. Problem II.
Let G be an absolutely almost simple algebraic group over a number field K , and S be afinite set of places of K containing all archimedean ones. Prove that if the S -rank rk S G := X v ∈ S rk K v G is ≥ , then S -arithmetic subgroups of G satisfy condition (PIG) ′ .We recall that according to Margulis’ Normal Subgroup Theorem [19, Ch. VIII], any noncentralnormal subgroup of a higher rank S -arithmetic subgroup Γ as above, hence in particular the subgroupΓ ( n ) for any n >
0, automatically has finite index.We conclude with one more problem which is at the meeting ground of Problems I and II.
Problem III.
Give a criterion or at least a verifiable sufficient condition for a free amalgamatedproduct
Γ = Γ ∗ Γ Γ to satisfy condition (PIG) ′ . References
1. H. Bass,
Theorems of Jordan and Burnside for algebraic groups , J. Algebra, , 245-254, 1983.2. A. Borel, Linear Algebraic Groups , 2nd edition, GTM , Springer, 1991.3. A. Borel, J. Tits,
Groupes r´eductifs , Publ. math. IHES (1965), 55-150.4. M. Burger, N. Monod, Bounded cohomology of lattices in higher rank Lie groups , J. Eur. Math. Soc. (1999),199-235 (Erratum: (1999), 338).5. D. Carter, G. Keller, Bounded elementary generation of SL n ( O ), Amer. J. Math. (1983), 673-687.6. P. Corvaja, U. Zannier, Applications of Diophantine approximation to integral points and transcendence , CambridgeTracts in Mathematics , Cambridge Univ. Press, 2018.7. J.D. Dixon, M.P.F. Du Sautoy, A. Mann, and D. Segal,
Analytic Pro-p Groups , 2nd edition, Cambridge Univ. Press,2009.8. I.V. Erovenko, A.S. Rapinchuk,
Bounded generation of S -arithmetic subgroups of isotropic orthogonal groups overnumber fields , J. Number Theory (2006), 28-48.9. A. Eskin, S. Mozes, H. Oh, On uniform exponential growth for linear groups, Invent. Math., 160(1):1-30, 2005.10. K. Fujiwara, The second bounded cohomology of a group acting on a Gromov-hyperbolic space , Proc. London Math.Soc. (1998), 70-94.11. K. Fujiwara, The second bounded cohomology of an amalgamated free product of groups , Trans. AMS (2000),1113-1129.12. K. Fujiwara,
On non-bounded generation of discrete subgroups in rank-1 Lie groups , Geometry, spectral theory,groups, and dynamics, 153-156, Contemp. Math. , Israel Math. Conf. Proc., AMS 2005.13. R.I. Grigorchuk,
Some results on bounded cohomology , Combinatorial and Geometric Group Theory (Edinburgh,1993), 111-163, London Math. Soc. Lecture Note Ser. , Cambridge Univ. Press 1995.14. R.I. Grigorchuk,
Bounded cohomology of group constructions , Math. Notes (1996), 392-394.15. M. Kassabov, Kazhdan constants for SL n ( Z ), Internat. J. Algebra Comput. (2005), 971-995.16. S. Lang, Algebra , volume 211 of
Graduate Texts in Mathematics . Springer-Verlag, New York, 3rd Edition, 2002.
OUNDED GENERATION 15
17. A. Lubotzky,
Subgroup growth and congruence subgroups , Invent. math. (1995), 267-295.18. A. Lubotzky, D. Segal,
Subgroup growth , volume 212 of
Progress in Mathematics . Birkh¨auser Verlag, Basel, 2003.19. G.A. Margulis,
Discrete Subgroups of Semisimple Lie Groups , Springer 1991.20. A.V. Morgan, A.S. Rapinchuk, and B. Sury,
Bounded generation of SL over rings of S -integers with infinitelymany units , Algebra & Number Theory (2018), 1949-1974.21. V.P. Platonov, A.S. Rapinchuk, Abstract properties of S -arithmetic groups and the congruence problem , RussianAcad. Sci. Izv. Math. (1993), 455-476.22. V.P. Platonov, A.S. Rapinchuk, Algebraic Groups and Number Theory , Academic Press, Boston, 1994.23. G. Prasad, A.S. Rapinchuk,
Existence of irreducible R -regular elements in Zariski-dense subgroups , Math. Res. Lett. (2003), 21-32.24. G. Prasad, A.S. Rapinchuk, Generic elements in Zariski-dense subgroups and isospectral locally symmetric spaces ,Thin subgroups and superstrong approximation, 211-252, Math. Sci. Res. Inst. Publ. , Cambridge Univ. Press,2014.25. G. Prasad, A.S. Rapinchuk, Generic elements of a Zariski-dense subgroup form an open subset , Trans. MoscowMath. Soc. (2017), 299-314.26. M.S. Raghunathan, Discrete subgroups of Lie groups , Springer, 1972.27. A.S. Rapinchuk,
The Congruence Subgroup Problem , Algebra, K-theory, groups, and education (New York 1997),175-188, Contemp. Math. , AMS, 1999.28. D. Segal,
Polycyclic groups , Cambridge Univ. Press, 1983.29. J.-P. Serre,
Le probl`eme des groupes de congruence pour SL , Ann. Math. (1970), 489-527.30. J.-P. Serre, Trees , Springer, 1980.31. Y. Shalom,
Bounded generation and Kazhdan’s property (T) , Publ. math. IHES (1999), 145-168.32. Y. Shalom, G.A. Willis, Commensurated subgroups of arithmetic groups, totally disconnected groups and adelicrigidity , Geom. Func. Anal. (2013), 1631-1683.33. O.I. Tavgen, Bounded generability of Chevalley groups over rings of S -integer algebraic numbers , Math. USSR-Izv. (1991), 101-128.34. M. Vsemirnov, Short unitriangular factorizations of SL ( Z [1 /p ]) , Q. J. Math. (2014), no. 1, 279-290. Dipartimento di Scienze Matematiche, Informatiche e Fisiche, via del le Scienze, 206, 33100 Udine,Italy
Email address : [email protected] Department of Mathematics, University of Virginia, Charlottesville, VA 22904-4137, USA
Email address : [email protected] Department of Mathematics, University of Virginia, Charlottesville, VA 22904-4137, USA
Email address : [email protected] Scuola Normale Superiore, Piazza dei Cavalieri, 7, 56126 Pisa, Italy
Email address ::