On products of conjugacy classes in general linear groups
aa r X i v : . [ m a t h . K T ] J un On products of conjugacy classes in general lineargroups
Raimund Preusser
Abstract
Let K be a field and n ≥
3. Let E n ( K ) ≤ H ≤ GL n ( K ) be an intermediategroup and C a noncentral H -class. Define m ( C ) as the minimal positive integer m such that ∃ i , . . . , i m ∈ {± } such that the product C i . . . C i m contains allnontrivial elementary transvections. In this article we obtain a sharp upperbound for m ( C ). Moreover, we determine m ( C ) for any noncentral H -class C under the assumption that K is algebraically closed or n = 3 or n = ∞ . In 1964, H. Bass [1] showed that if R is an (associative, unital) ring then ∀ H ≤ GL n ( R ) : H E n ( R ) ⊆ H ⇔ ∃ I ⊳ R : E n ( R, I ) ≤ H ≤ C n ( R, I ) (1)provided n is large enough with respect to the stable rank of R . Here E n ( R ) de-notes the elementary subgroup, E n ( R, I ) the relative elementary subgroup of level I and C n ( R, I ) the full congruence subgroup of level I . Bass’s result is known as Sandwich Classification Theorem . In the 1970’s and 80’s, the validity of this theoremwas extended by J. Wilson [10], I. Golubchik [4], L. Vaserstein [7, 8, 9] and others.Statement (1) holds true for example if R is a commutative ring and n ≥ σ is a matrix in GL n ( R )where R is a commutative ring and n ≥
3, then each of the elementary transvections t kl ( σ ij ) , t kl ( σ ii − σ jj ) ( i = j, k = l ) can be expressed as a finite product of E n ( R )-conjugates of σ and σ − . In 1960, J. Brenner [3] showed that in the case R = Z thereis a bound for the number of E n ( R )-conjugates needed for such an expression. In2018, the author [5] proved that indeed for any commutative ring there is a boundfor the number of E n ( R )-conjugates needed for such an expression. In 2020, theauthor [6] obtained bounds for different classes of noncommutative rings.Now consider the case that R = K is a field and n ≥
3. It follows from theSandwich Classification Theorem that for any noncentral σ ∈ GL n ( K ) the elementarytransvection t (1) can be expressed as a finite product of E n ( K )-conjugates of σ and Keywords and phrases. general linear groups, conjugacy classes, matrix identities. − . Rephrased, for any noncentral E n ( K )-class C (see beginning of Section 2) thereis an m ∈ N and i , . . . , i m ∈ {± } such that T ⊆ C i . . . C i m where T is the E n ( K )-class of t (1). What is the smallest m with this property? This paper aims to answerthis question. More generally, if E n ( K ) ≤ H ≤ GL n ( K ) is an intermediate group and C a noncentral H -class, we define m ( C ) := min { m ∈ N | ∃ i , . . . , i m ∈ {± } : T ⊆ C i . . . C i m } where T denotes the H -class of t (1). The Sandwich Classification Theorem impliesthat the minimum in the definition of m ( C ) exist. The goal of this paper is to compute m ( C ).The rest of the paper is organised as follows. In Section 2 we recall some stan-dard notation which is used throughout the paper. Moreover, we recall the Frobeniusform and the generalised Jordan form of a square matrix and some basic definitionsregarding general linear groups. In Section 3 we prove that m ( C ) ≤ H -class C and that this bound is sharp (see Proposition 22 and Example 24).Moreover, we show that if K is algebraically closed, then m ( C ) ≤
2. In Section 4 wecompute m ( C ) in the case n = 3, the main result of this section is Theorem 29. InSection 5 we consider the case n = ∞ . If G is a group and g, h ∈ G , we let g h := h − gh and [ g, h ] := ghg − h − . Let H be asubgroup of G and g, g ′ ∈ G . If there is an h ∈ H such that g h = g ′ , we write g ∼ H g ′ (or just g ∼ g ′ if H = G ). Clearly ∼ H is an equivalence relation on G . We call the ∼ H -equivalence class of an element g ∈ G the H -class of g and denote it by g H . Foran H -class C = g H we define C := C and C − := ( g − ) H .Throughout the paper K denotes a field and K ∗ the set of all nonzero elementsof K . N denotes the set of positive integers. For any m, n ∈ N , the set of all m × n matrices over K is denoted by M m × n ( K ). Instead of M n × n ( K ) we may write M n ( K ).The identity matrix in M n ( K ) is denoted by e or e n × n and the matrix with a one atposition ( i, j ) and zeros elsewhere is denoted by e ij . In this subsection n denotes a positive integer. We recall the Frobenius form and thegeneralised Jordan form of a matrix in M n ( K ). Definition 1.
Let P = X n + a n − X n − + · · · + a X + a ∈ K [ X ] be a monic polynomialof degree n . The companion matrix of P is the matrix[ P ] = − a − a . . . ...1 − a n − ∈ M n ( K ) . efinition 2. Let σ ∈ M m ( K ) and τ ∈ M n ( K ). The direct sum of σ and τ is thematrix σ ⊕ τ = (cid:18) σ τ (cid:19) ∈ M m + n ( K ) . Definition 3.
Let σ, τ ∈ M n ( K ). If there is an invertible ρ ∈ M n ( K ) such that σ = ρτ ρ − , then we write σ ∼ τ and call σ and τ similar . Theorem 4.
Let σ ∈ M n ( K ) . Then there are uniquely determined nonconstant,monic polynomials P , . . . , P r ∈ K [ X ] such that P | P | . . . | P r and σ ∼ [ P ] ⊕· · ·⊕ [ P r ] .Proof. See [2, Part V, Chapter 21, Theorem 4.4].The matrix [ P ] ⊕ · · · ⊕ [ P r ] in Theorem 4 is called the Frobenius form or rationalcanonical form of σ . We denote it by F ( σ ). The polynomials P , . . . , P r are calledthe invariant factors of σ .Recall that the characteristic matrix of a matrix σ ∈ M n ( K ) is the matrix Xe − σ ∈ M n ( K [ X ]). The characteristic polynomial of σ is the polynomial χ σ = det( Xe − σ ).Note that the characteristic polynomial of a companion matrix [ P ] is P . The invariantfactors of a matrix σ ∈ M n ( K ) are precisely the monic polynomials associated withthe nonconstant polynomials in the Smith normal form of the characteristic matrix Xe − σ . The Smith normal form of Xe − σ can be computed using the algorithmdescribed in [2, Part V, Chapter 20, Proof of Theorem 3.2].Let σ ∈ M n ( K ) and P , . . . , P r the invariant factors of σ . For each 1 ≤ i ≤ r wecan write P i = s i Q j =1 P q ij ij where P i , . . . , P is i ∈ K [ X ] are pairwise distinct irreducible,monic polynomials and q i , . . . , q is i ≥
1. This factorisation is unique up to the orderof the factors. The polynomials P q ij ij (1 ≤ i ≤ r, ≤ j ≤ s i ) form the system of elementary divisors of σ . Note that P q ij ij maybe equal to P q i ′ j ′ i ′ j ′ if i = i ′ .In order to define the generalised Jordan form of a matrix, we need the notionof the generalised Jordan block of a power P q of an irreducible polynomial P . Forpolynomials P of degree 1 one gets back the usual Jordan blocks. Definition 5.
Let q ∈ N and P ∈ K [ X ] an irreducible, monic polynomial of degree n . The generalised Jordan block corresponding to P q is the matrix J ( P q ) = [ P ] ξ [ P ]. . . . . . ξ [ P ] ∈ M qn ( K )where ξ ∈ M n ( K ) is the matrix that has a 1 at position (1 , n ) and zeros elsewhere. Theorem 6.
Let σ ∈ M n ( K ) and ( P q i i ) i ∈ Φ the system of elementary divisors of σ .Then σ ∼ L i ∈ Φ J ( P q i i ) where the order of direct summands maybe arbitrary.Proof. See [2, Part V, Chapter 21, Section 5].3he matrix L i ∈ Φ J ( P q i i ) in Theorem 6 is called the generalised Jordan form or primary rational canonical form of σ . It is uniquely determined up to the order ofthe generalised Jordan blocks. If the characteristic polynomial of σ splits into linearfactors, then one gets back the usual Jordan form of σ . GL n ( K ) In this subsection n denotes a positive integer. Definition 7.
The group GL n ( K ) consisting of all invertible elements of M n ( K ) iscalled the general linear group of degree n over K . Definition 8.
Let a ∈ K and 1 ≤ i = j ≤ n . Then the matrix t ij ( a ) = e + ae ij is called an elementary transvection . If a = 0, then t ij ( a ) is called nontrivial . Thesubgroup E n ( K ) of GL n ( K ) generated by the elementary transvections is called the elementary subgroup . Lemma 9.
For any σ ∈ GL n ( K ) there is an ǫ ∈ E n ( K ) such that σ = ǫd n (det( σ )) . Itfollows that E n ( K ) equals the subgroup SL n ( K ) of GL n ( K ) consisting of all matriceswith determinant .Proof. See for example [1, § Lemma 10.
The relations t ij ( a ) t ij ( b ) = t ij ( a + b ) , (R1)[ t ij ( a ) , t hk ( b )] = e and (R2)[ t ij ( a ) , t jk ( b )] = t ik ( ab ) (R3) hold where i = k, j = h in ( R and i = k in ( R . Definition 11.
Let a ∈ K ∗ and 1 ≤ i = j ≤ n . We define d i ( a ) := e + ( a − e ii ∈ GL n ( K )and d ij ( a ) := e + ( a − e ii + ( a − − e jj ∈ SL n ( K ) = E n ( K ) . Moreover, we define p ij := e + e ij + e ji − e ii − e jj ∈ GL n ( K )and ˆ p ij := e + e ij − e ji − e ii − e jj ∈ SL n ( K ) = E n ( K ) . .3 The stable general linear group GL ∞ ( K ) Definition 12.
The direct limit G ∞ ( K ) = lim −→ n GL n ( K ) with respect to the transitionhomomorphisms φ n,n + k : GL n ( K ) → GL n + k ( K ) , σ e k × k ⊕ σ is called the stablegeneral linear group over K .Let σ ∈ GL m ( K ) and τ ∈ GL n ( K ). If φ m, max { m,n } ( σ ) = φ n, max { m,n } ( τ ), we write σ ∼ ∞ τ . We identify GL ∞ ( K ) with the set S n GL n ( K ) / ∼ ∞ of all ∼ ∞ -equivalenceclasses made a group by defining [ σ ] ∞ [ τ ] ∞ = [ φ m, max { m,n } ( σ ) φ n, max { m,n } ( τ )] ∞ for any σ ∈ GL m ( K ) and τ ∈ GL n ( K ). Definition 13.
The subgroup E ∞ ( K ) of GL ∞ ( K ) generated by the elements [ t n,i,j ( a )] ∞ ( n ≥ , ≤ i = j ≤ n, a ∈ K ) is called the elementary subgroup . Here t n,i,j ( a ) de-notes the elementary transvection t ij ( a ) ∈ GL n ( K ). GL n ( K ) In this section n denotes an integer greater than 2. We set G := GL n ( K ) and E := E n ( K ). H denotes a subgroup of G containing E , and T denotes the H -classof t (1). If C is a noncentral H -class, then m ( C ) is defined as in Section 1.The lemma below shows that T = t (1) G = t (1) E . Lemma 14.
Let σ ∈ G . Then t (1) ∼ E σ ⇔ t (1) ∼ H σ ⇔ t (1) ∼ G σ .Proof. Clearly t (1) ∼ E σ ⇒ t (1) ∼ H σ ⇒ t (1) ∼ G σ . Hence it suffices to showthat t (1) ∼ G σ ⇒ t (1) ∼ E σ . Suppose that t (1) ∼ G σ . Then there is a ρ ∈ G such that t (1) = σ ρ . By Lemma 9 there is an ǫ ∈ E such that ρ = ǫd n (det( ρ )).Since n ≥
3, the matrix t (1) commutes with d n (det( ρ )) and hence t (1) = σ ǫ . Thus t (1) ∼ E σ .It follows from the next lemma that T contains all nontrivial elementary transvec-tion. Lemma 15.
Let t ij ( a ) and t kl ( b ) be nontrivial elementary transvections. Then t ij ( a ) ∼ E t kl ( b ) .Proof. It is an easy exercise to show that t ij ( a ) ǫ = t kl ( a ) for some ǫ ∈ E which is aproduct of ˆ p st ’s (see Definition 11). Hence t ij ( a ) ∼ E t kl ( a ). It remains to show that t kl ( a ) ∼ E t kl ( b ). But clearly t kl ( a ) d lm ( a − b ) = t kl ( b ) for any m = k, l .Let C be an H -class. Since similar matrices have the same characteristic polyno-mial (resp. determinant, trace, invariant factors, elementary divisors), we can define χ C (resp. det( C ), tr( C ), the invariant factors of C , the elementary divisors of C ,the Frobenius form F ( C )) in the obvious way. Below we compute F ( T ). Note that T = { g ∈ G | F ( σ ) = F ( T ) } by Lemma 14. Lemma 16. F ( T ) = [ X − ⊕ · · · ⊕ [ X − ⊕ [( X − ] . roof. By Lemma 15, we have F ( T ) = F ( t n,n − (1)). One checks easily that t n,n − (1) t n − ,n (1) = e ( n − × ( n − −
11 2 . Let C and D be noncentral H -classes. We write C ∼ D and call C and D conjugated if there is a ρ ∈ G such that C ρ = D . The lemma below implies that m ( C ) does only depend on the conjugacy class of C . Lemma 17.
Let C ∼ D be conjugated noncentral H -classes and suppose that T ⊆ C i . . . C i k where i , . . . , i k ∈ {± } . Then T ⊆ D i . . . D i k .Proof. Choose a σ ∈ C and a τ ∈ D . Since C ∼ D , there is a ρ ∈ G such that σ = τ ρ .By Lemma 9 there is an ǫ ∈ E and an a ∈ K ∗ such that ρ = ǫd n ( a ). Hence σ = τ ǫd n ( a ) . (2)Since T ⊆ C i . . . C i k , there are ρ , . . . , ρ k ∈ H such that t (1) = ( σ i ) ρ . . . ( σ i k ) ρ k . By Lemma 9 there are ǫ , . . . , ǫ k ∈ E and a , . . . , a k ∈ K such that ρ i = ǫ i d n ( a i ) (1 ≤ i ≤ k ). Note that d n ( a ) , . . . , d n ( a k ) ∈ H since H contains E . We have t (1) = ( σ i ) ǫ d n ( a ) . . . ( σ i k ) ǫ k d n ( a k ) . (3)It follows from Equations (2) and (3) that t (1) = ( τ i ) ǫd n ( a ) ǫ d n ( a ) . . . ( τ i k ) ǫd n ( a ) ǫ k d n ( a k ) . One easily checks that d n ( a ) commutes with elementary transvections modulo E .Hence there are ǫ ′ , . . . , ǫ ′ k ∈ E such that d n ( a ) ǫ i = ǫ ′ i d n ( a ) (1 ≤ i ≤ k ). We get t (1) = ( τ i ) ǫǫ ′ d n ( a ) d n ( a ) . . . ( τ i k ) ǫǫ ′ k d n ( a k ) d n ( a ) . Since n ≥
3, the matrix t (1) commutes with d n ( a ) and thus t (1) = ( τ i ) ǫǫ ′ d n ( a ) . . . ( τ i k ) ǫǫ ′ k d n ( a k ) ∈ D i . . . D i k . Lemma 18.
Let C be a noncentral H -class. Then m ( C ) = 1 iff C = T .Proof. Clear since T = T − by Lemma 15.Let C be a noncentral H -class. Proposition 19 below shows that if χ C has a rootin K (which is always true if K is algebraically closed), then m ( C ) ≤
2. But we willsee later that m ( C ) can be greater than 2 if χ C has no root.6 roposition 19. Let C be a noncentral H -class. If χ C has a root in K , then T ⊆ CC − .Proof. Let a ∈ K be a root of χ C . First note that a = 0 since the constant coefficientof χ C is ( − n det( C ) = 0. Since ( X − a ) divides χ C we get that ( X − a ) q is anelementary divisor of C for some q ∈ N (since χ C is the product of the elementarydivisors of C ). Choose a σ ∈ C . By Theorem 6 we have σ ∼ (cid:18) J (( X − a ) q ) τ (cid:19) =: ρ for some τ ∈ GL n − q ( K ). Recall that J (( X − a ) q ) = a a . . . . . .1 a ∈ GL q ( K ) . By Theorem 4 we may assume that τ is in Frobenius form, i.e. there are nonconstant,monic polynomials P , . . . , P r ∈ K [ X ] such that P | P | . . . | P r and τ = [ P ] ⊕· · ·⊕ [ P r ]. Case q = 1. Then ρ = (cid:18) a τ (cid:19) . Subcase P i ’s has degree 1. Since P | P | . . . | P r ,it follows that P = · · · = P r = X − b for some b ∈ K ∗ . Hence ρ = a b . . . b . One checks easily that [ ρ, t (1)] = t ( ab − − t ( ab − − ∈ DD − where D = ρ H . Since σ is noncentral,we have a = b and hence ab − − = 0. Hence T ⊆ DD − byLemma 15 and thus T ⊆ CC − by Lemma 17. Subcase i such that P i has degree t ≥
2. Write P i = X t + b t − X t − + · · · + b X + b . Then σ ∼ a − b − b − b . . . ...1 − b t − ∗ =: ξ.
7t follows that σ ∼ ξ t ( − a ) = a a ∗ ∗ ∗ ∗ − b . . . ...1 − b t − ∗ =: ζ . One checks easily that [ ζ , t (1)] = t ( a − ). It follows as inSubcase 1.1 that T ⊆ CC − . Case q = 2. Then ρ = a a τ . Subcase P i ’s has degree 1. Since P | P | . . . | P r ,it follows that P = · · · = P r = X − b for some b ∈ K ∗ . Hence ρ = a a b . . . b . Clearly σ ∼ ρ t ( a − b ) = b ∗ ∗ b . . . b =: ξ. One checks easily that [ ξ, t (1)] = t ( b − ). It follows as inSubcase 1.1 that T ⊆ CC − . Subcase i such that P i has degree t ≥
2. Write P i = X t + b t − X t − + · · · + b X + b . Then σ ∼ a a − b − b − b . . . ...1 − b t − ∗ =: ξ.
8t follows that σ ∼ ξ t ( − a ) = a a a ∗ ∗ ∗ ∗ − b . . . ...1 − b t − ∗ =: ζ . One checks easily that [ ζ , t (1)] = t ( a − ). It follows as inSubcase 1.1 that T ⊆ CC − . Case q ≥
3. Then ρ = a a a ∗ . One checks easily that [ ρ, t (1)] = t ( a − ). It follows as in Subcase 1.1 that T ⊆ CC − .Lemma 18 and Proposition 19 directly imply the following theorem. Theorem 20.
Suppose that K is algebraically closed. Let C be a noncentral H -class.Then m ( C ) = 1 if C = T respectively m ( C ) = 2 if C = T . The converse of Proposition 19 does not hold as the following example shows.Suppose that K = F and n = 4. Let C be the G -class of [ X + X + 1]. Set σ := ∈ G and τ := ∈ G. Clearly σ = t (1) τ and hence στ − = t (1). We leave it to the reader to check that F ( σ ) = F ( τ ) = [ X + X + 1] which implies that σ, τ ∈ C . It follows that T ⊆ CC − although χ C = X + X + 1 has no root in K . However, a “weak converse” ofProposition 19 does hold. Namely if C is a noncentral H -class such that T ⊆ CC − ,then χ C is reducible as Proposition 21 below shows.In Proposition 21 we use the following notation. If σ ∈ G , 1 ≤ i < · · · < i k ≤ n and 1 ≤ j < · · · < j k ≤ n , then σ i ,...,i k j ,...,j k denotes the k × k matrix whose entry at9osition ( s, t ) is σ i s ,j t . Recall that det( σ i ,...,i k i ,...,i k ), where 1 ≤ i < · · · < i k ≤ n , is calleda principal minor of size k . It is well-known that χ σ = n X k =0 ( − k a k X n − k where a k is the sum of all principal minors of size k . Proposition 21.
Let C be a noncentral H -class. If T ⊆ CC − , then χ C is reducible.Proof. Suppose there is a noncentral H -class C such that T ⊆ CC − and χ C isirreducible. We will show that this assumption leads to a contradiction. Let σ, τ ∈ C such that t (1) = στ − . It follows that σ = t (1) τ. (4)Clearly we may assume that τ ij = 0 for any i ≥ j ≥ i + 2 (5)(conjugate Equation (4) by appropriate elements of E commuting with t (1)). Since σ, τ ∈ C , we have tr( σ ) = tr( τ ). Hence tr( σ ) (4) = tr( t (1) τ ) = tr( τ ) + τ = tr( σ ) + τ whence τ = 0. Suppose that τ = 0. Then all nondiagonal entries of τ in the secondrow are 0. Therefore χ τ = χ C has a linear factor, which contradicts the assumptionthat χ C is irreducible. Hence τ = 0 and therefore we may assume that τ = 0(conjugate (4) by t ( − τ − τ ). Hence σ = τ τ τ . . . . . . τ n τ . . . τ τ τ . . . . . . ...... ... ... . . . . . . 0... ... ... . . . τ n − ,n τ n τ n τ n . . . . . . τ nn and, because of (4), τ = τ τ τ + τ . . . . . . τ n τ . . . τ τ τ . . . . . . ...... ... ... . . . . . . 0... ... ... . . . τ n − ,n τ n τ n τ n . . . . . . τ nn Consider the statement P ( k ) : τ k = 0 .
10e will show by induction on k that P ( k ) holds for any 3 ≤ k ≤ n . k = 3 Since σ, τ ∈ C , we have χ σ = χ τ . We compare the coefficients of X n − in χ σ and χ τ , namely the sums of all principal minors of size 2. Suppose det( σ i ,i i ,i ) = det( τ i ,i i ,i ).Then clearly i = 1 and i = 3. Moreover, det( σ , , ) − det( τ , , ) = − τ τ . Hence τ τ = 0. Since clearly τ = 0, we obtain τ = 0. Thus P (3) holds. k → k + 1 Suppose that P (3) , . . . , P ( k ) hold for some k ∈ { , . . . , n − } . We willshow that P ( k + 1) holds. We compare the coefficients of X n − k in χ σ and χ τ ,namely the sums of all principal minors of size k (multiplied by ( − k ). Supposedet( σ i ,...,i k i ,...,i k ) = det( τ i ,...,i k i ,...,i k ). Then clearly i = 1 and i = 3. Moreover,det( σ , ,i ...,i k , ,i ...,i k ) − det( τ , ,i ...,i k , ,i ...,i k ) = τ det( σ ,i ...,i k ,i ...,i k )(consider the Laplace expansion along the first row). In view of (5) we have σ ,i ...,i k ,i ...,i k = τ τ i . . . τ i τ i i τ i i . . . ... τ i τ i i τ i i . . . 0... ... ... . . . τ i k − i k τ i k τ i k i τ i k i . . . τ i k i k . Consider the statement Q ( j ) : i j = j + 1 . We will show by induction on j that Q ( j ) holds for any 3 ≤ j ≤ k . j = 3 Assume that i >
4. Then τ i = 0 by (5). But τ = 0 by P (3), whencedet( σ , ,i ...,i k , ,i ...,i k ) − det( τ , ,i ...,i k , ,i ...,i k ) = τ det( σ ,i ...,i k ,i ...,i k ) = 0. But this contradicts the as-sumption that det( σ i ,...,i k i ,...,i k ) = det( τ i ,...,i k i ,...,i k ). Hence i = 4 and thus Q (3) holds. j → j + 1 Suppose that Q (3) , . . . , Q ( j ) hold for some j ∈ { , . . . , k − } . Write σ , ,i ...,i k , ,i ...,i k = (cid:18) A BD E (cid:19) where A ∈ M ( j − × ( j − ( K ), B ∈ M ( j − × ( n − j +1) ( K ), D ∈ M ( n − j +1) × ( j − ( K ) and E ∈ M ( n − j +1) × ( n − j +1) ( K ). Then A = τ τ . . . τ τ τ . . . ... τ τ τ . . . 0... ... ... . . . τ j,j +1 τ j +1 , τ j +1 , τ j +1 , . . . τ j +1 ,j +1 and B is the matrix whose entry at position ( j − ,
1) is τ j +1 ,i j +1 and whose other entriesare zero. Assume that i j +1 > j + 2. Then τ j +1 ,i j +1 = 0 by (5). Hence det σ , ,i ...,i k , ,i ...,i k =11et( A ) det( E ). But τ = · · · = τ j +1 , = 0 by P (3) , . . . , P ( j + 1) whence det( A ) = 0.It follows that det( σ , ,i ...,i k , ,i ...,i k ) − det( τ , ,i ...,i k , ,i ...,i k ) = τ det( σ ,i ...,i k ,i ...,i k ) = 0. But this con-tradicts the assumption that det( σ i ,...,i k i ,...,i k ) = det( τ i ,...,i k i ,...,i k ). Hence i j +1 = j + 2 and thus Q ( j + 1) holds.We have shown that Q ( j ) holds for any 3 ≤ j ≤ k . Hence σ ,i ...,i k ,i ...,i k = σ , ...,k +11 , ...,k +1 = τ τ . . . τ τ τ . . . ... τ τ τ . . . 0... ... ... . . . τ k,k +1 τ k +1 , τ k +1 , τ k +1 , . . . τ k +1 ,k +1 . Since τ = · · · = τ k = 0 by P (3) , . . . , P ( k ), it follows that τ det( σ ,i ...,i k ,i ...,i k ) = ( − k τ τ k +1 , τ τ . . . τ k,k +1 . (6)Suppose that τ j,j +1 = 0 for some 3 ≤ j ≤ k . Then σ = A B C E F G I ∼ (cid:18) E ∗ ∗ (cid:19) where A ∈ M × ( K ), B ∈ M × ( j − ( K ), C ∈ M × ( n − j ) ( K ), E ∈ M ( j − × ( j − ( K ), F ∈ M ( n − j ) × ( K ), G ∈ M ( n − j ) × ( j − ( K ) and I ∈ M ( n − j ) × ( n − j ) ( K ). It follows that χ σ = χ C has a factor of degree j −
1, which contradicts the assumption that χ C is irreducible. Hence τ , . . . , τ k,k +1 = 0. It follows from (6) that τ k +1 , = 0. Thus P ( k + 1) holds.We have shown that P ( k ) holds for any 3 ≤ k ≤ n . Hence all nondiagonal en-tries of σ in the first column are zero and hence χ σ = χ C has a linear factor. But thiscontradicts the assumption that χ C is irreducible.The next proposition implies that m ( C ) ≤ H -class. Proposition 22.
Let C be a noncentral H -class. Then T ⊆ CC − CC − .Proof. Choose a σ ∈ C . By Theorem 4 and Lemma 17 we may assume that σ isin Frobenius form, i.e. there are nonconstant, monic polynomials P , . . . , P r ∈ K [ X ]such that P | P | . . . | P r and σ = [ P ] ⊕ · · · ⊕ [ P r ]. If one of the P i ’s has degree 1, then T ⊆ CC − ⊆ CC − CC − by Proposition 19 (since χ C = P . . . P r ). Hence we mayassume that the degree of each P i is at least 2. Case P i has degree 2. It follows that P = P = · · · = P r = X − a X − a for some a , a ∈ K since P | P | . . . | P r . Hence σ = a a a a ∗ σ − = − a − a a − − a − a a − ∗ . One checks easily that [ σ, t (1)] = t ( a − ) t ( −
1) which implies that [[ σ, t (1)] ,t (1)] = t ( − a − ). It follows from the formula[[ a, b ] , c ] = a ( a − ) b − a b − c − ( a − ) c − , (7)which holds for elements a, b, c of any group, that t ( − a − ) ∈ CC − CC − .Thus T ⊆ CC − CC − by Lemma 15. Case i such that P i has degree t ≥
3. By Lemma 17 wemay assume that i = 1. Write P = X t − a t − X t − − · · · − a X − a . Then σ = a a a . . . ...1 a t − ∗ and σ − = − a − a − a − a − a − a t − a − ∗ One checks easily that [ σ, t t − ,t (1)] = t t ( a − ) t t − ,t ( −
1) which implies that[[ σ, t t − ,t (1)] , t t − ,t (1)] = t t − , ( − a − ) (since t ≥ t t − , ( − a − ) ∈ CC − CC − . Thus T ⊆ CC − CC − by Lemma 15.Recall that if P = a n X n + · · · + a X + a ∈ K [ X ] is a polynomial of degree n ,then P ∗ = a X n + · · · + a n − X + a n ∈ K [ X ] is called the reciprocal polynomial of P .If σ ∈ G , then χ σ − = χ ∗ σ det( σ ) . It follows that χ σ is irreducible iff χ σ − is irreducible(note that ( P ∗ ) ∗ = P and ( P Q ) ∗ = P ∗ Q ∗ for any P, Q ∈ K [ X ] with nonzero constantcoefficient). 13 heorem 23. Let C be a noncentral H -class. If C = T , χ C is irreducible, det( C ) =1 and det( C ) = 1 , then m ( C ) = 4 .Proof. By Lemma 18 we have m ( C ) >
1. By Proposition 21 we have T CC − since χ C is irreducible. Similarly T C − C since χ C − is irreducible by the paragraphright before Theorem 23. Clearly T CC and T C − C − since det( C ) = 1.Similarly T C i C i C i for all i , i , i ∈ {± } since det( C ) = 1. Hence m ( C ) > m ( C ) = 4. Example 24.
Suppose K = Q . Let P = X n + 2 ∈ K [ X ]. Then P is irreducibleby Eisenstein’s criterion. Let C denote the H -class of [ P ]. Clearly C = T since F ( C ) = [ P ] but F ( T ) = [ X − ⊕ · · · ⊕ [ X − ⊕ [( X − ] by Lemma 16. Moreover, χ C = P and det( C ) ∈ {± } . It follows from Theorem 23 that m ( C ) = 4. n = 3 Set G := GL ( K ) and E := E ( K ). H denotes a subgroup of G containing E , and T denotes the H -class of t (1). We will determine m ( C ) for any noncentral H -class C . Proposition 25.
Let C be a noncentral H -class. Then the following are equivalent.(i) χ C has a root in K .(ii) T ⊆ CC − .(iii) T ⊆ C − C .Proof. The implication ( i ) ⇒ ( ii ) follows from Proposition 19 and the implication( ii ) ⇒ ( i ) from Proposition 21 (note that a polynomial P ∈ K [ X ] of degree 3 isreducible iff it has a root in K ). Applying the equivalence ( i ) ⇔ ( ii ) to C − weobtain χ C − has a root in K ⇔ T ⊆ C − C. It follows from the paragraph right before Theorem 23 that χ C − has a root in K ⇔ χ C has a root in K. Thus we have shown ( i ) ⇔ ( iii ).Next we will consider H -classes C such that χ C does not have a root in K . Proposition 26.
Let C be an H -class such that χ C does not have a root in K . Thenthe following are equivalent.(i) det( C ) = 1 .(ii) T ⊆ CC .(iii) T ⊆ C − C − . roof. We will first show ( i ) ⇔ ( ii ) and then ( i ) ⇔ ( iii ).The implication ( ii ) ⇒ ( i ) is obvious since det( T ) = 1. Suppose now that det( C ) =1. Choose a σ ∈ C . By Lemma 17 we may assume that σ is in Frobenius form. Since χ C does not have a root in K , it is irreducible (because n = 3). Hence the matrix σ has only one invariant factor and therefore σ = a b c for some a, b, c ∈ K . Note that a = det( σ ) and hence a = 1. Set σ = a − a . Then σ = a − − a . One checks easily that ( σ ) p t ( a ) = [ X − ⊕ [( X − ] = F ( t (1)) ∼ t (1). Hence σ ∼ t (1) and therefore σ ∼ E t (1) by Lemma 14. It follows that there is an ǫ ∈ E such that ( σ ) ǫ = t (1) . (8)Set ξ = − − b − − a + c . Clearly ξ = t ( − b − t ( a + c ) d ( −
1) = d ( − t ( b + 1) t ( − a − c ) ∈ E and ξ = e. (9)One checks easily that σ ξ = σ d ( − . (10)Clearly σ d ( − ǫ σ d ( − ǫξ ǫ (10) = ( σ ξ ) ǫ ( σ ξ ) ǫξ ǫ = σ ǫ ξ ǫ ξ − ǫ σ ǫ ξ ǫ ξ ǫ (9) = σ ǫ σ ǫ = t (1) . Hence t (1) ∈ CC and therefore T ⊆ CC . Thus we have shown ( i ) ⇔ ( ii ).Applying the equivalence ( i ) ⇔ ( ii ) to C − we obtaindet( C − ) = 1 ⇔ T ⊆ C − C − . C − ) = 1 ⇔ det( C ) = 1 . Thus we have shown ( i ) ⇔ ( iii ). Lemma 27.
Let a, d, f, x ∈ K ∗ and b, c ∈ K . Then ad bf c ∼ E df a − bfa c ∼ E ax dx − bxf x − c . Proof.
A straightforward computation shows that ad bf c ǫ = df a − bfa c where ǫ = ˆ p ˆ p t ( − a − c ) t ( a − b ) d ( − ∈ E . Moreover, ad bf c d ( x ) = ax dx − bxf x − c . Proposition 28.
Let C be an H -class such that χ C does not have a root in K . Thenthe following are equivalent.(i) det( C ) = 1 .(ii) T ⊆ CCC .(iii) T ⊆ C − C − C − .Proof. We will first show ( i ) ⇔ ( ii ) and then ( i ) ⇔ ( iii ).The implication ( ii ) ⇒ ( i ) is obvious since det( T ) = 1. Suppose now that det( C ) =1. Choose a σ ∈ C . By Lemma 17 we may assume that σ is in Frobenius form. Since χ C does not have a root in K , it is irreducible (because n = 3). Hence the matrix σ has only one invariant factor and therefore σ = a b c for some a, b, c ∈ K . Note that a = det( σ ) and hence a = 1. Case b = 0. Set σ = a . σ ) ˆ p = − a − a . It follows from Lemma 27 that there is an ǫ ∈ E such that( σ ) ǫ = a . (11)Set ξ = − − b − c . Clearly ξ = t ( − b ) t ( c ) d ( −
1) = d ( − t ( b ) t ( − c ) ∈ E and ξ = e. (12)One checks easily that σ ξ = σ d ( − . (13)Clearly σ d ( − ǫ σ d ( − ǫξ ǫ (13) = ( σ ξ ) ǫ ( σ ξ ) ǫξ ǫ = σ ǫ ξ ǫ ξ − ǫ σ ǫ ξ ǫ ξ ǫ (12) = σ ǫ σ ǫ = a . (14)Our next to goal is to show that there are ǫ ′ , ǫ ′′ ∈ E such that( t (1)( σ − ) ǫ ′ ) ǫ ′′ = a , which will finish Case 1 in view of (14). Clearly σ − = − ba − − ca − a − . A straightforward computation shows that( σ − ) ǫ ′ = − ba − + b − c − bba − b − c − b − − b − c ǫ ′ = ˆ p t ( b − c ) d ( − b ) ∈ E . It follows that( t (1)( σ − ) ǫ ′ ) t ( b − c ) = − bba − − b − whence, by Lemma 27, there is an ǫ ′′ ∈ E such that( t (1)( σ − ) ǫ ′ ) ǫ ′′ = a . (15)By (14) and (15) we have σ d ( − ǫ ( ǫ ′′ ) − σ d ( − ǫξ ǫ ( ǫ ′′ ) − σ ǫ ′ = t (1) . Thus T ⊆ CCC . Case b = 0 and c = 0. Let σ , ǫ and ξ be as in Case 1. We will showthat there are ǫ ′ , ǫ ′′ ∈ E such that( t (1)( σ − ) ǫ ′ ) ǫ ′′ = a , which will finish Case 2 in view of (14). Clearly σ − = − ca − a − and hence ( σ − ) ǫ ′ = c − c c − a − − c − a − where ǫ ′ = t ( ca − ) d ( c ) ∈ E . It follows that( t (1)( σ − ) ǫ ′ ) t (1)ˆ p = c − − c − a − − c whence, by Lemma 27, there is an ǫ ′′ ∈ E such that( t (1)( σ − ) ǫ ′ ) ǫ ′′ = a . (16)By (14) and (16) we have σ d ( − ǫ ( ǫ ′′ ) − σ d ( − ǫξ ǫ ( ǫ ′′ ) − σ ǫ ′ = t (1) . Thus T ⊆ CCC . 18 ase b = c = 0. Set σ = a . Then ( σ ) ˆ p t ( − a − ) = − a a − − a . It follows from Lemma 27 that there is an ǫ ∈ E such that( σ ) ǫ = a −
11 2 . (17)Set ξ = − − − . Clearly ξ = t ( − d ( −
1) = d ( − t (1) ∈ E and ξ = e. (18)One checks easily that σ ξ = σ d ( − . (19)Clearly σ d ( − ǫ σ d ( − ǫξ ǫ (19) = ( σ ξ ) ǫ ( σ ξ ) ǫξ ǫ = σ ǫ ξ ǫ ξ − ǫ σ ǫ ξ ǫ ξ ǫ (18) = σ ǫ σ ǫ = a −
11 2 . (20)Our next to goal is to show that there are ǫ ′ , ǫ ′′ ∈ E such that( t (1)( σ − ) ǫ ′ ) ǫ ′′ = a −
11 2 , which will finish Case 3 in view of (20). Clearly σ − = a − σ − ) ǫ ′ = − a − a − − a − a where ǫ ′ = d ( − a ) t ( − a − ) ∈ E . It follows that( t (1)( σ − ) ǫ ′ ) t (2 a − ) t (2 a − )ˆ p = − a − a − a − − a whence, by Lemma 27, there is an ǫ ′′ ∈ E such that( t (1)( σ − ) ǫ ′ ) ǫ ′′ = a −
11 2 . (21)By (20) and (21) we have σ d ( − ǫ ( ǫ ′′ ) − σ d ( − ǫξ ǫ ( ǫ ′′ ) − σ ǫ ′ = t (1) . Thus T ⊆ CCC .We have shown ( i ) ⇔ ( ii ). Applying the equivalence ( i ) ⇔ ( ii ) to C − we obtaindet( C − ) = 1 ⇔ T ⊆ C − C − C − . But clearly det( C − ) = 1 ⇔ det( C ) = 1 . Thus we have shown ( i ) ⇔ ( iii ).The theorem below follows from Lemma 18 and Propositions 22, 25, 26, 28. Theorem 29.
Let C be a noncentral H -class. Then the following holds.(i) If C = T , then m ( C ) = 1 .(ii) If C = T and χ C has a root in K , then m ( C ) = 2 . In this case T ⊆ CC − .(iii) If C = T , χ C has no root in K and det( C ) = 1 , then m ( C ) = 2 . In this case T ⊆ CC .(iv) If C = T , χ C has no root in K , det( C ) = 1 and det( C ) = 1 , then m ( C ) = 3 .In this case T ⊆ CCC .(v) If C = T , χ C has no root in K , det( C ) = 1 and det( C ) = 1 , then m ( C ) = 4 .In this case T ⊆ CC − CC − . xample 30. Suppose that K = F or K = F . Then m ( C ) ≤ H -class C since a = 1 for any a ∈ K ∗ . Example 31.
Suppose that K = F . Let C be the H -class of [ X − X + 2]. Then C = T by Lemma 16, χ C = X − X + 2 has no root in K , det( C ) = − = 1 anddet( C ) = 2 = 1. Hence m ( C ) = 4. Example 32.
Suppose that K = Q . Let C be the H -class of [ X − C = T by Lemma 16, χ C = X − K , det( C ) = 4 = 1 and det( C ) = 8 = 1.Hence m ( C ) = 4. Example 33.
Suppose that K = R . Then m ( C ) ≤ H -class C since any cubic polynomial with real coefficients has at least one real root. GL ∞ ( K ) Set G := GL ∞ ( K ) and E := E ∞ ( K ). H denotes a subgroup of G containing E ,and T denotes the H -class of [ t , , (1)] ∞ . We will determine m ( C ) for any noncentral H -class C ( m ( C ) is defined as in Section 1). Note that the center of G consists onlyof [ e ] ∞ . Hence there is only one central H -class.Our first goal is to define the Frobenius form of an element of G . Lemma 34.
Let n ∈ N and σ ∈ GL n ( K ) . If F ( σ ) = [ P ] ⊕ · · · ⊕ [ P r ] , then F (1 ⊕ σ ) = [ P ] ⊕ · · · ⊕ [ P i − ] ⊕ [ P i ( X − ⊕ [ P i +1 ] ⊕ · · · ⊕ [ P r ] where i = min { j ∈ { , . . . , r } | P j ( X −
1) divides P j +1 } . Here we set P := 1 and P r +1 := P r ( X − .Proof. Clearly the characteristic matrix of 1 ⊕ σ is the matrix A = ( X − ⊕ ( Xe n × n − σ ). Since the invariant factors of σ are P , . . . , P r , the matrix A is equivalent to thematrix B = ( X − ⊕ ⊕ · · · ⊕ ⊕ P ⊕ · · · ⊕ P r . By applying the algorithm describedin [2, Part V, Chapter 20, Proof of Theorem 3.2] we get that B is equivalent to thematrix C = 1 ⊕ · · · ⊕ ⊕ P ⊕ · · · ⊕ P i − ⊕ P i ( X − ⊕ P i +1 · · · ⊕ P r . Clearly C is the Smith normal form of A and hence P , . . . , P i − , P i ( X − , P i +1 , . . . P r are theinvariant factors of 1 ⊕ σ .For any σ ∈ G there is a minimal n σ ∈ N such that σ has a representativein GL n σ ( K ). For any n ≥ n σ we write σ ( n ) for the unique representative of σ inGL n ( K ). Proposition 35 below shows that the sequence F ( σ ( n σ ) ) , F ( σ ( n σ +1) ) , F ( σ ( n σ +2) ) . . . eventually stabilises (up to the equivalence relation ∼ ∞ ). Proposition 35.
Let σ ∈ G . Then there is a s ≥ n σ such that [ F ( σ ( s ) )] ∞ = [ F ( σ ( t ) ] ∞ for any t ≥ s . roof. If F ( σ ( n σ ) ) = [ P ] ⊕ · · · ⊕ [ P r ], then by Lemma 34, F ( σ ( n σ +1) ) = [ P ] ⊕ · · · ⊕ [ P i − ] ⊕ [ P i ( X − ⊕ [ P i +1 ] ⊕ · · · ⊕ [ P r ]where i = min { j ∈ { , . . . , r } | P j ( X −
1) divides P j +1 } . Since P i − ( X −
1) divides P i ( X −
1) but P j ( X −
1) does not divide P j +1 for any j ∈ { , . . . , i − } , we get F ( σ ( n σ +2) ) = [ P ] ⊕ · · · ⊕ [ P i − ] ⊕ [ P i − ( X − ⊕ [ P i ( X − ⊕ [ P i +1 ] ⊕ · · · ⊕ [ P r ] , again by Lemma 34. After repeating this step a finite number of times we arrive at F ( σ ( n σ + i ) ) = [ P ( X − ⊕ · · · ⊕ [ P i ( X − ⊕ [ P i +1 ] ⊕ · · · ⊕ [ P r ] . Hence, again by Lemma 34, F ( σ ( n σ + t ) ) = [ X − ⊕ · · · ⊕ [ X − ⊕ [ P ( X − ⊕ · · · ⊕ [ P i ( X − ⊕ [ P i +1 ] ⊕ · · · ⊕ [ P r ] . for any t ≥ i . Since [ X −
1] is the 1 × σ ∈ G . By Proposition 35 there is a s ≥ n σ such that [ F ( σ ( s ) )] ∞ = [ F ( σ ( t ) ] ∞ for any t ≥ s . We define the Frobenius form F ( σ ) of σ by F ( σ ) = [ F ( σ ( s ) )] ∞ . Clearly F ( σ ) is well-defined. Proposition 36.
Let σ, τ ∈ G . Then σ ∼ τ iff F ( σ ) = F ( τ ) .Proof. Choose an s ≥ n σ , n τ such that [ F ( σ ( s ) )] ∞ = [ F ( σ ( t ) )] ∞ and [ F ( τ ( s ) )] ∞ =[ F ( τ ( t ) )] ∞ for any t ≥ s . Then F ( σ ) = [ F ( σ ( s ) )] ∞ and F ( τ ) = [ F ( τ ( s ) )] ∞ . One checkseasily that σ ∼ τ ⇔∃ t ≥ s : σ ( t ) ∼ τ ( t ) in GL t ( K ) ⇔∃ t ≥ s : F ( σ ( t ) ) = F ( τ ( t ) ) ⇔ F ( σ ( s ) ) = F ( τ ( s ) ) ⇔ F ( σ ) = F ( τ ) . By Proposition 36 we can define the Frobenius form F ( C ) of an H -class C in theobvious way. Below we compute F ( T ). Note that T = { g ∈ G | F ( σ ) = F ( T ) } byLemma 14. Lemma 37. F ( T ) = [[( X − ]] ∞ .Proof. It is an easy exercise to show that F ( t , , (1)) = (cid:18) −
11 2 (cid:19) = [( X − ] . Hence F ( e n × n ⊕ t , , (1)) = [ X − ⊕ · · · ⊕ [ X − ⊕ [( X − ] for any n ∈ N , byLemma 34. The assertion of the lemma follows.22 emma 38. Let C be a noncentral H -class. Then m ( C ) = 1 iff C = T .Proof. Clear since T = T − by Lemma 15. Lemma 39.
Let C be a noncentral H -class. Then T ⊆ CC − .Proof. Choose a σ ∈ C and an n > n σ . Then clearly σ ( n ) is noncentral and χ σ ( n ) hasa linear factor. By Proposition 19 we get t n,n − ,n (1) ∈ C ( n ) ( C ( n ) ) − where C ( n ) is the E n ( K )-class of σ ( n ) . It follows that [ t , , (1)] ∞ ∈ CC − and thus T ⊆ CC − .Theorem 40 below follows directly from Lemmas 38 and 39. Theorem 40.
Let C be a noncentral H -class. Then m ( C ) = 1 if C = T respectively m ( C ) = 2 if C = T . References [1] H. Bass,
K-theory and stable algebra , Publ. Math. Inst. Hautes ´Etudes Sci. (1964), 5–60.[2] P.B. Bhattacharya, S.K. Jain, S.R. Nagpaul, Basic abstract algebra, 2nd ed.,Cambridge University Press, 1994.[3] J.L. Brenner, The linear homogeneous group, III , Ann. of Math. (1960), no.2, 210–223.[4] I.Z. Golubchik, On the general linear group over an associative ring , UspekhiMat. Nauk (1973), no. 3, 179–180 (Russian).[5] R. Preusser, Sandwich classification for GL n ( R ) , O n ( R ) and U n ( R, Λ) revisited ,J. Group Theory. (2018), no. 1, 21–44.[6] R. Preusser, Reverse decomposition of unipotents over noncommutative rings I:General linear groups , Linear Algebra Appl. (2020), 285–300.[7] L.N. Vaserstein,
On the normal subgroups of GL n over a ring , Lecture Notes inMath. (1981), 454–465.[8] L.N. Vaserstein, Normal subgroups of the general linear groups over Banach al-gebras , J. Pure Appl. Algebra (1986), 99–112.[9] L.N. Vaserstein, Normal subgroups of the general linear groups over von Neu-mann regular rings , Proc. Am. Math. Soc (1986), no. 2, 209–214.[10] J.S. Wilson, The normal and subnormal structure of general linear groups , Math.Proc. Cambridge Philos. Soc. (1972), 163–177. Chebyshev Laboratory, St. Petersburg State University, Russia
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