Sandwich classification for GL_n(R), O_{2n}(R) and U_{2n}(R,Λ) revisited
aa r X i v : . [ m a t h . K T ] M a y SANDWICH CLASSIFICATION FOR GL n ( R ) , O n ( R ) AND U n ( R, Λ) REVISITED
RAIMUND PREUSSER
Abstract.
Let n be a natural number greater or equal to 3, R a commutative ring and σ ∈ GL n ( R ). Weshow that t kl ( σ ij ) (resp. t kl ( σ ii − σ jj )) where i = j and k = l can be expressed as a product of 8 (resp.24) matrices of the form ǫ σ ± where ǫ ∈ E n ( R ). We prove similar results for the orthogonal groups O n ( R )and the hyperbolic unitary groups U n ( R, Λ) under the assumption that R is commutative and n ≥
3. Thisyields new, very short proofs of the Sandwich Classification Theorems for the groups GL n ( R ), O n ( R ) and U n ( R, Λ). Introduction
Let n be a natural number greater or equal to 3 and R a commutative ring. Let σ ∈ GL n ( R ) and set H := E n ( R ) σ , i.e. H is the smallest subgroup of GL n ( R ) which contains σ and is normalized by E n ( R ). Let I be the ideal of R defined by I := { x ∈ R | t ( x ) ∈ H } . Then clearly E n ( R, I ) ⊆ H . By the SandwichClassification Theorem (SCT) for GL n ( R ) one also has H ⊆ C n ( R, I ). It follows that σ ij , σ ii − σ jj ∈ I for any i = j , i.e. the matrices t ( σ ij ) and t ( σ ii − σ jj ) can be expressed as products of matrices of theform ǫ σ ± where ǫ ∈ E n ( R ). We show how one can use the theme of the paper [4] in order to find suchexpressions and give boundaries for the number of factors, see Theorem 12. This yields a new, very simpleproof of the SCT for GL n ( R ).Further we prove an orthogonal and a unitary version of Theorem 12 (cf. Theorem 27 and Theorem49). The proof of the orthogonal version is very simple. The proof of the unitary version is a bit morecomplicated, but still it is much shorter than the proof of the SCT for the groups U n ( R, Λ) given in [3](on the other hand, in [3] the ring R is only assumed to be quasi-finite and hence the result is a bit moregeneral). For the hyperbolic unitary groups U n ( R, Λ) this yields the first proof of the SCT which doesnot use localization.This paper is organized as follows. In Section 2 we recall some standard notation which will be usedthroughout the paper. In Section 3 we state two lemmas which will be used in the proofs of the maintheorems 12, 27 and 49. In Section 4 we recall the definitions of the general linear group GL n ( R ) andsome important subgroups, in Section 5 we prove Theorem 12. In Section 6 we recall the definitions ofthe (even-dimensional) orthogonal group O n ( R ) and some important subgroups, in Section 7 we proveTheorem 27. In Section 8 we recall the definitions of A. Bak’s hyperbolic unitary group U n ( R, Λ) andsome important subgroups and in the last section we prove Theorem 49.2.
Notation
By a natural number we mean an element of the set N := { , , , . . . } . If G is a group and g, h ∈ G ,we let h g := hgh − and [ g, h ] := ghg − h − . By a ring we will always mean an associative ring with 1 suchthat 1 = 0. Ideal will mean two-sided ideal. If X is a subset of a ring R , then we denote by I ( X ) theideal of R generated by X . If X = { x } , then we may write I ( x ) instead of I ( X ). The set of all invertibleelements in a ring R is denoted by R ∗ . If m and n are natural numbers and R is a ring, then the set ofall m × n matrices with entries in R is denoted by M m × n ( R ). If a ∈ M m × n ( R ), we denote the transposeof a by a t and the entry of a at position ( i, j ) by a ij . We denote the i -th row of a by a i ∗ and its j -th Department of Mathematics, University of Brasilia, Brazil
E-mail address : [email protected] . column by a ∗ j . We set M n ( R ) := M n × n ( R ). The identity matrix in M n ( R ) is denoted by e or e n × n andthe matrix with a 1 at position ( i, j ) and zeros elsewhere is denoted by e ij . If a ∈ M n ( R ) is invertible, theentry of a − at position ( i, j ) is denoted by a ′ ij , the i -th row of a − by a ′ i ∗ and the j -th column of a − by a ′∗ j . Further we denote by n R the set of all rows v = ( v , . . . , v n ) with entries in R and by R n the set of allcolumns u = ( u , . . . , u n ) t with entries in R . We consider n R as left R -module and R n as right R -module.3. Preliminaries
The following two lemmas are easy to check.
Lemma 1.
Let G be a group and a, b, c ∈ G . Then b − [ a, bc ] = [ b − , a ][ a, c ] . Lemma 2.
Let G be a group, E a subgroup and a ∈ G . Suppose that b ∈ G is a product of n elements ofthe form ǫ a ± where ǫ ∈ E . Then(i) ǫ ′ b is a product of n elements of the form ǫ a ± (ii) [ ǫ ′ , b ] is a product of n elements of the form ǫ a ± for any ǫ ′ ∈ E . Lemma 2 will be used in the proofs of the main theorems without explicit reference.4.
The general linear group GL n ( R )In this section n denotes a natural number, R a ring and I an ideal of R . We shall recall the definitionsof the general linear group GL n ( R ) and the following subgroups of GL n ( R ); the elementary subgroup E n ( R ), the preelementary subgroup E n ( I ) of level I , the elementary subgroup E n ( R, I ) of level I , theprincipal congruence subgroup GL n ( R, I ) of level I and the full congruence subgroup C n ( R, I ) of level I .4.1. The general linear group.Definition 3. GL n ( R ) := ( M n ( R )) ∗ is called general linear group .4.2. The elementary subgroup.Definition 4.
Let i, j ∈ { , . . . , n } such that i = j and x ∈ R . Then t ij ( x ) := e + xe ij is called an elementary transvection . The subgroup of GL n ( R ) generated by all elementary transvections is called elementary subgroup and is denoted by E n ( R ). An elementary transvection t ij ( x ) is called I - elementary if x ∈ I . The subgroup of GL n ( R ) generated by all I -elementary transvections is called preelementarysubgroup of level I and is denoted by E n ( I ). Its normal closure in E n ( R ) is called elementary subgroup oflevel I and is denoted by E n ( R, I ). Lemma 5.
The relations t ij ( x ) t ij ( y ) = t ij ( x + y ) , (R1)[ t ij ( x ) , t hk ( y )] = e and (R2)[ t ij ( x ) , t jk ( y )] = t ik ( xy ) (R3) hold where i = k, j = h in ( R and i = k in ( R .Proof. Straightforward computation. (cid:3)
Definition 6.
Let i, j ∈ { , . . . , n } such that i = j . Define p ij := e + e ij − e ji − e ii − e jj = t ij (1) t ji ( − t ij (1) ∈ E n ( R ). It is easy show that p − ij = p ji . Lemma 7.
Let x ∈ R and i, j, k ∈ { , . . . , n } be pairwise distinct indices. Then ANDWICH CLASSIFICATION REVISITED 3 (i) p ki t ij ( x ) = t kj ( x ) and(ii) p kj t ij ( x ) = t ik ( x ) .Proof. Follows from the relations in Lemma 5. (cid:3)
Congruence subgroups.Definition 8.
The kernel of the group homomorphism GL n ( R ) → GL n ( R/I ) induced by the canonicalmap R → R/I is called principal congruence subgroup of level I and is denoted by GL n ( R, I ). Obviously GL n ( R, I ) is a normal subgroup of GL n ( R ). Definition 9.
The preimage of
Center ( GL n ( R/I )) under the group homomorphism GL n ( R ) → GL n ( R/I )induced by the canonical map R → R/I is called full congruence subgroup of level I and is denoted by C n ( R, I ). Obviously GL n ( R, I ) ⊆ C n ( R, I ) and C n ( R, I ) is a normal subgroup of GL n ( R ). Theorem 10. If n ≥ and R is almost commutative (i.e. module finite over its center), then the equalities [ C n ( R, I ) , E n ( R )] = [ E n ( R, I ) , E n ( R )] = E n ( R, I ) hold.Proof. See [5], Corollary 14. (cid:3) Sandwich classification for GL n ( R )In this section n denotes a natural number greater or equal to 3 and R a commutative ring. Definition 11.
Let σ ∈ GL n ( R ). Then a matrix of the form ǫ σ ± where ǫ ∈ E n ( R ) is called an elementary σ -conjugate . Theorem 12.
Let σ ∈ GL n ( R ) , i = j and k = l . Then(i) t kl ( σ ij ) is a product of elementary σ -conjugates and(ii) t kl ( σ ii − σ jj ) is a product of elementary σ -conjugates.Proof. (i) Set τ := t ( − σ ) t ( σ ). One checks easily that the second row of στ − equals the second rowof σ and hence the second row of ξ := σ τ − is trivial. Set ζ := τ − [ t (1) , [ τ, σ ]] = τ − [ t (1) , τ ξ ] L.
1= [ τ − , t (1)][ t (1) , ξ ] . One checks easily that [ τ − , t (1)] = t ( − σ ) and [ t (1) , ξ ] = Q i =2 t i ( x i ) for some x , x , x , . . . , x n ∈ R .Hence ζ = t ( − σ ) Q i =2 t i ( x i ). It follows that [ t (1) , ζ ] = t ( σ ). Hence we have shown[ t (1) , t ( σ ) t ( − σ ) [ t (1) , [ t ( − σ ) t ( σ ) , σ ]]] = t ( σ ) . This implies that t ( σ ) is a product of 8 elementary σ -conjugates. It follows from Lemma 7 that t kl ( σ )is a product of 8 elementary σ -conjugates. Since one can bring σ ij to position (2 ,
3) by conjugating mono-mial matrices in E n ( R ) (see Definition 6) to σ , the assertion of (i) follows.(ii) Clearly the entry of t ji (1) σ at position ( j, i ) equals σ ii − σ jj + σ ji − σ ij . Applying (i) to t ji (1) σ we get that t kl ( σ ii − σ jj + σ ji − σ ij ) is a product of 8 elementary σ -conjugates (note that any elementary t ji (1) σ -conjugateis also an elementary σ -conjugate). Applying (i) to σ we get that t kl ( σ ij − σ ji ) = t kl ( σ ij ) t kl ( − σ ji ) is aproduct of 16 elementary σ -conjugates. It follows that t kl ( σ ii − σ jj ) = t kl ( σ ii − σ jj + σ ji − σ ij ) t kl ( σ ij − σ ji )is a product of 24 elementary σ -conjugates. (cid:3) As a corollary we get the Sandwich Classification Theorem for GL n ( R ). Note that if σ ∈ GL n ( R ) and I is an ideal of R , then σ ∈ C n ( R, I ) if and only if σ ij , σ ii − σ jj ∈ I for any i = j . SANDWICH CLASSIFICATION REVISITED
Corollary 13.
Let H be a subgroup of GL n ( R ) . Then H is normalized by E n ( R ) if and only if E n ( R, I ) ⊆ H ⊆ C n ( R, I ) (1) for some ideal I of R .Proof. First suppose that H is normalized by E n ( R ). Let I be the ideal of R defined by I := { x ∈ R | t ( x ) ∈ H } . Then clearly E n ( R, I ) ⊆ H . It remains to show that H ⊆ C n ( R, I ), i.e. that if σ ∈ H ,then σ ij , σ ii − σ jj ∈ I for any i = j . But that follows from the previous theorem. Suppose now that (1)holds for some ideal I . Then it follows from the standard commutator formulas in Theorem 10 that H isnormalized by E n ( R ). (cid:3) The even-dimensional orthogonal group O n ( R )In this section n denotes a natural number, R a commutative ring and I an ideal of R . We shall recall thedefinitions of the even-dimensional orthogonal group O n ( R ) and the following subgroups of O n ( R ); theelementary subgroup EO n ( R ), the preelementary subgroup EO n ( I ) of level I , the elementary subgroup EO n ( R, I ) of level I , the principal congruence subgroup O n ( R, I ) of level I , and the full congruencesubgroup CO n ( R, I ) of level I .6.1. The even-dimensional orthogonal group.Definition 14.
Set V := R n . We use the following indexing for the elements of the standard basis of V :( e , . . . , e n , e − n , . . . , e − ). That means that e i is the column whose i -th coordinate is one and all the othercoordinates are zero if 1 ≤ i ≤ n and the column whose (2 n + 1 + i )-th coordinate is one and all the othercoordinates are zero if − n ≤ i ≤ −
1. Let p ∈ M n ( R ) be the matrix with ones on the skew diagonal andzeros elsewhere. We define the quadratic form q : V → Rv v t (cid:18) p (cid:19) v. The subgroup O n ( R ) := { σ ∈ GL n ( R ) | q ( σv ) = q ( v ) ∀ v ∈ V } of GL n ( R ) is called (even-dimensional)orthogonal group . Remark 15.
The even-dimensional orthogonal groups are special cases of the hyperbolic unitary groups,cf. Example 32.
Definition 16.
We define Ω := { , ..., n, − n, ..., − } . Lemma 17.
Let σ ∈ GL n ( R ) . Then σ ∈ O n ( R ) if and only if(i) σ ′ ij = σ − j, − i ∀ i, j ∈ Ω and(ii) q ( σ ∗ j ) = 0 ∀ j ∈ Ω .Proof. See [2], p.167. (cid:3)
Lemma 18.
Let σ ∈ O n ( R ) , x ∈ R ∗ and k ∈ Ω . Then the statements below are true.(i) If the k -th column of σ equals e k x then the ( − k ) -th row of σ equals x − e t − k .(ii) If the k -th row of σ equals xe tk then the ( − k ) -th column of σ equals e − k x − .Proof. Follows from (i) in the previous lemma. (cid:3)
ANDWICH CLASSIFICATION REVISITED 5
The elementary subgroup.Definition 19. If i, j ∈ Ω such that i = ± j and x ∈ R , then the matrix T ij ( x ) := e + xe ij − xe − j, − i ∈ O n ( R )is called an elementary orthogonal transvection . The subgroup of O n ( R ) generated by all elementaryorthogonal transvections is called elementary orthogonal group and is denoted by EO n ( R ). An elementaryorthogonal transvection T ij ( x ) is called I - elementary if x ∈ I . The subgroup of O n ( R ) generated by all I -elementary orthogonal transvections is called preelementary subgroup of level I and is denoted by EO n ( I ).Its normal closure in EO n ( R ) is called elementary subgroup of level I and is denoted by EO n ( R, I ). Lemma 20.
The relations T ij ( x ) = T − j, − i ( − x ) , (R1) T ij ( x ) T ij ( y ) = T ij ( x + y ) , (R2)[ T ij ( x ) , T hk ( y )] = e, (R3)[ T ij ( x ) , T jk ( y )] = T ik ( xy ) , (R4)[ T ij ( x ) , T j, − i ( y )] = e (R5) hold where h = j, − i and k = i, − j in (R3) and i = ± k in (R4) .Proof. Straightforward calculation. (cid:3)
Definition 21.
Let i, j ∈ Ω such that i = ± j . Define P ij := e + e ij − e ji + e − i, − j − e − j, − i − e ii − e jj − e − i, − i − e − j, − j = T ij (1) T ji ( − T ij (1) ∈ EO n ( R ). It is easy show that ( P ij ) − = P ji . Lemma 22.
Let x ∈ R and i, j, k ∈ Ω such that i = ± j and k = ± i, ± j . Then(i) P ki T ij ( x ) = T kj ( x ) and(ii) P kj T ij ( x ) = T ik ( x ) .Proof. Follows from the relations in Lemma 20. (cid:3)
Congruence subgroups.Definition 23.
The kernel of the group homomorphism O n ( R ) → O n ( R/I ) induced by the canonicalmap R → R/I is called principal congruence subgroup of level I and is denoted by O n ( R, I ). Obviously O n ( R, I ) is a normal subgroup of O n ( R ). Definition 24.
The preimage of
Center ( O n ( R/I )) under the group homomorphism O n ( R ) → O n ( R/I )induced by the canonical map R → R/I is called full congruence subgroup of level I and is denoted by CO n ( R, I ). Obviously O n ( R, I ) ⊆ CO n ( R, I ) and CO n ( R, I ) is a normal subgroup of O n ( R ). Theorem 25. If n ≥ , then the equalities [ CO n ( R, I ) , EO n ( R )] = [ EO n ( R, I ) , EO n ( R )] = EO n ( R, I ) hold.Proof. See [2], Theorem 1.1 and Lemma 5.2. (cid:3) Sandwich classification for O n ( R )In this section n denotes a natural number greater or equal to 3 and R a commutative ring. Definition 26.
Let σ ∈ O n ( R ). Then a matrix of the form ǫ σ ± where ǫ ∈ EO n ( R ) is called an elementary (orthogonal) σ -conjugate . Theorem 27.
Let σ ∈ O n ( R ) , i = ± j and k = ± l . Then SANDWICH CLASSIFICATION REVISITED (i) T kl ( σ ij ) is a product of elementary orthogonal σ -conjugates,(ii) T kl ( σ i, − i ) is a product of elementary orthogonal σ -conjugates,(iii) T kl ( σ ii − σ jj ) is a product of elementary orthogonal σ -conjugates and(iv) T kl ( σ ii − σ − i, − i ) is a product of elementary orthogonal σ -conjugates.Proof. (i) Set τ := T ( − σ ) T ( σ ) T , − ( σ , − ). One checks easily that the second row of στ − equalsthe second row of σ and hence the second row of ξ := σ τ − is trivial. By Lemma 18 the second last columnof ξ also is trivial. Set ζ := τ − [ T (1) , [ τ, σ ]] = τ − [ T (1) , τ ξ ] L.
1= [ τ − , T (1)][ T (1) , ξ ] . One checks easily that [ τ − , T (1)] = T ( − σ ) and [ T (1) , ξ ] = Q i = ± T i ( x i ) for some x i ∈ R ( i = ± ζ = T ( − σ ) Q i = ± T i ( x i ). It follows that [ T (1) , ζ ] = T ( σ ). Hence we have shown[ T (1) , T ( σ ) T ( − σ ) T , − ( − σ , − ) [ T (1) , [ T ( − σ ) T ( σ ) T , − ( σ , − ) , σ ]]] = T ( σ ) . This implies that T ( σ ) is a product of 8 elementary σ -conjugates. It follows from Lemma 22 that T kl ( σ ) is a product of 8 elementary σ -conjugates. Since one can bring σ ij to position (2 ,
3) by conjugat-ing monomial matrices in EO n ( R ) (see Definition 21) to σ , the assertion of (i) follows.(ii) Clearly the entry of T ji (1) σ at position ( j, − i ) equals σ i, − i + σ j, − i . Applying (i) to T ji (1) σ we get that T kl ( σ i, − i + σ j, − i ) is a product of 8 elementary σ -conjugates (note that any elementary T ji (1) σ -conjugate isalso an elementary σ -conjugate). Applying (i) to σ we get that T kl ( σ j, − i ) is a product of 8 elementary σ -conjugates. It follows that T kl ( σ i, − i ) = T kl ( σ i, − i + σ j, − i ) T ji ( − σ j, − i ) is a product of 16 elementary σ -conjugates.(iii) Clearly the entry of T ji (1) σ at position ( j, i ) equals σ ii − σ jj + σ ji − σ ij . Applying (i) to T ji (1) σ we get that T kl ( σ ii − σ jj + σ ji − σ ij ) is a product of 8 elementary σ -conjugates (note that any elementary T ji (1) σ -conjugateis also an elementary σ -conjugate). Applying (i) to σ we get that T kl ( σ ij − σ ji ) = T kl ( σ ij ) T kl ( − σ ji ) is aproduct of 16 elementary σ -conjugates. It follows that T kl ( σ ii − σ jj ) = T kl ( σ ii − σ jj + σ ji − σ ij ) T kl ( σ ij − σ ji )is a product of 24 elementary σ -conjugates.(iv) Follows from (iii) since T kl ( σ ii − σ − i, − i ) = T kl ( σ ii − σ jj ) T kl ( σ jj − σ − i, − i ). (cid:3) As a corollary we get the Sandwich Classification Theorem for O n ( R ). Note that if σ ∈ O n ( R ) and I is an ideal of R , then σ ∈ CO n ( R, I ) if and only if σ ij , σ ii − σ jj ∈ I for any i = j . Corollary 28.
Let H be a subgroup of O n ( R ) . Then H is normalized by EO n ( R ) if and only if EO n ( R, I ) ⊆ H ⊆ CO n ( R, I ) (2) for some ideal I of R .Proof. First suppose that H is normalized by EO n ( R ). Let I be the ideal of R defined by I := { x ∈ R | T ( x ) ∈ H } . Then clearly EO n ( R, I ) ⊆ H . It remains to show that H ⊆ CO n ( R, I ), i.e. that if σ ∈ H ,then σ ij , σ ii − σ jj ∈ I for any i = j . But that follows from the previous theorem. Suppose now that (2)holds for some ideal I . Then it follows from the standard commutator formulas in Theorem 25 that H isnormalized by EO n ( R ) (cid:3) ANDWICH CLASSIFICATION REVISITED 7 Bak’s unitary group U n ( R, Λ)In order to classify the subgroups of a general linear group (resp. an even-dimensional orthogonalgroup) which are normalized by the elementary subgroup (resp. the elementary orthogonal group), thenotion of an ideal of a ring is sufficient. Bak’s dissertation [1] showed that the notion of an ideal by itselfwas not sufficient to solve the analogous classification problem for unitary groups, but that a refinementof the notion of an ideal, called a form ideal, was necessary. This led naturally to a more general notionof unitary group, which was defined over a form ring instead of just a ring and generalized all previousconcepts. We describe form rings ( R, Λ) and form ideals ( I, Γ) first, then the hyperbolic unitary group U n ( R, Λ) and its elementary subgroup EU n ( R, Λ) over a form ring ( R, Λ). For a form ideal ( I, Γ), werecall the definitions of the following subgroups of U n ( R, Λ); the preelementary subgroup EU n ( I, Γ) oflevel ( I, Γ), the elementary subgroup EU n (( R, Λ) , ( I, Γ)) of level ( I, Γ), the principal congruence subgroup U n (( R, Λ) , ( I, Γ)) of level ( I, Γ), and the full congruence subgroup CU n (( R, Λ) , ( I, Γ)) of level ( I, Γ).8.1.
Form rings and form ideals.Definition 29.
Let R be a ring and ¯: R → Rx ¯ x an involution on R , i.e. x + y = ¯ x + ¯ y , xy = ¯ y ¯ x and ¯¯ x = x for any x, y ∈ R . Let λ ∈ center ( R ) such that λ ¯ λ = 1 and set Λ min := { x − λ ¯ x | x ∈ R } and Λ max := { x ∈ R | x = − λ ¯ x } . An additive subgroup Λ of R such that(i) Λ min ⊆ Λ ⊆ Λ max and(ii) x Λ¯ x ⊆ Λ ∀ x ∈ R is called a form parameter for R . If Λ is a form parameter for R , the pair ( R, Λ) is called a form ring . Definition 30.
Let ( R, Λ) be a form ring and I an ideal such that ¯ I = I . Set Γ max = I ∩ Λ andΓ min = { x − λ ¯ x | x ∈ I } + h{ xy ¯ x | x ∈ I, y ∈ Λ }i . An additive subgroup Γ of I such that(i) Γ min ⊆ Γ ⊆ Γ max and(ii) x Γ¯ x ⊆ Γ ∀ x ∈ R is called a relative form parameter of level I . If Γ is a relative form parameter of level I , then ( I, Γ) iscalled a form ideal of ( R, Λ).Until the end of section 8 let n ∈ N , ( R, Λ) a form ring and ( I, Γ) a form ideal of ( R, Λ).8.2.
The hyperbolic unitary group.Definition 31.
Set V := R n . We use the following indexing for the elements of the standard basis of V :( e , . . . , e n , e − n , . . . , e − ). That means that e i is the column whose i -th coordinate is one and all the othercoordinates are zero if 1 ≤ i ≤ n and the column whose (2 n + 1 + i )-th coordinate is one and all the othercoordinates are zero if − n ≤ i ≤ −
1. Let p ∈ M n ( R ) be the matrix with ones on the skew diagonal andzeros elsewhere. We define the maps f : V × V → R h : V × V → R q : V → R/ Λ( v, w ) v t (cid:18) p (cid:19) w, ( v, w ) v t (cid:18) pλp (cid:19) w, v f ( v, v ) + Λwhere ¯ v is obtained from v by applying ¯ to each entry of v . For any v ∈ V , f ( v, v ) is called the valueof v and is denoted by | v | . The subgroup U n ( R, Λ) := { σ ∈ GL n ( R ) | ( h ( σu, σv ) = h ( u, v ) ∧ q ( σv ) = q ( v )) ∀ u, v ∈ V } of GL n ( R ) is called hyperbolic unitary group . SANDWICH CLASSIFICATION REVISITED
Example 32. If R is commutative, ¯ = id , λ = − max = R , then U n ( R, Λ) equals thesymplectic group Sp n ( R ). If R is commutative, ¯ = id , λ = 1 and Λ = Λ min = { } , then U n ( R, Λ) equalsthe orthogonal group O n ( R ). Definition 33.
We define Ω + := { , ..., n } , Ω − := {− n, ..., − } , Ω := Ω + ∪ Ω − and ǫ : Ω → {− , } i ǫ ( i ) := ( , if i ∈ Ω + , − , if i ∈ Ω − . Further if i, j ∈ Ω, we write i < j iff either i, j ∈ Ω + ∧ i < j or i, j ∈ Ω − ∧ i < j or i ∈ Ω + ∧ j ∈ Ω − . Lemma 34.
Let σ ∈ GL n ( R ) . Then σ ∈ U n ( R, Λ) if and only if(i) σ ′ ij = λ ( ǫ ( j ) − ǫ ( i )) / ¯ σ − j, − i ∀ i, j ∈ Ω and(ii) | σ ∗ j | ∈ Λ ∀ j ∈ Ω .Proof. See [2], p.167. (cid:3)
Lemma 35.
Let σ ∈ U n ( R, Λ) , x ∈ R ∗ and k ∈ Ω . Then the statements below are true.(i) If the k -th column of σ equals e k x then the ( − k ) -th row of σ equals x − e t − k .(ii) If the k -th row of σ equals xe tk then the ( − k ) -th column of σ equals e − k x − .Proof. Follows from (i) in the previous lemma. (cid:3)
Polarity map.Definition 36.
The map e : V −→ n Rv (cid:0) λ ¯ v − . . . λ ¯ v − n ¯ v n . . . ¯ v (cid:1) is called polarity map . One checks easily that h ( u, v ) = ˜ uv for any u, v ∈ V and that e is involutary linear ,i.e. ] u + v = ˜ u + ˜ v and f vx = ¯ x ˜ v for any u, v ∈ V and x ∈ R . Lemma 37. If σ ∈ U n ( R, Λ) and v ∈ V , then f σv = ˜ vσ − .Proof. See [2, Lemma 2.5]. (cid:3)
The elementary subgroup.Definition 38. If i, j ∈ Ω such that i = ± j and x ∈ R , then the matrix T ij ( x ) := e + xe ij − λ ( ǫ ( j ) − ǫ ( i )) / xe − j, − i ∈ U n ( R, Λ)is called an elementary short root transvection . If i ∈ Ω and y ∈ λ − ( ǫ ( i )+1) / Λ, then the matrix T i, − i ( y ) := e + ye i, − i ∈ U n ( R, Λ)is called an elementary long root transvection . If σ ∈ U n ( R, Λ) is an elementary short root transvectionor an elementary long root transvection, it is called an elementary unitary transvection . The subgroupof U n ( R, Λ) generated by all elementary unitary transvections is called elementary unitary group andis denoted by EU n ( R, Λ). An elementary unitary transvection T ij ( x ) is called ( I, Γ) -elementary if i = − j ∧ x ∈ I or i = − j ∧ x ∈ λ − ( ǫ ( i )+1) / Γ. The subgroup of U n ( R, Λ) generated by all ( I, Γ)-elementarytransvections is called preelementary subgroup of level ( I, Γ) and is denoted by EU n ( I, Γ). Its normalclosure in EU n ( R, Λ) is called elementary subgroup of level ( I, Γ) and is denoted by EU n (( R, Λ) , ( I, Γ)).
ANDWICH CLASSIFICATION REVISITED 9
Lemma 39.
The relations T ij ( x ) = T − j, − i ( − λ ( ǫ ( j ) − ǫ ( i )) / x ) , (R1) T ij ( x ) T ij ( y ) = T ij ( x + y ) , (R2)[ T ij ( x ) , T hk ( y )] = e, (R3)[ T ij ( x ) , T jk ( y )] = T ik ( xy ) , (R4)[ T ij ( x ) , T j, − i ( y )] = T i, − i ( xy − λ − ǫ ( i ) ¯ y ¯ x ) and (R5)[ T i, − i ( x ) , T − i,j ( y )] = T ij ( xy ) T − j,j ( − λ ( ǫ ( j ) − ǫ ( − i )) / ¯ yxy ) (R6) hold where h = j, − i and k = i, − j in (R3) , i, k = ± j and i = ± k in (R4) and i = ± j in (R5) and (R6) .Proof. Straightforward calculation. (cid:3)
Definition 40.
Let v ∈ V be isotropic (i.e. q ( v ) = 0) such that v − = 0. Then we denote the matrix − ¯ v − . . . − ¯ v − n − ¯ λ ¯ v n . . . − ¯ λ ¯ v v − ¯ λ ¯ v v . . . ...1 v n v − n . . . ...1 v − = e + ve t − − e ¯ λ ˜ v = T , − (¯ λ | v | + v − ¯ λ ¯ v ) − Y i =2 T i, − ( v i ) ∈ EU n ( R, Λ)by T ∗ , − ( v ). Clearly T ∗ , − ( v ) − = T ∗ , − ( − v ) (note that ˜ vv = 0 since v is isotropic) and σ T ∗ , − ( v ) = e + σvσ ′− , ∗ − σ ∗ ¯ λ ˜ vσ − L. e + σv f σ ∗ − σ ∗ ¯ λ f σv (3)for any σ ∈ U n ( R, Λ).
Definition 41.
Let i, j ∈ Ω such that i = ± j . Define P ij := e + e ij − e ji + λ ( ǫ ( i ) − ǫ ( j )) / e − i, − j − λ ( ǫ ( j ) − ǫ ( i )) / e − j, − i − e ii − e jj − e − i, − i − e − j, − j = T ij (1) T ji ( − T ij (1) ∈ EU n ( R, Λ). It is easy show that( P ij ) − = P ji . Lemma 42.
Let x ∈ R and i, j, k ∈ Ω such that i = ± j and k = ± i, ± j . Further let y ∈ λ − ( ǫ ( i )+1) / Λ .Then(i) P ki T ij ( x ) = T kj ( x ) ,(ii) P kj T ij ( x ) = T ik ( x ) and(iii) P − k, − i T i, − i ( y ) = T k, − k ( λ ( ǫ ( i ) − ǫ ( k )) / y ) .Proof. Follows from the relations in Lemma 39. (cid:3)
Lemma 43.
Let σ ∈ U n ( R, Λ) and i, j ∈ Ω such that i = ± j . Set ˆ σ := P ij σ . Then | ˆ σ ∗ i | = | σ ∗ j | , if ǫ ( i ) = ǫ ( j ) , | σ ∗ j | − ¯ σ ij σ − i,j + λ ¯ σ ij σ − i,j − ¯ σ − j,j σ jj + λ ¯ σ − j,j σ jj , if ǫ ( i ) = 1 , ǫ ( j ) = − , | σ ∗ j | − ¯ σ − i,j σ ij + λ ¯ σ − i,j σ ij − ¯ σ jj σ − j,j + λ ¯ σ jj σ − j,j , if ǫ ( i ) = − , ǫ ( j ) = 1 . Proof.
Straightforward computation. (cid:3)
Congruence subgroups.Definition 44.
The group consisting of all σ ∈ U n ( R, Λ) such that σ ≡ e mod I and f ( σv, σv ) ≡ f ( v, v ) mod Γ ∀ v ∈ V is called principal congruence subgroup of level ( I, Γ) and is denoted by U n (( R, Λ) , ( I, Γ)).By a theorem of Bak [1], 4.1.4, cf. [2], 4.4, it is a normal subgroup of U n ( R, Λ).
Lemma 45.
Let σ ∈ U n ( R, Λ) . Then σ ∈ U n (( R, Λ) , ( I, Γ)) if and only if(i) σ ≡ e mod I and(ii) | σ ∗ j | ∈ Γ ∀ j ∈ Ω .Proof. [2], p.174. (cid:3) Definition 46.
The subgroup { σ ∈ U n ( R, Λ) | [ σ, EU n ( R, Λ)] ⊆ U n (( R, Λ) , ( I, Γ)) } of U n ( R, Λ) is called full congruence subgroup of level ( I, Γ) and is denoted by CU n (( R, Λ) , ( I, Γ)). Obvi-ously U n (( R, Λ) , ( I, Γ)) ⊆ CU n (( R, Λ) , ( I, Γ)). If EU n ( R, Λ) is a normal subgroup of U n ( R, Λ) (whichfor example is true if n ≥ R is almost commutative, see [2, Theorem 1.1]), then CU n (( R, Λ) , ( I, Γ))is a normal subgroup of U n ( R, Λ).
Theorem 47. If n ≥ and R is almost commutative (i.e. module finite over its center), then the equalities [ CU n (( R, Λ) , ( I, Γ)) , EU n ( R, Λ)] = [ EU n (( R, Λ) , ( I, Γ)) , EU n ( R, Λ)] = EU n (( R, Λ) , ( I, Γ)) hold.Proof.
By [2, Theorem 1.1]), EU n (( R, Λ) , ( I, Γ)) is normal in U n ( R, Λ) and[ U n (( R, Λ) , ( I, Γ)) , EU n ( R, Λ)] ⊆ EU n (( R, Λ) , ( I, Γ)) (4)(note that in [2] the full congruence subgroup is defined a little differently). By [2, Lemma 5.2],[ EU n (( R, Λ) , ( I, Γ)) , EU n ( R, Λ)] = EU n (( R, Λ) , ( I, Γ)) . (5)Hence [ CU n (( R, Λ) , ( I, Γ)) , EU n ( R, Λ)]=[ EU n ( R, Λ) , CU n (( R, Λ) , ( I, Γ))]=[[ EU n ( R, Λ) , EU n ( R, Λ)] , CU n (( R, Λ) , ( I, Γ))] ⊆ EU n (( R, Λ) , ( I, Γ) (6)by the definition of CU n (( R, Λ) , ( I, Γ)), (4) and the three subgroups lemma. (5) and (6) imply the assertionof the theorem. (cid:3) Sandwich classification for U n ( R, Λ)In this section n denotes a natural number greater or equal to 3 and ( R, Λ) a form ring where R iscommutative. Definition 48.
Let σ ∈ U n ( R, Λ). Then a matrix of the form ǫ σ ± where ǫ ∈ EU n ( R, Λ) is called an elementary (unitary) σ -conjugate . Theorem 49.
Let σ ∈ U n ( R, Λ) , k = ± l and i = ± j . Then(i) T kl ( σ ij ) is a product of elementary unitary σ -conjugates,(ii) T kl ( σ i, − i ) is a product of elementary unitary σ -conjugates,(iii) T kl ( σ ii − σ jj ) is a product of elementary unitary σ -conjugates,(iv) T kl ( σ ii − σ − i, − i ) is a product of elementary unitary σ -conjugates and(v) T k, − k ( λ − ( ǫ ( k )+1) / | σ ∗ j | ) is a product of n + 4004 elementary unitary σ -conjugates. ANDWICH CLASSIFICATION REVISITED 11
Proof. (i) In step 1 below we show that T kl ( x ¯ σ σ , − ) where x ∈ R is a product of 16 elementary σ -conjugates. In step 2 we show that T kl ( x ¯ σ σ ) where x ∈ R is a product of 16 elementary σ -conjugates.In step 3 we show that T kl ( x ¯ σ σ ) is a product of 32 elementary σ -conjugates. In step 4 we use steps 1-3in order to prove (i).step 1 Set τ := T (¯ σ σ ) T ( − ¯ σ σ ) T , − (¯ σ σ , − ) T , − ( − ¯ σ σ , − + ¯ λ ¯ σ , − σ ). One checks eas-ily that the second row of στ − equals the second row of σ and hence the second row of ξ := σ τ − is trivial.By Lemma 35 the second last column of ξ also is trivial. Set ζ := τ − [ T − , (1) , [ τ, σ ]] = τ − [ T − , (1) , τ ξ ] L.
1= [ τ − , T − , (1)][ T − , (1) , ξ ] . One checks easily that [ τ − , T − , (1)] = T ( λ ¯ σ σ , − ) T − , ( z ) for some z ∈ Λ and [ T − , (1) , ξ ] = Q i =2 T i ( x i )for some x i ∈ R ( i = 2). Hence ζ = T ( λ ¯ σ σ , − ) T − , ( z ) Q i =2 T i ( x i ). It follows that [ T − , ( − x ¯ λ ) , [ T (1) , ζ ]] = T − , ( x ¯ σ σ , − ) for any x ∈ R . Hence we have shown[ T − , ( − x ¯ λ ) , [ T (1) , τ − [ T − , (1) , [ τ, σ ]]]] = T − , ( x ¯ σ σ , − ) . This implies that T − , ( x ¯ σ σ , − ) is a product of 16 elementary σ -conjugates. It follows from Lemma 42that T kl ( x ¯ σ σ , − ) is a product of 16 elementary σ -conjugates.step 2 Set τ := T , − (¯ σ σ ) T , − ( − ¯ σ σ ) T , − (¯ λ ¯ σ σ ) T , − (¯ σ σ − ¯ λ ¯ σ σ ). One checks easily thatthe second row of στ − equals the second row of σ and hence the second row of ξ := σ τ − is trivial. ByLemma 35 the second last column of ξ also is trivial. Set ζ := τ − [ T − , − (1) , [ τ, σ ]] = τ − [ T − , − (1) , τ ξ ] L.
1= [ τ − , T − , − (1)][ T − , − (1) , ξ ] . One checks easily that [ τ − , T − , − (1)] = T , − (¯ σ σ ) T , − ( z ) for some z ∈ ¯Λ and [ T − , − (1) , ξ ] = Q i =2 T i ( x i ) for some x i ∈ R ( i = 2). Hence ζ = T , − (¯ σ σ ) T , − ( z ) Q i =2 T i ( x i ). It follows that [ T − , ( − x ) , [ T − , (1) , ζ ]] = T − , ( x ¯ σ σ ) for any x ∈ R . Hence we have shown[ T − , ( − x ) , [ T − , (1) , τ − [ T − , − (1) , [ τ, σ ]]]] = T − , ( x ¯ σ σ ) . This implies that T − , ( x ¯ σ σ ) is a product of 16 elementary σ -conjugates. It follows from Lemma 42that T kl ( x ¯ σ σ ) is a product of 16 elementary σ -conjugates.step 3 Set τ := T ( − ¯ σ σ ) T (¯ σ σ ) T , − (¯ σ σ , − ) T , − ( − ¯ σ σ , − + ¯ λ ¯ σ , − σ ). One checks eas-ily that the second row of στ − equals the second row of σ and hence the second row of ξ := σ τ − is trivial.By Lemma 35 the second last column of ξ also is trivial. Set ζ := τ − [ T (1) , [ τ, σ ]] = τ − [ T (1) , τ ξ ] L.
1= [ τ − , T (1)][ T (1) , ξ ] . One checks easily that ψ := [ τ − , T (1)] = T ( − ¯ σ σ ) T , − ( y ) T , − ( z ) for some y ∈ ¯Λ and z ∈ R and θ := [ T (1) , ξ ] = Q i =2 T i ( x i ) for some x i ∈ R ( i = 2). Set χ := ψ − [ T (1) , ζ ] = ψ − [ T (1) , ψθ ] L.
1= [ ψ − , T (1)][ T (1) , θ ] . One checks easily that [ ψ − , T (1)] = T (¯ σ σ ) T , − ( a ) T , − ( b ) for some a ∈ ¯Λ and b ∈ R and[ T (1) , θ ] = T − , ( d ) for some d ∈ Λ. Hence χ = T (¯ σ σ ) T , − ( a ) T , − ( b ) T − , ( d ). It follows that[ T − , (¯ x ) , [ T , − (1) , χ ]] = T − , − ( − ¯ x ¯ σ σ ) ( R = T ( x ¯ σ σ ) for any x ∈ R . Hence we have shown[ T − , (¯ x ) , [ T , − (1) , ψ − [ T (1) , τ − [ T (1) , [ τ, σ ]]]]] = T ( x ¯ σ σ ) . This implies that T ( x ¯ σ σ ) is a product of 32 elementary σ -conjugates. It follows from Lemma 42 that T kl ( x ¯ σ σ ) is a product of 32 elementary σ -conjugates. step 4 Set I := I ( { ¯ σ σ , − , ¯ σ σ } ), J := I ( { ¯ σ σ , − , ¯ σ σ , ¯ σ σ } ) and τ := [ σ − , T ( − ¯ σ )] = ( e − σ ′∗ ¯ σ σ ∗ + σ ′∗ , − σ σ − , ∗ ) T (¯ σ ) . One checks easily that τ ≡ mod I and τ ≡ ¯ σ mod J . Set ζ := P P τ . Then ζ = τ and ζ = τ and hence ¯ ζ ζ ≡ σ mod I + ¯ J . Applying step 3 above to ζ , we get that T kl (¯ ζ ζ ) is a product of32 elementary ζ -conjugates. Since any elementary ζ -conjugate is a product of 2 elementary σ -conjugates,it follows that T kl (¯ ζ ζ ) is a product of 64 elementary σ -conjugates. Thus, by steps 1-3, T kl ( σ ) is aproduct of 64 + 16 + 16 + 16 + 16 + 32 = 160 elementary σ -conjugates. Since one can bring σ ij to position(3 ,
2) by conjugating monomial matrices in EU n ( R, Λ), the assertion of (i) follows.(ii)-(iv) See the proof of Theorem 27.(v) Set m:=160. In step 1 we show that T k, − k ( λ − ( ǫ ( k )+1) / ¯ x ¯ σ | σ ∗ | σ x ) where x ∈ R is a product of(2 n + 17) m + 4 elementary σ -conjugates. In step 2 we use step 1 in order to prove (v).step 1 Set v ′ := (cid:0) . . . σ ′− , − − σ ′− , − (cid:1) t = (cid:0) . . . σ − ¯ σ (cid:1) t ∈ V and v := σ − v ′ ∈ V .Then clearly v − = 0. Further q ( v ) = q ( σ − v ′ ) = q ( v ′ ) = 0 and hence v is isotropic. Set ξ := σ T ∗ , − ( − v ) ( ) = e − σv f σ ∗ + σ ∗ ¯ λ f σv = e − v ′ f σ ∗ + σ ∗ ¯ λ e v ′ . Then ξ = ∗ σ σ ∗ σ σ ∗ σ σ ∗ σ σ ∗ σ n σ ∗ σ − n, σ ∗ σ − , σ − σ − , σ σ − , σ ∗ α ∗ ∗ . . . ∗ ∗ . . . ∗ − ¯ σ ¯ σ ∗ ∗∗ β ∗ ∗ . . . ∗ ∗ . . . ∗ ¯ σ ¯ σ ∗ ∗ where α = σ − , σ − λ ¯ σ ¯ σ − , and β = σ − , σ + λ ¯ σ ¯ σ − , . Set τ := T − , ( σ − , σ ) T − , ( − σ − , σ ) . ANDWICH CLASSIFICATION REVISITED 13
It follows from (i) that τ is a product of 2 m elementary σ -conjugates. Clearly ξτ = ∗ σ σ ∗ σ σ ∗ σ σ ∗ σ σ ∗ σ n σ ∗ σ − n, σ ∗ σ − , σ
10 0 1 ∗ γ ∗ . . . ∗ ∗ . . . ∗ ∗ ∗ ∗∗ δ ∗ . . . ∗ ∗ . . . ∗ ∗ ∗ ∗ where γ = α + ¯ σ ¯ σ σ − , σ and δ = β − ¯ σ ¯ σ σ − , σ . Let x ∈ R and set ζ := T ∗ , − ( − v ) [ T , − ( − x ) , [ T ∗ , − ( v ) , σ ] τ ]= T ∗ , − ( − v ) [ T , − ( − x ) , T ∗ , − ( v ) ξτ ] L.
1= [ T ∗ , − ( − v ) , T , − ( − x )][ T , − ( − x ) , ξτ ] . Clearly ζ is a product of 4 m + 4 elementary σ -conjugates. One checks easily that[ T ∗ , − ( − v ) , T , − ( − x )]= T , − (¯ λ ( − ¯ xv − ¯ v − + λ ¯ xv − ¯ v − )) T , − (¯ λ ¯ x ¯ v − ) T , − ( − x ¯ v − )= T , − (¯ λ ( a + λ ¯ a )) T , − ( b + c ) T , − ( − x ( σ σ − σ σ ))for some a, b ∈ I ( σ ) , c ∈ I ( σ ). Further[ T , − ( − x ) , ξτ ] = ( − Y p =1 ,p =3 T p, − ( xσ p σ )) T − , − ( xγ ) T − , − ( xδ ) T , − ( y )where y = ¯ λ (¯ x ¯ σ | σ ∗ | σ x + d − λ ¯ d + e − λ ¯ e ) for some d ∈ I ( σ ) , e ∈ I ( σ − , ). Hence ζ = T , − (¯ λ ( a + λ ¯ a )) T , − ( b + c ) T , − ( − x ( σ ( σ − σ ) − σ σ )) ·· ( − Y p =2 ,p =3 T p, − ( xσ p σ )) T − , − ( xγ ) T − , − ( xδ ) T , − ( y ) . It follows from (i), (ii) and (iii) and relation (R5) in Lemma 39 that T , − ( y ) is a product of 4 m + 4 +2 m + 2 m + 4 m + (2 n − m + 3 m + 3 m = (2 n + 13) m + 4 elementary σ -conjugates. By (i) and relation(R5) in Lemma 39, T , − ( − ¯ λ ( d − λd )) and T , − ( − ¯ λ ( e − λe ))) each are a product of 2 m elementary σ -conjugates. Hence T , − (¯ λ (¯ x ¯ σ | σ ∗ | σ x )) = T , − ( y ) T , − ( − ¯ λ ( d − λd )) T , − ( − ¯ λ ( e − λe ))) is a product of(2 n + 17) m + 4 elementary σ -conjugates. It follows from Lemma 42 that T k, − k ( λ − ( ǫ ( k )+1) / ¯ x ¯ σ | σ ∗ | σ x )is a product of (2 n + 17) m + 4 elementary σ -conjugates. step 2 Clearly T k, − k ( λ − ( ǫ ( k )+1) / | σ ∗ | )= T k, − k ( λ − ( ǫ ( k )+1) / X q ∈ Ω σ ′ q σ q | σ ∗ | X r ∈ Ω σ ′ r σ r )= T k, − k ( λ − ( ǫ ( k )+1) / X q,r ∈ Ω ¯ σ ′ q ¯ σ q | σ ∗ | σ r σ ′ r )= T k, − k ( λ − ( ǫ ( k )+1) / X q ¯ σ ′ q ¯ σ q | σ ∗ | σ q σ ′ q ) T k, − k ( λ − ( ǫ ( k )+1) / X q 1. Hence A is a product of (2 n +17) m +4+(2 n − · m +6 m = (8 n +17) m +4 elementary σ -conjugates.On the other hand B = T k, − k ( λ − ( ǫ ( k )+1) / ( x − λ ¯ x )) where x ∈ I ( | σ ∗ | ). Since | σ ∗ | = P i ∈ Ω + ¯ σ i σ − i, , it followsfrom (i), (ii) and relation (R5) in Lemma 39 that B is a product of 4 m +( n − · m = (2 n +2) m elementary σ -conjugates. Hence T k, − k ( λ − ( ǫ ( k )+1) / | σ ∗ | ) is a product of (10 n + 19) m + 4 = 1600 n + 3044 elementary σ -conjugates. The assertion of (v) follows now from Lemma 43. (cid:3) As a corollary we get the Sandwich Classification Theorem for U n ( R, Λ). Corollary 50. Let H be a subgroup of U n ( R, Λ) . Then H is normalized by EU n ( R, Λ) if and only if EU n (( R, Λ) , ( I, Γ)) ⊆ H ⊆ CU n (( R, Λ) , ( I, Γ)) (7) for some form ideal ( I, Γ) of ( R, Λ) .Proof. First suppose that H is is normalized by EU n ( R, Λ). Let ( I, Γ) be the form ideal of ( R, Λ) definedby I := { x ∈ R | T ( x ) ∈ H } and Γ := { y ∈ Λ | T − , ( y ) ∈ H } . Then clearly EU n (( R, Λ) , ( I, Γ)) ⊆ H . It remains to show that H ⊆ CU n (( R, Λ) , ( I, Γ)), i.e. that if σ ∈ H and ǫ ∈ EU n ( R, Λ), then[ σ, ǫ ] ∈ U n (( R, Λ) , ( I, Γ)). By Lemma 45 it suffices to show that if σ ∈ H and ǫ ∈ EU n ( R, Λ), then[ σ, ǫ ] ≡ e mod I and | [ σ, ǫ ] ∗ j | ∈ Γ for any j ∈ Ω. But that follows from the previous theorem (applyingthe theorem to σ we get that σ ≡ diag ( x, . . . , x ) mod I for some x ∈ R and hence [ σ, ǫ ] ≡ e mod I ;applying it to [ σ, ǫ ] we get that | [ σ, ǫ ] ∗ j | ∈ Γ for any j ∈ Ω). Suppose now that (7) holds for some formideal ( I, Γ). Then it follows from the standard commutator formulas in Theorem 47 that H is normalizedby EU n ( R, Λ). (cid:3) References [1] A. Bak, The stable structure of quadratic modules , Thesis, Columbia University, 1969. 7, 10[2] A. Bak, N. Vavilov, Structure of hyperbolic unitary groups I: Elementary subgroups , Algebr. Colloq. (2000), no. 2,159-196. 4, 5, 8, 10[3] R. Preusser, Structure of hyperbolic unitary groups II: Classification of e-normal subgroups , Algebr. Colloq. (2017),no. 2, 195-232. 1[4] A. Stepanov, N. Vavilov, Decomposition of transvections: a theme with variations , K-Theory (2000), no. 2, 109-153. 1[5] L. N. Vaserstein, On the normal subgroups of GL n over a ring , Lecture Notes in Math.854