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Sandwich classification for O2n+1(R) and U2n+1(R,Δ) revisited

  • Raimund Preusser EMAIL logo
Published/Copyright: April 17, 2018

Abstract

In a recent paper, the author proved that if n3 is a natural number, R a commutative ring and σGLn(R), then tkl(σij) where ij and kl can be expressed as a product of 8 matrices of the form σ±1ε where εEn(R). In this article we prove similar results for the odd-dimensional orthogonal groups O2n+1(R) and the odd-dimensional unitary groups U2n+1(R,Δ) under the assumption that R is commutative and n3. This yields new, short proofs of the Sandwich Classification Theorems for the groups O2n+1(R) and U2n+1(R,Δ).

1 Introduction

The Sandwich Classification Theorem (SCT) for the lattice of subgroups of a linear group G that are normalized by the elementary subgroup E of G is one of the central points in the structure theory of linear groups. For general rings, not only semilocal or arithmetic ones, the SCT was first proved by H. Bass [4, 5] for GLn under the stable range condition. This condition allowed him to extract a transvection as well as to prove the standard commutator formula. The first proofs of the SCT for G=GLn over a commutative ring by J. Wilson [23] and I. Golubchik [7] used only direct calculations as neither localization techniques nor the standard commutator formula were known at that time. About ten years later the SCT was proved applying the standard commutator formula by L. Vaserstein [14] and, independently, by Z. Borewich and N. Vavilov [6]. In the latter article the authors introduced a trick to stabilize a column of a matrix whereas Vaserstein used localization. The next generation of the SCT proofs works for Chevalley groups. L. Vaserstein [15] and E. Abe [1] used localization methods whereas N. Vavilov, E. Plotkin and A. Stepanov [22] introduced the decomposition of unipotents, which in more detail was described in [13] and further developed in [18, 19, 21, 20, 16, 17] by Vavilov and his students and in [9] by V. Petrov. For generalized hyperbolic unitary groups the SCT was announced by A. Bak and N. Vavilov in [3], but the proof has never been published. It appeared later in [10] (which is essentially the author’s thesis). Recently A. Stepanov [12] proved the SCT for all Chevalley groups using the universal localization method.

Let n3 be a natural number and R a commutative ring. The SCT for GLn(R) states the following (it is also true over almost commutative rings, cf. [14]):

SCT.

Let H be a subgroup of GLn(R). Then H is normalized by En(R) iff

En(R,I)HCn(R,I)

for some ideal I of R.

The ideal I in the SCT is uniquely determined, namely

I=I(H)={xRt12(x)H}.

Let now σGLn(R) and set H:=σEn(R), i.e. H is the smallest subgroup of GLn(R) which contains σ and is normalized by En(R). Then, by the SCT, we have HCn(R,I) where I=I(H). It follows from the definition of Cn(R,I) that σij,σii-σjjI for any ij. Hence, by the definition of I(H), the matrices t12(σij) and t12(σii-σjj) can be expressed as products of matrices of the form σ±1ε where εEn(R). In [11] the author showed how one can use the theme of the paper [13] in order to find such expressions and gave boundaries for the number of factors. This yielded a new, very simple proof of the SCT for GLn(R). Further the author obtained similar results for the even-dimensional orthogonal groups O2n(R) and the even-dimensional unitary groups U2n(R,Λ).

In this article we prove similar results for the odd-dimensional orthogonal groups O2n+1(R) and the odd-dimensional unitary groups U2n+1(R,Δ) under the assumption that R is commutative and n3, cf. Theorems 5.2 and 7.2. The proof of the orthogonal version is quite simple. The proof of the unitary version is a bit more complicated, but still it is much shorter than the proof of the SCT for the groups U2n+1(R,Δ) given in [2] (on the other hand, in [2] the ring R is only assumed to be quasi-finite and hence the result is a bit more general). For the odd-dimensional unitary groups U2n+1(R,Δ) this yields the first proof of the SCT which does not use localization.

The rest of the paper is organized as follows. In Section 2 we recall some standard notation which will be used throughout the paper. In Section 3 we state two lemmas which will be used in the proofs of the main theorems 5.2 and 7.2. In Section 4 we recall the definitions of the odd-dimensional orthogonal group O2n+1(R) and some important subgroups. In Section 5 we prove Theorem 5.2. In Section 6 we recall the definitions of the odd-dimensional unitary group U2n+1(R,Δ) and some important subgroups. In Section 7 we prove Theorem 7.2.

2 Notation

By a natural number we mean an element of the set :={1,2,3,}. If G is a group and g,hG, we let gh:=hgh-1 and [g,h]:=ghg-1h-1. By a ring we will always mean an associative ring with 1 such that 10. Ideal will mean two-sided ideal. If X is a subset of a ring R, then we denote by I(X) the ideal of R generated by X. If X={x}, then we may write I(x) instead of I(X). If n is a natural number and R is a ring, then the set of all n×n matrices with entries in R is denoted by Mn(R). If aMn(R), we denote the entry of a at position (i,j) by aij, the i-th row of a by ai* and the j-th column of a by a*j. The group of all invertible matrices in Mn(R) is denoted by GLn(R) and the identity element of GLn(R) by e. If aGLn(R), then the entry of a-1 at position (i,j) is denoted by aij, the i-th row of a-1 by ai* and the j-th column of a-1 by a*j. Further we denote by Rn the set of all row vectors of length n with entries in R and by Rn the set of all column vectors of length n with entries in R. We consider Rn as left R-module and Rn as right R-module. If uRn (resp. uRn), we denote by ut its transpose in Rn (resp. in Rn).

3 Preliminaries

The following two lemmas are easy to check.

Lemma 3.1.

Let G be a group and a,b,cG. Then [a,bc]b-1=[b-1,a][a,c].

Lemma 3.2.

Let G be a group, E a subgroup and aG. Suppose that bG is a product of n elements of the form a±1ε where εE. Then for any εE

  1. bε is a product of n elements of the form a±1ε, and

  2. [ε,b] is a product of 2n elements of the form a±1ε.

Lemma 3.2 will be used in the proofs of the main theorems without explicit reference.

4 Odd-dimensional orthogonal groups

In this section n denotes a natural number and R a commutative ring. First we recall the definitions of the odd-dimensional orthogonal group O2n+1(R) and the elementary subgroup EO2n+1(R). For an admissible pair (I,J), we recall the definitions of the following subgroups of O2n+1(R); the preelementary subgroup EO2n+1(I,J) of level (I,J), the elementary subgroup EO2n+1(R,I,J) of level (I,J), the principal congruence subgroup O2n+1(R,I,J) of level (I,J), and the full congruence subgroup CO2n+1(R,I,J) of level (I,J).

4.1 The odd-dimensional orthogonal group

Definition 4.1.

Set M:=R2n+1. We use the following indexing for the elements of the standard basis of M: (e1,,en,e0,e-n,,e-1). That means that ei is the column whose i-th coordinate is one and all the other coordinates are zero if 1in, the column whose (n+1)-st coordinate is one and all the other coordinates are zero if i=0, and the column whose (2n+2+i)-th coordinate is one and all the other coordinates are zero if -ni-1. We define the quadratic form

Q:MR,uut(00p010000)u=u1u-1++unu-n+u02

where pMn(R) denotes the matrix with ones on the skew diagonal and zeros elsewhere. The subgroup

O2n+1(R):={σGL2n+1(R)Q(σu)=Q(u) for all uM}

of GL2n+1(R) is called (odd-dimensional) orthogonal group.

Remark 4.2.

The odd-dimensional orthogonal groups O2n+1(R) are special cases of the odd-dimensional unitary groups U2n+1(R,Δ), see [2, Example 15].

Definition 4.3.

Define Θ:={1,,n,0,-n,,-1} and Θhb:=Θ{0}.

Lemma 4.4.

Let σGL2n+1(R). Then σO2n+1(R) iff the conditions (i) and (ii) below hold:

    1. σij=σ-j,-i for all i,jΘhb,

    2. 2σ0j=σ-j,0 for all jΘhb,

    3. σi0=2σ0,-i for all iΘhb,

    4. 2σ00=2σ00.

  1. Q(σ*j)=δ0j for all jΘ.

Proof.

See [2, Lemma 17]. ∎

4.2 The polarity map

Definition 4.5.

The map

~:MMt,u(u-1u-n2u0unu1)

where Mt=R2n+1 is called the polarity map. Clearly ~ is linear, i.e. u+v~=u~+v~ and ux~=xu~ for any u,vM and xR.

Lemma 4.6.

If σO2n+1(R) and uM, then σu~=u~σ-1.

Proof.

Follows from Lemma 4.4. ∎

4.3 The elementary subgroup

If i,jΘ, let eij denote the matrix in M2n+1(R) with 1 in the (i,j)-th position and 0 in all other positions.

Definition 4.7.

If i,jΘhb, i±j and xR, the element

Tij(x):=e+xeij-xe-j,-i

of O2n+1(R) is called an (elementary) short root matrix. If iΘhb and xR, the element

Ti(x):=e+xe0,-i-2xei0-x2ei,-i

of O2n+1(R) is called an (elementary) extra short root matrix. If an element of O2n+1(R) is a short or extra short root matrix, then it is called an elementary matrix. The subgroup of O2n+1(R) generated by all elementary matrices is called the elementary subgroup and is denoted by EO2n+1(R).

Lemma 4.8.

The following relations hold for elementary matrices:

(S1)Tij(x)=T-j,-i(-x),
(S2)Tij(x)Tij(y)=Tij(x+y),
(S3)[Tij(x),Tkl(y)]=eif kj,-i and li,-j,
(S4)[Tij(x),Tjk(y)]=Tik(xy)if i±k,
(S5)[Tij(x),Tj,-i(y)]=e,
(E1)Ti(x)Ti(y)=Ti(x+y),
(E2)[Ti(x),Tj(y)]=Ti,-j(-2xy)if i±j,
(E3)[Ti(x),Ti(y)]=e,
(SE1)[Tij(x),Tk(y)]=eif kj,-i,
(SE2)[Tij(x),Tj(y)]=Tj,-i(-xy2)Ti(xy).

Proof.

Straightforward computation. ∎

Definition 4.9.

Let uM be such that u-1=0 and u is isotropic, i.e. Q(u)=0. Then we denote the matrix

(1-u-2-u-n-2u0-un-u201u21un1u01u-n1u-21)
  =e+ue-1t-e1u~
  =T1(u0)T2,-1(u2)Tn,-1(un)T-n,-1(u-n)T-2,-1(u-2)
  EO2n+1(R)

by T*,-1(u). Clearly T*,-1(u)-1=T*,-1(-u) (note that u~u=0 since u is isotropic) and

(4.1)T*,-1σ(u)=e+σuσ-1,*-σ*1u~σ-1=e+σuσ*1~-σ*1σu~

for any σO2n+1(R), the last equality by Lemma 4.6.

Definition 4.10.

Let i,jΘhb such that i±j. Define

Pij:=e-eii-ejj-e-i,-i-e-j,-j+eij-eji+e-i,-j-e-j,-i
=Tij(1)Tji(-1)Tij(1)EO2n+1(R).

It is easy to show that (Pij)-1=Pji.

Lemma 4.11.

Let i,j,kΘhb such that i±j and k±i,±j. Let xR. Then

  1. TijPki(x)=Tkj(x),

  2. TijPkj(x)=Tik(x), and

  3. TiP-k,-i(x)=Tk(x).

Proof.

Straightforward. ∎

4.4 Relative elementary subgroups

Definition 4.12.

An admissible pair is a pair (I,J) where I and J are ideals of R such that

2J,J2IJ

where J2={x2xJ}.

Definition 4.13.

Let (I,J) denote an admissible pair. A short root matrix Tij(x) is called (I,J)-elementary if xI. An extra short root matrix Ti(x) is called (I,J)-elementary if xJ. If an element of O2n+1(R) is an (I,J)-elementary short or extra short root matrix, then it is called an (I,J)-elementary matrix. The subgroup EO2n+1(I,J) of EO2n+1(R) generated by the (I,J)-elementary matrices is called the preelementary subgroup of level (I,J). Its normal closure EO2n+1(R,I,J) in EO2n+1(R) is called the elementary subgroup of level (I,J).

4.5 Congruence subgroups

In this subsection (I,J) denotes an admissible pair.

Definition 4.14.

The subgroup of O2n+1(R) consisting of all σO2n+1(R) such that

  1. σijδijmodI for any iΘhb,jΘ, and

  2. σ0jδ0jmodJ for any jΘ

is called principal congruence subgroup of level(I,J) and is denoted by O2n+1(R,I,J).

Theorem 4.15.

O2n+1(R,I,J) is a normal subgroup of O2n+1(R).

Proof.

Follows from [2, Corollary 35]. ∎

Definition 4.16.

The subgroup

{σO2n+1(R)[σ,EO2n+1(R)]O2n+1(R,I,J)}

of O2n+1(R) is called the full congruence subgroup of level (I,J) and is denoted by CO2n+1(R,I,J).

Lemma 4.17.

Let σO2n+1(R). Then σCO2n+1(R,I,J) iff

  1. σijI for any iΘhb,jΘ such that ij,

  2. σ0jJ for any jΘhb,

  3. σii-σjjI for any i,jΘhb, and

  4. σ00-σjjJ for any jΘhb.

Proof.

Straightforward. ∎

Theorem 4.18.

If n3, then the following equalities hold:

[CO2n+1(R,I,J),EO2n+1(R)]=[EO2n+1(R,I,J),EO2n+1(R)]
=EO2n+1(R,I,J).

Proof.

See [2, Theorem 39]. ∎

5 Sandwich classification for O2n+1(R)

In this section n denotes a natural number greater than or equal to 3 and R a commutative ring.

Definition 5.1.

Let σO2n+1(R). Then a matrix of the form σ±1ε where εEO2n+1(R) is called an elementary (orthogonal) σ-conjugate.

Theorem 5.2.

Let σO2n+1(R) and i,j,k,lΘhb such that i±j and k±l. Then

  1. Tkl(σij) is a product of 8 elementary orthogonal σ -conjugates,

  2. Tkl(σi,-i) is a product of 16 elementary orthogonal σ -conjugates,

  3. Tkl(σi0) is a product of 24 elementary orthogonal σ -conjugates,

  4. Tkl(2σ0j) is a product of 24 elementary orthogonal σ -conjugates,

  5. Tkl(σii-σjj) is a product of 24 elementary orthogonal σ -conjugates,

  6. Tkl(σii-σ-i,-i) is a product of 48 elementary orthogonal σ -conjugates,

  7. Tk(σ0j) is a product of 64n+148 elementary orthogonal σ -conjugates, and

  8. Tk(σ00-σjj) is a product of 192n+564 elementary orthogonal σ -conjugates.

Proof.

(i) Set τ:=T21(-σ23)T31(σ22)T2,-3(σ2,-1) and ξ:=τ-1σ. One checks easily that (στ-1)2*=σ2* and (τ-1σ-1)*,-2=σ*,-2. Hence ξ2*=e2t and ξ*,-2=e-2. Set

ζ:=[T32(1),[τ,σ]]τ-1=[T32(1),τξ]τ-1=[τ-1,T32(1)][T32(1),ξ],

the last equality by Lemma 3.1. One checks easily that [τ-1,T32(1)]=T31(-σ23) and [T32(1),ξ]=T-2(x-2)i0,±2Ti2(xi) for some xiR (i0,2). Hence ζ=T31(-σ23)T-2(x-2)i0,±2Ti2(xi). It follows that [T12(1),ζ]=T32(σ23). Hence we have shown

[T12(1),[T32(1),[τ,σ]]τ-1]=T32(σ23).

This implies that T32(σ23) is a product of 8 elementary σ-conjugates. It follows from Lemma 4.11 that Tkl(σ23) is a product of 8 elementary σ-conjugates. Since one can bring σij to position (2,3) conjugating by monomial matrices from EO2n+1(R) (see Definition 4.10), the assertion of (i) follows.

(ii) Clearly the entry of σTji(1) at position (j,-i) equals σi,-i+σj,-i. Applying (i) to σTji(1), we get that Tkl(σi,-i+σj,-i) is a product of 8 elementary σ-conjugates (note that any elementary σTji(1)-conjugate is also an elementary σ-conjugate). Applying (i) to σ, we get that Tkl(σj,-i) is a product of 8 elementary σ-conjugates. It follows that

Tkl(σi,-i)=Tkl(σi,-i+σj,-i)Tji(-σj,-i)

is a product of 16 elementary σ-conjugates.

(iii) Clearly the entry of σT-j(-1) at position (i,j) equals σi0+σij-σi,-j. Applying (i) to σT-j(-1), we get that Tkl(σi0+σij-σi,-j) is a product of 8 elementary σ-conjugates. Applying (i) to σ, we get that Tkl(-σij) and Tkl(σi,-j) each are a product of 8 elementary σ-conjugates. It follows that

Tkl(σi0)=Tkl(σi0+σij-σi,-j)Tkl(-σij)Tkl(σi,-j)

is a product of 24 elementary σ-conjugates.

(iv) Clearly the entry of σTi(-1) at position (i,j) equals 2σ0j+σij-σ-i,j. Applying (i) to σTi(-1), we get that Tkl(2σ0j+σij-σ-i,j) is a product of 8 elementary σ-conjugates. Applying (i) to σ, we get that Tkl(-σij) and Tkl(σ-i,j) each are a product of 8 elementary σ-conjugates. It follows that

Tkl(2σ0j)=Tkl(2σ0j+σij-σ-i,j)Tkl(-σij)Tkl(σ-i,j)

is a product of 24 elementary σ-conjugates.

(v) Clearly the entry of σTji(1) at position (j,i) equals σii-σjj+σji-σij. Applying (i) to σTji(1), we get that Tkl(σii-σjj+σji-σij) is a product of 8 elementary σ-conjugates. Applying (i) to σ, we get that Tkl(σij) and Tkl(-σji) each are a product of 8 elementary σ-conjugates. It follows that

Tkl(σii-σjj)=Tkl(σii-σjj+σji-σij)Tkl(σij)Tkl(-σji)

is a product of 24 elementary σ-conjugates.

(vi) Follows from (v) since

Tkl(σii-σ-i,-i)=Tkl(σii-σjj)Tkl(σjj-σ-i,-i).

(vii) Set m:=8. In Step 1 we show that for any xR the matrix Tk(xσ0jσjj) is a product of (2n+9)m+4 elementary σ-conjugates. In Step 2 we use Step 1 in order to prove (v).

Step 1. Set

u:=(00σ-1,-1-σ-1,-2)t=(00σ11-σ21)tM

and u:=σ-1uM. Then clearly u-1=0. Further Q(u)=Q(σ-1u)=Q(u)=0 and hence u is isotropic. Set

ξ:=T*,-1σ(-u)=e-σuσ*1~+σ*1σu~=e-uσ*1~+σ*1u~,

the first equality being a consequence of (4.1). Then

ξ=(1-σ11σ21σ11σ11-σ21σ211+σ21σ11-σ31σ21σ31σ111-σ41σ21σ41σ111-σn1σ21σn1σ111-σ01σ21σ01σ111-σ-n,1σ21σ-n,1σ111-σ-4,1σ21σ-4,1σ111-σ-3,1σ21σ-3,1σ111-α0******-σ31σ11**0α******σ31σ21**)

where α=σ-1,1σ11+σ-2,1σ21. Set

τ:=T-3,1(σ-3,1σ21)T-3,2(-σ-3,1σ11).

It follows from (i) that τ is a product of 2m elementary σ-conjugates. Clearly

ξτ=(1-σ11σ21σ11σ11-σ21σ211+σ21σ11-σ31σ21σ31σ111-σ41σ21σ41σ111-σn1σ21σn1σ111-σ01σ21σ01σ111-σ-n,1σ21σ-n,1σ111-σ-4,1σ21σ-4,1σ111001-α-βδ0********γα-β0********)

where β=σ-3,1σ21σ31σ11, γ=σ-3,1σ21σ31σ21 and δ=σ-3,1σ11σ31σ11. Let xR and set

ζ:=[T2,-3(-x),[T*,-1(u),σ]τ]T*,-1(-u)
=[T2,-3(-x),T*,-1(u)ξτ]T*,-1(-u)
=[T*,-1(-u),T2,-3(-x)][T2,-3(-x),ξτ],

the last equality by Lemma 3.1. Clearly ζ is a product of 4m+4 elementary σ-conjugates. One checks easily that

[T*,-1(-u),T2,-3(-x)]
  =T1,-2(xu-3)T1,-3(-xu-2)
  =T1,-2(x(σ23σ11-σ13σ21))T1,-3(-x(σ22σ11-σ12σ21)).

Further

[T2,-3(-x),ξτ]
  =(p=1,p3,0-4Tp,-3(xσp1σ11))T-2,-3(xδ)T-1,-3(x(α-β))T3(xσ01σ11).

Hence

ζ=T1,-2(x(σ23σ11-σ13σ21))T1,-3(-x((σ22-σ11)σ11-σ12σ21))
  (p=2,p3,0-4Tp,-3(xσp1σ11))T-2,-3(xδ)T-1,-3(x(α-β))T3(xσ01σ11).

It follows from (i), (ii) and (v) that T3(xσ01σ11) is a product of

4m+4+2m+4m+(2n-5)m+m+3m=(2n+9)m+4

elementary σ-conjugates. In view of Definition 4.10 and Lemma 4.11 we get that Tk(xσ0jσjj) is a product of (2n+9)m+4 elementary σ-conjugates.

Step 2. Clearly

Tk(σ0j)=Tk((sΘσjsσsj)σ0j)=sΘTk(σjsσsjσ0j).

By (i) and relation (SE2) in Lemma 4.8, Tk(σjsσsjσ0j) is a product of 3m elementary σ-conjugates if s±j,0. By (ii) and relation (SE2), Tk(σj,-jσ-j,jσ0j) is a product of 32m=6m elementary σ-conjugates. By (iv) and relation (SE2), Tk(σj0σ0jσ0j) is a product of 33m=9m elementary σ-conjugates (note that σj0 is a multiple of 2 by Lemma 4.4). By Step 1, Tk(σjjσjjσ0j) is a product of (2n+9)m+4 elementary σ-conjugates. Thus Tk(σ0j) is a product of

(2n-2)3m+6m+9m+(2n+9)m+4=8nm+18m+4=64n+148

elementary σ-conjugates.

(viii) One checks easily that the entry of σT-j(1) at position (0,j) equals σjj-σ00+σ0j-σj0-σ0,-j-σj,-j. Applying (vii) to σT-j(1), we get that Tk(σjj-σ00+σ0j-σj0-σ0,-j-σj,-j) is a product of 64n+148 elementary σ-conjugates. Applying (vii) to σ, we get that Tk(-σ0j) and Tk(σ0,-j) each are a product of 64n+148 elementary σ-conjugates. By (iii) and relation (SE2) in Lemma 4.8, Tk(σj0) is a product of 324=72 elementary σ-conjugates. By (ii) and relation (SE2) in Lemma 4.8, Tk(σj,-j) is a product of 316=48 elementary σ-conjugates. It follows that

Tk(σjj-σ00)=Tk(σjj-σ00+σ0j-σj0-σ0,-j-σj,-j)
  Tk(-σ0j)Tk(σj0)Tk(σ0,-j)Tk(σj,-j)

is a product of

3(64n+148)+72+48=192n+564

elementary σ-conjugates. ∎

As a corollary we get the Sandwich Classification Theorem for O2n+1(R).

Corollary 5.3.

Let H be a subgroup of O2n+1(R). Then H is normalized by EO2n+1(R) iff

(5.1)EO2n+1(R,I,J)HCO2n+1(R,I,J)

for some admissible pair (I,J).

Proof.

First suppose that H is normalized by EO2n+1(R). Let (I,J) be the admissible pair defined by

I:={xRT12(x)H}andJ:={xRT1(x)H}.

Then clearly EO2n+1(R,I,J)H. The inclusion HCO2n+1(R,I,J), i.e. that if σH, then the conditions (i)–(iv) in Lemma 4.17 are satisfied, follows from Theorem 5.2. Suppose now that (5.1) holds for some admissible pair (I,J). Then it follows from the standard commutator formula in Theorem 4.18 that H is normalized by EO2n+1(R). ∎

6 Odd-dimensional unitary groups

We describe Hermitian form rings (R,Δ) and odd form ideals (I,Ω) first, then the odd-dimensional unitary group U2n+1(R,Δ) and its elementary subgroup EU2n+1(R,Δ) over a Hermitian form ring (R,Δ). For an odd form ideal (I,Ω), we recall the definitions of the following subgroups of U2n+1(R,Δ):

  1. the preelementary subgroup EU2n+1(I,Ω) of level (I,Ω),

  2. the elementary subgroup EU2n+1((R,Λ),(I,Ω)) of level (I,Ω),

  3. the principal congruence subgroup U2n+1((R,Λ),(I,Ω)) of level (I,Ω),

  4. the normalized principal congruence subgroup NU2n+1((R,Λ),(I,Ω)) of level (I,Ω), and

  5. the full congruence subgroup CU2n+1((R,Λ),(I,Ω)) of level (I,Ω).

6.1 Hermitian form rings and odd form ideals

First we recall the definitions of a ring with involution with symmetry and a Hermitian ring.

Definition 6.1.

Let R be a ring and

¯:RR,xx¯

an anti-isomorphism of R (i.e. ¯ is bijective, x+y¯=x¯+y¯, xy¯=y¯x¯ for any x,yR and 1¯=1). Further let λR such that x¯¯=λxλ¯ for any xR. Then λ is called a symmetry for ¯, the pair (¯,λ) an involution with symmetry and the triple (R,¯,λ) a ring with involution with symmetry. A subset AR is called involution invariant iff x¯A for any xA. We call a quadruple (R,¯,λ,μ) where (R,¯,λ) is a ring with involution with symmetry and μR such that μ=μ¯λ a Hermitian ring.

Remark 6.2.

Let (R,¯,λ,μ) be a Hermitian ring.

  1. It is easy to show that λ¯=λ-1.

  2. The map

    ¯:RR,xx¯:=λ¯x¯λ

    is the inverse map of ¯. One checks easily that (R,¯,λ¯,μ¯) is a Hermitian ring.

Next we recall the definition of an R-module.

Definition 6.3.

If R is a ring, let R denote the underlying set of the ring equipped with the multiplication of the ring, but not the addition of the ring. A (right) R-module is a not necessarily abelian group (G,+.) equipped with a map

:G×RG,(a,x)ax

such that the following holds:

  1. a0=0 for any aG,

  2. a1=a for any aG,

  3. (ax)y=a(xy) for any aG and x,yR, and

  4. (+.ab)x=+.(ax)(bx) for any a,bG and xR.

A left R-module is defined analogously. An R-module is canonically an R-module, but not conversely. Let G and G be R-modules. A group homomorphism f:GG satisfying f(ax)=f(a)x for any aG and xR is called a homomorphism of R-modules. A subgroup H of G which is -stable (i.e. axH for any aH and xR) is called an R-submodule. Further, if AG and BR, then we denote by AB the subgroup of G generated by {abaA,bB}. We treat as an operator with higher priority than +..

An important example of an R-module is the Heisenberg group, which we define next. The odd form parameters Δ which are used to define the odd-dimensional unitary groups are certain R-submodules of the Heisenberg group.

Definition 6.4.

Let (R,¯,λ,μ) be a Hermitian ring. Define the map

+.:(R×R)×(R×R)R×R,
((x1,y1),(x2,y2))+.(x1,y1)(x2,y2):=(x1+x2,y1+y2-x¯1μx2).

Then (R×R,+.) is a group, which we call the Heisenberg group and denote by . Equipped with the map

:(R×R)×RR×R,
((x,y),a)(x,y)a:=(xa,a¯ya)

becomes an R-module.

Remark 6.5.

We denote the inverse of an element (x,y) by -.(x,y). One checks easily that -.(x,y)=(-x,-y-x¯μx) for any (x,y).

In order to define the odd-dimensional unitary groups we need the notion of a Hermitian form ring.

Definition 6.6.

Let (R,¯,λ,μ) be a Hermitian ring. Let (R,+) have the R-module structure defined by xa=a¯xa. Define the trace map

tr:R,(x,y)x¯μx+y+y¯λ.

One checks easily that tr is a homomorphism of R-modules. Set

Δmin:={(0,x-x¯λ)xR}andΔmax:=ker(tr).

An R-submodule Δ of lying between Δmin and Δmax is called an odd form parameter of (R,¯,λ,μ). Since Δmin and Δmax are R-submodules of , they are respectively the smallest and largest odd form parameters. A pair ((R,¯,λ,μ),Δ) is called a Hermitian form ring. We shall usually abbreviate it by (R,Δ).

Next we define an odd form ideal of a Hermitian form ring.

Definition 6.7.

Let (R,Δ) be a Hermitian form ring and I an involution invariant ideal of R. Set

J(Δ):={yRzR:(y,z)Δ},
I~:={xRJ(Δ)¯μxI}.

Obviously I~ and J(Δ) are right ideals of R, and II~. Further set

ΩminI:=+.{(0,x-x¯λ)xI}ΔI,
ΩmaxI:=Δ(I~×I).

An R-submodule Ω of lying between ΩminI and ΩmaxI is called a relative odd form parameter of level I. Since ΩminI and ΩmaxI are R-submodules of , they are respectively the smallest and the largest relative odd form parameters of level I. If Ω is a relative odd form parameter of level I, then (I,Ω) is called an odd form ideal of (R,Δ).

The following lemma is straightforward to check. It will be used in the proof of Theorem 7.2.

Lemma 6.8.

Let (R,¯,λ,μ) be a Hermitian ring, (a,b)Δmax, nN and x1,,xnR. Then

(a,b)i=1nxi=+.(+.i=1n(a,b)xi)(0,i,j=1,i>jnxibx¯j-xibx¯j¯λ).

6.2 The odd-dimensional unitary group

Let (R,Δ) be a Hermitian form ring and n. Let M, e1,,en,e0,e-n,,e-1 and p be defined as in Definition 4.1. If uM, then we call

(u1,,un,u-n,,u-1)tR2n

the hyperbolic part of u and denote it by uhb. Further we set

u*:=u¯tanduhb*:=u¯hbt.

Define the maps

b:M×MR,
(u,v)u*(00p0μ0pλ00)v=i=1nu¯iv-i+u¯0μv0+i=-n-1u¯iλv-i

and

q:M,
u(q1(u),q2(u)):=(u0,uhb*(0p00)uhb)=(u0,i=1nu¯iu-i).

Lemma 6.9.

  1. b is a λ-Hermitian form, i.e.

    1. b is biadditive,

    2. b(ux,vy)=x¯b(u,v)y for all u,vM, x,yR, and

    3. b(u,v)=b(v,u)¯λ for all u,vM.

  2. The map q has the following properties:

    1. q(ux)=q(u)x for all uM, xR,

    2. q(u+v)+.+.q(u)q(v)(0,b(u,v))modΔmin for all u,vM,

    3. tr(q(u))=b(u,u) for all uM.

Proof.

Straightforward computation. ∎

Definition 6.10.

The group

U2n+1(R,Δ):={σGL2n+1(R)b(σu,σv)=b(u,v) for all u,vM
and q(σu)q(u)modΔ for all uM}

is called the odd-dimensional unitary group.

Remark 6.11.

The odd-dimensional unitary groups U2n+1(R,Δ) are isomorphic to Petrov’s odd hyperbolic unitary groups with V0=R (see [8]). Namely U2n+1(R,Δ) is isomorphic to Petrov’s group U2l(R,𝔏) where Petrov’s pseudoinvolution ^ is defined by x^=-x¯λ, V0=R, B0(a,b)=a^1^-1μb,

𝔏={(x,y)R×R(x,-y)Δ}

and l=n.

Example 6.12.

The groups U2n+1(R,Δ) include as special cases the even-dimensional unitary groups U2n(R,Λ) where n and (R,Λ) is a form ring, the general linear groups GLn(R) where R is any ring and n2, and further the symplectic groups Spn(R) and the orthogonal groups On(R) where R is a commutative ring and n2. For details see [2, Example 15].

Definition 6.13.

Define

Θ+:={1,,n},Θ-:={-n,,-1},
Θ:=Θ+Θ-{0},Θhb:=Θ{0}

and the map

ε:Θhb{±1},i{1if iΘ+,-1if iΘ-.

Further if i,jΘ, we write i<j iff either iΘ+j=0 or i=0jΘ- or i,jΘ+i<j or i,jΘ-i<j or iΘ+jΘ-.

Lemma 6.14.

Let σGL2n+1(R). Then σU2n+1(R,Δ) iff the conditions (i) and (ii) below hold:

    1. σij=λ-(ε(i)+1)/2σ¯-j,-iλ(ε(j)+1)/2 for all i,jΘhb,

    2. μσ0j=σ¯-j,0λ(ε(j)+1)/2 for all jΘhb,

    3. σi0=λ-(ε(i)+1)/2σ¯0,-iμ for all iΘhb,

    4. μσ00=σ¯00μ.

  1. q(σ*j)(δ0j,0)modΔ for all jΘ.

Proof.

See [2, Lemma 17]. ∎

6.3 The polarity map

Definition 6.15.

The map

~:MM*,u(u¯-1λu¯-nλu¯0μu¯nu¯1)

where M*=R2n+1 is called the polarity map. Clearly ~ is involutary linear, i.e. u+v~=u~+v~ and ux~=x¯u~ for any u,vM and xR.

Lemma 6.16.

If σU2n+1(R,Δ) and uM, then σu~=u~σ-1.

Proof.

Follows from Lemma 6.14. ∎

6.4 The elementary subgroup

We introduce the following notation. In Definition 6.4 we defined an R-module structure on . Let (R,¯,λ¯,μ¯) be the Hermitian ring defined in Remark 6.2 (b). Let ¯ denote the Heisenberg group corresponding to (R,¯,λ¯,μ¯). The underlying set of both and ¯ is R×R. If we replace the Hermitian ring (R,¯,λ,μ) by the Hermitian ring (R,¯,λ¯,μ¯), then we get another R-module structure on R×R. We denote the group operation (resp. scalar multiplication) defined by (R,¯,λ,μ) on R×R by +.1 (resp. 1) and the group operation (resp. scalar multiplication) defined by (R,¯,λ¯,μ¯) on R×R by +.-1 (resp. -1). Further we set

Δ1:=ΔandΔ-1:={(x,y)R×R(x,y¯)Δ}.

One checks easily that ((R,¯,λ¯,μ¯),Δ-1) is a Hermitian form ring. Analogously, if (I,Ω) is an odd form ideal of (R,Δ), we set

Ω1:=ΩandΩ-1:={(x,y)R×R(x,y¯)Ω}.

One checks easily that (I,Ω-1) is an odd form ideal of (R,Δ-1).

If i,jΘ, let eij denote the matrix in M2n+1(R) with 1 in the (i,j)-th position and 0 in all other positions.

Definition 6.17.

If i,jΘhb, i±j and xR, the element

Tij(x):=e+xeij-λ(ε(j)-1)/2x¯λ(1-ε(i))/2e-j,-i

of U2n+1(R,Δ) is called an (elementary) short root matrix. If iΘhb and (x,y)Δ-ε(i), the element

Ti(x,y):=e+xe0,-i-λ-(1+ε(i))/2x¯μei0+yei,-i

of U2n+1(R,Δ) is called an (elementary) extra short root matrix. The extra short root matrices of the kind

Ti(0,y)=e+yei,-i

are called (elementary) long root matrices. If an element of U2n+1(R,Δ) is a short or extra short root matrix, then it is called elementary matrix. The subgroup of U2n+1(R,Δ) generated by all elementary matrices is called the elementary subgroup and is denoted by EU2n+1(R,Δ).

Lemma 6.18.

The following relations hold for elementary matrices:

(S1)Tij(x)=T-j,-i(-λ(ε(j)-1)/2x¯λ(1-ε(i))/2),
(S2)Tij(x)Tij(y)=Tij(x+y),
(S3)[Tij(x),Tkl(y)]=eif kj,-i and li,-j,
(S4)[Tij(x),Tjk(y)]=Tik(xy)if i±k,
(S5)[Tij(x),Tj,-i(y)]=Ti(0,xy-λ(-1-ε(i))/2y¯x¯λ(1-ε(i))/2),
(E1)Ti(x1,y1)Ti(x2,y2)=Ti(+.-ε(i)(x1,y1)(x2,y2)),
(E2)[Ti(x1,y1),Tj(x2,y2)]=Ti,-j(-λ-(1+ε(i))/2x¯1μx2)if i±j,
(E3)[Ti(x1,y1),Ti(x2,y2)]=Ti(0,-λ-(1+ε(i))/2(x¯1μx2-x¯2μx1)),
(SE1)[Tij(x),Tk(y,z)]=eif kj,-i,
[Tij(x),Tj(y,z)]=Tj,-i(zλ(ε(j)-1)/2x¯λ(1-ε(i))/2)
(SE2)    Ti(yλ(ε(j)-1)/2x¯λ(1-ε(i))/2,xzλ(ε(j)-1)/2x¯λ(1-ε(i))/2).

Proof.

Straightforward computation. ∎

Definition 6.19.

Let uM be such that u-1=0 and u is isotropic, i.e. q(u)Δ. Then we denote the matrix

(1-λ¯u¯-2λ-λ¯u¯-nλ-λ¯u¯0μ-λ¯u¯n-λ¯u¯2u1-λ¯u¯11u21un1u01u-n1u-21)
  =e+ue-1t-e1λ¯u~
  =T1(q1(u),λ¯(q2(u)-u¯1+λu1))i=2,i0-2Ti,-1(ui)EU2n+1(R,Δ)

by T*,-1(u). Clearly T*,-1(u)-1=T*,-1(-u) (note that u~u=tr(q(u))=0 since u is isotropic) and

(6.1)T*,-1σ(u)=e+σuσ-1,*-σ*1λ¯u~σ-1=e+σuσ*1~-σ*1λ¯σu~

for any σU2n+1(R,Δ), the last equality by Lemma 6.16.

Definition 6.20.

Let i,jΘhb such that i±j. Define

Pij:=e-eii-ejj-e-i,-i-e-j,-j+eij-eji
  +λ(ε(i)-ε(j))/2e-i,-j-λ(ε(j)-ε(i))/2e-j,-i
=Tij(1)Tji(-1)Tij(1)EU2n+1(R,Δ).

It is easy to show that (Pij)-1=Pji.

Lemma 6.21.

Let i,j,kΘhb such that i±j and k±i,±j. Let xR and (y,z)Δ-ε(i). Then

  1. TijPki(x)=Tkj(x),

  2. TijPkj(x)=Tik(x), and

  3. TiP-k,-i(y,z)=Tk(y,λ(ε(i)-ε(k))/2z).

Proof.

Straightforward. ∎

Lemma 6.22.

Let σU2n+1(R,Δ) and i,jΘhb such that i±j. Set σ^:=σPij. Then

q(σ^*i)={q(σ*j)if ε(i)=ε(j),+.q(σ*j)(0,-σ¯ijσ-i,j+σ¯ijσ-i,j¯λ-σ¯-j,jσjj+σ¯-j,jσjj¯λ)if ε(i)=1,ε(j)=-1,+.q(σ*j)(0,-σ¯-i,jσij+σ¯-i,jσij¯λ-σ¯jjσ-j,j+σ¯jjσ-j,j¯λ)if ε(i)=-1,ε(j)=1.

Proof.

Straightforward computation. ∎

6.5 Relative elementary subgroups

Definition 6.23.

Let (I,Ω) denote an odd form ideal of (R,Δ). A short root matrix Tij(x) is called (I,Ω)-elementary if xI. An extra short root matrix Ti(x,y) is called (I,Ω)-elementary if (x,y)Ω-ε(i). If an element of U2n+1(R,Δ) is an (I,Ω)-elementary short or extra short root matrix, then it is called an (I,Ω)-elementary matrix. The subgroup EU2n+1(I,Ω) of EU2n+1(R,Δ) generated by the (I,Ω)-elementary matrices is called the preelementary subgroup of level (I,Ω). Its normal closure EU2n+1((R,Δ),(I,Ω)) in EU2n+1(R,Δ) is called the elementary subgroup of level (I,Ω).

6.6 Congruence subgroups

In this subsection (I,Ω) denotes an odd form ideal of (R,Δ). If σM2n+1(R), we call the matrix (σij)i,jΘhbM2n(R) the hyperbolic part of σ and denote it by σhb. Further we define the submodule M(R,Δ):={uMu0J(Δ)} of M.

Definition 6.24.

The subgroup

U2n+1((R,Δ),(I,Ω)):={σU2n+1(R,Δ)σhbehbmodI and
  q(σu)q(u)modΩ for all uM(R,Δ)}

of U2n+1(R,Δ) is called the principal congruence subgroup of level (I,Ω).

Lemma 6.25.

Let σU2n+1(R,Δ). Then σU2n+1((R,Δ),(I,Ω)) iff the conditions (i) and (ii) below hold:

  1. σhbehbmodI.

    1. q(σ*j)Ω for all jΘhb, and

    2. (-.q(σ*0)(1,0))aΩ for all aJ(Δ).

Proof.

See [2, Lemma 28]. ∎

Remark 6.26.

Let σU2n+1(R,Δ). Define

I0:={xRxJ(Δ)I},I~0:={xRJ(Δ)¯μxI0}.

Then I0 is a left ideal of R and I~0 an additive subgroup of R. If J(Δ)=R then I0=I and I~0=I~. Lemma 6.25 implies that σU2n+1((R,Δ),(I,ΩmaxI)) iff σemodI,I~,I0,I~0, meaning that σhbehbmodI, σ0jI~ for any jΘhb, σi0I0 for any iΘhb and σ00-1I~0.

Definition 6.27.

The subgroup

NU2n+1((R,Δ),(I,Ω)):=NormalizerU2n+1(R,Δ)(U2n+1((R,Δ),(I,Ω)))

of U2n+1(R,Δ) is called the normalized principal congruence subgroup of level (I,Ω).

Remark 6.28.

  1. In many interesting situations, NU2n+1((R,Δ),(I,Ω)) equals U2n+1(R,Δ), for example the equality holds if Ω=ΩminI or Ω=ΩmaxI. But it can also happen that NU2n+1((R,Δ),(I,Ω))U2n+1(R,Δ), see [2, Example 38].

  2. EU2n+1(R,Δ)NU2n+1((R,Δ),(I,Ω)), see [2, Corollary 36].

Definition 6.29.

The subgroup

CU2n+1((R,Δ),(I,Ω)):=
{σNU2n+1((R,Δ),(I,Ω))[σ,EU2n+1(R,Δ)]U2n+1((R,Δ),(I,Ω))}

of U2n+1(R,Δ) is called the full congruence subgroup of level (I,Ω).

Lemma 6.30.

Let σU2n+1(R,Δ). Then σCU2n+1((R,Δ),(I,ΩmaxI)) iff

  1. σijI for any i,jΘhb such that ij,

  2. σi0I0 for any iΘhb,

  3. σ0jI~ for any jΘhb,

  4. σii-σjjI for any i,jΘhb, and

  5. σ00-σjjI~0 for any jΘhb.

Theorem 6.31.

If R is semilocal or quasifinite and n3, then the following equalities hold:

[CU2n+1((R,Δ),(I,Ω)),EU2n+1(R,Δ)]
  =[EU2n+1((R,Δ),(I,Ω)),EU2n+1(R,Δ)]
  =EU2n+1((R,Δ),(I,Ω)).

Proof.

See [2, Theorem 39]. ∎

7 Sandwich classification for U2n+1(R,Δ)

In this section n denotes a natural number greater than or equal to 3 and (R,Δ) a Hermitian form ring where R is commutative.

Definition 7.1.

Let σU2n+1(R,Δ). Then a matrix of the form σ±1ε where εEU2n+1(R,Δ) is called an elementary (unitary) σ-conjugate.

Theorem 7.2.

Let σU2n+1(R,Δ) and i,j,k,lΘhb such that k±l and i±j. Further let aJ(Δ). Then

  1. Tkl(σij) is a product of 160 elementary unitary σ -conjugates,

  2. Tkl(σi,-i) is a product of 320 elementary unitary σ -conjugates,

  3. Tkl(σi0a) is a product of 480 elementary unitary σ -conjugates,

  4. Tkl(a¯μσ0j) is a product of 480 elementary unitary σ -conjugates,

  5. Tkl(σii-σjj) is a product of 480 elementary unitary σ -conjugates,

  6. Tkl(σii-σ-i,-i) is a product of 960 elementary unitary σ -conjugates,

  7. Tk(q1(σ*j),λ-(ε(k)+1)/2q2(σ*j)) is a product of 1600n+5764 elementary unitary σ -conjugates, and

  8. Tk((σ00-σjj)a,x) is a product of 4800n+16812 elementary unitary σ -conjugates for some xR.

Proof.

(i) Let xR. In Step 1 we show that Tkl(xσ¯23σ21) is a product of 16 elementary σ-conjugates. In Step 2 we show that Tkl(xσ¯23σ2,-1) is a product of 16 elementary σ-conjugates. In Step 3 we show that Tkl(xσ¯23σ22) is a product of 32 elementary σ-conjugates. In Step 4 we use Steps 1–3 in order to prove (i).

Step 1. Set

τ:=T1,-2(σ¯23σ23)T3,-2(-σ¯23σ21)T3,-1(λ¯σ¯23σ22)T3(0,σ¯22σ21-λ¯σ¯21σ22)

and ξ:=τ-1σ. One checks easily that (στ-1)2*=σ2* and (τ-1σ-1)*,-2=σ*,-2. Hence ξ2*=e2t and ξ*,-2=e-2. Set

ζ:=[T-2,-1(1),[τ,σ]]τ-1
=[T-2,-1(1),τξ]τ-1
=[τ-1,T-2,-1(1)][T-2,-1(1),ξ],

the last equality by Lemma 3.1. One checks easily that

[τ-1,T-2,-1(1)]=T3,-1(σ¯23σ21)T1(z)

for some zΔ-1 and [T-2,-1(1),ξ]=T-2(x-2)i0,±2Ti2(xi) for some xiR (i0,2). Hence

ζ=T3,-1(σ¯23σ21)T1(z)T-2(x-2)i0,±2Ti2(xi).

It follows that [T-1,3(-x),[T-2,3(1),ζ]]=T-2,3(xσ¯23σ21) for any xR. Hence we have shown

[T-1,3(-x),[T-2,3(1),[T-2,-1(1),[τ,σ]]τ-1]]=T-2,3(xσ¯23σ21).

This implies that T-2,3(xσ¯23σ21) is a product of 16 elementary σ-conjugates. It follows from Lemma 6.21 that Tkl(xσ¯23σ21) is a product of 16 elementary σ-conjugates.

Step 2. Step 2 can be done similarly. Namely one can show that

[T-1,3(-xλ¯),[T12(1),[T-1,2(1),[τ,σ]]τ-1]]=T-1,2(xσ¯23σ2,-1)

for any xR where

τ=T21(σ¯23σ23)T31(-σ¯23σ22)T3,-2(σ¯23σ2,-1)T3(0,-σ¯22σ2,-1+λ¯σ¯2,-1σ22).

Step 3. Set

τ:=T21(-σ¯22σ23)T31(σ¯22σ22)T2,-3(σ¯22σ2,-1)T2(0,-σ¯23σ2,-1+λ¯σ¯2,-1σ23)

and ξ:=τ-1σ. One checks easily that (στ-1)2*=σ2* and (τ-1σ-1)*,-2=σ*,-2. Hence ξ2*=e2t and ξ*,-2=e-2. Set

ζ:=[T32(1),[τ,σ]]τ-1=[T32(1),τξ]τ-1=[τ-1,T32(1)][T32(1),ξ].

One checks easily that ψ:=[τ-1,T32(1)]=T31(-σ¯22σ23)T3(y)T3,-2(z) for some yΔ-1 and zR and θ:=[T32(1),ξ]=T-2(x-2)i0,±2Ti2(xi) for some xiR (i0,2). Set

χ:=[T12(1),ζ]ψ-1=[T12(1),ψθ]ψ-1=[ψ-1,T12(1)][T12(1),θ].

One checks easily that [ψ-1,T12(1)]=T32(σ¯22σ23)T3(a)T3,-1(b) for some aΔ-1 and bR and [T12(1),θ]=T-2(c) for some cΔ. Hence

χ=T32(σ¯22σ23)T3(a)T3,-1(b)T-2(c).

It follows that

[T-2,3(x¯),[T2,-1(1),χ]]=T-2,-1(-x¯σ¯22σ23)=T12(xσ¯23σ22)

for any xR, the last equality by (S1) of Lemma 6.18. Hence we have shown

[T-2,3(x¯),[T2,-1(1),[T12(1),[T32(1),[τ,σ]]τ-1]ψ-1]]=T12(xσ¯23σ22).

This implies that T12(xσ¯23σ22) is a product of 32 elementary σ-conjugates. It follows from Lemma 6.21 that Tkl(xσ¯23σ22) is a product of 32 elementary σ-conjugates.

Step 4. Set I:=I({σ¯23σ21,σ¯23σ2,-1¯}), J:=I({σ¯23σ21,σ¯23σ2,-1¯,σ¯23σ22}) and

τ:=[σ-1,T12(-σ¯23)]=(e-σ*1σ¯23σ2*+σ*,-2σ23σ-1,*)T12(σ¯23).

Clearly

τ11=1-σ11σ¯23σ21+σ1,-2σ23σ-1,1
=1-σ11σ¯23σ21I+λ¯σ¯2,-1σ23Iσ-1,1,

the last equality by Lemma 6.14, and

τ12=-σ11σ¯23σ22+σ1,-2σ23σ-1,2+τ11σ¯23
=-σ11σ¯23σ22J+λ¯σ¯2,-1σ23Jσ-1,2+τ11σ¯23,

again the last equality by Lemma 6.14. Hence τ111modI and τ12σ¯23modJ (note that IJ). Set ζ:=τP13P21. Then ζ22=τ11 and ζ23=τ12 and hence ζ¯23ζ22σ23modI+J¯. Applying Step 3 above to ζ, we get that Tkl(ζ¯23ζ22) is a product of 32 elementary ζ-conjugates. Since any elementary ζ-conjugate is a product of two elementary σ-conjugates, it follows that Tkl(ζ¯23ζ22) is a product of 64 elementary σ-conjugates. Thus, by Steps 1–3, Tkl(σ23) is a product of 64+16+16+16+16+32=160 elementary σ-conjugates. Since one can bring σij to position (3,2) conjugating by monomial matrices from EU2n+1(R,Δ), the assertion of (i) follows.

(ii)–(vi) See the proof of Theorem 5.2. In the proof of (iii) replace the matrix T-j(-1) by T-j(-.ε(j)(a,b)) where b is chosen such that (a,b)Δε(j). In the proof of (iv) replace the matrix Ti(-1) by Ti(-.-ε(i)(λ(ε(i)-1)/2a,b)) where b is chosen such that (a,b)Δ-ε(i).

(vii) By Lemma 6.21 it suffices to consider the case ε(k)=-1. Set m:=160. In Step 1 we show that for any xR the matrix Tk(q(σ*1)σ11x) is a product of (2n+19)m+4 elementary σ-conjugates. In Step 2 we use Step 1 in order to prove (vii).

Step 1. Set

u:=(00σ-1,-1-σ-1,-2)t=(00σ¯11-σ¯21)tM

and u:=σ-1uM. Then clearly u-1=0. Further

q(u)=q(σ-1u)q(u)=(0,0)modΔ

and hence u is isotropic. Set

ξ:=T*,-1σ(-u)=e-σuσ*1~+σ*1λ¯σu~=e-uσ*1~+σ*1λ¯u~,

the first equality being a consequence of (6.1). Then

ξ=(1-σ11σ21σ11σ11-σ21σ211+σ21σ11-σ31σ21σ31σ111-σ41σ21σ41σ111-σn1σ21σn1σ111-σ01σ21σ01σ111-σ-n,1σ21σ-n,1σ111-σ-4,1σ21σ-4,1σ111-σ-3,1σ21σ-3,1σ111-λβ¯α******-σ¯11σ¯31**γβ******σ¯21σ¯31**)

where

α=σ-2,1σ11-λσ¯11σ¯-2,1,
β=σ-1,1σ11+λσ¯21σ¯-2,1,
γ=-σ-1,1σ21+λσ¯21σ¯-1,1.

Set

τ:=T-3,1(σ-3,1σ21)T-3,2(-σ-3,1σ11).

It follows from (i) that τ is a product of 2m elementary σ-conjugates. Clearly

ξτ=(1-σ11σ21σ11σ11-σ21σ211+σ21σ11-σ31σ21σ31σ111-σ41σ21σ41σ111-σn1σ21σn1σ111-σ01σ21σ01σ111-σ-n,1σ21σ-n,1σ111-σ-4,1σ21σ-4,1σ111001*α+δ0*********β-ε0********)

where δ=σ¯11σ¯31σ-3,1σ11 and ε=σ¯21σ¯31σ-3,1σ11. Let xR and set

ζ:=[T2,-3(-x),[T*,-1(u),σ]τ]T*,-1(-u)
=[T2,-3(-x),T*,-1(u)ξτ]T*,-1(-u)
=[T*,-1(-u),T2,-3(-x)][T2,-3(-u),ξτ].

Clearly ζ is a product of 4m+4 elementary σ-conjugates. One checks easily that

[T*,-1(-u),T2,-3(-x)]
  =T1(0,λ¯(-x¯u-2u¯-3+λx¯u-2u¯-3¯))T1,-2(λ¯x¯u¯-3)T1,-3(-xu¯-2)
  =T1(0,λ¯(a+λa¯))T1,-2(b)T1,-3(-x(σ11σ22-σ12σ21))

for some a,bI({σ21,σ23}). Further

[T2,-3(-x),ξτ]
  =(p=1,p3,0-4Tp,-3(xσp1σ11))T-2,-3(x(α+δ))T-1,-3(x(β-ε))T3(y)

where

y=(q1(σ*1),λ¯q2(σ*1))σ11x+(0,c-λ¯c¯)

for some cI({σ31,σ-2,1}). Hence

ζ=T1(0,λ¯(a+λa¯))T1,-2(b)T1,-3(-x(σ11(σ22-σ11)-σ12σ21))
  (p=2,p3,0-4Tp,-3(xσp1σ11))T-2,-3(x(α+δ))T-1,-3(x(β-ε))T3(y).

It follows from (i), (ii) and (iii) and relation (S5) in Lemma 6.18 that T3(y) is a product of

4m+4+4m+2m+4m+(2n-5)m+3m+3m=(2n+15)m+4

elementary σ-conjugates. By (i) and relation (S5) in Lemma 6.18, T3(0,-c+λ¯c¯) is a product of 4m elementary σ-conjugates. Hence

T3((q1(σ*1),λ¯q2(σ*1))σ11x)=T3(y)T3(0,-c+λ¯c¯)

is a product of (2n+19)m+4 elementary σ-conjugates. It follows from Lemma 6.21 that Tk(q(σ*1)σ11x) is a product of (2n+19)m+4 elementary σ-conjugates.

Step 2. By Lemma 6.8 we have

Tk(q(σ*1))=Tk(q(σ*1)s=1-1σ1sσs1)
=Tk((+.s=1-1(q(σ*1)σ1sσs1))
  +.(0,s,t=1,s>t-1σ1sσs1q2(σ*1)σ1tσt1¯-σ1sσs1q2(σ*1)σ1tσt1¯¯λ))
=s=1-1Tk(q(σ*1)σ1sσs1)A:=
  Tk(0,s,t=1,s>t-1σ1sσs1q2(σ*1)σ1tσt1¯-σ1sσs1q2(σ*1)σ1tσt1¯¯λ)B:=

since q(σ*1)ΔΔmax. By Step 1, Tk(q(σ*1)σ11σ11) is a product of (2n+19)m+4 elementary σ-conjugates. By (i) and (SE2) in Lemma 6.18, Tk(q(σ*1)σ1sσs1) is a product of 3m elementary σ-conjugates if s±1,0. By (ii) and again (SE2), Tk(q(σ*1)σ1,-1σ-1,1) is a product of 6m elementary σ-conjugates. By (iv) and (SE2), Tk(q(σ*1)σ10σ01) is a product of 9m elementary σ-conjugates (note that σ10=λ¯σ¯0,-1μ by Lemma 6.14). Hence A is a product of

(2n+19)m+4+(2n-2)3m+6m+9m=(8n+28)m+4

elementary σ-conjugates. On the other hand B=Tk(0,x-λx¯) for some xI(q2(σ*1)). Since q2(σ*1)=r=1nσ¯r1σ-r,1, it follows from (i), (ii) and relation (S5) in Lemma 6.18 that B is a product of

4m+(n-1)2m=(2n+2)m

elementary σ-conjugates. Hence Tk(q(σ*1)) is a product of (10n+30)m+4=1600n+4804 elementary σ-conjugates. The assertion of (vii) follows now from Lemma 6.22.

(viii) By Lemma 6.21 it suffices to consider the case ε(k)=-1. Choose a bR such that (a,b)Δε(j). One checks easily that the entry of ξ:=σT-j(a,b) at position (0,j) equals (σjj-σ00)a+σ0j-σj0a2+σ0,-jb+σj,-jba2. Applying (vii) to ξ, we get that

Tk(q(ξ*j))=Tk((σjj-σ00)a+σ0j-σj0a2+σ0,-jb+σj,-jba2,x1)

is a product of 1600n+4804 elementary σ-conjugates where x1 is some ring element. Applying (vii) to σ, we get that

Tk(-.q(σ*j))=Tk(-σ0j,x2)andTk(-.q(σ*,-j)b)=Tk(-σ0,-jb,x3)

each are a product of 1600n+4804 elementary σ-conjugates where x2,x3 are some ring elements. By (iii) and (SE2) in Lemma 6.18, Tk(σj0a2,x4) is a product of 3480=1440 elementary σ-conjugates where x4 is some ring element. By (ii) and again (SE2), Tk(-σj,-jba2,x5) is a product of 3320=960 elementary σ-conjugates where x5 is some ring element. It follows that

Tk((σjj-σ00)a+σ0j-σj0a2+σ0,-jb+σj,-jba2,x1)
  Tk(-σ0j,x2)Tk(σj0a2,x4)Tk(-σ0,-jb,x3)Tk(-σj,-jba2,x5)
=Tk((σjj-σ00)a,x)

is a product of 3(1600n+4804)+1440+960=4800n+16812 elementary σ-conjugates where x is some ring element. ∎

As a corollary we get the Sandwich Classification Theorem for U2n+1(R,Δ).

Corollary 7.3.

Let H be a subgroup of U2n+1(R,Δ). Then H is normalized by EU2n+1(R,Δ) iff

(7.1)EU2n+1((R,Δ),(I,Ω))HCU2n+1((R,Δ),(I,Ω))

for some odd form ideal (I,Ω) of (R,Δ).

Proof.

First suppose that H is normalized by EU2n+1(R,Δ). Let (I,Ω) be the odd form ideal of (R,Δ) defined by

I:={xRT12(x)H}andΩ:={(y,z)ΔT-1(y,z)H}.

So the first inclusion of (7.1) clearly holds. It remains to show the second inclusion, i.e. that if σH and εEU2n+1(R,Δ), then [σ,ε]U2n+1((R,Δ),(I,Ω)). Since U2n+1((R,Δ),(I,Ω)) is normalized by EU2n+1(R,Δ) (see part (b) of Remark 6.28), we can assume that ε is an elementary matrix. By the previous theorem and Lemma 6.30, we have σCU2n+1((R,Δ),(I,ΩmaxI)) and therefore [σ,ε]U2n+1((R,Δ),(I,ΩmaxI)). Hence, by Lemma 6.25 it remains to show that q([σ,ε]*j)Ω for all jΘhb and (-.q([σ,ε]*0)(1,0))aΩ for all aJ(Δ). But applying the previous theorem to [σ,ε], we get that q([σ,ε]*j)Ω for all jΘhb. That (-.q([σ,ε]*0)(1,0))aΩ for all aJ(Δ) follows from the previous theorem and [2, Lemma 63].

Suppose now that (7.1) holds for some odd form ideal (I,Ω). Then it follows from the standard commutator formula in Theorem 6.31 that H is normalized by EU2n+1(R,Δ). ∎


Communicated by John S. Wilson


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Received: 2018-01-02
Published Online: 2018-04-17
Published in Print: 2018-07-01

© 2018 Walter de Gruyter GmbH, Berlin/Boston

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