Sandwich classification for $GL_n(R)$, $O_{2n}(R)$ and $U_{2n}(R,\Lambda)$ revisited

TL;DR

This paper provides concise proofs for the Sandwich Classification Theorems in classical groups by expressing certain matrix entries as products of a fixed number of elementary matrices, simplifying previous approaches.

Contribution

It introduces a new method to express matrix entries as products of elementary matrices, leading to shorter proofs of the Sandwich Classification Theorems for classical groups.

Findings

Expressed specific matrix entries as products of 8 or 24 elementary matrices.

Provided new, shorter proofs of the Sandwich Classification Theorems.

Extended results to orthogonal and hyperbolic unitary groups.

Abstract

Let $n$ be a natural number greater or equal to $3$, $R$ a commutative ring and $\sigma\in GL_n(R)$. We show that $t_{kl}(\sigma_{ij})$ (resp. $t_{kl}(\sigma_{ii}-\sigma_{jj}))$ where $i\neq j$ and $k\neq l$ can be expressed as a product of $8$ (resp. $24$) matrices of the form $^{\epsilon}\sigma^{\pm 1}$ where $\epsilon\in E_n(R)$. We prove similar results for the orthogonal groups $O_{2n}(R)$ and the hyperbolic unitary groups $U_{2n}(R,\Lambda)$ under the assumption that $R$ is commutative and $n\geq 3$. This yields new, very short proofs of the Sandwich Classification Theorems for the groups $GL_n(R)$, $O_{2n}(R)$ and $U_{2n}(R,\Lambda)$.

Equations

$$t_{ij}(x)t_{ij}(y)$$

$$[t_{ij}(x),t_{hk}(y)]$$

$$[t_{ij}(x),t_{jk}(y)]$$

$$[C_n(R,I),E_n(R)]=[E_n(R,I),E_n(R)]=E_n(R,I)$$

$$\zeta :={}^{{\tau }^{-1}}[t_{32}(1),[\tau ,\sigma ]]={}^{{\tau }^{-1}}[t_{32}(1),\tau \xi ]\stackrel{L.\text{ref}pre}{=}[{\tau }^{-1},t_{32}(1)][t_{32}(1),\xi ].$$

$$[t_{12}(1),{}^{t_{21}({\sigma }_{23})t_{31}(-{\sigma }_{22})}[t_{32}(1),[t_{21}(-{\sigma }_{23})t_{31}({\sigma }_{22}),\sigma ]]]=t_{32}({\sigma }_{23}).$$

$$E_n(R,I)\subseteq H\subseteq C_n(R,I)$$

$$q:V$$

$$v$$

$$T_{ij}(x):=e+x{e}^{ij}-x{e}^{-j,-i}\in O_{2n}(R)$$

$$T_{ij}(x)$$

$$T_{ij}(x)T_{ij}(y)$$

$$[T_{ij}(x),T_{hk}(y)]$$

$$[T_{ij}(x),T_{jk}(y)]$$

$$[T_{ij}(x),T_{j,-i}(y)]$$

$$[C{O}_{2n}(R,I),E{O}_{2n}(R)]=[E{O}_{2n}(R,I),E{O}_{2n}(R)]=E{O}_{2n}(R,I)$$

$$\zeta :={}^{{\tau }^{-1}}[T_{32}(1),[\tau ,\sigma ]]={}^{{\tau }^{-1}}[T_{32}(1),\tau \xi ]\stackrel{L.\text{ref}pre}{=}[{\tau }^{-1},T_{32}(1)][T_{32}(1),\xi ].$$

$$[T_{12}(1),{}^{T_{21}({\sigma }_{23})T_{31}(-{\sigma }_{22})T_{2,-3}(-{\sigma }_{2,-1})}[T_{32}(1),[T_{21}(-{\sigma }_{23})T_{31}({\sigma }_{22})T_{2,-3}({\sigma }_{2,-1}),\sigma ]]]=T_{32}({\sigma }_{23}).$$

$$E{O}_{2n}(R,I)\subseteq H\subseteq C{O}_{2n}(R,I)$$

$$\stackrel{ˉ}{}:R$$

$$x$$

$$f:V\times V$$

$$(v,w)$$

$$ϵ:\Omega$$

$$i$$

$$\tilde{}:V$$

$$v$$

$$T_{ij}(x):=e+x{e}^{ij}-{\lambda }^{(ϵ(j)-ϵ(i))/2}\overline{x}{e}^{-j,-i}\in U_{2n}(R,\Lambda )$$

$$T_{i,-i}(y):=e+y{e}^{i,-i}\in U_{2n}(R,\Lambda )$$

$$T_{ij}(x)$$

$$T_{ij}(x)T_{ij}(y)$$

$$[T_{ij}(x),T_{hk}(y)]$$

$$[T_{ij}(x),T_{jk}(y)]$$

$$[T_{ij}(x),T_{j,-i}(y)]$$

$$[T_{i,-i}(x),T_{-i,j}(y)]$$

$$\displaystyle\left(\begin{array}[]{cccc|cccc}1&-\bar{v}_{-2}&\dots&-\bar{v}_{-n}&-\bar{\lambda}\bar{v}_{n}&\dots&-\bar{\lambda}\bar{v}_{2}&v_{1}-\bar{\lambda}\bar{v}_{1}\\ &1&&&&&&v_{2}\\ &&\ddots&&&&&\vdots\\ &&&1&&&&v_{n}\\ \hline\cr&&&&1&&&v_{-n}\\ &&&&&\ddots&&\vdots\\ &&&&&&1&v_{-2}\\ &&&&&&&1\end{array}\right)$$

$$=$$

$${}^{\sigma }{T}_{\ast ,-1}(v)=e+\sigma v{\sigma }_{-1,\ast }^{\prime }-{\sigma }_{\ast 1}\bar{\lambda }\tilde{v}{\sigma }^{-1}\stackrel{L.\text{ref}35}{=}e+\sigma v\widetilde{{\sigma }_{\ast 1}}-{\sigma }_{\ast 1}\bar{\lambda }\widetilde{\sigma v}$$

$$|{\hat{\sigma }}_{\ast i}|=\{\begin{array}{l}|{\sigma }_{\ast j}|,\text{ if }ϵ(i)=ϵ(j),\\ |{\sigma }_{\ast j}|-{\bar{\sigma }}_{ij}{\sigma }_{-i,j}+\lambda \overline{{\bar{\sigma }}_{ij}{\sigma }_{-i,j}}-{\bar{\sigma }}_{-j,j}{\sigma }_{jj}+\lambda \overline{{\bar{\sigma }}_{-j,j}{\sigma }_{jj}},\text{ if }ϵ(i)=1,ϵ(j)=-1,\\ |{\sigma }_{\ast j}|-{\bar{\sigma }}_{-i,j}{\sigma }_{ij}+\lambda \overline{{\bar{\sigma }}_{-i,j}{\sigma }_{ij}}-{\bar{\sigma }}_{jj}{\sigma }_{-j,j}+\lambda \overline{{\bar{\sigma }}_{jj}{\sigma }_{-j,j}},\text{ if }ϵ(i)=-1,ϵ(j)=1.\end{array}$$

Full text

1. Introduction

Let $n$ be a natural number greater or equal to $3$ and $R$ a commutative ring. Let $\sigma\in GL_{n}(R)$ and set $H:={}^{E_{n}(R)}\sigma$, i.e. $H$ is the smallest subgroup of $GL_{n}(R)$ which contains $\sigma$ and is normalized by $E_{n}(R)$. Let $I$ be the ideal of $R$ defined by $I:=\{x\in R\mid t_{12}(x)\in H\}$. Then clearly $E_{n}(R,I)\subseteq H$. By the Sandwich Classification Theorem (SCT) for $GL_{n}(R)$ one also has $H\subseteq C_{n}(R,I)$. It follows that $\sigma_{ij},\sigma_{ii}-\sigma_{jj}\in I$ for any $i\neq j$, i.e. the matrices $t_{12}(\sigma_{ij})$ and $t_{12}(\sigma_{ii}-\sigma_{jj})$ can be expressed as products of matrices of the form ${}^{\epsilon}\sigma^{\pm 1}$ where $\epsilon\in E_{n}(R)$. We show how one can use the theme of the paper [4] in order to find such expressions and give boundaries for the number of factors, see Theorem 12. This yields a new, very simple proof of the SCT for $GL_{n}(R)$.

Further we prove an orthogonal and a unitary version of Theorem 12 (cf. Theorem 27 and Theorem 49). The proof of the orthogonal version is very 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 $U_{2n}(R,\Lambda)$ 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 more general). For the hyperbolic unitary groups $U_{2n}(R,\Lambda)$ this yields the first proof of the SCT which does not use localization.

This 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 12, 27 and 49. In Section 4 we recall the definitions of the general linear group $GL_{n}(R)$ and some important subgroups, in Section 5 we prove Theorem 12. In Section 6 we recall the definitions of the (even-dimensional) orthogonal group $O_{2n}(R)$ and some important subgroups, in Section 7 we prove Theorem 27. In Section 8 we recall the definitions of A. Bak’s hyperbolic unitary group $U_{2n}(R,\Lambda)$ and some important subgroups and in the last section we prove Theorem 49.

2. Notation

By a natural number we mean an element of the set $\mathbb{N}:=\{1,2,3,\dots\}$. If $G$ is a group and $g,h\in G$, we let ${}^{h}g:=hgh^{-1}$ and $[g,h]:=ghg^{-1}h^{-1}$. By a ring we will always mean an associative ring with $1$ such that $1\neq 0$. 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 $I(X)$. The set of all invertible elements in a ring $R$ is denoted by $R^{*}$. If $m$ and $n$ are natural numbers and $R$ is a ring, then the set of all $m\times n$ matrices with entries in $R$ is denoted by $M_{m\times n}(R)$. If $a\in M_{m\times n}(R)$, we denote the transpose of $a$ 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 column by $a_{*j}$. We set $M_{n}(R):=M_{n\times n}(R)$. The identity matrix in $M_{n}(R)$ is denoted by $e$ or $e^{n\times n}$ and the matrix with a $1$ at position $(i,j)$ and zeros elsewhere is denoted by $e^{ij}$. If $a\in M_{n}(R)$ is invertible, the entry of $a^{-1}$ at position $(i,j)$ is denoted by $a^{\prime}_{ij}$, the $i$-th row of $a^{-1}$ by $a^{\prime}_{i*}$ and the $j$-th column of $a^{-1}$ by $a^{\prime}_{*j}$. Further we denote by ${}^{n}R$ the set of all rows $v=(v_{1},\dots,v_{n})$ with entries in $R$ and by $R^{n}$ the set of all columns $u=(u_{1},\dots,u_{n})^{t}$ 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 $a,b,c\in G$. Then ${}^{b^{-1}}[a,bc]=[b^{-1},a][a,c]$.

Lemma 2**.**

Let $G$ be a group, $E$ a subgroup and $a\in G$. Suppose that $b\in G$ is a product of $n$ elements of the form ${}^{\epsilon}a^{\pm 1}$ where $\epsilon\in E$. Then

1. (i)

${}^{\epsilon^{\prime}}b$* is a product of $n$ elements of the form ${}^{\epsilon}a^{\pm 1}$* 2. (ii)

$[\epsilon^{\prime},b]$* is a product of $2n$ elements of the form ${}^{\epsilon}a^{\pm 1}$*

for any $\epsilon^{\prime}\in E$.

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 definitions of 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 $I$, the principal 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 $i,j\in\{1,\dots,n\}$ such that $i\neq j$ and $x\in R$. 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 $x\in I$. The subgroup of $GL_{n}(R)$ generated by all $I$-elementary transvections is called preelementary subgroup of level $I$ and is denoted by $E_{n}(I)$. Its normal closure in $E_{n}(R)$ is called elementary subgroup of level $I$ and is denoted by $E_{n}(R,I)$.

Lemma 5**.**

The relations

[TABLE]

hold where $i\neq k,j\neq h$ in $(R2)$ and $i\neq k$ in $(R3)$.

Proof.

Straightforward computation. ∎

Definition 6**.**

Let $i,j\in\{1,\dots,n\}$ such that $i\neq j$. Define $p_{ij}:=e+e^{ij}-e^{ji}-e^{ii}-e^{jj}=t_{ij}(1)t_{ji}(-1)t_{ij}(1)\in E_{n}(R)$. It is easy show that $p_{ij}^{-1}=p_{ji}$.

Lemma 7**.**

Let $x\in R$ and $i,j,k\in\{1,\dots,n\}$ be pairwise distinct indices. Then

1. (i)

${}^{p_{ki}}t_{ij}(x)=t_{kj}(x)$* and* 2. (ii)

${}^{p_{kj}}t_{ij}(x)=t_{ik}(x)$.

Proof.

Follows from the relations in Lemma 5. ∎

4.3. Congruence subgroups

Definition 8**.**

The kernel of the group homomorphism $GL_{n}(R)\rightarrow GL_{n}(R/I)$ induced by the canonical map $R\rightarrow R/I$ 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 $GL_{n}(R)\rightarrow GL_{n}(R/I)$ induced by the canonical map $R\rightarrow R/I$ is called full congruence subgroup of level $I$ and is denoted by $C_{n}(R,I)$. Obviously $GL_{n}(R,I)\subseteq C_{n}(R,I)$ and $C_{n}(R,I)$ is a normal subgroup of $GL_{n}(R)$.

Theorem 10**.**

If $n\geq 3$ and $R$ is almost commutative (i.e. module finite over its center), then the equalities

[TABLE]

hold.

Proof.

See [5], Corollary 14. ∎

5. 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 $\sigma\in GL_{n}(R)$. Then a matrix of the form ${}^{\epsilon}\sigma^{\pm 1}$ where $\epsilon\in E_{n}(R)$ is called an elementary $\sigma$-conjugate.

Theorem 12**.**

Let $\sigma\in GL_{n}(R)$, $i\neq j$ and $k\neq l$. Then

1. (i)

$t_{kl}(\sigma_{ij})$* is a product of $8$ elementary $\sigma$-conjugates and* 2. (ii)

$t_{kl}(\sigma_{ii}-\sigma_{jj})$* is a product of $24$ elementary $\sigma$-conjugates.*

Proof.

(i) Set $\tau:=t_{21}(-\sigma_{23})t_{31}(\sigma_{22})$. One checks easily that the second row of $\sigma\tau^{-1}$ equals the second row of $\sigma$ and hence the second row of $\xi:={}^{\sigma}\tau^{-1}$ is trivial. Set

[TABLE]

One checks easily that $[\tau^{-1},t_{32}(1)]=t_{31}(-\sigma_{23})$ and $[t_{32}(1),\xi]=\prod\limits_{i\neq 2}t_{i2}(x_{i})$ for some $x_{1},x_{3},x_{4},\dots,x_{n}\in R$. Hence $\zeta=t_{31}(-\sigma_{23})\prod\limits_{i\neq 2}t_{i2}(x_{i})$. It follows that $[t_{12}(1),\zeta]=t_{32}(\sigma_{23})$. Hence we have shown

[TABLE]

This implies that $t_{32}(\sigma_{23})$ is a product of $8$ elementary $\sigma$-conjugates. It follows from Lemma 7 that $t_{kl}(\sigma_{23})$ is a product of $8$ elementary $\sigma$-conjugates. Since one can bring $\sigma_{ij}$ to position $(2,3)$ by conjugating monomial matrices in $E_{n}(R)$ (see Definition 6) to $\sigma$, the assertion of (i) follows.

(ii) Clearly the entry of ${}^{t_{ji}(1)}\sigma$ at position $(j,i)$ equals $\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij}$. Applying (i) to ${}^{t_{ji}(1)}\sigma$ we get that $t_{kl}(\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij})$ is a product of $8$ elementary $\sigma$-conjugates (note that any elementary ${}^{t_{ji}(1)}\sigma$-conjugate is also an elementary $\sigma$-conjugate). Applying (i) to $\sigma$ we get that $t_{kl}(\sigma_{ij}-\sigma_{ji})=t_{kl}(\sigma_{ij})t_{kl}(-\sigma_{ji})$ is a product of $16$ elementary $\sigma$-conjugates. It follows that $t_{kl}(\sigma_{ii}-\sigma_{jj})=t_{kl}(\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij})t_{kl}(\sigma_{ij}-\sigma_{ji})$ is a product of $24$ elementary $\sigma$-conjugates. ∎

As a corollary we get the Sandwich Classification Theorem for $GL_{n}(R)$. Note that if $\sigma\in GL_{n}(R)$ and $I$ is an ideal of $R$, then $\sigma\in C_{n}(R,I)$ if and only if $\sigma_{ij},\sigma_{ii}-\sigma_{jj}\in I$ for any $i\neq j$.

Corollary 13**.**

Let $H$ be a subgroup of $GL_{n}(R)$. Then $H$ is normalized by $E_{n}(R)$ if and only if

[TABLE]

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 $I:=\{x\in R\mid t_{12}(x)\in H\}$. Then clearly $E_{n}(R,I)\subseteq H$. It remains to show that $H\subseteq C_{n}(R,I)$, i.e. that if $\sigma\in H$, then $\sigma_{ij},\sigma_{ii}-\sigma_{jj}\in I$ for any $i\neq j$. 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 $H$ is normalized by $E_{n}(R)$. ∎

6. The even-dimensional orthogonal group $O_{2n}(R)$

In this section $n$ denotes a natural number, $R$ a commutative ring and $I$ an ideal of $R$. We shall recall the definitions of the even-dimensional orthogonal group $O_{2n}(R)$ and the following subgroups of $O_{2n}(R)$; the elementary subgroup $EO_{2n}(R)$, the preelementary subgroup $EO_{2n}(I)$ of level $I$, the elementary subgroup $EO_{2n}(R,I)$ of level $I$, the principal congruence subgroup $O_{2n}(R,I)$ of level $I$, and the full congruence subgroup $CO_{2n}(R,I)$ of level $I$.

6.1. The even-dimensional orthogonal group

Definition 14**.**

Set $V:=R^{2n}$. We use the following indexing for the elements of the standard basis of $V$: $(e_{1},\dots,e_{n},e_{-n},\dots,e_{-1})$. That means that $e_{i}$ is the column whose $i$-th coordinate is one and all the other coordinates are zero if $1\leq i\leq n$ and the column whose $(2n+1+i)$-th coordinate is one and all the other coordinates are zero if $-n\leq i\leq-1$. Let $p\in M_{n}(R)$ be the matrix with ones on the skew diagonal and zeros elsewhere. We define the quadratic form

[TABLE]

The subgroup $O_{2n}(R):=\{\sigma\in GL_{2n}(R)\mid q(\sigma v)=q(v)\leavevmode\nobreak\ \forall v\in V\}$ of $GL_{2n}(R)$ 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 $\Omega:=\{1,...,n,-n,...,-1\}$.

Lemma 17**.**

Let $\sigma\in GL_{2n}(R)$. Then $\sigma\in O_{2n}(R)$ if and only if

1. (i)

$\sigma^{\prime}_{ij}=\sigma_{-j,-i}\leavevmode\nobreak\ \forall i,j\in\Omega$* and* 2. (ii)

$q(\sigma_{*j})=0\leavevmode\nobreak\ \forall j\in\Omega$.

Proof.

See [2], p.167. ∎

Lemma 18**.**

Let $\sigma\in O_{2n}(R)$, $x\in R^{*}$ and $k\in\Omega$. Then the statements below are true.

1. (i)

If the $k$-th column of $\sigma$ equals $e_{k}x$ then the $(-k)$-th row of $\sigma$ equals $x^{-1}e^{t}_{-k}$. 2. (ii)

If the $k$-th row of $\sigma$ equals $xe^{t}_{k}$ then the $(-k)$-th column of $\sigma$ equals $e_{-k}x^{-1}$.

Proof.

Follows from (i) in the previous lemma. ∎

6.2. The elementary subgroup

Definition 19**.**

If $i,j\in\Omega$ such that $i\neq\pm j$ and $x\in R$, then the matrix

[TABLE]

is called an elementary orthogonal transvection. The subgroup of $O_{2n}(R)$ generated by all elementary orthogonal transvections is called elementary orthogonal group and is denoted by $EO_{2n}(R)$. An elementary orthogonal transvection $T_{ij}(x)$ is called $I$-elementary if $x\in I$. The subgroup of $O_{2n}(R)$ generated by all $I$-elementary orthogonal transvections is called preelementary subgroup of level $I$ and is denoted by $EO_{2n}(I)$. Its normal closure in $EO_{2n}(R)$ is called elementary subgroup of level $I$ and is denoted by $EO_{2n}(R,I)$.

Lemma 20**.**

The relations

[TABLE]

hold where $h\neq j,-i$ and $k\neq i,-j$ in (R3) and $i\neq\pm k$ in (R4).

Proof.

Straightforward calculation. ∎

Definition 21**.**

Let $i,j\in\Omega$ such that $i\neq\pm j$. Define $P_{ij}:=e+e^{ij}-e^{ji}+e^{-i,-j}-e^{-j,-i}-e^{ii}-e^{jj}-e^{-i,-i}-e^{-j,-j}=T_{ij}(1)T_{ji}(-1)T_{ij}(1)\in EO_{2n}(R)$. It is easy show that $(P_{ij})^{-1}=P_{ji}$.

Lemma 22**.**

Let $x\in R$ and $i,j,k\in\Omega$ such that $i\neq\pm j$ and $k\neq\pm i,\pm j$. Then

1. (i)

${}^{P_{ki}}T_{ij}(x)=T_{kj}(x)$* and* 2. (ii)

${}^{P_{kj}}T_{ij}(x)=T_{ik}(x)$.

Proof.

Follows from the relations in Lemma 20. ∎

6.3. Congruence subgroups

Definition 23**.**

The kernel of the group homomorphism $O_{2n}(R)\rightarrow O_{2n}(R/I)$ induced by the canonical map $R\rightarrow R/I$ is called principal congruence subgroup of level $I$ and is denoted by $O_{2n}(R,I)$. Obviously $O_{2n}(R,I)$ is a normal subgroup of $O_{2n}(R)$.

Definition 24**.**

The preimage of $Center(O_{2n}(R/I))$ under the group homomorphism $O_{2n}(R)\rightarrow O_{2n}(R/I)$ induced by the canonical map $R\rightarrow R/I$ is called full congruence subgroup of level $I$ and is denoted by $CO_{2n}(R,I)$. Obviously $O_{2n}(R,I)\subseteq CO_{2n}(R,I)$ and $CO_{2n}(R,I)$ is a normal subgroup of $O_{2n}(R)$.

Theorem 25**.**

If $n\geq 3$, then the equalities

[TABLE]

hold.

Proof.

See [2], Theorem 1.1 and Lemma 5.2. ∎

7. Sandwich classification for $O_{2n}(R)$

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

Definition 26**.**

Let $\sigma\in O_{2n}(R)$. Then a matrix of the form ${}^{\epsilon}\sigma^{\pm 1}$ where $\epsilon\in EO_{2n}(R)$ is called an elementary (orthogonal) $\sigma$-conjugate.

Theorem 27**.**

Let $\sigma\in O_{2n}(R)$, $i\neq\pm j$ and $k\neq\pm l$. Then

1. (i)

$T_{kl}(\sigma_{ij})$* is a product of $8$ elementary orthogonal $\sigma$-conjugates,* 2. (ii)

$T_{kl}(\sigma_{i,-i})$* is a product of $16$ elementary orthogonal $\sigma$-conjugates,* 3. (iii)

$T_{kl}(\sigma_{ii}-\sigma_{jj})$* is a product of $24$ elementary orthogonal $\sigma$-conjugates and* 4. (iv)

$T_{kl}(\sigma_{ii}-\sigma_{-i,-i})$* is a product of $48$ elementary orthogonal $\sigma$-conjugates.*

Proof.

(i) Set $\tau:=T_{21}(-\sigma_{23})T_{31}(\sigma_{22})T_{2,-3}(\sigma_{2,-1})$. One checks easily that the second row of $\sigma\tau^{-1}$ equals the second row of $\sigma$ and hence the second row of $\xi:={}^{\sigma}\tau^{-1}$ is trivial. By Lemma 18 the second last column of $\xi$ also is trivial. Set

[TABLE]

One checks easily that $[\tau^{-1},T_{32}(1)]=T_{31}(-\sigma_{23})$ and $[T_{32}(1),\xi]=\prod\limits_{i\neq\pm 2}T_{i2}(x_{i})$ for some $x_{i}\in R\leavevmode\nobreak\ (i\neq\pm 2)$. Hence $\zeta=T_{31}(-\sigma_{23})\prod\limits_{i\neq\pm 2}T_{i2}(x_{i})$. It follows that $[T_{12}(1),\zeta]=T_{32}(\sigma_{23})$. Hence we have shown

[TABLE]

This implies that $T_{32}(\sigma_{23})$ is a product of $8$ elementary $\sigma$-conjugates. It follows from Lemma 22 that $T_{kl}(\sigma_{23})$ is a product of $8$ elementary $\sigma$-conjugates. Since one can bring $\sigma_{ij}$ to position $(2,3)$ by conjugating monomial matrices in $EO_{2n}(R)$ (see Definition 21) to $\sigma$, the assertion of (i) follows.

(ii) Clearly the entry of ${}^{T_{ji}(1)}\sigma$ at position $(j,-i)$ equals $\sigma_{i,-i}+\sigma_{j,-i}$. Applying (i) to ${}^{T_{ji}(1)}\sigma$ we get that $T_{kl}(\sigma_{i,-i}+\sigma_{j,-i})$ is a product of $8$ elementary $\sigma$-conjugates (note that any elementary ${}^{T_{ji}(1)}\sigma$-conjugate is also an elementary $\sigma$-conjugate). Applying (i) to $\sigma$ we get that $T_{kl}(\sigma_{j,-i})$ is a product of $8$ elementary $\sigma$-conjugates. It follows that $T_{kl}(\sigma_{i,-i})=T_{kl}(\sigma_{i,-i}+\sigma_{j,-i})T_{ji}(-\sigma_{j,-i})$ is a product of $16$ elementary $\sigma$-conjugates.

(iii) Clearly the entry of ${}^{T_{ji}(1)}\sigma$ at position $(j,i)$ equals $\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij}$. Applying (i) to ${}^{T_{ji}(1)}\sigma$ we get that $T_{kl}(\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij})$ is a product of $8$ elementary $\sigma$-conjugates (note that any elementary ${}^{T_{ji}(1)}\sigma$-conjugate is also an elementary $\sigma$-conjugate). Applying (i) to $\sigma$ we get that $T_{kl}(\sigma_{ij}-\sigma_{ji})=T_{kl}(\sigma_{ij})T_{kl}(-\sigma_{ji})$ is a product of $16$ elementary $\sigma$-conjugates. It follows that $T_{kl}(\sigma_{ii}-\sigma_{jj})=T_{kl}(\sigma_{ii}-\sigma_{jj}+\sigma_{ji}-\sigma_{ij})T_{kl}(\sigma_{ij}-\sigma_{ji})$ is a product of $24$ elementary $\sigma$-conjugates.

(iv) Follows from (iii) since $T_{kl}(\sigma_{ii}-\sigma_{-i,-i})=T_{kl}(\sigma_{ii}-\sigma_{jj})T_{kl}(\sigma_{jj}-\sigma_{-i,-i})$.

∎

As a corollary we get the Sandwich Classification Theorem for $O_{2n}(R)$. Note that if $\sigma\in O_{2n}(R)$ and $I$ is an ideal of $R$, then $\sigma\in CO_{2n}(R,I)$ if and only if $\sigma_{ij},\sigma_{ii}-\sigma_{jj}\in I$ for any $i\neq j$.

Corollary 28**.**

Let $H$ be a subgroup of $O_{2n}(R)$. Then $H$ is normalized by $EO_{2n}(R)$ if and only if

[TABLE]

for some ideal $I$ of $R$.

Proof.

First suppose that $H$ is normalized by $EO_{2n}(R)$. Let $I$ be the ideal of $R$ defined by $I:=\{x\in R\mid T_{12}(x)\in H\}$. Then clearly $EO_{2n}(R,I)\subseteq H$. It remains to show that $H\subseteq CO_{2n}(R,I)$, i.e. that if $\sigma\in H$, then $\sigma_{ij},\sigma_{ii}-\sigma_{jj}\in I$ for any $i\neq j$. 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 $H$ is normalized by $EO_{2n}(R)$ ∎

8. Bak’s unitary group $U_{2n}(R,\Lambda)$

In order to classify the subgroups of a general linear group (resp. an even-dimensional orthogonal group) which are normalized by the elementary subgroup (resp. the elementary orthogonal group), the notion of an ideal of a ring is sufficient. Bak’s dissertation [1] showed that the notion of an ideal by itself was not sufficient to solve the analogous classification problem for unitary groups, but that a refinement of the notion of an ideal, called a form ideal, was necessary. This led naturally to a more general notion of unitary group, which was defined over a form ring instead of just a ring and generalized all previous concepts. We describe form rings $(R,\Lambda)$ and form ideals $(I,\Gamma)$ first, then the hyperbolic unitary group $U_{2n}(R,\Lambda)$ and its elementary subgroup $EU_{2n}(R,\Lambda)$ over a form ring $(R,\Lambda)$. For a form ideal $(I,\Gamma)$, we recall the definitions of the following subgroups of $U_{2n}(R,\Lambda)$; the preelementary subgroup $EU_{2n}(I,\Gamma)$ of level $(I,\Gamma)$, the elementary subgroup $EU_{2n}((R,\Lambda),(I,\Gamma))$ of level $(I,\Gamma)$, the principal congruence subgroup $U_{2n}((R,\Lambda),(I,\Gamma))$ of level $(I,\Gamma)$, and the full congruence subgroup $CU_{2n}((R,\Lambda),(I,\Gamma))$ of level $(I,\Gamma)$.

8.1. Form rings and form ideals

Definition 29**.**

Let $R$ be a ring and

[TABLE]

an involution on $R$, i.e. $\overline{x+y}=\bar{x}+\bar{y}$, $\overline{xy}=\bar{y}\bar{x}$ and $\bar{\bar{x}}=x$ for any $x,y\in R$. Let $\lambda\in center(R)$ such that $\lambda\bar{\lambda}=1$ and set $\Lambda_{min}:=\{x-\lambda\bar{x}\mid x\in R\}$ and $\Lambda_{max}:=\{x\in R\mid x=-\lambda\bar{x}\}$. An additive subgroup $\Lambda$ of $R$ such that

1. (i)

$\Lambda_{min}\subseteq\Lambda\subseteq\Lambda_{max}$ and 2. (ii)

$x\Lambda\bar{x}\subseteq\Lambda\leavevmode\nobreak\ \forall x\in R$

is called a form parameter for $R$. If $\Lambda$ is a form parameter for $R$, the pair $(R,\Lambda)$ is called a form ring.

Definition 30**.**

Let $(R,\Lambda)$ be a form ring and $I$ an ideal such that $\bar{I}=I$. Set $\Gamma_{max}=I\cap\Lambda$ and $\Gamma_{min}=\{x-\lambda\bar{x}\mid x\in I\}+\langle\{xy\bar{x}\mid x\in I,y\in\Lambda\}\rangle$. An additive subgroup $\Gamma$ of $I$ such that

1. (i)

$\Gamma_{min}\subseteq\Gamma\subseteq\Gamma_{max}$ and 2. (ii)

$x\Gamma\bar{x}\subseteq\Gamma\leavevmode\nobreak\ \forall x\in R$

is called a relative form parameter of level $I$. If $\Gamma$ is a relative form parameter of level $I$, then $(I,\Gamma)$ is called a form ideal of $(R,\Lambda)$.

Until the end of section 8 let $n\in\mathbb{N}$, $(R,\Lambda)$ a form ring and $(I,\Gamma)$ a form ideal of $(R,\Lambda)$.

8.2. The hyperbolic unitary group

Definition 31**.**

Set $V:=R^{2n}$. We use the following indexing for the elements of the standard basis of $V$: $(e_{1},\dots,e_{n},e_{-n},\dots,e_{-1})$. That means that $e_{i}$ is the column whose $i$-th coordinate is one and all the other coordinates are zero if $1\leq i\leq n$ and the column whose $(2n+1+i)$-th coordinate is one and all the other coordinates are zero if $-n\leq i\leq-1$. Let $p\in M_{n}(R)$ be the matrix with ones on the skew diagonal and zeros elsewhere. We define the maps

[TABLE]

where $\bar{v}$ is obtained from $v$ by applying \leavevmode\nobreak\ \bar{}\leavevmode\nobreak\ to each entry of $v$. For any $v\in V$, $f(v,v)$ is called the value of $v$ and is denoted by $|v|$. The subgroup $U_{2n}(R,\Lambda):=\{\sigma\in GL_{2n}(R)\mid(h(\sigma u,\sigma v)=h(u,v)\wedge q(\sigma v)=q(v))\leavevmode\nobreak\ \forall u,v\in V\}$ of $GL_{2n}(R)$ is called hyperbolic unitary group.

Example 32**.**

If $R$ is commutative, $\leavevmode\nobreak\ \bar{}=id$, $\lambda=-1$ and $\Lambda=\Lambda_{max}=R$, then $U_{2n}(R,\Lambda)$ equals the symplectic group $Sp_{2n}(R)$. If $R$ is commutative, $\leavevmode\nobreak\ \bar{}\leavevmode\nobreak\ =id$, $\lambda=1$ and $\Lambda=\Lambda_{min}=\{0\}$, then $U_{2n}(R,\Lambda)$ equals the orthogonal group $O_{2n}(R)$.

Definition 33**.**

We define $\Omega_{+}:=\{1,...,n\}$, $\Omega_{-}:=\{-n,...,-1\}$, $\Omega:=\Omega_{+}\cup\Omega_{-}$ and

[TABLE]

Further if $i,j\in\Omega$, we write $i<j$ iff either $i,j\in\Omega_{+}\land i<j$ or $i,j\in\Omega_{-}\land i<j$ or $i\in\Omega_{+}\land j\in\Omega_{-}$.

Lemma 34**.**

Let $\sigma\in GL_{2n}(R)$. Then $\sigma\in U_{2n}(R,\Lambda)$ if and only if

1. (i)

$\sigma^{\prime}_{ij}=\lambda^{(\epsilon(j)-\epsilon(i))/2}\bar{\sigma}_{-j,-i}\leavevmode\nobreak\ \forall i,j\in\Omega$* and* 2. (ii)

$|\sigma_{*j}|\in\Lambda\leavevmode\nobreak\ \forall j\in\Omega$.

Proof.

See [2], p.167. ∎

Lemma 35**.**

Let $\sigma\in U_{2n}(R,\Lambda)$, $x\in R^{*}$ and $k\in\Omega$. Then the statements below are true.

1. (i)

*If the $k$-th column of $\sigma$ equals $e_{k}x$ then the $(-k)$-th row of $\sigma$ equals $\overline{x^{-1}}e^{t}_{-k}$. * 2. (ii)

If the $k$-th row of $\sigma$ equals $xe^{t}_{k}$ then the $(-k)$-th column of $\sigma$ equals $e_{-k}\overline{x^{-1}}$.

Proof.

Follows from (i) in the previous lemma. ∎

8.3. Polarity map

Definition 36**.**

The map

[TABLE]

is called polarity map. One checks easily that $h(u,v)=\tilde{u}v$ for any $u,v\in V$ and that \leavevmode\nobreak\ \widetilde{}\leavevmode\nobreak\ is involutary linear, i.e. $\widetilde{u+v}=\tilde{u}+\tilde{v}$ and $\widetilde{vx}=\bar{x}\tilde{v}$ for any $u,v\in V$ and $x\in R$.

Lemma 37**.**

If $\sigma\in U_{2n}(R,\Lambda)$ and $v\in V$, then $\widetilde{\sigma v}=\tilde{v}\sigma^{-1}$.

Proof.

See [2, Lemma 2.5]. ∎

8.4. The elementary subgroup

Definition 38**.**

If $i,j\in\Omega$ such that $i\neq\pm j$ and $x\in R$, then the matrix

[TABLE]

is called an elementary short root transvection. If $i\in\Omega$ and $y\in\lambda^{-(\epsilon(i)+1)/2}\Lambda$, then the matrix

[TABLE]

is called an elementary long root transvection. If $\sigma\in U_{2n}(R,\Lambda)$ is an elementary short root transvection or an elementary long root transvection, it is called an elementary unitary transvection. The subgroup of $U_{2n}(R,\Lambda)$ generated by all elementary unitary transvections is called elementary unitary group and is denoted by $EU_{2n}(R,\Lambda)$. An elementary unitary transvection $T_{ij}(x)$ is called $(I,\Gamma)$-elementary if $i\neq-j\leavevmode\nobreak\ \wedge\leavevmode\nobreak\ x\in I$ or $i=-j\wedge x\in\lambda^{-(\epsilon(i)+1)/2}\Gamma$. The subgroup of $U_{2n}(R,\Lambda)$ generated by all $(I,\Gamma)$-elementary transvections is called preelementary subgroup of level $(I,\Gamma)$ and is denoted by $EU_{2n}(I,\Gamma)$. Its normal closure in $EU_{2n}(R,\Lambda)$ is called elementary subgroup of level $(I,\Gamma)$ and is denoted by $EU_{2n}((R,\Lambda),(I,\Gamma))$.

Lemma 39**.**

The relations

[TABLE]

hold where $h\neq j,-i$ and $k\neq i,-j$ in (R3), $i,k\neq\pm j$ and $i\neq\pm k$ in (R4) and $i\neq\pm j$ in (R5) and (R6).

Proof.

Straightforward calculation. ∎

Definition 40**.**

Let $v\in V$ be isotropic (i.e. $q(v)=0$) such that $v_{-1}=0$. Then we denote the matrix

[TABLE]

by $T_{*,-1}(v)$. Clearly $T_{*,-1}(v)^{-1}=T_{*,-1}(-v)$ (note that $\tilde{v}v=0$ since $v$ is isotropic) and

[TABLE]

for any $\sigma\in U_{2n}(R,\Lambda)$.

Definition 41**.**

Let $i,j\in\Omega$ such that $i\neq\pm j$. Define $P_{ij}:=e+e^{ij}-e^{ji}+\lambda^{(\epsilon(i)-\epsilon(j))/2}e^{-i,-j}-\lambda^{(\epsilon(j)-\epsilon(i))/2}e^{-j,-i}-e^{ii}-e^{jj}-e^{-i,-i}-e^{-j,-j}=T_{ij}(1)T_{ji}(-1)T_{ij}(1)\in EU_{2n}(R,\Lambda)$. It is easy show that $(P_{ij})^{-1}=P_{ji}$.

Lemma 42**.**

Let $x\in R$ and $i,j,k\in\Omega$ such that $i\neq\pm j$ and $k\neq\pm i,\pm j$. Further let $y\in\lambda^{-(\epsilon(i)+1)/2}\Lambda$. Then

1. (i)

${}^{P_{ki}}T_{ij}(x)=T_{kj}(x)$, 2. (ii)

${}^{P_{kj}}T_{ij}(x)=T_{ik}(x)$* and* 3. (iii)

${}^{P_{-k,-i}}T_{i,-i}(y)=T_{k,-k}(\lambda^{(\epsilon(i)-\epsilon(k))/2}y)$.

Proof.

Follows from the relations in Lemma 39. ∎

Lemma 43**.**

Let $\sigma\in U_{2n}(R,\Lambda)$ and $i,j\in\Omega$ such that $i\neq\pm j$. Set $\hat{\sigma}:={}^{P_{ij}}\sigma$. Then

[TABLE]

Proof.

Straightforward computation. ∎

8.5. Congruence subgroups

Definition 44**.**

The group consisting of all $\sigma\in U_{2n}(R,\Lambda)$ such that $\sigma\equiv e\leavevmode\nobreak\ mod\leavevmode\nobreak\ I$ and $f(\sigma v,\sigma v)\equiv f(v,v)\leavevmode\nobreak\ mod\leavevmode\nobreak\ \Gamma\leavevmode\nobreak\ \forall v\in V$ is called principal congruence subgroup of level $(I,\Gamma)$ and is denoted by $U_{2n}((R,\Lambda),(I,\Gamma))$. By a theorem of Bak [1], 4.1.4, cf. [2], 4.4, it is a normal subgroup of $U_{2n}(R,\Lambda)$.

Lemma 45**.**

Let $\sigma\in U_{2n}(R,\Lambda)$. Then $\sigma\in U_{2n}((R,\Lambda),(I,\Gamma))$ if and only if

1. (i)

$\sigma\equiv e\leavevmode\nobreak\ mod\leavevmode\nobreak\ I$* and* 2. (ii)

$|\sigma_{*j}|\in\Gamma\leavevmode\nobreak\ \forall j\in\Omega$.

Proof.

[2], p.174. ∎

Definition 46**.**

The subgroup

[TABLE]

of $U_{2n}(R,\Lambda)$ is called full congruence subgroup of level $(I,\Gamma)$ and is denoted by $CU_{2n}((R,\Lambda),(I,\Gamma))$. Obviously $U_{2n}((R,\Lambda),(I,\Gamma))\subseteq CU_{2n}((R,\Lambda),(I,\Gamma))$. If $EU_{2n}(R,\Lambda)$ is a normal subgroup of $U_{2n}(R,\Lambda)$ (which for example is true if $n\geq 3$ and $R$ is almost commutative, see [2, Theorem 1.1]), then $CU_{2n}((R,\Lambda),(I,\Gamma))$ is a normal subgroup of $U_{2n}(R,\Lambda)$.

Theorem 47**.**

If $n\geq 3$ and $R$ is almost commutative (i.e. module finite over its center), then the equalities

[TABLE]

hold.

Proof.

By [2, Theorem 1.1]), $EU_{2n}((R,\Lambda),(I,\Gamma))$ is normal in $U_{2n}(R,\Lambda)$ and

[TABLE]

(note that in [2] the full congruence subgroup is defined a little differently). By [2, Lemma 5.2],

[TABLE]

Hence

[TABLE]

by the definition of $CU_{2n}((R,\Lambda),(I,\Gamma))$, (4) and the three subgroups lemma. (5) and (6) imply the assertion of the theorem. ∎

9. Sandwich classification for $U_{2n}(R,\Lambda)$

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

Definition 48**.**

Let $\sigma\in U_{2n}(R,\Lambda)$. Then a matrix of the form ${}^{\epsilon}\sigma^{\pm 1}$ where $\epsilon\in EU_{2n}(R,\Lambda)$ is called an elementary (unitary) $\sigma$-conjugate.

Theorem 49**.**

Let $\sigma\in U_{2n}(R,\Lambda)$, $k\neq\pm l$ and $i\neq\pm j$. Then

1. (i)

$T_{kl}(\sigma_{ij})$* is a product of $160$ elementary unitary $\sigma$-conjugates,* 2. (ii)

$T_{kl}(\sigma_{i,-i})$* is a product of $320$ elementary unitary $\sigma$-conjugates,* 3. (iii)

$T_{kl}(\sigma_{ii}-\sigma_{jj})$* is a product of $480$ elementary unitary $\sigma$-conjugates,* 4. (iv)

$T_{kl}(\sigma_{ii}-\sigma_{-i,-i})$* is a product of $960$ elementary unitary $\sigma$-conjugates and* 5. (v)

$T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}|\sigma_{*j}|)$* is a product of $1600n+4004$ elementary unitary $\sigma$-conjugates.*

Proof.

(i) In step 1 below we show that $T_{kl}(x\bar{\sigma}_{23}\sigma_{2,-1})$ where $x\in R$ is a product of $16$ elementary $\sigma$-conjugates. In step 2 we show that $T_{kl}(x\bar{\sigma}_{23}\sigma_{21})$ where $x\in R$ is a product of $16$ elementary $\sigma$-conjugates. In step 3 we show that $T_{kl}(x\bar{\sigma}_{23}\sigma_{22})$ is a product of $32$ elementary $\sigma$-conjugates. In step 4 we use steps 1-3 in order to prove (i).

step 1 Set $\tau:=T_{21}(\bar{\sigma}_{23}\sigma_{23})T_{31}(-\bar{\sigma}_{23}\sigma_{22})T_{3,-2}(\bar{\sigma}_{23}\sigma_{2,-1})T_{3,-3}(-\bar{\sigma}_{22}\sigma_{2,-1}+\bar{\lambda}\bar{\sigma}_{2,-1}\sigma_{22})$. One checks easily that the second row of $\sigma\tau^{-1}$ equals the second row of $\sigma$ and hence the second row of $\xi:={}^{\sigma}\tau^{-1}$ is trivial. By Lemma 35 the second last column of $\xi$ also is trivial. Set

[TABLE]

One checks easily that $[\tau^{-1},T_{-1,2}(1)]=T_{31}(\lambda\bar{\sigma}_{23}\sigma_{2,-1})T_{-1,1}(z)$ for some $z\in\Lambda$ and $[T_{-1,2}(1),\xi]=\prod\limits_{i\neq 2}T_{i2}(x_{i})$ for some $x_{i}\in R\leavevmode\nobreak\ (i\neq 2)$. Hence $\zeta=T_{31}(\lambda\bar{\sigma}_{23}\sigma_{2,-1})T_{-1,1}(z)\prod\limits_{i\neq 2}T_{i2}(x_{i})$. It follows that $[T_{-1,3}(-x\bar{\lambda}),[T_{12}(1),\zeta]]=T_{-1,2}(x\bar{\sigma}_{23}\sigma_{2,-1})$ for any $x\in R$. Hence we have shown

[TABLE]

This implies that $T_{-1,2}(x\bar{\sigma}_{23}\sigma_{2,-1})$ is a product of $16$ elementary $\sigma$-conjugates. It follows from Lemma 42 that $T_{kl}(x\bar{\sigma}_{23}\sigma_{2,-1})$ is a product of $16$ elementary $\sigma$-conjugates.

step 2 Set $\tau:=T_{1,-2}(\bar{\sigma}_{23}\sigma_{23})T_{3,-2}(-\bar{\sigma}_{23}\sigma_{21})T_{3,-1}(\bar{\lambda}\bar{\sigma}_{23}\sigma_{22})T_{3,-3}(\bar{\sigma}_{22}\sigma_{21}-\bar{\lambda}\bar{\sigma}_{21}\sigma_{22})$. One checks easily that the second row of $\sigma\tau^{-1}$ equals the second row of $\sigma$ and hence the second row of $\xi:={}^{\sigma}\tau^{-1}$ is trivial. By Lemma 35 the second last column of $\xi$ also is trivial. Set

[TABLE]

One checks easily that $[\tau^{-1},T_{-2,-1}(1)]=T_{3,-1}(\bar{\sigma}_{23}\sigma_{21})T_{1,-1}(z)$ for some $z\in\bar{\Lambda}$ and $[T_{-2,-1}(1),\xi]=\prod\limits_{i\neq 2}T_{i2}(x_{i})$ for some $x_{i}\in R\leavevmode\nobreak\ (i\neq 2)$. Hence $\zeta=T_{3,-1}(\bar{\sigma}_{23}\sigma_{21})T_{1,-1}(z)\prod\limits_{i\neq 2}T_{i2}(x_{i})$. It follows that $[T_{-1,3}(-x),$ $[T_{-2,3}(1),\zeta]]=T_{-2,3}(x\bar{\sigma}_{23}\sigma_{21})$ for any $x\in R$. Hence we have shown

[TABLE]

This implies that $T_{-2,3}(x\bar{\sigma}_{23}\sigma_{21})$ is a product of $16$ elementary $\sigma$-conjugates. It follows from Lemma 42 that $T_{kl}(x\bar{\sigma}_{23}\sigma_{21})$ is a product of $16$ elementary $\sigma$-conjugates.

step 3 Set $\tau:=T_{21}(-\bar{\sigma}_{22}\sigma_{23})T_{31}(\bar{\sigma}_{22}\sigma_{22})T_{2,-3}(\bar{\sigma}_{22}\sigma_{2,-1})T_{2,-2}(-\bar{\sigma}_{23}\sigma_{2,-1}+\bar{\lambda}\bar{\sigma}_{2,-1}\sigma_{23})$. One checks easily that the second row of $\sigma\tau^{-1}$ equals the second row of $\sigma$ and hence the second row of $\xi:={}^{\sigma}\tau^{-1}$ is trivial. By Lemma 35 the second last column of $\xi$ also is trivial. Set

[TABLE]

One checks easily that $\psi:=[\tau^{-1},T_{32}(1)]=T_{31}(-\bar{\sigma}_{22}\sigma_{23})T_{3,-3}(y)T_{3,-2}(z)$ for some $y\in\bar{\Lambda}$ and $z\in R$ and $\theta:=[T_{32}(1),\xi]=\prod\limits_{i\neq 2}T_{i2}(x_{i})$ for some $x_{i}\in R\leavevmode\nobreak\ (i\neq 2)$. Set

[TABLE]

One checks easily that $[\psi^{-1},T_{12}(1)]=T_{32}(\bar{\sigma}_{22}\sigma_{23})T_{3,-3}(a)T_{3,-1}(b)$ for some $a\in\bar{\Lambda}$ and $b\in R$ and $[T_{12}(1),\theta]=T_{-2,2}(d)$ for some $d\in\Lambda$. Hence $\chi=T_{32}(\bar{\sigma}_{22}\sigma_{23})T_{3,-3}(a)T_{3,-1}(b)T_{-2,2}(d)$. It follows that $[T_{-2,3}(\bar{x}),[T_{2,-1}(1),\chi]]=T_{-2,-1}(-\bar{x}\bar{\sigma}_{22}\sigma_{23})\overset{(R1)}{=}T_{12}(x\bar{\sigma}_{23}\sigma_{22})$ for any $x\in R$. Hence we have shown

[TABLE]

This implies that $T_{12}(x\bar{\sigma}_{23}\sigma_{22})$ is a product of $32$ elementary $\sigma$-conjugates. It follows from Lemma 42 that $T_{kl}(x\bar{\sigma}_{23}\sigma_{22})$ is a product of $32$ elementary $\sigma$-conjugates.

step 4 Set $I:=I(\{\overline{\bar{\sigma}_{23}\sigma_{2,-1}},\bar{\sigma}_{23}\sigma_{21}\})$, $J:=I(\{\overline{\bar{\sigma}_{23}\sigma_{2,-1}},\bar{\sigma}_{23}\sigma_{21},\bar{\sigma}_{23}\sigma_{22}\})$ and

[TABLE]

One checks easily that $\tau_{11}\equiv 1\leavevmode\nobreak\ mod\leavevmode\nobreak\ I$ and $\tau_{12}\equiv\bar{\sigma}_{23}\leavevmode\nobreak\ mod\leavevmode\nobreak\ J$. Set $\zeta:={}^{P_{13}P_{21}}\tau$. Then $\zeta_{22}=\tau_{11}$ and $\zeta_{23}=\tau_{12}$ and hence $\bar{\zeta}_{23}\zeta_{22}\equiv\sigma_{23}\leavevmode\nobreak\ mod\leavevmode\nobreak\ I+\bar{J}$. Applying step 3 above to $\zeta$, we get that $T_{kl}(\bar{\zeta}_{23}\zeta_{22})$ is a product of $32$ elementary $\zeta$-conjugates. Since any elementary $\zeta$-conjugate is a product of $2$ elementary $\sigma$-conjugates, it follows that $T_{kl}(\bar{\zeta}_{23}\zeta_{22})$ is a product of $64$ elementary $\sigma$-conjugates. Thus, by steps 1-3, $T_{kl}(\sigma_{23})$ is a product of $64+16+16+16+16+32=160$ elementary $\sigma$-conjugates. Since one can bring $\sigma_{ij}$ to position $(3,2)$ by conjugating monomial matrices in $EU_{2n}(R,\Lambda)$, the assertion of (i) follows.

(ii)-(iv) See the proof of Theorem 27.

(v) Set m:=160. In step 1 we show that $T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}\bar{x}\bar{\sigma}_{11}|\sigma_{*1}|\sigma_{11}x)$ where $x\in R$ is a product of $(2n+17)m+4$ elementary $\sigma$-conjugates. In step 2 we use step 1 in order to prove (v).

step 1 Set $v^{\prime}:=\begin{pmatrix}0&\dots&0&\sigma^{\prime}_{-1,-1}&-\sigma^{\prime}_{-1,-2}\end{pmatrix}^{t}=\begin{pmatrix}0&\dots&0&\bar{\sigma}_{11}&-\bar{\sigma}_{21}\end{pmatrix}^{t}\in V$ and $v:=\sigma^{-1}v^{\prime}\in V$. Then clearly $v_{-1}=0$. Further $q(v)=q(\sigma^{-1}v^{\prime})=q(v^{\prime})=0$ and hence $v$ is isotropic. Set

[TABLE]

Then

[TABLE]

where $\alpha=\sigma_{-2,1}\sigma_{11}-\lambda\bar{\sigma}_{11}\bar{\sigma}_{-2,1}$ and $\beta=\sigma_{-1,1}\sigma_{11}+\lambda\bar{\sigma}_{21}\bar{\sigma}_{-2,1}$. Set

[TABLE]

It follows from (i) that $\tau$ is a product of $2m$ elementary $\sigma$-conjugates. Clearly

[TABLE]

where $\gamma=\alpha+\bar{\sigma}_{11}\bar{\sigma}_{31}\sigma_{-3,1}\sigma_{11}$ and $\delta=\beta-\bar{\sigma}_{21}\bar{\sigma}_{31}\sigma_{-3,1}\sigma_{11}$. Let $x\in R$ and set

[TABLE]

Clearly $\zeta$ is a product of $4m+4$ elementary $\sigma$-conjugates. One checks easily that

[TABLE]

for some $a,b\in I(\sigma_{21}),c\in I(\sigma_{23})$. Further

[TABLE]

where $y=\bar{\lambda}(\bar{x}\bar{\sigma}_{11}|\sigma_{*1}|\sigma_{11}x+d-\lambda\bar{d}+e-\lambda\bar{e})$ for some $d\in I(\sigma_{31}),e\in I(\sigma_{-2,1})$. Hence

[TABLE]

It follows from (i), (ii) and (iii) and relation (R5) in Lemma 39 that $T_{3,-3}(y)$ is a product of $4m+4+2m+2m+4m+(2n-5)m+3m+3m=(2n+13)m+4$ elementary $\sigma$-conjugates. By (i) and relation (R5) in Lemma 39, $T_{3,-3}(-\bar{\lambda}(d-\lambda\overline{d}))$ and $T_{3,-3}(-\bar{\lambda}(e-\lambda\overline{e})))$ each are a product of $2m$ elementary $\sigma$-conjugates. Hence $T_{3,-3}(\bar{\lambda}(\bar{x}\bar{\sigma}_{11}|\sigma_{*1}|\sigma_{11}x))=T_{3,-3}(y)T_{3,-3}(-\bar{\lambda}(d-\lambda\overline{d}))T_{3,-3}(-\bar{\lambda}(e-\lambda\overline{e})))$ is a product of $(2n+17)m+4$ elementary $\sigma$-conjugates. It follows from Lemma 42 that $T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}\bar{x}\bar{\sigma}_{11}|\sigma_{*1}|\sigma_{11}x)$ is a product of $(2n+17)m+4$ elementary $\sigma$-conjugates.

step 2 Clearly

[TABLE]

since $|\sigma_{*1}|\in\Lambda\subseteq\Lambda_{max}$. By step 1, $T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}\bar{\sigma}^{\prime}_{11}\bar{\sigma}_{11}|\sigma_{*1}|\sigma_{11}\sigma^{\prime}_{11})$ is a product of $(2n+17)m+4$ elementary $\sigma$-conjugates. By (i), (ii) and relation (R6) in Lemma 39, $T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}\bar{\sigma}^{\prime}_{1q}\bar{\sigma}_{q1}|\sigma_{*1}|\sigma_{q1}\sigma^{\prime}_{1q})$ is a product of $3m$ elementary $\sigma$-conjugates if $q\neq\pm 1$ resp. a product of $6m$ elementary $\sigma$-conjugates if $q=-1$. Hence $A$ is a product of $(2n+17)m+4+(2n-2)\cdot 3m+6m=(8n+17)m+4$ elementary $\sigma$-conjugates. On the other hand $B=T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}(x-\lambda\bar{x}))$ where $x\in I(|\sigma_{*1}|)$. Since $|\sigma_{*1}|=\sum\limits_{i\in\Omega_{+}}\bar{\sigma}_{i1}\sigma_{-i,1}$, it follows from (i), (ii) and relation (R5) in Lemma 39 that $B$ is a product of $4m+(n-1)\cdot 2m=(2n+2)m$ elementary $\sigma$-conjugates. Hence $T_{k,-k}(\lambda^{-(\epsilon(k)+1)/2}|\sigma_{*1}|)$ is a product of $(10n+19)m+4=1600n+3044$ elementary $\sigma$-conjugates. The assertion of (v) follows now from Lemma 43. ∎

As a corollary we get the Sandwich Classification Theorem for $U_{2n}(R,\Lambda)$.

Corollary 50**.**

Let $H$ be a subgroup of $U_{2n}(R,\Lambda)$. Then $H$ is normalized by $EU_{2n}(R,\Lambda)$ if and only if

[TABLE]

for some form ideal $(I,\Gamma)$ of $(R,\Lambda)$.

Proof.

First suppose that $H$ is is normalized by $EU_{2n}(R,\Lambda)$. Let $(I,\Gamma)$ be the form ideal of $(R,\Lambda)$ defined by $I:=\{x\in R\mid T_{12}(x)\in H\}$ and $\Gamma:=\{y\in\Lambda\mid T_{-1,1}(y)\in H\}$. Then clearly $EU_{2n}((R,\Lambda),(I,\Gamma))\subseteq H$. It remains to show that $H\subseteq CU_{2n}((R,\Lambda),(I,\Gamma))$, i.e. that if $\sigma\in H$ and $\epsilon\in EU_{2n}(R,\Lambda)$, then $[\sigma,\epsilon]\in U_{2n}((R,\Lambda),(I,\Gamma))$. By Lemma 45 it suffices to show that if $\sigma\in H$ and $\epsilon\in EU_{2n}(R,\Lambda)$, then $[\sigma,\epsilon]\equiv e\leavevmode\nobreak\ mod\leavevmode\nobreak\ I$ and $|[\sigma,\epsilon]_{*j}|\in\Gamma$ for any $j\in\Omega$. But that follows from the previous theorem (applying the theorem to $\sigma$ we get that $\sigma\equiv diag(x,\dots,x)\leavevmode\nobreak\ mod\leavevmode\nobreak\ I$ for some $x\in R$ and hence $[\sigma,\epsilon]\equiv e\leavevmode\nobreak\ mod\leavevmode\nobreak\ I$; applying it to $[\sigma,\epsilon]$ we get that $|[\sigma,\epsilon]_{*j}|\in\Gamma$ for any $j\in\Omega$). Suppose now that (7) holds for some form ideal $(I,\Gamma)$. Then it follows from the standard commutator formulas in Theorem 47 that $H$ is normalized by $EU_{2n}(R,\Lambda)$. ∎