Hermitian Groups and K-Theory ^{∗}

Guoping Tang

Dept. of Math., University of Science and Technology of China Hefei, Anhui, 230026, P.R.China

Abstract This article provides fundemental constructions and results for Hermitian groups, which are necessary for an algebraic understanding of the functors K_{1} and K_{2} of Hermitian forms. These include a definition of the elementary subgroup EH of a general Hermitian group GH, the Hermitian Whitehead lemma for stabilized GH, a definition of the Hermitian Steinberg group, and the theorem that a stabilized Steinberg group is the universal central extension of its associated stabilized elementary group.

1. Introduction

The subject K-theory of forms was born in the late 60-ies and since then has undergone a broad and intensive development. An important landmark was the monograph K-theory of Forms of A. Bak. In recent years, several textbooks on the subject have appeared, beginning with the book The Classical Groups and K-Theory of A.J.Hahn and O.T.O’meara. However, most materials appearing in the 35 year span above are concerned with quadratic forms and not with Hermitian forms. Publications on Hermitian forms have been restricted to the functor K_{0}, with one notable exception in K-Theory of Forms where certain exact sequences involving other K-functors are constructed. One essential obstruction for the further development of the K- theory of Hermitian forms is insufficient knowledge of the structure of the general Hermitian groups GH. These groups are the analogues in the theory of Hermitian forms of the general linear groups GL in the theory of finitely generated projective modules. In fact, for general Hermitian groups not even the most basic construction, namely that of an elementary subgroup, is known.

∗1991 Mathematics Subject Classification. 19B99, 19C99, 19G38 Key words and Phrases. Metabolic forms, General Hermitian group, Elementary subgroup, Steinberg group The author gratefully acknowledges the support of the Deutsche Forschungsgemeischaft (DFG) and the hospitality of the University of Bielefeld during the preparation of the present article

The purpose of this paper is to fill the knowledge gap above. Accordingly, we define the elementary subgroup of a general Hermitian group and prove that a stabilized elementary subgroup is the commutator subgroup of its stabilized general Hermitian group. This permits one to interpret K_{1} as a quotient of a stabilized general Hermitian group modulo a stabilized elementary Hermitian group. For each elementary Hermitian group, we define a Steinberg group covering it and show that a stabilized Steinberg group is the universal central extension of the corresponding stabilized elementary group. This allows one to interpret K_{2} as the kernel of the canonical homomorphism from a stabilized Steinberg group to its corresponding stabilized elementary group. We establish the unexpected result that K_{2} can be strictly smaller than the center of the stabilized Steinberg group.

The rest of the paper is organized into two chapters. Each chapter has its own introduction. In chapter 1, we define the elementary subgroup of a general Hermitian group and prove that a stabilized elementary group is the commutator subgroup of the corresponding stabilized general Hermitian group. This result is deduced from the so-called Hermitian Whitehead Lemma. Some computations for Hermitian K_{1} are made at the end of the chapter. In chapter 2, we define the Hermitian Steinberg group and prove that a stabilized Steinberg group is a universal central extension of its stabilized elementary group.

I. Elementary Hermitian groups and the functor K_{1} of Hermitian forms

2. Introduction to I

Let R be an associative ring with involution a ↦ ā. Let λ ∈ center(R) such that λλ̄ = 1 and let Λ be a −λ-form parameter on R. For a_{1}, ⋯, a_{r} ∈ Λ and n ≥ r, let GH_{2n}(R, a_{1}, ⋯, a_{r}) denote the group of all automorphisms of the metabolic form defined by the 2n × 2n matrix

( \begin{matrix}A & λI \\ I & 0 \end{matrix} )

where A is the n × n diagonal matrix with diagonal (a_{1}, ⋯, a_{r}, 0, ⋯, 0), I is the n × n identity matrix, and 0 is the n × n zero matrix. GH_{2n}(R, a_{1}, ⋯, a_{r}) is called the n-th general Hermitian group of the elements a_{1}, …, a_{r}. For fixed a_{1}, ⋯, a_{r}, there is an obvious embedding GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟶ GH_{2(n+1)}(R, a_{1}, ⋯, a_{r}). A.Bak[Bak] defined

GH(R, a_{1}, ⋯, a_{r}) = lim_{n^{⟶}} GH_{2n}(R, a_{1}, ⋯, a_{r}),

showed that the commutator subgroup of this group is perfect, and posed the problem [Bak,p42] of finding a clear description of this subgroup in terms of functorially defined generators. In this chapter, we shall define in a functorial way generators for a subgroup EH_{2n}(R, a_{1}, ⋯, a_{r}) of the group GH_{2n}(R, a_{1}, ⋯, a_{r}), which will be called the elementary subgroup of GH_{2n}(R, a_{1}, ⋯, a_{r}). From the definition, it will be obvious that the stabilization map GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟶ GH_{2(n+1)}(R, a_{1}, ⋯, a_{r}) takes generators of EH_{2n}(R, a_{1}, ⋯, a_{r}) to generators of EH_{2(n+1)}(R, a_{1}, ⋯, a_{r}) and we define

EH(R, a_{1}, ⋯, a_{r}) = lim_{n^{⟶}} EH_{2n}(R, a_{1}, ⋯, a_{r}).

For n ≥ r + 3, a straightforward but somewhat tedious calculation shows that EH_{2n}(R, a_{1}, ⋯, a_{r}) is a perfect group. Our main result is the Hermitian Whitehead Lemma for GH(R, a_{1}, ⋯, a_{r}) , from which it follows that EH(R, a_{1}, ⋯, a_{r}) is the commutator subgroup of GH(R, a_{1}, ⋯, a_{r}) .

The organization of the rest of the chapter is as follows. In §3, we set up notation and develop some elementary facts concerning the group GH_{2n}(R, a_{1}, ⋯, a_{r}). In §4, we define the elementary Hermitian group EH_{2n}(R, a_{1}, ⋯, a_{r}) and show that it is perfect whenever n ≥ r + 3. In §5, we prove the Hermitian Whitehead Lemma and its corollary that EH(R, a_{1}, ⋯, a_{r}) is the commutator subgroup of GH(R, a_{1}, ⋯, a_{r}). In §6, we construct the K_{1}-group KH_{1}(R, Λ) of Λ-Hermitian forms and compare it with K_{1} of quadratic forms by the canonical natural transformation KQ_{1} ⟶ KH_{1}. Finally we show that our matrix description of KH_{1}(R, Λ) agrees with Bass’ categorical construction of K_{1} of a category with product, applied to the category of nonsingular Λ-Hermitian spaces under orthogonal sum.

3. General Hermitian Groups

Let R be a ring with involution, a ⟼ \overline{a}, and let λ ∈ center(R) such that λ\overline{λ} = 1. Set max^{λ}(R) = \{a ∈ R|a = λ\overline{a}\} and min^{λ}(R) = \{a + λ\overline{a}|a ∈ R\}. Recall that a λ-form parameter Λ is an additive subgroup of R such that min^{−λ}(R) ⊆ Λ ⊆ max^{−λ}(R) and aΛ\overline{a} ⊆ Λ for all a ∈ R. Form parameters were developed by A. Bak in order to unify the theories of quadratic and even λ-Hermitian forms and to provide a smooth transition from the classical notion of a quadratic form to that of an even λ-Hermitian form and from that of an even λ-Hermitian form to that of an arbitrary λ-Hermitian form, cf. [Bak, §1]. Since we shall be almost entirely interested in λ-Hermitan forms in this paper, our form parameters will always be (−λ)-form parameters. From the definitions of min^{λ}(R) and max^{λ}(R), it is clear that they are (−λ)-form parameters.

Throughout the article, we shall make the convention that all R-modules are right Rmodules. Let M be an R-module and Λ a −λ-form parameter on R; thus, min^{λ}(R) ⊂ Λ ⊂ max^{λ}(R). A sesquilinear form B on M is a biadditive map B : M × M ⟶ R such that B(ma, nb) = \overline{a}B(m, n)b for all m, n ∈ M and a, b ∈ R. A sesquilinear form B is called λ-Hermitian if B(m, n) = λ\overline{B(n, m)} for all m, n ∈ M. A λ-Hermitian form B is called Λ-Hermitian if B(m, m) ∈ Λ for all m ∈ M. Thus, max^{λ}(R)-Hermitian forms are arbitrary λ-Hermitian forms and min^{λ}(R)-Hermitian are, by definition, even λ-Hermitian forms.

The transition from quadratic forms to Hermitian forms is provided by the result [Bak, (1.1.)] of A. Bak which implies that the category of all max^{−λ}(R)-quadratic forms, cf. [Bak, §1], is the same as that of the category of all min^{λ}(R)-Hermitian forms. In the current paper, we shall be interested in quadratic forms only for the purpose of comparing the K-groups KQ_{i}(R, max^{−λ}(R)) [Bak, §4] of max^{−λ}(R)-quadratic forms with the K-groups KH_{i}(R, Λ) of Λ-Hermitian forms, under the canonical homomorphism KQ_{i}(R, max^{−λ}(R)) ⟶ KH_{i}(R, Λ). But since KQ_{i}(R, max^{−λ}(R)) ≅ KH_{i}(R, min^{λ}(R)), the comparision above is the same as that between KH_{i}(R, min^{λ}(R)) and KH_{i}(R, Λ).

A λ-Hermitian form B on an R-module M is called nonsingular if the induced map M ⟶ M^{⋆}, m ⟼ B(m, −), is an isomorphism, where M^{⋆} := Hom_{R}(M, R) has the usual right action of R defined by (f · r)(m) = \overline{r}(f(m)). A pair (M, B) where M is finitely generated and projective over R and B is nonsingular will be called a λ-Hermitian space. A Hermitian space (M, B) is called metabolic if it contains an R-direct summand U ⊆ M such that U = U^{⊥} := {m ∈ M | B(m, u) = 0 ∀u ∈ U}. By [Bak, (2.8)], (M, B) is metabolic on U if and only if it is isomorphic to a specifically constructed metabolic module Λ − M(U). For the rest of this paper, we shall be interested only in metabolic spaces on free modules U, because by [Bak, (2.10)] these form already a cofinal family of objects in the category of all λ-Hermitian spaces.

Let M_{n}(R) denote the ring of n × n matrices with coefficients in R. Give M_{n}(R) the involution (a_{ij}) ⟼ (\overline{a}_{ji}) called the conjugate transpose involution. If α ∈ M_{n}(R) then \overline{α} will denote its conjugate transpose. A matrix α is called λ-Hermitian if α = λ\overline{α}. If Λ is a (−λ)-form parameter on R and α ∈ M_{n}(R) then α is called Λ-Hermitian if it is λ-Hermitian and its diagonal coefficients lie in Λ.

Let M be a finitely generated, free R-module. After picking an R-basis m_{1}, …, m_{n} of M, there is a one to one correspondence between Λ-Hermitian forms B on M and n × n Λ-Hermitian matrices given by the rule B ⟼ (B(m_{i}, m_{j})). An R-basis m_{1}, ⋯, m_{2n} for a λ-Hermitian space (M, B) is called a metabolic basis if the 2n × 2n matrix

(B(m_{i}, m_{j})) = ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )

where A is the n × n diagonal matrix whose i-th diagonal coefficient is B(m_{i}, m_{i}), I is the n × n identity matrix, and 0 the n × n zero matrix. An R-basis m_{1}, …, m_{2n} for (M, B) is called a hyperbolic basis if it is metabolic and B(m_{i}, m_{i}) = 0 for all 1 ≤ i ≤ n. It follows from [Bak, (2,6)] that a λ-Hermitian space (M, B) is metabolic on a free module U of rank n if and only if M has a metabolic basis m_{1}, …, m_{2n} such that m_{1}, …, m_{n} is an R-basis for U.

Let m_{1}, …, m_{2n} be a metabolic basis for the λ-Hermitian space (M, B). Let

( \begin{matrix}A & λI \\ I & 0 \end{matrix} )

denote the λ-Hermitian matrix determined by B and the metabolic basis m_{1}, ⋯, m_{2n}. Let a_{1}, …, a_{n} denote the diagonal coefficients of A; thus, a_{i} = B(m_{i}, m_{i}). Using the basis m_{1}, …, m_{2n} we can identify the group Aut(M, B) of all automorphisms of the λ-Hermitian space (M, B) with a subgroup of the general linear group GL_{2n}(R) of rank 2n. Following [Bak, §3], we denote this subgroup by

GH_{2n}(R, a_{1}, …, a_{n})

and call it the general Hermitian group of a_{1}, …, a_{n}. Since each sequence of n elements a_{1}, …, a_{n} ∈ max^{λ}(R) arises from a metabolic basis of some λ-Hermitian space, it follows that GH_{2n}(R, a_{1}, …, a_{n}) is defined for any sequence of elements a_{1}, …, a_{n} ∈ max^{λ}(R). The next lemma shows that GH_{2n}(R, a_{1}, …, a_{n}) is uniquely defined by the sequence a_{1}, …, a_{n}.

Lemma 3.1 Let a_{1}, …, a_{n} ∈ max^{λ}(R). Then a 2n × 2n matrix

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ GL_{2n}(R)

belongs to GH_{2n}(R, a_{1}, …, a_{r}) if and only if anyone of the following equations hold:

\overline{( \begin{matrix}α & β \\ γ & δ \end{matrix} )} ( \begin{matrix}A & λI \\ I & 0 \end{matrix} ) ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )

( \begin{matrix}α & β \\ γ & δ \end{matrix} )^{−1} = ( \begin{matrix}δ̄ + β̄A & λβ̄ \\ ᾱĀ − Āβ̄A + λ̄γ̄ − Āδ̄ & ᾱ − Aβ̄ \end{matrix} ) .

Proof. The result is an immediate consequence of the definition of GH_{2n}(R, a_{1}, …, a_{n}). □

The following notation will be very useful. If 1 ≤ r ≤ n and a_{1}, …, a_{r} ∈ max^{λ}(R), set

GH_{2n}(R, a_{1}, …, a_{r}) = GH_{2n}(R, a_{1}, …, a_{r}, a_{r+1}, …, a_{n})

where a_{r+1} = a_{r+2} = ⋯ = a_{n} = 0. GH_{2n}(R, a_{1}, …, a_{r}) is called n-th general Hermitian group of a_{1}, …, a_{r}. A typical element of GH_{2n}(R, a_{1}, …, a_{r}) is denoted by a 2n × 2n matrix

( \begin{matrix}α & β \\ γ & δ \end{matrix} )

where α, β, γ, and δ are n × n block matrices. In order to deal effectively with technical difficulties caused by the elements a_{1}, …, a_{r}, we shall frequently partition the matrix above into the form

( \begin{matrix}α_{11} & α_{12} & β_{11} & β_{12} \\ α_{21} & α_{22} & β_{21} & β_{22} \\ γ_{11} & γ_{12} & δ_{11} & δ_{12} \\ γ_{21} & γ_{22} & δ_{21} & δ_{22} \end{matrix} )

where α_{11}, β_{11}, γ_{11}, δ_{11} are r × r matrices, α_{22}, β_{22}, γ_{22}, δ_{22} are (n − r) × (n − r) matrices, α_{12}, β_{12}, γ_{12}, δ_{12} are r × (n − r) matrices, and α_{21}, β_{21}, γ_{21}, δ_{21} are (n − r) × r matrices. The canonical homomorphism

GH_{2n}(R, a_{1}, …, a_{r}) ⟶ GH_{2(n+1)}(R, a_{1}, …, a_{r})

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ⟼ ( \begin{matrix}α & 0 & β & 0 \\ 0 & 1 & 0 & 0 \\ γ & 0 & δ & 0 \\ 0 & 0 & 0 & 1 \end{matrix} )

is called the stabilization homomorphism.

For convenience, we shall adopt the notation A =diag(a_{1}, ⋯, a_{r}, 0, ⋯, 0) and A_{1} =diag(a_{1}, ⋯, a_{r}) in the sequel.

Lemma 3.2 A 2n × 2n matrix ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ GL_{2n}(R) belongs to GH_{2n}(R, a_{1}, ⋯, a_{r}) if

and only if

ᾱAα + γ̄α + λᾱγ = A  (1)
ᾱAβ + γ̄β + λᾱδ = λI  (2)
β̄Aβ + δ̄β + λβ̄δ = 0.  (3)

Proof. The lemma is an immediate consequence of Lemma 3.1. □

To write down an element in GH_{2n}(R, a_{1}, ⋯, a_{r}) one has to solve equations in Lemma 3.2. To solve equations (1) and (3) in Lemma 3.2 we have to solve an equation of the kind X̄A_{1}X + Y + λȲ = 0. Of course, matrix equations are difficult to solve. The equation X̄A_{1}X + Y + λȲ = 0 above can be solved or not depending on whether or not the diagnonal elements of X̄A_{1}X belong to min^{λ}(R). This motivates making the following definition.

For any a_{1}, ⋯, a_{r} ∈ max^{λ}(R) let

C = { ^{t}(x_{1}, ⋯, x_{r}) ∈ R^{r} | ∑_{i=1}^{r} x̄_{i}a_{i}x_{i} ∈ min^{λ}(R)}.

Clearly, C is a R-module. The module C will play an important role in the subsequent development.

Lemma 3.3 For any r × m matrix X (m ≥ 1 arbitrary) there is a m × m matrix Y such that X̄A_{1}X + Y + λȲ = 0 if and only if the columns of X belong to C.

Proof. Consider the elements on the diagonal of X̄A_{1}X. The (i, i)-coefficient of X̄A_{1}X is x̄_{1}a_{1}x_{1} + ⋯ + x̄_{r}a_{r}x_{r}, where ^{t}(x_{1}, ⋯, x_{r}) is the i-th column of X. The Lemma then follows from the definition of C. □

Lemma 3.4 Suppose

σ = ( \begin{matrix}α_{11} & α_{12} & β_{11} & β_{12} \\ α_{21} & α_{22} & β_{21} & β_{22} \\ γ_{11} & γ_{12} & δ_{11} & δ_{12} \\ γ_{21} & γ_{22} & δ_{21} & δ_{22} \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}).

Then the columns of α_{11} − I, α_{12}, β_{11}, β_{12}, β̄_{11}, β̄_{21}, δ̄_{11} − I and δ̄_{21} belong to C.

Proof. By the equation (1) in Lemma 3.2 we have equations

ᾱ_{11}A_{1}α_{11} + γ̄_{11}α_{11} + γ̄_{21}α_{21} + λᾱ_{11}γ_{11} + λᾱ_{21}γ_{21} = A_{1}
ᾱ_{12}A_{1}α_{12} + γ̄_{12}α_{12} + γ̄_{22}α_{22} + λᾱ_{12}γ_{12} + λᾱ_{22}γ_{22} = 0.

It follows from Lemma 3.3 that the columns of α_{11} − I and α_{12} belong to C since A_{1} = λĀ_{1}. Similarly, by Lemma 3.2(3) the columns of β_{11} and β_{12} belong to C. Since σ^{−1} ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}), combining Lemma 3.1 and the result above it follows that the columns of δ̄_{11} + β̄_{11}Ā_{1} − I, δ̄_{12}, λβ̄_{11} and λβ̄_{21} belong to C . Now the lemma follows from the fact that C is R-module and λ ∈center(R).

Remark The Lemma above shows how the module C affects elements of GH_{2n}(R, a_{1}, ⋯, a_{r}). When a_{1}, ⋯, a_{r} ∈ min^{λ}(R), then C = R^{r}, and there is no restriction on elements of C. When C = 0, elements in GH_{2n}(R, a_{1}, ⋯, a_{r}) have a more rigid form by Lemma 3.4. This case can happen if R is a division ring.

The next Lemma is taken from [Bak, (2.12)] and will be very useful in the subsequent development.

Lemma 3.5 For any metabolic space (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) there is an isomorphism

(R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ⊥ (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ≅ (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ⊥ H(R^{n}).

Proof. The matrix

φ = ( \begin{matrix}I & I & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā & I & 0 \\ 0 & Ā & −I & I \end{matrix} )

defines such an isomorphism since

\overline{( \begin{matrix}I & I & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā & I & 0 \\ 0 & Ā & −I & I \end{matrix} )} ( \begin{matrix}A & & λI \\ & 0 & & λI \\ I & & 0 \\ & I & & 0 \end{matrix} ) ( \begin{matrix}I & I & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā & I & 0 \\ 0 & Ā & −I & I \end{matrix} ) = ( \begin{matrix}A & & λI \\ & A & & λI \\ I & & 0 \\ & I & & 0 \end{matrix} ) .

□

4. Elementary Hermitian Groups

The aim of this section is to describe elements in GH_{2n}(R, a_{1}, ⋯, a_{r}), which we shall use to construct our elementary Hermitian group. In what follows, we frequently write

an element ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) as a matrix

( \begin{matrix}α_{11} & α_{12} & β_{11} & β_{12} \\ α_{21} & α_{22} & β_{21} & β_{22} \\ γ_{11} & γ_{12} & δ_{11} & δ_{12} \\ γ_{21} & γ_{22} & δ_{21} & δ_{22} \end{matrix} )

as described after Lemma 3.1. By Lemma 3.2, if α and δ are invertible n × n matrices and γ is an n × n matrix then

( \begin{matrix}α & 0 \\ γ & δ \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟺ δ = ᾱ^{−1} and ᾱAα + γ̄α + λᾱγ = A.

In particular,

( \begin{matrix}I & 0 \\ γ & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟺ λγ + γ̄ = 0

i.e., γ is a −λ̄-hermitian matrix. Furthermore,

( \begin{matrix}I \\ α_{21} & I \\ & & I & δ_{12} \\ & & & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟺ δ_{12} = −ᾱ_{21},

and

( \begin{matrix}I \\ & α_{22} \\ & & I \\ & & & δ_{22} \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟺ δ_{22} = ᾱ_{22}^{−1}.

Suppose now that ( \begin{matrix}α & 0 \\ γ & δ \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) and α = ( \begin{matrix}α_{11} & α_{12} \\ α_{21} & α_{22} \end{matrix} ) has the

property that α_{22} is invertible. Then after multiplying ( \begin{matrix}α & 0 \\ γ & δ \end{matrix} ) on the left side by elements of the kind above, we can assume that α = ( \begin{matrix}α_{11} & α_{12} \\ 0 & I \end{matrix} ). The condition δ = ᾱ^{−1} implies that

δ = ( \begin{matrix}ᾱ_{11}^{−1} & 0 \\ −ᾱ_{12}ᾱ_{11}^{−1} & I \end{matrix} ) .

From the equation ᾱAα + γ̄α + λᾱγ = A, one gets the equation ᾱ_{11}A_{1}α_{11} + γ̄_{11}α_{11} + λᾱ_{11}γ_{11} = A_{1}. Thus, from Lemma 3.2, it follows that

( \begin{matrix}α_{11} & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ γ_{11} & 0 & ᾱ_{11}^{−1} & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r})

and that there is a decomposition

( \begin{matrix}α_{11} & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ γ_{11} & γ_{12} & ᾱ_{11}^{−1} & 0 \\ γ_{21} & γ_{22} & −ᾱ_{12}ᾱ_{11}^{−1} & I \end{matrix} ) = ( \begin{matrix}α_{11} & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ γ_{11} & 0 & ᾱ_{11}^{−1} & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ( \begin{matrix}I & α′_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & γ′_{12} & I & 0 \\ γ_{21} & γ′_{22} & −ᾱ′_{12} & I \end{matrix} )

in GH_{2n}(R, a_{1}, ⋯, a_{r}) for some α′_{12}, γ′_{12} and γ′_{22}. For the last factor in the decomposition above we can assume that γ_{21} = 0, after multiplying this factor on the right by

( \begin{matrix}I & 0 \\ γ & 0 \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) where γ = ( \begin{matrix}0 & λ̄γ̄_{21} \\ −γ_{21} & 0 \end{matrix} ). It follows from Lemma

3.2 that

( \begin{matrix}I & α′_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & γ′_{12} & I & 0 \\ 0 & γ′_{22} & −ᾱ′_{12} & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r})

⟺ γ′_{12} = −Ā_{1}α′_{12} and ᾱ′_{12}A_{1}α′_{12} − γ̄′_{22} − λγ′_{22} = 0.

The last equation implies that the columns of α′_{12} belong to C.

By Lemma 3.3, for an r × (n − r) matrix α_{12} there is an (n − r) × (n − r) matrix γ_{22} such that ᾱ_{12}A_{1}α_{12} − γ̄_{22} − λγ_{22} = 0 if and only if the columns of α_{12} in C. So for an r × (n − r) matrix α_{12} with columns in C,

( \begin{matrix}I & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & γ_{12} & I & 0 \\ 0 & γ_{22} & −ᾱ_{12} & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r})

⟺ γ_{12} = −Ā_{1}α_{12} and ᾱ_{12}A_{1}α_{12} − γ̄_{22} − λγ_{22} = 0.

If σ = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) with invertible α, then σ_{1} = ( \begin{matrix}α & 0 \\ γ & ᾱ^{−1} \end{matrix} ) ∈

GH_{2n}(R, a_{1}, ⋯, a_{r}). After multiplying σ on the left by the inverse of σ_{1}, we can reduce to the case α = I and γ = 0. For a matrix

( \begin{matrix}I & β \\ 0 & δ \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}),

Lemma 3.2 tells us that δ = I − Āβ and β̄Aβ − β − λβ̄ = 0, which implies

δ_{11} = I − Ā_{1}β_{11}, δ_{12} = −Ā_{1}β_{12}, δ_{21} = 0, δ_{22} = I and β̄_{11}A_{1}β_{11} − β_{11} − λβ̄_{11} = 0.

The invertibility of δ_{11} = I − Ā_{1}β_{11} follows from the invertibility of δ. Combining this with the equation β̄_{11}A_{1}β_{11} − β_{11} − λβ̄_{11} = 0, it follows from Lemma 3.2 that

( \begin{matrix}I & 0 & β_{11} & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I − Ā_{1}β_{11} & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}).

Thus, we have a decomposition

( \begin{matrix}I & 0 & β_{11} & β_{12} \\ 0 & I & β_{21} & β_{22} \\ 0 & 0 & I − Ā_{1}β_{11} & −Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} ) = ( \begin{matrix}I & 0 & β_{11} & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I − Ā_{1}β_{11} & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ( \begin{matrix}I & 0 & 0 & β′_{12} \\ 0 & I & β_{21} & β_{22} \\ 0 & 0 & I & −Ā_{1}β′_{12} \\ 0 & 0 & 0 & I \end{matrix} )

in GH_{2n}(R, a_{1}, ⋯, a_{r}) for some β′_{12}. The last matrix is in GH_{2n}(R, a_{1}, ⋯, a_{r}) if and only if β_{21} = −λβ′_{12} and β̄′_{12}A_{1}β′_{12} − β_{22} − λβ̄_{22} = 0.

Finally, it follows from Lemma 3.2 that

( \begin{matrix}I & 0 & 0 & 0 \\ 0 & α_{22} & 0 & β_{22} \\ 0 & 0 & I & 0 \\ 0 & γ_{22} & 0 & δ_{22} \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟺ ( \begin{matrix}α_{22} & β_{22} \\ γ_{22} & δ_{22} \end{matrix} ) ∈ GQ_{2(n−r)}(R, max^{−λ}(R)).

We have exhibited above centain special elements in GH_{2n}(R, a_{1}, ⋯, a_{r}). Using some of them, we define the elementary subgroup of GH_{2n}(R, a_{1}, ⋯, a_{r}) as follows.

Definition 4.1 Let EH_{2n}(R, a_{1}, ⋯, a_{r}) denote the subgroup of GH_{2n}(R, a_{1}, ⋯, a_{r}) generated by all matrices

( \begin{matrix}I \\ α_{21} & I \\ & & I & −ᾱ_{21} \\ & & & I \end{matrix} ) where α_{21} is arbitary,

( \begin{matrix}I \\ & I \\ γ_{11} & γ_{12} & I \\ γ_{21} & γ_{22} & & I \end{matrix} ) where ( \begin{matrix}γ_{11} & γ_{12} \\ γ_{21} & γ_{22} \end{matrix} ) is − λ̄ − hermitian,

( \begin{matrix}I \\ & ϵ \\ & & I \\ & ϵ̄^{−1} \end{matrix} ) where ϵ ∈ E_{n−r}(R),

( \begin{matrix}I & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā_{1}α_{12} & I & 0 \\ 0 & γ_{22} & −ᾱ_{12} & I \end{matrix} ) where ᾱ_{12}A_{1}α_{12} − γ̄_{22} − λγ_{22} = 0, and

( \begin{matrix}I & 0 & 0 & β_{12} \\ 0 & I & −λβ̄_{12} & β_{22} \\ 0 & 0 & I & −Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} ) where β̄_{12}A_{1}β_{12} − β_{22} − λβ̄_{22} = 0.

This subgroup is called the elementary Hermitian group of the elements a_{1}, ⋯, a_{r}.

The five types of matrices above are taken as generators of the elementary Hermitian group EH_{2n}(R, a_{1}, ⋯, a_{r}) for three reasons. First, each type is easy to construct and its inverse is easily determined. For example,

( \begin{matrix}I & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā_{1}α_{12} & I & 0 \\ 0 & γ_{22} & −ᾱ_{12} & I \end{matrix} )^{−1} = ( \begin{matrix}I & −α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & Ā_{1}α_{12} & I & 0 \\ 0 & λ̄γ̄_{22} & ᾱ_{12} & I \end{matrix} )

because ᾱ_{12}A_{1}α_{12} − γ̄_{22} − λγ_{22} = 0, and

( \begin{matrix}I & 0 & 0 & β_{12} \\ 0 & I & −λβ̄_{12} & β_{22} \\ 0 & 0 & I & −Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} )^{−1} = ( \begin{matrix}I & 0 & 0 & −β_{12} \\ 0 & I & λβ̄_{12} & λβ̄_{22} \\ 0 & 0 & I & Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} )

because β̄_{12}A_{1}β_{12} − β_{22} − λβ̄_{22} = 0. (It should be remarked that given a matrix of the kind

( \begin{matrix}α_{11} & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ γ_{11} & 0 & ᾱ_{11}^{−1} & 0 \\ 0 & 0 & 0 & I \end{matrix} ) or ( \begin{matrix}I & 0 & β_{11} & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I − Ā_{1}β_{11} & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}),

it is easy to determine its inverse, but such a matrix itself is difficult to construct.) Second, these matrices can be decomposed further as a product of so called elementary Hermitian matrices to be defined in Chapter 2. Third, the group they generate contains the elementary quadratic group EQ_{2(n−r)}(R, max^{−λ}(R)) and is perfect whenever n−r ≥ 3.

Theorem 4.2 If n ≥ r + 3, EH_{2n}(R, a_{1}, ⋯, a_{r}) is perfect.

Proof Denote by P the commutator subgroup of EH_{2n}(R, a_{1}, ⋯, a_{r}). Since n − r ≥ 3,

( \begin{matrix}I & 0 & 0 & 0 \\ 0 & ϵ & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & 0 & 0 & ϵ̄^{−1} \end{matrix} ) , ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & γ_{22} & 0 & I \end{matrix} ) , ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & β_{22} \\ 0 & 0 & I & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ∈ P

whenever they are in EH_{2n}(R, a_{1}, ⋯, a_{r}) ( see [Bak] p.30 ).

Now notice that there is a matrix e ∈ E_{n−r}(R) such that I − e is invertible. For example,

when m = n − r is even, take the m × m matrix e = ( \begin{matrix}I & −I \\ I & 0 \end{matrix} ) where I is the \frac{m}{2} ×

\frac{m}{2} identity matrix. When m = n − r is odd, take the block matrix

e = ( \begin{matrix}I & −I & 0 \\ I & 0 & x \\ o & y & 0 \end{matrix} )

where I is the \frac{m−1}{2} × \frac{m−1}{2} identity matrix, x = ^{t}(1, 0, ⋯, 0) is an \frac{m−1}{2} × 1 matrix and y = (−1, 0, ⋯, 0) is a 1 × \frac{m−1}{2} matrix. Let e ∈ E_{n−r}(R) such that I − e is invertible. Then

τ = ( \begin{matrix}I \\ & e \\ & & I \\ & ē^{−1} \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}).

For any α_{21},

[( \begin{matrix}I & 0 & 0 & 0 \\ (I − e)^{−1}eα_{21} & I & 0 & 0 \\ 0 & 0 & I & −\overline{(I − e)^{−1}eα_{21}} \\ 0 & 0 & 0 & I \end{matrix} ) , ( \begin{matrix}I \\ & e \\ & & I \\ & ē^{−1} \end{matrix} )] = ( \begin{matrix}I & 0 & 0 & 0 \\ α_{21} & I & 0 & 0 \\ 0 & 0 & I & −ᾱ_{21} \\ 0 & 0 & 0 & I \end{matrix} ) .

For any γ_{12},

[( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −γ_{12}(I − e)^{−1} & I & 0 \\ λ̄\overline{γ_{12}(I − e)^{−1}} & 0 & 0 & I \end{matrix} ) , ( \begin{matrix}I \\ & e \\ & & I \\ & ē^{−1} \end{matrix} )] = ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & γ_{12} & I & 0 \\ −λ̄γ̄_{12} & 0 & 0 & I \end{matrix} ) .

Suppose

σ = ( \begin{matrix}I & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā_{1}α_{12} & I \\ 0 & γ_{22} & −ᾱ_{12} & I \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}).

Then ᾱ_{12}A_{1}α_{12} − γ̄_{22} − λγ_{22} = 0 and it follows that

\overline{α_{12}(I − e)^{−1}}A_{1}α_{12}(I − e)^{−1} − \overline{(I − e)^{−1}}γ̄_{22}(I − e)^{−1} − λ\overline{(I − e)^{−1}}γ_{22}(I − e)^{−1} = 0, which implies by Lemma 3.2 that

σ_{1} = ( \begin{matrix}I & −α_{12}(I − e)^{−1} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & Ā_{1}α_{12}(I − e)^{−1} & I \\ 0 & \overline{(I − e)^{−1}}γ_{22}(I − e)^{−1} & \overline{α_{12}(I − e)^{−1}} & I \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r})

and

[σ_{1}, τ] = σ(I ⊥ ( \begin{matrix}I & 0 \\ c & I \end{matrix} ))

where c = λ̄\overline{γ_{22}(I − e)^{−1}} − γ_{22}(I − e)^{−1} is a −λ̄-hermitian matrix. Thus σ ∈ P. Suppose

σ_{2} = ( \begin{matrix}I & 0 & 0 & β_{12} \\ 0 & I & −λβ̄_{12} & β_{22} \\ 0 & 0 & I & −Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}).

Then β̄_{12}A_{1}β_{12} − β_{22} − λγ̄_{22} = 0 and it follows that

σ_{3} = ( \begin{matrix}I & 0 & 0 & −β_{12}(I − e)^{−1} \\ 0 & I & λ\overline{β_{12}(I − e)^{−1}} & \overline{(I − e)^{−1}}β_{22}(I − e)^{−1} \\ 0 & 0 & I & Ā_{1}β_{12}(I − e)^{−1} \\ 0 & 0 & 0 & I \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r})

and

[σ_{3}, τ̄^{−1}] = σ_{2}(I ⊥ ( \begin{matrix}I & d \\ 0 & I \end{matrix} ))

where d = −β_{22}(I − e)^{−1} + λ\overline{β_{22}(I − e)^{−1}} is a −λ-hermitian matrix. Thus σ_{2} ∈ P. Finally, for any −λ̄-hermitian matrix γ_{11} with zero diagonal, it follows from the identity

[( \begin{matrix}I & 0 & 0 & 0 \\ α_{21} & I & 0 & 0 \\ 0 & 0 & I & −ᾱ_{21} \\ 0 & 0 & 0 & I \end{matrix} ) , ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & γ_{12} & I & 0 \\ −λ̄γ̄_{12} & 0 & 0 & I \end{matrix} )] = ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ λ̄ᾱ_{21}γ̄_{12} − γ_{12}α_{21} & 0 & I & 0 \\ 0 & 0 & 0 & I \end{matrix} )

that ( \begin{matrix}I & 0 \\ γ_{11} & I \end{matrix} ) ⊥ I ∈ P whenever it is in EH_{2n}(R, a_{1}, ⋯, a_{r}). If the −λ̄-hermitian matrix γ_{11} is diagonal, it follows from the identity

[( \begin{matrix}I & 0 & 0 & 0 \\ α_{21} & I & 0 & 0 \\ 0 & 0 & I & −ᾱ_{21} \\ 0 & 0 & 0 & I \end{matrix} ) , ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & γ_{22} & 0 & I \end{matrix} )] = ( \begin{matrix}I & 0 & 0 & 0 \\ 0 & I & 0 & 0 \\ −ᾱ_{21}γ_{22}α_{21} & −ᾱ_{21}γ_{22} & I & 0 \\ −γ_{22}α_{21} & 0 & 0 & I \end{matrix} )

that ( \begin{matrix}I & 0 \\ γ_{11} & I \end{matrix} ) ⊥ I ∈ P whenever it is in EH_{2n}(R, a_{1}, ⋯, a_{r}). Thus P =

EH_{2n}(R, a_{1}, ⋯, a_{r}), i.e. EH_{2n}(R, a_{1}, ⋯, a_{r}) is perfect. □

5. Hermitian Whitehead Lemma

From the definition of EH_{2n}(R, a_{1}, ⋯, a_{r}) it is easy to see that the stabilization homomorphism

GH_{2n}(R, a_{1}, ⋯, a_{r}) ↪ GH_{2(n+1)}(R, a_{1}, ⋯, a_{r})

( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ⟼ ( \begin{matrix}α & 0 & β & 0 \\ 0 & 1 & 0 & 0 \\ γ & 0 & δ & 0 \\ 0 & 0 & 0 & 1 \end{matrix} )

induces a stabilization homomorphism

EH_{2n}(R, a_{1}, ⋯, a_{r}) ↪ EH_{2(n+1)}(R, a_{1}, ⋯, a_{r}).

Following [Bak,§3],we define stable groups as follows.

Definition 5.1 For any r elements a_{1}, a_{2}, ⋯, a_{r} ∈ max^{λ}(R) define

GH(R, a_{1}, ⋯, a_{r}) = lim_{n^{⟶}} GH_{2n}(R, a_{1}, ⋯, a_{r}),

EH(R, a_{1}, ⋯, a_{r}) = lim_{n^{⟶}} EH_{2n}(R, a_{1}, ⋯, a_{r}).

Lemma 5.2 Suppose a_{1}, a_{2}, ⋯, a_{r} ∈ max^{λ}(R). Then EH(R, a_{1}, ⋯, a_{r}) is perfect.

Proof The lemma is a direct consequence of Theorem 4.2 and the definition of EH(R, a_{1}, ⋯, a_{r}). □

From Lemma 3.5 and [Bak, Lemma 3.15], the commutator subgroup of GH(R, a_{1}, ⋯, a_{r}) is perfect. Thus EH(R, a_{1}, ⋯, a_{r}) is contained in this commutator subgroup. The goal of this section is to show that EH(R, a_{1}, ⋯, a_{r}) is the commutator subgroup. This will provide further justification for our definition of EH(R, a_{1}, ⋯, a_{r}).

Theorem 5.3 ( Hermitian Whitehead Lemma )

[GH_{2n}(R, a_{1}, ⋯, a_{r}), GH_{2n}(R, a_{1}, ⋯, a_{r})] ⊆ EH_{4n}(R, a_{1}, ⋯, a_{r}).

Proof. According to Lemma 3.5 there is an isomorphism

(R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ⊥ (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ≅ (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ⊥ H(R^{n}).

Many matrices can be taken to define this isomorphism. In the proof of Lemma 3.5 we wrote down such a matrix. To simplify our calculations, we shall use a different matrix, namely

φ = ( \begin{matrix}I & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & I & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 0 & 0 & −Ā_{1} & 0 & I & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & I & 0 & 0 \\ 0 & 0 & Ā_{1} & 0 & −I & 0 & I & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & I \end{matrix} ) .

This matrix can be obtained from the matrix in the proof of Lemma 3.5 by multiplying it on the right by the matrix

( \begin{matrix}I & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & I & 0 & −I & 0 & 0 & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & I & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & I & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & I & 0 \\ 0 & 0 & 0 & 0 & 0 & I & 0 & I \end{matrix} ) ∈ Aut((R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ⊥ (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} ))).

The isomorphism given above induces an isomorphism

Aut((R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} )) ⊥ (R^{2n}, ( \begin{matrix}A & λI \\ I & 0 \end{matrix} ))) ⟶ GH_{4n}(R, a_{1}, ⋯, a_{r})

X ⟼ φXφ^{−1}.

In particular, it induces an injection

GH_{2n}(R, a_{1}, ⋯, a_{r}) ⊥ GH_{2n}(R, a_{1}, ⋯, a_{r}) ⟶ GH_{4n}(R, a_{1}, ⋯, a_{r})
σ ⊥ σ′ ⟼ φ(σ ⊥ σ′)φ^{−1}.

To prove the theorem we need only to show that for any σ, τ ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}) στσ^{−1}τ^{−1} ⊥ I ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}). One might try to prove this by showing that σ ⊥ σ^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}). But this would be impossible because σ ⊥ σ^{−1} does not belong to GH_{4n}(R, a_{1}, ⋯, a_{r}) in general. Instead, we shall use the injection above and divide the proof into four steps.

Step 1 Suppose σ ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}). Then in the group GH_{4n}(R, a_{1}, ⋯, a_{r})

φ(σ ⊥ I)φ^{−1} = E_{1}(σ ⊥ I) for some E_{1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

Suppose

σ = ( \begin{matrix}α_{11} & α_{12} & β_{11} & β_{12} \\ α_{21} & α_{22} & β_{21} & β_{22} \\ γ_{11} & γ_{12} & δ_{11} & δ_{12} \\ γ_{21} & γ_{22} & δ_{21} & δ_{22} \end{matrix} ) ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}).

Notice that both φ(σ ⊥ I)φ^{−1} and σ ⊥ I belong to GH_{4n}(R, a_{1}, ⋯, a_{r}). A direct matrix calculation yields

φ(σ ⊥ I)φ^{−1} = ( \begin{matrix}α_{11} & α_{12} & I − α_{11} + β_{11}Ā_{1} & 0 & β_{11} & β_{12} & 0 & 0 \\ α_{21} & α_{22} & −α_{21} + β_{21}Ā_{1} & 0 & β_{21} & β_{22} & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ γ_{11} & γ_{12} & δ_{11}Ā_{1} − Ā_{1} − γ_{11} & 0 & δ_{11} & δ_{12} & 0 & 0 \\ γ_{21} & γ_{22} & −γ_{21} + δ_{21}Ā_{1} & 0 & δ_{21} & δ_{22} & 0 & 0 \\ −γ_{11} & −γ_{12} & γ_{11} − δ_{11}Ā_{1} + Ā_{1} & 0 & I − δ_{11} & −δ_{12} & I & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & I \end{matrix} ) = E_{1}(σ ⊥ I)

where

E_{1} = ( \begin{matrix}I & 0 & I − α_{11} + β_{11}Ā_{1} & 0 & 0 & 0 & 0 & 0 \\ 0 & I & −α_{21} + β_{21}Ā_{1} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 0 & 0 & (δ_{11} − I)Ā_{1} − γ_{11} & 0 & I & 0 & 0 & 0 \\ 0 & 0 & −γ_{21} + δ_{21}Ā_{1} & 0 & 0 & I & 0 & 0 \\ γ′_{11} & γ′_{12} & γ_{11} − (δ_{11} − I)Ā_{1} & 0 & δ′_{11} & δ′_{12} & I & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & I \end{matrix} )

for some γ′_{11}, γ′_{12}, δ′_{11} and δ′_{12}. E_{1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}) because it can be expressed as a product of three generators of EH_{4n}(R, a_{1}, ⋯, a_{r}) .

Step 2 Suppose σ ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}). Then in GH_{4n}(R, a_{1}, ⋯, a_{r})

φ(I ⊥ σ)φ^{−1} = UE_{2}(σ ⊥ I)U^{−1} for some U, E_{2} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

With the same assumption on σ we have φ(I ⊥ σ)φ^{−1} ∈ GH_{4n}(R, a_{1}, ⋯, a_{r}). A direct matrix calculation shows that

φ(I ⊥ σ)φ^{−1} = ( \begin{matrix}I & 0 & α_{11} − I & α_{12} & β_{11} & 0 & β_{11} & β_{11} \\ 0 & I & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & α_{11} & α_{12} & β_{11} & 0 & β_{11} & β_{12} \\ 0 & 0 & α_{21} & α_{22} & β_{21} & 0 & β_{21} & β_{22} \\ 0 & 0 & −Ā_{1}(α_{11} − I) & −Ā_{1}α_{12} & I − Ā_{1}β_{11} & 0 & −Ā_{1}β_{11} & −Ā_{1}β_{12} \\ 0 & 0 & 0 & 0 & 0 & I & 0 & 0 \\ 0 & 0 & x_{1} & x_{2} & x_{3} & 0 & x_{4} & x_{5} \\ 0 & 0 & γ_{21} & γ_{22} & δ_{21} & 0 & δ_{21} & δ_{22} \end{matrix} )

where x_{1} = γ_{11} + Ā_{1}(α_{11} − I), x_{2} = γ_{12} + Ā_{1}α_{12} , x_{3} = δ_{11} − I + Ā_{1}β_{11}, x_{4} = δ_{11} + Ā_{1}β_{11}, and x_{5} = δ_{12} + Ā_{1}β_{22}. Now, notice that adding the third column to the first column and interchanging the second and fourth column we obtain a matrix with α_{11} as (1,1)-entry and α_{12} as (1,2)-entry. Since we want to accomplish these manipulations by multiplying φ(I ⊥ σ)φ^{−1} by elements in GH_{4n}(R, a_{1}, ⋯, a_{r}), the definition of the following matrix U becomes natural.

Set

U = ( \begin{matrix}I & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & I & 0 & 0 & 0 & 0 & 0 & 0 \\ I & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & I & 0 & −I & 0 \\ 0 & 0 & 0 & 0 & 0 & I & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & I & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & I \end{matrix} ) ( \begin{matrix}I & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & −I & 0 & 0 & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & I & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & I & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & −I \\ 0 & 0 & 0 & 0 & 0 & 0 & I & 0 \\ 0 & 0 & 0 & 0 & 0 & I & 0 & 0 \end{matrix} ) .

Then U ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}) since the first factor of U is a generator of EH_{4n}(R, a_{1}, ⋯, a_{r}) and the second factor is in EH_{4n}(R, a_{1}, ⋯, a_{r}) as shown by the identity

( \begin{matrix}0 & 0 & −I \\ 0 & I & 0 \\ I & 0 & 0 \end{matrix} ) = ( \begin{matrix}I & 0 & 0 \\ 0 & I & 0 \\ I & 0 & I \end{matrix} ) ( \begin{matrix}I & 0 & −I \\ 0 & I & 0 \\ 0 & 0 & I \end{matrix} ) ( \begin{matrix}I & 0 & 0 \\ 0 & I & 0 \\ −I & 0 & I \end{matrix} ) .

Thus U^{−1}φ(I ⊥ α)φ^{−1}U ∈ GH_{4n}(R, a_{1}, ⋯, a_{r}) and direct matrix calculation shows

U^{−1}φ(I ⊥ α)φ^{−1}U = ( \begin{matrix}α_{11} & α_{12} & α_{11} − I & 0 & β_{11} & β_{12} & 0 & 0 \\ α_{21} & α_{22} & α_{21} & 0 & β_{21} & β_{22} & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ γ_{11} & γ_{12} & γ_{11} & 0 & δ_{11} & δ_{12} & 0 & 0 \\ γ_{21} & γ_{22} & γ_{21} & 0 & δ_{21} & δ_{22} & 0 & 0 \\ y_{1} & y_{2} & y_{3} & 0 & y_{4} & y_{5} & I & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & I \end{matrix} )

where y_{1} = γ_{11}+Ā_{1}(α_{11}−I), y_{2} = γ_{12}+Ā_{1}α_{12}, y_{3} = γ_{11}+Ā_{1}(α_{11}−I), y_{4} = δ_{11}+Ā_{1}β_{11}−I, and y_{5} = δ_{12} + Ā_{1}β_{22}. Hence it follows that

U^{−1}φ(I ⊥ α)φ^{−1}U = E_{2}(α ⊥ I)

where

E_{2} = ( \begin{matrix}I & 0 & α_{11} − I & 0 & 0 & 0 & 0 & 0 \\ 0 & I & α_{21} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & I & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & I & 0 & 0 & 0 & 0 \\ 0 & 0 & γ_{11} & 0 & I & 0 & 0 & 0 \\ 0 & 0 & γ_{21} & 0 & 0 & I & 0 & 0 \\ γ′′_{11} & γ′′_{12} & y_{3} & 0 & δ′′_{11} & δ′′_{12} & I & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & I \end{matrix} ) .

for some γ′′_{11}, γ′′_{12}, δ′′_{11} and δ′′_{12}. It is also easy to see that E_{2} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}). Thus

φ(I ⊥ σ)φ^{−1} = UE_{2}(σ ⊥ I)U^{−1}.

Step 3 Suppose σ ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}). Then in GH_{4n}(R, a_{1}, ⋯, a_{r})

φ(I ⊥ σ)φ^{−1} = E_{3}(σ ⊥ I) for some E_{3} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

From Step 2 it is sufficient to prove that

(σ ⊥ I)U^{−1}(σ ⊥ I)^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r})

or equivalently

(σ ⊥ I)U(σ ⊥ I)^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

Recall that U can be expressed as a product of four matrices of the form

X = ( \begin{matrix}I & 0 & 0 & 0 \\ x & I & 0 & 0 \\ 0 & 0 & I & −x̄ \\ 0 & 0 & 0 & I \end{matrix} ) or Y = ( \begin{matrix}I & y & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & 0 & I & 0 \\ 0 & 0 & −ȳ & I \end{matrix} )

where I, x and y are n × n matrices, and both X and Y are in EH_{4n}(R, a_{1}, ⋯, a_{r}). Now, write σ and σ^{−1} as follows

σ = ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) and σ^{−1} = ( \begin{matrix}α′ & β′ \\ γ′ & δ′ \end{matrix} ) .

Then

(σ ⊥ I)X(σ ⊥ I)^{−1} = ( \begin{matrix}α & 0 & β & 0 \\ 0 & I & 0 & 0 \\ γ & 0 & δ & 0 \\ 0 & 0 & 0 & I \end{matrix} ) ( \begin{matrix}I & 0 & 0 & 0 \\ x & I & 0 & 0 \\ 0 & 0 & I & −x̄ \\ 0 & 0 & 0 & I \end{matrix} ) ( \begin{matrix}α′ & 0 & β′ & 0 \\ 0 & I & 0 & 0 \\ γ′ & 0 & δ′ & 0 \\ 0 & 0 & 0 & I \end{matrix} )

= ( \begin{matrix}I & 0 & 0 & −βx̄ \\ xα′ & I & xβ′ & 0 \\ 0 & 0 & I & −δx̄ \\ 0 & 0 & 0 & I \end{matrix} ) = ( \begin{matrix}I & 0 & 0 & 0 \\ xα′ & I & 0 & 0 \\ 0 & 0 & I & −\overline{xα′} \\ 0 & 0 & 0 & I \end{matrix} ) ( \begin{matrix}I & 0 & 0 & −βx̄ \\ 0 & I & xβ′ & xα′βx̄ \\ 0 & 0 & I & δx̄ − \overline{xα′} \\ 0 & 0 & 0 & I \end{matrix} ) .

Obviously, the first factor is in EH_{4n}(R, a_{1}, ⋯, a_{r}) and the second factor is also in EH_{4n}(R, a_{1}, ⋯, a_{r}) since it is in GH_{4n}(R, a_{1}, ⋯, a_{r}). Thus

(σ ⊥ I)X(σ ⊥ I)^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

A similar calculation shows

(σ ⊥ I)Y (σ ⊥ I)^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

Hence

(σ ⊥ I)U^{−1}(σ ⊥ I)^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}),

which completes the proof of Step 3. Step 4 Suppose σ, τ ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}). Then in GH_{4n}(R, a_{1}, ⋯, a_{r})

(στσ^{−1}τ^{−1}) ⊥ I ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

From Step 1 and Step 3 we have

φ(σ ⊥ σ^{−1})φ^{−1} = φ(σ ⊥ I)φ^{−1}φ(I ⊥ σ^{−1})φ^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}). Thus for any σ, τ ∈ GH_{2n}(R, a_{1}, ⋯, a_{r}), it follows from

στσ^{−1}τ^{−1} ⊥ I = (σ ⊥ σ^{−1})(τ ⊥ τ^{−1})((τσ)^{−1} ⊥ τσ)

that

φ(στσ^{−1}τ^{−1} ⊥ I)φ^{−1} ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}),

and further from Step 1 that

στσ^{−1}τ^{−1} ⊥ I ∈ EH_{4n}(R, a_{1}, ⋯, a_{r}).

This completes the proof of the theorem. □ Theorem 5.4 For any elements a_{1}, ⋯, a_{r} ∈ max^{λ}(R), EH(R, a_{1}, ⋯, a_{r}) is the commutator subgroup of GH(R, a_{1}, ⋯, a_{r}) and is perfect. Proof. The theorem is an immediate consequence of Theorem 5.3 and Lemma 5.2. □

Remark Suppose Λ is a −λ-form parameter such that Λ/ min^{λ}(R) is finite and let a_{1}, ⋯, a_{N} ∈ Λ be a set of coset representatives for Λ/ min^{λ}(R). Bak defined

GH(R, Λ) = lim_{n^{⟶}} GH_{2n}(R, a_{1}, ⋯, a_{N})

and posed after [Bak, (3.14)] the problem of finding reasonable generators for the commutator subgroup of GH(R, Λ), which are functorial in (R, Λ) and can be used as generators for a presentation of the universal central extension of the commutator subgroup. This was the starting point for the investigations of the current paper and Theorem 5.4 above together with the definition of EH(R, a_{1}, …, a_{r}) provides the desired generators.

6. KH_{1}-groups

We develop some machinery and use it to define KH_{1}-groups. The same machinery will be used in section 11 to define KH_{2}-groups. At the end of the section we compare various K_{1}-groups and show that Bass’ categorical definition of K_{1} of Λ-Hermitian spaces coincides with the one here.

Let 𝒞 be a category. Call morphisms f : A ⟶ B and f′ : A′ ⟶ B′ equivalent if there are isomorphisms A ⟶ A′ and B ⟶ B′ such that the diagram

\begin{matrix}A & ⟶^{f} & B \\ ↓ & & ↓ \\ A′ & ⟶_{f′} & B′ \end{matrix}

commutes. Let 𝒞^{∼} denote the category whose objects are the isomorphism classes [A] of objects A of 𝒞 and whose morphisms are the equivalence classes [f] of morphisms f of 𝒞. For each object [A] ∈ 𝒞^{∼}, pick an object A^{∼} of 𝒞 such that A^{∼} is a member of the class [A], i.e A^{∼} ≅ A, and for each morphism [f] : [A] ⟶ [B] in 𝒞^{∼}, pick a morphism f^{∼} of 𝒞 such that f^{∼} ∈ [f] and f^{∼} : A^{∼} ⟶ B^{∼}. Let F : 𝒞 ⟶ 𝒟 be a functor. Suppose that for each object A of 𝒞, F (Aut(A)) = {1_{F (A)}}. Then F induces a functor F^{∼} : 𝒞^{∼} ⟶ 𝒟, [A] ⟼ F (A^{∼}) and [f] ⟼ F (f^{∼}). Clearly if 𝒞 satisfies the condition that Aut(A) = {1_{A}} for every object A of 𝒞 then F^{∼} is defined. Suppose that 𝒟= ((groups)) and the objects of 𝒞^{∼} form a set. Then colim F^{∼} exists. In this situation, we define colim F = colimF^{∼}. If additionally the category 𝒞^{∼} is isomorphic to a directed partially ordered set then we define \underrightarrow{lim}_{𝒞^{∼}}F = \underrightarrow{lim}_{𝒞^{∼}}F^{∼} = colimF^{∼}.

Let

Δ^{−λ}(R)

denote the category whose objects are finite sequences (a_{1}, … a_{r}) of elements a_{i} of max^{−λ}(R) such that a_{i} ≠ a_{j} whenever i ≠ j. A morphism (a_{1}, …, a_{r}) ⟶ (a′_{1}, …, a′_{r′}) is any injective function f : {1, …, r} ⟶ {1, …, r′} such that a_{i} = a′_{f(i)}. Clearly Aut (a_{i}, …, a_{r}) = 1 and

Δ^{−λ}(R)^{∼}

is isomorphic to the directed partially ordered set whose objects are the nonempty finite subsets of max^{−λ}(R) and whose inequalities are the natural inclusions among such subsets. For any functor F : Δ^{−λ}(R) ⟶ ((groups)) and any form parameter Λ ⫅ max^{−λ}(R), \underrightarrow{lim}_{Λ}F

is defined (where the latter Λ refers to the full subcategory of Δ^{−λ}(R)^{∼} whose objects are all finite nonempty subsets of Λ).

Consider the functor

Δ^{−λ}(R) ⟶ ((groups))
(a_{1}, …, a_{r}) ⟼ GH_{2r}(R, a_{1}, …, a_{r}).

For each morphism (a_{1}, …, a_{r}) ⟶ (a′_{1}, …, a′_{r′}), the stabilization of GH_{2r}(R, a_{1}, …, a_{r}) ⟶ GH_{2r}(R, a′_{1}, …, a′_{r′}) to a homomorphism GH(R, a_{1}, …, a_{r}) ⟶ GH(R, a′_{1}, …, a′_{r′}) defines a functor

Δ^{−λ}(R) ⟶ ((groups))
(a_{1}, …, a_{r}) ⟼ GH(R, a_{1}, …, a_{r}).

This induces a functor

Δ^{−λ}(R) ⟶ ((groups))
(a_{1}, …, a_{r}) ⟼ EH(R, a_{1}, …, a_{r}).

Definition 6.1 Define

KH_{1}(R, a_{1}, …, a_{r}) = GH(R, a_{1}, …, a_{r})/EH(R, a_{1}, …, a_{r}).

By the Hermitian Whitehead lemma, KH_{1}(R, a_{1}, …, a_{r}) is an abelian group. Definition 6.2 Suppose Λ is a -λ-form parameter on R. Define

GH(R, Λ) = \underrightarrow{lim}_{S⫅Λ}GH(R, S)
EH(R, Λ) = \underrightarrow{lim}_{S⫅Λ}EH(R, S)
KH_{1}(R, Λ) = GH(R, Λ)/EH(R, Λ)

= \underrightarrow{lim}_{S⫅Λ}KH_{1}(R, S).

Obviously GH, EH, and KH_{1} define functors ((form rings)) ⟶ ((groups)) and KH_{1} takes its values in ((abelian groups)). Definition 6.3 Define

KH_{1}^{λ}(R) = KH_{1}(R, max^{−λ}(R)).

Corollary 6.4 EH(R, Λ) is the commutator subgroup of GH(R, Λ) and is perfect. Proof This follows immediately from Lemma 5.2 and Theorem 5.3.

Let Λ be a −λ-form parameter on R and Λ_{1} a λ-form parameter on R. Next we compare the K_{1}-groups K_{1}(R), KQ_{1}(R, Λ_{1}) and KH_{1}(R, Λ). For any a_{1}, ⋯, a_{r} ∈ Λ, there is a group homomorphism

KQ_{1}(R, Λ_{1}) ⟶ KH_{1}(R, a_{1}, ⋯, a_{r})

induced by the embedding GQ_{2n}(R, Λ_{1}) ↪ GH_{2(n+r)}(R, a_{1}, ⋯, a_{r}), and a group homomorphism

KH_{1}(R, a_{1}, ⋯, a_{r}) ⟶ K_{1}(R)

induced by the embedding GH_{2n}(R, a_{1}, ⋯, a_{r}) ↪ GL_{2n}(R). Taking direct limits, one obtains group homomorphisms

KQ_{1}(R, Λ_{1}) ⟶ KH_{1}(R, Λ)
KH_{1}(R, Λ) ⟶ K_{1}(R).

Usually, the homomorphisms above are not bijective. We give an example such that KQ_{1}(R, Λ_{1}) = 1, KH_{1}(R, Λ) = 1 + J where J is the Jacobson radical of R, and K_{1}(R) = units (R).

Let F be a finite field of characteristic 2. Let R = F [x, y]/(x^{2}, y^{2}, xy, yx) with trivial involution and λ = 1. Since R is a commutative local ring, K_{1}(R) = units (R). Let Λ_{1} = max^{−λ}(R). Clearly max^{−λ}(R) = R. Thus for any uϵR,

( \begin{matrix}1 & u \\ 0 & 1 \end{matrix} ) and ( \begin{matrix}1 & 0 \\ u & 1 \end{matrix} ) ∈ EQ(R, max^{−λ}(R)) since max^{−λ}(R) = R. If u ∈ units (R)

then from the decomposition

( \begin{matrix}u & 0 \\ 0 & u^{−1} \end{matrix} ) = ( \begin{matrix}1 & u − 1 \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ) ( \begin{matrix}1 & −u^{−1}(u − 1) \\ 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 \\ −u & 1 \end{matrix} )

it follows that ( \begin{matrix}u & 0 \\ 0 & u^{−1} \end{matrix} ) ∈ EQ(R, max^{−λ}(R)). But since R is local and max^{−λ}(R) = R,

the matrices ( \begin{matrix}u & 0 \\ 0 & u^{−1} \end{matrix} ) where u ranges over units (R) generate KQ_{1}(R, max^{−λ}(R)).

Thus KQ_{1}(R, max^{−λ}(R)) = 1.

Consider the group GH_{2n}(R, 1). For this group, the module C = { x ∈ R | x^{2} = 0 } = J where J = Jacobson radical (R). For an element u ≠ 1 ∈ R, the matrix

σ = ( \begin{matrix}1 & u \\ 0 & 1 − u \end{matrix} ) ⊥ I ∈ GH_{2n}(R, 1) if and only if u = ax + by ∈ J for a, b ∈ F. This

means the determinate of σ is not 1 when 0 ≠ u ∈ R. But every element in EH(R, 1) has determinate 1. Consequently KH_{1}(R, 1) is not equal to KQ_{1}(R, max^{−λ}(R)) = 1. In fact, if Λ is the smallest −λ-form parameter containing 1 then KH_{1}(R, Λ) = 1 + J. The last assertion is left as an exercise. This completes the example.

Drop for the moment the assumption that modules we consider are free. Let Λ be a −λform parameter on R. Let H(R, Λ) denote as in [Bak, p. 13] the category of non-singular Λ-Hermitian spaces on finitely generated projective R-modules and isomorphisms. Let M(R, Λ)_{free} denote the full subcategory of H(R, Λ) of all Λ-metabolic spaces Λ − M(P) on finitely generated free R-modules P. For a description of Λ − M above as a functor, see [Bak, p. 165]. It follows from [Bak, (2.9)] and the fact that the Λ-metabolic functor Λ − M has the property Λ − M(P) ⊥ Λ − M(Q) ≅ Λ − M(P ⊕ Q) that the category M(R, Λ)_{free} is cofinal in H(R, Λ). Both categories above are categories with product in the sense of Bass, under orthogonal sum ⊥. Thus one can use the procedure in [Bass, VII 1.4] to define their K_{1}-groups.

Theorem 6.5 Suppose Λ is a −λ-form parameter on R. Then

KH_{1}(R, Λ) ≅ K_{1}(M(R, Λ)_{free}) ≅ K_{1}(H(R, Λ))

where K_{1} is taken in the sense of [Bass, VII 1.4].

Proof The second isomorphism follows from [Bak, Lemma 6.23]. The first isomorphism follows from Lemma 3.5 and Theorem 5.4. □

II. Hermitian Steinberg groups and the functor K_{2} of Hermitian forms

7. Introduction to II

Let R be a ring with involution a ↦ ā. Let λ ∈ center(R) such that λλ̄ = 1 and let Λ be a −λ-form parameter on R. For a_{1}, ⋯, a_{r} ∈ Λ and n ≥ r, let GH_{2n}(R, a_{1}, ⋯, a_{r}) denote the general Hermitian group and EH_{2n}(R, a_{1}, ⋯, a_{r}) denote its elementary subgroup, as in chapter 1. In the current chapter, we define the metabolic or Hermitian Steinberg group StH_{2n}(R, a_{1}, ⋯, a_{r}) as a certain extension of EH_{2n}(R, a_{1}, ⋯, a_{r}). From the definition, it is easy to show that StH_{2n}(R, a_{1}, ⋯, a_{r}) is a perfect group and that there is a stabilization homomorphism StH_{2n}(R, a_{1}, ⋯, a_{r}) ⟶ StH_{2(n+1)}(R, a_{1}, ⋯, a_{r}) covering the stabilization homomorphism EH_{2n}(R, a_{1}, ⋯, a_{r}) ⟶ EH_{2(n+1)}(R, a_{1}, ⋯, a_{r}). Let

StH(R, a_{1}, ⋯, a_{r}) = lim_{n^{⟶}} StH_{2n}(R, a_{1}, ⋯, a_{r}).

The main theorems of the chapter are that the stabilized Steinberg group StH(R, a_{1}, ⋯, a_{r}) is a universal central extension of the stabilized elementary group EH(R, a_{1}, ⋯, a_{r}) and that the action of GH(R, a_{1}, …, a_{r}) on KH_{2}(R, a_{1}, …, a_{r}) := Ker (StH(R, a_{1}, …, a_{r}) ⟶ EH(R, a_{1}, …, a_{r})), which is induced by lifting to StH(R, a_{1}, …, a_{r}) the conjugation action of GH(R, a_{1}, …, a_{r}) on EH(R, a_{1}, …, a_{r}), is trivial. The latter result is a K_{2}-analog of the Whitehead lemma. We establish the unexpected result that KH_{2}(R, a_{1}, …, a_{r}) does not coincide in general with the center (StH(R, a_{1}, …, a_{r})), but can be strictly smaller.

The remainder of the chapter is organized as follows. In §8, we define in a functorial way generators for the group EH_{2n}(R, a_{1}, ⋯, a_{r}) and establish a long list (Lemma (8.2)) of relations for the generators. The relations provide an alternative demonstration of the fact established already in chapter 1 that EH_{2n}(R, a_{1}, ⋯, a_{r}) is perfect when n ≥ r + 3. In §9, we define StH_{2n}(R, a_{1}, ⋯, a_{r}) as the group presented by the generators and relations above and establish some preliminary results for this group and its stabilization StH(R, a_{1}, ⋯, a_{r}). From the defining relations, it follows immediately that these groups are perfect. In §10, we prove the result that StH(R, a_{1}, ⋯, a_{r}) is a universal central extension of EH(R, a_{1}, ⋯, a_{r}). In §11, we construct the Steinberg group StH(R, Λ) of Λ-Hermitian spaces and the K_{2}-group KH_{2}(R, Λ) of Λ-Hermitian spaces and show that StH(R, Λ) is a GH(R, Λ)-perfect, universal, GH(R, Λ)-central extension of EH(R, Λ). Then we show that KH_{2}(R, Λ) can be strictly smaller than center (StH(R, Λ)). Finally we show that our Steinberg group description of KH_{2}(R, Λ) agrees with Bak’s categorical construction of K_{2} of a category with product, applied to the category of nonsingular Λ-Hermitian spaces under orthogonal sum.

8. Functorial Generators of Elementary Hermitian Groups

Recall Definition 4.1 of the elementary subgroup EH_{2n}(R, a_{1}, ⋯, a_{r}) of the general Hermitian group GH_{2n}(R, a_{1}, ⋯, a_{r}). We decompose now the generators of this group as products of so called elementary matrices in EH_{2n}(R, a_{1}, ⋯, a_{r}). This is well known, cf. [Bak,§3], except for the last two kinds of generators in Definition 4.1

Let ζ ∈ C = {^{t}(x_{1}, x_{2}, ⋯, x_{r}) ∈ R^{r} | ∑_{i=1}^{r} x̄_{i}a_{i}x_{i} ≡ 0 mod min^{λ}(R)}. By definition, there is an element y ∈ R such that ∑_{i=1}^{r} x̄_{i}a_{i}x_{i} = y + λȳ. Fix such an element y for each ζ ∈ C and denote it by ζ_{f}. We shall always take 0_{f} = 0 and (−ζ)_{f} = ζ_{f}.

Define the 2n × 2n matrix

Hm_{i}(ζ) = ( \begin{matrix}I & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā_{1}α_{12} & I & 0 \\ 0 & γ_{22} & −ᾱ_{12} & I \end{matrix} ) ,

where α_{12} is the r × (n − r) matrix whose (i − r)-th column is ζ and other columns are 0, and γ_{22} is the (n − r) × (n − r) matrix whose (i − r, i − r)-entry is ζ̄_{f} and other entries are 0. Then Hm_{i}(ζ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}) since ᾱ_{12}A_{1}α_{12} = γ̄_{22} + λγ_{22}.

Define the 2n × 2n matrix

r_{i}(ζ) = ( \begin{matrix}I & 0 & 0 & β_{12} \\ 0 & I & −λβ̄_{12} & β_{22} \\ 0 & 0 & I & −Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} ) ,

where β_{12} is the r × (n − r) matrix whose (i − r)-th column is ζ and other columns are 0, and β_{22} is the (n − r) × (n − r) matrix whose (i − r, i − r)-entry is λζ̄_{f} and other entries are 0. Then r_{i}(ζ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}) since β̄_{12}A_{1}β_{12} = β_{22} + λβ̄_{22}.

Let Hϵ_{ij}(a), l_{ij}(a), and r_{ij}(a) have the same meaning as in [Bak, p.43]. The matrices Hϵ_{ij}(a), l_{ij}(a), r_{ij}(a), Hm_{i}(ζ) and r_{i}(ζ) are called elementary matrices of GH_{2n}(R, a_{1}, ⋯, a_{r}).

Remark. In the definition of the elementary matrix r_{i}(ζ), if we use β′_{22} instead of β_{22}, where β′_{22} is the (n − r) × (n − r) matrix whose (i − r, i − r)-entry is ζ_{f} and other entries are 0, the resulting matrix r′_{i}(ζ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}) since β̄_{12}A_{1}β_{12} = β′_{22} + λβ̄′_{22}. It might seem better to use r′_{i}(ζ) rather than r_{i}(ζ). However, this would cause trouble in Lemma 9.9 and so we shall not use r′_{i}(ζ).

Lemma 8.1 EH_{2n}(R, a_{1}, ⋯, a_{r}) has generators

Hϵ_{ij}(a) r + 1 ≤ i ≤ n; 1 ≤ j ≤ n; i ≠ j
l_{ij}(a) 1 ≤ i, j ≤ n
r_{ij}(a) r + 1 ≤ i, j ≤ n
Hm_{j}(ζ) r + 1 ≤ j ≤ n
r_{j}(ζ) r + 1 ≤ j ≤ n

where ζ ∈ C = {^{t}(x_{1}, x_{2}, ⋯, x_{r}) ∈ R^{r} | ∑_{i=1}^{r} x̄_{i}a_{i}x_{i} ≡ 0 mod min^{λ}(R)}.

Proof. It is obvious that all the matrices above are in EH_{2n}(R, a_{1}, ⋯, a_{r}). Conversely, any matrix

( \begin{matrix}I & α_{12} & 0 & 0 \\ 0 & I & 0 & 0 \\ 0 & −Ā_{1}α_{12} & I & 0 \\ 0 & γ_{22} & −ᾱ_{12} & I \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r})

is easily decomposed as a product

∏_{i} Hm_{i}(ζ_{i}) ∏_{i≤j} l_{ij}(a_{ij})

where ζ_{i} is the (i − r)-th column of α_{12}, i = r + 1, ⋯, n.

Similarly, any matrix

( \begin{matrix}I & 0 & 0 & β_{12} \\ 0 & I & −λβ̄_{12} & β_{22} \\ 0 & 0 & I & −Ā_{1}β_{12} \\ 0 & 0 & 0 & I \end{matrix} ) ∈ EH_{2n}(R, a_{1}, ⋯, a_{r})

is easily decomposed as a product

∏_{i} r_{i}(ζ_{i}) ∏_{i≤j} r_{ij}(a_{ij})

where ζ_{i} is the (i − r)-th column of β_{12}, i = r + 1, ⋯, n. □ The subscripts of Hϵ_{ij}(a), l_{ij}(a), r_{ij}(a), Hm_{j}(ζ) and r_{j}(ζ) have certain restrictions in the Lemma 8.1. In the sequel, we shall always assume these restriction are satisfied. Next we provide the list of identities promised for EH_{2n}(R, a_{1}, ⋯, a_{r}). Some of them are taken from [Bak, p.43 ]. Lemma 8.2 Let Hϵ_{ij}(a), l_{ij}(a), r_{ij}(a), Hm_{j}(ζ) and r_{j}(ζ) be as in the Lemma 8.1, and ζ = (x_{1}, ⋯, x_{r}) ∈ C. Then the following equations hold.

1. [r_{ij}(a), l_{kl}(b)] = 1 ({i, j} ∩ {k, l} = ∅)

2. [r_{ij}(a), l_{ik}(b)] = Hϵ_{jk}(−λāb) (i, j, and k distinct)

3. [r_{ij}(a), l_{jj}(b)] = Hϵ_{ij}(ab)r_{ii}(−λabā) (i ≠ j)

4. [r_{jj}(b), l_{ji}(a)] = Hϵ_{ji}(ba)l_{ii}(λ̄āba) (i ≠ j)

5. [r_{i}(ζ), l_{jk}(a)] = 1 (i ≠ j, k and j, k ≥ r + 1)

6. [r_{i}(ζ), l_{ik}(a)] = Hm_{k}(ζa)Hϵ_{ik}(−ζ_{f}a)l_{kk}(āζ̄_{f}a − \overline{(ζa)}_{f}) (i ≠ k; k ≥ r + 1)

7. [r_{i}(ζ), l_{jk}(a)] = Hϵ_{ij}(x̄_{k}ā) (i ≠ j; j ≥ r + 1; k ≤ r)

8. [r_{i}(ζ), l_{jk}(a)] = Hϵ_{ij}(x̄_{k}ā)Hϵ_{ik}(−λx̄_{j}a)r_{ii}(x̄_{k}āx_{j} − λx̄_{j}ax_{k}) (j ≠ k; j, k ≤ r)

9. [r_{i}(ζ), l_{jj}(a)] = Hϵ_{ij}(−λx̄_{j}a)r_{ii}(−λx̄_{j}ax_{j}) (j ≤ r) E1. Hϵ_{ij}(a + b) = Hϵ_{ij}(a)Hϵ_{ij}(b) (i ≠ j)

E2. [Hϵ_{ij}(a), Hϵ_{kl}(b)] = 1 (i ≠ l and j ≠ k)

E3. [Hϵ_{ij}(a), Hϵ_{jk}(b)] = Hϵ_{ik}(ab) (i, j, and k distinct )

E4. [Hm_{i}(ζ), Hm_{j}(ξ)] = l_{ij}(ζ̄Ā_{1}ξ) (i ≠ j)

E5. Hm_{i}(ζ)Hm_{i}(ξ) = Hm_{i}(ζ + ξ)l_{ii}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ − \overline{(ζ + ξ)}_{f})

E6. [Hm_{i}(ζ), Hϵ_{jk}(a)] = Hϵ_{ji}(−ax_{k}) (i ≠ j; k ≤ r)

E7. [Hm_{i}(ζ), Hϵ_{jk}(a)] = 1 (i ≠ j; k ≥ r + 1)

E8. [Hm_{i}(ζ), Hϵ_{ik}(a)] = Hm_{k}(ζa)l_{ki}(āζ̄_{f})l_{kk}(āζ̄_{f}a − (\overline{ζa})_{f}) (k ≥ r + 1)

L1. l_{ij}(a + b) = l_{ij}(a)l_{ij}(b)

L2. l_{ij}(a)l_{kl}(b) = l_{kl}(b)l_{ij}(a)

L3. l_{ij}(a) = l_{ji}(−λ̄ā)

L4. [l_{ij}(a), Hϵ_{kl}(b)] = 1 (k ≠ i, j, l)

L5a. [l_{ij}(a), Hϵ_{jk}(b)] = l_{ik}(ab) (i, j, and k distinct )

L5b. [l_{ij}(a), Hϵ_{ji}(b)] = l_{ii}(ab − λ̄b̄ā) (i ≠ j)

L6. [l_{ii}(a), Hϵ_{ik}(b)] = l_{ik}(ab)l_{kk}(b̄ab) (i ≠ k)

L7a. [Hm_{i}(ζ), l_{jk}(a)] = l_{ji}(−ax_{k})l_{ki}(λ̄āx_{j})l_{ii}(−x̄_{j}ax_{k} + λ̄x̄_{k}āx_{j}) (j ≠ k, j, k ≤ r)

L7b. [Hm_{i}(ζ), l_{jj}(a)] = l_{ji}(−ax_{j})l_{ii}(−x̄_{j}ax_{j}) (j ≤ r)

L8a. [Hm_{i}(ζ), l_{jk}(a)] = l_{ik}(−x̄_{j}a) (k ≥ r + 1; j ≤ r; i ≠ k)

L8b. [Hm_{i}(ζ), l_{jk}(a)] = l_{ij}(λ̄x̄_{k}ā) (k ≤ r; j ≥ r + 1; i ≠ j)

L9. [Hm_{i}(ζ), l_{ji}(a)] = l_{ii}(−x̄_{j}a + λ̄āx_{j}) (j ≤ r)

L10. [Hm_{i}(ζ), l_{jk}(a)] = 1 (j, k ≥ r + 1)

R1. r_{ij}(a + b) = r_{ij}(a)r_{ij}(b)

R2. r_{ij}(a)r_{kl}(b) = r_{kl}(b)r_{ij}(a)

R3. r_{ij}(a) = r_{ji}(−λā) R4. [r_{ij}(a), Hϵ_{kl}(b)] = 1 (l ≠ i, j, k) R5a. [r_{ij}(a), Hϵ_{kj}(b)] = r_{ik}(−ab̄) (i, j, and k distinct ) R5b. [r_{ij}(a), Hϵ_{ij}(b)] = r_{ii}(λbā − ab̄) (i ≠ j) R6. [r_{ii}(a), Hϵ_{ki}(b)] = r_{ik}(−ab̄)r_{kk}(bab̄) (i ≠ k) R7. [r_{i}(ζ), r_{j}(ξ)] = r_{ij}(ζ̄A_{1}ξ) (i ≠ j) R8. r_{i}(ζ)r_{i}(ξ) = r_{i}(ζ + ξ)r_{ii}((ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ) R9. [Hm_{i}(ζ), r_{jk}(a)] = 1 (i ≠ j, k) R10. [Hm_{i}(ζ), r_{ik}(a)] = r_{k}(ζa)Hϵ_{ki}(λāζ̄_{f})r_{kk}((ζa)_{f} − āζ_{f}a) (i ≠ k) R11. [Hm_{i}(ζ), r_{ji}(a)] = r_{j}(−λζā)Hϵ_{ji}(−aζ̄_{f})r_{jj}(λaζ̄_{f} ā + \overline{(ζā)}_{f}) (i ≠ j) R12. [r_{i}(ζ), r_{jk}(a)] = 1 R13. [r_{k}(ζ), Hϵ_{ij}(a)] = r_{ik}(−ax_{j}) (i ≠ k; j ≤ r) R14. [r_{i}(ζ), Hϵ_{ij}(a)] = r_{ii}(−ax_{j} + λx̄_{j}ā)) (j ≤ r) R15. [r_{k}(ζ), Hϵ_{ij}(a)] = 1 (j ≥ r + 1; j ≠ i, k) R16. [r_{j}(ζ), Hϵ_{ij}(a)] = r_{i}(−ζā)r_{ji}(ζ_{f} ā)r_{ii}((ζā)_{f} − aζ_{f} ā) (j ≥ r + 1; i ≠ j) R17. [Hm_{i}(ζ), r_{j}(ξ)] = Hϵ_{ji}(−ξ̄A_{1}ζ) (i ≠ j)

Proof A tedious calculation. □ Remark For a λ-form parameter Λ_{1}, [H-O’M] uses seven relations and one family of generators for EQ_{2n}(R, Λ_{1}) to take the place of the 21 (resp. 28) relations and three families of generators used in [K-M 1,2] (resp. [Bak]). We prefer using three families of generators Hϵ_{ij}(a), l_{ij}(a) and r_{ij}(a) and 28 relations as in [Bak] instead of using the 7 relations in [H-O’M], because the subscripts of Bak fit better to our approach. Lemma 8.3 If n − r ≥ 1 then 1) EH_{2n}(R, a_{1}, ⋯, a_{r}) is generated by the elementary matrices having

n as one of their subscripts, and 2) GH_{2(n−1)}(R, a_{1}, ⋯, a_{r}) normalizes EH_{2n}(R, a_{1}, ⋯, a_{r}).

Proof 1) follows from L5a), L6, E3, E8, R5a), R6 and R10 in Lemma 8.2.

2) follows from 1) and direct matrix calculation. □

Theorem 8.4 If n − r ≥ 3 then EH_{2n}(R, a_{1}, ⋯, a_{r}) and EH(R, a_{1}, ⋯, a_{r}) are perfect.

Proof. This follows directly from E3, L5a), L6, E8, R5a), R6 and R16. (In chapter 1, we proved this theorem by using different calculations. However, the current calculations will be valid in Hermitian Steinberg groups). □

9. Hermitian Steinberg Groups

Since any perfect group has a (unique up to isomorphism) perfect, universal, central extension, it is natural to ask what the universal central extensions of EH_{2n}(R, a_{1}, ⋯, a_{r}) and EH(R, a_{1}, ⋯, a_{r}) are ? In this section, we define Hermitian Steinberg groups StH_{2n}(R, a_{1}, ⋯, a_{r}) and StH(R, a_{1}, ⋯, a_{r}) and derive some results which will be used in §10 to prove that StH(R, a_{1}, ⋯, a_{r}) is a perfect, universal, central extension of EH(R, a_{1}, ⋯, a_{r}).

Definition 9.1 Let

StH_{2n}(R, a_{1}, ⋯, a_{r})

denote the free group generated by the symbols l̃_{ij}(a), r̃_{ij}(a), Hϵ̃_{ij}(a), Hm̃_{i}(ζ) and r̃_{i}(ζ) modulo the relations given in Lemma 8.2. Thus, by definition, in the case of r̃_{ij}(a) (resp. l̃_{ij}(a) ) where i = j, the element a ∈ max^{−λ}( resp. a ∈ \overline{max^{−λ}}), and in the case of H̃m_{i}(ζ) and r̃_{i}(ζ), the element ζ ∈ C. StH_{2n}(R, a_{1}, ⋯, a_{r}) is called the 2n-th Hermitian Steinberg group of the elements a_{1}, ⋯, a_{r}. It is obvious that there is a natural homomorphism

StH_{2n}(R, a_{1}, ⋯, a_{r}) ⟶ StH_{2(n+1)}(R, a_{1}, ⋯, a_{r})

which covers the natural homomorphism

EH_{2n}(R, a_{1}, ⋯, a_{r}) ↪ EH_{2(n+1)}(R, a_{1}, ⋯, a_{r}).

Define

StH(R, a_{1}, ⋯, a_{r}) = lim_{n^{⟶}} StH_{2n}(R, a_{1}, ⋯, a_{r}).

StH(R, a_{1}, ⋯, a_{r}) is called the Hermitian Steinberg group of the elements a_{1}, ⋯, a_{r}.

The major task in this chapter is to prove that StH(R, a_{1}, ⋯, a_{r}) is a perfect, universal, central extension of EH(R, a_{1}, ⋯, a_{r}). We shall use the concepts G-perfect and G-central extension used in [Bak, p.49]. We shall also use the commutator formulas in Lemma 9.1 below, without explicit mention.

Lemma 9.2 a) If x, y, and z are elements of a group, let x^{y} = y^{−1}xy and [x, y] = x^{−1}y^{−1}xy. Then [y, x] = [x, y]^{−1}, [xy, z] = ([x, z]^{y})[y, z], and [x, y]^{z} = [x^{z}, y^{z}] = [x^{z}, [z, y^{−1}]y]. In particular, if z commutes with y then [zx, y] = [x, y].

b) Let M and G be groups together with a fixed group homomorphism G ⟶ Aut(M) which provides M with a G-action m ⟼ m^{g}. We call M a G-group. If m, n ∈ M and f, g ∈ G, let [m, g] = m^{−1}m^{g}. Then [mn, f] = ([m, f]^{n})[n, f], [m, fg] = [m, g][m, f]^{g}, [m, f]^{g} = [m^{g}, f^{g}], and [m, f]^{n} = [m^{n}, [n, f^{−1}]f]. In particular, if n is fixed by f then [nm, f] = [m, f].

c) Let M and N be G-groups and let ϕ : M ⟶ N be a G-central extension of N. If x, y ∈ M such that ϕ(x) = ϕ(y) and g ∈ G then [x, g] = [y, g].

It is not too difficult to prove the centrality of extension StH(R, a_{1}, ⋯, a_{r}) ⟶ EH(R, a_{1}, ⋯, a_{r}). But its universality is more difficult to establish and its proof is more complicated, than that of the corresponding result in the quadratic case StQ(R, Λ) ⟶ EQ(R, Λ). In the Hermitian case, 5 families of generators and 48 relations take the place of 3 families of generators and 28 relations in the quadratic case. Another complication is that certain symmetries enjoyed by the generators of EQ(R, Λ) are not valid for those of EH(R, a_{1}, ⋯, a_{r}). For example, r_{ij}(a) ∉ EH(R, a_{1}, ⋯, a_{r}) but l_{ij}(a) ∈ EH(R, a_{1}, ⋯, a_{r}) for i ≤ r.

To prove the universality, it suffices to show that for any central extension f : F ⟶ EH(R, a_{1}, ⋯, a_{r}), we can find a set of pre-images in F of all the generators Hϵ_{ij}(a), r_{ij}(a), l_{ij}(a), Hm_{i}(ξ) and r_{i}(ξ) of EH(R, a_{1}, ⋯, a_{r}), which satisfy the relations in Lemma 8.2. We shall use the universality of the extension St(R) ⟶ E(R) to find the pre-images of the matrices Hϵ_{ij}(a), where i, j ≥ r + 1. We shall use these pre-images to find pre-images of the Hϵ_{ij}(a), where 1 ≤ j ≤ r (Lemma 9.2). We shall use all of the preimages above to find the pre-images of the l_{ij}(a), where i, j ≥ 1 (Lemma 9.4), pre-images of the Hm_{i}(ξ) (Lemma 9.3) and pre-images of the r_{i}(ξ) and r_{ij}(a) (Lemma 9.5). Finally, we check that all of these pre-images satisfy the relations mentioned in Lemma 8.2.

Let

EH_{n,r}(R) = ⟨ϵ_{ij}(a) | r + 1 ≤ i ≤ n; 1 ≤ j ≤ n; i ≠ j; a ∈ R ⟩.

This group is a subgroup of the usual elementary subgroup E_{n}(R) of GL_{n}(R). Define an action of EH_{n,r}(R) on the elementary Hermitian group EH_{2n}(R, a_{1}, ⋯, a_{r}) by putting

u^{x} = ( x^{−1} ; x̄ ) u ( \begin{matrix}x \\ & x̄^{−1} \end{matrix} ) , where x ∈ EH_{n,r}(R), u ∈ EH_{2n}(R, a_{1}, ⋯, a_{r}).

Then EH_{2n}(R, a_{1}, ⋯, a_{r}) is an EH_{n,r}(R)-group.

Denote by E_{n,r}(R) the subgroup of EH_{n,r}(R) generated by all ϵ_{ij}(a), where r + 1 ≤ i, j ≤ n; i ≠ j and set

EH_{∞,r}(R) = lim_{n→∞} EH_{n,r}(R)

E_{∞,r}(R) = lim_{n→∞} E_{n,r}(R)

Lemma 9.3 If n − r ≥ 5 then H_{n} = ⟨Hϵ_{ij}(a) | r + 1 ≤ i ≤ n; 1 ≤ j ≤ r; a ∈ R ⟩ is an E_{n,r}(R)-perfect, universal, E_{n,r}(R)-central extension of itself.

Proof. The proof is essentially the same, but a bit easier, than that of a similar result in [Bak, (3.28)]. Details are left to the reader. □

Corollary 9.4 Let H_{n} be as in Lemma 9.3. Set H_{∞} = lim_{n→∞} H_{n}. Then H_{∞} is an E_{∞,r}(R)-perfect, universal, E_{∞,r}(R)-central extension of itself.

Proof This is a direct consequence of Lemma 9.2. □

Lemma 9.5 If n − r ≥ 5 then Hm_{n} = ⟨ Hm_{i}(ζ), l_{ij}(a) | r + 1 ≤ i, j ≤ n; ζ ∈ C⟩ is an E_{n,r}(R)-perfect, universal, E_{n,r}(R)-central extension of itself.

Proof: The generators Hm_{i}(ζ) and l_{ij}(a) satisfy the following relations:

E4. [Hm_{i}(ζ), Hm_{j}(ξ)] = l_{ij}(ζ̄Ā_{1}ξ)  (i ≠ j)
E5. Hm_{i}(ζ)Hm_{i}(ξ) = Hm_{i}(ζ + ξ)l_{ii}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ − \overline{(ζ + ξ)}_{f})
E7. [Hm_{i}(ζ), ϵ_{jk}(a)] = 1  (i ≠ j; k ≥ r + 1)
E8. [Hm_{i}(ζ), ϵ_{ik}(a)] = Hm_{k}(ζa)l_{ki}(āζ̄_{f})l_{kk}(āζ̄_{f}a − (\overline{ζa})_{f}) (k ≥ r + 1)
L1. l_{ij}(a + b) = l_{ij}(a)l_{ij}(b)
L2. l_{ij}(a)l_{kl}(b) = l_{kl}(b)l_{ij}(a)
L3. l_{ij}(a) = l_{ji}(−λ̄ā)
L4. [l_{ij}(a), ϵ_{kl}(b)] = 1  (k ≠ i, j, l)
L5a. [l_{ij}(a), ϵ_{jk}(b)] = l_{ik}(ab)  (i, j, and k distinct )
L5b. [l_{ij}(a), ϵ_{ji}(b)] = l_{ii}(ab − λ̄b̄ā)  (i ≠ j)
L6. [l_{ii}(a), ϵ_{ik}(b)] = l_{ik}(ab)l_{kk}(b̄ab)  (i ≠ k)
L10. [Hm_{i}(ζ), l_{jk}(a)] = 1  (j, k ≥ r + 1)

Relations E8, L5a), and L6 show that Hm_{n} is E_{n,r}(R)-perfect. Suppose ϕ : F ⟶ Hm_{n} is an E_{n,r}(R)-central extension of Hm_{n}. Since L_{n} = ⟨ l_{ij}(a) | r + 1 ≤ i, j ≤ n ⟩ is the E_{n,r}(R)-perfect, universal, E_{n,r}(R)-central extension of itself (see[Bak, (3.28) ]), there is an E_{n,r}(R)-equivariant homomorphism h : L_{n} ⟶ ϕ^{−1}(L) suth that

[figure: L_{n}; Hm_{n}; ϕ^{−1}(L_{n}); L_{n}; h; ϕ; id]

commutes. Let L_{ij}(a) = h(l_{ij}(a)). Then ϕ(L_{ij}(a)) = l_{ij}(a) and L_{ij}(a) satisfy L1, L2, L3, L4, L5a, L5b, and L6.

In what follows we will determine elements H_{i}(ζ) in F which satisfy a version of E4—E8 and L10 where Hm_{i}(ζ) and l_{ij}(a) are replaced respectively by H_{i}(ζ) and L_{ij}(a). Obviously, we can use E8 to define such elements H_{i}(ζ) in F as follows.

For each Hm_{i}(ζ) ∈ Hm_{n}, pick a lifting x_{i}(ζ) and define, for i ≠ p, p ≥ r + 1, H_{i}^{p}(ζ) = [x_{p}(ζ), ϵ_{pi}(1)]L_{ip}(−ζ̄_{f}).

Thus H_{i}^{p}(ζ) does not depend on the choice of x_{p}(ζ) and H_{i}^{p}(ζ) = z_{1}H_{i}^{l}(ζ) = z_{2}x_{i}(ζ), where z_{1}, z_{2} ∈ ker ϕ. It follows that whether E4, E5, E7, and L10 hold or not is independent of the choice of p in the definition of H_{i}^{p}(ζ). So we can choose any p ≥ r + 1 which is convenient for the proof of E4, E5, E7, and L10. To simplify notation, we shall write H_{i}(ζ) in place of H_{i}^{p}(ζ) while verifying E4, E5, E7, and L10.

L10). Choose p ≠ i, k, and p ≥ r+1. Then [H_{i}(ζ), L_{jk}(a)] = [[x_{p}(ζ), ϵ_{pi}(1)]L_{ip}(−ζ̄_{f}), L_{jk}(a)] = [[x_{p}(ζ), ϵ_{pi}(1)], L_{jk}(a)]^{L_{ip}(−ζ̄_{f})}[L_{ip}(−ζ̄_{f}), L_{jk}(a)]. Since [x_{p}(ζ), L_{jk}(a)] ∈ ker ϕ, [ϵ_{pi}(1), L_{jk}(a)] = 1, it follows that [[x_{p}(ζ), ϵ_{pi}(1)], L_{jk}(a)] = 1. Thus [H_{i}(ζ), L_{jk}(a)] = [L_{ip}(−ζ̄_{f}), L_{jk}(a)] = 1.

Next we want to prove a fact needed in the calculations below.

(A) x and L_{ij}(a) commute for any x ∈ F, r + 1 ≤ i, j ≤ n.

It is easy to see that any element in Hm_{n} can be uniquely expressed as a product

Hm_{r+1}(ζ_{r+1})Hm_{r+2}(ζ_{r+2}) ⋯ Hm_{n}(ζ_{n}) ∏_{i≥j} l_{ij}(a_{ij}).

Thus

x = zH_{r+1}(ζ_{r+1})H_{r+2}(ζ_{r+2}) ⋯ H_{n}(ζ_{n}) ∏_{i≥j} L_{ij}(a_{ij})

where z ∈ ker ϕ. It follows now from L10 and L2, which have already been verified, that [x, L_{ij}(a)] = 1.

E7). Choose p ≠ i, j, or k, and p ≥ r + 1. Then [H_{i}(ζ), ϵ_{jk}(a)] = [[x_{p}(ζ), ϵ_{pi}(1)]L_{ip}(−ζ̄_{f}), ϵ_{jk}(a)] = ( using (A) )[[x_{p}(ζ), ϵ_{pi}(1)], ϵ_{jk}(a)][L_{ip}(−ζ̄_{f}), ϵ_{jk}(a)] = 1, since [x_{p}(ζ), ϵ_{jk}(a)] ∈ ker ϕ, [ϵ_{pi}(1), ϵ_{jk}(a)] = 1 and [L_{ip}(−ζ̄_{f}), ϵ_{jk}(a)] = 1.

E4). Choose p ≠ i, j, and p ≥ r + 1. Then

[H_{i}(ζ), H_{j}(ξ)] = [[x_{p}(ζ), ϵ_{pi}(1)]L_{ip}(−ζ̄_{f}), H_{j}(ξ)] = ( using (A) )[[x_{p}(ζ), ϵ_{pi}(1)], H_{j}(ξ)]

= [x_{p}(ζ), ϵ_{pi}(1)]^{−1}[x_{p}(ζ), ϵ_{pi}(1)]^{H_{j}(ξ)} = (E7) [x_{p}(ζ), ϵ_{pi}(1)]^{−1}[x_{p}(ζ)^{H_{j}(ξ)}, ϵ_{pi}(1)].

Since ϕ(x_{p}(ζ)^{H_{j}(ξ)}) = ϕ(x_{p}(ζ))l_{pj}(ζ̄Ā_{1}ξ)), it follows that

[x_{p}(ζ)^{H_{j}(ξ)}, ϵ_{pi}(1)] = [x_{p}(ζ)L_{pj}(ζ̄Ā_{1}ξ), ϵ_{pi}(1)]
= ( using (A) )[x_{p}(ζ), ϵ_{pi}(1)][L_{pj}(ζ̄Ā_{1}ξ), ϵ_{pi}(1)] = (L3, L5)[x_{p}(ζ), ϵ_{pi}(1)]L_{ij}(ζ̄Ā_{1}ξ).

This completes the proof of E4.

E5). Choose p ≠ i and p ≥ r + 1. Then

H_{i}(ζ)H_{i}(ξ) = [x_{p}(ζ), ϵ_{pi}(1)]L_{ip}(−ζ̄_{f})[x_{p}(ξ), ϵ_{pi}(1)]L_{ip}(−ξ̄_{f})
= ( using (A) and L1 )[x_{p}(ζ), ϵ_{pi}(1)][x_{p}(ξ), ϵ_{pi}(1)]L_{ip}(−ζ̄_{f} − ξ̄_{f}).

By a direct computation it follows that

[x_{p}(ζ), ϵ_{pi}(1)][x_{p}(ξ), ϵ_{pi}(1)] = [[x_{p}(ζ), ϵ_{pi}(1)]^{−1}, x_{p}(ξ)][x_{p}(ζ)x_{p}(ξ), ϵ_{pi}(1)].

Since

ϕ([x_{p}(ζ), ϵ_{pi}(1)]^{−1}) = [Hm_{p}(ζ), Hϵ_{pi}(1)]^{−1} = Hm_{i}(−ζ)l_{ip}(−ζ̄_{f})l_{ii}(ζ̄A_{1}ζ),

it follows from (A) and E4 that

[[x_{p}(ζ), ϵ_{pi}(1)]^{−1}, x_{p}(ξ)] = [x_{i}(−ζ), x_{p}(ξ)] = [H_{i}(−ζ), H_{p}(ξ)] = L_{ip}(−ζ̄Ā_{1}ξ).

Similarly, using (A) and L6 it follows that

[x_{p}(ζ)x_{p}(ξ), ϵ_{pi}(1)] = [x_{p}(ζ + ξ)L_{pp}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ − \overline{(ζ + ξ)}_{f}), ϵ_{pi}(1)]
= [x_{p}(ζ + ξ), ϵ_{pi}(1)][L_{pp}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ̄− \overline{(ζ + ξ)}_{f}), ϵ_{pi}(1)]

= H_{i}(ζ + ξ)L_{ip}(\overline{(ζ + ξ)}_{f})L_{pi}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ − \overline{(ζ + ξ)}_{f})L_{ii}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ − \overline{(ζ + ξ)}_{f}).

Noting that λ(ζ + ξ)_{f} + \overline{(ζ + ξ)}_{f} = \overline{(ζ + ξ)}Ā_{1}(ζ + ξ), λζ_{f} + ζ̄_{f} = ζ̄Ā_{1}ζ, λξ_{f} + ξ̄_{f} = ξ̄Ā_{1}ξ and using (A), L10, L1, L2, and L3, we deduce from the above that

H_{i}(ζ)H_{i}(ξ) = H_{i}(ζ + ξ)L_{ii}(ζ̄_{f} + ξ̄_{f} + ζ̄Ā_{1}ξ − \overline{(ζ + ξ)}_{f}).

This completes the proof of E5.

To prove E8) and that H_{i}^{p}(ζ) is independent of the choice of p, we only need to prove for p, q ≥ r + 1 and p ≠ q,

(B) [x_{p}(ζ), ϵ_{pi}(a)]L_{iq}(āζ̄_{f}a) = [x_{q}(ζa), ϵ_{qi}(1)]L_{ip}(āζ̄_{f})L_{ii}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{qi}(āζ̄_{f}a − \overline{(ζa)}_{f}).

To start the verification of (B), set e = ϵ_{qi}(1) and f = ϵ_{pq}(a). Then [e, f] = [f, e]^{−1} = ϵ_{pi}(a)^{−1} = g^{−1}, where g = ϵ_{pi}(a). Let y = [x_{p}(ζ), f]. Then

y^{e} = [x_{p}(ζ)^{e}, f^{e}] = [x_{p}(ζ), fg] = [x_{p}(ζ), gf] = [x_{p}(ζ), f][x_{p}(ζ), g]^{f} = y[x_{p}(ζ), g]^{f}.

Thus

[y, e] = [x_{p}(ζ), g]^{f}.

Using ϕ(y) = [Hm_{p}(ζ), Hϵ_{pq}(a)] = Hm_{q}(ζa)l_{qp}(āζ̄_{f})l_{qq}(āζ̄_{f}a − \overline{(ζa)}_{f}), we obtain

[y, e] = [x_{q}(ζa)L_{qp}(āζ̄_{f})L_{qq}(āζ̄_{f}a − \overline{(ζa)}_{f}), ϵ_{qi}(1)]
= ( using (A) )[x_{q}(ζa), ϵ_{qi}(1)][L_{qp}(āζ̄_{f}), ϵ_{qi}(1)][L_{qq}(āζ̄_{f}a − \overline{(ζa)}_{f}), ϵ_{qi}(1)]
=(L5a, L3, L6)[x_{q}(ζa), ϵ_{qi}(1)]L_{ip}(āζ̄_{f})L_{qi}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ii}(āζ̄_{f}a − \overline{(ζa)}_{f}).

On the other hand, from

ϕ([x_{p}(ζ), g]) = [Hm_{p}(ζ), Hϵ_{pi}(a)] = Hm_{i}(ζa)l_{ip}(āζ̄_{f})l_{ii}(āζ̄_{f}a − \overline{(ζa)}_{f})

and (A), we obtain

[x_{p}(ζ), g]^{f} = [x_{p}(ζ), g][[x_{p}(ζ), g], f] = [x_{p}(ζ), ϵ_{pi}(a)][x_{i}(ζa)L_{ip}(āζ̄_{f})L_{ii}(āζ̄_{f}a − \overline{(ζa)}_{f}), ϵ_{pq}(a)]

=(E7, L5a, L4) [x_{p}(ζ), ϵ_{pi}(a)]L_{iq}(āζ̄_{f}a).

Hence we have established (B). Setting a = 1 in (B) shows that H_{i}^{p}(ζ) does not depend on the choice of p. E8) follows easily from (B) by using the identity ζ̄A_{1}ζ = ζ_{f} + λζ̄_{f}. □

Corollary 9.6 Keep the notion Hm_{n} as in Lemma 9.5. Let Hm_{∞} = lim_{n→∞} Hm_{n}. Then Hm_{∞} is an E_{∞,r}(R)-perfect, universal, E_{∞,r}(R)-central extension of itself.

Proof This is a direct consequence of Lemma 9.5. □

Lemma 9.7 If n − r ≥ 5 then L_{n} = ⟨ l_{ij}(a) | 1 ≤ i, j ≤ n ⟩ is an EH_{n,r}(R)-perfect, universal, EH_{n,r}(R)-central extension of itself.

Proof. The generators l_{ij}(a) of L_{n} satisfy the following relations:

1. l_{ij}(a + b) = l_{ij}(a)l_{ij}(b)
2. l_{ij}(a)l_{kl}(b) = l_{kl}(b)l_{ij}(a)
3. l_{ij}(a) = l_{ji}(−λ̄ā)
4. [l_{ij}(a), ϵ_{kl}(b)] = 1  (k ≠ i, j, l)
5a. [l_{ij}(a), ϵ_{jk}(b)] = l_{ik}(ab)  (i, j, and k distinct )
5b. [l_{ij}(a), ϵ_{ji}(b)] = l_{ii}(ab − λ̄b̄ā)  (i ≠ j)
6. [l_{ii}(a), ϵ_{ik}(b)] = l_{ik}(ab)l_{kk}(b̄ab)  (i ≠ k)

where for ϵ_{kl}(b), we always assume that r + 1 ≤ k ≤ n.

Relations 5a) and 6) show that L_{n} is EH_{n,r}(R)-perfect.

Suppose f : F ⟶ L_{n} is an EH_{n,r}(R)-central extension of L_{n}. First, we prove that F is abelian.

For each l_{ij}(a) ∈ L_{n}, pick a lifting y_{ij}(a) ∈ F. Then F is generated by the y_{ij}(a)’s and ker f. We need only to prove that y_{ij}(a) and y_{kl}(b) commute for any 1 ≤ i, j, k, l ≤ n. If i ≠ j, pick p ≥ r + 1 such that p ≠ i, j, k, and l. Then

y_{ij}(a) = z[y_{ip}(a), ϵ_{pj}(1)] for some z ∈ ker f.

It follows that

y_{ij}(a)^{y_{kl}(b)} = z[y_{ip}(a)^{y_{kl}(b)}, [y_{kl}(b), ϵ_{pj}(1)^{−1}]ϵ_{pj}(1)] = z[y_{ip}(a), ϵ_{pj}(1)] = y_{ij}(a)

because [y_{ip}(a), y_{kl}(b)], [y_{kl}(b), ϵ_{pj}(1)^{−1}] ∈ ker f. If i = j, pick p ≥ r + 1 such that p ≠ i, j, k, l. Then

y_{ii}(a) = zy_{ip}(a)[y_{pp}(a), ϵ_{pi}(1)] for some z ∈ ker f.

Since it has been shown that y_{ip}(a) and y_{kl}(b) commute, it follows that

y_{ii}(a)^{y_{kl}(b)} = zy_{ip}(a)[y_{pp}(a), ϵ_{pi}(1)]^{y_{kl}(b)}
= zy_{ip}(a)[y_{pp}(a)^{y_{kl}(b)}, [y_{kl}(b), ϵ_{pi}(1)^{−1}]ϵ_{pi}(1)] = zy_{ip}(a)[y_{pp}(a), ϵ_{pi}(1)] = y_{ii}(a)

because [y_{pp}(a), y_{kl}(b)], [y_{kl}(b), ϵ_{pi}(1)^{−1}] ∈ ker f. Hence F is abelian.

For the remainder of this proof we will make without explicit mention frequent use of this fact, and in particular of the equation [xy, e] = [x, e][y, e], where x, y ∈ F and e ∈ EH_{n,r}(R).

Similarly, we can prove that

(C) [y_{ij}(a), ϵ_{kl}(b)] = 1 for any k ≠ i, j, or l.

Next we establish that

(D) [y_{ip}(a), ϵ_{pj}(b)] = [y_{ik}(ab), ϵ_{kj}(1)],

for any distinct i, j, p, and k such that p, k ≥ r + 1. Since

[y_{ip}(a), ϵ_{pk}(b)]^{ϵ_{kj}(1)} = ( using (C) )[y_{ip}(a), ϵ_{pk}(b)ϵ_{pj}(b)] = [y_{ip}(a), ϵ_{pj}(b)ϵ_{pk}(b)]
= [y_{ip}(a), ϵ_{pk}(b)][y_{ip}(a), ϵ_{pj}(b)]^{ϵ_{pk}(b)} = ( using (C) )[y_{ip}(a), ϵ_{pk}(b)][y_{ip}(a), ϵ_{pj}(b)],

it follows that

[y_{ip}(a), ϵ_{pj}(b)] = [[y_{ip}(a), ϵ_{pk}(b)], ϵ_{kj}(1)] = [y_{ik}(ab), ϵ_{kj}(1)].

This establishes (D).

For a ∈ R, i ≠ j, p ≠ i or j and p ≥ r + 1, define

L_{ij}(a) = [y_{ip}(a), ϵ_{pj}(1)].

Then L_{ij}(a) is independent of the choice of y_{ip}(a) and p by (D). That the L_{ij}(a)’s satisfy 5a) is also a consequence of (D). To define the elements L_{ii}(a), we need to establish

(E) [y_{ii}(a), ϵ_{ik}(b)]L_{jk}(b̄ab) = [y_{jj}(b̄ab), ϵ_{jk}(1)]L_{ik}(ab) where a = −λ̄ā.

This is verified as follows:

[y_{ii}(a), ϵ_{ij}(b)]^{ϵ_{jk}(1)} = [y_{ii}(a), ϵ_{ij}(b)ϵ_{ik}(b)] = [y_{ii}(a), ϵ_{ij}(b)][y_{ii}(a), ϵ_{ik}(b)]^{ϵ_{ij}(b)}.

Thus

[[y_{ii}(a), ϵ_{ij}(b)], ϵ_{jk}(1)] = [y_{ii}(a), ϵ_{ik}(b)]^{ϵ_{ij}(b)}.

[[y_{ii}(a), ϵ_{ij}(b)], ϵ_{jk}(1)] = [y_{ij}(ab)y_{jj}(b̄ab), ϵ_{jk}(1)] = [y_{ij}(ab), ϵ_{jk}(1)][y_{jj}(b̄ab), ϵ_{jk}(1)]
= L_{ik}(ab)[y_{jj}(b̄ab), ϵ_{jk}(1)] = [y_{jj}(b̄ab), ϵ_{jk}(1)]L_{ik}(ab)
[y_{ii}(a), ϵ_{ik}(b)]^{ϵ_{ij}(b)} = [y_{ii}(a)^{ϵ_{ij}(b)}, ϵ_{ik}(b)] = [y_{ii}(a)y_{ij}(ab)y_{jj}(b̄ab), ϵ_{ik}(b)]
= [y_{ii}(a), ϵ_{ik}(b)][y_{ij}(ab), ϵ_{ik}(b)] = [y_{ii}(a), ϵ_{ik}(b)][y_{ji}(−λ̄b̄ā), ϵ_{ik}(b)]
= [y_{ii}(a), ϵ_{ik}(b)]L_{jk}(−λ̄b̄āb) = [y_{ii}(a), ϵ_{ik}(b)]L_{jk}(b̄ab)

since a = −λ̄ā. This establishes E).

For a = −λ̄ā, p ≠ i and p ≥ r + 1, define

L_{ii}(a) = [y_{pp}(a), ϵ_{pi}(1)]L_{pi}(−a).

Then L_{ii}(a) is independent of the choice of y_{pp}(a) and p by (E). It follows from (E) that 6) holds.

To complete the proof of Lemma 9.7, we have to verify that the elements L_{ij}(a) satisfy all the relations 1)— 6). Relations 5a) and 6) have been proved already. Since F is abelian,

2) holds and 4) follows from (C). Relation 1) is easy to check. It remains to check relations 5b) and 3).

Consider relation 3) first. When i = j it is obvious, since a = −λ̄ā. If i ≠ j, choose p ≥ r + 1 and k ≥ r + 1 such that i, j, k, and p are distinct. From

L_{ki}(a)^{ϵ_{kj}(1)} = [y_{kp}(a), ϵ_{pi}(1)]^{ϵ_{kj}(1)} = [y_{kp}(a)y_{jp}(a), ϵ_{pi}(1)]
= [y_{kp}(a), ϵ_{pi}(1)][y_{jp}(a), ϵ_{pi}(1)] = L_{ki}(a)L_{ji}(a)

it follows that

L_{ji}(a) = [L_{ki}(a), ϵ_{kj}(1)] = [y_{ik}(−λ̄ā), ϵ_{kj}(1)] = L_{ij}(−λ̄ā).

We turn to the verification of 5b). Choose p ≥ r + 1 and k ≥ r + 1 such that i, k, and p are distinct. Note that

[y_{pk}(a), ϵ_{kp}(b)]^{ϵ_{pi}(1)} = [y_{pk}(a)y_{ik}(a), ϵ_{kp}(b)ϵ_{ki}(b)]
= [y_{pk}(a), ϵ_{kp}(b)ϵ_{ki}(b)][y_{ik}(a), ϵ_{kp}(b)ϵ_{ki}(b)](F is abelian )

= [y_{pk}(a), ϵ_{ki}(b)ϵ_{kp}(b)][y_{ik}(a), ϵ_{kp}(b)ϵ_{ki}(b)]
= [y_{pk}(a), ϵ_{kp}(b)][y_{pk}(a), ϵ_{ki}(b)]^{ϵ_{kp}(b)}[y_{ik}(a), ϵ_{ki}(b)][y_{ik}(a), ϵ_{kp}(b)]^{ϵ_{ki}(b)}
= [y_{pk}(a), ϵ_{kp}(b)]L_{pi}(ab)^{ϵ_{kp}(b)}[y_{ik}(a), ϵ_{ki}(b)]L_{ip}(ab)^{ϵ_{ki}(b)}
= [y_{pk}(a), ϵ_{kp}(b)][y_{ik}(a), ϵ_{ki}(b)]L_{ip}(ab − λ̄b̄ā)( usingF is abelian and 1) ).

It follows that

[y_{ik}(a), ϵ_{ki}(b)]L_{ip}(ab − λ̄b̄ā) = [[y_{pk}(a), ϵ_{kp}(b)], ϵ_{pi}(1)]
= [y_{pp}(ab − λ̄b̄ā), ϵ_{pi}(1)] = L_{ii}(ab − λ̄b̄ā)L_{ip}(ab − λ̄b̄ā).

This implies 5b). □

Corollary 9.8 Keep the notion L_{n} as in Lemma 9.7. Let L_{∞} = lim_{n→∞} L_{n}. Then L_{∞} is an EH_{∞,r}(R)-perfect, universal, EH_{∞,r}(R)-central extension of itself.

Proof This is a direct consequence of Lemma 9.7. □ Lemma 9.9 If n − r ≥ 5 then K_{n} = ⟨ r_{i}(ζ), r_{ij}(a) | r + 1 ≤ i, j ≤ n; ζ ∈ C ⟩ is an EH_{n,r}(R)-perfect, universal, EH_{n,r}(R)-central extension of itself. Proof. The generators r_{i}(ζ) and r_{ij}(a) of K_{n} satisfy the following relations:

R1. r_{ij}(a + b) = r_{ij}(a)r_{ij}(b)
R2. r_{ij}(a)r_{kl}(b) = r_{kl}(b)r_{ij}(a)
R3. r_{ij}(a) = r_{ji}(−λā)
R4. [r_{ij}(a), ϵ_{kl}(b)] = 1  (l ≠ i, j, k)
R5a. [r_{ij}(a), ϵ_{kj}(b)] = r_{ik}(−ab̄)  (i, j, and k distinct )
R5b. [r_{ij}(a), ϵ_{ij}(b)] = r_{ii}(λbā − ab̄)  (i ≠ j)
R6. [r_{ii}(a), ϵ_{ki}(b)] = r_{ik}(−ab̄)r_{kk}(bab̄)  (i ≠ k)
R7. [r_{i}(ζ), r_{j}(ξ)] = r_{ij}(ζ̄A_{1}ξ)  (i ≠ j)
R8. r_{i}(ζ)r_{i}(ξ) = r_{i}(ζ + ξ)r_{ii}((ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ)
R12. [r_{i}(ζ), r_{jk}(a)] = 1
R13. [r_{k}(ζ), ϵ_{ij}(a)] = r_{ik}(−ax_{j})  (i ≠ k; j ≤ r)
R14. [r_{i}(ζ), ϵ_{ij}(a)] = r_{ii}(λx̄_{j}ā − ax_{j})  (j ≤ r)
R15. [r_{k}(ζ), ϵ_{ij}(a)] = 1  (j ≥ r + 1; j ≠ i, k)
R16. [r_{j}(ζ), ϵ_{ij}(a)] = r_{i}(−ζā)r_{ji}(ζ_{f} ā)r_{ii}((ζā)_{f} − aζ_{f} ā) (j ≥ r + 1, i ≠ j)

Relations R5a), R6, and R16 show that K_{n} is EH_{n,r}(R)-perfect.

Suppose ϕ : F ⟶ K_{n} is an EH_{n,r}(R)-central extension of K_{n}. Since R_{n} = ⟨ r_{ij}(a) | r + 1 ≤ i, j ≤ n ⟩ is an E_{n,r}(R)-perfect, universal, E_{n,r}(R)-central extension of itself (see[Bak, (3.28)]), there is an E_{n,r}(R)-equivariant homomorphism h : R_{n} ⟶ ϕ^{−1}(R_{n}) such that

[figure: R_{n}; K_{n}; ϕ^{−1}(R_{n}); R_{n}; h; ϕ; id]

commutes. Let R_{ij}(a) = h(r_{ij}(a)) (i, j ≥ r + 1). Then clearly ϕ(R_{ij}(a)) = r_{ij}(a) and the elements R_{ij}(a) together with the elements ϵ_{kl}(b) (k, l ≥ r + 1) satisfy R1, R2, R3, R5a, R5b, and R6. Furthermore it is clear that the elements R_{ij}(a) together with the elements ϵ_{kl}(b) (k ≥ r + 1; l ≤ r) satisfy R1, R2, R3, R5a, R5b, and R6. It is easy to check that they also satisfy R4. This detail is left to the reader.

Next, we define a pre-image R_{i}(ζ) ∈ F for each r_{i}(ζ) ∈ K_{n} and show that the elements R_{i}(ζ) satisfy R7—R16. It is natural to define them using R16 as follows.

For each generator r_{i}(ζ) ∈ K_{n}, pick a lifting x_{i}(ζ) ∈ F. For i ≠ p, p ≥ r + 1, define

R_{i}^{p}(ζ) = [x_{p}(ζ), ϵ_{ip}(−1)]R_{pi}(ζ_{f}).

Clearly R_{i}^{p}(ζ) does not depend on the choice of x_{p}(ζ) for a fixed p.

Remark. In the definition of r_{i}(ζ), if one took β_{22} to be the matrix whose (i-r,i-r)-entry is ζ_{f} then the formula R16 would look like

R16′. [r_{j}(ζ), ϵ_{ij}(a)] = r_{i}(−ζā)r_{ji}(λζ̄_{f}a)r_{ii}((ζā)_{f} − λaζ̄_{f} ā) (j ≥ r + 1, i ≠ j).

In this case, R_{i}^{p}(ζ) must be defined as

R_{i}^{p}(ζ) = [x_{p}(ζ), ϵ_{ip}(−1)]R_{pi}(−λζ̄_{f})R_{ii}(λζ̄_{f} − ζ_{f}).

But this definition would obviously cause more trouble in the following computation.

Since R_{i}^{p}(ζ) = z_{1}R_{i}^{l}(ζ) = z_{2}x_{i}(ζ) = z_{3}R_{pi}(ζ_{f})[x_{p}(ζ), ϵ_{ip}(−a)] where z_{1}, z_{2}, z_{3} ∈ ker ϕ, it follows easily that whether R7, R8, R12, R13, R14, and R15 hold or not is independent of the choice of p in the definition of R_{i}^{p}(ζ). So we can choose any p ≥ r + 1, which is convenient for verifying R7, R8, R12, R13, R14, and R15. To simplify notation, we shall write R_{i}(ζ) in place of R_{i}^{p}(ζ) in the verifications of R7, R8, R12, R13, R14, and R15. R12). Choose p ≠ i, j, or k and p ≥ r + 1. Then [R_{i}(ζ), R_{jk}(a)] = [R_{pi}(ζ_{f})[x_{p}(ζ), ϵ_{ip}(−1)], R_{jk}(a)] = ( R2 )[[x_{p}(ζ), ϵ_{ip}(−1)], R_{jk}(a)] = 1, since [x_{p}(ζ), R_{jk}(a)] ∈ ker ϕ and [R_{jk}(a), ϵ_{ip}(−1)] = 1. This completes the proof of R12. Next we want to prove a fact needed in the calculations below. (C) x and R_{ij}(a) commute for any x ∈ F, r + 1 ≤ i, j ≤ n. It is easy to see that any element in K_{n} can be uniquely expressed as a product

r_{r+1}(ζ_{r+1})r_{r+2}(ζ_{r+2}) ⋯ r_{n}(ζ_{n}) ∏_{i≥j} r_{ij}(a_{ij}).

Thus

x = zR_{r+1}(ζ_{r+1})R_{r+2}(ζ_{r+2}) ⋯ R_{n}(ζ_{n}) ∏_{i≥j} R_{ij}(a_{ij}),

where z ∈ ker ϕ. It follows now from R12 and R2 that [x, R_{ij}(a)] = 1. In particular, R_{i}^{p}(ζ) = [x_{p}(ζ), ϵ_{ip}(−1)]R_{pi}(ζ_{f}) = R_{pi}(ζ_{f})[x_{p}(ζ), ϵ_{ip}(−1)].

R15). Suppose p ≠ i, j, or k and p ≥ r + 1. Then [R_{k}(ζ), ϵ_{ij}(a)] = [R_{pk}(ζ_{f})[x_{p}(ζ), ϵ_{kp}(−1)], ϵ_{ij}(a)] = ( R2, R4 )[[x_{p}(ζ), ϵ_{kp}(−1)], ϵ_{ij}(a)] = 1, because [x_{p}(ζ), ϵ_{ij}(a)] ∈ ker ϕ and [ϵ_{ij}(a), ϵ_{kp}(−1)] = 1. R7). Choose p ≠ i, j, and p ≥ r + 1. Then

[R_{i}(ζ), R_{j}(ξ)] = [R_{pi}(ζ_{f})[x_{p}(ζ), ϵ_{ip}(−1)], R_{j}(ξ)] = ( R12 )[[x_{p}(ζ), ϵ_{ip}(−1)], R_{j}(ξ)] = [x_{p}(ζ), ϵ_{ip}(−1)]^{−1}[x_{p}(ζ), ϵ_{ip}(−1)]^{R_{j}(ξ)} = ( R15 )[x_{p}(ζ), ϵ_{ip}(−1)]^{−1}[x_{p}(ζ)^{R_{j}(ξ)}, ϵ_{ip}(−1)].

Since ϕ(x_{p}(ζ)^{R_{j}(ξ)}) = r_{p}(ζ)r_{pj}(ζ̄A_{1}ξ), it follows that

[R_{i}(ζ), R_{i}(ξ)] = [x_{p}(ζ), ϵ_{ip}(−1)]^{−1}[x_{p}(ζ)R_{pj}(ζ̄A_{1}ξ), ϵ_{ip}(−1)]

=( using ( C ) )[R_{pj}(ζ̄A_{1}ξ), ϵ_{ip}(−1)]] = R_{ij}(ζ̄A_{1}ξ).

R8). Choose p ≠ i, and p ≥ r + 1. Then

R_{i}(ζ)R_{i}(ξ) = [x_{p}(ζ), ϵ_{ip}(−1)]R_{pi}(ζ_{f})[x_{p}(ξ), ϵ_{ip}(−1)]R_{pi}(ξ_{f})

= [x_{p}(ζ), ϵ_{ip}(−1)][x_{p}(ξ), ϵ_{ip}(−1)]R_{pi}(ζ_{f} + ξ_{f})

using ( C ) and R1. Now by a direct computation, one gets that

[x_{p}(ζ), ϵ_{ip}(−1)][x_{p}(ξ), ϵ_{ip}(−1)] = [[x_{p}(ζ), ϵ_{ip}(−1)]^{−1}, x_{p}(ξ)][x_{p}(ζ)x_{p}(ξ), ϵ_{ip}(−1)]. From

ϕ(x_{p}(ζ)x_{p}(ξ)) = r_{p}(ζ)r_{p}(ξ) = r_{p}(ζ + ξ)r_{pp}(ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ),

it follows that [x_{p}(ζ)x_{p}(ξ), ϵ_{ip}(−1)] = [x_{p}(ζ + ξ)R_{pp}(ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ζ̄A_{1}ξ), ϵ_{ip}(−1),]

= ( using (C) )[x_{p}(ζ + ξ), ϵ_{ip}(1)][R_{pp}(ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ), ϵ_{ip}(−1)]

= ( R6 )R_{i}(ζ + ξ)R_{pi}(−(ζ + ξ)_{f})R_{pi}(ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ)R_{ii}(ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ)

= R_{i}(ζ + ξ)R_{pi}(−ζ_{f} − ξ_{f} − ξ̄A_{1}ζ)R_{ii}(ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ).

From ϕ([x_{p}(ζ), ϵ_{ip}(−1)]^{−1}) = [r_{p}(ζ), Hϵ_{ip}(−1)]^{−1} = r_{i}(ζ)^{−1}r_{pi}(−ζ_{f})^{−1} = r_{i}(−ζ)r_{ii}(ζ_{f}−λζ̄_{f})r_{pi}(ζ_{f}), it follows that

[[x_{p}(ζ), ϵ_{ip}(−1)]^{−1}, x_{p}(ξ)] = [x_{i}(−ζ)R_{ii}(ζ_{f} − λζ̄_{f})R_{pi}(ζ_{f}), x_{p}(ξ)]

=( using (C) )[x_{i}(−ζ), x_{p}(ξ)] = [R_{i}(−ζ), R_{p}(ξ)] = ( R7 )R_{ip}(−ζ̄A_{1}ξ) = R_{pi}(ξ̄A_{1}ζ), using R3) and the equation A_{1} = λĀ_{1}. Hence R_{i}(ζ)R_{i}(ξ) = R_{pi}(ξ̄A_{1}ζ)R_{i}(ζ+ξ)R_{pi}(−ζ_{f}−ξ_{f}−ξ̄A_{1}ζ)R_{ii}((ζ+ξ)_{f}−ζ_{f}−ξ_{f}−ξ̄A_{1}ζ)R_{pi}(ζ_{f}+ξ_{f})

= R_{i}(ζ + ξ)R_{ii}((ζ + ξ)_{f} − ζ_{f} − ξ_{f} − ξ̄A_{1}ζ).

Thus R8 is proved. R13). Choose p ≠ i, j, k and p ≥ r + 1. Then [R_{k}(ζ), ϵ_{ij}(a)] = [[x_{p}(ζ), ϵ_{kp}(−1)]R_{pk}(ζ_{f}), ϵ_{ij}(a)]=( (C) and R4 )[[x_{p}(ζ), ϵ_{kp}(−1)], ϵ_{ij}(a)]

= [x_{p}(ζ), ϵ_{kp}(−1)]^{−1}[x_{p}(ζ), ϵ_{kp}(−1)]^{ϵ_{ij}(a)} = [x_{p}(ζ), ϵ_{kp}(−1)]^{−1}[x_{p}(ζ)^{ϵ_{ij}(a)}, ϵ_{kp}(−1)]

= [x_{p}(ζ), ϵ_{kp}(−1)]^{−1}[x_{p}(ζ)R_{ip}(−ax_{j}), ϵ_{kp}(−1)]=( (C) )[R_{ip}(−ax_{j}), ϵ_{kp}(−1)] = R_{ik}(−ax_{j}). R14). Choose p ≠ i, j, and p ≥ r + 1. Then

[R_{i}(ζ), ϵ_{ij}(a)] = [[x_{p}(ζ), ϵ_{ip}(−1)]R_{pi}(ζ_{f}), ϵ_{ij}(a)]=( (C) and R4)[[x_{p}(ζ), ϵ_{ip}(−1)], ϵ_{ij}(a)]

= [x_{p}(ζ), ϵ_{ip}(−1)]^{−1}[x_{p}(ζ), ϵ_{ip}(−1)]^{ϵ_{ij}(a)} = [x_{p}(ζ), ϵ_{kp}(−1)]^{−1}[x_{p}(ζ)^{ϵ_{ij}(a)}, ϵ_{ip}(−1)] = [x_{p}(ζ), ϵ_{ip}(−1)]^{−1}[x_{p}(ζ)R_{ip}(−ax_{j}), ϵ_{ip}(−1)] = [R_{ip}(−ax_{j}), ϵ_{ip}(−1)] = R_{ii}(λx̄_{j}ā − ax_{j}). R16). To prove R16, we only need to show (D) [x_{p}(ζ), ϵ_{ip}(a)] = [x_{k}(ζā), ϵ_{ik}(1)]R_{pi}(ζ_{f} ā)R_{ki}(−(ζā)_{f})R_{ii}((ζā)_{f} − aζ_{f} ā) where p, q ≥ r + 1 and p ≠ q. To begin the verification of this equation, set e = ϵ_{iq}(1) and f = ϵ_{qp}(−a). Then [e, f] = ϵ_{ip}(−a) = g^{−1}, where g = ϵ_{ip}(a). fg = ϵ_{kp}(−a)ϵ_{ip}(a) = gf. Let y = [x_{p}(ζ), f]. Since

ϕ(y) = [r_{p}(ζ), Hϵ_{qp}(−a)] = r_{q}(ζā)r_{pq}(−ζ_{f} ā)r_{qq}((ζā)_{f} − aζ_{f} ā),

it follows that

y = zx_{q}(ζā)R_{pq}(−ζ_{f} ā)R_{qq}((ζā)_{f} − aζ_{f} ā)

where z ∈ ker ϕ. From

y^{e} = [x_{p}(ζ)^{e}, f^{e}] = (R15)[x_{p}(ζ), fg] = [x_{p}(ζ), gf] = [x_{p}(ζ), f][x_{p}(ζ), g]^{f} = y[x_{p}(ζ), g]^{f}, it follows that

( E ) [y, e] = [x_{p}(ζ), g]^{f}. Consider the left hand side of equation ( E ) above.

[y, e] = [zx_{q}(ζā)R_{pq}(−ζ_{f} ā)R_{qq}((ζā)_{f} − aζ_{f} ā), ϵ_{iq}(1)]

=( using ( C ))[x_{q}(ζā), ϵ_{iq}(1)][R_{pq}(−ζ_{f} ā), ϵ_{iq}(1)][R_{qq}((ζā)_{f} − aζ_{f} ā), ϵ_{iq}(1)]
=( R5a), R6 )[x_{q}(ζā), ϵ_{iq}(1)]R_{pi}(ζ_{f} ā)R_{qi}(−(ζā)_{f} + aζ_{f} ā)R_{ii}((ζā)_{f} − aζ_{f} ā),

which is the right hand of (D) multiplied by R_{qi}(aζ_{f} ā) on the right. Consider the right hand side of the equation ( E ) above. Since

ϕ([x_{p}(ζ), g]) = ϕ([x_{p}(ζ), ϵ_{ip}(a)]) = [r_{p}(ζ), Hϵ_{ip}(a)] = r_{i}(−ζā)r_{pi}(ζ_{f} ā)r_{ii}((ζā)_{f} − aζ_{f} ā), it follows that

[x_{p}(ζ), g] = zx_{i}(−ζā)R_{pi}(ζ_{f} ā)R_{ii}((ζā)_{f} − aζ_{f} ā)

where z ∈ ker ϕ. Thus

[x_{p}(ζ), g]^{f} = zx_{i}(−ζā)^{f}R_{pi}(ζ_{f} ā)^{f}R_{ii}((ζā)_{f} − aζ_{f} ā)^{f}
=(R15, R4, R5a) )zx_{i}(−ζā)R_{pi}(ζ_{f} ā)R_{qi}(aζ_{f} ā)R_{ii}((ζā)_{f} − aζ_{f} ā)
= [x_{p}(ζ), g]R_{qi}(aζ_{f} ā) = [x_{p}(ζ), ϵ_{ip}(a)]R_{qi}(aζ_{f} ā),

which is the left hand of ( D ) multiplied by R_{qi}(aζ_{f} ā) on the right. Thus, ( D ) is proved. Setting a = 1 in ( D ) shows that the element R_{i}^{p}(ζ) does not depend on the choice of p and that R16 holds. □

Corollary 9.10 Keep the notation K_{n} as in Lemma 9.9. Let K_{∞} = lim_{n→∞} K_{n}. Then K_{∞} is an EH_{∞,r}(R)-perfect, universal, EH_{∞,r}(R)-central extension of itself.

Proof This is a direct consequence of Lemma 9.9. □

Lemma 9.11 Let ϕ : F ⟶ G be a central extension of G. Let H and K be subgroups of G such that [K, H] ⊆ K. Let K̃ = ϕ^{−1}(K). Define an action of H on K̃ by setting

x^{e} = u^{−1}xu, where x ∈ K̃, e ∈ H, and u ∈ ϕ^{−1}(H) is a pre-image of e.

Then ϕ : K̃ ⟶ K is an H-central extension of K.

Proof. Since [K, H] ⊆ K, K is normalized by H, i.e. K is a H-group under conjugation by elements of H. Since ϕ : F ⟶ G is a central extension of G, K̃ is also an H-group under the action defined above and ϕ : K̃ ⟶ K is an H-central extension of K.

10. Universality of Stabilized Steinberg Groups

Our goal is to prove that StH(R, a_{1}, ⋯, a_{r}) is the perfect, universal, central extension of EH(R, a_{1}, ⋯, a_{r}).

Theorem 10.1 The canonical homomorphism

StH(R, a_{1}, ⋯, a_{r}) ⟶ EH(R, a_{1}, ⋯, a_{r}), x̃_{ij}(a) ⟼ x_{ij}(a), x̃_{i}(ζ) ⟼ x_{i}(ζ)

where x_{ij} = r_{ij}, l_{ij}, or Hϵ_{ij} and x_{i} = Hm_{i} or r_{i}, is a perfect, universal, central extension of EH(R, a_{1}, ⋯, a_{r}).

Proof. Perfectness follows directly from relations E3, R5a), R6, L5a), L6, E8 and R16 in Lemma 8.2.

Centrality is established as follows . Let ρ : StH(R, a_{1}, ⋯, a_{r}) ⟶ EH(R, a_{1}, ⋯, a_{r}) denoted the canonical homomorphism. Let x ∈ ker ρ. The element x can be written as a product of the generators x̃_{kl}(a) and x̃_{i}(ζ) of StH(R, a_{1}, ⋯, a_{r}). Fix an integer q such that q ≥ all k, l and i appearing in the product above. Let S denote the subgroup of StH(R, a_{1}, ⋯, a_{r}) generated by all Hϵ̃_{iq}(a), Hm̃_{q}(ζ) and l̃_{qj}(a) such that a ∈ R, ζ ∈ C, i ≥ r + 1, j ≥ 1. Let T denote the subgroup of StH(R, a_{1}, ⋯, a_{r}) generated by all Hϵ̃_{qi}(a), r̃_{q}(ζ) and r̃_{jq}(a) such that a ∈ R, ζ ∈ C, i ≥ 1, j ≥ r + 1. Let N denote the subgroup of StH(R, a_{1}, ⋯, a_{r}) generated by S and T. To prove centrality, it sufficies to show that N = StH(R, a_{1}, ⋯, a_{r}), and that x fixes each element of S and T.

Relation 2) in Lemma 8.2 shows that N contains all Hϵ̃_{kl}(a) such that k ≥ r+1 and l ≥ 1. Combining this fact with relations L5a) and L6 in Lemma 8.2, one easily deduces that N contains all l̃_{kl}(a) such that k, l ≥ 1. Similarly, using relations R5a) and R6, one deduces that N contains all r̃_{kl}(a) such that k, l ≥ r + 1. Combining these facts with relations E8 and R16, one derives that N contains all Hm̃_{i}(ζ) and r̃_{i}(ζ) such that i ≥ r + 1 and ζ ∈ C. Thus N = StH(R, a_{1}, ⋯, a_{r}).

Next we show that conjugation by x fixes each element of S and T. We handle first the case of S.

We claim that every element of S can be written as a product Hm̃_{q}(ζ) ∏_{i} Hϵ̃_{iq}(a_{i}) ∏_{j} l̃_{jq}(b_{j}) such that the indices i and j are taken in increasing order. Relations E2, L2, L4, E7, and L10 in Lemma 8.2 show that all pairs of generators of S commute except for pairs (l̃_{qi}(a), Hϵ̃_{iq}(b)), ( Hm̃_{q}(ζ), l̃_{qj}(a)) (j ≤ r), and (Hm̃_{q}(ζ), Hm̃_{q}(ξ)). But relation L5b) says that [l̃_{qi}(a), Hϵ̃_{iq}(b)] = l̃_{qq}(ab − λ̄b̄ā). Relation L9 says that [Hm̃_{q}(ζ), l̃_{qj}(a)] = l̃_{qq}(ax_{j} − λ̄x̄_{j}ā) (j ≤ r). Thus every element of S can be written as product ∏_{j} Hm̃_{q}(ζ_{j}) ∏_{i}(∏_{j} Hϵ̃_{iq}(a_{ij})) ∏_{i}(∏_{j} l̃_{qi}(b_{ij})). By E5 and L10, ∏_{j} Hm̃_{q}(ζ_{j}) = Hm̃_{q}(∑_{j} ζ_{j})l̃_{qq}(a) for some a ∈ R. By E1, ∏_{j} Hϵ̃_{iq}(a_{ij}) = Hϵ̃_{iq}(∑_{j} a_{ij}). By L1, ∏_{j} l̃_{qi}(b_{ij}) = l̃_{qi}(∑_{j} b_{ij}). Now the claim follows from L4.

It is easy to see that ρ(Hm̃_{q}(ζ) ∏_{i} Hϵ̃_{iq}(a_{i}) ∏_{j} l̃_{qj}(b_{j})) = 1 ⟺ ζ = 0 and a_{i} = b_{j} = 0 for all i ≥ r + 1 and j ≥ 1. Thus ρ|_{S} is injective. Since ρ(x) = 1, it follows that conjugation by ρ(x) fixes each element of ρ(S). Thus conjugation by x fixes each element of S ⟺ conjugation by x normalizes S, because ρ|_{S} is injective. If k and l < q then it follows from relations E2, E3, L4, L5, E6, and E7 that Hϵ̃_{kl}(a) normalizes S. From relations L2, L4, L5a), L7,L8a), L8b) and L10, it follows that l̃_{kl}(a) normalizes S. Using E7, E8, E4, L8b), and L10, we obtain that Hm̃_{k}(ζ) (k < q) normalizes S. Using R4, R9, 1, 2, and 4, we obtain that r̃_{kl}(a) normalizes S. By R15, R17, 5, 6, and 7, r̃_{j}(ζ) normalize S. Thus x normalizes S.

We handle now the case of T. As in the case of S, we begin by showing that every element of T can be written as a product r̃_{q}(ζ) ∏_{i} Hϵ̃_{qi}(a_{i}) ∏_{j} r̃_{jq}(b_{j}) such that the indices i and j are taken in increasing order. Relations E2, R2, R4, R12, and R15 in Lemma 8.2 show that all pairs of generators of T commute except for pairs (r̃_{iq}(a), Hϵ̃_{qi}(b)), ( r̃_{q}(ζ), Hϵ̃_{qj}(a))(j ≤ r), and (r̃_{q}(ζ), r̃_{q}(ξ)). But relations R5b) and R3 say that [r̃_{iq}(a), Hϵ̃_{qi}(b)] = r̃_{qq}(−ba + λāb̄). Relation R14 says that [Hϵ̃_{qj}(a), r̃_{q}(ζ)] = r̃_{qq}(ax_{j} − λx̄_{j}ā) (j ≤ r). Thus every element of T can be written as product ∏_{j} r̃_{q}(ζ_{j}) ∏_{i}(∏_{j} Hϵ̃_{qi}(a_{ij})) ∏_{i}(∏_{j} r̃_{iq}(b_{ij})). By R8 and R4, ∏_{j} r̃_{q}(ζ_{j}) = r̃_{q}(∑_{j} ζ_{j})r̃_{qq}(a) for some a ∈ R. By E1, ∏_{j} Hϵ̃_{qi}(a_{ij}) = Hϵ̃_{qi}(∑_{j} a_{ij}). By R1, ∏_{j} r̃_{iq}(b_{ij}) = r̃_{iq}(∑_{j} b_{ij}). It follows that every element of T can be expressed as a required product.

It is easy to see that ρ(r̃_{q}(ζ) ∏_{i} Hϵ̃_{qi}(a_{i}) ∏_{j} r̃_{jq}(b_{j})) = 1 ⟺ ζ = 0 and a_{i} = b_{j} = 0 for all i ≥ 1 and j ≥ r + 1. Thus ρ|_{T} is injective. Since ρ(x) = 1, it follows that conjugation by ρ(x) fixes each element of ρ(T). Thus conjugation by x fixes each element of T ⟺ conjugation by x normalizes T, because ρ|_{T} is injective. If k and l < q then it follows from relations E2, E3, R4, R5, R13, and R15 that Hϵ̃_{kl}(a) normalizes T. From relations L4, L5, 1,2,3, and 5, it follows that l̃_{kl}(a) normalizes T. Using E6, E7, R9, R11, and R17, we obtain that Hm̃_{k}(ζ) (k < q) normalizes T. Using R4, R5, R2, and R12, we obtain that r̃_{kl}(a) normalizes T. By R13, R15, R16, R12 and R7, r̃_{j}(ζ) normalize T. Thus x normalizes T.

Next we establish universality.

Let f : F ⟶ EH(R, a_{1}, ⋯, a_{r}) be a central extension of EH(R, a_{1}, ⋯, a_{r}). Since the elementary Steinberg group St(R) is a universal central extension of the elementary group E(R) (cf. [Milnor, 5.10]), there are elements Hϵ̃_{ij}(a) ∈ F (i, j ≥ r + 1; i ≠ j; a ∈ R) such that f(Hϵ̃_{ij}(a)) = Hϵ_{ij}(a) and the elements Hϵ̃_{ij}(a) satisfy E1, E2, and E3. Recall the groups E_{∞,r}(R) and EH_{∞,r}(R) defined prior to 9.3. For any element x ∈ EH_{∞,r}(R), the action of x on EH(R, a_{1}, ⋯, a_{r}) is by definition conjugation by

the element ( \begin{matrix}x \\ & x̄^{−1} \end{matrix} ) ∈ EH(R, a_{1}, ⋯, a_{r}). Thus we can identify EH_{∞,r}(R) with its

image under the canonical homomorphism

EH_{∞,r}(R) ⟶ EH(R, a_{1}, ⋯, a_{r}), x ⟼ ( \begin{matrix}x \\ & x̄^{−1} \end{matrix} ) .

It is clear that H_{∞} = ⟨Hϵ_{kl}(b) | k ≥ r + 1, 1 ≤ l ≤ r, b ∈ R⟩ is an E_{∞,r}(R)-group where E_{∞,r}(R) acts on H_{∞} by conjugation. By Lemma 9.11

f : f^{−1}(H_{∞}) ⟶ H_{∞}

is an E_{∞,r}(R)-central extension. The action of E_{∞,r}(R) is explained in Lemma 9.11. It follows from Corollary 9.4 that there is an E_{∞,r}(R)-splitting

s_{1} : H_{∞} ⟶ f^{−1}(H_{∞}).

Let Hϵ̃_{ij}(a) = s_{1}(Hϵ_{ij}(a)) (i ≥ r + 1, 1 ≤ j ≤ r). Then the elements Hϵ̃_{ij}(a) (i ≥ r + 1, j ≥ 1) satisfy E1, E2, and E3.

Similarly, by Lemma 9.11

f : f^{−1}(Hm_{∞}) ⟶ Hm_{∞}

is an E_{∞,r}(R)-central extension. It follows from Corollary 9.6 that there is an E_{∞,r}(R)- splitting

s_{2} : Hm_{∞} ⟶ f^{−1}(Hm_{∞}).

Again, by Lemma 9.11

f : f^{−1}(L_{∞}) ⟶ L_{∞}

is an EH_{∞,r}(R)-central extension. It follows from Corollary 9.8 that there is an EH_{∞,r}(R)- splitting

s_{3} : L_{∞} ⟶ f^{−1}(L_{∞}).

Since Hm_{∞} ∩ L_{∞} = ⟨L_{ij}(a) | i, j ≥ r + 1⟩ is E_{∞,r}(R)-perfect and the homomorphism f is E_{∞,r}(R)- central, the splittings s_{2} and s_{3} are agree on ⟨L_{ij}(a) | i, j ≥ r + 1⟩. Now, define Hm̃_{i}(ζ) = s_{2}(Hm_{i}(ζ)) (i ≥ r + 1) and l̃_{kl}(a) = s_{3}(l_{kl}(a)) (k, l ≥ 1). Then Hm̃_{i}(ζ) and l̃_{kl}(a) satisfy E4, E5, E7, L1 — L6 and L10.

Using Corollary 9.10, we can find elements r̃_{i}(ζ) and r̃_{ij}(a) (i ≥ r + 1) in F such that f(r̃_{i}(ζ)) = r_{i}(ζ), f(r̃_{ij}(a)) = r_{ij}(a) and which satisfy R1—R8, R12—R16.

Let StQ(R) = StQ(R, max^{−λ}(R)) denote the stabilized quadratic Steinberg group with maximal form parameter max^{−λ}(R) and EQ(R) = EQ(R, max^{−λ}(R)) the stabilized elementary quadratic group with maximal form parameter max^{−λ}(R). These groups are defined in [Bak, §3]. By [Bak, (3.17)], StQ(R) ⟶ EQ(R) is a perfect universal central extension. Thus the natural embedding EQ(R) ⟶ EH_{∞,r}(R, a_{1}, ⋯, a_{r}) induces a unique homomorphism StQ(R) ⟶ F covering the one above. It follows now from the generators and relations used in [Bak, p. 43-44] to define StQ(R) that the elements r̃_{ij}(a) and l̃_{kl}(b) (k, l ≥ r + 1) satisfy 1—4.

Above, we have picked a counterpart in F for each generator of EH(R, a_{1}, ⋯, a_{r}). Morever we have shown that the counterparts satisfy all the relations in Lemma 8.2, except possibly relations 1, 2, 4, 5, 6, 7, 8, 9, E6, L7a), L7b), L8a), L8b), R9, R10, R11, and R17. We shall check in order that these relations are satisfied. This will complete the proof of the theorem.

To simplify notation, we shall write H_{ij}, L_{ij}, R_{ij}, Hm_{i}, and R_{i} in place of Hϵ̃_{ij}, l̃_{ij}, r̃_{ij}, Hm̃_{i}, and r̃_{i}. The letter p will always refer to an integer such that r + 1 ≤ p.

1). Choose p ≠ i, j, k, or l. Then L_{kl}(−b)^{R_{ij}(a)}=(L5a)[L_{kp}(−b), H_{pl}(1)]^{R_{ij}(a)} =(R4)[L_{kp}(−b)^{R_{ij}(a)}, H_{pl}(1)] = [L_{kp}(−b), H_{pl}(1)] = L_{kl}(−b), since [R_{ij}(a), L_{kp}(b)] ∈ ker f.

2). If k ≥ r + 1 then 2) holds. If k ≤ r, pick p ≠ i, j, k. Then L_{ik}(−b)^{R_{ij}(a)}=(L5a)[L_{ip}(−b), H_{pk}(1)]^{R_{ij}(a)} =(R4, 2) for p ≥ r + 1)[L_{ip}(−b)H_{jp}(−λāb), H_{pk}(1)] = [L_{ip}(−b), H_{pk}(1)]^{H_{jp}(−λāb)}[H_{jp}(−λāb), H_{pk}(1)] =(L5a, L4, E3) L_{ik}(−b)H_{jk}(−λāb).

4). If i ≥ r + 1 then 4) holds. If i ≤ r, choose p ≠ i, j. Then

L_{ji}(−a)^{R_{jj}(b)} = ( L5a )[L_{jp}(−a), H_{pi}(1)]^{R_{jj}(b)}
=( 4) for p ≥ r + 1, R4 ) [H_{jp}(ba)L_{pp}(λ̄āba)L_{jp}(−a), H_{pi}(1)].

By E3, [H_{jp}(ba), H_{pi}(1)] = H_{ji}(ba) which commutes with L_{pp}(λ̄āba) by L4. Furthermore

[L_{pp}(λ̄āba)L_{jp}(−a), H_{pi}(1)] = (L6, L2, L5a) L_{pi}(λ̄āba)L_{ii}(λ̄āba)L_{ji}(−a).

Thus

L_{ji}(−a)^{R_{jj}(a)} = H_{ji}(ba)^{L_{jp}(−a)}L_{pi}(λ̄āba)L_{ii}(λ̄āba)L_{ji}(−a) = H_{ji}(ba)L_{ii}(λ̄āba)L_{ji}(−a).

E6). Choose p ≠ i, j, or k. Then

H_{jk}(−a)^{Hm_{i}(ζ)} = (E3)[H_{jp}(a), H_{pk}(−1)]^{Hm_{i}(ζ)}=(E7)[H_{jp}(a), H_{pk}(−1)^{Hm_{i}(ζ)}]

= [H_{jp}(a), H_{pi}(−x_{k})H_{pk}(−1)] = Hϵ_{pi}(−x_{k})Hϵ_{pk}(−1) )=(E3, E2)H_{ji}(−ax_{k})H_{jk}(−a),

since f(H_{pk}(−1)^{Hm_{i}(ζ)}) = Hϵ_{pi}(−x_{k})Hϵ_{pk}(−1).

L8a). Pick p ≠ i, j, or k where k ≥ r + 1, j ≤ r. Then L_{jk}(−a)^{Hm_{i}(ζ)} =(L5a)[L_{jp}(−a), H_{pk}(1)]^{Hm_{i}(ζ)} =(E7)[L_{jp}(−a)^{Hm_{i}(ζ)}, H_{pk}(1)] = [L_{ip}(−x̄_{j}a)L_{jp}(−a), H_{pk}(1)], since f(L_{jp}(−a)^{Hm_{i}(ζ)}) = l_{ip}(−x̄_{j}a)l_{jp}(−a). Now, it is easy to complete the proof of L8a).

L8b) is a direct consequence of L8a) and L3).

L9). Choose p ≠ i, j where j ≤ r. Then L_{ji}(−a)^{Hm_{i}(ζ)} =( L5a) )[L_{jp}(−a), H_{pi}(1)]^{Hm_{i}(ζ)} =( L8a), E7 ) [L_{ip}(−x̄_{j}a)L_{jp}(−a), H_{pi}(1)] =( L5b), L2, L5a) ) L_{ii}(λ̄āx_{j} − x̄_{j}a)L_{ji}(−a).

L7a). Choose p ≠ i, j or k where j ≠ k, j, k ≤ r. Then

L_{jk}(−a)^{Hm_{i}(ζ)} = ( L5a) )[L_{jp}(−a), H_{pk}(1)]^{Hm_{i}(ζ)}
=( L8a), E6 )[L_{ip}(−x̄_{j}a)L_{jp}(−a), H_{pi}(x_{k})H_{pk}(1)]
= [L_{ip}(−x̄_{j}a), H_{pi}(x_{k})H_{pk}(1)]^{L_{jp}(−a)}[L_{jp}(−a), H_{pi}(x_{k})H_{pk}(1)] = u^{L_{jp}(−a)}v,

where

u = [L_{ip}(−x̄_{j}a), H_{pi}(x_{k})H_{pk}(1)] = [L_{ip}(−x̄_{j}a), H_{pk}(1)][L_{ip}(−x̄_{j}a), H_{pi}(x_{k})]^{H_{pk}(1)}
=( L5a), L5b) )L_{ik}(−x̄_{j}a)L_{ii}(λ̄x̄_{k}āx_{j} − x̄_{j}ax_{k})^{H_{pk}(1)}
=(L4)L_{ik}(−x̄_{j}a)L_{ii}(λ̄x̄_{k}āx_{j} − x̄_{j}ax_{k}),

which commutes with L_{jp}(−a). Thus

u^{L_{jp}(−a)} = u.

Similarly, one gets that

v = [L_{jp}(−a), H_{pi}(x_{k})H_{pk}(1)] = L_{ji}(−ax_{k})L_{jk}(−a).

Hence

L_{jk}(−a)^{Hm_{i}(ζ)} = L_{ik}(−x̄_{j}a)L_{ii}(λ̄x̄_{k}āx_{j} − x̄_{j}ax_{k})L_{ji}(−ax_{k})L_{jk}(−a).

L7b). Choose p ≠ i, j where j ≤ r. For any a = −λ̄ā, it follows from L6 that L_{jj}(−a) = L_{pj}(a)[L_{pp}(−a), H_{pj}(1)]. By L8b), L_{pj}(a)^{Hm_{i}(ζ)} = L_{pi}(−ax_{j})L_{pj}(a). Since

[L_{pp}(−a), H_{pj}(1)]^{Hm_{i}(ζ)}=( L10, E6 )[L_{pp}(−a), H_{pi}(x_{j})H_{pj}(1)]

= [L_{pp}(−a), H_{pj}(1)]^{H_{pi}(x_{j})}[L_{pp}(−a), H_{pi}(x_{j})]
=(L6)(L_{pj}(−a)L_{jj}(−a))^{H_{pi}(x_{j})}L_{pi}(ax_{j})L_{ii}(−x̄_{j}ax_{j})

=(L4, L5)L_{pj}(−a)L_{ji}(−ax_{j})L_{jj}(−a)L_{pi}(ax_{j})L_{ii}(−x̄_{j}ax_{j}),

it follows that

L_{jj}(−a)^{Hm_{i}(ζ)} = L_{ji}(−ax_{j})L_{ii}(−x̄_{j}ax_{j})L_{jj}(−a).

R9). Choose p ≠ i, j, or k. Then R_{jk}(a)^{Hm_{i}(ζ)} = (R5a) [R_{jp}(a), H_{kp}(−1)]^{Hm_{i}(ζ)} =(E7)[R_{jp}(a)^{Hm_{i}(ζ)}, H_{kp}(−1)] = [R_{jp}(a), H_{kp}(−1)] = R_{jk}(a), because f(R_{jp}(a)^{Hm_{i}(ζ)}) = r_{jp}(a).

R10). Choose p ≠ i, or k. Then

R_{ik}(−a)^{Hm_{i}(ζ)}=(R5a)[R_{ip}(a), H_{kp}(1)]^{Hm_{i}(ζ)}=(E7)[R_{ip}(a)^{Hm_{i}(ζ)}, H_{kp}(1)].

Since we always take (−ζ)_{f} = ζ_{f}, we have that

f(R_{ip}(a)^{Hm_{i}(ζ)}) = r_{p}(−ζa)Hϵ_{pi}(−λāζ̄_{f})r_{pp}((ζa)_{f} − āζ_{f}a)r_{ip}(a)
=(R15, R2, R12 )Hϵ_{pi}(−λāζ̄_{f})r_{ip}(a)r_{pp}((ζa)_{f} − āζ_{f}a)r_{p}(−ζa).

Thus

R_{ik}(−a)^{Hm_{i}(ζ)} = [H_{pi}(−λāζ̄_{f})R_{ip}(a)R_{pp}((ζa)_{f} − āζ_{f}a)R_{p}(−ζa), H_{kp}(1)].

From E3, R2, R12, R4, R15, R3 and R5a), it follows that

[H_{pi}(−λāζ̄_{f}), H_{kp}(1)]^{R_{ip}(a)R_{pp}((ζa)_{f} −āζ_{f} a)R_{p}(−ζa)}

= H_{ki}(λāζ̄_{f})^{R_{ip}(a)R_{pp}((ζa)_{f} −āζ_{f} a)R_{p}(−ζa)} = H_{ki}(λāζ̄_{f})^{R_{ip}(a)} = R_{pk}(−āζ_{f}a)H_{ki}(λāζ̄_{f}).

By R5a), [R_{ip}(a), H_{kp}(1)] = R_{ik}(−a) and R_{ik}(−a) commutes with any R_{j}(ξ) and R_{jl}(b) according to R12 and R2. By R6,

[R_{pp}((ζa)_{f} − āζ_{f}a), H_{kp}(1)] = R_{pk}(−(ζa)_{f} + āζ_{f}a)R_{kk}((ζa)_{f} − āζ_{f}a),

and the right hand side commutes with any R_{j}(ξ) and R_{jl}(b) according to R12 and R2. By R16,

[R_{p}(−ζa), H_{kp}(1)] = R_{k}(ζa)R_{pk}((ζa)_{f}).

Thus

R_{ik}(−a)^{Hm_{i}(ζ)}

= R_{pk}(−āζ_{f}a)H_{ki}(λāζ̄_{f})R_{ik}(−a)R_{pk}(−(ζa)_{f} + āζ_{f}a)R_{kk}((ζa)_{f} − āζ_{f}a)R_{k}(ζa)R_{pk}((ζa)_{f})

=( R4, R2, R12, R15) R_{k}(ζa)H_{ki}(λāζ̄_{f})R_{kk}((ζa)_{f} − āζ_{f}a)R_{ik}(−a).

This completes the proof of R10).

R11) follows directly from R10).

5) Choose p ≠ i, j, or k. Then L_{jk}(−a)^{R_{i}(ζ)} = (L5a)[L_{jp}(−a), H_{pk}(1)]^{R_{i}(ζ)} = (R15)[L_{jp}(−a)^{R_{i}(ζ)}, H_{pk}(1)] = [L_{jp}(−a), H_{pk}(1)] = L_{jk}(−a), since [R_{i}(ζ), L_{jk}(a)] ∈ ker f.

6) Choose p ≠ i, j, or k. Then

L_{ik}(−a)^{R_{i}(ζ)} = (L5a)[L_{ip}(−a), H_{pk}(1)]^{R_{i}(ζ)} = (R15)[L_{ip}(−a)^{R_{i}(ζ)}, H_{pk}(1)]

= [Hm_{p}(ζa)H_{ip}(−ζ_{f}a)L_{pp}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ip}(−a), H_{pk}(1)],

because

f(L_{ip}(−a)^{R_{i}(ζ)}) = Hm_{p}(ζa)Hϵ_{ip}(−ζ_{f}a)l_{pp}(āζ̄_{f}a − \overline{(ζa)}_{f})l_{ip}(−a).

It follows from E8 that

[Hm_{p}(ζa), H_{pk}(1)] = Hm_{k}(ζa)L_{kp}(\overline{(ζa)}_{f}),

and the right hand side commutes with

H_{ip}(−ζ_{f}a)L_{pp}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ip}(−a)

according to E7, L4, and L2. It follows from E3, L4, L3, and L5a) that

[H_{ip}(−ζ_{f}a), H_{pk}(1)]^{L_{pp}(āζ̄_{f} a−\overline{(ζa)}_{f})L_{ip}(−a)} = H_{ik}(−ζ_{f}a)^{L_{ip}(−a)} = L_{pk}(λ̄āζ_{f}a)H_{ik}(−ζ_{f}a). Using L6, L5a), and L2, we obtain that

[L_{pp}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ip}(−a), H_{pk}(1)] = L_{pk}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{kk}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ik}(−a). Thus

L_{ik}(−a)^{R_{i}(ζ)}

= Hm_{k}(ζa)L_{kp}(\overline{(ζa)}_{f})L_{pk}(λ̄āζ_{f}a)H_{ik}(−ζ_{f}a)L_{pk}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{kk}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ik}(−a)

=(L4, L2, L1)Hm_{k}(ζa)H_{ik}(−ζ̄_{f}a)L_{kk}(āζ̄_{f}a − \overline{(ζa)}_{f})L_{ik}(−a),

because ζ_{f} + λζ̄_{f} = ζ̄A_{1}ζ. This completes the proof of 6). 7). Choose p ≠ i, j, or k. Then L_{jk}(−a)^{R_{i}(ζ)} = (L5a)[L_{jp}(−a), H_{pk}(1)]^{R_{i}(ζ)} =(5, R13)[L_{jp}(−a), H_{pk}(1)R_{pi}(x_{k})] = [L_{jp}(−a), R_{pi}(x_{k})][L_{jp}(−a), H_{pk}(1)]^{R_{pi}(x_{k})} =(L3, 2, L5a), 1)H_{ij}(x̄_{k}ā)L_{jk}(−a). 8). Choose p ≠ i. Then

L_{jk}(−a)^{R_{i}(ζ)}=(L5a)[L_{jp}(−a), H_{pk}(1)]^{R_{i}(ζ)} = [L_{jp}(−a)^{R_{i}(ζ)}, H_{pk}(1)^{R_{i}(ζ)}]

=(7, R13)[H_{ip}(−λx̄_{j}a)L_{jp}(−a), H_{pk}(1)R_{pi}(x_{k})]
= [H_{ip}(−λx̄_{j}a), H_{pk}(1)R_{pi}(x_{k})]^{L_{jp}(−a)}[L_{jp}(−a), H_{pk}(1)R_{pi}(x_{k})] = u^{L_{jp}(−a)}v,

where

u = [H_{ip}(−λx̄_{j}a), H_{pk}(1)R_{pi}(x_{k})] = [H_{ip}(−λx̄_{j}a), R_{pi}(x_{k})][H_{ip}(−λx̄_{j}a), H_{pk}(1)]^{R_{pi}(x_{k})}

=(R3, R5b), E3, R4)R_{ii}(x̄_{k}āx_{j} − λx̄_{j}ax_{k})H_{ik}(−λx̄_{j}a).

It follows from 1 and L4 that u commutes with L_{jp}(−a). Furthermore

v = [L_{jp}(−a), H_{pk}(1)R_{pi}(x_{k})] = [L_{jp}(−a), R_{pi}(x_{k})][L_{jp}(−a), H_{pk}(1)]^{R_{pi}(x_{k})}

=(L3, 2, L4, 1)H_{ij}(x̄_{k}ā)L_{jk}(−a).

Thus L_{jk}(−a)^{R_{i}(ζ)} = u^{L_{jp}(−a)}v = uv = R_{ii}(x̄_{k}āx_{j} − λx̄_{j}ax_{k})H_{ik}(−λx̄_{j}a)H_{ij}(x̄_{k}ā)L_{jk}(−a)

=(R4, E2)H_{ij}(x̄_{k}ā)H_{ik}(−λx̄_{j}a)R_{ii}(x̄_{k}āx_{j} − λx̄_{j}ax_{k})L_{jk}(−a).

9). Choose p ≠ i, j. Using L6, we deduce that

L_{jj}(a) = L_{pj}(−a)[L_{pp}(a), H_{pj}(1)].

Thus

L_{jj}(−a)^{R_{i}(ζ)} = L_{pj}(a)^{R_{i}(ζ)}[L_{pp}(−a)^{R_{i}(ζ)}, H_{pj}(1)^{R_{i}(ζ)}]
=(7, 5, R13)L_{pj}(a)H_{ip}(−x̄_{j}ā)[L_{pp}(−a), H_{pj}(1)R_{pi}(x_{j})]
= L_{pj}(a)H_{ip}(−x̄_{j}ā)[L_{pp}(−a), R_{pi}(x_{j})][L_{pp}(−a), H_{pj}(1)]^{R_{pi}(x_{j})}

=(R3, 3, L6)L_{pj}(a)H_{ip}(−x̄_{j}ā)H_{ip}(−λx̄_{j}a)R_{ii}(−λx̄_{j}ax_{j})L_{pj}(−a)^{R_{pi}(x_{j})}L_{jj}(−a)^{R_{pi}(x_{j})}. Since L_{jj}(a) ∈ EH(R, a_{1}, ⋯, a_{r}), it follows that a = −λ̄ā. Using relations 2 and 1 we obtain that

L_{pj}(−a)^{R_{pi}(x_{j})}L_{jj}(−a)^{R_{pi}(x_{j})} = H_{ij}(−λx̄_{j}a)L_{pj}(−a)L_{jj}(−a).

Thus

L_{jj}(−a)^{R_{i}(ζ)} = L_{pj}(a)R_{ii}(−λx̄_{j}ax_{j})H_{ij}(−λx̄_{j}a)L_{pj}(−a)L_{jj}(−a)
=(1, L4, R4)H_{ij}(−λx̄_{j}a)R_{ii}(−λx̄_{j}ax_{j})L_{jj}(−a).

R17). Choose p ≠ i, j. It follows from E8 that

Hm_{i}(ζ) = [Hm_{p}(ζ), H_{pi}(1)]L_{ip}(−ζ̄_{f}).

From

f(Hm_{p}(ζ)^{R_{j}(ξ)}) = Hm_{p}(ζ)Hϵ_{jp}(−ξ̄A_{1}ζ) = Hϵ_{jp}(−ξ̄A_{1}ζ)Hm_{p}(ζ),

it follows that

Hm_{i}(ζ)^{R_{j}(ξ)} = (R15, 5)[Hm_{p}(ζ)^{R_{j}(ξ)}, H_{pi}(1)]L_{ip}(−ζ̄_{f})
= [H_{jp}(−ξ̄A_{1}ζ)Hm_{p}(ζ), H_{pi}(1)]L_{ip}(−ζ̄_{f})
= [H_{jp}(−ξ̄A_{1}ζ), H_{pi}(1)]^{Hm_{p}(ζ)}[Hm_{p}(ζ), H_{pi}(1)]L_{ip}(−ζ̄_{f})
= (E3, E7)H_{ji}(−ξ̄A_{1}ζ)Hm_{i}(ζ)=(E7)Hm_{i}(ζ)H_{ji}(−ξ̄A_{1}ζ).

This completes the proof of R17 and of the theorem.

11. KH_{2}-groups and a K_{2}-analog of the Whitehead lemma

We use the notation and machinery developed in section 6 to define KH_{2}-groups of Λ-Hermitian spaces. Then we prove that the action of GH-groups on KH_{2}-groups is trivial. This is the K_{2}-analog of the Whitehead lemma. We show next that Bak’s categorical definition of K_{2} of Λ-Hermitian spaces coincides with the one here. Finally we show that KH_{2} can be strictly smaller than the center of the Hermitian Steinberg group.

Definition 11.1 Define

KH_{2}(R, a_{1}, …, a_{r}) = Ker (StH(R, a_{1}, …, a_{r}) ⟶ EH(R, a_{1}, …, a_{r}).

By Theorem 10.1, KH_{2}(R, a_{1}, …, a_{r}) is abelian.

Definition 11.2 Suppose Λ is a -λ-form parameter on R. Define

StH(R, Λ) = \underrightarrow{lim}_{S⫅Λ}StH(R, S)
KH_{2}(R, Λ) = Ker (StH(R, Λ) ⟶ EH(R, Λ))

= \underrightarrow{lim}_{S⫅Λ}KH_{2}(R, Λ).

Obviously StH and KH_{2} define functors ((form rings)) ⟶ ((groups)) and KH_{2} takes its values in ((abelian groups)).

Definition 11.3 Define

KH_{2}^{λ}(R) = KH_{2}(R, max^{−λ}(R)).

By the Hermitian Whitehead lemma 5.3, EH(R, a_{1}, …, a_{r}) is the commutator subgroup of GH(R, a_{1}, …, a_{r}) and therefore normal in the latter. By the universality 10.1 of the extension StH(R, a_{1}, …, a_{r}) ⟶ EH(R, a_{1}, …, a_{r}), the action of GH(R, a_{1}, …, a_{r}) on EH(R, a_{1}, …, a_{r}) by conjugation lifts uniquely to an action of GH(R, a_{1}, …, a_{r}) on StH(R, a_{1}, …, a_{r}), which respects multiplication in StH(R, a_{1}, …, a_{r}). Since the extension above is clearly equivariant with respect to the action of GH(R, a_{1}, …, a_{r}), there is an induced action of GH(R, a_{1}, …, a_{r}) on KH_{2}(R, a_{1}, …, a_{r}).

Recall the terms G-perfect and G-central extension given in [Bak, p.49].

Theorem 11.4 The homomorphism StH(R, a_{1}, …, a_{r}) ⟶ EH(R, a_{1}, …, a_{r}) is a GH(R, a_{1} …, a_{r})-perfect, universal, GH(R, a_{1}, …, a_{r})-central extension. In particular, the action of GH(R, a_{1}, …, a_{r}) on KH_{2}(R, a_{1}, …, a_{r}) is trivial.

Taking limits, one obtains from the theorem above the following result.

Theorem 11.5 Suppose Λ is a -λ-form parameter on R. Then the homomorphism StH(R, Λ) ⟶ EH(R, Λ) is a GH(R, Λ)-perfect, universal, GH − (R, Λ)-central extension. In particular, the action of GH(R, Λ) on KH_{2}(R, Λ) is trivial.

Proof of Theorem 11.4. Let

Θ_{2k} : GH_{2r}(R, a_{1}, …, a_{r}) ⊥ GH_{2r}(R, a_{1}, …, a_{r}) → GH_{4r}(R, a_{1}, …, a_{r})
σ ⊥ σ′ ⟼ φ_{2k}(σ ⊥ σ′)φ_{2r}^{−1}

denote the injection defined in the proof of the Hermitian Whitehead lemma 5.3. The stabilizes to an injection.

Θ : GH(R, a_{1}, …, a_{r}) ⊥ GH(R, a_{1}, …, a_{r}) ⟶ GH(r, a_{1}, …, a_{r})
σ ⊥ σ′ ⟼ φ^{(σ⊥σ′)}φ^{−1}.

By the universality of Steinberg groups, Θ lifts uniquely to a homomorphism

Θ^{∼} : StH(R, a_{1}, …, a_{r}) ⊥ StH(R, a_{1}, …, a_{r}) ⟶ StH(R, a_{1}, …, a_{r}).

Again by universality, Θ^{∼} is equivariant with respect to the action of GH(R, a_{1}, …, a_{r}) ⊥ GH(R, a_{1}, …, a_{r}) on StH(R, a_{1}, …, a_{r}) ⊥ StH(R, a_{1}, …, a_{r}) and of GH(R, a_{1}, …, a_{r}) on StH(R, a_{1}, …, a_{r}). Θ^{∼} induces an equivariant homomorphism

Θ^{∼} : KH_{2}(R, a_{1}, …, a_{r}) ⊥ KH_{2}(R, a_{1}, …, a_{r}) ⟶ KH_{2}(R, a_{1}, …, a_{r}).

By the centrality of KH_{2}(R, a_{1}, …, a_{r}) in StH(R, a_{1}, …, a_{r}), the action of GH(R, a_{1}, …, a_{r}) on KH_{2}(R, a_{1}, …, a_{r}) factors through KH_{1}(R, a_{1}, …, a_{r}). Clearly

Θ^{∼} : KH_{2}(R, a_{1}, …, a_{r}) ⊥ KH_{2}(R, a_{1}, …, a_{r}) ⟶ KH_{2}(R, a_{1}, …, a_{r})

is equivariant with respect to

Θ : KH_{1}(R, a_{1}, …, a_{r}) ⊥ KH_{1}(R, a_{1}, …, a_{r}) ⟶ KH_{1}(R, a_{1}, …, a_{r}).

Let

ι_{ij} : KH_{i}(R, a_{1}, …, a_{r}) ⟶ KH_{i}(R, a_{1}, …, a_{r}) ⊥ KH_{i}(R, a_{1}, …, a_{r})(i = 1, 2)

denote the canonical embedding on the first (resp. second) coordinate when j = 1 (resp. j = 2). By Step 1 (resp. Step 2) in the proof of Theorem 5.3, Θι_{1,1} (resp. Θι_{1,2}) is the identity isomorphism. These arguments lifts to KH_{2} because in the KH_{2} situation, the elements in KH_{2} corresponding to the matrices E_{1} and E_{2} in Steps 1 and 2, respectively, are trivial. Thus Θ^{∼}ι_{2,1} and Θ^{∼}ι_{2,2} are the identity morphisms on KH_{2}(R, a_{1}, …, a_{r}). Clearly the action of 0 ⊥ KH_{1}(R, a_{1}, …, a_{r}) on KH_{2}(R, a_{1}, …, a_{r}) ⊥ 0 is trivial. By applying Θ and Θ^{∼} and the fact that Θ^{∼} is equivariant over Θ, we obtain that the action of KH_{1}(R, a_{1}, …, a_{r}) on KH_{2}(R, a_{1}, …, a_{r}) is trivial. □

Recall the definitions of the categories H H H(R, Λ) and M M M(R, Λ)_{free} given prior to (6.5).

Theorem 11.6 Suppose Λ is a −λ− form parameter on R. Then

KH_{2}(R, Λ) ≅ K_{2}(M M M(R, Λ)_{free}) ≅ K_{2}(H H H(R, Λ))

where K_{2} is taken in the sense of [Bak, p. 64].

Proof The second isomorphism is from [Bak, Lemma 6.23]. The first follows from Lemma 3.5. □

We close by giving an example where KH_{2}(R, a_{1}, ⋯, a_{r}) does not coincide with the center of StH(R, a_{1}, ⋯, a_{r}). Let F be a field of characteristic 2 and trivial involution. λ = 1. Then min^{λ}(R) = 0, max^{λ}(R) = max^{−λ}(R) = F. Take a_{1} = 1 then C = { x ∈ F | x^{2} = 0} = 0. Thus EH(F, 1) has three kinds of generators l_{ij}(a), r_{st}(b) and Hϵ_{it}(c) where i, j ≥ 1 and s, t ≥ 2. It follows from 1), L2 and L4 that l_{11}(a) (a ∈ F) commutes with all generators of EH(F, 1), hence the pre-image l̃_{11}(a) of l_{11}(a) in StH(F, 1) is in the center of StH(F, 1) but not in the kernel of StH(F, 1) ⟶ EH(F, 1).

Acknowledgement

This paper formed part of the author’s the dissertation. His sincere appreciation goes to Professor A. Bak for his encouragement, many useful discussions and patient guidance.

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