UNITARY SK_{1} FOR A GRADED DIVISION RING AND ITS QUOTIENT DIVISION RING

A. R. WADSWORTH AND V. I. YANCHEVSKIĬ

Abstract. Let E be a graded division ring finite-dimensional over its center with torsion-free abelian grade group, and let q(E) be its quotient division ring. Let τ be a degree-preserving unitary involution on E. We prove that SK_{1}(E, τ) ≅ SK_{1}(q(E), τ).

1. Introduction

Let D be a division algebra finite-dimensional over its center K. Let τ be a unitary involution on D, i.e., τ is an antiautomorphism of D with τ^{2} = id such that τ|_{K} ≠ id. By definition,

SK_{1}(D, τ) = Σ′_{τ} (D)/Σ_{τ} (D),

where

Σ_{τ} (D) = ⟨{a ∈ D^{∗} | a = τ(a)}⟩  and  Σ′_{τ} (D) = {a ∈ D^{∗} | Nrd_{D}(a) = τ(Nrd_{D}(a))}.

Analogously, suppose E = ⨁_{γ∈Γ_{E}} E_{γ} is a graded division algebra (i.e., a graded ring in which all nonzero homogeneous elements are units) of finite rank over its center Z, and with torsion-free abelian grade group Γ_{E}. Suppose E has a unitary graded involution τ, which is a ring antiautomorphism with τ(E_{γ}) = E_{γ} for each γ, such that τ^{2} = id and τ|_{Z} ≠ id. Then there is a reduced norm map Nrd_{E} : E → Z, and one can define Σ_{τ} (E), Σ′_{τ} (E), and SK_{1}(E, τ) just as we did above for D. Any such E has a ring of (central) quotients, q(E) = E ⊗_{Z} q(Z), where q(Z) is the quotient field of the integral domain Z, and it is known that q(E) is a division ring finite-dimensional over its center, which is q(Z). The unitary graded involution τ on E extends canonically to a unitary involution on q(E), also called τ. In this paper we will prove the following:

Theorem 1.1. Let E be a graded division algebra (with torsion-free abelian grade group) finite-dimensional over its center, and let τ be a unitary graded involution on E. Let q(E) be the quotient division ring of E and let τ denote also the extension of τ to a unitary involution of q(E). Then,

SK_{1}(E, τ) ≅ SK_{1}(q(E), τ).

This theorem complements a result in [HaW_{2}]: Suppose D is a division algebra finite-dimensional over its center K = Z(D), and suppose K has a Henselian valuation v. It is well-known that v has a unique extension to a valuation on D. The filtration on D induced by the valuation yields an associated graded ring gr(D) which is a graded division algebra of finite rank over its center. If τ is a unitary involution on D which is compatible with the valuation, there is an induced graded involution τ̃ on gr(D). It was shown in [HaW_{2}, Th. 3.5] that if the restriction of v to the τ-fixed field K^{τ} is Henselian and K is tamely ramified over K^{τ}, then τ is compatible with v, τ̃ is unitary, and

SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃).  (1.1)

This theorem served to focus attention on unitary SK_{1} for graded division algebras. The graded division algebra gr(D) has a significantly simpler structure than the valued division algebra D; notably gr(D) has a much more tractable group of units (as they are all homogeneous). Consequently, SK_{1} calculations are often substantially easier for gr(D) than for D. This was demonstrated in [HaW_{2}] and [W], where many formulas for SK_{1}(gr(D), τ̃) were proved. By that approach new results on SK_{1}(D, τ) were obtained, as well as new and simpler proofs of results that had previously been proved by calculations that were often quite complicated. Now, by virtue of the theorem proved here, all the results proved in [HaW_{2}] and [W] on SK_{1}(E, τ) for a graded division algebra E carry over to yield corresponding results for SK_{1}(q(E), τ).

One of the key results for unitary SK_{1} is the “Stability Theorem,” which says that the unitary SK_{1} is unchanged on passage from a division algebra D to a rational division algebra over D. (This was originally proved in [Y_{3}, §23].) We will show at the end of § 4 that the Stability Theorem is a quick corollary of our theorem.

When the torsion-free abelian grade group Γ_{E} of a graded division algebra E is finitely-generated (hence a free abelian group), E has a concrete description as an iterated twisted Laurent polynomial ring over the division ring E_{0}, as follows: Take any homogeneous x_{1}, …, x_{n} in the group E^{∗} of units of E such that Γ_{E} = ℤ deg(x_{1}) ⊕ … ⊕ ℤ deg(x_{n}). Then E = E_{0}[x_{1}, x_{1}^{−1}, …, x_{n}, x_{n}^{−1}; σ_{1}, …, σ_{n}], i.e., E = E_{n}, where for i = 1, 2, …, n, E_{i} = E_{i−1}[x_{i}, x_{i}^{−1}; σ_{i}], which is a Laurent polynomial ring with multiplication twisted by the relation x_{i}c = σ_{i}(c)x_{i} for all c ∈ E_{i−1}. Each σ_{i} is a graded (i.e., degree-preserving) automorphism of E_{i−1}, and σ_{i} is completely determined by its action on E_{0} and on x_{1}, x_{2}, …, x_{i−1}. To assure that E is finite-dimensional over its center, it is assumed that some power of σ_{i} is an inner automorphism of E_{i−1}. The quotient division ring q(E) is the iterated twisted rational division algebra

A_{n} = E_{0}(x_{1}, …, x_{n}; σ_{1}, …, σ_{n}),  (1.2)

which is also the quotient division algebra of the iterated twisted polynomial ring E_{0}[x_{1}, …, x_{n}; σ_{1}, …, σ_{n}].

In [Y_{5}] the second author gave formulas for SK_{1}(A_{n}, τ_{n}) for the iterated twisted rational division algebras A_{n} as in (1.2) above, for a unitary involution τ_{n} on A_{n} arising from a unitary graded involution on the iterated twisted polynomial ring E_{0}[x_{1}, …, x_{n}; σ_{1}, …, σ_{n}]. The crucial identity

Σ′_{τ_{n}}(A_{n}) = (Σ′_{τ_{n}}(A_{n}) ∩ E_{0}) · Σ_{τ_{n}}(A_{n}).  (1.3)

is given in that paper, with a much too brief and incomplete sketch of proof. (More detailed proofs of this identity and other results in [Y_{5}] are given in this author’s habilitation, which, regrettably, was never published. A complete proof of the n = 1 case was given in [Y_{3}].) As will be seen in § 4 below, the proof of this identity, stated there as (4.1), constitutes the bulk of the proof of our Theorem. Our proof of this identity is by induction on n, as is done in [Y_{5}], but the proof differs susbstantially, in that it is carried out as much as possible by calculations in the divisor group Div(T) of the twisted polynomial ring in one variable T = E_{0}[x_{1}; σ_{1}]. This Div(T) is a free abelian group, and calculations with it are significantly simpler than with the quotient division ring q(T). The style of proof here is analogous to the approach in [HaW_{1}, § 5] to proving the corresponding result for the nonunitary SK_{1}(E) of a graded division algebra.

Th. 1.1 above, when combined with (1.1), provides a unifying perspective for understanding why there is such a great similarity between the formulas for SK_{1}(D, τ) for D a division algebra over a Henselian field, as in [Y_{2}] and [Y_{6}] and the formulas for SK_{1}(E_{0}(x_{1}, …, x_{n}; σ_{1}, …, σ_{n}), τ_{n}) given in [Y_{3}] and[Y_{5}]: The formulas in each case coincide with formulas for SK_{1}(E, τ) for a related graded division algebra E.

Theorem 1.1 and formula (1.1) are unitary analogues to results for nonunitary SK_{1} for division algebras over Henselian fields, graded division algebras, and their associated quotient division algebras given in [HaW_{1}, Th. 4.8, Th. 5.7]. This is another manifestation of the philosophy that results about SK_{1} ought to have corresponding results for the unitary SK_{1}. From the perspective of algebraic groups, this pholosophy is motivated by the fact that division algebras are associated with algebraic groups of inner type A_{n}, while division algebras with unitary involution are associated algebraic groups of outer type A_{n} (see, e.g., [KMRT, Ch. VI]).

2. Graded division algebras and unitary involutions

In this section we recall some basic known facts about graded division algebras and unitary involutions which will be used in the proof of Th. 1.1. A good reference for the properties of graded division algebras stated here without proof is [HwW].

Let Γ be a torsion-free abelian group. A ring E is a graded division ring (with grade group in Γ) if E has additive subgroups E_{γ} for γ ∈ Γ such that E = ⨁_{γ∈Γ} E_{γ} and E_{γ} · E_{δ} = E_{γ+δ} for all γ, δ ∈ Γ, and each E_{γ} \ {0} lies in E^{∗}, the group of units of E. For background on graded division rings and proofs of their properties mentioned here, see [HwW]. The grade group of E is

Γ_{E} = {γ ∈ Γ | E_{γ} ≠ {0} },

a subgroup of Γ. For a ∈ E_{γ}\{0} we write deg(a) = γ. A significant property is that E^{∗} = ⋃_{γ∈Γ_{E}} E_{γ}\{0}, i.e., every unit of E is actually homogeneous. (Γ_{E} torsion-free is needed for this.) Thus, E is not a division ring if |Γ_{E}| > 1. But, E has no zero divisors. (This also depends on having Γ_{E} torsion-free.) However, E_{0} is a division ring, and each E_{γ} (γ ∈ Γ_{E}) is a 1-dimensional left- and right- E_{0}-vector space.

Let M be any graded left E-module, i.e., M is a left M-module with additive subgroups M_{γ} such that M = ⨁_{γ∈Γ} M_{γ} and E_{γ} · M_{ε} ⊆ M_{γ+ε} for all γ, ε ∈ Γ. Then, M is a free E-module with homogeneous base, and any two such bases have the same cardinality, which is called the dimension, dim_{E}(M); M is therefore said to be a graded vector space over E.

Let Z = Z(E), the center of E, which is a graded subring of E. Indeed, Z is a graded field, i.e., a commutative graded division ring. Then E is a left (and right) graded Z-vector space, and we write [E:Z] for dim_{Z}(E). In this paper we work exclusively with finite-dimensional graded division algebras, i.e., those E with [E:Z] < ∞. Clearly Z_{0} is a field, and E_{0} is a finite-dimensional Z_{0}-algebra. Moreover, Γ_{Z} is a subgroup of Γ_{E}, and it is easy to verify the “fundamental equality”

[E:Z] = [E_{0}:Z_{0}] |Γ_{E}:Γ_{Z}|.  (2.1)

We have Z_{0} ⊆ Z(E_{0}) ⊆ E_{0}. Let G(Z(E_{0})/Z_{0}) be the Galois group for the finite-degree field extension Z(E_{0}) of Z_{0}. There is a well-defined canonical map

θ_{E} : Γ_{E} → G(Z(E_{0})/Z_{0})  given by  θ_{E}(deg(a)): c ↦ aca^{−1} for all a ∈ E^{∗}, c ∈ E_{0}.

clearly Γ_{Z} ⊆ ker(θ_{E}), so | im(θ_{E})| ≤ |Γ_{E}:Γ_{Z}| < ∞. Moreover, the fixed field of im(θ_{E}) is Z_{0}. Hence, Z(E_{0}) is Galois over Z_{0} with abelian Galois group G(Z(E_{0})/Z_{0}) = im(θ_{E}).

Since Z is a commutative ring with no zero divisors, it has a quotient field q(Z). Then E has its ring of central quotients

q(E) = E ⊗_{Z} q(Z).

Because E is a free, hence torsion-free Z-module, the canonical map E → q(E), a ↦ a ⊗ 1, is injective. Therefore, we view E as a subring of q(E). Note that q(E) has no zero divisors since E has none. Furthermore, q(E) is a q(Z)-algebra with [q(E):q(Z)] = [E:Z] < ∞. Hence, q(E) is a division ring, called the quotient division algebra of E. Clearly, Z(q(E)) = q(Z). The index of E is defined to be ind(E) = \sqrt{[E:Z]} = ind(q(E)) ∈ ℤ.

It is known that E is an Azumaya algebra over Z, and hence from general principles that there is a reduced norm map Nrd_{E} : E → Z. In fact, by [HaW_{1}, Prop. 3.2(i)], Nrd_{E} coincides with the restriction to E of the usual reduced norm Nrd_{q(E)} on q(E). Also, by [HaW_{1}, Prop. 3.2(iv)], for a ∈ E_{0}, we have

Nrd_{E}(a) = N_{Z(E_{0})/Z_{0}}(Nrd_{E_{0}}(a))^{λ}  where λ = ind(E)/( ind(E_{0}) · [Z(E_{0}:Z_{0}]),  (2.2)

where Nrd_{E_{0}} is the reduced norm for E_{0} and N_{Z(E_{0})/Z_{0}} is the field norm from Z(E_{0}) to Z_{0}.

A graded involution on the graded division algebra E is a ring antiautomorphism τ: E → E such that τ^{2} = id_{E} and τ(E_{γ}) = E_{γ} for each γ ∈ Γ. Such a τ is said to be unitary (or of the second kind) if τ|_{Z} ≠ id_{Z} where Z = Z(E). Assuming τ is unitary, let F = Z^{τ} = {c ∈ Z | τ(c) = c}, which is a graded subfield of Z with [Z:F] = 2. It follows from the fundamental equality that either [Z_{0}:F_{0}] = 2 and Γ_{Z} = Γ_{F} or Z_{0} = F_{0} and |Γ_{Z}:Γ_{F}| = 2. The second case, where the involution induced by τ on E_{0} is not unitary, tends to be uninteresting, but it can occur. We write τ also for the induced involution τ ⊗ id_{q(F)} on q(E) = E ⊗_{F} q(F). Recall that for all a ∈ E we have

Nrd_{E}(τ(a)) = τ(Nrd_{E}(a)),  (2.3)

since this equality holds for Nrd_{q(E)}. (For, if a ∈ q(E), then Nrd_{q(E)}(a) is determined by the minimal polynomial p_{a} of a over Z(q(E)), and p_{τ(a)} = τ(p_{a}).)

If τ′ is another unitary graded involution on E, we write τ ∼ τ′ if τ|_{Z} = τ′|_{Z}. In particular, for any c ∈ E^{∗}, if τ(c)c^{−1} ∈ Z^{∗}, then τ′ = int(c) ∘ τ is a unitary graded involution on E, and τ′ ∼ τ. Here int(c) denotes the inner automorphism a ↦ cac^{−1} of E. Since c is homogeneous, int(c) is clearly a graded (i.e., degree-preserving) automorphism of E.

For a unitary graded involution τ on E, set S_{τ} (E) = {a ∈ E^{∗} | τ(a) = a}, the set of symmetric units of E, and set

Σ_{τ} (E) = ⟨S_{τ} (E)⟩ and Σ′_{τ} (E) = {a ∈ E^{∗} | Nrd_{E}(a) ∈ S_{τ} (E)}.

Then, by definition, SK_{1}(E, τ) = Σ′_{τ} (E)/Σ_{τ} (E).

We recall a few fundamental facts about unitary involutions on ungraded division algebras which have analogues for graded division aglebras:

Lemma 2.1. Let D be a division algebra finite-dimensional over its center, and let τ and τ′ be unitary involutions on D.

(a) If τ ∼ τ′ (i.e., τ|_{Z(D)} = τ′|_{Z(D)}), then Σ′_{τ′}(D) = Σ′_{τ} (D) and Σ_{τ′}(D) = Σ_{τ} (D), so SK_{1}(D, τ′) = SK_{1}(D, τ).

(b) [D^{∗}, D^{∗}] ⊆ Σ_{τ} (D), where [D^{∗}, D^{∗}] = ⟨aba^{−1}b^{−1} | a, b ∈ D^{∗}⟩.

For a proof of (a), see [Y_{1}, Lemma 1], and for (b) see [KMRT, Prop. 17.26, p. 267]. Part (b) was originally proved by Platonov and Yanchevskiĭ. See [HaW_{2}, Remark 4.1(ii), Lemma 2.3(iv)] for the graded versions of (a) and (b).

3. The divisor group of a twisted polynomial ring

The proof of Th. 1.1, both in the case Γ_{E} ≅ ℤ and also in the induction argument for Γ_{E} ≅ ℤ^{n} will use properties of twisted polynomial rings in one variable over a division ring. In this section we give the properties we need about the divisor group of such a twisted polynomial ring.

Let D be a division ring finite dimensional over its center K, and let σ be an automorphism of D whose restriction to K has finite order, and let T be the twisted polynomial ring

T = D[x; σ],

consisting of polynomials ∑_{i=0}^{k} a_{i}x^{i} with a_{i} ∈ D, with the usual addition of polynomials, but multiplication twisted by σ, so that

(ax^{i})(bx^{j}) = aσ^{i}(b)x^{i+j} for all a, b ∈ D, i, j ≥ 0.

For the factorization theory of such rings T, [J, Ch. 1] is an excellent reference. Let A = q(T) = D(x; σ), the quotient division ring of T, which is a twisted rational function division algebra in one variable. We will make fundamental use of the “divisor group” Div(T) described in [HaW_{1}, § 5]: Let S be the set of isomorphism classes [S] of simple left T -modules S; then

Div(T) = ⨁_{[S]∈S} ℤ[S],

the free abelian group on S. (Note that in the commutative case when D is a field and σ = id, then T is a polynomial ring, and Div(T) is its usual divisor group. This is the source of the terminology.) For simple T -modules S, S′, we have ann_{T} (S) and ann_{T} (S′) are maximal two-sided ideals of T, and S ≅ S′ iff ann_{T} (S) = ann_{T} (S′). Thus, we could have indexed Div(T) by the maximal two-sided ideals of T; but, the indexing by simple modules is more natural for our purposes. We call an element α = ∑ n_{[S]}[S] of Div(T) a divisor, and call it an effective divisor if every n_{[S]} ≥ 0.

Also, there is a degree homomorphism

deg : Div(T) → ℤ given by deg ( ∑ n_{[S]}[S]) = ∑ n_{[S]} dim_{D}(S).

Note that if M is any left T -module of finite length (equivalently, finite-dimensional as a D-vector space), then M determines an effective Jordan-Hölder divisor

jh(M) = ∑_{i=1}^{k}[M_{i}/M_{i−1}],

where {0} = M_{0} ⫋ M_{1} ⫋ … ⫋ M_{k} = M is a chain of T -submodules of M with each M_{i}/M_{i−1} simple. The Jordan-Hölder Theorem shows that jh(M) is well-defined.

Every simple left T -module has the form T/T p for some p ∈ T with p irreducible, i.e., p has no factorization into a product of terms of positive degree. For nonzero f ∈ T with deg(f) > 0, the division algorithm shows that every nonzero element of T/T f has the form s + T f for some s ∈ T with deg(s) < deg(f). Using this, it is easy to check (cf. [HaW_{1}, Lemma 5.2]) that for nonzero f, g ∈ T of positive degree

T/T f ≅ T/T g iff deg(f) = deg(g) and there exist nonzero s, t ∈ T with
deg(s) = deg(t) < deg(f) and ft = sg.

(3.1)

There is a divisor function

δ: T \ {0} → Div(T) given by δ(f) = jh(T/T f).

It is easy to check that δ(fg) = δ(f) + δ(g), hence δ extends to a well-defined map δ: A^{∗} → Div(T) given by δ(fz^{−1}) = δ(f) − δ(z) for all f ∈ T \ {0}, z ∈ Z(T) \ {0}. Clearly, δ(T) is the monoid of effective divisors in Div(T). Note also that deg(δ(f)) = deg(f) for all f ∈ T \ {0}. It is proved in [HaW_{1}, Prop. 5.3] that there is an exact sequence:

1 ⟶ [A^{∗}, A^{∗}]D^{∗} ⟶ A^{∗} ⟶^{δ} Div(T) ⟶ 0  (3.2)

Let R = Z(T); so R is the polynomial ring K^{σ}[y], where K^{σ} = {b ∈ K | σ(b) = b} and y = c^{−1}x^{m}, with m minimal such that σ^{m}|_{K} = id and c ∈ D^{∗} satisfying σ^{m} = int(c) on D. Note that q(R) = Z(A). This R has its own divisor group Div(R), definable just as with Div(T), and there is a corresponding divisor map δ_{R}: q(R)^{∗} → Div(R). The reduced norm map Nrd_{A}: A → Z(A) maps T to R, (as T is integral over R, and R is integrally closed), and it is shown in [HaW_{1}, Prop. 5.4] that there is a corresponding “reduced norm” map DNrd : Div(T) → Div(R) which is injective, and such that the following diagram commutes:

A^{∗} ⟶^{δ} Div(T)
Nrd_{A}↓  _{DNrd}↓
q(R)^{∗} ⟶^{δ_{R}} Div(R)

(3.3)

Of course, T is a graded ring, with T_{i} = Dx^{i} for all nonnegative i ∈ ℤ. Suppose ρ: T → T is a graded automorphism of T, i.e., a ring automorphism such that ρ(Dx^{i}) = Dx^{i} for all i. Then, ρ restricts to a graded automorphism of R and also determines ring automorphisms of A and q(R), all denoted ρ.

Lemma 3.1. The graded automorphism ρ of T determines an automorphism of Div(T) mapping S to itself, and also determines an automorphism of Div(R). The automorphisms determined by ρ on each of the terms in diagram (3.3) are compatible with the maps in that diagram.

Proof. Since ρ(D^{∗}) = D^{∗} and ρ([A^{∗}, A^{∗}]) = [A^{∗}, A^{∗}], the exact sequence (3.2) shows that ρ determines a well-defined automorphism ρ of Div(T) given by ρδ(a) = δ(ρ(a)) for all a ∈ A^{∗}. Any class in S has the form [T/T p] for some irreducible p in T. Then, ρ(p) is also irreducible in F; so, ρ[T/T p] = [T/T ρ(p)] ∈ S. For Div(R), ρ is defined analogously by ρ(δ_{R}(q)) = δ_{R}(ρ(q)) for all q ∈ q(R)^{∗}. Since ρ is an automorphism of A, we have ρ(Nrd_{A}(a)) = Nrd_{A}(ρ(a)) for all a ∈ A. For, Nrd_{A}(a) is determined by the constant term of the minimal polynomial p_{a} ∈ Z(A)[X] of a over Z(A), and p_{ρ(a)} = ρ(p_{a}). Thus, we have ρ ∘ δ = δ ∘ ρ, ρ ∘ δ_{R} = δ_{R} ∘ ρ, and ρ ∘ Nrd_{A} = Nrd_{A} ∘ρ. Since diagram (3.3) commutes and δ is surjective, it follows that ρ ∘ DNrd = DNrd ∘ρ. □

Suppose τ: T → T is a unitary graded involution. That is, τ is a ring antiautomorphism of T with τ^{2} = id_{T}, τ|_{R} ≠ id, and τ(Dx^{i}) = Dx^{i} for all i. (The last condition is equivalent to: τ(D) = D and τ(x) = dx for some d ∈ D^{∗}.) This τ extends to a unitary involution on A given by τ(fr^{−1}) = τ(r)^{−1}τ(f) for all f ∈ T, r ∈ R \ {0}. Since τ([A^{∗}, A^{∗}]) = [A^{∗}, A^{∗}] and τ(D^{∗}) = D^{∗}, the exact sequence (3.2) above shows that τ determines an automorphism of order at most 2 of Div(T), also denoted τ. This map τ: Div(T) → Div(T) is given by τ(δ(a)) = δ(τ(a)) for all a ∈ A^{∗}. (For a simple left T -module S, we have S ≅ T/T p for some irreducible p ∈ T. Then, τ(p) is irreducible in T, and τ[S] = [T/T τ(p)]. It may seem surprising that this is well-defined, independent of the choice of p, but the well-definition follows easily from (3.1). In terms of two-sided ideals, τ[S] is the isomorphism class of simple left T -modules with annihilator τ(ann_{T} (S)).) Let Div(T)^{τ} = {α ∈ Div(T) | τ(α) = α}.

Lemma 3.2. Let τ be a unitary graded involution on T. Suppose τ|_{D} ≠ id or T is noncommutative. Then,

δ(Σ_{τ} (A)) = Div(T)^{τ} .

Proof. Suppose first that τ|_{D} ≠ id.

Let Ω = δ(Σ_{τ} (A)) ⊆ Div(T). Since δ maps generators of Σ_{τ} (A) into Div(T)^{τ}, we have Ω ⊆ Div(T)^{τ}. We must prove that this inclusion is equality. Suppose that Ω ⫋ Div(T)^{τ}.

Note that if α ∈ Div(T), say α = δ(a), then α + τ(α) = δ(aτ(a)) ∈ Ω. Take any η ∈ Div(T)^{τ} \ Ω. We can write η = α − β, where α and β are effective divisors. Then,

α + τ(β) = η + (β + τ(β)) ≡ η (mod Ω).

So, α + τ(β) is an effective divisor in Div(T)^{τ} \ Ω.

Let ξ be an effective divisor in Div(T)^{τ} \ Ω of minimal degree. Necessarily deg(ξ) > 0, as ξ ≠ 0. Say ξ = δ(z) for some z ∈ T. So, deg(z) = deg(ξ) > 0. Factor z into irreducibles in T, say z = p_{1} … p_{ℓ}, and let π_{i} = δ(p_{i}). So,

π_{1} + … + π_{ℓ} = η = τ(η) = τ(π_{1}) + … + τ(π_{ℓ}).

Because the π_{i} and τ(π_{i}) are all part of the ℤ-base of the free abelian group Div(T), we must have π_{1} = τ(π_{j}) for some index j. Suppose j > 1. Then, π_{1} + π_{j} = π_{1} + τ(π_{j}) ∈ Ω. Let ξ′ = ξ − (π_{1} + π_{j}). Then, ξ′ ≡ ξ (mod Ω). But, ξ′ is an effective divisor with deg(ξ′) < deg(ξ). This contradicts the minimality of deg(ξ). So, we must have j = 1, i.e., τ(π_{1}) = π_{1}. If π_{1} ∈ Ω, then ξ − π_{1} is an effective divisor in Div(T)^{τ} \ Ω of degree less than deg(ξ). Hence π_{1} ∈ Div(T)^{τ} \ Ω, so in fact ξ = π_{1} by the minimality of deg(ξ).

To simplify notation, let p = p_{1} and π = π_{1}. Since p and τ(p) are irreducible, we have π = [T/T p] and τ(π) = [T/T τ(p)]. Hence, the equality π = τ(π) implies that T/T p ≅ T/T τ(p). Therefore, by (3.1), there exist f, g ∈ T \ {0} with deg(f) = deg(g) < deg(p) and

pf = gτ(p).  (3.4)

Suppose first that τ(f) = g. Then, pf = τ(f)τ(p) = τ(pf), so pf ∈ Σ_{τ} (A), hence δ(pf) ∈ Ω. Then, δ(f) = δ(pf) − π ≡ −π (mod Ω). Therefore, δ(f) is an effective divisor in Div(T)^{τ} \ Ω with

deg(δ(f)) = deg(f) < deg(p) = deg(π) = deg(ξ).

This contradicts the minimality of deg(ξ). So, we must have τ(f) ≠ g.

Now suppose that τ(f) ≠ −g. Then, let f′ = f + τ(g) ∈ T \ {0}, and g′ = τ(f′) = τ(f) + g. By applying τ to (3.4), we have pτ(g) = τ(f)τ(p), which when added to (3.4) yields pf′ = g′τ(p). But then the preceding argument for the case τ(f) = g applies for f′, as f′ ≠ 0 and τ(f′) = g′; it shows that δ(f′) ∈ Div(T)^{τ} \ Ω. Since deg(f′) ≤ max ( deg(f), deg(τ(g)) = deg(f) < deg(ξ), this contradicts the minimality of deg(ξ).

We thus have τ(f) = −g while τ(f) ≠ g. Hence, char(D) ≠ 2. Since we have assumed τ|_{D} ≠ id, there is c ∈ D^{∗} with τ(c) = −c. Then,

pfc = gτ(p)c = c^{−1}cgτ(p)c = c^{−1}τ(pfc)c.  (3.5)

Since τ(c) = −c, the map τ′ = int(c^{−1}) ∘ τ is a unitary involution on A with τ′ ∼ τ. Hence, by Lemma 2.1(a), Σ_{τ′}(A) = Σ_{τ} (A). Since equation (3.5) shows pfc = τ′(pfc), we thus have pfc ∈ Σ_{τ′}(A) = Σ_{τ} (A), and hence δ(pfc) ∈ Ω. This shows that δ(fc) ≡ −π (mod Ω), and hence δ(fc) ∈ Div(T)^{τ} \ Ω. But, δ(fc) is an effective divisor, with deg(δ(fc)) = deg(f) < deg(π) = deg(ξ). Thus, we have a contradiction to the minimality of deg(ξ). The contradiction arose from the assumption that Ω ⫋ Div(T)^{τ}. So, we must have Div(T)^{τ} = Ω = δ(Σ_{τ} (A)), as desired.

We have thus far assumed that τ|_{D} ≠ id. Now, suppose that τ_{D} = id. Then, D must be commutative, since τ is an antiautomorphism. Our hypothesis now is that T is noncommutative; therefore, σ = int(x)|_{D} ≠ id. Since, τ(Dx) = Dx there is a nonzero d ∈ D with τ(dx) = ±dx. Then let τ̃ = int(dx) ∘ τ, which is a unitary graded involution of T with τ̃ ∼ τ as unitary involutions on A. Hence, Στ̃(A) = Σ_{τ} (A) by Lemma 2.1(a). Also, for any a ∈ A^{∗},

τ̃(δ(a)) = δ(τ̃(a)) = δ(dx τ(a)(dx)^{−1}) = δ(dx) + δ(τ(a)) − δ(dx) = δ(τ(a)) = τ(δ(a)).

Thus, τ̃ and τ have the same action on Div(T). Furthermore, τ̃|_{D} = int(dx)|_{D} = σ ≠ id. Therefore, the preceding argument shows that the Lemma holds for τ̃. This yields for τ,

δ(Σ_{τ} (A)) = δ(Στ̃(A)) = Div(T)τ̃ = Div(T)^{τ} .

Thus, the lemma holds in all cases. □

Remark. The assumption that τ|_{D} ≠ id or T is noncommutative is definitely needed for Lemma 3.2. For example, suppose that T is commutative and τ is defined by τ|_{D} = id and τ(x) = −x. Then, δ(x) ∈ Div(T)^{τ} \ δ(Σ_{τ} (A)).

Now we combine the action of the graded unitary involution τ on Div(T) with a finite group action. Let H be a finite abelian group which acts on the set S of isomorphism classes of simple left T -modules. This action induces an action of H on Div(T), making Div(T) into a permutation module for H. There is an associated norm map N_{H}: Div(T) → Div(T) given by N_{H}(α) = ∑_{h∈H} hα. Suppose the actions of H and τ on Div(T) are related by

τ(hα) = h^{−1}τ(α)  for all h ∈ H, α ∈ Div(T).  (3.6)

Lemma 3.3. {α ∈ Div(T) | N_{H}(α) = N_{H}(τ(α))} = I_{H}(Div(T)) + ∑_{h∈H} Div(T)^{hτ} ,

where I_{H}(Div(T)) = ⟨hα − α | h ∈ H, α ∈ Div(T)⟩ and Div(T)^{hτ} = {α ∈ Div(T) | hτ(α) = α}.

Proof. Let G be the semidirect product group H⋊_{ψ}⟨τ⟩ built using the homomorphism ψ: ⟨τ⟩ → Aut(H) given by ψ(τ)(h) = h^{−1}. (This is well-defined as H is abelian.) More explicitly, G = H ∪ Hτ (disjoint union), where the multiplication is determined by that in H together with τh = h^{−1}τ and τ^{2} = 1. Note that G is a generalized dihedral group in the terminology of [HaW_{2}, § 2.4] since every element of G \ H has order 2. The hypothesis (3.6) shows that the actions of τ and H on Div(T) combine to yield a ℤ-linear group action of G on Div(T). Also, G sends S to S, since this is true for H and τ. Thus, Div(T) is a permutation G-module. From this group action we build a new action of G on Div(T), denoted ∗, given by

h ∗ α = hα and (hτ) ∗ α = −hτα for all h ∈ H, α ∈ Div(T).

Let { \atop Div(T)} denote Div(T) with this twisted G-action, and let Ñ_{G}: { \atop Div(T)} → { \atop Div(T)} be the associated norm map, given by

Ñ_{G}(α) = ∑_{h∈H} hα − ∑_{h∈H} hτα = N_{H}(α) − N_{H}(τα) for all α ∈ Div(T).

Thus, the Lemma describes ker(Ñ_{G}).

Note that while Div(T) is a permutation G-module, the twisted G-module { \atop Div(T)} need not be a permutation G-module, as the twisted action of G does not map S to itself. Write S = ⋃_{j∈J} O_{j}, where the O_{j} are the distinct G-orbits of S (for the original G-action). Then, Div(T) = ⨁_{j∈J} M_{j}, where each M_{j} = ⨁_{s∈O_{j}} ℤs, which is a G-submodule of Div(T). Let O be one of the orbits, and let M = ⨁_{s∈O} ℤs. Take any s ∈ O, and let V = H · s, which is an H-orbit within O. Then, τV = τH · s = Hτ · s, which is also an H-orbit in O, with V ∪ τV = Hs ∪ τHs = Gs = O. Let V = {s_{1}, …, s_{n}}. There are two possible cases:

Case I. V ∩ τV = ∅. Then, {s_{1}, …, s_{n}, −τ(s_{1}), …, −τ(s_{n})} is a ℤ-base of M which is mapped to itself by the ∗-action of G. Thus, M̃ (i.e., M with the twisted G-action) is a permutation G-module. Consequently, Ĥ^{−1}(G, M̃) = 0 (as is true for all G-modules). This means that

ker(Ñ_{G}) ∩ M = I_{G}(M̃) = ⟨g ∗ m − m | g ∈ G, m ∈ M⟩ = ⟨hm − m, −hτm − m | h ∈ H, m ∈ M⟩.

Each hm − m ∈ I_{H}(M) ⊆ I_{H}(Div(T)), while −hτm − m ∈ M^{hτ} ⊆ Div(T)^{hτ}. Thus,

ker(Ñ_{G}) ∩ M ⊆ I_{H}(Div(T)) + ∑_{h∈H} Div(T)^{hτ} .

Case II. V ∩ τV ≠ ∅. Then, V = τV, since they are each H-orbits. So, for each t ∈ V there is k ∈ H with t = τk^{−1}t = kτ t; so, t ∈ M^{kτ}. Thus,

ker(Ñ_{G}) ∩ M ⊆ M = ∑_{h∈H} M^{hτ} ⊆ ∑_{h∈H} Div(T)^{hτ} .

By combining the two cases, we obtain

ker(Ñ_{G}) = ∑_{j∈J} ( ker(Ñ_{G}) ∩ M_{j}) ⊆ I_{H}(Div(T)) + ∑_{h∈H} Div(T)^{hτ} .

This proves ⊆ in the Lemma, and the reverse inclusion is clear. □

4. Proof of the theorem

We can now prove Theorem 1.1.

Proof. Let E be a graded division ring (finite-dimensional over its center) with a unitary graded involution τ, and let Q = q(E). Because Nrd_{Q} |_{E} = Nrd_{E}, as noted in § 2, we have Σ′_{τ} (E) ⊆ Σ′_{τ} (Q). Also, clearly Σ_{τ} (E) ⊆ Σ_{τ} (Q). Therefore, there is a canonical map φ_{E} : SK_{1}(E, τ) → SK_{1}(Q, τ). We will show that φ_{E} is an isomorphism.

To see that φ_{E} is injective, give Γ_{E} a total ordering making it into a totally ordered abelian group. Then define a function λ: E \ {0} → E^{∗} as follows: For c ∈ E \ {0}, write c = ∑_{γ∈Γ_{E}} c_{γ} where each c_{γ} ∈ E_{γ}. Then set λ(c) = c_{ε}, where ε is minimal such that c_{ε} ≠ 0. It is easy to check that λ(cd) = λ(c)λ(d) for all c, d ∈ E \ {0}. It follows that λ extends to a well-defined group epimorphism λ: Q^{∗} → E^{∗} given by λ(cz^{−1}) = λ(c)λ(z)^{−1} for all c ∈ E \ {0}, z ∈ Z(E) \ {0}. Clearly λ|_{E^{∗}} = id. Note that as τ is a graded involution, we have λ(τ(c)) = τ(λ(c)) for all c ∈ E \ {0}, and hence for all c ∈ Q^{∗}. Now take any u ∈ Σ_{τ} (Q) ∩ E^{∗}. Then, u = s_{1} … s_{m} for some s_{i} ∈ Q^{∗} with each τ(s_{i}) = s_{i}. Then, u = λ(u) = λ(s_{1}) … λ(s_{m}) with τ(λ(s_{i})) = λ(τ(s_{i})) = λ(s_{i}); so, u ∈ Σ_{τ} (E). This shows that Σ_{τ} (Q) ∩ E^{∗} ⊆ Σ_{τ} (E). Therefore, the canonical map φ_{E} is injective.

To show that φ_{E} is surjective, we will prove the following:

Σ′_{τ} (Q) = (Σ′_{τ} (Q) ∩ E_{0}^{∗}) · Σ_{τ} (Q)  (4.1)

Note that surjectivity of φ_{E} follows immediately from (4.1) because Σ′_{τ} (Q) ∩ E_{0}^{∗} ⊆ Σ′_{τ} (E).

We next reduce to the case of a finitely-generated grade group. For any graded division algebra E, observe that if Δ is a subgroup of Γ_{E}, then E has the graded division subalgebra E_{Δ} = ⨁_{ε∈Δ} E_{ε}, which is finite-dimensional over its center with Γ_{E_{Δ}} = Δ. Let Q_{Δ} = q(E_{Δ}) ⊆ Q. Take any γ ∈ Γ_{Z(E)} such that τ|_{Z(E)_{γ}} ≠ id and any c_{1}, …, c_{m} ∈ E^{∗} such that the c_{i} span E as a graded Z(E)-vector space. Then, the c_{i} also span Q as a Z(Q)-vector space. So, if Δ is any finitely-generated subgroup of Γ_{E} such that γ ∈ Δ and deg(c_{1}), …, deg(c_{m}) ∈ Δ, then τ|_{E_{Δ}} is a unitary graded involution and the map Q_{Δ} ⊗_{Z(Q_{Δ})} Z(Q) → Q is surjective. This map is also injective, as its domain is simple since Q_{Δ} is central simple and finite-dimensional over Z(Q_{Δ}). Therefore, [Q_{Δ}:Z(Q_{Δ})] = [Q:Z(Q)], and hence Nrd_{Q_{Δ}} = Nrd_{Q} |_{Q_{Δ}}, so Σ′_{τ} (Q_{Δ}) = Σ′_{τ} (Q) ∩ Q_{Δ}^{∗}. Now, E = \underrightarrow{lim} E_{Δ} and Σ′_{τ} (Q) = \underrightarrow{lim} Σ′_{τ} (Q_{Δ}) as Δ ranges over finitely-generated subgroups of Γ_{E} containing γ and deg(c_{1}), …, deg(c_{m}); therefore, equality (4.1) holds for E if it holds for each E_{Δ}. Thus, it suffices to prove (4.1) for E with Γ_{E} finitely-generated.

Henceforward, suppose Γ_{E} is a finitely generated, hence free, abelian group, say Γ_{E} = ℤγ_{1} ⊕…⊕ℤγ_{n}. Take any nonzero x_{i} ∈ E_{γ_{i}} for i = 1, 2, …, n. Then, E = E_{0}[x_{1}, x_{1}^{−1}, ….x_{n}, x_{n}^{−1}], an iterated twisted Laurent polynomial ring over the division ring E_{0}. We prove (4.1) by induction on n.

Suppose first that n = 1. Let D = E_{0} and x = x_{1}, and let T = D[x], which is a graded subring of E. So, T is the twisted polynomial ring D[x; σ], where σ = int(x)|_{D}. Then, σ|_{Z(D)} has finite order since [Z(D):Z(E)_{0}] < ∞. To be consistent with the notation of § 3, let A = q(T) = q(E) = Q. Since τ is a graded unitary involution on E, its restriction τ|_{T} is a graded unitary involution on T. Let Div(T) and δ: A^{∗} → Div(T) be as in § 3. If T is commutative, then A is also commutative, so SK_{1}(A) = 1 and (4.1) holds trivially. Thus, we may assume that T is noncommuative. Take any a ∈ Σ′_{τ} (A). So, Nrd_{A}(a) = τ(Nrd_{A}(a)) = Nrd_{A}(τ(a)) by (2.3). Recall from [HaW_{1}, Remark 5.1(iv)] that δ(Nrd_{A}(a)) = rδ(a), where r = ind(A). Hence,

r τ(δ(a)) = r δ(τ(a)) = δ(Nrd_{A}(τ(a)) = δ(Nrd_{A}(a)) = r δ(a).

Then, τ(δ(a)) = δ(a), as Div(T) is torsion-free. By Lemma 3.2, we have δ(a) = δ(b) for some b ∈ Σ_{τ} (A). It then follows from the exact sequence (3.2) that a = bcd for some c ∈ [A^{∗}, A^{∗}] and d ∈ D^{∗}. Recall that [A^{∗}, A^{∗}] ⊆ Σ_{τ} (A) by Lemma 2.1(b). So, bc ∈ Σ_{τ} (A) ⊆ Σ′_{τ} (A). Hence, d = (bc)^{−1}a ∈ Σ′_{τ} (A) ∩ D^{∗}, so a = (bc)d ∈ Σ_{τ} (A)(Σ′_{τ} (A) ∩ D^{∗}). This proves the inclusion ⊆ in (4.1). The reverse inclusion is clear. Thus, (4.1) holds when n = 1.

Now assume n > 1. By induction (4.1) holds for all graded division algebras with grade group of rank less than n. Let x = x_{1} as above, and let

Z = Z(E);
D = E_{0};
T = D[x];
R = Z(T);
A = q(T);
B = A[x_{2}, x_{2}^{−1}, …, x_{n}, x_{n}^{−1}] ⊆ Q;
C = Z(B).

For a clearer picture of B and C, let U = E_{0}[x_{1}, x_{1}^{−1}], which is a graded division subalgebra of E with U_{0} = E_{0}, Γ_{U} = ℤγ_{1}, and q(U) = A. Let F = U∩Z, which is a graded field with F_{0} = Z_{0} and Γ_{F} = Γ_{Z}∩Γ_{U}. So,

|Γ_{U}:Γ_{F}| = |Γ_{E} ∩ ℤγ_{1} : Γ_{Z} ∩ ℤγ_{1}| ≤ |Γ_{E}:Γ_{Z}| < ∞,

and hence [U:F] = [U_{0}:F_{0}] |Γ_{U}:Γ_{F}| = [E_{0}:Z_{0}] |Γ_{U}:Γ_{F}| < ∞. Since F is central in U, we have A = q(U) = q(F) ⊗_{F} U. Hence, B = q(F) ⊗_{F} E, so C = Z(B) = q(F) ⊗_{F} Z, from which it is clear that the rank of B as a free C-module equals [E:Z], which is finite. Let E′ = U[x_{2}, x_{2}^{−1}, …, x_{n}, x_{n}^{−1}] = E, but we grade E′ so that E′_{0} = U, Γ_{E′} = ℤγ_{2} ⊕ … ⊕ ℤγ_{n}, and E′_{ε} = ⨁_{j∈ℤ} E_{jγ_{1}+ε}, for each ε ∈ Γ_{E′}. Now, B is obtained from E′ by inverting nonzero elements of F, which are all central in E′ and homogeneous of degree 0. So, the grading on E′ extends to a grading on B with B_{0} = A and Γ_{B} = Γ_{E′} = ℤγ_{2} ⊕ … ⊕ ℤγ_{n}; this grading restricts to a grading on C = Z(B). Note that B is a graded division ring since B_{0} = A is a division ring and each homogeneous component B_{ε} of B contains a unit of B (namely any nonzero homogeneous element of E in B_{ε}). We saw above that [B:C] < ∞. Also, q(B) = Q, as E ⊆ B ⊆ Q = q(E). Our unitary involution τ is a graded involution for E, so also a graded involution for E′, hence also a graded involution for B. Moreover, τ is a unitary involution for B, as τ|_{Z} ≠ id and Z = Z(E) ⊆ B∩Z(Q) ⊆ Z(B). Since Γ_{B} has ℤ-rank n − 1, (4.1) holds for τ on B by induction, i.e.

Σ′_{τ} (Q) = (Σ′_{τ} (Q) ∩ A^{∗}) · Σ_{τ} (Q).

Therefore, to prove (4.1) for E, it suffices to prove

Σ′_{τ} (Q) ∩ A^{∗} ⊆ (Σ′_{τ} (q) ∩ D^{∗}) · Σ_{τ} (Q).  (4.2)

For this, we will work with Div(T) and the action on it by τ and various automorphisms. Note that while τ is unitary as an involution on B, its restriction to A need not be unitary. But, if τ|_{A} is not unitary, then τ|_{Z(A)} = id, so τ|_{C_{0}} = id, hence C is totally ramified over C^{τ}, i.e., C_{0} = (C^{τ})_{0}. But then SK_{1}(B, τ) = 1 by [HaW_{2}, Th. 4.5]. So, SK_{1}(Q, τ) = SK_{1}(B, τ) = 1 since (4.1) holds for B, whence (4.2) and (4.1) for E obviously hold. Thus, we will assume from now on that τ|_{A} is unitary.

Assume that T is not commutative.

Since B is a graded division algebra and C = Z(B), it is known (see § 2) that Z(B_{0}) is Galois over C_{0}, and there is a well-defined epimorphism θ_{B} : G(Z(B_{0})/C_{0}) given by ε ↦ int(b_{ε})|_{Z(B_{0})} for any nonzero b_{ε} in B_{ε}. Here, B_{0} = A. Let H = G(Z(A)/C_{0}), which is abelian, as θ_{B} is surjective. We next define a group action of H on Div(T) so as to be able to use Lemma 3.3.

We use the notation associated to T and Div(T) as in § 3, including the epimormphism δ: A^{∗} → Div(T). Observe that conjugation by an element of E^{∗} yields a degree preserving automorphism of T, and hence a well-defined automorphism of Div(T). That is, for any y ∈ E^{∗} and a ∈ A^{∗}, define

y · δ(a) = δ(yay^{−1}).  (4.3)

This is well-defined because ker(δ) = [A^{∗}, A^{∗}]D^{∗} (see (3.2)), which int(y) maps to itself. For any irreducible p of the twisted polynomial ring T, its conjugate ypy^{−1} is also irreducible in T, and y · [T/T p] = y · δ(p) = [T/T ypy^{−1}]. Hence, the action of y maps the distinguished ℤ-base S of Div(T) to itself. Therefore, the group action of E^{∗} on Div(T) given by (4.3) makes Div(T) into a permutation E^{∗}-module. Note that if u ∈ E^{∗} and int(u)|_{Z(A)} = id, then by Skolem-Noether there is c ∈ A^{∗} with int(u)|_{A} = int(c). Then, for a ∈ A,

u · δ(a) = δ(cac^{−1}) = δ(c) + δ(a) − δ(c) = δ(a),  (4.4)

so u acts trivially on Div(T). Now, for any y ∈ E^{∗}, we have τ ∘ int(y) ∘ τ ∘ int(y) = int(τ(y)^{−1}y). Since τ preserves degrees, we have τ(y)^{−1}y ∈ E_{0} = D ⊆ A, hence int(τ(y)^{−1}y)|_{Z(A)} = id. Therefore, (4.4) with u = τ(y)^{−1}y shows that τyτy · α = α for all α ∈ Div(T), i.e.,

τy · α = y^{−1}τ · α.  (4.5)

We obtain from the E^{∗}-action an induced action of H on Div(T): For any h ∈ H, choose any y_{h} ∈ E^{∗} such that int(y_{h})|_{Z(A)} = h. (Such a y_{h} exists because the composition E^{∗} → B^{∗} → Γ_{B} is surjective, as is θ_{B} : Γ_{B} → H.) Then, for α ∈ Div(T), set h · α = y_{h} · α. If we had made a different choice of y_{h}, say y′_{h}, then as int(y_{h}^{′−1}y_{h})|_{Z(A)} = h^{−1}h = id, the calculation in (4.4) shows that y′_{h} · α = y_{h} · α for all α ∈ Div(T). Thus, the action of H on Div(T) is well-defined, independent of the choice of the y_{h}. So, Div(T) is a permutation H-module, and from (4.5), we have τh · α = h^{−1}τ · α for all h ∈ H and α ∈ Div(T). Therefore, Lemma 3.3 applies for the norm map N_{H} of H on Div(T). We interpret the terms appearing in that lemma: First,

I_{H}(Div(T)) = δ([E^{∗}, A^{∗}]).  (4.6)

For, to see ⊆, take any generator h · β − β of I_{H}(Div(T)), and take any b ∈ A^{∗} with δ(b) = β. Then,

h · β − β = δ(y_{h}by_{h}^{−1}) − δ(b) = δ(y_{h}by_{h}^{−1}b^{−1}) ∈ δ([E^{∗}, A^{∗}]).

For the reverse inclusion, take any z ∈ E^{∗} and a ∈ A^{∗}. Let h = int(z)|_{Z(A)}. Since we could have chosen z for y_{h}, we have

δ(zaz^{−1}a^{−1}) = δ(zaz^{−1}) − δ(a) = h · δ(a) − δ(a) ∈ I_{H}(Div(T)).

This proves (4.6).

For any h ∈ H, we have selected y_{h} ∈ E^{∗} so that int(y_{h})|_{Z(A)} = h. For any d ∈ D = E_{0}, we have int(dy_{h})|_{Z(A)} = int(y_{h})|_{Z(A)}. Since E_{deg(y_{h})} = E_{0}y_{h}, the graded involution τ maps E_{0}y_{h} to itself. Hence, there is d_{h} ∈ E_{0} with τ(d_{h}y_{h}) = ±d_{h}y_{h}. Then, replace y_{h} by d_{h}y_{h}, and set τ_{h} = int(y_{h}) ∘ τ. Since τ(y_{h}) = ±y_{h}, the map τ_{h} is a unitary graded involution on E with τ_{h} ∼ τ on E. So, τ_{h} restricts to a graded involution on T. If τ_{h}|_{T} is not unitary, then τ_{h}|_{A} is not unitary, so SK_{1}(Q, τ_{h}) = 1 by the same argument as given above for τ. But, SK_{1}(Q, τ_{h}) = SK_{1}(Q, τ) as τ_{h} ∼ τ by Lemma 2.1(a), and the triviality of SK_{1}(Q, τ) implies (4.2) and (4.1). Thus, we are done if any τ_{h}|_{A} is not unitary. We therefore assume that each τ_{h}|_{A} is unitary, so τ_{h}|_{T} is a graded unitary involution. Since the action of τ_{h} on Div(T) coincides with that of hτ, Lemma 3.2 yields

Div(T)^{hτ} = δ(Σ_{τ_{h}}(A))  for each h ∈ H.  (4.7)

(The Lemma applies because we are assuming that T is noncommutative.)

We now prove (4.2). For this, take any a ∈ Σ′_{τ} (Q)∩A^{∗}, and let α = δ(a). Since A = B_{0} and Q = q(B), the norm formula (2.2) yields

Nrd_{Q}(a) = N_{Z(A)/C_{0}}(Nrd_{A}(a))^{λ} where λ = ind(Q)/([Z(A):C_{0}] · ind(A)).

Note that N_{Z(A)/C_{0}} is just the norm map N_{H} for the H-module Z(A). Here, Div(T) and Div(R) are also H-modules, and the maps δ_{R}: Z(A) → Div(R) and DNrd : Div(T) → Div(R) in diagram (3.3) are H-equivariant. This follows from Lemma 3.1 since for h ∈ H the action of h on Z(A), Div(R), and Div(T) is induced via int(y_{h}), which is a graded automorphism of T. Therefore, δ_{R} and DNrd commute with the norm maps N_{H} on Z(A), Div(R), and Div(T). Thus, for α = δ(a) we have, using commutative diagram (3.3),

δ_{R}(Nrd_{Q}(a)) = δ_{R}(N_{Z(A)/C_{0}}(Nrd_{A}(a))^{λ} = λδ_{R}N_{H}(Nrd_{A}(a)) = λN_{H}δ_{R}(Nrd_{A}(a))
= λN_{H}(DNrd δ(a)) = λ DNrd(N_{H}(α)).

Since a ∈ Σ′_{τ} (Q), we have Nrd_{Q}(a) = Nrd_{Q}(τ(a)). Thus,

λ DNrd(N_{H}(α)) = δ_{R}(N_{Q}(a)) = δ_{R}(N_{Q}(τ(a)) = λ DNrd(N_{H}(τ(α))).  (4.8)

But, Div(R) is a torsion-free ℤ-module, and DNrd is injective by [HaW_{1}, Prop. 5.4]. Therefore, equation (4.8) yields N_{H}(α) = N_{H}(τ(α)). By Lemma 3.3 and (4.6) and (4.7), we thus have

α ∈ I_{H}(Div(T)) + ∑_{h∈H} Div(T)^{hτ} = δ([E^{∗}, A^{∗}]) + ∑_{h∈H} δ(Σ_{τ_{h}}(A)).

Hence, there exist b ∈ [E^{∗}, A^{∗}] and c_{h} ∈ Σ_{τ_{h}}(A) for each h ∈ H with

δ(a) = δ(b) + ∑_{h∈H} δ(c_{h}) = δ(b ∏_{h∈H} c_{h}).

Then, as ker(δ) = [A^{∗}, A^{∗}]D^{∗}, there exist d ∈ D^{∗} and e ∈ [A^{∗}, A^{∗}] such that a = deb ∏_{h∈H} c_{h}. Let q = eb ∏_{h∈H} c_{h}. Since [A^{∗}, A^{∗}] ⊆ Σ_{τ} (Q) by Lemma 2.1(b), we have eb ∈ Σ_{τ} (Q). But also, each c_{h} ∈ Σ_{τ_{h}}(A) ⊆ Σ_{τ_{h}}(Q) = Σ_{τ} (Q), with the last equality given by Lemma 2.1(a), since τ_{h} ∼ τ as involutions on Q. Hence, q ∈ Σ_{τ} (Q) ⊆ Σ′_{τ} (Q). So, d = aq^{−1} ∈ D^{∗} ∩ Σ′_{τ} (Q), whence a ∈ (Σ′_{τ} (Q) ∩ D^{∗}) · Σ_{τ} (Q). This proves (4.2), under the assumption that T is noncommutative.

There remains the case where T is commutative. Suppose that int(x_{j})|_{D} ≠ id for some j > 1. Then, we can rearrange the order of the x_{i}, viewing the order of generators of E as x_{j}, x_{2}, …, x_{j−1}, x_{1}, x_{j+1}, …, x_{n} (and their inverses). Then, we have T = D[x_{j}], which is not commutative. So, we are back to the previously proved case; hence, (4.2) holds.

The final possibility is that each D[x_{j}] is commutative for j = 1, 2, …, n, i.e., D ⊆ Z(E). So, for each h ∈ H, we have τ_{h}|_{D} = int(y_{h})|_{D} ∘ τ|_{D} = τ|_{D}. In the basic argument for (4.2), we used the assumption that T is commutative only to be able to apply Lemma 3.2 for each τ_{h}. Now T is commutative, but if τ|_{D} ≠ id then also each τ_{h}|_{D} ≠ id, so Lemma 3.2 still applies for each τ_{h}; then the preceding argument for (4.2) again goes through. Thus, we may assume that τ|_{D} = id. (Hence, E_{0} = (Z(E^{τ}))_{0}, which implies that SK_{1}(E, τ) = 1 by [HaW_{2}, Th. 4.5]. But, we are not finished because we do not know that SK_{1}(Q, τ) = 1.) For any y ∈ E^{∗}, we have τ(y) = cy for some c ∈ E_{0}^{∗} = D^{∗}, as each E_{γ} is a τ-stable 1-dimensional E_{0}-vector space. Then y = τ(cy) = c^{2}y, so c^{2} = 1. Thus, τ(y) = ±y for each y ∈ E^{∗}. If τ(x_{1}) = 1, then τ|_{A} = id, which we saw earlier implies SK_{1}(Q, τ) = SK_{1}(B, τ) = 1, which trivially implies (4.2). Thus, we may assume that τ(x_{1}) = −x_{1}. Likewise, we may assume that τ(x_{j}) = −x_{j} for each j. For, if τ(x_{j}) = x_{j}, then after reordering the generators of E by interchanging x_{1} and x_{j}, we have A = D(x_{j}), so again τ|_{A} = id and SK_{1}(Q, τ) = 1. Similarly, we are done if τ(x_{1}x_{2}) = x_{1}x_{2}. For, we can then replace x_{1} by x_{1}x_{2} as a generator of E, as Γ_{E} = ℤ(γ_{1} + γ_{2}) ⊕ ℤγ_{2} ⊕ … ⊕ ℤγ_{n}. Then A = D(x_{1}x_{2}) and τ|_{A} = id, showing SK_{1}(Q, τ) = 1, as before. Thus, we may assume that τ(x_{1}x_{2}) = −x_{1}x_{2}. Then, x_{2}x_{1} = (−1)^{2}τ(x_{1}x_{2}) = −x_{1}x_{2}. Likewise, by reordering the x_{i}, we are done if τ(x_{i}x_{j}) = x_{i}x_{j} for any distinct i and j. So, we may assume that τ(x_{i}x_{j}) = −x_{i}x_{j}, and hence x_{j}x_{i} = −x_{i}x_{j}, whenever i ≠ j. If n ≥ 3, we then find that τ(x_{1}x_{2}x_{3}) = x_{1}x_{2}x_{3}. In this case, we can replace x_{1} by x_{1}x_{2}x_{3} as a generator of E, as Γ_{E} = ℤ(γ_{1} + γ_{2} + γ_{3}) ⊕ ℤγ_{2} ⊕ … ⊕ ℤγ_{n}. Then, A = D(x_{1}x_{2}x_{3}), and τ|_{A} = id, so SK_{1}(Q, τ) = 1, as before. The remaining case is that n = 2, D is central in E, τ|_{D} = id, τ(x_{i}) = −x_{i} for i = 1, 2, and x_{2}x_{1} = −x_{1}x_{2}. But then, Z(E) = D[x_{1}^{2}, x_{2}^{2}] and τ|_{Z(E)} = id, which is a contradiction as τ is unitary. Thus, in all cases that can occur (4.2) holds, so also (4.1). Hence, SK_{1}(Q, τ) = SK_{1}(E, τ), as desired. □

Here is a quick consequence of the Theorem:

Corollary 4.1 (Stability Theorem [Y_{3}]). Let D be a division algebra finite-dimensional over Z(D), and let τ be a unitary involution on D. Then, SK_{1}(D, τ) ≅ SK_{1}(D(x), τ′), where D(x) is the rational division algebra over D and τ′ is the canonical extension of τ to D(x), with τ′(x) = x.

Proof. Let E = D[x, x^{−1}], the (untwisted) Laurent polynomial ring in one variable over D. Then, E is a graded division algebra with E_{0} = D, Γ_{E} = ℤ, and E_{j} = Dx^{j} for all j ∈ ℤ. Also, q(E) = D(x) and Z(E) = Z(D)[x, x^{−1}], so [E:Z(E)] = [D:Z(D)] < ∞. The extension τ′ of τ to D(x) clearly restricts to a unitary graded involution on E. Note that E^{∗} = ⋃_{j∈ℤ} D^{∗}x^{j}. For d ∈ D^{∗} and j ∈ ℤ, we have τ′(dx^{j}) = τ(d)x^{j} and Nrd_{E}(dx^{j}) = Nrd_{D}(d)x^{mj}, where m = ind(D(x)) = ind(D). Hence, Σ′_{τ′}(E) = ⋃_{j∈ℤ} Σ′_{τ} (D) x^{j} and Σ_{τ′}(E) = ⋃_{j∈ℤ} Σ_{τ} (D) x^{j}, showing that SK_{1}(E, τ′) ≅ SK_{1}(D, τ). (This is a special case of the result in [HaW_{2}, Cor. 4.10] that if E is a graded division algebra with unitary graded involution τ, and Γ_{E} = Γ_{Z(E)^{τ}}, then SK_{1}(E, τ) ≅ SK_{1}(E_{0}, τ|_{E_{0}}).) Since SK_{1}(E, τ′) ≅ SK_{1}(D(x), τ′) by Th. 1.1, we have SK_{1}(D(x), τ′) ≅ SK_{1}(D, τ). □

5. Examples

Here are a few examples of SK_{1}(Q, τ) that follow from known results about SK_{1}(E, τ).

For a field K containing a primitive n-th root of unity ω (n ≥ 2) and any a, b ∈ K^{∗}, let (\frac{a,b}{K})_{ω} denote the degree n symbol algebra over K with generators i, j and relations i^{n} = a, j^{n} = b, and ij = ωji. Note that if K has a nonidentity automorphism η such that η^{2} = id and η(ω) = ω^{−1} and if η(a) = a and η(b) = b, then (\frac{a,b}{K})_{ω} has a unitary involution τ satisfying τ(ci^{k}j^{ℓ}) = j^{ℓ}i^{k}η(c) for all c ∈ K, k, ℓ ∈ ℤ.

Example 5.1. Let r_{1}, …, r_{m} be integers with each r_{i} ≥ 2, and let s = lcm(r_{1}, …, r_{m}) and n = r_{1} … r_{m}. Let L be a field containing a primitive s-th root of unity ω, and suppose L has an automorphism η of order 2 such that η(ω) = ω^{−1}. (For example, take L = ℂ and η to be complex conjugation.) Let L = K(x_{1}, …, x_{2m}), a rational function field over L. For k = 1, 2, …, m, let ω_{k} = ω^{s/r_{k}}, which is a primitive r_{k}-th root of unity in L. Let

Q = (\frac{x_{1},x_{2}}{K} )_{ω_{1}} ⊗_{K} … ⊗_{K} (\frac{x_{2m−1},x_{2m}}{K} )_{ω_{m}}.

So, Q is a division algebra over K of exponent s and index n. Extend η to an automorphism of order 2 of K by setting η(x_{ℓ}) = x_{ℓ} for ℓ = 1, 2, …, 2m. Each symbol algebra (\frac{x_{2k−1},x_{2k}}{K} )_{ω_{k}} has a unitary involution τ_{k} as described above, with τ_{k}|_{K} = η. Let τ = τ_{1} ⊗ … ⊗ τ_{m}: Q → Q. Since the τ_{k} all agree on K, this τ is a well-defined unitary involution on Q. Let μ_{ℓ}(L) denote the group of all ℓ-th roots of unity in L. Then,

\begin{aligned}SK_{1}(Q, τ) & ≅ \{c ∈ L^{∗} | η(c^{n}) = c^{n}\} / \{c ∈ L^{∗} | η(c^{s}) = c^{s}\} \\ & ≅ \{ξ ∈ μ_{n}(L) | η(ξ) = ξ^{−1}\}/μ_{s}(L). \end{aligned}  (5.1)

For, let Z = L[x_{1}, x_{1}^{−1}, …, x_{2m}, x_{2m}^{−1}], the (commutative) iterated Laurent polynomial ring over L, with its usual multigrading in x_{1}, …, x_{2m}; that is, Γ_{Z} = ℤ^{2m} with Z_{(k_{1},...,k_{2m})} = Lx_{1}^{k_{1}} … x_{2m}^{k_{2m}}. Then, Z is a graded field. Let E be the tensor product of graded symbol algebras, E = (\frac{x_{1},x_{2}}{Z} )_{ω_{1}} ⊗_{Z} … ⊗_{Z} (\frac{x_{2m−1},x_{2m}}{Z} )_{ω_{m}}. Then, the grading on Z extends to a grading on E, with deg(i_{k}) = \frac{1}{r_{k}} deg(x_{2k−1}) and deg(j_{k}) = \frac{1}{r_{k}} deg(x_{2k}), where i_{k} and j_{k} are the standard generators for (\frac{x_{2k−1},x_{2k}}{Z} ). Then, E is a graded division algebra with center Z, and E is totally ramified over Z, i.e., E_{0} = L = Z_{0}. Clearly q(Z) = K and q(E) = E⊗_{Z}q(Z) = Q, and τ restricts to a unitary graded involution on E. Thus, SK_{1}(Q, τ) ≅ SK_{1}(E, τ) by Th. 1.1 , and formula (5.1) follows from the corresponding formula for SK_{1}(E, τ) given in [HaW_{2}, Th. 5.1], as E is totally ramified over Z (i.e., E_{0} = Z_{0}) and Z is unramified over Z^{τ}.

Let F ⊆ K ⊆ N be fields with N Galois over K, K Galois over F, and [K:F] = 2. Let Br(K) denote the Brauer group of K, and Br(N/K) = ker ( Br(K) → Br(N)), the relative Brauer group. Let Br(N/K; F) = {[A] ∈ Br(N/K) | cor_{K→F} [A] = 1}, where cor_{K→F} is the corestriction mapping Br(K) to Br(F). Recall that the theorem of Albert-Riehm says that Br(N/K; F) consists of the classes of central simple K -algebras A such that N splits A and A has a unitary involution τ such that K^{τ} = F (see [KMRT, Th. 3.1, p. 31]). Suppose N is a cyclic Galois extension of K with [N:K] = n and G(N/K) = ⟨σ⟩. For b ∈ K^{∗}, let (N/K, σ, b) denote the cyclic algebra

(N/K, σ, b) = ⨁_{i=0}^{n−1} Ny^{i}, where yc = σ(c)y for all c ∈ K and y^{n} = b.

Let η be the nonidentity F -automorphism of K. Suppose further that N is Galois over F and N/F is dihedral. That is, suppose there exists ρ ∈ G(N/F) with ρ|_{K} = η, ρ^{2} = id_{N}, and ρσρ^{−1} = σ^{−1}. So, G(N/F) = ⟨σ, ρ⟩, and this group is dihedral if n ≥ 3. Observe that when N/F is dihedral, if b ∈ F^{∗}, then (N/K, σ, b) has a unitary involution τ given by τ(cy^{i}) = y^{i}ρ(c) for all c ∈ K, i ∈ ℤ. Note that τ|_{K} = η, so K^{τ} = F and [(N/K, σ, b)] ∈ Br(N/K; F).

Example 5.2. Let F ⊆ L be fields with [L:F] = 2 and L Galois over F with G(L/F) = {id_{L}, η}. Let N_{1} and N_{2} be cyclic Galois extensions of L which are linearly disjoint over L with each N_{i} dihedral over F as just described. Let n_{j} = [N_{j}:L] and let G(N_{j}/L) = ⟨σ_{j}⟩; extend each σ_{j} to N_{1}N_{2} so that σ_{1}|_{N_{2}} = id_{N_{2}} and σ_{2}|_{N_{1}} = id_{N_{1}}. Let K = L(x_{1}, x_{2}), a rational function field over L, and extend η to K by η(x_{i}) = x_{i} for i = 1, 2 and likewise extend the σ_{j} to N_{1}N_{2}K by σ_{j}(x_{i}) = x_{i} for j = 1, 2, i = 1, 2. Let

Q = (N_{1}K/K, σ_{1}, x_{1}) ⊗_{K} (N_{2}K/K, σ_{2}, x_{2}),

which is a division algebra over K with exponent lcm(n_{1}, n_{2}) and degree n_{1}n_{2}. As noted above, each (N_{j}K/K, σ_{j}, x_{j}) has a unitary involution τ_{j} with τ_{j}|_{K} = η. Let τ = τ_{1} ⊗τ_{2}, which is a unitary involution on Q. Then,

SK_{1}(Q, τ) ≅ Br(N_{1}N_{2}/L; F) /[Br(N_{1}/L; F) · Br(N_{2}/L; F)].  (5.2)

It was shown in [Y_{4}] that the right expression in (5.2) can be made into any finite abelian group by choosing L to be an algebraic number field and suitably choosing N_{1} and N_{2}. To view Q as a ring of quotients, first take Z = L[x_{1}, x_{1}^{−1}, x_{2}, x_{2}^{−1}] ⊆ K, so Z is a commutative twice-iterated Laurent polynomial ring over L, and we give Z its usual grading by multi-degree in x_{1} and x_{2}, as in the preceding example. Let E = N_{1}N_{2}[y_{1}, y_{1}^{−1}, y_{2}, y_{2}^{−1}] ⊆ Q, where y_{1} and y_{2} are the standard generators of the symbol algebras of Q. This E is a twisted iterated Laurent polynomial ring with y_{1}^{n_{1}} = x_{1}, y_{2}^{n_{2}} = x_{2}, y_{1}y_{2} = y_{2}y_{1} and for all c ∈ N_{1}N_{2}, y_{1}c = σ_{1}(c)y_{1} and y_{2}c = σ_{2}(c)y_{2}. We extend the grading on Z = Z(E) to E by setting deg(y_{1}) = (\frac{1}{n_{1}} , 0) and deg(y_{2}) = (0, \frac{1}{n_{2}} ); so, Γ_{E} = \frac{1}{n_{1}} ℤ × \frac{1}{n_{2}} ℤ. We can see that E is a graded division algebra by noting that E_{0} = N_{1}N_{2}, a field, and each homogeneous component E_{γ} of E is a 1-dimensional E_{0}-vector space containing a unit of E. This E is semiramified since [E_{0}:Z_{0}] = [N_{1}N_{2}:L] = n_{1}n_{2} = |Γ_{E}:Γ_{Z}|; indeed, it is decomposably semiramified in the terminology of [W], since E = (N_{1}Z/Z, σ_{1}, x_{1}) ⊗_{Z} (N_{2}Z/Z, σ_{2}, x_{2}) which expreses E as a tensor product of semiramified graded cyclic algebras. Since τ on Q clearly restricts to a unitary graded involution on E (recall that τ(y_{1}) = y_{1} and τ(y_{2}) = y_{2}), Th. (1.1) shows that SK_{1}(Q, τ) ≅ SK_{1}(E, τ). But further, let K′ = L((x_{1}))((x_{2})), a twice-iterated Laurent power series field over L, and let D = (N_{1}K′/K′, σ_{1}, x_{1}) ⊗_{K′} (N_{2}K′/K′, σ_{2}, x_{2}), which is a central simple division algebra over K′. Then, the standard rank 2 Henselian valuation v on K′ has associated graded ring gr(K′) = Z, and for the unique extension of v to D we have gr(D) = E. Because each N_{j}K′ is dihedral over F ((x_{1}))((x_{2})), there is a unitary involution τ̂ on D built just as for τ on Q. This τ̂ is compatible with the valuation on D, and the involution on E induced by τ̂ is clearly τ. Thus, SK_{1}(E, τ) ≅ SK_{1}(D, τ̂) by (1.1) above. But SK_{1}(D, τ̂) was computed in [Y_{2}, Th. 5.6] (with another proof given in [W, Th. 7.1(ii)]), and the formula given there combined with the isomorphisms stated here yield (5.2).

References

[J] N. Jacobson, Finite-dmensional division algebras over fields, Springer-Verlag, Berlin, 1996.

[HaW_{1}] R. Hazrat, A. R. Wadsworth, SK_{1} of graded division algebras, Israel J. Math., to appear, preprint available

        (No. 318) at: http://www.math.uni-bielefeld.de/LAG/ .

[HaW_{2}] R. Hazrat, A. R. Wadsworth, Unitary SK_{1} of graded and valued division algebras, Proc. London Math. Soc., to

        appear, preprint available (No. 369) at: http://www.math.uni-bielefeld.de/LAG/ .

[HwW] Y.-S. Hwang and A. R. Wadsworth, Correspondences between valued division algebras and graded division algebras, J. Algebra, 220 (1999), 73–114.

[KMRT] M.-A. Knus, A. S. Merkurjev, M. Rost, J.-P. Tignol, The Book of Involutions, Amer. Math. Soc. Coll. Pub. 44, Providence, RI, 1998.

[W] A. R. Wadsworth, Unitary SK_{1} of semiramified graded and valued division algebras, preprint available at arXiv:1009.3904.

[Y_{1}] V. I. Yanchevskiĭ, Simple algebras with involution, and unitary groups, Mat. Sb. (N.S.), 1974, 93 (135), 3, 368–380, (in Russian); English transl., Math. of the USSR-Sbornik, 22:3 (1974), 372–385.

[Y_{2}] V. I. Yanchevskiĭ, Reduced unitary K-Theory and division rings over discretely valued Hensel fields, Izv. Akad. Nauk SSSR Ser. Mat., 42 (1978), 879–918 (in Russian); English transl., Math. USSR Izvestiya, 13 (1979), 175–213.

[Y_{3}] V. I. Yanchevskiĭ, Reduced Whitehead unitary groups of skew-fields of noncommutative rational functions, Rings and modules, 2. Zap. Nauchn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 94 (1979), 142–148, 152 (in Russian); English transl: J. Soviet Math., 19 (1982), 1067–1071.

[Y_{4}] V. I. Yanchevskiĭ, The inverse problem of reduced K-theory, Mat. Zametki, 26 (1979), 475–482 (in Russian); English transl., A converse problem in reduced unitary K-theory, Math. Notes, 26 (1979), 728–731.

[Y_{5}] V. I. Yanchevskiĭ, Reduced unitary Whitehead groups and noncommutative rational functions, Dokl. Akad. Nauk BSSR, 24 (1980), 588–591, 667 (in Russian); English transl: Amer. Math. Soc. Transl. (2), 154 (1992), 63–67.

[Y_{6}] V. I. Yanchevskiĭ, Reduced unitary K-theory. Applications to algebraic groups. Mat. Sb. (N.S.), 110 (152) (1979), 579–596 (in Russian); English transl., Math. USSR Sbornik. 38 (1981) 533–548.

Department of Mathematics, University of California, San Diego, 9500 Gilman Drive, La Jolla, CA 92093-0112, USA

   E-mail address: arwadsworth@ucsd.edu

Institute of Mathematics of the National Academy of Sciences of Belarus, Ul. Surganova 11, 220072 Minsk, Belarus

   E-mail address: yanch@im.bas-net.by