UNITARY SK_{1} OF GRADED AND VALUED DIVISION ALGEBRAS, I

R. HAZRAT AND A. R. WADSWORTH

Abstract. The reduced unitary Whitehead group SK_{1} of a graded division algebra equipped with a unitary involution (i.e., an involution of the second kind) and graded by a torsion-free abelian group is studied. It is shown that calculations in the graded setting are much simpler than their nongraded counterparts. The bridge to the non-graded case is established by proving that the unitary SK_{1} of a tame valued division algebra wih a unitary involution over a henselian field coincides with the unitary SK_{1} of its associated graded division algebra. As a consequence, the graded approach allows us not only to recover results available in the literature with substantially easier proofs, but also to calculate the unitary SK_{1} for much wider classes of division algebras over henselian fields.

1. Introduction

Motivated by Platonov’s striking work on the reduced Whitehead group SK_{1}(D) of valued division algebras D, see [P_{2}, P_{4}], V. Yanchevskiĭ, considered the unitary analogue, SK_{1}(D, τ), for a division algebra D with unitary (i.e., second kind) involution τ, see [Y_{1}, Y_{2}, Y_{3}, Y_{4}]. Working with division algebras over a field with henselian discrete (rank 1) valuation whose residue field also contains a henselian discrete valuation, and carrying out formidable technical calculations, he produced remarkable analogues to Platonov’s results. By relating SK_{1}(D, τ) to data over the residue algebra, he showed not only that SK_{1}(D, τ) could be nontrivial but that it could be any finite abelian group, and he gave a formula in the bicyclic case expressing SK_{1}(D, τ) as a quotient of relative Brauer groups. Over the years since then several approaches have been given to understanding and calculating the (nonunitary) group SK_{1} using different methods, notably by Ershov [E], Suslin [S_{1}, S_{2}], Merkurjev and Rost [Mer] (For surveys on the group SK_{1}, see [P_{4}], [G], [Mer] or [W_{2}, §6].) However, even after the passage of some 30 years, there does not seem to have been any improvement in calculating SK_{1} in the unitary setting. This may be due in part to the complexity of the formulas in Yanchevskiĭ’s work, and the difficulty in following some of his arguments.

This paper is a sequel to [HaW] where the reduced Whitehead group SK_{1} for a graded division algebra was studied. Here we consider the reduced unitary Whitehead group of a graded division algebra with unitary graded involution. As in our previous work, we will see that the graded calculus is much easier and more transparent than the non-graded one. We calculate the unitary SK_{1} in several important cases. We also show how this enables one to calculate the unitary SK_{1} of a tame division algebra over a henselian field, by passage to the associated graded division algebra. The graded approach allows us not only to recover most of Yanchevskiĭ’s results in [Y_{2}, Y_{3}, Y_{4}], with very substantially simplified proofs, but also extend them to arbitrary value groups and to calculate the unitary SK_{1} for wider classes of division algebras. There is a significant simplification gained by considering arbitrary value groups from the outset, rather than towers of discrete valuations. But the greatest gain comes from passage to the graded setting, where the reduction to arithmetic considerations in the degree 0 division subring is quicker and more transparent.

We briefly describe our principal results. Let E be a graded division algebra, with torsion free abelian grade group Γ_{E}, and let τ be a unitary graded involution on E. “Unitary” means that the action of τ on the center T = Z(E) is nontrivial (see §2.3). The reduced unitary Whitehead group for τ on E is defined as

The first author acknowledges the support of EPSRC first grant scheme EP/D03695X/1. The second author would like to thank the first author and Queen’s University, Belfast for their hospitality while the research for this paper was carried out.

SK_{1}(E, τ) = {a ∈ E^{∗} | Nrd_{E}(a^{1−τ}) = 1}/⟨a ∈ E^{∗} | a^{1−τ} = 1⟩,

where Nrd_{E} is the reduced norm map Nrd_{E} : E^{∗} → T^{∗} (see [HaW, §3]). Here, a^{1−τ} means aτ(a)^{−1}. Let R = T^{τ} = {t ∈ T | τ(t) = t} ⫋ T (see §2.3). Let E_{0} be the subring of homogeneous elements of degree 0 in E; likewise for T_{0} and R_{0}. For an involution ρ on E_{0}, S_{ρ}(E_{0}) denotes {a ∈ E_{0} | ρ(a) = a} and Σ_{ρ}(E_{0}) = ⟨S_{ρ}(E_{0})∩E_{0}^{∗}⟩. Let n be the index of E, and e the exponent of the group Γ_{E}/Γ_{T}. Since [T : R] = 2, there are just two possible cases: either (i) T is unramified over R, i.e., Γ_{T} = Γ_{R}; or (ii) T is totally ramified over R, i.e., |Γ_{T} : Γ_{R}| = 2 . We will prove the following formulas for the unitary SK_{1}:

(i) Suppose T/R is unramified:

• If E/T is unramified, then SK_{1}(E, τ) ≅ SK_{1}(E_{0}, τ|_{E_{0}}) (Prop. 4.10).

• If E/T is totally ramified, then (Th. 5.1):

SK_{1}(E, τ) ≅ {a ∈ T_{0}^{∗} | a^{n} ∈ R_{0}^{∗}}/{a ∈ T_{0}^{∗} | a^{e} ∈ R_{0}^{∗}}
≅ {ω ∈ μ_{n}(T_{0}) | τ(ω) = ω^{−1}}/μ_{e}.

• If Γ_{E}/Γ_{T} is cyclic, and σ is a generator of Gal(Z(E_{0})/T_{0}), then (Prop. 4.13):

∘ SK_{1}(E, τ) ≅ {a ∈ E_{0}^{∗} | N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a)) ∈ R_{0}}/(Σ_{τ} (E_{0}) · Σ_{στ} (E_{0})).

∘ If E_{0} is a field, then SK_{1}(E, τ) = 1.

• If E has a maximal graded subfield M unramified over T and another maximal graded subfield L totally ramified over T, with τ(L) = L, then E is semiramified and (Cor. 4.11)

SK_{1}(E, τ) ≅ {a ∈ E_{0} | N_{E_{0}/T_{0}}(a) ∈ R_{0}} / ∏_{h∈Gal(E_{0}/T_{0})}E_{0}^{∗hτ} .

(ii) If T/R is totally ramified, then SK_{1}(E, τ) = 1 (Prop. 4.5).

The bridge between the graded and the non-graded henselian setting is established by Th. 3.5, which shows that for a tame division algebra D over a henselian valued field with a unitary involution τ, SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃) where gr(D) is the graded division algebra associated to D by the valuation, and τ̃ is the graded involution on gr(D) induced by τ (see §3). Thus, each of the results listed above for graded division algebras yields analogous formulas for valued division algebras over a henselian field, as illustrated in Example 5.3 and Th. 5.4. This recovers existing formulas, which were primarily for the case with value group ℤ or ℤ × ℤ, but with easier and more transparent proofs than those in the existing literature. Additionally, our results apply for any value groups whatever. The especially simple case where E/T is totally ramified and T/R is unramified is entirely new.

In the sequel to this paper [W_{3}], the very interesting special case will be treated where E/T is semiramified (and T/R is unramified) and Gal(E_{0}/T_{0}) is bicyclic. This case was the setting of essentially all of Platonov’s specifically computed examples with nontrivial SK_{1}(D) [P_{2}, P_{3}], and likewise Yanchevskiĭ’s unitary examples in [Y_{3}] where the nontrivial SK_{1}(D, τ) was fully computed. This case is not pursued here because it requires some more specialized arguments. For such an E, it is known that [E] decomposes (nonuniquely) as [I ⊗_{T} N] in the graded Brauer group of T, where I is inertial over T and N is nicely semiramified, i.e., semiramified and containing a maximal graded subfield totally ramified over T. Then a formula will be given for SK_{1}(E) as a factor group of the relative Brauer group Br(E_{0}/T_{0}) modulo other relative Brauer groups and the class of I_{0}. An exactly analogous formula will be proved for SK_{1}(E, τ) in the unitary setting.

UNITARY SK_{1} OF DIVISION ALGEBRAS 3

2. Preliminaries

Throughout this paper we will be concerned with involutory division algebras and involutory graded division algebras. In the non-graded setting, we will denote a division algebra by D and its center by K; this D is equipped with an involution τ, and we set F = K^{τ} = {a ∈ K | τ(a) = a}. In the graded setting, we will write E for a graded division algebra with center T, and R = T^{τ} where τ is a graded involution on E. (This is consistent with the notation used in [HaW].) Depending on the context, we will write τ(a) or a^{τ} for the action of the involution on an element, and K^{τ} for the set of elements of K invariant under τ. Our convention is that a^{στ} means σ(τ(a)).

In this section, we recall the notion of graded division algebras and collect the facts we need about them in §2.1. We will then introduce the unitary and graded reduced unitary Whitehead groups in §2.2 and §2.3.

2.1. Graded division algebras. In this subsection we establish notation and recall some fundamental facts about graded division algebras indexed by a totally ordered abelian group. For an extensive study of such graded division algebras and their relations with valued division algebras, we refer the reader to [HW_{2}]. For generalities on graded rings see [NvO].

Let R = ⨁_{γ∈Γ} R_{γ} be a graded ring, i.e., Γ is an abelian group, and R is a unital ring such that each R_{γ} is a subgroup of (R, +) and R_{γ} · R_{δ} ⊆ R_{γ+δ} for all γ, δ ∈ Γ. Set

Γ_{R} = {γ ∈ Γ | R_{γ} ≠ 0}, the grade set of R;
R^{h} = ⋃_{γ∈Γ_{R}} R_{γ}, the set of homogeneous elements of R.

For a homogeneous element of R of degree γ, i.e., an r ∈ R_{γ} \ {0}, we write deg(r) = γ. Recall that R_{0} is a subring of R and that for each γ ∈ Γ_{R}, the group R_{γ} is a left and right R_{0}-module. A subring S of R is a graded subring if S = ⨁_{γ∈Γ_{R}}(S∩R_{γ}). For example, the center of R, denoted Z(R), is a graded subring of R. If T = ⨁_{γ∈Γ} T_{γ} is another graded ring, a graded ring homomorphism is a ring homomorphism f: R → T with f(R_{γ}) ⊆ T_{γ} for all γ ∈ Γ. If f is also bijective, it is called a graded ring isomorphism; we then write R ≅_{gr} T.

For a graded ring R, a graded left R-module M is a left R-module with a grading M = ⨁_{γ∈Γ′} M_{γ}, where the M_{γ} are all abelian groups and Γ′ is a abelian group containing Γ, such that R_{γ} · M_{δ} ⊆ M_{γ+δ} for all γ ∈ Γ_{R}, δ ∈ Γ′. Then, Γ_{M} and M^{h} are defined analogously to Γ_{R} and R^{h}. We say that M is a graded free R-module if it has a base as a free R-module consisting of homogeneous elements.

A graded ring E = ⨁_{γ∈Γ} E_{γ} is called a graded division ring if Γ is a torsion-free abelian group and every non-zero homogeneous element of E has a multiplicative inverse in E. Note that the grade set Γ_{E} is actually a group. Also, E_{0} is a division ring, and E_{γ} is a 1-dimensional left and right E_{0} vector space for every γ ∈ Γ_{E}. Set E_{γ}^{∗} = E_{γ} \ {0}. The requirement that Γ be torsion-free is made because we are interested in graded division rings arising from valuations on division rings, and all the grade groups appearing there are torsion-free. Recall that every torsion-free abelian group Γ admits total orderings compatible with the group structure. (For example, Γ embeds in Γ ⊗_{ℤ} ℚ which can be given a lexicographic total ordering using any base of it as a ℚ-vector space.) By using any total ordering on Γ_{E}, it is easy to see that E has no zero divisors and that E^{∗}, the multiplicative group of units of E, coincides with E^{h} \ {0} (cf. [HW_{2}, p. 78]). Furthermore, the degree map

deg: E^{∗} → Γ_{E}  (2.1)

is a group epimorphism with kernel E_{0}^{∗}.

By an easy adaptation of the ungraded arguments, one can see that every graded module M over a graded division ring E is graded free, and every two homogenous bases have the same cardinality. We thus call M a graded vector space over E and write dim_{E}(M) for the rank of M as a graded free E-module. Let S ⊆ E be a graded subring which is also a graded division ring. Then we can view E as a graded left S-vector space, and we write [E : S] for dim_{S}(E). It is easy to check the “Fundamental Equality,”

[E : S] = [E_{0} : S_{0}] |Γ_{E} : Γ_{S}|,  (2.2)

where [E_{0} : S_{0}] is the dimension of E_{0} as a left vector space over the division ring S_{0} and |Γ_{E} : Γ_{S}| denotes the index in the group Γ_{E} of its subgroup Γ_{S}.

A graded field T is a commutative graded division ring. Such a T is an integral domain (as Γ_{T} is torsion free), so it has a quotient field, which we denote q(T). It is known, see [HW_{1}, Cor. 1.3], that T is integrally closed in q(T). An extensive theory of graded algebraic field extensions of graded fields has been developed in [HW_{1}].

If E is a graded division ring, then its center Z(E) is clearly a graded field. The graded division rings considered in this paper will always be assumed finite-dimensional over their centers. The finite-dimensionality assures that E has a quotient division ring q(E) obtained by central localization, i.e., q(E) = E ⊗_{T} q(T), where T = Z(E). Clearly, Z(q(E)) = q(T) and ind(E) = ind(q(E)), where the index of E is defined by ind(E)^{2} = [E : T] (see [HW_{2}, p. 89]). If S is a graded field which is a graded subring of Z(E) and [E : S] < ∞, then E is said to be a graded division algebra over S. We recall a fundamental connection between Γ_{E} and Z(E_{0}): The field Z(E_{0}) is Galois over T_{0}, and there is a well-defined group epimorphism

Θ_{E} : Γ_{E} → Gal(Z(E_{0})/T_{0}) given by deg(e) ↦ (z ↦ eze^{−1}),  (2.3)

for any e ∈ E^{∗}. (See [HW_{2}, Prop. 2.3] for a proof).

Let E = ⨁_{α∈Γ_{E}} E_{α} be a graded division algebra with a graded center T (with, as always, Γ_{E} a torsion-free abelian group). After fixing some total ordering on Γ_{E}, define a function

λ: E \ {0} → E^{∗} by λ(∑ c_{γ}) = c_{δ}, where δ is minimal among the γ ∈ Γ_{E} with c_{γ} ≠ 0.

Note that λ(a) = a for a ∈ E^{∗}, and

λ(ab) = λ(a)λ(b) for all a, b ∈ E \ {0}.  (2.4)

Let Q = q(E) = E ⊗_{T} q(T), which is a division ring as E has no zero divisors and is finite-dimensional over T. We can extend λ to a map defined on all of Q^{∗} = Q \ {0} as follows: for q ∈ Q^{∗}, write q = ac^{−1} with a ∈ E \ {0}, c ∈ Z(E) \ {0}, and set λ(q) = λ(a)λ(c)^{−1}. It follows from (2.4) that λ: Q^{∗} → E^{∗} is well-defined and is a group homomorphism. Since the composition

E^{∗} ↪ Q^{∗} ⟶^{λ} E^{∗}  (2.5)

is the identity, λ is a splitting map for the injection E^{∗} ↪ Q^{∗}.

For a graded division algebra E over its center T, there is a reduced norm map Nrd_{E} : E^{∗} → T^{∗} (see [HaW, §3]) such that for a ∈ E one has Nrd_{E}(a) = Nrd_{q(E)}(a). The reduced Whitehead group, SK_{1}(E), is defined as E^{(1)}/E′, where E^{(1)} denotes the set of elements of E^{∗} with reduced norm 1, and E′ is the commutator subgroup [E^{∗}, E^{∗}] of E^{∗}. This group was studied in detail in [HaW]. We will be using the following facts which were established in that paper:

Remarks 2.1. Let n = ind(E).

(i) For γ ∈ Γ_{E}, if a ∈ E_{γ} then Nrd_{E}(a) ∈ E_{nγ}. In particular, E^{(1)} is a subset of E_{0}. (ii) If S is any graded subfield of E containing T and a ∈ S, then Nrd_{E}(a) = N_{S/T}(a)^{n/[S:T]}. (iii) Set

∂ = ind(E)/( ind(E_{0}) [Z(E_{0}):T_{0}]).  (2.6)

If a ∈ E_{0}, then,

Nrd_{E}(a) = N_{Z(E_{0})/T_{0}}Nrd_{E_{0}}(a)^{∂} ∈ T_{0}.  (2.7)

(iv) If N is a normal subgroup of E^{∗}, then N^{n} ⊆ Nrd_{E}(N)[E^{∗}, N].

For proofs of (i)-(iv) see [HaW, Prop. 3.2 and 3.3].

(v) SK_{1}(E) is n-torsion.

Proof. By taking N = E^{(1)}, the assertion follows from (iv). □

A graded division algebra E with center T is said to be inertial (or unramified) if Γ_{E} = Γ_{T}. From (2.2), it then follows that [E : T] = [E_{0} : T_{0}]; indeed, E_{0} is central simple over T_{0} and E ≅_{gr} E_{0} ⊗_{T_{0}} T. At the other extreme, E is said to be totally ramified if E_{0} = T_{0}. In an intermediate case E is said to be semiramified if E_{0} is a field and [E_{0} : T_{0}] = |Γ_{E} : Γ_{T}| = ind(E). These definitions are motivated by analogous definitions for valued division algebras ([W_{2}]). Indeed, if a tame valued division algebra is unramified, semiramified, or totally ramified, then so is its associated graded division algebra. Likewise, a graded field extension L of T is said to be inertial (or unramified) if L ≅_{gr} L_{0} ⊗_{T_{0}} T and the field L_{0} is separable over T_{0}. At the other extreme, L is totally ramified over T if [L : T] = |Γ_{L} : Γ_{T}|. A graded division algebra E is said to be inertially split if E has a maximal graded subfield L with L inertial over T. When this occurs, E_{0} = L_{0}, and ind(E) = ind(E_{0}) [Z(E_{0}) : T_{0}] by Lemma 2.2 below. In particular, if E is semiramified then E is inertially split, E_{0} is abelian Galois over T_{0}, and the canonical map Θ_{E} : Γ_{E} → Gal(E_{0}/T_{0}) has kernel Γ_{T} (see (2.3) above and [HW_{2}, Prop. 2.3]).

Lemma 2.2. Let E be a graded division algebra with center T. For the ∂ of (2.6), ∂^{2} = | ker(Θ_{E})/Γ_{T}|. Also, ∂ = 1 iff E is inertially split.

Proof. Since Θ_{E} is surjective, Γ_{T} ⊆ ker(Θ_{E}), and Z(E_{0}) is Galois over T_{0}, we have

∂^{2} = ind(E)^{2} / ( ind(E_{0})^{2} [Z(E_{0}) : T_{0}]^{2}) = [E : T] /([E_{0} : Z(T_{0})] [Z(E_{0}) : T_{0}] | Gal(Z(E_{0})/T_{0})|)
= [E_{0} : T_{0}] |Γ_{E}/Γ_{T}| / ([E_{0} : T_{0}] | im(Θ_{E})|) = | ker(Θ_{E})/Γ_{T}|.

Now, suppose M is a maximal subfield of E_{0} with M separable over T_{0}. Then, M ⊇ Z(E_{0}) and [M : Z(E_{0})] = ind(E_{0}). Let L = M ⊗_{T_{0}} T which is a graded subfield of E inertial over T, with L_{0} = M. Then,

[L : T] = [L_{0} : T_{0}] = [L_{0} : Z(E_{0})] [Z(E_{0}) : T_{0}] = ind(E)/∂.

Thus, if ∂ = 1, then E is inertially split, since L is a maximal graded subfield of E which is inertial over T. Conversely, suppose E is inertially split, say I is a maximal graded subfield of E with I inertial over T. So, [I_{0} : T_{0}] = [I : T] = ind(E). Since I_{0}Z(E_{0}) is a subfield of E_{0}, we have

[I_{0} : T_{0}] ≤ [I_{0}Z(E_{0}) : T_{0}] = [I_{0}Z(E_{0}) : Z(E_{0})] [Z(E_{0}) : T_{0}]
≤ ind(E_{0}) [Z(E_{0}) : T_{0}] = ind(E)/∂ = [I_{0} : T_{0}]/∂.

So, as ∂ is a positive integer, ∂ = 1. □

2.2. Unitary SK_{1} of division algebras. We begin with a description of unitary K_{1} and SK_{1} for a division algebra with an involution. The analogous definitions for graded division algebras will be given in §2.3.

Let D be a division ring finite-dimensional over its center K of index n, and let τ be an involution on D, i.e., τ is an antiautomorphism of D with τ^{2} = id. Let

S_{τ} (D) = {d ∈ D | τ(d) = d};
Σ_{τ} (D) = ⟨S_{τ} (D) ∩ D^{∗}⟩ .

Note that Σ_{τ} (D) is a normal subgroup of D^{∗}. For, if a ∈ S_{τ} (D), a ≠ 0, and b ∈ D^{∗}, then bab^{−1} = [baτ(b)][bτ(b)]^{−1} ∈ Σ_{τ} (D), as baτ(b), bτ(b) ∈ S_{τ} (D).

Let φ be an isotropic m-dimensional, nondegenerate skew-Hermitian form over D with respect to an involution τ on D. Let ρ be the involution on M_{m}(D) adjoint to φ, let U_{m}(D) = {a ∈ M_{m}(D) | aρ(a) = 1} be the unitary group associated to φ, and let EU_{m}(D) denote the normal subgroup of U_{m}(D) generated by the unitary transvections. For m > 2, the Wall spinor norm map Θ: U_{m}(D) → D^{∗}/Σ_{τ} (D)D′ was developed in [Wa], where it was shown that ker(Θ) = EU_{m}(D). Here, D′ denotes the multiplicative commutator group [D^{∗}, D^{∗}]. Combining this with [D, Cor. 1 of §22] one obtains the commutative diagram:

\xymatrix{U_{m}(D)/ EU_{m}(D) \ar[r]_{≅}^{Θ} \ar[d] & D^{∗}/(Σ_{τ} (D)D′) \ar[d]^{1−τ} \\ GL_{m}(D)/E_{m}(D) \ar[r]^{det} \ar[d]_{Nrd} & D^{∗}/D′ \ar[d]^{Nrd} \\ K^{∗} \ar[r]^{id} & K^{∗}}  (2.8)

where the map det is the Dieudonné determinant and 1 − τ: D^{∗}/(Σ_{τ} (D)D′) ⟶ D^{∗}/D′ is defined as xΣ_{τ} (D)D′ ↦ x^{1−τ} D′, where x^{1−τ} means xτ(x)^{−1} (see [HM, 6.4.3]).

From the diagram, and parallel to the “absolute” case, one defines the unitary Whitehead group,

K_{1}(D, τ) = D^{∗}/(Σ_{τ} (D)D′).

For any involution τ on D, recall that

Nrd_{D}(τ(d)) = τ(Nrd_{D}(d)),  (2.9)

for any d ∈ D. For, if p ∈ K[x] is the minimal polynomial of d over K, then τ(p) is the minimal polynomial of τ(d) over K (see also [D, §22, Lemma 5]).

We consider two cases:

2.2.1. Involutions of the first kind. In this case the center K of D is elementwise invariant under the involution, i.e., K ⊆ S_{τ} (D). Then S_{τ} (D) is a K-vector space. The involutions of this kind are further subdivided into two types: orthogonal and symplectic involutions (see [KMRT, Def. 2.5]). By ([KMRT, Prop. 2.6]), if char(K) ≠ 2 and τ is orthogonal then, dim_{K}(S_{τ} (D)) = n(n + 1)/2, while if τ is symplectic then dim_{K}(S_{τ} (D)) = n(n − 1)/2. However, if char(K) = 2, then dim_{K}(S_{τ} (D)) = n(n + 1)/2 for each type.

If dim_{K}(S_{τ} (D)) = n(n + 1)/2, then for any x ∈ D^{∗}, we have xS_{τ} (D) ∩ S_{τ} (D) ≠ 0 by dimension count; it then follows that D^{∗} = Σ_{τ} (D), and thus K_{1}(D, τ) = 1. However, in the case dim_{K}(S_{τ} (D)) = n(n−1)/2, Platonov showed that K_{1}(D, τ) is not in general trivial, settling Dieudonné’s conjecture in negative [P_{1}]. Note that whenever τ is of the first kind we have Nrd_{D}(τ(d)) = Nrd_{D}(d) for all d ∈ D, by (2.9). Thus, K_{1}(D, τ) is sent to the identity under the composition Nrd∘(1−τ). This explains why one does not consider the kernel of this map, i.e., the unitary SK_{1}, for involutions of the first kind. If char(K) ≠ 2 and τ is symplectic, then as the m-dimensional form φ over D is skew-Hermitian, its associated adjoint involution ρ on M_{m}(D) is of orthogonal type, so there is an associated spin group Spin(M_{m}(D), ρ). For any a ∈ S(D) one then has Nrd_{D}(a) ∈ K^{∗2} ([KMRT, Lemma 2.9]). One defines K_{1} Spin(D, τ) = R(D)/(Σ_{τ} (D)D′), where R(D) = {d ∈ D^{∗} | Nrd_{D}(d) ∈ K^{∗2}}. This group is related to Spin(M_{m}(D), ρ), and has been studied in [MY], parallel to the work on absolute SK_{1} groups and unitary SK_{1} groups for unitary involutions.

2.2.2. Involutions of the second kind (unitary involutions). In this case K ⊈ S_{τ} (D). Then, let F = K^{τ} (= K ∩ S_{τ} (D) ), which is a subfield of K with [K : F] = 2. It was already observed by Dieudonné that U_{m}(D) ≠ EU_{m}(D). An important property proved by Platonov and Yanchevskiĭ, which we will use frequently, is that

D′ ⊆ Σ_{τ} (D).  (2.10)

(For a proof, see [KMRT, Prop. 17.26].) Thus K_{1}(D, τ) = D^{∗}/Σ_{τ} (D), which is not trivial in general. The kernel of the map Nrd ∘(1−τ) in diagram (2.8), is called the reduced unitary Whitehead group, and denoted by SK_{1}(D, τ). Using (2.9), it is straightforward to see that

SK_{1}(D, τ) = Σ′_{τ} (D)/Σ_{τ} (D), where Σ′_{τ} (D) = {a ∈ D^{∗} | Nrd_{D}(a) ∈ F^{∗}}.

Note that we use the notation SK_{1}(D, τ) for the reduced unitary Whitehead group as opposed to Draxl’s notation USK_{1}(D, τ) in [D, p. 172] and Yanchevskiĭ’s notation SUK_{1}(D, τ) [Y_{2}] and the notation USK_{1}(D) in [KMRT].

Before we define the corresponding groups in the graded setting, let us recall that all the groups above fit in Tits’ framework [T] of the Whitehead group W (G, K) = G_{K}/G_{K}^{+} where G is an almost simple, simply connected linear algebraic group defined over an infinite field K, with char(K) ≠ 2, and G is isotropic over K. Here, G_{K} is the set of K-rational points of G and G_{K}^{+}, is the subgroup of G_{K}, generated by the unipotent radicals of the minimal K-parabolic subgroups of G. In this setting, for G_{K} = SL_{n}(D), n > 1, we have W (G, K) = SK_{1}(D); for τ an involution of first or second kind on D and F = K^{τ}, for G_{F} = SL_{n}(D, τ) := SL_{n}(D)∩U_{n}(D) we have W (G, F) = SK_{1}(D, τ); and for τ a symplectic involution on D and ρ the adjoint involution of an m-dimensional isotropic skew-Hermitian form over D with m ≥ 3, for the spinor group G_{K} = Spin(M_{m}(D), ρ) we have W (G, K) is a double cover of K_{1} Spin(D, τ) (see [MY]).

2.3. Unitary SK_{1} of graded division algebras. We will now introduce the unitary K_{1} and SK_{1} in the graded setting. Let E = ⨁_{γ∈Γ_{E}} E_{γ} be a graded division ring (with Γ_{E} a torsion-free abelian group) such that E has finite dimension n^{2} over its center T, a graded field. Let τ be a graded involution of E, i.e., τ is an antiautomorphism of E with τ^{2} = id and τ(E_{γ}) = E_{γ} for each γ ∈ Γ_{E}. We define S_{τ} (E) and Σ_{τ} (E), analogously to the non-graded cases, as the set of elements of E which are invariant under τ, and the multiplicative group generated by the nonzero homogenous elements of S_{τ} (E), respectively. We say the involution of the first kind if all the elements of the center T are invariant under τ; it is of the second kind (or unitary) otherwise. If τ is of the first kind then, parallel to the non-graded case, either dim_{T}(S_{τ} (E)) = n(n+1)/2 or dim_{T}(S_{τ} (E)) = n(n−1)/2. Indeed, one can show these equalities by arguments analogous to the nongraded case as in the proof of [KMRT, Prop. 2.6(1)], as E is split by a graded maximal subfield and the Skolem–Noether theorem is available in the graded setting ([HW_{2}, Prop. 1.6]). (These equalities can also be obtained by passing to the quotient division algebra as is done in Lemma 2.3(i) below.)

Define the unitary Whitehead group

K_{1}(E, τ) = E^{∗}/(Σ_{τ} (E)E′),

where E′ = [E^{∗}, E^{∗}]. If τ is of the first kind, char(T) ≠ 2, and dim_{T}(S_{τ} (E)) = n(n − 1)/2, a proof similar to [KMRT, Prop. 2.9], shows that if a ∈ S_{τ} (E) is homogeneous, then Nrd_{E}(a) ∈ T^{∗2} (This can also be verified by passing to the quotient division algebra, then using Lemma 2.3(i) below and invoking the corresponding result for ungraded division algebras.) For this type of involution, define the spinor Whitehead group

K_{1} Spin(E, τ) = {a ∈ E^{∗} | Nrd_{E}(a) ∈ T^{∗^{2}}} /(Σ_{τ} (E)E′).

When the graded involution τ on E is unitary, i.e., τ|_{T} ≠ id, let R = T^{τ}, which is a graded subfield of T with [T : R] = 2. Furthermore, T is Galois over R, with Gal(T/R) = {id, τ|_{T}}. (See [HW_{1}] for Galois theory for graded field extensions.) Define the reduced unitary Whitehead group

SK_{1}(E, τ) = Σ′_{τ} (E) / (Σ_{τ} (E) E′) = Σ′_{τ} (E) / Σ_{τ} (E),  (2.11)

where

Σ′_{τ} (E) = {a ∈ E^{∗} | Nrd_{E}(a^{1−τ}) = 1} = {a ∈ E^{∗} | Nrd_{E}(a) ∈ R^{∗}}

and

Σ_{τ} (E) = ⟨a ∈ E^{∗} | a^{1−τ} = 1⟩ = ⟨S_{τ} (E) ∩ E^{∗}⟩.

Here, a^{1−τ} means aτ(a)^{−1}. See Lemma 2.3(iv) below for the second equality in (2.11). The group SK_{1}(E, τ) will be the main focus of the rest of the paper.

We will use the following facts repeatedly:

Lemma 2.3.

(i) Any graded involution on E extends uniquely to an involution of the same kind (and type) on Q = q(E). (ii) For any graded involution τ on E, and its extension to Q = q(E), we have Σ_{τ} (Q) ∩ E^{∗} ⊆ Σ_{τ} (E). (iii) If τ is a graded involution of the first kind on E with dim_{T}(S_{τ} (E)) = n(n + 1)/2, then Σ_{τ} (E) = E^{∗}. (iv) If τ is a unitary graded involution on E, then E′ ⊆ Σ_{τ} (E).

(v) If τ is a unitary graded involution on E, then SK_{1}(E, τ) is a torsion group of bounded exponent dividing n = ind(E).

Proof.

(i) Let τ be a graded involution on E. Then q(E) = E ⊗_{T} q(T) = E ⊗_{T} (T ⊗_{T^{τ}} q(T^{τ})) = E ⊗_{T^{τ}} q(T^{τ}). The unique extension of τ to q(E) is τ ⊗ id_{q(T^{τ})}, which we denote simply as τ. It then follows that S_{τ} (q(E)) = S_{τ} (E) ⊗_{T^{τ}} q(T^{τ}). Since q(T^{τ}) = q(T)^{τ}, the assertion follows.

(ii) Note that for the map λ in the sequence (2.5) we have τ(λ(a)) = λ(τ(a)) for all a ∈ Q^{∗}. Hence, λ(Σ_{τ} (Q)) ⊆ Σ_{τ} (E). Since λ|_{E^{∗}} is the identity, we have Σ_{τ} (Q) ∩ E^{∗} ⊆ Σ_{τ} (E).

(iii) The extension of the graded involution τ to Q = q(E), also denoted τ, is of the first kind with dim_{Q}(S_{τ} (Q)) = n(n+1)/2 by (i). Therefore Σ_{τ} (Q) = Q^{∗} (see §2.2.1). Using (ii) now, the assertion follows.

(iv) Since τ is a unitary graded involution, its extension to Q = q(E) is also unitary, by (i). But Q′ ⊆ Σ_{τ} (Q), as noted in (2.10). From (2.5) it follows that Q′ ∩ E^{∗} = E′. Hence, using (ii), E′ ⊆ E^{∗} ∩ Q′ ⊆ E^{∗} ∩ Σ_{τ} (Q) ⊆ Σ_{τ} (E).

(v) Setting N = Σ′_{τ} (E), Remark 2.1(iv) above, coupled with the fact that E′ ⊆ Σ_{τ} (E) (iv), implies that SK_{1}(E, τ) is an n-torsion group. This assertion also follows by using (ii) which implies the natural map SK_{1}(E, τ) → SK_{1}(Q, τ) is injective and the fact that unitary SK_{1} of a division algebra of index n is n-torsion ([Y_{2}, Cor. to 2.5]). □

2.4. Generalized dihedral groups and field extensions. The nontrivial case of SK_{1}(E, τ) for τ a unitary graded involution turns out to be when T = Z(E) is unramified over R = T^{τ} (see §4.2). When that occurs, we will see in Lemma 4.6(ii) below that Z(E_{0}) is a so-called generalized dihedral extension over R_{0}. We now give the definition and observe a few easy facts about generalized dihedral groups and extensions.

Definition 2.4.

(i) A group G is said to be generalized dihedral if G has a subgroup H such that [G : H] = 2 and every

τ ∈ G\H satisfies τ^{2} = id.

Note that if G is generalized dihedral and H the distinguished subgroup, then H is abelian and (hτ)^{2} = id, for all τ ∈ G\H and h ∈ H. Thus, τ^{2} = id and τhτ^{−1} = h^{−1} for all τ ∈ G\H, h ∈ H. Furthermore, every subgroup of H is normal in G. Clearly every dihedral group is generalized dihedral, as is every elementary abelian 2-group. More generally, if H is any abelian group and χ ∈ Aut(H) is the map h ↦ h^{−1}, then the semi-direct product H ⋊_{i} ⟨χ⟩ is a generalized dihedral group, where i: ⟨χ⟩ → Aut(H) is the inclusion map. It is easy to check that every generalized dihedral group is isomorphic to such a semi-direct product.

(ii) Let F ⊆ K ⊆ L be fields with [L : F] < ∞ and [K : F] = 2. We say that L is generalized dihedral for K/F if L is Galois over F and every element of Gal(L/F)\ Gal(L/K) has order 2, i.e., Gal(L/F) is a generalized dihedral group. Note that when this occurs, L is compositum of fields L_{i} containing K with each L_{i} generalized dihedral for K/F with Gal(L_{i}/K) cyclic, i.e., L_{i} is Galois over F with Gal(L_{i}/F) dihedral (or a Klein 4-group). Conversely, if L and M are generalized dihedral for K/F then so is their compositum.

Example 2.5. Let n ∈ ℕ, n ≥ 3, and let F ⊆ K be fields with [K : F] = 2 and K = F (ω), where ω is a primitive n-th root of unity (so char(F) ∤ n). Suppose the non-identity element of Gal(K/F) maps ω to ω^{−1}. For any c_{1}, …, c_{k} ∈ F^{∗}, if ω ∉ F ( \sqrt[n]{c_{1}}, …, \sqrt[n]{c_{k}}), then K( \sqrt[n]{c_{1}}, …, \sqrt[n]{c_{k}}) is generalized dihedral for K/F.

3. Henselian to graded reduction

The main goal of this section is to prove an isomorphism between the unitary SK_{1} of a valued division algebra with involution over a henselian field and the graded SK_{1} of its associated graded division algebra. We first recall how to associate a graded division algebra to a valued division algebra.

Let D be a division algebra finite dimensional over its center K, with a valuation v: D^{∗} → Γ. So, Γ is a totally ordered abelian group, and v satisifies the conditions that for all a, b ∈ D^{∗},

(1) v(ab) = v(a) + v(b);

(2) v(a + b) ≥ min{v(a), v(b)} (b ≠ −a).

Let

V_{D} = {a ∈ D^{∗} | v(a) ≥ 0} ∪ {0}, the valuation ring of v;
M_{D} = {a ∈ D^{∗} | v(a) > 0} ∪ {0}, the unique maximal left (and right) ideal of V_{D};
\overline{D} = V_{D}/M_{D}, the residue division ring of v on D; and
Γ_{D} = im(v), the value group of the valuation.

Now let K be a field with a valuation v, and suppose v is henselian; that is, v has a unique extension to every algebraic field extension of K. Recall that a field extension L of K of degree n < ∞ is said to be tamely ramified or tame over K if, with respect to the unique extension of v to L, the residue field \overline{L} is separable over \overline{K} and char(\overline{K}) ∤ (n/[\overline{L} : \overline{K}]). Such an L is necessarily defectless over K, i.e., [L : K] = [\overline{L} : \overline{K}] |Γ_{L} : Γ_{K}|, by [EP, Th. 3.3.3] (applied to N/K and N/L, where N is a normal closure of L over K). Along the same lines, let D be a division algebra with center K (so, by convention, [D : K] < ∞); then the henselian valuation v on K extends uniquely to a valuation on D ([W_{1}]). With respect to this valuation, D is said to be tamely ramified or tame if the center Z(\overline{D}) is separable over \overline{K} and char(\overline{K}) ∤ [ ind(D)/( ind(\overline{D})[Z(\overline{D}) : \overline{K}])]. Recall from [JW, Prop. 1.7], that whenever the field extension Z(\overline{D})/\overline{K} is separable, it is abelian Galois. It is known that D is tame if and only if D is split by the maximal tamely ramified field extension of K, if and only if char(\overline{K}) = 0 or char(\overline{K}) = p ≠ 0 and the p-primary component of D is inertially split, i.e., split by the maximal unramified extension of K ([JW, Lemma 6.1]). We say D is strongly tame if char(\overline{K}) ∤ ind(D). Note that strong tameness implies tameness. This is clear from the last characterization of tameness, or from (3.1) below. Recall also from [Mor, Th. 3], that for a valued division algebra D finite dimensional over its center K (here not necessarily henselian), we have the “Ostrowski theorem”

[D : K] = q^{k} [\overline{D} : \overline{K}] |Γ_{D} : Γ_{K}|,  (3.1)

where q = char(\overline{D}) and k ∈ ℤ with k ≥ 0 (and q^{k} = 1 if char(\overline{D}) = 0). If q^{k} = 1 in equation (3.1), then D is said to be defectless over K. For background on valued division algebras, see [JW] or the survey paper [W_{2}].

Remark 3.1. If a field K has a henselian valuation v and L is a subfield of K with [K : L] < ∞, then the restriction w = v|_{L} need not be henselian. But it is easy to see that w is then “semihenselian,” i.e., w has more than one but only finitely many different extensions to a separable closure L_{sep} of L. See [En] for a thorough analysis of semihenselian valuations. Notably, Engler shows that w is semihenselian iff the residue field \overline{L}_{w} is algebraically closed but there is a henselian valuation u on L such that u is a proper coarsening of w and the residue field \overline{L}_{u} is real closed. When this occurs, char(L) = 0, L is formally real, w has exactly two extensions to L_{sep}, the value group Γ_{L,w} has a nontrivial divisible subgroup, and the henselization of L re w is L(\sqrt{−1}), which lies in K. For example, if we take any prime number p, let w_{p} be the p-adic discrete valuation on ℚ, and let L = {r ∈ ℝ | r is algebraic over ℚ}; then any extension of w_{p} to L is a semihenselian valuation. Note that if v on K is discrete, i.e., Γ_{K} ≅ ℤ, then w on L cannot be semihenselian, since Γ_{L} has no nontrivial divisible subgroup; so, w on L must be henselian. This preservation of the henselian property for discrete valuations was asserted in [Y_{2}, Lemma, p. 195], but the proof given there is invalid.

One associates to a valued division algebra D a graded division algebra as follows: For each γ ∈ Γ_{D}, let

D^{≥γ} = {d ∈ D^{∗} | v(d) ≥ γ} ∪ {0}, an additive subgroup of D;
D^{>γ} = {d ∈ D^{∗} | v(d) > γ} ∪ {0}, a subgroup of D^{≥γ}; and
gr(D)_{γ} = D^{≥γ}/D^{>γ}.

Then define

gr(D) = ⨁_{γ∈Γ_{D}} gr(D)_{γ}.

Because D^{>γ}D^{≥δ} + D^{≥γ}D^{>δ} ⊆ D^{>(γ+δ)} for all γ, δ ∈ Γ_{D}, the multiplication on gr(D) induced by multiplication on D is well-defined, giving that gr(D) is a graded ring, called the associated graded ring of D. The multiplicative property (1) of the valuation v implies that gr(D) is a graded division ring. Clearly, we have gr(D)_{0} = \overline{D} and Γ_{gr(D)} = Γ_{D}. For d ∈ D^{∗}, we write d̃ for the image d + D^{>v(d)} of d in gr(D)_{v(d)}. Thus, the map given by d ↦ d̃is a group epimorphism ρ : D^{∗} → gr(D)^{∗} with kernel 1 + M_{D}, giving us the short exact sequence

1 ⟶ 1 + M_{D} ⟶ D^{∗} ⟶ gr(D)^{∗} ⟶ 1,  (3.2)

which will be used throughout. For a detailed study of the associated graded algebra of a valued division algebra refer to [HW_{2}, §4]. As shown in [HaW, Cor. 4.4], the reduced norm maps for D and gr(D) are related by

{ \atop Nrd_{D}(a)} = Nrd_{gr(D)}(ã) for all a ∈ D^{∗}.  (3.3)

Now let K be a field with a henselian valuation v and, as before, let D be a division algebra with center K. Then v extends uniquely to a valuation on D, also denoted v, and one obtains associated to D the graded division algebra gr(D) = ⨁_{γ∈Γ_{D}} D_{γ}. Further, suppose D is tame with respect to v. This implies that [gr(D) : gr(K)] = [D : K], gr(K) = Z(gr(D)) and D has a maximal subfield L with L tamely ramified over K ([HW_{2}, Prop. 4.3]). We can then associate to an involution τ on D, a graded involution τ̃ on gr(D). First, suppose τ is of the first kind on D. Then v ∘ τ is also a valuation on D which restricts to v on K; then, v ∘ τ = v since v has a unique extension to D. So, τ induces a well-defined map τ̃: gr(D) → gr(D), defined on homogeneous elements by τ̃(ã) = τ̃(a) for all a ∈ D^{∗}. Clearly, τ̃ is a well-defined graded involution on gr(D); it is of the first kind, as it leaves Z(gr(D)) = gr(K) invariant.

If τ is a unitary involution on D, let F = K^{τ}. In this case, we need to assume that the restriction of the valuation v from K to F induces a henselian valuation on F, and that K is tamely ramified over F. Since (v ∘ τ)|_{F} = v|_{F}, an argument similar to the one above shows that v ∘ τ coincides with v on K and thus on D, and the induced map τ̃ on gr(D) as above is a graded involution. That K is tamely ramified over F means that [K : F] = [gr(K) : gr(F)], \overline{K} is separable over \overline{F}, and char(\overline{F}) ∤ |Γ_{K} : Γ_{F} |. Since [K : F] = 2, K is always tamely ramified over F if char(\overline{F}) ≠ 2. But if char(\overline{F}) = 2, K is tamely ramified over F if and only if [\overline{K} : \overline{F}] = 2, Γ_{K} = Γ_{F}, and \overline{K} is separable (so Galois) over \overline{F}. Since K is Galois over F, the canonical map Gal(K/F) → Gal(\overline{K}/\overline{F}) is surjective, by [EP, pp. 123–124, proof of Lemma 5.2.6(1)]. Hence, τ induces the nonidentity \overline{F} -automorphism \overline{τ} of \overline{K}. Also τ̃ is unitary, i.e., τ̃|_{gr(K)} ≠ id. This is obvious if

char(\overline{F}) ≠ 2, since then K = F (\sqrt{c}) for some c ∈ F^{∗}, and τ̃(\sqrt{ c̃}) = { \atop τ(\sqrt{c})} = −\sqrt{ c̃} ≠ \sqrt{ c̃}. If char(\overline{F}) = 2, then K is unramified over F and τ̃|_{gr(K)_{0}} = \overline{τ} (the automorphism of \overline{K} induced by τ|_{K}) which is nontrivial as Gal(K/F) maps onto Gal(\overline{K}/\overline{F}); so again τ̃|_{gr(K)} ≠ id. Thus, τ̃ is a unitary graded involution in any characteristic. Moreover, for the graded fixed field gr(K)τ̃ we have gr(F) ⊆ gr(K)τ̃ ⫋ gr(K) and [gr(K) : gr(F)] = 2, so gr(K)τ̃ = gr(F).

Theorem 3.2. Let (D, v) be a tame valued division algebra over a henselian field K, with char(\overline{K}) ≠ 2. If τ is an involution of the first kind on D, then

K_{1}(D, τ) ≅ K_{1}(gr(D), τ̃),

and if τ is symplectic, then

K_{1} Spin(D, τ) ≅ K_{1} Spin(gr(D), τ̃).

Proof. Let ρ: D^{∗} → gr(D)^{∗} be the group epimorphism given in (3.2). Clearly ρ(S_{τ} (D)) ⊆ Sτ̃(gr(D)), so ρ(Σ_{τ} (D)) ⊆ Στ̃(gr(D)). Consider the following diagram:

\xymatrix{1 \ar[r] & (1 + M_{D}) ∩ Σ_{τ} (D)D′ \ar[r] \ar[d] & Σ_{τ} (D)D′ \ar[r]^{ρ} \ar[d] & Στ̃(gr(D)) gr(D)′ \ar@{.>}[r] \ar[d] & 1 \\ 1 \ar[r] & (1 + M_{D}) \ar[r] & D^{∗} \ar[r]^{ρ} & gr(D)^{∗} \ar[r] & 1.}  (3.4)

The top row of the diagram is exact. To see this, note that ρ(D′) = gr(D)′. Thus, it suffices to show that ρ maps S_{τ} (D) ∩ D^{∗} onto Sτ̃(gr(D)) ∩ gr(D)^{∗}. For this, take any d ∈ D^{∗} with d̃ = τ̃(d̃). Let b = \frac{1}{2}(d + τ(d)) ∈ S_{τ} (D). Since v(b) = v(τ(b)) and d̃+ τ̃(d) = 2d̃≠ 0, b̃ = \frac{1}{2}({ \atop d + τ(d)}) = \frac{1}{2}(d̃+ τ̃(d)) = d̃. Since τ on D is an involution of the first kind, the index of D is a power of 2 ([D, Th. 1, §16]). As char(\overline{K}) ≠ 2, it follows that the valuation is strongly tame, and by [Ha, Lemma 2.1],

1 + M_{D} = (1 + M_{K})[D^{∗}, 1 + M_{D}] ⊆ Σ_{τ} (D)D′.

Therefore, the left vertical map is the identity map. It follows (for example using the snake lemma) that K_{1}(D, τ) ≅ K_{1}(gr(D), τ̃). The proof for K_{1} Spin when τ is of symplectic type is similar. □

The key to proving the corresponding result for unitary involutions is the Congruence Theorem:

Theorem 3.3 (Congruence Theorem). Let D be a tame division algebra over a field K with henselian valuation v. Let D^{(1)} = {a ∈ D^{∗} | Nrd_{D}(a) = 1}. Then,

D^{(1)} ∩ (1 + M_{D}) ⊆ [D^{∗}, D^{∗}].

This theorem was proved by Platonov in [P_{2}] for v a complete discrete valuation, and it was an essential tool in all his calculations of SK_{1} for division rings. The Congruence Theorem was asserted by Ershov in [E] in the generality given here. A full proof is given in [HaW, Th. B.1].

Proposition 3.4 (Unitary Congruence Theorem). Let D be a tame division algebra over a field K with henselian valuation v, and let τ be a unitary involution on D. Let F = K^{τ}. If F is henselian with respect to v|_{F} and K is tamely ramified over F, then

(1 + M_{D}) ∩ Σ′_{τ} (D) ⊆ Σ_{τ} (D).

Proof. The only published proof of this we know is [Y_{2}, Th. 4.9], which is just for the case v discrete rank 1; that proof is rather hard to follow, and appears to apply for other valuations only if D is inertially split. Here we provide another proof, in full generality.

We use the well-known facts that

Nrd_{D}(1 + M_{D}) = 1 + M_{K} and N_{K/F} (1 + M_{K}) = 1 + M_{F} .  (3.5)

(The second equation holds as K is tamely ramified over F.) See [E, Prop. 2] or [HaW, Prop. 4.6, Cor. 4.7] for a proof.

Now, take m ∈ M_{D} with Nrd_{D}(1 + m) ∈ F. Then Nrd_{D}(1 + m) ∈ F ∩ (1 + M_{K}) = 1 + M_{F}. By (3.5) there is c ∈ 1 + M_{K} with Nrd_{D}(1 + m) = N_{K/F} (c) = cτ(c), and there is b ∈ 1 + M_{D} with Nrd_{D}(b) = c. Then,

Nrd_{D}(bτ(b)) = cτ(c) = N_{K/F} (c) = Nrd_{D}(1 + m).

Let s = (1 + m)(bτ(b))^{−1} ∈ 1 + M_{D}. Since Nrd_{D}(s) = 1, by the Congruence Theorem for SK_{1}, Th. 3.3 above, s ∈ [D^{∗}, D^{∗}] ⊆ Σ_{τ} (D), (recall (2.10)) . Since bτ(b) ∈ S_{τ} (D), we have 1+m = s(bτ(b)) ∈ Σ_{τ} (D). □

Theorem 3.5. Let D be a tame division algebra over a field K with henselian valuation v. Let τ be a unitary involution on D, and let F = K^{τ}. If F is henselian with respect to v|_{F} and K is tamely ramified over F, then τ induces a unitary graded involution τ̃ of gr(D) with gr(F) = gr(K)τ̃, and

SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃).

Proof. That τ̃ is a unitary graded involution on gr(D) and gr(F) = gr(K)τ̃ was already observed (see the discussion before Th. 3.2). For the canonical epimorphism ρ: D^{∗} → gr(D)^{∗}, a ↦ ã, it follows from (3.3) that ρ(Σ′_{τ} (D) ⊆ Σ′ ; τ̃(gr(D)). Also, clearly ρ(S_{τ} (D)) ⊆ Sτ̃(gr(D)), so ρ(Σ_{τ} (D)) ⊆ Στ̃(gr(D)). Thus, there is a commutative diagram

\xymatrix{1 \ar[r] & (1 + M_{D}) ∩ Σ_{τ} (D) \ar[d] \ar[r] & Σ_{τ} (D) \ar[d] \ar[r]^{ρ} & Στ̃(gr(D)) \ar[d] \ar@{.>}[r] & 1 \\ 1 \ar[r] & (1 + M_{D}) ∩ Σ′_{τ} (D) \ar[r] & Σ′_{τ} (D) \ar[r]^{ρ} & Σ′ ; τ̃(gr(D)) \ar@{.>}[r] & 1,}  (3.6)

where the vertical maps are inclusions, and the left vertical map is bijective, by Prop. 3.4 above.

To see that the bottom row of diagram (3.6) is exact at Σ′ ; τ̃(gr(D)), take b ∈ D with Nrd_{gr(D)}(b̃) ∈ gr(F). Let c = Nrd_{D}(b) ∈ K^{∗}. Then c̃ = Nrd_{gr(D)}(b̃) ∈ gr(F), so c̃ = t̃ for some t ∈ F^{∗}. Let u = c^{−1}t ∈ 1 + M_{K}. By (3.5) above, there is d ∈ 1 + M_{D} with Nrd_{D}(d) = u. So, Nrd_{D}(bd) = cu = t ∈ F^{∗}. Thus, bd ∈ Σ′_{τ} (D) and ρ(bd) = b̃d = b̃. This gives the claimed exactness, and shows that the bottom row of diagram (3.6) is exact.

To see that the top row of diagram (3.6) is exact at Στ̃(gr(D)), it suffices to show that ρ maps S_{τ} (D) ∩ D^{∗} onto Sτ̃(gr(D)) ∩ gr(D)^{∗}. For this, take any d ∈ D^{∗} with d̃ = τ̃(d̃). If char(\overline{F}) ≠ 2, as in the proof of Th. 3.2, let b = \frac{1}{2}(d + τ(d)) ∈ S_{τ} (D). Since v(b) = v(τ(b)) and d̃+ τ̃(d) = 2d̃≠ 0, we have b̃ = \frac{1}{2}({ \atop d + τ(d)}) = \frac{1}{2}(d̃+ τ̃(d)) = d̃. If char(\overline{F}) = 2, then K is unramified over F, so \overline{K} is Galois over \overline{F} with [\overline{K} : \overline{F}] = 2, and the map \overline{τ}: \overline{K} → \overline{K} induced by τ is the nonidentity \overline{F} -automorphism of \overline{K}. Of course, \overline{K} = gr(K)_{0} and \overline{τ} = τ̃|_{gr(K)_{0}}. Because \overline{K} is separable over \overline{F}, the trace tr_{\overline{K}/\overline{F}} is surjective, so there is r ∈ V_{K} with r̃+ τ̃(r̃) = 1 ∈ gr(F)_{0}. Let c = rd + τ(rd) ∈ S_{τ} (D). We have r̃d = r̃d̃ and

{ \atop τ(rd)} = τ̃(r̃d) = τ̃(r̃d̃) = τ̃(d̃)τ̃(r̃) = τ̃(r̃)d̃.

Since v(rd) = v(τ(rd)) and r̃d + { \atop τ(rd)} = r̃d̃+ τ̃(r̃)d̃= d̃≠ 0, we have c̃ = r̃d + { \atop τ(rd)} = d̃. So, in all cases ρ(S_{τ} (D) ∩ D^{∗}) = Sτ̃(gr(D)) ∩ gr(D)^{∗}, from which it follows that the bottom row of diagram (3.6) is exact.

Since each row of (3.6) is exact, we have a right exact sequence of cokernels of the vertical maps, which yields the isomorphism of the theorem. □

Having established the bridge between the unitary K-groups in the graded setting and the non-graded henselian case (Th. 3.2, Th. 3.5), we can deduce known formulas in the literature for the unitary Whitehead group of certain valued division algebras, by passing to the graded setting. The proofs are much easier than those previously available. We will do this systematically for unitary involutions in Section 4. Before we turn to that, here is an example with an involution of the first kind: Example 3.6. Let E be a graded division algebra over its center T with an involution τ of the first kind. If E is unramified over T, then, by using E^{∗} = E_{0}^{∗}T^{∗}, it follows easily that

K_{1}(E, τ) ≅ K_{1}(E_{0}, τ|_{E_{0}}),  (3.7)

and, if char(E) ≠ 2 and τ is symplectic,

K_{1} Spin(E, τ) ≅ K_{1} Spin(E_{0}, τ|_{E_{0}}).  (3.8)

Now if D is a tame and unramified division algebra over a henselian valued field and D has an involution τ of the first kind, then the associated graded division ring gr(D) is also unramified with the corresponding graded involution τ̃ of the first kind; then Th. 3.2 and (3.7) above show that

K_{1}(D, τ) ≅ K_{1}(gr(D), τ̃) ≅ K_{1}(gr(D)_{0}, τ|_{gr(D)_{0}}) = K_{1}(\overline{D}, \overline{τ}),

yielding a theorem of Platonov-Yanchevskiĭ [PY, Th. 5.11] (that K_{1}(D, τ) ≅ K_{1}(\overline{D}, \overline{τ}) when D is unramified over K and the valuation is henselian and discrete rank 1.) Similarly, when char(\overline{D}) ≠ 2 and τ is symplectic,

K_{1} Spin(D, τ) ≅ K_{1} Spin(gr(D), τ̃) ≅ K_{1} Spin(gr(D)_{0}, τ|_{gr(D)_{0}}) = K_{1} Spin(\overline{D}, \overline{τ}).

Remark 3.7. We have the following commutative diagram connecting unitary SK_{1} to non-unitary SK_{1}, where SH^{0}(D, τ) and SH^{0}(D) are the cokernels of Nrd ∘ (1 − τ) and Nrd respectively (see diagram (2.8)).

\xymatrix{1 \ar[r] & SK_{1}(D, τ) \ar[r] \ar[d] & D^{∗}/Σ(D) \ar[r]^{Nrd∘(1−τ)} \ar[d]^{1−τ} & K^{∗} \ar[r] \ar[d]^{id} & SH^{0}(D, τ) \ar[r] \ar[d] & 1 \\ 1 \ar[r] & SK_{1}(D) \ar[r] & D^{∗}/D′ \ar[r]^{Nrd} & K^{∗} \ar[r] & SH^{0}(D) \ar[r] & 1.}  (3.9)

Now, let D be a tame valued division algebra with center K and with a unitary involution τ, such that the valuation restricts to a henselian valuation on F = K^{τ}. By Th. 3.5, SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃) and by [HaW, Th. 4.8, Th. 4.12], SK_{1}(D) ≅ SK_{1}(gr(D)) and SH^{0}(D) ≅ SH^{0}(gr(D)). However, SH^{0}(D, τ) is not stable under “valued filtration”, i.e., SH^{0}(D, τ) ≅ S̸H^{0}(gr(D), τ̃). In fact using (3.2), we can build a commutative diagram with exact rows,

\xymatrix{1 \ar[r] & (1 + M_{K}) ∩ Nrd(D^{∗})^{1−τ} \ar[r] \ar@{^{(}->}[d] & Nrd(D^{∗})^{1−τ} \ar[r] \ar@{^{(}->}[d] & Nrd( gr(D)^{∗})^{1−τ̃} \ar[r] \ar@{^{(}->}[d] & 1 \\ 1 \ar[r] & 1 + M_{K} \ar[r] & K^{∗} \ar[r] & gr(K)^{∗} \ar[r] & 1,}

which induces the exact sequence

1 ⟶ (1 + M_{K})/((1 + M_{K}) ∩ Nrd(D^{∗})^{1−τ} ) ⟶ SH^{0}(D, τ) ⟶ SH^{0}(gr(D), τ̃) ⟶ 1.

By considering the norm N_{K/F} : K^{∗} → F^{∗}, we clearly have Nrd(D^{∗})^{1−τ} ⊆ ker N_{K/F}. However, by (3.5), N_{K/F} : 1 + M_{K} → 1 + M_{F} is surjective, which shows that 1 + M_{K}/((1 + M_{K}) ∩ Nrd(D^{∗})^{1−τ} ) is not trivial and thus SH^{0}(D, τ) ≅ S̸H^{0}(gr(D), τ̃).

4. Graded Unitary SK_{1} Calculus

Let E be a graded division algebra over its center T with a unitary graded involution τ, and let R = T^{τ}. Since [T : R] = 2 = [T_{0} : R_{0}] |Γ_{T} : Γ_{R}|, there are just two possible cases:

• T is totally ramified over R, i.e., |Γ_{T} : Γ_{R}| = 2

• T is unramfied over R, i.e., |Γ_{T} : Γ_{R}| = 1.

We will consider SK_{1}(E, τ) in these two cases separately in §4.1 and §4.2.

The following notation will be used throughout this section and the next: Let τ′ be another involution on E. We write τ′ ∼ τ if τ′|_{Z(E)} = τ|_{Z(E)}. For t ∈ E^{∗}, let φ_{t} denote the map from E to E given by conjugation by t, i.e., φ_{t}(x) = txt^{−1}. Let Σ_{0} = Σ_{τ} ∩ E_{0}^{∗} and Σ′_{0} = Σ′_{τ} ∩ E_{0}^{∗}.

We first collect some facts which will be used below. They all follow by easy calculations.

Remarks 4.1.

(i) We have τ′ ∼ τ if and only if there is a t ∈ E^{∗} with τ(t) = t and τ′ = τφ_{t}. (The proof is analogous to the ungraded version given, e.g. in [KMRT, Prop. 2.18].)

(ii) If τ′ ∼ τ, then Σ_{τ′} = Σ_{τ} and Σ′_{τ′} = Σ′_{τ}; thus SK_{1}(E, τ′) = SK_{1}(E, τ). (See [Y_{1}, Lemma 1] for the

analogous ungraded result.) (iii) For any s ∈ E^{∗}, we have τφ_{s} = φ_{τ(s)^{−1}}τ. Hence, τφ_{s} is an involution (necessarily ∼ τ) if and only if

τφ_{s} = φ_{s^{−1}}τ if and only if τ(s)/s ∈ T. (iv) If s ∈ E_{γ}^{∗} and τ(s) = s, then Σ′_{τ} ∩ E_{γ} = sΣ′_{0} and S_{τ} ∩ E_{γ} = s(S_{τ_{s}} ∩ E_{0}) where τ_{s} = τφ_{s}.

4.1. T/R totally ramified. Let E be a graded division algebra with a unitary graded involution τ such that T = Z(E) is totally ramified over R = T^{τ}. In this section we will show that SK_{1}(E, τ) = 1. Note that the assumption that T/R is totally ramified implies that char(T) ≠ 2. For, if char(T) = 2 and T is totally ramified over a graded subfield R with [T : R] = 2, then for any x ∈ T^{∗}\R^{∗}, we have deg(x^{2}) ∈ Γ_{R}, so x^{2} ∈ R; thus, T is purely inseparable over R. That cannot happen here, as τ|_{T} is a nontrivial R-automorphism of T.

Lemma 4.2. If T is totally ramified over R, then τ ∼ τ′ for some graded involution τ′, where τ′|_{E_{0}} is of the first kind.

Proof. Let Z_{0} = Z(E_{0}). Since T is totally ramified over R, T_{0} = R_{0}, so τ|_{Z_{0}} ∈ Gal(Z_{0}/T_{0}). Since the map Θ_{E} : Γ_{E} → Gal(Z_{0}/T_{0}) is surjective (see (2.3)), there is γ ∈ Γ_{E} with Θ_{E}(γ) = τ|_{Z_{0}}. Choose y ∈ E_{γ}^{∗} with τ(y) = ±y. Then set τ′ = τφ_{y^{−1}}. □

Example 4.3. Here is a construction of examples of graded division algebras E with unitary graded involution τ with E totally ramified over Z(E)^{τ}. We will see below that these are all such examples. Let R be any graded field with char(R) ≠ 2, and let A be a graded division algebra with center R, such that A is totally ramified over R with exp(Γ_{A}/Γ_{R}) = 2. Let T be a graded field extension of R with [T : R] = 2, T totally ramified over R, and Γ_{T} ∩ Γ_{A} = Γ_{R}. Let E = A ⊗_{R} T, which is a graded central simple algebra over T, as A is graded central simple over R, by [HW_{2}, Prop. 1.1]. But because Γ_{T} ∩ Γ_{A} = Γ_{R}, we have E_{0} = A_{0} ⊗_{R_{0}} T_{0} = R_{0} ⊗_{R_{0}} R_{0} = R_{0}. Since E_{0} is a division ring, E must be a graded division ring, which is totally ramified over R, as E_{0} = R_{0}. Now, because A is totally ramified over R, we have exp(A) = exp(Γ_{A}/Γ_{R}) = 2, and A = Q_{1} ⊗_{R} … ⊗_{R} Q_{m}, where each Q_{i} is a graded symbol algebra of degree at most 2, i.e., a graded quaternion algebra. Let σ_{i} be a graded involution of the first kind on Q_{i} (e.g., the canonical symplectic graded involution), and let ρ be the nonidentity R-automorphism of T. Then, σ = σ_{1} ⊗ … ⊗ σ_{m} is a graded involution of the first kind on A, so σ ⊗ ρ is a unitary graded involution on E, with T^{τ} = R.

Proposition 4.4. If E is totally ramified over R, and E ≠ T, then Σ_{τ} = E^{∗}, so SK_{1}(E, τ) = 1. Furthermore, E and τ are as described in Ex. 4.3.

Proof. We have E_{0} = T_{0} = R_{0}. For any γ ∈ Γ_{E}, there is a nonzero a ∈ E_{γ} with τ(a) = ϵa where ϵ = ±1. Then, for any b ∈ E_{γ}, b = ra for some r ∈ E_{0} = R_{0}. Since r is central and symmetric, τ(b) = ϵb. Thus, every element of E^{∗} is symmetric or skew-symmetric. Indeed, fix any t ∈ T^{∗} \ R^{∗}. Then τ(t) ≠ t, as t ∉ R^{∗}. Hence, τ(t) = −t. Since t is central and skew-symmetric, every a ∈ E^{∗} is symmetric iff ta is skewsymmetric. Thus, E^{∗} = S_{τ}^{∗} ∪ tS_{τ}^{∗}. To see that Σ_{τ} = E^{∗}, it suffices to show that t ∈ Σ_{τ}. To see this, take any c, d ∈ E^{∗} with dc ≠ cd. (They exist, as E ≠ T.) By replacing c (resp. d) if necessary by tc (resp. td), we may assume that τ(c) = c and τ(d) = d. Then, dc = τ(cd) = ϵcd, where ϵ = ±1; since dc ≠ cd, ϵ = −1; hence τ(tcd) = tcd. Thus, t = (tcd)c^{−1}d^{−1} ∈ Σ_{τ} (E), completing the proof that Σ_{τ} (E) = E^{∗}.

For γ ∈ Γ_{E}, let \overline{γ} = γ +Γ_{T} ∈ Γ_{E}/Γ_{T}. To see the structure of E, recall that as E is totally ramified over T there is a well-defined nondegenerate ℤ-bilinear symplectic pairing β: (Γ_{E}/Γ_{T}) × Γ_{E}/Γ_{T}) → E_{0}^{∗} given by β(\overline{γ}, \overline{δ}) = y_{γ}y_{δ}y_{γ}^{−1}y_{δ}^{−1} for any nonzero y_{γ} ∈ E_{γ}, y_{δ} ∈ E_{δ}. The computation above for c and d shows that im(β) = {±1}. Since the pairing β is nondegenerate by [HW_{2}, Prop. 2.1] there is a symplectic base of Γ_{E}/Γ_{T}, i.e., a subset \{\overline{γ}_{1}, \overline{δ}_{1}, …, \overline{γ}_{m}, \overline{δ}_{m}\} of Γ_{E}/Γ_{T} such that β(\overline{γ}_{i}, \overline{δ}_{i}) = −1 while β(\overline{γ}_{i}, \overline{γ}_{j}) = β(\overline{δ}_{i}, \overline{δ}_{j}) = 1 for all i, j, and β(\overline{γ}_{i}, \overline{δ}_{j}) = 1 whenever i ≠ j, and Γ_{E} = ⟨γ_{1}, δ_{1}, …, γ_{m}, δ_{m}⟩ + Γ_{T}. Choose any nonzero i_{i} ∈ E_{γ_{i}} and j_{i} ∈ E_{δ_{i}}. The properties of the \overline{γ}_{i}, \overline{δ}_{i} under β translate to: i_{i}j_{i} = −j_{i}i_{i} while i_{i}i_{j} = i_{j}i_{i} and j_{i}j_{j} = j_{j}j_{i} for all i, j, and i_{i}j_{j} = j_{j}i_{i} whenever i ≠ j. Since β(\overline{2γ_{i}}, \overline{η}) = 1 for all i and all η ∈ Γ_{E}, each i_{i}^{2} is central in E. But also τ(i_{i}^{2}) = i_{i}^{2}, as τ(i_{i}) = ±i_{i}. So, each i_{i}^{2} ∈ R^{∗}, and likewise each j_{i}^{2} ∈ R^{∗}. Let Q_{i} = R-span{1, i_{i}, j_{i}, i_{i}j_{i}} in E. The relations on the i_{i}, j_{i} show that each Q_{i} is a graded quaternion algebra over R, and the distinct Q_{i} centralize each other in E. Since each Q_{i} is graded central simple over R, Q_{1} ⊗_{R} … ⊗_{R} Q_{m} is graded central simple over R by [HW_{2}, Prop. 1.1]. Let A = Q_{1} … Q_{m} ⊆ E. The graded R-algebra epimorphism Q_{1} ⊗_{R} … ⊗_{R} Q_{m} → A must be an isomorphism, as the domain is graded simple. If Γ_{T} ⊆ Γ_{A}, then T ⊆ A, since E is totally ramified over R. But this cannot occur, as T centralizes A but T ⫌ R = Z(A). Hence, as |Γ_{T} : Γ_{R}| = 2, we must have Γ_{T}∩Γ_{A} = Γ_{R}. The graded R-algebra homomorphism A ⊗_{R} T → E is injective since its domain is graded simple, by [HW_{2}, Prop. 1.1]; it is also surjective, since E_{0} = R_{0} ⊆ A ⊗_{R} T and Γ_{A⊗_{R}T} ⊇ ⟨γ_{1}, δ_{1}, …, γ_{m}, δ_{m}⟩ + Γ_{T} = Γ_{E}. Clearly, τ = τ|_{A} ⊗ τ|_{T}. □

Proposition 4.5. If E ≠ T and T is totally ramified over R, then Σ_{τ} = E^{∗}, so SK_{1}(E, τ) = 1.

Proof. The case where E_{0} = T_{0} was covered by Prop. 4.4. Thus, we may assume that E_{0} ⫌ T_{0}. By Lemma 4.2 and Remark 4.1(ii), we can assume that τ|_{E_{0}} is of the first kind. Further, we can assume that E_{0}^{∗} = Σ_{τ|_{E_{0}}} (E_{0}). For, if τ|_{E_{0}} is symplectic, take any a ∈ E_{0}^{∗} with τ(a) = −a, and let τ′ = τφ_{a}. Then, τ′ ∼ τ (see Remark 4.1(iii)). Also, τ′|_{Z(E_{0})} = τ|_{Z(E_{0})}, as a ∈ E_{0} and so φ_{a}|_{Z(E_{0})} = id. Therefore, τ′|_{E_{0}} is of the first kind. But as τ(a) = −a, τ′|_{E_{0}} is orthogonal. Thus E_{0}^{∗} = Σ_{τ′|_{E_{0}}} (E_{0}), as noted at the beginning of §2.2.1. Now replace τ by τ′.

We consider two cases.

Case I. Suppose for each γ ∈ Γ_{E} there is x_{γ} ∈ E_{γ}^{∗} such that τ(x_{γ}) = x_{γ}. Then, E^{∗} = ⋃_{γ∈Γ_{E}} E_{0}^{∗}x_{γ} ⊆ Σ_{τ} (E), as desired.

Case II. Suppose there is γ ∈ Γ_{E} with E_{γ} ∩ S_{τ} = 0. Then τ(d) = −d for each d ∈ E_{γ}. Fix t ∈ E_{γ}^{∗}. For any a ∈ E_{0}, we have ta ∈ E_{γ}; so, −ta = τ(ta) = τ(a)τ(t) = −τ(a)t. That is,

τ(a) = φ_{t}(a) for all a ∈ E_{0}.  (4.1)

Let τ′′ = τφ_{t}, which is a unitary involution on E with τ′′ ∼ τ (see Remark 4.1(iii)). But, τ′′(a) = a for all a ∈ E_{0}, i.e., τ′′|_{E_{0}} = id. This implies that E_{0} is a field. Replace τ by τ′′. The rest of the argument uses this new τ. So τ|_{E_{0}} = id. If we are now in Case I for this τ, then we are done by Case I. So, assume we are in Case II. Take any γ ∈ Γ_{E} with E_{γ} ∩ S_{τ} = 0. For any nonzero t ∈ E_{γ}, equation (4.1) applies to t, showing φ_{t}(a) = τ(a) = a for all a ∈ E_{0}; hence for the map Θ_{E} of (2.3), Θ_{E}(γ) = id_{E_{0}}. But recall that E_{0} is Galois over T_{0} and Θ_{E} : Γ_{E} → Gal(E_{0}/T_{0}) is surjective. Since E_{0} ≠ T_{0}, there is δ ∈ Γ_{E} with Θ_{E}(δ) ≠ id. Hence, there must be some s ∈ E_{δ}^{∗} ∩ S_{τ}. Likewise, since Θ_{E}(γ − δ) = Θ_{E}(γ)Θ_{E}(δ)^{−1} ≠ id, there is some r ∈ E_{γ−δ}^{∗} ∩ S_{τ}. Then, as rs ∈ E_{γ}^{∗}, we have E_{γ}^{∗} = E_{0}^{∗}rs ⊆ Σ_{τ}. This is true for every γ with E_{γ} ∩ S_{τ} = 0. But for any other γ ∈ Γ_{E}, there is an x_{γ} in E_{γ}^{∗} ∩ S_{τ}; then E_{γ}^{∗} = E_{0}^{∗}x_{γ} ⊆ Σ_{τ}. Thus, E^{∗} = ⋃_{γ∈Γ_{E}} E_{γ}^{∗} ⊆ Σ_{τ}. □

4.2. T/R unramified. Let E be a graded division algebra with a unitary involution τ such that T = Z(E) is unramified over R = T^{τ}. In this subsection, we will give a general formula for SK_{1}(E, τ) in terms of data in E_{0}.

Lemma 4.6. Suppose T is unramified over R. Then,

(i) Every E_{γ} contains both nonzero symmetric and skew symmetric elements. (ii) Z(E_{0}) is a generalized dihedral extension for T_{0} over R_{0} (see Def. 2.4). (iii) If T is unramified over R, then SK_{1}(E, τ) = Σ′_{0}/Σ_{0}.

Proof.

(i) If char(E) = 2, it is easy to see that every E_{γ} contains a symmetric element (which is also skew symmetric) regardless of any assumption on T/R. Let char(E) ≠ 2. Since [T_{0} : R_{0}] = 2 and R_{0} = T_{0}^{τ}, there is c ∈ T_{0} with τ(c) = −c. Now there is t ∈ E_{γ}, t ≠ 0, with τ(t) = ϵt where ϵ = ±1. Then τ(ct) = −ϵct.

(ii) Let G = Gal(Z(E_{0})/R_{0}) and H = Gal(Z(E_{0})/T_{0}). Note that [G : H] = 2. Since τ is unitary, τ|_{Z(E_{0})} ∈ G \ H. We will denote τ|_{Z(E_{0})} by \overline{τ} and will show that for any h ∈ H, (\overline{τ}h)^{2} = 1. By (2.3), Θ_{E} : Γ_{E} → Gal(Z(E_{0})/T_{0}) is onto, so there is γ ∈ Γ_{E}, such that Θ_{E}(γ) = h. Also by (i), there is an x ∈ E_{γ}^{∗} with τ(x) = x. Then τφ_{x} is an involution, where φ_{x} is conjugation by x; therefore, τφ_{x}|_{Z(E_{0})} ∈ G has order 2. But φ_{x}|_{Z(E_{0})} = Θ_{E}(γ) = h. Thus (\overline{τ}h)^{2} = 1.

(iii) By (i), for each γ ∈ Γ_{E}, there is s_{γ} ∈ E_{γ}, s_{γ} ≠ 0, with τ(s_{γ}) = s_{γ}. By Remark 4.1(iv), Σ′_{τ} = ⋃_{γ∈Γ_{E}} s_{γ}Σ′_{0}. Since each s_{γ} ∈ S_{τ} ⊆ Σ_{τ}, the injective map Σ′_{0}/Σ_{0} → Σ′_{τ} /Σ_{τ} is an isomorphism. □

To simplify notation in the next theorem, let \overline{τ} = τ|_{Z(E_{0})} ∈ Gal(Z(E_{0})/R_{0}), and for any h ∈ Gal(Z(E_{0})/T_{0}), write Σ_{h\overline{τ}} (E_{0}) for Σ_{ρ}(E_{0}) for any unitary involution ρ on E_{0} such that ρ|_{Z(E_{0})} = h\overline{τ}. This is well-defined, independent of the choice of ρ, by the ungraded analogue of Remark(4.1)(ii).

Theorem 4.7. Let E be a graded division algebra with center T, with a unitary graded involution τ, such that T is unramified over R = T^{τ}. For each γ ∈ Γ_{E} choose a nonzero x_{γ} ∈ S_{τ} ∩E_{γ}. Let H = Gal(Z(E_{0})/T_{0}). Then,

SK_{1}(E, τ) ≅ (Σ′_{τ} ∩ E_{0})/(Σ_{τ} ∩ E_{0}),

with

Σ′_{τ} ∩ E_{0} = {a ∈ E_{0}^{∗} | N_{Z(E_{0})/T_{0}}Nrd_{E_{0}}(a)^{∂} ∈ R_{0}}, where ∂ = ind(E)/( ind(E_{0}) [Z(E_{0}) : T_{0}]) (4.2)

and

Σ_{τ} ∩ E_{0} = P · X, where P = ∏_{h∈H}Σ_{h\overline{τ}} (E_{0}) and X = ⟨x_{γ}x_{δ}x_{γ+δ}^{−1} | γ, δ ∈ Γ_{E}⟩ ⊆ E_{0}^{∗}. (4.3)

Furthermore, if H = ⟨h_{1}, …, h_{m}⟩, then P = ∏_{(ε_{1},...,ε_{m})∈\{0,1\}^{m}} Σ_{h_{1}^{ε_{1}}...h_{m}^{εm} \overline{τ}} (E_{0}).

Before proving the theorem, we record the following:

Lemma 4.8. Let A be a central simple algebra over a field K, with an involution τ and an automorphism or anti-automorphism σ. Then,

(i) στσ^{−1} is an involution of A of the same kind as τ, and

S_{στσ^{−1}} = σ(S_{τ}), so Σ_{στσ^{−1}} = σ(Σ_{τ}).

(ii) Suppose A is a division ring. If σ and τ are each unitary involutions, then (writing S_{τ}^{∗} = S_{τ} ∩ A^{∗}),

S_{τ}^{∗} ⊆ S_{σ}^{∗} · σ(S_{τ}^{∗}) = S_{σ}^{∗} · S_{στσ^{−1}}^{∗}, so Σ_{τ} ⊆ Σ_{σ} · Σ_{στσ^{−1}}.

Proof.

(i) This follows by easy calculations.

(ii) Observe that if a ∈ S_{τ}^{∗}, then a = (aσ(a))σ(a^{−1}) with aσ(a) ∈ S_{σ}^{∗} and σ(a^{−1}) ∈ σ(S_{τ}^{∗}) = S_{στσ^{−1}}^{∗} by (i). Thus, (ii) follows from (i) and the fact that A′ ⊆ Σ_{τ} ∩ Σ_{σ} (see (2.10)). □

Proof of Theorem 4.7. First note that by Lemma 4.6(iii) the canonical map

(Σ′_{τ} ∩ E_{0}) / (Σ_{τ} ∩ E_{0}) ⟶ Σ′_{τ} /Σ_{τ} = SK_{1}(E, τ)

is an isomorphism. The description of Σ′_{τ} ∩ E_{0} in (4.2) is immediate from the fact that for a ∈ E_{0}, Nrd_{E}(a) = N_{Z(E_{0})/T_{0}}Nrd_{E_{0}}(a)^{∂} ∈ T_{0} (see Remark 2.1(iii)).

For Σ_{τ} ∩ E_{0}, note that for each γ ∈ Γ_{E}, if a ∈ E_{0}, then ax_{γ} ∈ S_{τ} if and only if x_{γ}τ(a)x_{γ}^{−1} = a. That is, S_{τ} ∩ E_{γ} = S(φ_{x_{γ}}τ; E_{0})x_{γ}, where S(φ_{x_{γ}}τ; E_{0}) denotes the set of symmetric elements in E_{0} for the unitary involution φ_{x_{γ}}τ|_{E_{0}}. Therefore,

Σ_{τ} ∩ E_{0} = ⟨S(φ_{x_{γ}}τ; E_{0})^{∗}x_{γ} | γ ∈ Γ_{E}⟩ ∩ E_{0}.

Take a product a_{1}x_{1} … a_{k}x_{k} in Σ_{τ} ∩ E_{0} where each x_{i} = x_{γ_{i}} for some γ_{i} ∈ Γ_{E} and a_{i} ∈ S(φ_{x_{i}}τ; E_{0})^{∗}. Then,

a_{1}x_{1} … a_{k}x_{k} = a_{1}φ_{x_{1}}(a_{2}) … φ_{x_{1}...x_{i−1}}(a_{i}) … φ_{x_{1}...x_{k−1}}(a_{k})x_{1} … x_{k} ∈ E_{γ_{1}+...+γ_{k}}.  (4.4)

So, γ_{1} + … + γ_{k} = 0. Now, as a_{i} ∈ S(φ_{x_{i}}τ; E_{0}) and τφ_{x_{j}}^{−1} = φ_{x_{j}}τ for all j, by Lemma 4.8(i) we obtain

φ_{x_{1}...x_{i−1}}(a_{i}) ∈ S(φ_{x_{1}} … φ_{x_{i−1}}(φ_{x_{i}}τ)φ_{x_{i−1}}^{−1} … φ_{x_{1}}^{−1}; E_{0})^{∗} = S(φ_{x_{1}...x_{i−1}x_{i}x_{i−1}...x_{1}}τ; E_{0})^{∗} ⊆ Σ_{h\overline{τ}} (E_{0}) ⊆ P,

(4.5)

where h = φ_{x_{1}...x_{i−1}x_{i}x_{i−1}...x_{1}}|_{Z(E_{0})} ∈ H. Note also that if k = 1, then x_{1} ∈ S_{τ} ∩ E_{0}^{∗} ⊆ Σ\overline{τ} (E_{0}) ⊆ P.

If k > 1, then

x_{1} … x_{k} = x_{γ_{1}} … x_{γ_{k}} = (x_{γ_{1}}x_{γ_{2}}x_{γ_{1}+γ_{2}}^{−1})(x_{γ_{1}+γ_{2}}x_{γ_{3}} … x_{γ_{k}}),

with (γ_{1} + γ_{2}) + γ_{3} + … + γ_{k} = 0. It follows by induction on k that x_{1} … x_{k} ∈ X. With this and (4.4) and (4.5), we have a_{1}x_{1} … a_{k}x_{k} ∈ P · X (which is a group, as E′_{0} ⊆ Σ\overline{τ} (E_{0}) ⊆ P by (2.10)), showing that Σ_{τ} ∩ E_{0} ⊆ P · X. For the reverse inclusion, take any h ∈ H and choose γ ∈ Γ_{E} with φ_{x_{γ}}|_{Z(E_{0})} = h. Then, x_{γ} ∈ S_{τ}^{∗} ⊆ Σ_{τ} and S(φ_{x_{γ}}τ; E_{0})^{∗}x_{γ} = S_{τ}^{∗} ∩ E_{γ} ⊆ Σ_{τ}, so Σ_{h\overline{τ}} (E_{0}) = Σ_{φ_{xγ} τ} (E_{0}) = ⟨S(φ_{x_{γ}}τ; E_{0})^{∗}⟩ ⊆ Σ_{τ} ∩ E_{0}. Thus, P ⊆ Σ_{τ} ∩ E_{0}, and clearly also X ⊆ Σ_{τ} ∩ E_{0}. Hence, Σ_{τ} ∩ E_{0} = P · X.

The final equality for P in the Theorem follows from Lemma 4.9 below by taking U = E_{0}^{∗}, A = H, and W_{h} = Σ_{h\overline{τ}} (E_{0}) for h ∈ H. To see that the lemma applies, note that each Σ_{h\overline{τ}} (E_{0}) contains E′_{0} by (2.10). Furthermore, take any h, ℓ ∈ H, and choose x, y ∈ E^{∗} ∩ S_{τ} with φ_{x}|_{Z(E_{0})} = h and φ_{y}|_{Z(E_{0})} = ℓ. Then,

(φ_{y}τ)(φ_{x}τ)(φ_{y}τ)^{−1} = φ_{y}τφ_{x}φ_{y}^{−1} = φ_{yx^{−1}y}τ,

and φ_{yx^{−1}y}|_{Z(E_{0})} = ℓh^{−1}ℓ = ℓ^{2}h^{−1}. Hence, by Lemma 4.8(ii), Σ_{h\overline{τ}} (E_{0}) ⊆ Σ_{ℓ\overline{τ}} (E_{0})Σ_{ℓ^{2}h^{−1}\overline{τ}} (E_{0}). This shows that hypothesis (4.6) of Lemma 4.9 below is satisfied here. □

Lemma 4.9. Let U be a group, A an abelian group, and {W_{a} | a ∈ A} a family of subgroups of U with each W_{a} ⊇ [U, U]. Suppose

W_{a} ⊆ W_{b}W_{2b−a} for all a, b ∈ A.  (4.6)

If A = ⟨a_{1}, …, a_{m}⟩, then

∏_{a∈A}W_{a} = ∏_{(ε_{1},...,ε_{m})∈{0,1}^{m}}W_{ε_{1}a_{1}+...+ε_{m}a_{m}}.

Proof. Since each W_{a} ⊇ [U, U], we have W_{a}W_{b} = W_{b}W_{a}, and this is a subgroup of U, for all a, b ∈ A. Let

Q = ∏_{(ε_{1},...,ε_{m})∈{0,1}^{m}}W_{ε_{1}a_{1}+...+ε_{m}a_{m}}.

We prove by induction on m that each W_{a} ⊆ Q. The lemma then follows, as Q is a subgroup of U. Note that condition (4.6) can be conveniently restated,

if a + b = 2d ∈ A, then W_{a} ⊆ W_{d}W_{b}.  (4.7)

Take any c ∈ A. Then, (4.7) shows that W_{−c} ⊆ W_{0}W_{c}. Take any i ∈ ℤ, and suppose W_{ic} ⊆ W_{0}W_{c}. Then by (4.7) W_{−ic} ⊆ W_{0}W_{ic} ⊆ W_{0}W_{c}. So, by (4.7) again, W_{(i+2)c} ⊆ W_{c}W_{−ic} ⊆ W_{0}W_{c} and W_{(i−2)c} ⊆ W_{−c}W_{−ic} ⊆ W_{0}W_{c}. Hence, by induction (starting with j = 0 and j = 1), W_{jc} ⊆ W_{0}W_{c} for every j ∈ ℤ. This proves the lemma when m = 1.

Now assume m > 1 and let B = ⟨a_{1}, …, a_{m−1}⟩ ⊆ A. By induction, for all b ∈ B,

W_{b} ⊆ ∏_{(ε_{1},...,ε_{m−1})∈{0,1}^{m−1}}W_{ε_{1}a_{1}+...+ε_{m−1}a_{m−1}} ⊆ Q.

Also, by the cyclic case done above, W_{ja_{m}} ⊆ W_{0}W_{a_{m}} ⊆ Q for all j ∈ ℤ. So, for any b ∈ B, j ∈ ℤ, using (4.7),

W_{2b+ja_{m}} ⊆ W_{b}W_{−ja_{m}} ⊆ Q,  (4.8)

and

W_{b+2ja_{m}} ⊆ W_{ja_{m}}W_{−b} ⊆ Q.  (4.9)

Let d = a_{i_{1}} + … + a_{i_{ℓ}} for any indices 1 ≤ i_{1} < i_{2} < … < i_{ℓ} ≤ m − 1. Since W_{d+a_{m}} ⊆ Q by hypothesis, from (4.7) and (4.9) it follows that

W_{d+3a_{m}} ⊆ W_{d+2a_{m}}W_{d+a_{m}} ⊆ Q.  (4.10)

Now, take any element of A; it has the form b + ja_{m} for some b ∈ B and j ∈ ℤ. If b ∈ 2B or if j is even, then (4.8) and (4.9) show that W_{b+ja_{m}} ⊆ Q. The remaining case is that j is odd and b ∉ 2B, so b = 2c + d, where c ∈ B and d = a_{i_{1}} + … + a_{i_{ℓ}} for some indices with 1 ≤ i_{1} < i_{2} < … < i_{ℓ} ≤ m − 1. Set q = 1 if j ≡ 3 (mod 4) and q = 3 if j ≡ 1 (mod 4). Then, W_{d+qa_{m}} ⊆ Q by definition if q = 1 or by (4.10) if q = 3. Hence, by (4.7),

W_{b+ja_{m}} = W_{(2c+d)+ja_{m}} ⊆ W_{(c+d)+((j+q)/2)a_{m}}W_{d+qa_{m}} ⊆ Q,  (4.11)

using (4.9) as c + d ∈ B and (j + q)/2 is even. Thus, W_{a} ⊆ Q, for all a ∈ A. □

Corollary 4.10. If E is unramified over R, then SK_{1}(E, τ) ≅ SK_{1}(E_{0}, τ|_{E_{0}}).

Proof. Since E is unramified over R, we have T is unramified over R, Z(E_{0}) = T_{0}, and Γ_{E} = Γ_{R}, so we can choose all the x_{γ}’s to lie in R. The assertion thus follows immediately from Th. 4.7, as P = Σ_{τ|_{E_{0}}} (E_{0}) and X ⊆ R_{0}^{∗} ⊆ P. (Alternatively, more directly, one can observe that Σ′_{0} = Σ′_{τ|_{E_{0}}} (E_{0}) and Σ_{0} = Σ_{τ|_{E_{0}}} (E_{0}) and so deduce the Corollary by Lemma 4.6(iii).) □

Corollary 4.11. If T is unramified over R and E has a maximal graded subfield M unramified over T and another maximal graded subfield L totally ramified over T with τ(L) = L, then E is semiramified with E_{0} = M_{0} (a field) and Γ_{E} = Γ_{L}, and

SK_{1}(E, τ) ≅ \{a ∈ E_{0} | N_{E_{0}/T_{0}}(a) ∈ R_{0}\} / ∏_{h∈Gal(E_{0}/T_{0})}E_{0}^{∗h\overline{τ}} .  (4.12)

Proof. Let n = ind(E). Since [M_{0} : T_{0}] = n and |Γ_{L} : Γ_{T}| = n, it follows from the Fundamental Equality (2.2) for E/T, M/T, and L/T that [E_{0} : T_{0}] = [Γ_{E} : Γ_{T}] = n, E_{0} = M_{0}, which is is a field, and Γ_{E} = Γ_{L}. Thus E is semiramified, so for the ∂ of (2.6), ∂ = 1. Now, L^{τ} is a graded subfield of L with [L : L^{τ}] = 2. Since L_{0} = T_{0} while (L^{τ})_{0} = (L_{0})^{τ} = R_{0}, L must be unramified over L^{τ}; hence, Γ_{E} = Γ_{L} = Γ_{L^{τ}}. Therefore, one can choose all the x_{γ}’s in Th. 4.7 to lie in L^{τ}. Then each x_{γ}x_{δ}x_{γ+δ}^{−1} ∈ (L^{τ})_{0}^{∗} = R_{0}^{∗} = E_{0}^{∗\overline{τ}}. Hence, X ⊆ P = ∏_{h∈Gal(E_{0}/T_{0})} E_{0}^{∗h\overline{τ}}, so the formula for SK_{1}(E, τ) in Th. 4.7 reduces to (4.12). □

Remark 4.12. In a sequel to this paper [W_{3}], the following will be shown: With the hypotheses of Th. 4.7, suppose E is semiramified with a graded maximal subfield L totally ramified over T such that τ(L) = L, and suppose Gal(E_{0}/T_{0}) is bicyclic, say E_{0} = N ⊗_{T_{0}} N′ with N and N′ each cyclic Galois over T_{0}. Then,

SK_{1}(E, τ) ≅ Br(E_{0}/T_{0}; R_{0})/( Br(N/T_{0}; R_{0}) + Br(N′/T_{0}; R_{0})),

where Br(E_{0}/T_{0}) is the relative Brauer group ker ( Br(T_{0}) → Br(E_{0})) and Br(E_{0}/T_{0}; R_{0}) is the kernel of cor_{E_{0}→R_{0}} : Br(E_{0}/T_{0}) → Br(E_{0}/R_{0}). (Compare this with [Y_{2}, Th. 5.6].) A further formula will be given assuming only that E is semiramified over T with Gal(E_{0}/T_{0}) bicyclic.

In his construction of division algebras D with nontrivial SK_{1}, Platonov worked originally in [P_{2}, §4] with a division algebra D where Z(D) is a Laurent power series field; he gave an exact sequence relating SK_{1}(D) with SK_{1}(\overline{D}) and what he called the “group of projective conorms.” Yanchevskiĭ gave in [Y_{2}, 4.11] an analogous exact sequence in the unitary case. Their results and proofs are valid whenever Z(D) has a henselian discrete (rank 1) valuation. We show here that their results hold more generally whenever Z(D) has a henselian valuation with Γ_{D}/Γ_{Z(D)} cyclic. We work in the equivalent graded setting where the arguments are more transparent.

As before, let E be a graded division algebra finite dimensional over its center T with a unitary graded involution τ, and let R = T^{τ}. Assume that T is unramified over R and that Γ_{E}/Γ_{T} is cyclic group. (This cyclicity holds, e.g., whenever Γ_{T} ≅ ℤ.) It follows that the surjective map Θ_{E} : Γ_{E} → Gal(Z(E_{0})/T_{0}) has kernel Γ_{T}. (For, by [HW_{2}, Prop. 2.1, (2.3), Remark 2.4(i)], ker(Θ_{E})/Γ_{T} has a nondegenerate symplectic pairing, and hence has even rank as a finite abelian group. But here ker(Θ_{E})/Γ_{T} is a cyclic group.) Hence, ∂ = 1 by Lemma 2.2, so E is inertially split. Invoking Lemma 4.6(i), choose any s ∈ E^{∗} with deg(s) + Γ_{T} a generator of Γ_{E}/Γ_{T}, such that τ(s) = s. Let σ = φ_{s} ∈ Aut_{T}(E); so στ is another T/R-graded involution of E, and τσ = σ^{−1}τ (see Remark 4.1(iii)). By the choice of s, σ|_{Z(E_{0})} is a generator of the cyclic group Gal(Z(E_{0})/T_{0}). Note that Gal(Z(E_{0})/R_{0}) = ⟨σ|_{Z(E_{0})}, τ|_{Z(E_{0})}⟩ is a dihedral group. Recall our convention that c^{στ} means σ(τ(c)). Let

𝒮 = {(β, b) ∈ Z(E_{0})^{∗} × E_{0}^{∗} | β^{σ−1} = Nrd_{E_{0}}(b)};
𝒩 = π_{1}(𝒮) (projection into the first component)
= {β ∈ Z(E_{0})^{∗} | β^{σ−1} = Nrd_{E_{0}}(b) for some b ∈ E_{0}^{∗}};
𝒲 = T_{0}^{∗} · Nrd_{E_{0}}(E_{0}^{∗}) ⊆ Z(E_{0})^{∗};
𝒫 = 𝒩/𝒲, which is Platonov’s group of projective conorms for E [P_{2}, §4].
𝒮_{τ} = {(α, a) ∈ Z(E_{0})^{∗} × E_{0}^{∗} | α^{σ−1} = Nrd_{E_{0}}(a)^{1−στ} };
𝒩_{τ} = π_{1}(𝒮_{τ}) (projection into the first component)
= {α ∈ Z(E_{0})^{∗} | α^{σ−1} = Nrd_{E_{0}}(a)^{1−στ} for some a ∈ E_{0}^{∗}};
𝒲_{τ} = T_{0}^{∗} · Nrd_{E_{0}}(Σ_{τ} (E_{0})) ⊆ Z(E_{0})^{∗};
𝒫U_{τ} = 𝒩_{τ} /𝒲_{τ} , which is Yanchevskiĭ’s group of unitary projective conorms for (E, τ) [Y_{2}, 4.11].

Proposition 4.13. If T is unramified over R and Γ_{E}/Γ_{T} is cyclic, then for any generator σ of the cyclic group Gal(Z(E_{0})/T_{0}), we have

(i) SK_{1}(E) ≅ {a ∈ E_{0}^{∗} | N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a)) = 1} / ([E_{0}^{∗}, E_{0}^{∗}] · {c^{σ−1} | c ∈ E_{0}^{∗}}). (ii) SK_{1}(E, τ) ≅ {a ∈ E_{0}^{∗} | N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a)) ∈ R_{0}} / (Σ_{τ} (E_{0}) · Σ_{στ} (E_{0})).

(iii) The following sequence is exact:

SK_{1}(E_{0}, στ) ⟶ SK_{1}(E, τ) ⟶^{f} 𝒫U_{τ} ⟶ 1,  (4.13)

where the map f: SK_{1}(E, τ) → 𝒫U_{τ} is the composition of a Σ_{τ} (E_{0}) · Σ_{στ} (E_{0}) ↦ (α, a) ∈ 𝒮_{τ} and (α, a) ↦ α𝒲_{τ} ∈ 𝒫U_{τ}.

(iv) There is a commutative diagram with exact rows:

\xymatrix{ & SK_{1}(E_{0}, στ) \ar[r] \ar[d]_{\begin{matrix}a \\ ↓ \\ a^{1−στ} \end{matrix}} & SK_{1}(E, τ) \ar[r]^{f} \ar[d]_{\begin{matrix}a \\ ↓ \\ a^{1−στ} \end{matrix}} & 𝒫U_{τ} \ar[r] \ar[d]_{\begin{matrix}α \\ ↓ \\ α \end{matrix}} & 1 \\ 1 \ar[r] & SK_{1}(E_{0}) \ar[r] \ar[d]_{\begin{matrix}b \\ ↓ \\ b \end{matrix}} & SK_{1}(E) \ar[r]^{g} \ar[d]_{\begin{matrix}b \\ ↓ \\ b \end{matrix}} & 𝒫 \ar[r] \ar[d]_{\begin{matrix}β \\ ↓ \\ β^{1+τ} \end{matrix}} & 1 \\  & SK_{1}(E_{0}, στ) \ar[r] & SK_{1}(E, τ) \ar[r] & 𝒫U_{τ} \ar[r] & 1}

where the map g: SK_{1}(E) → 𝒫 is the composition of b [E_{0}^{∗}, E_{0}^{∗}]⟨c^{σ−1}⟩ ↦ (β, b) ∈ 𝒮 and (β, b) ↦ β𝒲 ∈ 𝒫.

(v) If E_{0} is a field, then SK_{1}(E, τ) = 1.

Proof.

(i) This formula was given by Suslin [S_{1}, Prop. 1.7] for a division algebra over a field with a complete discrete valuation. In order to prove it in the graded setting we need two exact sequences which were given in [HaW, Th. 3.4]:

Γ_{E}/Γ_{T} ∧ Γ_{E}/Γ_{T} ⟶ E^{(1)}/[E_{0}^{∗}, E^{∗}] ⟶ SK_{1}(E) ⟶ 1,
1 ⟶ ker Ñ/[E_{0}^{∗}, E^{∗}] ⟶ E^{(1)}/[E_{0}^{∗}, E^{∗}] ⟶ μ_{∂}(T_{0}) ∩ Ñ(E_{0}^{∗}) ⟶ 1,

where E^{(1)} = {a ∈ E^{∗} | Nrd_{E}(a) = 1} ⊆ E_{0} and Ñ = N_{Z(E_{0})/T_{0}} ∘ Nrd_{E_{0}} : E_{0}^{∗} → T_{0}^{∗}. Since ∂ = 1 (see the paragraph prior to the Proposition) and the wedge product of a cyclic group with itself is trivial, these exact sequences yield

SK_{1}(E) ≅ {a ∈ E_{0}^{∗} | N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a)) = 1}/[E_{0}^{∗}, E^{∗}].

We are left to show that [E_{0}^{∗}, E^{∗}] = [E_{0}^{∗}, E_{0}^{∗}]·{c^{σ−1} | c ∈ E_{0}^{∗}}. This follows from the fact that E^{∗}/T^{∗}E_{0}^{∗} ≅ Γ_{E}/Γ_{T} is cyclic together with the following observation, which is easily verified using the standard commutator identities: If G is a group and N is a normal subgroup of G such that G/Z(G)N is a cyclic group generated by, say, xZ(G)N, then [N, G] = [N, N][x, N] where [x, N] = {[x, n] | n ∈ N}. (Here, take G = E^{∗}, N = E_{0}^{∗}, and for x take any s ∈ E_{γ}^{∗} for any γ ∈ Γ_{E} such that Θ_{E}(γ + Γ_{T}) = σ.)

(ii) By Th. 4.7, taking into account that ∂ = 1 and Gal(Z(E_{0})/T_{0}) = ⟨σ⟩, we have,

SK_{1}(E, τ) ≅ (Σ′_{τ} ∩ E_{0})/(Σ_{τ} ∩ E_{0})

= {a ∈ E_{0}^{∗} | N_{Z(E_{0})/T_{0}}Nrd_{E_{0}}(a) ∈ R_{0}} / (Σ_{τ} (E_{0}) · Σ_{στ} (E_{0}) · ⟨x_{γ}x_{δ}x_{γ+δ}^{−1} | γ, δ ∈ Γ_{E}⟩), (4.14)

where for each γ ∈ Γ_{E}, x_{γ} is chosen in E_{γ}^{∗} with x_{γ} = τ(x_{γ}) and x_{γ} ≠ 0, using Lemma 4.6(i). Let L = R[s], where s is chosen in E^{∗} with φ(s)|_{E_{0}} = σ, which is possible as Θ_{E} : Γ_{E} → Gal(Z(E_{0})/T_{0}) is surjective (see (2.3)). Moreover, s can be chosen with τ(s) = s. Since ker(Θ_{E}) = Γ_{T} = Γ_{R}, we have Γ_{E} = ⟨deg(s)⟩ + Γ_{R}. Thus, L is a graded subfield of E with Γ_{L} = Γ_{E} and τ|_{L} = id. For each γ ∈ Γ_{E} we can choose x_{γ} ∈ L_{γ}^{∗}; then for all γ, δ ∈ Γ_{E}, we have x_{γ}x_{δ}x_{γ+δ}^{−1} ∈ L_{0}^{∗} ⊆ Σ_{τ} (E_{0}). Thus, the ⟨x_{γ}x_{δ}x_{γ+δ}^{−1}⟩ term in (4.14) is redundant, yielding the formula in (ii).

(iii) We first check that f is well-defined: Take any a ∈ E_{0}^{∗} with N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a)) ∈ R_{0}^{∗}. Let c = Nrd_{E_{0}}(a). Then, as R_{0} = T_{0}^{τ}, 1 = N_{Z(E_{0})/T_{0}}(c)^{1−τ} = N_{Z(E_{0})/T_{0}}(c)^{1−στ} = N_{Z(E_{0})/T_{0}}(c^{1−στ}). By Hilbert 90, there is α ∈ Z(E_{0})^{∗} with α^{σ−1} = c^{1−στ} = Nrd_{E_{0}}(a)^{1−στ}. Hence (α, a) ∈ 𝒮_{τ}, so α ∈ 𝒩_{τ}, and the choice of α is unique up to T_{0}^{∗} ⊆ 𝒲_{τ}. Thus, the image of a in 𝒫U_{τ} is independent of the choice of α. Suppose further that a = pq for some p ∈ Σ_{τ} (E_{0}), q ∈ Σ_{στ} (E_{0}), say, p = s_{1} … s_{k} with each s_{i} ∈ S_{τ} (E_{0}). Then,

Nrd_{E_{0}}(p)^{τ} = Nrd_{E_{0}}(s_{1})^{τ} … Nrd_{E_{0}}(s_{k})^{τ} = Nrd_{E_{0}}(s_{1}^{τ}) … Nrd_{E_{0}}(s_{k}^{τ})
= Nrd_{E_{0}}(s_{1}) … Nrd_{E_{0}}(s_{k}) = Nrd_{E_{0}}(p);

(4.15)

likewise, Nrd_{E_{0}}(q)^{στ} = Nrd_{E_{0}}(q). So,

α^{σ−1} = Nrd_{E_{0}}(pq)^{1−στ} = Nrd_{E_{0}}(p)^{1−στ} Nrd_{E_{0}}(q)^{1−στ} = Nrd_{E_{0}}(p)^{1−σ}.

Hence, (αNrd_{E_{0}}(p))^{σ−1} = 1, showing that αNrd_{E_{0}}(p) ∈ T_{0}; Thus α ∈ 𝒲_{τ}. This proves that f is welldefined.

For the subjectivity of f, take any α ∈ 𝒩_{τ}. Then, there is a ∈ E_{0}^{∗} with α^{σ−1} = Nrd_{E_{0}}(a)^{1−στ}. So, N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a))^{1−στ} = N_{Z(E_{0})/T_{0}}(α^{σ−1}) = 1, which shows that N_{Z(E_{0})/T_{0}}(Nrd_{E_{0}}(a)) ∈ T_{0}^{στ} = T_{0}^{τ} = R_{0}, and hence a ∈ Σ′_{τ} (E) ∩ E_{0}^{∗}. Since f(aΣ_{τ} (E_{0})Σ_{στ} (E_{0})) = α𝒲_{τ}, f is surjective.

Finally, we determine ker(f): The image of SK_{1}(E_{0}, στ) in SK_{1}(E, τ) is Σ′_{στ} (E_{0})Σ_{τ} (E_{0})/Σ_{στ} (E_{0})Σ_{τ} (E_{0}). An element in this image is represented by some a ∈ Σ′_{στ} (E_{0}). For such an a, Nrd_{E_{0}}(a)^{1−στ} = 1. Then (1, a) ∈ 𝒮_{τ}, so that f maps the image of a to 1 in 𝒫U_{τ}. Conversely, suppose aΣ_{τ} (E_{0})Σ_{στ} (E_{0}) ∈ ker(f). That is, Nrd_{E_{0}}(a)^{1−στ} = α^{σ−1}, where α ∈ 𝒲_{τ}, so α = cNrd_{E_{0}}(d) with c ∈ T_{0}^{∗} and d ∈ Σ_{τ} (E_{0}). So, Nrd_{E_{0}}(d) = Nrd_{E_{0}}(d)^{τ} by the argument of (4.15) above, and hence

Nrd_{E_{0}}(a)^{1−στ} = α^{σ−1} = (cNrd_{E_{0}}(d))^{σ−1} = Nrd_{E_{0}}(d)^{σ−1} = Nrd_{E_{0}}(d)^{στ−1}.

Thus, Nrd_{E_{0}}(ad)^{1−στ} = 1, i.e., ad ∈ Σ′_{στ} (E_{0}). Hence, a = (ad)d^{−1} ∈ Σ′_{στ} (E_{0})Σ_{τ} (E_{0}). This shows that ker(f) coincides with the image of SK_{1}(E_{0}, στ) in SK_{1}(E, τ), completing the proof of exactness of the sequence.

(iv) Exactness of the middle row is proved by an analogous but easier argument to that for (iii). Commutativity of the left rectangles of the diagram is evident. Commutativity of the top right rectangle is clear from the definitions. Commutativity of the bottom right rectangle is easy to check using the identity

(1 − στ) ∘ (σ − 1) = (σ − 1) ∘ (1 + τ),  (4.16)

which follows from (στ)^{2} = id. Note that for each column of the diagram, the composition of the two maps is the squaring map.

(v) For this part, the proof follows closely Yanchevskiĭ’s proof in [Y_{2}, 4.13]. (But our notational convention for products of functions is fg = f ∘ g, whereas his appears to be fg = g ∘ f.) Suppose E_{0} is a field. For simplicity we denote \overline{τ} = τ|_{E_{0}} by τ. Take a ∈ Σ′_{τ} (E)∩E_{0}. So, N_{E_{0}/T_{0}}(a^{1−τ}) = 1. We will show that a ∈ E_{0}^{τ}E_{0}^{στ}. It then follows by (ii) above that SK_{1}(E, τ) = 1. But since E_{0} is cyclic over T_{0}, by Hilbert 90 there is a b ∈ E_{0}^{∗} such that a^{τ−1} = b^{σ−1} where ⟨σ⟩ = Gal(E_{0}/T_{0}). So, 1 = a^{(τ+1)(τ−1)} = b^{(τ+1)(σ−1)}. Analogously to (4.16), we have (τ + 1)(σ − 1) = (σ − 1)(1 − τσ). So b^{(σ−1)(1−τσ)} = 1. Setting c = b^{(1−τσ)}, we have c^{σ−1} = 1, so c ∈ T_{0}. But, N_{T_{0}/R_{0}}(c) = c^{1+τσ} = b^{(1+τσ)(1−τσ)} = 1. By Hilbert 90 we have c = d^{τσ−1} for some d ∈ T_{0}^{∗}. Let t = bd ∈ E_{0}^{∗}. Then, t^{1−τσ} = b^{(1−τσ)}d^{(1−τσ)} = d^{(τσ−1)}d^{(1−τσ)} = 1, i.e., t ∈ E_{0}^{τσ}. So, σ(t) = τ(t) ∈ E_{0}^{στ}. Thus, a^{τ−1} = b^{σ−1} = (t/d)^{σ−1} = t^{σ−1} = t^{τ−1}, as d ∈ T_{0}. This shows that (aτ(t))^{τ−1} = 1, i.e., aτ(t) ∈ E^{τ}; hence a = (aτ(t))τ(t)^{−1} ∈ E_{0}^{τ}E_{0}^{στ}. □

5. Totally ramified algebras

For a graded division algebra E totally ramified over its center T with a unitary graded involution τ, two possible cases can arise: either T is totally ramified over R = T^{τ}, or T is unramified over R. In the first case, we showed in Prop. 4.4 that SK_{1}(E, τ) is trivial. We now obtain an easily computable explicit formula for SK_{1}(E, τ) in the second case. For a field K and for n ∈ ℕ, we write μ_{n} for the group of all n-th roots of unity in an algebraic closure of K. Then set μ_{n}(K) = μ_{n} ∩ K^{∗}.

Theorem 5.1. If E is totally ramified over T of index n and T is unramified over R, then

SK_{1}(E, τ) ≅ {a ∈ T_{0}^{∗} | a^{n} ∈ R_{0}^{∗}} / {a ∈ T_{0}^{∗} | a^{e} ∈ R_{0}^{∗}}  (5.1)
≅ {ω ∈ μ_{n}(T_{0}) | τ(ω) = ω^{−1}}/μ_{e},  (5.2)

where e is the exponent of Γ_{E}/Γ_{T}. In particular,

(i) The restriction of the map K_{1}(E, τ) → K_{1}(E) given by aΣ_{τ} ↦ a^{1−τ} E′, induces an injective map

α: SK_{1}(E, τ) ⟶ SK_{1}(E) ≅ μ_{n}(T_{0})/μ_{e}.

(ii) If the exponent e of E is odd, then α is an isomorphism. (iii) If e > 2 then T_{0} = R_{0}(μ_{e}), and τ acts on μ_{e} by ω ↦ ω^{−1}.

Proof. Since T is unramified over R and E_{0} = T_{0}, the formulas of Th. 4.7 for SK_{1}(E, τ) reduce to ∂ = n and

SK_{1}(E, τ) ≅ {a ∈ T_{0}^{∗} | a^{n} ∈ R_{0}^{∗}} / (R_{0}^{∗} ⟨x_{γ}x_{δ}x_{γ+δ}^{−1} | γ, δ ∈ Γ_{E}⟩),  (5.3)

where each x_{γ} ∈ E_{γ}^{∗} with τ(x_{γ}) = x_{γ}. Recall that as E/T is totally ramified, the canonical pairing E^{∗} × E^{∗} → μ_{e}(T_{0}) given by (s, t) ↦ [s, t] is surjective ([HW_{2}, Prop. 2.1]), and μ_{e}(T_{0}) = μ_{e}, i.e., T_{0} contains all e-th roots of unity. Since each E_{γ} = T_{0}x_{γ} with T_{0} central, it follows that {[x_{δ}, x_{γ}] | γ, δ ∈ Γ_{E}} = μ_{e}. Now consider c = x_{γ}x_{δ}x_{γ+δ}^{−1} for any γ, δ ∈ Γ_{E}. Then, τ(c) = x_{γ+δ}^{−1}x_{δ}x_{γ}. Note that x_{δ}x_{γ} and x_{γ+δ} each lie in E_{γ+δ} = T_{0}x_{γ+δ}, so they commute. Hence,

τ(c)c^{−1} = x_{γ+δ}^{−1}(x_{δ}x_{γ})x_{γ+δ}x_{δ}^{−1}x_{γ}^{−1} = [x_{δ}, x_{γ}].  (5.4)

Since [x_{δ}, x_{γ}] ∈ μ_{e}, this shows that c ∈ {a ∈ T_{0}^{∗} | a^{e} ∈ R_{0}^{∗}}. For the reverse inclusion, take any d in T_{0}^{∗} such that d^{e} ∈ R_{0}^{∗}. So τ(d)d^{−1} ∈ μ_{e}. Thus, τ(d)d^{−1} = [x_{δ}, x_{γ}], for some γ, δ ∈ Γ_{E}. Taking c = x_{γ}x_{δ}x_{γ+δ}^{−1}, we have τ(d)d^{−1} = τ(c)c^{−1} by (5.4), which implies that dc^{−1} is τ-stable, so lies in R_{0}^{∗}; thus, d ∈ R_{0}^{∗} ⟨x_{γ}x_{δ}x_{γ+δ}^{−1} | γ, δ ∈ Γ_{E}⟩. Therefore, R_{0}^{∗} ⟨x_{γ}x_{δ}x_{γ+δ}^{−1} | γ, δ ∈ Γ_{E}⟩ = {a ∈ T_{0}^{∗} | a^{e} ∈ R_{0}^{∗}}. Inserting this in (5.3) we obtain (5.1).

(i) Consider the well-defined map α: SK_{1}(E, τ) → SK_{1}(E) given by aΣ_{τ} ↦ a^{1−τ} E′ (see diagram (3.9) for the non-graded version). By [HaW, Cor. 3.6(ii)], SK_{1}(E) ≅ μ_{n}(T_{0})/μ_{e}. Taking into account formula (5.1) for SK_{1}(E, τ), it is easy to see that α is injective.

We now verify that

im(α) = {ω ∈ μ_{n}(T_{0}) | τ(ω) = ω^{−1}} /μ_{e},  (5.5)

and thus obtain (5.2). Indeed, since μ_{e} = {[x_{δ}, x_{γ}] | γ, δ ∈ Γ_{E}}, by setting c = x_{γ}x_{δ}x_{γ+δ}^{−1} we have [x_{δ}, x_{γ}] = τ(c)c^{−1} by (5.4). This shows that μ_{e} ⊆ {ω ∈ μ_{n}(T_{0}) | τ(ω) = ω^{−1}}. Now for any ω ∈ μ_{n}(T_{0}) with τ(ω) = ω^{−1}, we have N_{T_{0}/R_{0}}(ω) = 1, so Hilbert 90 guarantees that ω = c^{1−τ} for some c ∈ T_{0}^{∗}. Then, (c^{n})^{1−τ} = ω^{n} = 1, so c^{n} ∈ R_{0}^{∗}. Thus, c ∈ Σ′_{τ}, and clearly α(cΣ_{τ}) = ωμ_{e}. This shows ⊇ in (5.5); the reverse inclusion is clear from the definition of α.

(ii) Suppose e is odd. Let m = |μ_{n}(T_{0})|. So, μ_{n}(T_{0}) = μ_{m}, with m | n. Also, e | m, as μ_{e} ⊆ T_{0}. Since e and n have the same prime factors, this is also true for e and m. Recall that Aut(μ_{m}) ≅ (ℤ/mℤ)^{∗}, the multiplicative group of units of the ring ℤ/mℤ; so, | Aut(μ_{m})| = φ(m), where φ is Euler’s φ-function. Since e | m and e and m have the same prime factors (all odd), the canonical map ψ: Aut(μ_{m}) → Aut(μ_{e}) given by restriction is surjective with kernel of order φ(m)/φ(e) = m/e, which is odd. Therefore, ψ induces an isomorphism on the 2-torsion subgroups, _{2}Aut(μ_{m}) ≅ _{2}Aut(μ_{e}). Now, τ|_{μ_{m}} ∈ _{2}Aut(μ_{m}) and we saw for (i) that τ|_{μ_{e}} is the inverse map ω ↦ ω^{−1}. The inverse map on μ_{m} also lies in _{2}Aut(μ_{m}) and has the same restriction to μ_{e} as τ. Hence, τ|_{μ_{m}} must be the inverse map. That is, {ω ∈ μ_{n}(T_{0}) | τ(ω) = ω^{−1}} = μ_{n}(T_{0}). Therefore, (5.5) above shows that im(α) = μ_{n}(T_{0})/μ_{e}, which we noted above is isomorphic to SK_{1}(E).

(iii) We saw in the proof of part (i) that τ acts on μ_{e} by the inverse map. So, if e > 2, then μ_{e} ⊈ R_{0}. Since [T_{0} : R_{0}] = 2, it then follows that T_{0} = R_{0}(μ_{e}). □

Remark 5.2. The isomorphism SK_{1}(E, τ) ≅ SK_{1}(E) of part (ii) of the above theorem can be obtained under the milder condition that E_{0} = T_{0}E′ provided that the exponent of E is a prime power. The proof is similar.

Example 5.3. Let r_{1}, …, r_{m} be integers with each r_{i} ≥ 2. Let e = lcm(r_{1}, …, r_{m}), and let n = r_{1} … r_{m}. Let C be any field such that μ_{e} ⊆ C and C has an automorphism θ of order 2 such that θ(ω) = ω^{−1} for all ω ∈ μ_{e}. Let R be the fixed field C^{θ}. Let x_{1}, …, x_{2m} be 2m independent indeterminates, and let K be the iterated Laurent power series field C((x_{1})) … ((x_{2m})). This K is equipped with its standard valuation v: K^{∗} → ℤ^{2m} where ℤ^{2m} is given the right-to-left lexicographical ordering. With this valuation K is henselian (see [W_{2}, p. 397]). Consider the tensor product of symbol algebras

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

where for 1 ≤ i ≤ m, ω_{i} is a primitive r_{i}-th root of unity in C. Using the valuation theory developed for division algebras, it is known that D is a division algebra, the valuation v extends to D, and D is totally ramified over K (see [W_{2}, Ex. 4.4(ii)] and [TW, Ex. 3.6]) with

Γ_{D}/Γ_{K} ≅ ∏_{i=1}^{m}(ℤ/r_{i}ℤ) × (ℤ/r_{i}ℤ),

and \overline{D} = \overline{K} ≅ C. Extend θ to an automorphism θ′ of order 2 on K in the obvious way, i.e., acting by θ on the coefficients of a Laurent series, and with θ′(x_{i}) = x_{i} for 1 ≤ i ≤ 2m. On each of the symbol algebras (\frac{x_{2i−1},x_{2i}}{K} )_{ω_{i}} with its generators i_{i} and j_{i} such that i_{i}^{r_{i}} = x_{2i−1}, j_{i}^{r_{i}} = x_{2i}, and

i_{i}j_{i} = ω_{i}j_{i}i_{i}, define an involution τ_{i} as follows: τ_{i}(c i_{i}^{k}j_{i}^{l}) = θ′(c) j_{i}^{l}i_{i}^{k}, where c ∈ K and 0 ≤ l, k < r_{i}. Clearly K^{τ_{i}} = K^{θ′} = R((x_{1})) … ((x_{2m})), and therefore τ_{i} is a unitary involution. Since the τ_{i} agree on K for 1 ≤ i ≤ m, they yield a unitary involution τ = ⊗_{i=1}^{m}τ_{i} on D. Now by Th. 3.5, SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃). Since D is totally ramified over K, which is unramified over K^{τ}, we have correspondingly that gr(D) is totally ramified over gr(K), which is unramified over gr(K)τ̃. Also, gr(K)_{0} ≅ \overline{K} ≅ C. We have exp(gr(D)) = exp(D) = exp(Γ_{D}/Γ_{K}) = lcm(r_{1}, …, r_{m}) = e and ind(gr(D)) = ind(D) = r_{1} … r_{m} = n. By Th. 5.1,

SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃) ≅ {ω ∈ μ_{n}(C) | θ(ω) = ω^{−1}}/μ_{e},

while by [HaW, Th. 4.8, Cor. 3.6(ii)],

SK_{1}(D) ≅ SK_{1}(gr(D)) ≅ μ_{n}(C)/μ_{e}.

Here are some more specific examples:

(i) Let C = ℂ, the complex numbers, and let θ be complex conjugation, which maps every root of unity to its inverse. So, R = C^{θ} = ℝ. Then, SK_{1}(D, τ) ≅ SK_{1}(D) ≅ μ_{n}/μ_{e} ≅ ℤ/(n/e)ℤ.

(ii) Let r_{1} = r_{2} = 4, so e = 4 and n = 16. Let ω_{16} be a primitive sixteenth root of unity in ℂ, and let C = ℚ(ω_{16}), the sixteenth cyclotomic extension of ℚ. Recall that Gal(C/ℚ) ≅ Aut(μ_{16}) ≅ (ℤ/4ℤ) × (ℤ/2ℤ), Let θ: C → C be the automorphism which maps ω_{16} ↦ (ω_{16})^{7}. Then, θ^{2} = id_{C}, as 7^{2} ≡ 1 (mod 16), and {ω ∈ μ_{16} | θ(ω) = ω^{−1}} = μ_{8}. Thus, SK_{1}(D, τ) ≅ μ_{8}/μ_{4} ≅ ℤ/2ℤ, while SK_{1}(D) ≅ μ_{16}/μ_{4} ≅ ℤ/4ℤ. So, here the injection SK_{1}(D, τ) → SK_{1}(D) is not surjective.

(iii) Let r_{1} = … = r_{m} = 2, so e = 2 and n = 2^{m}. Here, C could be any quadratic extension of any field R with char(R) ≠ 2. Take θ to be the unique nonidentity R-automorphism of C. The resulting D is a tensor product of m quaternion algebras over C((x_{1})) … ((x_{2m})), and SK_{1}(D, τ) ≅ {ω ∈ μ_{2^{m}}(C) | θ(ω) = ω^{−1}}/μ_{2}, while SK_{1}(D) ≅ μ_{2^{m}}(C)/μ_{2}.

Ex. 5.3 gives an indication how to use the graded approach to recover results in the literature on the unitary SK_{1} in a unified manner and to extend them from division algebras with discrete valued groups to arbitrary valued groups. While SK_{1}(D) has long been known for the D of Ex. 5.3, the formula for SK_{1}(D, τ) is new.

Here is a more complete statement of what the results in the preceding sections yield for SK_{1}(D, τ) for valued division algebras D.

Theorem 5.4. Let (D, v) be a tame valued division algebra over a field K with v|_{K} henselian, with a unitary involution τ; let F = K^{τ}, and suppose v|_{F} is henselian and that K is tamely ramified over F. Let \overline{τ} be the involution on \overline{D} induced by τ. Then,

(1) Suppose K is unramified over F.

(i) If D is unramified over K, then SK_{1}(D, τ) ≅ SK_{1}(\overline{D}, \overline{τ}).

(ii) If D is totally ramified over K, let e = exp(D) and n = ind(D); then,

SK_{1}(D, τ) ≅ \{ω ∈ μ_{n}(\overline{K}) | τ(ω) = ω^{−1}\}/μ_{e},

while SK_{1}(D) ≅ μ_{n}(\overline{K})/μ_{e}. (iii) If D has a maximal graded subfield M unramified over K and another maximal graded subfield

L totally ramified over K, with τ(L) = L, then D is semiramified and

SK_{1}(D, τ) = \{a ∈ \overline{D}^{∗} | N_{\overline{D}/\overline{K}}(a) ∈ \overline{F} \} / ∏_{h∈Gal(\overline{D}/\overline{K})}\overline{F}^{∗h\overline{τ}} .

(iv) Suppose Γ_{D}/Γ_{K} is cyclic. Let σ be a generator of Gal(Z(\overline{D})/\overline{K}). Then,

SK_{1}(D, τ) ≅ \{a ∈ \overline{D}^{∗} | N_{Z(\overline{D})/\overline{K}}(Nrd\overline{D}(a)) ∈ \overline{F} \} / (Σ\overline{τ} (\overline{D}) · Σ_{σ\overline{τ}} (\overline{D})).

(v) If D is inertially split, \overline{D} is a field and Gal(\overline{D}/\overline{K}) is cyclic, then SK_{1}(D, τ) = 1.

(2) If K is totally ramified over F, then SK_{1}(D, τ) = 1.

Proof. Let gr(D) be the associated graded division algebra of D. The tameness assumptions assure that gr(K) is the center of gr(D) with [gr(D) : gr(K)] = [D : K] and that the graded involution τ̃ on gr(D) induced by τ is unitary with gr(K)τ̃ = gr(K^{τ}). In each case of Th. 5.4, the conditions on D yield analogous conditions on gr(D). Since by Th. 3.5, SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃), (2) and (1)(v) follow immediately from Prop. 4.5 and Prop. 4.13(v), respectively. Part (1)(i), also follows from Th. 3.5, and Cor. 4.10 as follows:

SK_{1}(D, τ) ≅ SK_{1}(gr(D), τ̃) ≅ SK_{1}(gr(D)_{0}, τ|_{gr(D)_{0}}) = SK_{1}(\overline{D}, \overline{τ}).

Parts (1)(ii), (1)(iii), and (1)(iv) follow similarly using Th. 5.1, Cor. 4.11, and Prop. 4.13(ii) respectively. □

In the special case that the henselian valuation on K is discrete (rank 1), Th. 5.4 (1)(i), (iii), (iv), (v) and (2) were obtained by Yanchevskiĭ [Y_{2}]. In this discrete case, the assumption that v on K is henselian already implies that v|_{F} is henselian (see Remark 3.1).

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Department of Pure Mathematics, Queen’s University, Belfast BT7 1NN, United Kingdom

   E-mail address: r.hazrat@qub.ac.uk

Department of Mathematics, University of California at San Diego, La Jolla, California 92093-0112, U.S.A.

   E-mail address: arwadsworth@ucsd.edu