Multiples of trace forms and algebras with involution

Eva Bayer–Fluckiger

Introduction

Let k be a field of characteristic ≠ 2, and let L be a Galois extension of k with group G. Let

q_{L} : L × L → k
q_{L}(x, y) = Tr_{L/k}(xy)

be the trace form. It is well–known that L has a normal basis over k, in other words there exists x ∈ L such that {gx}_{g∈G} is a basis of L as a k-vector space. Such a basis is called a self–dual normal basis if q_{L}(gx, hx) = δ_{g,h} for all g, h ∈ G.

The following question was studied in [1], [2], [3], [5], [7] :

Question. Which Galois extensions have a self–dual normal basis ?

This question is settled in some cases, for instance when G has odd order [3], when the 2–Sylow subgroups of G are elementary abelian [7], but it is open in general.

Note that the existence of a normal basis is equivalent with the fact that L is a free k[G]–module of rank one. A similar reformulation can be given for the self–dual normal basis question. Indeed, remark that the quadratic form q_{L} is invariant by G, that is q_{L}(gx, gy) = q_{L}(x, y) for all x, y ∈ L and for all g ∈ G. In other words, q_{L} is a G–quadratic form (cf. 1.3). Let us define the unit G–quadratic form as being q_{0} : k[G] × k[G] → k characterized by q_{0}(g, h) = δ_{g,h}. Then the existence of a self–dual normal basis is equivalent with the isomorphism of q_{L} and q_{0} as G–quadratic forms.

It is more natural to work in the category of G–Galois algebras instead of Galois extensions with group G. Let us denote by L_{0} the split G–Galois algebra; then q_{L_{0}} ≃_{G} q_{0}. This leads us to the following question :

Question. Let L and L′ be two G–Galois algebras. When are the G–forms q_{L} and q_{L′} isomorphic ?

The results of [3] and [7] apply to this more general situation. However, a complete answer to the question seems out of reach at this point. For this reason, a weaker question was raised in [1]. Indeed, if ϕ is a non–degenerate quadratic form and q is a G–quadratic form, then the tensor product ϕ ⊗ q is a G–quadratic form. One can ask the following

Question. Let L and L′ be two G–Galois algebras, and let ϕ be a non–degenerate quadratic form. When are the G–forms ϕ ⊗ q_{L} and ϕ ⊗ q_{L′} isomorphic ?

If ϕ is an odd dimensional form, then this question is equivalent with the previous one. Let W (k) be the Witt ring of k, and let I be the ideal of W (k) consisting of the Witt classes of the even dimensional forms. Let k_{s} be a separable closure of k, and set Γ_{k} = Gal(k_{s}/k). For every positive integer n, let us denote by e_{n} : I^{n}/I^{n+1} → H^{1}(Γ_{k}, Z/2Z) the Milnor– Voevodsky isomorphism (see 1.2). Let cd_{2}(Γ_{k}) be the 2–cohomological dimension of Γ_{k} (cf. 1.1), and let d ≥ 0 be an integer. The following two statements are easy consequences of the above isomorphisms (cf. 1.2.2, 1.2.5) :

I. Suppose that cd_{2}(Γ_{k}) ≤ d. Let q and q′ be two quadratic forms with dim(q) = dim(q′), and let ϕ ∈ I^{d}. Then

ϕ ⊗ q ≃ ϕ ⊗ q′.

II. Suppose that cd_{2}(Γ_{k}) ≤ d. Let q and q′ be two quadratic forms with dim(q) = dim(q′), and let ϕ ∈ I^{d−1}. Then

ϕ ⊗ q ≃ ϕ ⊗ q′ if and only if e_{d−1}(ϕ) ∪ (disc(q)) = e_{d−1}(ϕ) ∪ (disc(q′)) in H^{d}(k).

It is natural to look for similar statements concerning trace forms of G–Galois algebras, as proposed in [1]. As an analog of I, we have the following :

Theorem. (Chabloz, [9]) Suppose that cd_{2}(Γ_{k}) ≤ d. Let L and L′ be two G–Galois algebras, and let ϕ ∈ I^{d}. Then ϕ ⊗ q_{L} ≃_{G} ϕ ⊗ q_{L′}.

In order to go further, we need some invariants defined in [7]. Let f_{L} : Γ_{k} → G be a continuous homomorphism corresponding to the G–Galois algebra L. The homomorphism f_{L} induces f_{L}^{∗} : H^{1}(G, Z/2Z) → H^{1}(Γ_{k}, Z/2Z). For all x ∈ H^{1}(G, Z/2Z), set x_{L} = f_{L}^{∗}(x). Then x_{L} is an invariant of the G–quadratic form q_{L} (cf. [7], 2.2.3).

The following statement is inspired by II, and is proved in §3 :

Theorem. Suppose that cd_{2}(Γ_{k}) ≤ d. Let L and L′ be two G–Galois algebras, and let ϕ ∈ I^{d−1}. Then the G–quadratic forms ϕ ⊗ q_{L} and ϕ ⊗ q_{L′} are isomorphic if and only if e_{d−1}(ϕ) ∪ x_{L} = e_{d−1}(ϕ) ∪ x_{L′} for all x ∈ H^{1}(G, Z/2Z).

This was conjectured in [1], and proved in special cases by Chabloz, Monsurro, Morales, Parimala and Schoof (see [4], [5], [6], [9] and [12]).

The proof uses results concerning hermitian forms over algebras with involution (see 1.5 and §2 for details). Let (D, σ) be a division algebra with involution over k, and let W (D, σ) be the Witt group of hermitian forms over (D, σ). Then W (D, σ) is a W (k)– module. Let us denote by J the W (k)–submodule of W (D, σ) consisting of even dimensional hermitian forms over D.

Theorem. Suppose that cd_{2}(Γ_{k}) ≤ d. Then (a) We have I^{d}J = 0.

(b) If σ is of the second kind, then I^{d−1}J = 0.

(c) If σ is of the first kind and of the symplectic type, then I^{d−2}J = 0.

Part (a) is due to Chabloz [9]. Parts (b) and (c) are proved in §2, and are used in the proof of the main result of this paper in §3. In order to deal with involutions of the first kind and of the orthogonal type, we need the following :

Theorem. Suppose that cd_{2}(Γ_{k}) ≤ d. Let (D, σ) be a quaternion algebra with an orthogonal involution, and let (V, h) and (V′, h′) be two hermitian forms over (D, σ) with dim_{D}(V) = dim_{D}(V′). Let ϕ ∈ I^{d−1}. Then ϕ ⊗ h ≃ ϕ ⊗ h′ if and only if

e_{d−1}(ϕ) ∪ (disc(h)) = e_{d−1}(ϕ) ∪ (disc(h′)).

This follows from results of Parimala, Sridharan and Suresh [14] and of Berhuy [8].

§1. Definitions, notation and basic facts

1.1. Galois cohomology

Let k_{s} be a separable closure of k, and set Γ_{k} = Gal(k_{s}/k). For any discrete Γ_{k}– module C, set H^{i}(k, C) = H^{i}(Γ_{k}, C). We say that the 2–cohomological dimension of Γ_{k} is at most d, denoted by cd_{2}(Γ_{k}) ≤ d, if H^{i}(k, C) = 0 for all i > d and for every finite 2–primary Γ_{k}–module C.

Set H^{i}(k) = H^{i}(k, Z/2Z), and recall that H^{1}(k) ≃ k^{∗}/k^{∗2}. For all a ∈ k^{∗}, let us denote by (a) ∈ H^{1}(k) the corresponding cohomology class. We use the additive notation for H^{1}(k). If a_{1}, …, a_{n} ∈ k^{∗}, we denote by (a_{1}) ∪ … ∪ (a_{n}) ∈ H^{n}(k) their cup product.

If U is a linear algebraic group defined over k, let H^{1}(k, U) be the pointed set H^{1}(Γ_{k}, U(k_{s})) (cf. [16], [17] Chap. 10).

1.2. Quadratic forms

All quadratic forms are supposed to be non–degenerate. We denote by W (k) the Witt ring of k, and by I = I(k) the fundamental ideal of W (k). For all a_{1}, …, a_{n} ∈ k^{∗}, let us denote by << a_{1}, …, a_{n} >>=< 1, −a_{1} > ⊗ … ⊗ < 1, −a_{n} > the associated n–fold Pfister form. It is well–known that I^{n} is generated by the n-fold Pfister forms. The following has been conjectured by Milnor, and proved by Voevodsky (see also Orlov–Vishik–Voevodsky [13], [18], [19], and the survey paper [10]) :

Theorem 1.2.1. (Voevodsky) : For every positive integer n, there exists an isomorphism

e_{n} : I^{n}/I^{n+1} → H^{n}(k)

such that

e_{n}(<< a_{1}, …, a_{n} >>) = (a_{1}) ∪ … ∪ (a_{n})

for all a_{1}, …, a_{n} ∈ k^{∗}.

It is easy to see that the above theorem has the following consequences : Corollary 1.2.2. Suppose that cd_{2}(Γ_{k}) ≤ d. Let q and q′ be two quadratic forms with dim(q) = dim(q′), and let ϕ ∈ I^{d}. Then

ϕ ⊗ q ≃ ϕ ⊗ q′.

Proof. Note that by 1.2.1, cd_{2}(Γ_{k}) ≤ d implies that I^{d+1} = 0. As dim(q) = dim(q′), we have q ⊕ (−q′) ∈ I. Therefore ϕ ⊗ (q ⊕ (−q′)) ∈ I^{d+1} = 0. This implies that ϕ ⊗ q ≃ ϕ ⊗ q′.

For every quadratic form q, let us denote by disc(q) ∈ H^{1}(k) its discriminant. Recall that if n = dim(q), then disc(q) = (−1)^{\frac{n(n−1)}{2}} det(q). We need the following proposition : Proposition 1.2.3. Let q and q′ be two quadratic forms with dim(q) = dim(q′), and let ϕ ∈ I^{d−1}. Then

e_{d}(ϕ ⊗ (q ⊕ (−q′)) = e_{d−1}(ϕ) ∪ (disc(q)) + e_{d−1}(ϕ) ∪ (disc(q′)).

Proof. Set Q = q ⊕ (−q′), and let m = dim(q) = dim(q′), n = 2m = dim(Q). Note that (−1)^{\frac{n(n−1)}{2}} = (−1)^{m}, hence (disc(Q)) = (disc(q)) + (disc(q′)). We have Q ∈ I, and e_{1}(Q) = (disc(Q)) = (disc(q)) + (disc(q′)). Therefore

e_{d}(ϕ ⊗ Q) = e_{d−1}(ϕ) ∪ e_{1}(Q) = e_{d−1}(ϕ) ∪ (disc(q)) + e_{d−1}(ϕ) ∪ (disc(q′)), hence the proposition is proved. Corollary 1.2.4. Let q and q′ be two quadratic forms with dim(q) = dim(q′), and let ϕ ∈ I^{d−1}. If ϕ ⊗ q ≃ ϕ ⊗ q′, then e_{d−1}(ϕ) ∪ (disc(q)) = e_{d−1}(ϕ) ∪ (disc(q′)) ∈ H^{d}(k). Proof. As ϕ ⊗ q ≃ ϕ ⊗ q′, the quadratic form ϕ(⊗(q ⊕ (−q′)) is hyperbolic. Hence e_{d}(ϕ ⊗ (q ⊕ (−q′)) = 0. By prop. 1.2.3, we have

e_{d}(ϕ ⊗ (q ⊕ (−q′)) = e_{d−1}(ϕ) ∪ (disc(q)) + e_{d−1}(ϕ) ∪ (disc(q′)),

therefore e_{d−1}(ϕ) ∪ (disc(q)) = e_{d−1}(ϕ) ∪ (disc(q′)), as claimed. Corollary 1.2.5. Suppose that cd_{2}(Γ_{k}) ≤ d. Let q and q′ be two quadratic forms with dim(q) = dim(q′), and let ϕ ∈ I^{d−1}. Then

ϕ ⊗ q ≃ ϕ ⊗ q′ if and only if e_{d−1}(ϕ) ∪ (disc(q)) = e_{d−1}(ϕ) ∪ (disc(q′)) ∈ H^{d}(k).

Proof. Set Q = q ⊕ (−q′). By prop. 1.2.3, we have

e_{d}(ϕ ⊗ Q) = e_{d−1}(ϕ) ∪ (disc(q)) + e_{d−1}(ϕ) ∪ (disc(q′)).

Hence e_{d}(ϕ ⊗ Q) = 0 is equivalent with e_{d−1}(ϕ) ∪ (disc(q)) = e_{d−1}(ϕ) ∪ (disc(q′)). But by 1.2.1, e_{d}(ϕ ⊗ Q) = 0 is equivalent with ϕ ⊗ Q hyperbolic, hence with ϕ ⊗ q ≃ ϕ ⊗ q′.

1.3. G–quadratic forms

Let G be a finite group, and let us denote by k[G] the associated group ring. A G–quadratic form is a pair (M, q), where M is a k[G]–module that is a finite dimensional k–vector space, and q : M × M → k is a non–degenerate symmetric bilinear form such that

q(gx, gy) = q(x, y)

for all x, y ∈ M and all g ∈ G. We say that two G–quadratic forms (M, q) and (M′, q′) are isomorphic if there exists an isomorphism of k[G]–modules f : M → M′ such that q(f(x), f(y)) = q′(x, y) for all x, y ∈ M. If this is the case, we write (M, q) ≃_{G} (M′, q′), or q ≃_{G} q′.

If ϕ is a quadratic form over k, and q a G–quadratic form, then the tensor product ϕ ⊗ q is a G–quadratic form.

1.4. Trace forms

Let L be a G–Galois algebra, and let

q_{L} : L × L → k, q_{L}(x, y) = Tr_{L/k}(xy),

be its trace form. Then q_{L} is a G–quadratic form.

Let : k[G] → k[G] be the canonical involution of the group ring k[G], in other words the k–linear involution of k[G] characterized by \overline{g} = g^{−1} for all g ∈ G. Let U_{G} be the linear algebraic group defined over k such that for every commutative k–algebra A, we have U_{G}(A) = \{x ∈ A[G] | x\overline{x} = 1\}. Recall that we denote by H^{1}(k, U_{G}) the pointed set H^{1}(Γ_{k}, U(k_{s})).

Let f_{L} : Γ_{k} → G be a continuous homomorphism corresponding to L. The composition of f_{L} with the inclusion of G in U_{G}(k_{s}) is a 1–cocycle Γ_{k} → U_{G}(k_{s}). Let us denote by u(L) its class in the cohomology set H^{1}(k, U_{G}). The following is proved in [7], prop. 1.5.1

Proposition 1.4.1 Let L and L′ be two G–Galois algebras. Then the G–quadratic forms q_{L} and q_{L′} are isomorphic if and only if u(L) = u(L′) ∈ H^{1}(k, U_{G}).

The trace form of a G–Galois algebra, considered as a G–form, determines the trace forms of all of its subalgebras of fixed points (se [7], 1.4). We have a similar result for multiples of trace forms, as follows :

Proposition 1.4.2 Let L and L′ be two G–Galois algebras. Let H be a subgroup of G, and set E = L^{H}, E′ = L^{′H}. Let ϕ be a quadratic form over k. Suppose that ϕ ⊗ q_{L} ≃_{G} ϕ ⊗ q_{L′}. Then we have

(a) The quadratic forms ϕ ⊗ q_{E} and ϕ ⊗ q_{E′} are isomorphic.

(b) If moreover H is a normal subgroup of G, then the (G/H)–quadratic forms ϕ ⊗ q_{E} and ϕ ⊗ q_{E′} are isomorphic.

Proof. The proof of this statement is similar to the proof of [7], 1.5.1.

The homomorphism f_{L} : Γ_{k} → G induces f_{L}^{∗} : H^{1}(G, Z/2Z) → H^{1}(k). For any x ∈ H^{1}(G, Z/2Z), set x_{L} = f_{L}^{∗}(x). Then the elements x_{L} are invariants of the G–quadratic form q_{L} (cf. [7], 2.2.3). Let x ∈ H^{1}(G, Z/2Z), and let χ : G → Z/2Z the corresponding homomorphism. Let H be the kernel of χ, and let E_{χ} = L^{H} be the invariant subalgebra; it is a quadratic subalgebra of L. The discriminant of the quadratic algebra E_{χ} is equal to x_{L}.

The following is a generalization of [7], 2.2.3 :

Proposition 1.4.3 Let L and L′ be two G–Galois algebras. Let ϕ ∈ I^{d−1}. Suppose that ϕ ⊗ q_{L} ≃_{G} ϕ ⊗ q_{L′}. Then e_{d−1}(ϕ) ∪ x_{L} = e_{d−1}(ϕ) ∪ x_{L′} for all x ∈ H^{1}(G, Z/2Z).

Proof. Let x ∈ H^{1}(G, Z/2Z), and let χ : G → Z/2Z the corresponding homomorphism. Let H be the kernel of χ, and set E_{χ} = L^{H}, E′_{χ} = L^{′H}. By prop. 1.4.2, the quadratic forms ϕ ⊗ q_{E} and ϕ ⊗ q_{E′} are isomorphic. Using 1.2.4, we have e_{d−1}(ϕ) ∪ disc(q_{E}) = e_{d−1}(ϕ)∪disc(q_{E′}). As the discriminant of a quadratic algebra is equal to the discriminant of its trace form, we obtain e_{d−1}(ϕ) ∪ x_{L} = e_{d−1}(ϕ) ∪ x_{L′}, so the proposition is proved.

1.5. Hermitian forms over division algebras with involution

Let D be a division algebra over k. An involution of D is a k–linear anti–automorphism σ : D → D of order 2. Let K be the center of D. We say that (D, σ) is a division algebra with involution over k if the fixed field of σ in K is equal to k. If K = k, then σ is said to be of the first kind. After extension to k_{s}, the involution σ is determined by a symmetric or a skew–symmetric form. In the first case, σ is said to be of the orthogonal type, in the second one, of the symplectic type. If K ≠ k, then K is a quadratic extension of k and the restriction of σ to K is the non–trivial automorphism of K over k. In that case, the involution is said to be of the second kind, or a unitary involution, or a K/k–involution. See for instance [11] or [15], chap 7, for more details on algebras with involution.

Let (D, σ) be a division algebra with involution over k. A hermitian form over (D, σ) is by definition a pair (V, h), where V is a finite dimensional D–vector space, and h : V × V → D is hermitian with respect to σ. We say that (V, h) is hyperbolic if there exists a sub D–vector space W of V with dim(V) = 2dim(W) and such that h(x, y) = 0 for all x, y ∈ W. This leads to a notion of Witt group W (D, σ) (cf. for instance [15], Chap 7. §2). Note that the tensor product of a quadratic form over k with a hermitian form over (D, σ) is a hermitian form over (D, σ), hence W (D, σ) is a W (k)–module.

Let (V, h) be a hermitian form over (D, σ), as above. Let n = dim_{D}(V), and let H be the matrix of h with respect to some D–basis of V. Let us denote by Nrd : M_{n}(D) → k the reduced norm. The discriminant of h is by definition disc(h) = (−1)^{\frac{n(n−1)}{2}} Nrd(H) ∈ k^{∗}/k^{∗2}.

§2. Multiples of hermitian forms

Let (D, σ) a division algebra with involution over k. Let us denote by J the sub W (k)–module of W (D, σ) consisting of the hermitian forms (V, h) with dim_{D}(V) even. Suppose that cd_{2}(Γ_{k}) ≤ d.

Theorem 2.1.1. (a) We have I^{d}J = 0.

(b) If σ is of the second kind, then I^{d−1}J = 0.

(c) If σ is of the first kind and of the symplectic type, then I^{d−2}J = 0.

Part (a) was proved by Chabloz in [9]. We need the following lemma :

Lemma 2.1.2. Let a ∈ D^{∗} such that σ(a) = a. We have :

(a) If ϕ ∈ I^{d}, then ϕ⊗ < 1, a > is hyperbolic.

(b) If ϕ ∈ I^{d−1} and σ is of the second kind, then ϕ⊗ < 1, a > is hyperbolic.

(c) If ϕ ∈ I^{d−2} and σ is of the first kind and of the symplectic type, then ϕ⊗ < 1, a > is hyperbolic.

Proof.

(a) Let F = k(a), and let f : W (F) → W (D, σ) be the base change homomorphism. We have f(ϕ⊗ < 1, a >) = ϕ⊗ < 1, a >, hence it suffices to check that ϕ⊗ < 1, a >= 0 in W (F). Note that ϕ⊗ < 1, a >∈ I^{d+1}(F). As cd_{2}(k) ≤ d, we have cd_{2}(F) ≤ d, therefore by Theorem 1.2.1, we have I^{d+1}(F) = 0. Hence ϕ⊗ < 1, a > is hyperbolic, and this concludes the proof of (a).

(b) Suppose that σ is a K/k–involution, and set E = K(a), F = k(a). Let us denote by τ : E → E the restriction of σ : D → D to E. Then τ is an E/F –involution, an involution of the second kind. Let f : W (E, τ) → W (D, σ) be the base change homomorphism. We have f(ϕ⊗ < 1, a >) = ϕ⊗ < 1, a >, hence it suffices to check that ϕ⊗ < 1, a >= 0 in W (E). Let E = F (\sqrt{δ}), for some δ ∈ F. Let tr_{E/F} : W (E, τ) → W (F) be the W (F)– homomorphism given by the trace of hermitian forms. It is well–known that tr_{E/F} is injective, and its image is equal to < 1, −δ > W (F) (cf. for instance [15], Chap 10, §1). We have

tr_{E/F} (< 1, a >)) =< 1, −δ > ⊗ < 1, a >,

hence tr_{E/F} (ϕ⊗ < 1, a >) = ϕ⊗ < 1, −δ > ⊗ < 1, a >. This implies that

tr_{E/F} (ϕ⊗ < 1, a >) ∈ I^{d+1}(F).

As in (a), we see that I^{d+1}(F) = 0. Therefore ϕ⊗ < 1, a >= 0 in W (E, τ), and (b) is proved.

(c) Suppose that σ is symplectic. Then the degree of D is even. Set deg(D) = 2m, and let us prove the statement by induction on m. If m = 1, then D is a quaternion algebra and σ is the canonical involution of D. As σ(a) = a, we have a ∈ k^{∗}. Let trd : W (D, σ) → W (k) be the W (k)–homomorphism given by the reduced trace of hermitian forms. It is well– known that this homomorphism is injective, and its image is equal to n_{D}W (k), where n_{D} is the norm form of the quaternion algebra D (cf. [15], chap. 10, §1). We have trd(< 1, a >) = n_{D}⊗ < 1, a >∈ I^{3}. Let ϕ ∈ I^{d−2}. Then

trd(ϕ⊗ < 1, a >) = ϕ ⊗ n_{D}⊗ < 1, a >∈ I^{d+1} = 0.

This implies that trd(ϕ⊗ < 1, a >) = 0 in W (k), hence ϕ⊗ < 1, a >= 0 in W (D, σ).

Suppose that m > 1. If a ∉ k, set F = k(a). If a ∈ k, take any b ∈ D^{∗} such that σ(b) = b and that b ∉ k (this is possible as m > 1) and set F = k(b). Let D′ = Z_{D}(F) be the centralizer of F in D. Note that F is invariant by σ. By [11], 2.9. we know that [F : k] ≤ m. The F –algebra D′ is also invariant by σ. We have [F : k] > 1, hence degD′ < deg(D). Let deg(D′) = 2m′. As m′ < m, we can apply the induction hypothesis, hence ϕ⊗ < 1, a >= 0 in W (D′, σ). Let f : W (D′, σ) → W (D, σ) be the base change homomorphism. We have f(ϕ⊗ < 1, a >) = ϕ⊗ < 1, a >. Hence ϕ⊗ < 1, a >= 0 in W (D, σ), as claimed.

Proof of Theorem 2.1.1 Let q ∈ J. We have q =< a_{1}, …, a_{n} >, with a_{i} ∈ D^{∗}, σ(a_{i}) = a_{i}. Note that n is even, as q ∈ J. Set m = \frac{n}{2}. Let H =< 1, −1 >, and let us denote by [m]H the orthogonal sum of m copies of H. Then in W (D, σ), we have q = q ⊕ [m]H =< 1, a_{1} > ⊕ < −1, a_{2} > ⊕ … ⊕ < −1, a_{n} >. By the lemma, ϕ⊗ < 1, a_{i} > and ϕ⊗ < −1, a_{i} > are hyperbolic for all i whenever ϕ ∈ I^{d}, or ϕ ∈ I^{d−1} and σ is unitary, or ϕ ∈ I^{d−2} and σ is symplectic. Hence ϕ ⊗ q is hyperbolic in these cases too, so the theorem is proved.

The following is a consequence of results of Parimala, Sridharan and Suresh [14] and of Berhuy [8].

Theorem 2.1.3 Suppose that D is a quaternion algebra, and that σ is of the first kind and of the orthogonal type. Let h ∈ J, and let ϕ ∈ I^{d−1}. Then ϕ ⊗ h is hyperbolic if and only if e_{d−1}(ϕ) ∪ (disc(h)) = 0.

Proof. By Berhuy [8], Theorem 13, it suffices to show that e_{d−1}(ϕ) ∪ (disc(h)) = 0 if and only if e_{n,D}(ϕ ⊗ h) = 0 for all n ≥ 0 (cf. [8], 2.2 for the definition of the invariant e_{n,D}). As cd_{2}(Γ_{k}) ≤ d, we have e_{n,D}(ϕ ⊗ h) = 0 for n > d, so it suffices to check that e_{d−1}(ϕ) ∪ (disc(h)) = 0 is equivalent with e_{n,D}(ϕ ⊗ h) = 0 for all n = 0, …, d. Let k(D) be the function field of the quadric associated to D. Then D ⊗ k(D) ≃ M_{2}(k(D)), and h_{k(D)} corresponds via Morita equivalence to a quadratic form q_{h} over k(D). Note that disc(q_{h}) = disc(h). Similarly, the hermitian form (ϕ ⊗ h)_{k(D)} corresponds to a quadratic form q_{ϕh} over k(D), and we have q_{ϕh} ≃ ϕ ⊗ q_{h}.

For all n = 0, …, d, we have by construction that e_{n,D}(ϕ ⊗ h) = 0 if and only if e_{n}(q_{ϕh}) = 0 (cf. [8], 2.2). But q_{ϕh} ≃ ϕ ⊗ q_{h}, hence

e_{n}(q_{ϕh}) = e_{n}(ϕ ⊗ q_{h}) = e_{n−1}(ϕ) ∪ (disc(q_{h})) = e_{n−1}(ϕ) ∪ (disc(h)).

If n < d, then e_{n}(ϕ) = 0 as ϕ ∈ I^{d−1}. We have e_{d}(q_{ϕh}) = e_{d−1}(ϕ) ∪ (disc(h)). Hence e_{n}(q_{ϕh}) = 0 for all n ≥ 0 if and only if e_{d−1}(ϕ) ∪ (disc(h)) = 0. This concludes the proof.

Corollary 2.1.4 Suppose that D is a quaternion algebra, and that σ is of the first kind and of the orthogonal type. Let h and h′ be two hermitian forms over (D, σ), and let ϕ ∈ I^{d−1}. Then ϕ ⊗ h ≃ ϕ ⊗ h′ if and only if e_{d−1}(ϕ) ∪ (disc(h)) = e_{d−1}(ϕ) ∪ (disc(h′)).

Proof. The hermitian forms ϕ ⊗ h and ϕ ⊗ h′ are isomorphic if and only if ϕ ⊗ (h ⊕ (−h′)) is hyperbolic. By 2.1.3, this is equivalent with e_{d−1}(ϕ) ∪ disc(h ⊕ (−h′)) = 0. Note that as dim_{k}(D) is even, disc(−h′) = disc(h′). Therefore

e_{d−1}(ϕ) ∪ (disc(h ⊕ (−h′))) = e_{d−1}(ϕ) ∪ (disc(h)) + e_{d−1}(ϕ) ∪ (disc(h′)),

hence the corollary is proved.

Let us denote by J_{2} the sub W (k)–module of J consisting of the classes of the hermitian forms h such that (disc(h)) = 0.

Corollary 2.1.5 Suppose that D is a quaternion algebra, and that σ is of the first kind and of the orthogonal type. Then I^{d−1}J_{2} = 0.

Proof. This is an immediate consequence of 2.1.4.

§3. Multiples of trace forms

Let L and L′ be two G–Galois algebras. The aim of this section is to prove a result concerning multiples of trace forms (see 3.1.2) that was conjectured in [1], and to derive some consequences for generalized self–dual normal bases. Suppose that cd_{2}(Γ_{k}) ≤ d.

Theorem 3.1.1. Let ϕ ∈ I^{d−1}. Then ϕ ⊗ q_{L} ≃_{G} ϕ ⊗ q_{L′} if and only if e_{d−1}(ϕ) ∪ x_{L} = e_{d−1}(ϕ) ∪ x_{L′} for all x ∈ H^{1}(G, Z/2Z).

Special cases of this have been proved in [1], [4], [5], [6], [9] and [12].

Proof. The condition is necessary by 1.4.3. Let us prove that it is also sufficient. By [3], 4.1 and [7], 2.3.1 we can assume that k is perfect. Set A = k[G], and let us denote by σ_{A} : A → A the canonical involution. Let R_{A} be the radical of the algebra A, and set \overline{A} = A/R. Then the projection A → \overline{A} induces a bijection of pointed sets H^{1}(k, U_{A}) → H^{1}(k, U\overline{A}). We have

\overline{A} ≃ A_{1} × … × A_{s} × (A_{s+1} × A′_{s+1}) × … × (A_{m} × A′_{m}),

where A_{i} is a simple algebra for all i = 1, …, m, with σ(A_{i}) = A_{i} for i = 1, …, s and σ(A_{i}) = A′_{i} for i = s + 1, …, m. Let σ_{i} : A_{i} → A_{i} be the restriction of σ_{A} to A_{i} for i = 1, …, s, and let us denote by σ_{i} : A_{i} × A′_{i} → A_{i} → A_{i} × A′_{i} the restriction of σ_{A} to A_{i}×A′_{i} if i = s+1, …, m. Let F_{i} be the maximal subfield of the center of A_{i} such that σ_{i} is F_{i}–linear if i = 1, …, s, and let U_{i} be the norm–one–group of (A_{i}, σ_{i}). For i = s+1, …, m, let F_{i} be the center of A_{i}, and let U_{i} be the norm–one–group of ((A_{i} × A_{i}), σ_{i}). Then U_{i} is a linear algebraic group defined over F_{i} for all i = 1, …, m. We have a bijection of pointed sets

H^{1}(k, U_{A}) → ∏_{i=1,...,m} H^{1}(F_{i}, U_{i}).

If i = s + 1, …, m, then U_{i} is a general linear group, hence H^{1}(F_{i}, U_{i}) = 0. Hence we have a bijection of pointed sets

H^{1}(k, U_{A}) → ∏_{i=1,...,s} H^{1}(F_{i}, U_{i}).

Let us denote by u_{i}, u′_{i} ∈ H^{1}(F_{i}, U_{i}), i = 1, …, s, the images of u(L), u(L′) ∈ H^{1}(k, U_{A}).

For all i = 1, …, s, the simple algebra A_{i} is a matrix algebra over a division algebra with involution D_{i}, and the classes u_{i}, u′_{i} correspond to isomorphism classes of hermitian forms h_{i}, h′_{i} over D_{i}.

Let r = dim(ϕ), and set B = M_{r}(A). Let us denote by σ_{B} : B → B the involution induced by σ_{A} and the transposition, i.e. σ_{B}(a_{i,j}) = (σ_{A}(a_{j,i})) for all a_{i,j} ∈ A. Let R_{B} be the radical of B, and set \overline{B} = B/R_{B}.

We have

\overline{B} ≃ M_{r}(A_{1}) × … × M_{r}(A_{m}).

As above, we get a bijection of pointed sets

H^{1}(k, U_{B}) → ∏_{i=1,...,s} H^{1}(F_{i}, U_{M_{r}(A_{i})}).

Sending a G–quadratic form to its tensor product with the quadratic form ϕ gives us a map f : H^{1}(k, U_{A}) → H^{1}(k, U_{B}). The map f induces \overline{f} : H^{1}(k, U\overline{A}) → H^{1}(k, U\overline{B}), and

f_{i} : H^{1}(F_{i}, U_{i}) → H^{1}(k, U_{M_{r}(A_{i})})

for all i = 1, …, s. The image of the isomorphism class of the hermitian form h_{i} is the hermitian form ϕ ⊗ h_{i}.

Let us show that for all i = 1, …, s, we have ϕ ⊗ h_{i} ≃ ϕ ⊗ h′_{i}. This is equivalent with proving that ϕ⊗(h_{i} ⊕(−h′_{i})) is hyperbolic. If U_{i} is unitary or symplectic, then this follows from Theorem 2.1.1. (b) and (c). Suppose that U_{i} is orthogonal. Then A_{i} = M_{n_{i}}(D_{i}), where D_{i} = F_{i} or D_{i} is a quaternion field with center F_{i} (cf. [15], Chap. 8, 13.5. (ii) ). We have (U_{i}/U_{i}^{0})(k_{s}) ≃ Z/2Z. Let ι : Γ_{F_{i}} → Γ_{k} be the inclusion, and let us consider

δ_{i} : Γ_{F_{i}} →^{ι}Γ_{k}⟶^{ϕ_{L}}G → U_{G}(k_{s}) → U_{i}(k_{s}) ≃ Z/2Z

and

δ′_{i} : Γ_{i}→^{ι}Γ_{k}⟶^{ϕ_{L′}}G → U_{G}(k_{s}) → U_{i}(k_{s}) ≃ Z/2Z.

Then δ_{i}, δ′_{i} are 1–cocycles that define elements (δ_{i}), (δ′_{i}) ∈ H^{1}(F_{i}, Z/2Z) corresponding to the relative discriminants disc(h_{i}) and disc(h′_{i}) of the hermitian forms h_{i} and h′_{i} with respect to the unit hermitian form h_{0}.

Note that δ_{i} = ιx_{L} and δ_{i} = ιx_{L′} for all x ∈ H^{1}(G, Z/2Z). By hypothesis, we have e_{d−1}(ϕ) ∪ x_{L} = e_{d−1}(ϕ) ∪ x_{L′} for all x ∈ H^{1}(G, Z/2Z). This implies that

e_{d−1}(ϕ) ∪ (disc(h_{i})) = e_{d−1}(ϕ) ∪ (disc(h′_{i}))

for all i = 1, …, s. By cor. 2.1.4, we conclude that the hermitian forms ϕ ⊗ h_{i} and ϕ ⊗ h′_{i} are isomorphic, and hence the theorem is proved.

Recall that for any G–quadratic form q, we denote by [m]q the orthogonal sum of m copies of q, in other words the quadratic form < 1, …, 1 > ⊗q. Let us denote by ϵ_{d−1} ∈ H^{d−1}(k) the cup product of d − 1 copies of (−1) ∈ H^{1}(k). The following is an immediate consequence of 3.1.1 :

Corollary 3.1.2 Let L and L′ be two G–Galois algebras. Then [2^{d−1}]q_{L} ≃_{G} [2^{d−1}]q_{L′} if and only if ϵ_{d−1} ∪ x_{L} = ϵ_{d−1} ∪ x_{L′}.

Let L_{0} be the split G–Galois algebra, and let q_{0} = q_{L_{0}} be its trace form. Recall that a G–Galois algebra is said to have a self–dual normal basis if q_{L} ≃_{G} q_{0}.

For any positive integer m, we denote by [m]L the product of m copies of the G–Galois algebra L. We say that [m]L has a self–dual normal basis if [m]q_{L} ≃_{G} [m]q_{0}. A subalgebra E of L is said to be a subalgebra of invariants if there exists a subgroup H of G such that E = L^{H}.

Corollary 3.1.3 Let L be a G–Galois algebra. Then the algebra [2^{d−1}]L has a self–dual normal basis if and only if the discriminant of every quadratic subalgebra of invariants is a sum of 2^{d−1} squares.

Proof. Let us denote by q the 2^{d−1}–dimensional unit form, and let D(q) be the set of non–zero elements of k represented by q. Then a ∈ D(q) if and only if a is a sum of 2^{d−1} squares in k. It is well–known that a ∈ D(q) if and only if the quadratic form q⊕ < −a > is isotropic. This is equivalent with the 2^{d}–fold Pfister form q⊗ < 1, −a > being hyperbolic, hence by 1.2.1 with ϵ_{d−1} ∪ (a) = 0.

Let x ∈ H^{1}(G, Z/2Z). The argument above shows that x_{L} is a sum of 2^{d−1} squares if and only if ϵ_{d−1} ∪ x_{L} = 0. Let χ : G → Z/2Z the corresponding homomorphism. Let H be the kernel of χ, and let E_{x} = L^{H} be the invariant subalgebra. Then E_{x} is a quadratic subalgebra of L, and its discriminant is equal to x_{L}.

By 3.1.2, the algebra [2^{d−1}]L has a self–dual normal basis if and only if ϵ_{d−1} ∪ x_{L} = 0 for all x ∈ H^{1}(G, Z/2Z). We have just seen that this is equivalent with the discriminant of the quadratic subalgebra E_{x} being a sum of 2^{d−1} squares for all x ∈ H^{1}(G, Z/2Z), hence the corollary is proved.

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Address E. Bayer–Fluckiger, Ecole Polytechnique Fédérale de Lausanne, Mathématiques, CSAG– IMB–FSB, Station 8, 1015 Lausanne, Switzerland, eva.bayer@epfl.ch