Galois Algebras, Hasse Principle

and Induction–Restriction Methods

E. Bayer-Fluckiger and R. Parimala

Received: May 9, 2011 Revised: August 29, 2011

Communicated by Ulf Rehmann

Let k be a field of characteristic ≠ 2, and let L be a Galois extension of k with group G. Let us denote by q_{L} : L × L → k the trace form, defined by q_{L}(x, y) = Tr_{L/k}(xy). Let (gx)_{g∈G} be a normal basis of L over k. We say that this is a self–dual normal basis if q_{L}(gx, hx) = δ_{g,h}. If the order of G is odd, then L always has a self–dual normal basis over k (cf. [1]). This is no longer true in general if the order of G is even; some partial results are given in [2].

If k is a global field, then it is natural to ask whether a local–global principle holds for this problem. In order to make this question precise, we have to consider G–Galois algebras and not only field extensions. Moreover, it is useful to note that q_{L} is a G–quadratic form, in other words q_{L}(gx, gy) = q_{L}(x, y) for all x, y ∈ L and g ∈ G. The G–Galois algebra has a self–dual normal basis if and only if the G–form q_{L} is isomorphic to the unit G–form. This leads to the following question:

Question. Suppose that k is a global field, and let L and L′ be two G–Galois algebras. Assume that for all places v of k, the G–forms q_{L_{v}} and q_{L′_{v}} are isomorphic over k_{v}. Are the G–forms q_{L} and q_{L′} isomorphic over k ?

Note that a similar Hasse principle does not hold for arbitrary G–forms, cf. Morales [5]. In the context of trace forms of G–Galois algebras, positive results are obtained in [2] in some special cases. However, the problem is open in general.

The starting point of this paper is to investigate this question. The main tool, which is of independent interest, is to develop induction–restriction methods for arbitrary G–forms, generalizing some results of [2] and of Lequeu in [4]. The key ingredient is an odd determinant property of the group G (cf. §2) which is shown to hold for instance if the normalizer of a 2–Sylow subgroup S controls the fusion of S in G. We obtain the following :

Theorem. Suppose that k is a global field of characteristic ≠ 2. Let G be a finite group, and suppose that G has the odd determinant property if char(k) = 0. Let L and L′ be two G–Galois algebras such that for all places v of k, the G–forms q_{L_{v}} and q_{L′_{v}} are isomorphic over k_{v}. Then the G–forms q_{L} and q_{L′} are isomorphic over k.

Corollary. Suppose that k and G are as above. Then a G–Galois algebra has a self–dual normal basis over k if and only if such a basis exists over all the completions of k.

We thank Jean–Pierre Serre for many fruitful conversations and correspondence. We also thank Emmanuel Lequeu for very interesting conversations, and for showing us a preliminary version of his paper [4]. The authors acknowledge hospitality of Emory University and the École Polytechnique Fédérale de Lausanne, institutions in which the work was carried out. The first named author was partially supported by the Swiss National Science Fundation, grant 200020-109174/1, and the second named author was partially supported by NSF grant DMS 1001872.

§1. Definitions and basic facts

Let k be a field of characteristic ≠ 2, let G be a finite group, and let k[G] be the associated group ring. We refer to [7] for basic facts on k[G]–modules.

Group ring and involution

Let ι : k[G] → k[G] be the canonical involution of the group ring, in other words the k–linear involution of k[G] characterized by ι(g) = g^{−1} for all g ∈ G. Let R be the radical of k[G]. Then k[G]/R is a semi–simple k–algebra, hence we have a decomposition k[G]/R = ∏_{i=1,...,r} M_{n_{i}}(D_{i}), where D_{1}, …, D_{r} are division algebras. Let us denote by K_{i} the center of D_{i}, and let D_{i}^{op} be the opposite algebra of D_{i}.

Galois Algebras, Hasse Principle … 679

Note that ι(R) = R, hence ι induces an involution ι : k[G]/R → k[G]/R. Therefore k[G]/R decomposes into a product of involution invariant factors. These can be of two types: either an involution invariant matrix algebra M_{n_{i}}(D_{i}), or a product M_{n_{i}}(D_{i})×M_{n_{i}}(D_{i}^{op}), with M_{n_{i}}(D_{i}) and M_{n_{i}}(D_{i}^{op}) exchanged by the involution. We say that a factor is unitary if the restriction of the involution to its center is not the identity: in other words, either an involution invariant M_{n_{i}}(D_{i}) with ι|K_{i} not the identity, or a product M_{n_{i}}(D_{i}) × M_{n_{i}}(D_{i}^{op}) . Otherwise, the factor is said to be of the first kind. In this case, the component is of the form M_{n_{i}}(D_{i}) and the restriction of ι to K_{i} is the identity. We say that the component is orthogonal if after base change to a separable closure ι is given by the transposition, and symplectic otherwise.

We say that a component M_{n_{i}}(D_{i}) is split if D_{i} is a commutative field.

G–quadratic forms

A G–quadratic form is a pair (V, q), where V is a k[G]–module that is a finite dimensional k–vector space, and q : V × V → k is a non–degenerate symmetric bilinear form such that

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

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

Let S be a subgroup of G. We have two operations, induction and restriction (see for instance [2], 1.2 for details):

If (V, q) is an S–quadratic form, then Ind_{S}^{G}(V, q) is a G–quadratic form;

If (V, q) is a G–quadratic form, then Res_{S}^{G}(V, q) is an S–quadratic form.

The following result will be used in the sequel

Theorem 1.1. (see [1], th. 4.1) Let q and q′ be two G–quadratic forms. If they become isomorphic over an odd degree extension, then they are isomorphic.

It is well–known that S–quadratic forms correspond bijectively to k[S]– hermitian forms with respect to the involution ι : k[S] → k[S]. We will use the same notation for the S–quadratic form and the corresponding hermitian form.

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 us recall a result from [2] that will be basic for the proof of the main theorem:

Lemma 1.2. (cf. [2], 2.1.1.): Let S be a 2–Sylow subgroup of G. For any G–Galois algebra L, there exists an odd degree field extension k′/k and an S– Galois algebra M over k′ such that the G–form (L, q_{L}) ⊗_{k} k′ is isomorphic to the G–form Ind_{S}^{G}(q_{M}).

§2. The induction-restriction functor and the odd determinant property

The aim of this section is to introduce the odd determinant property, and to state a result (th. 2.2), which will be used in the proof of the Hasse principle result of §3.

Let G be a finite group, let S be a 2–Sylow subgroup of G, and let N = N_{G}(S) be the normalizer of S in G. Then N acts on S, and we denote by Σ the set of orbits of S under the action of N.

Let X be the Z–module of Z–valued functions on S invariant under conjugation by N, and let Φ : X → X be Res_{S}^{G} Ind_{S}^{G} considered as an endomorphism of X (cf. [7], 7.2).

Definition 2.1 We say that G has the odd determinant property if the determinant of Φ : X → X is an odd integer.

One of the main results of this paper is the following

Theorem 2.2 Suppose that G has the odd determinant property. Let (V_{1}, q_{1}) and (V_{2}, q_{2}) be two S–quadratic spaces. Suppose that

Res_{S}^{G} Ind_{S}^{G}(V_{1}, q_{1}) ≃_{S} Res_{S}^{G} Ind_{S}^{G}(V_{2}, q_{2}).

Then

Ind_{S}^{G}(V_{1}, q_{1}) ≃_{G} Ind_{S}^{G}(V_{2}, q_{2}).

This result is used in the proof of the Hasse principle stated in the introduction, see th. 3.1. The proof relies on an analysis of the odd determinant property, and is the subject matter of sections 4-11. The structure of the proof of th. 2.2 is as follows. Sections 5 and 6 study induction and restriction properties of S– quadratic forms. Section 7 is concerned with the odd determinant property in the special case where all the characters of S over k are absolutely irreducible. Using a filtration introduced in §9 and the quadratic descent argument of §8, we obtain a general result (see th. 10.1) based on the case considered in §7. This is then used in §11 to prove th. 2.2.

We next show that the odd determinant property holds if N controls the fusion of S in G.

Definition 2.3 We say that N controls the fusion of S in G if for all subsets T and T′ of S, if there exists g ∈ G with gT g^{−1} = T′ then there exists n ∈ N such that nT n^{−1} = T′.

There are many examples of groups G in which the normalizer controls the fusion of the 2–Sylow subroups; see for instance Thévenaz [8] for a survey.

Remark. Note that we only use the following property, which is clearly satisfied if N controls the fusion of S in G :

(*) For all s, t ∈ S, if there exists g ∈ G with gsg^{−1} = t then there exists n ∈ N such that nsn^{−1} = t.

It does not seem to be known whether there exist groups G having property (*) where N does not control the fusion of S in G.

Proposition 2.4 Suppose that N controls the fusion of S in G. Then G has the odd determinant property.

In order to prove this proposition, we need the following lemma:

Lemma 2.5 Suppose that N controls the fusion of S in G, and let x ∈ S. Then C_{S}(x) is a 2–Sylow subgroup of C_{G}(x).

Proof. Let S_{0} be a 2–Sylow subgroup of C_{G}(x) containing x and let S_{1} be a 2–Sylow subgroup of G containing S_{0}. Let g ∈ G be such that gS_{1}g^{−1} = S. In view of the fusion hypothesis, there exists n ∈ N such that ngxg^{−1}n^{−1} = x. Let us consider Int(ng) : G → G. Then, as Int(ng)(x) = x, we have Int(ng)(C_{G}(x)) = C_{G}(x). We have Int(ng)(S_{1}) = S, hence Int(ng)(S_{0}) = Int(ng)(S_{1} ∩ C_{G}(x)) = S ∩ C_{G}(x) = C_{S}(x). This implies that C_{S}(x) is a 2–Sylow subgroup of C_{G}(x), as claimed.

Proof of prop. 2.4 For σ ∈ Σ, let q_{σ} be the function on S which is equal to 1 on σ and 0 otherwise. Note that the set (q_{σ})_{σ∈Σ} is a basis of the Z–module X. Let σ, σ′ ∈ Σ, and fix x ∈ σ′. By definition, the coefficient of q_{σ} in Φ(q_{σ′}) is equal to

\frac{1}{#S} #\{g ∈ G | gxg^{−1} ∈ σ \}.

As N controls the fusion of S in G, we have gxg^{−1} ∈ σ if and only if x ∈ σ. Therefore the coefficient of q_{σ} in Φ(q_{σ′}) is equal to 0 if σ ≠ σ′.

The coefficient of q_{σ} in Φ(q_{σ}) is equal to

\frac{1}{#S} #C_{G}(x) #σ = \frac{1}{#S} #C_{G}(x) \frac{#N}{#C_{N}(x)} = \frac{#N}{#S} \frac{#C_{G}(x)}{#C_{S}(x)} \frac{#C_{S}(x)}{#C_{N}(x)}.

Therefore it suffices to check that \frac{#C_{G}(x)}{#C_{S}(x)} is odd, and this follows from lemma 2.5.

§3. Hasse principle

In this section, we suppose that k is a global field of characteristic ≠ 2. Let G be a finite group, and let us denote by k[G] the associated group ring. One of the main results of this paper is the following

Theorem 3.1 Suppose that G has the odd determinant property if char(k) = 0, and let L and L′ be two G–Galois algebras. Then q_{L} ≃_{G} q_{L′} over k if and only if q_{L} ≃_{G} q_{L′} over all the completions of k.

As an immediate consequence, we get

Corollary 3.2 Suppose that G has the odd determinant property if char(k) = 0. Then a G–Galois algebra has a self–dual normal basis over k if and only if it has a self–dual normal basis over every completion of k.

By prop. 2.3, we know that G has the odd determinant property whenever for a 2–Sylow subgroup S, the normalizer N_{G}(S) controls the fusion of S in G. Hence we have

Corollary 3.3 Suppose that for a 2–Sylow subgroup S of G, the normalizer N_{G}(S) controls the fusion of S in G. Then the trace forms of two G–Galois algebras are G–isomorphic over k if and only if they are G–isomorphic over each completion of k. In particular, a G–Galois algebra has a self–dual normal basis over k if and only if it has a self–dual normal basis over every completion of k.

Corollary 3.4 Suppose that G has a normal 2–Sylow subgroup. Then the trace forms of two G–Galois algebras are isomorphic over k if and only if they are isomorphic over each completion of k. In particular, a G–Galois algebra has a self–dual normal basis over k if and only if it has a self–dual normal basis over every completion of k.

Proof. This is an immediate consequence of 3.3.

The proof of th. 3.1 relies on th. 2.2, and on some properties of group rings and of quadratic and hermitian forms that we recall in this section. Let us first note that the Hasse principle holds for any G–form provided the orthogonal components of the group ring are split:

Theorem 3.5 Suppose that all the orthogonal components of k[G] are split, and let q, q′ be two G–forms. Then q ≃_{G} q′ over k if and only if q ≃_{G} q′ over all the completions of k.

Proof. This follows from the Hasse principle for unitary and symplectic forms, as well as the Hasse principle for quadratic forms over global fields (see for instance [6], chap. 10).

Therefore th. 3.1 is new for number fields only – indeed, if char(k) > 0, then all the orthogonal components of k[G] are split.

Proposition 3.6 Let S be a 2–group. Then the orthogonal and unitary components of k[S] are split, and the symplectic components of k[S] are either split, or of the form M_{n}(H) where H is a quaternion division algebra over its center.

Proof. Note that k[S] = Q[S] ⊗_{Q} k if char(k) = 0, and k[S] = F_{p}[S] ⊗_{F_{p}} k if char(k) = p ≠ 0. Therefore it is sufficient to prove the proposition when k = Q or k = F_{p}. As the Brauer group of a finite field is trivial, every component is split if k = F_{p}.

Suppose that k = Q. Then each component of Q[S] is invariant under ι (cf. [6], Chap 8, 13.2.).

Let M_{n}(D) be a symplectic component of Q[S]. This implies that the algebra M_{n}(D) is of order one or two in the Brauer group of Q, and it is well–known that this can only happen if D is a commutative field or a quaternion algebra.

Let us now show that the orthogonal and unitary components of Q[S] are split. Let v be a non–dyadic place of Q, and let O_{v} be the ring of integers of Q_{v}. Since #S is invertible in O_{v}, it follows that O_{v}[S] is Azumaya over its center. This implies that this algebra is split mod π, where π is a uniformizer at v, therefore it is split over O_{v}. In particular every component of Q_{v}[S] is split.

If v is the real place of Q, then every orthogonal and unitary component of Q_{v}[S] = R[S] is split (cf. [6], Chap 8, 13.5).

Let M_{n}(D) be an orthogonal or unitary component of Q[S], and let Z(D) = K. As S is a 2–group, K is a subfield of a 2–cyclotomic field, hence K admits a unique dyadic place. Since D is split at all the other places, D is split at the dyadic place as well, hence D is split.

Corollary 3.7 Let S be a 2–group, and let q, q′ be two S–forms. Then q ≃_{S} q′ over k if and only if q ≃_{S} q′ over all the completions of k.

Proof. This follows from 3.5 and 3.6.

We are now ready to prove 3.1. The proof uses th. 2.2, which will be proved in section 11.

Proof of th. 3.1 Suppose first that char(k) > 0. Then all the components of k[G] are split, hence th. 3.5 implies the desired result.

Suppose now that char(k) = 0, in other words that k is an algebraic number field. By lemma 1.2, there exists an odd degree field extension k′/k and S– Galois algebras M and M′ over k′ such that (L, q_{L}) ⊗_{k} k′ ≃_{G} Ind_{S}^{G}(M, q_{M}), and (L,′ q_{L′}) ⊗_{k} k′ ≃_{G} Ind_{S}^{G}(M′, q_{M′}). Recall that by hypothesis the G–forms (L, q_{L}) and (L′, q_{L′}) are isomorphic over all the completions of k. This implies that the G–forms (L, q_{L}) ⊗_{k} k′ and (L,′ q_{L′}) ⊗_{k} k′ are isomorphic over all the completions of k′. Hence the S–forms Res_{S}^{G}(L, q_{L})⊗_{k}k′ ≃_{S} Res_{S}^{G} Ind_{S}^{G}(M, q_{M}), and Res_{S}^{G}(L′, q_{L′}) ⊗_{k} k′ ≃_{S} Res_{S}^{G} Ind_{S}^{G}(M′, q_{M′}) are isomorphic over all the completions of k′. By corollary 3.7, this implies that the S forms Res_{S}^{G}(L, q_{L})⊗_{k} k′ ≃_{S} Res_{S}^{G} Ind_{S}^{G}(M, q_{M}), and Res_{S}^{G}(L′, q_{L′}) ⊗_{k} k′ ≃_{S} Res_{S}^{G} Ind_{S}^{G}(M′, q_{M′}) are isomorphic over k′. As G has the odd determinant property, th. 2.2 implies that the G–forms Ind_{S}^{G}(M, q_{M}) and Ind_{S}^{G}(M′, q_{M′}) are isomorphic. As (L, q_{L}) ⊗_{k} k′ ≃_{G} Ind_{S}^{G}(M, q_{M}) and (L,′ q_{L′}) ⊗_{k} k′ ≃_{G} Ind_{S}^{G}(M′, q_{M′}), we get (L, q_{L}) ⊗_{k} k′ ≃_{G} (L,′ q_{L′}) ⊗_{k} k′. By th. 1.1, this implies that (L, q_{L}) ≃_{G} (L,′ q_{L′}), and this completes the proof of th. 3.1.

§4. Properties of determinants in characteristic 2

This section is concerned with properties of determinants of linear transformations over rings of characteristic 2 that will be needed in the following sections. Let F be a field of characteristic 2, and let R = F [X]/(X^{2} + 1). We start by recalling a result of linear algebra :

Proposition 4.1 Let M = R^{n} be the free R–module of rank n, and let f : M → M be an R–linear map. Then

N_{R/F} (det(f)) = det(f_{F}),

where det(f_{F}) is the determinant of f considered as an F –linear map.

Corollary 4.2 Let

A = ( \begin{matrix}a_{1,1} & b_{1,1} & … & a_{1,n} & b_{1,n} \\ b_{1,1,} & a_{1,1} & … & b_{1,n} & a_{1,n} \\ … & … & … & … & … \\ a_{n,1} & b_{n,1} & … & a_{n,n} & b_{n,n} \\ b_{n,1} & a_{1,n} & … & b_{n,n} & a_{n,n} \end{matrix} )

and

B = ( \begin{matrix}a_{1,1} + b_{1,1} & … & a_{1,n} + b_{1,n} \\ … & … & … \\ a_{n,1} + b_{n,1} & … & a_{n,n} + b_{n,n} \end{matrix} )

with a_{i,j}, b_{i,j} ∈ F. Then

det(B)^{2} = det(A).

Proof. Let f : R^{n} → R^{n} be defined by f(e_{j}) = Σ_{1≤i≤n}(a_{i,j} + b_{i,j}X)e_{i}, where e_{1}, …, e_{n} is the standard basis of R^{n}. The matrix of f with respect to the basis e_{1}, …, e_{n} is (a_{i,j} + b_{i,j}X). We have

N_{R/F} (det(a_{i,j} + b_{i,j}X)) = (det(a_{i,j} + b_{i,j}))^{2}, which is equal to det(B)^{2}. By 4.1, this is the determinant of f as an F –linear map. On the other hand, the determinant of f in the basis e_{1}, e_{1}X, e_{2}, e_{2}X, …, e_{n}, e_{n}X is equal to det(A); hence we have det(B)^{2} = det(A), as claimed.

We also need the following observation:

Lemma 4.3 Let n ∈ N, and suppose that the group {1, ι} of order 2 acts on the set {1, …, n} in such a way that {1, …, r} is the set of fixed points. Let (d_{i,j})_{1≤i,j≤n} be an integral matrix such that d_{ι(i),ι(j)} = d_{i,j} for all i, j. Then

det(d_{i,j})_{1≤i,j≤n} ≡ det(d_{i,j})_{1≤i,j≤r}det(d_{i,j})_{r+1≤i,j≤n} (mod 2).

Proof. Let S be the set of permutations of {1, 2, ...n}. For s ∈ S and 1 ≤ i ≤ n, set ι ∗ s(i) = ιsι(i). We have

det(d_{i,j})_{1≤i,j≤n} ≡ Σ_{s∈S}( ∏_{1≤i≤n} d_{si,i}) mod 2.

Set

H = {s ∈ S | s(i) ≤ r for i ≤ r}.

Then

Σ_{s∈S}( ∏_{1≤i≤n} d_{si,i}) = Σ_{s∈H}( ∏_{1≤i≤n} d_{si,i}) + Σ_{s∉H}( ∏_{1≤i≤n} d_{si,i}).

For s ∉ H, we have ι ∗ s ∉ H and s ≠ ι ∗ s. In view of d_{i,j} = d_{ι(i),ι(j)} for all i, j, we get

Σ_{s∉H}( ∏_{1≤i≤n} d_{si,i}) ≡ 0 mod 2.

Let

S^{1} = {s ∈ S | s(i) = i for i ≥ r + 1},

and

S^{2} = {s ∈ S | s(i) = i for i ≤ r}.

Then we have

Σ_{s∈H}( ∏_{1≤i≤n} d_{si,i}) = [Σ_{s∈S^{1}}( ∏_{1≤i≤r} d_{si,i})].[Σ_{s∈S^{2}}( ∏_{r+1≤i≤n} d_{si,i})].

This completes the proof of the lemma.

Lemma 4.3 is used in the next sections, in particular in the proofs of 7.1, 8.4 and 8.5.

§5. Group rings of 2–groups and decomposition of S–quadratic forms

The aim of this section is to introduce some tools and notation that will be used in the sequel. In particular, we set up a decomposition of the quadratic forms invariant by a 2–group, generalizing the approach of [2], §5.

Group rings of 2–groups

Let k be a field of characteristic ≠ 2, and let S be a 2–group. Recall that ι : k[S] → k[S] is the canonical involution of the group ring.

As the characteristic of k is not 2, the algebra k[S] is semi–simple. We have a decomposition of k[S] into simple factors, corresponding to the irreducible representations of S over k, hence also to the irreducible characters of S over k. Let us denote by S′_{k} the set of these irreducible characters. Each of them determines a component M_{n_{x}}(Δ_{x}) of k[S], where Δ_{x} is a division algebra. Let K_{x} = Z(Δ_{x}) be the center of Δ_{x}. Recall that the orthogonal and unitary components are split, and that the symplectic components are either split, or of the form M_{n}(H) where H is a quaternion division algebra (see prop. 3.6).

Let us denote by U_{x} the simple k[S]–module associated to the irreducible character x ∈ S′_{k}. Note that it is isomorphic to Δ_{x}^{n_{x}}. Let Y_{k} be the free Z–module generated by S′_{k}.

Note that ι acts on S′_{k} by ι(x)(s) = x(s^{−1}) for all x ∈ S′_{k} and s ∈ S. We say that x ∈ S′_{k} is self–dual if ι(x) = x. This is equivalent to requiring that the corresponding component M_{n_{x}}(Δ_{x}) is stable by ι. If x ∈ S′_{k} is not self–dual, then there exists x′ ∈ S′_{k} such that x′ ≠ x and ι(x) = x′. In this case, set \overline{x} = (x, x′). If x is self–dual, then set \overline{x} = x. Let us denote by \overline{S}′_{k} the set of \overline{x} for x ∈ S′_{k}.

Set M_{n_{x}}(Δ\overline{x}) = M_{n_{x}}(Δ_{x}) if x is self–dual, and M_{n_{x}}(Δ\overline{x}) = M_{n_{x}}(Δ_{x}) × M_{n_{x′}} (Δ_{x′}) if ι(x) = x′ ≠ x. Similarly, set K\overline{x} = K_{x} is x is self–dual and K\overline{x} = K_{x} × K_{x′} if \overline{x} = (x, x′). Note that K\overline{x} is an étale algebra, but not necessarily a field. Let K0 ; \overline{x} = \{a ∈ K\overline{x} | ι(a) = a\} be the invariants of ι in K\overline{x}. When x is not self–dual, then we have K_{x} ≃ K_{x′} ≃ K0 ; \overline{x}.

The involution ι of k[S] restricts to the factors M_{n_{x}}(Δ\overline{x}), and it is adjoint to a hermitian or skew–hermitian form, which we fix in the different cases as follows.

If x is orthogonal, then Δ\overline{x} = K_{x}. In this case, we set D\overline{x} = K_{x}, and we chose the involution τ\overline{x} : D\overline{x} → D\overline{x} to be the identity. The restriction of the involution ι to this factor is adjoint to a symmetric form on Dn_{x} ; \overline{x} which we denote by ρ\overline{x}. We define m_{x} = n_{x}, and the symmetric form is supported on the simple module U_{x}.

If x is symplectic, then Δ\overline{x} = K_{x} or a quaternion division algebra. We set D\overline{x} = M_{2}(K_{x}) in the first case, and D\overline{x} = Δ_{x} in the second case. In both cases, we choose the involution τ\overline{x} : D\overline{x} → D\overline{x} to be the standard symplectic involution of D\overline{x}. In this case, the restriction of the involution ι to this factor is adjoint to a hermitian form over Dm_{x} ; \overline{x} with respect to the involution τ\overline{x} which we denote by ρ\overline{x}. The form ρ\overline{x} is supported on the module U_{x} ⊕ U_{x} and n_{x} = 2m_{x} if D\overline{x} is not division, it is supported on the module U_{x} and m_{x} = n_{x} if D\overline{x} is division.

If x is unitary, then Δ\overline{x} = K\overline{x}, and K\overline{x} is a quadratic algebra over K0 ; \overline{x}. We set D\overline{x} = K\overline{x}, and we fix the involution τ\overline{x} : D\overline{x} → D\overline{x} to be the non–trivial automorphism of this quadratic algebra. Then the restriction of the involution ι to this factor is adjoint to a hermitian form on Dn_{x} ; \overline{x} with respect to the involution τ\overline{x} which we denote by ρ\overline{x}. We set m_{x} = n_{x} in this case. The form ρ\overline{x} is supported on U_{x} if x is self–dual, and on U_{x_{1}} ⊕ U_{x_{2}} if \overline{x} = (x_{1}, x_{2}) with x_{1} ≠ x_{2} and ι(x_{1}) = x_{2}.

Set U\overline{x} = U_{x} ⊕ U_{x} if x is symplectic and D\overline{x} not division, U\overline{x} = U_{x_{1}} ⊕ U_{x_{2}} if \overline{x} is unitary with \overline{x} = (x_{1}, x_{2}) such that ι(x_{1}) = x_{2} and x_{1} ≠ x_{2}, and U\overline{x} = U_{x} in all other cases. Note that U\overline{x} ≃ Dm_{x} ; \overline{x}. Therefore in all cases we have a hermitian form ρ\overline{x} : U\overline{x} × U\overline{x} → D\overline{x} which we fix throughout. We denote the hermitian form (U\overline{x}, ρ\overline{x}) by ρ\overline{x}.

We also fix a quadratic form n\overline{x} : D\overline{x} → K0 ; \overline{x} to be the one–dimensional unit form if x is orthogonal, the reduced norm form of the quaternion algebra D\overline{x} if x is symplectic, and the norm form of the quadratic algebra D\overline{x} if x is unitary.

Decomposition of S–quadratic forms

Let (V, q) be an S–quadratic form. Then (V, q) decomposes as an orthogonal sum of hermitian forms (M\overline{x}, Q\overline{x}) for x ∈ S′_{k}, over M_{m_{x}}(D\overline{x}) with respect to the restriction of ι to this factor. By Morita theory, fixing ρ\overline{x}, the hermitian form (M\overline{x}, Q\overline{x}) is uniquely determined up to isomorphism by a hermitian form h\overline{x} over a free D\overline{x}–module W\overline{x} of finite rank with respect to the involution τ\overline{x}, and conversely the hermitian form (W\overline{x}, h\overline{x}) is uniquely determined up to isomorphism by (M\overline{x}, Q\overline{x}). Moreover, by Jacobson’s theorem the hermitian form (W\overline{x}, h\overline{x}) corresponds to a quadratic form (V\overline{x}, g\overline{x}) over K0 ; \overline{x} with the property that (V\overline{x}, g\overline{x})⊗n\overline{x} is uniquely determined by (W\overline{x}, h\overline{x}) (cf. [6], 10.1.1 and 10.1.7) . We have (M\overline{x}, Q\overline{x}) ≃ ρ\overline{x}⊗_{K0 ; \overline{x}} (V\overline{x}, g\overline{x}), and (V\overline{x}, g\overline{x})⊗n\overline{x} is uniquely determined by (M\overline{x}, Q\overline{x}), hence by (V, q). In other words, we have

(V, q) ≃ ⨁_{\overline{x}∈\overline{S}′_{k}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V\overline{x}, g\overline{x}),

and if (V_{1}, q_{1}) and (V_{2}, q_{2}) are two S–quadratic forms with

(V_{1}, q_{1}) ≃ ⨁_{\overline{x}∈\overline{S}′_{k}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V 1 ; \overline{x}, g1 ; \overline{x}) and (V_{2}, q_{2}) ≃ ⨁_{\overline{x}∈\overline{S}′_{k}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V 2 ; \overline{x}, g2 ; \overline{x}),

then

(V_{1}, q_{1}) ≃_{S} (V_{2}, q_{2})

if and only if

n\overline{x} ⊗ g1 ; \overline{x} ≃ n\overline{x} ⊗ g2 ; \overline{x}

for all \overline{x} ∈ S′_{k}.

§6. Induction of S–forms

Let k be a field of characteristic ≠ 2. Let G be a finite group, and let S be a 2–Sylow subgroup of G, We use the notation introduced in §5. In particular, S′_{k} is the set of irreducible characters of S over k. Recall that ι : k[S] → k[S] is the standard involution, and that for x ∈ S′_{k} we set \overline{x} = x if x is selfdual, and \overline{x} = (x, x′) if ι(x) = x′ ≠ x.

Let N = N_{G}(S) be the normalizer of S in G. Then N acts on S′_{k} by n(x)(s) = x(nsn^{−1}) for all n ∈ N, x ∈ S′_{k} and s ∈ S. Note that the actions of N and ι commute. We need the following lemmas:

Lemma 6.1 The orbits of S′_{k} under N have odd cardinality.

Proof. Let x ∈ S′_{k} and let ω be the orbit of x under N. We have ♯(ω) = ♯(N/Stab_{N}(x)). As S ⊂ Stab_{N}(x), we see that ♯(N/Stab_{N}(x)) is odd.

Lemma 6.2 Let x, x′ ∈ S′_{k} such that ι(x) = x′ ≠ x. Let ω, ω′ be the orbits of x, respectively x′. Then ω ≠ ω′.

Proof. Indeed, suppose that ω = ω′. As the actions of N and ι commute, we see that for every n ∈ N, we have ιn(x) ≠ n(x). This implies that ιy ≠ y for every y ∈ ω, and therefore ω has even cardinality, contradicting lemma 6.1. Therefore ω ≠ ω′.

Let us denote by Ω_{k} the set of orbits of S′_{k} under N. There is an induced action of N on the free Z–module generated by S′_{k} and the set of orbits under this action is the free Z–module generated by Ω_{k}.

Let us define an action of ι on Ω_{k} by letting ιω to be the orbit of ι(x) for any x ∈ ω; this is well–defined as the actions of N and ι on S′_{k} commute. For any

Galois Algebras, Hasse Principle … 689

ω ∈ Ω_{k}, set \overline{ω} = ω if ιω = ω, and \overline{ω} = (ω_{1}, ω_{2}) with ιω_{1} = ω_{2} and ω_{1} ≠ ω_{2}. Let \overline{Ω}_{k} be the set of all \overline{ω} with ω ∈ Ω_{k}. Let us fix a field extension K0 ; \overline{ω} of k such that K0 ; \overline{ω} ≃ K0 ; \overline{x} for all \overline{x} ∈ \overline{ω}. Let (V, q) be an S–quadratic form. Then we have an orthogonal decomposition

(V, q) ≃ ⨁_{\overline{x}∈\overline{S}′_{k}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V\overline{x}, g\overline{x}),

where (V\overline{x}, g\overline{x}) is a quadratic form over K0 ; \overline{x}, and (V\overline{x}, g\overline{x}) ⊗ n\overline{x} is uniquely determined by (V, q) (cf. §5). For all \overline{ω} ∈ \overline{Ω}_{k}, let us consider the orthogonal sum

(V\overline{ω}, g\overline{ω}) = ⨁_{\overline{x}∈\overline{ω}}(V\overline{x}, g\overline{x}).

Then (V\overline{ω}, g\overline{ω}) is a quadratic form over K0 ; \overline{ω}. Note that Ind_{S}^{G}(ρ\overline{x}) does not depend on the choice of \overline{x} ∈ \overline{ω}. Set

I(\overline{ω}) = Ind_{S}^{G}(ρ\overline{x})

where \overline{x} is any element of \overline{ω}. Therefore we have

Ind_{S}^{G}(V, q) = ⨁_{\overline{ω}∈\overline{Ω}_{k}} I(\overline{ω}) ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω}).

Set

A(V, q) = Res_{S}^{G} Ind_{S}^{G}(V, q).

Then we have

A(V, q) = ⨁_{\overline{ω}∈\overline{Ω}_{k}} Res_{S}^{G}(I(\overline{ω})) ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω}).

Let \overline{y} ∈ \overline{S}′_{k}, and let us take the \overline{y}–component of the equation above. We get

A(V, q)\overline{y} = ⨁_{\overline{ω}∈\overline{Ω}_{k}} Res_{S}^{G}(I(\overline{ω}))\overline{y} ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω}).

Let \overline{ω}′ ∈ \overline{Ω}_{k} such that \overline{y} ∈ \overline{ω}′. Note that the S–quadratic spaces A(V, q)\overline{y} and Res_{S}^{G}(I\overline{ω})\overline{y} do not depend on the choice of \overline{y} ∈ \overline{ω}′. Set

A(V, q)\overline{ω}′ = A(V, q)\overline{y}

Documenta Mathematica 16 (2011) 677–707

and

Res_{S}^{G}(I(\overline{ω}))\overline{ω}′ = Res_{S}^{G}(I(\overline{ω}))\overline{y}

for any \overline{y} ∈ \overline{ω}′.

Then we have

Res_{S}^{G}(I(\overline{ω}))\overline{ω}′ = ρ\overline{y} ⊗_{K_{\overline{ω}′}^{0}} F_{\overline{ω},\overline{ω}′}

for \overline{y} ∈ \overline{ω}′, where F_{\overline{ω},\overline{ω}′} is a quadratic form over K_{\overline{ω}′}^{0}.

Hence

A(V, q)\overline{ω}′ = ρ\overline{y} ⊗_{K_{\overline{ω}′}^{0}} ⨁_{\overline{ω}∈\overline{Ω}_{k}}[F_{\overline{ω},\overline{ω}′} ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω})].

Notation. Let ω, ω′ ∈ Ω_{k} be such that K0 ; \overline{ω} = K_{\overline{ω}′}^{0} = k. We define d_{\overline{ω},\overline{ω}′} to be the dimension of the k–vector space underlying the quadratic form F_{\overline{ω},\overline{ω}′}.

Note that d_{\overline{ω},\overline{ω}′} is the number of times ρ\overline{y} occurs in Res_{S}^{G} Ind_{S}^{G}(ρ\overline{x}) for any \overline{x} ∈ \overline{ω}, \overline{y} ∈ \overline{ω}′. As U\overline{x} is the underlying module of ρ\overline{x}, the integer d_{\overline{ω},\overline{ω}′} can also be seen as the number of times U\overline{y} occurs in Res_{S}^{G} Ind_{S}^{G}(U\overline{x}) for any \overline{x} ∈ \overline{ω}, \overline{y} ∈ \overline{ω}′.

Let (V_{1}, q_{1}) and (V_{2}, q_{2}) be two S–quadratic forms. If A(V_{1}, q_{1}) ≃ A(V_{2}, q_{2}), then A(V_{1}, q_{1})\overline{ω}′ ≃ A(V_{2}, q_{2})\overline{ω}′ for all \overline{ω}′ ∈ \overline{Ω}_{k}. Hence, if we have

(V_{1}, q_{1}) ≃ ⨁_{\overline{x}∈\overline{S}′_{k}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V 1 ; \overline{x}, g1 ; \overline{x}) and (V_{2}, q_{2}) ≃ ⨁_{\overline{x}∈\overline{S}′_{k}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V 2 ; \overline{x}, g2 ; \overline{x}),

then, for each \overline{ω}′ ∈ \overline{Ω}_{k},

⨁_{\overline{ω}∈\overline{Ω}_{k}} n\overline{ω}′ ⊗_{K_{\overline{ω}′}^{0}} [F_{\overline{ω},\overline{ω}′} ⊗_{K0 ; \overline{ω}} (V 1 ; \overline{ω}, g1 ; \overline{ω})] ≃ ⨁_{\overline{ω}∈\overline{Ω}_{k}} n\overline{ω}′ ⊗_{K_{\overline{ω}′}^{0}} [F_{\overline{ω},\overline{ω}′} ⊗_{K0 ; \overline{ω}} (V 2 ; \overline{ω}, g2 ; \overline{ω})].

§7. Odd determinant property – a special case

The aim of this section and the next ones is to establish some technical results relative to the odd determinant property. These will be used in §11 to prove th. 2.2.

We keep the notation of the previous sections, and we suppose that all the characters in S′_{k} are absolutely irreducible.

Recall that Ω_{k} is the set of N–orbits of S′_{k}. The following notation will be important in the sequel:

Notation. Let us define d_{ω,ω′} as being the number of times U_{y} occurs in

Res_{S}^{G} Ind_{S}^{G}(U_{x})

for x ∈ ω, y ∈ ω′.

The standard involution ι : k[S] → k[S] acts on Ω_{k}. Note that d_{ιω,ιω′} = d_{ω,ω′} for all ω, ω′ ∈ Ω_{k}. Let us define

Ω^{1} = {ω ∈ Ω_{k} | ιω = ω}

and

Ω^{2} = {ω ∈ Ω_{k} | ιω ≠ ω}.

Since all the characters in S′_{k} are absolutely irreducible and in view of Lemma 6.2, Ω^{1} is precisely the set of orbits of irreducible orthogonal and symplectic characters.

Proposition 7.1 Suppose that

det_{ω,ω′∈Ω_{k}}(d_{ω,ω′}) ≡ 1 (mod 2).

Then

det_{ω,ω′∈Ω^{1}}(d_{ω,ω′}) ≡ 1 (mod 2)

and

det_{ω,ω′∈Ω^{2}}(d_{ω,ω′}) ≡ 1 (mod 2).

Proof. Since the group {1, ι} acts on Ω with fixed points precisely Ω^{1}, it follows from lemma 4.3 that

det_{ω,ω′∈Ω_{k}}(d_{ω,ω′}) ≡ det_{ω,ω′∈Ω^{1}}(d_{ω,ω′})det_{ω,ω′∈Ω^{2}}(d_{ω,ω′}) (mod 2).

Hence we have

det_{ω,ω′∈Ω^{1}}(d_{ω,ω′}) ≡ 1 (mod 2),

and

det_{ω,ω′∈Ω^{2}}(d_{ω,ω′}) ≡ 1 (mod 2).

This completes the proof of the proposition.

We define Ω^{1,o} = {ω ∈ Ω^{1} | ω orthogonal}, and Ω^{1,s} = {ω ∈ Ω^{1} | ω symplectic}.

Proposition 7.2 Suppose that det_{ω,ω′∈Ω^{1}}(d_{ω,ω′}) ≡ 1 (mod 2). Then

det_{ω,ω′∈Ω^{1,o}}(d_{ω,ω′}) ≡ 1 (mod 2),

and

det_{ω,ω′∈Ω^{1,s}}(d_{ω,ω′}) ≡ 1 (mod 2).

Proof. Let ω be orthogonal and ω′ symplectic. For x ∈ ω, y ∈ ω′, recall that U_{x} and U_{y} are the simple k[S]–modules associated to x and y respectively. Then ρ\overline{y} is supported on U\overline{y} = U_{y} ⊕ U_{y} and ρ\overline{x} is supported on U\overline{x} = U_{x}, hence the \overline{y}–component of Res_{S}^{G} Ind_{S}^{G}(U_{x}, ρ\overline{x}) is isomorphic to (U_{y} ⊕ U_{y}, ρ\overline{y}) ⊗_{k} F_{\overline{ω},\overline{ω}′}. Thus the module U_{y} occurs with even multiplicity in Res_{S}^{G} Ind_{S}^{G}(U_{x}), so that d_{ω,ω′} ≡ 0 (mod 2). Therefore the matrix (d_{ω,ω′})_{ω,ω′∈Ω^{1}} has the shape

( \begin{matrix}A & 0 \\ ∗ & B \end{matrix} ) .

mod 2, where A = det_{ω,ω′∈Ω^{1,o}}(d_{ω,ω′}) and B = det_{ω,ω′∈Ω^{1,s}}(d_{ω,ω′}). This completes the proof of the proposition.

For any ω ∈ Ω_{k}, recall that \overline{ω} = ω if ιω = ω, and \overline{ω} = (ω_{1}, ω_{2}) with ιω_{1} = ω_{2} and ω_{1} ≠ ω_{2}. Let \overline{Ω}_{k} be the set of all \overline{ω} with ω ∈ Ω_{k}. Let

\overline{Ω}^{2} = \{\overline{ω} = (ω_{1}, ω_{2}) ∈ \overline{Ω} | ιω_{1} = ω_{2} and ω_{1} ≠ ω_{2}\}.

Proposition 7.3 Suppose that det_{ω,ω′∈Ω^{2}}(d_{ω,ω′}) ≡ 1 (mod 2). Then we have

det_{\overline{ω},\overline{ω}′∈\overline{Ω}^{2}}(d_{\overline{ω},\overline{ω}′}) ≡ 1 (mod 2).

Proof. Let \overline{ω} = (ω_{1}, ω_{2}) and \overline{ω}′ = (ω′_{1}, ω′_{2}), and let d_{ω_{1},ω′_{1}} = a, d_{ω_{1},ω′_{2}} =

b. Then d_{\overline{ω},\overline{ω}′} = a + b. For a suitable ordering of the orbits \overline{Ω}^{2}, and the corresponding ordering of Ω^{2}, the matrices d_{\overline{ω},\overline{ω}′} and d_{ω,ω′} are of the shape B and A as in corollary 4.2. Hence det(B)^{2} ≡ det(A) (mod 2). This gives the desired result.

We have the following

Proposition 7.4 Suppose that G has the odd determinant property. Then

det_{ω,ω′∈Ω_{k}}(d_{ω,ω′}) ≡ 1 (mod 2).

For the proof of prop. 7.4, we need the following lemma Lemma 7.5 Let K be a field of characteristic 0, and assume that all the characters in S′_{K} are absolutely irreducible. Suppose that G has the odd determinant property. Then

det_{ω,ω′∈Ω_{K}} (d_{ω,ω′}) ≡ 1 (mod 2).

Proof. Let X_{K} = X ⊗_{Z} K be the vector space of K–valued functions on S invariant under conjugation by N. For all ω ∈ Ω_{K}, set p_{ω} = Σ_{x∈ω}x. Note that as all the characters in S′_{K} are absolutely irreducible, the set (p_{ω})_{ω∈Ω_{K}} is a basis of X_{K}.

Let Φ : X_{K} → X_{K} be Res_{S}^{G} Ind_{S}^{G} considered as an endomorphism of X_{K}. Note that we have

Φ(p_{ω}) = (♯ω)Σ_{ω∈Ω_{K}} d_{ω,ω′}p_{ω′}.

This implies that the matrix of Φ in the basis (p_{ω})_{ω∈Ω_{K}} is equal to ((♯ω)d_{ω,ω′}).

On the other hand, the odd determinant property implies that the determinant of Φ : X_{Z} → X_{Z} is odd (cf. §2). Hence the determinant of Φ : X_{K} → X_{K} is also odd. Note that ♯ω is odd for all ω ∈ Ω (see lemma 6.1). This implies that det_{ω,ω′∈Ω_{K}} (d_{ω,ω′}) is odd, hence the lemma is proved.

Proof of prop. 7.4 Note that for any field E and any ω, ω′ ∈ Ω_{E}, we have

d_{ω,ω′} = ⟨x, Res_{S}^{G} Ind_{S}^{G}x′⟩_{S} = ⟨Ind_{S}^{G}x, Ind_{G}^{S}x′⟩_{G}

for any x ∈ ω, x′ ∈ ω′.

If char(k) = 0, then the proposition follows from lemma 7.5. Suppose that char(k) > 0. Let A be a complete discrete valuation ring of characteristic 0 with residue field k, and let π be a uniformizer of A. Let K be the field of fractions of A. Then all the characters in S′_{K} are absolutely irreducible. Indeed, we have k[S] = ∏_{1≤i≤r} M_{n_{i}}(k), where r is the number of irreducible representations of S over k. Since A[S] is complete with respect to the ideal πA[S], the isomorphism A[S]/πA[S] → ∏_{1≤i≤r} M_{n_{i}}(k) can be lifted to an isomorphism A[S] ≃ ∏_{1≤i≤r} M_{n_{i}}(A), hence we have K[S] ≃ ∏_{1≤i≤r} M_{n_{i}}(K). Thus every character of S′_{K} is absolutely irreducible, hence by lemma 7.5 we have

det_{ω,ω′∈Ω_{K}} (d_{ω,ω′}) ≡ 1 (mod 2).

Let us show that the matrices (d_{ω,ω′})_{ω,ω′∈Ω_{k}} and (d_{ω,ω′})_{ω,ω′∈Ω_{K}} are equal for suitable orderings of the sets Ω_{k} and Ω_{K}. As S is a 2–group and char(k) ≠ 2, every k[S]–module is projective. If P is a projective k[S]–module, then Ind_{S}^{G}(P) is projective as well.

Let P be a projective k[S]–module. Since A[S] is π–adically complete, there is a projective A[S]–module P̃ such that P̃/πP̃ ≃ P. Then P̃_{K} = P̃ ⊗_{A} K is a projective K[S]–module. Moreover, P is simple if and only if P̃_{K} is simple. Note that if P and Q are simple k[S]–modules, then we have

⟨Ind_{S}^{G}(P), Ind_{S}^{G}(Q)⟩_{G} = ⟨Ind_{S}^{G}P̃_{K}, Ind_{S}^{G}Q̃_{K}⟩_{G}.

Therefore the matrices (d_{ω,ω′})_{ω,ω′∈Ω_{k}} and (d_{ω,ω′})_{ω,ω′∈Ω_{K}} are equal for suitable orderings of the sets Ω_{k} and Ω_{K}, and this completes the proof of the proposition.

§8. Odd determinant property–behavior under quadratic extension

This section contains a quadratic descent argument. Together with a filtration introduced in §9, this quadratic descent will enable us to reduce to the case where all the characters are absolutely irreducible, cf. §7. Putting these informations together in §10, we obtain a result (th. 10.1) that will be used in §11 to prove th. 2.2. We start by recalling and introducing some notation that will be needed in this section and the next ones.

Let G be a finite group and let S be a 2–Sylow subgroup of G. For any field E with char(E) ≠ 2, we denote by S′_{E} the set of irreducible characters of S over E, and by Ω_{E} be the set of orbits of S′_{E} under the action of N = N_{G}(S). Recall that ι : E[S] → E[S] is the standard involution, and that for x ∈ S′_{E} we denote \overline{x} = x if x is selfdual, and \overline{x} = (x, x′) if ι(x) = x′ ≠ x.

For any ω ∈ Ω_{E}, recall that \overline{ω} = ω if the characters of ω are invariant under ι, and \overline{ω} = (ω_{1}, ω_{2}) if there exist x_{1} ∈ ω_{1} and x_{2} ∈ ω_{2} such that ι(x_{1}) = x_{2} with x_{1} ≠ x_{2}. Let \overline{Ω}_{E} be the set of all \overline{ω} with ω ∈ Ω_{E}, and let K0 ; \overline{ω} = K0 ; \overline{x} for \overline{x} ∈ \overline{ω}.

Let us define d_{ω,ω′} as being the number of times U_{y} occurs in

Res_{S}^{G} Ind_{S}^{G}(U_{x})

for x ∈ ω, y ∈ ω′.

Let us recall that for all ω, ω′ ∈ Ω_{k} such that K0 ; \overline{ω} = K_{\overline{ω}′}^{0} = k, we denote by d_{\overline{ω},\overline{ω}′} the dimension of the k–vector space underlying the quadratic form F_{\overline{ω},\overline{ω}′} (see §6).

Set

Ω_{E}^{0} = \{ω ∈ Ω_{E} | K0 ; \overline{ω} = E\}

\overline{Ω}_{E}^{0} = \{\overline{ω} ∈ \overline{Ω}_{E} | K0 ; \overline{ω} = E\}

Ω_{E}^{1} = {ω ∈ Ω_{E}^{0} | ω orthogonal or symplectic}

Ω_{E}^{1,o} = {ω ∈ Ω_{E}^{0} | ω orthogonal}

Ω_{E}^{1,s} = {ω ∈ Ω_{E}^{0} | ω symplectic}

Ω_{E}^{2} = {ω ∈ Ω_{E}^{0} | ω unitary}

\overline{Ω}_{E}^{2} = \{\overline{ω} ∈ \overline{Ω}_{E}^{0} | \overline{ω} unitary\}

and

δ_{E}^{1} = det_{ω,ω′∈Ω_{E}^{1}}(d_{ω,ω′})

δ_{E}^{2} = det_{ω,ω′∈Ω_{E}^{2}}(d_{ω,ω′})

\overline{δ}_{E}^{2} = det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{E}^{2}}(d_{\overline{ω},\overline{ω}′})

δ_{E}^{1,o} = det_{ω,ω′∈Ω_{E}^{1,o}}(d_{ω,ω′})

δ_{E}^{1,s} = det_{ω,ω′∈Ω_{E}^{1,s}}(d_{ω,ω′})

Let L/K be a quadratic extension, and let τ : L → L be the non–trivial automorphism of L/K. Then τ acts on S′_{L} by (τx)(s) = τ(x(s)) for all s ∈ S and x ∈ S′_{L}. This induces an action of τ on Ω_{L}.

Proposition 8.1 Let ω ∈ Ω_{L}. Then τω = ω if and only if there is a character x ∈ S′_{L} with x ∈ ω such that τx = x.

Proof. If there exists x ∈ ω such that τx = x, then we have τω = ω. Conversely, suppose that ω ∈ Ω_{L} is such that τω = ω. If we had τx ≠ x for every x ∈ ω, then ♯(ω) would be even, contradicting lemma 6.1. This implies that there exists x ∈ ω with τx = x, hence the proposition is proved.

Note that τ acts on the center of L[S], and that the action of τ on S′_{L} can be described in terms of this action. This leads to the following observation, which will be used in the sequel:

Lemma 8.2 Let x ∈ S′_{L} be an orthogonal or symplectic character such that τx = x. Let L = K_{x}. Then

(i) There exists x_{0} ∈ S′_{K} such that K_{x_{0}} = K.

(ii) If x is orthogonal (resp. symplectic) then x_{0} is orthogonal (resp. symplectic).

(iii) If x_{0} is orthogonal, then (x_{0})_{L} = x

(iv) If x_{0} is symplectic, then (x_{0})_{L} = x or (x_{0})_{L} = 2x.

Proof. For any field E, let us denote by Z(E[S])) the center of E[S].

The Galois automorphism τ : L → L over K acts on L[S], hence also on Z(L[S]). Then the subalgebra of Z(L[S]) fixed by τ is equal to Z(K[S]). The hypothesis implies that L is one of the factors in the decomposition of Z(L[S]). Note that the restriction of τ to the factor L in Z(L[S]) is non–trivial, and that the fixed field is equal to K. This corresponds to a factor in the decomposition of K[S], and hence to a character x_{0} of S′_{K}. This proves (i). Noting that the base change to L of the factor corresponding to x_{0} in K[S] is the factor corresponding to (x_{0})_{L}, points (ii) and (iii) are immediate. Suppose now that x is symplectic. Then the same reasoning proves that if Δ_{x_{0}} is a quaternion division algebra and Δ_{x} = L, then (x_{0})_{L} = 2x; if both Δ_{x_{0}} and Δ_{x} are quaternion division algebras, or if Δ_{x_{0}} = K and Δ_{x} = L, then (x_{0})_{L} = x. This proves (iv).

Corollary 8.3 Let ω ∈ Ω_{L}^{1} be such that τω = ω.

(i) If ω ∈ Ω_{L}^{1,o}, there exists ω_{0} ∈ Ω_{K}^{1,o} such that (ω_{0})_{L} = ω.

(ii) If ω ∈ Ω_{L}^{1,s}, then there exists ω_{0} ∈ Ω_{K}^{1,s} such that (ω_{0})_{L} = ω or (ω_{0})_{L} = 2ω.

Proof. By prop. 8.1 we can choose x ∈ ω such that τx = x. Let x_{0} ∈ S′_{K} such that K_{x_{0}} = K (see lemma 8.2 (i)). Hence we have K_{(x_{0})} ⊗_{K} L = L = K_{x}.

(i) Suppose that ω ∈ Ω_{L}^{1,o}. Then x is orthogonal. Hence x_{0} is orthogonal, and (x_{0})_{L} = x (cf. 8.2 (ii) and (iii)). Let ω_{0} be the orbit of x_{0}; then ω_{0} ∈ Ω_{K}^{1,o} and (ω_{0})_{L} = ω. (ii) Suppose that ω ∈ Ω_{L}^{1,s}. Then x is symplectic. Hence x_{0} is symplectic, and (x_{0})_{L} = x or (x_{0})_{L} = 2x (cf. 8.2 (ii) and (iv)). Let ω_{0} be the orbit of x_{0}. Then ω_{0} ∈ Ω_{K}^{1,s} has the required property.

Proposition 8.4 Suppose that

δ_{L}^{1,o} ≡ 1 (mod 2).

Then

δ_{K}^{1,o} ≡ 1 (mod 2).

Proof. The automorphism τ of L/K induces a permutation of Ω_{L}^{1,o}. Set

Ω_{L}^{1,1} = {ω ∈ Ω_{L}^{1,o} |τω = ω}

and

Ω_{L}^{1,2} = {ω ∈ Ω_{L}^{1,o} |τω ≠ ω}.

Let S_{L} be the group of permutations of Ω_{L}^{1,o}, and let S_{L}^{1} respectively S_{L}^{2} be the group of permutations of Ω_{L}^{1,1}, respectively Ω_{L}^{1,2}, regarded as subgroups of S_{L}. Set

α = Σ_{s∈S_{L}^{1}}( ∏_{ω∈Ω_{L}^{1,1}} d_{sω,ω}),

β = Σ_{s∈S_{L}^{2}}( ∏_{ω∈Ω_{L}^{1,2}} d_{sω,ω}),

By lemma 4.3, we have

δ_{L}^{1,o} ≡ αβ (mod 2).

On the other hand, we have

δ_{K}^{1,o} ≡ α (mod 2).

Indeed, by cor. 8.3 (i) the map ω_{0} ↦ (ω_{0})_{L} induces a bijection between Ω_{K}^{1,o} and Ω_{L}^{1,1} with d_{ω_{0},ω′_{0}} = d_{(ω_{0})_{L},(ω′_{0})_{L}} for ω_{0}, ω′_{0} ∈ Ω_{K}^{1,o}. It follows that δ_{K}^{1,o} ≡ α (mod 2). This completes the proof of the proposition.

Proposition 8.5 Suppose that

δ_{L}^{1,s} ≡ 1 (mod 2).

Then

δ_{K}^{1,s} ≡ 1 (mod 2).

Proof. The automorphism τ of L/K induces a permutation of Ω^{1,s}. Let

Ω_{L}^{1,1} = {w ∈ Ω_{L}^{1,s} |τω = ω},
Ω_{L}^{1,2} = {w ∈ Ω_{L}^{1,s} |τω ≠ ω}.

Let S_{L} be the group of permutations of Ω_{L}^{1,s}, and let S_{L}^{1} respectively S_{L}^{2} be the group of permutations of Ω_{L}^{1,1}, respectively Ω_{L}^{1,2}, regarded as subgroups of S_{L}. Set

α = Σ_{s∈S_{L}^{1}}( ∏_{ω∈Ω_{L}^{1,1}} d_{sω,ω}),

γ = Σ_{s∈S_{L}^{2}}( ∏_{ω∈Ω_{L}^{1,2}} d_{sω,ω}).

Arguing as in 8.4, we get

δ_{L}^{1,s} ≡ γα (mod 2).

Claim. We have

δ_{K}^{1,s} ≡ α (mod 2).

Let us write

Ω_{L}^{1,1} = Ω_{L/K}^{s} ∪ Ω_{L/K}^{ns},

where

Ω_{L/K}^{s} = {ω ∈ Ω_{L}^{1,1} | there exists ω_{0} ∈ Ω_{K}^{1,s} with (ω_{0})_{L} = 2ω},

and

Ω_{L/K}^{ns} = {ω ∈ Ω_{L}^{1,1} | there exists ω_{0} ∈ Ω_{K}^{1,s} with (ω_{0})_{L} = ω}.

By corollary 8.3 (ii) the above is a disjoint union decomposition.

For ω ∈ Ω_{L/K}^{s} with 2ω = (ω_{0})_{L} and ω′ ∈ Ω_{L/K}^{ns} with ω′ = (ω′_{0})_{L}, ω_{0}, ω′_{0} ∈ Ω_{K}, we have: If d_{ω′_{0},ω_{0}} = r, then d_{ω′,ω} = 2r. Thus the matrix (d_{ω,ω′})_{ω,ω′∈Ω_{L}^{1}} is congruent to the matrix

( \begin{matrix}A & ∗ \\ 0 & B \end{matrix} )

where A = (d_{ω,ω′})_{ω,ω′∈Ω_{L/K}^{s}}, and B = (d_{ω,ω′})_{ω,ω′∈Ω_{L/K}^{ns}}. Therefore

α ≡ det(A)det(B) (mod 2).

We next determine δ_{K}^{1,s} (mod 2). Let us write

Ω_{K}^{1,s} = Ω_{K/L}^{s} ∪ Ω_{K/L}^{ns},

with

Ω_{K/L}^{s} = {ω ∈ Ω_{K} | there exists ω_{0} ∈ Ω_{L}^{1,1} with (ω)_{L} = 2ω_{0}},

and

Ω_{K/L}^{ns} = {ω ∈ Ω_{K} | there exists ω_{0} ∈ Ω_{L}^{1,1} with (ω)_{L} = ω_{0}}.

Documenta Mathematica 16 (2011) 677–707

For ω, ω′ ∈ Ω_{K/L}^{s}, if ω_{L} = 2ω_{0}, ω′_{L} = 2ω′_{0} for some ω_{0}, ω′_{0} ∈ Ω_{L/K}^{s}, we have d_{ω,ω′} = d_{ω_{0},ω′_{0}}. Also, for ω, ω′ ∈ Ω_{K/L}^{ns}, if ω_{L} = ω_{0}, ω′_{L} = ω′_{0} for some ω_{0}, ω′_{0} ∈ Ω_{L/K}^{ns}, we have d_{ω,ω′} = d_{ω_{0},ω′_{0}}. Thus the matrix (d_{ω,ω′})_{ω,ω′∈Ω_{K/L}^{s}} is equal to A, and the matrix (d_{ω,ω′})_{ω,ω′∈Ω_{K/L}^{ns}} is equal to B. Further, if ω ∈ Ω_{K/L}^{s}, then

ω_{L} = 2ω_{0} for some ω_{0} ∈ Ω_{L}^{1,1}, and if ω′ ∈ Ω_{K/L}^{ns}, then ω′_{L} = ω′_{0} for some ω′_{0} ∈ Ω_{L}^{1,1}. If d_{ω_{0},ω′_{0}} = r, then d_{ω,ω′} = 2r. Thus the matrix (d_{ω,ω′})_{ω,ω′∈Ω_{K}^{1,s}} is congruent to

( \begin{matrix}A & 0 \\ ∗ & B \end{matrix} ) .

mod 2. Therefore

δ_{K}^{1,s} = det(A)det(B) ≡ α (mod 2),

and this completes the proof of the claim. Therefore we see that δ_{L}^{1,s} ≡ 1 (mod 2) implies δ_{K}^{1,s} ≡ 1 (mod 2), hence prop. 8.5 is proved.

Let us recall that for any field E of characteristic ≠ 2, we denote by \overline{S}′_{E} the set of \overline{x} for x ∈ S′_{E}.

The Galois automorphism τ : L → L over K induces an action on \overline{S}′_{L} which we denote by \overline{y} ↦ τ\overline{y}.

Lemma 8.6 Let \overline{y} ∈ \overline{S}′_{L} with \overline{y} unitary such that τ\overline{y} = \overline{y}. Then there exists \overline{x} ∈ \overline{S}′_{K} with \overline{x} unitary such that \overline{x}_{L} = \overline{y}. Moreover, if K0 ; \overline{y} = L, then K0 ; \overline{x} = K.

Proof. Suppose that \overline{y} = y and τy = y. Then by the method of lemma 8.2 we see that there is a unitary character x ∈ S′_{K} such that x_{L} = y. Moreover, τ restricts to a non–trivial automorphism of K_{y} which commutes with ι. If E = (K_{y})^{ι}, then τ|E is non–trivial and K_{x} = E. Since K_{x_{L}} = EL = K_{y}, we have x_{L} = y. Further, if K_{y}^{0} = L, then K_{x}^{0} = K.

Suppose that \overline{y} = (y, ιy) with ιy ≠ y and τ\overline{y} = \overline{y}. Then τy = y or τιy = y. Set M = K_{y} × K_{ιy}.

Suppose first that τy = y. Then there is an x ∈ S′_{K} such that x_{L} = y. Further, τ induces an automorphism on M which takes each factor K_{y} and K_{ιy} in itself. Moreover, we have M^{τ} = K_{x} × K_{ιx}. Thus \overline{x} = (x, ιx) ∈ \overline{S}′_{K} is unitary with \overline{x}_{L} = \overline{y}. Moreover, if K_{y}^{0} = M^{ι} = L, then K_{x}^{0} = K.

Suppose now that τιy = y. Then τy = ιy, and τ switches the factors K_{y} and K_{ιy} of M. Let E = M^{τ}. Then E is a field which is a factor of the center of K[S] and ι restricted to E is non–trivial. Let x ∈ S′_{K} be the character associated to E. Then x is unitary, K_{x} = E and K_{x_{L}} = EL = M = K_{y} × K_{ιy}. Thus x_{L} = (y, ιy). Further, if K0 ; \overline{y} = M^{ι} = L, then K0 ; \overline{x} = E^{ι} = K, where \overline{x} = x.

The automorphism τ induces an action on \overline{Ω}_{L}^{2} that we denote by \overline{ω} ↦ τ\overline{ω}.

Corollary 8.7 Let \overline{ω} ∈ \overline{Ω}_{L}^{2} be a unitary orbit with τ\overline{ω} = \overline{ω}. Then there is a unitary orbit \overline{ω}_{0} ∈ \overline{Ω}_{K}^{2} such that (\overline{ω}_{0})_{L} = \overline{ω}.

Proof. Suppose that \overline{ω} = ω and τω = ω. By proposition 8.1, there is a character y ∈ S′_{L} belonging to ω with τy = y. In this case, the proposition follows from lemma 8.6.

Suppose that \overline{ω} = (ω, ιω) with ιω ≠ ω. Then τ\overline{ω} = \overline{ω} implies that τω = ω or τιω = ω.

Suppose first that τω = ω. Then by proposition 8.1 there is a y ∈ ω such that τy = y. Further, \overline{y} = (y, ιy) is unitary with τ\overline{y} = \overline{y}. In this case, we appeal to lemma 8.6 to conclude the proof.

Suppose now that τιω = ω and that τω ≠ ω. Then τι induces an action on the characters in ω. As ♯ω is odd by lemma 6.1, there exists y ∈ ω such that τιy = y and τy ≠ y, since τω ≠ ω. Then \overline{y} = (y, ιy) is a unitary pair with τ\overline{y} = \overline{y} and the proposition follows from lemma 8.6.

Proposition 8.8 If det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{L}^{2}}(d_{\overline{ω},\overline{ω}′}) ≡ 1 (mod 2), then det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{K}^{2}} (d_{\overline{ω},\overline{ω}′}) ≡ 1 (mod 2).

Proof. Recall that τ : L → L is the non–trivial automorphism of L/K. Let us write \overline{Ω}_{L}^{2} = Ω_{L}^{2,1} ∪ Ω_{L}^{2,2}, where

Ω_{L}^{2,1} = \{\overline{ω} ∈ \overline{Ω}_{L} |τ\overline{ω} = \overline{ω}\},

and

Ω_{L}^{2,2} = \{\overline{ω} ∈ \overline{Ω}_{L} |τ\overline{ω} ≠ \overline{ω}\}.

Arguing as in 8.4 and using 4.3, we get

det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{L}^{2}}(d_{\overline{ω},\overline{ω}′}) = [det_{\overline{ω},\overline{ω}′∈Ω_{L}^{2,1}}(d_{\overline{ω},\overline{ω}′})][det_{\overline{ω},\overline{ω}′∈Ω_{L}^{2,2}}(d_{\overline{ω},\overline{ω}′})].

Note that scalar extension induces a bijection between \overline{Ω}_{K}^{2} and Ω_{L}^{2,1}, and we have d_{\overline{ω},\overline{ω}′} = d_{\overline{ω}_{L},\overline{ω}′_{L}} for \overline{ω}, \overline{ω}′ ∈ \overline{Ω}_{K}^{2}. This proves the proposition.

§9. A filtration

Let k be a field of characteristic ≠ 2, let G be a finite group and let S be a 2–Sylow subgroup of G. In this section, we introduce a quadratic filtration of the field k that will be needed in the next two sections.

Let κ be the prime field of k, that is, κ = Q if char(k) = 0 and κ = F_{p} if char(k) = p > 0. Note that k[S] = κ[S] ⊗_{κ} k, hence it is interesting to investigate the structure of κ[S] in both cases.

Suppose first that κ = Q. We have the following lemma:

Lemma 9.1 Let S be a 2–group, and let Q[S] = ∏_{i=1,...,r} M_{n_{i}}(D_{i}) where the D_{i}’s are division algebras, and let Z(D_{i}) = K_{i}. Let ι : Q[S] → Q[S] be the standard involution. Then each component of Q[S] is invariant under ι. Let us denote by K_{i}^{0} the invariant subfield of K_{i} under the restriction of ι to K_{i}. Then there exists m ∈ N such that K_{i}^{0} is a subfield of the real 2–cyclotomic subfield Q(ζ_{2^{m}} + ζ_{2^{m}}^{−1}).

Proof. The fact that each component of Q[S] is invariant under ι follows from [6], Chap. 8, 13.2. We know that as S is a 2–group, there exists m ∈ N such that for all i = 1, …, r the field K_{i} is a subfield of the cyclotomic field Q(ζ_{2^{m}}). The standard involution ι : Q[S] → Q[S] is positive definite, hence its restriction to each component is positive definite as well. This implies (cf [6], Chap 8, 13.5) that K_{i}^{0} ⊂ R for all i. Hence for all i = 1, …, r, we have K_{i}^{0} ⊂ Q(ζ_{2^{m}} + ζ_{2^{m}}^{−1}) as claimed.

With the notation of lemma 9.1, let L = Q(ζ_{2^{m}} + ζ_{2^{m}}^{−1}). Since L/Q is cyclic of degree a power of 2, it has a unique set of subfields which fit into a filtration

L_{0} = Q ⊂ L_{1} ⊂ L_{2} ⊂ … ⊂ L_{s} = L

with all inclusions being strict, and L_{i}/L_{i−1} of degree 2.

Suppose now that κ = F_{p} for some prime number p. We have

F_{p}[S] = ∏_{i=1,...,r} M_{n_{i}}(K_{i})

where the K_{i}’s are finite degree extensions of F_{p}. As S is a 2–group, the degrees of these extensions are powers of 2. There exists a finite extension L/F_{p} of degree a power of 2 containing all the K_{i}^{0}’s. Note that as F_{p} is a finite field, the extension L/F_{p} is cyclic. Hence in this case too, we have a unique set of subfields of L which fit into a filtration

L_{0} = Q ⊂ L_{1} ⊂ L_{2} ⊂ … ⊂ L_{s} = L

with all inclusions being strict, and L_{i}/L_{i−1} of degree 2.

Let

k_{0} = k ⊂ k_{1} ⊂ k_{2} ⊂ … ⊂ k_{t} = Lk

be the induced strict filtration of Lk/k. Note that every subfield of Lk containing k is one of the fields k_{i}. Let k_{r} be the smallest of these fields containing

K0 ; \overline{x} for all \overline{x} ∈ S′_{k}.

§10. The odd determinant property revisited

For any field E, set

\overline{δ}_{E}^{1,o} = det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{E}^{1,o}}(d_{\overline{ω},\overline{ω}′})

\overline{δ}_{E}^{1,s} = det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{E}^{1,s}}(d_{\overline{ω},\overline{ω}′})

\overline{δ}_{E}^{2} = det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{E}^{2}}(d_{\overline{ω},\overline{ω}′})

d_{E}^{0} = det_{\overline{ω},\overline{ω}′∈\overline{Ω}_{E}^{0}}(d_{\overline{ω},\overline{ω}′}).

The result below will be instrumental in the proof of th. 2.2 in the next section:

Theorem 10.1 Let G be a finite group having the odd determinant property. Then for any field K of characteristic not 2, we have

d_{K}^{0} ≡ 1 (mod 2).

Proof. We first treat the case where all the characters in S′_{K} are absolutely irreducible. The reduction to this case is via the filtration introduced in §9, and the quadratic descent of §8.

Suppose first that all the characters in S′_{K} are absolutely irreducible. For x ∈ S′_{K}, the form ρ\overline{x} is supported on U_{x} if x is orthogonal, on U_{x} ⊕ U_{x} if x is symplectic, and U_{x_{1}} ⊕ U_{x_{2}} if \overline{x} = (x_{1}, x_{2}) with ι(x_{1}) = x_{2} and x_{1} ≠ x_{2}. Noting that for a general K, the integers d_{\overline{ω},\overline{ω}′} can be computed after base changing to an algebraic closure of K, we get the following:

1) d_{\overline{ω},\overline{ω}′} = d_{ω,ω′} if ω, ω′ ∈ Ω_{K}^{1,s};

2) d_{\overline{ω},\overline{ω}′} = 2d_{ω,ω′} if ω is symplectic and ω′ is not symplectic;

3) d_{\overline{ω},\overline{ω}′} = 2d_{ω,ω′} if ω is unitary and ω′ is orthogonal;

4) d_{\overline{ω},\overline{ω}′} = d_{ω,ω′} if ω, ω′ are orthogonal.

Thus the matrix (d_{\overline{ω},\overline{ω}′})_{\overline{ω},\overline{ω}′∈\overline{Ω}_{K}^{0}} has the following shape modulo 2

( \begin{matrix}A & ∗ & ∗ \\ 0 & B & ∗ \\ 0 & 0 & C \end{matrix} ) ,

where

A = (d_{\overline{ω},\overline{ω}′})_{\overline{ω},\overline{ω}′∈\overline{Ω}_{K}^{1,o}}
B = (d_{\overline{ω},\overline{ω}′})_{\overline{ω},\overline{ω}′∈\overline{Ω}_{K}^{2}}
C = (d_{\overline{ω},\overline{ω}′})_{\overline{ω},\overline{ω}′∈\overline{Ω}_{K}^{1,s}}.

Thus d_{K}^{0} ≡ 1 (mod 2) if and only if \overline{δ}_{K}^{1,0} = det(A) ≡ 1 (mod 2), \overline{δ}_{K}^{1,s} = det(C) ≡ 1 (mod 2), and \overline{δ}_{K}^{2} = det(B) ≡ 1 (mod 2). We also note that for ω, ω′ ∈ Ω_{K}^{1,o}, or for ω, ω′ ∈ Ω_{K}^{1,s}, we have d_{\overline{ω},\overline{ω}′} = d_{ω,ω′}. Therefore \overline{δ}_{K}^{1,o} = δ_{K}^{1,o}, and \overline{δ}_{K}^{1,s} = δ_{K}^{1,s}. Thus d_{K}^{0} ≡ 1 (mod 2) if and only if δ_{K}^{1,o} ≡ 1 (mod 2), δ_{K}^{1,s} ≡ 1 (mod 2), and \overline{δ}_{K}^{2} ≡ 1 (mod 2). There exists a field extension L/K and a filtration by quadratic extensions

K ⊂ K_{2} ⊂ … ⊂ K_{n} = L

such that all characters in S′_{L} are absolutely irreducible (cf. §9). By prop. 7.2 and 7.3, we have δ_{L}^{1,o} ≡ 1 (mod 2), δ_{L}^{1,s} ≡ 1 (mod 2), and \overline{δ}_{L}^{2} ≡ 1 (mod 2). By the quadratic descent results 8.4, 8.5 and 8.8, we get \overline{δ}_{K}^{1,0} = det(A) ≡ 1 (mod 2), \overline{δ}_{K}^{1,s} = det(C) ≡ 1 (mod 2), and \overline{δ}_{K}^{2} = det(B) ≡ 1 (mod 2). Therefore d_{K}^{0} ≡ 1 (mod 2).

§11. Proof of the induction–restriction result The aim of this section is to prove th. 2.2. Let

k_{0} = k ⊂ k_{1} ⊂ k_{2} ⊂ … ⊂ k_{r}

be the filtration introduced in §9, k_{r} being the smallest of these fields containing

K0 ; \overline{x} for all \overline{x}.

Proof of theorem 2.2 Let (V, h) be an S–quadratic form. We have a decomposition (cf. §5)

(V, h) ≃ ⨁_{\overline{x}} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V\overline{x}, g\overline{x}),

Documenta Mathematica 16 (2011) 677–707

where (V\overline{x}, g\overline{x}) is a quadratic form over K0 ; \overline{x} (cf. §5). Recall that the Witt class

(V\overline{x}, g\overline{x}) ⊗ n\overline{x} ∈ W (K0 ; \overline{x})

is uniquely determined by (V, h), where n\overline{x} is the reduced norm of D\overline{x} over K0 ; \overline{x} if D\overline{x} is a quaternion algebra, the norm of K\overline{x} over K0 ; \overline{x} if K\overline{x} is a quadratic algebra, and n\overline{x} = 1 otherwise. We have

Ind_{S}^{G}(V, h) = ⨁_{\overline{ω}∈\overline{Ω}} I(\overline{ω}) ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω}),

where

(V\overline{ω}, g\overline{ω}) = ⨁_{\overline{x}∈\overline{w}}(V\overline{x}, g\overline{x})

is a quadratic space determined up to multiplication by n\overline{ω} = n\overline{x}. We have I(\overline{ω}) = Ind_{S}^{G}(ρ\overline{x}), which does not depend on the choice of \overline{x} ∈ \overline{ω}.

We have

Res_{S}^{G} Ind_{S}^{G}(V, h) = ⨁_{\overline{ω}∈\overline{Ω}} Res_{S}^{G}(I(\overline{ω})) ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω}).

For \overline{y} ∈ \overline{S}′_{k}, the \overline{y}–component of Res_{S}^{G}(I\overline{ω}) is ρ\overline{y} ⊗_{K_{\overline{ω}′}^{0}} F_{\overline{ω},\overline{ω}′}, where \overline{y} ∈ \overline{ω}′, and where F_{\overline{ω},\overline{ω}′} is a quadratic space over K_{\overline{ω}′}^{0}, determined up to multiplication by n\overline{ω}′.

Let (V_{1}, h_{1}) and (V_{2}, h_{2}) be two S–quadratic forms such that

Res_{S}^{G} Ind_{S}^{G}(V_{1}, h_{1}) ≃_{S} Res_{S}^{G} Ind_{S}^{G}(V_{2}, h_{2}).

Let

(V_{1}, h_{1}) ≃ ⨁_{\overline{x}∈\overline{S}′} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V 1 ; \overline{x}, g1 ; \overline{x}) and (V_{2}, h_{2}) ≃ ⨁_{\overline{x}∈\overline{S}′} ρ\overline{x} ⊗_{K0 ; \overline{x}} (V 2 ; \overline{x}, g2 ; \overline{x})

and

(V i ; \overline{ω}, gi ; \overline{ω}) = ⨁_{\overline{x}∈\overline{w}}(V i ; \overline{x}, gi ; \overline{x})

for i = 1, 2.

Note that as the k[S]–modules Res_{S}^{G} Ind_{S}^{G}(V_{1}) and Res_{S}^{G} Ind_{S}^{G}(V_{2}) are isomorphic, the k[G]–modules Ind_{S}^{G}(V_{1}) and Ind_{S}^{G}(V_{2}) are also isomorphic (see for instance [3], cor. 6.8). This implies that dim(V 1 ; \overline{ω}) = dim(V i ; \overline{ω}) for all \overline{ω} ∈ \overline{Ω}.

Claim. We have

n\overline{ω} ⊗_{k} (V 1 ; \overline{ω}, g1 ; \overline{ω}) ≃ n\overline{ω} ⊗_{k} (V 2 ; \overline{ω}, g2 ; \overline{ω}).

For the proof, we distinguish two cases

Case 1. Suppose that K0 ; \overline{x} = k for all \overline{x} ∈ S′_{k}.

Then we have Res_{S}^{G}(I(\overline{ω}))\overline{y} = ρ\overline{y} ⊗_{k} F_{\overline{ω},\overline{ω}′}, where \overline{y} ∈ \overline{ω}′, where F_{\overline{ω},\overline{ω}′} is a quadratic form over k, and n\overline{ω}′ ⊗_{k} F_{\overline{ω},\overline{ω}′} is determined by Res_{S}^{G}(I(\overline{ω}))\overline{y}. Hence

Res_{S}^{G} Ind_{S}^{G}(V_{i}, h_{i}) = ⨁_{\overline{y}∈\overline{S}′_{k}} ρ\overline{y} ⊗_{k} [ ⨁_{\overline{ω}∈\overline{Ω}_{k}} F_{\overline{ω},\overline{ω}′} ⊗_{k} (V i ; \overline{ω}, gi ; \overline{ω})]

Suppose that Res_{S}^{G} Ind_{S}^{G}(V_{1}, h_{1}) ≃ Res_{S}^{G} Ind_{S}^{G}(V_{2}, h_{2}), and set gi ; \overline{ω} = (V i ; \overline{ω}, gi ; \overline{ω}) for i = 1, 2. Then

n\overline{ω}′ ⊗_{k} [⨁_{\overline{ω}∈\overline{Ω}} F_{\overline{ω},\overline{ω}′} ⊗_{k} g1 ; \overline{ω}] ≃ n\overline{ω}′ ⊗_{k} [⨁_{\overline{ω}∈\overline{Ω}} F_{\overline{ω},\overline{ω}′} ⊗_{k} g2 ; \overline{ω}].

Let us denote by f_{\overline{ω},\overline{ω}′} the element of W (k) determined by the quadratic form F_{\overline{ω},\overline{ω}′}, and let (f̃_{\overline{ω},\overline{ω}′}) be the matrix of cofactors of the matrix (f_{\overline{ω},\overline{ω}′}) in the Witt ring W (k). Then the product (f̃_{\overline{ω},\overline{ω}′})(n\overline{ω}′ ⊗_{k} f_{\overline{ω},\overline{ω}′}) is equal to

φ ( \begin{matrix}n_{\overline{ω}_{1}} & 0 & … & 0 \\ 0 & n_{\overline{ω}_{2}} & … & 0 \\ 0 & … & … & n_{\overline{ω}_{n}} \end{matrix} ) ,

a diagonal matrix with diagonal entries φ.n_{\overline{ω}_{j}}, where φ ∈ W (k) is the determinant of the matrix (f_{\overline{ω},\overline{ω}′}). Let vi ; \overline{ω} be the element of W (k) determined by the quadratic form gi ; \overline{ω} = (V i ; \overline{ω}, gi ; \overline{ω}) for i = 1, 2. Then we get

φ.n\overline{ω} ⊗_{k} (v1 ; \overline{ω} − v2 ; \overline{ω}) = 0

in W (k), for every \overline{ω} ∈ \overline{Ω}.

Note that det(dim((f_{\overline{ω},\overline{ω}′})) = dim(det((f_{\overline{ω},\overline{ω}′})), and that

det(dim((f_{\overline{ω},\overline{ω}′})) = det(d_{\overline{ω},\overline{ω}′}) = d_{k}^{0}.

Since G has the odd determinant property, by prop. 10.1 we have d_{k}^{0} ≡ 1 (mod 2). Therefore dim(φ) is odd, hence φ is not a zero divisor in W (k) (see for instance [6], 2.6.5). Therefore we have

n\overline{ω} ⊗_{k} (v1 ; \overline{ω} − v2 ; \overline{ω}) = 0

in W (k), for all \overline{ω} ∈ \overline{Ω}, and hence n\overline{ω} ⊗_{k} (V 1 ; \overline{ω}, g1 ; \overline{ω}) and n\overline{ω} ⊗_{k} (V 2 ; \overline{ω}, g2 ; \overline{ω}) are in the same Witt class. Recall that dim(V 1 ; \overline{ω}) = dim(V 2 ; \overline{ω}) for all \overline{ω} ∈ \overline{Ω}. Hence the quadratic forms n\overline{ω} ⊗_{k} (V 1 ; \overline{ω}, g1 ; \overline{ω}) and n\overline{ω} ⊗_{k} (V 2 ; \overline{ω}, g2 ; \overline{ω}) have the same dimension and are in the same Witt class, therefore we have

n\overline{ω} ⊗_{k} (V 1 ; \overline{ω}, g1 ; \overline{ω}) ≃ n\overline{ω} ⊗_{k} (V 2 ; \overline{ω}, g2 ; \overline{ω})

for all \overline{ω} ∈ \overline{Ω}. This completes the proof of the claim in case 1. General case. Let us consider (V_{i}, h_{i}) ⊗_{k} k_{r}. We have

Res_{S}^{G} Ind_{S}^{G}(V_{1}, h_{1}) ⊗_{k} k_{r} ≃_{S} Res_{S}^{G} Ind_{S}^{G}(V_{2}, h_{2}) ⊗_{k} k_{r}.

Moreover, K0 ; \overline{ω} ⊗_{k} k_{r} ≃ ∏_{α∈Gal(K0 ; \overline{ω}/k)} k_{r}^{α}. The orbit \overline{ω} splits into distinct conjugate orbits over k_{r}. Each \overline{ω} ∈ \overline{Ω}_{k} with K0 ; \overline{ω} = k_{r} occurs as one of the conjugate orbits over k_{r}. Using case 1, we get, for orbits \overline{ω} with K0 ; \overline{ω} = k_{r},

n\overline{ω} ⊗ (V 1 ; \overline{ω}, g1 ; \overline{ω}) ≃ n\overline{ω} ⊗ (V 2 ; \overline{ω}, g2 ; \overline{ω}).

Cancelling these factors, we may assume that

Ind_{S}^{G}(V, h) = ⨁_{\overline{ω}∈\overline{Ω}} I(\overline{ω}) ⊗_{K0 ; \overline{ω}} (V\overline{ω}, g\overline{ω})

with K0 ; \overline{ω} ⊂ k_{r−1} for all \overline{ω} in the above decomposition. Inductively we get, for all \overline{ω}, that

n\overline{ω} ⊗ (V 1 ; \overline{ω}, g1 ; \overline{ω}) ≃ n\overline{ω} ⊗ (V 2 ; \overline{ω}, g2 ; \overline{ω}).

This completes the proof of the theorem.

Bibliography

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[2] E. Bayer–Fluckiger and J–P. Serre, Torsions quadratiques et bases normales autoduales, Amer. J. Math. 116 (1994) 1–64.

[3] D.J. Benson, R.A. Parker, The Green ring of a finite group, J. Algebra 87 (1984), 290-331.

[4] E. Lequeu, Hermitian G-Forms and Induction, in preparation.

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E. Bayer–Fluckiger École Polytechnique Fédérale de Lausanne Epfl/Fsb/Mathgeom/Csag Station 8 1015 Lausanne, Switzerland eva.bayer@epfl.ch R. Parimala Department of Mathematics & Computer Science Emory University Atlanta, GA 30322, USA. parimala@mathcs.emory.edu 708

Documenta Mathematica 16 (2011)