Documenta Math. 677 Galois Algebras, Hasse Principle and Induction–Restriction Methods

Keywords: Dual Normal Bases ; Forms Reference EPFL-ARTICLE-176135View record in Web of Science Record created on 2012-04-11, modified on 2017-10-19

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 Lv 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 ni (D i ), where D 1 , . . ., D r are division algebras.Let us denote by K i the center of D i , and let D op i be the opposite algebra of D i .
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 ni (D i ), or a product M ni (D i )×M ni (D op i ), with M ni (D i ) and M ni (D op i ) 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 ni (D i ) with ι|K i not the identity, or a product M ni (D i ) × M ni (D op i ) .Otherwise, the factor is said to be of the first kind.In this case, the component is of the form M ni (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 ni (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 G S (V, q) is a G-quadratic form; If (V, q) is a G-quadratic form, then Res G S (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 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 G S (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 G S Ind G S 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 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 Squadratic 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 ′ .

Documenta Mathematica 16 (2011) 677-707
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.
In view of the fusion hypothesis, there exists n ∈ N such that ngxg 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 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 Therefore it suffices to check that #CG(x) #CS(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.
. 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 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 If v is the real place of Q, then every orthogonal and unitary component of [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 Recall that by hypothesis the G-forms (L, q L ) and (L ′ , q L ′ ) are isomorphic over all the completions of k.This implies are isomorphic over all the completions of k ′ .By corollary 3.7, this implies that the S forms Res ), 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 : where det(f F ) is the determinant of f considered as an F -linear map.

Corollary 4.2 Let
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 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 Proof.Let S be the set of permutations of {1, 2, ...n}.For s ∈ S and 1 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 and Then we have This completes the proof of the lemma.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 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 nx (∆ x ) of k[S], where ∆ x is a division algebra.Let K x = Z(∆ x ) be the center of ∆ 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 ∆ nx 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 The involution ι of k[S] restricts to the factors M nx (∆ 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 ∆ x = K x .In this case, we set D x = K x , and we chose the involution τ x : D x → D x to be the identity.The restriction of the involution ι to this factor is adjoint to a symmetric form on D nx x which we Documenta Mathematica 16 (2011) 677-707 denote by ρ x .We define m x = n x , and the symmetric form is supported on the simple module U x .
If x is symplectic, then ∆ x = K x or a quaternion division algebra.We set D x = M 2 (K x ) in the first case, and D x = ∆ x in the second case.In both cases, we choose the involution τ x : D x → D x to be the standard symplectic involution of D x .In this case, the restriction of the involution ι to this factor is adjoint to a hermitian form over D mx x with respect to the involution τ x which we denote by ρ x .The form ρ x is supported on the module and K x is a quadratic algebra over K 0 x .We set D x = K x , and we fix the involution τ x : D x → D 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 D nx x with respect to the involution τ x which we denote by ρ x .We set m x = n x in this case.The form ρ x is supported on U x if x is self-dual, and on Therefore in all cases we have a hermitian form ρ x : U x × U x → D x which we fix throughout.We denote the hermitian form (U x , ρ x ) by ρ x .
We also fix a quadratic form n x : D x → K 0 x to be the one-dimensional unit form if x is orthogonal, the reduced norm form of the quaternion algebra D x if x is symplectic, and the norm form of the quadratic algebra D 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 x , Q x ) for x ∈ S ′ k , over M mx (D x ) with respect to the restriction of ι to this factor.By Morita theory, fixing ρ x , the hermitian form (M x , Q x ) is uniquely determined up to isomorphism by a hermitian form h x over a free D x -module W x of finite rank with respect to the involution τ x , and conversely the hermitian form (W x , h x ) is uniquely determined up to isomorphism by (M x , Q x ).Moreover, by Jacobson's theorem the hermitian form (W x , h x ) corresponds to a quadratic form (V x , g x ) over K 0 x with the property that (V x , g x )⊗n x is uniquely determined by (W x , h x ) (cf. [6], 10.1.1 and 10.1.7) .We have (M x , Q x ) ≃ ρ x ⊗ K 0 x (V x , g x ), and (V x , g x )⊗n x is uniquely determined by (M x , Q x ), hence by (V, q).In other words, we have Documenta Mathematica 16 (2011) 677-707 and if (V 1 , q 1 ) and (V 2 , q 2 ) are two S-quadratic forms with if and only if 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, is the standard involution, and that for x ∈ S ′ k we set x = x if x is selfdual, and 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 Let ω, ω ′ be the orbits of x, respectively x ′ .Then 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 ω ∈ Ω k , set ω = ω if ιω = ω, and ω = (ω 1 , ω 2 ) with ιω 1 = ω 2 and ω 1 = ω 2 .Let Ω k be the set of all ω with ω ∈ Ω k .Let us fix a field extension K 0 ω of k such that K 0 ω ≃ K 0 x for all x ∈ ω.
Let (V, q) be an S-quadratic form.Then we have an orthogonal decomposition where (V x , g x ) is a quadratic form over K 0 x , and (V x , g x ) ⊗ n x is uniquely determined by (V, q) (cf.§5).
For all ω ∈ Ω k , let us consider the orthogonal sum Then (V ω , g ω ) is a quadratic form over K 0 ω .
Note that Ind G S (ρ x ) does not depend on the choice of x ∈ ω.Set where x is any element of ω.

Therefore we have
Set A(V, q) = Res G S Ind G S (V, q).Then we have Let y ∈ S ′ k , and let us take the y-component of the equation above.We get Let ω ′ ∈ Ω k such that y ∈ ω ′ .Note that the S-quadratic spaces A(V, q) y and Res G S (I ω ) y do not depend on the choice of y ∈ ω ′ .Set A(V, q) ω ′ = A(V, q) y Documenta Mathematica 16 (2011) 677-707 and Res G S (I(ω)) ω ′ = Res G S (I(ω)) y for any y ∈ ω ′ .

Then we have
Res G S (I(ω)) .
We define d ω,ω ′ to be the dimension of the k-vector space underlying the quadratic form F ω,ω ′ .Note that d ω,ω ′ is the number of times ρ y occurs in Res G S Ind G S (ρ x ) for any x ∈ ω, y ∈ ω ′ .As U x is the underlying module of ρ x , the integer d ω,ω ′ can also be seen as the number of times 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 and 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.
Proof.Since the group {1, ι} acts on Ω with fixed points precisely Ω 1 , it follows from lemma 4.3 that Hence we have This completes the proof of the proposition.
We define Ω 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 y is supported on U y = U y ⊕ U y and ρ x is supported on Thus the module U y occurs with even multiplicity in Res G S Ind G S (U x ), so that d ω,ω ′ ≡ 0 (mod 2).Therefore the matrix (d ω,ω ′ ) ω,ω ′ ∈Ω 1 has the shape ).This completes the proof of the proposition.
We have the following Proposition 7.4 Suppose that G has the odd determinant property.Then For the proof of prop.7.4, we need the following lemma Documenta Mathematica 16 (2011) 677-707 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 ′ ∈ΩK (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∈ΩK is a basis of X K .
Let Φ : X K → X K be Res G S Ind G S considered as an endomorphism of X K .Note that we have This implies that the matrix of Φ in the basis (p ω ) ω∈ΩK 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 ′ ∈ΩK (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 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 Let P be a projective k[S]-module.Since A[S] is π-adically complete, there is a projective A[S]-module P such that P /π P ≃ P .Then PK = P ⊗ A K is a projective K[S]-module.Moreover, P is simple if and only if PK is simple.Note that if P and Q are simple k[S]-modules, then we have Therefore the matrices (d ω,ω ′ ) ω,ω ′ ∈Ω k and (d ω,ω ′ ) ω,ω ′ ∈ΩK 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 x = x if x is selfdual, and For any ω ∈ Ω E , recall that ω = ω if the characters of ω are invariant under ι, and 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 Ω E be the set of all ω with ω ∈ Ω E , and let K 0 ω = K 0 x for x ∈ ω.
Let us define d ω,ω ′ as being the number of times U y occurs in Let us recall that for all ω, ω ′ ∈ Ω k such that K 0 ω = K 0 ω ′ = k, we denote by d ω,ω ′ the dimension of the k-vector space underlying the quadratic form F ω,ω ′ (see §6).
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 (τ x)(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 τ ω = ω if and only if there is a character x ∈ S ′ L with x ∈ ω such that τ x = x.Proof.If there exists x ∈ ω such that τ x = x, then we have τ ω = ω.Conversely, suppose that ω ∈ Ω L is such that τ ω = ω.If we had τ x = x for every x ∈ ω, then ♯(ω) would be even, contradicting lemma 6.1.This implies that there exists x ∈ ω with τ x = 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: L be an orthogonal or symplectic character such that τ Proof.For any field E, let us denote by The Galois automorphism τ : L → L over K acts on L[S], hence also on ).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 ∆ x0 is a quaternion division algebra and ∆ x = L, then (x 0 ) L = 2x; if both ∆ x0 and ∆ x are quaternion division algebras, or if ∆ x0 = K and ∆ x = L, then (x 0 ) L = x.This proves (iv).
Proposition 8.4 Suppose that Proof.The automorphism τ of L/K induces a permutation of Ω On the other hand, we have Indeed, by cor.8.3 (i) the map ω 0 → (ω 0 ) L induces a bijection between Ω 1,o ).This completes the proof of the proposition.Proof.The automorphism τ of L/K induces a permutation of Ω 1,s .Let Let S L be the group of permutations of Ω For ω, ω Proof.Suppose that y = y and τ y = 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 xL = EL = K y , we have x L = y.Further, if K 0 y = L, then K 0 x = K.
Suppose first that τ y = y.Then there is an x ∈ S ′ K such that x L = y.Further, τ an automorphism on M which takes each factor K y and K ιy in itself.Moreover, we have 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 is positive definite, hence its restriction to each component is positive definite as well.This implies (cf [6], Chap 8, 13.5) that K 0 i ⊂ R for all i.Hence for all i = 1, . . ., r, we have With the notation of lemma 9.1, let Since L/Q is cyclic of degree a power of 2, it has a unique set of subfields which fit into a filtration 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 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 0 i '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 with all inclusions being strict, and L i /L i−1 of degree 2.

Let
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 K 0 x for all x ∈ S ′ k .§10.The odd determinant property revisited For any field E, set 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 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 x is supported on U Noting that for a general K, the integers d ω,ω ′ can be computed after base changing to an algebraic closure of K, we get the following: Thus the matrix (d ω,ω ′ ) ω,ω ′ ∈Ω 0 K has the following shape modulo 2 Documenta Mathematica 16 (2011) 677-707 where (V x , g x ) is a quadratic form over K 0 x (cf.§5).Recall that the Witt class is uniquely determined by (V, h), where n x is the reduced norm of D x over K 0 x if D x is a quaternion algebra, the norm of K x over K 0 x if K x is a quadratic algebra, and n x = 1 otherwise.We have where is a quadratic space determined up to multiplication by n ω = n x .We have I(ω) = Ind G S (ρ x ), which does not depend on the choice of x ∈ ω.

We have Res
For y ∈ S ′ k , the y-component of Res G S (I ω ) is ρ y ⊗ K 0 ω ′ F ω,ω ′ , where y ∈ ω ′ , and where F ω,ω ′ is a quadratic space over K 0 ω ′ , determined up to multiplication by n ω ′ .Let (V 1 , h 1 ) and (V 2 , h 2 ) be two S-quadratic forms such that and for i = 1, 2.
Note that as the k[S]-modules Res G S Ind G S (V 1 ) and Res G S Ind G S (V 2 ) are isomorphic, the k[G]-modules Ind G S (V 1 ) and Ind G S (V 2 ) are also isomorphic (see for instance [3], cor.6.8).This implies that dim(V 1 ω ) = dim(V i ω ) for all ω ∈ Ω.
Claim.We have .
Documenta Mathematica 16 (2011) 677-707 For the proof, we distinguish two cases Case 1. Suppose that K 0 x = k for all x ∈ S ′ k .Then we have Res G S (I(ω)) y = ρ y ⊗ k F ω,ω ′ , where y ∈ ω ′ , where F ω,ω ′ is a quadratic form over k, and n ω ′ ⊗ k F ω,ω ′ is determined by Res G S (I(ω)) y .Hence Suppose that Res G S Ind G S (V 1 , h 1 ) ≃ Res G S Ind G S (V 2 , h 2 ), and set g i ω = (V i ω , g i ω ) for i = 1, 2. Then Let us denote by f ω,ω ′ the element of W (k) determined by the quadratic form F ω,ω ′ , and let ( fω,ω ′ ) be the matrix of cofactors of the matrix (f ω,ω ′ ) in the Witt ring W (k). Then the product ( fω,ω ′ )(n ω ′ ⊗ k f ω,ω ′ ) is equal to a diagonal matrix with diagonal entries ϕ.n ω j , where ϕ ∈ W (k) is the determinant of the matrix (f ω,ω ′ ).Let v i ω be the element of W (k) determined by the quadratic form g i ω = (V i ω , g i ω ) for i = 1, 2. Then we get ϕ.n ω ⊗ k (v 1 ω − v 2 ω ) = 0 in W (k), for every ω ∈ Ω.
General case.Let us consider (V i , h i ) ⊗ k k r .We have Moreover, K 0 ω ⊗ k k r ≃ α∈Gal(K 0 ω /k) k α r .The orbit ω splits into distinct conjugate orbits over k r .Each ω ∈ Ω k with K 0 ω = k r occurs as one of the conjugate orbits over k r .Using case 1, we get, for orbits ω with K 0 ω = k r , . Cancelling these factors, we may assume that with K 0 ω k r−1 for all ω in the above decomposition.Inductively we get, for all ω, that . This completes the proof of the theorem.

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

mod 2 .
Therefore δ 1,s K = det(A)det(B) ≡ α (mod 2), and this completes the proof of the claim.Therefore we see that δ 1,s L ≡ 1 (mod 2) implies δ 1,s K ≡ 1 (mod 2), hence prop.8.5 is proved.Let us recall that for any field E of characteristic = 2, we denote by S ′ E the set of x for x ∈ S ′ E .The Galois automorphism τ : L → L over K induces an action on S ′ L which we denote by y → τ y.Lemma 8.6 Let y ∈ S ′ L with y unitary such that τ y = y.Then there exists x ∈ S ′ K with x unitary such that x L = y.Moreover, if K 0 y = L, then K 0 x = 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 isomorphismA[S]/πA[S] → 1≤i≤r M ni (k) can be lifted to an isomorphism A[S] ≃ 1≤i≤r M ni (A), hence we have K[S] ≃ 1≤i≤r M ni (K).Thus every character of S ′ K is absolutely irreducible, hence by lemma 7.5 we have det ω,ω ′ ∈ΩK (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 ′ ∈ΩK 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 G S (P ) is projective as well.