# Hasse principle for $G$-quadratic forms

## Abstract

Introduction. Let k be a global field of characteristic = 2. The classical Hasse-Minkowski theorem states that if two quadratic forms become isomorphic over all the completions of k, then they are isomorphic over k as well. It is natural to ask whether this is true for G-quadratic forms, where G is a finite group. In the case of number fields the Hasse principle for G-quadratic forms does not hold in general, as shown by J. Morales [M 86]. The aim of the present paper is to study this question when k is a global field of positive characteristic. We give a sufficient criterion for the Hasse principle to hold (see th. 2.1.), and also give counter-examples. These counter-examples are of a different nature than those for number fields : indeed, if k is a global field of positive characteristic, then the Hasse principle does hold for G-quadratic forms on projective k[G]- modules (see cor. 2.3), and in particular if k[G] is semi-simple, then the Hasse principle is true for G-quadratic forms, contrarily to what happens in the case of number fields. On the other hand, there are counter-examples in the non semi-simple case, as shown in §3. Note that the Hasse principle holds in all generality for G-trace forms (cf. [BPS 13]).The third named author is partially supported by National Science Foundation grant DMS-1001872. §1. Definitions, notation and basic facts Let k be a field of characteristic = 2. All modules are supposed to be left modules. Documenta Mathematica 18 (2013) 383-392 E. Bayer-Fluckiger, N. Bhaskhar, and R. Parimala G-quadratic spaces Let G be a finite group, and let k[G] be the associated group ring. A G- quadratic space is a pair (V, q), where V is a k[G]-module that is a finite dimensional k-vector space, and q : V $\times V \rightarrow k$ is a non-degenerate symmetric bilinear form such that $q(gx, gy) = q(x, y)$ for all x, y $\in V$ and all g $\in G$. Two G-quadratic spaces (V, q) and (V $^{\prime}$, q$^{\prime}$) are isomorphic if there exists an isomorphism of k[G]-modules f : V $\rightarrow V^{\prime}$ such that q$^{\prime}$(f (x), f (y)) = $q(x, y)$ for all x, y $\in V$ . If this is the case, we write (V, q) $\simeq G$ (V $^{\prime}$, q$^{\prime}$), or simply q $\simeq G$ q$^{\prime}$.Hermitian forms Let R be a ring endowed with an involution r $\rightarrow r$. For any R-module M , we denote by M * its dual $HomR(M, R)$. Then M * has an R-module structure given by (rf )$(x) = f$ (x)r for all r $\in R$, x $\in M$ and f $\in M$ * . If M and N are two R-modules and if f : M $\rightarrow N$ is a homomorphism of R-modules, then f induces a homomorphism f * : $N * \rightarrow M * defined$ by f * (g) = gf for all g $\in N$ * , called the adjoint of f . A hermitian form is a pair (M, h) where M is an R-module and h : M $\times M \rightarrow R$ is biadditive, satisfying the following two conditions: (1.1) $h(rx, sy) = rh(x, y)$s and $h(x, y) = h(y, x)$ for all x, y $\in M$ and all r, s $\in R$. (1.2) The homomorphism h : M $\rightarrow M * given$ by y $\rightarrow h( , y)$ is an isomorphism. Note that the existence of h implies that M is self-dual, i.e. isomorphic to its dual. If G is a finite group, then the group algebra R = k[G] has a natural k-linear involution, characterized by the formula g = g - 1 for every g $\in G$. We have the following dictionary (see for instance [BPS 13, 2.1, Example] a) R-module M ?? k-module M with a k-linear action of G; b) R-dual M * ?? k-dual of M , with the contragredient (i.e. dual) action of G. c) hermitian space (M, h) ?? symmetric bilinear form on M , which is G- invariant and defines an isomorphism of M onto its k-dual. Therefore a hermitian space over k[G] corresponds to a G-quadratic space, as defined above. Hermitian elements Let E be a ring with an involution $\sigma$: E $\rightarrow E$ and put E0 = z $\in E\times | \sigma (z) = z$.Documenta Mathematica 18 (2013) 383-392 If z $\in E0$, the map hz : E$\times E \rightarrow E$ defined by $hz(x, y) = x$.z.$\sigma (y)$ is a hermitian space over E; conversely, every hermitian space over E with underlying module E is isomorphic to hz for some z $\in E0$. Define an equivalence relation on E0 by setting z $\equiv$z$^{\prime}$ if there exists e $\in E\times$with z$^{\prime} = \sigma (e)$ze; this is equivalent to (E, hz) $\simeq$(E, hz$^{\prime}$). Let $H(E, \sigma )$ be the quotient of E0 by this equivalence relation. If z $\in E0$, we denote by [z] its class in $H(E, \sigma )$. Classifying hermitian spaces via hermitian elements Let (M, h0) be a hermitian space over R. Set $EM = End(M )$. Let $\tau$: EM $\rightarrow$EM be the involution of EM induced by h0, i.e. $\tau (e) = h - 1 0$ e* h0, for e $\in$EM , where e* is the adjoint of e. If (M, h) is a hermitian space (with the same underlying module M ), we have $\tau$(h - 1 0 h) = h - 1 0 (h - 1 0 h)* h0 = h - 1 0 h* (h - 1 0 )* h0 = h - 1 0 h. Hence h - 1 0 h is a hermitian element of $(EM , \tau )$; let [h - 1 0 h] be its class in $H(EM , \tau )$. Lemma 1.1. (see for instance [BPS 13, lemma 3.8.1]) Sending a hermitian space (M, h) to the element [h - 1 0 h] of H $(EM , \tau )$ induces a bijection between the set of isomorphism classes of hermitian spaces (M, h) and the set $H(EM , \tau )$. Components of algebras with involution Let A be a finite dimensional k-algebra, and let $\iota$: A $\rightarrow A$ be a k-linear involution. Let RA be the radical of A. Then A/RA is a semi-simple k- algebra, hence we have a decomposition A/RA = M (D i=1,$\dots$,r ni i), where D1, . . . , Dr are division algebras. Let us denote by Ki the center of Di, and let Dop be the opposite algebra of D i i. Note that $\iota (RA) = RA$, hence $\iota$induces an involution $\iota$: A/RA $\rightarrow$A/RA. Therefore A/RA decomposes into a product of involution invariant factors. These can be of two types : either an involution invariant matrix algebra Mn (D (D (Dop), with M (D (Dop) i i), or a product Mni i) $\times$Mni i ni i) and Mni 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 Mn (D (D i i) with $\iota$|Ki not the identity, or a product Mni i)$\times$Mn (Dop) . Otherwise, the factor is said to be of the first kind. In this case, i i the component is of the form Mn (D i i) and the restriction of $\iota$to Ki is the identity. We say that the component is orthogonal if after base change to a separable closure $\iota$is given by the transposition, and symplectic otherwise. A component Mn (D i i) is said to be split if Di is a commutative field.Completions Documenta Mathematica 18 (2013) 383-392 E. Bayer-Fluckiger, N. Bhaskhar, and R. Parimala If k is a global field and if v is a place of k, we denote by kv the completion of k at v. For any k-algebra E, set Ev = E $\otimes k$ kv. If K/k is a field extension of finite degree and if w is a place of K above v, then we use the notation w|v. §2. Hasse principle In this section, k will be a global field of characteristic = 2. Let us denote by $\Sigma k$ the set of all places of k. The aim of this section is to give a sufficient criterion for the Hasse principle for G-quadratic forms to hold. All modules are left modules, and finite dimensional k-vector spaces. Theorem 2.1. Let V be a k[G]-module , and let $E = End(V )$. Let RE be the radical of E, and set E = E/RE. Suppose that all the orthogonal components of E are split, and let (V, q), (V, q$^{\prime}$) be two G-forms. Then q $\simeq G$ q$^{\prime}$ over k if and only if q $\simeq G$ q$^{\prime}$ over all the completions of k. This is announced in [BP 13], and replaces th. 3.5 of [BP 11]. The proof of th. 2.1 relies on the following proposition Proposition 2.2. Let E be a finite dimensional k-algebra endowed with a k- linear involution $\sigma$: E $\rightarrow E$. Let RE be the radical of E, and set E = E/RE. Suppose that all the orthogonal components of E are split. Then the canonical map $H(E, \sigma ) \rightarrow H(E v\in \Sigma v , \sigma v )$ is injective. k Proof. The case of a simple algebra. Suppose first that E is a simple k- algebra. Let K be the center of E, and let F be the fixed field of $\sigma$in K. Let $\Sigma F$ denote the set of all places of F . For all v $\in \Sigma k$, set Ev = E $\otimes k$ kv, and note that Ev = E H(E $H(E w|v w , therefore v\in \Sigma v , \sigma v ) = w , \sigma w )$. k w$\in \Sigma F$ By definition, $H(E, \sigma )$ is the set of isomorphism classes of one dimensional hermitian forms over E. Moreover, if $\sigma$is orthogonal, then the hypothesis implies that E is split, in other words we have E $\simeq Mn(F )$. Therefore the conditions of [R 11, th. 3.3.1] are fulfilled, hence the Hasse principle holds for hermitian forms over E with respect to $\sigma$. This implies that the canonical map $H(E, \sigma ) \rightarrow H(E v\in \Sigma v , \sigma v )$ is injective. k The case of a semi-simple algebra. Suppose now that E is semi-simple. Then E $\simeq E1 \times . . . \times$Er $\times A \times$Aop, where E1, . . . , Er are simple algebras which are stable under the involution $\sigma$, and where the restriction of $\sigma$to A $\times$Aop exchanges the two factors. Applying [BPS 13, lemmas 3.7.1 and 3.7.2] we are reduced to the case where E is a simple algebra, and we already know that the result is true in this case. Documenta Mathematica 18 (2013) 383-392 General case. We have E = E/RE. Then E is semi-simple, and $\sigma$induces a k-linear involution $\sigma$: E $\rightarrow E$. We have the following commutative diagram f $H(E, \sigma ) - \rightarrow H(E v\in \Sigma v , \sigma )$ k $\downarrow \downarrow f H(E, \sigma ) - \rightarrow H(E v\in \Sigma v , \sigma )$, k where the vertical maps are induced by the projection E $\rightarrow E$. By [BPS 13, lemma 3.7.3], these maps are bijective. As E is semi-simple, the map f is injective, hence f is also injective. This concludes the proof. Proof of th. 2.1. It is clear that if q $\simeq G$ q$^{\prime}$ over k, then q $\simeq G$ q$^{\prime}$ over all the completions of k. Let us prove the converse. Let (V, h) be the k[G]-hermitian space corresponding to (V, q), and let $\sigma$: E $\rightarrow E$ be the involution induced by (V, h) as in §1. Let (V, h$^{\prime}$) be the k[G]-hermitian space corresponding to (V, q$^{\prime}$), and set u = h - 1h$^{\prime}$. Then u $\in E0$, and by lemma 1.1. the element [u] $\in H(E, \sigma )$ determines the isomorphism class of (V, q$^{\prime}$); in other words, we have q $\simeq G$ q$^{\prime}$ if and only if [u] = [1] in $H(E, \sigma )$. Hence the theorem is a consequence of proposition 2.2.Corollary 2.3 Suppose that $char(k) =$p > 0, and let V be a projective k[G]- module.. Let (V, q), (V, q$^{\prime}$) be two G-forms. Then q $\simeq G$ q$^{\prime}$ over k if and only if q $\simeq G$ q$^{\prime}$ over all the completions of k. Proof. Since V is projective, there exists a k[G]-module W and n $\in N$ such that V $\oplus W \simeq$k[G]n. The endomorphism ring of k[G]n is $Mn(k[G])$, and as $char(k) =$p > 0, we have k[G] = Fp[G] $\otimes F$ k. Hence M p $n(k[G])$ is isomorphic to $Mn(Fp[G]) \otimes F$ k. Let $E = End(V )$, let R p E be the radical of E, and let E = E/RE. Let us show that all the components of E are split. Let e be the idempotent endomorphism of V $\oplus W$ which is the identity of V . Set $\Lambda = End(V \oplus W )$ and let R$\Lambda$be the radical of $\Lambda$. Then e$\Lambda e = E$ and eR$\Lambda e = RE$. Set $\Lambda = \Lambda /R\Lambda$, and and let e be the image of e in $\Lambda$. Set k[G] = k[G]/$rad(k[G])$. Then we have E $\simeq e\Lambda$e $\simeq eMn(k[G])$e. This implies that E is a component of the semi-simple algebra $Mn(k[G])$. Let us show that all the components of $Mn(k[G])$ are split. As Fp is a finite field, Fp[G]/$(rad(Fp[G])$ is a product of matrix algebras over finite fields. Moreover, for any finite field F of characteristic p, the tensor product F $\otimes F$ k is a product of fields. This p shows that (Fp[G]/$(rad(Fp[G]))\otimes F$ k is a product of matrix algebras over finite p extensions of k; in particular, it is semi-simple. The natural isomorphism Fp[G] $\otimes F$ k $\rightarrow$k[G] induces an isomorphism [F k $\rightarrow p$ p[G]/$(rad(Fp[G]))] \otimes$Fp k[G]/$(rad(Fp[G])$.k[G]). Therefore $rad(Fp[G].k[G])$ is the radical of k[G], and we have an isomorphism [Fp[G]/$(rad(Fp[G]))]\otimes F$ k $\rightarrow$k[G]/$(rad(k[G]))$. Hence p all the components of k[G]/$(rad(k[G]))$ are split. This implies that all the components of E are split as well. Therefore the corollary follows from th. 2.1. The following corollary is well-known (see for instance [R 11, 3.3.1 (b)]). Documenta Mathematica 18 (2013) 383-392 E. Bayer-Fluckiger, N. Bhaskhar, and R. Parimala Corollary 2.4 Suppose that $char(k) =$p > 0, and that the order of G is prime to p. Then two G-quadratic forms are isomorphic over k if and only if they become isomorphic over all the completions of k. Proof. This follows immediately from cor. 2.3. §3. Counter-examples to the Hasse principle Let k be a field of characteristic p > 0, let Cp be the cyclic group of order p, and let G = Cp $\times$Cp $\times$Cp. In this section we give counter-examples to the Hasse principle for G$\times$G-quadratic forms over k in the case where k is a global field. We start with some constructions that are valid for any field of positive characteristic. 3.1 A construction Let D be a division algebra over k. It is well-known that there exist indecomposable k[G]-modules such that their endomorphism ring modulo the radical is isomorphic to D. We recall here such a constuction, brought to our attention by R. Guralnick, in order to use it in 3.2 in the case of quaternion algebras. The algebra D can be generated by two elements (see for instance [J 64, Chapter VII, §12, th. 3, p. 182]). Let us choose i, j $\in D$ be two such elements.Let us denote by Dop the opposite algebra of D, and let d be the degree of D. Then we have D $\otimes k$ Dop $\simeq Md2(k)$. Let us choose an isomorphism f : D $\otimes k$ Dop $\simeq Md2(k)$, and set a1 = f ($1 \otimes 1$) = 1, a2 = f (i $\otimes 1$) and a3 = f (j $\otimes 1$).Let g1, g2, g3 $\in G$ be three elements of order p such that the set ${g1, g2, g3}$ generates G.and let us define a representation G $\rightarrow GL2d2(k)$ by sending gm to the matrix I am 0 I for all m = 1, 2, 3. Note that this is well-defined because $char(k) = p$. This endowes k2d2 with a structure of k[G]-module. Let us denote by N this k[G]- module, and let EN be its endomorphism ring. Then x y EN = | x $\in$Dop $\subset M 0$ x d2 (k), y $\in$Md2 (k) , and its radical is 0 y RN = | y $\in M 0$ 0 d2 (k) , hence EN /RN $\simeq$Dop. Documenta Mathematica 18 (2013) 383-392 3.2. The case of a quaternion algebra Let H be a quaternion algebra over k. Then by 3.1, we get a k[G]-module N = NH with endomorphism ring EN such that EN /RN $\simeq$Hop, where RN is the radical of EN . We now construct a G-quadratic form q over N in such a way that the involution it induces on EN /RN $\simeq$Hop is the canonical involution. Let i, j $\in H$ such that i2, j2 $\in k\times$and that ij = - ji. Let $\tau$: H $\rightarrow H$ be the orthogonal involution of H obtained by composing the canonical involution of H with $Int(ij)$. Let $\sigma$: Hop $\rightarrow$Hop be the canonical involution of Hop. Let us consider the tensor product of algebras with involution $(H, \tau ) \otimes (Hop, \sigma ) = (M4(k), \rho )$. Then $\rho$is a symplectic involution of $M4(k)$ satisfying $\rho (am) = am$ for all m = 1, 2, 3, since $\tau (i)$ = (ij)( - i)$(ij) - 1 = i, \tau (j)$ = (ij)( - j)$(ij) - 1 = j$. Let $\alpha \in M4(k)$ be a skew-symmetric matrix such that for all x $\in M4(k)$, we have $0 \alpha \rho (x) = \alpha - 1xT \alpha$, where xT denotes the transpose of x. Set A = . - $\alpha 0$ Then AT = A. Let q : N $\times N \rightarrow k$ be the symmetric bilinear form defined by A : $q(v, w) = vT$ Aw for all v, w $\in N$ . Let $\gamma : M8(k) \rightarrow M8(k)$ be the involution adjoint to q, that is $\gamma (X) = A - 1XT$ A for all X $\in M8(k)$, i.e. $q(f v, w) = q(v, \gamma (f )$w) for all f $\in M8(k)$ and all v, w $\in N$ . The involution $\gamma$restricts to an involution of EN , as for all x, y $\in M4(k)$, we have x y $\alpha - 1xT \alpha - \alpha - 1yT \alpha \gamma$= .0 x $0 \alpha - 1xT \alpha$It also sends RN to itself, and induces an involution $\gamma$on Hop $\simeq$EN /RN that coincides with the canonical involution of Hop. We claim that q : N $\times N \rightarrow k$ is a G-quadratic form. To check this, it suffices to show that $q(gmv, gmw) = q(v, w)$ for all v, w $\in N$ and for all m = 1, 2, 3. Since $\rho (am) = am$ for all m = 1, 2, 3, we have - 1 I a I a $\gamma m = m 0$ I 0 I and hence $q(gmv, gmw) = q(v, \gamma (gm)$gmw) = $q(v, w)$ for all m = 1, 2, 3 and all v, w $\in N$ . Thus q is a G-quadratic form, and by construction, the involution of EN induced by q is the restriction of $\gamma$to EN . Documenta Mathematica 18 (2013) 383-392 E. Bayer-Fluckiger, N. Bhaskhar, and R. Parimala 3.3. Two quaternion algebras Let H1 and H2 be two quaternion algebras over k. By the construction of 3.2, we obtain two indecomposable k[G]-modules N1 and N2. Set E1 = EN and

## Cite this article

Eva Bayer-Fluckiger, Nivedita Bhaskhar, Hasse principle for $G$-quadratic forms. Doc. Math. 18 (2013), pp. 383–392

DOI 10.4171/DM/402