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Four-fold Massey products in Galois cohomology

Published online by Cambridge University Press:  17 August 2018

Pierre Guillot
Affiliation:
Université de Strasbourg & CNRS, Institut de Recherche Mathématique Avancée, UMR 7501, F-67000 Strasbourg, France email guillot@math.unistra.fr
Ján Mináč
Affiliation:
Department of Mathematics, Western University, London, Ontario, N6A 5B7, Canada email minac@uwo.ca
Adam Topaz
Affiliation:
Mathematical Institute, University of Oxford, Andrew Wiles Building, Radcliffe Observatory Quarter, Woodstock Road, Oxford OX2 6GG, UK email topaz@maths.ox.ac.uk
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Abstract

In this paper, we develop a new necessary and sufficient condition for the vanishing of $4$-Massey products of elements in the modulo-$2$ Galois cohomology of a field. This new description allows us to define a splitting variety for $4$-Massey products, which is shown in the appendix to satisfy a local-to-global principle over number fields. As a consequence, we prove that, for a number field, all such $4$-Massey products vanish whenever they are defined. This provides new explicit restrictions on the structure of absolute Galois groups of number fields.

Type
Research Article
Copyright
© The Authors 2018 

1 Introduction

Let $F$ be a field and $\overline{F}$ a separable closure of $F$ . The absolute Galois group of $F$ , denoted $G_{F}:=\operatorname{Gal}(\overline{F}/F)$ is an object of great interest in algebra and number theory. Many aspects of modern Galois theory, in one way or another, aim to understand the structural properties of $G_{F}$ . Recent major results in Galois cohomology show that such absolute Galois groups are extremely rare among all profinite groups. The most notable restriction on absolute Galois groups arises from the Bloch–Kato conjecture, which is now a theorem due to Rost and Voevodsky; see [Reference VoevodskyVoe11, Reference RostRos98, Reference Suslin and JoukhovitskiSJ06, Reference Haesemeyer and WeibelHW09, Reference WeibelWei09]. In particular, if $F$ contains a primitive $p$ th root of unity, then $\operatorname{H}^{\ast }(G_{F},\mathbb{Z}/p)$ is a quadratic algebra. More explicitly, this means that $\operatorname{H}^{\ast }(G_{F},\mathbb{Z}/p)$ is generated by elements of $\operatorname{H}^{1}(G_{F},\mathbb{Z}/p)$ , and the relations are generated only by those relations appearing in degree $2$ . This is a very strong restriction on the group-theoretical structure of $G_{F}$ . Recently, other explicit structural restrictions on absolute Galois groups started to arise, based on the notion of Massey products in the context of Galois cohomology.

We will recall the definition of Massey products below, but we briefly note that, given $x_{1},\ldots ,x_{n}\in \operatorname{H}^{1}(G_{F},\mathbb{Z}/p)$ where $p$ is a prime, the $n$ -Massey product, denoted $\langle x_{1},\ldots ,x_{n}\rangle$ , is a (possibly empty) subset of $\operatorname{H}^{2}(G_{F},\mathbb{Z}/p)$ . In the case $n=2$ , one has a simple description in terms of the cup-product as $\langle x_{1},x_{2}\rangle =\{x_{1}\cup x_{2}\}$ . Just as the cup-product $x_{1}\cup x_{2}$ provides an obstruction to the existence of Heisenberg extensions of $F$ (of degree $p^{3}$ ), the $n$ -Massey product provides as obstruction for the existence of higher $\mathbb{Z}/p$ -unipotent extensions. In this respect, we are primarily interested in situations where the $n$ -Massey product $\langle x_{1},\ldots ,x_{n}\rangle$ contains $0$ . In this article, when $\langle x_{1},\ldots ,x_{n}\rangle$ contains $0$ , we will simply say that ‘ $\langle x_{1},\ldots ,x_{n}\rangle$ vanishes.’

In the breakthrough paper on the subject, Hopkins and Wickelgren [Reference Hopkins and WickelgrenHW15] proved that, given $x_{1},x_{2},x_{3}\in \operatorname{H}^{1}(G_{F},\mathbb{Z}/p)$ , the triple Massey product $\langle x_{1},x_{2},x_{3}\rangle$ always vanishes whenever it is non-empty, in the case where $F$ is a number field and $p=2$ . This result was later extended by Mináč and Tân to arbitrary fields and $p=2$  [Reference Mináč and TânMT17c] partially based on ideas appearing in [Reference Gao, Leep, Mináč and SmithGLMS03], and to arbitrary primes $p$ with $F$ a global field [Reference Mináč and TânMT15b].

The end of 2014 and early 2015 saw a surge of activity on triple Massey products of elements of $\operatorname{H}^{1}(G_{F},\mathbb{Z}/p)$ , significantly extending the results mentioned above. Matzri [Reference MatzriMat11] was first to announce his proof, extending these results to all primes $p$ and all fields $F$ which contain $\unicode[STIX]{x1D707}_{p}$ . Shortly thereafter, the arguments from [Reference MatzriMat11] were refined by Efrat and Matzri, and this work was eventually published in [Reference Efrat and MatzriEM17]. At around the same time as when [Reference Efrat and MatzriEM17] was posted, Mináč and Tân [Reference Mináč and TânMT16] also released their proof that triple Massey products of elements of $\operatorname{H}^{1}(G_{F},\mathbb{Z}/p)$ always vanish when defined, while also removing the condition that $F$ must contain $\unicode[STIX]{x1D707}_{p}$ . Motivated by these results, Mináč and Tân [Reference Mináč and TânMT17c, Reference Mináč and TânMT16, Reference Mináč and TânMT15a] eventually formulated the so-called $n$ -Massey vanishing conjecture, which states that for an arbitrary field $F$ , the $n$ -Massey products of elements of $\operatorname{H}^{1}(G_{F},\mathbb{Z}/p)$ always vanish whenever they are non-empty.

The case of number fields has always played a particularly important role in this context, as will be outlined in the historical discussion below. In this respect, the present paper presents the first significant result concerning vanishing of $4$ -Massey products in Galois cohomology. Namely, this paper proves the following result.

Main Theorem. Let $F$ be a number field, and let $x_{1},x_{2},x_{3},x_{4}\in \operatorname{H}^{1}(G_{F},\mathbb{Z}/2)$ be given. If the Massey product $\langle x_{1},x_{2},x_{3},x_{4}\rangle$ is non-empty, then it vanishes.

To achieve this, we construct a ‘splitting variety’ for the problem at hand, which works over any field of characteristic not $2$ , and which is compatible with base-change. More precisely, given a field $F$ of characteristic not $2$ and $x_{1},x_{2},x_{3},x_{4}\in \operatorname{H}^{1}(G_{F},\mathbb{Z}/2)$ , we construct an $F$ -variety $\mathscr{X}_{F}$ such that the following are equivalent:

  1. (1) the set of  $F$ -rational points $\mathscr{X}_{F}(F)$ is non-empty;

  2. (2) the $4$ -Massey product $\langle x_{1},x_{2},x_{3},x_{4}\rangle$ vanishes.

Furthermore, this construction is compatible with base-change, in the sense that for all $L/F$ , one has $\mathscr{X}_{F}\otimes _{F}L=\mathscr{X}_{L}$ .

The study of the geometry and arithmetic of this variety falls within the reach of the recent results in [Reference Harpaz and WittenbergHW16]. Hence, when $F$ is a number field, it turns out to be possible to prove that (a variant of) $\mathscr{X}_{F}$ satisfies a local-to-global principle for the existence of rational points, as soon as the corresponding Brauer–Manin obstruction vanishes. In the proof of Theorem 6.1, we arrange the existence of local points satisfying certain additional conditions which force the corresponding Brauer–Manin obstruction to vanish; this proves the modulo- $2$ case of the $4$ -Massey vanishing conjecture over number fields. Moreover, it turns out that the Brauer–Manin obstruction always vanishes in ‘generic’ situations, which allows us to prove a stronger version of our result in such cases (see Theorem C).

The overall strategy we take in this paper is similar to the approach taken by Wickelgren and Hopkins [Reference Hopkins and WickelgrenHW15], primarily because of the fact that we use splitting varieties. However, their methods are highly specialized to the case of triple Massey products. Indeed, working with $n$ -Massey products for $n\geqslant 4$ is substantially harder, especially with respect to their definability and indeterminacy (see the discussion below). Moreover, there is a technical, yet fundamental difference between the two approaches arising from the additional conditions one must arrange for the local points of the splitting variety. The details and new tools we develop here are therefore completely different and more technically involved than those in [Reference Hopkins and WickelgrenHW15].

We want to say more about the context, and the history of the subject. Massey products were first introduced in the context of algebraic topology by Massey [Reference MasseyMas58], as a collection of ‘higher-order cohomology operations’ defined in terms of the cochain algebra. For example, Massey products of elements of $\operatorname{H}^{1}$ play a central role in (rational) homotopy theory, as being the most basic obstruction to 1-formality of manifolds – see the work of Deligne, Griffiths, Morgan and Sullivan [Reference Deligne, Griffiths, Morgan and SullivanDGMS75, Reference SullivanSul77, Reference MorganMor78, Reference MorganMor86]. Massey products in Galois cohomology were first systematically considered over number fields by Morishita [Reference MorishitaMor02, Reference MorishitaMor04], Vogel [Reference VogelVog05] and Sharifi [Reference SharifiSha07]. These papers were primarily focused on Galois groups of extensions with restricted ramification over number fields, where Massey products have interesting connections with topics from Iwasawa theory, Milnor invariants, and Rédei symbols. Understanding Massey products in the Galois cohomology of number fields is particularly important, as they also show up in the context of Grothendieck’s section conjecture; see Wickelgren [Reference WickelgrenWic09, Reference WickelgrenWic12, Reference WickelgrenWic12]. More generally, Massey products in Galois cohomology play an important role in understanding the structure of nilpotent quotients of absolute Galois groups. Therefore, a detailed understanding of Massey products in Galois cohomology could lead to significant generalizations of results in [Reference Efrat and MináčEM11, Reference Mináč and Spira.Min96] from the two-step nilpotent setting to more general settings.

The investigation of Massey products in Galois cohomology of arbitrary fields has recently started progressing very rapidly. This surge started with the work of Hopkins and Wickelgren [Reference Hopkins and WickelgrenHW15], and further progressed by Mináč and Tân [Reference Mináč and TânMT17b, Reference Mináč and TânMT17c, Reference Mináč and TânMT16, Reference Mináč and TânMT17a] and Efrat and Matzri [Reference Efrat and MatzriEM15, Reference Efrat and MatzriEM17, Reference EfratEfr14, Reference MatzriMat11]. In fact, ideas related to vanishing of modulo- $2$ triple Massey products already appeared in 2003 by Gao, Leep, Mináč and Smith [Reference Gao, Leep, Mináč and SmithGLMS03], albeit using different terminology. However, it is important to note that all of the results mentioned above were restricted to studying triple Massey products.

Until now, the cases of the $n$ -Massey vanishing conjecture for $n\geqslant 4$ have remained completely open. Already in the case $n=4$ , having a non-empty $4$ -Massey product forces the vanishing of a new invariant defined by Isaksen [Reference IsaksenIsa15]. In particular, prior to the present paper there was not even a strategy for approaching the $4$ -Massey vanishing conjecture.

We now introduce the basic concepts, and the precise notation, needed to state the main results of the paper.

1.1 Basic notation

Throughout the paper, $F$ will denote a field of characteristic not $2$ . We will use the usual notation $\operatorname{H}^{\ast }(F,A):=\operatorname{H}^{\ast }(G_{F},A)$ for the Galois cohomology of $F$ with coefficients in the $G_{F}$ -module $A$ .

Recall that Kummer theory yields a canonical isomorphism

$$\begin{eqnarray}F^{\times }/F^{\times 2}\xrightarrow[{}]{\cong }\operatorname{H}^{1}(F,\mathbb{F}_{2}).\end{eqnarray}$$

For an element $x\in F^{\times }$ , we write $[x]$ for its class in $F^{\times }/F^{\times 2}$ , and $\unicode[STIX]{x1D712}_{x}\in \operatorname{H}^{1}(F,\mathbb{F}_{2})$ for the image of $[x]$ under the Kummer isomorphism. We will usually consider $\unicode[STIX]{x1D712}_{x}$ as a (continuous) homomorphism

$$\begin{eqnarray}\unicode[STIX]{x1D712}_{x}:G_{F}\rightarrow \mathbb{F}_{2}\end{eqnarray}$$

via the canonical identification $\operatorname{H}^{1}(F,\mathbb{F}_{2})=\operatorname{Hom}^{\text{cont}}(G_{F},\mathbb{F}_{2})$ .

Given $a,b\in F^{\times }$ , we will usually write $(a,b)_{F}$ (or $(a,b)$ when $F$ is understood) for the cup-product $\unicode[STIX]{x1D712}_{a}\,\cup \,\unicode[STIX]{x1D712}_{b}\in \operatorname{H}^{2}(F,\mathbb{F}_{2})$ . This notation borrows from the fact that $\operatorname{H}^{2}(F,\mathbb{F}_{2})$ is canonically isomorphic to the $2$ -torsion of $\operatorname{Br}(F)$ , and that the class of the quaternion algebra $(a,b)_{F}$ corresponds to $\unicode[STIX]{x1D712}_{a}\cup \unicode[STIX]{x1D712}_{b}$ via this identification.

1.2 The groups $\mathbb{U}_{n}(\mathbb{F}_{2})$

The group $\mathbb{U}_{n}(\mathbb{F}_{2})$ , for $n\geqslant 2$ , is composed of the $n\times n$ upper-triangular matrices with entries in $\mathbb{F}_{2}$ with $1$ s along the diagonal. The group $\mathbb{U}_{n}(\mathbb{F}_{2})$ is endowed with $n-1$ homomorphisms

$$\begin{eqnarray}s_{1},\ldots ,s_{n-1}:\mathbb{U}_{n}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}\end{eqnarray}$$

defined as $s_{i}(g)=g_{i,i+1}$ (the $i$ th near-diagonal component of $g$ ).

The center ${\mathcal{Z}}(\mathbb{U}_{n}(\mathbb{F}_{2}))$ of $\mathbb{U}_{n}(\mathbb{F}_{2})$ consists of those matrices whose only possibly non-zero coefficient above the diagonal is in the top-right corner. In particular, the map $g\mapsto g_{1,n}$ induces an isomorphism ${\mathcal{Z}}(\mathbb{U}_{n}(\mathbb{F}_{2}))\cong \mathbb{F}_{2}$ . We write $\overline{\mathbb{U}}_{n}(\mathbb{F}_{2}):=\mathbb{U}_{n}(\mathbb{F}_{2})/{\mathcal{Z}}(\mathbb{U}_{n}(\mathbb{F}_{2}))$ , and consider $\mathbb{U}_{n}(\mathbb{F}_{2})$ as an extension of $\overline{\mathbb{U}}_{n}(\mathbb{F}_{2})$ by $\mathbb{F}_{2}$ . Furthermore, we denote by $\unicode[STIX]{x1D709}_{n}$ the element of $\operatorname{H}^{2}(\overline{\mathbb{U}}_{n}(\mathbb{F}_{2}),\mathbb{F}_{2})$ associated to this extension.

1.3 Massey products

Let $\unicode[STIX]{x1D6E4}$ be a profinite group, and let $x_{1},\ldots ,x_{n}\in \operatorname{H}^{1}(\unicode[STIX]{x1D6E4},\mathbb{F}_{2})$ be given. In this context, we say that the $n$ -Massey product $\langle x_{1},\ldots ,x_{n}\rangle$ is defined provided that there exists a homomorphism $\unicode[STIX]{x1D711}:\unicode[STIX]{x1D6E4}\rightarrow \overline{\mathbb{U}}_{n+1}(\mathbb{F}_{2})$ such that $x_{i}=s_{i}\circ \unicode[STIX]{x1D711}$ for $i=1,\ldots ,n$ . Furthermore, in this case we say that $\unicode[STIX]{x1D711}$ is a defining system for the $n$ -Massey product $\langle x_{1},\ldots ,x_{n}\rangle$ .

The $n$ -Massey product associated to the defining system $\unicode[STIX]{x1D711}$ , denoted by $\langle x_{1},\ldots ,x_{n}\rangle _{\unicode[STIX]{x1D711}}$ , is defined to be $\unicode[STIX]{x1D711}^{\ast }\unicode[STIX]{x1D709}_{n+1}\in \operatorname{H}^{2}(\unicode[STIX]{x1D6E4},\mathbb{F}_{2})$ , the pull-back of $\unicode[STIX]{x1D709}_{n+1}$ along $\unicode[STIX]{x1D711}$ . Note that $\langle x_{1},\ldots ,x_{n}\rangle _{\unicode[STIX]{x1D711}}=0$ if and only if the map $\unicode[STIX]{x1D711}:\unicode[STIX]{x1D6E4}\rightarrow \overline{\mathbb{U}}_{n+1}(\mathbb{F}_{2})$ lifts to a homomorphism $\widetilde{\unicode[STIX]{x1D711}}:\unicode[STIX]{x1D6E4}\rightarrow \mathbb{U}_{n+1}(\mathbb{F}_{2})$ .

Finally, the $n$ -Massey product $\langle x_{1},\ldots ,x_{n}\rangle$ is defined as the set

$$\begin{eqnarray}\langle x_{1},\ldots ,x_{n}\rangle :=\{\langle x_{1},\ldots ,x_{n}\rangle _{\unicode[STIX]{x1D711}}\},\end{eqnarray}$$

where $\unicode[STIX]{x1D711}$ varies over all defining systems for $\langle x_{1},\ldots ,x_{n}\rangle$ . In particular, the $n$ -Massey product $\langle x_{1},\ldots ,x_{n}\rangle$ is non-empty if and only if it is defined. As mentioned above we will be primarily interested in situations where the $n$ -Massey product $\langle x_{1},\ldots ,x_{n}\rangle$ contains $0$ , and we say that ‘ $\langle x_{1},\ldots ,x_{n}\rangle$ vanishes’ in such situations. Note that, when we say ‘ $\langle x_{1},\ldots ,x_{n}\rangle$ vanishes’ we are also implying that $\langle x_{1},\ldots ,x_{n}\rangle$ is defined (as $\langle x_{1},\ldots ,x_{n}\rangle$ is non-empty).

Remark.

We have presented the definition of defining systems and Massey products in the context of group cohomology from the point of view of embedding problems. This is nevertheless equivalent to the classical (highly technical) definitions, by the work of Dwyer [Reference DwyerDwy75]. For our purposes, Massey products are defined as above.

We will simplify the notation somewhat in the context of Galois cohomology. Namely, given $a_{1},\ldots ,a_{n}\in F^{\times }$ , we write $\langle a_{1},\ldots ,a_{n}\rangle$ instead of $\langle \unicode[STIX]{x1D712}_{a_{1}},\ldots ,\unicode[STIX]{x1D712}_{a_{n}}\rangle$ . We will follow this convention when talking about defining systems as well as Massey products themselves.

1.4 Main results

We are now prepared to state our main theorems which characterize the vanishing of $4$ -Massey products in modulo- $2$ Galois cohomology.

Theorem A. Let $F$ be a field of characteristic not $2$ . Let $a,b,c,d\in F^{\times }$ be given, choose square roots $\sqrt{a}$ respectively $\sqrt{d}$ of $a$ respectively $d$ in an algebraic closure of $F$ , and put $E:=F[\sqrt{a},\sqrt{d}]$ . Then the following are equivalent.

  1. (1) The $4$ -Massey product $\langle a,b,c,d\rangle$ vanishes (i.e. it is defined and contains $0$ ).

  2. (2) There exist $B\in F[\sqrt{a}]$ , $C\in F[\sqrt{d}]$ and $z_{1},z_{2}\in F^{\times }$ such that the following conditions hold:

    1. (a) one has $\operatorname{N}_{F[\sqrt{a}]/F}(B)=b\cdot z_{1}^{2}$ and $\operatorname{N}_{F[\sqrt{d}]/F}(C)=c\cdot z_{2}^{2}$ ;

    2. (b) one has $(B,C)_{E}=0$ , $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ and $(b,c)_{F}=0$ .

  3. (3) There exist $B\in F[\sqrt{a}]$ , $C\in F[\sqrt{d}]$ and $z_{1},z_{2}\in F^{\times }$ such that the following conditions hold:

    1. (a) one has $\operatorname{N}_{F[\sqrt{a}]/F}(B)=b\cdot z_{1}^{2}$ and $\operatorname{N}_{F[\sqrt{d}]/F}(C)=c\cdot z_{2}^{2}$ ;

    2. (b) one has $(B,C)_{E}=(B,c)_{E}=(b,C)_{E}=(b,c)_{E}=0$ .

Note that condition (2) of Theorem A can be readily described in terms of polynomial equations over $F$ , hence defining an (affine) $F$ -variety. Theorem A then shows that this variety has an $F$ -point if and only if $\langle a,b,c,d\rangle$ vanishes. Note, however, that these equations depend on whether $a$ , $d$ , and/or $ad$ are squares in $F$ (the definition involves a Weil-restriction from $F[\sqrt{a},\sqrt{d}]$ to $F$ ); in other words, the variety is not compatible with base-change to extensions of $F$ . This is undesirable, as compatibility with base-change will be an important property towards the end of the paper. With some additional work, we are able to obtain the following characterization theorem which provides us with our desired uniform polynomial equations.

Theorem B. Let $F$ be a field of characteristic not $2$ and let $a,b,c,d\in F^{\times }$ be given. Consider the finite étale $F$ -algebra

$$\begin{eqnarray}{\mathcal{E}}:=F[X,Y]/(X^{2}-a,Y^{2}-d).\end{eqnarray}$$

Then the following are equivalent.

  1. (1) The $4$ -Massey product $\langle a,b,c,d\rangle$ vanishes.

  2. (2) There exist $x_{1},y_{1},x_{2},y_{2}\in F$ , $z_{1},z_{2}\in F^{\times }$ and $u,v\in {\mathcal{E}}$ such that the following equations are satisfied:

    1. (a) one has $x_{1}^{2}-y_{1}^{2}\cdot a=b\cdot z_{1}^{2}$ and $x_{2}^{2}-y_{2}^{2}\cdot d=c\cdot z_{2}^{2}$ ;

    2. (b) one has $u^{2}-\widetilde{B}v^{2}=\widetilde{C}$ in ${\mathcal{E}}$ , where $\widetilde{B}=x_{1}+y_{1}\cdot X\in {\mathcal{E}}$ and $\widetilde{C}=x_{2}+y_{2}\cdot Y\in {\mathcal{E}}$ .

Note that the polynomial equations described by condition (2) of Theorem B actually have the same shape over any field which contains $a,b,c,d$ . The $F$ -variety defined by these equations is what we will eventually call the splitting variety for $\langle a,b,c,d\rangle$ .

It is important to note that both Theorems A and B will play a key role in this paper. Indeed, condition (2) of Theorem A has an immediate and direct formulation involving cup-products in modulo-2 Galois cohomology (we exhibit some direct applications, over any field, in § 4). On the other hand, condition (2) of Theorem B defines a splitting variety whose geometry is remarkably simple. In generic situations, it satisfies the Hasse principle for the existence of rational points; in all cases, the local-to-global principle is governed by the Brauer–Manin obstruction, which takes a simple form here. See Theorem A.1 in the appendix for a detailed statement. We thereby obtain a more precise version of the Theorem announced above.

Theorem C. Let $F$ be a number field, and let $a,b,c,d\in F^{\times }$ be given. Then the following are equivalent:

  1. (1) the $4$ -Massey product $\langle a,b,c,d\rangle$ vanishes;

  2. (2) the $4$ -Massey product $\langle a,b,c,d\rangle$ is defined.

If furthermore $ad$ , $ab$ , $cd$ are all non-squares in $F$ , then the above conditions are further equivalent to the following:

  1. (3) one has $(a,b)_{F}=(b,c)_{F}=(c,d)_{F}=0$ .

Theorem C will be proved in Theorems 6.1 and 6.2 below. It is natural to ask whether the implication $(3)\;\Longrightarrow \;(1)$ holds in general. It turns out that this implication fails in general, even over number fields. See Remark 6.3, Example A.15, and the surrounding discussions for more details.

Organization of the paper

After some preliminaries in the next section, we prove Theorem A in § 3. In § 4, we give some first applications of Theorem A by proving a few cases of the 4-Massey vanishing conjecture by hand, over arbitrary fields. Then in § 5 we introduce the splitting variety $\mathscr{X}_{F}$ , as well as a variant $X_{F}$ which will simplify some calculations. The next section, that is § 6, gives a proof of Theorem C. Finally in § 7 we make some of our constructions explicit, and explain concretely how to get a Galois extension with group  $\mathbb{U}_{5}(\mathbb{F}_{2})$ when  $\langle a,b,c,d\rangle$ vanishes and  $a,b,c,d$ are linearly independent modulo squares; incidentally, this gives an alternative, more pedestrian proof for the implication $(3)\;\Longrightarrow \;(1)$ in Theorem A in this case.

An appendix by Wittenberg shows that the variety $X_{F}$ satisfies the local-to-global principle alluded to above, which is of course a crucial ingredient for Theorem C.

2 Preliminaries

2.1 The groups  $\mathbb{U}_{3}(\mathbb{F}_{2})$ and $\mathbb{U}_{5}(\mathbb{F}_{2})$

We shall need special notation for these two groups. First note that we define the dihedral group of order  $8$ to be  $\mathbb{U}_{3}(\mathbb{F}_{2})$ , and we may write  $D_{4}=\mathbb{U}_{3}(\mathbb{F}_{2})$ . An element  $g\in D_{4}$ is a matrix of the form

$$\begin{eqnarray}g=\left(\begin{array}{@{}ccc@{}}1 & s_{1}(g) & t(g)\\ 0 & 1 & s_{2}(g)\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Thus  $D_{4}$ is equipped with maps  $s_{1},s_{2},t:D_{4}\rightarrow \mathbb{F}_{2}$ . (The letter  $t$ is for ‘top’.) Note that the first two are group homomorphisms, but  $t$ is not. Our favourite generators are the involutions  $\unicode[STIX]{x1D70E}_{1}$ and  $\unicode[STIX]{x1D70E}_{2}$ , with  $s_{i}(\unicode[STIX]{x1D70E}_{i})=1$ and  $s_{j}(\unicode[STIX]{x1D70E}_{i})=t(\unicode[STIX]{x1D70E}_{i})=0$ for  $j\neq i$ .

Similarly, an element  $g\in \mathbb{U}_{5}(\mathbb{F}_{2})$ will be written

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}1 & s_{1}(g) & t_{1}(g) & u_{1}(g) & z(g)\\ 0 & 1 & s_{2}(g) & u_{3}(g) & u_{2}(g)\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & s_{4}(g)\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

This endows  $\mathbb{U}_{5}(\mathbb{F}_{2})$ with maps  $s_{1},\ldots ,z:\mathbb{U}_{5}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}$ , and  $s_{i}$ is a group homomorphism for $1\leqslant i\leqslant 4$ .

More generally the group  $\mathbb{U}_{n}(\mathbb{F}_{2})$ has homomorphisms  $s_{i}:\mathbb{U}_{n}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}$ for  $1\leqslant i\leqslant n-1$ , already mentioned in the Introduction, obtained by looking at the entries on what we call the near-diagonal. If we define elements  $\unicode[STIX]{x1D70E}_{i}$ by requiring  $s_{i}(\unicode[STIX]{x1D70E}_{i})=1$ while all the other entries of  $\unicode[STIX]{x1D70E}_{i}$ above the diagonal are  $0$ , then each  $\unicode[STIX]{x1D70E}_{i}$ is an involution, and these generate  $\mathbb{U}_{n}(\mathbb{F}_{2})$ .

We note that  $\mathbb{U}_{n}(\mathbb{F}_{2})$ has an automorphism which exchanges  $\unicode[STIX]{x1D70E}_{i}$ with  $\unicode[STIX]{x1D70E}_{n-i}$ . Most of our considerations respect this symmetry, and this motivates the notation above for  $\mathbb{U}_{5}(\mathbb{F}_{2})$ . (The automorphism is given by ‘the transpose but along the other diagonal’, followed by  $g\mapsto g^{-1}$ .)

2.2 Around the group  $D_{4}$

We write  $s=(s_{1},s_{2}):D_{4}\rightarrow C_{2}\times C_{2}$ , where we have identified  $\mathbb{F}_{2}$ with the cyclic group of order  $2$ in multiplicative notation, written  $C_{2}$ . There is an exact sequence

$$\begin{eqnarray}1\longrightarrow \mathbb{F}_{2}\longrightarrow D_{4}\stackrel{s}{\longrightarrow }C_{2}\times C_{2}\longrightarrow 1,\end{eqnarray}$$

the kernel of  $s$ being generated by  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]$ (which is the element  $g$ with  $t(g)=1$ and  $s_{i}(g)=0$ , $i=1,2$ ).

The quotient group  $C_{2}\times C_{2}$ is generated by the images of  $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}$ , written  $\overline{\unicode[STIX]{x1D70E}}_{1},\overline{\unicode[STIX]{x1D70E}}_{2}$ . The cohomology group  $\operatorname{H}^{1}(C_{2}^{2},\mathbb{F}_{2})=\operatorname{Hom}(C_{2}^{2},\mathbb{F}_{2})$ is endowed with the dual basis  $\overline{s}_{1},\overline{s}_{2}$ . The next lemma is very well known.

Lemma 2.1. The cohomology class of the above extension is  $\overline{s}_{1}\overline{s}_{2}\in \operatorname{H}^{2}(C_{2}^{2},\mathbb{F}_{2})$ .

We introduce the two elementary abelian subgroups  $E_{1},E_{2}$ , where  $E_{1}$ is the kernel of  $s_{2}$ and  $E_{2}$ is the kernel of  $s_{1}$ (the switch is justified by the next lemma). A very useful observation is that  $t$ , when restricted to either of these, is a group homomorphism.

Lemma 2.2. The corestriction

$$\begin{eqnarray}\operatorname{cores}:\operatorname{H}^{1}(E_{i},\mathbb{F}_{2})\longrightarrow \operatorname{H}^{1}(D_{4},\mathbb{F}_{2})\end{eqnarray}$$

carries  $t|_{E_{i}}$ to  $s_{i}$ , for  $i=1,2$ . More generally if  $\unicode[STIX]{x1D6E4}$ is any profinite group with a continuous homomorphism  $\unicode[STIX]{x1D711}:\unicode[STIX]{x1D6E4}\rightarrow D_{4}$ , and if  $H=\unicode[STIX]{x1D711}^{-1}(E_{i})$ is assumed to have index  $2$ in  $\unicode[STIX]{x1D6E4}$ , then the corestriction

$$\begin{eqnarray}\operatorname{cores}:\operatorname{H}^{1}(H,\mathbb{F}_{2})\longrightarrow \operatorname{H}^{1}(\unicode[STIX]{x1D6E4},\mathbb{F}_{2})\end{eqnarray}$$

carries  $t\circ \unicode[STIX]{x1D711}$ to  $s_{i}\circ \unicode[STIX]{x1D711}$ , for  $i=1,2$ .

Proof. We recall some properties of the corestriction

$$\begin{eqnarray}\operatorname{cores}:\operatorname{H}^{i}(N,M)\longrightarrow \operatorname{H}^{i}(G,M),\end{eqnarray}$$

where  $N$ is a subgroup of finite index of the profinite group  $G$ , and  $M$ is a  $G$ -module. In fact, we only need to consider the case when  $M$ has a trivial action, $N$ is closed of index  $2$ (and thus is normal) in  $G$ , and  $i=1$ , so that

$$\begin{eqnarray}\operatorname{cores}:\operatorname{Hom}(N,M)\longrightarrow \operatorname{Hom}(G,M).\end{eqnarray}$$

Here if  $f:N\rightarrow M$ , then the map  $\operatorname{cores}(f):G\rightarrow M$ is characterized as follows. Pick  $\unicode[STIX]{x1D70F}\in G\smallsetminus N$ . Then (i) $\operatorname{cores}(f)(n)=f(n)+f(\unicode[STIX]{x1D70F}^{-1}n\unicode[STIX]{x1D70F})$ for $n\in N$ , and (ii) $\operatorname{cores}(f)(\unicode[STIX]{x1D70F})=f(\unicode[STIX]{x1D70F}^{2})$ . This follows from the material in [Reference TateTat76, § 2], for example.

Let us use this for  $N=E_{1}$ and  $G=D_{4}$ . If

$$\begin{eqnarray}n=\left(\begin{array}{@{}ccc@{}}1 & a & b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right)\quad \text{and}\quad \unicode[STIX]{x1D70F}=\left(\begin{array}{@{}ccc@{}}1 & c & d\\ 0 & 1 & 1\\ 0 & 0 & 1\end{array}\right),\end{eqnarray}$$

then

$$\begin{eqnarray}\unicode[STIX]{x1D70F}^{-1}n\unicode[STIX]{x1D70F}=\left(\begin{array}{@{}ccc@{}}1 & a & a+b\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Moreover

$$\begin{eqnarray}\unicode[STIX]{x1D70F}^{2}=\left(\begin{array}{@{}ccc@{}}1 & 0 & c\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Thus  $t(n)+t(\unicode[STIX]{x1D70F}^{-1}n\unicode[STIX]{x1D70F})=b+a+b=a=s_{1}(n)$ , and $t(\unicode[STIX]{x1D70F}^{2})=c=s_{1}(\unicode[STIX]{x1D70F})$ . Hence the first statement of the lemma for  $i=1$ . The other cases are treated similarly.◻

2.3 $D_{4}$ -extensions of fields

We proceed to apply the above observations in a Galois-theoretic context, but a couple of comments are in order. First, the group  $D_{4}$ has an automorphism exchanging  $\unicode[STIX]{x1D70E}_{1}$ and  $\unicode[STIX]{x1D70E}_{2}$ , but the proposition below is not ‘symmetric’ in this way; it involves the subgroup  $E_{2}$ and not  $E_{1}$ , for example (so that one could get a new proposition by exchanging the roles of various players). Second, when asked for a basis for  $E_{2}$ , the reader would probably offer  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}],\unicode[STIX]{x1D70E}_{2}$ ; however, later considerations with  $\mathbb{U}_{5}(\mathbb{F}_{2})$ compel us to work with  $\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}],\unicode[STIX]{x1D70E}_{2}$ instead (specifically, we want Lemma 2.5 to have the simple form given below). This is reflected in the proposition below, since the dual basis of  $\operatorname{H}^{1}(E_{2},\mathbb{F}_{2})$ is  $t,s_{2}+t$ (or, in more complete notation, $t|_{E_{2}},s_{2}|_{E_{2}}+t|_{E_{2}}$ ).

Proposition 2.3. Let $F$ be a field of characteristic not $2$ and let $a,b\in F^{\times }$ be given. Then the following are equivalent:

  1. (1) $(a,b)_{F}=0$ ;

  2. (2) there exists a continuous homomorphism  $\unicode[STIX]{x1D711}:G_{F}\longrightarrow D_{4}$ such that  $s_{1}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{a}$ and  $s_{2}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{b}$ ;

  3. (3) there exist  $x,y,z\in F$ such that  $x^{2}-ay^{2}=bz^{2}$ , with  $z\neq 0$ .

When the equivalent conditions hold and $B:=x+y\sqrt{a}$ , we will say that $\unicode[STIX]{x1D711}$ from (2) and $x,y,z$ from (3) are consistent, provided that $\unicode[STIX]{x1D712}_{bB}=t\circ \unicode[STIX]{x1D711}$ and $\unicode[STIX]{x1D712}_{B}=(s_{2}+t)\circ \unicode[STIX]{x1D711}$ as elements of $\operatorname{H}^{1}(F[\sqrt{a}],\mathbb{F}_{2})$ . Then given any $\unicode[STIX]{x1D711}$ as in (2), we can choose $x,y,z$ as in (3) which are consistent with $\unicode[STIX]{x1D711}$ . Conversely, given any $x,y,z$ as in (3), we can choose $\unicode[STIX]{x1D711}$ as in (2) which is consistent with $x,y,z$ .

Again, this is essentially known, but we need the precise version given here. Note that it is necessary to deal with the cases when either  $a$ or  $b$ is a square, and that the proof below gives more concrete information in some situations. Also note that, as promised, the elements  $t,s_{2}+t$ , related to  $E_{2}$ , make an uncanny appearance.

Proof. We can combine  $\unicode[STIX]{x1D712}_{a}$ and  $\unicode[STIX]{x1D712}_{b}$ into a homomorphism  $G_{F}\rightarrow C_{2}\times C_{2}$ . The obstruction to lifting it to  $D_{4}$ is the cohomology class of the extension, so that Lemma 2.1 gives immediately the equivalence of (1) and (2).

We first conduct the rest of the proof under the following assumption.

Assumption.

Assume for the moment that  $a$ is not a square in  $F$ .

Suppose (2) holds. Note that  $\unicode[STIX]{x1D711}^{-1}(E_{2})=G_{F[\sqrt{a}]}$ . Let  $B^{\prime }=x^{\prime }+y^{\prime }\sqrt{a}\in F[\sqrt{a}]$ be such that $\unicode[STIX]{x1D712}_{B^{\prime }}=t\circ \unicode[STIX]{x1D711}|_{G_{F[\sqrt{a}]}}$ . Since  $a$ is not a square in  $F$ , the subgroup  $G_{F[\sqrt{a}]}$ has index  $2$ in  $G_{F}$ . Lemma 2.2 shows then that  $\operatorname{cores}(\unicode[STIX]{x1D712}_{B^{\prime }})=s_{2}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{b}$ . Now recall that under the identifications of  $\operatorname{H}^{1}(F,\mathbb{F}_{2})$ with  $F^{\times }\!/F^{\times 2}$ and of  $\operatorname{H}^{1}(F[\sqrt{a}],\mathbb{F}_{2})$ with  $F[\sqrt{a}]^{\times }\!/F[\sqrt{a}]^{\times 2}$ , the corestriction becomes the usual norm  $N_{F[\sqrt{a}]/F}$ . It follows that

$$\begin{eqnarray}N_{F[\sqrt{a}]/F}(B^{\prime })=(x^{\prime })^{2}-a(y^{\prime })^{2}=b\text{ mod squares}.\end{eqnarray}$$

This gives (3), clearly, but we need to modify  $B^{\prime }$ to get the consistency statement. And indeed, we put  $B=bB^{\prime }$ , so that  $B^{\prime }=bB$ modulo squares, and the result follows.

Next, we must prove that (3) implies (2), or equivalently (1). We may as well suppose that  $a$ and  $b$ are both not squares, for (1) holds trivially otherwise. The assumption is that  $b$ is a norm from  $F[\sqrt{a}]$ , or in more cohomological terms, that  $\unicode[STIX]{x1D712}_{b}$ is the corestriction of an element from the subgroup  $G_{F[\sqrt{a}]}$ . That the cup product  $\unicode[STIX]{x1D712}_{a}\unicode[STIX]{x1D712}_{b}=0$ then follows from the Arason exact sequence [Reference ArasonAra75].

However, to prove the claimed consistency, a more explicit argument is needed. Assume  $b$ is not a square. The element  $B=x+y\sqrt{a}$ , where  $x,y$ are as in (3), is fixed up to squares by  $\operatorname{Gal}(F[\sqrt{a},\sqrt{b}]/F)$ , as is readily checked. It follows that  $K=F[\sqrt{a},\sqrt{b},\sqrt{B}]$ is Galois over  $F$ (by equivariant Kummer theory, if you will).

We distinguish two cases, and assume first that  $a$ and  $b$ are not equal modulo squares. We now introduce elements  $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}\in \operatorname{Gal}(K/F)$ which are dual to  $\sqrt{a},\sqrt{b}$ in the obvious sense. Direct computation shows that  $\unicode[STIX]{x1D70E}_{1}^{2}=\unicode[STIX]{x1D70E}_{2}^{2}=1$ , and that  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}](\sqrt{B})=-\sqrt{B}$ , so that  $[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]\neq 1$ . From this one draws readily that  $\operatorname{Gal}(K/F)\cong D_{4}$ and (again!) that (2) holds. Also, one computes that  $\unicode[STIX]{x1D70E}_{2}(\sqrt{B})=\pm \sqrt{B}$ . We want to ensure that  $\unicode[STIX]{x1D70E}_{2}(\sqrt{B})=-\sqrt{B}$ , and to achieve this we replace  $\unicode[STIX]{x1D70E}_{2}$ by  $\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]$ if needed. Having done this, the elements  $\sqrt{B}$ , $\sqrt{bB}$ are dual to  $\unicode[STIX]{x1D70E}_{2}$ , $\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}]$ , and one checks that the corresponding map  $\unicode[STIX]{x1D711}:G_{F}\rightarrow D_{4}$ has precisely the required consistency.

If  $a=b$ modulo squares, one sees that  $\operatorname{Gal}(K/F)$ has order 4, so is abelian, and if  $\unicode[STIX]{x1D70E}$ is the non-trivial element of  $\operatorname{Gal}(F[\sqrt{a}]/F)$ extended to  $\operatorname{Gal}(K/F)$ , another direct calculation shows that  $\unicode[STIX]{x1D70E}$ does not have order  $2$ . So  $\operatorname{Gal}(K/F)\cong C_{4}$ can be identified with the subgroup of  $D_{4}$ generated by  $\unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{2}$ , and (2) follows. Consistency is automatic.

Finally, if  $b$ is a square in  $F$ , we use the same extension  $K=F[\sqrt{a},\sqrt{B}]$ of  $F$ , but compute that  $\operatorname{Gal}(K/F)\cong C_{2}^{2}$ . We identify this group with  $E_{1}$ appropriately, yielding a consistent  $\unicode[STIX]{x1D711}$ .

The case when  $a$ is a square.

In this situation (1) holds trivially, and thus (2) also holds. As for (3), if  $a=u^{2}$ then put

$$\begin{eqnarray}x_{0}=\frac{b+1}{2}\quad \text{and}\quad y_{0}=\frac{1-b}{2u}\end{eqnarray}$$

and compute that  $x_{0}^{2}-ay_{0}^{2}=b$ .

Let us see how we can adjust  $\unicode[STIX]{x1D711}$ from  $x,y,z$ . Put  $B=x+y\sqrt{a}\in F$ , and consider the characters  $\unicode[STIX]{x1D712}_{bB}$ and  $\unicode[STIX]{x1D712}_{b}$ , together defining a homomorphism  $G_{F}\rightarrow C_{2}\times C_{2}$ . Identifying Klein’s group with  $E_{2}$ sitting in  $D_{4}$ appropriately, we obtain  $\unicode[STIX]{x1D711}:G_{F}\rightarrow D_{4}$ satisfying our requirements.

Our very last step is to see how one can adjust  $x,y,z$ from  $\unicode[STIX]{x1D711}$ . First put  $B_{0}=x_{0}+y_{0}\sqrt{a}\in F$ where  $x_{0}$ and  $y_{0}$ are as above, which is a non-zero element since  $B_{0}(x-y\sqrt{a})=b\neq 0$ . For  $f\in F$ , put  $x=(f/B_{0})x_{0}$ and  $y=(f/B_{0})y_{0}$ , so that  $x^{2}-ay^{2}=bz^{2}$ for some  $z\in F^{\times }$ , while  $B=x+y\sqrt{a}=f$ , an arbitrary element of  $F$ . Of course  $\unicode[STIX]{x1D711}$ lands in the subgroup  $E_{2}$ , so we only need to pick  $f$ so that  $\unicode[STIX]{x1D712}_{f}=(s_{2}+t)\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{B}$ ; we have then (2) and (3) simultaneously and consistently.◻

2.4 The group  $\mathbb{U}_{5}(\mathbb{F}_{2})$ and its subquotients

Let  $S$ (for ‘square’) be the subgroup of matrices of the form

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & y_{1} & y_{2}\\ 0 & 1 & 0 & y_{3} & y_{4}\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Then  $S\cong C_{2}^{4}$ , and our favourite $\mathbb{F}_{2}$ -basis, denoted $e_{1},e_{2},e_{3},e_{4}$ , will be given by

$$\begin{eqnarray}\displaystyle & \displaystyle e_{1}=e=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right),\quad e_{2}=\unicode[STIX]{x1D70E}_{1}e\unicode[STIX]{x1D70E}_{1}^{-1}=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right), & \displaystyle \nonumber\\ \displaystyle & \displaystyle e_{3}=\unicode[STIX]{x1D70E}_{4}e\unicode[STIX]{x1D70E}_{4}^{-1}=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right),\quad e_{4}=(\unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4})e(\unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4})^{-1}=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 1 & 1\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right). & \displaystyle \nonumber\end{eqnarray}$$

The centralizer of  $S$ in  $\mathbb{U}_{5}(\mathbb{F}_{2})$ , which we denote by  $\mathscr{C}(S)$ , is easily seen to be composed of the elements  $g$ for which  $s_{1}(g)=s_{4}(g)=0$ , that is  $\mathscr{C}(S)=\ker s_{1}\cap \ker s_{4}$ . In particular  $\unicode[STIX]{x1D70E}_{2}$ and  $\unicode[STIX]{x1D70E}_{3}$ centralize  $S$ , and from the formulae above we see that  $S$ is normal in  $\mathbb{U}_{5}(\mathbb{F}_{2})$ . We shall write  $G=\mathbb{U}_{5}(\mathbb{F}_{2})/S$ , which we identify with  $D_{4}\times D_{4}$ , as we visibly may. The image of  $\mathscr{C}(S)$ in  $G$ , that is  $\mathscr{C}(S)/S$ , will be denoted by  $N$ , a normal subgroup of  $G$ .

One has  $G/N=\mathbb{U}_{5}(\mathbb{F}_{2})/\mathscr{C}(S)=\langle \unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{4}\rangle \cong C_{2}^{2}$ (we shall often write  $\unicode[STIX]{x1D70E}_{i}$ for the image of this element in various quotients, whenever no confusion can arise). The next observation is now clear, but it is crucial.

Lemma 2.4. The action of  $G/N$ on  $S$ , induced by conjugation, turns it into a free $\mathbb{F}_{2}[G/N]$ -module of rank $1$ . A specific isomorphism  $\mathbb{F}_{2}[G/N]\rightarrow S$ is given by $1\mapsto e$ , for example.

The group  $N$ itself also has a simple structure: one has  $N\cong C_{2}^{4}$ , a basis being 

$$\begin{eqnarray}\unicode[STIX]{x1D70E}_{2}[\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2}],\unicode[STIX]{x1D70E}_{2},\unicode[STIX]{x1D70E}_{3},\unicode[STIX]{x1D70E}_{3}[\unicode[STIX]{x1D70E}_{4},\unicode[STIX]{x1D70E}_{3}].\end{eqnarray}$$

(A choice which respects the ambient ‘symmetry’ already alluded to.) The corresponding dual basis of $\operatorname{H}^{1}(N,\mathbb{F}_{2})$ will be denoted  $x_{1},x_{2},x_{3},x_{4}$ . To bridge the notation with that of the previous sections, we regard  $N$ as sitting in  $D_{4}\times D_{4}$ , which itself possesses six maps  $s_{1},s_{2},t_{1},s_{3},s_{4},t_{2}$ to  $\mathbb{F}_{2}$ , using names adapted from § 2.1. With this notation, one has  $x_{1}=t_{1}$ , $x_{2}=s_{2}+t_{1}$ , $x_{3}=s_{3}+t_{2}$ , and $x_{4}=t_{2}$ (where restrictions to  $N$ are implicit).

Next we introduce some subgroups of  $S$ , and use them to produce extensions of  $N$ . These extensions turn out to control the entire situation, as will be explained. So we let

$$\begin{eqnarray}S_{1}:=\langle e_{2},e_{3},e_{4}\rangle ,\quad S_{2}:=\langle e_{1},e_{3},e_{4}\rangle ,\quad S_{3}:=\langle e_{1},e_{2},e_{4}\rangle .\end{eqnarray}$$

(The group  $S_{4}$ which could be defined using the same logic will not play any role, as it happens. Also note that among these three, only  $S_{1}$ respects the ambient ‘symmetry’.)

Lemma 2.5. Using the notation above, the following hold:

  1. (1) the cohomology class of the extension

    $$\begin{eqnarray}0\longrightarrow S/S_{1}\cong \mathbb{F}_{2}\longrightarrow \mathscr{C}(S)/S_{1}\longrightarrow N\longrightarrow 1\end{eqnarray}$$
    is  $x_{2}x_{3}$ ;
  2. (2) the cohomology class of the extension

    $$\begin{eqnarray}0\longrightarrow S/S_{2}\cong \mathbb{F}_{2}\longrightarrow \mathscr{C}(S)/S_{2}\longrightarrow N\longrightarrow 1\end{eqnarray}$$
    is  $x_{1}x_{3}$ ;
  3. (3) the cohomology class of the extension

    $$\begin{eqnarray}0\longrightarrow S/S_{3}\cong \mathbb{F}_{2}\longrightarrow \mathscr{C}(S)/S_{3}\longrightarrow N\longrightarrow 1\end{eqnarray}$$
    is  $x_{2}x_{4}$ .

Proof. An element of  $\mathscr{C}(S)$ has the form

$$\begin{eqnarray}g=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}(g) & u_{1}(g) & z(g)\\ 0 & 1 & s_{2}(g) & u_{3}(g) & u_{2}(g)\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Let us multiply two of these, say  $g$ and  $g^{\prime }$ , using the shorthand  $s_{2}=s_{2}(g)$ and  $s_{2}^{\prime }=s_{2}(g^{\prime })$ , and so on:

We shall use the set-theoretic section  $\sec :N\rightarrow \mathscr{C}(S)$ given by

$$\begin{eqnarray}\sec (g)=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}(g) & 0 & 0\\ 0 & 1 & s_{2}(g) & 0 & 0\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Let us prove (1). Using the section  $N\rightarrow \mathscr{C}(S)/S_{1}$ induced by  $\sec$ , we end up with the bijection of sets $\unicode[STIX]{x1D6F7}:S/S_{1}\times N\rightarrow \mathscr{C}(S)/S_{1}$ given by

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}(x,g)=x\sec (g)=\left(\begin{array}{@{}ccccc@{}}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & x & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right)\times \sec (g)=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}(g) & 0 & 0\\ 0 & 1 & s_{2}(g) & x & 0\\ 0 & 0 & 1 & s_{3}(g) & t_{2}(g)\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Here we have used (∗) to perform the calculation. A caveat: in these expressions, we have identified  $S/S_{1}$ with  $\mathbb{F}_{2}$ (this can be done uniquely!), so that  $x\in S/S_{1}$ can be seen as an entry ( $0$ or $1$ ) of a matrix. A second caveat is that the matrix displayed is understood modulo  $S_{1}$ only.

From the theory of group extensions, we have $\unicode[STIX]{x1D6F7}(x,g)\unicode[STIX]{x1D6F7}(y,g^{\prime })=\unicode[STIX]{x1D6F7}(x+y+c(g,g^{\prime }),gg^{\prime })$ , where the expression $c(g,g^{\prime })$ , is what we are after, that is, it is a two-cocycle representing the cohomology class of the extension under scrutiny. So we compute, from (∗), that

$$\begin{eqnarray}\unicode[STIX]{x1D6F7}(x,g)\unicode[STIX]{x1D6F7}(y,g^{\prime })=\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}+t_{1}^{\prime } & t_{1}s_{3}^{\prime } & t_{1}t_{2}^{\prime }\\ 0 & 1 & s_{2}+s_{2}^{\prime } & s_{2}s_{3}^{\prime }+x+y & s_{2}t_{2}^{\prime }\\ 0 & 0 & 1 & s_{3}+s_{3}^{\prime } & t_{2}+t_{2}^{\prime }\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

Another useful computational remark is that, in $S$ identified with the additive group of  $2\times 2$ -matrices, we have

$$\begin{eqnarray}\left(\begin{array}{@{}cc@{}}a & b\\ c & d\end{array}\right)=(a+b+c+d)e_{1}+(a+b)e_{2}+(b+d)e_{3}+be_{4}\equiv \left(\begin{array}{@{}cc@{}}0 & 0\\ a+b+c+d & 0\end{array}\right)\text{ mod }S_{1}.\end{eqnarray}$$

Thus the last matrix displayed, viewed in  $\mathscr{C}(S)/S_{1}$ , is also

$$\begin{eqnarray}\left(\begin{array}{@{}ccccc@{}}1 & 0 & t_{1}+t_{1}^{\prime } & 0 & 0\\ 0 & 1 & s_{2}+s_{2}^{\prime } & x+y+s_{2}s_{3}^{\prime }+t_{1}s_{3}^{\prime }+s_{2}t_{2}^{\prime }+t_{1}t_{2}^{\prime } & 0\\ 0 & 0 & 1 & s_{3}+s_{3}^{\prime } & t_{2}+t_{2}^{\prime }\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

We conclude that, as predicted, $c(g,g^{\prime })=s_{2}(g)s_{3}(g^{\prime })+t_{1}(g)s_{3}(g^{\prime })+s_{2}(g)t_{2}(g^{\prime })+t_{1}(g)t_{2}(g^{\prime })=x_{2}(g)x_{3}(g^{\prime })$ , and  $c$ is indeed the cup-product of  $x_{2}$ and  $x_{3}$ .

The proofs of (2) and (3) are similar. ◻

Remark 2.6. Note that in each case  $\mathscr{C}(S)/S_{i}\cong C_{2}^{2}\times D_{4}$ .

3 The fundamental cup-products

In this section, we prove Theorem A, as stated in the Introduction. We shall see that we are led naturally to another statement first, in which the following notation is used. When  $a\in F$ , we put  $F_{a}=F[X]/(X^{2}-a)$ . When  $e+fX\in F_{a}$ , we define its norm to be $\operatorname{N}_{F_{a}/F}(e+fX)=e^{2}-af^{2}$ . When a square root  $\sqrt{a}$ has been chosen in some field containing  $F$ , there is a homomorphism $F_{a}\rightarrow F[\sqrt{a}]$ mapping  $X$ to  $\sqrt{a}$ . Of course when  $a$ is not a square in  $F$ , this map is an isomorphism of extensions of  $F$ , and the norm just introduced coincides with the usual norm map between fields. However, it is useful to work with  $F_{a}$ for those occasions when  $a$ is already a square in  $F$ .

We start by the implication $(1)\;\Longrightarrow \;(2)$ from Theorem A.

3.1 Proving that the four cup-products vanish

Theorem 3.1. Let  $F$ be a field of characteristic not  $2$ , let $a,b,c,d\in F^{\times }$ be given and put $E:=F[\sqrt{a},\sqrt{d}]$ . Suppose there exist a continuous homomorphism  $\overline{\unicode[STIX]{x1D711}}:G_{F}\rightarrow \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})$ such that  $\unicode[STIX]{x1D712}_{a}=s_{1}\circ \overline{\unicode[STIX]{x1D711}}$ , $\unicode[STIX]{x1D712}_{b}=s_{2}\circ \overline{\unicode[STIX]{x1D711}}$ , $\unicode[STIX]{x1D712}_{c}=s_{3}\circ \overline{\unicode[STIX]{x1D711}}$ , and $\unicode[STIX]{x1D712}_{d}=s_{4}\circ \overline{\unicode[STIX]{x1D711}}$ . (In other words, we assume that $\langle a,b,c,d\rangle$ is defined.) Then there exist  $\widetilde{B}\in F_{a}$ and  $\widetilde{C}\in F_{d}$ such that the following hold, where $B$ denotes the image of $\widetilde{B}$ under $F_{a}\rightarrow F[\sqrt{a}]$ and $C$ denotes the image of $\widetilde{C}$ under $F_{d}\rightarrow F[\sqrt{d}]$ .

  1. (1) One has $\operatorname{N}_{F_{a}/F}(\widetilde{B})=b$ modulo squares and $\operatorname{N}_{F_{d}/F}(\widetilde{C})=c$ modulo squares.

  2. (2) One has $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ , and $(b,c)_{F}=0$ .

  3. (3) There is a class  $u\in \operatorname{H}^{2}(F,\mathbb{F}_{2})$ whose image under the canonical restriction map $\operatorname{H}^{2}(F,\,\mathbb{F}_{2})\longrightarrow \operatorname{H}^{2}(E,\mathbb{F}_{2})$ is $(B,C)_{E}$ .

  4. (4) Suppose that  $\overline{\unicode[STIX]{x1D711}}$ can be lifted to  $\unicode[STIX]{x1D711}:G_{F}\rightarrow \mathbb{U}_{5}(\mathbb{F}_{2})$ . (In other words, assume that $\langle a,b,c,d\rangle$ vanishes.) Then one has  $(B,C)_{E}=0$ .

Note that the mere existence of  $\widetilde{B}$ and  $\widetilde{C}$ implies also that  $(a,b)_{F}=0$ and  $(c,d)_{F}=0$ , by Proposition 2.3.

Proof. First we consider the composition

$$\begin{eqnarray}\overline{\overline{\unicode[STIX]{x1D711}}}:G_{F}\longrightarrow \overline{U}_{5}(\mathbb{F}_{2})\longrightarrow \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})/S=G=D_{4}\times D_{4}.\end{eqnarray}$$

Projecting further onto the left factor, we make a first use of Proposition 2.3. We draw the existence of  $x,y,z\in F$ satisfying $x^{2}-ay^{2}=bz^{2}$ , that is  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=bz^{2}$ , with  $\widetilde{B}=x+yX\in F_{a}$ . If  $B=x+y\sqrt{a}$ is the corresponding element, then the proposition says that we can arrange to have  $\unicode[STIX]{x1D712}_{bB}=t_{1}\circ \overline{\unicode[STIX]{x1D711}}=x_{1}\circ \overline{\unicode[STIX]{x1D711}}$ , $\unicode[STIX]{x1D712}_{B}=(s_{2}+t_{1})\circ \overline{\unicode[STIX]{x1D711}}=x_{2}\circ \overline{\unicode[STIX]{x1D711}}$ . (We fully use the notation from § 2.4.)

Using the right factor, we find elements  $\widetilde{C}$ and  $C$ similarly, such that  $\unicode[STIX]{x1D712}_{C}=x_{3}\circ \overline{\unicode[STIX]{x1D711}}$ , $\unicode[STIX]{x1D712}_{cC}=x_{4}\circ \overline{\unicode[STIX]{x1D711}}$ . This also proves the first assertion.

We prove that the cup-products vanish as announced in the second assertion, starting with $(b,c)_{F}=0$ (which of course does not depend on the choices for  $B$ and  $C$ ). For this, consider the map

$$\begin{eqnarray}\mathbb{U}_{5}(\mathbb{F}_{2})\longrightarrow \mathbb{U}_{3}(\mathbb{F}_{2})=D_{4}\end{eqnarray}$$

which discards the top row and the rightmost column of an element of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ ; this factors through  $\overline{U}_{5}(\mathbb{F}_{2})$ . Postcomposing  $\overline{\unicode[STIX]{x1D711}}$ with this, we draw from Proposition 2.3 that  $(b,c)_{F}=0$ , as required.

Next we turn to the proof of  $(B,c)_{F[\sqrt{a}]}=0$ , the cup-product  $(b,C)_{F[\sqrt{d}]}$ being treated in a similar way. Define a group homomorphism  $\unicode[STIX]{x1D70B}:\ker (s_{1})\subset \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})\rightarrow \mathbb{U}_{3}(\mathbb{F}_{2})=D_{4}$ by

$$\begin{eqnarray}g\mapsto \left(\begin{array}{@{}ccc@{}}1 & t_{1}(g) & u_{1}(g)\\ 0 & 1 & s_{3}(g)\\ 0 & 0 & 1\end{array}\right).\end{eqnarray}$$

One checks that  $\unicode[STIX]{x1D70B}$ is well defined. Note that  $\unicode[STIX]{x1D70B}\circ \overline{\unicode[STIX]{x1D711}}:G_{F[\sqrt{a}]}\rightarrow D_{4}$ is a lift for  $(t_{1}\circ \overline{\unicode[STIX]{x1D711}},s_{3}\circ \overline{\unicode[STIX]{x1D711}}):G_{F[\sqrt{a}]}\rightarrow \mathbb{F}_{2}\times \mathbb{F}_{2}$ (using that  $G_{F[\sqrt{a}]}=\overline{\unicode[STIX]{x1D711}}^{-1}(\ker (s_{1}))$ ). Therefore, we see from Proposition 2.3 that the cup product of  $t_{1}\circ \overline{\unicode[STIX]{x1D711}}$ and  $s_{3}\circ \overline{\unicode[STIX]{x1D711}}$ is zero. This means, in alternative notation, that  $(Bb,c)_{F[\sqrt{a}]}=0$ , so  $(B,c)_{F[\sqrt{a}]}=0$ .

We now turn to the third assertion. Let  $\unicode[STIX]{x1D6FC}$ be the cohomology class of the extension

$$\begin{eqnarray}0\longrightarrow \mathbb{F}_{2}\longrightarrow \mathbb{U}_{5}(\mathbb{F}_{2})\longrightarrow \overline{\mathbb{U}}_{5}(\mathbb{F}_{2})\longrightarrow 1.\end{eqnarray}$$

Let us write down an explicit two-cocycle  $\unicode[STIX]{x1D6FE}$ representing  $\unicode[STIX]{x1D6FC}$ . First, recall the functions $s_{1},t_{1},\ldots :\mathbb{U}_{5}(\mathbb{F}_{2})\rightarrow \mathbb{F}_{2}$ introduced in § 2.1. Multiplying two matrices  $g,h\in \mathbb{U}_{5}(\mathbb{F}_{2})$ , the top-right coefficient must be  $z(g)+z(h)+\unicode[STIX]{x1D6FE}(g,h)$ , and we deduce

$$\begin{eqnarray}\unicode[STIX]{x1D6FE}(g,h)=s_{1}(g)u_{2}(h)+t_{1}(g)t_{2}(h)+u_{1}(g)s_{4}(h).\end{eqnarray}$$

To obtain a two-cocycle representing the pull-back  $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})\in \operatorname{H}^{2}(F,\mathbb{F}_{2})$ , we only need compose with  $\overline{\unicode[STIX]{x1D711}}$ . Restricting to the subgroup  $G_{E}$ where  $s_{1}\circ \overline{\unicode[STIX]{x1D711}}$ and  $s_{4}\circ \overline{\unicode[STIX]{x1D711}}$ both vanish, we obtain that  $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})_{E}$ is represented by the two-cocyle

$$\begin{eqnarray}\unicode[STIX]{x1D70E},\unicode[STIX]{x1D70F}\mapsto (t_{1}\circ \overline{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D70E}))(t_{2}\circ \overline{\unicode[STIX]{x1D711}}(\unicode[STIX]{x1D70F})).\end{eqnarray}$$

It follows that  $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})_{E}=t_{1}\circ \overline{\unicode[STIX]{x1D711}}\cup t_{2}\circ \overline{\unicode[STIX]{x1D711}}$ , the cup-product of the classes  $t_{i}\circ \overline{\unicode[STIX]{x1D711}}\in \operatorname{H}^{1}(E,\mathbb{F}_{2})$ . Or in other words $\overline{\unicode[STIX]{x1D711}}^{\ast }(\unicode[STIX]{x1D6FC})_{E}=(bB,cC)_{E}$ .

Given that  $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ , and  $(b,c)_{F}=0$ , it follows that one has $u=\unicode[STIX]{x1D711}^{\ast }(\unicode[STIX]{x1D6FC})\in \operatorname{H}^{2}(F,\mathbb{F}_{2})$ satisfies  $u_{E}=(B,C)_{E}$ .

Finally, suppose that  $\unicode[STIX]{x1D711}$ exists as in the fourth assertion. We note that  $G_{E}=\unicode[STIX]{x1D711}^{-1}(\mathscr{C}(S))$ (again the notation $\mathscr{C}(S)$ for the centralizer of  $S$ is from § 2.4, and we had noted  $\mathscr{C}(S)=\ker (s_{1})\,\cap \,\ker (s_{4})$ ). The composition

$$\begin{eqnarray}f:G_{E}\longrightarrow \mathscr{C}(S)\longrightarrow N\end{eqnarray}$$

factors via  $\mathscr{C}(S)/S_{i}$ (for  $i=1,2,3$ ), so from the Lemma 2.5, we must have  $f^{\ast }(x_{2}x_{3})=0$ , $f^{\ast }(x_{1}x_{3})=0$ and  $f^{\ast }(x_{2}x_{4})=0$ . But these translate as  $(B,C)_{E}=0$ , which we were after, and $(bB,C)_{E}=0$ ,   $(B,cC)_{E}=0$ , consistently with the above.◻

3.2 Shapiro’s lemma and the converse

Let  $G$ be a finite group, let  $N$ be a subgroup, and let  $k$ be a field. For any  $kN$ -module  $A$ , we let  $\operatorname{Coind}_{N}^{G}(A)$ denote  $\operatorname{Hom}_{N}(kG,A)$ , which is a (left) $kG$ -module with action  $(\unicode[STIX]{x1D70E}\cdot f)(x)=f(x\unicode[STIX]{x1D70E})$ . (We are thinking of  $G$ and  $N$ as being the groups bearing those names in the discussion above, with  $k=\mathbb{F}_{2}$ , and  $A$ having trivial action.) The well-known Shapiro’s lemma states the existence of an isomorphism

$$\begin{eqnarray}sh:\operatorname{H}^{2}(G,\operatorname{Coind}_{N}^{G}(A))\longrightarrow \operatorname{H}^{2}(N,A).\end{eqnarray}$$

More precisely, the map is obtained using  $\operatorname{ev}:\operatorname{Coind}_{N}^{G}(A)\rightarrow A$ which evaluates at  $1\in G$ , followed by restriction (in cohomology) to  $N$ . Note, if the class  $\unicode[STIX]{x1D6FC}\in \operatorname{H}^{2}(G,\operatorname{Coind}_{N}^{G}(A))$ describes the extension

$$\begin{eqnarray}0\longrightarrow \operatorname{Coind}_{N}^{G}(A)\longrightarrow \unicode[STIX]{x1D6E4}\stackrel{p}{\longrightarrow }G\longrightarrow 1,\end{eqnarray}$$

then  $sh(\unicode[STIX]{x1D6FC})$ corresponds to

$$\begin{eqnarray}0\longrightarrow A\longrightarrow \frac{p^{-1}(N)}{\ker (\operatorname{ev})}\longrightarrow N\longrightarrow 1,\end{eqnarray}$$

as is easily verified.

In order to recognize that a given $G$ -module, say  $S$ , is isomorphic to  $\operatorname{Coind}_{N}^{G}(A)$ for some  $A$ , let us merely consider the case where  $A=k^{r}$ with trivial  $N$ -action, and assume that  $N$ is normal in  $G$ to boot. Then  $\operatorname{Coind}_{N}^{G}(A)=(k[G/N]^{\ast })^{r}$ . The dual module  $(k[G/N])^{\ast }$ is free of rank one, that is, it is isomorphic to  $k[G/N]$ . (Consider the map taking  $1\in G$ to $\unicode[STIX]{x1D6FF}_{1}$ , the Dirac delta function at  $1$ . Thus, $\unicode[STIX]{x1D6FF}_{1}(N)=1$ and $\unicode[STIX]{x1D6FF}_{1}(g\cdot N)=0$ if $g\notin N$ .) We conclude that  $S$ is isomorphic to  $\operatorname{Coind}_{N}^{G}(k^{r})$ if and only if  $N$ acts trivially and we can find a basis for  $S$ as a free  $k[G/N]$ of rank  $r$ . If this basis is  $\unicode[STIX]{x1D700}_{1},\ldots ,\unicode[STIX]{x1D700}_{r}$ , then  $\ker (ev)$ is the  $k$ -vector space spanned by  $g\unicode[STIX]{x1D700}_{i}$ , for  $g\in G$ , $g\neq 1$ , $i=1,\ldots ,r$ .

For example, let  $G,N,S$ recover their concrete meanings as in § 2.4 (all the accompanying notation will be used, too). Then Lemma 2.4 asserts that  $S\cong \operatorname{Coind}_{N}^{G}(\mathbb{F}_{2})$ , in such a way that  $\ker (ev)$ is identified with  $S_{1}$ . As a result, the cohomology class of

corresponds via  $sh$ to the cohomology class of the first extension treated in Lemma 2.5, that is, $x_{2}x_{3}$ .

But  $S$ can be regarded in another way. Consider the subgroup  $G^{\prime }\subset G$ of elements mapping into  $\langle \unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4}\rangle$ under  $G\rightarrow G/N$ (equivalently, $g\in G^{\prime }$ if  $s_{1}(g)=s_{4}(g)$ ), and view  $S$ as a  $G^{\prime }$ -module. Lemma 2.4 shows that  $S$ is a free  $\mathbb{F}_{2}[\langle \unicode[STIX]{x1D70E}_{1}\unicode[STIX]{x1D70E}_{4}\rangle ]=\mathbb{F}_{2}[G^{\prime }/N]$ -module, with basis  $e_{1},e_{2}$ (or alternatively  $e_{1},e_{3}$ ). Thus we also have  $S\cong \operatorname{Coind}_{N}^{G^{\prime }}(\mathbb{F}_{2}\oplus \mathbb{F}_{2})$ , and  $\ker (ev)$ is spanned by  $e_{3},e_{4}$ (or  $e_{2},e_{4}$ in the alternative). Now the cohomology class of

$$\begin{eqnarray}0\longrightarrow S\longrightarrow \mathbb{U}_{5}(\mathbb{F}_{2})^{\prime }\longrightarrow G^{\prime }\longrightarrow 1,\end{eqnarray}$$

where  $\mathbb{U}_{5}(\mathbb{F}_{2})^{\prime }$ is the preimage of  $G^{\prime }$ , is taken by Shapiro to that of the extension

$$\begin{eqnarray}0\longrightarrow \mathbb{F}_{2}e_{1}\oplus \mathbb{F}_{2}e_{2}\longrightarrow \frac{\mathscr{C}(S)}{\langle e_{3},e_{4}\rangle }\longrightarrow N\longrightarrow 1.\end{eqnarray}$$

This extension is described by two classes in  $\operatorname{H}^{2}(N,\mathbb{F}_{2})$ , corresponding to the exact sequences obtained by factoring out  $e_{1}$ and  $e_{2}$ respectively. From Lemma 2.5, these are  $x_{1}x_{3}$ and  $x_{2}x_{3}$ respectively. With the alternative choice of basis for  $S$ , this discussion ends with  $x_{2}x_{4}$ and  $x_{2}x_{3}$ .

There is a well-known version of Shapiro’s lemma for profinite groups (cf. [Reference Neukirch, Schmidt and WingbergNSW08, ch. 1, § 6]), which can be deduced from the version mentioned above using a straightforward limit argument. We record this version below in the context of Galois cohomology, since we will use it later on.

Lemma 3.2. Let  $E/F$ be a finite Galois extension, let  $A$ be a trivial, discrete  $G_{E}$ -module, and consider  $\operatorname{Coind}_{G_{E}}^{G_{F}}(A)$ , a discrete  $G_{F}$ -module. Then Shapiro’s map

$$\begin{eqnarray}\operatorname{H}^{2}(F,\operatorname{Coind}_{G_{E}}^{G_{F}}(A))\longrightarrow \operatorname{H}^{2}(E,A),\end{eqnarray}$$

defined as above, is an isomorphism.

With the preparations above, we can now prove our primary converse to Theorem 3.1.

Theorem 3.3. Let  $F$ be a field of characteristic not  $2$ , let  $a,b,c,d\in F^{\times }$ be given, and put  $E:=F[\sqrt{a},\sqrt{d}]$ . Assume that there exist  $\widetilde{B}\in F_{a}$ such that  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=b$ modulo squares, and  $\widetilde{C}\in F_{d}$ such that $\operatorname{N}_{F_{d}/F}(\widetilde{C})=c$ modulo squares, with the following additional property: if  $B$ denotes the image of  $\widetilde{B}$ under  $F_{a}\rightarrow F[\sqrt{a}]$ and  $C$ denotes the image of  $\widetilde{C}$ under  $F_{d}\rightarrow F[\sqrt{d}]$ , then one has $(B,C)_{E}=(B,c)_{E}=(b,C)_{E}=(b,c)_{E}=0$ .

Then there exist a continuous homomorphism  $\unicode[STIX]{x1D711}:G_{F}\rightarrow \mathbb{U}_{5}(\mathbb{F}_{2})$ such that  $s_{1}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{a}$ , $s_{2}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{b}$ , $s_{3}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{c}$ and  $s_{4}\circ \unicode[STIX]{x1D711}=\unicode[STIX]{x1D712}_{d}$ . In other words, the Massey product  $\langle a,b,c,d\rangle$ is defined and vanishes.

Proof. We start by assuming that neither  $a$ nor  $d$ is a square in  $F$ (we identify  $F_{a}$ and  $F_{d}$ with  $F[\sqrt{a}]$ and  $F[\sqrt{d}]$ respectively). From Proposition 2.3 (applied twice), we obtain a homomorphism

$$\begin{eqnarray}f:G_{F}\longrightarrow D_{4}\times D_{4}=G\end{eqnarray}$$

such that  $s_{i}\circ f=\unicode[STIX]{x1D712}_{a},\unicode[STIX]{x1D712}_{b},\unicode[STIX]{x1D712}_{c},\unicode[STIX]{x1D712}_{d}$ according as  $i=1,2,3,4$ , and also  $f^{\ast }(x_{1})=\unicode[STIX]{x1D712}_{bB}$ , $f^{\ast }(x_{2})=\unicode[STIX]{x1D712}_{B}$ , $f^{\ast }(x_{3})=\unicode[STIX]{x1D712}_{C}$ , $f^{\ast }(x_{4})=\unicode[STIX]{x1D712}_{cC}$ . If  $\unicode[STIX]{x1D6FC}\in \operatorname{H}^{2}(D_{4}\times D_{4},S)$ is the class of the extension

$$\begin{eqnarray}0\longrightarrow S\longrightarrow \mathbb{U}_{5}(\mathbb{F}_{2})\longrightarrow D_{4}\times D_{4}\longrightarrow 1,\end{eqnarray}$$

then its pull-back  $f^{\ast }(\unicode[STIX]{x1D6FC})\in \operatorname{H}^{2}(F,S)$ is represented by the fibred-product of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ with  $G_{F}$ over  $D_{4}\times D_{4}$ (via $f$ ). To conclude the proof of the theorem, we will show that one has  $f^{\ast }(\unicode[STIX]{x1D6FC})=0$ , so that this fibred-product is a split extension of $G_{F}$ by $S$ . The composition of such a splitting with the projection to $\mathbb{U}_{5}(\mathbb{F}_{2})$ provides the necessary homomorphism $\unicode[STIX]{x1D711}$ .

Note that  $G_{E}=f^{-1}(N)$ . Suppose first that  $[E:F]=4$ . The  $G_{F}$ -module  $S$ is isomorphic to  $\operatorname{Coind}_{G_{E}}^{G_{F}}(\mathbb{F}_{2})$ , the trivial module of  $G_{E}$ (co)induced to  $G_{F}$ . Shapiro’s lemma applies, yielding the isomorphism

$$\begin{eqnarray}\operatorname{H}^{2}(F,S)\longrightarrow \operatorname{H}^{2}(E,\mathbb{F}_{2}).\end{eqnarray}$$

From the comments above and the naturality of Shapiro’s isomorphism, we see that  $f^{\ast }(\unicode[STIX]{x1D6FC})$ is taken to  $f^{\ast }(x_{2}x_{3})=f^{\ast }(x_{2})f^{\ast }(x_{3})=(B,C)_{E}=0$ , so we are done in this case.

We turn to the case where $a=d$ mod squares and $[E:F]=2$ ; another way to phrase this is by saying that the image of  $f$ lies within  $G^{\prime }$ . Now the  $G_{F}$ -module  $S$ is isomorphic, although not canonically, to  $\operatorname{Coind}_{G_{E}}^{G_{F}}(\mathbb{F}_{2}\oplus \mathbb{F}_{2})$ : we have at least the two possibilities given in the discussion preceding the proof, and for definiteness say we pick the basis  $e_{1},e_{2}$ .

Now Shapiro’s lemma gives an isomorphism

$$\begin{eqnarray}\operatorname{H}^{2}(F,S)\longrightarrow \operatorname{H}^{2}(E,\mathbb{F}_{2}\oplus \mathbb{F}_{2})=\operatorname{H}^{2}(E,\mathbb{F}_{2})\oplus \operatorname{H}^{2}(E,\mathbb{F}_{2}).\end{eqnarray}$$

This takes  $f^{\ast }(\unicode[STIX]{x1D6FC})$ to a pair of cohomology classes, and again from naturality, they are  $f^{\ast }(x_{2}x_{3})=(B,C)_{E}$ and  $f^{\ast }(x_{1}x_{3})=(bB,C)_{E}$ . These are both zero by assumption, and  $f^{\ast }(\unicode[STIX]{x1D6FC})=0$ also in this case.

Finally, suppose that  $a$ is a square in  $F$ (a symmetric argument deals with the case when  $d$ is a square). A neat way to handle this is to use the Massey vanishing conjecture for  $n=3$ , (which is now a theorem, see [Reference MatzriMat11, Reference Efrat and MatzriEM17, Reference Mináč and TânMT17c, Reference Mináč and TânMT16]). First we claim that  $(b,c)_{F}=0$ . Indeed, $(b,c)_{E}=0$ by assumption; if  $d$ is also a square in  $F$ then  $E=F$ , and otherwise  $E=F[\sqrt{d}]$ and  $\operatorname{N}_{E/F}(C)=c$ mod squares, so  $(b,C)_{E}=0$ implies  $(b,c)_{F}=0$ by the projection formula in group cohomology. We also have  $(c,d)_{F}=0$ by Proposition 2.3. We deduce that the Massey product  $\langle b,c,d\rangle$ vanishes, and so that there is a continuous  $\unicode[STIX]{x1D713}:G_{F}\rightarrow \mathbb{U}_{4}(\mathbb{F}_{2})$ compatible with  $b,c,d$ . However, if we see  $\mathbb{U}_{4}(\mathbb{F}_{2})$ as the subgroup of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ consisting of those matrices whose first row is that of the identity matrix, we see that we have built  $G_{F}\rightarrow \mathbb{U}_{5}(\mathbb{F}_{2})$ compatible with  $1,b,c,d$ , as required.◻

Remark 3.4. It is possible to avoid relying on the Massey vanishing conjecture for  $n=3$ if one wishes to do so. Here is a sketch. When  $d$ is not a square, use another argument based on Shapiro’s lemma, this time replacing  $G^{\prime }$ by  $G^{\prime \prime }$ , the subgroup of  $G$ of elements mapping to  $\langle \unicode[STIX]{x1D70E}_{d}\rangle \subset G/N$ . When  $d$ is a square, define  $f$ as in the first part of the proof; this time  $f$ takes values in  $N$ . The exact sequence

$$\begin{eqnarray}0\longrightarrow S\longrightarrow \mathscr{C}(S)\longrightarrow N\longrightarrow 1\end{eqnarray}$$

is central, and controlled by four cohomology classes in  $\operatorname{H}^{2}(N,\mathbb{F}_{2})$ , pulling back to  $(B,C)$ , $(b,C)$ , $(B,c)$ and  $(b,c)$ in  $\operatorname{H}^{2}(F,\mathbb{F}_{2})$ under  $f^{\ast }$ .

The proof of Theorem A is now almost complete. In fact, what we have is the following.

Theorem 3.5. Let  $F$ be a field of characteristic not  $2$ , and let  $a,b,c,d\in F^{\times }$ be given. Then the following statements are equivalent.

  1. (1) The Massey product $\langle a,b,c,d\rangle$ is defined and vanishes.

  2. (2) There exist  $\widetilde{B}\in F_{a}$ such that  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=b$ modulo squares, and  $\widetilde{C}\in F_{d}$ such that $\operatorname{N}_{F_{d}/F}(\widetilde{C})=c$ modulo squares, with the following extra property. If  $B$ is the image of  $\widetilde{B}$ under  $F_{a}\rightarrow F[\sqrt{a}]$ and  $C$ is the image of  $\widetilde{C}$ under  $F_{d}\rightarrow F[\sqrt{d}]$ , then $(B,C)_{F[\sqrt{a},\sqrt{d}]}=0$ , $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ , and $(b,c)_{F}=0$ .

  3. (3) There exist  $\widetilde{B}\in F_{a}$ such that  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=b$ modulo squares, and  $\widetilde{C}\in F_{d}$ such that $\operatorname{N}_{F_{d}/F}(\widetilde{C})=c$ modulo squares, with the following additional property: If  $B$ denotes the image of  $\widetilde{B}$ under  $F_{a}\rightarrow F[\sqrt{a}]$ and  $C$ denotes the image of  $\widetilde{C}$ under  $F_{d}\rightarrow F[\sqrt{d}]$ , then one has $(B,C)_{E}=(B,c)_{E}=(b,C)_{E}=(b,c)_{E}=0$ .

Proof. Theorem 3.1 shows the implication (1) $\;\Longrightarrow \;$ (2), while (2) $\;\Longrightarrow \;$ (3) is trivial, and Theorem 3.3 shows (3) $\;\Longrightarrow \;$ (1).◻

The only difference with Theorem A is the presence of the elements  $\widetilde{B}$ and  $\widetilde{C}$ instead of just  $B,C$ . Clearly, if neither  $a$ nor  $d$ is a square in  $F$ , then the two results are the same. On the other hand, when one of these two elements is a square, in fact when one of  $a,b,c,d$ is a square, things become very easy, as we proceed to show in the following subsection.

3.3 Some trivial cases

Lemma 3.6. Let  $F$ be a field of characteristic different from  $2$ and let  $a,b,c,d\in F^{\times }$ be given. Suppose that $(a,b)_{F}=(b,c)_{F}=(c,d)_{F}=0$ . Finally, assume that one of  $a,b,c,d$ is a square in  $F$ . Then the Massey product $\langle a,b,c,d\rangle$ is defined and vanishes.

Proof. First assume that  $a$ is a square, or equivalently  $a=1$ . Then the argument given in the last paragraph of the proof of Theorem 3.3 shows that $0\in \langle a,b,c,d\rangle$ . The case when  $d=1$ is treated similarly.

On the other hand, suppose that  $b=1$ . From  $(c,d)_{F}=0$ , we draw the existence of $G_{F}\rightarrow \mathbb{U}_{3}(\mathbb{F}_{2})$ compatible with  $c$ and  $d$ , by Proposition 2.3. The element  $a$ itself defines  $\unicode[STIX]{x1D712}_{a}:G_{F}\rightarrow \mathbb{F}_{2}\cong C_{2}$ . Since the subgroup of  $\mathbb{U}_{5}(\mathbb{F}_{2})$ generated by  $\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{3},\unicode[STIX]{x1D70E}_{4}$ is isomorphic to  $C_{2}\times \mathbb{U}_{3}(\mathbb{F}_{2})$ , we see immediately that we may combine our two homomorphisms into one of the form  $G_{F}\rightarrow \mathbb{U}_{5}(\mathbb{F}_{2})$ , showing that $0\in \langle a,1,b,c\rangle$ . The case when  $c=1$ is treated similarly.◻

Remark 3.7. The above proof, via the reference to the proof of Theorem 3.3, uses the Massey vanishing conjecture for  $n=3$ . Without this, it is still a general fact about Massey products that $\langle a,b,c,d\rangle$ vanishes when it is defined and one of them is a square [Reference FennFen83, Lemma 6.2.4]. However, the statement just given is stronger.

We proceed to show how we can improve the statement of Theorem 3.5 to that of Theorem A from the Introduction, so that the two are in fact equivalent. We have already mentioned that this is obvious when neither  $a$ nor  $d$ is a square in  $F$ .

Let us call (1A), (2A), (3A) the conditions of Theorem A, and keep (1), (2), (3) for those of Theorem 3.5, which we know are equivalent. Note (1) $=$ (1A).

Suppose  $a$ is a square in  $F$ , but not  $d$ . Assume condition (1A). We must show that condition (2A) holds. Indeed, from condition (2), we have the element  $C$ with  $\operatorname{N}_{F[\sqrt{d}]/F}(C)=c$ mod squares, and satisfying  $(b,C)_{F[\sqrt{d}]}=0$ , and moreover  $(b,c)_{F}=0$ . Now put  $B=b\in F=F[\sqrt{a}]$ , so that  $\operatorname{N}_{F[\sqrt{a}]/F}(B)=B=b$ . Then  $(B,C)_{E}=(b,C)_{E}=0$ , while  $(B,c)_{F[\sqrt{a}]}=(b,c)_{F}=0$ . We do have condition (2A), and so also (3A).

Conversely, condition (3A) is enough to ensure that  $(a,b)_{F}=(1,b)_{F}=0$ , $(b,c)_{F}=0$ (apply the projection formula to  $(b,C)_{E}=0$ ) and  $(c,d)_{F}=0$ (merely because  $C$ exists, cf. Proposition 2.3). So Lemma 3.6 applies and shows that condition (1A) holds.

The situation when  $a$ is not a square, but  $d$ is, is clearly similar.

Now suppose  $a$ and  $d$ are both squares. Suppose condition (1A) holds, and so also (2), and we have  $(a,b)=(c,d)=0$ (because  $\widetilde{B}$ and  $\widetilde{C}$ exist, cf. remark after Theorem 3.1) and  $(b,c)=0$ . Thus condition (2A) holds with  $B=b$ and  $C=c$ , and (2A) $=$ (3A) here. Conversely, condition (3A) contains the statement $(b,c)=0$ , while $(a,b)=(1,b)=0$ and  $(c,d)=(c,1)=0$ , and we see by the last lemma that condition (1A) holds. We have therefore just proven Theorem A.

Remark 3.8. Theorem 3.5 is heavier on notation than Theorem A, so we have deemed it unfit for the Introduction. However, the extra case-by-case considerations needed to establish the latter, as just given, are perhaps an indication that it is less natural. (Also, it relies more seriously on the Massey vanishing conjecture for  $n=3$ , see Remark 3.4.) From now on, we shall refer to Theorem 3.5, rather than to Theorem A.

3.4 Maps from profinite groups into  $\mathbb{U}_{5}(\mathbb{F}_{2})$

A routine modification of the arguments given above produces the next result.

Theorem 3.9. Let  $\unicode[STIX]{x1D6E4}$ be a profinite group, and let  $\unicode[STIX]{x1D712}_{1},\unicode[STIX]{x1D712}_{2},\unicode[STIX]{x1D712}_{3},\unicode[STIX]{x1D712}_{4}\in \operatorname{H}^{1}(\unicode[STIX]{x1D6E4},\mathbb{F}_{2})$ be given. Put  $\unicode[STIX]{x1D6E4}_{1}:=\ker (\unicode[STIX]{x1D712}_{1})$ , $\unicode[STIX]{x1D6E4}_{4}:=\ker (\unicode[STIX]{x1D712}_{4})$ , and $\unicode[STIX]{x1D6E4}_{14}:=\unicode[STIX]{x1D6E4}_{1}\cap \unicode[STIX]{x1D6E4}_{4}$ . The the following statements are equivalent.

  1. (1) There exists a continuous homomorphism $\unicode[STIX]{x1D711}:\unicode[STIX]{x1D6E4}\rightarrow \mathbb{U}_{5}(\mathbb{F}_{2})$ such that  $\unicode[STIX]{x1D712}_{i}=s_{i}\circ \unicode[STIX]{x1D711}$ for  $i=1,2,3,4$ . In other words, $\langle \unicode[STIX]{x1D712}_{1},\unicode[STIX]{x1D712}_{2},\unicode[STIX]{x1D712}_{3},\unicode[STIX]{x1D712}_{4}\rangle$ is defined and vanishes.

  2. (2) There exist a continuous homomorphism  $\unicode[STIX]{x1D6E4}\rightarrow D_{4}$ given by

    $$\begin{eqnarray}\unicode[STIX]{x1D6FE}\mapsto \left(\begin{array}{@{}ccc@{}}1 & \unicode[STIX]{x1D712}_{1}(\unicode[STIX]{x1D6FE}) & \unicode[STIX]{x1D701}_{1}(\unicode[STIX]{x1D6FE})\\ 0 & 1 & \unicode[STIX]{x1D712}_{2}(\unicode[STIX]{x1D6FE})\\ 0 & 0 & 1\end{array}\right),\end{eqnarray}$$
    and another one given by
    $$\begin{eqnarray}\unicode[STIX]{x1D6FE}\mapsto \left(\begin{array}{@{}ccc@{}}1 & \unicode[STIX]{x1D712}_{3}(\unicode[STIX]{x1D6FE}) & \unicode[STIX]{x1D701}_{2}(\unicode[STIX]{x1D6FE})\\ 0 & 1 & \unicode[STIX]{x1D712}_{4}(\unicode[STIX]{x1D6FE})\\ 0 & 0 & 1\end{array}\right),\end{eqnarray}$$
    such that  $\unicode[STIX]{x1D701}_{1}|_{\unicode[STIX]{x1D6E4}_{14}}\cup \unicode[STIX]{x1D701}_{2}|_{\unicode[STIX]{x1D6E4}_{14}}=0$ , $\unicode[STIX]{x1D701}_{1}|_{\unicode[STIX]{x1D6E4}_{1}}\cup \unicode[STIX]{x1D712}_{3}|_{\unicode[STIX]{x1D6E4}_{1}}=0$ , $\unicode[STIX]{x1D712}_{2}|_{\unicode[STIX]{x1D6E4}_{4}}\cup \unicode[STIX]{x1D701}_{2}|_{\unicode[STIX]{x1D6E4}_{4}}=0$ , $\unicode[STIX]{x1D712}_{2}\cup \unicode[STIX]{x1D712}_{3}=0$ .
  3. (3) $\unicode[STIX]{x1D701}_{1}$ and  $\unicode[STIX]{x1D701}_{2}$ exist as above and satisfy  $\unicode[STIX]{x1D701}_{1}|_{\unicode[STIX]{x1D6E4}_{14}}\cup \unicode[STIX]{x1D701}_{2}|_{\unicode[STIX]{x1D6E4}_{14}}=\unicode[STIX]{x1D701}_{1}|_{\unicode[STIX]{x1D6E4}_{14}}\cup \unicode[STIX]{x1D712}_{3}|_{\unicode[STIX]{x1D6E4}_{14}}=\unicode[STIX]{x1D712}_{2}|_{\unicode[STIX]{x1D6E4}_{14}}\cup \unicode[STIX]{x1D701}_{2}|_{\unicode[STIX]{x1D6E4}_{14}}=\unicode[STIX]{x1D712}_{2}|_{\unicode[STIX]{x1D6E4}_{14}}\cup \unicode[STIX]{x1D712}_{3}|_{\unicode[STIX]{x1D6E4}_{14}}=0$ .

We shall have no use for this theorem in the sequel, so we leave the proof to the reader.

4 First applications

While our main objective in this paper is to prove the $4$ -Massey vanishing conjecture for number fields, there are a few cases which can be treated over any field. Usually, we use the following trick when working ‘by hand’. It will have a more theoretical use below, too.

Proposition 4.1. Let  $F$ be a field of characteristic not  $2$ , and let  $a,b,c,d\in F^{\times }$ . Let  $\widetilde{B}_{0}\in F_{a}$ be any initial element such that  $\operatorname{N}_{F_{a}/F}(\widetilde{B}_{0})=b$ , and likewise let  $\widetilde{C}_{0}\in F_{d}$ be any initial element such that  $\operatorname{N}_{F_{d}/F}(\widetilde{C}_{0})=c$ . The following statements are equivalent:

  1. (1) there exist  $\widetilde{B}\in F_{a}$ with  $\operatorname{N}_{F_{a}/F}(\widetilde{B})=bz_{1}^{2}$ , and  $\widetilde{C}\in F_{d}$ with  $\operatorname{N}_{F_{d}/F}(\widetilde{C})=cz_{2}^{2}$ , for some  $z_{1},z_{2}\in F^{\times }$ , such that we have simultaneously $(B,C)_{F[\sqrt{a},\sqrt{d}]}=0$ , $(B,c)_{F[\sqrt{a}]}=0$ , $(b,C)_{F[\sqrt{d}]}=0$ , $(b,c)_{F}=0$ , where  $B\in F[\sqrt{a}]$ and  $C\in F[\sqrt{d}]$ correspond to  $\widetilde{B},\widetilde{C}$ respectively;

  2. (2) there exist  $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\in F^{\times }$ such that  $\widetilde{B}=\unicode[STIX]{x1D6FD}\widetilde{B}_{0}$ and  $\widetilde{C}=\unicode[STIX]{x1D6FE}\widetilde{C}_{0}$ satisfy the previous condition.

Of course, by Theorem 3.5 these conditions are also equivalent to the vanishing of  $\langle a,b,c,d\rangle$ , but the point of the proposition is to show that it makes sense to start with any  $\widetilde{B}_{0},\widetilde{C}_{0}$ and then look for the ‘modifiers’ $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}$ .

Proof. It is plain that (2) implies (1), so here is the non-trivial part. Assume that $\widetilde{B},\widetilde{C}$ exist as in the first part. We claim that we can in fact find $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\in F$ and $\unicode[STIX]{x1D706}\in F[\sqrt{a}]$ , $\unicode[STIX]{x1D707}\in F[\sqrt{d}]$ such that $B=\unicode[STIX]{x1D6FD}B_{0}\unicode[STIX]{x1D706}^{-2}$ and $C=\unicode[STIX]{x1D6FE}C_{0}\unicode[STIX]{x1D707}^{-2}$ , where $B_{0}\in F[\sqrt{a}]$ and $C_{0}\in F[\sqrt{d}]$ correspond to $\widetilde{B}_{0},\widetilde{C}_{0}$ respectively. Clearly this will give the result.

To prove the claim, we suppose first that  $a$ is not a square in the field  $F$ , and write $\operatorname{N}_{F[\sqrt{a}]/F}(Bz_{1}^{-1})=b=\operatorname{N}_{F[\sqrt{a}]/F}(B_{0})$ . Since we have $\operatorname{N}_{F[\sqrt{a}]/F}(Bz_{1}^{-1}B_{0}^{-1})=1$ , by Hilbert’s Theorem 90 we have

$$\begin{eqnarray}\frac{Bz_{1}^{-1}}{B_{0}}=\frac{\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D706})}{\unicode[STIX]{x1D706}}\end{eqnarray}$$

for some  $\unicode[STIX]{x1D706}\in F[\sqrt{a}]$ , where  $\unicode[STIX]{x1D70E}$ is the non-trivial element of  $\operatorname{Gal}(F[\sqrt{a}]/F)$ . Rewrite this  $B=\unicode[STIX]{x1D6FD}B_{0}\unicode[STIX]{x1D706}^{-2}$ , with  $\unicode[STIX]{x1D6FD}=z_{1}\unicode[STIX]{x1D706}\unicode[STIX]{x1D70E}(\unicode[STIX]{x1D706})\in F$ , and we are done.

If on the other hand  $a$ is a square in  $F$ , then  $B_{0},B\in F^{\times }$ , so we may put  $\unicode[STIX]{x1D6FD}=B/B_{0}$ and  $\unicode[STIX]{x1D706}=1$ .

A similar argument works with  $C,C_{0}$ .◻

We give a series of examples. These will demonstrate how, in practice, one looks for  $\unicode[STIX]{x1D6FD}$ and  $\unicode[STIX]{x1D6FE}$ rather than  $B$ and  $C$ directly. Logically speaking, only the trivial implication of the last proposition is used, at this point (although knowing that the converse holds gives us confidence in the whole approach).

The elements  $a$ and  $d$ are always assumed not to be squares in  $F$ , so we identify  $F_{a}$ and  $F[\sqrt{a}]$ , as well as  $F_{d}$ and  $F[\sqrt{d}]$ , and  $\widetilde{B}_{0}=B_{0}$ , $\widetilde{C}_{0}=C_{0}$ . As you have guessed, the characteristic is always not $2$ . We will make some forward references to the results of the next section, which reduce the number of conditions to check.

Example 4.2 (The abaa case).

In this example, we show that when  $(a,b)=(a,a)=0$ , the Massey product  $\langle a,b,a,a\rangle$ is defined and vanishes, under the assumption that  $a\neq b$ modulo squares, and that  $a$ and  $b$ are not squares.

Clearly we can find  $B_{0}$ and  $C_{0}$ as in the Proposition, by our assumption (and using Proposition 2.3, of course). We claim that  $B_{0}$ and  $C_{0}$ form a basis for  $F[\sqrt{a}]$ as an  $F$ -vector space. Indeed, if we had  $\unicode[STIX]{x1D6FD}B_{0}=\unicode[STIX]{x1D6FE}C_{0}$ for  $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\in F^{\times }$ , then by taking norms to  $F$ we would find that  $b=a$ modulo squares, a contradiction. Therefore there must exist  $\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE}\in F$ such that

$$\begin{eqnarray}\unicode[STIX]{x1D6FD}B_{0}+\unicode[STIX]{x1D6FE}C_{0}=1.\end{eqnarray}$$

This implies  $(\unicode[STIX]{x1D6FD}B_{0},\unicode[STIX]{x1D6FE}C_{0})_{F[\sqrt{a}]}=0$ . Moreover  $(\unicode[STIX]{x1D6FD}B_{0},a)_{F[\sqrt{a}]}=(\unicode[STIX]{x1D6FD}B_{0},1)_{F[\sqrt{a}]}=0$ . By Lemma 5.5 below, the other cup-products vanish automatically (the reader will also enjoy looking for a direct argument). By Theorem 3.5, the Massey product is defined and vanishes.

Example 4.3 (The aaaa case).

Now we complete the discussion of the previous example, and turn to the case when  $a=b$ modulo squares (still assuming that  $a$ is not a square). We show that  $(a,a)=0$ implies that $\langle a,a,a,a\rangle$ is defined and vanishes.

Since  $(a,a)=0$ , we draw the existence of  $B_{0}$ with  $\operatorname{N}_{F[\sqrt{a}]/F}(B_{0})=a$ from Proposition 2.3. It is always true that  $(a,-a)=0$ , so we have  $(a,-1)=0$ , implying that  $a$ is a sum of two squares in  $F$ . We invoke the ‘norm principle’ from [Reference Elman and LamEL76]: from the fact that  $\operatorname{N}_{F[\sqrt{a}]/F}(B_{0})$ is the product of two elements, namely  $a$ and  $1$ , which are each the sum of two squares in  $F$ , this principle implies that there exists  $\unicode[STIX]{x1D6FD}\in F^{\times }$ such that  $B=\unicode[STIX]{x1D6FD}B_{0}$ is the sum of two squares in  $F[\sqrt{a}]$ . In turn, this means that  $(B,-1)=0$ and so  $(B,B)=0$ .

Obviously  $(B,a)=(B,1)=0$ in the cohomology of  $F[\sqrt{a}]$ since  $a$ is a square there, so Theorem 3.5 applies (with