On the Tits–Weiss conjecture and the Kneser–Tits conjecture for $\mathrm {E}^{78}_{7,1}$ and $\mathrm {E}^{78}_{8,2}$ (With an Appendix by R. M. Weiss)

Abstract

We prove that the structure group of any Albert algebra over an arbitrary field is R-trivial. This implies the Tits–Weiss conjecture for Albert algebras and the Kneser–Tits conjecture for isotropic groups of type $\mathrm {E}_{7,1}^{78}, \mathrm {E}_{8,2}^{78}$ . As a further corollary, we show that some standard conjectures on the groups of R-equivalence classes in algebraic groups and the norm principle are true for strongly inner forms of type $^1\mathrm {E}_6$ .

1. Introduction

The primary aim of this article is to prove the long standing Tits–Weiss conjecture on U-operators in Albert algebras and the Kneser–Tits conjecture for algebraic groups of type $\mathrm {E}^{78}_{7,1}$ and $\mathrm {E}^{78}_{8,2}$ .

The Tits–Weiss conjecture asserts that the structure group ${\mathrm{Str}}(A)$ of an arbitrary Albert algebra A is generated by the inner structure group, formed by the so-called U-operators, and the central homotheties. This problem was raised by Tits and Weiss in their 2002 book [26], where they studied spherical buildings and the corresponding generalised polygons attached to isotropic groups of relative rank $2$ . Despite many efforts, this problem has remained unsolved.

If $\mathbf{G}$ is an isotropic simple simply connected group over a field K of relative rank $\geq 2$ , then by [18] the group ${\mathbf{G}}(K)$ is generated by the K-points of isotropic subgroups of ${\mathbf{G}}$ of relative rank $1$ . This result allows one to reduce many problems for ${\mathbf{G}}(K)$ to groups of relative rank $1$ . For instance, this is the case for the Kneser–Tits problem (see the paragraph below). Note also that isotropic groups of relative rank $1$ give rise to important examples of more general groups of rank 1. The latter were introduced by Tits in the early 1990s, who called them Moufang sets. They have proved to be important in the classification of simple groups, incidence geometry, the theory of buildings and other areas. Further still, rank 1 groups are useful in studying isotropic groups of exceptional types, where algebraic groups and their associated root subgroups are typically parametrised by a nonassociative structure and, as emphasised in [6], a rich interplay emerges between rank 1 groups, nonassociative algebras and linear algebraic groups.

The Kneser–Tits conjecture for a simple simply connected isotropic group ${\mathbf{G}}$ over a field K asserts that the abstract group ${\mathbf{G}}(K)$ of K-points of ${\mathbf{G}}$ coincides with its normal subgroup ${\mathbf{G}}(K)^+$ generated by the unipotent radicals of the minimal parabolic K-subgroups of ${\mathbf{G}}$ . We refer to [9] for a survey of the history and recent results on this conjecture. Its importance comes from the fact that the group ${\mathbf{G}}(K)^+$ has a natural $BN$ -pair structure and hence is projectively simple (i.e., simple modulo its centre) by a celebrated theorem of Tits. So if ${\mathbf{G}}(K)={\mathbf{G}}(K)^+$ , we would have many more new examples of projectively simple abstract groups. In this way, we would obtain analogues of finite simple groups of Lie type in the case of infinite fields. It is also worth mentioning that the information about the normal subgroup structure of ${\mathbf{G}}(K)$ is crucial in the arithmetic of algebraic groups for studying, among other things, congruence subgroups, discrete subgroups, lattices and locally symmetric spaces. In general, the Kneser–Tits conjecture does not hold and the first counterexample was constructed by V. Platonov in 1975 [17]. However, specialists believe that the conjecture does hold for many isotropic groups of exceptional type, including those of type $\mathrm {E}_{7,1}^{78}$ and $\mathrm {E}_{8,2}^{78}$ .

The bridge connecting the Tits–Weiss conjecture and the Kneser–Tits conjecture for the abovementioned forms of type $\mathrm {E}_7$ and $\mathrm {E}_8$ is provided by a theorem of Tits and Weiss (see the Appendix), which states that the two conjectures are equivalent. ¹ It is interesting to mention that the proof in the Appendix is characteristic free. Furthermore, using P. Gille’s results on Whitehead groups [9], one can easily see that

1. 1. the Kneser–Tits conjecture for the abovementioned groups reduces to the R-triviality of structure groups of Albert algebras and

2. 2. the conjecture holds in arbitrary characteristic once it is established in characteristic zero.

Our main result is the following.

Theorem. Let A be an Albert algebra over a field K. Then the structure group ${\mathbf{Str}}(A)$ of A is R-trivial; that is, for any field extension $F/K$ the group of R-equivalence classes ${\mathbf{Str}}(A)(F)/R$ is trivial.

As explained above, this implies that the Tits–Weiss conjecture on U-operators holds for Albert algebras over any field and that the same is true for the Kneser–Tits conjecture for groups of type $\mathrm {E}^{78}_{7,1}$ and $\mathrm {E}^{78}_{8,2}$ . Our proof is geometric in nature. We carefully analyse the properties of the natural action of the structure group ${\mathbf{Str}}(A)$ on the corresponding Albert algebra A. The information that we need is encoded in the Galois cohomology of the stabilisers of subalgebras of A. We compute the Galois cohomology of all these stabilisers and, using this information, we explicitly construct a system of generators of ${\mathbf{Str}}(A)(K)$ , which we prove are R-trivial.

An important step of the proof of our main result is the weak Skolem–Noether theorem for isomorphic embeddings due to S. Garibaldi and H. Petersson (for the terminology, definitions and the precise statement we refer to [7]). In Section 5 we prove Theorem 5.3, which can be viewed as a weaker version of S. Garibaldi and H. Petersson’s result.

We would also like to mention that it follows directly from our main theorem that two standard conjectures hold for simple simply connected strongly inner forms of type $\mathrm {E}_6$ : the abelian nature of the group of R-equivalence classes and the existence of transfers for the functor of R-equivalence classes. For these groups, the norm principle holds as well. For details we refer to the last section of the article.

Lastly, we mention that our results were proved independently by M. Thakur [24]. His proof differs from ours and is based on some explicit formulas for automorphisms of subalgebras of Albert algebras and extensions of these automorphisms.

Conventions and Notation

Throughout, K will denote a fixed field. A K-ring is a morphism $K \to R$ in the category of unital commutative associative rings. As is customary, by an abuse of language, the target R of such a morphism is also referred to as a K-ring.

By an algebraic K-group, or simply a K-group for convenience, we will understand a group scheme ${\mathbf{G}}$ of finite type over ${\textrm {Spec}}(K).$ ² The connected component of the identity of an algebraic K-group ${\mathbf{G}}$ will be denoted by ${\mathbf{G}}^\circ .$

Let A be a finite-dimensional algebra over K (not assumed associative or commutative). For any K-ring R the R-module $A \otimes _K R$ has a natural R-algebra structure. We denote by ${\mathbf {Aut}}(A)$ the algebraic K-group whose functor of points is given by

$$ \begin{align*}{\mathbf{Aut}}(A) : R \to {\textrm{Aut}}_{\textrm{R-alg}}(A \otimes_K R).\end{align*} $$

Let V be a K-subspace of $A.$ We let ${\mathbf {Aut}}(A,V)$ and ${\mathbf {Aut}}(A/V)$ be the closed algebraic subgroups of ${\mathbf {Aut}}(A)$ whose functor of points are given by

$$ \begin{align*}{\mathbf{Aut}}(A, V) : R \to \{ g \in {\textrm{Aut}}_{\textrm{R-alg}}(A \otimes_K R) : g(V \otimes_K R) = V \otimes_K R \}\end{align*} $$

and

$$ \begin{align*}{\mathbf{Aut}}(A/V) : R \to \{ g \in {\textrm{Aut}}_{\textrm{R-alg}}(A \otimes_K R) : g _{|(V \otimes_K R)} = {\textrm{id}} \},\end{align*} $$

respectively.

If $\bf G$ is an algebraic K-group, we will denote $H_{\textrm {fppf}}^1(K, {\mathbf{G}})$ simply by $H^1(K, {\mathbf{G}}).$ Whenever ${\mathbf{G}}$ is smooth, one knows that $H_{\textrm {fppf}}^1(K, {\mathbf{G}}) = H_{\acute{e}t}^1(K, {\mathbf{G}})$ and that this last is nothing but the usual (nonabelian) Galois cohomology.

In view of point (2) above, in order to prove our main results we may assume that K is of characteristic zero. That said, many of the preliminary results are of independent interest and hold with less restrictions on the nature of the base field. For this reason, we will henceforth assume, unless specifically stated otherwise, that K of characteristic different from 2 and 3.

2. Preliminaries

For later use we record some facts about Albert algebras and algebraic groups.

2.1. Albert Algebras

A Jordan algebra over K is a unital, commutative, not necessarily associative K-algebra A in which the Jordan identity

$$ \begin{align*} {(xy)(xx)=x(y(xx))} \end{align*} $$

is satisfied. In particular, A is power associative. Given an associative algebra B with multiplication $\cdot $ , the anticommutator $\frac {1}{2}(x\cdot y+y\cdot x)$ defines on B the structure of a Jordan algebra, denoted by $B^+$ . A Jordan algebra A is said to be special if it is isomorphic to a Jordan subalgebra of $B^+$ for an associative algebra B and exceptional otherwise. An Albert algebra is by definition a simple, exceptional Jordan algebra. It is known that any Albert algebra has dimension 27 and that, if the field K is separably closed, then all Albert algebras over K are isomorphic, as follows from [13, Proposition 37.11]. Thus, all Albert algebras over K are twisted forms of each other.

If A is an Albert algebra over $K,$ then ${\mathbf{H}}={\mathbf {Aut}}(A)$ is a simple K-group of type ${\textrm {F}}_4$ . It is known that any such group arises in this fashion and that two Albert algebras are isomorphic if and only if their automorphism groups are. Moreover, A is equipped with a cubic form $N:A\to K$ , called the norm of A. Our main object of study is the structure group ${\mathbf{Str}}(A)$ of A. This is the algebraic K-group whose functor of points is as follows: if R is a K-ring,

$$ \begin{align*}{\mathbf{Str}}(A)(R)=\{ x\in {\textbf{GL}}(A)(R)\ | \ N_R(x(a))=\nu(x) N_R(a)\ \ \forall \,a\in A\otimes_K R\ \},\end{align*} $$

where $N_R$ is the base change of N to $A\otimes _K R$ and the multiplier $\nu (x)\in R^\times $ .

The derived subgroup

$$ \begin{align*}{\mathbf{G}}=[{\mathbf{Str}}(A),{\mathbf{Str}}(A)]\end{align*} $$

of ${\mathbf{Str}}(A)$ is known to be a strongly inner form of a split simple simply connected K-group of type ${\mathrm{E}}_6$ . Moreover, ${\mathbf{Str}}(A)$ is an almost direct product of ${\mathbf{G}}_m$ and ${\mathbf{G}}$ (the intersection being the centre of $\bf {G}$ ). Hence,

$$ \begin{align*}\overline{{\mathbf{G}}}={\mathbf{Str}}(A)/{\mathbf{G}}_m\end{align*} $$

is an adjoint K-group of type ${\mathrm{E}}_6$ . Since in the split and even isotropic case, ${\mathbf{G}}$ and ${\mathbf{Str}}(A)$ are R-trivial (see [5]), we may, in the proof of the main theorem, assume that $\bf G$ is K-anisotropic. This amounts to the Albert algebra A being a division algebra, which is equivalent to the norm map N being anisotropic; that is, the equation $N(a)=0$ having no nonzero solutions over K.

We denote the group of K-points of ${\mathbf{Str}}(A)$ (respectively ${\mathbf{G}},\, {\overline {\mathbf{G}}},\, {\mathbf{H}}$ ) by ${\mathrm{Str}}(A)$ (respectively $G,\,\overline {G},\,H$ ). The group ${\mathbf{H}}$ coincides with the stabiliser in ${\mathbf{Str}}(A)$ of $1\in A$ ; see e.g. [20, 5.9.4].

2.2. R-equivalence in Algebraic Groups

Let ${\mathbf{G}}$ be an affine algebraic K-group. Two K-points $x,y\in {\mathbf{G}}(K)$ are called R-equivalent if there is a path from x to y; that is, if there exists a rational map $f:\mathbb A_K^1\dashrightarrow {\mathbf{G}}$ defined at $0,1$ and mapping $0$ to x and $1$ to y. One can easily verify that this is indeed an equivalence relation on ${\mathbf{G}}(K)$ and that, moreover, ${\mathbf{G}}$ induces a group structure on the set ${\mathbf{G}}(K)/R$ of all R-equivalence classes. We will denote the set of elements in ${\mathbf{G}}(K)$ equivalent to $1$ by $R{\mathbf{G}}(K)$ .

2.3. R-triviality of Cohomology Classes and the Norm Principle

Let ${\mathbf{G}}$ be a semisimple K-group, ${\mathbf{Z}}\subset {\mathbf{G}}$ a central subgroup and let

(1) $$ \begin{align} [\xi]\in {\textrm{Ker}}\,[H^1(K,{\textbf{Z}})\longrightarrow H^1(K,{\mathbf{G}})]. \end{align} $$

Definition. We say that $[\xi ]$ is R-trivial if there exists

$$ \begin{align*}c(t)=[\xi(t)]\in {\textrm{Ker}}\,[H^1(K(t),{\textbf{Z}}_{K(t)})\longrightarrow H^1(K(t),{\mathbf{G}}_{K(t)})], \end{align*} $$

with $K(t)$ a purely transcendental extension of K, such that $c(t)$ is defined at $t=0, 1$ and $c(0)=1$ and $c(1)=[\xi ]$ .

Remark 2.1. The above definition requires some clarification. Here and below, if ${\mathbf{G}}$ is an algebraic K-group, then an element in ${\mathbf{G}}(K(t))$ (respectively a class in $H^1(K(t),{\mathbf{G}}_{K(t)})$ ), where t is a variable over K, is said to be defined at 0 and 1 if it is in the image of ${\mathbf{G}}(\mathcal O)$ (respectively $H^1(\mathcal O,{\mathbf{G}}_{\mathcal O})$ ), where $\mathcal O$ is the intersection in $K(t)$ of the localisations $K[t]_{(t)}$ and $K[t]_{(t-1)}$ . In particular, via the evaluation maps $\varepsilon _0,\varepsilon _1:\mathcal O\to K$ , we can evaluate such an element or a class at 0 and at 1.

Example 2.2. Let D be a central simple algebra of degree n over K and set ${\mathbf{G}}={\mathbf{SL}}(1,D)$ . The centre ${\mathbf{Z}}$ of ${\mathbf{G}}$ is isomorphic to $\boldsymbol {\mu }_n$ and thus $H^1(K,{\mathbf{Z}})\simeq K^\times /K^{\times n}$ . Moreover, $H^1(K,{\mathbf{G}})\simeq K^\times /{\textrm {Nrd}}\,(D^\times )$ . Hence,

$$ \begin{align*}{\textrm{Ker}} [H^1(K,{\textbf{Z}})\longrightarrow H^1(K,{\mathbf{G}})]\simeq {\textrm{Nrd}}(D^\times)/K^{\times n}, \end{align*} $$

and as D is the affine space ${\mathbb A}_K^{n^2}$ , any element of the above kernel is R-trivial.

Example 2.3. Let f be a Pfister form over K, set ${\mathbf{G}}={\mathbf{Spin}}(f)$ and let again ${\mathbf{Z}}\subset {\mathbf{G}}$ be its centre. Then all cohomology classes in (1) are R-trivial; indeed, by [14, Proposition 7] the group ${\mathbf{G}}/{\mathbf{Z}}={\mathbf{PGO}}^+(f)$ is stably rational, and hence R-trivial, and since the canonical map

$$ \begin{align*}({\mathbf{G}}/{\textbf{Z}})(K)\to {\textrm{Ker}}\,[H^1(K,{\textbf{Z}})\to H^1(K,{\mathbf{G}})] \end{align*} $$

is surjective, any element in the kernel is R-trivial.

Example 2.4. Let f be an n-fold Pfister form over K, and let g be a nondegenerate subform of $f\oplus \mathbb H$ of codimension $2$ , where $\mathbb H$ is the hyperbolic plane. If d is the determinant of g, then we have a decomposition

(2) $$ \begin{align} g\oplus a\langle 1, -d\rangle \simeq f\oplus \mathbb H \end{align} $$

for some scalar $a\in K^\times $ . We claim that the group ${\mathbf{PGO}}^+(g)$ is R-trivial or, equivalently (see [14]), that the multiplier of any similitude with respect to g is R-trivial. In particular, if ${\mathbf{G}}={\mathbf{Spin}}(g)$ and $\bf {Z}$ is its centre, then arguing as in the previous example, one finds that every element in the kernel (1) is R-trivial.

To compute the group of multipliers of g we first recall that every multiplier of g is contained in the set $N_{K(\sqrt {d})/K}(K(\sqrt {d})^\times )$ (see [13, Theorem 13.38]). Therefore, it follows from (2) that a multiplier m with respect to g is a multiplier with respect to f as well and hence is contained in the value group $D(f)$ of f. Conversely, (2) implies that every element

$$ \begin{align*}m\in N_{K(\sqrt{d})/K}(K(\sqrt{d})^\times)\cap D(f)\end{align*} $$

is a multiplier of g. Let now $U\subset K(\sqrt {d})$ be the open subvariety consisting of all elements with nonzero norm and let $X\subset U\times \mathbb {A}_K^{2^n}$ be the K-variety consisting of the elements $(x,y)$ satisfying $N_{K(\sqrt {d})/K}(x)=f(y)$ . Consider the map $X\to {\mathbf{G}}_m$ given by $(x,y) \to N_{K(\sqrt {d})/K}(x)$ . Then the group of multipliers of g is the image of the K-points of X. Since X is K-rational, the group of multipliers of g is R-trivial.

Let ${\mathbf{G}}$ be a semisimple K-group and let ${\mathbf{Z}}\subset {\mathbf{G}}$ be a central subgroup. For any finite extension $L/K$ we have the restriction map

$$ \begin{align*}res^L_K: H^1(K,{\textbf{Z}}) \to H^1(L,{\textbf{Z}}) \end{align*} $$

and the corestriction map

$$ \begin{align*}cor^L_K: H^1(L,{\textbf{Z}})\to H^1(K,{\textbf{Z}}). \end{align*} $$

Definition. Let $L/K$ be a finite field extension. We say that the norm principle holds for a cohomology class

$$ \begin{align*}[\eta]\in {\textrm{Ker}}\,[H^1(L, {\textbf{Z}})\longrightarrow H^1(L,{\mathbf{G}})] \end{align*} $$

if

$$ \begin{align*}cor^L_K([\eta]) \in {\textrm{Ker}}\,[H^1(K,{\textbf{Z}}) \longrightarrow H^1(K,{\mathbf{G}})]. \end{align*} $$

We also say that the norm principle holds for the pair $({\mathbf{Z}},{\mathbf{G}})$ if it holds for every $[\eta ]\in {\textrm {Ker}}\,[H^1(L, {\mathbf{Z}})\longrightarrow H^1(L,{\mathbf{G}})]$ whenever $L/K$ is a finite extension.

Theorem 2.5 P. Gille [8]

Let K be a field of arbitrary characteristic and let $L/K$ be a finite field extension. The norm principle holds for all R-trivial elements $[\eta ]\in {\textrm {Ker}}\,[H^1(L, {\mathbf{Z}})\longrightarrow H^1(L,{\mathbf{G}})]$ . Moreover, $cor^L_K([\eta ])$ is R-trivial.

2.4. Groups of Type $\mathrm D_4$

First we recall a statement about groups of type $\mathrm D_4$ inside split groups of type $\mathrm {F}_4$ , proved in [1].

Proposition 2.6. Let ${\mathbf{F}}$ be a split K-group of type $\mathrm {F}_4$ . Then any K-subgroup ${\mathbf{M}}\subset {\mathbf{F}}$ of type $\mathrm {D}_4$ is quasi-split.

For later use we need one more fact about groups of type $^{3,6}\mathrm {D}_4$ inside $\mathrm {F}_4$ . Let thus ${\mathbf{M}}\subset {\mathbf{F}}$ be a split subgroup of type $\mathrm {D}_4$ and consider its normaliser ${\mathbf{N}}=N_{\mathbf{F}}({\mathbf{M}})$ . The quotient group ${\mathbf{N}}/{\mathbf{M}}$ is isomorphic to the group of outer automorphisms ${\mathbf{Out}}({\mathbf{M}})$ of ${\mathbf{M}}$ . This is the symmetric group $S_3$ , which we view as a constant finite group scheme over K.

Let $ [\xi ]\in H^1(K,{\mathbf{N}})$ be an arbitrary cohomology class and consider its image $[\overline {\xi }]\in H^1 (K,{\mathbf{Out}}({\mathbf{M}}))$ . Since ${\mathbf{Out}}({\mathbf{M}})$ is a constant group scheme, any cocycle $\overline {\xi }$ representing it corresponds to a homomorphism $\phi _{\xi }:{\textrm {Gal}}(K^{sep}/K) \to S_3$ . The image ${\textrm {Im}}\,\phi _{\xi }$ is then isomorphic to the Galois group of the minimal Galois extension $F/K$ over which the twisted group $^{\xi }{\mathbf{M}}$ becomes a group of inner type. It follows that ${\textrm {Im}}\,\phi _{\xi }$ is generated by the cycle $(123)$ if $^{\xi }{\mathbf{M}}$ has type $^3\mathrm {D}_4$ and is equal to $S_3$ if $^{\xi }{\mathbf{M}}$ has type $^6\mathrm {D}_4$ .

Lemma 2.7. Assume that $^{\xi }{\mathbf{M}}$ has type $^{3,6}{\mathrm {D}}_4$ . Then the natural map $\phi :{\mathbf {}^{\xi }\mathbf{N}}(K) \to {^{\overline {\xi }}(S_3)(K)}$ is surjective.

Proof. If ${\textrm {Im}}\,\phi _{\xi }=S_3$ , we have $^{\overline {\xi }}(S_3)(K)=1$ and there is nothing to prove. Assume next that ${\textrm {Im}}\,\phi _{\xi }=\langle (123)\rangle $ . In this case ${^{\overline {\xi }}(S_3)(K)} $ consists of three elements. Note that $\phi $ is the composition of the two natural maps ${\mathbf {}^{\xi }\mathbf {N}}(K) \to {\mathbf {Aut}}({^\xi }{\mathbf{M}})(K)$ and ${\mathbf {Aut}}({^\xi }{\mathbf{M}})(K)\to {^{\overline {\xi }}(S_3)(K)}$ . By [7], the group ${\mathbf {Aut}}({^\xi }{\mathbf{M}} )(K)$ contains an outer automorphism, say a, and it suffices to lift it modulo inner automorphisms to ${\mathbf {}^{\xi }\mathbf {N}}(K)$ .

The exact sequence

$$ \begin{align*}1 \longrightarrow {\textbf{Z}} \longrightarrow {\mathbf{}^{\xi}\mathbf {N}}\longrightarrow {\mathbf{Aut}}({^\xi}{\textbf{M}} ) \longrightarrow 1, \end{align*} $$

where ${\mathbf{Z}}$ is the centre of ${\mathbf {}^{\xi }\mathbf {N}} $ , induces the connecting map

$$ \begin{align*}\psi: {\mathbf{Aut}}({^\xi}{\textbf{M}} )(K)\longrightarrow H^1(K,{\textbf{Z}}). \end{align*} $$

Let $\eta :=\psi (a)$ . Let $L/K$ be the cubic field extension over which $^\xi {\mathbf{M}}$ becomes an inner form. By construction, we can say even more: it is a strongly inner form of type ${\textrm {D}}_4$ ; hence, $^\xi {\mathbf{M}}_L\simeq {\textrm {Spin}}(f_L)$ where $f_L$ is a $3$ -fold Pfister form. Therefore, a viewed over L can be lifted modulo inner automorphisms to an element of $^\xi {\mathbf{N}}(L)$ . In other words, there is an element $a'\in ({\mathbf {Aut}}({^\xi }{\mathbf{M}}))^\circ (L)$ such that its image under the connecting map $\psi _L$ is $\eta _L$ .

According to Example 2.3, $\eta _L$ is R-trivial. This means that if t is a variable over K, then there exists a class

$$ \begin{align*}[\eta(t)]\in {\textrm{Ker}}[H^1(L(t),{\textbf{Z}})\to H^1(L(t),{^\xi}{\textbf{M}})] \end{align*} $$

such that $\eta (t)$ is defined at $t=0,1$ and $\eta (0)=1, \eta (1)=\eta _L$ . We now pass to $[\widetilde {\eta (t)}]:={\textrm {cor}}^L_K([\eta (t)]$ . By a result of P. Gille (see Theorem 2.5), the class $[\widetilde {\eta (t)}]$ belongs to

$$ \begin{align*}{\textrm{Ker}}[H^1(K(t),{\textbf{Z}}) \to H^1(K(t),({\mathbf{Aut}}({^\xi}{\textbf{M}}))^\circ] \end{align*} $$

and can be specialised at $t=0,1$ . Since $L/K$ has degree $3$ and since $H^1(K,{\mathbf{Z}})$ is a $2$ -group, we have $\widetilde {\eta (t)}(1)=\eta $ .

Choose an element $b(t)\in ({\mathbf {Aut}}({^\xi }{\mathbf{M}}))^\circ (K(t))$ which maps to $\widetilde {\eta (t)}$ under $\psi _{K(t)}$ and which can be specialised at $t=0,1$ . By construction, $\psi (b(1))=[\eta ]$ . This implies that $\psi (ab(1)^{-1})=1$ . In other words, $ab(1)^{-1}$ has a lifting to $^\xi {\mathbf{N}}(K)$ , as required.

2.5. Conjugacy of Maximal Tori

Let ${\mathbf{G}}$ be an absolutely simple semisimple K-group. Let ${\mathbf{T}}$ and ${\mathbf{T}}'$ be maximal tori in ${\mathbf{G}}.$ Since all maximal tori become conjugate upon extension to $K^{sep}$ , there exists $g\in {\mathbf{G}}(K^{sep})$ such that ${\mathbf{T}}'=g{\mathbf{T}}g^{-1}$ . Since ${\mathbf{T}}$ and ${\mathbf{T}}'$ are K-subgroups, we have $(g^{-1})^\tau g\in N_{\mathbf{G}}({\mathbf{T}})(K^{sep})$ for all $\tau \in {\textrm {Gal}}(K^{sep}/K)$ . Thus, the class of the cocycle $(\xi _\tau )=((g^{-1})^\tau g)$ with coefficients in $N_{\mathbf{G}}({\mathbf{T}})$ is a cohomological obstruction to the conjugacy of ${\mathbf{T}}$ and ${\mathbf{T}}'$ . Note that since ${\mathbf{T}}'=g{\mathbf{T}}g^{-1}$ , the twisted tori $^{(\xi _\tau )}{\mathbf{T}}$ and ${\mathbf{T}}'$ are isomorphic K-groups.

Next we will show that under some additional assumptions, one can choose g such that the cocycle $(\xi _\tau )$ takes values in ${\mathbf{T}}(K^{sep})\subset N_{\mathbf{G}}({\mathbf{T}})(K^{sep})$ . Note that for such a choice of g we have $^{(\xi _\tau )}{\mathbf{T}}\simeq {\mathbf{T}}$ . Therefore, a necessary condition for this is that ${\mathbf{T}}$ and ${\mathbf{T}}'$ be isomorphic over K, since $^{(\xi _\tau )}{\mathbf{T}}\simeq {\mathbf{T}}'$ . Furthermore, our claim will hold true in ${\mathbf{G}}$ if it does in ${\mathbf{G}}/{\mathbf{Z}}$ for some central subgroup ${\mathbf{Z}}$ (because ${\mathbf{Z}}$ is contained in ${\mathbf{T}}$ ). This reduces the problem to the adjoint case.

Assume thus that ${\mathbf{T}}\simeq {\mathbf{T}}'$ and that ${\mathbf{G}}$ is adjoint. Let $F/K$ be the minimal splitting field of ${\mathbf{T}}$ (and hence of ${\mathbf{T}}'$ ) and let $\Gamma ={\textrm {Gal}}(F/K)$ . The group $\Gamma $ acts naturally on the character lattices $X({\mathbf{T}})_{\ast }$ and $X({\mathbf{T}}')_{\ast }$ and these actions preserve the root systems $\Sigma =\Sigma (G,{\mathbf{T}})$ and $\Sigma '=\Sigma (G,{\mathbf{T}}')$ . Thus, we have two canonical embeddings $\rho _1:\Gamma \hookrightarrow {\mathrm{Aut}}(\Sigma )$ and $\rho _2:\Gamma \hookrightarrow {\mathrm{Aut}}(\Sigma ')$ . Since ${\mathbf{G}}$ is adjoint, $\Sigma $ and $\Sigma '$ generate $X({\mathbf{T}})_{\ast }$ and $X({\mathbf{T}}')_{\ast }$ , respectively. Since $\Sigma $ and $\Sigma '$ are root systems of the same type we may identify them, which in turn gives rise to an identification $X({\mathbf{T}})_{\ast }=X({\mathbf{T}}')_{\ast }$ . After all of these identifications we obtain two actions of $\Gamma $ on each of $\Sigma $ and $X({\mathbf{T}})_{\ast }$ through $\rho _1$ and $\rho _2$ .

Lemma 2.8. Assume that there is an inner automorphism $\rho : {\mathrm{Aut}}(\Sigma )\to {\mathrm{Aut}}(\Sigma )$ such that $\rho |_{{\textrm {Im}}\,\rho _1}=\rho _2\circ \rho _1^{-1}$ . Then there is a $\Gamma $ -equivariant automorphism $X({\mathbf{T}})_{\ast }\to X({\mathbf{T}})_{\ast }$ preserving the root system $\Sigma $ , where $\Gamma $ acts on the domain through $\rho _1$ and on the codomain through $\rho _2$ .

Proof. Let $\rho ={\textrm {Int}}(a)$ where $a\in {\mathrm{Aut}}(\Sigma )$ . The map $a:\Sigma \to \Sigma $ can be extended uniquely to an automorphism $a_{X({\mathbf{T}})_\ast }:X({\mathbf{T}})_\ast \to X({\mathbf{T}})_\ast $ preserving roots. It is straightforward to check that it is $\Gamma $ -equivariant.

We are now ready to conclude this section with the following theorem. Since we will mainly be concerned with outer forms of type $\mathrm A_2$ , it is stated for groups of outer type.

Theorem 2.9. Let ${\mathbf{G}}$ be an absolutely simple semisimple K-group of outer type with $|{\textrm {Out}}({\mathbf{G}})|=2$ and let ${\mathbf{T}}$ and ${\mathbf{T}}'$ be two isomorphic maximal tori in ${\mathbf{G}}$ , with corresponding root systems $\Sigma $ and $\Sigma '$ , respectively. Assume that there is an inner automorphism $\rho : {\mathrm{Aut}}(\Sigma )\to {\mathrm{Aut}}(\Sigma )$ such that $\rho |_{{\textrm {Im}}\,\rho _1}=\rho _2\circ \rho _1^{-1}$ , where $\rho _1$ and $\rho _2$ are the above embeddings of $\Gamma $ into ${\mathrm{Aut}}(\Sigma )$ . If there is $f\in {\mathrm{Aut}}({\mathbf{G}})(K)\setminus {\textrm {Int}}({\mathbf{G}})(K)$ such that $f({\mathbf{T}})={\mathbf{T}}$ , then there is $g\in {\mathbf{G}}(K^{sep})$ such that $g{\mathbf{T}}g^{-1}={\mathbf{T}}'$ and $(g^{-1})^\tau g\in {\mathbf{T}}(K^{sep})$ for all $\tau \in {\textrm {Gal}}(K^{sep}/K)$ .

Proof. Without loss of generality, we may assume that ${\mathbf{G}}$ is adjoint. The assumptions of Lemma 2.8 are satisfied. Let thus

$$ \begin{align*}a_{X({\mathbf{T}})_\ast}:X({\mathbf{T}})_\ast\to X({\mathbf{T}})_\ast\end{align*} $$

be the $\Gamma $ -equivariant map constructed in that lemma. Using the identification of $X({\mathbf{T}})_\ast $ and $X({\mathbf{T}}')_\ast $ , we obtain a $\Gamma $ -equivariant map $X({\mathbf{T}})_\ast \to X({\mathbf{T}}')_\ast $ , which can be extended to a K-group isomorphism $a:{\mathbf{T}}\to {\mathbf{T}}'$ that induces an isomorphism between $\Sigma $ and $\Sigma '$ . By [11, Theorem 32.1] the map $a_{\mathbf{T}}$ can be further extended to an automorphism $a_{\mathbf{G}}:{\mathbf{G}} \to {\mathbf{G}}$ . Replacing $a_{\mathbf{G}}$ with $a_{\mathbf{G}}\circ f$ , if necessary, we may assume that $a_{\mathbf{G}}$ is inner, say $a_{\mathbf{G}}={\textrm {Int}}(g)$ , where $g\in {\mathbf{G}}(K^{sep})$ . Since ${\textrm {Int}}(g)|_{\mathbf{T}}: {\mathbf{T}} \to {\mathbf{T}}'$ is a K-group isomorphism and ${\textrm {Int}}((g^{-1})^\tau g)$ fixes $\Sigma $ , it follows that $(g^{-1})^\tau g\in {\mathbf{T}}(K^{sep})$ for all $\tau \in {\textrm {Gal}}(K^{sep}/K)$ .

Example 2.10. We keep the above notation. Let $E/K$ be a quadratic étale extension and let B be a central simple algebra of degree $3$ over E equipped with an involution $\sigma $ of the second kind. Consider two isomorphic cubic subfields $L,L'\subset B_\sigma $ where $B_\sigma \subset B$ is the subset consisting of all $\sigma $ -invariant elements. Since the maximal subfields $L\cdot E$ and $L'\cdot E$ of B are $\sigma $ -stable, they give rise to two maximal K-tori ${\mathbf{T}}$ and ${\mathbf{T}}'$ in ${\mathbf{G}}={\mathbf{SU}}(B,\sigma )$ , given by

$$ \begin{align*}{\mathbf{T}}=\{ x\in L\cdot E \ | \ \sigma(x)x=1,\ \ {\textrm{Nrd}}(x)=1\}; \end{align*} $$

$$ \begin{align*}{\mathbf{T}}'=\{ x\in L'\cdot E \ | \ \sigma(x)x=1,\ \ {\textrm{Nrd}}(x)=1\}. \end{align*} $$

Clearly, ${\mathbf{T}}\simeq {\mathbf{T}}'$ and the Galois group $\Gamma $ of the minimal splitting field of ${\mathbf{T}}$ (and hence of ${\mathbf{T}}'$ ) are of order divisible by $6$ . Now ${\mathbf{SU}}(B,\sigma )$ is of type $\mathrm {A}_2$ , with $W(\mathrm {A}_2)\simeq S_3$ and the automorphism group of its root system $\Sigma $ ,

$$ \begin{align*}{\textrm{Aut}}(\Sigma)\simeq W(\mathrm{A}_2)\times \mathbb{Z}/2\simeq S_3\times \mathbb{Z}/2, \end{align*} $$

is of order $12$ . Thus, $\Gamma $ has order $6$ or $12$ .

Case $1$ : $|\Gamma |=12$ . Then ${\textrm {Im}}\,\rho _1$ and ${\textrm {Im}}\,\rho _2$ coincide with ${\mathrm{Aut}}(\Sigma )$ . Note that

$$ \begin{align*}\rho_2\circ \rho_1^{-1}:{\textrm{Aut}}(\Sigma)\to {\textrm{Aut}}(\Sigma)\end{align*} $$

preserves the Weyl group $W(\mathrm {A}_2)$ since $\rho ^{-1}_1(W(\mathrm {A}_2))$ (respectively $\rho _2^{-1}(W(\mathrm {A}_2))$ ) coincides with ${\textrm {Gal}}(F/E) < {\textrm {Gal}}(F/K)$ , where $F/K$ is the Galois closure of $L\cdot E/K$ . Hence, $\rho _2\circ \rho _1^{-1}$ obviously satisfies all of the assumptions in Theorem 2.9 and the map $f(x)=\sigma (x)^{-1}$ is an outer automorphism of ${\mathbf{G}}$ preserving ${\mathbf{T}}$ .

Case $2$ : $|\Gamma |=6$ . The automorphism group ${\mathrm{Aut}}(\Sigma )$ has 3 subgroups of order $6$ , namely, $\Gamma _1=W(\mathrm {A}_2)=S_3\times 0$ , the subgroup $\Gamma _2\subset S_3\times \mathbb {Z}/2$ generated by the two elements $((123),0)$ and $((12),1)$ , where $(123)$ and $(12)$ are standard cycles in $S_3$ and the cyclic subgroup $\Gamma _3\subset S_3\times \mathbb {Z}/2$ of order $6$ generated by the two elements $((123),0)$ and $({\textrm {Id}},1)$ .

Since ${\mathbf{G}}$ has outer type, we know from [19, Lemma 4.1] that $\Gamma $ does not embed into $\Gamma _1=W(\mathrm {A}_2)\simeq S_3$ . If $\rho _1(\Gamma )= \Gamma _2\simeq S_3$ , then ${\textrm {Im}}\,\rho _1={\textrm {Im}}\,\rho _2$ and since every automorphism of $\Gamma _2$ is obviously inner, the automorphism $\rho _2\circ \rho _1^{-1}$ of $\Gamma _2$ can be extended to an inner automorphism of ${\mathrm{Aut}}(\Sigma )$ . If instead $\rho _1(\Gamma )= \Gamma _3\simeq \mathbb {Z}/3\times \mathbb {Z}/2$ , then again ${\textrm {Im}}\,\rho _1={\textrm {Im}}\,\rho _2$ . The group $\Gamma _3$ has a unique nontrivial automorphism given by $x\mapsto x^{-1}$ and one easily checks that it is the restriction of an inner automorphism of ${\mathrm{Aut}}(\Sigma )$ .

Thus, in all cases, the hypothesis of Theorem 2.9 is satisfied.

3. Subgroups of the Automorphism Group of an Albert Algebra

In this section, we will study automorphisms of Albert algebras related to 9-dimensional subalgebras. Recall that for our purposes it suffices to consider Albert algebras that are division algebras. Therefore, throughout this section, A is an arbitrary division Albert algebra over K. The main result of this section is the rationality, hence R-triviality, of the group of all automorphisms stabilising a 9-dimensional subalgebra.

3.1. $9$ -dimensional Subalgebras and Their Automorphisms

By [13, Theorem 37.12 (2)], any proper nontrivial subalgebra of A is either a cubic field extension $K\subset L\subset A$ or a $9$ -dimensional subalgebra $K\subset S \subset A$ . Furthermore, S is of the form $S=D^+$ where D is a central simple algebra of degree $3$ over K or $S=B_\sigma ^+$ where B is a central division algebra of degree $3$ over a quadratic field extension $E/K$ equipped with an involution $\sigma $ of the second kind. For later use we record some facts related to automorphism groups of $D^+,B_\sigma ^+$ and their extensions to automorphisms of A.

First, let $S=D^+$ . By [13, Theorem 39.14 (2)], the algebra A has a presentation $A=D\oplus D\oplus D$ (as a vector space) where the subalgebra S coincides with the first component. By [13, formula (37.7)], we have an exact sequence

$$ \begin{align*}1\longrightarrow {\mathbf{Aut}}(D) \longrightarrow {\mathbf{Aut}}(D^+) \longrightarrow \mathbb{Z}/2\longrightarrow 1. \end{align*} $$

This sequence is split if and only if D is split. Since D is a division algebra, any K-automorphism of $D^+$ thus comes from ${\mathbf {Aut}}(D)(K)={\mathrm{Aut}}(D)$ ; that is, is given by conjugation $x\mapsto dxd^{-1}$ for some $d\in D^\times $ . Moreover, such an automorphism can be extended to A by the formula

$$ \begin{align*}(x,y,z) \mapsto ( gxg^{-1}, gyg^{-1}, gzg^{-1}). \end{align*} $$

Thus, the sequence implies that

$$ \begin{align*}{\mathbf{Aut}}(D^+)^\circ\simeq {\textbf{PGL}}(1,D).\end{align*} $$

Note that ${\mathbf{PGL}}(1,D)$ is rational and hence, in particular, R-trivial.

Next, let $S=B_\sigma ^+$ . Here the situation is completely analogous to that of $S=D^+$ . Namely, by [13, Theorem 39.18 (2)], the algebra A admits the presentation $A=B_\sigma ^+\oplus B$ as a vector space, with the first component a subalgebra. By [13, Section 37.B], the algebraic K-group ${\mathbf {Aut}}(B^+_\sigma )$ is smooth and

$$ \begin{align*}{\mathbf{Aut}}(B_\sigma^+)^\circ\simeq {\textbf{PGU}}(B,\sigma).\end{align*} $$

Passing to the quadratic field extension $E/K$ , we conclude that

$$ \begin{align*}{\mathbf{Aut}}(B^+_\sigma)(K)={\mathbf{Aut}}(B^+_\sigma)^\circ(K). \end{align*} $$

Thus, any K-automorphism of $B_\sigma ^+$ is given by conjugation $x\mapsto bxb^{-1}$ for some $b\in B^\times $ satisfying $b\sigma (b)\in K$ . Since ${\mathbf{PGU}}(B,\sigma )$ has rank $2$ it is rational and hence R-trivial. In Corollary 3.2 we shall see that any K-automorphism of $B_\sigma ^+$ can be extended to a K-automorphism of A.

3.2. The Group ${\mathbf {Aut}}(A/B_\sigma ^+)$

Note that if $E=K\times K$ and $B=D\otimes _K E$ with the flip involution $\sigma $ , then $B_\sigma ^+$ is equal to $D^+$ embedded diagonally into B. This provides a unified treatment of both kinds of 9-dimensional subalgebras. Therefore, here and in what follows in this section, we will let $E/K$ be a quadratic étale extension, including the possibility of E being split; doing so, any 9-dimensional subalgebra of A is of the form $B_\sigma ^+$ .

As above, let $A=B_\sigma ^+\oplus B$ . Recall the algebraic K-group

$$ \begin{align*}{\mathbf{Aut}}(A/B_\sigma^+)\subset {\mathbf{Aut}}(A)={\textbf{H}}\end{align*} $$

(see Introduction). One knows (see [13, Section 39.B]) that ${\mathbf {Aut}}(A/B_\sigma ^+)$ is simple simply connected of type $\mathrm {A}_2$ (hence connected). Thus,

$$ \begin{align*}{\mathbf{Aut}}(A/B_\sigma^+)\simeq {\textbf{SU}}(B,\tau)\end{align*} $$

where $\tau $ is some involution of the second kind on B, which in general is different from $\sigma $ . Being an algebraic group of rank $2$ , this group is rational and hence R-trivial.

3.3. The Group ${\mathbf {Aut}}(A,B_\sigma ^+)$

Recall from the Introduction the K-group

$$ \begin{align*}{\mathbf{Aut}}(A,B_\sigma^+)\subset {\mathbf{Aut}}(A)={\textbf{H}}.\end{align*} $$

By [13, Proposition 39.16], we have an exact sequence

$$ \begin{align*}1\longrightarrow {\mathbf{Aut}}(A/B_\sigma^+)\longrightarrow {\mathbf{Aut}}(A,B_\sigma^+)\longrightarrow {\mathbf{Aut}}(B_\sigma^+) \longrightarrow 1. \end{align*} $$

Moreover, by [13, Corollary 39.12],

$$ \begin{align*}{\mathbf{Aut}}(A,B_\sigma^+)^\circ(K)={\mathbf{Aut}}(A,B_\sigma^+)(K). \end{align*} $$

Furthermore, the above sequence induces the exact sequence

(3) $$ \begin{align} 1\longrightarrow {\mathbf{Aut}}(A/B_\sigma^+)\longrightarrow {\mathbf{Aut}}(A,B_\sigma^+)^{\circ}\longrightarrow {\mathbf{Aut}}(B_\sigma^+)^{\circ} \longrightarrow 1. \end{align} $$

Thus,

$$ \begin{align*}{\mathbf{Aut}}(A,B_\sigma^+)^\circ/{\mathbf{Aut}}(A/B_\sigma^+)\simeq {\mathbf{Aut}}(B_\sigma^+)^{\circ}\simeq {\textbf{PGU}}(B,\sigma)\simeq {\textbf{SU}}(B,\sigma)/{\textbf{Z}}, \end{align*} $$

where ${\mathbf{Z}}\subset {\mathbf{SU}}(B,\sigma )$ is the centre.

From the point of view of algebraic groups, (3) implies that the algebraic K-group ${\mathbf{G}}:={\mathbf {Aut}}(A,B_\sigma ^+)^\circ $ is semisimple and is an almost direct product of two simple simply connected groups of type $\mathrm {A}_2$ : the first is

$$ \begin{align*}{\mathbf{G}}_1:={\mathbf{Aut}}(A/B_\sigma^+)={\textbf{SU}}(B,\tau) \end{align*} $$

and the second is isomorphic to ${\mathbf{G}}_2:={\mathbf{SU}}(B,\sigma )$ . The centres ${\mathbf{Z}}_1$ and ${\mathbf{Z}}_2$ of ${\mathbf{G}}_1$ and ${\mathbf{G}}_2$ , respectively, are both isomorphic to ${\mathbf{Z}}=R_{E/K}^{(1)}(\boldsymbol {\mu }_3)$ and ${\mathbf{G}}_1\cap {\mathbf{G}}_2={\mathbf{Z}}$ (see [13, Corollary 39.12]). Thus, we have the exact sequence

(4) $$ \begin{align} 1\longrightarrow {\textbf{Z}} \longrightarrow {\mathbf{G}}_1\times {\mathbf{G}}_2 \stackrel{\phi}{\longrightarrow} {\mathbf{G}} \longrightarrow 1, \end{align} $$

where ${\mathbf{Z}}$ is embedded codiagonally; that is, via $z\mapsto (z,z^{-1})$ . Identifying the image $\phi ({\mathbf{G}}_1\times 1)\subset {\mathbf{G}}$ with ${\mathbf{G}}_1$ , we recover the sequence (3) in the form

(5) $$ \begin{align} 1\longrightarrow {\mathbf{G}}_1 \longrightarrow {\mathbf{G}} \stackrel{\psi}{\longrightarrow} {\mathbf{G}}/{\mathbf{G}}_1 \longrightarrow 1. \end{align} $$

Note that ${\mathbf{G}}/{\mathbf{G}}_1\simeq {\mathbf{G}}_2/{\mathbf{Z}}$ .

3.4. Rationality of ${\mathbf {Aut}}(A,B_\sigma ^+)^\circ $

We keep the notation introduced in Subsection 3.3.

Proposition 3.1. The K-group ${\mathbf{G}}={\mathbf {Aut}}(A,B_\sigma ^+)^\circ $ is rational and hence R-trivial.

Proof. Consider the exact sequences (4) and (5). The two groups ${\mathbf{G}}_1$ and ${\mathbf{G}}_2/{\mathbf{Z}}$ in (5), being groups of rank $2$ , are rational over K. Therefore, it suffices to show that $\psi $ has a rational section. This is equivalent to proving that

$$ \begin{align*}{\textrm{Ker}}\,[H^1(F,{\mathbf{G}}_1) \longrightarrow H^1(F,{\mathbf{G}})]=1 \end{align*} $$

for all field extensions $F/K$ .

Fix a field extension $F/K$ and let $[\xi ]\times 1\in H^1(F,{\mathbf{G}}_1\times 1)$ be a class whose image in $H^1(F,{\mathbf{G}})$ is trivial. From (4) it follows that there is $[\lambda ]\in H^1(F,{\mathbf{Z}})$ whose image in $H^1(F, {\mathbf{G}}_1\times {\mathbf{G}}_2)$ is $[\xi ]\times 1$ . Since ${\mathbf{Z}}$ is embedded codiagonally into ${\mathbf{G}}_1\times {\mathbf{G}}_2$ , the image of $[\lambda ]$ under the natural map

$$ \begin{align*}H^1(F,{\textbf{Z}})\to H^1(F,{\mathbf{G}}_2)\end{align*} $$

is trivial. We distinguish two cases.

Case $1$ : The quadratic étale extension $E/K$ is split; that is, $E=K\times K$ . Then up to K-isomorphism we may assume that ${\mathbf{G}}_1={\mathbf{SL}}(1,D_1)$ and ${\mathbf{G}}_2={\mathbf{SL}}(1,D_2)$ , where $D_1$ and $D_2$ are central simple algebras over K of degree $3$ and either $D_2=D_1$ or $D_2=D_1^{\textrm {op}}$ . In both cases their centres are $\boldsymbol {\mu }_3$ , whence $H^1(F,{\mathbf{Z}})\simeq F^\times /F^{\times 3}$ , so that the class $[\lambda ]\in H^1(F,{\mathbf{Z}})$ is represented by some $f\in F^\times $ . The fact that

$$ \begin{align*}H^1(F,{\mathbf{G}}_2)\simeq F^\times/ {\textrm{Nrd}}(D_2^\times)\end{align*} $$

together with the image of $[\lambda ]$ in $H^1(F,{\mathbf{G}}_2)$ being trivial then imply that f is a reduced norm in $D_2$ , and hence also in $D_1$ . This implies that the image $[\xi ]$ of $[\lambda ]$ in $H^1(F,{\mathbf{G}}_1)$ is trivial, which completes the proof in this case.

Case $2$ : $E/K$ is a separable field extension. Let $F'=F\cdot E$ and consider the class $[\lambda ]^2\in H^1(F,{\mathbf{Z}})$ . From Case $1$ we know that $res_F^{F'}([\lambda ]^2)$ is contained in ${\textrm {Ker}}\,[H^1(F',{\mathbf{Z}})\to H^1(F',{\mathbf{G}}_1)]$ . It follows from Example 2.2 and the norm principle (Theorem 2.5) that

$$ \begin{align*}[\lambda]=[\lambda]^4=cor^{F'}_F(res_F^{F'}([\lambda]^2) \end{align*} $$

is contained in ${\textrm {Ker}}\,[H^1(F,{\mathbf{Z}})\to H^1(F,{\mathbf{G}}_1)]$ and the proof is complete.

As a direct consequence of our proof, we have the following.

Corollary 3.2. For any field extension $F/K$ , the canonical map ${\mathbf{G}}(F) \to ({\mathbf{G}}_2/{\mathbf{Z}})(F)$ is surjective. Hence, if $A_F$ is a division Albert algebra, then

$$ \begin{align*}{\mathbf{Aut}}(A,B_\sigma^+)(F) \longrightarrow {\mathbf{Aut}}(B_\sigma^+)(F)\end{align*} $$

is surjective.

4. Subgroups of the Structure Group of an Albert Algebra

We now turn to the structure group and consider, along the same lines as in the previous section, subgroups of it related to 9-dimensional subalgebras. Throughout this section, as in the previous, A denotes an arbitrary division Albert algebra over K. We moreover continue using the convention that the quadratic étale algebra E involved in the definition of $B_\sigma ^+$ may be split.

4.1. The Group ${\mathbf{Str}}(B_\sigma ^+)$

The structure group ${\mathbf{Str}}(B_\sigma ^+)$ of the Jordan algebra $B_\sigma ^+$ is a closed subgroup of the algebraic K-group ${\mathbf{GL}}(B_\sigma ^+)$ consisting of all similitudes. More precisely, for any K-ring R,

$$ \begin{align*}{\mathbf{Str}}(B_\sigma^+)(R) = \{x \in {\textbf{GL}}(B_\sigma^+)(R) \ | \ {\textrm{Nrd}}_R(x(b)) = \nu(x){\textrm{Nrd}}_R(b) \ \ \forall b\in B_\sigma^+\otimes_K R\} \end{align*} $$

where ${\textrm {Nrd}}_R$ is the base change of ${\textrm {Nrd}}$ to $B\otimes _K R$ and the multiplier $\nu (x)\in R^\times $ . By [12, Chap. V, Thm. 5.12.10], the group ${\mathrm{Str}}(B_\sigma ^+)={\mathbf{Str}}(B_\sigma ^+)(K)$ consists of the linear maps of the form

$$ \begin{align*}B_\sigma^+\longrightarrow B_\sigma^+,\ \ x \mapsto \lambda bx\sigma(b), \end{align*} $$

where $b\in B^\times $ and $\lambda \in K^\times $ . It follows that we have a surjective map of algebraic K-groups

$$ \begin{align*}{\mathbf{G}}_m \times R_{E/K}({\textbf{GL}}(1,B)) \longrightarrow {\mathbf{Str}}(B_\sigma^+)^\circ\end{align*} $$

and that ${\mathbf{Str}}(B_\sigma ^+)^\circ (K)={\mathbf{Str}}(B_\sigma ^+)(K)$ . The kernel of this map is the torus $R_{E/K}({\mathbf{G}}_{m,E})$ , where the embedding

$$ \begin{align*}R_{E/K}({\mathbf{G}}_{m,E})\hookrightarrow {\mathbf{G}}_m \times R_{E/K}({\textbf{GL}}(1,B)) \end{align*} $$

is given by $x\mapsto (N_{E/K}(x^{-1}),x)$ . Thus, we have the exact sequence

(6) $$ \begin{align} 1\longrightarrow R_{E/K}({\mathbf{G}}_{m,E}) \longrightarrow {\mathbf{G}}_m \times R_{E/K}({\textbf{GL}}(1,B)) \stackrel{\phi}{\longrightarrow} {\mathbf{Str}}(B_\sigma^+)^\circ \longrightarrow 1. \end{align} $$

Lemma 4.1. The algebraic K-group ${\mathbf{Str}}(B_\sigma ^+)^\circ $ is rational and hence, in particular, R-trivial.

Proof. We identify $\phi ({\mathbf{G}}_m\times 1)\subset {\mathbf{Str}}(B_\sigma ^+)^\circ $ with ${\mathbf{G}}_m$ . Since for any field extension $F/K$ one has $H^1(F,{\mathbf{G}}_m)=1$ , the canonical map

$$ \begin{align*}{\mathbf{Str}}(B_\sigma^+)^\circ\to {\mathbf{Str}}(B_\sigma^+)^\circ/{\mathbf{G}}_m\end{align*} $$

has a rational section. Therefore, it suffices to establish the rationality of ${\mathbf{Str}}(B_\sigma ^+)^\circ /{\mathbf{G}}_m$ . By (6) we have

$$ \begin{align*}{\mathbf{Str}}(B_\sigma^+)^\circ/{\mathbf{G}}_m\simeq R_{E/K}({\textbf{GL}}(1,B^\times))/R_{E/K}({\mathbf{G}}_{m,E}), \end{align*} $$

which is clearly rational.

4.2. The Group ${\mathbf{Str}}(A,B_\sigma ^+)$

Let ${\mathbf{Str}}(A,B_\sigma ^+)$ (respectively ${\mathbf{Str}}(A/B_\sigma ^+)$ ) be the closed subgroup of ${\mathbf{Str}}(A)$ consisting of those elements stabilising $B_\sigma ^+$ (respectively fixing $B_\sigma ^+$ pointwise). Note that since the elements of ${\mathbf{Str}}(A/B_\sigma ^+)$ fix the identity element of A, it follows from [20, Proposition 5.9.4] that ${\mathbf{Str}}(A/B_\sigma ^+)={\mathbf {Aut}}(A/B_\sigma ^+)$ . This group is the kernel of the canonical restriction map

$$ \begin{align*}{\mathbf{Str}}(A,B_\sigma^+)^\circ \to {\mathbf{Str}}(B_\sigma^+)^\circ.\end{align*} $$

Note that the restriction map is well defined because the restriction to $B^+_\sigma $ of the norm on A is the norm on $B^+_\sigma $ . Since, by [7, Proposition 7.2.4], this map is surjective on the level of $K^{sep}$ -points, we have the exact sequence

(7) $$ \begin{align} 1\longrightarrow {\mathbf{Aut}}(A/B_\sigma^+) \longrightarrow {\mathbf{Str}}(A,B_\sigma^+)^\circ \stackrel{\phi}{\longrightarrow} {\mathbf{Str}}(B_\sigma^+)^\circ \longrightarrow 1 \end{align} $$

of algebraic K-groups.

Proposition 4.2. The group ${\mathbf{Str}}(A,B_\sigma ^+)^\circ $ is rational and hence R-trivial.

Proof. By [7], for any field extension $F/K$ the map

$$ \begin{align*}{\mathbf{Str}}(A,B_\sigma^+)^\circ(F) \stackrel{\phi_F}{\longrightarrow} {\mathbf{Str}}(B_\sigma^+)^\circ(F) \end{align*} $$

is surjective, implying that $\phi $ has a rational section. Thus, ${\mathbf{Str}}(A,B_\sigma ^+)^\circ $ is birationally isomorphic to ${\mathbf {Aut}}(A/B_\sigma ^+) \times {\mathbf{Str}}(B_\sigma ^+)^\circ $ . It remains to be noted that, being a group of rank $2$ , the group ${\mathbf {Aut}}(A/B_\sigma ^+)$ is rational and by Lemma 4.1 the group ${\mathbf{Str}}(B_\sigma ^+)^\circ $ is rational.

Corollary 4.3. The natural map ${\mathrm{Str}}(A,B_\sigma ^+)\to {\mathrm{Str}}(B_\sigma ^+)$ is surjective.

Proof. From the proof of Proposition 4.2 it follows that the map

$$ \begin{align*}{\mathbf{Str}}(A,B_\sigma^+)^\circ(K)\to {\mathbf{Str}}(B_\sigma^+)^\circ(K)\end{align*} $$

is surjective, and from 4.1 we know that ${\mathbf{Str}}(B_\sigma ^+)(K)={\mathbf{Str}}(B_\sigma ^+)^\circ (K)$ . The assertion follows.

5. The Weak Skolem–Noether Property for Isomorphic Embeddings

Let A be an Albert algebra over a field K. Let $K\subset L\subset A$ and $K\subset L'\subset A$ be two isomorphic separable cubic field extensions; one says that they are weakly equivalent if there exists an element $g\in {\mathrm{Str}}(A)$ such that $g(L)=L'$ . That this property holds follows from Theorem B (Skolem–Noether theorem for Albert algebras) due to S. Garibaldi and H. Petersson [7].

The goal of this section is to give a self-contained proof of the validity of weak equivalence based on the technique of conjugacy of maximal tori detailed in Section 2.

We start with the intermediate step of a $9$ -dimensional Jordan algebra $B^+_\sigma $ .

Proposition 5.1. Let L and $L'$ be two isomorphic separable cubic field extensions of the base field K contained in the subalgebra $B_\sigma ^+$ . Then there exists an element $s\in {\mathrm{Str}}(B_\sigma ^+)$ such that $s(L)=L'$ .

Proof. Let $E/K$ be the étale quadratic extension over which B is defined. The two cubic fields L and $L'$ give rise to the two $4$ -dimensional (maximal) tori

$$ \begin{align*}{\mathbf{T}}=\{ x\in R_{L\cdot E/K}({\mathbf{G}}_{m,L\cdot E})\ | \ \sigma(x)x\in {\mathbf{G}}_{m,K}\} \end{align*} $$

and

$$ \begin{align*}{\mathbf{T}}'=\{ x\in R_{L'\cdot E/K}({\mathbf{G}}_{m,L'\cdot E})\ | \ \sigma(x)x\in {\mathbf{G}}_{m,K}\} \end{align*} $$

of the algebraic K-group ${\mathbf{Sim}}(B,\sigma )$ . ³ From the point of view of algebraic groups, ${\mathbf{Sim}}(B,\sigma )$ is a reductive group which is the almost direct product of the central $2$ -dimensional torus ${\mathbf{P}}=R_{E/K}({\mathbf{G}}_{m,E})$ and the simple simply connected group ${\mathbf{SU}}(B,\sigma )$ of outer type $\mathrm {A}_2$ . Note that ${\mathbf{P}}$ is contained in both ${\mathbf{T}}$ and ${\mathbf{T}}'$ . Let ${\mathbf{T}}_{ss}={\mathbf{T}}\cap {\mathbf{SU}}(B,\sigma )$ and ${\mathbf{T}}^{\prime }_{ss}={\mathbf{T}}'\cap {\mathbf{SU}}(B,\sigma )$ . Then ${\mathbf{T}}={\mathbf{P}}\cdot {\mathbf{T}}_{ss}$ and ${\mathbf{T}}'={\mathbf{P}}\cdot {\mathbf{T}}^{\prime }_{ss}$ .

By Theorem 2.9 and Example 2.10, there exists $b\in {\mathbf{SU}}(B,\sigma )(K^{sep})$ such that $b{\mathbf{T}}_{ss}b^{-1}={\mathbf{T}}^{\prime }_{ss}$ and $b^{-\tau +1} \in {\mathbf{T}}_{ss}(K^{sep})$ for all $\tau \in {\textrm {Gal}}(K^{sep}/K)$ . Since ${\mathbf{P}}\subset {\mathbf{Sim}}(B,\sigma )$ is a central torus, we also have $b{\mathbf{T}}b^{-1}={\mathbf{T}}'$ .

Consider the following commutative diagram with exact rows:

Here ${\mathbf{J}}={\mathbf{G}}_{m,K}\times R_{E/K}({\mathbf{GL}}(1,B))$ , so that the lower sequence is the exact sequence (6), $\phi _1$ is given by conjugation,

$$ \begin{align*}\phi_2(x,y): B_\sigma^+\longrightarrow B_\sigma^+,\ \ a\mapsto xya\sigma(y) \end{align*} $$

and

$$ \begin{align*}\lambda_2(x)=(\nu(x^{-1}),x)=(x^{-1}x^{-\sigma},x)\end{align*} $$

where $\nu (x)$ is the multiplier of x. The map $\lambda _2$ takes ${\mathbf{T}}$ and ${\mathbf{T}}'$ into the quasi-trivial tori

$$ \begin{align*}\widetilde{{\mathbf{T}}}={\mathbf{G}}_{m,K}\times R_{L\cdot E/K}({\mathbf{G}}_{m,L\cdot E}) \end{align*} $$

and

$$ \begin{align*}\widetilde{{\mathbf{T}}}'={\mathbf{G}}_{m,K}\times R_{L'\cdot E/K}({\mathbf{G}}_{m,L'\cdot E}), \end{align*} $$

respectively. Next we need the following.

Lemma 5.2. The torus $\widetilde {{\mathbf{T}}}$ (respectively $\widetilde {{\mathbf{T}}'}$ ) is the centraliser of ${\mathbf{T}}$ (respectively ${\mathbf{T}}'$ ) in ${\mathbf{G}}_{m,K}\times R_{E/K}({\mathbf{GL}}(1,B))$ .

Proof. We consider the case of the torus ${\mathbf{T}}$ only. The torus ${\mathbf{T}}'$ is handled analogously. It suffices to show that

$$ \begin{align*}C_{R_{E/K}({\textbf{GL}}(1,B))}({\mathbf{T}})=R_{L\cdot E/K}({\mathbf{G}}_{m,L\cdot E}).\end{align*} $$

Without loss of generality, we may assume that the base field K is algebraically closed. Then B may be identified with the algebra $M_3 \times M_3$ , the involution $\sigma $ switches the components and L with diagonal matrices fixed by $\sigma $ . With this identification we have

$$ \begin{align*}R_{E/K}({\textbf{GL}}(1,B))={\textbf{GL}}_3\times {\textbf{GL}}_3\end{align*} $$

and

$$ \begin{align*}R_{L\cdot E/K}({\mathbf{G}}_{m,L\cdot E})=\{(t_1,t_2)\ | \ t_1,t_2 \mbox{ are diagonal matrices}\} \end{align*} $$

and, moreover,

$$ \begin{align*}{\mathbf{T}}=\{(t_1, ut_1) \ |\ t_1 \mbox{ is a diagonal matrix and } u\in K^\times\}. \end{align*} $$

It follows that

$$ \begin{align*}C_{{\textbf{GL}}_3\times {\textbf{GL}}_3}({\mathbf{T}})=\{ (t_1,t_2)\ | \ t_1,t_2 \mbox{ are diagonal matrices}\} \end{align*} $$

and we are done.

We now return to proving the proposition. By the above lemma the equality $b{\mathbf{T}}b^{-1}={\mathbf{T}}'$ implies $b\widetilde {{\mathbf{T}}}b^{-1}=\widetilde {{\mathbf{T}}}'$ and

$$ \begin{align*}\lambda_2(b^{-\tau+1})=(\lambda_2(b))^{-\tau+1}=(\nu(b^{-1})^{-\tau+1},b^{-\tau+1}) \in \widetilde{{\mathbf{T}}}(K^{sep}).\end{align*} $$

Since $H^1(K,R_{L\cdot E/K}({\mathbf{G}}_{m,L\cdot E}))=1$ , we can pick an element

$$ \begin{align*}a\in R_{L\cdot E/K}({\mathbf{G}}_{m,L\cdot E})(K^{sep})=((L\cdot E)\otimes_K K^{sep})^\times \end{align*} $$

such that $b^{-\tau +1}=a^{-\tau +1}$ . Clearly, $c:=ba^{-1}$ is ${\textrm {Gal}}(K^{sep}/K)$ -invariant, hence, $c\in B^\times $ , and, defining s as the map $x\mapsto cx\sigma (c)$ , the following claim completes the proof of the proposition.

Claim. $cL\sigma (c)=L'$ .

The claim is equivalent to $c(L\cdot E)\sigma (c)=L'\cdot E$ . Moreover, it suffices to show that

$$ \begin{align*}c((L\cdot E)\otimes_K K^{sep})\sigma(c)=(L'\cdot E)\otimes_K K^{sep}.\end{align*} $$

But $\sigma (c)=\sigma (a^{-1})\sigma (b)$ and both $a^{-1}$ and $\sigma (a^{-1})$ are in $(L\cdot E)\otimes _K K^{sep}$ . Therefore, it suffices to show that

$$ \begin{align*}b((L\cdot E)\otimes_K K^{sep})\sigma(b)=(L'\cdot E)\otimes_K K^{sep}.\end{align*} $$

Recall that, by construction, b is a similitude; hence, $\sigma (b)=\nu (b)b^{-1}$ , where $\nu (b)$ is the multiplier of b. Since $b{\mathbf{T}}b^{-1}={\mathbf{T}}'$ , the claim follows upon noting that the centraliser of ${\mathbf{T}}$ (respectively ${\mathbf{T}}'$ ) in $B\otimes _K K^{sep}$ is $(L\cdot E)\otimes _K K^{sep}$ (respectively $(L'\cdot E)\otimes _K K^{sep}$ ).

The proposition in conjunction with Corollary 4.3 yields the following.

Theorem 5.3. Let A be a division Albert algebra over K and let $L\subset A$ and $L'\subset A$ be isomorphic separable cubic field extensions of K. Then there exists a 9-dimensional subalgebra $B_\sigma ^+$ of A and an element $s\in {\mathrm{Str}}(A,B_\sigma ^+)$ such that $s(L)=L'=f(L)$ .

Proof. If $L'=L$ , we can take any any $9$ -dimensional subalgebra $B^+_\sigma $ containing L and choose $s=\mathrm {Id}$ . Otherwise, the cubic subfields $L,L'$ generate a subalgebra in A of the form $B_\sigma ^+$ and we can take any lift of the element s constructed in the above proposition.

Remark 5.4. To be able to approach the Skolem–Noether isomorphic embedding problem using torsor techniques is of independent interest. It seems plausible that this approach can shed some light on the still open and much more difficult Skolem–Noether problem for isotopic embeddings.

6. Reduction to $\mathrm {F}_4$

Throughout this section, we let A be a division Albert algebra over K. We will show that an arbitrary element in ${\mathrm{Str}}(A)$ can be written as a product of R-trivial elements and elements in ${\mathbf{H}}(K)$ , thereby reducing the problem to subgroups of type $\mathrm {F}_4$ . To begin with, we recall from [1] how to associate, to any $a\in A^\times $ , a subgroup of type $\mathrm {D}_4$ in ${\mathbf{G}} =[{\mathbf{Str}}(A),{\mathbf{Str}}(A)]$ and a $2$ -dimensional torus. Let $L\subset A$ be the K-subalgebra generated by a if a is not a scalar multiple of the identity of A and by any element that is not such a scalar multiple if a is. Since A is a division algebra, it is known that L is a cubic subfield. Let ${\mathbf{G}}^L$ be the algebraic K-group whose functor of points is given by

$$ \begin{align*}{\mathbf{G}}^L(R) =\{ x\in {\mathbf{G}}(R) \ | \ x(l \otimes 1)=l \otimes 1 \ \ \forall l\in L\} \end{align*} $$

for any K-ring R. Since ${\mathbf{G}}^L$ stabilises $1\in L\subset A$ , we have ${\mathbf{G}}^L\subset {\mathbf{H}}\subset {\mathbf{G}}$ . It is known that over a separable closure of K the group ${\mathbf{G}}^L$ is conjugate to the standard subgroup in ${\mathbf{G}}$ of type $\mathrm {D}_4$ generated by roots $\alpha _2,\alpha _3,\alpha _4, \alpha _5$ . The lemma below shows that the centraliser ${\mathbf{S}}^{'L}=C_{\mathbf{G}}({\mathbf{G}}^L)$ of ${\mathbf{G}}^L$ in ${\mathbf{G}}$ is a $2$ -dimensional torus over K and that

$$ \begin{align*}{\textbf{Z}}^L:={\textbf{S}}^{'L}\cap {\textbf{H}}={\textbf{S}}^{'L}\cap {\mathbf{G}}^L\end{align*} $$

is the centre of ${\mathbf{G}}^L$ .

Lemma 6.1. Let K be an arbitrary field. ${\mathbf{G}}$ be a split simple simply connected K-group of type ${\mathrm{E}}_6.$ Let ${\mathbf{T}}\subset {\mathbf{G}}$ be a maximal split torus. Denote by $\Pi =\{\alpha _1,\ldots ,\alpha _6\}$ a basis of the root system $\Sigma ({\mathbf{G}},{\mathbf{T}})$ and by $\Sigma _1$ the root subsystem in $\Sigma $ generated by $\alpha _2,\alpha _3,\alpha _4,\alpha _5$ . Then ${\mathbf{S}}=C_{\mathbf{G}}({\mathbf{G}}_{\Sigma _1})$ is a $2$ -dimensional subtorus in ${\mathbf{T}}$ and ${\mathbf{S}}\cap {\mathbf{G}}_{\Sigma _1}$ is the centre of ${\mathbf{G}}_{\Sigma _1}$ .

Proof. Let ${\mathbf{T}}_{\Sigma _1}={\mathbf{G}}_{\Sigma _1}\cap {\mathbf{T}}$ . This is a maximal split torus of ${\mathbf{G}}_{\Sigma _1}$ . Clearly, one has $C_{\mathbf{G}}({\mathbf{G}}_{\Sigma _1})\subset C_{\mathbf{G}}({\mathbf{T}}_{\Sigma _1})$ . By properties of reductive groups, $C_{{\mathbf{G}}}({\mathbf{T}}_{\Sigma _1})$ is a reductive group whose derived subgroup $[C_{{\mathbf{G}}}({\mathbf{T}}_{\Sigma _1}),C_{{\mathbf{G}}}({\mathbf{T}}_{\Sigma _1})]$ is generated by roots in $\Sigma $ orthogonal to $\alpha _2,\alpha _3,\alpha _4,\alpha _5$ . Since $\Sigma $ is of type ${\mathrm{E}}_6$ there are no such roots, and this implies that $C_{{\mathbf{G}}}({\mathbf{T}}_{\Sigma _1})={\mathbf{T}}$ . Thus, $S=C_{\mathbf{G}}({\mathbf{G}}_{\Sigma _1})$ is a subtorus in ${\mathbf{T}}$ . To describe it explicitly we will use the canonical cocharacters $\check {\alpha }_i: {\mathbf{G}}_m\to {\mathbf{T}}$ .

Since ${\mathbf{G}}$ is simply connected, every element in $t\in {\mathbf{T}}(K)$ can be written uniquely in the form $t=\prod _i \check {\alpha }_i(t_i)$ where $t_i\in K$ . For a root $\alpha \in \Sigma $ let ${\mathbf{U}}_{\alpha }$ be the associated root subgroup in ${\mathbf{G}}$ . Using Steinberg’s notation (introduced in [21]), every element in ${\mathbf{U}}_{\alpha }$ can be written as $x_{\alpha }(u)$ where $u\in K$ . According to the Steinberg relations [21], we have

$$ \begin{align*}\check{\alpha}_i(t_i) x_{\alpha}(u)\check{\alpha}_i(t_i)^{-1}=x_\alpha(t_i^{<\alpha,\alpha_i>}u). \end{align*} $$

Using this relation, it is straightforward to verify that ${\mathbf{S}}$ consists of the elements in ${\mathbf{T}}$ of the form

$$ \begin{align*}\check{\alpha}_1\left(\frac{b^2}{a^2}\right) \check{\alpha}_3(b) \check{\alpha}_4(a^2) \check{\alpha}_2(a) \check{\alpha}_5\left(\frac{a^3}{b}\right) \check{\alpha}_6\left(\frac{a^4}{b^2}\right) \end{align*} $$

where $a,b\in K^\times $ . Lastly, we note that the centre of ${\mathbf{G}}_{\Sigma _1}$ is generated by the two elements $\check {\alpha }_3(-1)\check {\alpha }_2(-1)$ and $\check {\alpha }_5(-1)\check {\alpha }_2(-1)$ which are contained in ${\mathbf{S}}$ .

Let ${\mathbf{S}}^L\subset {\mathbf{Str}}(A)$ be the $3$ -dimensional torus in ${\mathbf{Str}}(A)$ generated by ${\mathbf{S}}^{'L}$ and ${\mathbf{G}}_m$ , where the latter is embedded in ${\mathbf{Str}}(A)$ as the subgroup of homotheties. The following was proved in [1].

Lemma 6.2. With the notation above, ${\mathbf{S}}^L\simeq R_{L/K}({\mathbf{G}}_{m,L})$ and ${\mathbf{S}}^{'L}\simeq R_{L/K}^{(1)}({\mathbf{G}}_{m,L})$ . Moreover,

$$ \begin{align*}\{ a\in A \ | \ x(a)=a\ \ \forall x\in {\mathbf{G}}^L(K)\}=L,\end{align*} $$

and the natural action of ${\mathbf{Str}}(A)$ on A induces an action of ${\mathbf{S}}^L$ on L, which is transitive on the level of the $K^{sep}$ -points of the open subset

$$ \begin{align*}L^\times=\{x\in L\ | \ N(x)\not=0\} \end{align*} $$

of L and gives rise to an exact sequence

$$ \begin{align*}1\longrightarrow {\textbf{Z}}^L \longrightarrow {\textbf{S}}^L \longrightarrow {\textbf{L}}^\times \longrightarrow 1.\end{align*} $$

From the exact sequence in cohomology associated to the sequence in the lemma we have a map $L^\times \to H^1(K,{\mathbf{Z}}^L)$ , which is surjective since $H^1(K,{\mathbf{S}}^L)$ is trivial. Using this map we can attach, to any $a\in L^\times $ , a class $[\xi _a]=(a_\tau )\in H^1(K,{\mathbf{Z}}^L)$ . From [1] we moreover know that the image of this class in $H^1(K,{\mathbf{H}})$ is trivial if a is in the ${\mathrm{Str}}(A)$ -orbit of $1$ .

Let now $g\in {\mathrm{Str}}(A)$ be an arbitrary element and set $a:=g(1)$ . As detailed above, we attach to a a subfield L in A and with it the closed subgroups ${\mathbf{G}}^L$ , ${\mathbf{Z}}^L$ and ${\mathbf{S}}^L$ of ${\mathbf{Str}}(A)$ , as well as the class $[\xi _a]=(a_\tau )\in H^1(K,{\mathbf{Z}}^L)$ . Since the image of this class in $H^1(K,{\mathbf{H}})$ is trivial, there exists $f\in {\mathbf{H}}(K^{sep})$ such that $a_{\tau }=(f^{-1})^\tau f$ for all $\tau \in {\textrm {Gal}}(K^{sep}/K)$ .

Lemma 6.3. The subset $f(L)\subset A_{K^{sep}}=A\otimes _K K^{sep}$ is contained in $A=A\otimes _K K$ .

Proof. Let $l\in L$ . We need to show that $f(l)$ is ${\textrm {Gal}}(K^{sep}/K)$ -invariant. Take any $\tau \in {\textrm {Gal}}(K^{sep}/K)$ . Then

$$ \begin{align*}\tau(f(l))=(f^\tau)(\tau(l))=f^\tau(l)=f (f^{-1}f^\tau(l))=f a^{-1}_{\tau}(l)=f(l), \end{align*} $$

since ${\mathbf{Z}}^L\subset {\mathbf{G}}^L$ fixes L pointwise and the statement follows.

Since ${\mathbf{H}} = {\mathbf {Aut}}(A)$ , the above lemma implies that the map $L\to L'=f(L)$ given by $l \mapsto f(l)$ is a field isomorphism over K. Assume that $L\not =L'$ . Let $B_\sigma ^+\subset A$ be the $9$ -dimensional subalgebra generated by L and $L'$ . Recall from [1] that $\xi _a$ is constructed explicitly as follows. Choose $t\in {\mathbf{S}}^L(K^{sep})$ such that $t(1)=a=g(1)$ ; this is possible by Lemma 6.2. Then $a_\tau =t^{-\tau +1}$ , from which we conclude that $ft^{-1}$ is defined over K and that

$$ \begin{align*}ft^{-1}(g(1))=ft^{-1}(a)=f(1)=1, \end{align*} $$

which implies that $ft^{-1}g\in {\mathbf{H}}(K)$ . Thus, modulo ${\mathbf{H}}(K)$ , we may assume that $g=ft^{-1}$ .

Let now $s\in {\mathrm{Str}}(A,B_\sigma ^+)$ be the element constructed in Theorem 5.3. It is R-trivial, since so is ${\mathbf{Str}}(A,B_\sigma ^+)^\circ $ and it satisfies $s(L)=f(L)$ . Furthermore,

$$ \begin{align*}L=s^{-1}(f(L))=s^{-1}ft^{-1}(L)=s^{-1}g(L), \end{align*} $$

since $t(L)=L$ . It follows that modulo R-trivial elements we may assume that $g(L)=L$ ; that is, $g\in {\mathbf{Str}}(A,L)(K)$ , where ${\mathbf{Str}}(A,L)\subset {\mathbf{Str}}(A)$ is the subgroup of all elements stabilising L.

Passing to a separable closure of K one can easily check that the connected component of ${\mathbf{Str}}(A,L)$ is ${\mathbf{S}}^L\cdot {\mathbf{G}}^L$ . Hence,

$$ \begin{align*}{\mathbf{Str}}(A,L)=N_{{\mathbf{Str}}(A)}({\textbf{S}}^L\cdot {\mathbf{G}}^L) \end{align*} $$

and

$$ \begin{align*}{\mathbf{Str}}(A,L)/{\textbf{S}}^L\cdot {\mathbf{G}}^L\simeq N_{{\textbf{H}}}({\mathbf{G}}^L)/{\mathbf{G}}^L \simeq {\textbf{Out}}({\mathbf{G}}^L). \end{align*} $$

By Lemma 2.7 we can, if necessary, multiply g by an element from $N_{\mathbf{H}}({\mathbf{G}}^L)(K)$ to obtain an element $g'\in ({\mathbf{S}}^L\cdot {\mathbf{G}}^L)(K)$ . To complete our reduction to subgroups of type $\mathrm {F}_4$ , it remains to show that $g'$ is R-trivial modulo elements from ${\mathbf{H}}(K)$ . This is the content of the following result.

Proposition 6.4. Let $g\in ({\mathbf{S}}^L\cdot {\mathbf{G}}^L)(K)$ . Then there exists an R-trivial element j in $({\mathbf{S}}^L\cdot {\mathbf{G}}^L)(K)$ such that $gj\in {\mathbf{G}}^L(K)\subset {\mathbf{H}}(K)$ .

Proof. Our argument is based on the consideration of the exact sequence

(8) $$ \begin{align} 1\longrightarrow {\textbf{Z}}^L\longrightarrow {\textbf{S}}^L\times {\mathbf{G}}^L \longrightarrow {\textbf{S}}^L\cdot {\mathbf{G}}^L \longrightarrow 1. \end{align} $$

In the corresponding exact sequence in cohomology, the element g is mapped to a class $[\eta ]$ in $H^1(K,{\mathbf{Z}}^L)$ . Since $H^1(K,{\mathbf{S}}^L)$ is trivial, this class belongs to

$$ \begin{align*}{\textrm{Ker}}\,[H^1(K,{\textbf{Z}}^L)\longrightarrow H^1(K,{\mathbf{G}}^L)]. \end{align*} $$

We first prove that it is R-trivial. Since ${\mathbf{Z}}^L$ is a group of exponent $2$ , by the norm principle it suffices to prove that $[\eta ]$ becomes R-trivial after extending scalars to $L/K$ . Two cases are possible.

If $L/K$ is a Galois extension, then ${\mathbf{G}}^L_L$ , being a strongly inner form of type $^1\mathrm {D}_4$ , is of the form ${\mathbf{G}}^L_L\simeq {\mathbf{Spin}}(h)$ , where h is a $3$ -fold Pfister form. Hence, by Example 2.3, the class ${res}^L_K([\eta ])$ is R-trivial.

If $L/K$ is not a Galois extension, then ${\mathbf{G}}^L_L\simeq {\mathbf{Spin}}(h)$ for some h satisfying all conditions in Example 2.4. Therefore, ${res}^L_K([\eta ])$ is also R-trivial.

Now the sequence (8) induces the exact sequence

$$ \begin{align*}({\textbf{S}}^L\times {\mathbf{G}}^L)(K(x)) \longrightarrow ({\textbf{S}}^L \cdot {\mathbf{G}}^L)(K(x))\longrightarrow H^1(K(x),{\textbf{Z}}^L)\longrightarrow H^1(K(x),{\mathbf{G}}^L), \end{align*} $$

where x is a variable over K. Let

$$ \begin{align*}\eta(x)\in {\textrm{Ker}}\,[H^1(K(x),{\textbf{Z}}^L) \longrightarrow H^1(K(x),{\mathbf{G}}^L)] \end{align*} $$

be defined at 0 and 1 and such that $\eta (0)=1$ and $\eta (1)=\eta $ ; such an element exists since $\eta $ is R-trivial. Take any element $g(x)\in ({\mathbf{S}}^L \cdot {\mathbf{G}}^L)(K(x))$ that is defined at 0 and 1 and whose image in $H^1(K(x), {\mathbf{Z}}^L)$ is $\eta (x)$ . Note that $g(0)^{-1}g(1)$ is R-trivial in ${\mathbf{S}}^L \cdot {\mathbf{G}}^L$ . By construction, $g(0)=u_0$ and $g(1)=u_1g$ for some $u_0,u_1\in ({\mathbf{S}}^L\times {\mathbf{G}}^L)(K)$ . Hence,

$$ \begin{align*}g=u_1^{-1}g(1)=u_1^{-1}g(0)\left(g(0)^{-1}g(1)\right)=u_1^{-1}u_0\left(g(0)^{-1}g(1)\right). \end{align*} $$

Writing $u_1^{-1}u_0=us$ for some $u\in {\mathbf{G}}^L(K)$ and $s\in {\mathbf{S}}^L(K)$ and noting that ${\mathbf{S}}^L$ is a rational torus, the proof is complete upon setting $j=\left (sg(0)^{-1}g(1)\right )^{-1}$ .

We have thus altogether proved the following.

Theorem 6.5. Let $L\subset A$ be a cubic subfield and let $g\in {\mathrm{Str}}(A)$ be such that $g(L)=L$ . Then modulo R-trivial elements, g can be written as a product $g=g_1g_2$ where $g_1(L)=g_2(L)=L$ , $g_1\in N_{\mathbf{H}}({\mathbf{G}}^L)(K)$ and $g_2\in {\mathbf{G}}^L(K)$ .

7. End of the Proof

We now finish the proof that ${\mathbf{Str}}(A)$ is R-trivial. To begin we recall the following known fact.

Theorem 7.1. Let K be a perfect field. Let ${\mathbf{G}}$ be a semisimple K-group and $g\in {\mathbf{G}}(K)$ . Then the semisimple and unipotent components $g_s,g_u$ of the Jordan decomposition $g=g_sg_u$ are defined over K. Moreover, if ${\mathbf{G}}$ is K-anisotropic, then $g_u=1$ .

Proof. For the first statement see [2, Page 81, Corollary 1] and for the second statement see [3].

Next we have the following.

Lemma 7.2. Let A be a division Albert algebra over K and let $g\in {\mathbf{H}}(K)$ . Then g fixes a cubic subfield in A pointwise.

Proof. Since ${\mathbf{H}}$ is K-anisotropic, g is semisimple. Indeed, let $g=g_sg_u$ be its Jordan decomposition over an algebraic closure $\overline {K}$ of K. Let $\widetilde {K}=K^{p^{-\infty }}$ . Here p is the characteristic of K. Since $p\not =2,3$ , the extension $\widetilde {K}/K$ cannot kill the cohomological invariant $g_3$ attached to ${\mathbf{H}}$ . By [13, Theorem 40.8, part (2)], A is still a division algebra over $\widetilde {K}$ , whence ${\mathbf{H}}_{\widetilde {K}}$ is still anisotropic. It follows from the above theorem that $g_u=1$ . ⁴

Hence, g is contained in a maximal K-torus ${\mathbf{T}}\subset {\mathbf{H}}$ . Using an explicit reduced model of the split Albert algebra $A\otimes _K K^{sep}$ , one can easily check that over $K^{sep}$ , every element in ${\mathbf{T}}$ fixes a commutative subalgebra

$$ \begin{align*}W\simeq K^{sep}\times K^{sep}\times K^{sep}\end{align*} $$

of $ A\otimes _K K^{sep}$ pointwise. It follows that g fixes a vector $v\in A$ which is not a scalar multiple of the identity, whence it fixes the cubic subfield $K(v)\subset A$ generated by v.

In order to finish our proof we need one final ingredient. Let A be an Albert algebra over K and let $p\in A^\times $ . Recall that this is equivalent to $N(p)\neq 0$ , where N is the cubic norm of A. The isotope $A^{(p)}$ of A is the algebra with underlying linear space A and multiplication

$$ \begin{align*}x\bullet_p y=x(yp)+y(xp)-(xy)p,\end{align*} $$

where juxtaposition denotes the multiplication of A. It is known that $A^{(p)}$ is an Albert algebra and that the cubic norm of $A^{(p)}$ is $\nu N$ , where $\nu =N(p)$ . From this it follows that $A^{(p)}$ is a division algebra if and only if A is and that ${\mathbf{Str}}(A)\simeq {\mathbf{Str}}(A^{(p)})$ .

Proposition 7.3 [25]. Let A be a division Albert algebra over K. Then there exists $p\in A^\times $ such that $A^{(p)}$ contains a cubic cyclic subfield.

Remark 7.4. If the ground field F contains a cubic root of unity, then due to a result of H. Petersson and M. Racine [16], one can take $p=1$ .

We are now ready to prove our main result.

Theorem 7.5. Let A be a division Albert algebra over K. Then the algebraic K-group ${\mathbf{Str}}(A)$ is R-trivial.

Proof. By the above discussion, we may replace A by any isotope $A^{(p)}$ . Thus, by the preceding proposition we may assume that A contains a cubic cyclic subfield L. Let $g\in {\mathrm{Str}}(A)$ . By the results of Section 6, summarised in Theorem 6.5, we may assume that $g\in {\mathbf{H}}(K)$ . Three cases are possible.

Case $1$ : g fixes L but not pointwise. Let $F\subset A$ be a cubic field extension of K pointwise fixed by g; such a field exists by Lemma 7.2. Since $F\not =L$ , L and F generate a 9-dimensional subalgebra $B_\sigma ^+\subset A$ that is stabilised by g. In the course of the proof of Corollary 4.3 we saw that

$$ \begin{align*}{\mathbf{Str}}(A,B_\sigma^+)(K)={\mathbf{Str}}(A,B_\sigma^+)^\circ(K).\end{align*} $$

By Proposition 4.2, the algebraic K-group ${\mathbf{Str}}(A,B_\sigma ^+)^\circ $ is R-trivial and therefore g is R-trivial.

Case $2$ : g fixes L pointwise. Hence, g is in the group of K-points of the algebraic K-group ${\mathbf{Str}}(A/L)$ whose functor of points is as follows: if R is a K-ring,

$$ \begin{align*}{\mathbf{Str}}(A/L)(R)=\{x\in {\mathbf{Str}}(A)(R)\ | \ x(l)=l \ \ \mbox{for all}\ l\in L\}={\mathbf{G}}^L(R). \end{align*} $$

Since L is cyclic, by Lemma 2.7 there is an element $h_1\in N_{\mathbf{H}}({\mathbf{G}}^L)(K)$ that stabilises L but does not fix it pointwise. The same is true for $h_2:=h_1^{-1}g$ , since g fixes L pointwise. From Case (1) we know that $h_1$ and $h_2$ are R-trivial and hence so is g.

Case 3: $g(L)\not =L$ . Since $g\in {\mathrm{Aut}}(A)$ , the field $g(L)$ is a cubic cyclic subfield of A isomorphic to L. The subfields L and $g(L)$ generate a 9-dimensional subalgebra $B_\sigma ^+$ of A. By Theorem 5.3 there exists $h_1\in {\mathrm{Str}}(A,B_\sigma ^+)$ such that $h_1(g(L))=L$ and by Proposition 4.2, $h_1$ is R-trivial. Let $h_2:=h_1g$ . By construction, it belongs to

$$ \begin{align*}{\mathbf{Str}}(A,L)(K)=\{ x\in {\mathrm{Str}}(A)\ | \ x(L)=L\}. \end{align*} $$

By Theorem 6.5, $h_2$ can be written, modulo R-trivial elements, as a product $h_2=h_3h_4$ with $h_3\in {\mathbf{G}}^L(K)$ and $h_4\in N_{\mathbf{H}}({\mathbf{G}}^L)(K)$ . In particular, $h_3$ and $h_4$ are in ${\mathbf{H}}(K)$ and stabilise L. By Cases (1) and (2), the elements $h_3$ and $h_4$ are R-trivial and hence so is $h_2$ and g. This completes the proof.

Theorem 7.6. Let $\overline {\mathbf{G}}$ be an adjoint strongly inner form of type ${\mathrm{E}}_6$ . Then $\overline {\mathbf{G}}$ is R-trivial.

Proof. Indeed, the canonical map ${\mathbf{Str}}(A) \to \overline {\mathbf{G}}$ has a rational section, since its kernel ${\mathbf{G}}_m$ has trivial Galois cohomology in dimension 1. We conclude by the above theorem.

8. Applications

8.1. The Kneser–Tits Problem for $\mathrm {E}_{7,1}^{78}$ and $\mathrm {E}_{8,2}^{78}$

Theorem 8.1. Let K be an arbitrary field. Let ${\mathbf{G}}$ be a simple simply connected K-group of type $\mathrm {E}_{7,1}^{78}$ or $\mathrm {E}_{8,2}^{78}.$ Then the Kneser–Tits conjecture for ${\mathbf{G}}$ holds.

Proof. According to [9, Remarque 7.4] we may, without loss of generality, assume that ${\textrm {char}}(K)=0$ . The Tits index of ${\mathbf{G}}$ is of the form

or

In both cases the semisimple anisotropic kernel ${\mathbf{H}}$ of ${\mathbf{G}}$ is a strongly inner form of type $\mathrm {E}_6$ . If ${\mathbf{S}}\subset {\mathbf{G}}$ is a maximal split torus whose centraliser is ${\mathbf{H}}$ , then arguing as in [5] one easily verifies that over an algebraic closure of K the intersection ${\mathbf{S}}\cap {\mathbf{H}}$ is the centre of ${\mathbf{H}}$ .

Furthermore, by [9, Théorème 7.2], the Kneser–Tits problem has an affirmative answer if and only if ${\mathbf{G}}(K)/R=1$ . It follows from the Bruhat–Tits decomposition that

$$ \begin{align*}{\mathbf{G}}(K)/R\simeq C_{\textbf{S}}({\mathbf{G}})/R \simeq ({\textbf{H}}/(C_{\textbf{S}}({\mathbf{G}})\cap {\textbf{H}}))/R \simeq \overline{{\textbf{H}}}/R \end{align*} $$

where $\overline {{\mathbf{H}}}$ is the corresponding adjoint group. It remains to note that by Theorem 7.6, the group $\overline {{\mathbf{H}}}$ is R-trivial.

8.2. The Tits–Weiss Conjecture

Theorem 8.2. Let A be an Albert algebra defined over an arbitrary field $K.$ Then the group ${\mathbf{Str}}(A)(K)$ is generated by the U-operators $U_a$ with $a\in A^\times $ and the scalar homotheties.

Proof. By the main result of the Appendix, the Tits–Weiss conjecture holds if and only if the Kneser–Tits conjecture holds for groups of type $\mathrm {E}_{7,1}^{78}$ and $\mathrm {E}_{8,2}^{78}$ . The result follows.

Remark 8.3. From this theorem, which we have now established in arbitrary characteristic, the R-triviality of ${\mathbf{Str}}(A)$ is immediate. Thus, Theorem 7.5 holds in arbitrary characteristic as well.

8.3. Properties of the Functor of R-Equivalence Classes for Strongly Inner Forms of Type $^1\mathrm {E}_6$

To a reductive K-group ${\mathbf{G}}$ one can attach the functor of R-equivalence classes

$$ \begin{align*}\mathcal{G}/R: Fields/K \longrightarrow Groups,\ \ \ F/K \to {\mathbf{G}}(F)/R \end{align*} $$

where $Fields/K$ is the category of field extensions of K and $Groups$ is the category of abstract groups. The experts expect that the following conjectures hold:

Conjecture 1. The functor $\mathcal {G}/R$ factors through the subcategory of abelian groups of the category $Groups$ ; that is, for all field extensions $F/K$ the group ${\mathbf{G}}(F)/R$ is abelian.

Conjecture 2. If F is a finitely generated field over its prime subfield, then ${\mathbf{G}}(F)/R$ is finite.

Conjecture 3. The functor $\mathcal {G}/R$ has transfers; that is, there is a functorial collection of maps ${\textrm {tr}}^F_K: {\mathbf{G}}(F)/R \to {\mathbf{G}}(K)/R$ for all finite field extensions $F/K$ .

Furthermore, one expect that the norm principle holds for all semisimple K-groups. Of course, all of these conjectures are obviously true for R-trivial K-groups. In particular, they hold for rational K-groups. For instance, this is the case for groups of type $\mathrm {G}_2$ . In [1] we investigated the case of groups of type $\mathrm {F}_4$ arising from the first Tits construction. Here we consider the next case of simple simply connected strongly inner forms of type $\mathrm {E}_6$ .

Theorem 8.4. Let K be an arbitrary field and let ${\mathbf{G}}$ be a simple simply connected K-group which is a strongly inner form of type $\mathrm {E}_6$ . Then

1. (i) ${\mathbf{G}}(K)/R$ is an abelian group.

2. (ii) the functor $\mathcal {G}/R$ has transfers.

3. (iii) if ${\mathbf{Z}}$ is the centre of ${\mathbf{G}}$ , the norm principle holds for $({\mathbf{Z}},{\mathbf{G}})$ .

Proof.

1. (i) The group ${\mathbf{G}}$ is the derived subgroup of the R-trivial K-group ${\mathbf{Str}}(A)$ for an Albert algebra A. It follows that $[{\mathrm{Str}}(A),{\mathrm{Str}}(A)]\subset R{\mathbf{G}}(K)$ and hence ${\mathbf{G}}(K)/R$ is abelian.

2. (ii) Apply Theorem $3.4$ in [4] to ${\mathbf{G}}'=R_{F/K}({\mathbf{G}})$ .

3. (iii) Since the corresponding adjoint group $\overline {\mathbf{G}}$ is R-trivial, the result follows from Theorem 2.5.