%%% metamail: Ignoring invalid mailcap entry: 	x-mozilla-flags=deleted
%%% metamail: Ignoring invalid mailcap entry: 	x-mozilla-flags=deleted
%%% Envelope-to: rehmann@localhost
%%% X-Authentication-Warning: hopf.math.ethz.ch: chernous owned process doing -bs
%%% Date: Sun, 13 Jan 2002 14:46:19 +0100 (CET)
%%% From: "Prof. Vladimir Chernousov" <chernous@math.ethz.ch>
%%% To: Ulf Rehmann <rehmann@mathematik.Uni-Bielefeld.DE>
%%% Subject: Re: Paper
%%% In-Reply-To: <E16LpOy-0002Cd-00@dea.ur.de>
%%% Content-Type: MULTIPART/MIXED; BOUNDARY="-2071885934-1982192816-1010929579=:5694"
%%% X-UIDL: 6bd246f8a38f3d886cdbbce57d786c01
%%% 
%%% Content-Type: TEXT/PLAIN; charset=US-ASCII
%%% 
%%% metamail: Ignoring invalid mailcap entry: 	x-mozilla-flags=deleted
%%% Content-Type: APPLICATION/x-tex; name="rsh.tex"
%%% Content-Transfer-Encoding: BASE64
%%% Content-ID: <Pine.LNX.4.33.0201131446190.5694@hopf.math.ethz.ch>
%%% Content-Description: 
%%% Content-Disposition: attachment; filename="rsh.tex"
%%% 
%%% 
%%% This message contains data in an unrecognized format, application/x-tex,
%%% which is being decoded and written to the file named "/tmp/rsh.tex".
%%% If you do not want this data, you probably should delete that file.
%%% Wrote file /tmp/rsh.tex
%%% 
\documentclass[11pt]{article}
\usepackage{amsfonts,amssymb,eucal}
\usepackage{amsbsy}
\usepackage[all]{xy}
\makeindex \voffset = -18mm \textheight 230mm \textwidth 150mm
\oddsidemargin 4mm
%\usepackage[notcite,notref]{showkeys}
%%%%%%%%%%%% arrows %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\font\rus=wncyr10
\newcommand{\SH}{\hbox{\rus\char88}}
\newcommand \ra {\rightarrow } 
\newcommand \lra {\longrightarrow }
\newcommand \hra {\hookrightarrow }
\newcommand \mt {\mapsto }
\newcommand \Lra {\Leftrightarrow }
\newcommand \Ra {\Rightarrow }
%%%%%%%%%%% algebraic-geometry symbols %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%% arithmetical symbols %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{sublemma}[theorem]{Sublemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
%%%%%%%%%%%%% specific symbols %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\palka }{\mathop {\! \! \mid }\nolimits }
\newcommand{\qed}{$\ \Box$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newenvironment{pf}{\par\noindent{\em Proof}.}{\hfill\framebox(6,6)
\par\medskip}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\title{\bf The kernel of the Rost invariant, Serre's Conjecture II and the 
Hasse principle for quasi-split groups $^{3,6}D_4,E_6,E_7$}
\author{}

\author{V. Chernousov
\thanks{The author gratefully acknowledge the support of 
TMR ERB FMRX CT-97-0107 and Forschungsinstitut f\"ur Mathematik,
ETH in Z\"urich.}
%and the hospitality of the University of Bielefeld.}
\vspace{0mm}\\
{\small Forschungsinstitut f\"ur Mathematik }\\
{\small ETH--Zentrum}\\
{\small  CH-8092 Z\"urich}\\
{\small Switzerland}}
%\vspace{0mm}\\
%V. Guletski\v{\i}
%\vspace{0mm}\\
%{\small Institute of Mathematics,}\\ 
%{\small Surganova str. 11, Minsk 220072, Belarus}}
\date{}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}

\maketitle

%\baselineskip = 5mm
%
\begin{abstract}
We prove that for a simple simply connected quasi-split group of type 
$^{3,6}D_4,E_6, E_7$ defined
over a perfect field $F$ of characteristic $\not= 2,3$ the Rost invariant has 
trivial kernel. In certain cases we give a formula for the Rost
invariant. It follows immediately from the result above 
that if ${\rm cd}\,F\le 2$ (resp. ${\rm vcd}\,F\le 2$) then Serre's 
Conjecture II (resp. 
the Hasse principle) holds for such a group. For a $(C_2)$-field,
in particular $\mathbb{C}(x,y)$, we prove the stronger result
that Serre's Conjecture II holds for all (not necessary
quasi-split) exceptional groups of type $^{3,6}D_4,E_6,E_7$.  
\end{abstract}


%\tableofcontents


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{\it Introduction}

This paper grew out of the letters~\cite{[Ch98],[Ch00]} where we 
sketched how Harder's proof~\cite{[H65],[H66]} of the Hasse principle
for exceptional groups $^{3,6}D_4,E_6,E_7$
over number fields can be carried over to the case
of quasi-split groups defined over a perfect field
of cohomological dimension  $\le 2$  and how the same ideas can be 
applied to describe in particular the kernel of the 
Rost invariant for $^2E_6$. 

In this paper we give full proofs of all these results. The main
ones are the following. Let $G_0$ be a quasi-split simple simply 
connected exceptional group of type $^{3,6}D_4,E_6,E_7$
defined over a perfect field $F$ of characteristic $\not=2,3$. Then
\begin{itemize}
\item the kernel of the Rost invariant of $G_0$ is trivial;

\item if ${\rm cd}\,F\le 2$, then Serre's Conjecture II holds for $G_0$;

\item if ${\rm vcd}\,F\le 2$, then the Hasse principle Conjecture II holds 
for $G_0$;

\item if $F$ is a $(C_2)$-field, then Serre's Conjecture II holds for an 
arbitrary simple simply connected group (not necessary quasi-split) of 
the same type as $G_0$.
\end{itemize}

In ~\cite{[G97]} Gille proved that a group $G$ of inner type
$E_6$ or $E_7$ defined over $F$ is $F$-split iff there exist finite field
extensions $E_1,\ldots, E_s$ of $F$ splitting $G$ and such that the 
$$ g.c.d.\{\,[E_1:F],\ldots,[E_s:F]\,\}=1.$$ As an easy corollary
of our results we obtain that the same quasi-splitting criterion is true
for outer forms of type $^{3,6}D_4$ or $^2E_6$.  

Another corollary says that for any $F$-group $G$ (not necessary quasi-split)
of type $^{3,6}D_4,\,^{1,2}E_6$ or $E_7$ there exists a chain of cyclic
extensions
\[
L_0=F\subset L_1\subset \cdots \subset L_n
\]
of degrees $2$ or $3$
such that $G$ splits over $L_n$. 

Recall that for all classical groups and groups of type $G_2,F_4$
Serre's Conjecture II and the Hasse principle Conjecture II
were proved by E. Bayer-Fluckiger and R. Parimala~\cite{[BP95],[BP98]}.
J. Ferrar \cite{[Fer]} essentially proved the Hasse principle
Conjecture II for inner groups $E_6$ with the Tits algebras of index $\le 3$. 
Note also that our results related to Serre's Conjecture II
(items $2,4$) were obtained independently by P. Gille~\cite{[G01]} using 
different methods.


S. Garibaldi~\cite{[Gar]} also proved independently the triviality
of the kernel of the Rost invariant for quasi-split groups $E_6,E_7$.
His argument is based on consideration of explicit geometric realizations of 
quasi-split $E_6,\,E_7$ and studying properties of algebra structures 
which occur in his constructions.  
Since items $2,3$ immediately follow from the first one, Garibaldi's 
result gives another proof of Serre's Conjecture II and the Hasse principle 
Conjecture II for quasi-split groups $E_6,E_7$.

Our approach is based on different
ideas which, as we have already mentioned, come back to G. Harder. 
In contrast to Garibaldi's paper 
we focus on studying intrinsic properties of groups splitting over small 
extensions of the ground field $F$.
The main body of the paper are Sections $4,5$ where we 
study properties of
groups splitting over an extension $K/F$ 
of degree $p=2,3$. In this part $F$ is an arbitrary field of
characteristic $\not= 2,3$. We show
that the $F$-structure of such groups can be described completely
by certain numerical invariants. In the case of quadratic extensions
we follow the author proof \cite{[Ch89]} of the Platonov-Margulis conjecture
on the projective simplicity of groups of rational points splitting
over a quadratic extension. The results obtained in these two sections
allow us
to give a formula for the Rost invariant of strongly inner or outer forms
of type $E_6,E_7$ splitting over an extension of degree $p=2,3$
in the case where the Galois group ${\rm Gal}(F^s/F)$ is a pro-$p$-group
(see \ref{formula} and 6.2.3).

Gille's splitting criterion~\cite{[G97]} for groups of inner type $E_6,E_7$ 
combined with the results described in Sections 4,5 gives 
proofs of 
the main theorems more or less quickly. This is done in Sections 6,7,8.

Finally we note that the same methods together with the results of
\cite{[Ch94],[Ch]} prove Serre's Conjecture II
for $E_8$ in each of the following cases:

\begin{itemize}
\item the nilpotent closure of the basic field $F$ 
has cohomological dimension $1$;

\item Gille's splitting criterion~\cite{[G97]} holds for $E_8$.
\end{itemize}

\medskip

\noindent
{\bf Acknowledgements.} 
%I gratefully acknowledge the support of
%TMR ERB FMRX CT-97-0107 and Forschungsinstitut f\"ur Mathematik,
%ETH in Z\"urich.
I am grateful to J.-P. Serre and S. Garibaldi 
for useful comments and discussions of
the first version of the paper. 

\medskip


\begin{center} {\it \Large{Notation}}
\end{center}

The letter $F$ denotes a perfect field and
$F^s$ denotes a separable closure of $F$. 
If $L/F$ is a separable extension, then $R_{L/F}$ denotes the Weil functor of 
restriction of scalars.

We say that a reductive $F$-subgroup 
$G'\subset G$ is standard if there is a maximal $F$-torus $T$ of $G$
normalizing $G'$. 


If $G$ is a reductive algebraic group and $T\subset G$ is a maximal
torus, we let $\Sigma(G,T)$ denote the root system of $G$ with
respect to $T$.


If $S\subset G$ is a maximal $F$-split torus then the semisimple part
of the centralizer $C_G(S)$ is called the semisimple $F$-anisotropic
kernel of $G$.
 

We number the simple roots of exceptional groups as in \cite{[B68]}.



\section{\it The Rost invariant and its properties}





In the $90$s, for a simple simply connected linear algebraic group $G$ defined
over a field $F$, M.~Rost~\cite{[R]} constructed a cohomological 
$H^3(\mathbb{Q}/\mathbb{Z}(2))$--invariant of $G$ which
nowadays is called  the Rost invariant of the group $G$. 
Thus, for any field extension $K/F$ the Rost invariant associates
a canonical map of pointed sets
   $$   
     R_G^K:H^1(K,G)\to H^3(K,\mathbb{Q}/\mathbb{Z}(2)).
   $$
By abuse of language, instead of $R_G^K$ we will use the abbreviated 
notation $R_G$ for the Rost invariant whenever there is no danger
of confusion.

In this section we will state (without proofs) properties 
of the Rost invariant
used in Section~\ref{rost}. For the proofs we refer  to \cite{[R]},  
\cite{[KMRT]} (\S 31), \cite{[EKLV98]}, \cite{[G00]}, \cite{[M01]} (Appendix
A and B).
We note that
an explicit formula for $R_G$ is known only in a few cases. But 
fortunately
that will do for us to prove triviality of the kernel of
the Rost invariant for quasi-split groups $E_6,E_7$.

\subsection{Inner type $A_n$}

Let $A$ be a central simple algebra over $F$ of exponent 
$e$ and degree $n+1$. Let $G=SL_1(A)$. Assume that the characteristic
of $F$ doesn't divide $n+1$. Then 
$H^1(F,G)=F^{\times}/\,{\rm Nrd}\,(A^{\times})$ and
the formula for the Rost invariant $R_G$ is given (up to a sign) by
$$
R_{G}: F^{\times}/{\rm Nrd}\,(A^{\times})\lra H^3(F,\mu_e^{\otimes 2}),\quad
R_G\,(x\,{\rm Nrd}\,(A^{\times}))=(x)\cup [A].
$$ 
    \begin{theorem}\label{ms}{\rm (\cite{[MS]}, Theorem 12.2)}
     If $n+1$ is square-free, then $R_G$ is injective.
     \end{theorem}




\subsection{Spinor groups}\label{sg} 




Let $f$ be a quadratic form over a field $F$ of characteristic $\not=2$
of dimension at least $5$ and let $G= Spin(f)$. The set $H^1(F,G)$ fits 
into the exact
sequence
$$
H^1(F,Spin(f))\stackrel{\pi}{\lra} H^1(F,SO(f)) \lra\, _2Br F.
$$
%The set $H^1(K,SO(f))$ classifies quadratic forms of the same dimension and
%discriminant as $f$. The connecting map $\delta$ takes such a form $f'$
%to the Hasse-Witt invariant $e_2([f']-[f])$. 
%Thus the image of $\pi$
It follows that ${\rm Im}\,\pi$
consists of classes of quadratic forms having the same dimension, discriminant
and Hasse-Witt invariant as $f$. Therefore, for any cocycle
$\xi\in Z^1(F,Spin(f))$ we obtain that $\pi([\xi])-[f]\in I^3F$ and
the formula for the Rost invariant is given by
\begin{equation}\label{11111}
R_{Spin(f)}: H^1(F,Spin(f))\lra  H^3(F,\mu_2^{\otimes 2}), \quad
R_{Spin(f)}([\xi])=e_3(\pi([\xi])-[f]).
\end{equation}
Here $e_3$ is the Arason invariant~\cite{[Ar75]}.
   \begin{proposition}\label{d5}
    Let $f$ be a quadratic form of dimension $\le 12$
    over a field $F$ such that $G=Spin(f)$ is quasi-split over $F$,
    i.e. $f$ is of maximal Witt index. Then $R_G$ has trivial kernel.
   \end{proposition}
\begin{pf} The statement follows from (\ref{11111}) and the Arason-Pfister 
Hauptsatz~\cite{[L]}, X.3.1.
\end{pf}




\subsection{Outer forms $A_n$}

Let $K/F$ be a quadratic extension, $V$ a vector space of dimension
$n+1$ over $K$ and $h$ a hermitian form on $V$. The group $SU(V,h)$ of 
isometries with determinant $1$ is an almost simple simply connected group
of type $A_n$ defined over $F$ and splitting over $K$.
The hermitian form $h$ corresponds to the quadratic form $q$ on
the vector space $V$ viewed as an $F$-vector space: $q$ is given by the
formula $q(v)=h(v,v)$. It easily follows that we have a natural embedding
$SU(V,h)\hookrightarrow SO(V,q)$ which can be lifted to an
embedding $SU(V,h)\hookrightarrow Spin(V,q)$, since $SU(V,h)$
is simply connected. It turns out that the Rost invariant for $SU(V,h)$
is just the restriction of $R_{Spin}$.
     \begin{proposition}\label{a5} 
      Let $h$ be a quasi-split hermitian form of dimension
      $\le 6$. Then the Rost invariant $R_{SU(V,h)}$ has trivial kernel.
      \end{proposition}
\begin{pf} Apply the same argument as in Proposition~\ref{d5}.
\end{pf}
 
\subsection{The Rost numbers}




Let $G,H$ be almost simple simply connected linear algebraic groups over $F$
and let $\rho: H\hookrightarrow G$ be an $F$-embedding. The 
restriction $R_G$ at 
$\rho(H)$ gives a cohomological invariant of $H$, hence is equal to
$n_{\rho}R_H$ for a positive integer $n_{\rho}$ (see~\cite{[KMRT]}, \S 31). 
The smallest such integer is called the Rost number of embedding $\rho$.
    \begin{proposition}\label{number}
     Let $\rho: H\hookrightarrow G$ be an $F$-embedding such that
     $\rho(H)$ is a standard subgroup. Assume that
     all roots of the root system $\Sigma=\Sigma(G,T)$ of $G$ with respect 
     to a maximal torus $T$ have the same length. Then $n_{\rho}=1$; in 
     particular, for any
     $[\xi]\in H^1(F,H)$ one has $R_G(\rho([\xi]))=R_H([\xi])$.
    \end{proposition}
\begin{pf} Let $T$ be a maximal $F$-torus of $G$ normalizing $\rho(H)$.
Then $\rho(H)\cdot T$ is a reductive subgroup of $G$. Since $T$ is a maximal torus
of $\rho(H)\cdot T$ and all maximal tori are conjugate, we easily obtain
that the connected component $S$ of the intersection $T\cap \rho(H)$ is
a maximal torus of $\rho(H)$. Since $T$ normalizes $\rho(H)$, there is a natural
embedding 
   \[
     \Sigma(\rho(H),S)\hookrightarrow \Sigma(G,T).
   \]
Hence the coroots
of $\rho(H)$ are also the coroots of $G$ (we used the fact that all roots
of the root system $\Sigma=\Sigma(G,T)$ have the same length). The rest of the proof
follows from \cite{[M01]}, Appendix B.
\end{pf}     



\section{\it Steinberg's theorem}




We state now two theorems which are due to Steinberg~\cite{[St65]}.
Although they are not formulated explicitly in~\cite{[St65]},
their proofs can be easily obtained from arguments contained in
\cite{[St65]}, \S $10$ (see also \cite{[PR]}, Propositions $6.18,\,6.19$,
p. 338--339). 

Let $G_0$ be a simple (not necessary
simply connected) linear algebraic group split or quasi-split over $F$. Let 
$\xi\in Z^1(F,G_0)$ be a cocycle and let $G=\,^{\xi}G_0$ be the 
corresponding twisted group.
     \begin{theorem}\label{steinberg1} 
      For any maximal torus $S\subset G$ over $F$ there is an $F$-embedding 
      $S\hookrightarrow G_0$ such that the class $[\xi]$
      lies in the image of $H^1(F,S)\to H^1(F,G_0)$.
     \end{theorem}
     \begin{theorem}\label{steinberg2}
      In the notation above assume
      that $G_0$ is a simple simply connected group split over $F$ and that
      $G$ is isotropic over $F$.
      Then $\xi$ is equivalent to a cocycle with coefficients
      in a proper semisimple simply connected $F$-split subgroup of $G_0$
      which is standard and isomorphic over $F^s$ to a semisimple 
      anisotropic kernel of $G$. 
     \end{theorem}
\begin{pf} Let $S_1\subset G$ be a maximal $F$-split torus. Then
the semisimple part of its centralizer $C_G(S_1)$ is a semisimple simply 
connected $F$-anisotropic subgroup of $G$ (see~\cite{[T66]}). 
Let $S\subset G$ be a maximal torus over $F$ containing 
$S_1$. By Theorem~\ref{steinberg1}, there is an $F$-embedding
$\phi:S\hookrightarrow G_0$ such that 
     $$
       [\xi]\in {\rm Im}\,[H^1(F,S)\to H^1(F,G_0)].
     $$
The centralizer $C_{G_0}(\phi(S_1))\subset G_0$ is a reductive subgroup
of $G_0$ and
it follows from the construction of $\phi$ (see the proof of 
Proposition $6.18$ in \cite{[PR]}, p. 339) that the groups 
$C_{G_0}(\phi(S_1))$ and 
$C_{G}(S_1)$ are isomorphic over $F^s$. In particular,
$$H=[C_{G_0}(\phi(S_1)),C_{G_0}(\phi(S_1))]$$
is a simply connected semisimple algebraic group over $F$.

The group $H$ is 
clearly standard. It is split over $F$, since any $F$-split subtorus in 
$G_0$, 
and in particular $\phi(S_1)$,
lies in a maximal $F$-split torus. Furthermore, from the exact sequence
$$
H^1(F,H)\lra  H^1(F,C_{G_0}(\phi(S_1))) \lra H^1(F,C_{G_0}(\phi(S_1))/H)=1
$$
we see that $\xi$ is equivalent to a cocycle with
coefficients in $H$.
\end{pf} 



\section{\it Groups splitting over a quadratic extension}\label{africa}




Throughout this section $F$ denotes a field of 
characteristic $\not= 2$, $K=F(\sqrt{d})$ its quadratic extension and 
$\tau$ the non-trivial automorphism of $K/F$.

Let $G$ be a simple simply 
connected algebraic group of rank $n$ defined over $F$ and splitting over $K$.
By Lemma 6.17 in~\cite{[PR]}, p. 329, there is a Borel 
%It is known that the intersection of two Borel subgroups of $G$ which are
%in general position is a maximal torus.
%Let $B\subset G$ be a Borel 
subgroup over $K$ such that $B\cap \tau(B)=T$ is
a maximal torus. We remark that this lemma in~\cite{[PR]} is proved 
under the condition 
that ${\rm char}(F)=0$. However the same proof with trivial modifications
works in the case of positive characteristic.
 
The torus $T$ is $F$-defined and splitting over $K$, 
since any $K$-torus in $B$ splits over $K$.
For our purposes it suffices to treat only the case
where $T$ is an $F$-anisotropic torus. 



\subsection{The structure of $G(K)$}\label{chevalley}





Let $\mathfrak{g}$ be the Lie algebra of $G$ and
let $\Sigma = \Sigma(T,G)$ be the root system of $G$ relative to $T$. 
The Borel subgroup $B$ determines an ordering of $\Sigma$, hence 
the system of simple roots $\Pi = \{\alpha_1,\ldots,\alpha_n \}$.
Let $\Sigma^+$ (resp. $\Sigma^-$) be the set of positive (resp. negative) 
roots and let $B^-$ be the Borel subgroup opposite to $B$ with respect to $T$.
We pick a Chevalley basis \cite{[St67]}
$$
\lbrace H_{\alpha_1},\ldots H_{\alpha_n},\ X_{\alpha},\ \alpha\in\Sigma\rbrace
$$  
in  $\mathfrak{g}$ corresponding to the pair $(T,B)$. 
This basis is unique up to signs 
and automorphisms of $\mathfrak{g}$ which preserve $B$ and $T$ (see 
\cite{[St67]}, \S 1, Remark $1$). 

Since $G$ is a Chevalley (in other words, quasi-split) group over $K$,  
its $K$-structure is well known. For  details and proofs
of all standard facts about $G(K)$ used in this section we refer  to 
\cite{[St67]}. In particular, we refer to \cite{[St67]} for the definition
of the operators $X_{\alpha}^n\left/n!\right.$ in the case
${\rm char}(F)=p > 0$ (see also~\cite{[SGA3]}).
Recall only that $G(K)$ is generated by the so-called root 
subgroups $U_{\alpha} = \langle x_{\alpha}(u)\mid u \in K\rangle$, 
where $\alpha\in\Sigma$ and
$$
x_{\alpha}(u)=\sum_{n=0}^{\infty}u^nX_{\alpha}^n\left/n!\right.\,,
$$ 
and $T$ is generated by the one-parameter subgroups 
   $$
     T_{\alpha} = T\cap G_{\alpha} = \langle\, h_{\alpha}(t)\mid t\in 
     K^*\,\rangle.
   $$ 
Here $G_{\alpha}$ is the subgroup generated by $U_{\pm\alpha}$ and  
  $$
    h_{\alpha}(t) = w_{\alpha}(t)\,w_{\alpha}(1)^{-1},\quad
    w_{\alpha}(t) = x_{\alpha}(t)\,x_{-\alpha}(-t^{-1})\,x_{\alpha}(t). 
  $$
Furthermore, since $G$ is a simply connected group, the following relations 
hold in $G$ (cf. \cite{[St67]}, Lemma 28 b), Lemma 20 c)\,):

\medskip
\noindent
(A) \ $T = T_{\alpha_1}\times\cdots\times T_{\alpha_n};$

\medskip
\noindent
(B) \ for any two roots \ $\alpha,\,\beta\in \Sigma\;$  we have
$$ 
h_{\alpha}(t)\,x_{\beta}(u)\,h_{\alpha}(t)^{-1} = 
x_{\beta}(t^{\langle\beta,\alpha\rangle}u)
$$
\ \ \ \ \ where $\langle\beta,\alpha\rangle\,= 2\,(\beta,\alpha)/(\alpha,
\alpha)$.

\medskip

If $\Delta\subset\Sigma^+$ is a subset, we let $G_{\Delta}$ denote the subgroup
generated by $U_{\pm\alpha},\,\alpha\in\Delta$. 



\subsection{Galois descent data}\label{basicnotation}



We shall now describe explicitly the $F$-structure of $G$, 
i.e. the action of $\tau$ on the generators $\{x_{\alpha}(u),\,\alpha\in
\Sigma\}$. 
\begin{lemma}\label{mole1}
$\tau(\alpha)=-\alpha$ for any $\alpha\in\Sigma.$
\end{lemma}
\begin{pf} The character $\alpha+\tau(\alpha)$ of $T$ is $F$-defined, hence
it is zero, since by our assumption $T$ is $F$-anisotropic.
\end{pf} 
    \begin{lemma}\label{mole2}
    $\tau(B)=B^-$.
    \end{lemma}
\begin{pf}
The statement follows immediately from Lemma~\ref{mole1}.
\end{pf}
    \begin{lemma}\label{mole20}
    $T_{\alpha}\simeq R_{K/F}^{(1)}(G_{m,K})$.
    \end{lemma}
\begin{pf}
$T_{\alpha}$ is a one-dimensional $F$-torus splitting over $K$. According to 
Lemma~\ref{mole1}, $\tau$ acts on its character lattice by multiplication by 
$-1$. So the result follows. \end{pf}


Let $\alpha\in\Sigma$. Since, by Lemma~\ref{mole1},
$\tau(\alpha)=-\alpha$, 
there exists a constant $c_{\alpha}\in K^{\times}$ such that 
$\tau(X_{\alpha})=c_{\alpha}X_{-\alpha}$. It follows that the action of 
$\tau$ on $G(K)$ is determined completely
by the family $\{c_{\alpha},\,\alpha\in\Sigma\}$.
We call these constants structure constants of $G$ with respect
to $T$. We summarize their properties in the following two lemmas.
\begin{lemma}\label{hen0}
Let $\alpha\in\Sigma$. Then we have

\smallskip

\noindent
{\rm (i)} $c_{-\alpha}=c_{\alpha}^{-1}$;

\smallskip

\noindent {\rm (ii)} $c_{\alpha}\in F^{\times}$;

\smallskip

\noindent {\rm (iii)} if $\beta\in\Sigma$ is a root such that 
$\alpha+\beta\in\Sigma$, then 
$c_{\alpha+\beta}=\pm c_{\alpha}\,c_{\beta}$; in 
particular, the family
$\{c_{\alpha},\,\alpha\in\Sigma\}$ is determined completely by its
subfamily $\{c_{\alpha_1},\ldots,c_{\alpha_n}\}$.
\end{lemma}
\begin{pf} 
(i) Apply $\tau$ to the equality $[X_{\alpha},X_{-\alpha}]=H_{\alpha}$
and use the fact that $\tau(H_{\alpha})=-H_{\alpha}$.

\smallskip

\noindent
(ii) One has $X_{\alpha}=\tau^2(X_{\alpha})=\tau(c_{\alpha})c_{-\alpha}
X_{\alpha}$, hence $\tau(c_{\alpha})c_{-\alpha}=1$. Substituting 
$c_{-\alpha}=c_{\alpha}^{-1}$ we obtain $\tau(c_{\alpha})=c_{\alpha}$,
as required.

\smallskip


\noindent
(iii) Apply $\tau$ to the equality $[X_{\alpha},X_{\beta}]=\pm (r+1)
X_{\alpha+\beta}.$ Here $r$ is an integer depending on $\alpha,\beta$
only.
\end{pf}
\begin{remark}
{\rm The number $r$ above is the biggest integer such that
\[
\beta,\,\beta-\alpha,\ldots,\beta-r\alpha
\]
are roots, but $\beta-(r+1)\alpha$ is not a root. Therefore,  this number
is equal to $0$ or $1$, since, by our assumption, $\alpha+\beta$ is a root. 
It follows, in particular, that $r+1\not=0$, since we assumed that
${\rm char}(F)\not=2$.}
\end{remark}
    \begin{lemma}\label{hen1}
    $\tau[\,x_{\alpha}(u)\,]=x_{-\alpha}(c_{\alpha}\tau(u))$ for any 
    $u\in K$ and $\alpha\in\Sigma$. 
    \end{lemma}
\begin{pf} This follows from the equality $x_{\alpha}(u)={\rm exp\,}
(uX_{\alpha})$. \end{pf}



\subsection{Comparison of different Galois descent data}\label{tool}


The family $\{c_{\alpha},\,\alpha\in\Sigma\}$ determining the action of $\tau$
on $G(K)$ depends on the chosen Borel subgroup $B$ and the corresponding
Chevalley basis. Given another Borel subgroup and Chevalley basis
we get another family of constants and we want now to describe the relation 
between the old ones and the new ones.
 
Let $B'\subset G$ be a Borel subgroup over $K$ such that
the intersection $T'=B'\cap\tau(B')$ is a maximal and $F$-anisotropic torus. 
Both  tori $T$ and $T'$ are isomorphic
over $F$ since, by Property (A) in \ref{chevalley} and Lemma \ref{mole20}, 
they are the direct products of $n$ copies of the torus
$R_{K/F}^{(1)}(G_{m,K})$. Furthermore, there exists
an $F$-isomorphism $T\to T'$ preserving
positive roots, i.e. which takes
$(\Sigma')^+=\Sigma(G,T')^+$ into $\Sigma^+=\Sigma(G,T)^+$. Any such 
isomorphism can be extended to an inner automorphism 
  $$
    i_g: G\lra G,\quad x\to g\,x\,g^{-1}
  $$ 
for some $g\in G(F^s)$, which takes $B$ into $B'$ (\,see \cite{[Hum]}, 
Theorem 32.1\,). 
    \begin{lemma}\label{lion0}
    $g$ can be chosen in $G(K)$.
    \end{lemma}
\begin{pf}
Since the restriction 
$i_g\, |_{T}$ is an $F$-defined isomorphism we easily get
that $t_{\sigma}=g^{-1+\sigma}\in T(F^s)$ for any $\sigma\in
{\rm Gal}\,(F^s/F)$. Consider the cocycle $\xi=(t_{\sigma})\in Z^1(F,T)$.
Since $T$ splits over $K$, 
$res_K(\xi)$ viewed as a cocycle in $T$ is trivial, by Hilbert's Theorem $90$. 
It follows that there is $z\in T(F^s)$ such that
$t_{\sigma}=z^{1-\sigma},\,\sigma\in{\rm Gal}\,(F^s/K)$.
Then the element $g'=gz$ is stable under ${\rm Gal}\,(F^s/K)$ and 
clearly $g'B(g')^{-1}=B'$.
\end{pf}



Let $g$ be  an element from Lemma~\ref{lion0} and let $t=g^{-1+\tau}$.
Since $t$ belongs to $T(K)$, it can be written (cf.~\ref{chevalley})  
as a product
$t=h_{\alpha_1}(t_1)\cdots h_{\alpha_n}(t_n)$,
where $t_1,\ldots,t_n\in K^{\times}$ are some parameters. Using the equality
$t\,\tau(t)=1$ and the fact that $\tau$ acts on characters of $T$
as multiplication by $-1$ one can easily see that $t_1,\ldots,t_n\in 
F^{\times}$. Consider the set 
    $$
      \{H_{\alpha_1}'=gH_{\alpha_1}g^{-1},\ldots ,H_{\alpha_n}'=gH_{\alpha_n}
      g^{-1},\,X'_{\alpha}=gX_{\alpha}g^{-1},\,\alpha\in\Sigma\}
    $$
which is a Chevalley basis related to the pair $(T',B')$.
Let $\{c'_{\alpha},\,\alpha\in \Sigma\}$ be the corresponding
structure constants.
   \begin{lemma}\label{lion1}
   For each $\alpha\in\Sigma$ one has  
   $c'_{\alpha}=t_1^{-\langle\alpha,\alpha_1\rangle}\cdots 
   t_n^{-\langle\alpha,\alpha_n\rangle}\cdot c_{\alpha}.$
   \end{lemma}
\begin{pf} Apply $\tau$
to the equality $X'_{\alpha}=gX_{\alpha}g^{-1}$ and use relation
(B) in \ref{chevalley}.
\end{pf}

\medskip

The converse is also true.
     \begin{lemma}\label{tiger1}
     Let $g\in G(K)$ be an element such that $t=g^{-1+\tau}\in T(K)$. Then 
     $T'=gTg^{-1}$ is an $F$-defined maximal torus splitting over $K$
     and the restriction of the inner automorphism $i_g$ to $T$ is an
     $F$-defined isomorphism.
     The structure constants $\{c'_{\alpha}\}$ related to
     $T'$ are given by the formulas in Lemma \ref{lion1}.
     \end{lemma}
\begin{pf}
This is clear.
\end{pf}
    \begin{remark}
     {\rm The elements $g\in G(K)$ such that $g^{-1+\tau}\in T(K)$
     correspond to the elements of the kernel of $\mu: H^1(K/F,T(K))\to 
     H^1(K/F,G(K))$. The exact sequence in Galois cohomology attached 
     to the exact sequence
    $$
      1\lra T(K)\lra G(K)\lra (G/T)(K)\lra 1
    $$
shows that they come from
$F$-points of the homogeneous variety $G/T$. It was shown in 
{\rm \cite{[Ch89]}}
that $G/T$ is an $F$-rational variety, hence all such elements can be
parametrized by points from an open subset
of an affine space.}
\end{remark}
\begin{remark}
{\rm The restriction of $i_g$ to $T$ induces a bijection between
$H^1(K/F,T(K))$ and  $H^1(K/F,T'(K))$. If we identify these two sets
(recall that $T$ and $T'$ are isomorphic over $F$), 
then this bijection is the translation
by the class of the cocycle $\xi=(t_{\tau})\in Z^1(K/F,T(K))$, where 
$t_{\tau}=g^{-1+\tau}.$}
\end{remark}






\subsection{Type $A_n$}\label{computations}







It follows from Lemma \ref{hen1} that, for each root $\alpha\in\Sigma$,
the subgroup $G_{\alpha}$ of  $G$ is defined over $F$.
Since it is a simple simply connected group of rank $1$, we obtain that 
$G_{\alpha} \simeq  SL(1,D_{\alpha})$, where $D_{\alpha}$ is
a quaternion algebra  over $F$. 
     \begin{lemma}\label{lion3}
     $D_{\alpha}\simeq (d,c_{\alpha})$.
     \end{lemma}
\begin{pf}  Since $T_{\alpha}= R_{K/F}^{(1)}(G_{m,K})
\subset SL(1,D_{\alpha})$, the algebra $D_{\alpha}$ splits over $K$. Hence 
it is of the form $D_{\alpha}=(d,f)$, where $f\in F^{\times}$,
and we need to prove that $$c_{\alpha}\equiv f\pmod{N_{K/F}(K^{\times})}.$$ 

Let $i,j\in D^{\times}$ be elements such that $i^2=d,\,j^2=f,\,ij=-ji.$
We may identify $T_{\alpha}$ with $R_{K/F}^{(1)}(F(i)^{\times})$.
A straightforward computation shows that the triple 
    $$
      H'_{\alpha}=\frac{1}{\sqrt{d}}\,i,\quad X'_{\alpha}=
      \frac{-\sqrt{d}+i}{-2f\sqrt{d}}\,j,\quad X'_{-\alpha}=
      \frac{\sqrt{d}+i}{2\sqrt{d}}\,j
    $$
belonging to $D\otimes_F K$ is a Chevalley 
basis with 
respect to $T_{\alpha}$. Note that under the standard identification 
$D\otimes_F K\simeq M_2(K)$ the triple $H'_{\alpha},\,X'_{\alpha},\,
X'_{-\alpha}$ corresponds to the matrices
\[
\left(
\begin{array}{cc}
1&0\\
0&-1
\end{array}\right),\quad
\left(
\begin{array}{cc}
0&1\\
0&0
\end{array}\right),\quad
\left(
\begin{array}{cc}
0&0\\
1&0
\end{array}\right).
\]
From the above formulas for $X_{\alpha}$ and $X_{\alpha}'$ we easily get
$\tau(X_{\alpha}')= fX'_{-\alpha}$. 

Let us now
come back to our former Chevalley basis $H_{\alpha},X_{\alpha},X_{-\alpha}$.
Recall that a Chevalley basis
is unique up to an automorphism of the corresponding Lie algebra
$sl_2(K)$. It follows that there is an inner automorphism $i_d$ of
$D_K$ which takes $H_{\alpha}',X_{\alpha}',
X_{-\alpha}'$ into $H_{\alpha},X_{\alpha},X_{-\alpha}$.
Since $H_{\alpha}$ and $H_{\alpha}'$ are proportional (being elements
of the Lie algebra of $T_{\alpha}$), $d$ lies in the Weyl group
of $T_{\alpha}$. Then $X_{\alpha}$ is proportional to either
$X'_{\alpha}$ or $X'_{-\alpha}$. Without loss of generality we may assume
that $X_{\alpha}$ is proportional to $X'_{\alpha}$. Then we have
$X_{\alpha}=zX_{\alpha}',\,
X_{-\alpha}=z^{-1}X_{-\alpha}'$
for some $z\in K^{\times}$ and hence
 $$c_{\alpha}X_{-\alpha}=\tau(X_{\alpha})=
\tau(z)\tau(X_{\alpha}')=\tau(z)fX_{-\alpha}'=z\tau(z)fX_{-\alpha},$$
implying $c_{\alpha}=z\tau(z)f$, as required.
\end{pf}

\medskip

In conclusion of this subsection we consider outer forms of type 
$^{2\!}A_{n-1}$ splitting over a quadratic extension. Let  
   $$
     h=a_1x_1x_1^{\tau}-a_2x_2x_2^{\tau}+\ldots + (-1)^{n-1}a_{n}x_nx_n^{\tau}
   $$ 
be a hermitian form of dimension $n$ relative to the quadratic extension
$K/F$, where $a_1,\ldots,a_{n}$ are elements in $F^{\times}$.
Let $T\subset {\rm SU}(h)$ be 
a maximal $F$-torus consisting of all diagonal matrices 
in ${\rm SU}(h)$. Clearly, $T$ is $F$-anisotropic and splits over
$K$. As above, we denote the corresponding structure constants related to
$T$ and the standard Chevalley basis of ${\rm SU}(h)_K\simeq {\rm SL}_{n,K}$ by 
$\{c_{\alpha},\,\alpha\in \Sigma=\Sigma(SU(h),T)\}.$
    \begin{lemma}\label{lion2} One has the following:
     $$
       c_{\alpha_1}\equiv a_1/a_2,\ c_{\alpha_2}\equiv a_2/a_3,\,\ldots,\
       c_{\alpha_{n-1}}\equiv a_{n-1}/a_{n}\pmod{N_{K/F}(K^{\times})}.
     $$
\end{lemma}
\begin{pf}
The subgroup $G_{\alpha_i}\subset {\rm SU}(h)$ coincides with 
${\rm SU}(h_i)\subset {\rm SU}(h)$, where
$$h_i=a_ix_ix_i^{\tau}-a_{i+1}x_{i+1}x_{i+1}^{\tau}.$$
Since ${\rm SU}\,(h_i)\simeq
{\rm SL}\,(1,D_{\alpha_i})$ and $D_{\alpha_i}
=(d,a_{\alpha_i}/a_{\alpha_{i+1}}),$
the result follows from Lemma~\ref{lion3}.
\end{pf}


\subsection{A cohomological property}


    \begin{proposition}\label{mole3}
     The natural map $H^1(K/F,T(K))\to H^1(K/F,G(K))$ is surjective.
    \end{proposition}
\begin{pf}
We are under the conditions of 
Lemma 6.28 in \cite{[PR]}, p.~369, so the result follows from that lemma.
\end{pf}





\section{\it Groups $E_6,E_7$ splitting over cubic extensions.}





Throughout this section $F$ denotes a field of 
characteristic $\not= 3$, $K/F$ a cyclic extension of degree $3$ and 
$\tau$ a non-trivial automorphism of $K/F$. We also assume that
$\Gamma= Gal(F^s/F)$ is a pro-$3$-group. We let $G_0$ 
denote a simple simply connected split group over $F$ 
of type $E_6,E_7$. 
\begin{theorem}\label{eagle1}
Any cocycle $\xi\in Z^1(K/F,G_0)$ is equivalent to a cocycle
with coefficients in a standard simple simply connected $F$-subgroup $H< G_0$
 of inner type $A_2$.
\end{theorem} 
\begin{pf}
We split our proof into a sequence of simple observations.
The main ingredients are the classification of Tits 
indices
\cite{[T66]}, Steinberg theorem~\cite{[St65]} and the fact
that $\Gamma$ is a pro-$3$-group; in particular there are no simple 
$F$-anisotropic groups of types $A_1,A_3,B_n,C_n,\,^{1,2}D_n,G_2$. 
We let $G=\,^{\xi}G_0$ denote the corresponding twisted group.
Consider first the key case where $G$ is an $F$-anisotropic group of 
type $E_6$.

%\subsubsection{Construction of a simple subgroup of type $D_4$}



\subsection{Type $E_6$: anisotropic case}
\subsubsection{Construction of a special torus}
    \begin{proposition}\label{tor}
    $G$ contains a maximal $F$-defined torus $S$ splitting over $K$.
    \end{proposition}
\begin{pf} We follow Harder's arguments \cite{[H65],[H66]} (cf. 
\cite{[PR]}, Chapter $6$). In this part $G$ is not necessary a strongly
inner form.
Let us start with the construction of a simple 
simply connected $F$-subgroup of $G$ of type
$^3D_4$ splitting over $K$. Since $G$ splits over $K$, we choose 
a $K$-split maximal torus $T$.
Let $\Sigma=\Sigma(G,T)$ and let $\Pi=
\{\alpha_1,\ldots,\alpha_6\}\subset\Sigma$ be a basis.
Let ${\cal{P}}\subset G$ be the standard parabolic subgroup over 
$K$ corresponding to $\alpha_6$. 

Consider the connected component
$C$ of the intersection ${\cal{P}}\cap \tau({\cal{P}})\cap 
\tau^2({\cal{P}})$. 
It is defined over $F$, hence reductive,
since $G$ is $F$-anisotropic. 
Denote the central torus of $C$ by $S_1$ and let
$C^{(1)}=[C,C]=C_1\cdot...\cdot C_s$  be the decomposition of the semisimple 
part of $C$ into an almost direct product of the simple components over $F^s$.
It easily follows from ${\rm dim}\,C\ge 30$ that $C^{(1)}$ contains
a simple component, say $C_1$, of type not $A_n$ (see Lemma~$6.32$ 
in \cite{[PR]}, p. 380).
       \begin{lemma} $C_1$ is defined over $F$.\label{eagle2}
       \end{lemma}
\begin{pf}
If $C_1$ is defined over $L/F$, then there is a canonical embedding 
$H_1=R_{L/F}(C_1)\hookrightarrow C$, since $C$ is 
$F$-defined. Consider a Levi subgroup $\cal{L}\subset P$ over $K$ 
containing $C$. The parabolic subgroup $\cal{P}$ corresponds to $\alpha_6$, 
so the semisimple part $[{\cal{L,L}}]$ of $\cal{L}$
has type $D_5$. It follows then from the inclusion $[C,C]\subset 
[{\cal{L,L}}]$ that 
    $$
      {\rm rank}\, H_1= [L:F]\cdot {\rm rank}\, C_1\le 5.
    $$ 
But $[L:F]$ is a power of $3$ and $C_1$ is not of type $A_n$, hence $[L:F]=1$.
\end{pf}
    \begin{lemma} $C_1$ has type $^{3}D_4$.
    \end{lemma}
\begin{pf} The conditions $C_1$ is $F$-anisotropic and ${\rm rank}\,C_1\le 5$ 
imply that $C_1$ is of type either $F_4$ or $D_4$. A group of type $F_4$ 
being a group of dimension $52$ can not be embedded  into the group 
$[{\cal{L},{L}}]$ of type $D_5$, since ${\rm dim}\,[{\cal{L},{L}}]=45$.
So the result follows. 
\end{pf}

%\subsubsection{Description of the structure of $C$}

     \begin{lemma} $C^{(1)}=C_1$.
     \end{lemma} 
\begin{pf} If $C^{(1)}$ contains another simple component
$C_2$, then its rank is $1$, since $C^{(1)}$ has rank at most $5$.
It follows that
$C_2$ is defined over $F$ and hence $F$-isotropic being a group of type
$A_1$ --- a contradiction. 
\end{pf}
    \begin{lemma}\label{tor1} $S_1\simeq R_{K/F}^{(1)}(G_{m,K})$.
    \end{lemma}
\begin{pf} We have ${\rm dim}\,C\ge 30$. On the other hand, $C^{(1)}=[C,C]$ is 
a group of dimension $28$ and rank $4$. Hence ${\rm dim}\,S_1=2$. 
According to the classification of 
$2$-dimensional tori \cite{[V]} it suffices now to show that $S_1$
is $K$-split or, what is the same, $K$-isotropic, since $\Gamma$ is a 
pro-$3$-group. 

Let $\cal{L}$ be a Levi subgroup of $\cal{P}$ over $K$ containing
$C$. Both groups $\cal{L}$ and $C$ are reductive and have
the same rank $6$. It follows that the central torus $A$ of $\cal{L}$
is contained in $C$. Since it commutes with $[\cal{L},\cal{L}]$
which contains $[C,C]=C_1$, we conclude that $A\subset S_1$.
It remains to note that $A$ is $K$-split.
\end{pf}
     \begin{lemma}\label{tor2} $C_1$ 
     contains a maximal torus $S_2$ isomorphic to $R_{K/F}^{(1)}(G_{m,K})
     \times  R_{K/F}^{(1)}(G_{m,K})$.
     \end{lemma}  
\begin{pf} Consider the following chain of inclusions:
$$C=S_1\cdot C_1\subset C_G(S_1)\subset C_G(A)= \cal{L}.$$
It gives
\[
[C,C]\subset [C_G(S_1),C_G(S_1)]\subset [\cal{L},\cal{L}].
\]
Since $[C,C]$ and $[\cal{L},\cal{L}]$ have type $D_4$ and $D_5$ respectively
and $[C_G(S_1),C_G(S_1)]$ is a group of rank $4$, we conclude that
$[C_G(S_1),C_G(S_1)]=[C,C]=C_1$. 

Since $S_1$ splits over $K$, so are
$C_G(S_1)$ and $[C_G(S_1),C_G(S_1)]=C_1$. Thus 
$C_1$ is an $F$-group
of type $D_4$ splitting over $K$. Then arguing just as in \cite{[PR]}, 
p. 371--372, we obtain that $C_1$ contains a torus of the required form.
\end{pf}

\medskip

\noindent
Lemmas~\ref{tor1} and \ref{tor2} show that $S=S_1\cdot S_2$ is an 
$F$-defined maximal torus of $G$ splitting over $K$. 
Proposition~\ref{tor} is proved. 
\end{pf}

\medskip

Let $S$ be the torus constructed in Proposition~\ref{tor}.
According to Theorem~\ref{steinberg1} there exists an $F$-embedding $S
\hookrightarrow G_0$  such that
    $$
    [\xi]\in {\rm Im}\,[H^1(F,S)\lra H^1(F,G_0)].
    $$
We fix this embedding and starting from this point we may
forget about the way it was constructed. 
The only fact we need to know is
that $S$ is an almost direct product of three copies of $R_{K/F}^{(1)}
(G_{m,K})$. Based on this fact we are going to show that there exists a 
decomposition
     \begin{equation}\label{dec}
      S\simeq R_{K/F}^{(1)}(G_{m,K})\times R_{K/F}^{(1)}(G_{m,K})\times 
      R_{K/F}^{(1)}(G_{m,K})
     \end{equation}
and a new $F$-embedding 
$\eta: S\hookrightarrow G_0$ with the following properties: 

\medskip

\noindent
(a) {\it the images of the first
two components in decomposition~(\ref{dec})
lie in an $F$-split standard subgroup of $G_0$ isomorphic to 
$SL_3\times SL_3$,}

\smallskip
 
\noindent
(b) {\it $\xi$ is equivalent to a cocycle with coefficients in $\eta(S)$.} 




\subsubsection{The Weyl group $W_{E_6}$}



Let $\Sigma=\Sigma(G_0,S)$ be the root system of $G_0$ with respect to 
$S$, $W=W(\Sigma)$ the Weyl group of $\Sigma$ and $\Pi=\{\alpha_1,\ldots,
\alpha_6\}$ a fixed basis of $\Sigma$. 
    \begin{lemma}\label{sylow} Let $W_3\subset W$ be a $3$-Sylow subgroup.
    Then $$W_3\simeq (\mathbb{Z}/3\times \mathbb{Z}/3\times \mathbb{Z}/3)
    \,\rtimes\,\mathbb{Z}/3,$$ where 
    $\mathbb{Z}/3$ acts by permuting the components of $\mathbb{Z}/3\times 
     \mathbb{Z}/3\times \mathbb{Z}/3$.
    \end{lemma}  
\begin{pf} We have $\#\,(W_3)=3^4.$ Let $\Sigma_1,\Sigma_2$ and $\Sigma_3$
be the three subroot systems of $\Sigma$ of type
$A_2$ generated by roots
$\langle\alpha_1,\alpha_3\rangle,\,\langle\alpha_5,\alpha_6\rangle\,$
and $\langle\alpha_2,-\beta\rangle$ respectively, where 
$\beta\in \Sigma$ is the positive root of maximal length
with respect to the basis $\Pi$.
Let $w_0,w_1\in W$ 
be the elements of maximal length with respect to the basis
$\{\alpha_1,\ldots,
\alpha_6\}$ and $\{\alpha_1,\alpha_3,\alpha_4,\alpha_2,-\beta,\alpha_5\}$
respectively.

It is easy to see that $w=w_0\cdot w_1$ has order $3$ and takes the roots
$\alpha_1,\alpha_3,\alpha_5,\alpha_6,\alpha_2,-\beta$ into
$\alpha_6,\alpha_5,\alpha_2,-\beta,\alpha_3,\alpha_1$ respectively.
Therefore $w$ permutes the components of the subsystem $\Sigma'=\Sigma_1
\times\Sigma_2\times\Sigma_3\subset\Sigma$ and their Weyl groups.
Let $v_1\in W(\Sigma_1)$ be an arbitrary element  of order $3$. Then 
$v_2=wv_1w^{-1}\in W(\Sigma_2)$ and  $v_3=wv_2w^{-1}\in W(\Sigma_3)$. 
It follows that 
$v_1,v_2,v_3$ commutes and $w$ permutes them;
in particular, $v_1,v_2,v_3,w$ generate a subgroup $W_3\subset W$ of order
$3^4$. 
\end{pf}  

\medskip

Since $S$ is an $F$-defined maximal torus splitting over $K$, the
group ${\rm Gal}\,(K/F)$ acts on the root system $\Sigma$. Thus we have
a natural embedding ${\rm Gal}\,(K/F)\hookrightarrow W$. Choosing an 
appropriate 
basis of $\Sigma$ we may assume that the image of ${\rm Gal}\,(K/F)$
lies in the $3$-Sylow subgroup $W_3$ constructed in Lemma~\ref{sylow}.
    \begin{lemma}\label{action} 
     The image of ${\rm Gal}\,(K/F)$ lies in a subgroup 
     $\langle v_1,v_2,v_3\rangle$ of $W_3$.
     \end{lemma}
\begin{pf} Let 
$\tau=v_1^{i_1}v_2^{i_2}v_3^{i_4}w^{i_4}$, where $i_1,
\ldots, i_4$ are integers.
Since $S$ is $F$-anisotropic, we have $\alpha_1+\tau(\alpha_1)+
\tau^2(\alpha_1)=0$. On the other hand, assuming that $i_4\not=0$
we easily obtain that $\tau(\alpha_1)\in \Sigma_2$ and $\tau^2(\alpha_1)
\in \Sigma_3$.
But then $\alpha_1+\tau(\alpha_1)+\tau^2(\alpha_1)\not=0$ --- a contradiction.
\end{pf}
  


\subsubsection{Special embedding $S\hookrightarrow G_0$}\label{special}



Lemma~\ref{action} shows that $\Sigma_1,\Sigma_2,\Sigma_3$ are stable
under the action of
${\rm Gal}\,(K/F)$, hence the subgroup $G_{\Sigma_1}\cdot G_{\Sigma_2}\cdot
G_{\Sigma_3}\subset G_0$ is $F$-defined and of type $A_2\times A_2
\times A_2$. Note also that
the three intersections $S\cap G_{\Sigma_i}$ are $2$-dimensional tori
isomorphic to $R_{K/F}^{(1)}(G_{m,K})$. This follows from the facts
that they are $F$-anisotropic, splitting over $K$ and from the classification
of $2$-dimensional tori. Clearly, the tori $S\cap G_{\Sigma_i}$, $i=1,2,3,$ 
generate $S$. 

We are now ready to
 construct a new embedding $S\hookrightarrow G_0$ with the required 
properties (a) and (b). Let $T\subset G_0$ be
a maximal $F$-split torus and let $\{\beta_1,
\ldots,\beta_6\}$ be a basis of the root
system $\Sigma(G_0,T)$. We consider the subsystems
$\Sigma_1',\Sigma_2',\Sigma_3'$ generated by $\{\beta_1,
\beta_3\},\,\{\beta_5,\beta_6\},\,\{\beta_2,-\beta'\}$ respectively.
Here $\beta'$ is the root of maximal length with respect to the
basis $\{\beta_1,\ldots,\beta_6\}$.    
Since $G_{\Sigma_1'},G_{\Sigma_2'}$ and $G_{\Sigma_3'}$ are isomorphic to 
$SL_3$ over $F$, there are $F$-embeddings $\phi_i:R_{K/F}^{(1)}(G_{m,K})
\hookrightarrow G_{\Sigma_i'},\,i=1,2,3$. Then the torus 
$$S'= {\rm Im}\,\phi_1\cdot {\rm Im}\,\phi_2\cdot {\rm Im}\,\phi_3 $$ 
is maximal in $G_0$ isomorphic to $S$ over $F$.
      \begin{proposition}\label{canon}
      $[\xi]\in {\rm Im}\,[H^1(F,S')\lra H^1(F,G_0)].$
      \end{proposition}
\begin{pf} It follows from the constructions of $S$ and $S'$
that there is an $F$-defined isomorphism $S\to S'$ which takes
$\Sigma(G_0,S)$ into $\Sigma(G_0,S')$. Any such 
isomorphism can be extended to an inner automorphism $i_g:G_0\to G_0$ over
$F^s$, by Theorem $32.1$ in \cite{[Hum]}. Since the restriction $i_g|_{S}$ is 
$F$-defined, we obtain that  
$\sigma(g)^{-1}\cdot g\in S$
for any $\sigma\in {\rm Gal}\,(F^s/F)$. Let $\xi=(a_{\sigma})$, 
where $a_{\sigma}\in 
S$. Then the equivalent cocycle $$\xi'=(g\cdot a_{\sigma}\cdot 
\sigma(g)^{-1})=
(g\cdot a_{\sigma}\cdot (\sigma(g)^{-1}g)\cdot g^{-1})$$ 
takes values in $S'$ and we are done. 
\end{pf}


\subsubsection{The direct product decomposition}

In view of Proposition \ref{canon} the torus $S'$ satisfies
condition (b). Let us prove that it satisfies condition (a)
as well. We do not need the torus $S$ any more and to ease notation we 
will denote $S'$ by $S$.
We also denote $\Sigma(G_0,S')$ by $\Sigma$ and let $\Pi=\{\alpha_1,\ldots,
\alpha_6\}$ be the corresponding basis of $\Sigma$.
% be the corresponding root system with the 
%basis $\Pi=\{\alpha_1,\ldots,\alpha_6\}$. 



For any root $\alpha\in\Sigma$ 
we have $\alpha+\tau(\alpha)+\tau^2(\alpha)=0$, since $S$ is $F$-anisotropic.
It follows that the set $\{\pm\alpha,\pm\tau(\alpha)\}$ generates an 
$F$-defined subsystem of $\Sigma$ of
rank at most $2$. This subsystem has the automorphism $\tau$,
which is of order $3$,
hence it has type $A_2$. In particular, the subgroup $G_{\{\alpha,
\tau(\alpha)\}}$ is an $F$-defined inner form of type $A_2$
splitting over $K$.
For $\alpha\in\Sigma$ we let 
   $$
     S_{\{\alpha,\tau(\alpha)\}}=S\cap G_{\{\alpha,
     \tau(\alpha)\}}.
    $$ 
Since  $S_{\{\alpha,\tau(\alpha)\}}$ splits over $K$, we
obtain that $S_{\{\alpha,\tau(\alpha)\}}\simeq R_{K/F}^{(1)}(G_{m,K})$.


Recall also that, by construction, the
three subsystems of $\Sigma$ generated by the roots $\{\alpha_1,
\alpha_3\},$ $\{\alpha_5,\alpha_6\}$ and $\{\alpha_2,-\beta\}$ are stable 
under the action
of ${\rm Gal}\,(K/F)$ and have type $A_2$. Hence 
\[
S_{\{\alpha_1,\tau(\alpha_1)\}}=S_{\{\alpha_1,\alpha_3\}}, \quad
S_{\{\alpha_5,\tau(\alpha_5)\}}=S_{\{\alpha_5,\alpha_6\}}.
\]
    \begin{proposition}\label{splitting} The product morphism
          $S_{\{\alpha_1,\alpha_3\}}\times S_{\{\alpha_5,\alpha_6\}}\times 
           S_{\{\alpha_4,\tau(\alpha_4)\}}\lra S$ is 
           an $F$-isomorphism. 
    \end{proposition}
\begin{pf} It suffices to show that it
is an isomorphism
over $F^s$. We use notation from 
Steinberg's book \cite{[St67]}. Recall that
given a root $\alpha\in\Sigma$, one can associate to it the one-parameter
subgroup $S_{\alpha}=\langle\, h_{\alpha}(t)\mid t\in 
{\overline F}^{\times}\,\rangle \leq S$.
By~\ref{chevalley} (A), $S$ is the direct product of such subgroups 
corresponding to $\alpha_1,\ldots,\alpha_6$.
If $\alpha=m_1\alpha_1+\cdots+m_6\alpha_6$, then 
   \begin{equation}\label{raz}
    h_{\alpha}(t)= h_{\alpha_1}(t)^{m_1}\cdots h_{\alpha_6}(t)^{m_6},
   \end{equation} 
since all roots have the same length.

It is clear that $\tau(\alpha_4)\not=\pm\beta $, 
since otherwise in view of the fact that the root subsystem 
$\Sigma_3=\langle\,\alpha_2,-\beta\,
\rangle$ is $F$-defined we would have $\alpha_4\in \Sigma_3$. Therefore 
$\tau(\alpha_4)$ is of the form
$$\tau(\alpha_4)=\pm(m_1\alpha_1+
m_3\alpha_3+\cdots+m_6\alpha_6+\alpha_2),$$
where $m_1,m_3,\ldots,m_6$ are positive integers. 
Then it follows from (\ref{raz}) that 
\[
S=S_{\alpha_1}\times S_{\alpha_3}\times\cdots\times S_{\alpha_6}
\times S_{\tau(\alpha_4)}=S_{\{\alpha_1,\alpha_3\}}\times
S_{\{\alpha_5,\alpha_6\}}\times S_{\{\alpha_4,\tau(\alpha_4)\}},
\]
as required.
\end{pf} 



\subsubsection{Concluding argument}


\begin{em} Proof of Theorem \ref{eagle1} in anisotropic case. \end{em}
%   \begin{proposition} There is a simple simply connected $F$-defined
%     subgroup $H\leq G_0$ of inner type $A_2$ such that 
%     $$[\xi]\in Im\,[H^1(F,H)\lra H^1(F,G_0)].$$
%    \end{proposition}
By Proposition~\ref{canon}
we may assume that $\xi\in Z^1(F,S)$. Let $\xi=(a_{\sigma})$,
where $a_{\sigma}\in S$.
Applying Proposition~\ref{splitting}
we split $\xi$ into the product of three cocycles
$\xi=\xi_1\cdot\xi_2\cdot\xi_3$
 with coefficients
in $S_{\{\alpha_1,\alpha_3\}},S_{\{\alpha_5,\alpha_6\}}$
and $S_{\{\alpha_4,\tau(\alpha_4)\}}$ respectively. 
Since $G_{\Sigma_1}$
and $G_{\Sigma_2}$ are isomorphic to $SL_3$, $\xi_1$ and
$\xi_2$ are trivial cocycles. 

Let $g_1\in G_{\Sigma_1}$ and $g_2\in G_{\Sigma_2}$ be
such that $\xi_1=(g_1^{-1+\sigma})$ and $\xi_2=(g_2^{-1+\sigma})$.
Clearly, $g_1$ and $g_2$ commutes. It follows that
the equivalent cocycle $$\xi'=((g_1g_2)\,a_{\sigma}\,
(g_1g_2)^{-\sigma})=(g_1g_2)^{\sigma}\,[(g_2^{-\sigma}g_2)
(g_1^{-\sigma}g_1)a_{\sigma}]\,(g_1g_2)^{-\sigma}$$ has 
coefficients in 
$$(g_1g_2)^{\sigma}\, S_{\{\alpha_4,\tau(\alpha_4)\}}\,(g_1g_2)^{-\sigma}
\leq (g_1g_2)^{\sigma}\, G_{\{\alpha_4,\tau(\alpha_4)\}}\,
(g_1g_2)^{-\sigma}.$$
The condition $(g_1g_2)^{-1}\,(g_1g_2)^{\sigma}\in S$
easily implies that 
   \begin{equation}\label{cel}
    H=(g_1g_2)^{\sigma}\, G_{\{\alpha_4,\tau(\alpha_4)\}}\,(g_1g_2)^{-\sigma}
   \end{equation}  
is an $F$-defined subgroup of type $A_2$ splitting over $K$, hence we are 
done. 
\begin{remark}\label{formula}
{\rm Let $S\hookrightarrow G_0$ be the embedding constructed in~\ref{special}.
Proposition~\ref{canon} says that the canonical map
$H^1(K/F,S)\to H^1(K/F,G_0)$ is surjective. By Proposition~\ref{splitting},
any cocycle $\xi\in Z^1(K/F,G_0)$ can be written as a triple
\[
\xi=(a_1N_{K/F}(K^{\times}),a_2N_{K/F}(K^{\times}),a_3N_{K/F}(K^{\times})),
\]
where $a_1,a_2,a_3\in F^{\times}$.
One can show that the above subgroup~(\ref{cel}) is isomorphic
to $SL(1,T)$, where $T$ is a cubic central simple algebra
of the form $T=(K/F,a_1^{n_1}a_2^{n_2}c_{\alpha_4})$ for certain
integers $n_1,n_2$ and a constant $c_{\alpha_4}\in F^{\times}$.
These integers and the constant can be computed explicitly, but we omit details.
As a result, we obtain the following formula for the Rost invariant:
\[
R_{G_0}([\xi])= T\cup (a_3).
\]}
\end{remark} 


 
 
\subsubsection{Moving Lemma}\label{move}






The above argument will be used below several times. Let us formulate it 
for the future reference as follows.

\medskip

\noindent
{\bf Moving Lemma} {\it Let $G$ be a simple group over $F$ and let $T$
be a maximal torus of $G$ over $F$. Let $\xi\in Z^1(F,T)$ be a cocycle.
Assume that $\xi=\xi_1\cdot\xi_2$ is the product of two cocycles
with coefficients in $T$ such that $\xi_1$ takes values in $T\cap H$, where
$H$ is a proper $F$-subgroup of $G$
normalizing by $T$, and $\xi_2$ viewed as a cocycle with 
coefficients in $G$
is trivial. Then $\xi$ is equivalent to a cocycle with coefficients
in a proper $F$-subgroup of $G$ of the same type as $H$.} 
\begin{pf}
Let $\xi_1=(a_{\sigma}),\,\xi_2=(b_{\sigma})$, where $\sigma\in 
{\rm Gal}\,(F^s/F),\,a_{\sigma}\in T(F^s)\cap H(F^s)$ and
$b_{\sigma}\in T(F^s)$. Since
$\xi_2$ is trivial, there is $g\in G(F^s)$ such that $b_{\sigma}=g^{-1+\sigma}$
for all $\sigma\in {\rm Gal}\,(F^s/F)$.
Then the equivalent cocycle $\xi'=(c_{\sigma})$, where
\[
c_{\sigma}= g(a_{\sigma}b_{\sigma})g^{-\sigma}= ga_{\sigma}g^{-1},
\]
takes values in a proper $F$-subgroup $gHg^{-1}$ which has the same type
as $H$.
\end{pf}




\subsection{Type $E_6$: isotropic case}





This case is much easier. 
\begin{lemma}
Let the twisted group $G=\,^{\xi}G_0$ be $F$-isotropic. 
Then it splits over $F$; in particular $\xi$ is trivial. 
\end{lemma}
\begin{pf}
Assume the 
contrary. According to~\cite{[T66]}
all admissible Tits' indices of type $^1E_6$ corresponding to
isotropic groups are as follows:

\begin{equation}\label{p}
\begin{array}{ll}
\mbox{(i)}
\qquad 
\begin{picture}(85,30)
\put(00,02){\line(1,0){80}}
\put(40,02){\line(0,1){20}}
\put(00,02){\circle*{3}}
\put(20,02){\circle*{3}}
\put(40,02){\circle*{3}}
\put(60,02){\circle*{3}}
\put(80,02){\circle*{3}}
\put(40,22){\circle*{3}}
\put(00,02){\circle{10}}
\put(80,02){\circle{10}}
\put(-5,-11){$\alpha_1$}
\put(15,-11){$\alpha_3$}
\put(35,-11){$\alpha_4$}
\put(55,-11){$\alpha_5$}
\put(75,-11){$\alpha_6$}
\put(35,30){$\alpha_2$}
\end{picture}
&
\qquad
\mbox{(ii)}
\qquad 
\begin{picture}(80,30)
\put(00,02){\line(1,0){80}}
\put(40,02){\line(0,1){20}}
\put(00,02){\circle*{3}}
\put(20,02){\circle*{3}}
\put(40,02){\circle*{3}}
\put(60,02){\circle*{3}}
\put(80,02){\circle*{3}}
\put(40,22){\circle*{3}}
%\put(00,02){\circle{10}}
%\put(20,02){\circle{10}}
%\put(60,02){\circle{10}}
%\put(80,02){\circle{10}}
\put(40,02){\circle{10}}
\put(40,22){\circle{10}}
\put(-5,-11){$\alpha_1$}
\put(15,-11){$\alpha_3$}
\put(35,-11){$\alpha_4$}
\put(55,-11){$\alpha_5$}
\put(75,-11){$\alpha_6$}
\put(35,30){$\alpha_2$}
\end{picture}
\end{array}
\end{equation}

\vspace{10mm}

In case (i) the semisimple anisotropic kernel of $G$ is a classical group
of type $D_4$. Since $\Gamma$ is a pro-3-group, any such a group is $F$-split.

Consider case (ii). Applying Theorem~\ref{steinberg2}
we obtain that $\xi$ is equivalent to a cocycle with coefficients
in $SL_3\times SL_3$. Hence $\xi$ is trivial and $G$ is $F$-split. 
\end{pf}


 
\subsection{Type $E_7$}



     \begin{proposition}\label{funny1} There exists no $F$-anisotropic 
       groups of type $E_7$
       splitting over $K$.
     \end{proposition}
\begin{pf} Assume the contrary. Let $G$ be an $F$-anisotropic simple group 
of type $E_7$ splitting over $K$.
To get a contradiction we can proceed as 
in case $E_6$. Namely,
let ${\cal{P}}\subset G$ be the parabolic subgroup over 
$K$ corresponding to the root $\alpha_7$. Then we consider the connected 
component $C$ of the intersection ${\cal{P}}\cap \tau({\cal{P}})\cap 
\tau^2({\cal{P}})$. 
As in case $E_6$, it is easy to conclude that $C$ is a reductive group of 
dimension at least $52$ whose semisimple part $[C,C]$ contains an
$F$-defined simple component $C_1$ of type not $A_n$. 


Let $\cal{L}\subset \cal{P}$ be a Levi subgroup containing $C_1$.
Its semisimple part has type $E_6$, hence 
$C_1$ is a group of type either $^3D_4, F_4,$ or  $E_6$.
By a dimension
argument, $C_1$ cannot be of type $^3D_4$. If $C_1$ is of type $E_6$,
then $C_1$ is the semisimple part of $\cal{L}$.
It follows that $C_G(C_1)$ is defined over $F$. But it coincides with the 
central torus $A$ of $\cal{L}$ whose dimension is $1$. Hence $A$ is $F$-defined
and $F$-isotropic being a torus of dimension $1$.


Assume now that $C_1$ is a group of type $F_4$ and let us consider 
$C_G(C_1)$. Since it contains $A$, $C_G(C_1)$
is a nontrivial reductive group over $F$ whose rank
is either $1,2$ or $3$. Let $S\subset
C_G(C_1)$ be a maximal torus over $F$. It commutes with $C_1$, by 
construction, hence $C_1\subset [C_G(S),C_G(S)]$. 

If ${\rm dim}\,S=1$, then $S$ is $F$-isotropic.
If ${\rm dim}\, S=3$, then $C_1=[C_G(S),C_G(S)]$ would be a standard subgroup of 
$G$ of type $F_4$. But a root system of type $E_7$ does not contain a subsystem
of type $F_4$.

Thus ${\rm dim}\,S=2$
and $C_G(S)$ is a reductive group whose semisimple part $[C_G(S),C_G(S)]$
has rank $5$ and contains the subgroup $C_1$ of type $F_4$. It follows that
$[C_G(S),C_G(S)]$
 contains another simple component over $F$ of rank $1$. But any such
a group is $F$-isotropic. 
\end{pf}

We are now ready to complete the proof of Theorem~\ref{eagle1} in the
case of $E_7$
by reducing it to $E_6$. By Proposition~\ref{funny1}, $G$ is 
$F$-isotropic. Looking at Tits tables~\cite{[T66]} and taking into 
consideration the fact that
$\Gamma$ is a pro-$3$- group we obtain that the only possibility
for the semisimple $F$-anisotropic
kernel of $G$ is to be a group of type $E_6$.  Applying Theorem~\ref{steinberg2}
we get that
$\xi$ is equivalent to a cocycle with coefficients in an $F$-split
standard subgroup of $G_0$ of type $E_6$. So the result follows.
\end{pf}





\section{{\it The triviality of the kernel of $R_G$ for 
$E_6,E_7$}}\label{rost}







\begin{theorem}\label{rrost} Let $G_0$ be a quasi-split simple simply 
connectedgroup of type $E_6,E_7$
defined over a field $F$ of characteristic $\not=2,3$. Then ${\rm Ker}\,R_{G_0}=1.$
\end{theorem}

We split the proof into three parts considering inner forms of type $E_6$,
outer forms of type $E_6$ and type $E_7$ separately. 
Let $G_0$ be the given quasi-split group and let 
$[\xi]$ be an element of ${\rm Ker}\,R_{G_0}$.
Let $G=\,^{\xi}G_0$ be the corresponding twisted group. The triviality of $\xi$
is clearly equivalent to saying that $G$
is quasi-split over $F$.

\subsection{Inner forms of type $E_6$}\label{inner}

Our strategy is to reduce to the case where $G$ is an $F$-isotropic group
and then to apply Theorem~\ref{steinberg2} and Proposition~\ref{number}. 

\subsubsection{Reduction to a pro-$p$-case}\label{p-case} 

It is entirely based on the following
theorem which is due to P. Gille~\cite{[G93]}, \cite{[G97]}. 
\begin{theorem}\label{gille}
Let $G$ be a simple simply connected group of type $E_6$ or $E_7$ over $F$.
Assume that there exist finite field extensions $E_1,\ldots,
E_s$ of $F$ splitting $G$ and such that the 
$$ g.c.d.\{\,[E_1:F],\ldots,[E_s:F]\,\}=1.$$ Then $G$ splits over $F$.
\end{theorem}

Let $\Gamma_p\subset \Gamma$ be a Sylow $p$-subgroup and let
$F_p=(F^s)^{\Gamma_p}$ be
fixed points of $\Gamma_p$. Assuming that Theorem~\ref{rrost}
is proved over $F_p$ we can find for each $p$ a finite extension
$E_p/F$  contained in $F_p$ and splitting $G$. Since the $g.c.d$
of all degrees $[E_p: F]$ is clearly $1$, we obtain from
Gille's theorem that $G$ splits over $F$.

\subsubsection{Reduction to the case $p=2,3$}



By Theorem~\ref{steinberg1}, we may assume
that $\xi$ takes values in an $F$-defined maximal torus $T\subset G_0$.
   \begin{lemma}\label{poryadok}
    Any element of the group $H^1(F,T)$ has order of the form $2^n3^m$.
   \end{lemma}
\begin{pf} See Proposition~$6.21$ in \cite{[PR]}, p. 375. 
\end{pf}

Lemma~\ref{poryadok} implies that in the case $p\not=2,3$ any
cocycle from $Z^1(F,T)$ is trivial. 

\subsubsection{The case $p=3$}


Theorem~\ref{eagle1} says that $\xi$ is equivalent
to a cocycle with coefficients in a standard simple simply connected
$F$-subgroup of $G_0$ of type $A_2$. Then, by Theorem~\ref{ms} and 
Proposition~\ref{number}, $\xi$ is trivial. 

\subsubsection{The case $p=2$}

Assume first that $G$ be $F$-isotropic. According to~\cite{[T66]}, we have
two possibilities for an $F$-anisotropic kernel of $G$:

\smallskip

\noindent
(i) an inner form of type $A_2\times A_2$. But
$\Gamma$ is a pro-$2$-group, hence there are no anisotropic inner forms
of type $A_2$ over $F$;
\smallskip

\noindent
(ii) an inner form of type $D_4$. In this case, by 
Theorem~\ref{steinberg2}, $\xi$ is equivalent to a cocycle with coefficients 
in a standard simple simply connected $F$-split subgroup of $G_0$ of type
$D_4$, hence it is trivial, by Propositions~\ref{d5} and ~\ref{number}.


\smallskip

Let $G$ be now $F$-anisotropic. Let $E/F$ be a finite Galois extension which 
splits $G$ and is minimal with this property.  
Since $\Gamma$ is a pro-$2$-group, there is a chain of 
subfields 
$$F=E_0\subset E_1\subset \ldots \subset E_s=E$$
such that $[E_{i+1} : E_i ] = 2$ and $s\ge 1$.               
Since $G$ is not split over $E_{s - 1}$ and the $F$-isotropic case
has been already studied, $G$ is anisotropic over that 
field.  Thus it suffices to treat the case where $G$ splits 
over a quadratic extension $K/F$. 
Then $G$ has a maximal
torus $T$ defined over $F$ and splitting over $K$ (see \S~\ref{africa}). 
Since $G$ is a group of inner type,
${\rm Gal}\,(K/F)$ has the natural embedding into the Weyl group. We also know
that the image of the nontrivial element of ${\rm Gal}\,(K/F)$
acts on the corresponding
root system $\Sigma(G,T)$ by multiplication by $-1$. But the Weyl group
of type $E_6$ does not contain $-1$: a contradiction.



\subsection{Outer forms of type $E_6$}

Let $G_0$ be a quasi-split group of type $E_6$ and let $K=F(\sqrt{d})$
be the quadratic extension over which $G_0$ becomes split.
Since the inner case has been already treated, we have ${\rm Res_K}([\xi])=1$,
hence $G$ splits over $K$.

\subsubsection{Isotropic case}

\begin{proposition}\label{gul10}
If $G$ is $F$-isotropic, then $[\xi]=1.$
\end{proposition}
\begin{pf}
According to Tits' classification 
\cite{[T66]} the $F$-index of $G$ is one of the following:
$$
%\begin{equation}
\begin{array}{lll}
\mbox{(a)}
\quad 
\begin{picture}(80,20)
\put(00,02){\line(1,0){25}}
\put(65,02){\oval(80,14)[l]}
\put(0,02){\circle*{3}}
\put(25,02){\circle*{3}}
\put(65,-5){\circle*{3}}
\put(65,9){\circle*{3}}
\put(45,9){\circle*{3}}
\put(45,-5){\circle*{3}}
\put(0,02){\circle{10}}
\put(25,02){\circle{10}}
\put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$}
\put(40,15){$\alpha_3$}
\put(62,15){$\alpha_1$}
\put(62,-16){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}       
&
\mbox{(b)}
\quad 
\begin{picture}(80,20)
\put(00,02){\line(1,0){25}}
\put(65,02){\oval(80,14)[l]}
\put(0,02){\circle*{3}}
\put(25,02){\circle*{3}}
\put(65,-5){\circle*{3}}
\put(65,9){\circle*{3}}
\put(45,9){\circle*{3}}
\put(45,-5){\circle*{3}}
\put(65,02){\oval(10,25)}
\put(0,02){\circle{10}}
\put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$}
\put(40,14){$\alpha_3$}
\put(62,20){$\alpha_1$}
\put(62,-19){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}
&
\mbox{(c)}
\quad 
\begin{picture}(80,20)
\put(00,02){\line(1,0){25}}
\put(65,02){\oval(80,14)[l]}
\put(0,02){\circle*{3}}
\put(25,02){\circle*{3}}
\put(65,-5){\circle*{3}}
\put(65,9){\circle*{3}}
\put(45,9){\circle*{3}}
\put(45,-5){\circle*{3}}
\put(0,02){\circle{10}}
\put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$}
\put(40,14){$\alpha_3$}
\put(62,14){$\alpha_1$}
\put(62,-16){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}\\
\bigskip
\mbox{(d)}
\quad 
\begin{picture}(80,50)
\put(00,02){\line(1,0){25}}
\put(65,02){\oval(80,14)[l]}
\put(0,02){\circle*{3}}
\put(25,02){\circle*{3}}
\put(65,-5){\circle*{3}}
\put(65,9){\circle*{3}}
\put(45,9){\circle*{3}}
\put(45,-5){\circle*{3}}
\put(65,02){\oval(10,25)}
\put(-5,14){$\alpha_2$}
\put(20,14){$\alpha_4$}
\put(40,14){$\alpha_3$}
\put(62,20){$\alpha_1$}
\put(62,-19){$\alpha_6$}
\put(40,-16){$\alpha_5$}
\end{picture}\\
\end{array}
%\end{equation}
$$
\bigskip

\begin{lemma}\label{gull2}
The Tits index of $G$ is not of the form {\rm (a),\,(b) or (c)}. 
\end{lemma}
\begin{pf}
Let $T\subset G$ be a maximal $F$-torus containing a maximal $F$-split
torus of $G$. We fix a basis $\Pi=\{\alpha_1,\ldots,\alpha_6\}$ of
$\Sigma(G,T)$ corresponding to the above pictures. According
to~\cite{[T66]} the maximal $F$-split subtorus of $T$ is the identity 
component of 
the subgroup of $T$ defined by the following system of equations:
\[
\left\{
\begin{array}{lcll}
\gamma(t) & = & \sigma^*(\gamma)(t)\qquad &\mbox{ for all $\gamma\in \Pi$ 
and all $\sigma \in Gal(F^s/F)$,}\\
\alpha_i(t) & = & 1 & \mbox{ for all 
non-distinguished vertices $\alpha_i,$}\\ 
\end{array}
\right.  
\]

\noindent
where $t \in T$ and $\sigma^*$ denotes the $*$-action of 
$\Gamma$ on $\Pi$ (see~\cite{[T66]}).
It follows from these equations and the above pictures that in all
cases (a),\,(b) and (c) the maximal 
$F$-split subtorus of  $T$ contains the $F$-split 
one-dimensional subtorus $S_1=\langle h_{\beta}(u)\mid u\in (F^s)^{\times}
\rangle$ corresponding to the root $\beta$ of maximal length. 
Then the semisimple part of the centralizer $C_G(S_1)$ is $F$-defined and
coincides with the subgroup $G_{\Sigma_1}\subset G$
generated by roots $\Sigma_1=\{\alpha_1,\alpha_3,\ldots,\alpha_6\}$.
Let $S_2\subset G_{\Sigma_1}$ be an arbitrary maximal
$F$-defined torus and let $S=S_1\cdot S_2$.

Arguing as in Theorem~\ref{steinberg2} it is easy to see that there is an 
$F$-embedding $S\hookrightarrow G_0$
such that $[\xi]\in {\rm Im}\,[H^1(F,S)\to H^1(F,G_0)]$
and the image of $S_2$ lies in a standard quasi-split simple simply connected
$F$-subgroup $H$ of type $^{2\!}A_5$.
Since $S/S_2$ is an $F$-split torus, any cocycle in $Z^1(F,S)$
is equivalent to a cocycle with coefficients in $S_2$. Thus
$\xi$ is equivalent to a cocycle with coefficients in $H$,
hence $\xi$ is trivial, by Propositions~\ref{a5} and \ref{number}.
\end{pf}
     \begin{lemma}\label{gull3}
      A group with the Tits index {\rm (d)} can not split over a 
      quadratic extension. In particular,
      the Tits index of $G$ is not of the form {\rm (d)}. 
      \end{lemma}
\begin{pf} Let $H$ be an $F$-isotropic group with the Tits index of the form
(d). If $H$ splits over a quadratic extension $K/F$, then so is
its semisimple $F$-anisotropic kernel $L$.  
In case (d) $L$ is an almost simple simply connected $F$-anisotropic group of 
type $^2D_4$. Since $L$ splits over a quadratic extension of the ground
field, there are
two possibilities.

\medskip

Case I. $L\simeq {\rm Spin}\,(f)$, where $f$ is an $F$-anisotropic quadratic
form of dimension $8$. Since $L$ has type $^2D_4$,
the discriminant of $f$ is nontrivial. On the other hand, $L$ is 
$F$-anisotropic and splits over $K$, hence $f$ is of the form
$$
f\simeq a_1(x_1^2-dx_2^2)+a_2(x_1^3-dx_4^2)+ a_3(x_5^2-dx_6^2)+
a_4(x_7^2-dx_8^2),
$$
where $a_1,\ldots,a_4\in F^{\times}$. It follows that $f$ has trivial 
discriminant --- a contradiction.

\medskip

Case II. $L\simeq {\rm Spin}\,(D,f)$, where $D$ is a quaternion
algebra over $F$ and $f$ is a skew-hermitian form over $D$ (with
respect to the standard involution on $D$). Let $E$ be the function field
of the Severi-Brauer variety corresponding to $D$. By a
result of Parimala, Sridharan and Suresh~\cite{[PSS]}, $L$ is still 
anisotropic over $E$ and clearly has type $^2D_4$ over $E$. 
Then we can proceed to Case I to get a contradiction.
\end{pf}

Lemmas~\ref{gull2} and \ref{gull3} complete the proof of 
Proposition~\ref{gul10}. 
\end{pf}



\subsubsection{Anisotropic case}

By Theorem \ref{steinberg1}, there is a maximal $F$-anisotropic
torus $S\subset G_0$ splitting over $K$ and such that 
   \begin{equation}\label{obraz}
    [\xi] \in {\rm Im}\,[H^1(F,S)\to H^1(F,G_0)].
   \end{equation} 
Then according to Proposition~\ref{mole3} for any maximal $F$-anisotropic
torus of $G_0$ splitting over $K$, condition (\ref{obraz}) still holds.
We keep the notation introduced in Section~\ref{africa}. 
In particular, given a Chevalley basis of the Lie 
algebra of $G_0$ with respect to $S$ we denote the corresponding
structure constants by $\{c_{\alpha},\,\alpha\in\Sigma=\Sigma(G_0,S)\}$.
\begin{lemma}\label{snake3} 
There exists a maximal $F$-anisotropic torus $S\subset G_0$ splitting over $K$
and such that the structure constants are of the form
$$
       c_{\alpha_1}\equiv 1,\ c_{\alpha_2}\equiv 1,\,\ldots,\
       c_{\alpha_6}\equiv 1\pmod{N_{K/F}(K^{\times})}.
     $$
\end{lemma}
\begin{pf}  
Let $T\subset G_0$ be a centralizer of a maximal $F$-split
torus in $G_0$. The Tits $F$-index of $G_0$ is of the form
\begin{eqnarray}\label{diagram1}
\begin{picture}(80,25)
\put(00,02){\line(1,0){25}}
\put(65,02){\oval(80,14)[l]}
\put(0,02){\circle*{3}}
\put(25,02){\circle*{3}}
\put(65,-5){\circle*{3}}
\put(65,9){\circle*{3}}
\put(45,9){\circle*{3}}
\put(45,-5){\circle*{3}}
\put(65,02){\oval(10,25)}
\put(0,02){\circle{10}}
\put(25,02){\circle{10}}
\put(45,02){\oval(10,25)}
\put(-5,14){$\alpha_2$}
\put(20,20){$\alpha_4$}
\put(40,20){$\alpha_3$}
\put(62,20){$\alpha_1$}
\put(62,-19){$\alpha_6$}
\put(40,-19){$\alpha_5$}
\end{picture}
\end{eqnarray}



\vspace{8mm}

\noindent
Here $\{\alpha_1,\ldots,\alpha_6\}$ is a basis
of $\Sigma(G_0,T)$. 
Let $\beta\in\Sigma(G_0,T)$ be the root of maximal length with respect
to the basis $\{\alpha_1,\ldots,\alpha_6\}$. It follows from (\ref{diagram1})
that the subgroup $G_{\beta}\subset G_0$
corresponding to  $\beta$
is an $F$-split group of rank $1$, hence $G_{\beta}\simeq {\rm SL}_2$.
Let $S_1\subset G_{\beta}$ be a maximal $F$-anisotropic torus of dimension
$1$ which is isomorphic to $R^{(1)}_{K/F}(G_{m,K})$.

The subgroup $H=G_{\Sigma_1}\subset G_0$ corresponding
to the set $\Sigma_1=\{\alpha_1,\alpha_3,\ldots\alpha_6\}$ is a quasi-split 
$F$-group of type $^{2\!}A_5$, hence it is $F$-isomorphic to ${\rm SU}\,(h)$,
where 
\begin{equation}\label{equation1}
    h=x_1x_1^{\tau}-x_2x_2^{\tau}+x_3x_3^{\tau}-x_4x_4^{\tau}+
    x_5x_5^{\tau}-x_6x_6^{\tau}.
\end{equation}
We choose the $F$-anisotropic torus 
$S_2\subset H=SU(h)$ consisting of all diagonal isometries of $h$ 
and we let 
$S=S_1\cdot S_2$. Since $G_{\beta}$ commutes with $H$,
$S$ is a maximal $F$-anisotropic torus in $G_0$ 
splitting over $K$. 


Let $\Sigma$ be the root system of $G_0$ with respect to $S$.
We may now forget about $T$ and we work with the root system
$\Sigma$ related to $S$. In order to ease notation we denote a 
basis in
$\Sigma$ by the same letters $\{\alpha_1,\ldots,\alpha_6\}$. 
We choose a Chevalley basis (with respect to $S$) of the Lie algebra of 
$G_0$
and a basis $\{\alpha_1,\ldots,\alpha_6\}$ of $\Sigma$ in such a way that
its subset $\{\alpha_1,\alpha_3,\ldots,\alpha_6\}$ is a basis of the root 
system of ${\rm SU}\,(h)$ with respect to $S_2$.
Then by Lemma~\ref{lion2},  
     $$
       c_{\alpha_1}\equiv 1,\ c_{\alpha_3}\equiv 1,\,\ldots,\
       c_{\alpha_6}\equiv 1\pmod{N_{K/F}(K^{\times})}.
     $$
We now want to modify $S$ (if necessary) in 
such a way that $c_{\alpha_2}\equiv 1\pmod{N_{K/F}(K^{\times})}$ as well. 

Assume that $c_{\alpha_2}\not\equiv 1$. Consider the cocycle
$\zeta=(b_{\tau})\in 
Z^1(K/F,S_1(K))$, where $b_{\tau}=h_{\beta}(c_{\alpha_2})$ and
 $\beta$ is the root of $\Sigma$ of maximal length with respect to the 
ordering determined by the basis $\{\alpha_1,\ldots,\alpha_6\}$.
Since $G_{\beta}\simeq {\rm SL}_2$, 
$\zeta$ viewed
as a cocycle with coefficients in $G_{\beta}$ is trivial.

Let $b_{\tau}=g^{-1+\tau},$ where $g\in G_{\beta}(K)$. 
We claim that
the torus $S'=gSg^{-1}$ is as required. Indeed, by Lemma~\ref{tiger1},
$S'$ is defined over $F$ and splits over $K$. Since
$g$ commutes with
 $H$, Lemma~\ref{lion1} shows that the constants 
$c_{\alpha_1},c_{\alpha_3},\ldots,c_{\alpha_6}$ related to the tori $S$ and 
$S'$ coincide, hence are as required. 
Since $\langle{\beta},\alpha_2\rangle=1,$ 
Lemma~\ref{lion1} again shows that the constant  
$c_{\alpha_2}$ related to $S'$ equals $1$.
\end{pf}
%\begin{remark}\label{snake4}
%Conjugation $S$ by $g$ does not change the subtorus $S_2$ 
%and the subgroup $H$. So for the corresponding new basis $\{\alpha_1,\ldots,
%\alpha_6\}$ of the root system $\Sigma(S',G_0)$ the subgroup of $G_0$
%generated by $\alpha_1,\alpha_3,\ldots,\alpha_6$ still coincides with $H$,
%hence isomorphic to $SU(h),$ where $h$ is the hermitian form given 
%by~{\rm (\ref{equation1})}.
%\end{remark} 


Applying the Moving Lemma~\ref{move} we want next to modify our cocycle $\xi$, 
i.e. take an equivalent one,
in such a way that the new one has coefficients in a simple simply connected
$F$-group of type $D_5$. 

To do so, let $S$ be the torus constructed
in Lemma \ref{snake3}. Let $L\subset G_0$ 
be a subgroup generated by the simple roots $\alpha_2,\ldots,\alpha_6$
from the root system $\Sigma(G_0,S)$. Clearly, $L$ is an $F$-defined group
of type $D_5$ splitting over $K$. 
Since $S$ splits over $K$, we have $H^1(F,S)=H^1(K/F,S(K))$. Hence
$\xi$ can be written in the form $\xi=(s_{\tau})$, where
$$
s_{\tau}=h_{\alpha_1}(u_1)\cdots h_{\alpha_6}(u_6),
$$
and $\tau$ is the nontrivial automorphism of $K/F$. It easily follows
from $s_{\tau}\tau(s_{\tau})=1$ that
$u_1,\ldots,u_6\in F^{\times}$.

Since $c_{\alpha_1}\equiv 1$, we have $G_{\alpha_1} \simeq SL_2$ over $F$, 
by Lemma~\ref{lion3}. Hence there is
$g\in G_{\alpha_1}(K)$ such that $g^{-\tau+1}=h_{\alpha_1}(u_1)$.
Consider the equivalent cocycle $\tilde{\xi}=(\widetilde{s_{\tau}})$, where
$$\widetilde{s_{\tau}}=g^{\tau} h_{\alpha_1}(u_1)\cdots h_{\alpha_6}(u_6) 
g^{-1}=
g\,h_{\alpha_2}(u_2)\cdots h_{\alpha_6}(u_6)\,g^{-1}.
$$
It has coefficients in ${\widetilde L}=g\,L\,g^{-1}$.

The group $\widetilde{L}$ is a simple
simply connected group of type $D_5$ defined over $F$. It contains
the maximal torus $\widetilde{S_2}=g\,S_2\,g^{-1}$ splitting over $K$.
In order to stress that
$\widetilde{L}$ sits inside $G_0$, we number simple roots in 
$\Sigma(\widetilde{L},\widetilde{S_2})$ as follows.
$$
\begin{picture}(95,40)
\put(00,02){\line(1,0){90}}
\put(30,02){\line(0,1){30}}
\put(00,02){\circle*{3}}
\put(30,02){\circle*{3}}
\put(60,02){\circle*{3}}
\put(90,02){\circle*{3}}
%\put(80,02){\circle*{3}}
\put(30,32){\circle*{3}}
%\put(00,02){\circle{10}}
%\put(80,02){\circle{10}}
%\put(-5,-10){$\alpha_1$}
\put(-5,-10){$\alpha_3$}
\put(25,-10){$\alpha_4$}
\put(55,-10){$\alpha_5$}
\put(85,-10){$\alpha_6$}
\put(25,40){$\alpha_2$}
\end{picture}
$$

\bigskip


\begin{lemma}\label{mod}
The structure constants of $\widetilde{L}$ with respect to the torus
$\widetilde{S_2}$ are as follows:
$$
c_{\alpha_2}\equiv c_{\alpha_4} \equiv
c_{\alpha_5}\equiv c_{\alpha_6}\equiv 1 \pmod{N_{K/F}(K^{\times})},\quad
c_{\alpha_3}\equiv u_1 \pmod{N_{K/F}(K^{\times})};
$$
in particular, $\widetilde{L}$ has $F$-rank at least $2$.
\end{lemma}
\begin{pf} The first statement follows from Lemma~\ref{lion1},
the second one from the fact that the subgroups 
$G_{\alpha_2}$ and $G_{\alpha_5}$ of $\widetilde{L}$ corresponding to 
$\alpha_2$ and $\alpha_5$ split over $F$, by Lemma~\ref{lion3},
and commutes.
\end{pf}

Since $\widetilde{L}$ splits over the quadratic extension
$K/F$, there are two possibilities for $\widetilde{L}$.
   \begin{lemma} Assume that
    $\widetilde{L}\simeq {\rm Spin}\,(f)$, where $f$ is a quadratic
    form over $F$ of dimension $10$. Then $\widetilde{L}$ is quasi-split.
   \end{lemma}
\begin{pf}
We have to exclude the 
possibilities for the Witt index $i$ of $f$ to be $i=2,3$.

\smallskip
 
Assume that $i=2$. Then $f$ is the direct sum of two hyperbolic planes
and a certain $F$-anisotropic quadratic form of dimension $6$. It follows
that the dimension of a maximal $F$-split
torus in $\widetilde{L}$ equals $2$ and the semisimple part of its
centralizer has type $A_3=D_3$. 

On the other hand, all maximal $F$-split tori in
$\widetilde{L}$ are conjugate over $F$. Since $G_{\alpha_2}$
and $G_{\alpha_5}$ are split over $F$, they contain $F$-split tori
of dimension $1$, say $T_2$ and $T_5$. Then simple calculations
with roots show that the
semisimple part of the centralizer $C_{\widetilde{L}}(T_2\cdot T_5)$
has type $A_1\times A_1$ --- a contradiction.

\smallskip

Assume that $i=3$. Let $f_{an}$ be the anisotropic part of $f$.
It has dimension $4$ and splits over $K$, in particular
$f_{an}$ has trivial discriminant. It follows that the function field
$E$ of the projective quadric $f_{an}=0$
splits $f$. Hence $G_0$ has rank at least $5$
over $E$ which is impossible, since $E$ does not contain $K$.
\end{pf}

\medskip

Since $\widetilde{L}$ is quasi-split and $R_{G_0}([\xi])=
R_{\widetilde{L}}([\tilde\xi])=1$,
Proposition~\ref{d5} gives $[\tilde\xi]=1$. 


\medskip

It remains to consider the case $\widetilde{L}\simeq {\rm Spin}\,(D,f)$, 
where $D$ is a quaternion algebra over $F$ and $f$ is a skew-hermitian form 
over $D$ (with respect to the standard involution on $D$). Let $E$
be the function field of the Severi-Brauer variety of $D$. The
extension $E/F$ splits $D$ and so we can
reduce to the previous case if we show that the twisted
group $G=\,^{\xi}G_0$ is still anisotropic over $E$.
\begin{lemma}
$G$ is anisotropic over $E$.
\end{lemma}
\begin{pf}
Assume the contrary, i.e. $G$ is $E$-isotropic. Then 
$G$ is quasi-split over $E$,
since the isotropic case has been already treated.
Let $U$ be the unipotent radical
of a Borel subgroup $B\subset G$ over $E$. 
The twisted group $^{\tilde{\xi}}(\widetilde{L})$ being $F$-anisotropic  is 
still anisotropic over $E$, by \cite{[PSS]}.
Hence the intersection $U\cap\, 
^{\tilde{\xi}}({\widetilde L})$
is trivial. On the other hand, we have 
${\rm dim}\,^{\tilde{\xi}}({\widetilde L})=45$,
${\rm dim}\,U=36$, so
${\rm dim}\,U\cap\,^{\tilde{\xi}}({\widetilde L})$ is at least $3$. 
\end{pf}


\subsubsection{A formula for the Rost invariant}\label{cat1}

We keep the notation from the previous subsection. Using
the above material one can easily  produce a formula
for the Rost invariant for any cocycle $\xi\in Z^1(K/F,G_0(K))$. Namely,
let $S$ be the torus from Lemma~\ref{snake3}. By Proposition~\ref{mole3},
we have the surjection $$H^1(K/F,S(K))\to H^1(K/F,G_0(K)).$$ Hence
we may assume that $\xi$ has coefficients in $S$.

Let $\xi=(a_{\tau})$ and let $a_{\tau}=h_{\alpha_1}(t_1)
\cdots h_{\alpha_6}(t_6)$, where $t_1,\ldots,t_6\in F^{\times}$. 
We write $\xi$ as the product $\xi=\xi_1\cdot\xi_2$,
where 
$$\xi_1=(h_{\alpha_1}(t_1)h_{\alpha_4}(t_4)h_{\alpha_6}(t_6)),\quad
\xi_2=(h_{\alpha_3}(t_3)h_{\alpha_2}(t_2)h_{\alpha_5}(t_5)).
$$
Since, by Lemma~\ref{lion3}, the subgroup of $G_0$ generated by
the roots $\alpha_2,\alpha_3,\alpha_5$ splits over $F$, $\xi_2$
is a trivial cocycle. Then, by the Moving Lemma~\ref{move}, $\xi$ 
is equivalent
to a cocycle with coefficients in an $F$-subgroup $H\subset G_0$
of type $A_1\times A_1\times A_1$. 

According to Lemma~\ref{lion1} the three components of $H$ correspond
to the quaternion algebras $D_1=(d,t_3),\,D_4=(d,t_2t_3t_5),
\,D_6=(d,t_5)$. Then, by Proposition~\ref{number} and from the formula
for the Rost invariant for groups of inner type $A_n$,
we obtain that the cocycle
$R_{G_0}(\xi)$ is given by
$$
 R_{G_0}(\xi)=(d)\cup (t_3)\cup (t_1) + (d)\cup (t_2t_3t_5)\cup (t_4)
+(d)\cup (t_5)\cup (t_6).
$$









\subsection{Type $E_7$}








This case is entirely similar to case $E_6$. By Gille's Theorem~\ref{gille}
and Lemma~\ref{poryadok}, we may assume that ${\rm Gal}\,(F^s/F)$
is a pro-$p$-group, where $p=2,3$. If $p=3$, then in view of
Theorem~\ref{eagle1} $\xi$ is equivalent to a cocycle with coefficients
in a subgroup of $G_0$ of type $A_2$. Therefore
$\xi$ is trivial, by Theorem~\ref{ms}.

Let now $p=2$. Then we may assume that $G$ is split over a
quadratic extension $K/F$. It is easy to see that $G$, and hence
$G_0$, contains a maximal $F$-anisotropic torus $S$ splitting
over $K$. By Theorem~\ref{steinberg1}, there is an embedding $S\hookrightarrow
G_0$ such that $\xi$ is equivalent to a cocycle with coefficients
in the image of $S$. Arguing as in $E_6$ we may assume that all structure
constants of $S$ corresponding to all simple roots are $1$ modulo norms 
$N_{K/F}(K^{\times})$.
After that we can modify $\xi$ (as in Lemma~\ref{mod})
in such a way that the new cocycle
lies in a simple simply connected $F$-subgroup $H$ of $G_0$ of type
$E_6$ with the following structure constants:
$$
c_{\alpha_1}\equiv c_{\alpha_2} \equiv
c_{\alpha_3}\equiv c_{\alpha_4}\equiv c_{\alpha_5}\equiv 
1 \pmod{N_{K/F}(K^{\times})},\quad
c_{\alpha_6}\equiv u_1 \pmod{N_{K/F}(K^{\times})}.
$$
It follows  that $F$-rank of $H$ is at least $3$.
Looking at the tables~\cite{[T66]}, we see that $H$ is either quasi-split
or split over $F$. Hence the result follows from Proposition~\ref{number}, 
since $H$ is standard.



\subsection{Type $^{3,6}D_4$}





This case is known (see \cite{[KMRT]}). However in order to get 
a self-contained proof of Serre's Conjecture II 
for groups of exceptional types over $(C_2)$--fields (except $E_8$)
let us show how this case follows from the above.

Let $G_0$ be a simple simply connected quasi-split trialitarian group over 
$F$. Denote
the minimal extension of $F$ over which $G_0$ becomes split by $L$
and  let $\Gamma ={\rm Gal}\,(L/F)$. If $G_0$ is a group 
of type $^3D_4$, then $[L:F]=3$ and $\Gamma\simeq\mathbb{Z}/3$; otherwise we 
have 
$[L:F]=6$ and $\Gamma\simeq S_3$. We start with the following 
    \begin{proposition}\label{dd4}
      Let $\xi\in Z^1(F,G_0)$ be a cocycle such that the twisted group
      $G=\,^{\xi}G_0$ is quasi-split over a quadratic extension $K/F$.
      Assume that $K$ and $L$ are linearly disjoint over $F$. Then
      $\xi$ is equivalent to a cocycle with coefficients in a standard
      simple simply connected $F$-subgroup of $G_0$ of type $A_1$.
    \end{proposition}
\begin{pf}
Let $\sigma\in\Gamma$ be an arbitrary element of order $3$. We fix a 
subfield $P\subseteq L$ 
of degree $3$ over $F$. If $[L:F]=3$, the field $P$ 
coincides with $L$. If $[L:F]=6,$ let $\lambda$ be the non-trivial 
automorphism $L/P$. We have clearly $\lambda\sigma\lambda^{-1}=\sigma^2$. 
Let $\tau$ be the nontrivial automorphism of $K/F$.
Since $K$ and $L$ are linearly disjoint over $F$, we
can extend the automorphisms 
$$
\sigma,\,\lambda\in {\rm Gal}\,(L/F),\quad\tau\in {\rm Gal}\,(K/F)
$$ 
in a natural way to the  
extension $E/F$, where $E=K\!\cdot\! L$. For simplicity we denote 
these extensions by the same letters $\sigma,\,\lambda,\,\tau$.

Let $B\subset G$ be a Borel subgroup over $K$ such that $S=B\cap\tau(B)$
is a maximal torus. $S$ is split over $E$.
Let $\Sigma =\Sigma(G,S)$ and let 
$\Pi=\{\alpha_1,\ldots,\alpha_4\}\subset\Sigma$ be the basis corresponding 
to the Borel subgroup $B$. 
The action of $\Gamma$ on root subgroups was described
in~\cite{[ChT]} and is given by the formulas  
\begin{equation}\label{e38}
\begin{array}{rcl}
\vphantom{\left[\prod\limits_A^B\right]}
\sigma [x_{\pm\alpha_1}(u)]=x_{\pm\alpha_3}[\sigma (u)],&\qquad & 
\sigma [x_{\pm\alpha_3}(u)]=x_{\pm\alpha_4}[\sigma (u)],\\
\vphantom{\left[\prod\limits_A^B\right]}
\sigma [x_{\pm\alpha_4}(u)]=x_{\pm\alpha_1}[\sigma (u)],& \qquad & 
\sigma [x_{\pm\alpha_2}(u)]=x_{\pm\alpha_2}[\sigma (u)],\\
\vphantom{\left[\prod\limits_A^B\right]}
\lambda [x_{\pm\alpha_1}(u)]=x_{\pm\alpha_1}[\lambda (u)],& \qquad & 
\lambda [x_{\pm\alpha_2}(u)]=x_{\pm\alpha_2}[\lambda (u)],\\
\vphantom{\left[\prod\limits_A^B\right]}
\lambda [x_{\pm\alpha_3}(u)]=x_{\pm\alpha_4}[\lambda (u)], & \qquad &
\lambda [x_{\pm\alpha_4}(u)]=x_{\pm\alpha_3}[\lambda (u)],\\
\vphantom{\left[\prod\limits_A^B\right]}
\tau([x_{\alpha_i}(u)]=x_{-\alpha_i}(\tau(u)).&\qquad  &  
\end{array}
\end{equation}
Formulas (\ref{e38}) show that $S\simeq S_1\times S_2$, where 
  $$
     S_1=S\cap G_{\alpha_2}=\langle\, h_{\alpha_2}(u)\,\rangle\simeq
     R_{K/F}^{(1)}(G_{m,K}),
  $$ 
   $$
     S_2=S\cap G_{\{\alpha_1,\alpha_3,\alpha_4\}}=\langle\, h_{\alpha_1}(u_1)
     h_{\alpha_3}(u_3)h_{\alpha_4}(u_4)\,\rangle\simeq
     R_{P/F}\,[R^{(1)}_{P\cdot K/P}(G_{m,P\cdot K})].
  $$

\smallskip

The rest of the proof is the same as in Section~\ref{special}.
Namely, we construct the following embedding $\psi:S\hookrightarrow G_0$.
Let $T\subset G_0$ be the centralizer of a maximal $F$-split torus
of $G_0$. The Tits index of $G_0$ over $F$
is of the form:
\begin{equation}\label{e33}
\begin{picture}(40,25)
\put(00,00){\line(1,0){40}}
\put(40,0){\oval(80,30)[l]}
\put(0,0){\circle*{3}}
\put(40,0){\circle*{3}}
\put(40,15){\circle*{3}}
\put(40,-15){\circle*{3}}
\put(00,0){\circle{10}}
\put(40,0){\oval(15,40)}
\put(-20,0){$\alpha_2$}
\put(50,0){$\alpha_3$}
\put(50,15){$\alpha_1$}
\put(50,-15){$\alpha_4$}
\end{picture}       
\end{equation}

\vspace{10mm}

\noindent
This picture shows that the subgroup $H_1\subset G_0$ (respectively $H_2$) 
generated by 
$\alpha_2$ (respectively $\alpha_1,\alpha_3,\alpha_4$)
from the root system $\Sigma(G_0,T)$ is isomorphic
to $SL_{2,F}$ (respectively $R_{P/F}(SL_{2,P})$). Hence there are
natural embeddings $S_1\hookrightarrow H_1,\,S_2\hookrightarrow H_2$
which can be extended to $\psi:S\hookrightarrow G_0$.
Arguing as in Proposition~\ref{canon} we obtain that $\xi$
is equivalent to a cocycle with coefficients in $\psi(S)$.
Then it can be written as a product of two cocycles $\xi=\xi_1\cdot\xi_2$
with coefficients in $\psi(S_1)$ and $\psi(S_2)$ respectively. Since $\xi_2$
has coefficients in $H_2$ which has trivial Galois cohomology, 
the Moving Lemma~\ref{move} completes the proof.
\end{pf} 
     
    \begin{theorem}\label{trialitarian}
    If $\xi\in Z^1(F,G_0)$ is a cocycle such that $R_{G_0}([\xi])=1$,
      then $\xi$ is trivial.
    \end{theorem}
\begin{pf}
The triviality of $\xi$ is equivalent to saying that the twisted group
$G=\,^{\xi}G_0$ is quasi-split over $F$.
As usual we distinguish two cases: isotropic and anisotropic.

Let first $G$ be $F$-isotropic. We claim that then $G$ is quasi-split over $F$.
Assume the contrary. Then the $F$-index of $G$ is of the form 
\begin{equation}\label{d4}
\begin{picture}(40,15)
\put(00,00){\line(1,0){30}}
\put(30,0){\oval(60,20)[l]}
\put(0,0){\circle*{3}}
\put(30,0){\circle*{3}}
\put(30,10){\circle*{3}}
\put(30,-10){\circle*{3}}
\put(00,0){\circle{10}}
\end{picture}
\end{equation}

\vspace{4mm}

\noindent
Let $S_1$ be a maximal $F$-split torus in $G$.
It follows from the picture that 
the semisimple part of the centralizer $C_G(S_1)$ 
coincides with the subgroup $G_{\Sigma_2}\subset G$
generated by the non-distinguished roots $\Sigma_2=
\{\alpha_1,\alpha_3,\alpha_4\}$. 
Let $S_2\subset G_{\Sigma_2}$ be an arbitrary maximal
$F$-defined torus and let $S=S_1\cdot S_2$.

Arguing as in Theorem~\ref{steinberg2} we obtain that there is an 
$F$-embedding $S\hookrightarrow G_0$
such that $[\xi]\in {\rm Im}\,[H^1(F,S)\to H^1(F,G_0)]$
and the image of $S_2$ lies in a standard quasi-split simple simply connected
$F$-subgroup $H\subset G_0$ of type $A_1\times A_1\times A_1$.
% i.e. $H=R_{P/F}(SL_{2,P})$.
Since $S/S_2$ is a split torus over $F$, any cocycle in $Z^1(F,S)$
is equivalent to a cocycle with coefficients in $S_2$. It follows
that 
$\xi$ is equivalent to a cocycle with coefficients in $H$ and $H$ has trivial 
Galois cohomology.


Let $G$ be $F$-anisotropic. Consider first the case $[L:F]=3$.
Since ${\rm Ker}\, R_{G_0}^L=1$, we obtain that $[\xi]\in H^1(L/F,G_0(L))$. 
Then the argument on p. 372--373 in \cite{[PR]}
shows that $\xi$ up to equivalence lies in
a standard simple simply connected subgroup of $G_0$ of inner type $A_2$. 
Hence $\xi$ is trivial, by
Theorem~\ref{ms} and Proposition~\ref{number}.

The last case is $[L:F]=6$. Let $K_1=F(\sqrt{d})$ be the quadratic 
extension contained in $L$. Since we have 
already proved that ${\rm Ker}\,R_{G_0}^{K_1}=1$, we obtain 
that $\xi\in Z^1(K_1/F,G_0(K_1))$, in particular $G$ is quasi-split over
$K_1$. Then we can easily find as above (see also \cite{[PR]}, p. 354)
an $F$-subgroup $H$ of $G$ of type $A_1$. 
Let $K/F$ be a quadratic extension splitting $H$ and linearly disjoint with
$L$. Then $G$ is isotropic over $K$ and hence quasi-split over $K$,
for the isotropic case has been already treated.
Since $L$ and $K$ are linearly disjoint, Proposition~\ref{dd4}
and Theorem~\ref{ms} show that $\xi$ is trivial.
\end{pf}  



\section{\it Serre's Conjecture {\rm II} and the Hasse principle Conjecture 
{\rm II}}




{\textbf{Conjecture II}} (Serre \cite{[S94]}) Let $F$ be a perfect field 
with $cd\,(F)\le 2$
and let $G$ be a simply connected semisimple linear algebraic group 
over $F$. Then  $H^1(F,G)=1$.

\medskip

\noindent
{\textbf{The Hasse Principle Conjecture {\rm II}}} (\,Colliot-Th\'el\`ene, 
Scheiderer \cite{[CT96],[Sch96]}) Let 
$F$ be a field with $vcd\,(F)\le 2$ and let $G$ be a  simply connected
semisimple linear algebraic group over $F$. Then the canonical map
$$
\theta :\ H^1(F,G)\longrightarrow \prod_{\xi\in \Omega_F} H^1(F_{\xi},G)
$$
induced by the restriction maps has trivial kernel. Here $\Omega_F$ 
denotes the set of all orderings of $F$ and
$F_{\xi}$ is a real closure of $F$ at the ordering $\xi$.

        \begin{theorem}\label{serre} Let $G_0$ be a quasi-split simple 
         simply connected group over a perfect field $F$ which is of type 
         $^{3,6}D_4,E_6,E_7$. Assume 
         that $cd_p\,(F)\le 2$, where $p=2,3$. Then $H^1(F,G_0)=1$.
         \end{theorem}
\begin{pf} Let $\zeta\in Z^1(F,G_0)$. Since $cd_p\,(F)\le 2,\,p=2,3,$
we get $R_{G_0}([\zeta])=1$. It follows then from Theorem~\ref{rrost}
that $[\zeta]=1$.
\end{pf}
     \begin{theorem}\label{hhasse}
      Let $G_0$ be a quasi-split simple simply connected
      algebraic group over $F$ which is of type $^{3,6}D_4,E_6,E_7$. Assume that 
      $vcd\,(F)\le 2$. Then
      $$
        \theta :\ H^1(F,G_0)\longrightarrow \prod_{\xi\in \Omega_F} 
        H^1(F_{\xi},G_0)
      $$
      has trivial kernel. 
      \end{theorem}
{\em Proof.} 
Let $\zeta\in Z^1(F,G_0)$ be a locally trivial cocycle.
Let $K=F(\sqrt{-1})$. By Theorem~\ref{serre}, the restriction of $\zeta$
at $K$ is trivial. It follows that $R_{G_0}([\zeta])$ has coefficients
in $\mathbb{Z}/2$.
Hence the result follows from Theorem~\ref{rrost} and 
the following theorem which is due to Arason.
    \begin{theorem}\label{arason} The Hasse principle holds for 
     $H^3(F,\mathbb{Z}/2)$.
    \end{theorem}
\begin{pf} Let $\eta\in Z^3(F,\mathbb{Z}/2)$ be a locally trivial cocycle. 
Then by Arason's theorem~\,\cite{[A75]}, Satz 
3,  there is an integer $r$ such that $\eta\,\cup 
(-1)^r=0$. On
the other hand from the exact sequence 
$$
H^i(K,\mathbb{Z}/2)\stackrel{cor}\lra H^i(F,\mathbb{Z}/2)
\stackrel{\cup(-1)}\lra
H^{i+1}(F,\mathbb{Z}/2)\stackrel{res}\lra H^{i+1}(K,\mathbb{Z}/2)
$$
(\,\cite{[Ar75]}, Corollary 4.6\,) and from the equalities 
$$H^i(K,\mathbb{Z}/2)=H^{i+1}(K,\mathbb{Z}/2)=0,\quad i\ge 3
$$ 
we conclude that the product $\cup(-1)$ is an isomorphism. Therefore, 
$\eta=0$.
\end{pf}



\section{\it $(C_2)$-fields}





A field $F$ has $(C_2)$-property if every homogeneous equation
$f(x_1,\ldots,x_n)=0$ of degree $d$ with coefficients in $F$ has a 
nontrivial solution in $F^n$ if $n > d^2$. Property $(C_2)$ implies
$cd(F)\le 2$, but the converse is not true in general case.
The main property of $(C_2)$-fields we are going to use is
     \begin{theorem}{\rm (\cite{[Art82]})}
       Every central simple algebra $A$ over a $(C_2)$-field $F$ of exponent 
       $p$ has index $p$ if $p=2,3$.
       \end{theorem}
     \begin{theorem}\label{c2} Let $G$ be a simple simply connected algebraic
      group of type $^{3,6}D_4,E_6,E_7$ defined over a $(C_2)$-field $F$.
      Then $H^1(F,G)=1$.
     \end{theorem}
\begin{pf}
Denote a simple simply connected (\,resp. adjoint\,)
$F$-quasi-split group of the same type as $G$ by $G_0$ (\,resp. ${\overline 
G_0}$\,). In the case $^{3,6}D_4$, as in $6.4$, let $L/F$ be the minimal 
Galois extension
over which $G_0$ becomes split and $P/F$ a cubic extension contained
in $L$.
We have the natural map $\varphi: H^1(F,{\overline G_0})\lra 
H^2(F,Z)$, where $Z$ is the center of $G_0$.
Let $\xi\in Z^1(F,{\overline G_0})$ be such that $G=\,^{\xi}G_0$.
Then there is the natural bijection (see~\cite{[S94]})
$$
\varphi^{-1}(\varphi([\xi]))\lra
H^1(F,G)/\sim,
$$ 
where $\sim$ is the equivalence relation given by
multiplication on elements from $H^1(F,Z)$. It follows that we need only
to show that the fiber  $\varphi^{-1}(\varphi([\xi]))$ consists of 
one element, namely $[\xi]$, 
and any cocycle with coefficients in the centre of $G$ is trivial, viewed as
a cocycle in $G$. 
%$\eta\in Z^1(F,Z)$ is trivial in $H^1(F,G)$.

\subsection{Type $^{3,6}D_4$} 

Let $[\xi_1]\in \varphi^{-1}(\varphi([\xi]))$
and let $G_1=\,^{\xi_1}G_0$ be the corresponding twisted group.
It is known that
\[
H^2(F,Z)\simeq {\rm Ker}\,[\,{\rm _2Br}\,P\lra\, {\rm _2Br} F\,].
\]
Since exponent of an algebra over $P$ coincides with its index,
$\varphi([\xi])$ can be represented by a quaternion algebra over $P$. 
Then, by Proposition 43.9 in \cite{[KMRT]}, there is a quadratic extension
$K/F$ which kills $\varphi([\xi])=\varphi([\xi_1])$. This implies,
by Theorem~\ref{serre}, that $G$ and $G_1$ are quasi-split over $K$. 
Arguing as in Proposition~\ref{dd4} (see also~\cite{[PR]}, p. 354)
we can easily see that each of $G$ and $G_1$ contains a simple simply 
connected subgroup
of outer type $A_2$ splitting over $K$. Since $cd_2(F)\le 2$, any such  
subgroup is isotropic over $F$. Hence both groups $G$ and $G_1$ are 
$F$-isotropic.


If one of $G,G_1$ is quasi-split over $F$, there is nothing to prove.
Otherwise, both of them have the same Tits index (\ref{d4}).
Denote by $H_0$ the subgroup in $G_0$ generated by the root subgroups
corresponding to the roots $\alpha_1,\,\alpha_3,\,\alpha_4$ from 
diagram~(\ref{e33}). The group $H_0$ is a semisimple simply connected 
group of type $A_1\times A_1\times 
A_1$ containing the center $Z$. Let ${\overline H_0}\subset {\overline G_0}$ 
be its image in the adjoint group. 
Taking into consideration the diagram~(\ref{d4}) and
arguing as in Theorem~\ref{steinberg2}
we get immediately that up to 
equivalence the cocycles
$\xi$ and $\xi_1$ lies in $Z^1(F,{\overline H_0})$. Hence $[\xi]=[\xi_1]$,
since for groups of classical types Conjecture II holds by
\cite{[BP95]}.

To show that
$$
{\rm Im}\,[\,H^1(F,Z)\lra H^1(F,G)\,]=1
$$
it suffices to note that the center $Z$ lies in the semisimple 
$F$-anisotropic kernel of $G$ which
has type $A_1\times A_1\times A_1$ (see diagram~(\ref{d4})) and hence 
has trivial Galois cohomology.

\medskip

The above argument shows that we can apply the same proof for the other types 
$E_6,E_7$ if we find an $F$-split torus $S\subset G_0$
satisfying the following conditions:

\medskip
{\it
\noindent
(a) $[\xi],\,[\xi_1]\in {\rm Im}\,[\,H^1(F,C_{{\overline G_0}}
({\overline S}))
\lra H^1(F,{\overline G_0})\,]$, where ${\overline S}$ is the image of
$S$ in ${\overline G_0}$,


\medskip

\noindent
(b) $H_0=[\,C_{G_0}(S),C_{G_0}(S)\,]$ is a semisimple simply connected 
group
of classical type containing the center $Z$ of $G_0$,


\medskip

\noindent
(c) the centralizer $C_{G_0}(S)$ is an almost direct product of $S$
and $H_0$.}

\subsection{Type $^1E_6$} 

We have $Z\simeq \mu_3$, hence there is a cubic cyclic extension
$E/F$ killing 
$\varphi([\xi])=\varphi([\xi_1])$.
   \begin{lemma} $G$ and $G_1$ are $F$-isotropic.
    \end{lemma}
\begin{pf} Assume the contrary. Let $C_1$ be a subgroup constructed in
Lemma~\ref{eagle2}. It has type $D_4$, hence is $F$-isotropic by the argument 
above --- a contradiction.
\end{pf}

All admissible $F$-isotropic Tits indices are given by diagrams~(\ref{p}).
Since $G,G_1$ are split over a cubic extension $E/F$ we conclude
that the only case which can occur is represented by diagram~(\ref{p}),
index~(ii). 
Then one can easily see that the torus $S_1$ constructed in 
Theorem~\ref{steinberg2} satisfies properties $(a),\,(b)$ and $(c)$.

\subsection{Type $^2E_6$} Let $K/F$ be the quadratic extension over which $G_0$
becomes a group of inner type and let $\tau$ be the nontrivial
automorphism $K/F$.
    \begin{lemma} $G$ and $G_1$ are $F$-isotropic.
    \end{lemma}
\begin{pf} Assume the contrary. We have
already proved that $G$ is $K$-isotropic. If it splits over 
$K$, then it contains an $F$-defined subgroup of type $A_2$
splitting over the quadratic extension $K/F$. Since $F$ has $(C_2)$-property,
this subgroup is $F$-isotropic.

Thus $G$ is not $K$-split, hence its $K$-Tits index is of the form (ii) in 
diagram (\ref{p}) (note that the index (i) can not occur since its 
anisotropic kernel
is a classical group of type $D_4$ and any such group is $F$-isotropic
over a $(C_2)$-field). Let $P$ be a minimal parabolic subgroup of $G$ over 
$K$ in generic position and let $C$ be the connected component of 
$P\cap\tau(P)$. Since, by our assumption, $G$ is $F$-anisotropic,
$C$ is a reductive $F$-group. By dimension argument, $C_K$ is a
Levi subgroup of $P$.
%whose semisimple part $[C,C]$ is the semisimple part of the Levi subgroup
%of $P$ over $K$. 
This implies that $[C,C]$ is a semisimple $K$-anisotropic 
kernel of $G$. It follows then from picture (ii) in diagram (\ref{p})
that the centraliser $C_G([C,C])$ is an $F$-defined subgroup of $G$
of type $A_2$ splitting over $K$. But any such group is $F$-isotropic.
\end{pf}
 

\medskip

All admissible $F$-isotropic Tits indices of type $^2E_6$
are given by the diagrams
in Proposition~\ref{gul10}. Since the $K$-Tits indices of $G,G_1$ are of the form
(ii) in (\ref{p}), we conclude that over $F$ they
are of the form either (a) or (c) from the diagrams 
in Proposition~\ref{gul10}. In both
cases the subgroups in $G,G_1$ corresponding to the root of maximal length
are isomorphic to $SL_2$ over $F$, hence contains a $1$-dimensional split
torus $S$. The rest of the proof is the same as for inner forms of type $E_6$. 

\subsection{Type $E_7$} As above one can easily see that both $G,G_1$
are $F$-isotropic. Looking at the Tits tables~\cite{[T66]}
and taking into the consideration the fact (which has been already proved
above) that all simple groups of type not $A_n$ are $F$-isotropic
we obtain that $G,G_1$ have the same Tits indices of the form
$$
\begin{picture}(100,30)
\put(00,02){\line(1,0){100}}
\put(60,02){\line(0,1){20}}
\put(00,02){\circle*{3}}
\put(20,02){\circle*{3}}
\put(40,02){\circle*{3}}
\put(60,02){\circle*{3}}
\put(80,02){\circle*{3}}
\put(100,02){\circle*{3}}
\put(60,22){\circle*{3}}
\put(20,02){\circle{10}}
\put(60,02){\circle{10}}
\put(80,02){\circle{10}}
\put(100,02){\circle{10}}
\end{picture}
$$
The rest of the proof is the same as for inner forms of type $E_6$.
Thus we complete the proof of Theorem~\ref{c2}.
\end{pf} 
\begin{remark} {\rm Note that under the proof of Theorem~\ref{c2}
we have showed that any simple group of type $^{3,6}D_4,E_6,E_7$
is isotropic and also described all admissible Tits indices
of such groups.}
\end{remark} 






\begin{thebibliography}{Z}



\bibitem[A75]{[A75]}
J. Arason, Primideale im graduierten Wittring und im mod 2 Cohomologiering,
Math. Z. {\bf 145} (1975), 139--143.

\bibitem[Ar75]{[Ar75]}
J. Arason, Cohomologische Invarianten quadratischer Formen, J. Algebra 
{\bf 36} (1975), 448--491.


\bibitem[Art82]{[Art82]}
M. Artin, Brauer--Severi Varieties (Notes by A. Verschoren),
Lecture Notes in Math.{\bf 917} (1982), 194--210.

 
\bibitem[BP95]{[BP95]}
E. Bayer-Fluckiger, R. Parimala, Galois cohomology of the classical groups 
over fields of cohomological dimension $\leq 2$, Invent. Math. {\bf 122}
(1995), no. 2, 195--229. 


\bibitem[BP98]{[BP98]}
E. Bayer-Fluckiger, R. Parimala, Classical groups and the Hasse principle,
Ann. of Math. {\bf 147} (1998), 1--43. 


\bibitem[Bourb68]{[B68]}
N. Bourbaki, Groupes et Alg\`ebres de Lie, Chapitres 4,5 et 6,
Hermann, Paris 1968.

\bibitem[Ch89]{[Ch]} V. Chernousov,
On the Hasse principle for groups of type $E_8$, Dokl. Akad. Nauk SSSR,
{\bf 306} (1989), N\,25, 1059-1063.

\bibitem[Cher89]{[Ch89]}
V. Chernousov, On projective simplicity of groups of rational points of 
some algebraic groups over algebraic number fields, Izvestiya Akad. Nauk SSSR,
Ser. Math., {\bf 53} (1989),  no 2, 398--410.

\bibitem[Ch94]{[Ch94]} V. Chernousov, Remark on $({\rm mod }\,5)$--Serre's 
invariant for groups of type $E_8$, Math. Zametki, {\bf 56} (1994), 
N\,1, 116-121.

\bibitem[Ch98]{[Ch98]}
V. Chernousov, Letter to P. Gille, November, 1998.


\bibitem[ChT99]{[ChT]}
V. Chernousov, L. Timoshenko, On the group of $R$-equivalence classes
of semisimple groups over arithmetic fields, Algebra and Analis,
{\bf 11} (1999), no 6, 191-221.

\bibitem[Ch00]{[Ch00]}
V. Chernousov, Letter to Garibaldi, October 2000.


\bibitem[CT96]{[CT96]} 
J--L. Colliot-Th\'el\`ene, Groupes lin\'eaires sur les corps de fonctions de 
courbes r\'eelles, J. reine angew. Math. {\bf 474} (1996), 139--167.


\bibitem[EKLV98]{[EKLV98]}
H. Esnault, B. Kahn, M. Levine, E. Viehweg, The Arason invariant and mod 
$2$ algebraic cycles, J. Amer. Math. Soc. 11 (1998), no. 1, 73--118. 


\bibitem[Fer69]{[Fer]}
J. C. Ferrar, Lie algebras of type $^1E_6$, J. of Algebra 13 (1969),
57--72.

\bibitem[Gar01]{[Gar]}
R. S. Garibaldi, The Rost invariant has trivial kernel for quasi-split
groups of low rank, To appear in Commentarii Mathematici Helvetici. 


\bibitem[G93]{[G93]}
P. Gille, $R$-\'equivalence et principe de norme en cohomologie 
galoisienne, C. R. Acad. Sci. Paris, S\'er. I Math., {\bf 316} (1993), 
no. 4, 315--320.


\bibitem[G97]{[G97]}
P. Gille, La $R$-\'equivalence sur les groupes alg\'ebriques r\'eductifs 
d\'efinis sur un corps global,
Inst. Hautes \'Etudes Sci. Publ. Math., No. 86 (1997), 199--235 (1998).


\bibitem[G00]{[G00]}
P. Gille, Invariants cohomologiques de Rost en caract\'eristique positive, 
$K$-Theory, {\bf 21} (2000), no. 1, 57--100.



\bibitem[G01]{[G01]}
P. Gille, Cohomologie galoisienne des groupes quasi-d\'eploy\'es sur des 
corps de 
dimension cohomologique $\leq 2$, Composito Math. {\bf 125} (2001), no. 3,
283--325. 




\bibitem[H65]{[H65]}
G. Harder, \"Uber die Galoiskohomologie halbeinfacher 
Matrizengruppen I, Math. Z. {\bf 90} (1965), no 5, 404--424.

\bibitem[H66]{[H66]}
G. Harder, \"Uber die Galoiskohomologie halbeinfacher 
Matrizengruppen II, Math. Z. {\bf 92} (1966), no 5, 396--415.

\bibitem[Hum75]{[Hum]}
J. Humphreys, Linear Algebraic Groups, Springer-Verlag 1975.


\bibitem[KMRT98]{[KMRT]} 
M. Knus, A. Merkurjev, M. Rost, J--P. Tignol, 
The Book of Involutions, {\bf 44} Colloquium publications: American 
Mathematical Society 1998.

\bibitem[L73]{[L]}
T.-Y. Lam, The Algebraic Theory of Quadratic Forms,
Benjamin, MA, 1973.

\bibitem[MS82]{[MS]}
A. Merkurjev, A. Suslin, $K$-cohomology of Severi-Brauer varieties
and the norm residue homomorphism, Izv. Akad. Nauk SSSR {\bf 46} (1982), 
1011--1046.

\bibitem[M01]{[M01]}
A. Merkurjev, Unramified cohomology of classifying varieties
for classical simply connected groups, Preprint 2001.

\bibitem[PSS99]{[PSS]}
R. Parimala, R. Sridharan, V. Suresh, Hermitian analogue of a theorem
of Springer, Preprint 1999.


\bibitem[PR94]{[PR]}
V. Platonov, A. Rapinchuk,  Algebraic Groups and Number Theory,
Academic Press 1994.

\bibitem[R]{[R]}
M. Rost, Cohomological invariants, in preparation. 


\bibitem[S94]{[S94]}
J.--P. Serre, Cohomologie Galoisienne, Springer, 1994.

\bibitem[S93]{[S93]}
J.--P. Serre, Cohomologie galoisienne: progr\`es et probl\`emes,
S\'em. Bourbaki 1993--1994, expos\'e 783.

\bibitem[Sch96]{[Sch96]}
C. Scheiderer, Hasse principles and approximation theorems for homogeneous
spaces over fields of virtual cohomological dimension one, Invent. math. 
{\bf 125} (1996), 307--365.

\bibitem[SGA3]{[SGA3]} S\'eminaire de G\'eom\'etrie alg\'ebrique de l'I.H.E.S.,
1963--1964, Sch\'emas en groupes, dirig\'e par M. Demazure et A. Grothendieck,
Lecture Notes in Math. 151--153, Springer (1970).

\bibitem[St65]{[St65]}
R. Steinberg, Regular elements of semisimple groups, Publ. Math. {\rm
I.H.E.S.} {\bf 25}, 281--312 (1965).

\bibitem[St67]{[St67]}
R. Steinberg, Lectures on Chevalley Groups, Yale University 1967.

\bibitem[T66]{[T66]} 
J. Tits, Classification of algebraic semisimple groups,        
In Algebraic Groups and discontinuous Subgroups,
(eds. A.Borel and G.D.Mostow), Proc. Symp. Pure Math. {\bf 9}, 33--62 (1966).  

\bibitem[V98]{[V]} V. Voskresenskii, Algebraic Groups and their Birational
Invariants, Translated from Russian by Boris E. Kunyavskii, 
Amer. Math. Soc., Providence, RI, 1998



\end{thebibliography}


\end{document}

