\documentclass[12pt]{amsart}
\usepackage{amsfonts,amssymb,eucal,amsmath}
\usepackage{amsbsy}
\usepackage{amscd}
\usepackage{amsthm}
\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 }
\newcommand{\End}{\operatorname{End}}
\newcommand{\Hom}{\operatorname{Hom}}
\newcommand{\CH}{\operatorname{CH}}
\newcommand{\Corr}{\operatorname{\mathbf{Corr}}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\A}{\mathbb{A}}
\renewcommand{\P}{\mathbb{P}}
\newcommand{\Spec}{\operatorname{Spec}}
\newcommand{\inv}{^{-1}}
\renewcommand{\(}{\bigl(}
\renewcommand{\)}{\bigr)}
\newcommand{\GL}{\operatorname{GL}}
\newcommand{\PGL}{\operatorname{PGL}}
\newcommand{\PSO}{\operatorname{PSO}}
\newcommand{\SO}{\operatorname{SO}}
\newcommand{\Aut}{\operatorname{Aut}}
\newcommand{\cH}{\mathbb H}
\newcommand{\ga}{\mathbb{G}_a}
\newcommand{\id}{\operatorname{id}}



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



\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{sublemma}[theorem]{Sublemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\theoremstyle{definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newcommand{\iso}{\stackrel{\sim}{\to}}





%%%%%%%%%%%%% specific symbols %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



%\newcommand{\qed}{$\ \Box$}



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



%\newenvironment{pf}{\par\noindent{\em Proof}.}{\hfill\framebox(6,6)

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



\title[Another proof of Totaro's theorem on $E_8$-torsors]
{Another proof of Totaro's theorem on splitting fields of $E_8$-torsors}


\date{}

\author[V. Chernousov]{Vladimir Chernousov*}

\thanks{*\,Supported by the Canada Research Chairs
Program, and by NSERC's Grant G121210944}


\address
{Department of mathematical sciences, University of Alberta,
Edmonton, Alberta, Canada T6G 2G1}


\email {chernous@math.ualberta.ca}



%\thanks{Supported by the Canada Research Chairs
%Program, and by NSERC's Grant G121210944}



\subjclass[2000]{Primary 11E72; Secondary 14M17, 20G15}


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



\begin{document}







%\maketitle



\begin{abstract}
We give a short proof of Totaro's theorem that every
$E_8$-torsor over a field $k$ becomes trivial over a finite
separable extension of $k$
of degree dividing $d(E_8)=2^63^25$.
\end{abstract}



\maketitle



\section{Introduction}


In the paper we give a short proof of the following theorem
due to B. Totaro~\cite{Tot1}.
\begin{theorem}\label{main}
Let $k$ be an arbitrary field. Then every $E_8$-torsor defined
over $k$ becomes trivial over a finite separable extension of $k$
of degree dividing $d(E_8)=2^63^25$.
\end{theorem}

Note that in a second paper on $E_8$-torsors~\cite{Tot2}, Totaro
showed that the bound $2^63^25$ is exact, i.e. there is an
$E_8$-torsor that can not be split by an extension whose degree is a
proper divisor of $2^63^25$.
%that the so-called torsion index for $E_8$
%introduced by Grothendieck~\cite{gro} is equal to $2^63^25$. As a
%consequence it follows that the bound $2^63^25$ in the above
%theorem is exact.

The original proof of Theorem~\ref{main} is based on an analysis of
the subgroup structure of the Weyl group of type $E_8$, Brauer's
theory of blocks, Aschbacher's theorem on the maximal subgroups of
the classical groups over finite fields, and the classification of
solvable primitive linear groups. Moreover, some of the computations
in ~\cite{Tot1} were made with the aid of a computer. The aim of the
present paper is to simplify the proof. Eventually following the
main Totaro's idea on considering Galois orbits in the corresponding
root system $\Sigma(E_8)$ we give a short straightforward proof of
Theorem~\ref{main}.

\section{Generic case and possible bad cases}
Let $G_0$ be a split group of type $E_8$ over $k$. Let $\xi\in
Z^1(k,G_0)$, and let $G={^{\xi}G_0}$ be the corresponding twisted
group. Consider a maximal $k$-defined torus $T\subset G$. Let $E/k$
be a minimal finite extension splitting $T$. The  extension $K/k$ is
necessarily Galois, and its Galois group $\Gamma$ acts in a natural
way on the root system $\Sigma=\Sigma(G,T)$ of $G$ with respect to
$T$. This gives rise to a canonical embedding $\Gamma\hra W$ where
$W=W(E_8)$ is the corresponding Weyl group. If we choose a base of
$\Sigma$, then the action of $\Gamma$ on $\Sigma$ induces an action
of $\Gamma$ on the set $R=\Sigma/(\pm 1)$. This set  has $120$
elements, and we always choose positive roots as representatives of
the elements of $R$.

The case of ``generic" $E_8$-torsors is easy.
\begin{lemma} Assume that $\Gamma$ has an orbit on $R$ of size
dividing $120=2^3\cdot 3\cdot 5$. Then there is a finite separable
extension $L/k$ of degree dividing $d(E_8)$ such that $G$ splits
over $L$.
\end{lemma}
\begin{proof} Let $\alpha\in R$ be such that $|\Gamma(\alpha)|$
divides $120$. Let ${\rm Stab}_{\Gamma}(\alpha)$ be the stabilizer
of $\alpha$ in $\Gamma$, and consider the subfield $L_1\subset E$
corresponding to ${\rm Stab}_{\Gamma}(\alpha)$. Taking an extension
$L_2/L_1$ of degree $2$ if necessary, we may assume that $\Sigma$
has a root $\alpha$ stable with respect to an (absolute) Galois
group of $L_2$.
%Note that $[L_2:k]$ divides $2^4\cdot 3\cdot 5$.
The centralizer $\Sigma'$ of $\alpha$ in $\Sigma$ is the subsystem
of type $E_7$ which is stable with respect to the Galois group of
$L_2$. If $H\subset G$ is the subgroup in $G$ of type $E_7$
corresponding to $\Sigma'$, then $H$ is $L_2$-defined and, by a
result of Tits~\cite{tits}, splits over a separable extension
$L_3/L_2$ of degree dividing $2^23$. Clearly $L_3$ also splits $G$,
and $[L_3:k]=[L_3:L_2][L_2:L_1][L_1:k]$ divides
$(2^23)2(120)=2^63^25$, as required.
\end{proof}

If $\Sigma$ contains a proper subroot system stable with respect to
$\Gamma$, then using known results on groups of classical types and
Tits results~\cite{tits} on splitting fields of groups of types
$G_2,F_4,E_6,E_7$, it is easy to conclude that $G$ splits over a
finite separable extension of $k$ of degree dividing $d(E_8)$. Thus,
we may henceforth assume without loss of generality that $\Sigma$
does not contain root subsystems stable with respect to $\Gamma$. In
this case, possible ``bad" orbit decompositions are given by the
following:
\begin{lemma}{\rm (~\cite{Tot1}, Lemma $4.1$)}\label{bad} If $\Gamma$
has no orbits on $R$ of size dividing $120$, then the orbit sizes
of $\Gamma$ are either

\noindent {\rm (a)} $64+$ ( multiples of $7$ summing to $56$);

\noindent {\rm (b)} $50+$ (multiples of $7$ summing to $70$);

\noindent {\rm (c)} $45+$ (multiples of $25$ summing to $75$);

\noindent {\rm (d)} $36+$ (multiples of $7$ summing to $84$) or

\noindent {\rm (e)} (multiples of $16$ summing to $48$) +
(multiples of $9$ summing to $72$).
\end{lemma}
For convenience of the reader we give a sketch of the proof due to
Totaro. It is based on the following result.
\begin{lemma}\label{sylow}
{\rm (i)} A $7$-Sylow subgroup of $W$ has only one fixed point in
$R$.

\noindent {\rm (ii)} A $5$-Sylow subgroup of $W$ has $4$ orbits of
size $25$ and $4$ orbits of size $5$ in $R$.
\end{lemma}
\begin{proof} This is easy to check by direct inspection.
\end{proof}

\noindent {\em Proof of Lemma~\ref{bad}}. Let us first assume that
$7$ divides $|\Gamma|$. Then, by Lemma~\ref{sylow}, all orbits of
$\Gamma$ in $R$ have sizes divisible by $7$ except for one whose
size is $\equiv 1$ modulo $7$. The size of this exceptional orbit
is either $36,\ 50$ or $64$, since by our assumption there is no
orbit of size dividing $120$. Thus, assuming that $|\Gamma|$ is a
multiple of $7$ we have cases {\rm (a), (b), (d)}.

Assume next that $|\Gamma|$ is not divisible by $7$, but divisible
by $25$. Since the sum of sizes of all orbits of $\Gamma$ in $R$
is $120$, and sizes of orbits do not divide $120$ we find, by
Lemma~\ref{sylow}, that all orbits of $\Gamma$ have size divisible
by $25$ except for one whose size is $45$. Hence we have case (c).

Finally, assume that the order of $\Gamma$ is divisible by neither
$7$ nor $25$. Recall that $|W|=2^{14}3^55^27$. Since there is no
orbit of $\Gamma$ whose size divide $120$, all of them have sizes
a multiple of $16$ or $9$. The only way it can happen is case $\rm
(e)$. Lemma~\ref{bad} is proved. \qed

\medskip

By \cite{Tot1}, Lemma $6.1$, cases ${\rm (b), (c)}$ are
impossible. By \cite{Tot1}, Lemma $4.2$, in case ${\rm (a)}$ the
complementary subset to the orbit of size $64$ forms a subsystem
of type $D_8$. The remaining cases ${\rm (d)}$ and ${\rm (e)}$,
which caused most of the complications in~\cite{Tot1}, will be
dealt with in a simple fashion in the following two sections.

For later use, we need the following fact related to the Rost
invariant for $E_7$. For the definition and properties of the Rost
invariant $R_G$ of an algebraic group $G$ we refer to~\cite{gms}.
\begin{proposition}\label{e7}
Let $H_0$ be a split simple simply connected algebraic group of type
$E_7$ defined over an arbitrary field $K$, and let
$$R_{H_0}:H^1(K,H_0)\to H^3(K,{\bf Q}/{\bf Z}(2))$$ be the Rost
invariant of $H_0$. Let $\xi\in H^1(K,H_0)$ be such that the
$3$-component of $R_{H_0}(\xi)$ is trivial. Then there is a
separable extension $L/K$ of degree dividing $4$ such that $\xi$
is trivial over $L$.
\end{proposition}
\begin{proof} By~\cite{tits}, there is a quasi-split subgroup
$H'\subset H_0$ of type $E_6$ such that $\xi$ is in the image of
$H^1(K,H')\to H^1(K,H_0)$. Taking a proper quadratic extension
$E/K$ if necessary, we may assume that $H'$ is split over $E$. One
knows that for a split group $H'_E$ of type $E_6$ the $2$-component
of $R_{H'}(\xi_E)$, where $\xi_E$ is the image of $\xi$ under the
restriction map $H^1(K,H_0)\to H^1(E,H_0)$, is a symbol. Taking
again a separable quadratic extension $L/E$ killing this symbol we
may assume that the $2$-component of $R_{H'}(\xi_L)$ is trivial over
$L$. Then $\xi_L\in {\rm Ker}\,R_{H'}$.
%Here $\xi_L$ is the image of $\xi$ under the
%restriction map $H^1(K,H_0)\to H^1(L,H_0)$.
It remains to observe that ${\rm Ker}\,R_{H'}=1$, by ~\cite{gar}
(see also~\cite{ch}).
\end{proof}

\section{An orbit of size $36$}\label{36}

Let $R_1\subset R$ be an orbit of $\Gamma$ of size $36$, and let
$R_2=R\setminus R_1$. Take a positive root $\alpha\in R_1$ and
consider $\Gamma_1=Stab_{\Gamma}(\alpha)$. Note that in the
definition of $\Gamma_1$, $\alpha$ is viewed as an element of $R$,
but not of $\Sigma$. Let $E_1'\subset E$ be the subfield
corresponding to $\Gamma_1$. Taking a proper quadratic extension
$E_1/E'_1$ if necessary, we may assume that $\alpha$ viewed as a
root in $\Sigma$ is stable with respect to an (absolute) Galois
group of $E_1$. Since $|R_1|=36$, the index $[E_1:k]$ is either
$2^23^2$ or $2^33^2$.
\begin{lemma}\label{3component}
If the $3$-component of $R_{G_0}([\xi])$ is trivial over $E_1$,
then there is a separable extension $E_2/k$ of degree dividing
$2^53^2$ which kills $\xi$.
\end{lemma}
\begin{proof} Let $\Sigma'$ be the root subsystem of $\Sigma$
consisting of roots orthogonal to $\alpha$. Consider the subgroup
$H$ of $G$ corresponding to $\Sigma'$. It has type $E_7$ and is
defined over $E_1$ since so is $\alpha$. Since $H$ contains a
semisimple anisotropic $E_1$-kernel of $G$, by a result due to R.
Steinberg (cf.~\cite{ch}, Theorem $3.2$), there is a cocycle
$\xi_1\in Z^1(E_1,H_0)$, where $H_0\subset G_0$ is a canonical
$E_1$-split subgroup of type $E_7$, such that $\xi$ is equivalent
to $\xi_1$ over $E_1$. Note that $R_{G_0}(\xi)=R_{H_0}(\xi_1)$.
Then, by Proposition~\ref{e7}, there is a separable extension
$E_2/E_1$ of degree dividing $4$ which kills $\xi_1$, and hence
$\xi$. Its degree over $k$ divides $4(2^33^2)$, as required.
\end{proof}

By Lemma~\ref{3component}, we may henceforth assume without loss
of generality that the $3$-component of $R_{G_0}([\xi])$ is
nontrivial over $E_1$.
\begin{lemma}\label{4roots} Let $\beta\in R_2$. Then $|\Gamma_1(\beta)|$ is
multiple of $21$.
\end{lemma}
\begin{proof}
Since $\Gamma_1$ contains a $7$-Sylow subgroup of $W$, the size of
$\Gamma_1(\beta)$ is divisible by $7$ by Lemma~\ref{sylow} (i).
Assume that $|\Gamma_1(\beta)|$ is not divisible by $3$. Take the
extension $E_2/E_1$ of degree prime to $3$ corresponding to the
stabilizer $\Gamma_2={\rm Stab}_{\Gamma_1}(\beta)$. By a counting
argument, there are at least two roots in $R_2$ different from
$\beta$ whose $\Gamma_2$-orbits have sizes not divisible by $3$.
Repeating the above construction $2$ times, we can find a finite
extension $E/E_1$ of degree prime to $3$ with the property that an
(absolute) Galois group of $E$ stabilizers $\alpha$ and at least
$3$ roots in $R_2$. Then it follows from Tits'
classification~\cite{t66} that the $E$-rank of $G$ is at most $5$.
Again, by Tits' classification, all simple groups which could
appear in a semisimple $E$-anisotropic kernel of $G$ have trivial
$3$-components of the Rost invariant, implying therefore that
$R_{G_0}(\xi_E)$ has also trivial $3$-component. On the other
hand, since $[E:E_1]$ is prime to $3$, the $3$-component of
$R_{G_0}(\xi_E)$ is still nontrivial -- a contradiction.
\end{proof}

Recall that we assumed that $\Sigma$ has no subroot systems stable
with respect to $\Gamma$; in particular we may assume that $R_1$ is
not a subroot system. It follows that there is $\delta\in R_1$ such
that either $\alpha+\delta$ or $\alpha-\delta$ is a root, call it
$\beta=\alpha\pm\delta$, belonging to $R_2$. Since the size of
$\Gamma_1(\beta)$ is divisible by $21$, so is $|\Gamma_1(\delta)|$.
Since $R_1$ consists of $36$ elements, the size of
$\Gamma_1(\delta)$,  hence that of $\Gamma_1(\beta)$, is exactly
$21$.
% The orbit $\Gamma_1(\beta)$ of $\beta$ can not have an order
%bigger $21$, since $\beta$ is a sum of $\alpha$ and $\pm\delta$
%and the $\Gamma_1$-orbit of $\delta$ is in $R_1$ whose order is
%$36$.

Let $R'_1=\Gamma_1(\delta)$, $R''_1=R_1\setminus R'_1$,
$R'_2=\Gamma_1(\beta)$, $R''_2=R_2\setminus R'_2$. Recall that we
denote the subsystem of $\Sigma$ of type $E_7$ consisting of all
roots in $\Sigma$ orthogonal to $\alpha$ by $\Sigma'$.
\begin{lemma}
$\pm R''_2$ coincides with $\Sigma'$.
\end{lemma}
\begin{proof} Since $(\alpha,\beta)=\pm 1$ and $(\alpha,\delta)=\pm 1$,
the intersection of $\Sigma'/\pm 1$ with $R'_1$ and $R'_2$ is
empty, hence $$(\Sigma'/\pm 1)=((\Sigma'/\pm 1)\cap R''_1)\cup
((\Sigma'/\pm 1)\cap R''_2.)$$ The order of $(\Sigma'/\pm 1)\cap
R''_2$ being $\Gamma_1$-stable is divisible by $21$. Since $R''_1$
has order $16$ and $|\Sigma'/\pm 1|=63$, we have $(\Sigma'/\pm
1)\cap R''_1=\emptyset$.
\end{proof}
As a direct consequence of the above lemma we have
\begin{corollary}\label{system}
{\rm (i)} $(\alpha,\gamma)=\pm 1$, if $\gamma\in R_1$ and
$\gamma\not=\alpha$.

\noindent {\rm (ii)} $\alpha\pm\gamma_1\in R''_1$, if $\gamma_1\in
R''_1$.

\noindent {\rm (iii)} $(\gamma_1,\gamma_2)=\pm 1$, if
$\gamma_1,\gamma_2\in R_1$, $\gamma_1\not=\gamma_2$.
\end{corollary}
\begin{proof}
Properties (i) and (ii) are clear since $(\Sigma'/\pm 1)\subset
R_2$. Property (iii) follows from (i), since $\alpha$ was an
arbitrary root in $R_1$.
\end{proof}
\begin{lemma} $\pm R''_1$ is a  subroot system of $\Sigma$.
\end{lemma}
\begin{proof}
Let $\gamma\in R''_1$. We have to show that $\gamma\pm\gamma'\in
R''_1$ for all $\gamma'\in R''_2$ different from $\gamma$. Arguing
as above, we see that there exists a subset $R'_{1,\gamma}$  of
$R_1$, with $21$ elements, comprised of roots whose sum with
$\gamma$ is in $R_2$. By Corollary~\ref{system}, the remaining $14$
roots in $R_1 \setminus R'_{1,\gamma}$ have sum with $\gamma$  in
$R_1\setminus R'_{1,\gamma}$. We will be finished if we show that
$R'_{1,\gamma}=R'_1$.

Let $\delta\in R'_1$. By Corollary~\ref{system} (iii), either
$\gamma+\delta$ or $\gamma-\delta$ is a root. Call it $\beta$.
Since $(\alpha,\beta)\equiv{0}$ modulo $2$, we have either
$\alpha=\pm\beta$ or $\beta\in \Sigma'=R''_2$. The first case is
impossible, since the $\Gamma_1$-orbits of $\delta$ and $\gamma$
consist of $21$ and at most $14$ elements respectively. Then
$\beta\in R_2$, so that $\delta\in R'_{1,\gamma}$.
\end{proof}
 To finish the consideration of orbits of size $36$, it remains to note
that the subroot system $R''_1$ is $\Gamma_1$-stable, hence it has
an automorphism of order $7$. However the minimal simple root
system having an automorphism of order $7$ has type $A_6$ and
consists of $42$ elements.

\section{An orbit of size a multiple of $16$}

We start with an explicit description of a $3$-Sylow subgroup of
$W$, denoted below by $\Psi$, and its action on the root system
$\Sigma$. Recall that $|\Psi|=3^5$. Let $\Pi=\{\alpha_1,\ldots,
\alpha_8\}$ be a fixed basis of $\Sigma$. Here and below we label
roots as in~\cite{bourbaki}. Consider the subroot system of type
$E_6\times A_2$ in $\Sigma$ generated by
$\Sigma_1=\langle\,\alpha_1,\ldots,\alpha_6\,\rangle$ and
$\Sigma_2=\langle\,\alpha_8,-\alpha\,\rangle$ where $\alpha$ is the
highest root of $\Sigma^+$. Comparing the orders of the Weyl groups
of type $E_6,A_2,E_8$, we find that the direct product
$\Psi=\Psi_1\times\Psi_2$ of $3$-Sylow subgroups $\Psi_1$ of
$W(E_6)$ and $\Psi_2$ of $W(A_2)$ is a $3$-Sylow subgroup of $W$.

Recall that $\Psi_2$ has order $3$. As for $\Psi_2$, we choose the
subgroup in $W(A_2)$ generated by the element $e$ which takes
$\alpha_8$ into $-\alpha$ and $-\alpha$ into $-(\alpha_8-\alpha)$.

The root system $\Sigma_1$ contains a subroot system $\Sigma_3$ of
type $A_2\times A_2\times A_2$ generated by the roots
$\langle\alpha_1,\alpha_3\rangle,\,\langle\alpha_5,\alpha_6\rangle\,$
and $\langle\alpha_2,-\beta\rangle$ respectively, where $\beta$ is
the positive root of maximal length in $\Sigma_1$ with respect to
the basis $\alpha_1,\ldots,\alpha_6$. Let $w_0,w_1\in W(E_6)$ be
the elements of maximal length with respect to the bases
$\{\alpha_1,\ldots, \alpha_6\}$ and
$\{\alpha_1,\alpha_3,\alpha_4,\alpha_2,-\beta,\alpha_5\}$
respectively. Let $d=w_0 w_1$. It is easy to see that $d$ 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 $d$ permutes the components of $\Sigma_3$
and their Weyl groups.

Let $a$ be an arbitrary element  of order $3$ in the Weyl group of
the first component of $\Sigma_3$. Denote $b=dad^{-1}$ and
$c=dbd^{-1}$. Clearly, $a,b,c$ commute and $d$ permutes them.
Consider the subgroup $\Psi_1$ in $W(E_6)$ generated by $a,b,c,d$.
Since $\Psi_1$ has order $3^4$, it is a $3$-Sylow subgroup of
$W(E_6)$.

One easily checks that there are $4$ orbits of $\Psi$ on $R$ which
are as follows. The $\Psi$-orbit of $\alpha_7$ consists of $81$
elements in $\Sigma^+\setminus \{\Sigma_1^+\cup \Sigma_2^+\}$. The
$\Psi$-orbit of $\alpha_1$ consists of $9$ elements and coincides
with $\Sigma_3^+$. The $\Psi$-orbit of $\alpha_8$ consists of $3$
elements in $\Sigma_2^+=\{\alpha_8,\alpha,\alpha-\alpha_8\}$.
Lastly, the $\Psi$-orbit of $\alpha_4$ consists of the remaining
$27$ elements in $\Sigma_1^+\setminus\Sigma_3^+$.

We also need information about  the stabilizer ${\rm
Stab}_{\Psi}(\beta)$ of a root $\beta\in R$. It is easy to see
that for each root $\beta\in \Psi(\alpha_7)=\Sigma^+\setminus
\{\Sigma_1^+\cup \Sigma_2^+\}$ one has ${\rm
Stab_{\Psi}(\beta})\subset \langle\,a\,\rangle\cup
\langle\,b\,\rangle\cup \langle\,c\,\rangle$. Furthermore, for
each $\beta\in \Psi(\alpha_4)$, ${\rm Stab}_{\Psi_1}(\beta)$ has
order $3$ and is generated by an element of the form
$da^{\epsilon_1}b^{\epsilon_2}c^{\epsilon_3}$ where $\epsilon_i$
is $0,1$ or $2$.

Let $R_1$ and $R_2$ be unions of orbits of $\Gamma$ whose sizes
are divisible by $16$ and $9$ respectively.
%the subsets
%$R_1,R_2$ which are $\Gamma$-stable and have orders $48,\,72$
%respectively. Recall that all $\Gamma$-orbits in $R_1,\,R_2$ have
%orders divisible by $16$ and $9$ respectively.
Let
$\Gamma_3\le\Gamma$ be a $3$-Sylow subgroup. Without loss of
generality we may assume that $\Gamma_3$ is a subgroup of $\Psi$.
\begin{lemma}\label{order}  $|\Gamma_3|\le 3^3$.
\end{lemma}
\begin{proof} If $|\Gamma_3|=3^5$, then $\Gamma_3=\Psi$ and hence
$\Gamma_3$ has the orbit $\Gamma_3(\alpha_7)=\Psi(\alpha_7)$ of size
$81$; which is impossible.

Assume that $|\Gamma_3|=3^4=81$. Then $\Gamma_3$ is a normal
subgroup in $\Psi$ and hence $\Psi$ acts in a natural way on
$\Gamma_3$-orbits. Since $\Psi$ has the orbit $\Psi(\alpha_7)$ of
size $81$, $\Gamma_3$ has at least three orbits of size $27$.
Since $R_1$ and $R_2$ contain at most one and two orbits of size
$27$ respectively, we find that $\Gamma_3$ has exactly $3$ orbits
of size $27$ and their union is necessarily $\Sigma^+\setminus
\{\Sigma_1^+\cup \Sigma_2^+\}$. It follows that for each
$\beta\in\Sigma^+\setminus \{\Sigma_1^+\cup \Sigma_2^+\}$ we have
${\rm Stab}_{\Psi}(\beta)\subset \Gamma_3$ and this implies
$\langle\,a,b,c\,\rangle\subset\Gamma_3$. But then the orbit
$\Gamma_3(\alpha_4)$ contains at least $27$ elements giving thus
the fourth orbit of size $27$ -- a contradiction.
\end{proof}

We are ready to finish the proof. Since $|\Gamma_3|\le 27$, the
$\Gamma_3$-orbits of roots in $R_2$ have sizes divisible by $9$ or
$27$. Since $|R_2|=72$, there is at least one $\beta\in R_2$ such
that the size of its $\Gamma_3$-orbit is not divisible by $27$. As
in \S\,\ref{36}, consider $\Gamma'={\rm Stab}_{\Gamma}(\beta)$ and
let $E_1\subset E$ be the subfield corresponding to $\Gamma'$. If
the $3$-component of $R_{G_0}(\xi)$ is trivial over $E_1$, then
the same argument as in Lemma~\ref{3component} completes the
proof. Thus we may assume without loss of generality that
$|\Gamma_3|=27$, and that for each root $\beta\in R_2$, whose
$\Gamma_3$-orbit has size divisible by $9$ but not by $27$, the
$3$-component of $R_{G_0}(\xi)$ is nontrivial over the
corresponding field $E_1$.

Note that in this possible ``bad" case we have that ${\rm
Stab}_{\Gamma_3}(\beta)$, being a group of order $3$, is a $3$-Sylow
subgroup of $\Gamma'$. By arguing as in Lemma~\ref{4roots}, we may
therefore additionally assume that a nontrivial $x\in{\rm
Stab}_{\Gamma_3}(\beta)$ has at most $3$ invariant positive roots
with respect to the canonical action of $\Gamma_3\subset W$ on
$\Sigma$. In particular, this assumption implies that for each root
in $R_2\cap (\Sigma^+\setminus \{\Sigma_1^+\cup \Sigma_2^+\})$ its
$\Gamma_3$-orbit has size $27$,  hence that $\beta$ with the above
property is in $\Sigma_1^+$. We also have $e\not\in \Gamma_3$, since
each root in $\Sigma_1$ is stable with respect to $e$.

Consider the canonical morphism $$f:\Psi\to
\Psi/\langle\,e\,\rangle\simeq\Psi_1= \langle\,a,b,c,d\,\rangle.$$
Since $e\not\in \Gamma_3$, the image $f(\Gamma_3)$ has order $27$,
hence it is a normal subgroup in $\Psi_1$. As in
Lemma~\ref{order}, we find that $\Psi_1$ acts on $\Gamma_3$-orbits
of $\Gamma_3$ on $\Sigma_1^+$. Thus
$\Sigma_1^+\setminus\Sigma_3^+$, being a unique $\Psi_1$-orbit of
size $27$, is a disjoint union of $3$ $\Gamma_3$-orbits of size
$9$. Then for each root $\beta\in\Sigma_1^+\setminus\Sigma_3^+$,
${\rm Stab}_{\Psi_1}(\beta)$, being a group of order $3$, is
contained in $\Gamma_3$. However it is easy to see that all such
stabilizers generate $\Psi_2$, whose order is $3^4$. This
contradicts our assumption that $|\Gamma_3|=27$.

\medskip

\begin{center}
{\large{\it   Acknowledgement}}
\end{center}

Work on the present paper started in 2002 at the Centre Bernoulli,
E.P.F.L., Lausanne, and continued at Bielefeld University in 2003.
The author thanks both centres for their hospitality and their
stimulating atmosphere. The author would also like to thank
Forschergruppe ``Spektrale Analysis, asymptotische Verteilungen und
stochastische Dynamik'' for its support.



\begin{thebibliography}{10}



%\bibitem{Borel91}
%A.~Borel, \emph{Linear algebraic groups}, second ed., Graduate
%Texts in Mathematics, vol. 126, Springer-Verlag, New York, 1991.



%\bibitem{BorelTits}
%Armand Borel and Jacques Tits, \emph{Groupes r\'eductifs}, Inst.\
%Hautes \'Etudes Sci.\ Publ.\ Math.\ \textbf{27} (1965), 55--150.



\bibitem{bourbaki}
N.~Bourbaki, \emph{\'{E}l\'ements de math\'ematique}, Masson,
Paris, 1981, Groupes et alg\`ebres de Lie, Chapitres 4, 5 et 6.

\bibitem{ch}
V. Chernousov, The kernel of the Rost invariant, Serre's
Conjecture II and the Hasse principle for quasi-split groups
$^{3,6}D_4,E_6,E_7$, Math. Ann. {\bf 326} (2003), 297--330.

\bibitem{gar}
R. S. Garibaldi, The Rost invariant has trivial kernel for
quasi-split groups of low rank, Comment. Math. Helv. {\bf 76}
(2001), no. 4, 684--711.

\bibitem{gms} S. Garibaldi, A. Merkurjev, J.-P. Serre,
Cohomological invariants in Galois Cohomology, American
Mathematical Society, Providence, RI, 2003.

%\bibitem{gro}
%A. Grothendieck, Torsion homologique et sections rationnelles,
%Expos\'e {\bf 5} in Anneaux de Chow et applications, S\'eminaire
%C.\~Chevalley 1958, Paris (1958).

%\bibitem{Steinberg67}
%R.~Steinberg, \emph{Lectures on Chevalley groups}, Yale University
%1967

\bibitem{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} (1966), 33--62.


\bibitem{tits}
J.~Tits, Sur les degr\'es des extensions de corps d\'eployant les
groupes alg\'ebriques simples, C. R. Acad. Sci. {\bf 315} (1992),
1131--1138.

\bibitem{Tot1} B. Totaro, Splitting fields for $E_{8}$-torsors, Duke
Math. J. {\bf 121} (2004), no. 3, 425--455.

\bibitem{Tot2} B. Totaro, The torsion index of $E_8$ and other
groups, Duke Math.J. {\bf 129} (2005), 219--248

% preprint (2004).


\end{thebibliography}





\end{document}