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\title[Zero cycles] % colontitle
{Connectedness of classes of fields and zero cycles on projective
homogeneous varieties}


\date{}

\author
[V. Chernousov] {Vladimir Chernousov}


\author
[A. Merkurjev] {Alexander Merkurjev}


\address
{Department of Mathematical Sciences, University of Alberta,
Edmonton, Alberta, Canada T6G 2G1}

\address{Department of Mathematics,
         University of California,
         Los Angeles, CA 90095-1555}

\email {chernous@math.ualberta.ca}

\email {merkurev@math.ucla.edu}

\thanks{The first author was supported in part by Canada Research Chairs
Program and NSERC Canada Grant G121210944.}

\thanks{The second author was supported in part by NSF Grant \#0355166.}

\subjclass[2000]{Primary 20G15, 14M15, 14M17}

\begin{document}
\maketitle

\section{Introduction}




Let $X$ be a proper scheme of finite type over a field $F$. A {\it
zero-cycle} on $X$ is the formal sum $\sum n_i[x_i]$ where $n_i\in
\mathbb{Z}$ and $x_i$ are closed (zero-dimensional) points of the
variety $X$. The factor group of the group of zero-cycles modulo
rational equivalence is called {\it Chow group of dimension zero}
and is denoted by $\CH_0(X)$. The assignment $x\mapsto \deg(x)$
extends to the \emph{degree homomorphism}
\[
\deg:\CH_0(X)\ra\Z.
\]
%Note that $\deg$ is the push-forward
%homomorphism for the structure morphism $X\ra\Spec F$.
The image of $\deg$ coincides with $n(X)\Z$ where $n(X)$ is the
greatest common divisor of the degrees $\deg(x)=[F(x):F]$ over all
closed points $x\in X$. We denote the kernel of $\deg$ by
$\overline{\CH}_0(X)$.

The main purpose of the paper is to present a characteristic free
uniform method of computing the group $\overline{\CH}_0(X)$ for
projective homogeneous varieties of semisimple algebraic groups. The
method is based on the idea of parametrization of fields over which
$X$ has a point. We illustrate the method by proving  that in many
cases the group $\overline{\CH}_0(X)$ is trivial and give examples
of varieties when this group is not trivial. The main results of the
paper can be summarized as follows.

%In the present paper is devoted we study studying the group $\cat
%C(X)$ for projective homogeneous varieties of semisimple algebraic
%groups over a given field $F$. We prove that in many cases the group
%$\overline{\CH}_0(X)$ is trivial and give examples of varieties when this
%group is not trivial. The main results of the paper can be
%summarized as follows.

Let $X$ be a scheme over $F$. We denote by $\cA(X)$ the class of all
field extensions $L/F$ such that $X(L)\not=\emptyset$. We say that
two fields $L_0,L_1\in\cA(X)$ of the same degree $n$ over $F$ are
simply $X$-equivalent if they are members of a continuous family of
fields $L_t\in \cA(X)$, $t\in\A^1$, of degree $n$ over $F$ (for
precise definition see Section~\ref{mainequivalence}). We say that
$L$ and $L'$ are $X$-equivalent if they can be connected by a chain
of fields $L_0=L, L_1,\ldots, L_r=L'$ such that $L_i$ and $L_{i+1}$
are simply $X$-equivalent for $i=0,\ldots,r-1$. Furthermore we say
that the class $\cA(X)$ is \emph{connected} if every two fields in
$\cA(X_E)$ of degree $n(X_E)$ over $E$ are $X_E$-equivalent over any
special extension $E/F$ (see Section~\ref{mainequivalence}).

Our first result (Theorem \ref{maincor}) asserts that if $X$ is an
arbitrary proper scheme over $F$ such that the class $\cA(X)$ is
connected and $\overline{\CH}_0(X_L)=0$ for any field
$L\in\cA(X)$, then $\overline{\CH}_0(X)=0$. Note that the
condition $\overline{\CH}_0(X_L)=0$ always holds for projective
homogeneous varieties $X$. Thus the connectedness of the class
$\cA(X)$ for such $X$ implies $\overline{\CH}_0(X)=0$.

We prove the connectedness of $\cA(X)$ for various classes of
projective homogeneous varieties. These include: Severi-Brauer
varieties, certain generalized Severi-Brauer varieties, quadrics,
involution varieties, projective homogeneous varieties related to
groups of exceptional types $^{3,6}{\cat D}_4,{\cat G}_2,{\cat
F}_4,{^{1,2}}{\cat E}_6, {\cat E}_7$ with trivial Tits algebras. As
an application we get that $\overline{\CH}_0(X)=0$, i.e. the Chow
group $\CH_0(X)$ is infinite cyclic for all above mentioned
varieties. We borrow from \cite{Krashen06} the idea of using
symplectic involutions in the case of generalized Severi-Brauer
varieties and involution varieties.

Some of our results were known before under certain restrictions on
characteristic of the ground field $F$. Triviality of
$\overline{\CH}_0(X)$ in the case of the Severi-Brauer variety
$X=SB(A)$ was proven by Panin in \cite{Panin84} if $\ch F$ does not
divide $\ind(A)$. Quadrics over fields of characteristic $\not=2$
were considered by Swan~\cite{swan} and Karpenko~\cite{karpenko}.
The cases of certain generalized Severi-Brauer varieties and
involution varieties were treated by Krashen in \cite{Krashen06}
under the assumption $\ch F=0$. Involution varieties of algebras of
index at most 2 were considered in \cite{Merkurjev95b} by the second
author under assumption $\ch F\neq 2$. Petrov, Semenov and
Zainoulline~\cite{psz}, independently, have recently shown that
$\overline{\CH}_0(X)=0$ for projective homogeneous varieties $X$
related to groups of types ${\cat G}_2,{\cat F}_2, {^1{\cat E}_6}$
over fields of characteristic $0$ with trivial Tits algebras.

The notion of $X$-equivalence used in the paper as the main
technical tool is formulated in terms of field extensions and
discrete valuations, so that we avoid symmetric power constructions
used in \cite{Krashen06} to describe closed points. Flexibility of
the notion of $X$-equivalence allows us not to impose any
characteristic restrictions on $F$ (except for the trialitarian
${\cat D}_4$ and ${\cat E}_6,\,{\cat E}_7$ where we assume that
characteristic is $\not=2,3$). Another advantage of our method is
its transparency and shortness. Even for those varieties where the
result was already known our proofs are simpler.

%at least in the case of classical groups.
%Note that triviality of $\overline{\CH}_0(X)$ in the case of
%the Severi-Brauer variety $X=SB(A)$ was proven by Panin in
%\cite{Panin84} if $\ch F$ does not divide $\ind(A)$ and in the cases
%of certain generalized Severi-Brauer varieties and involution
%varieties by Krashen in \cite{Krashen06} under the assumption $\ch
%F=0$.

Finally we remark that most likely our results on algebraic groups
of exceptional types are close to optimal. It looks hopeless to
weaken restrictions on Tits algebras and prove that $\CH_0(X)$ is
infinite cyclic for larger classes of projective homogeneous
varieties of exceptional groups.

At the end of the paper we give two examples of projective
homogeneous varieties $X$ with $\overline{\CH}_0(X)\not=0$ related
to algebraic groups of types ${\cat A}_1+{\cat A}_1+ {\cat A}_1$ and
${\cat B}_3$ with nontrivial Tits algebras. Note that our first
result is minimal possible since $\overline{\CH}_0(X)=0$ for all
projective homogeneous varieties $X$ of dimension at most $2$ (see
Proposition~\ref{dim2again}).

% and this suggests that the restriction of
%considering exceptional groups with trivial Tits algebras seems to
%be essential.





\section{Preliminary facts on algebraic groups}

\subsection{Parabolic subgroups}
Let $G$ be a simple simply connected algebraic group over a field
$F$. Fix a maximal torus $T\subset G$ over $F$ and a basis $\Delta$
of the root system $\Sigma=\Sigma(G,T)$ of $G$ with respect to $T$.
Recall that for each subset $S\subset\Delta$ one can associate the
parabolic subgroup $P_S$ in $G$, whose semisimple part is generated
by the corresponding root subgroups $U_{\pm\alpha}$ of $G$ for all
roots $\alpha\in S$. It is defined over a separable closure
$F_{sep}$ of $F$ and is called the \emph{standard parabolic
subgroup} of type $S$. Every parabolic subgroup $P$ in $G$ over
$F_{sep}$ is conjugate to a unique standard parabolic subgroup
$P_S$. We say that \emph{$P$ is of type $S$}.

Let $X_S$ be the variety of all parabolic subgroups of $G$ of type
$S$. If $S$ is stable with respect to the so-called $*$-action of
the Galois group ${\rm Gal}\,(F_{sep}/F)$ (see~\cite{t66}), the
variety $X_S$ is defined over $F$. It has a rational point if and
only if $G$ contains an $F$-defined parabolic subgroup of type $S$.
If $X_S(F)\not=\emptyset$, then we say that $X_S$ is
\emph{isotropic}.

\subsection{Tits indices}
Recall~\cite{t66} that given $G$ one can associate a geometric
picture called the {\it Tits index} of $G$. It consists of the
corresponding Dynkin diagram of $G$ with some vertices circled.
The set $S_0$ of uncircled vertices corresponds to an $F$-subgroup
of $G$ called a \emph{semisimple anisotropic kernel}. This
subgroup is the semisimple part of the centralizer $C_G(T_1)$
where $T_1\subset G$ is a maximal $F$-split subtorus in $G$. All
maximal $F$-split subtori are conjugate over $F$. Hence a
semisimple anisotropic kernel of $G$ is defined uniquely up to
conjugation.

All vertices in Tits index of $G$ are circled if and only if $G$
is quasi-split over $F$ and none of them is circled if and only if
$G$ is $F$-anisotropic. The last occurs if and only if $G$
contains no nontrivial $F$-split subtorus.

Let $S$ be stable with respect to the $*$-action of ${\rm
Gal}\,(F_{sep}/F)$. Then the variety $X_S$ of parabolic subgroups
in $G$ of type $S$ is defined over $F$ and it has a rational point
if and only if $S$ contains the subset $S_0$ of all uncircled
vertices. Hence an $F$-defined parabolic subgroup $P\subset G$ of
type $S$ is minimal if and only if $S=S_0$; in particular all
minimal parabolic subgroups in $G$ have the same type and they are
conjugate over $F$.

\subsection{Basic types}

We say that $S$ is a \emph{basic} type for $G$ if there is a form
$H$ of $G$ over a field extension $L/F$ such that $H$ is
$L$-isotropic and its minimal parabolic subgroups have type $S$.
%(recall~\cite{t66} that all minimal $L$-defined parabolic
%subgroups of $H$ are conjugate over $L$ and hence they have the
%same type).
In other words, a type $S$ is basic if in the classification
table~\cite{t66} of Tits indices there is one whose set of all
uncircled vertices coincides with $S$.

\section{Strongly inner forms}

One knows that $G$ is a form of a quasi-split simply connected
group $G_0$, i.e.  there exists a cocycle $\xi\in Z^1(F,{\rm
Aut}\,G_0)$ such that $G$ is isomorphic over $F$ to the twisted
group $^{\xi}G_0$. One says that $G$ is of the same \emph{inner
type} as $G_0$ or an \emph{inner form} of $G_0$ if the class
$[\xi]$ is contained in ${\rm Im}\,[H^1(F,{\overline G}_0)\to
H^1(F,{\rm Aut}\,G_0)]$ where ${\overline G}_0=G_0/Z$ is the
corresponding adjoint group and $Z\subset G_0$ is the center of
$G_0$. Otherwise $G$ is an \emph{outer form} of $G_0$. One says
that $G$ is a \emph{strongly inner form} of $G_0$ if $[\xi]$ is
contained in ${\rm Im}\,[H^1(F,G_0)\to H^1(F,{\rm Aut}\,G_0)]$.
%More generally, one says that $G$ is a strongly inner form of an
%$F$-group $H$ if $G\simeq{^{\xi}}H$ where $\xi\in Z^1(F,H)$.

\begin{remark} The notion of strongly inner forms was
first introduced by Tits in~\cite{tits90}.
\end{remark}

For future reference we need some structure facts on strongly
inner forms of a quasi-split simple simply connected group $G_0$.
In what follows $\xi\in Z^1(F,G_0)$ and $G={^{\xi}}G_0$ is the
corresponding twisted group.

\subsection{ Quasi-splitness criterion}

\begin{lemma}\label{fact1}
$G$ is quasi-split if and only if $[\xi]=1$.
\end{lemma}
\begin{proof} Let $G$ be quasi-split. Since
$H^1(F,{\rm Aut}\,G_0)$ classifies all forms of $G_0$,
$$[\xi]\in {\rm Ker}\,[H^1(F,G_0)\to H^1(F,{\rm Aut}\,G_0)].
$$
The group ${\rm Aut}\,G_0$ is the semi-direct product of the
adjoint group ${\overline G}_0$ and the automorphism group ${\rm
Aut}\,D$ of the Dynkin diagram $D$ of $G_0$. It follows that ${\rm
Aut}\,G_0(F)\to {\rm Aut}\,D(F)$ is surjective, hence the mapping
$H^1(F,{\overline G_0})\to H^1(F,{\rm Aut}\,G_0)$ has trivial
kernel. This implies that $[\xi]$ is contained in the kernel of $
H^1(F,G_0)\to H^1(F,{\overline G}_0)$. Consider now an exact
sequence $1\to Z\to G_0\to{\overline G}_0\to 1$. It gives rise to
an exact sequence
$$
H^1(F,Z)\to H^1(F,G_0)\to H^1(F,{\overline G_0}).
$$
It follows that $\xi$ up to equivalence takes values in $Z$. As
$G_0$ is simply connected and quasi-split, it contains a maximal
quasi-split torus $T\subset G_0$. Since $Z\subset T$ and
$H^1(F,T)=1$, by Hilbert Theorem $90$, the result follows.
\end{proof}

\subsection{Application of Steinberg's Theorem}

Let $C_1\subset G$ be a maximal $F$-split torus. The centralizer
$C_G(C_1)$ is a reductive group over $F$ whose central torus $C$
contains $C_1$. Note that if $G_0$ is split, we have $C=C_1$.
However in general case $C\not=C_1$. The semisimple part
$H=[C_G(C_1),C_G(C_1)]$ of $C_G(C_1)$ is an $F$-anisotropic simply
connected group, not necessarily simple. Its Tits index coincides
with the subindex of $G$ generated by uncircled vertices. The
group $C_G(C_1)$ is an almost direct product of $H$ and $C$, i.e
$H\cap C$ is a finite central subgroup of $H$ and $C_G(S_1)=H\cdot
C$.
\begin{theorem}\label{fact2}
Assume that $H^1(F,C/H\cap C)=1$. Then $H$ is a strongly inner
form of a quasi-split simply connected group $H_0$ over $F$.
\end{theorem}
\begin{proof} Let $C_2\subset H$ be a maximal torus over $F$ and
let $T=C\cdot C_2$. By dimension count, $T$ is a maximal torus in
$G$ over $F$. It easily follows from arguments contained
in~\cite[\S\,10]{[St65]} (see also~\cite[Propositions $6.18, 6.19$,
p. 338--339]{[PR]}) that there is an $F$-embedding $\phi: T\to G_0$
such that the class $[\xi]$ is contained in ${\rm Im}\,[H^1(F,T)\to
H^1(F,G_0)]$. Since $C\cap H=C\cap C_2$ and $H^1(F,C/C\cap H)=1$,
the mapping $H^1(F,C_2)\to H^1(F,T)$ is surjective. Thus, up to
equivalence, we may assume that $\xi$ takes values in $C_2$.

The centralizer $C_{G_0}(C_1)$ is a reductive $F$-subgroup in $G_0$
and from the construction of $\phi$ (see the proof of
in~\cite[Proposition~$6.18$, p. 339]{[PR]}) it follows that
$C_{G_0}(C_1)$ and $C_G(C_1)$ are isomorphic over $F_{sep}$. Hence
their semisimple parts $H_0$ and $H$ are isomorphic over $F_{sep}$.
As all maximal $F$-split tori in $G_0$ are conjugate, $C_{G_0}(C_1)$
(and hence $H_0=[C_{G_0}(C_1),C_{G_0}(C_1)]$) is a quasi-split
group. By construction $H_0$ contains $C_2$, hence $\xi$ can be
viewed as an element in $Z^1(F,H_0)$ and then clearly we have
$H\simeq {^{\xi}}H_0$.
\end{proof}

\subsection{Groups isotropic over a quadratic extension}

Assume that $F$ is infinite and perfect. Let $L/F$ be a quadratic
extension making $G$ isotropic. We denote the generator of ${\rm Gal}(L/F)$
by $\sigma$. If $P\subset G$ is an $L$-defined
standard parabolic subgroup of type $S\subset\Delta$ we denote by
$P^-$ the opposite parabolic subgroup (it is generated by root subgroups
$U_{-\alpha},\, \alpha\in \Delta\setminus S,$ and
$U_{\pm\alpha},\, \alpha\in S$). The class $\mathcal{P}$ of
parabolic subgroups conjugate to $P$ is called self-opposite if it
contains $P^-$. Clearly, $P\cap P^-$ is a reductive part of $P$.

\begin{lemma}\label{opposite} If $\mathcal{P}$ is self-opposite, then it
contains an $L$-parabolic subgroup $Q\in\mathcal{P}$ such that
$Q\cap \sigma(Q)$ is a reductive part of $Q$.
\end{lemma}
\begin{proof} See \cite[Lemma $17'$, p. 383]{[PR]}.
\end{proof}



%\section{Equivalences of field extensions}

\section{Comparison of two homogeneous varieties}
%\section{Reduction to basic types}\label{reductionadmissible}

%\subsection{Projective homogeneous varieties}




%Let $X$ be a proper scheme over $F$. The assignment $x\mapsto
%\deg(x)$ extends to the \emph{degree homomorphism}
%\[
%\deg:\CH_0(X)\ra\Z.
%\]
%Note that $\deg$ is the push-forward homomorphism for the structure
%morphism $X\ra\Spec F$.

%Write $n(X)$ for the greatest common divisor of the degrees
%$\deg(x)=[F(x):F]$ over all closed points $x\in X$. The image of
%$\deg$ coincides with $n(X)\Z$. We denote the kernel of $\deg$ by
%$\overline{\CH}_0(X)$.

%In the present paper we prove that in certain cases the group $\cat
%C(X)$ is trivial and give examples of varieties when this group is
%not trivial. All the varieties we consider in the paper are
%projective homogeneous under an action of a semi-simple algebraic
%group. A projective homogeneous variety is called \emph{isotropic}
%if it has a rational point.




\begin{proposition}\label{twohomo}
Let $X$ be a scheme over $F$ and $Y$ be a projective homogeneous
variety over $F$ such that $Y_{F(x)}$ is isotropic for every $x\in
X$. Then the projection $f:X\times Y\ra X$ induces an isomorphism
$f_*:\CH_0(X\times Y)\ra \CH_0(X)$.
\end{proposition}

\begin{proof}
Consider the spectral sequence  \cite[\S 8]{Rost98a}
\[
E^1_{p,q}=\coprod_{x\in X_{(p)}} A_q(Y_{F(x)}, K_p)\Rightarrow
A_{p+q}(X\times Y,K_0)
\]
for the projection $f$, where $A_*(X\times Y,K_*)$ are the
$K$-homology groups defined in \cite{Rost98a}. It gives an exact
sequence
\[
E^1_{1,0}\xra{\partial} E^1_{0,0}\ra \CH_0(X\times Y)\ra 0.
\]
We shall identify the terms $E^1_{n,0}$. It is proven in \cite[Th.
7.5]{CGM05} that the Chow motive of the isotropic variety $Y_{F(x)}$
is isomorphic over $F(x)$ to a direct sum of the motive $\Z$ and
twisted motives of the form $Z(r)$ for some projective homogeneous
variety $Z$ and $r>0$. By the dimension consideration,
$A_0(Z(r),K_n)=A_{-r}(Z,K_{n+r})=0$, hence
\[
A_0(Y_{F(x)}, K_n)\simeq A_0(\Spec F(x),K_n)=K_n F(x)
\]
where the isomorphism is the push-forward map with respect to the
structure morphism $Y_{F(x)}\ra \Spec F(x)$. Thus
\[
E^1_{n,0}=\coprod_{x\in X_{(0)}}K_n F(x),
\]
and therefore $\CH_0(X\times Y)=\Coker(\partial)\simeq \CH_0(X)$.
\end{proof}

\begin{corollary}\label{equivalence}
Let $X$ and $Y$ be two projective homogeneous varieties over $F$
such that each of them is isotropic over residue field of every
point of another. Then $\CH_0(X)\simeq \CH_0(Y)$ and
$\overline{\CH}_0(X)\simeq \overline{\CH}_0(Y)$.
\end{corollary}

Taking $Y=\Spec F$ in Corollary \ref{equivalence} we get

\begin{corollary}\label{equiv2}
Let $X$ be an isotropic projective homogeneous variety. Then
$\overline{\CH}_0(X)=0$.
\end{corollary}

Let $X$ be a variety over a field $F$. Consider the
Brown-Gersten-Quillen spectral sequence
\begin{equation}\label{bgq}
E_2^{p,q}=A^p(X,K_{-q})\Rightarrow K_{-p-q}(X),
\end{equation}
where $A^p(X,K_{-q})$ is the $K$-cohomology group and $K_n(X)$ is
the $K$-group of $X$ (see \cite[\S 7]{Quillen73}). If $X$ is
projective, the group $E_2^{0,-1}=A^0(X,K_1)$ of regular invertible
functions on $X$ is equal to $K_1(F)=F\m$. Since the composite
$K_1(F)\ra K_1(X)\xra{e} A^0(X,K_1)$ is the identity, the edge
homomorphism $e$ is surjective. In particular, all the differentials
going from $E_*^{0,-1}$ are trivial. If $d=\dim X\leq 2$, all
differentials coming to $E_*^{d,-d}$ are trivial, in particular, the
edge homomorphism $e':\CH_0(X)\ra K_0(X)$ is injective.

If $X$ is projective, the degree homomorphism for $\CH_0(X)$ is the
composition
\[
\CH_0(X)\xra{e'} K_0(X)\xra{f_*} K_0(\Spec F)=\Z,
\]
where the $f_*$ is the push-forward homomorphism with respect to the
structure morphism $f:X\ra\Spec(F)$. Therefore, $\Ker(e')\subset
\overline{\CH}_0(X)$. If $X$ is a projective homogeneous variety,
the group $K_0(X)$ is torsion free (see \cite{Panin94}), hence
$\overline{\CH}_0(X)\subset \Ker(e')$ since $\overline{\CH}_0(X)$ is
a torsion group. Thus $\overline{\CH}_0(X)= \Ker(e')$. In particular
we have proven

\begin{proposition}\label{dim2again}
Let $X$ be a projective homogeneous variety of dimension at most
$2$. Then $\overline{\CH}_0(X)=0$.
\end{proposition}


%We shall need the following

%\begin{lemma}\label{sameresidue}
%Let $X$ be a proper scheme over $F$ such that $\overline{\CH}_0(X_L)=0$ for
%any field $L\in \cA(X)$ (e.g., $X$ is a projective homogeneous
%variety by Corollary \ref{equiv2}). Let $x$ and $x'$ be two closed
%points of $X$ such that the residue fields $F(x)$ and $F(x')$ are
%isomorphic over $F$. Then $[x]=[x']$ in $\CH_0(X)$.
%\end{lemma}

%\begin{proof}
%Let $L=F(x)$. By assumption there are rational points $y$ and $y'$
%of $X_L$ over $x$ and $x'$ respectively. Since $[y]=[y']$ in
%$\CH_0(X_L)$, taking the push-forward homomorphism with respect to
%$X_L\ra X$ we get $[x]=f_*([y])=f_*([y'])=[x']$.
%\end{proof}




\subsection{Reduction to a basic
type}\label{reductionadmissible}


Let $G$ be a simple simply connected group over $F$. We fix a
maximal torus $T\subset G$ over $F$, the corresponding root system
$\Sigma=\Sigma(G,T)$, a basis $\Delta\subset \Sigma$ and a subset
$S\subset\Delta$. Consider the variety $X_S$ of parabolic
subgroups in $G$ of type $S$. Let $L=F(X_S)$ be the field of
rational functions of $X_S$. Since $X_S(L)\not=\emptyset$, there
is an $L$-defined parabolic subgroup $P$ in $G$ of type $S$. It
contains a minimal parabolic subgroup $P'\subset P$ of $G$ over
$L$ whose type $S'$ is a subset in $S$.

\begin{proposition}\label{admissible}
There are canonical isomorphisms $\CH_0(X_S)\simeq\CH_0(X_{S'})$ and
$\overline{\CH}_0(X_S)\simeq \overline{\CH}_0(X_{S'})$.
\end{proposition}
\begin{proof}
We show that $X_S$ and $X_{S'}$ satisfy conditions of Corollary
\ref{equivalence}. As $S'\subset S$, there is a natural map
$f:X_{S'}\to X_S$ over $F$. Therefore $X_S$ is isotropic over
$F(x')$ for every $x'\in X_{S'}$.

Conversely, let $x\in X_S$ and set $E=F(x)$. The set $S'$ consists
of all uncircled vertices of the Tits index of $G$ over $F(X)$.
Therefore $S'$ contains all uncircled vertices of the Tits
index of $G$ over $E(X)$. Since $E(X)/E$ purely transcendental,
and the Tits index does not change under such extensions, the set
$S'$ contains all uncircled vertices of the Tits index of $G$ over
$E$, i.e., $X_{S'}(E)\neq\emptyset$.
\end{proof}

\begin{remark}\label{innerform} The proposition allows us to
consider basic subsets $S\subset\Delta$ only while computing
$\CH_0(X_S)$. In fact we proved even more: it is sufficient to
compute $\CH_0(X_S)$ for those basic subsets $S$ that appear as
sets of uncircled vertices of the Tits indices of $G$ over all
field extensions $L/F$. Thus we can narrow the set of basic types
by considering Tits indices of strongly inner forms of $G_0$ only.
\end{remark}




\section{Special fields}

Let $p$ be a prime integer. A field $F$ is called \emph{$p$-special}
if the degree of every finite field extension of $F$ is a power of
$p$. We say that $F$ is \emph{special} if $F$ is $p$-special for
some $p$. The following properties of special fields follow directly
from the definition.
\begin{enumerate}
    \item An algebraic extension of a $p$-special field is
    $p$-special.
    \item For every field $F$ and every prime $p$ there is an algebraic field
extension $F_p/F$ such that the field $F_p$ is $p$-special and the
degree of any finite sub-extension $L/F$ of $F_p/F$ is prime to $p$.
\end{enumerate}



If $X$ is a scheme over a $p$-special field $F$, then $n(X)$ is a
power of $p$ and there is a closed point $x\in X$ such that $\deg
x=n(X)$.

The following statement reduces the problem of computing $\CH_0(X)$
to the case of schemes
over $p$-special fields.

\begin{lemma}\label{special}
Let $X$ be a proper scheme over $F$. Assume that for any $p$-special
field extension $L/F$, the group $\overline{\CH}_0(X_L)$ is trivial.
Then $\overline{\CH}_0(X)=0$.
\end{lemma}

\begin{proof}
Let $a\in \overline{\CH}_0(X)$ and fix a prime integer $p$. As the
image of $a$ in $\overline{\CH}_0(X_{F_p})$ is trivial, $a$ is
trivial over some finite sub-extension $L/F$ in $F_p/F$. Applying
the push-forward homomorphism with respect to the morphism $X_L\ra
X$, we get $[L:F]\cdot a=0$. In particular, the order of $a$ is not
divisible by $p$. Since this holds for every prime $p$, we have
$a=0$.
\end{proof}

The following statement enables us to consider only closed points of
minimal degree.

\begin{lemma}\label{reduction}
Let $X$ be a proper scheme over a $p$-special field $F$. Assume that
for any finite field extension $L/F$, every two closed points of
$X_L$ of the smallest degree $n(X_L)$ are rationally equivalent.
Then $\overline{\CH}_0(X)=0$.
\end{lemma}

\begin{proof}
Let $x$ be a closed point of $X$ of the smallest degree $p^a=n(X)$
and let $y$ be a closed point of $X$ of degree $p^b$. We claim that
$[y]=p^{b-a}[x]$ in $\CH_0(X)$.

We shall prove the claim by induction on $p^b=\deg(y)$. We may
assume that $p^b>p^a$. Choose an intermediate field $F\subset
L\subset F(y)$ such  that $[L:F]=p$ and a point $y'\in X_L$ over $y$
of degree $p^{b-1}$. Note that $n(X_L)$ divides $n(X)$ and $n(X)$
divides $p\cdot n(X_L)$. Therefore we have the following two cases.

Case 1: $n(X_L)=n(X)=p^a$. Then there is a point $x'\in X_L$ over
$x$ of degree $p^a$. By the induction hypothesis applied to $X_L$,
$[y']=p^{b-1-a}[x']$ in $\CH_0(X_L)$. Taking the push-forward
homomorphism with respect to $f:X_L\ra X$ we get
\[
[y]=f_*([y'])=p^{b-1-a}f_*([x'])=p^{b-1-a}\cdot p\ [x]=p^{b-a}[x].
\]

Case 2: $n(X_L)=p^{a-1}$. Choose a closed point $z'\in X_L$ of
degree $p^{a-1}$ and denote by $z$ its image in $X$. The degree of
$z$ is $p^a$, in particular, $[z]=[x]$ in $\CH_0(X)$. By the
induction hypothesis applied to $X_L$, $[y']=p^{(b-1)-(a-1)}[z']$ in
$\CH_0(X_L)$. Therefore
\[
[y]=f_*([y'])=p^{(b-1)-(a-1)}f_*([z'])=p^{b-a}[z]=p^{b-a}[x].
\]
The claim is proven. It follows that every $0$-cycle $\alpha$ on
$X$ is equivalent to $m[x]$ for some $m\in\Z$. If $\deg \alpha=0$,
then $m=0$, i.e., $\alpha=0$ in $\CH_0(X)$.
\end{proof}


\section{Equivalences of field extensions}\label{mainequivalence}

\subsection{$X$-equivalence}


Let $F$ be a field and let $\cA$ be the class of all field
extensions of $F$. For any subclass $\cB\subset \cA$ and every field
$L\in\cA$ we write $\cB_L$ for the subclass of $\cB$ of all fields
containing $L$. Let $n(\cB)$ denote the greatest common divisor of
the degrees $[E:F]$ over all fields $E\in\cB$ of finite degree over
$F$. (We set $n(\cB)=0$ if $\cB$ has no finite degree extensions.)
In particular, $n(\cA)=1$. If $F$ is $p$-special then $n(\cB)$ is a
power of $p$ or $0$ and if $n(\cB)>0$ then there is a field
$E\in\cB$ such that $[E:F]=n(\cB)$.


For every $a\in F$, let $v_a$ be the discrete valuation of the
rational function field $F(t)$ over $F$ corresponding to the
irreducible polynomial $f(t)=t-a$.


Let $\cB\subset \cA$ be a subclass of field extensions of $F$. Two
fields $L$ and $L'$ in $\cB$ of the same degree $n$ over $F$ are
called \emph{simply $\cB$-equivalent} if there is a degree $n$
field extension $E/F(t)$ such that $E\in\cB$ and two discrete
valuations of $E$ over $v_0$ and $v_1$ with residue fields
isomorphic to $L$ and $L'$ respectively over $F$. Two fields $L$
and $L'$ in $\cB$ are called \emph{$\cB$-equivalent} if there is a
chain of fields $L_0=L, L_1,\dots, L_r=L'$ in $\cB$ such that
$L_i$ is simply $\cB$-equivalent to $L_{i+1}$ for all
$i=0,1,\dots, r-1$.

Note that if two fields $L$ and $L'$ in $\cB$ contain a common
subfield $K\in\cA$, i.e., $L,L'\in \cB_K$, then $L\sim_{\cB_K} L'$
implies $L\sim_{\cB} L'$.

\begin{example}\label{splitfields}
Let $X$ be a scheme of finite type over $F$. Denote by $\cA(X)$ the
class of all fields $E\in\cA$ such that $X(E)\neq\emptyset$. The
number $n\(\cA(X)\)$ coincides with $n(X)$, and $\cA(X)$-equivalent
fields will be called \emph{$X$-equivalent}.
\end{example}


\begin{example}
In the notation of Proposition \ref{admissible} the classes
$\cA(X_S)$ and $\cA(X_{S'})$ coincide.
\end{example}

We shall need the following

\begin{lemma}\label{sameresidue}
Let $X$ be a proper scheme of finite type over $F$ such that
$\overline{\CH}_0(X_L)=0$ for any field $L\in \cA(X)$ (e.g., $X$ is
a projective homogeneous variety by Corollary \ref{equiv2}). Let $x$
and $x'$ be two closed points of $X$ such that residue fields $F(x)$
and $F(x')$ are isomorphic over $F$. Then $[x]=[x']$ in $\CH_0(X)$.
\end{lemma}

\begin{proof}
Let $L=F(x)$. By assumption there are rational points $y$ and $y'$
of $X_L$ over $x$ and $x'$ respectively. As $[y]=[y']$ in
$\CH_0(X_L)$, taking the push-forward homomorphism with respect to
$X_L\ra X$ we get $[x]=f_*([y])=f_*([y'])=[x']$.
\end{proof}




\begin{lemma}\label{equivalent}
Let $X$ be a proper scheme of finite type over $F$ such that
$\overline{\CH}_0(X_L)=0$ for any field $L\in\cA(X)$. Let $x$ and
$x'$ be two rational points of $X$ of degree $n(X)$ such that
residue fields $F(x)$ and $F(x')$ are $X$-equivalent. Then
$[x]=[x']$ in $\CH_0(X)$.
\end{lemma}


\begin{proof}
We may assume that $F(x)$ and $F(x')$ are simply $X$-equivalent. Let
$E$ be the field extension of $F(t)$ of degree $n=n(X)$ in the
definition of the simple equivalence between $F(x)$ and $F(x')$. We
have the dominant morphism $\Spec E\ra \Spec F(t)\ra \P_F^1$. Since
$E\in\cA(X)$, there is a morphism $\Spec E\ra X$ over $F$. Denote by
$C$ the closure of the image of the diagonal morphism $\Spec E\ra
X\times \P_F^1$. Thus $C$ is an integral proper curve with a
surjective morphism $f:C\ra \P_F^1$. By assumption there are two
discrete valuations $v$ and $v'$ of $E$ over the valuations $v_0$
and $v_1$ of $F(t)$ respectively. Let $v$ and $v'$ dominate points
$c$ and $c'$ on $C$ respectively, so that $f(c)=0$ and $f(c')=1$.
Note that the residue field of $c$ is isomorphic to a subfield of
the residue field of $v$, which in its turn is isomorphic to the
field $F(x)$ of degree $n$ over $F$. On the other hand, $C\subset
X\times \P_F^1$, hence $n(C)\geq n(X)=n$. Therefore, $c$ is a closed
point of degree $n$, $f\inv(0)=\Spec F(c)$ and $F(c)\simeq F(x)$.
Similarly, $f\inv(1)=\Spec F(c')$ and $F(c')\simeq F(x')$.

The morphism $f$ is flat, hence $[c]=f^*([0])=f^*([1])=[c']$ in
$\CH_0(C)$. Consider the projection $g:C\ra X$. Let $z=g(c)$ and
$z'=g(c')$. As $F(z)$ is isomorphic to a subfield of $F(c)$ and
$n(X)=n$, we have $F(z)\simeq F(c)\simeq F(x)$. Similarly
$F(z')\simeq F(c')\simeq F(x')$. Therefore,
$[z]=g_*([c])=g_*([c'])=[z']$. By Lemma \ref{sameresidue},
$[x]=[z]=[z']=[x']$.
\end{proof}

A class $\cB\subset\cA$ is called \emph{connected} if for any
special field $E\in\cA$ every two fields in $\cB_E$ of degree
$n(\cB_E)$ over $E$ are $\cB_E$-equivalent. For example, the whole
class $\cA$ is connected, and therefore for every $X$ with $X(F)\neq
0$ the class $\cA(X)=\cA$ is connected.


\begin{theorem}\label{maincor}
Let $X$ be a proper scheme of finite type over $F$ such that
$\overline{\CH}_0(X_L)=0$ for any field $L\in\cA(X)$. Suppose that
the class $\cA(X)$ is connected. Then $\overline{\CH}_0(X)=0$.
\end{theorem}

\begin{proof}
By Lemma \ref{special} we may assume that $F$ is $p$-special for
some prime $p$. It follows from Lemma \ref{equivalent} that for any
finite field extension $L/F$, every two closed point of $X_L$ of
degree $n(X_L)$ are rationally equivalent. By Lemma \ref{reduction},
$\overline{\CH}_0(X)=0$.
\end{proof}






\section{Severi-Brauer varieties}\label{severibrauer}

Let $A$ be a central simple algebra of degree $m$ over a field $F$
and let $X$ be the corresponding Severi-Brauer variety of right
ideals of $A$ of dimension $\deg(A)$ (see \cite[\S 1]{Book}). The
variety $X$ is projective homogeneous of the group $\gSL_1(A)$ of
type $S=\{\alpha_2,\ldots,\alpha_{m-1}\}$ where simple roots
$\alpha_1,\ldots,\alpha_{m-1}$ of the Dynkin diagram ${\cat
A}_{m-1}$ are numbered as follows:
$$
\begin{array}{c}
\begin{picture}(140,25)
\put(20,00){\line(1,0){40}}%
\put(70,00){\dots}%
\put(90,00){\line(1,0){20}}%
\put(20,0){\circle*{3}}%
\put(40,0){\circle*{3}}%
\put(110,0){\circle*{3}}%
%\put(230,0){\circle*{3}}%
\put(15,8){$\alpha_1$}%
\put(35,8){$\alpha_2$}%
\put(100,8){$\alpha_{m-1}$}%
\end{picture}
\end{array}
$$

\bigskip

\noindent Here $m-1$ is the rank of $\gSL_1(A)$. A field extension
$L/F$ belongs to $\cA(X)$ if and only if the algebra $A_L=A\tens_F
L$ is split. In particular $n(X)=\ind(A)$. If $A$ is split, i.e.,
$A=\End(V)$ for some vector space $V$, the Severi-Brauer variety $X$
is the projective space $\P(V)$.

Let $D$ be a (unique up to isomorphism) central division algebra
$F$-algebra Brauer equivalent to $A$. Every maximal subfield of $D$
has degree $n=n(X)$ over $F$ and it is a splitting field for $D$ and
$A$ and it belongs to $\cA(X)$. Conversely, every field in $\cA(X)$
of the smallest degree $n$ is isomorphic over $F$ to a maximal
subfield of $D$ \cite[\S 4.4]{Herstein94}.

For every element $d\in D$ denote by $\Prd_d=\Prd_d(s)\in F[s]$ the
reduced characteristic polynomial of $d$. The polynomial $\Prd_d$ is
monic of degree $n$ and $\Prd_d$ is irreducible if and only if $d$
generates a maximal subfield of $D$. In this case $F(d)\simeq
F[s]/(\Prd_d)$.

Recall that a finite field extension $L/F$ is called \emph{simple}
if $L$ is generated over $F$ by one element. In particular every
separable finite extension is simple.

\begin{lemma}\label{twosubfields}
Let $L\subset D$ be a simple maximal subfield and let $d\in D$. Then
there is a maximal subfield $K\subset D$ containing $d$ such that
$L\sim_X K$.
\end{lemma}

\begin{proof}
We have $L=F(a)$ for some $a\in D$. In particular the polynomial
$\Prd_a$ is irreducible. Consider the element $c=td+(1-t)a\in D[t]$
and its reduced characteristic polynomial $\Prd_c\in F[s,t]$. Note
that $\Prd_c(s,0)=\Prd_a$ and $\Prd_c(s,1)=\Prd_d$. In particular
$\Prd_c$ is irreducible.

Set $R=F[s,t]/(\Prd_c)$. Since $\Prd_c$ is irreducible, $R$ is a
domain. The quotient field $E$ of $R$ is the maximal subfield of
$D_{F(t)}$ generated by $c$ and therefore belongs to $\cA(X)$. Let
$Q$ be the principal ideal of $R$ generated by $t$. The factor ring
$R/Q$ is isomorphic to $F[s]/(\Prd_a)\simeq L$. In particular, $Q$
is a maximal ideal of $R$. As the localization $R_Q$ is a Noetherian
local domain of dimension one with principal maximal ideal, $R_Q$ is
a discrete valuation ring. The corresponding discrete valuation of
$E$ with residue field $L$ restricts to $v_0$ on $F(t)$.

Let $v$ be a discrete valuation of $E$ restricting to $v_1$. Since
$R$ is integral over $F[t]$, $R$ is contained in the valuation ring
$O$ of $v$. As $R/(1-t)=F[s]/(\Prd_d)$ and $\Prd_d$ is a power of
the minimal polynomial $h=h(s)$ of $d$, the intersection of $R$ with
the maximal ideal of $O$ is generated by $1-t$ and $h$. In
particular residue field $K$ of $O$ contains $F[s]/(h)=F(d)$.

On the other hand, since $X$ is proper, a morphism $\Spec E\ra X$
extends to $\Spec O\ra X$. Therefore $K$ contains $F(x)$ for some
$x\in X$. But $[K:F]\leq [E:F(t)]=n$, therefore $K=F(x)$ and
$[K:F]=n$. Thus $K$ is isomorphic to a maximal subfield of $D$
containing $d$ and $L\sim_X K$.
\end{proof}

\begin{theorem}\label{equiv}
Let $A$ be a central simple $F$-algebra and let $X$ be the
Severi-Brauer variety $SB(A)$. Then the class $\cA(X)$ is connected.
\end{theorem}

\begin{proof}
It is sufficient to show that every two maximal subfields of $D$
are $X$-equivalent. We proceed by induction on $\deg D$ and may
assume that $\deg (D)>1$. Choose a separable maximal subfield $L$
of $D$. It is sufficient to show that $L$ is $X$-equivalent to any
maximal subfield $E\subset D$. Choose an element $d\in E\setminus
F$. By Lemma \ref{twosubfields}, $L\sim_X K$ for some maximal
subfield $K$ containing $d$. Let $M=F(d)$ and consider the
centralizer $D'$ of $M$ in $D$. The fields $E$ and $K$ are maximal
subfields of $D'$. By the induction hypothesis, $K\sim_{X_M}E$ and
therefore, $K\sim_X E$. Finally, $L\sim_X K\sim_X E$.
\end{proof}

Theorem \ref{maincor} yields

\begin{corollary}\label{severi}{\rm (Cf., \cite{Panin84})}
Let $X$ be a Severi-Brauer variety. Then $\overline{\CH}_0(X)=0$.
\end{corollary}



\section{Generalized Severi-Brauer varieties}

Let $A$ be a central simple $F$-algebra.
% of even degree over $F$.
Denote by $X=SB(2,A)$ the generalized Severi-Brauer variety of
right ideals of $A$ of dimension $2\deg(A)$ (see \cite[\S
1]{Book}). It is a projective homogeneous variety of the algebraic
group $\gSL_1(A)$ of type
$S=\{\alpha_1,\alpha_3,\ldots,\alpha_m\}$ where simple roots are
numbered as in Section~\ref{severibrauer}. A field extension $L/F$
belongs to $\cA(X)$ if and only if $\ind A_L\leq 2$. If $A$ is
split, i.e., $A=\End(V)$ for some vector space $V$, the variety
$SB(2,A)$ is the Grassmannian variety of planes in $V$.

If index of $A$ is odd, then a field extension $L/F$ belongs to
$\cA(X)$ if and only if $L$ splits $A$. Therefore the class $\cA(X)$
coincides with $\cA(SB(A))$. Hence the class $\cA(X)$ is connected,
by Theorem~\ref{equiv}.

Suppose now that $\ind A=2n$ is even. We have $n(X)=n$. Let $D$ be a
central division $F$-algebra of degree $2n$ Brauer equivalent to
$A$. In particular $\cA(X)=\cA\(SB(2,D)\)$. Every finite extension
of $F$ of degree $n$ in $\cA(X)$ is isomorphic to a subfield of $D$
and conversely every subfield of $D$ of degree $n$ is contained in
$\cA(X)$.

Suppose that $D$ has a symplectic involution $\sigma$. Denote by
$\Symd(D,\sigma)$ the subspace of $D$ of all elements $d+\sigma(d)$
for $d\in D$. The characteristic polynomial of an element $a\in
\Symd(D,\sigma)$ is the square of the monic \emph{pfaffian
characteristic polynomial} $\Prp_a(s)$ of degree $n$ \cite[Prop.
2.9]{Book}. In particular, $[F(a):F]\leq n$, moreover the polynomial
$\Prp_a(s)$ is irreducible if and only if $[F(a):F]= n$.

\begin{lemma}\label{symd}
For any subfield $K$ of $D$ of degree $n$ there is a symplectic
involution $\tau$ of $D$ such that $K\subset \Symd(D,\tau)$.
\end{lemma}

\begin{proof}
The centralizer $Q$ of $K$ in $D$ is a quaternion algebra over
$K$. By \cite[Th. 4.14]{Book}, the canonical involution of $Q$ can
be extended to a symplectic involution $\tau$ on $D$. Choose an
element $y\in Q$ such that $y+\tau(y)=1$. Then $x=xy+\tau(xy)$ for
every $x\in K$, i.e., $K\subset  \Symd(D,\tau)$.
\end{proof}





\begin{lemma}\label{symplectic}
Let $\sigma$ and $\tau$ be two distinct symplectic involutions on
$D$ and let $L$ be a simple subfield of $\Symd(D,\sigma)$ of degree
$n$. Then
\begin{enumerate}
    \item There exists a subfield $K$ of $D$ of degree $n$ such that $L\sim_X K$
    and $K\cap \Symd(D,\tau)$ contains a non-constant element.
    \item For every $d\in \Symd(D,\sigma)$ there exists a subfield
    $K$ of $D$ of degree $n$ such that $d\in K$ and $L\sim_X K$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $L=F(a)$ for some $a\in\Symd(D,\sigma)$.

(1) Choose $b\in D\m$ such that $\sigma=\Int(b)\circ\tau$
\cite[Prop. 2.7]{Book}. We have $ba\in\Symd(D,\tau)$. Replacing $a$
by $a+1$ if necessary we assume that $ba\neq F$.

Consider the element $c=t+(1-t)b\in D[t]$ and the involution
$\sigma_t=Int(c)\circ \sigma_{F(t)}$ on $D_{F(t)}$. Then
$ca\in\Symd(D_{F(t)},\sigma_t)$. The pfaffian characteristic
polynomial $\Prp_{ca}$ belongs to $F[s,t]$. We have
$\Prp_{ca}(s,0)=\Prp_{ba}$ and $\Prp_{ca}(s,1)=\Prp_a$, in
particular $\Prp_{ca}$ is irreducible. Considering the quotient
field $E$ of the factor ring $F[s,t]/(\Prp_{ca})$ as in the proof of
Lemma \ref{twosubfields} we find two discrete valuations of $E$ over
$v_1$ and $v_0$ respectively with residue fields isomorphic to $L$
and a subfield $K$ of $D$ containing the nonconstant element $ba$.
Thus $L\sim_X K$.

(2) Let $E$ be the degree $n$ subfield of $D_{F(t)}$ generated by
$ta+(1-t)d$. As in the proof of Lemma \ref{twosubfields} we see that
there are two discrete valuations of $E$ with residue fields
$L=F(a)$ and $K$ respectively, where $K$ is a subfield of $D$ of
degree $n$ and $d\in K$.
\end{proof}

\begin{theorem}\label{gensb}
Let $X$ be the generalized Severi-Brauer variety $SB(2,A)$ of a
central simple algebra $A$ of exponent not divisible by $4$. Then
the class $\cA(X)$ is connected.
\end{theorem}

\begin{proof}
By definition of connectedness we may assume that $F$ is a
$p$-special field. If $p$ is odd and we already know that the class
$\cA(X)$ is connected. Thus we may assume that $p=2$, $\exp(A)=2$
and $\ind(A)=2n$ is even. By \cite[Th. 3.8]{Book}, the algebra $A$
has a symplectic involution.

We use the notation above. It is sufficient to show that every two
subfields of $D$ of degree $n$ are $X$-equivalent.

Let $L$ and $E$ be subfields of $D$ of degree $n$. We may assume
that $L/F$ is separable (see \cite[Prop. 5]{Tits}) and therefore is
simple. We prove that $L\sim_X E$ by induction on $n$. By Lemma
\ref{symd}, we can choose symplectic involutions $\sigma$ and $\tau$
of $D$ such that $L\subset \Symd(A,\sigma)$ and $E\subset
\Symd(A,\tau)$.

Case 1: $\sigma=\tau$. Choose an element $d\in E\setminus F$. By
Lemma \ref{symplectic}(2), there is a subfield $K$ of $D$ of
degree $n$ containing $d$ such that $L\sim_X K$. Let $D'$ be the
centralizer of $M=F(d)$ in $D$. By the induction hypothesis
applied to $D'$ with the involution $\tau|_{D'}$ we have
$K\sim_{X_M} E$, therefore $K\sim_X E$ and $L\sim_X K\sim_X E$.

Case 2: $\sigma\neq\tau$. By Lemma \ref{symplectic}(1), there
exists a subfield $K$ of $D$ of degree $n$ such that $L\sim_X K$
and $K\cap \Symd(D,\tau)$ contains a non-constant element $d$.
Choose a subfield $H\subset \Symd(D,\tau)$ of degree $n$
containing $d$. Let $D'$ be the centralizer of $M=F(d)$ in $D$. By
the induction hypothesis applied to $D'$, $K\sim_{X_M} H$ and
therefore $K\sim_X H$. By the first case, $H\sim_X E$, hence
$L\sim_X K \sim_X H\sim_X E$.
\end{proof}

\begin{corollary}\label{genseveri}{\rm (Cf., \cite{Krashen06})}
Let $X$ be the generalized Severi-Brauer variety $SB(2,A)$ of a
central simple algebra $A$ of exponent not divisible by $4$. Then
$\overline{\CH}_0(X)=0$.
\end{corollary}

\section{Quadrics}

Let $q$ be a quadratic form on an $F$-vector space $V$ of finite
dimension. Denote by $X$ the projective quadric hypersurface in
$\P(V)$ defined by the equation $q=0$. If $X$ is smooth, it is a
projective homogeneous variety of the special orthogonal group
$\gO^+(q)$ of type $S=\{\alpha_2,\ldots,\alpha_m\}$ where simple
roots $\alpha_1,\ldots \alpha_m$ of the Dynkin diagrams ${\cat
D}_m$ and ${\cat B}_m$ are numbered as follows:
$$
\begin{array}{cc}
\begin{picture}(120,30)
\put(00,00){\line(1,0){20}} \put(30,00){\dots}
\put(50,00){\line(1,0){20}} \put(70,0){\line(2,-1){20}}
\put(70,0){\line(2,1){20}} \put(0,0){\circle*{3}}
\put(20,0){\circle*{3}} \put(50,0){\circle*{3}}
\put(70,0){\circle*{3}} \put(90,10){\circle*{3}}
\put(90,-10){\circle*{3}} \put(-5,8){$\alpha_1$}
\put(15,8){$\alpha_2$}
%\put(65,8){$\alpha_d$} \put(127,8){$\alpha_{id}$}
\put(58,8){$\alpha_{m-2}$} \put(95,10){$\alpha_{m-1}$}
\put(95,-13){$\alpha_{m}$}
\end{picture} &
\begin{picture}(100,20)
\put(30,00){\line(1,0){20}} \put(60,00){\dots}
\put(80,00){\line(1,0){20}} \put(100,1.1){\line(1,0){20}}
\put(100,-1.2){\line(1,0){20}} \put(106,-2.5){$>$}
\put(30,0){\circle*{3}} \put(50,0){\circle*{3}}
%\put(80,0){\circle*{3}} \put(100,0){\circle*{3}}
%\put(120,0){\circle*{3}}
\put(80,0){\circle*{3}} \put(100,0){\circle*{3}}
\put(120,0){\circle*{3}} \put(25,8){$\alpha_1$}
\put(45,8){$\alpha_2$} \put(90,8){$\alpha_{m-1}$}
\put(120,8){$\alpha_m$}
\end{picture}
\end{array}
$$


\bigskip

\noindent Here the first and second diagrams correspond to the cases
$\dim V=2m$ and $\dim V=2m+1$ respectively.

The quadric $X$ has a rational point if and only if $q$ is
isotropic. If $q$ is anisotropic, then $n(X)=2$ by Springer's
theorem.

\begin{example}\label{dim2}
Let $q$ be a nonzero $2$-dimensional quadratic form. Then we have $\dim
X=0$, moreover $X=\Spec C_0(q)$, where $C_0(q)$ is the even
Clifford algebra of $q$. The form $q$ is non-degenerate (resp.
anisotropic) if and only if $C_0(q)$ is an \'etale quadratic
$F$-algebra (resp. a quadratic field extension of $F$).
\end{example}


\begin{theorem}\label{quadforms}
The class $\cA(X)$ for a projective quadric $X$ is connected.
\end{theorem}

\begin{proof}
We may assume that $q$ is anisotropic. Let $L/F$ be a quadratic
field extension making $q$ isotropic and let $x\in X$ be a point
with $F(x)\simeq L$. Let $\{1,\alpha\}$ be a basis of $L$ over $F$
and let $w=u+\alpha v$, $u,v\in V$,  be an isotropic vector in
$V_L$. Then $w\in U_L$, where $U$ is the $2$-dimensional subspace of
$V$ spanned by $v$ and $u$. Moreover, by Example \ref{dim2},
$F(x)\simeq C_0(q|_{U})$.

Conversely, if $U\subset V$ is a $2$-dimensional subspace, then
\[
X\cap \P(U)=\Spec F(x)\simeq C_0(q|_{U}),
\]
where $x$ is a closed point of degree $2$. Thus it is sufficient to
show that for every two $2$-dimensional subspaces $U$ and $U'$, the
quadratic field extensions $L=C_0(q|_{U})$ and $L'=C_0(q|_{U'})$ are
$X$-equivalent. We may assume that $U\cap U'\neq 0$.

Let $\{v,u\}$ and $\{v,u'\}$ be bases of $U$ and $U'$
respectively. Consider the $F[t]$-submodule $W$ of $V_{F[t]}$
generated by $v$ and $tu+(1-t)u'$. Let $R=C_0(q_{F[t]}|_{W})$. The
quotient field $E$ of $R$ belongs to $\cA(X)$. Since $R/(t)=L'$ is
a field, the principal ideal $Q$ generated by $t$ is maximal.
Therefore the localization $R_Q$ is a Noetherian local domain of
dimension one with principal maximal ideal, i.e., $R_Q$ is a
discrete valuation ring. The corresponding discrete valuation of
$E$ with residue field $L'$ restricts to $v_0$ on $F(t)$.
Similarly, there is a discrete valuation of $E$ with residue field
$L$ restricting to $v_1$, i.e., $L$ and $L'$ are simply
$X$-equivalent.
\end{proof}

\begin{corollary}\label{quadric}{\rm (Cf., \cite{karpenko}, \cite{swan})}
Let $X$ be a smooth projective quadric. Then
$\overline{\CH}_0(X)=0$.
\end{corollary}







\section{Involution varieties}

Let $A$ be a central simple algebra of degree $2m$ with a quadratic
pair $(\sigma, f)$, where $\sigma$ is an involution on $A$ of the
first kind and $f:\Sym(A,\sigma)\ra F$ is a linear map satisfying
certain properties \cite[\S 5.B]{Book}. If $A$ is split, i.e.,
$A\simeq \End(V)$ for a $2n$-dimensional vector space $V$ over $F$,
the space $\Sym(A,\sigma)$ is identified with the subspace of
symmetric elements in $V\tens_F V$ and $f(v\tens v)=q(v)$ for a
non-degenerate quadratic form $q$ on $V$.

Consider the morphism
\[
\varphi: SB(A)\ra \P\(\Sym(A,\sigma)\)
\]
taking an ideal $I$ in $A$ to the (1-dimensional) subspace $I\cdot
\sigma(I)$ of $\Sym(A,\sigma)$ \cite[p. 122]{Book}. Let $W=\Ker(f)$.
The inverse image $X=\varphi\inv\P(W)$ is called the
\emph{involution variety of the quadratic pair}. Thus $X$ is a
hypersurface in the Severi-Brauer variety $SB(A)$. In the split case
as above, $X$ is a smooth projective quadric hypersurface in $\P(V)$
given by the quadratic form $q$. In other words, $X$ is a twisted
form of a projective quadric. It is a projective homogeneous variety
of the special orthogonal group $\gO^+(A,\sigma,f)$ \cite[\S
26]{Book} of type $S=\{\alpha_2,\ldots,\alpha_m\}$, where $S$ is the
subset of vertices of the Dynkin diagram ${\cat D}_m$. In fact, any
twisted form of a smooth projective quadric is isomorphic to an
involution variety.

If $A$ is not split, then $\ind A$ is even. Consider the case
$\ind(A)=2$. Let $Q$ be a quaternion division algebra Brauer
equivalent to $A$. Then $A\simeq \End_Q(V)$ for some right
$Q$-module $V$. The Severi-Brauer variety $SB(A)$ can be identified
with the variety $\P_Q(V)$ of reduced rank 1 $Q$-submodules in $V$.
Note that for any $Q$-submodule $U\subset V$ of rank 1, the
subvariety $\P_Q(U)$ of $\P_Q(V)$ is a conic curve. Then either the
conic $\P_Q(U)$ is contained in $X$ or the intersection of $\P_Q(U)$
with $X$ is a point of degree $2$. Therefore $n(X)=2$.




\begin{theorem}\label{invindex2}
If $\ind(A)=2$ then the class $\cA(X)$ is connected.
\end{theorem}

\begin{proof}
As $n(X)=2$ it is sufficient to show that every two closed points
$x$ and $x'$ of degree $2$ in $X$ are $X$-equivalent. By \cite[Lemma
1]{Merkurjev95b}, there are $Q$-submodules $U\subset V$ and
$U'\subset V$ of rank 1 such that $x\in \P_Q(U)$ and $x'\in
\P_Q(U')$. In fact $F(x)$ and $F(x')$ are isomorphic to the even
Clifford algebras of the restrictions of the quadratic pair to $U$
and $U'$ respectively. Let $v$ and $v'$ be nonzero vectors of $U$
and $U'$ respectively. Consider $w=tv+(1-t)v'\in V_{F[t]}$ and let
$W=Q[t]w$. Let $R$ be the even Clifford algebra of the restriction
of the quadratic pair $(\sigma,f)\tens F[t]$ on $W$. As in the proof
of Theorem \ref{quadforms} we see that the quotient field $E$ of $R$
is a quadratic extension of $F(t)$ equipped with two discrete
valuations over $v_0$ and $v_1$ respectively and residue fields
isomorphic to $F(x)$ and $F(x')$, i.e., $F(x)$ and $F(x')$ are
simply $X$-equivalent.
\end{proof}

Consider the general case when $\ind(A)=2n$. Assume that $F$ is a
$2$-special field. We claim that $n(X)=2n$. Indeed, $n(X)\geq 2n$
since $X$ is a hypersurface in $SB(A)$ and $n(SB(A))=2n$. On the
other hand, choose a field extension $K/F$ of degree $n$ such that
$\ind A_K=2$. As $n(X_K)=2$, we conclude that $n(X)\leq 2n$.


Set $Y=SB(2,A)$. We know that $n(Y)=n$.

\begin{lemma}\label{xandy}
Let $y_0$ and $y_1$ be two closed points of $Y$ of degree $n$ such
that $F(y_0)$ and $F(y_1)$ are simply $Y$-equivalent. Then there
exist points $x_0$ and $x_1$ of $X$ of degree $2n$ such that
$F(x_0)$ and $F(x_1)$ are simply $X$-equivalent and $F(y_i)$ is
isomorphic to a subfield of $F(x_i)$ for $i=0,1$.
\end{lemma}

\begin{proof}
Let $E/F(t)$ be an extension of degree $n$ such that
$Y(E)\neq\emptyset$ and let $w_i$, $i=0,1$, be two discrete
valuations of $E$ over the valuations $v_i$ of $F(t)$ with
$F(w_i)\simeq F(y_i)$.

Since $\ind A_E=2$, there is a quadratic field extension $K/E$ such
that $X(K)\neq \emptyset$. Let $u_i$ be a discrete valuation of $K$
extending $w_i$ and let $L_i$ be the residue field of $u_i$. As $X$
is projective, a morphism $\Spec K\ra X$ factors through a morphism
$\Spec O_i\ra X$, where $O_i$ is the valuation ring of $u_i$. In
particular $X(L_i)\neq \emptyset$, therefore
\[
2n=[K:F(t)]\geq [L_i:F]\geq n(X)=2n.
\]
Hence there is a closed point $x_i\in X$ of degree $2n$ with
$F(x_i)\simeq L_i$. Note that $L_0$ and $L_1$ are simply
$X$-equivalent and $F(y_i)\simeq F(w_i)\subset L_i$.
\end{proof}




\begin{theorem}\label{inv}
Let $A$ be a central simple algebra of even degree with a
quadratic pair and let $X$ be the corresponding involution
variety. Then the class $\cA(X)$ is connected.
\end{theorem}

\begin{proof}
We may assume that $F$ is $p$-special. If $p\not=2$, then
$X(F)\not=\emptyset$ and the class $\cA(X)$ is connected. Let $F$ be
a $2$-special field. If $A$ splits, then $X$ is a smooth projective
quadric, and the result follows from Theorem \ref{quadforms}. Thus
we may assume that $\ind A=2n$ for some $n\ge 1$.

Let $x$ and $x'$ be two closed points of $X$ of degree $2n$. We
would like to show that $F(x)$ and $F(x')$ are $X$-equivalent. As
$F$ is $2$-special, we can choose closed points $y$ and $y'\in Y$ of
degree $n$ such that $F(y)\subset F(x)$ and $F(y')\subset F(x')$. By
Theorem \ref{gensb}, $F(y)$ and $F(y')$ are $Y$-equivalent, i.e.,
there are closed points $y_0=y,y_1,\dots, y_r=y'$ of degree $n$ such
that $F(y_i)$ and $F(y_{i+1})$ are simply $Y$-equivalent for
$i=0,\dots, r-1$. By Lemma \ref{xandy} applied to each pair
$(y_i,y_{i+1})$ there are closed points $x'_i$, $x_{i+1}$ of $X$ of
degree $2n$ such that $F(x'_i)$ and $F(x_{i+1})$ are simply
$X$-equivalent and $F(y_i)\subset F(x'_i)$, $F(y_{i+1})\subset
F(x_{i+1})$.

Set for convenience $x_0=x$ and $x'_r=x'$. For every $i=0,\dots,r$,
the fields $F(x_i)$ and $F(x'_i)$ contain $F(y_i)$. The index of $A$
over $F(y_i)$ is equal to $2$. By Theorem \ref{invindex2}, the
fields $F(x_i)$ and $F(x'_i)$ are $X_{F(y_i)}$-equivalent, and
therefore are $X$-equivalent. Finally we have
\[
F(x)=F(x_0)\sim F(x'_0)\sim F(x_1)\sim F(x'_1)\sim\dots \sim
F(x_r)\sim F(x'_r)=F(x').\qedhere
\]
\end{proof}



\begin{corollary}\label{involutionvarieties}{\rm (Cf., \cite{Krashen06})}
Let $A$ be a central simple algebra of even degree with a quadratic
pair and let $X$ be the corresponding involution variety. Then
$\overline{\CH}_0(X)=0$.
\end{corollary}

\section{Cayley-Dickson and Jordan algebras}

Let $C$ be a Cayley-Dickson algebra over $F$. Denote by $N$ the norm
form of $C$ given by $N(x)=x\bar x$, where $x\mapsto\bar x$ is the
canonical involution of $C$. It is known
\cite[Prop. 33.18, Th. 33.19]{Book}
that $N$ is a 3-fold Pfister form, and $C$ is split
if and only if $N$ is split. Therefore, Theorem \ref{quadforms}
applied to the quadric given by $N$ yields

\begin{theorem}\label{cayley}
The class of splitting fields of $C$ is connected.
\end{theorem}

\begin{remark}
One can give a direct proof of Theorem \ref{cayley} along the lines
of the proof of Lemma \ref{twosubfields} using the fact that a
quadratic field extension of $L/F$ splits $C$ if and only if $L$ can
be embedded into $C$.
\end{remark}

Let $J$ be an exceptional simple $27$-dimensional Jordan algebra
over a field $F$ of degree $3$ arising from the first Tits
construction \cite[Ch. IX, \S 12]{jacobson}.
Recall that there is an associated
central simple $F$-algebra $A$ of degree $3$ and an element $a\in
F\m$. The algebra $J$ is split if and only if $a$ is a reduced
norm for $A$.
%has zero divisors if and only if $a$ is a
%reduced norm for $A$. If $F$ is $3$-special then $J$ is split if
%and only if $J$ has zero divisors.

\begin{theorem}\label{jordan}
Let $J$ be an exceptional simple $27$-dimensional Jordan algebra
over a $3$-special field $F$ of degree $3$ arising from the first
Tits construction. Then the class of all splitting fields of $J$ is
connected.
\end{theorem}

\begin{proof}
We may assume that $J$ is not split. Therefore for the class $\cB$
of all splitting fields of $J$ we have $n(\cB)=3$. Let $L$ and $L'$
be two cubic extensions of $F$ splitting $J$. We shall show that
$L\sim_{\cB} L'$. By \cite[Cor. 3]{PeRaproc84} we may assume that
$L$ and $L'$ are subfields of $J$. Choose generators $x$ and $x'$ of
$L$ and $L'$ over $F$ respectively and let $E$ be the subfield of
$J_{F(t)}$ generated by $tx+(1-t)x'$. As in the proof of Lemma
\ref{twosubfields} one shows using the cubic minimal polynomials of
$x$ and $x'$ that there are two discrete valuations of $E$ over
$v_0$ and $v_1$ with residue fields $L$ and $L'$ respectively, i.e.,
$L$ and $L'$ are simply $\cB$-equivalent.
\end{proof}





\section{Groups of exceptional types}
% ${\cat G}_2,{\cat F}_4,{^{3,6}{\cat D}_4},{{^1}{\cat E}_6}$}





In what follows $G_0$ denotes split or quasi-split simple simply
connected algebraic group over $F$ of one of the following types:
${\cat G}_2,{\cat F}_4,{\cat D}_4,{\cat E}_6,{\cat E}_7$. Let
$\xi\in Z^1(F,G_0)$ be a cocycle and let $G={^{\xi}G_0}$ be the
corresponding strongly inner form of $G_0$.

\begin{theorem}\label{exceptional}
Let $X=X_S$ be a projective homogeneous variety of the group $G$ as
above. If $G$ is an outer form of type ${\cat E}_6$ of of type
${\cat E}_7$ we assume that $S\not=\emptyset$. If $G$ is a
trialitarian group or of types ${\cat E}_6,{\cat E}_7$ we assume
that ${\rm char}(F)\not=2,3$. Then the class $\cA(X)$ is connected.
In particular, we have $\overline{\CH}_0(X)=0$.
\end{theorem}
%\begin{proof}
If $S=\Delta$, then $X_S={\rm Spec}\,F$ and there is nothing to
prove. Thus, by Proposition~\ref{admissible} we may assume that
$S$ is a proper basic subset of $\Delta$. By Lemma~\ref{special},
we may also assume that $F$ is $p$-special.


If $p\not=2,3$, then we have $H^1(F,G_0)=1$ for all types under the
consideration. Indeed, by Steinberg's theorem~\cite{[St65]}, $\xi$
is equivalent to a cocycle with coefficients in a maximal torus
$T\subset G_0$ and, by~\cite[Proposition $6.21$, p. 375]{[PR]}, we
have $H^1(F,T)=1$ for $p$-special fields with $p\not=2,3$. It
follows that $G$ is split or quasi-split and hence
$X_S(F)\not=\emptyset$. Thus we may assume that $p=2,3$.

To finish the proof we shall either show directly that $\cA(X)$ is
connected or construct a projective homogeneous variety $Y$ (under
some algebraic simple group) such that

\medskip

\noindent (i) {\it The class $\cA(Y)$ is connected;}

\noindent (ii) $\cA(X_S)=\cA(Y)$.

\medskip


We shall construct $Y$ using case-by-case consideration.

\section{Types ${\cat G}_2$ and ${\cat F}_4$}\label{typef4}

By~\cite{t66}, each isotropic group of type ${\cat G}_2$ is split.
Hence the only basic type for $G$ is $S=\emptyset$. Since
$X_{\emptyset}$ is a variety of Borel subgroups of $G$, the class
$\cA(X_{\emptyset})$ consists of all field extensions making $G$
split.

Recall that the correspondence $C\mapsto G=\Aut(C)$ gives rise to a
bijection between the set of isomorphism classes of Cayley-Dickson
algebras and the set of isomorphism classes of simple groups of type
${\cat G}_2$ \cite[Theorem 26.19]{Book}. In particular the classes
of splitting fields for $C$ and $G$ coincide. The connectedness of
$\cA(X_{\emptyset})$ follows from Theorem \ref{cayley}.




%\subsection{Type ${\cat F}_4$}\label{typef4}

We pass to type ${\cat F}_4$. According to~\cite{t66}, if $G$ is
isotropic over a field extension $L/F$ but not $L$-split, then its
Tits index  is of the form

$$
%\begin{equation}\label{f4}
\begin{picture}(100,5)
\put(00,00){\line(1,0){20}} \put(20,1.1){\line(1,0){30}}
\put(20,-1.2){\line(1,0){30}} \put(30,-2.5){$>$}
\put(50,00){\line(1,0){20}} \put(00,0){\circle*{3}}
\put(20,0){\circle*{3}} \put(50,0){\circle*{3}}
\put(70,0){\circle*{3}} \put(70,0){\circle{10}}
\put(-5,8){$\alpha_1$} \put(15,8){$\alpha_2$} \put(40,8){$\alpha_3$}
\put(60,8){$\alpha_4$}
\end{picture}
%\end{equation}
$$

\medskip

\noindent It follows that there exist two basic types only:
$$S_1=\emptyset \mbox{\ \ and\ \ }
S_2=\{\alpha_1,\alpha_2,\alpha_3\}.$$ The class $\cA(X_{S_1})$
consists of all field extensions $L/F$ making $G$ split and
$\cA(X_{S_2})$ consists of all field extensions $L/F$ making $G$
isotropic.

The correspondence $J\mapsto G=\Aut(J)$ gives rise to a bijection
between the set of isomorphism classes of exceptional simple
$27$-dimensional Jordan algebra of degree $3$ and the set of
isomorphism classes of simple groups of type ${\cat F}_4$
\cite[Theorem 26.18]{Book}. In particular the classes of splitting
fields for $J$ and $G$ coincide.


If $p=3$, then $J$ arises from the first Tits construction
(cf.~\cite[Ch. IX, \S 12]{jacobson}).
%Note that since there are no anisotropic groups of
%type ${\cat B}_3$,
We claim that the diagram above cannot appear as the Tits index of
$G$ over any extension $L/F$ of $F$. Indeed, assume the contrary.
Since $G$ is not split over $L$, so is $J$. The anisotropic kernel
of $G$ over $L$ has type ${\cat B}_3$, hence there is a finite
extension $E/L$ of degree $[E:L]=2^a$ for some positive integer $a$
making $G$ (and hence $J$) split. On the other side, the extension
$E/L$ does not split $J$ since it arises from the first Tits
construction -- a contradiction.
%(remind that if $A$ and $c$ are the simple
%central algebra of index $3$ and the element in $F$ corresponding
%to the first Tits construction, then $J$ is split over $E$ if and
%only if $c$ is the reduced norm for $A_E$).
Thus we showed that the classes $\cA(X_{S_1})$, $\cA(X_{S_2})$ and
the class of all splitting fields of $J$ coincide and we are done
by Theorem \ref{jordan}.


If $p=2$, then $J$ is reduced. Recall
(cf.~\cite{[GMS]},~\cite{petersson}) that given $J$ one can
associate a $3$-Pfister form $f_3(J)$ and a $5$-Pfister form
$f_5(J)$ with the following properties (cf. \cite{springer}, \cite{racine}):

\medskip

\noindent ({\it a}) {\it The group $G$ is split over a field
extension $L/F$ if and only if $f_3(J)$ is hyperbolic over $L$;}

\noindent ({\it b}) {\it The group $G$ is isotropic over a field
extension $L/F$ if and only if $f_5(J)$ is hyperbolic over $L$.}

\medskip

%Note that $X_{S_2}$ is split if and only if $G$ is isotropic and
%$X_{S_1}$ is split if and only if $G$ is split.

Let $Y_1$ and $Y_2$ be the projective quadric defined by the
equations $f_3(J)=0$ and $f_5(J)=0$ respectively. By (a) and (b),
we have $\cA(X_{S_1})=\cA(Y_1)$ and $\cA(X_{S_2})=\cA(Y_2)$. The
connectedness of $\cA(X_{S_1})$ and $\cA(X_{S_2})$ follows from
Theorem \ref{quadforms}.




\section{Type ${\cat D}_4$}

Suppose first that $G$ is anisotropic of type $^1{\cat D}_4$ or
$^2{\cat D}_4$. Then $G\simeq {\gSpin}(f)$, where $f$ is an
$8$-dimensional quadratic form with trivial even Clifford algebra.
Let $dF^{\times 2}$ be the discriminant of $f$. The form
$f\perp\langle 1,-d\rangle$ has trivial discriminant and Clifford
algebra, therefore it is represented by a $3$-Pfister form $g$ in
the Witt ring $W(F)$.


As $g$ can have Witt index $0$ or $4$ over a field extension $L/F$,
it follows that if $f$ becomes isotropic over $L/F$, then its Witt
index over $L$ is either $1$ or $f$ is quasi-split (i.e., its Witt
index is either 3 or 4). Hence for the Tits index of $G$ over $L$
there are the following possibilities only:
$$
%\begin{equation}
%\label{222}
\begin{array}{ccc}
\begin{picture}(70,30)
\put(00,00){\line(1,0){30}} \put(30,0){\line(2,-1){30}}
\put(30,0){\line(2,1){30}} \put(0,0){\circle*{3}}
\put(30,0){\circle*{3}} \put(60,15){\circle*{3}}
\put(60,-15){\circle*{3}} \put(00,0){\circle{10}}
\put(-5,10){$\alpha_1$} \put(20,10){$\alpha_2$}
\put(70,15){$\alpha_3$} \put(70,-18){$\alpha_4$}
\end{picture}
%\end{equation}
\qquad \qquad

\begin{picture}(70,30)
\put(00,00){\line(1,0){30}} \put(30,0){\line(2,-1){30}}
\put(30,0){\line(2,1){30}} \put(0,0){\circle*{3}}
\put(30,0){\circle*{3}} \put(60,15){\circle*{3}}
\put(60,-15){\circle*{3}} \put(30,0){\circle{10}}
\put(60,-15){\circle{10}} \put(60,15){\circle{10}}
\put(00,00){\circle{10}} \put(-5,10){$\alpha_1$}
\put(20,10){$\alpha_2$} \put(70,15){$\alpha_3$}
\put(70,-18){$\alpha_4$}
\end{picture}

\qquad \qquad

\begin{picture}(70,30)
\put(00,00){\line(1,0){30}} \put(60,0){\oval(60,20)[l]}
\put(0,0){\circle*{3}} \put(30,0){\circle*{3}}
\put(60,10){\circle*{3}} \put(60,-10){\circle*{3}}
\put(00,0){\circle{10}} \put(30,0){\circle{10}}
\put(60,0){\oval(10,30)} \put(-10,10){$\alpha_1$}
\put(15,10){$\alpha_2$} \put(70,8){$\alpha_3$}
\put(70,-8){$\alpha_4$}
\end{picture}
\end{array}
$$

\bigskip

\bigskip

By Remark~\ref{innerform}, we need to consider the following basic
types only: $$S_1=\{\alpha_2,\alpha_3,\alpha_4\} \mbox{\ \ and \ \
} S_2=\emptyset.$$ The variety $X_{S_1}$ is isomorphic to the
quadric $Y$ defined by $f=0$. The class $\cA(Y)$ is connected by
Theorem~\ref{quadforms}.

The second case $S_2$ corresponds to the variety of Borel
subgroups of $G$. A field extension $L/F$ makes $X_{\emptyset}$
isotropic if and only if it splits $g$. Let $Z$ be the projective
quadric defined by $g=0$. Then the classes $\cA(X_{\emptyset})$
and $\cA(Z)$ coincide, and we are done again by
Theorem~\ref{quadforms}.

We now turn to trialitarian cases $^3{\cat D}_4$ and $^6{\cat D}_4$.
We need to consider the case $p=3$ only. Let $E/F$ be a cubic
extension over which $G$ is a classical group.
\begin{lemma}\label{trial}
Let $F$ be $3$-special. If $L/F$ is an extension such that $G$ is
$L$-isotropic, then $G$ is quasi-split over $L$.
%In particular,
%the only basic type for $G$ is $S=\emptyset$.
\end{lemma}
\begin{proof} If $G$ is not quasi-split over $L$, then, by~\cite{t66}, its
Tits index over $L$ is of the form
$$
\begin{picture}(40,15)
\put(00,00){\line(1,0){30}} \put(30,0){\oval(60,20)[l]}
\put(0,0){\circle*{3}} \put(30,0){\circle*{3}}
\put(30,10){\circle*{3}} \put(30,-10){\circle*{3}}
\put(00,0){\circle{10}}
\end{picture}
$$
\medskip

\noindent Hence the semisimple $L$-anisotropic kernel of $G$ is
isomorphic to ${\rm R_{E'/L}}({\rm  \gSL}_1(D))$, where $D$ is a
quaternion algebra over $E'$ and $E'=E\cdot L$.  Note that such a
group is still anisotropic over $E'$. On the other side, since $F$
is $3$-special, $G$ is quasi-split over $E$ and hence over
$E'=E\cdot L$ -- a contradiction.
\end{proof}

Lemma~\ref{trial} implies that we need to consider the basic type
$S=\emptyset$ only. Let $R_{G_0}:H^1(F,G_0)\to H^3(F,\mathbb{Z}/3)$
be the Rost invariant. We refer to~\cite{[GMS]} for the definition
and properties of the Rost invariant. It takes values in
$\mathbb{Z}/3$ since $F$ is $3$-special. By~\cite[Theorem 6.14]{ch},
${\rm Ker}\,R_{G_0}=1$
%and by Lemma~\ref{fact1}, $G$ is
%quasi-trivial if and only if $[\xi]=1$. It follows that two
%classes of field extensions making $G$ quasi-split and killing
%$R_{G_0}([\xi])$ coincide.
%and the image $R_{G_0}(\xi)$ of $\xi$ is a $3$-symbol, say
%$(a)\cup(b)\cup(c)$, where $a,b,c\in F^{\times}$. Then this symbol
%By our assumption, $G$ is not quasi-split over $L$, hence
%$R_{G_0}([\xi])$ is also nontrivial over $L$. By~\cite{chpl},
%Lemma~$8$, $D$ has a maximal subfield of the form $E(\sqrt{a})$
%for some $a \in L$. This implies that $L$-anisotropic kernel of
%$G$ (and hence $G$) is quasi-split over a quadratic extension
%$L(\sqrt{a})$ of $L$. However any quadratic extension of $L$ does
%not kill $R_{G_0}([\xi])$ since it has order $3$ -- a
%contradiction.
%\end{proof}
%By~\cite{ch}, Theorem $6.14$,
and $R_{G_0}([\xi])$ is a $3$-symbol, say $(a)\cup(b)\cup(c)$, where
$a,b,c\in F^{\times}$.

Let $J$ be the Jordan algebra arising from the first Tits
construction corresponding to the central simple algebra $A=(a,b)$
over $F$ of degree $3$ and element $c$. Recall that it is split
over a field extension $L/F$ of $F$ if and only if $c\in {\rm
Nrd}\,A_L$ or equivalently if and only if $R_{G_0}([\xi])$ is
trivial over $L$, by \cite[Th. 12.2]{MerkSus82}. As ${\rm
Ker}\,R_{G_0}=1$, the last holds if and only if $[\xi_L]=1$. Thus,
by Lemma~\ref{fact1}, the classes of splitting fields of $J$ and $
\cA(X_{\emptyset})$ coincide and the connectedness of
$X_{\emptyset}$ follows from Theorem~\ref{jordan}.


\section{Inner forms of type  $^1{\cat E}_6$}\label{1e6}

According to~\cite{t66} all admissible Tits' indices of type
$^1{\cat E}_6$ corresponding to isotropic (not split) groups are as
follows:

$$
%\begin{equation}\label{p}
\begin{array}{ll}
\mbox{(a)} \qquad
\begin{picture}(85,30)
\put(00,02){\line(1,0){80}} \put(40,02){\line(0,2){20}}
\put(00,02){\circle*{3}} \put(20,02){\circle*{3}}
\put(40,02){\circle*{3}} \put(60,02){\circle*{3}}
\put(80,02){\circle*{3}} \put(40,22){\circle*{3}}
\put(00,02){\circle{10}} \put(80,02){\circle{10}}
\put(-5,-11){$\alpha_1$} \put(15,-11){$\alpha_3$}
\put(35,-11){$\alpha_4$} \put(55,-11){$\alpha_5$}
\put(75,-11){$\alpha_6$} \put(35,30){$\alpha_2$}
\end{picture}
& \qquad \mbox{(b)} \qquad
\begin{picture}(80,30)
\put(00,02){\line(1,0){80}} \put(40,02){\line(0,1){20}}
\put(00,02){\circle*{3}} \put(20,02){\circle*{3}}
\put(40,02){\circle*{3}} \put(60,02){\circle*{3}}
\put(80,02){\circle*{3}} \put(40,22){\circle*{3}}
%\put(00,02){\circle{10}}
%\put(20,02){\circle{10}}
%\put(60,02){\circle{10}}
%\put(80,02){\circle{10}}
\put(40,02){\circle{10}} \put(40,22){\circle{10}}
\put(-5,-11){$\alpha_1$} \put(15,-11){$\alpha_3$}
\put(35,-11){$\alpha_4$} \put(55,-11){$\alpha_5$}
\put(75,-11){$\alpha_6$} \put(35,30){$\alpha_2$}
\end{picture}
\end{array}
%\end{equation}
$$
\vspace{5mm}

\noindent In case (b) the corresponding anisotropic kernel has type
${\cat A}_2+ {\cat A}_2$. Since ${\rm \gSL}_3\times{\rm \gSL}_3$ has
no nontrivial strongly inner forms, Theorem~\ref{fact2} implies that
diagram (b) can not appear as the Tits index of a strongly inner
form of type $^1{\cat E}_6$ (note that in the notation of that
theorem the condition $H^1(F,C/C\cap H)=1$ holds, since $C=C_1$ is
an $F$-split torus). Thus, by Remark~\ref{innerform}, the only basic
types to be consider are
$$S_1=\emptyset \mbox{\ \ and\ \ } S_2=\{\alpha_2,\alpha_3,\alpha_4,
\alpha_5\}.$$

\begin{lemma}\label{spin8} Assume that $G$ has diagram (a) over an extension
$L/F$. Then the anisotropic $L$-kernel of $G$ is isomorphic to ${\rm
\gSpin}(f)$ where $f$ is a $3$-Pfister form over $L$.
\end{lemma}\label{kernel}
\begin{proof} The anisotropic $L$-kernel is a strongly inner form
of type $^1{\cat D}_4$, by Theorem~\ref{fact2}. So the result is
clear.
\end{proof}

%Note that in case (a) the corresponding anisotropic kernel is a
%strongly inner form of a split group ${\rm Spin}_8$, by
%Theorem~\ref{fact2}, hence it is isomorphic to ${\rm Spin}(f)$
%where $f$ is a $3$-Pfister form. In particular, $G$ is split over
%a field extension $L/F$ if and only if $f$ is hyperbolic over $L$.
\begin{lemma}\label{three} Let $F$ be $3$-special. Then
$\cA(X_{S_1})=\cA(X_{S_2})$.
\end{lemma}
\begin{proof} The inclusion $\cA(X_{S_1})\subset \cA(X_{S_2})$ is
clear. Conversely, let $L\in \cA(X_{S_2})$. If
$X_{S_1}(L)=\emptyset$, then $G$ has Tits index (a) over $L$. The
corresponding anisotropic kernel of $G$ over $L$ is isomorphic to
${\rm \gSpin}(f)$ where $f$ is a $3$-Pfister form, by
Lemma~\ref{spin8}. Hence it can be split by a quadratic extension
$E/L$. On the other side, the Rost invariant $R_{G_0}$ has trivial
kernel (cf.~\cite{ch}, \cite{gar}), takes values in
$H^3(F,\mathbb{Z}/3)$ and $R_{G_0}([\xi])$ is a symbol, by
\cite{ch}. Let $J$ be a Jordan algebra corresponding to the symbol
$R_{G_0}([\xi])$. As ${\rm Ker}\,R_{G_0}=1$, $J$ is nontrivial
over $L$ and remains so over its quadratic extension $E/L$ since
$J$ arises from the first Tits construction. However $J$, $[\xi]$
(and hence $G$) have the same splitting fields -- a contradiction.
\end{proof}

If $p=3$, then by Lemma~\ref{three}, $\cA(X_{S_1})=\cA(X_{S_2})$
and the classes of splitting fields of $J$ and $\cA(X_{S_1})$
coincide where $J$ is the Jordan algebra constructed in
Lemma~\ref{three}. The connectedness of
$\cA(X_{S_1}),\,\cA(X_{S_2})$ follows from Theorem~\ref{jordan}.

Let now $p=2$.
\begin{lemma}\label{isotropic} If $F$ is $2$-special, then $G$ is $F$-isotropic.
%and its Tits index has form {\rm (i)}.
\end{lemma}
\begin{proof}
%We need only to show that $G$ is $F$-isotropic. Let $H_0$ be a split
%group of type ${\cat F}_4$.
There exists the canonical embedding $H_0\to G_0$ where $H_0$ is a
split group of type ${\cat F}_4$. It induces the mapping
$H^1(F,H_0)\to H^1(F,G_0)$ which is surjective, by~\cite[page 51,
Exercise 22.9]{[GMS]}, since $F$ is $2$-special. Each class in
$H^1(F,H_0)$ is split over a quadratic extension $E/F$, by
property (a) in Section~\ref{typef4}, hence so is $[\xi]$.
By~\cite[Lemma $6.4$]{ch}, $G$ is isotropic over $F$.
\end{proof}
Lemma~\ref{isotropic} implies that $X_{S_2}(F)\not=\emptyset$ and
hence $\cA(X_{S_2})$ is connected. Furthermore, if $G$ is not
split, then, by Lemma~\ref{spin8}, its anisotropic kernel is
isomorphic to ${\rm Spin}(f)$ where $f$ is a $3$-Pfister form $f$.
Hence $\cA(X_{S_1})=\cA(Y)$ where $Y$ is the quadric defined by
$f=0$.

\section{Outer forms of type ${\cat E}_6$}

%In this section  $G_0$ denotes a simple simply connected
%quasi-split group of type $^2{\cat E}_6$ over a field $F$ of
%characteristic $\not=2$ and $G={^{\xi}}G_0$ where $\xi\in
%Z^1(F,G_0)$.
%We may assume that $F$ is $p$-special.
If $p\not=2$, then $G_0$ and $G$ are inner forms of type ${\cat
E}_6$. So we may assume that $p=2$. Then $F$ is perfect since by
our assumption ${\rm char}(F)\not=2$.


\subsection{Basic types}

Let $K=F(\sqrt{d})$ be the quadratic extension which makes $G_0$
(and hence $G$) an inner form of type ${\cat E}_6$. We denote by
$\sigma$ a unique nontrivial automorphism of $K/F$. According to
Tits' classification \cite{t66} if $G$ is $F$-isotropic but not
quasi-split, then its $F$-index 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)}.
\end{lemma}
\begin{proof}
In case (a) the anisotropic kernel of $G$ has type ${\cat A}_2+
{\cat A}_2$ and a maximal $F$-split torus $C_1$ has dimension $2$.
By dimension count $C_1$ coincides with the center $C$ of its
centralizer; in particular $C$ and all its quotients have trivial
cohomology in dimension $1$. Hence, by Theorem~\ref{fact2}, the
anisotropic kernel of $G$ is a strongly inner form of
$R_{K/F}(\gSL_3)$. As $H^1(F,R_{K/F}(\gSL_3))=1$, the group
$R_{K/F}(\gSL_3)$ has no nontrivial strongly inner forms.
\end{proof}
Thus there are four basic types only:
$$
S_1=\{\alpha_2,\alpha_3,\alpha_4,\alpha_5\}, \ S_2=\{
\alpha_1,\alpha_3,\alpha_4,\alpha_5,\alpha_6\}, \
S_3=\{\alpha_3,\alpha_4,\alpha_5\}, \  S_4=\emptyset.
$$
According to Lemma~\ref{isotropic}, $G$ is isotropic over $K$. If it
is not $K$-split, then its Tits index is given by diagram (a) in
Section~\ref{1e6} and hence its anisotropic kernel over $K$ is
isomorphic to $\gSpin(f)$ where $f$ is a $3$-Pfister form over $K$.
Since $f$ can be split by a quadratic extension of $K$, we obtain
that $n(X_{S_4})$ divides $4$.

To compute the numbers $n(X_S)$ we need information on groups of
type ${\cat D}_n$.

\subsection{Classical groups of type ${\cat D}_n$}

Recall that a simple simply connected $F$-group of type $^{1}{\cat
D}_n$ can be realized as a spinor group ${\rm {\bf Spin}}(A,\tau)$
associated to a central simple algebra $A$ over $F$ equipped with an
involution $\tau$ of the first kind, orthogonal type and trivial
discriminant. If $A$ is represented by a skew field $D$ in the
Brauer group ${\rm Br}\,(F)$, then $A\simeq {\rm End}_D(V)$ where
$V$ is a right $D$-module, and $\tau$ corresponds to a
non-degenerate skew-hermitian form $h$ of trivial discriminant on
$V$ with respect to a suitable involution $\tau'$ on $D$ of
symplectic type.

${\rm {\bf Spin}}(A,\tau)$ is isotropic over $F$ if and only if
$h$ is isotropic, i.e. there is a vector $v\in V$ such that
$h(v,v)=0$. Tits index of ${\rm {\bf Spin}}(A,\tau)$ over $F$ is
of the form
$$
%\begin{equation}
%\label{7}
\begin{picture}(250,20)
\put(00,00){\line(1,0){20}} \put(30,00){\dots}
\put(50,00){\line(1,0){40}} \put(100,00){\dots}
\put(120,00){\line(1,0){40}} \put(170,00){\dots}
\put(190,00){\line(1,0){20}} \put(210,0){\line(2,-1){20}}
\put(210,0){\line(2,1){20}} \put(0,0){\circle*{3}}
\put(20,0){\circle*{3}} \put(50,0){\circle*{3}}
\put(70,0){\circle*{3}} \put(90,0){\circle*{3}}
\put(120,0){\circle*{3}} \put(140,0){\circle*{3}}
\put(160,0){\circle*{3}} \put(190,0){\circle*{3}}
\put(210,0){\circle*{3}} \put(230,10){\circle*{3}}
\put(230,-10){\circle*{3}} \put(70,0){\circle{10}}
\put(140,0){\circle{10}} %\put(230,-10){\circle{10}}
\put(-5,8){$\alpha_1$} \put(15,8){$\alpha_2$}
\put(65,8){$\alpha_d$} \put(127,8){$\alpha_{id}$}
\put(240,10){$\alpha_{rd-1}$} \put(240,-13){$\alpha_{rd}$}
\end{picture}
%\end{equation}
$$

\bigskip

\noindent Here $d={\rm deg}(D)$,  $i$ is the Witt index of $h$ and
$rd=n$.
%\begin{proposition}\label{subform}
%Let $D$ be a quaternion algebra over $F$ and $h$ be a
%skew-hermitian form on a vector space $V$ over $D$ with respect to
%the canonical involution on $D$. Assume that $L/F$ is a quadratic
%extension killing $D$ and making $h$ isotropic over $L$. Then $h$
%can be written as the direct sum $h=h_1\oplus h_2$ such that $h_2$
%is a $1$-dimensional subform isotropic over $K$.
%\end{proposition}
%\begin{proof} ????
%\end{proof}

\subsection{Trialitarian effect}\label{affect} We keep the above
notation. Assume that we are given a pair $(A,\tau)$ such that $A$
has index $2$ and degree $8$. Consider the even Clifford algebra of
$(A,\tau)$. It is of the form $A_1\times A_2$ where $A_1,A_2$ are
central simple algebras of degree $8$ equipped with involutions
$\tau_1,\tau_2$ of orthogonal type and $A\cdot A_1\cdot A_2=1$ in
the Brauer group ${\rm Br}\,(F)$ (see \cite{Book}). The center $Z$
of ${\rm {\bf Spin}}(A,\tau)$ is isomorphic to
$Z\simeq\mathbb{Z}/2\times \mathbb{Z}/2$. Three nontrivial elements
in $Z$ give rise to the quotient morphisms $\phi_i:
\gSpin(A,\tau)\to \gO^+(A_i,\tau_i),\,i=1,2$ and $\phi:
\gSpin(A,\tau)\to \gO^+(A,\tau)$. If ${\rm {\bf Spin}}(A,\tau)$ is
isotropic over $F$, there are two possibilities for its Tits index:
$$
%\begin{equation}
%\label{222}
\begin{array}{cc}
\begin{picture}(70,30)
\put(-5,00){\line(1,0){35}} \put(30,0){\line(2,-1){30}}
\put(30,0){\line(2,1){30}} \put(-5,0){\circle*{3}}
\put(30,0){\circle*{3}} \put(60,15){\circle*{3}}
\put(60,-15){\circle*{3}} \put(30,0){\circle{10}}
\put(-5,8){$\alpha_1$} \put(20,8){$\alpha_2$}
\put(70,15){$\alpha_3$} \put(70,-18){$\alpha_4$}
\end{picture}
%\end{equation}
\qquad \qquad & \qquad

\begin{picture}(70,30)
\put(-5,00){\line(1,0){35}} \put(30,0){\line(2,-1){30}}
\put(30,0){\line(2,1){30}} \put(-5,0){\circle*{3}}
\put(30,0){\circle*{3}} \put(60,15){\circle*{3}}
\put(60,-15){\circle*{3}} \put(30,0){\circle{10}}
\put(60,-15){\circle{10}} \put(-5,8){$\alpha_1$}
\put(20,8){$\alpha_2$} \put(70,15){$\alpha_3$}
\put(70,-18){$\alpha_4$}
\end{picture}
\end{array}
$$

\bigskip

\bigskip

\noindent In the first case all algebras $A,A_1,A_2$ have index
$2$ and in the second case one of the algebras $A_1,A_2$ is split
and a quadratic form $f$ corresponding to its involution is
isotropic. Note that according to the above picture isotropy of
$f$ implies automatically that its Witt index is $2$.

\begin{lemma}\label{rank2} Assume that $A,A_1,A_2$ have indices at
most $2$. Then there is a quadratic extension $L/F$ such that
${\rm {\bf Spin}}(A,\tau)$ has rank at least $2$ over $L$.
\end{lemma}
\begin{proof} If all algebras $A,A_1,A_2$ are matrix algebras, then
$G\simeq\gSpin(f)$ where $f$ is a $3$-Pfister form. Such a form can
be split by a quadratic extension.

Assume that $A$ has index $2$. Consider the pair $(A_1,\tau_1)$ and
the corresponding involution variety $Y$. As $n(Y)=2$, there is a
quadratic extension $L/F$ belonging to $\cA(Y)$. The remark above
the lemma shows that $L$-rank of ${\rm {\bf Spin}}(A,\tau)$ is at
least $2$.
%Fixing an orthogonal base in the corresponding vector space $V$
%over $D_1$, me can identify $A_1=M_4(D_1)$ and $h_1$ with a
%diagonal matrix $h_1={\rm diag}(a_1,a_2,a_3,a_4)$ where
%$a_1,\ldots,a_4$ are pure quaternions in $D_1$. The involution
%$\tau_1$ is given by $\tau_1(x)=h{\overline x}h^{-1}$ where
%$x=(x_{ij})\in M_4(D_1)$ and ${\overline x}=({\overline x_{ji}})$.
%Up to isogeny we have $G\simeq {\rm SO}(D_1,h_1)$. Consider a
%maximal subfield $K\subset D_1$ generated by $a_1$ and a
%$1$-dimensional torus $T_1=R^{(1)}_{K/F}(G_{m,K})$. We have an
%embedding $T_1\to {\rm SO}(D_1,h_1)$ given by $t\to {\rm
%diag}(t,1,1,1)$. Passing from $F$ to $K$ we split $D_1$, split $T_1$
%and hence make $G$ isotropic over $K$. The remark above the lemma
%shows that $K$-rank of $G$ is at least $2$.
\end{proof}



\subsection{Computing $n(X_S)$}\label{splittingnumber}

\begin{proposition}\label{number}
We have $n(X_{S_1})=n(X_{S_2})=n(X_{S_3})=2$.
\end{proposition}
\begin{proof} We consider the most difficult case when $G$ is
anisotropic over $F$. Isotropic cases are reduced to groups of
classical types and can be treated much easier.

We may assume that $G$ is not split over $K$, since otherwise
$X_{S_i}(K)\not=\emptyset,\,i=1,\ldots,4$ and we are done. Then
its Tits $K$-index is given by diagram (a) in Section~\ref{1e6}
and so the variety of parabolic subgroups in $G$ of type
$S_1=\{\alpha_2,\alpha_3,\alpha_4,\alpha_5\}$ contains a
$K$-point. By Lemma~\ref{opposite}, there is a parabolic subgroup
$P$ in $G$ over $K$ of type $S_1$ such that $H_1=P\cap \sigma(P)$
is a reductive part of $P$.

$H_1$ is defined over $F$ and is an almost direct product of a
simple simply connected $F$-group of classical type ${\cat D}_4$
and a $2$-dimensional central torus $T_1$. Since $T_1$ is
$F$-anisotropic and split over $K$, $\sigma$ acts on the character
lattice of $T_1$ by multiplication $-1$. Hence each
$1$-dimensional subtorus in $T_1$ is $F$-defined.

Let ${\tilde \alpha}$ be the highest root in $\Sigma$ with respect
to the ordering given on diagram (a) in Section \ref{1e6}. The
restriction mapping ${\tilde \alpha}|_{T_1}:T_1\to G_m$ is a
nontrivial homomorphism. Let $T_2$ be its kernel. It has dimension
$1$ and, by construction, $T_2$ commutes with $U_{\pm{\tilde
\alpha}}$. As $T_2$ is defined over $F$, so is $H_2=C_G(T_2)$ which
is a reductive group whose semisimple part $[H_2,H_2]$ is a simple
simply connected group of classical type ${\cat D}_5$ generated by
roots $\alpha_2,\alpha_3,\alpha_4,\alpha_5,{\tilde \alpha}$.

Let us describe the structure of $[H_2,H_2]$. Note first that it
has $K$-rank $1$ (and hence $K$-isotropic), since $T_1$ is
contained in $H_2=C_G(T_2)$ and $T_1$ is a maximal $K$-split torus
in $G$ of dimension $2$.
%Note that diagram (a) in \ref{1e6} implies that the group $\langle
%U_{\pm{\tilde \alpha}} \rangle$ is split over $K$.
Furthermore, up to isogeny, we have $[H_2,H_2]\simeq \gSU(D,h)$
where $D$ is a skew field over $F$ equipped with a symplectic
involution $\tau$ and $h$ is a skew hermitian form over $D$ with
respect to $\tau$. As $[H_2,H_2]$ is $K$-isotropic and its
anisotropic semisimple kernel over $K$ coincides with that of $G$
(it is a group of type ${\cat D}_4$ generated by
$\{\alpha_2,\alpha_3,\alpha_4,\alpha_5\}$), Theorem~\ref{fact2}
applied to $G$ over $K$ shows that this anisotropic kernel has
trivial Tits algebras over $K$. It follows that the discriminant
of $h$ is trivial over $K$, the algebra $D$ splits over $K$ and
hence $D$ is either $F$ or a quaternion skew field.

We claim that the discriminant of $h$ is $d$. Indeed, we know that
it is trivial over $K$, hence we need only to exclude the
possibility for it to be trivial over $F$. Let $Z$ be the variety
of Borel subgroups of $[H_2,H_2]$. If the discriminant of $h$ is
trivial over $F$, then $[H_2,H_2]$ is a group of inner type and
hence it has rank $5$ over a field extension $F(Z)/F$. This
implies automatically that $G$ is split over $F(Z)$. But $F$ is
algebraically closed in $F(Z)$, hence $K$ is not contained in
$F(Z)$. It follows that $G$ is still a group of outer type over
$F(Z)$ -- a contradiction.


Assume now that $D$ is a quaternion skew field. By \cite[Lemma
1]{Merkurjev95b}, $h$ can be written in the form $h=h_1\oplus h_2$
where $h_2$ is a $1$-dimensional skew-hermitian form over $D$
isotropic over $K$ and $h_1$ is a $4$-dimensional form with
trivial discriminant. Lemma~\ref{rank2} applied to $(D,h_1)$ shows
that there is a quadratic extension $L/F$ such that $L$-rank of
$\gSU(D,h_1)$ is at least $2$. Note that the conditions of this
lemma are satisfied since the anisotropic kernel of $\gSU(D,h_1)$
over $K$ is a group of classical type $D_4$ with trivial Tits
algebras. An inspection of possible Tits indices of $G$ over $L$
shows that $G$ is either quasi-split over $L$ or has Tits index
(b) in the above list of Tits indices of outer type ${\cat E}_6$.
In all cases we have
$X_{S_1}(L)\not=\emptyset,\,X_{S_2}(L)\not=\emptyset,\,
X_{S_3}(L)\not=\emptyset$ and hence
$n(X_{S_1})=n(X_{S_2})=n(X_{S_3})=2$.

If $D=F$ and $h$ is a $10$-dimensional quadratic form over $F$, we
can apply the same argument as above. Namely, since $h$ is
isotropic over $K$, we can write it in the form $h=h_1\oplus h_2$
where $h_2$ is a $2$-dimensional quadratic form splitting over $K$
and $h_1$ is an $8$-dimensional quadratic form with trivial
discriminant. Arguing as in Lemma~\ref{rank2} we easily see that
$h_1$ has rank at least $2$ over a proper quadratic extension
$L/F$. So the result follows.
\end{proof}



\subsection{Basic type $S_1$}

Let $L\in \cA(X_{S_1})$ and $[L:F]=2$. As $K\in \cA(X_{S_1})$, it
suffices to show that $L$ and $K$ are $X_{S_1}$-equivalent.
%We may assume that $L\not=K$ since otherwise there is nothing to prove.
%To show that $L\sim_{X_{S_1}} K$
We shall construct an $F$-subgroup $H\subset G$ of classical type
${\cat D}_5$ with the properties:

\medskip

\noindent 1) {\it there is an $F$-embedding $Y\to X_{S_1}$, where
$Y$ is the involution variety corresponding to $H$;}

\noindent 2) {\it $L,K\in \cA(Y)$}.

\medskip

\noindent Property 2) implies that $L$ and $K$ are $Y$-equivalent,
by Theorem~\ref{inv}, and property 1) implies that $L$ and $K$ are
$X_{S_1}$-equivalent.

To construct $H$ with the required properties we repeat the
argument in \ref{splittingnumber} by verbatim. Namely, $G$
contains a parabolic subgroup $P$ of type $S_1$ over $L$. We may
assume that $P$ is standard in the corresponding ordering of the
root system of $G$. Let $\tau$ be the nontrivial automorphism of
$L/F$. Without lost of generality we may assume that $H_1=P\cap
\tau(P)$ is a reductive part of $P$, by Lemma~\ref{opposite}. It
is an $F$-reductive group whose semisimple part is a simple simply
connected group of type ${\cat D}_4$ generated by roots
$\alpha_2,\alpha_3,\alpha_4,\alpha_5$ and whose central torus
$T_1$ has dimension $2$. This torus is isotropic over $L$ and
splits over $K\cdot L$.

Let $T_2\subset T_1$ be the kernel of the restriction mapping
${\tilde \alpha}|_{T_1}$ where $\tilde \alpha$ is the highest root
in the chosen ordering of the root system of $G$. Below we shall
show that $T_2$ is $F$-defined. Then the centralizer
$H_2=C_G(T_2)$ is a reductive $F$-group whose semisimple part
$H=[H_2,H_2]$ is a simple simply connected group of type ${\cat
D}_5$ over $F$. We claim that it has the required properties.

Let us check property 1). The Dynkin diagram of $H$ is of the form
$$
\begin{picture}(150,25)
\put(00,00){\line(1,0){60}}
%\put(20,00){\line(1,0){40}}
%\put(120,00){\line(1,0){40}} \put(190,00){\line(1,0){20}}
\put(85,0){\oval(50,15)[l]}
%\put(30,00){\dots} \put(100,00){\dots}
%\put(170,00){\dots}
\put(0,0){\circle*{3}} \put(30,0){\circle*{3}}
\put(60,0){\circle*{3}}
%\put(70,0){\circle*{3}}
%\put(90,0){\circle*{3}} \put(120,0){\circle*{3}}
%\put(140,0){\circle*{3}} \put(160,0){\circle*{3}}
%\put(190,0){\circle*{3}} \put(210,0){\circle*{3}}
\put(85,7){\circle*{3}} \put(85,-8){\circle*{3}}
%\put(70,0){\circle{10}} \put(140,0){\circle{10}}
\put(-10,8){$-\tilde \alpha$} \put(20,8){$\alpha_2$}
\put(45,8){$\alpha_4$}
%\put(133,8){$\alpha_{rd}$}
\put(95,8){$\alpha_3$} \put(95,-8){$\alpha_5$}
\end{picture}
$$

\medskip

\noindent The corresponding involution variety $Y$ is the variety
of parabolic subgroups in $H$ of type
$S_1=\{\alpha_2,\alpha_4,\alpha_3,\alpha_5\}$. The subgroup
$P_H=P\cap H$ of $H$ is parabolic of type $S_1$. Then the required
embedding is given by $H/P_H\to G/P,\ hP_H\to hP$. It is well
defined and it is easy to check that it is $F$-defined using the
twisted argument.

To check 2) we need information about the structure of $T_1$. It is
proven in \cite[Theorem $2.12$]{cht}, that $T_1$ viewed over $L$ is
isomorphic to $T_1\simeq R_{E/L}({\bf G}_{m,E})$ where $E=L\cdot K$.
Thus $T_1$ viewed over $L$ is an almost direct product of a
$1$-dimensional split $L$-subtorus, say $V_1$, and a $1$-dimensional
$L$-anisotropic torus $V_2=R^{(1)}_{E/L}({\bf G}_{m,E})$. It follows
that ${\rm Gal}(L/F)$ preserves $V_1,V_2$ and hence both of them are
$F$-defined.
%As $V_1$ splits over $L$, it is of the form
%$V_1=R^{(1)}_{L/F}({\bf G}_{m,L})$.
Note that computation in \cite{cht} shows that $V_1$ is contained
in $H$ and commutes with the subgroup of $H$ generated by
$\alpha_2,\alpha_4,\alpha_3,\alpha_5$. The last implies that $H$
is $L$-isotropic and $L\in\cA(Y)$.
\begin{lemma}
We have $V_2=T_2$ and $T_2$ splits over $K$.
\end{lemma}
\begin{proof} The $2$-torsion part of $T_1$ is of the form
$R_{E/L}(\mu_2)$. One easily checks the centralizer of $-1\in
R_{E/L}(\mu_2)$ in $G$ is $H\cdot T_2$, hence $T_2$ is
$L$-defined. But $T_1$ contains only two $L$-defined subtori,
namely $V_1$ and $V_2$. As $V_1\subset H$, we get $T_2=V_2$.

Assume that $T_2$ is not split over $K$. Let $Z$ be the variety
$Z$ of Borel subgroups of $H$ over $K$. The group $H$ being
quasi-split over $K(H/B)$ contains a split torus over $K(H/B)$,
say $T_3$, of dimension at least $4$. One easily checks that
$C_G(T_3)$ is a maximal torus in $G$ containing $T_2$. As $K$ is
algebraically closed in $K(Z)$, the torus $T_2$ is still
anisotropic over $K(Z)$ and hence $C_G(T_3)$ does not split over
$K(Z)$.

On the other side, as rank of $G$ over $K(H/B)$ is at least $4$,
Tits classification implies that $G$ is split over $K(H/B)$ (see
the diagrams in Section~\ref{1e6}) and hence so is the torus
$C_G(T_3)$ because any $K(Z)$-split torus in $G$ is contained in a
maximal $K(Z)$-split torus of $G$. This contradicts our assumption
that $T_2$ is anisotropic over $K(H/B)$.
\end{proof}

As $T_2$ is $K$-split, the $K$-anisotropic kernel of $G$ is
contained in $H_2=C_G(T_2)$, hence $G$ and $H$ have the same
$K$-anisotropic kernel. Then Tits classification of groups of
inner type ${\cat E}_6$ shows that $K\in \cA(Y)$.


\subsection{Basic type $S_2$}\label{types2}


Assume that $L\in\cA(X_{S_2})$ and $[L:F]=2$. Let $\tau$ be the
nontrivial automorphism of $L/F$. Choose a parabolic subgroup
$P\subset G$ over $L$ of type $S_2$ such that $H'=P\cap\tau(P)$ is a
reductive part of $P$. It is a reductive subgroup of $G$ over $F$
whose semisimple part $[H',H']$ has type $A_5$ and is generated by
roots $\alpha_1,\alpha_3,\ldots,\alpha_6$ in a proper ordering of
the corresponding root system of $G$. The centralizer
$H''=C_G([H',H'])$ is an $F$-defined subgroup of $G$ of type ${\cat
A}_1$ containing the central torus of $H'$. It is generated by the
root subgroups $U_{\pm{\tilde \alpha}}$ where $\tilde\alpha$ is the
highest root.

Consider $H=H''\cdot [H',H']$. It is an $F$-defined semisimple
subgroup of $G$ of type ${\cat A}_1+ {\cat A}_5$ whose component of
type ${\cat A}_1$ is split over $L$ and is of the form ${\rm
SL}_1(D)$ where $D$ is a quaternion algebra over $F$. It follows
that $L\in \cA(SB(D))$. Conversely, any splitting field of $D$ is
contained in $\cA(X_{S_2})$.

Let $L_1$ be another quadratic extension of $F$ such that
$L_1\in\cA(X_{S_2})$.  We are going to show that $L_1\in
\cA(SB(D))$. Theorem~\ref{equiv} then implies $L_1\sim_{SB(D)} L$
and hence $L_1\sim_{X_{S_2}} L$.

Applying the above construction to $L_1$ we can construct an
$F$-subgroup $H_1$ of $G$ of type ${\cat A}_1+ {\cat A}_5$. Its
component of type ${\cat A}_1$ correspond to quaternion algebras
$D_1$ splitting over $L_1$ and it suffices to show that $D_1=D$.

$H_1$ and $H$ are related to subsystems of type ${\cat A}_1+ {\cat
A}_5$ in the root system ${\cat E}_6$. Since every two roots in the
root system of type ${\cat E}_6$ are conjugate by an element in the
corresponding Weyl group $W({\cat E}_6)$, so are every two root
subsystems of type ${\cat A}_1+ {\cat A}_5$. Then $H_1,H$ are
conjugate over $F_{sep}$. Let $H_1=gHg^{-1}$ where $g\in
G(F_{sep})$. Fixing $F$-defined maximal tori $T_1,T$ in $H_1,H$
respectively, we may assume additionally that $T_1=gTg^{-1}$. As
$H_1,H,T_1,T$ are $F$-defined, for each $\sigma\in{\rm
Gal}(F_{sep}/F)$ we have $a_{\sigma}=g^{-\sigma+1}\in N_G(H)\cap
N_G(T)$. It follows that $H_1={^{\lambda}}H$ where
$\lambda=(a_{\sigma})$; in particular we have $D_1={^{\lambda}}D$.

The torus $T$ can be decomposed as $T=T'T''$ where $T',T''$ are
intersections of $T$ with simple components of $H$ of types ${\cat
A}_1$ and ${\cat A}_5$ respectively. Since groups of type ${\cat
A}_1$ have no nontrivial outer automorphisms, one checks that
$a_{\sigma}$ can be decomposed as
$a_{\sigma}=a_{\sigma}'a_{\sigma}''$ where $a_{\sigma}'$ and
$a_{\sigma}''$ are contained in the normalizers of $T'$ and $T''$ in
the simple components of $H$ respectively. Thus twisting of $D$ is
given actually by the cocycle $\lambda'=(a_{\sigma}')$. This cocycle
takes values in $\gSL_1(D)$ and hence twisting does not change $D$
since any cocycle in $Z^1(F,\gSL_1(D))$ is equivalent to a cocycle
with coefficients in the center $\mu_2$ of $\gSL_1(D)$.

\subsection{Basic type $S_3$} Let $L_1,L_2\in \cA(X_{S_3})$ be
two quadratic extensions of $F$ contained in $\cA(X_{S_3})$. As
$\cA(X_{S_3})\subset \cA(X_{S_2})$, the above construction applied
to $L_1$ or $L_2$ gives us an $F$-defined subgroup $H$ of $G$ of
type ${\cat A}_1+ {\cat A}_5$. Its simple components, say $H'$ and
$H''$, are of the form $H'=\gSL_1(D)$ and $H''=\gSU(A,\tau)$ where
$D$ is a quaternion algebra over $F$ splitting over $L_1,L_2$ and
$A$ is a central simple algebra of degree $6$ over $K$ equipped with
an involution $\tau$ of the second kind.

Any field extension $L/F$ splitting $D$ makes $H''$ a strongly inner
form of a quasi-split group of type ${\cat A}_5$, by
Theorem~\ref{fact2}, hence it also splits $A$. Then
$[A]=[D\otimes_F K]$ in the Brauer group ${\rm Br}(K)$ and hence
$H''\simeq \gSU(T,h)$ where $T=D\otimes_F K$ and $h$ is a
skew-hermitian form of dimension $3$ given on a vector space $V$
over $T$ equipped with the standard involution, say $\sigma$, of the
second kind.

We are also given that $H''$ is isotropic over $L_1,L_2$, i.e. $h$
represents zero over $L_1,L_2$. Since $L_1,L_2$ splits $T$ and have
degree $2$ over $F$, there are $1$-dimensional $T$-submodules
$U_1,U_2$ in $V$ such that $h$ restricted to $U_1,U_2$ is isotropic
over $L_1$ and $L_2$ respectively. This fact can be reformulated as
follows. Let $v_1\in U_2$ and $v_2\in U_2$ be two nontrivial
vectors. Let $\sigma_1,\sigma_2$ be involutions on $T$ corresponding
to the elements $a_1=h(v_1,v_1)$ and $a_2=h(v_2,v_2)$, i.e.
$\sigma_1,\sigma_2$ are compositions of $\sigma$ with the inner
conjugation given by $a_1,a_2$. Then $\gSU(T,\sigma_1)$ and
$\gSU(T,\sigma_2)$ are split over $L_1$ and $L_2$ respectively.

Clearly $\gSU(T,\sigma_1)$ and $\gSU(T,\sigma_2)$ are subgroups in
$H''={\rm SU}(T,h)$ and being groups of type ${\cat A}_1$ they are
of the form $\gSL_1(D_1)$ and $\gSL_1(D_2)$ where $D_1,D_2$ are
quaternion algebras over $F$ contained in $T$ such that $D_1\otimes
K=D_2\otimes K=T$. Recall that the pure part $D_i^{\circ}$ of $D_i$
is given by
$$D_i^{\circ}=\{\,x\in T\mid \sigma_i(x)=-x\}=\{\,x\in D^{\circ}\otimes K \mid
a_i\sigma(x)a_i^{-1}=-x\}.$$

We now note that $v_i$ is defined up to $t_i\in T$, hence $D_i$ is
defined up to conjugation in $T$. Since by our construction, $D_i$
and $D$ are split over $L_i$, we can modify  $v_i,a_i$ and $D_i$
(by replacing them with $t_iv_i$, $t_ia_i\sigma(t_i)$ and
$t_iD_it_i^{-1}$ for a proper element $t_i\in T$) such that $D\cap
D_i$ contains a common subfield isomorphic to $L_i$.

More precisely, this common subfield can be obtained as follows. As
$a_i$ is skew with respect to $\sigma$ we can write it in
 the form $a_i=a_i'+a_i''$ where $a_i'\in K$ and $a_i''\in
 D^{\circ}$. Modifying $a_i$, if necessary, we may assume that
 $a_i''\not=0$. Then we have
 $$
 D^{\circ}\cap D_i^{\circ}=\{\,d\in D^{\circ}\mid
 \sigma_i(d)=-d\}=\{\,d\in D^{\circ}\mid a_ida_i^{-1}=d\}.
$$
It follows that $D^{\circ}\cap D_i^{\circ}$ has dimension $1$ and
is generated by $a_i''$.

We are ready to finish the proof. Consider a vector
$v(t)=(1-t)v_1+tv_2$ where $t$ is an indeterminate. Let
$a(t)=h(v(t),v(t))$. Write $a(t)=a(t)'+a(t)''$ as above. Here
$a(t)'\in K\otimes_F F(t)$ and $a(t)''\in D^{\circ}\otimes_F
F(t)$. Clearly, $a(t)''$ generate a maximal subfield $L(t)$ in
$D\otimes_F F(t)$ splitting $D\otimes_F F(t)$ and making $H''$
isotropic. This implies that $L(t)\in \cA(X_{S_3})$. Since
$v(0)=v_1$ and $v(1)=v_2$, arguing as in Theorem~\ref{quadforms}
we see that $L(t)$ has two discrete valuations with residue fields
isomorphic to $L_1,L_2$.

\section{Type ${\cat E}_7$}
By Tits classification~\cite{t66}, if $G$ is isotropic over an
extension $L/F$ but not split, then its Tits index is one of the
following:
$$
\begin{array}{ll}
\mbox{a)} \qquad
\begin{picture}(105,45)
\put(00,02){\line(1,0){100}} \put(60,02){\line(0,1){20}}
\put(00,02){\circle*{3}} \put(20,02){\circle*{3}}
\put(40,02){\circle*{3}} \put(60,02){\circle*{3}}
\put(80,02){\circle*{3}} \put(100,02){\circle*{3}}
\put(60,22){\circle*{3}} \put(00,02){\circle{10}}
%\put(60,02){\circle{10}}\put(80,02){\circle{10}}
%\put(100,02){\circle{10}}
\put(-5,-11){$\alpha_7$}
\put(15,-11){$\alpha_6$} \put(35,-11){$\alpha_5$}
\put(55,-11){$\alpha_4$} \put(75,-11){$\alpha_3$}
\put(95,-11){$\alpha_1$} \put(55,30){$\alpha_2$}
\end{picture}
\qquad & \mbox{b)} \qquad
\begin{picture}(125,45)
\put(00,02){\line(1,0){100}} \put(60,02){\line(0,1){20}}
\put(00,02){\circle*{3}} \put(20,02){\circle*{3}}
\put(40,02){\circle*{3}} \put(60,02){\circle*{3}}
\put(80,02){\circle*{3}} \put(100,02){\circle*{3}}
\put(60,22){\circle*{3}} %\put(00,02){\circle{10}}
%\put(20,02){\circle{10}}
\put(100,02){\circle{10}}
\put(-5,-11){$\alpha_7$} \put(15,-11){$\alpha_6$}
\put(35,-11){$\alpha_5$} \put(55,-11){$\alpha_4$}
\put(75,-11){$\alpha_3$} \put(95,-11){$\alpha_1$}
\put(55,30){$\alpha_2$}
\end{picture}\\
\mbox{c)} \qquad
\begin{picture}(105,45)
\put(00,02){\line(1,0){100}} \put(60,02){\line(0,1){20}}
\put(00,02){\circle*{3}} \put(20,02){\circle*{3}}
\put(40,02){\circle*{3}} \put(60,02){\circle*{3}}
\put(80,02){\circle*{3}} \put(100,02){\circle*{3}}
\put(60,22){\circle*{3}} \put(00,02){\circle{10}}
\put(20,02){\circle{10}}\put(100,02){\circle{10}}
%\put(100,02){\circle{10}}
\put(-5,-11){$\alpha_7$} \put(15,-11){$\alpha_6$}
\put(35,-11){$\alpha_5$} \put(55,-11){$\alpha_4$}
\put(75,-11){$\alpha_3$} \put(95,-11){$\alpha_1$}
\put(70,20){$\alpha_2$}
\end{picture}
\end{array}
$$
\medskip

\noindent Thus we need to consider the following basic types only:
$$
S_1=\{\alpha_1,\ldots,\alpha_6 \}, \ \
S_2=\{\alpha_2,\ldots,\alpha_7\}, \ \
S_3=\{\alpha_2,\ldots,\alpha_5\}, \ \ S_4=\emptyset.
$$
If $p=3$, then we can repeat the same argument as in the case of
type $^1{\cat E}_6$. Consider the case $p=2$.
\begin{lemma} Let $F$ be $2$-special. Then we have $n(S_1)=n(S_2)=n(S_3)=2.$
\end{lemma}
\begin{proof}
By~\cite{tits}, $G$ contains an $F$-defined subgroup $H$ of type
${\cat E}_6$. It is a strongly inner form of a quasi-split group of
type ${\cat E}_6$, since $F$ is $2$-special. We know that any such a
group (and hence $G$) has rank at least $2$ over a proper quadratic
extension $E/F$. The above diagrams show that
$X_{S_i}(E)\not=\emptyset$ for $i=1,2,3$.
\end{proof}
By Lemma~\ref{isotropic}, if $E/F$ is a quadratic extension such
that $X_{S_1}(E)\not=\emptyset$, then $X_{S_3}(E)\not=\emptyset$.
Hence it suffices to consider types $S_2,S_3$ only.

\subsection{Basic type $S_2$}\label{s2} Repeating verbatim the argument
in~\ref{types2} we find that $G$ contains an $F$-defined subgroup
$H$ of type ${\cat A}_1+ {\cat D}_6$ such that
$\cA(X_{S_2})=\cA(SB(D))$ where $D$ is a quaternion algebra over $F$
corresponding to the simple component of $H$ of type ${\cat A}_1$.
The connectedness of $\cA(S_2)$ follows from Theorem~\ref{equiv}.

\subsection{Basic type $S_3$} Let $L/F$ be a quadratic
extension belonging to $\cA(X_{S_3})$. Let $H_1$ and $H_2$ be the
simple components of types ${\cat A}_1$ and ${\cat D}_6$
respectively of the group $H$ constructed in~\ref{s2}. We have
$H_1\simeq \gSL_1(D)$ where $D$ is a quaternion algebra over $F$.
The above Tits diagrams show that the anisotropic $L$-kernel of
$H_2$ is of the form $\gSpin(f)$ where $f$ is a $4$-Pfister form. It
follows that up to isogeny $H$ is of the form $\gSU(T,h)$ where $T$
is a quaternion algebra over $F$ equipped with the standard
symplectic involution $\tau$ and $h$ is a skew hermitian form with
respect to $\tau$ defined over a $6$-dimensional vector space $V$
over $T$.
\begin{lemma} If $T$ is nontrivial, then we have $D=T$.
\end{lemma}
\begin{proof} If $E/F$ is an extension splitting $D$, then $H_2$
viewed over $E$ is a strongly inner form of a split group of type
${\cat D}_6$. Hence $H_2$ is of the form $\gSpin(g)$ where $g$ is a
$6$-dimensional quadratic form from $I^3$. It follows that $T$ is
split over $E$. As this is true for all splitting fields of $D$, the
quaternion algebras $D$ and $T$ coincide.
\end{proof}
Let $Y$ be the involution variety corresponding to ${\rm
SO}(D,h)$. By our construction, $L\in \cA(Y)$. Conversely, let
$E\in\cA(Y)$. We noted above that $H_2$ viewed over $E$ is related
to the quadratic form $g$ of dimension $12$ from $I^3$. Since $g$
is isotropic over $E$ and since $10$-dimensional quadratic forms
from $I^3$ are isotropic, we conclude that $E$-rank of $H_2$ is at
least $2$. This implies that $E$-rank of $G$ is at least $2$. An
inspection of the above Tits diagrams shows that $E\in
\cA(X_{S_3})$. Thus we showed that $\cA(Y)=\cA(X_{S_3})$ and the
connectedness of $\cA(Y)$ follows from Theorem~\ref{inv}.

\section{Two examples of a nontrivial $\overline{\CH}_0(X)$}

In this section we give two examples of projective homogeneous
varieties $X$ with nontrivial group $\overline{\CH}_0(X)$.

\subsection{First Example}

We owe to Vishik the observation that \cite{KarpMerk90} essentially
contains an example. Let $F$ be a field of characteristic different
from 2 and let $a,b\in F\m$. Let $Q_{a,b}$ be the quadric of the
2-fold Pfister form $\langle\!\langle a,b\rangle\!\rangle=\langle
1,-a,-b,ab\rangle$. In the split case (over a quadratic extension
$E/F$), $Q_{a,b}$ is isomorphic to $\P_F^1\times\P_F^1$, in
particular, there are two families of lines on $Q_{a,b}$. Let
$l_{a,b}$ be one of these lines and denote by $u_{a,b}\in
\CH_1(Q_{a,b})$ the push-forward of the class of $l_{a,b}$ under the
morphism $(Q_{a,b})_E\ra Q_{a,b}$.

Let $X_{a,b}$ be the variety of isotropic planes of
$\langle\!\langle a,b\rangle\!\rangle$. Over the field
$E\in\cA(X_{a,b})$, the variety $X_{a,b}$ is the disjoint union of
two copies of the projective line. We can view $l_{a,b}$ as a
rational point of $X_{a,b}$ over $E$. Denote the push-forward of the
class of $l_{a,b}$ in $\CH_0 (X_{a,b})$ by $x_{a,b}$, so $x_{a,b}$
is a $0$-cycle of degree 2. The incidence correspondence between
$Q_{a,b}$ and $X_{a,b}$ induces a map $\CH_0(X_{a,b})\ra
\CH_1(Q_{a,b})$ taking $u_{a,b}$ to $x_{a,b}$.


Let $q=\langle 1,-a,-b,ab,-c,-d,cd\rangle$ for some $a,b,c,d\in F\m$
and let $Q$ be the corresponding $5$-dimensional quadric. We have
natural embeddings $Q_{a,b}\hookrightarrow Q$ and
$Q_{c,d}\hookrightarrow Q$. Denote the images of $u_{a,b}$ and
$u_{c,d}$ in $\CH_1(Q)$ by the same symbols.

It is shown in \cite[Cor. 5.2]{KarpMerk90} that in the case $-1\in
F^{\times 2}$, the classes $u_{a,b}$ and $u_{c,d}$ are equal if and
only if the Pfister form $\langle\!\langle a,b,c,d\rangle\!\rangle$
is isotropic.

Denote by $X$ the variety of isotropic planes of $q$. The images of
$x_{a,b}$ and $x_{c,d}$ in $\CH_0(X)$ will be denoted by the same
symbols. We have the incidence correspondence between $Q$ and $X$
that induces a homomorphism $\CH_0(X)\ra \CH_1(Q)$ taking
$x_{a,b}-x_{c,d}$ to $u_{a,b}-u_{c,d}$.

If $F$ is a field such that $-1\in F^{\times 2}$ and the Pfister
form $\langle\!\langle a,b,c,d\rangle\!\rangle$ is anisotropic, the
class $u_{a,b}-u_{c,d}$ is nontrivial and therefore
$x_{a,b}-x_{c,d}\neq 0$. As $\deg(x_{a,b}-x_{c,d})=0$, the class
$x_{a,b}-x_{c,d}$ represents a nontrivial element of
$\overline{\CH}_0(X)$. Note that $X$ corresponds to the subset
$\{\alpha_1,\alpha_3\}$ of the set of vertices of the Dynkin diagram
${\cat B}_3$.

\subsection{Second Example}


We shall give an example of a 3-dimensional projective homogeneous
variety $X$ with a nontrivial group $\overline{\CH}_0(X)$. Note that
this is an example of the smallest possible dimension in view of
Proposition \ref{dim2again}.

Let $C_i$, $i=1,2,3$, be three conic curves corresponding to
quaternion algebras $Q_i$ over $F$ and let $X=C_1\times C_2\times
C_3$. For every subset $S\subset \{1,2,3\}$ let $Q_S$ be the tensor
product of $Q_i$ for all $i\in S$ (in particular,
$Q_{\emptyset}=F$). We define the complex
\begin{equation}\label{complex}
\coprod_{S\neq\emptyset} K_1(Q_S)\xra{\alpha} (F\m)^3\xra{\beta}
H^3(F,\Z/2\Z)
\end{equation}
by the following formulas:
\[
\alpha(x)=(a_1,a_2,a_3),\quad a_i=\prod_{S\ni i} \Nrd_S(x_S),
\]
\[
\beta(a_1,a_2,a_3)=(a_1)\cup [Q_1]+ (a_2)\cup [Q_2]+(a_3)\cup [Q_3],
\]
where $\Nrd_S:K_1(Q_S)\ra K_1(F)=F\m$ is the reduced norm
homomorphism, $(a)$ denotes the class in $H^1(F,\Z/2\Z)$
corresponding to $a\in F\m$ under the Kummer isomorphism and $[Q]$
denotes the class of $Q$ in $H^2(F,\Z/2\Z)$.

\begin{proposition}\label{3conics}
The group $\overline{\CH}_0(X)$ is isomorphic to the homology group
of the complex $(\ref{complex})$.
\end{proposition}

\begin{proof}
In the spectral sequence (\ref{bgq}) for $X$ the only possibly
non-trivial differential coming to $E_*^{3,-3}$ is
$d_2:A^1(X,K_2)=E_2^{1,-2}\ra E_2^{3,-3}=\CH_0(X)$ and all the
differentials coming to $E_*^{1,-2}$ are trivial. Therefore, the
sequence
\[
K_1(X)^{(1)}\ra A^1(X,K_2)\xra{d_2}\CH_0(X)\xra{e'} K_0(X)
\]
is exact. As the image of $d_2$ coincides with
$\Ker(e')=\overline{\CH}_0(X)$, it is sufficient to identify the
cokernel of the first homomorphism in the sequence with the homology
group of the complex $(\ref{complex})$.

The group $A^1(X,K_2)$ was computed in \cite{Merkurjev95}. If all
$Q_i$ are split, we have
\[
A^1(X,K_2)=K_1(F)\tens\CH^1(X)\simeq F\m\tens \Z^3=(F\m)^3,
\]
where we identify $\CH^1(X)$ with $\Pic(X)=\Z h_1\oplus \Z h_2\oplus
\Z h_3\simeq\Z^3$ (here $h_i$ is the class of a rational point in
$C_i$). In the general case the group $A^1(X,K_2)$ is canonically
isomorphic to a subgroup of $(F\m)^3$, namely to the kernel of the
homomorphism $\beta$.

The group $K_1(X)$ was computed in \cite[\S 8, Th. 4.1]{Quillen73}:
\begin{equation}\label{k1}
K_1(X)=\coprod_S K_1(Q_S).
\end{equation}
The first term of the topological filtration $K_1(X)^{(1)}$ is the
kernel of the natural homomorphism $K_1(X)\ra K_1 F(X)$. We claim
that for every $S$, the corresponding homomorphism $K_1(Q_S)\ra K_1
F(X)$ is the reduced norm map $\Nrd_S$ followed by the natural
homomorphism $K_1 F\ra K_1 F(X)$. Since the group $K_1$ of a field
injects when the field gets extended, we may assume that all $Q_i$
are split. In this case the reduced norm $\Nrd_S$ is an isomorphism
and (\ref{k1}) identifies $K_1(X)$ with the direct sum of $K_1(F)
L_S$, where $ L_S$ is the tensor product of the canonical line
bundles $L_i$ on the $C_i\simeq \P_F^1$ such that $i\in S$. As $
L_S$ maps to $1$ in $K_0 F(X)$, the claim follows.

We have shown that $K_1(X)^{(1)}$ consists of all $x=(x_S)$ such
that $\sum \Nrd_S(x_S)=0$. We claim that the image of $x$ in
$A^1(X,K_2)\subset (F\m)^3$ is equal to $(a_1,a_2,a_3)$, where $a_i$
is the product of $\Nrd_S(x_S)$ for all $S\ni i$. We can assume that
all $Q_i$ are split. Under the identification of $K_1(Q_S)$ with
$K_1(F)$, the element $x$ corresponds to $\sum b_S
L_S=\sum_{S\neq\emptyset} b_S( L_S-1)$, where $b_S=\Nrd_S(x_S)$.
Note that the image of $ L_S-1$ under the natural map
$K_0(X)^{(1)}\ra \CH^1(X)$ is equal to the first Chern class $c_1(
L_S)=\sum_{i\in S}c_1( L_i)=\sum_{i\in S}h_i$. Therefore, the image
of $x$ in $A^1(X,K_2)=\Z h_1\oplus \Z h_2\oplus \Z h_3$ is equal to
$a_1h_1\oplus a_2h_2\oplus a_3h_3$, where $a_i$ is the product of
$b_S$ for all $S\ni i$.

It follows from the claim that the image of the homomorphism
$K_1(X)^{(1)}\ra A^1(X,K_2)\subset (F\m)^3$ is equal to $\Im\alpha$.
\end{proof}

\begin{example}\label{second}
Let $F=k(\!(t)\!)$ be the Laurent power series field over a field
$k$ of characteristic different from 2, $Q_i=(b_i,-t)_F$ for some
$b_i\in k\m$. Let $b_S$ be the product of $b_i$ for all $i\in S$,
$k_S=k(b_S^{1/2})$ and $\Norm_S:k\m_S\ra k\m$ be the norm
homomorphism. Since $(t^n,t^m,t^p)\in \Im(\alpha)$ for all $n,m,p$,
$H^p(F,\Z/2\Z)$ is equal to $H^p(k,\Z/2\Z)\oplus
H^{p-1}(k,\Z/2\Z)\cup (-t)$ and $\Im \Nrd_S=\Im\Norm_S\cdot
F^{\times 2}\cup \Im\Norm_S\cdot t\cdot F^{\times 2}$, the homology
group of (\ref{complex}) is isomorphic to the homology group of the
complex
\[
\coprod_{S\neq\emptyset}k_S\m\xra{\alpha'} (k\m)^3\xra{\beta'}
H^2(k,\Z/2\Z),
\]
where
\[
\alpha'(x)=(a_1,a_2,a_3),\quad a_i=\prod_{S\ni i} \Norm_S(x_S),
\]
\[
\beta'(a_1,a_2,a_3)=(a_1)\cup (b_1)+(a_2)\cup (b_2)+(a_3)\cup (b_3).
\]
There is a field $k$ and the elements $b_i$ so that the complex is
not acyclic (see \cite[\S 5]{ShapiroTignol82}) and therefore
$\overline{\CH}_0(X)\neq 0$ in this case.
\end{example}



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\end{document}
