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%%% %Brauer equivalence in a  homogeneous space
%%% %with connected stabilizer.
%%% 
%%% % Mikhail Borovoi and Boris Kunyavskii
%%% 
%%% 
%%% %27.09.2000
%%%   
%%% 
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\documentstyle[12pt,amscd,amssymb,verbatim]{amsart}


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%\topmatter
\begin{document}
\title[Brauer equivalence]{Brauer equivalence in a homogeneous space with 
connected stabilizer}
\author{Mikhail Borovoi and Boris Kunyavski\u\i }
\thanks{This research was supported by    
the Israel Science Foundation founded by the Israel Academy of Sciences
and Humanities --- Center of Excellence Program.}
\thanks{The first named author was partially supported by the Hermann Minkowski
Center for Geometry.} 
\thanks{The second
named author was partially supported by the Ministry of Absorption (Israel) 
and the Minerva Foundation through the Emmy Noether Research Institute of 
Mathematics.}

\date{}
\address{Borovoi: Raymond and Beverly Sackler School of Mathematical 
Sciences,
Tel Aviv University, 69978 Tel Aviv, Israel} 
\email{borovoi@@math.tau.ac.il} 
\address{Kunyavski\u\i :  Department of Mathematics and Computer Science, 
Bar-Ilan University, 52900 Ramat Gan, Israel} 
\email{kunyav@@macs.biu.ac.il} 






%\begin{comment}





 
\begin{abstract}
Let $G$ be a simply connected algebraic group over a field $k$ of
characteristic 0, $H$ a connected $k$-subgroup of $G$, 
$X=H\backslash G$. 
  When $k$ is a local field or a number field, we compute the set of 
Brauer equivalence classes in $X(k)$.
\end{abstract}

\maketitle

%\endtopmatter



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\def\Bra{\text{Br}_a\,}









\section{Introduction}
\label{subsec:intro}
In this note we investigate the Brauer equivalence in a homogeneous space
$X=H\bks G$, where $G$ is a simply connected  algebraic group 
over a local
field or a number field, and $H$ is a connected subgroup of $G$.
 
 In more detail, 
let $k$ be a field of characteristic 0, and let $\kb$ be a fixed algebraic
closure of $k$.
  For a smooth algebraic variety $Y$ over $k$, 
set $\Yb=Y_\kb=Y\times_k \kb$.
Let $\Br Y$ denote the cohomological Brauer group of $Y$,
$\Br Y=H^2_{\text{\'et}}(Y,\Gm)$.
Set $\Bro Y=\ker[\Br Y\to \Br\Yb]$.
There is a canonical pairing
\begin{equation}
Y(k)\times\Bro Y\to \Br k, \qquad (y,b)\mapsto b(y)%\tag{0.1}
\end{equation}
called the Manin pairing.
 We define the Brauer equivalence on $Y(k)$
as follows: $y_1\sim y_2$ if $(y_1, b)=(y_2, b)$ for all $b\in\Bro Y$.
We denote the set of classes of Brauer equivalence in $Y(k)$ 
by $Y(k)/\Br$.
  Note that we define the Brauer equivalence in terms of $\Bro Y$,
not in terms of $\Bro Y^c$ or $\Br Y^c$, where $Y^c$ is a smooth 
compactification of $Y$.


  

The notion of $B$-equivalence for a subgroup $B$ of the Brauer group 
$\Br Y$ was introduced by Manin [Ma1], [Ma2].
  Colliot-Th\'el\`ene and Sansuc [CT/Sa1] investigated the 
Brauer equivalence in algebraic tori (they defined the Brauer equivalence
in terms of the Brauer group of a smooth compactification). The Brauer 
equivalence in reductive groups was studied in [Th]. 

 
Let $G$ be a simply connected semisimple algebraic group over $k$.
  Let $H$ be a connected subgroup of $G$. 
  We denote by $H\tor$ the biggest toric quotient group of $H$.
 We are interested in the Brauer equivalence in the set
$X(k)$ where $X=H\bks G$.
%  We denote by $X(k)/\Bro$ the set of $\Bro$-equivalence classes in $X(k)$. 
%and by $X(k)/\text{R}$ the set of R-equivalence classes.
 
  We compute $X(k)/\Br$ when $k$ is a local field. 
Namely, we prove that there is a bijection
$$
X(k)/\Br\toeq\im[\ker[H^1(k,H)\to H^1(k,G)]\to H^1(k,H\tor)]
$$
(Theorem 2.1).
  Moreover, when $k$ is a nonarchimedean local field, we prove that there is
a bijection $X(k)/\Br\to H^1(k, H\tor)$ (Theorem 2.2).

We also compute $X(k)/\Br$ when $k$ is a number field.
We prove that there is a bijection
$$
X(k)/\Br\toeq \im\left[\ker[H^1(k,H)\to H^1(k,G)]\to
\bigoplus_v H^1(k_v, H\tor)\right]
$$
(Theorem  3.1), where $v$ runs over the set of places of $k$.
  Moreover when $k$ is a totally  imaginary number field, 
we prove that there is a bijection
$$
X(k)/\Br \toeq H^1(k,H\tor)/\Sh^1(k,H\tor)
$$ 
(Theorem 3.6), where $\Sh^1$ denotes the Shafarevich--Tate kernel.

In Example 3.11 we compute $X(k)/\Br$ when $X$ is a symmetric space of 
a simply connected almost simple group over a totally imaginary
number field.

Note that all our results remain true when $G$ is any simply connected
algebraic group (an extension of a simply connected semisimple group
by a unipotent group), not necessarily a simply connected semisimple group.

\medskip

\noindent
{\it Acknowledgements}. The paper was finished while the authors 
were visiting Sonderforschungsbereich 343 ``Diskrete Strukturen 
in der Mathematik'' (University of Bielefeld) whose hospitality 
and support are gratefully appreciated. 
We thank D.~Harari and A.~N.~Skorobogatov 
for help in proving Theorem  1.5. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Generalities over an arbitrary field}
\begin{subsec}
We introduce some notation.
% Let $k$ be a field of characteristic 0, and let $\kb$ be a fixed algebraic 
% closure of $k$. 
  For a smooth algebraic variety $Y$ over a field $k$ of characteristic 0
%, set $\Yb=Y\times_k \kb$. 
let $U(Y)=k[Y]^\times/k^\times$. 
%Denote by $\gg$ the absolute Galois group of $k$. 
  Let $\Pic Y$ denote the Picard group of $Y$.
  Let $\Br Y$ and $\Bro Y$ be as in the Introduction.
%  Let $\Br Y$ denote the cohomological Brauer group of $Y$,
%$\Br Y=H^2_{\text{\'et}}(Y,\Gm)$.
Set $\Bra Y =\coker[\Br k \to\Bro Y]$.
Assume that $Y$ has a $k$-rational point $y$, and define
$$
\Brr_{y} Y=\ker[\Bro Y @>{y^*}>> \Br k]
$$
where  $y^*$ is the specialization map.

  We prove that  $\Brr_{y}Y\simeq\Bra Y$.
Consider the composed map
$$
\Br k\to\Bro Y @>{y^*}>>\Br k\,,
$$
it is the identity of $\Br k$.
  It follows that the exact sequence 
$$
0\to\Brr_y Y\to\Bro Y @>{y^*}>>\Br k\to 0
$$
splits, and we obtain an isomorphism $\Brr_y Y\oplus\Br k\simeq\Bro Y$.
  Thus we obtain an isomorphism $\Brr_y Y\to\Bra Y$ and a splitting
$\Bra Y\to \Bro Y$ of the exact sequence 
$$
0\to\Br k\to\Bro Y\to\Bra Y\to 0.
$$


%There is a canonical pairing
%$$
%Y(k)\times\Bro Y\to \Br k, \qquad (y,b)\mapsto b(x)\tag1.1
%$$
%called the Manin pairing, cf. [Sa], 6.2.
% We define the Brauer equivalence on $Y(k)$
%as follows: $y_1\sim y_2$ if $(y_1, b)=(y_2, b)$ for all $b\in\Bro Y$.
 % Note that we define the Brauer equivalence in terms of $\Bro Y$,
%not in terms of $\Br Y^c$, where $Y^c$ is a smooth compactification of a
%smooth variety $Y$.
\end{subsec}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{subsec}
We wish to investigate the Brauer equivalence in homogeneous spaces.
 Let $G$ be a simply connected semisimple algebraic group over
a field $k$ of characteristic  0.
 Let $H\subset G$ be a connected $k$-subgroup.
Set $X=H\bks G$; then $X$ is a right homogeneous space of $G$ with connected
stabilizer.
  The  variety $X$ has a distinguished $k$-rational point $x_0$, the image
in $X(k)$ of the unit element $e\in G(k)$.

%  Let $H\tor$ denote the biggest quotient group of $H$ which is an algebraic
%torus.
  We construct a map $X(k)\to H^1(k,H\tor)$ taking $x_0$ to 1.
  From an  exact sequence in Galois cohomology of nonabelian groups
(cf. [Se], I-5.4, Prop.~36) we obtain a canonical bijection
$$
X(k)/G(k) @>\sim>> \ker[H^1(k,H)\to H^1(k,G)]\,.
$$
  Thus we obtain a map $X(k)\to H^1(k,H)$.
  Composing this map with the canonical map $H^1(k,H)\to H^1(k,H\tor)$
induced by the homomorphism $H\to H\tor$, we obtain a map 
\begin{equation}
\alp\colon X(k)\to H^1(k,H\tor)\,.%\tag{2.1}
\end{equation}
By construction this map is constant on the orbits of $G(k)$ in $X(k)$.

Let $\X(H)$ denote the group of $k$-characters of  $H$, i.e.
$\X(H)=\Hom_k(H,\Gm)$.
  We have $\X(H)=\X(H\tor)$.
\end{subsec}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{prop}
  There is a canonical isomorphism $\bet\colon\X(H)\toeq \Pic X$.
\end{prop}

\begin{pf}
Let $\chi\in \X(H)$.
Set $L(\chi)=H\bks(\Ga\times G)$, where $H$ acts on $\Ga\times G$ by 
$h*(u,g)=(\chi(h)u, hg)$ (here $h\in H,\ u\in \Ga, \ g\in G$).
  There is a canonical map $L(\chi)\to H\bks G=X$, and one can easily see 
that $L(\chi)$ is a linear bundle over $X$.
  We denote its class in $\Pic X$ by $\bet(\chi)$.
 One can easily see that $\bet\colon\X(H)\to \Pic X$ is a homomorphism.

We wish to prove that $\bet$ is an isomorphism. 
  We need the following fact:
$U(G)=1$.
Indeed, by Rosenlicht's theorem [Ro] $U(G)=\X(G)$ and clearly $\X(G)=1$
because $G$ is semisimple.
  We need also the fact that $\Pic G=0$, cf. [Sa], 6.9(iv).

We construct a map which is inverse to $\bet$.
We follow an idea of Mumford [Mu], Ch.~I, \S 2.
  Let $\pi$ denote the canonical map $G\to X=H\bks G$.
  Let $L\to X$ be a line bundle over $X$. 
  Consider the pullback $\pi^*L$, it is a line bundle over $G$.
Since $\Pic G=0$, the line bundle $\pi^*L$ is trivial.
 This means that there exists an isomorphism
$$
\mu\colon\pi^*L\to \Ga\times G\,.
$$
  There is a canonical action of $H$ on $\pi^*L$, and using $\mu$ we obtain
an action of $H$ on $\Ga\times G$ extending the standard action on $G$,
$h*g=hg$.
  The action of $H$ on $\Ga\times G$ can be written as
$$
h*(u,g)=(\chi_g(h)u,hg) \text{\quad where } h\in H, g\in G, u\in\Ga, 
$$
and $\chi\colon G\times H\to \Gm$ is a regular map.
  If we fix $h$, then $\chi_g(h)\colon G\to \Gm$ is a regular map, in other
words it is a nowhere zero regular function on $G$.
  Since $U(G)=1$, we conclude that $\chi_g(h)$ is constant in $g$. 
  We may therefore write $\chi(h)$ instead of $\chi_g(h)$.

Since $h_1*(h_2*(u,g))=(h_1h_2)*(u,g)$, we obtain that 
$\chi(h_1)\chi(h_2)=\chi(h_1h_2)$.
  Hence $\chi\colon H\to \Gm$ is a regular homomorphism, $\chi\in\X(H)$.

If we start from another isomorphism 
$$
\mu'\colon\pi^*L\to\Ga\times G\,,
$$
then $\mu'$ differs from $\mu$ by a nowhere zero regular function $f$ on
$G$, and this $f$ must be a constant because $U(G)=1$.
  Then the homomorphism $\chi$ is replaced by $\chi'$ where 
$\chi'_g(h)=\chi_g(h)f(hg)f(g)\min$, i.e. $\chi'=\chi$.
  Thus $\chi$ is defined uniquely by $L$.
  One can easily check that we have constructed a homomorphism 
$\del\colon\Pic X\to\X(H)$ and that this homomorphism is inverse to $\bet$.
  Thus $\bet$ is an isomorphism.
\end{pf}


\noindent {\bf 1.3.1.} {\it Remark.}  
  We will use Proposition 1.3 only in the case when $k$ is algebraically
closed. 
  In this case Proposition 1.3  was proved in [Po], Cor.~to Thm.~4.


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


\begin{subsec}
We have seen in the proof of Proposition 1.3 that $U(G)=1$. 
  Hence $U(\Gb)=1$.
It follows that $U(\Xb)=1$.

Since $X(k)\ne\emptyset$ and $U(\Xb)=1$, we have by [Sa], 6.3(iii)
$$
\Bra X=H^1(k,\Pic \Xb)\,.
$$
  We have  $\Brr_{x_0}\,X\simeq\Bra X$.
 By Proposition 1.3, $\Pic \Xb=\X(\Hb)$.
  We obtain
\begin{equation}
\Brr_{x_0}\,X=H^1(k,\X(\Hb))=H^1(k,\X(\Hb\tor))\,.%\tag{1.2}
\end{equation}

There is a canonical cup product pairing
\begin{equation}
H^1(k,H\tor)\times H^1(k,\X(\Hb\tor))\to\Br k\,.%\tag{1.3}
\end{equation}
The pairing (1.3) together with the map (1.1) $X(k)\to H^1(k,H\tor)$ 
and the isomorphism (1.2) defines a pairing
\begin{equation}
X(k)\times\Brr_{x_0}\,X\to\Br k\,.%\tag{1.4}
\end{equation}
\end{subsec}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{theorem} The pairing (1.4) up to sign coincides with the restriction
of the Manin pairing (0.1) to 
$X(k)\times\Brr_{x_0}X\subset X(k)\times\Bro X$. 
\end{theorem}

This theorem will be proved in the Appendix.

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

\section{Brauer equivalence over a local field}

\begin{theorem} 
Let $G,\ H,\ X$ be as in 1.2.
Assume that $k$ is a local field (archimedean or not). 
Then the map (1.1) $\alp\colon X(k)\to H^1(k,H\tor)$ induces a bijection
$$ 
X(k)/\Br\toeq\im[\ker[H^1(k,H)\to H^1(k,G)]\to H^1(k,H\tor)]\,.
$$
\end{theorem}

\begin{pf}
It follows from Theorem 1.5 that two points $x_1,x_2\in X(k)$ are 
Brauer-equi\-va\-lent if and only if $(\alp(x_1),\eta )=
(\alp(x_2),\eta )$
for every $\eta\in H^1(k, \X(\Hb\tor))$.
Since $k$ is a local field, the cup product pairing (1.3)
is perfect (Tate--Nakayama duality, cf. [Mi], Cor.~I.2.4),
and it follows that $x_1$ and $x_2$ are Brauer-equivalent if and only if
$\alp(x_1)=\alp(x_2)$.
  Thus the set of classes of Brauer equivalence is in a bijective 
correspondence with $\im \alp$.
  We see that we must only describe the image of $X(k)$ in $H^1(k,H\tor)$.
  But the image of $X(k)$ in $H^1(k,H)$ is the same as the image of 
$X(k)/G(k)$ and it equals $\ker[H^1(k,H)\to H^1(k,G)]$.
  Hence the image of $X(k)$ in $H^1(k,H\tor)$ is 
$$
\im[\ker[H^1(k,H)\to H^1(k,G)]\to H^1(k,H\tor)]\,,
$$
and the assertion of the theorem follows.
\end{pf}

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


\begin{theorem} Let $G,H,X$ be as in 1.2, and assume that $k$ is a 
non\-archimedean local field. 
 Then the map (1.1) $\alp$ induces a bijection $$X(k)/\Br\to H^1(k,
H\tor).$$
\end{theorem}

\begin{pf}
  Since $G$ is a simply connected group, by Kneser's theorem  (see
 [Pl/Ra], 6.1, Thm.~4) $H^1(k,G)=1$.
  We see now from Theorem 2.1 that $X(k)/\Br$ is in a bijective 
correspondence with $\im[H^1(k,H)\to H^1(k,H\tor)]$. 
  Let $H\ssu$ denote $\ker[H\to H\tor]$, it is an extension of a semisimple
group by a unipotent group. 
Since $k$ is local non\-archimedean and $(H\ssu)\tor=1$,
the map $H^1(k,H)\to H^1(k, H\tor)$ is surjective, cf. [Bo], Cor.~6.4.
This proves the theorem.
\end{pf}


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

\section{Brauer equivalence over a number field}

\begin{theorem} 
Let $k$ be a number field, and let $G,H,X$ be as  in 1.2. 
  Then the map
$$
X(k)\to X(k)/G(k)\toeq \ker[H^1(k,H)\to H^1(k,G)]
\to \bigoplus_vH^1(k_v, H\tor)
$$
induces a bijection 
$$
X(k)/\Br\toeq\im\left[\ker[H^1(k,H)\to H^1(k,G)]\to\bigoplus_v 
H^1(k_v,H\tor)\right]
$$
where $v$ runs over the set of places of $k$.
\end{theorem}


\begin{subsec}
Before proving Theorem 3.1 we note that $\Bro G=\Br k$ (cf. [Sa], 6.9(iv)),
hence every orbit of $G(k)$ in $X(k)$ is contained in one class of Brauer
equivalence.
  It follows that the map $X(k)\to X(k)/\Br$ factors through $X(k)/G(k)$:
$$
X(k)\to X(k)/G(k)\to X(k)/\Br
$$
and these maps are surjective.

For a place $v$ of $k$ consider the map
\begin{equation}
X(k)\to X(k)/\Br\to X(k_v)/\Br\to H^1(k_v, H\tor)%\tag{3.1}
\end{equation}
where the last arrow is defined by Theorem 2.1.
\end{subsec}

\begin{lemma}The map (3.1) $X(k)\to H^1(k_v,H\tor)$
factors
$$
X(k)\to X(k)/G(k)\toeq\ker[H^1(k,H)\to H^1(k,G)]\to H^1(k_v,H\tor)
$$
with the obvious arrows.
\end{lemma}

\begin{pf}
The statement of the lemma shows how to define a map
from $X(k)/G(k)$ to $H^1(k,H\tor)$. 
We can define a map from $X(k_v)/G(k_v)$ to $H^1(k_v,H\tor)$ 
in a similar way. 
Consider the diagram
 %%
\setlength{\unitlength}{0.7mm}
\begin{center}
\begin{picture}(160,86)
\put(25,7){\makebox(0,0){$X(k_v)/G(k_v)$}}
\put(25,52){\makebox(0,0){$X(k)/G(k)$}}
\put(78,27){\makebox(0,0){$X(k_v)/\Br$}}
\put(78,75){\makebox(0,0){$X(k)/\Br$}}
\put(130,7){\makebox(0,0){$H^1(k_v,H\tor)$}}
\put(130,52){\makebox(0,0){$H^1(k,H\tor)$}}
\put(45,7){\vector(1,0){66}}
\put(42,52){\vector(1,0){68}}
\put(25,45){\vector(0,-1){30}}
\put(78,67){\vector(0,-1){30}}
\put(130,45){\vector(0,-1){30}}
\put(42,55){\vector(3,2){23}}
\put(45,10){\vector(3,2){18}}
\put(94,22){\vector(3,-2){17}}
\end{picture}
\end{center}
%%%
It is commutative (the lower triangle is commutative by Theorem 2.1).
  It follows that the diagram
%%
\begin{center}
\begin{picture}(160,60)
\put(25,7){\makebox(0,0){$X(k_v)/G(k_v)$}}
\put(25,52){\makebox(0,0){$X(k)/G(k)$}}
\put(78,52){\makebox(0,0){$X(k)/\Br$}}
%\put(78,75){\makebox(0,0){$X(k)/\Br$}}
\put(130,7){\makebox(0,0){$H^1(k_v,H\tor)$}}
\put(130,52){\makebox(0,0){$X(k_v)/\Br$}}
\put(45,7){\vector(1,0){66}}
\put(42,52){\vector(1,0){19}}
\put(94,52){\vector(1,0){19}}
\put(25,45){\vector(0,-1){30}}
%\put(78,67){\vector(0,-1){30}}
\put(130,45){\vector(0,-1){30}}
%\put(42,55){\vector(3,2){23}}
%\put(45,10){\vector(3,2){18}}
%\put(94,22){\vector(3,-2){17}}
\end{picture}
\end{center}
%%
is commutative, which proves the lemma.
\end{pf}

\begin{lemma}{\rm (cf. [Ma/Ts], 4.5).}
Let $Y$ be a variety over a global field $k$.
Then the map $Y(k)/\Br\to\prod_v Y(k_v)/\Br$ is injective.
\end{lemma}

\begin{pf}
Let $y_1,y_2\in Y(k)$, and assume that $y_1$ and $y_2$ are 
Brauer-equivalent in $Y(k_v)$ for all places $v$ of $k$.
 This means that for every $b_v\in\Bro Y_{k_v}$, $(y_1,b_v)=(y_2,b_v)$.
  Let now $b\in\Bro Y$.
  We wish to compare $(y_1,b)$ and $(y_2, b)$.
  Consider $\loc_v(y_i,b)\in \Br k_v$, $i=1,2$, where $\loc$ means 
localization.
  We have $\loc_v(y_i,b)=(y_i,\loc_v b)$, 
where $\loc_v b \in\Bro Y_{k_v}$.
  By assumption we have $(y_1,\loc_v b)=(y_2,\loc_v b)$.
  We see that $\loc_v(y_1,b)=\loc_v(y_2,b)$ for all $v$.
  It follows that $(y_1,b)=(y_2,b)$, because the map 
$\loc\colon\Br k\to\prod_v\Br k_v$ is injective.
  Thus $y_1$ and $y_2$ are Brauer-equivalent in $Y(k)$.
\end{pf}


\begin{subsec} {\em Proof of Theorem 3.1.}
By Lemma 3.3 the obvious map
\begin{equation}
\ker[H^1(k,H)\to H^1(k,G)]\to\prod_v H^1(k_v, H\tor)%\tag{3.2}
\end{equation}
has the same image as the map
\begin{equation}
X(k)/\Br\to \prod_v X(k_v)/\Br\to\prod_v H^1(k_v,H\tor),%\tag{3.3}
\end{equation}
where the last arrow is described in Theorem 2.1.
  But the image of map (3.2) is contained in $\bigoplus_v H^1(k_v, H\tor)$
(because the image of $H^1(k,H\tor)$ in $\prod_v H^1(k_v, H\tor)$
is contained in $\bigoplus_v H^1(k_v, H\tor)$, cf. e.g. [Vo], 11.3, Cor.~1
of Prop.~1).
  Thus the image of map (3.3) is contained in $\bigoplus_v H^1(k_v, H\tor)$
and is equal to the image of $\ker[H^1(k,H)\to H^1(k,G)]$.

The first arrow in (3.3) is injective by Lemma 3.4, and the second arrow
is injective by Theorem 2.1, and so the composition (3.3) is injective.
Since (3.2) and (3.3) have the same images, we obtain a bijection
$$
X(k)/\Br\toeq\im\left[\ker[H^1(k,H)\to H^1(k,G)]\to\bigoplus_v 
H^1(k_v,H\tor)\right].
$$  
This proves Theorem 3.1.
\qed
\end{subsec}

\begin{theorem} 
In Theorem 3.1 assume that $k$ is a totally imaginary number field.
  Then the bijection of Theorem 3.1 induces a bijection
$$
X(k)/Br\to H^1(k, H\tor)/\Sh^1(k,H\tor).
$$
\end{theorem}

To prove Theorem 3.6 we need a proposition and two corollaries.

\begin{proposition}
Let $k$ be a totally imaginary number field, and $L=(\Fb, \kap)$
a $k$-kernel ($k$-lien) (see [Sp], [Bo], [F/S/S] for a definition), 
where $\Fb$ is a connected linear $\kb$-group such that $\Fb\tor=1$.
   Then every element of $H^2(k,L)$ is neutral.
\end{proposition}

\begin{pf}
The proposition follows from [Bo], Thm.~6.8(iii) and Thm.~6.3(ii).
Note that in the case when $\Fb$ is semisimple, the proposition was
proved by Douai ([Do], Cor.~5.1), see also [Bo], Cor.~6.9.
  The proposition follows also from Douai's result and [Bo], Prop.~4.1.
\end{pf}

\begin{corollary}
Let $k$ be a totally imaginary number field and let
$$
1\to G_1 \to G_2\to G_3\to 1
$$
be an exact sequence of linear $k$-groups.
 If $G_1$ is connected and $G_1\tor=1$, then the map
$H^1(k,G_2)\to H^1(k, G_3)$ is surjective.
\end{corollary}

\begin{pf}
We argue as in [Bo], the proof of Cor.~6.4. Let $\xi\in H^1(k,G_3)$, 
and let $\psi\in Z^1(k,G_3)$ be a cocycle from the class $\xi$. 
According to Springer ([Sp], 1.20), one can associate to $\psi$ 
a $k$-kernel $L_{\psi }=(G_{1\kb }, \kap _{\psi })$ and 
a cohomology class $\del (\psi )\in H^2(k,L_{\psi })$ which is 
the obstruction to lifting $\xi$ to $H^1(k,G_2)$. Since $G\tor _{1\kb }=1$, 
by Prop. 3.7 the class $\del (\psi )$ is neutral, and hence $\xi$ comes 
from $H^1(k,G_2)$.  
\end{pf}

\begin{corollary}
Let $F$ be a connected linear group over a totally imaginary number field
$k$.
  Then the map $H^1(k,F)\to H^1(k, F\tor)$ is surjective.
\end{corollary}

\begin{pf}
We have an exact sequence
$$
1\to F\ssu\to F\to F\tor\to 1
$$
where $(F\ssu)\tor=1$. 
  Now the corollary follows from Corollary 3.8.
\end{pf}

\begin{subsec} {\em Proof of Theorem 3.6}.
Since  $G$ is simply connected and $k$ is a totally imaginary
number field, $H^1(k,G)=1$  (Kneser--Harder--Cher\-nou\-sov, see
[Pl/Ra] \S 6.1, Thm.~6). 
Thus $\ker[H^1(k,H)\to H^1(k,G)]=H^1(k,H)$.
  By Theorem 3.1 $X(k)/\Br$ is in a bijective correspondence with
$$
\im\left[H^1(k,H)\to H^1(k, H\tor)\to\bigoplus_v H^1(k_v,H\tor)\right].
$$
By Corollary 3.9 the map $H^1(k,H)\to H^1(k,H\tor)$ is surjective.
We see that $X(k)/\Br$ is in a bijective correspondence with
$$
\im\left[H^1(k,H\tor)\to\bigoplus_vH^1(k_v,H\tor)\right]=
H^1(k,H\tor)/\Sh^1(k,H\tor).
$$
\qed
\end{subsec}

\begin{example} \rm
Let $G$ be a simply connected almost simple group over a number field
$k$, $H\subset G$ a connected $k$-subgroup, $X=H\bs G$. 
Assume that $X$ is a symmetric space, i.e. $H$ is the group of invariants
of an involution of $G$. 
  From the classification of involutions of simple Lie algebras
(see e.g. [He], X-5, p.~514) it follows that $\dim H\tor\leq 1$.

If $H\tor=1$ or $H\tor$ is a one-dimensional split torus,
then $H^1(k_v, H\tor)$ $=1$ for all $v$, and by Theorem 3.1 
$X(k)/\Br$ consists of one element.

If $H\tor$ is a one-dimensional nonsplit torus, then $H\tor$ splits
over a quadratic extension $K$ of $k$.
Assume in addition that $k$ is totally imaginary.
  Then by Theorem 3.6 $X(k)/\Br=H^1(k,H\tor)/\Sh^1(k,H\tor)$.
  Since $K/k$ is cyclic, we have $\Sh^1(k,H\tor)=1$ ([Vo], 11.6, Cor.3), 
and we see that 
$$
X(k)/\Br=H^1(k,H\tor)=k^\times/N_{K/k}K^\times
$$
where $N_{K/k}$ denotes the norm map.
\end{example}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Appendix} \label{app}

In this appendix, we prove Theorem 1.5. 

We use the description of the Manin pairing with the help of torsors given 
in [CT/Sa2], \S 2. We first recall some generalities. 

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

\def\Gmkb{{\Bbb G}_{m,\kb}}
\def\Sb{{\overline S}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{subsec} {\em Torsors.} 
Let $X$ be a smooth geometrically integral $k$-variety such that 
$U(\Xb )=1$ and $\Pic\Xb$ is a finitely generated $\gg$-module,
where $\gg=\Gal(\kb/k)$. 
  Let $S$ be a  $k$-torus.
  Let $(Y, p_Y \colon Y\to X$) be a right $X$-torsor under $S$.

\begin{comment}
 By a (right) $X$-torsor under $S$ we mean a pair $(Y,p_Y)$ where 
$Y$ is an $X$-scheme, $p_Y\colon Y\to X$ is a $k$-morphism locally trivial in 
the \'etale topology, and $S$ acts on $Y$ so that this action extends the 
trivial action of $S$ on $X$, i.e. the canonical morphism 
$Y\times _XS\to Y\times _XY$ given by $(y,s)\mapsto (y,y\cdot s)$ is 
an isomorphism.
\end{comment}

 The set of isomorphism classes of $X$-torsors under $S$ 
is  the first \'etale cohomology group $H^1(X,S)$. We have a canonical 
map 
$$
\chi\colon H^1(X,S)\to\Hom _{\gg }(\XX (\Sb),\Pic\Xb )
$$
defined as follows. To any $X$-torsor $Y$ under $S$ we associate a 
homomorphism $\chi _Y$ sending a character 
$\lam\in\XX(\Sb)$, $\lam\colon \Sb \to \Gmkb$ 
to the push-forward $\lam _*(Y)$ which is an 
$X$-torsor under $\Gmkb$; since $\Pic\Xb =H^1(\Xb,\Gm )$, 
we identify the isomorphism class of $\lam _*(Y)$ with an element of 
$\Pic\Xb$. We call $\chi _Y$ the type of $Y$. 
If $S$ is the N\'eron--Severi  
torus of $X$ (i.e. $\XX (\Sb)=\Pic\Xb$) and $\chi _Y$ is the identity map, the 
torsor $Y$ is called universal. 
\end{subsec}

\begin{subsec} {\em Evaluation map.} 
Let $Y$ be an $X$-torsor under $S$. We then have a natural evaluation 
map $\theta _Y\colon X(k)\to H^1(k,S)$ which associates to every rational 
point $x\in X(k)$ the isomorphism class of the fibre $Y_x$ of $Y$ at $x$. 
 The map $\theta _Y$ allows us to identify the Manin 
pairing with a cup-product. We use the following construction. 
%Consider a natural sequence of $\gg$-modules 
%$$
%1\to\kb ^\times\to\kb (X)^\times\to\kb (X)^\times /\kb ^\times\to 1.
%$$
%If $X(k)\neq\emptyset$, this sequence splits ([CT/Sa2], Prop.~2.2.2). 
To $x\in X(k)$ we can associate a homomorphism of $\gg$-modules 
$\sig\colon \kb (X)^\times\to\kb ^\times$, $\sig(f)=f(x)$ for 
$f\in\kb(X)^\times$. 
There is a canonical isomorphism 
$H^1(k ,\Pic\Xb )\toeq\Bra X$, cf. [Sa], 6.3(iii).
Together with the isomorphism $\Bra X\toeq\Brr_x X$ (see 1.1)
 this yields an embedding 
$t_x\colon H^1(k ,\Pic\Xb )\to\Bro X$ (in  [CT/Sa2], p.~449, this embedding
is denoted by $t_\sig$). 
\end{subsec}

Let $Y^x$ denote an $X$-torsor under $S$ 
of type $\chi _Y\colon\XX (\Sb)\to\Pic\Xb$ which is trivial 
at $x$ (such a 
torsor exists and is unique up to isomorphism, cf. [CT/Sa2], p.~449). 

\begin{theorem} {\text{\rm{([CT/Sa2], Prop. 2.7.10)}}} \label{duke}
Let  
$$
\rho\colon H^1(k,\XX (\Sb))\to H^1(k ,\Pic\Xb ) @>{t_x}>> \Bro X
$$
be the composed map where the left arrow is induced by $\chi _Y$.  
Then the following diagram is 
commutative up to sign:

\begin{equation}
\CD
X(k)   @. \times @. \Bro X @>>> \Br k \\
@V{\theta _{Y^x}}VV @. @AA{\rho }A @\vert \\
H^1(k,S) @. \times @. H^1(k, \XX (\Sb)) @>>> \Br k 
\endCD
\tag{4.1}
\end{equation}
Here the top row is the Manin pairing and the bottom row is the 
cup-product. 
\end{theorem}

\begin{subsec} {\em Abelianization of torsors.}
Let us now suppose that $X$ is as in 1.2, i.e. $X=H\bks G$.
We have a canonical map $G\to X$. 
Define a right action of $H$ on $G$ by $g*h=h^{-1} g$, 
where $g\in G$, $h\in H$.
 With this action the map $G\to H$ is a right
(non-abelian) $X$-torsor under $H$. 
%Denote this torsor by $Y_0$. 
  Set $S=H\tor$.
  We consider the natural homomorphism 
$\psi\colon H\to S$.
  Let $H\ssu$ denote the kernel of $\psi$. 
Let  $\psi _*\colon H^1(k,H)\to H^1(k,S)$ and 
$\psi _*^X\colon H^1(X,H)\to H^1(X,S)$ 
be the induced push-forward maps, they send non-abelian torsors under 
$H$ to abelian torsors under $S$. 
Explicitly, to any right $X$-torsor $Z$ under $H$ we associate an 
$X$-torsor $\psi _*^X(Z)=Z/H\ssu$ under $S$ (abusing notation, we use the
same symbol $\psi _*$ for maps of torsors and their cohomology classes).  
%(In other words, we factorize the action of $H$ by the kernel of $\psi$.)
Denote by $Y$ the torsor under $S$ 
obtained by applying $\psi _*^X$ to $G\to X$, i.e. $Y=H\ssu\bks G$.
Note that by Proposition 1.3, 
we have an isomorphism $\XX(\Sb)\toeq\Pic\Xb$ so that $Y$ is a universal 
torsor. With the notation of Theorem \ref{duke}, $Y=Y^{x_0}$ 
where $x_0\in X(k)$ is the image of the unit element 
of $G$. %under the canonical projection $G\to X$.% Below (see 4.6) 
%we assume that the point $x_0$ is used in the definition of $\rho$.   
\end{subsec}

\begin{lemma} \label{lem:app}
The map $\theta _Y\colon X(k)\to H^1(k,H\tor)$ 
coincides with the map $\alp$ 
defined by (1.1). 
\end{lemma}
\begin{pf}
Indeed, $\alp$ is the composition 
$$
\psi _*\circ\del\colon X(k) @>\del>> H^1(k,H) @>{\psi_*}>>H^1(k,H\tor)
$$ 
where $\del$ is the connecting map.
  Note that the 
image of $x\in X(k)$ under $\del$ coincides with the (isomorphism class of 
the) fibre of $G\to X$ at $x$ (cf. [Se], I.5.4, Prop.~36, [Ha/Sk], 4.1). 
 Furthermore, push-forward commutes with 
specialization, i.e. for any $X$-torsor $Z$ under $H$ and any $x\in X(k)$ 
the fibre $[\psi _*^X(Z)]_x$ coincides with $\psi _*(Z_x)$. 
Hence $\psi _*(\del (x))$ coincides with the 
fibre of the abelian $X$-torsor $Y$ under $H\tor$ at $x$, which is equal to 
$\theta _Y(x)$ (by the definition of $\theta _Y$). Thus $\alp =\theta _Y$. 
\end{pf}

\begin{subsec} 
 To finish the proof of Theorem 1.5, it only remains to apply diagram 
(4.1). 
Indeed, denote by $i$ the isomorphism $\Brr_{x_0}X\toeq H^1(k,\XX (\Sb ))$. 
Then the pairing (1.4) $X(k)\times\Brr_{x_0}X\to\Br k$
is given by 
\begin{equation}
(x,b)\mapsto \alp (x)\cup i(b),
\tag{4.2}
\end{equation}
whereas the Manin pairing (0.1) $X(k)\times\Bro X\to\Br k$
 restricted to $X(k)\times\Brr_{x_0}X$ is given by 
\begin{equation}
(x,b)\mapsto b(x).
\tag{4.3}
\end{equation}
Any $b\in \Brr_{x_0}X$ can be written as $b=\rho (z)$ with 
$z\in H^1(k,\XX (\Sb ))$, where in the definition of $\rho$ (in Theorem 
4.3) we use $x_0$ instead of $x$. 
On applying diagram (4.1) to the Manin pairing (4.3), we get an equality 
(up to sign) 
$$
(x,\rho (z))=\theta _Y(x)\cup z,
$$
where $Y=Y^{x_0}$ in $\theta_Y$.
On the other hand, taking into account that 
$\alp =\theta _Y$ (Lemma \ref{lem:app}) and $i(\rho (z))=z$, 
formula (4.2) gives the same value $\theta _Y(x)\cup z$ 
for $(x,\rho (z))$. Thus formulas (4.2) and (4.3) coincide (up to sign),  
and the Manin pairing restricted to $\Brr_{x_0}X$ coincides 
(up to sign) with pairing (1.4). The theorem is proved.  
 
\end{subsec}

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

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Authors: Mikhail Borovoi and Boris Kunyavskii 

Title: Brauer equivalence in a homogeneous space with connected stabilizer 

Email: borovoi@math.tau.ac.il , kunyav@macs.biu.ac.il 

1991 Mathematics Subject Classification: 14G25; 14M17

Keywords and Phrases: linear algebraic group; homogeneous space; 
Brauer equivalence 

Abstract Text: 
Let G be a simply connected algebraic group over a field k of 
characteristic 0, H a connected k-subgroup of G, X=H\G. When k 
is a local field or a number field, we compute the set of Brauer 
equivalence classes in X(k).


