%\input a4newnew.tex

\hsize=13.50cm    
\vsize=18cm       
\parindent=12pt   \parskip=0pt     
\pageno=1 

% deux jeux d'offsets. Le premier concerne la \magnification=1000
% (si l'image n'est pas agrandie)

\hoffset=15mm    % offset horizontal en \magnification=1000
\voffset=1cm    % offset vertical en \magnification=1000
 
% Le deuxieme concerne  la \magnification=1200
% (lorsque l'image est agrandie de 20%)

\ifnum\mag=\magstep1
\hoffset=-2mm   % offset horizontal en \magnification=1200
\voffset=.8cm   % offset horizontal en \magnification=1200
\fi

% ��������������������������� Reglages ���������������������������
% ������������ auquels il ne vaut mieux pas toucher ���������������

\pretolerance=500 \tolerance=1000  \brokenpenalty=5000


% ��������������������� Debut des macros privees �������������������
\catcode`\@=11
% ��������������������� Les fontes ���������������������������������

\font\eightrm=cmr8         \font\eighti=cmmi8
\font\eightsy=cmsy8        \font\eightbf=cmbx8
\font\eighttt=cmtt8        \font\eightit=cmti8
\font\eightsl=cmsl8        \font\sixrm=cmr6
\font\sixi=cmmi6           \font\sixsy=cmsy6
\font\sixbf=cmbx6

% Fontes AMS

\font\tengoth=eufm10       \font\tenbboard=msbm10
\font\eightgoth=eufm10 at 8pt      \font\eightbboard=msbm10 at 8 pt
\font\sevengoth=eufm7      \font\sevenbboard=msbm7
\font\sixgoth=eufm7 at 6 pt        \font\fivegoth=eufm5

 \font\tencyr=wncyr10       \font\tenbcyr=wncyb10
\font\eightcyr=wncyr10 at 8 pt      \font\eightbcyr=wncyb10 at 8 pt
\font\sevencyr=wncyr10 at 7 pt      \font\sevenbcyr=wncyb10 at 7 pt
\font\sixcyr=wncyr10 at 6 pt        


% Pour que les accents se placent correctement en mode math en corps 8 et 6

\skewchar\eighti='177 \skewchar\sixi='177
\skewchar\eightsy='60 \skewchar\sixsy='60

% Nouvelles familles pour les maths

\newfam\gothfam           \newfam\bboardfam
\newfam\cyrfam

\def\tenpoint{%
  \textfont0=\tenrm \scriptfont0=\sevenrm \scriptscriptfont0=\fiverm
  \def\rm{\fam\z@\tenrm}%
  \textfont1=\teni  \scriptfont1=\seveni  \scriptscriptfont1=\fivei
  \def\oldstyle{\fam\@ne\teni}\let\old=\oldstyle
  \textfont2=\tensy \scriptfont2=\sevensy \scriptscriptfont2=\fivesy
  \textfont\gothfam=\tengoth \scriptfont\gothfam=\sevengoth
  \scriptscriptfont\gothfam=\fivegoth
  \def\goth{\fam\gothfam\tengoth}%
  \textfont\bboardfam=\tenbboard \scriptfont\bboardfam=\sevenbboard
  \scriptscriptfont\bboardfam=\sevenbboard
  \def\bb{\fam\bboardfam\tenbboard}%
 \textfont\cyrfam=\tencyr \scriptfont\cyrfam=\sevencyr
  \scriptscriptfont\cyrfam=\sixcyr
  \def\cyr{\fam\cyrfam\tencyr}%
  \textfont\itfam=\tenit
  \def\it{\fam\itfam\tenit}%
  \textfont\slfam=\tensl
  \def\sl{\fam\slfam\tensl}%
  \textfont\bffam=\tenbf \scriptfont\bffam=\sevenbf
  \scriptscriptfont\bffam=\fivebf
  \def\bf{\fam\bffam\tenbf}%
  \textfont\ttfam=\tentt
  \def\tt{\fam\ttfam\tentt}%
  \abovedisplayskip=12pt plus 3pt minus 9pt
  \belowdisplayskip=\abovedisplayskip
  \abovedisplayshortskip=0pt plus 3pt
  \belowdisplayshortskip=4pt plus 3pt 
  \smallskipamount=3pt plus 1pt minus 1pt
  \medskipamount=6pt plus 2pt minus 2pt
  \bigskipamount=12pt plus 4pt minus 4pt
  \normalbaselineskip=12pt
  \setbox\strutbox=\hbox{\vrule height8.5pt depth3.5pt width0pt}%
  \let\bigf@nt=\tenrm       \let\smallf@nt=\sevenrm
  \normalbaselines\rm}

\def\eightpoint{%
  \textfont0=\eightrm \scriptfont0=\sixrm \scriptscriptfont0=\fiverm
  \def\rm{\fam\z@\eightrm}%
  \textfont1=\eighti  \scriptfont1=\sixi  \scriptscriptfont1=\fivei
  \def\oldstyle{\fam\@ne\eighti}\let\old=\oldstyle
  \textfont2=\eightsy \scriptfont2=\sixsy \scriptscriptfont2=\fivesy
  \textfont\gothfam=\eightgoth \scriptfont\gothfam=\sixgoth
  \scriptscriptfont\gothfam=\fivegoth
  \def\goth{\fam\gothfam\eightgoth}%
  \textfont\cyrfam=\eightcyr \scriptfont\cyrfam=\sixcyr
  \scriptscriptfont\cyrfam=\sixcyr
  \def\cyr{\fam\cyrfam\eightcyr}%
  \textfont\bboardfam=\eightbboard \scriptfont\bboardfam=\sevenbboard
  \scriptscriptfont\bboardfam=\sevenbboard
  \def\bb{\fam\bboardfam}%
  \textfont\itfam=\eightit
  \def\it{\fam\itfam\eightit}%
  \textfont\slfam=\eightsl
  \def\sl{\fam\slfam\eightsl}%
  \textfont\bffam=\eightbf \scriptfont\bffam=\sixbf
  \scriptscriptfont\bffam=\fivebf
  \def\bf{\fam\bffam\eightbf}%
  \textfont\ttfam=\eighttt
  \def\tt{\fam\ttfam\eighttt}%
  \abovedisplayskip=9pt plus 3pt minus 9pt
  \belowdisplayskip=\abovedisplayskip
  \abovedisplayshortskip=0pt plus 3pt
  \belowdisplayshortskip=3pt plus 3pt 
  \smallskipamount=2pt plus 1pt minus 1pt
  \medskipamount=4pt plus 2pt minus 1pt
  \bigskipamount=9pt plus 3pt minus 3pt
  \normalbaselineskip=9pt
  \setbox\strutbox=\hbox{\vrule height7pt depth2pt width0pt}%
  \let\bigf@nt=\eightrm     \let\smallf@nt=\sixrm
  \normalbaselines\rm}

\tenpoint


% Definition des petites capitales qui reagissent au \tenpoint et \eightpoint
% La syntaxe est celle d'un changement de fonte :
% {\pc FERMAT}, {\pc EUCLIDE} et {\pc G\"ODEL}.

\def\pc#1{\bigf@nt#1\smallf@nt}         \def\pd#1 {{\pc#1} }


%���������������������� dactylographie francaise ������������������

\catcode`\;=\active
\def;{\relax\ifhmode\ifdim\lastskip>\z@\unskip\fi
\kern\fontdimen2 \font\kern -1.2 \fontdimen3 \font\fi\string;}

\catcode`\:=\active
\def:{\relax\ifhmode\ifdim\lastskip>\z@\unskip\fi\penalty\@M\ \fi\string:}

\catcode`\!=\active
\def!{\relax\ifhmode\ifdim\lastskip>\z@
\unskip\fi\kern\fontdimen2 \font \kern -1.1 \fontdimen3 \font\fi\string!}

\catcode`\?=\active
\def?{\relax\ifhmode\ifdim\lastskip>\z@
\unskip\fi\kern\fontdimen2 \font \kern -1.1 \fontdimen3 \font\fi\string?}

\def\^#1{\if#1i{\accent"5E\i}\else{\accent"5E #1}\fi}
\def\"#1{\if#1i{\accent"7F\i}\else{\accent"7F #1}\fi}

\frenchspacing

% ���������������� Le format de sortie ����������������������������
% Haut et bas de page 

\newtoks\auteurcourant      \auteurcourant={\hfil}
\newtoks\titrecourant       \titrecourant={\hfil}

\newtoks\hautpagetitre      \hautpagetitre={\hfil}
\newtoks\baspagetitre       \baspagetitre={\hfil}

\newtoks\hautpagegauche     
\hautpagegauche={\eightpoint\rlap{\folio}\hfil\the\auteurcourant\hfil}
\newtoks\hautpagedroite     
\hautpagedroite={\eightpoint\hfil\the\titrecourant\hfil\llap{\folio}}

\newtoks\baspagegauche      \baspagegauche={\hfil} 
\newtoks\baspagedroite      \baspagedroite={\hfil}

\newif\ifpagetitre          \pagetitretrue  

% \nopagenumbers : c'est un peu violent, mais a marche. Alors ...

\def\nopagenumbers{\def\folio{\hfil}}  

\headline={\ifpagetitre\the\hautpagetitre
\else\ifodd\pageno\the\hautpagedroite\else\the\hautpagegauche\fi\fi}

\footline={\ifpagetitre\the\baspagetitre\else
\ifodd\pageno\the\baspagedroite\else\the\baspagegauche\fi\fi
\global\pagetitrefalse}

% Redefinition de \raggedbottom pour avoir plus de mou en bas de page
% (necesssaire quand il y a beaucoup de grumeaux, des grosses
% formules centrees et pas beaucoup de texte entre)

\def\raggedbottom{\topskip 10pt plus 36pt\r@ggedbottomtrue}


% ������������������ Macros de mise en page ���������������������

% Un point-tiret

\def\pointir{\unskip . --- \ignorespaces}

% Macros Bigbreak et \Medbreak pour que les blancs verticaux ne s'ajoutent pas

\def\Bigbreak{\vskip-\lastskip\bigbreak}
\def\Medbreak{\vskip-\lastskip\medbreak}

% Texte centre dans une boite : le resultat est le plus petit rectangle
% qui contient le texte. Ce resultat est
% place dans une boite centree.
% La syntaxe est celle d'un tableau \`a une colonne

\def\ctexte#1\endctexte{%
  \hbox{$\vcenter{\halign{\hfill##\hfill\crcr#1\crcr}}$}}

% Titres centres (en gras)

\long\def\ctitre#1\endctitre{%
    \ifdim\lastskip<24pt\vskip-\lastskip\bigbreak\bigbreak\fi
  		\vbox{\parindent=0pt\leftskip=0pt plus 1fill
          \rightskip=\leftskip
          \parfillskip=0pt\bf#1\par}
    \bigskip\nobreak}

\long\def\section#1\endsection{%
\vskip 0pt plus 3\normalbaselineskip
\penalty-250
\vskip 0pt plus -3\normalbaselineskip
\Bigbreak
\message{[section \string: #1]}{\bf#1\unskip}\pointir}

\long\def\sectiona#1\endsection{%
\vskip 0pt plus 3\normalbaselineskip
\penalty-250
\vskip 0pt plus -3\normalbaselineskip
\Bigbreak
\message{[sectiona \string: #1]}%
{\bf#1}\medskip\nobreak}

\long\def\subsection#1\endsubsection{%
\Medbreak
{\it#1\unskip}\pointir}

\long\def\subsectiona#1\endsubsection{%
\Medbreak
{\it#1}\par\nobreak}

\def\rem#1\endrem{%
\Medbreak
{\it#1\unskip} : }

\def\remp#1\endrem{%
\Medbreak
{\pc #1\unskip}\pointir}

\def\rema#1\endrem{%
\Medbreak
{\it #1}\par\nobreak}

\def\newparwithcolon#1\endnewparwithcolon{
\Medbreak
{#1\unskip} : }

\def\newparwithpointir#1\endnewparwithpointir{
\Medbreak
{#1\unskip}\pointir}

\def\newpara#1\endnewpar{
\Medbreak
{#1\unskip}\smallskip\nobreak}


\def\remarque{\newparwithcolon
              \pc REMARQUE
              \endnewparwithcolon}


\def\remarques{\newpara
               \pc REMARQUES
               \endnewpar}

\def\exemple{\newparwithcolon
              \pc EXEMPLE
              \endnewparwithcolon}


\def\exemples{\newpara
               \pc EXEMPLES
               \endnewpar}

\def\definition{\newparwithpointir
                \pc D\'EFINITION
                \endnewparwithpointir}

\def\definitions{\newpara
                 \pc D\'EFINITIONS
                 \endnewpar}

\def\demonstration{\newparwithcolon
                   \it D\'emonstration
                   \endnewparwithcolon}

%���������������������������������������������
% enonces de theoremes avec numerotation apres 
% #1 = THEOREME, COROLLAIRE, etc.
% #2 = numero (par exemple 3, 3.1, etc.)
% #3 = l'enonce du th proprememnt dit.

\long\def\th#1 #2\enonce#3\endth{%
   \Medbreak
   {\pc#1} {#2\unskip}\pointir{\it #3}\medskip}

\long\def\tha#1 #2\enonce#3\endth{%
   \Medbreak
   {\pc#1} {#2\unskip}\par\nobreak{\it #3}\medskip}

%���������������������������������������������
% enonces de theoremes avec numerotation d'abord
% #1 = numero (par exemple 3, 3.1, etc.)
% #2 = THEOREME, COROLLAIRE, etc.
% #3 = l'enonce du th proprememnt dit.

\long\def\Th#1 #2 #3\enonce#4\endth{%
   \Medbreak
   #1 {\pc#2} {#3\unskip}\pointir{\it #4}\medskip}

\long\def\Tha#1 #2 #3\enonce#4\endth{%
   \Medbreak
   #1 {\pc#2} #3\par\nobreak{\it #4}\medskip}
%���������������������������������������������

% les differents retraits, voir aussi \item

\def\decale#1{\smallbreak\hskip 28pt\llap{#1}\kern 5pt}
\def\decaledecale#1{\smallbreak\hskip 34pt\llap{#1}\kern 5pt}
\def\puce{\smallbreak\hskip 6pt{$\scriptstyle\bullet$}\kern 5pt}

% ������������������������������������������������������������������
% ��������������� ce que Knuth n'a pas fait ������������������������
% ������������������������������������������������������������������

% Un \displaylines qui numerote  droite.
% La syntaxe est la meme que celle de \eqalignno

\def\displaylinesno#1{\displ@y\halign{
\hbox to\displaywidth{$\@lign\hfil\displaystyle##\hfil$}&
\llap{$##$}\crcr#1\crcr}}

% Un \displaylines qui numerote a gauche.
% La syntaxe est la meme que celle de \leqalignno

\def\ldisplaylinesno#1{\displ@y\halign{ 
\hbox to\displaywidth{$\@lign\hfil\displaystyle##\hfil$}&
\kern-\displaywidth\rlap{$##$}\tabskip\displaywidth\crcr#1\crcr}}

% Un \eqalign qui accepte plusieurs alignements verticaux
% Motif : \hfil ** & ** \hfil & \hfil ** & ** \hfill, etc.

\def\eqalign#1{\null\,\vcenter{\openup\jot\m@th\ialign{
\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
&&\quad\strut\hfil$\displaystyle{##}$&$\displaystyle{{}##}$\hfil
\crcr#1\crcr}}\,}

% Systeme d'equations precede d'une accolade.
% Copi\'e sur \eqalign, on s'en sert comme une matrice
% syntaxe : signe & coef & inconnue
% les coef sont justifi\'es \`a droite (\hfil coef)
% et les inconnues \`a gauche (inconnue\hfil)
% attention : un seul & ou deux && avant le signe =
% selon la justification choisie !
% Exemple : $$\system{
%             &2 &x &- &3 & y & = &&  -5 \cr
%            -&  &x &+ &  & y & = &&   6 \cr
%           }$$

\def\system#1{\left\{\null\,\vcenter{\openup1\jot\m@th
\ialign{\strut$##$&\hfil$##$&$##$\hfil&&
        \enskip$##$\enskip&\hfil$##$&$##$\hfil\crcr#1\crcr}}\right.}

% pour avoir des messages raisonnables avec les lettres accentu\'ees

\let\@ldmessage=\message

\def\message#1{{\def\pc{\string\pc\space}%
                \def\'{\string'}\def\`{\string`}%
                \def\^{\string^}\def\"{\string"}%
                \@ldmessage{#1}}}


% ������������������ Divers gadgets ���������������������������������

% Pour se rendre la vie facile : \up{er}, \up{i\`eme}, n\up{0}, etc.

\def\up#1{\raise 1ex\hbox{\smallf@nt#1}}

% Utilisation : \cf. \etc.

\def\cf{{\it cf}} \def\etc{{\it etc}}

\def\qed{\raise -2pt\hbox{\vrule\vbox to 10pt{\hrule width 4pt
                 \vfill\hrule}\vrule}}

\def\cqfd{\unskip\penalty 500\quad\qed\medbreak}

\def\virg{\raise .4ex\hbox{,}}   % virgule aprs une fraction

\def\pv{\ ;\enspace} % point-virgule de ponctuation en maths

\def\build#1_#2^#3{\mathrel{
\mathop{\kern 0pt#1}\limits_{#2}^{#3}}}

% Entoure #2 d'un filet. Le filet est ecarte tout autour de #1
% Syntaxe \boxit{5pt}{...}. La ligne de base n'est pas perdue.

\def\boxit#1#2{%
\setbox1=\hbox{\kern#1{#2}\kern#1}%
\dimen1=\ht1 \advance\dimen1 by #1 \dimen2=\dp1 \advance\dimen2 by #1 
\setbox1=\hbox{\vrule height\dimen1 depth\dimen2\box1\vrule}%  
\setbox1=\vbox{\hrule\box1\hrule}%
\advance\dimen1 by .6pt \ht1=\dimen1 
\advance\dimen2 by .6pt \dp1=\dimen2  \box1\relax}

% ������������������������������������������������������������������
% fin des macros privees
% ������������������������������������������������������������������

\catcode`\@=12

% pour qu'il la ferme
\showboxbreadth=-1  \showboxdepth=-1





% \setbox1=\ctexte
% \bf NOUVEAU FORMAT a4\cr
% \noalign{\medskip}
% \bf \date\cr
% \endctexte

% $$\boxit{7pt}{\boxit{7pt}{\box1}}$$


\input amssym.def
\input amssym.tex
%\input grille

% (le reglage se fait a la fin du fichier)

\def\Grille{\setbox13=\vbox to 5\unitlength{\hrule width 109mm\vfill} 
\setbox13=\vbox to 65mm{\offinterlineskip\leaders\copy13\vfill\kern-1pt\hrule} 
\setbox14=\hbox to 5\unitlength{\vrule height 65mm\hfill} 
\setbox14=\hbox to 109mm{\leaders\copy14\hfill\kern-2mm\vrule height 65mm}
\ht14=0pt\dp14=0pt\wd14=0pt \setbox13=\vbox to
0pt{\vss\box13\offinterlineskip\box14} \wd13=0pt\box13}

\def\grille{\noalign{\Grille}}

%

\def\fleche(#1,#2)\dir(#3,#4)\long#5{%
\noalign{\leftput(#1,#2){\vector(#3,#4){#5}}}}


\def\ligne(#1,#2)\dir(#3,#4)\long#5{%
\noalign{\leftput(#1,#2){\lline(#3,#4){#5}}}}


\def\put(#1,#2)#3{\noalign{\setbox1=\hbox{%
    \kern #1\unitlength
    \raise #2\unitlength\hbox{$#3$}}%
    \ht1=0pt \wd1=0pt \dp1=0pt\box1}}

%

\def\diagram#1{\def\normalbaselines{\baselineskip=0pt\lineskip=5pt}
\matrix{#1}}

\def\hfl#1#2#3{\smash{\mathop{\hbox to#3{\rightarrowfill}}\limits
^{\scriptstyle#1}_{\scriptstyle#2}}}

\def\gfl#1#2#3{\smash{\mathop{\hbox to#3{\leftarrowfill}}\limits
^{\scriptstyle#1}_{\scriptstyle#2}}}

\def\vfl#1#2#3{\llap{$\scriptstyle #1$}
\left\downarrow\vbox to#3{}\right.\rlap{$\scriptstyle #2$}}

\def\ufl#1#2#3{\llap{$\scriptstyle #1$}
\left\uparrow\vbox to#3{}\right.\rlap{$\scriptstyle #2$}}

\def\pafl#1#2#3{\llap{$\scriptstyle #1$}
\left\Vert\vbox to#3{}\right.\rlap{$\scriptstyle #2$}}

\def\vl#1#2#3{\llap{$\scriptstyle #1$}
\left\vert\vbox to#3{}\right.\rlap{$\scriptstyle #2$}}

%


 \message{`lline' & `vector' macros from LaTeX}
 \catcode`@=11
\def\{{\relax\ifmmode\lbrace\else$\lbrace$\fi}
\def\}{\relax\ifmmode\rbrace\else$\rbrace$\fi}
\def\newcount{\alloc@0\count\countdef\insc@unt}
\def\newdimen{\alloc@1\dimen\dimendef\insc@unt}
\def\newwrite{\alloc@7\write\chardef\sixt@@n}

\newwrite\@unused
\newcount\@tempcnta
\newcount\@tempcntb
\newdimen\@tempdima
\newdimen\@tempdimb
\newbox\@tempboxa

\def\@spaces{\space\space\space\space}
\def\@whilenoop#1{}
\def\@whiledim#1\do #2{\ifdim #1\relax#2\@iwhiledim{#1\relax#2}\fi}
\def\@iwhiledim#1{\ifdim #1\let\@nextwhile=\@iwhiledim
        \else\let\@nextwhile=\@whilenoop\fi\@nextwhile{#1}}
\def\@badlinearg{\@latexerr{Bad \string\line\space or \string\vector
   \space argument}}
\def\@latexerr#1#2{\begingroup
\edef\@tempc{#2}\expandafter\errhelp\expandafter{\@tempc}%
%% error help message pieces.
\def\@eha{Your command was ignored.
^^JType \space I <command> <return> \space to replace it
  with another command,^^Jor \space <return> \space to continue without
it.} 
\def\@ehb{You've lost some text. \space \@ehc}
\def\@ehc{Try typing \space <return>
  \space to proceed.^^JIf that doesn't work, type \space X <return> \space to
  quit.}
\def\@ehd{You're in trouble here.  \space\@ehc}

\typeout{LaTeX error. \space See LaTeX manual for explanation.^^J
 \space\@spaces\@spaces\@spaces Type \space H <return> \space for
 immediate help.}\errmessage{#1}\endgroup}
\def\typeout#1{{\let\protect\string\immediate\write\@unused{#1}}}

% line & circle fonts
\font\tenln    = line10
\font\tenlnw   = linew10
%\font\tencirc  = circle10
%\font\tencircw = circlew10


\newdimen\@wholewidth
\newdimen\@halfwidth
\newdimen\unitlength 

\unitlength =1pt

%\newbox\@picbox
%\newdimen\@picht

\def\thinlines{\let\@linefnt\tenln \let\@circlefnt\tencirc
  \@wholewidth\fontdimen8\tenln \@halfwidth .5\@wholewidth}
\def\thicklines{\let\@linefnt\tenlnw \let\@circlefnt\tencircw
  \@wholewidth\fontdimen8\tenlnw \@halfwidth .5\@wholewidth}

\def\linethickness#1{\@wholewidth #1\relax \@halfwidth .5\@wholewidth}



\newif\if@negarg

\def\lline(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@linelen=#3\unitlength
\ifnum\@xarg =0 \@vline
  \else \ifnum\@yarg =0 \@hline \else \@sline\fi
\fi}

\def\@sline{\ifnum\@xarg< 0 \@negargtrue \@xarg -\@xarg \@yyarg -\@yarg
  \else \@negargfalse \@yyarg \@yarg \fi
\ifnum \@yyarg >0 \@tempcnta\@yyarg \else \@tempcnta -\@yyarg \fi
\ifnum\@tempcnta>6 \@badlinearg\@tempcnta0 \fi
\setbox\@linechar\hbox{\@linefnt\@getlinechar(\@xarg,\@yyarg)}%
\ifnum \@yarg >0 \let\@upordown\raise \@clnht\z@
   \else\let\@upordown\lower \@clnht \ht\@linechar\fi
\@clnwd=\wd\@linechar
\if@negarg \hskip -\wd\@linechar \def\@tempa{\hskip -2\wd\@linechar}\else
     \let\@tempa\relax \fi
\@whiledim \@clnwd <\@linelen \do
  {\@upordown\@clnht\copy\@linechar
   \@tempa
   \advance\@clnht \ht\@linechar
   \advance\@clnwd \wd\@linechar}%
\advance\@clnht -\ht\@linechar
\advance\@clnwd -\wd\@linechar
\@tempdima\@linelen\advance\@tempdima -\@clnwd
\@tempdimb\@tempdima\advance\@tempdimb -\wd\@linechar
\if@negarg \hskip -\@tempdimb \else \hskip \@tempdimb \fi
\multiply\@tempdima \@m
\@tempcnta \@tempdima \@tempdima \wd\@linechar \divide\@tempcnta \@tempdima
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\def\rightheadline{\tenrm \hfill \quad UNIPOTENT SUBGROUPS }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%




{UNIPOTENT SUBGROUPS OF REDUCTIVE GROUPS}

{IN  CHARACTERISTIC $p>0$ }

\vskip4mm

{PHILIPPE GILLE}

\vskip4mm


\vskip4mm


\noindent{\bf Abstract \pointir }
We prove two conjectures of Tits on the unipotent elements of semisimple
algebraic groups defined over a field with positive characteristic.
\footnote{}{2000 Mathematics Subject Classification :  20G15,  12G05.}

\vskip8mm

\noindent {\bf Summary }
\vskip2mm
{\def\page{\leaders\hbox to 5mm{\hfil.\hfil}\hfill} \sl
\par 1 \pointir Introduction \page \par

\qquad 1.1 \pointir Good finite subgroups \page \par
\qquad 1.2 \pointir The conjectures of Tits\page \par

\qquad 1.3 \pointir Complements on finite subgroups \page \par

2 \pointir Finite unipotent subgroups  and Galois cohomology \page\par
\qquad 2.1 \pointir A Galois cohomology class
\page\par
\qquad 2.2 \pointir  How to see if a finite  unipotent subgroup is $k$--good
\page\par
\qquad 2.3 \pointir The case $[k:k^p] \leq p$
\page\par
\qquad 2.4 \pointir Bad unipotents and torsors on the affine line
\page\par


3 \pointir Conjugacy classes
of $k$--anisotropic automorphisms of split groups
\page \par


4 \pointir  Anisotropic automorphisms are special
\page \par
\qquad 4.1  \pointir Special automorphisms \page\par
\qquad 4.2  \pointir Triality \page\par
\qquad 4.3 \pointir Lifting in characteristic $0$ \page\par
\qquad 4.4  \pointir End of the proof of Theorem 3\page\par
}


\vskip8mm

\noindent{\bf 1 \pointir  Introduction}

\noindent{\bf 1.1 \pointir Good finite  subgroups.}
Let $k$ be a field   of
characteristic
$p\geq 0$, and  then $k_s$ be a separable closure of  $k$. Let $G/k$ be a
reductive
connected
algebraic group. For the notions on algebraic groups  used here, we refer to
the paper by Borel and Tits [BT1] and the book of Demazure and Gabriel
 [DG]. If $G$ is
semisimple and absolutely almost simple, we denote by
$S(G)$ the finite set of torsion primes of $G$ as defined by Serre
in [Se2]. In the case  of $G$  simply connected, the   the Dynkin index
will be denoted by
$d_G$ (cf. [LS], \pa 2). Note taht  the prime factors of
$d_G$ lie in $S(G)$. We recall  Tits' definition of good  unipotent elements.


\th DEFINITION 1 [T2]
\enonce Let  $u$ be a unipotent element of $G(k)$.
 The element $u$ is $k$--good
 if $u$  lies in the unipotent radical of a $k$--parabolic
subgroup of $G$, otherwise
$u$  is $k$--bad.
\endth

In characteristic $0$, all unipotent elements are $k$--good, so the definition
is relevant only  if $\car(k)=p>0$; in that case,  a unipotent element $u$
has finite order
$q=p^r$ and can be viewed as a morphism $u^\sharp: {\bb Z}/q{\bb Z} \rg G$
sending $1$ to $u$. This viewpoint leads to the following definition of
goodness for
morphisms $M \rg G$ where  $M/k$ is a finite \'etale group.



\th DEFINITION 1'
\enonce  Let  $M/k$ be a finite \'etale group and $\phi: M/k \rg G$ be a
morphism.
 The morphism  $\phi$ is $k$--good if $\phi(M(k_s))$  lies in the radical
of a $k$--parabolic
subgroup of $G$, otherwise $\phi$  is $k$--bad.
\endth

If $\phi$ is injective, we also say
that $M$ is a $k$--good subgroup of $G$.
We say that $\phi:M\rg G$ is unipotent if $\phi(M(k_s))$ consists of
unipotent elements. Let
us prove that the two definitions are compatible.

\th LEMMA 1
\enonce Assume  $\car(k)=p>0$.  Let $u$ be a unipotent element of
 order $q$ and
$u^\sharp: {\bb Z}/q{\bb Z} \rg G$ be  the associated morphism. Then $u$ is
$k$--good if and
only if
$u^\sharp$ is $k$--good.
\endth

\rem Proof
\endrem
If $u$ is $k$--good,  it  is obvious that the morphism $u^\sharp$ is $k$--good.
Conversely, assume that the morphism $u^\sharp: {\bb Z}/q {\bb Z} \rg G$ is
$k$--good.
Then  $u=u^\sharp(1)$  lies in the  radical $R(P)$ of a
$k$--parabolic subgroup $P$ of $G$,  which is an extension of a $k$--torus
$S$ by the
unipotent radical $R_uP$ of $P$, i.e. we have an exact sequence
$1 \rg R_u(P) \rg R(P) \rg S \rg 1$. As $S(k)$ has no element of order $p$,
$u$ lies in   $R_u(P)(k)$, and $u$ is $k$--good.
\hfill\hfill\cqfd


\th DEFINITION 2 [T2]
\enonce Let $\alpha$ be a $k$--automorphism of $G$ of finite order.
The morphism $\alpha$ is $k$--isotropic if $\alpha$ normalizes some proper
$k$--parabolic subgroup of $G$, otherwise $\phi$  is $k$--anisotropic.
\endth


As above, we can extend this definition
to isotropic  morphisms $M \rg \Aut(G)$ which will yield the previous one
in the case
$M={\bb Z}/n{\bb Z}$.


\th DEFINITION 2'
\enonce Let $\phi \in \Hom_{k-gr}(M, \Aut(G))$.
The morphism $\phi$ is $k$--isotropic if $\phi(M(k_s))$ normalizes some proper
$k$--parabolic subgroup of $G$, otherwise $\phi$  is $k$--anisotropic.
\endth


We say that $\phi \in \Hom_{k-gr}(M, G)$ is $k$--isotropic if the
composite $Ad\circ\phi : M \rg G$ is $k$--isotropic. Obviously,
$k$--anisotropic implies
$k$--bad.

\vskip2mm


\headline={\ifodd\pageno\rightheadline \else\leftheadline\fi}


\noindent{\bf 1.2\pointir The conjectures of Tits.}
 Our first result is the proof of the following conjecture (see Theorem 2
of \pa 2).

\th CONJECTURE 1  [T3]
\enonce Assume $p>0$, $[k:k^p] \leq p$ and that $G/k$ is semisimple and simply
connected. Then every unipotent subgroup of $G(k)$ (i.e a subgroup
consisting of
unipotent elements) is
$k$--embeddable into
the unipotent radical of a $k$-parabolic subgroup of $G$.
\endth

The case
 $p=2$ is due to Tits [T2, \pa 4.5].  Further known cases are when
$p$ is not a
torsion prime of $G$, and  types $A_n$ and $C_n$ ({\it loc.
cit.}, \pa 3.5 and 4.4).
In  conjecture  1,
the  condition $[k:k^p]\leq p$ is necessary
as the following theorem of Tits shows.

\th THEOREM 1 [T4, prop. S1] and  [T5, th. 7]
\enonce Assume that $G$ is split simply connected and almost simple.
If  $[k :k^p]\geq p^2$ and $p$ divides $d_G$, then  the group  $G(k)$
contains a
$k$--bad unipotent
element of order $p$.
\hfill\hfill\cqfd
\endth

Our proof of conjecture 1 is based on
the reduction to a problem in Galois cohomology
for the group $G_{k((t))}$, which is done in Proposition 3 (\pa 2.2), and on
known cases of Serre's
conjecture \deux\enskip
in Galois cohomology [Gi1].
 Our second result is the proof of the following
conjecture (see Theorem 3 of \pa 4), which is related to the shape of
conjugacy  classes
of $k$--anisotropic
automorphisms.

\th CONJECTURE 2 [T2]
\enonce Assume $p>0$ and that  $G/k$ is split and almost simple. Let
$\alpha$ be a
$k$--anisotropic automorphism of order $p$. Then $\alpha$ normalizes a
maximal $k$--split
torus  of $G$.
\endth

Let us recall first the known cases handled by
Tits.
\begingroup\parindent=1,5cm
\item{1)} $p=2$, all types,
\item{2)} $p=3$, $G$ of  type $D_4$ or $E_6$, using a triality argument
due to Harder [H],
\item{3)} type $A_n$.
\par\endgroup

It turns out that conjecture 2 and a large part of the present  paper make
sense in
a broader
setting,  including the characteristic zero case. To this end,  we
formulate the
following  conjecture,
which extends the preceeding one.


\th CONJECTURE 2'
\enonce Assume  $G/k$ is split almost simple. Let $l$ be a prime of $S(G)$.
We assume
that $l=\car(k)$ or either  that $k$ contains a primitive $l$--root of unity.
 Let $\alpha$ be a $k$--anisotropic automorphism of order $l$. Then $\alpha$
normalizes a maximal $k$--split torus  of $G$.
\endth

Such anisotropic automorphisms are called {\it special} automorphisms.
 In section 4, we prove conjecture  2'  except for
the  $E_8$ case. However, the case of $E_8$ and $l=p=5$ is proven in [Gi3].
Tits'
original conjecture is thus true.

The proof  of conjecture 2' requires several steps. First
we work in characteristic zero and  prove by a
case-by-case analysis conjecture 2'  (except for the $E_8$ case) by using
methods
 inspired by the proof of  Hasse principle (\pa 4.2).
The tame case $l\not =p$ is then not hard and we can
concentrate on the wild case
$l=p=\car(k)$. We consider a complete discrete valuation ring $A$ with
residue field
$k$ and fraction field $F_A$ of characteristic zero. We remark that
special elements lift in
characteristic zero, i.e. if $\alpha \in \Aut(G)(k)$ is a special
automorphism of order $p$,
then there exists $\widetilde \alpha  \in \Aut(G)(A)$ of order $p$
 specializing to $\alpha$. We show that anisotropic automorphisms lift in
characteristic $0$
 (\pa 4.3). The proof  concludes by a specialization argument.


\vskip2mm

\noindent{\bf 1.3 \pointir  Complements on finite subgroups.}
In this section, we remark on the extension of some known results to our
present
setting.
The  argument as used in the proof of proposition 3.2 of [T2] yields in
fact the
following.

\th PROPOSITION 1
\enonce Let $\phi \in \Hom_{k-gr}(M, G)$,
and let $P/k$ be a $k$--parabolic subgroup normalized by $\phi$.
Let $L$ be a Levi subgroup of $P$
defined over $k$, so that $P=R_u(P) \rtimes L$ and let $\phi'$ be the
composition of $M
\rg P \rg L$.

\decale{$a)$} The following are equivalent :
\begingroup\parindent=2,5cm
\item{i)} $\phi$ is $k$--good in $G$
\item{ii)} $\phi'$ is $k$--good in $G$
\item{iii)} $\phi'$ is $k$--good in $L$.
\par\endgroup
\decale{$b)$} If $P$ is minimal among all $k$--parabolic subgroups
normalized by
$\phi$, then $\phi'$ is $k$--anisotropic in $L$.
\hfill\hfill\cqfd
\endth


Similarly, we can extend lemma  3.5  of [T2].

\th LEMMA 2
\enonce Let $H/k$ be a reductive subgroup of $G/k$. Let $\phi: M \rg
\Aut(G,H) \subset
\Aut(G)$ be a $k$--anisotropic morphism.
Then the morphism $\phi: M \rg \Aut(G,H) \rg \Aut(H)$ is $k$--anisotropic.
\hfill\hfill\cqfd
\endth

We also recall:


\th PROPOSITION  2 ([BT2], prop. 3.6)
\enonce
Let $U \subset G(k)$ be a unipotent subgroup such that  every
element of $U$ is $k_s$--embeddable in a Borel subgroup. Then $U$
is $k$--embeddable in the unipotent radical of a $k$--parabolic subgroup of
$G$.
\hfill\hfill\cqfd
\endth

\vskip2mm


In particular and in contrast to the tame case, a unipotent element $u$ is
$k$--good iff it
 is $k_s$--good.

\th LEMMA 3
\enonce  Let $g \in G(k)$ be an element of finite order
invertible in $k$. Then $g$ lies in a maximal $k$--torus
of $G$. In particular $g$ is $k_s$--good.
\endth


\rem Proof
\endrem  Let $g \in G(k)$ be an element of finite order
invertible in $k$. Then $g$ is semisimple and
$g$ lies in some maximal torus $T/k$. As  $T \times_k k_s$ is a split
maximal torus,
$g$ is $k_s$--good.
\hfill\hfill\cqfd


More generally, if $\car(k)=0$, Borel-Mostow's theorem asserts that
an automorphism of $G$ of  finite order normalizes a maximal torus of $G$
[BM] (see also [P]).

\vskip2mm


\noindent{\bf 2\pointir Finite unipotent subgroups  and Galois cohomology}

 \vskip2mm
\noindent{\bf 2.1 \pointir  A Galois cohomology class.}
Let us set $K=k((t))$ and denote by $K_{mod}$ a maximal tamely ramified
extension
of $K$.
Let $I=I_w \rtimes \mu_m$ be a $k$--\'etale group, semi--direct product of
$\mu_m$  ($m \in k^\times$) by  a constant $p$--group $I_w$.
Throughout this paper, we  assume that
\vskip2mm

{\it there exists a  totally ramified field extension $L/K$ such that
$\Spec(L) \rg
\Spec(K)$ is an
$I$--torsor.}

\vskip2mm
This hypothesis is satisfied
in the two extreme cases $I=\mu_m$ (obviously)
 and $I=I_w$  since the  maximal pro--$p$ quotient of
$\Gal(K_s/K)$ is a free pro--$p$ group [Se1, \pa \deux.2.2, Cor. 1]. We
denote by
$\widetilde L=L.K(\mu_m)$, which is a Galois extension of $K$ with Galois
group
$\widetilde I$. So, there is a diagram of field extensions
$$
\diagram{
&&\widetilde L \cr
 &\diagup&& \diagdown\cr
L&&&&K(\mu_m) ,\cr
 &\diagdown&& \diagup \cr
&& L\cap K(\mu_m)   \cr
&& \mid \cr
&& K}
$$
and the residue field of $\widetilde L$ is $k(\mu_m)$.
We denote by $f_\theta \in Z^1(\widetilde I, I(\widetilde L))$ the $1$--cocycle
defined by the $I$--torsor $\Spec(L) \rg \Spec(K)$ and by $\theta \in
H^1(K,I)$ its class.
For simplicity, the reader  can think of $I={\bb Z}/p{\bb Z}$; this case,
which is of main
interest for us, was treated  in [Gi2].

We consider the set $\Hom_{k-gr}(I, G)$ and pick $\phi \in \Hom_{k-gr}(I, G)$.
The key point, due to Serre,  is to associate to  $\phi:
I \rg G$ the  $1$--cocycle
$$
f_\phi=\phi_*( f_\theta) \hbox{\quad in \quad} Z^1(\widetilde
I,G(\widetilde L))
$$
 and the cohomology class
$$
\gamma(\phi)=\phi_*( \theta) = [f_\phi] \hbox{\quad in \quad} H^1(K,G).
$$

\rem Remark 1
\endrem We shall use functoriality  for a surjective
base change
$\lambda: I' \rg I$
and an extension $L'/L/K$ of monodromy group $I'$ :  if
there exists  $\theta'$ such that $\lambda_*(\theta')=\theta$, then
$\phi_*( \theta)_{L'}=
\lambda^*(
\phi\circ \lambda)$ in $H^1(L',G)$.
\vskip2mm

\noindent{\bf 2.2\pointir  How to see if a finite unipotent subgroup is
$k$--good }

\th PROPOSITION 3
\enonce  $a)$ If $\phi$ is $k$--good, then $\gamma(\phi)=1$ in $H^1(K,G)$.

\decale{$b)$} Assume that $\phi$ is unipotent. Then the following
assertions are
equivalent:
\begingroup\parindent=2cm
\item{$i)$} $\phi$ is $k$--good,
\item{$ii)$} $\gamma(\phi)=1$ in $H^1(K,G)$,
\item{$iii)$} $\gamma(\phi)_{K_{mod}} =1 \in H^1(K_{mod},G)$.
\par \endgroup
\endth


\rem Proof
\endrem
\noindent $a):$ Let $P/k$ be a parabolic subgroup of $G/k$ such that
the radical $R(P)$ contains $\Im(\phi)$. The group $R(P)$ is split and we
know that
$H^1(K,R(P))=1$.
Thus
$\gamma(\phi)=1$.


$b):$ As we consider unipotent morphisms, we can assume that
$I=I_w$, i.e.
$I$ is a finite $p$--group.
There is only the implication $iii) \Longrightarrow i) $ to prove.   By
hypothesis,
there exists a tamely ramified extension  $K'/K$ of valuation ring $O'/O$
and residue
extension $k'/k$ such that $\gamma(\phi)_{K'}=1$ in $H^1(K',G)$. One can assume
$K'=k'((t'))$,  $(t')^e=t$ and  $(e,p)=1$, and that $G_{k'}$ is split.
Let $B/k'$ be a
Borel subgroup of $G/k'$. We consider the extension
$L'=L.K'/K'$, which is Galois  of group $I=\Gal(L'/K')$ and
valuation ring $O_{L'}$.
Let us denote by $X/k'=G/B$ the variety of Borel subgroups
of $G$ and  by $\pi: G/k' \rg X/k'$ the canonical map. This variety is
projective, the morphism
$\pi$ is smooth and one knows that the map on $k'$--points $G(k') \rg
X(k')$ is onto.
Using Hensel's lemma and the valuative criterion of properness,
one obtains easily that the map $\pi:  G(O_{L'}) \rg X(O_{L'})$ is onto and
the map
$G(O_{L'}) /B(O_{L'}) \simlgr G(k')/B(k')$ is bijective.
We have the following commutative diagram of pointed sets (cf. [Se1], \pa
1.5.4, prop.
36)
$$
\diagram{
\Bigl[ G(O_{L'}) /B(O_{L'}) \Bigr]^I &\hfl{}{}{4mm}&
H^1(I,B(O_{L'}))&\hfl{}{}{4mm}& H^1(I,G(O_{L'}) ) \cr
\vfl{}{\wr}{4mm} &&\vfl{}{}{4mm} && \vfl{}{}{4mm} \cr
\Bigl[ G(L') / B(L')\Bigr]^I &\hfl{}{}{4mm}&
H^1(I,B(L')) &\hfl{}{}{4mm}& H^1(I, G(L') ). \cr
}
$$
Taking the fixed points of the twisted $I$--sets by the cocycle $f_\phi$,
one gets
$$
\Bigl[ _{f_\phi}\Bigl(G(O_{L'}) / B(O_{L'})\Bigr) \Bigr]^I
\simlgr  \Bigl[ _{f_\phi}\Bigl(G(L') / B(L') \Bigr) \Bigr]^I,
$$
so the class $[f_\phi] \in H^1(I,G(O_{L'}) )$ has a reduction
in $H^1(I,B(O_{L'}) )$. Thus
there exists $h \in G(O_{L'})$ such that  $h \phi(\sigma) ({^\sigma \cskip
h \cskip ^{-1}}) \in
B(O_{L'})$ for any $\sigma \in I$. Let   $\overline h$ be the image of $h$
in $G(k')$. Then
$\overline h \phi(\sigma) {(\overline h)}\cskip^{-1} \in B(k')$ for any $\sigma
\in I$
(we used here that
$L/K$ is totally ramified) and $\phi$ is $k'$--good.
By Proposition 2, because $k'/k$ is separable, $\phi$ is $k$--good.
\hfill\hfill\cqfd


The wild inertia group $\Gal(K_s/K_{mod})$ is a pro--$p$ group. If $p$ is
a good prime for $G$,
i.e.  if $p$ is not a torsion prime of  $G$, we know   that $H^1(K_{mod},G)=1$
[Se2, th 4"]. The last proposition  together with Proposition 2  gives then
another
 proof of Tits' result [T2, cor. 2.6].

\th COROLLARY 1
\enonce Assume $p=\car(k) \not \in S(G)$. Then any unipotent subgroup $U$
of $G(k)$
is $k$--embeddable in the unipotent radical of a $k$--parabolic subgroup of
$G$.
\endth


\vskip2mm

\noindent{\bf 2.3 \pointir The case $[k:k^p] \leq p$ }

\th THEOREM 2
\enonce Assume $\car(k)=p>0$,  $[k:k^p]\leq p$ and that $G$ is semisimple
simply
connected.  Then every unipotent subgroup of $G(k)$ is
$k$--embeddable in the unipotent radical of a $k$-parabolic subgroup of $G$.
\endth

We recall Kato's definition [K] of the $p$--dimension $dim_p(k)$ of $k$ by
means of
 cohomology groups $H^i_p(k)$. Let $\Omega_k$ be the  $k$-vector
space of the $1$--differential forms
 of the  ${\bb Z}$-algebra $k$.
For any nonnegative integer $i$, we set
 $\Omega^i_k=\bigwedge^i\Omega_k$ and the exterior differential
  $d$ maps $\Omega_k^i$
to $\Omega_k^{i-1}$. There exists an unique additive $p$--linear application
$\gamma: \Omega^i_k\rg \Omega^i_k/d \Omega^{i-1}_k$ such that
$\gamma(x\omega)=x^pw$ for any logarithmic differential form
$\omega=dy_1/y_1 \wedge\cdots \wedge dy_i/y_i$. The operator
$\omega$ is inverse of the Cartier operator. We set
$$
H^{i+1}_p(k)=Coker\Bigl(\gamma-1:\Omega^i_k \rg \Omega^i_k/d
\Omega_k^{i-1}\Bigr).
$$

\vskip2mm

\noindent If $[k:k^p]=\infty$,  one sets
$\dim_p(k)=\infty$. If $[k: k^p]=p^r$, one sets :
\begingroup\parindent=2cm
\item{-} $\dim_p(k)=r$ if
$H^{r+1}_p(k')=0$  for any finite extension $k'/k$,
\item{-} $\dim_p(k)=r+1$ otherwise.
\par\endgroup

\vskip2mm

\rem Proof of Theorem 2
\endrem
Assume $[k:k^p] \leq p$. By Proposition 2, one can assume
that $k=k_s$,  that $G$ is split and almost simple and that $U$ is
generated by a
unipotent element of order $q$. One has $_pBr(k')=H^2_p(k')=0$ for any
finite  extension
$k'/k$ because $k$ is separably closed, so $\dim_p(k) \leq 1$.  Then
$\dim_p(K) \leq 2$
by [K, corollary to th. 3] where $K=k((t))$ as in \pa 2. By  definition,
one has
$H^3_p(K')=0$ for any finite field extension  $K'/K$.  As the wild inertia
group
$\Gal(K_s/K_{mod})$ is a pro--$p$ group, the main result of [Gi1] gives then
$$
H^1(K_{mod}, G)=1.
$$
Then Proposition 3.b shows that $u$ is $k$--good.
\hfill\hfill\cqfd

\vskip2mm


\noindent{\bf 2.4 \pointir Bad unipotents and torsors on the affine line.}
Raghunathan has conjectured that if $G$ is semisimple simply connected,  then
$G$--torsors on the affine line
${\bb A}^1_k$ are constant, i.e. come from $G$--torsors on $\Spec(k)$ [R,
p. 189]. It turns
out that  bad unipotent elements yield counterexamples to this conjecture.

\th PROPOSITION 4
\enonce Assume that $G/k$ admits a bad unipotent element $u$ of order $p$.
Let ${\cal P} : {\bb A}^1_k \rg  {\bb A}^1_k$ be the Artin--Schreier covering,
which is Galois of group ${\bb F}_p$ and
defined by $t= {\cal P} (x)=x^p-x$.
Let us define the $1$--cocycle $h=(h_\sigma)_{\sigma \in {\bb F}_p}$ for
this covering by
$$
h_\sigma = u^\sigma \in G(k[x])   \enskip  (\sigma \in {\bb F}_p)
$$
Then  the cocycle $h$ defines a $G$--torsor on ${\bb A}^1_k$ which  is not
isomorphic
to a constant torsor.
\endth


\rem Proof
\endrem
One can assume  $k=k_s$. So $H^1(k,G)=1$ and we have to show that
the torsor defined by $h$ is not trivial.
We lift our cocycle $h$ in $G\bigl(k(({1 \over x}))\bigr)$ and we observe
that  $k(x)/k(t)$
is wildly ramified at $\infty$. As $u$ is a
bad unipotent element, Proposition 3.b shows  that $[ h]_{k(({1 \over t}))}$
is not trivial in $H^1(k(({1 \over t})),G)$, and a  fortiori $[h]$ is not
trivial
in $H^1({\bb A}^1_k,G)$.
\hfill\hfill\cqfd

\rem Remark 2
\endrem Theorem 1 (\pa \un.2) yields non constant $G$--torsors on the
affine line for
$G$  split semisimple simply connected provided $p$ divides $d_G$ and
$[k:k^p]\geq
p^2$.

\vskip2mm



 \noindent{\bf 3 \pointir   Conjugacy classes
of $k$--anisotropic automorphisms of split groups}

From now on, we assume that $G/k$ is split and semisimple.  So $G$ is the
extension
of ${\bb Z}$ to $k$ of  the  corresponding Chevalley
group scheme $G_{/ \Spec({\bb Z})}$. For any ring $A$, we denote by
simplicity $G
\times_{\bb Z} A =  G_J\times_{\Spec({\bb Z})} \Spec(A)$.

The goal of this section is
to give a powerful cohomological criterion analogous to Proposition 3
for distinguishing conjugacy classes of $k$--anisotropic automorphisms.
We denote by
$T/{\bb Z}$ a maximal split torus of $G/{\bb Z}$ and by $B/{\bb Z}$ a Borel
subgroup containing $T$.
Let $J\subset \Aut(G,B,T)$ be a subgroup and let us consider the semi-direct
product
$G_J/{\bb Z}= G/{\bb Z} \rtimes J$.  We have a natural map $G_J/{\bb Z} \rg
\Aut(G)/{\bb Z}$ with kernel the center of $G/{\bb Z}$
and we note that if $G/{\bb Z}$ is adjoint and  $J =\Aut(G,B,T)$, one has
$G_J/{\bb Z}=\Aut(G)/{\bb Z}$.
Let us introduce the following notation :
\begingroup\parindent=1cm
\item{-} $W=N_G(T)/T$ and $W_J=N_{G_J}(T)/T=W \rtimes J$ (the Weyl groups),
\item{-} $\widehat T= \Hom_{{\bb Z}-gr}(T, {\bb G}_m)$, $\widehat T^0=
\Hom_{{\bb Z}-gr}( {\bb G}_m,T)$.
\par\endgroup

\vskip2mm


\th PROPOSITION 5
\enonce
Let $\phi \in \Hom_{gr}\bigl(I, G_J(k)\bigr)$.

\decale{$a)$} The following assertions are equivalent

\begingroup\parindent=2,5cm
\item{$i)$} $\phi$ is $k$--anisotropic,
\item{$ii)$} $\Bigl[_{f_\phi} G_J(O_{\widetilde L}) \Bigr]^{\widetilde I}=
\bigl(_{f_\phi}
G_J\bigr)(K)$,
\item{$iii)$} the group $_{f_\phi} G_J/K$ is $K$--anisotropic.
\par\endgroup

\decale{$b$)} Assume  $\phi$ is $k$--anisotropic. Let $ \phi': I \rg G_J$
be another group morphism.  The following
assertions are equivalent :
\begingroup\parindent=2,5cm
\item{$i)$} $\phi'$ is conjugate to $\phi$ under $G_J(k)$,
\item{$ii)$} $\gamma(\phi)=\gamma(\phi')$ in $H^1(K,G_J)$.
\par \endgroup

\decale{$c$)} Let $k'/k$ be a Galois extension such that $\phi/{k'}:I \rg
G_J$ is a
$k'$--anisotropic morphism. With the notation
$Z_G(\phi)=Z_G(\Im(\phi))$, one has an
injection
$$
H^1(k'/k,Z_{G_J}(\phi)) \hookrightarrow H^1(K.k'/K, _{f_\phi}G_J).
$$
\endth

\rem Proof
\endrem  Let us begin by the second assertion.

{$b)$} The implication $i) \Longrightarrow ii)$ is
obvious. Conversely, assume that $\gamma(\phi)=\gamma(\phi')$ in
$H^1(\widetilde
L/K,G)$. Then there exists $g \in G(L)$ such that
$$
g^{-1} \phi(\sigma) \eskip{^\sigma \cskip g}=\phi'(\sigma)  \qquad (\sigma \in
\widetilde I).
$$
Let  ${\cal B}$ be the Bruhat--Tits building of the group
$G_{\widetilde L}$ [BrT,\pa \un.7.4] which is a metric space. This building
is equipped
with an
(isometric) left action of the group  $G_J(\widetilde L)$ and an  action of the
Galois group
$\widetilde I =\Gal(\widetilde L/K)$ denoted by
$x \mapsto {^\sigma \cskip  x}$ for any $\sigma \in \widetilde I$. We
denote by  $c$
the center of the building, i.e. the  vertex of
${\cal B}$ fixed by  $G(O_{\widetilde L})$.
One defines a twisted action of $\widetilde I$  on  ${\cal B}$  by
$$
\sigma. x =\phi(\sigma). ^\sigma \cskip x \qquad (\sigma \in \widetilde I).
$$
This action is isometric and simplicial (i.e. maps a facet on a facet).
Clearly, because
$\Im(\phi) \subset G_J(k)$, the center $c$ is fixed under the twisted action of
$\widetilde I$. Due to
$$
\sigma . (g.c)=\phi(\sigma) (^\sigma \cskip g). c=g \phi'(\sigma).c=g.c,
$$
we see that  $g.c$ is also fixed by $\widetilde I$ for the twisted action.
Assume  that $g.c \not=c$. As
the  segment $[c,g.c]$ is fixed pointwise by
$\widetilde I$,
there exists a facet of ${\cal B}$ containing strictly $c$ which is  stabilized
by the twisted action of $\widetilde I$. Hence there exists (by the
isomorphim between
the link of $c$ in ${\cal B}$ and the spherical building of $G_J/k(\mu_m)$)
a facet  of the
spherical building of $G_J/k(\mu_m)$  which is stabilized  by $\widetilde
I$. In
other words,
there exists a proper $k$--parabolic  subgroup $P$ of $G$ such that
$$
\Bigl[ _{f_\phi}\Bigl(G_J(k(\mu_m)) / N_{G_J}(P)(k(\mu_m))\Bigr)
\Bigr]^{\widetilde I}\not
=\emptyset, \enskip i.e
$$
$$
\Bigl\{ [\varphi] \in G_J(k(\mu_m)) / N_{G_J}(P)(k(\mu_m))\enskip \mid \enskip
\varphi^{-1}
\Im(\phi) \varphi \in  N_{G_J}(P)(k(\mu_m)) \Bigr\}^{ \Gal(k(\mu_m)/k)}\not
=\emptyset,
$$
which means that $\phi$ normalizes the $k$--parabolic subgroup
$\varphi(P)$. This
contradicts  the anisotropy of $\phi$. We conclude that
$g.c=c$, i.e. $g
\in G_J(O_L)$. Reducing the identity $g^{-1}\phi(\sigma) (^\sigma \cskip
g)=\phi'(\sigma)$ to
$G_J(k(\mu_m))$, one gets $\overline g^{-1}\phi  \overline g=\phi'$ with
$\overline g \in
G(k)$.

{ $a$)} $i) \Longrightarrow ii)$: Assume  $\phi$ is
$k$--anisotropic. The
same argument as before with
$\phi'=\phi$ shows that
$$
\bigl(_{f_\phi} G_J\bigr)(K)= \{ g \in G_J(\widetilde L) \enskip \mid \enskip
\phi(s) (^\sigma \cskip g) \phi(\sigma)^{-1}=g \enskip \forall \sigma \in
\widetilde I \} \subset
G_J(O_{\widetilde L}),
$$
so $\bigl(_{f_\phi} G_J\bigr)(K)= \Bigl[_{f_\phi} G_J(O_{\widetilde L})
\Bigr]^{\widetilde
I}$.

\noindent $ii) \Longrightarrow iii)$: As
 $\bigl(_{f_\phi} G\bigr)(K) \subset
 G_J(O_{\widetilde L})$ is bounded, the group $_{f_\phi} G/K$ does not
contain any non trivial $K$--split torus, hence
it  is $K$--anisotropic.

\noindent $iii) \Longrightarrow i)$: We prove  $not \enskip  i)
\Longrightarrow$ not
$\enskip iii)$. Assume
 $\phi$  is $k$--isotropic, i.e. there exists a $k$--parabolic subgroup $P$
such that
$\Im(\phi) \subset  N_{G_J}(P)(k)$.
 Then the class $\gamma(\phi)$ in $H^1(K,G_J)$ comes from $H^1(K,
N_{G_J}(P))$, and the twisted group $_{f_\phi}G/K$ is isotropic.

 {$c)$} We set $K'=K \otimes_k k'$, $O'=L \otimes_k k'$, etc.. The
first assertion
shows that $\bigl(_{f_\phi} G_J\bigr)(K')= \Bigl[_{f_\phi}
G_J(O_{\widetilde L'}) \Bigr]^{\widetilde I}$. But the map of $\Gal(k'/k)$
groups
$Z_{G_J}(\phi)(k')
\rg
\Bigl[_{f_\phi} G_J(O_{\widetilde L'})\Bigr]^{\widetilde I}$ is split, so
the map
$H^1(k'/k,Z_{G_J}(\phi))
\rg H^1(K.k'/K, {_{f_\phi}G_J})$ is injective.
\hfill\hfill\cqfd

\vskip2mm

First, we give the following corollary of the proposition and of  Lemma 3
(\pa 1.3).

 \th COROLLARY 2
 \enonce Assume that the adjoint group $G_{ad}$ of $G$  is simple.
 Assume $m$ is an integer such that $k$ contains a primitive $m$-th root
of unity and such that there exists a $k$--anisotropic
 automorphism  of order $m$. Then $m \in S(G)$.
 \endth

 \rem Proof
 \endrem
 One can assume that $G=G_{ad}$. Let $\alpha$ be a $k$--anisotropic
 automorphism  of order $m$ and assume that $m \not \in S(G)$.
 The exact sequence $1 \rg G \rg \Aut(G) \rg \Out(G) \rg 1$ and
the fact that $\Out(G)$ is a finite group such that the
prime divisors of $\sharp \Out(G)$ are bad primes
shows that $\alpha$ is an inner
automorphism. By Lemma 3, there exists a maximal torus $S$ which contains
$\alpha$. The choice of a primitive
$m$--root of unity $\zeta$ induces
an isomorphim ${\bb Z}/m{\bb Z} \simlgr \mu_m$; taking
$L=k(\root m \of t)/K$ and  $\theta=(t) \in H^1(K,\mu_m)$ as in section
$2$, we see
that  $\gamma(\alpha) \in \Im\bigl( H^1(L/K,S) \rg H^1(L/K,G) \bigr)$.
But $ H^1(L/K,S)$ is an $S(G)$-primary torsion group, so $H^1(L/K,S)=1$ and
$\gamma(\alpha)=1$. By Proposition 5.a,  $\alpha$ is then
$k$--isotropic.
Contradiction.
 \hfill\hfill\cqfd



\rem   Questions
\endrem
Assume  $p>0$. We lack an analogue of Proposition 2 (\pa 3.1)
for $k$--anisotropic automorphisms of  order $p$. More precisely, we ask
the following :
\vskip2mm
\begingroup\parindent=1,5cm
\item{1)} Let $\alpha$ be a $k$--anisotropic automorphism of $G$ of
order $p$. Is
$\alpha_{k_s}$ $k_s$--anisotropic ?
\item{2)} Let $\alpha,\alpha'$ be $k$--anisotropic automorphisms of $G$
of order $p$
such that $\alpha$ and $\alpha'$ are conjugate under
$\Aut(G)(k_s)$.
Are $\alpha$ and $\alpha'$ conjugate under $\Aut(G)(k)$ ?
\par\endgroup


\vskip4mm




\noindent{\bf 4 \pointir Anisotropic automorphisms are  special }

In this section, we prove conjecture 2'; as mentionned earlier, this proves
Tits original conjecture is proven  as the case $E_8$, $p=5$ has already
been done
in [Gi3].

\th THEOREM 3
\enonce  Conjecture 2' is true for all types excepting possibly
 case  $E_8$, $l=5 \not=p$.
\endth

The  steps  have been described in section 1.2 and we take into
account  cases done by Tits
in the wild case.
Their proof works  also in the tame case, which is much simpler.

\vskip2mm
\noindent{\bf 4.1 \pointir Special automorphisms}

\th DEFINITION 3
\enonce A $k$--automorphism $f$ of a torus $S/k$
is anisotropic if the $k$--group $S^f$ of fixed points of $S$ by $f$
is finite.
\endth

\th DEFINITION 3'
\enonce $a)$ An element $w$ of $W_J$ is special if  the automorphism $w : T
\rg T $ defined
by $t \mapsto w.t$ is anisotropic.

\decale{$b)$} An element  of $G_J(k)$ of order $n$ is special (relative to
the split torus
$T$) if it lies in
$N_{G_J}(T)(k)$ and its image in $W_J$ is a special element of order $n$.
\endth

Some special elements of $W_J$ are listed in [Sp] for any simple type.


\vfill\eject


\th LEMMA 4
\enonce Let $w$ be a special element of $W_J$ of order $n$.

\decale{$a)$} The homomorphism  $N(w)=1+w+...+ w^{n-1} : T \rg T$ is trivial.

\decale{$b)$} Let $g$ be a special element relative to $T$ mapping to $w$.
Then  for any $\tau \in T(k)$, $\tau g$ is a special element of $G_J(k)$
of order $n$
and is $N_G(T)(\overline k)$--conjugate to $g$.

\decale{$c)$} Let $g$ be an element in $N_G(T)(k)$ such that $g$ is
$k$--anisotropic in
$G$.  Then  $g$ acts anisotropically on $T$.
Moreover, if $\car(k)=p>0$ and $g$ has order $p^r$, then  $g$  is special
(relative to $T$).

\decale{$d)$} Assume that $n$ is odd. Let $A$ be a complete
discrete
valuation ring
with residue
field $k$ and fraction field $F_A$. Let $g$ be a special element of
$G_J(k)$ (relative to
$T$) mapping to $w$. Then there exists $\widetilde g\in N_{G_J \times_{\bb
Z} A}(T)(A)$
of order $n$  mapping to $g$ in $G_J(k)$ such that $\widetilde g$ is a
special element of
$G_J(F_A)$ (relative to $T$) and mapping to $w$ in $W_J$.
\endth

\rem Proof
\endrem
$a)$  As $(1-w).N(w)=0$ and $1-w$ is an invertible automorphism of
$\widehat T \otimes_{\bb Z} {\bb Q}$, then $N(w)=0$ as
an
automorphism of
$\widehat T \otimes_{\bb Z} {\bb Q}$, hence $N(w):T \rg T$ is trivial.

 $b)$ Let us denote by $w$ the image of $g$ in $W_J$. We have to
show that $\tau
g$ has order $n$; it is given by the computation
$$
(\tau g)^n=\tau \times g \tau g^{-1} \times g^2\tau g^{-2} \cdots  \times
g^{n-1}  \tau
g^{(-n+1)}=\tau \times (w.\tau) \times \cdots \times  (w^{n-1}. \tau)
=N(w).\tau=1.
$$
So the order  of $\tau g$ divides $n$ and as $w$ has order $n$,
the order of $\tau g$ is exactly $n$.
 We consider the equation $$
\tau g=x g x^{-1}    \qquad (x \in T(\overline k)),
$$
or equivalently
$$
\tau=xgx^{-1}g^{-1}=x  \times w.x^{-1} \qquad (x \in T(\overline k)).
$$
As the group morphism $T \rg T$ given by $t \mapsto t \times w.t^{-1}$ has
finite
kernel, it is surjective, so there exists $x\in  T(\overline k)$ such that
$\tau g=x g x^{-1}$.

 $c)$ Let $g$ be an element of $N_{G_J}(T)(k)$
such that $g$ is $k$--anisotropic. Assume that $g$ acts isotropically on
$T$, then
$T^{\langle g \rangle} =\Ker(1-g: T \rg T)$ has dimension greater than $1$,
so there exists
a $k$--split subtorus $S$ of $T$ such that $g \in Z_G(S)(k)$. As $Z_G(S)$
is a $k$--Levi
subgroup of a proper $k$--parabolic subgroup, $g$ is $k$--isotropic. So $g$
acts anisotropically on $T$. Assume moreover  that $g$ has order $p^r$; we have
to show that
the image of $g$ in $W_J$, say
$w_0$, has order $p^r$. If not, $g^{p^{r-1}}$ is an element of order $p$
of $T(k)$, and as
$\mu_p(k)=1$, this leads to a contradiction.

 $d)$ As $G_J$ is a Chevalley group defined over ${\bb Z}$,
the sequence $1 \rg T \rg N_{G_J}(T_J) \rg W_J \rg 1$ can be defined over
${\bb Z}$, and
as  $H^1_{\acute et}({\bb Z},T)\simlgr \bigl(\Pic({\bb
Z})\bigr)^{rank(G)}=1$, one has the
exact sequence
$$
1 \rg T({\bb Z}) \rg N_{G_J}(T_J) ({\bb Z}) \rg W_J \rg 1
$$
and an isomorphism  $T({\bb Z})\approx ({\bb Z} / 2 {\bb Z})^{rank(G)}$
(see [T1]).
So the class of the preceeding extension is killed by $2$.
Let $g\in N_{G_J}(T_J) (k)$ be of odd order $n$. Then there exists $g_0 \in
N_{G_J}(T_J) ({\bb Z})$ of order $n$ with same image in $W_J$ as
$g$. There thus  exists
$\tau \in T(k)$ such that $\alpha=\tau \overline \alpha_0$, where
$\overline \alpha_0$
is the reduction in $k$ of $\alpha_0$ via the map $N_{G_J}(T_J) ({\bb Z})
 \rg N_{G_J}(T_J) (A) \rg N_{G_J}(T_J) (k)$. As $T\times_{\bb Z} A$ is a
smooth group
 scheme,  the map $T(A) \rg T(k)$ is surjective by Hensel's
lemma. We pick
 $\widetilde \tau \in T(A)$ mapping to $\tau$. The element $\widetilde
g:=\widetilde \tau
\alpha_0$ of  $N_{G_J}(T_J) (A)$ lifts $g=\tau \overline \alpha_0$, maps to
$w$ in $W_J$ and by assertion $b)$, $\widetilde g$ is a special element of
$G_J(F_A)$
relative to $T$.
\hfill\hfill\cqfd


 \vskip2mm


 \noindent{\bf 4.2  \pointir Triality}


In this section, we handle the case $l=3$  assuming then that $k$ contains
a primitive third root of unity.
First, let us remark that we can assume that  $G$ is adjoint, so the group
$\Aut(G)=G \rtimes \Aut(G,B,T)$ is of the type we studied before.
Let $\alpha \in \Aut(G)(k)$ be a $k$--anisotropic automorphism of order $3$.
We shall use the following triality argument.

\th LEMMA 5
\enonce Assume that $G$ has  type $D_4$ (resp.  $F_4,E_6, E_7,E_8$).  Then
$\alpha$
normalizes  a $k$--Levi subgroup of a $k$--parabolic subgroup  of $G$ of
type $A_2$
(resp. $A_2$, $A_2 \times A_2$, $E_6$, $E_7$).
\endth

\rem Proof
\endrem
Let $P$ be a $k$--parabolic subgroup of $G$ such that $P_{red}$ has type
$A_3$ (resp.
$C_3$, $D_5$, $E_6$, $E_7$). Then $\codim_G(P)= 6$ (resp. $15, 16, 27,57$).
Let us define the $k$--group
$$
C=P \cap \alpha(P) \cap \alpha^2(P).
$$
Then $codim_G(C) \leq  3 \times codim_G(P)$, so $dim_G(C)\geq 28-10$
(resp. $7$, $30$, $52$, $77$) (cf. [PR] p. 380).
We claim that $C$ is a $k$--Levi subgroup of some $k$--parabolic subgroup
of $G$. Let $Q$ be a $k$--parabolic subgroup such that $C \subset Q \subset
P$ and
minimal for this property. Then the $k$--parabolic subgroup
$R_u(Q). \bigl( Q \cap \alpha(Q) \bigr)$ contains $C$ and is contained in
$Q$, so
$R_u(Q). \bigl( Q \cap \alpha(Q)\bigr)= Q$ and $M:=Q \cap \alpha(Q)$ is a
$k$--Levi
subgroup
of $G$ ([BT1], proposition 4.10). But $C= M \cap \alpha^2(Q)$ is a
$k$--parabolic subgroup of $M$; if
$C\not= M$, then
$\alpha$ normalizes the split $k$--unipotent group $R_u(C)$, and by
Proposition 3.1 of
[BT2], $\alpha$ normalizes a $k$--proper parabolic subgroup of $G$, which
contradicts the assumption of anisotropy. So $C$ is a $k$--Levi subgroup of
$Q/k$ and
the restriction  of $\alpha$ is still $k$--anisotropic (Lemma 2). As
$dim_k(C) \geq 10$, the
only possibility is that $C$ has type $A_2$. Other
 cases 	are considered on a case-by-case basis using Corollary 2 and we
leave the details of the argument to the reader.
\hfill\hfill\cqfd

Let us recall that Theorem 3 is true for $A_2$ and prove it inductively in
the cases
considered in Lemma 5.
 Lemma 5 gives a $k$--Levi subgroup $C$ of
a $k$--parabolic proper subgroup of $G$ normalized by $\alpha$. So the
connected
center $Z_G(C)^0$ is a $k$--split torus normalized by $\alpha$ and, by
induction, the derived group ${\cal D}C$ contains a $k$--split maximal
torus $S$ normalized by
$\alpha$. It  follows that $Z_G(C)^0.S$ is a maximal $k$-split torus of $C$
(and $G$)
normalized by $\alpha$.


 \vskip2mm

\noindent{\bf  4.3 \pointir Lifting in characteristic $0$.}


\th LEMMA 6
\enonce  Assume  $p=\car(k)>0$. Let $A$ be a complete discrete valuation
ring with
residue field $k$ and fraction field $F_A$. Let
$\phi: I \rg G_J$ be a group homomorphism which lifts to $\widetilde \phi :
I \rg
G_J\times_{\Spec({\bb Z})} \Spec(A)$.

\decale{$a)$} If $\widetilde \phi: I \rg G_J\times_{\bb Z} F_A$ is a
$F_A$--isotropic
morphism, then $\phi$ is a $k$--isotropic morphism.

\decale{$b)$} Assume  $\phi$ is $k$--anisotropic.   If $\widetilde \phi$
normalizes some maximal $F_A$--split torus of  $G_{J,F_A}$, then
$\phi$ normalizes some maximal $k$--split torus of $G_J/k$.
 Moreover,
let $\phi': I \rg G_J$ be a group homomorphism which lifts to
 $\widetilde \phi : I \rg G_J\times_{\Spec({\bb Z})} \Spec(A)$ such that
$\widetilde \phi$
and $\widetilde \phi'$ are  conjugate under
$G_J(F_A)$. Then $\phi$ and $\phi'$ are conjugate under $G_J(k)$ and
$\phi'$ normalizes some maximal $k$--split torus of $G_J/k$.
\endth


\rem Proof
\endrem
$a)$ Assume that   $\widetilde \phi$ is a $k$--isotropic element of
$G_J(F_A)$. This means
 that there exists a standard parabolic subgroup $P/{\bb Z}$ of
$G/{\bb Z}$ such that  $I$ (through $\widetilde \phi$) has a fixed point in
$X(F_A)$, where
$X=G_J /N_{G_J}(P)$ is the projective ${\bb Z}$--scheme of parabolic subgroups
of $G_J$ of the same type as $P$.  As $X(A)=X(F_A)$, then $I$ has a fixed
point in
$X(k)$, and $\phi$ is $k$--isotropic.

\noindent $b)$ Assume that $\phi$ is $k$--anisotropic and that $\widetilde
\phi$ normalizes a $F_A$--maximal split torus, i.e. there exists $g \in
G_J(F_A)$ such
 that
$$
g^{-1} \widetilde \phi(\sigma) g =\widetilde f(\sigma) \in N_{G_J}(T)(F_A)
\enskip (\sigma \in I),\leqno(*)
$$
where $\widetilde f: I \rg N_{G_J}(T)(F_A)$ is a group homomorphism.
We consider the Bruhat--Tits building ${\cal B}$ of the group $G_{J,F_A}$
with center $c$,  which is
the vertex stabilized by $G_J(A)$. We denote by
${\cal B}^I$ the fixed point of $X$ by $I$ (through $ \widetilde \phi$),
which is convex
and contains $c$. The torus $T$ defines an apartment ${\cal A}= \widehat T
\otimes_{\bb
Z} {\bb R}
\subset X$.  From $(*)$, we get $g. {\cal A}  \subset {\cal B}^I$. Let us
denote by $\pi(c)$
the projection of $c$ on the apartment $g. {\cal A}$.

\rem Claim $\pi(c)=g.c=c$
\endrem
 We write $c=g.a $ with $a \in {\cal A}$.  For any $\sigma \in I$, one has
 $$
 g.a =\phi(\sigma) g.a=g \widetilde  f(\sigma) .a,
 $$
 so $\widetilde f(\sigma).a =a$. As $I$ acts on ${\cal A}$ trough
$\widetilde f$,
 we get $a \in ({\widehat T}^0\otimes_{\bb Z} {\bb R} \bigr)^I ={\cal A}^I$.
But $({\widehat T}^0\otimes_{\bb Z} {\bb R} \bigr)^I=
({\widehat T}^0\bigr)^I \otimes_{\bb Z} {\bb R}$, so if $a \not =c$,
it means then
 $\widetilde f$ is $k$--isotropic, which gives a contradiction
to the anisotropy
of $\phi$ by $a)$. We conclude that $\pi(c)=g.c$.

As ${\cal B}^I$ is convex, the segment $[c, \pi(c)]$ is fixed pointwise by
$\alpha$.  If $c \not=\pi(c) $, there exists a facet $F$ of ${\cal B}$
strictly containing $c$  which is stabilized by $I$. By the isomorphism
between the link of $c$ and the spherical building of $G_J/k$, there exists
a proper $k$--parabolic subgroup $P$ of $G$ such that $\phi(P)=P$, which again
gives a contradiction to the anisotropy of $\phi$. We conclude that
$\pi(c)=c$, i.e.
$c \in g. {\cal A}$.
So $g.c=c$, i.e. $g \in G_J(A)$ and $\widetilde f(\sigma) \in N_{G_J}(T)(A)$.
Reducing $(*)$ in $k$, it yields
$\overline g^{-1} \phi(\sigma) \overline g \in N_{G_J}(T)(k)$, so
$\phi$ normalizes the maximal $k$--split torus $\overline g.T$. One  can
then assume that
$\phi$ and $\widetilde \phi$  normalizes $T$.

Now, let  $\phi': I
\rg G_J$ be a group homomorphism which lifts to $\widetilde \phi : I \rg
G_J\times_{\bb Z} A$ and such that $\widetilde \phi$ and $\widetilde \phi'$
are conjugate
under
$G_J(F_A)$.   Let $g \in G_J(F_A )$ be an element such
that
$\widetilde\phi'= g \widetilde \phi g^{-1}$. Then
$\widetilde \phi'$ normalizes the $F_A$--torus $gTg^{-1}$. But $\widetilde
\phi$ is anisotropic by
assertion a), so $\widetilde \phi'$ is anisotropic and the proof of
assertion b)
shows that $g \in G_J(A )$. We have then $\widetilde \phi'= g \widetilde
\phi g^{-1}$ with $g \in G_J(A)$,
so $\phi= \overline g \phi {(\overline g)}^{-1}$ with  $\overline g \in
G_J(k)$, and
$\phi'$ normalizes the  maximal $k$--split torus  $\overline gT\overline
g^{-1}$.
\hfill\hfill\cqfd



\vskip2mm

\th LEMMA 7
\enonce  Let $\phi \in \Hom_{gr} (I, G_J(k))$ be a $k$--anisotropic
morphism. Then  there
exists $g \in G_J(k_s)$  such that the following hold

\begingroup\parindent=2,5cm
\item{$i)$}  $g^{-1} \Im(\phi) g \subset N_{G_J}(T)(k_s)$,
\item{$ii)$}  ${g \cskip ^{-1}}  {^s \cskip g} \subset  N_{G_J}(T)(k_s)
\enskip \forall  s
\in \Gal(k_s/k)$.
\par \endgroup
The torus $g.T$ is defined over $k$ and normalized by $\phi$.
Moreover, in the case $I$ is cyclic, one can replace condition $i)$ by
\begingroup\parindent=2,5cm
\item{$i')$}  $g^{-1} \Im(\phi) g \in  N_{G_J}(T)(k)$,
\par \endgroup
\endth



\rem Remark 3
\endrem  Lemma 7 yields conjecture 2 of Tits by an uniform argument in the
case $k$ is
separably closed.



\rem Proof of Lemma 7
\endrem It is well--known that the map
$$
H^1(K, N_{G_J}(T)) \rg H^1(K,G_J)
$$
is surjective. In other words, there exists a finite Galois extension $M/K$
such  that $K \subset
L
\subset M \subset K_s$, an element $g \in G_J(M)$ and a cocycle $h \in
Z^1\bigl(\Gal(M/K),
N_{G_J}(T)(K_s)\bigr)$ such that
$$
g \cskip ^{-1}  \phi(s){^s\cskip g} =  h_s  \qquad (s\in \Gal(M/K)).\leqno(*)
$$
Let us denote by $K'/K$ the maximal unramified  subextension of $M$, by
$O'$ its valuation ring and by $k'$ its residue field
Set $L'=L.K$. We may  assume  $K'=k'((t))$.
$$
\diagram{
&&M \cr
&&\mid \cr
&&L'  \cr
 &\diagup&& \diagdown\cr
K'&&&&L \cr
 &\diagdown&& \diagup \cr
&&K \cr}
$$
Let  ${\cal B}$ be the Bruhat--Tits building of the group $G_M$ [BrT1, \pa
\un.7.4].
This building is equipped
with  an  action of the group
$G_J(M)$ and an  action of the Galois group $\Gal(M/K)$ denoted by
$x \mapsto {^s \cskip x}$  for any $s \in \Gal(M/K)$. We denote by  $c$
the center of the building, i.e. the  vertex of
${\cal B}$ fixed by  $G_J(O_M)$, where $O_M$ is the valuation ring of $M$.
The torus $T \times_k M$ defines an apartment
${\cal A}=\widehat T^0 \otimes_{\bb Z} {\bb R} \subset {\cal B}$,
which contains the center $c$, and which is fixed pointwise by $\Gal(M/K)$.
One defines
a twisted action of
$\Gal(M/K)$  on  ${\cal B}$  by
$$
s. x =\phi(s) . {^s \cskip x} \qquad (s \in \Gal(M/K)).
$$
This action is isometric and simplicial (i.e. maps a facet on a facet).
Clearly, since
$\Im(\phi) \subset  G_J(k')$, the center $c$ is fixed under the twisted
action of $\Gal(M/K)$.
We claim that the apartment  $g{\cal A}$ is stabilised by
$\Gal(M/K)$ for the twisted action. This
follows from the following
computation; for any $x
=g.a
\in {\cal A}$, one has
$$
s.x= \phi(s) . {^s \cskip (g.a)}= g. (h_s. a) \in g {\cal A}.
$$
Let us denote by $\pi(c)=g.a$ the projection of $c$ to
 the apartment $g {\cal A}$.

\rem Claim: $\pi(c)= g.c=c$
\endrem
As $c$ is fixed under  the twisted action of
$\Gal(L/K)$, the projection $\pi(c)$ is also fixed.
Moreover (observe $^s \cskip a=a$), one has
$\pi(c)= g.a=s.(g.a)= \phi(s). {^s \cskip g}. {^s \cskip a}= g h_s.a \qquad
(s \in \Gal(L/K))$,
so
$$
a =h_s . a \qquad (s \in \Gal(M/K) ).
$$
Now, we consider the maximal torus $$
{_nT}/K \subset {_nG}/K \approx { _{f_\phi}G/K}.
$$
The group  $_nG/K \approx _{f_\phi}G/K$
is anisotropic by  Proposition 5, so the torus $_nT$ is anisotropic and
 $0=\widehat {_nT}^0(K) \otimes_{\bb Z} {\bb R}= \{ x \in {\cal A} \enskip
\mid \enskip h_s.  x = x \}$. One deduces that $a=c$, i.e. $\pi(c)=h.c$.

Let us assume that $\pi(c)\not =c$. Then the segment $[c, \pi(c)]$ is fixed
pointwise
by the twisted action of $\Gal(M/K)$. So there exists a facet of ${\cal B}$
containing
strictly $c$ which is stabilized by the twisted action of $\Gal(M/K)$
and there exists (by the isomorphim between the link of $c$ in ${\cal B}$
and the spherical building of $G/k$)  a facet  of the spherical building of
$G/k$
which is pointwise fixed by $\Gal(M/K)$. In other words, there exists
a proper $k$--parabolic subgroup of $G$ such that
$$
\Bigl[ _{f_\phi}\Bigl(G_J(k') / N_{G_J}(P)(k')\Bigr) \Bigr]^{\Gal(M/K)} \not
=\emptyset.
$$
The group $\Gal(M/K)$ acts on the previous set through $\Gal(L'/K)=I \times
\Gal(k'/k)$. One has
$$
\displaylines{
\Bigl[ _{f_\phi}\Bigl(G_J(k') / N_{G_J}(P)(k')\Bigr) \Bigr]^I
= \{ [d] \in G_J(k') / N_{G_J}(P)(k')\enskip \mid \enskip
d^{-1} \Im(\phi) d \in  N_{G_J}(P)(k') \} ,\cr
\Bigl[ _{f_\phi}\Bigl(G_J(k') / N_{G_J}(P)(k')\Bigr) \Bigr]^{\Gal(k'/k)} \cr
= \{ [d] \in G_J(k') / N_{G_J}(P)(k')\enskip \mid \enskip d^{-1} (^s \cskip
d) \in
N_{G_J}(P)(k') \enskip  \forall s \in \Gal(k'/k) \} .\cr
}
$$
Let $d \in  N_{G_J}(P)(k')$ be a coset in the intersection of the two
previous sets and let us
denote by $z \in Z^1(k'/k, N_{G_J}(P))$ the cocycle defined by
$z_s=d^{-1}(^s \cskip d)$.
The element $d$ gives rise to a  trivialization $\beta_d:  _z G_J \simlgr G_J$
such that $\Im(\phi) \subset   \beta_d( _zN_{G_J}(P))$, which gives a
contradiction to
the anisotropy of $\phi$. We conclude that $\pi(c)=c$.
Let us denote by $a \mapsto \overline a$ the map $G_J(O_M) \rg G_J(k')$.
The identity $(*)$ holds in $G_J(O_M)$ and replacing $d$ by $\overline d \in
G_J(k')$, one can assume that $g \in G_J(k') \subset G_J(L')$, that $L'=M$
and $h_s \in
G_J(k')
\subset G_J(L')$ for every $s \in \Gal(L'/K)$. So we get $i)$ and $ii)$.


Assume now that $I$ is cyclic with a generator $\sigma$.
Identity $(*)$ for $\sigma$ yields
$$
g^{-1} \phi(\sigma) g =h_\sigma \enskip \in \enskip  N_{G_J}(T)(k'). \leqno(**)
$$
Since $\Gal(k'/k)$ acts trivially on the Weyl group $W_J$, the map
$s \mapsto h_\sigma^{-1} (^s \cskip  h_\sigma)$ defines a
$\Gal(k'/k)$--cocycle in
$T(k')$ and
by Hilbert's theorem  90, there exists $\tau \in T(k')$ such that
$h_\sigma^{-1} (^s \cskip
h_\sigma)=\tau^{-1} (^s \cskip \tau)$. Replacing $g$ by $g\tau$ in $(**)$,
one can
assume
that  $h_\sigma \in  N_{G_J}(k)$ and taking $h=h_\sigma$, the proposition
is proved.
\hfill\hfill\cqfd


\th LEMMA 8
\enonce Assume that $p=\car(k)$ is an odd prime; let $A$ be a complete discrete
valuation ring with
residue field $k$ and fraction field $F_A$.
Let $H/k$ be a $k$--form of $G_J$,  ${\goth H}$ a $A$--form of $G_J \times_{\bb
Z}A$  with special fiber $H/k$ Let $S$ be a maximal $k$--torus of $H$,
and let ${\goth S}$ be a maximal $A$--torus of
${\goth H}$ with special fiber $S$.
Let $h \in N_H(S)(k)$ be an element of finite order $q=p^r$. Then $h$
lifts to
 an element $\widetilde h \in N_{\goth H}({\goth S})(A)$ of
order $q$.
\endth


\rem Remark 4
\endrem
Such an
${\goth H}$ exists
by Hensel isomorphism
$H^1_{\acute et}(A, \Aut(G_J)) \simlgr H^1(k, \Aut(G_J))$ and such an
$A$--torus $\goth S$ exists by  [SGA3, exp.
{\uppercase\expandafter{\romannumeral 15}}, \pa 8].

\rem Proof
\endrem
Let $A_s/A$
be the \'etale extension associated to $k_s/k$.
We remark first that we can assume that $h$ generates $H/H^0$.
Let us denote by $w$ the image of  $h$ in $(N_H(S)/S)(k)$, which has order $q$.

\rem First case  : $w$ acts anisotropically on $S \times_k k_s$
\endrem
First,  remark that the lifting assumption in  $N_{\goth H}({\goth
S})(A_s)$ is satisfied by Lemma 4.d (\pa 4.1, we use here that $p$ is odd).
Let us denote  the Weyl group scheme $N_{\goth H}({\goth S}) /{\goth S}$ by
$W({\goth S})/A$; that is a
twisted finite group,  so isomorphic to $N_G(S)/S$. We also denote  by
$\langle w \rangle$ the  cyclic subgroup of $W({\goth S})$ generated  by
$w$ and
by $N_w/A$ the preimage of  $\langle w \rangle$ in $N_{\goth H}( {\goth S})$.
By Hensel's lemma, one has the following exact commutative diagram of
pointed sets
$$
\diagram{
1&\hfl{ }{}{4mm}& {\goth S}(A)& \hfl{ }{}{4mm}&N_w(A)
& \hfl{ }{}{4mm}& \langle w \rangle& \hfl{ }{}{4mm}&  H^1(A, {\goth S}) \cr
&& \vfl{ }{}{4mm}&& \vfl{}{}{4mm}&& \pafl{ }{}{4mm}&& \vfl{ }{\wr}{4mm} \cr
1 & \hfl{ }{}{4mm}& S(k)& \hfl{ }{}{4mm}&N_w(k)
& \hfl{ }{}{4mm}& \langle w \rangle & \hfl{ }{}{4mm}&  H^1(k,S) , \cr
}
$$
and
$$
\diagram{
&&1&& 1 \cr
&& \vfl{ }{}{2mm}&& \vfl{ }{}{2mm} \cr
&&{\bb S}&=& {\bb S} \cr
&& \vfl{ }{}{2mm}&& \vfl{ }{}{2mm} \cr
1& \hfl{ }{}{4mm}& {\goth S}(A)& \hfl{ }{}{4mm}&N_w(A)
& \hfl{ }{}{4mm}& \langle w \rangle& \hfl{ }{}{4mm}& 1 \cr
&& \vfl{sp}{}{4mm}&& \vfl{sp}{}{4mm}&& \pafl{ }{}{4mm} \cr
1& \hfl{ }{}{4mm}& S(k)& \hfl{ }{}{4mm}&N_w(k)
& \hfl{ }{}{4mm}& \langle w \rangle& \hfl{ }{}{4mm}&  1,\cr
&& \vfl{ }{}{2mm}&& \vfl{ }{}{2mm} \cr
&&1&& 1\cr
}
$$
where ${\bb S}:= \Ker\bigl({\goth S}(A) \rg S(k)\bigr)$. So, the
obstruction to lifting $h$ to
 an element of order $q$ in
$N_{\goth H}({\goth S})(A)$  is the class, say
$\eta$ in $H^2(\langle h \rangle, {\bb S})$, of  the restriction to the
cyclic group
$\langle h \rangle$ of the vertical extension. We put
 $\mu/A=\Ker(1-w: {\goth S} \rg {\goth S})$ which is
a finite
$A$--multiplicative group, and we have
$$
H^2(\langle h \rangle, {\bb S})= \bigl({\bb S}\bigr)^{\langle w \rangle } /
N(w) . {\bb
S} = \Ker\bigl( \mu(A) \rg \mu(k) \bigr),
$$
because $N(w)=0$ (Lemma 4.a).
The same approach for $k=k_s$ gives an obstruction $\eta_s$ in $\mu(A_s)$ which
vanishes. As $\mu(A) \subset \mu(A_s)$, one has $\eta=0$, and $h$
lifts to $N_{\goth H}({\goth S})(A)$.

\rem Second case : $h$ acts isotropically on $S\times_k k_s$
\endrem
This means that $\dim(S^{\langle h \rangle}) \geq 1$, i.e. there exists a
non--trivial
subtorus $S_0$ of $S$ such that $h \in Z_H(S_0)(k)$. We have
$Z_{H^0}(S_0)=Z_H(S_0)^0$.  As $h$ generates $H/H^0$,
we have an exact sequence $1 \rg Z_{H^0}(S_0) \rg Z_{H}(S_0) \rg H/H^0 \rg 1$
which is split by $h$. The element  $h$ acts trivially on the coradical
torus (cf. [SGA3],
exp. {\uppercase\expandafter{\romannumeral 22}}, \pa 6.2)
$C=:corad(Z_{H^0}(S_0))$ of $Z_{H^0}(S_0)$, and it defines an extension of  the
canonical morphism $Z_{H^0}(S_0) \rg  C$
 to a  morphism $Z_H(S_0)^0 \rg C$. So one can define  the
group
$H'/k$ by the following diagram
$$
\diagram{
&&1&&1\cr
&&\ufl{}{}{2mm}&&\ufl{}{}{2mm} \cr
&&  C &=& C  \cr
&&\ufl{ }{}{4mm}&&\ufl{}{}{4mm} \cr
1& \hfl{ }{}{4mm}& Z_{H^0}(S_0) & \hfl{ }{}{4mm}& Z_H(S_0) & \hfl{ }{}{4mm}&
H/H^0 & \hfl{ }{}{4mm}& 1\cr
&&\ufl{}{}{4mm}&&\ufl{}{}{4mm}&&\pafl{}{}{4mm} \cr
1& \hfl{ }{}{4mm}& DZ_{H^0}(S_0) & \hfl{ }{}{4mm}& H' & \hfl{ }{}{4mm}&
H/H^0 & \hfl{ }{}{4mm}& 1.\cr
&&\ufl{}{}{2mm}&&\ufl{}{}{2mm} \cr
&&1&&1\cr
}
$$
So $H'$ is a $k$--form of some  $G'_{J'}$ and $h \in H(k)$ because
$C(k)$ has no $k$--points of $p$--power order. Denoting $S'=\Ker(S \rg
C) \subset H'$, $h$ acts anisotropically on the maximal torus $S' \times_k
k_s$ of
$H'$. Defining ${\goth H}'/A$ and ${\goth S}'/A$ by Hensel's lift, the
first case
shows
that $h$ lifts in
$N_{{\goth H}'}({\goth S'})(A)=N_{{\goth H}}({\goth S})(A) \cap {\goth
H}'(A)$, so $h$
lifts in
$N_{{\goth H}}({\goth S})(A)$.
\hfill\hfill\cqfd



\noindent{\bf 4.4 \pointir End of the proof of Theorem 3}

Recall that  $\car(k)=p>0$ and we consider a $k$--anisotropic automorphism
$\alpha$ of prime order $l$ of $G$.
We can assume that $l$ is odd. Let $A$ be a
complete
discrete valuation ring with residue field  $k$ and with fraction field
$F_A$ such that $F_A$
contains a primitive $p$--root of unity.  First, we show that $\alpha$
lifts to $\Aut(G)(A)$, i.e.
there exists $\widetilde \alpha \in
\Aut(G)(F_A)$ of order
$l$ lifting $\alpha$.
If $l=p$, this is a consequence of  Lemmas 7 and 8.
If $l\not=p$, then  ${\bb Z}/l {\bb Z}$ is a finite group
of multiplicative type; as $\Aut(G)$ is smooth, the Grothendieck rigidity
theorem [SGA3,
exp. {\uppercase\expandafter{\romannumeral 6}}, Cor. 7.3] says that the map
$$
\Hom_{A-gr}\bigl( {\bb Z}/l {\bb Z}, \Aut(G) \bigr) \rg
\Hom_{k-gr}\bigl( {\bb Z}/l {\bb Z}, \Aut(G) \bigr)
$$
is surjective, so $\alpha$ lifts in characteristic zero.
By Lemma 6.a, $\widetilde  \alpha$ is an $F_A$--anisotropic automorphism
of $G/F_A$,  the characteristic zero  case shows that $\widetilde \alpha$ is
special
and  finally Lemma 6.b shows  that $\alpha$ is special.


\vskip8mm

\noindent{\bf Acknowledgments :}
It is a pleasure to thank Tam\'as Szamuely for his careful reading of a
preliminary version
of this paper.

I wish to express my hearty thanks to Jean--Pierre Serre and Jacques
Tits, who introduced me to this subject.


\vskip12mm


\noindent{\bf References}
{\leftskip=8mm
%\input biblio
\parindent=0pt


\newtoks\ref
\newtoks\auteur
\newtoks\titre
\newtoks\editeur
\newtoks\annee
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\newtoks\tome
\newtoks\pages
\newtoks\reste
\newtoks\autre



\def\livre{\leavevmode
\llap{[\the\ref]\enspace}%
\the\auteur\pointir
{\sl \the\titre},
\the\editeur, 
{\the\annee}.
\smallskip
\filbreak}

\def\article{\leavevmode
\llap{[\the\ref]\enspace}%
\the\auteur\pointir
\the\titre,
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{\bf\the\tome},
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\the\pages. 
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\llap{[\the\ref]\enspace}%
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\let\+=\tabalign

\def\signature#1\endsignature{\vskip 15mm minus 5mm\rightline{\vtop{#1}}}




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\vskip8mm

\signature
\+ Philippe GILLE\cr
\+ D\'epartement de Math\'ematiques, CNRS \cr
\+ B\^at. 425, Universit\'e de Paris--Sud\cr
\+ F--91405 ORSAY CEDEX\cr
\+ Courrier \'electronique: gille@math.u-psud.fr\cr
\endsignature
}


\bye
