%%%% format a4
\catcode'32=9
\magnification=1200

\voffset=1cm
\hoffset=0cm
%\hoffset=1cm
\font\tenpc=cmcsc10
%\font\eightpc=cmcsc8

% Charge des fontes de 8 et 6 points :
\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

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

% Chargement des fontes AMS

\font\tengoth=eufm10
\font\tenbboard=msbm10
\font\eightgoth=eufm8
\font\eightbboard=msbm8
\font\sevengoth=eufm7
\font\sevenbboard=msbm7
\font\sixgoth=eufm6
\font\fivegoth=eufm5

\newfam\gothfam
\newfam\bboardfam

\catcode`\@=11

\def\raggedbottom{\topskip 10pt plus 36pt
\r@ggedbottomtrue}
\def\pc#1#2|{{\bigf@ntpc #1\penalty
\@MM\hskip\z@skip\smallf@ntpc #2}}

\def\tenpoint{%
  \textfont0=\tenrm \scriptfont0=\sevenrm \scriptscriptfont0=\fiverm
  \def\rm{\fam\z@\tenrm}%
  \textfont1=\teni \scriptfont1=\seveni \scriptscriptfont1=\fivei
  \def\oldstyle{\fam\@ne\teni}%
  \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\bboard{\fam\bboardfam}%
  \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
  \abovedisplayshortskip=0pt plus 3pt
  \belowdisplayskip=12pt plus 3pt minus 9pt
  \belowdisplayshortskip=7pt plus 3pt minus 4pt
  \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@ntpc=\tenrm \let\smallf@ntpc=\sevenrm
  \let\petcap=\tenpc
  \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}%
  \textfont2=\eightsy \scriptfont2=\sixsy \scriptscriptfont2=\fivesy
  \textfont\gothfam=\eightgoth \scriptfont\gothfam=\sixgoth
  \scriptscriptfont\gothfam=\fivegoth
  \def\goth{\fam\gothfam\eightgoth}%
  \textfont\bboardfam=\eightbboard \scriptfont\bboardfam=\sevenbboard
  \scriptscriptfont\bboardfam=\sevenbboard
  \def\bboard{\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 2pt minus 6pt
  \abovedisplayshortskip=0pt plus 2pt
  \belowdisplayskip=9pt plus 2pt minus 6pt
  \belowdisplayshortskip=5pt plus 2pt minus 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@ntpc=\eightrm \let\smallf@ntpc=\sixrm
  \normalbaselines\rm}

\let\bb=\bboard

\tenpoint

%���������������������� dactylographie fran�aise �����������������

\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

%���������������������� format de sortie �����������������

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

\def\appeln@te{}
\def\vfootnote#1{\def\@parameter{#1}\insert\footins\bgroup\eightpoint
  \interlinepenalty\interfootnotelinepenalty
  \splittopskip\ht\strutbox % top baseline for broken footnotes
  \splitmaxdepth\dp\strutbox \floatingpenalty\@MM
  \leftskip\z@skip \rightskip\z@skip
  \ifx\appeln@te\@parameter\indent \else{\noindent #1\ }\fi
  \footstrut\futurelet\next\fo@t}

\pretolerance=500 \tolerance=1000 \brokenpenalty=5000
\newdimen\hmargehaute \hmargehaute=0cm
\newdimen\lpage \lpage=13.3cm
\newdimen\hpage \hpage=20cm
\newdimen\lmargeext \lmargeext=1cm
\hsize=11.25cm
\vsize=18cm
\parskip 0pt
\parindent=12pt

\def\margehaute{\vbox to \hmargehaute{\vss}}%
\def\margebasse{\vss}

\output{\shipout\vbox to \hpage{\margehaute\nointerlineskip
  \corpsdepage\margebasse}
  \advancepageno \global\pagetitrefalse
  \ifnum\outputpenalty>-20000 \else\dosupereject\fi}

\def\corpsdepage{\hbox to \lpage{\hss\pagetexte\hskip\lmargeext}}
\def\pagetexte{\vbox{\makeheadline\pagebody\makefootline}}
\headline={\ifpagetitre\titleheadline \else
  \ifodd\pageno\rightheadline \else\leftheadline\fi\fi}
\def\leftheadline{\eightpoint\hfil\the\auteurcourant\hfil}
\def\rightheadline{\eightpoint\hfil\the\titrecourant\hfil}
\def\titleheadline{\hfill}
\pagetitretrue

\def\footnoterule{\kern-6\p@
  \hrule width 2truein \kern 5.6\p@} % the \hrule is .4pt high



\let\rmpc=\sevenrm
\def\pd#1#2 {\pc#1#2| }

\def\pointir{\discretionary{.}{}{.\kern.35em---\kern.7em}\nobreak
\hskip 0em plus .3em minus .4em }

\def\resume#1{\vbox{\eightpoint \pc R\'ESUM\'E|\pointir #1}}
\def\abstract#1{\vbox{\eightpoint \pc ABSTRACT|\pointir #1}}

\def\titre#1|{\message{#1}
              \par\vskip 30pt plus 24pt minus 3pt\penalty -1000
              \vskip 0pt plus -24pt minus 3pt\penalty -1000
              \centerline{\bf #1}
              \vskip 5pt
              \penalty 10000 }

\def\section#1|{\par\vskip .3cm
                {\bf #1}\pointir}

\def\ssection#1|{\par\vskip .2cm
                {\it #1}\pointir}

\long\def\th#1|#2\finth{\par\medskip
              {\petcap #1\pointir}{\it #2}\par\smallskip}

\long\def\tha#1|#2\fintha{\par\medskip
                    {\petcap #1.}\par\nobreak{\it #2}\par\smallskip}
\def\cf{{\it cf}}

\def\rem#1|{\par\medskip
            {{\it #1}\pointir}}

\def\rema#1|{\par\medskip
             {{\it #1.}\par\nobreak }}

%
\def\ieme{\raise 1ex\hbox{\pc{}i\`eme|}}
\def\omini{\raise 1ex\hbox{\pc{}o|}}
\def\emini{\raise 1ex\hbox{\pc{}e|}}
\def\ermini{\raise 1ex\hbox{\pc{}er|}}
\def\remini{\raise 1ex\hbox{\pc{}re|}}

%reference pour un article :
\def\article#1|#2|#3|#4|#5|#6|#7|
    {{\leftskip=7mm\noindent
     \hangindent=2mm\hangafter=1
     \llap{[#1]\hskip.35em}{#2}\pointir
     #3, {\sl #4}, t.\nobreak\ {\bf #5}, {\oldstyle #6},
     p.\nobreak\ #7.\par}}
%reference pour un livre :
\def\livre#1|#2|#3|#4|
    {{\leftskip=7mm\noindent
    \hangindent=2mm\hangafter=1
    \llap{[#1]\hskip.35em}{#2}\pointir
    {\sl #3}\pointir #4.\par}}
%reference complementaire :
\def\divers#1|#2|#3|
    {{\leftskip=7mm\noindent
    \hangindent=2mm\hangafter=1
     \llap{[#1]\hskip.35em}{#2}\pointir
     #3.\par}}
%
\mathchardef\conj="0365
\def\dem{\par{\it D\'emonstration}\pointir}
\def\qed{\quad\raise -2pt\hbox{\vrule\vbox to 10pt{\hrule width 4pt
\vfill\hrule}\vrule}}

\def\virg{\raise 2pt\hbox{,}}   % virgule apr�s une fraction

\def\cqfd{\penalty 500 \hbox{\qed}\par\smallskip}


%les differents retraits, voir aussi \item

\def\decale#1|{\par\noindent\hskip 28pt\llap{#1}\kern 5pt}
\def\decaledecale#1|{\par\noindent\hskip 34pt\llap{#1}\kern 5pt}
% pour les titres en deux lignes et les sections sans point-tiret :
\def\titrea#1|#2|{\message{#1 #2}
  \par\vskip.5cm plus .1cm minus .1cm\penalty -1000
  \centerline{\bf #1}
  \centerline{\bf #2}
  \vskip 5pt
  \penalty 10000 }
\def\sectiona#1|{\par\vskip .3cm
  {\bf #1.}
  \par\nobreak\vskip 3pt }
\def\ssectiona#1|{\par\vskip .2cm
  {\it #1.}
  \par\nobreak\vskip 2pt }

\def\rest#1{\ifinner\setbox1=\hbox{$\textstyle{#1}$}
            \else\setbox1=\hbox{$\displaystyle{#1}$}\fi
            \dimen1=\ht1
            \advance\dimen1 by\dp1
            \divide\dimen1 by 2
            \box1\lower 2pt\hbox{$\left|\vbox to\dimen1{}\right.$}}

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

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

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

\def\lfq{\leavevmode\raise.3ex\hbox{$\scriptscriptstyle
\langle\!\langle$}\thinspace}
\def\rfq{\leavevmode\thinspace\raise.3ex\hbox{$\scriptscriptstyle
\rangle\!\rangle$}}


\catcode`\@=12

%format to draw oblique lines

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

%������������������������������������������������������������������

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

%������������������������������������������������������������������

\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{$\scriptstyle #2$}}

%������������������������������������������������������������������


 \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
\@tempdima \ht\@linechar \multiply\@tempdima \@tempcnta
\divide\@tempdima \@m
\advance\@clnht \@tempdima
\ifdim \@linelen <\wd\@linechar
   \hskip \wd\@linechar
  \else\@upordown\@clnht\copy\@linechar\fi}

%\def\@hline{\ifnum \@xarg <0 \hskip -\@linelen \fi
%\vrule \@height \@halfwidth \@depth \@halfwidth \@width
%\@linelen \ifnum \@xarg <0 \hskip -\@linelen \fi}


\def\@hline{\ifnum \@xarg <0 \hskip -\@linelen \fi
\vrule height \@halfwidth depth \@halfwidth width \@linelen
\ifnum \@xarg <0 \hskip -\@linelen \fi}

\def\@getlinechar(#1,#2){\@tempcnta#1\relax\multiply\@tempcnta 8
\advance\@tempcnta -9 \ifnum #2>0 \advance\@tempcnta #2\relax\else
\advance\@tempcnta -#2\relax\advance\@tempcnta 64 \fi
\char\@tempcnta}

\def\vector(#1,#2)#3{\@xarg #1\relax \@yarg #2\relax
\@linelen=#3\unitlength
\ifnum\@xarg =0 \@vvector
  \else \ifnum\@yarg =0 \@hvector \else \@svector\fi
\fi}

\def\@hvector{\@hline\hbox to 0pt{\@linefnt
\ifnum \@xarg <0 \@getlarrow(1,0)\hss\else
    \hss\@getrarrow(1,0)\fi}}

\def\@vvector{\ifnum \@yarg <0 \@downvector \else \@upvector \fi}

\def\@svector{\@sline
\@tempcnta\@yarg \ifnum\@tempcnta <0 \@tempcnta=-\@tempcnta\fi
\ifnum\@tempcnta <5
  \hskip -\wd\@linechar
  \@upordown\@clnht \hbox{\@linefnt  \if@negarg
  \@getlarrow(\@xarg,\@yyarg) \else \@getrarrow(\@xarg,\@yyarg) \fi}%
\else\@badlinearg\fi}

\def\@getlarrow(#1,#2){\ifnum #2 =\z@ \@tempcnta='33\else
\@tempcnta=#1\relax\multiply\@tempcnta \sixt@@n \advance\@tempcnta
-9 \@tempcntb=#2\relax\multiply\@tempcntb \tw@
\ifnum \@tempcntb >0 \advance\@tempcnta \@tempcntb\relax
\else\advance\@tempcnta -\@tempcntb\advance\@tempcnta 64
\fi\fi\char\@tempcnta}

\def\@getrarrow(#1,#2){\@tempcntb=#2\relax
\ifnum\@tempcntb < 0 \@tempcntb=-\@tempcntb\relax\fi
\ifcase \@tempcntb\relax \@tempcnta='55 \or
\ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta
24 \advance\@tempcnta -6 \else \ifnum #1=3 \@tempcnta=49
\else\@tempcnta=58 \fi\fi\or
\ifnum #1<3 \@tempcnta=#1\relax\multiply\@tempcnta
24 \advance\@tempcnta -3 \else \@tempcnta=51\fi\or
\@tempcnta=#1\relax\multiply\@tempcnta
\sixt@@n \advance\@tempcnta -\tw@ \else
\@tempcnta=#1\relax\multiply\@tempcnta
\sixt@@n \advance\@tempcnta 7 \fi\ifnum #2<0 \advance\@tempcnta 64 \fi
\char\@tempcnta}



\def\@vline{\ifnum \@yarg <0 \@downline \else \@upline\fi}


\def\@upline{\hbox to \z@{\hskip -\@halfwidth \vrule
  width \@wholewidth height \@linelen depth \z@\hss}}

\def\@downline{\hbox to \z@{\hskip -\@halfwidth \vrule
  width \@wholewidth height \z@ depth \@linelen \hss}}

\def\@upvector{\@upline\setbox\@tempboxa\hbox{\@linefnt\char'66}\raise
     \@linelen \hbox to\z@{\lower \ht\@tempboxa\box\@tempboxa\hss}}

\def\@downvector{\@downline\lower \@linelen
      \hbox to \z@{\@linefnt\char'77\hss}}

%INITIALIZATION
\thinlines

\newcount\@xarg
\newcount\@yarg
\newcount\@yyarg
\newcount\@multicnt
\newdimen\@xdim
\newdimen\@ydim
\newbox\@linechar
\newdimen\@linelen
\newdimen\@clnwd
\newdimen\@clnht
\newdimen\@dashdim
\newbox\@dashbox
\newcount\@dashcnt
 \catcode`@=12


% macros supplementaires (J.D.)

\newbox\tbox
\newbox\tboxa

%\def\picbox(#1,#2,#3)#4{\setbox\tbox=\hbox{#4}%
%     \ht\tbox=#1 \wd\tbox=#2 \dp\tbox=#3 \box\tbox}

\def\leftzer#1{\setbox\tbox=\hbox to 0pt{#1\hss}%
     \ht\tbox=0pt \dp\tbox=0pt \box\tbox}

\def\rightzer#1{\setbox\tbox=\hbox to 0pt{\hss #1}%
     \ht\tbox=0pt \dp\tbox=0pt \box\tbox}

\def\centerzer#1{\setbox\tbox=\hbox to 0pt{\hss #1\hss}%
     \ht\tbox=0pt \dp\tbox=0pt \box\tbox}

% sytaxe: \image(hauteur totale reservee, distance
%    verticale de l'origine par rapport au bas de
%    la place reservee){materiel a inserer}
% L'origine est toujours centre horizontalement
%
\def\image(#1,#2)#3{\vbox to #1{\offinterlineskip
    \vss #3 \vskip #2}}

% \leftput place le materiel avec son point de reference
%    en #1,#2 unites \unitlength
% De meme pour \centerput et \rightput

\def\leftput(#1,#2)#3{\setbox\tboxa=\hbox{%
    \kern #1\unitlength
    \raise #2\unitlength\hbox{\leftzer{#3}}}%
    \ht\tboxa=0pt \wd\tboxa=0pt \dp\tboxa=0pt\box\tboxa}

\def\rightput(#1,#2)#3{\setbox\tboxa=\hbox{%
    \kern #1\unitlength
    \raise #2\unitlength\hbox{\rightzer{#3}}}%
    \ht\tboxa=0pt \wd\tboxa=0pt \dp\tboxa=0pt\box\tboxa}

\def\centerput(#1,#2)#3{\setbox\tboxa=\hbox{%
    \kern #1\unitlength
    \raise #2\unitlength\hbox{\centerzer{#3}}}%
    \ht\tboxa=0pt \wd\tboxa=0pt \dp\tboxa=0pt\box\tboxa}

\unitlength=1mm

\def\cput(#1,#2)#3{\noalign{\nointerlineskip\centerput(#1,#2){#3}
                             \nointerlineskip}}

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

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  \noindent{#1}\enspace{\it #2}\par
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              {\petcap #1\pointir}{\it #2}\par
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%a macro for grille.tex

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%%%%%%%%%%%%%%%%%%%% Start of the paper

\pageno=131
\let\findem=\cqfd
\auteurcourant={I.G. MACDONALD}
\titrecourant={A NEW CLASS OF SYMMETRIC FUNCTIONS}

\eightpoint
\leftline{Publ. I.R.M.A. Strasbourg, $\oldstyle 1988$, 372/S--20}
\leftline{Actes 20\emini\ S\'eminaire Lotharingien, p. 131--171}

\tenpoint

\vskip 1.5cm
\centerline{{\bf A NEW CLASS OF SYMMETRIC FUNCTIONS}}
\vskip 2.8mm
\centerline{\sevenrm BY}
\vskip 2.8mm
\centerline{I. G. MACDONALD}
\vskip 1cm
\noindent 
{\bf Contents.}

{\parindent=1cm
\item{1.} Introduction
\item{2.} The symmetric functions $P_\lambda (q,t)$
\item{3.} Duality
\item{4.} Skew $P$ and $Q$ functions
\item{5.} Explicit formulas 
\item{6.} The Kostka matrix
\item{7.} Another scalar product
\item{8.} Conclusion
\item{9.} Appendix

}
\titre 1. Introduction|
I will begin by reviewing briefly some aspects of the theory
of symmetric functions. This will serve to fix notation
and to provide some motivation for the subject of these lectures.

Let $x_1$, \dots, $x_n$ be independent indeterminates. The symmetric
group ${\goth S}_n$ acts on the polynomial ring
${\bboard Z}[x_1,\ldots,x_n]$ by permuting the $x$'s, and we shall
write
  $$\Lambda _n={\bb Z}[x_1,\ldots,x_n]^{{\goth S}_n}$$
for the subring of symmetric polynomials in
$x_1$, \dots, $x_n$. If $f\in  \Lambda _n$, we may write
  $$f=\sum _{r\geq 0} f^{(r)}$$
where $f^{(r)}$ is the homogeneous component of $f$ of degree~$r$ ;
each $f^{(r)}$ is itself symmetric, 
and so $\Lambda _n$ is a {\it graded} ring :
  $$\Lambda _n=\bigoplus_{r\geq 0} \Lambda _n^r,$$
where $\Lambda _n^r$ is the additive group of symmetric polynomials
of degree~$r$ in $x_1$, \dots, $x_n$. (By convention, $0$ is homogeneous
of every degree.)

If we now adjoin another indeterminate $x_{n+1}$, we can form
$\Lambda _{n+1}={\bb Z}[x_1,\ldots,x_{n+1}]^{{\goth S}_{n+1}}$,
and we have a surjective homomorphism (of graded rings)
  $$\Lambda _{n+1}\rightarrow  \Lambda _n$$
defined by setting $x_{n+1}=0$. The mapping
$\Lambda _{n+1}^r\rightarrow \Lambda _n^r$ 
is surjective for all $r\geq 0$, and bijective
if and only if $r\leq n$. 

Often it is convenient to pass to the limit. Let
  $$\displaylines{\Lambda ^r=\limproj_n \Lambda _n^r\cr
\noalign{\hbox{for each $r\geq 0$, and let}}
\Lambda =\bigoplus_{r\geq 0} \Lambda ^r.\cr}$$
By the definition of inverse (or projective) limits, an element of
$\Lambda _n^r$ is a sequence $(f_n)_{n\geq 0}$ where
$f_n\in  \Lambda _n^r$ for each~$n$, and $f_n$ is obtained from
$f_{n+1}$ by setting $x_{n+1}=0$. We may therefore regard the
$f_n$ as the partial sums of an infinite series $f$ of monomials
of degree~$r$ in infinitely many indeterminates
$x_1$, $x_2$, \dots\  For example, if $f_n=x_1+\cdots+x_n$,
then $f=\sum _{i=1}^\infty  x_i$. Thus the elements of $\Lambda $ are no longer
polynomials, and we call them symmetric {\it functions}.
(Of course, they aren't functions either, but they have to be
called something !)

For each $n$ there is a surjective homomorphism 
$\Lambda \rightarrow \Lambda _{n}$,
obtained by setting $x_{n+1}=x_{n+2}=\cdots=0$.
The graded ring $\Lambda $ is the {\it ring of symmetric functions}.
If $R$ is any commutative ring, we write
  $$\Lambda _R=\Lambda \otimes _{\lower 2pt\hbox{$\scriptstyle \bb Z$}} R,\qquad 
\Lambda _{n,R}=\Lambda _n\otimes _{\lower2pt\hbox{$\scriptstyle \bb Z$}} R$$
for the ring of symmetric functions (resp. symmetric
polynomials in $n$ indeterminates) with coefficients in~$R$.

There are various ${\bb Z}$-bases of the ring $\Lambda $, some of which
we shall review. They all are indexed by {\it partitions}. A partitions~$\lambda $
is a (finite or infinite) sequence
  $$\lambda =(\lambda _1,\lambda _2,\lambda _3,\ldots\,)$$
of non-negative integers, such that $\lambda _1\geq \lambda _2\geq \cdots$ and
  $$\left|\lambda \right|=\sum \lambda _i<\infty ,$$
so that from a certain point onwards (if the sequence 
$\lambda $ is infinite) all the $\lambda _i$ are zero. We shall not distinguish
between two such sequences which differ only by a string of zeros
at the end. Thus $(2,1)$, $(2,1,0)$, $(2,1,0,0,\ldots\,)$
are all to be regarded as the same partition.

The nonzero $\lambda _i$ are called the 
{\it parts} of $\lambda $, and the number
of parts is the {\it length} $\ell(\lambda )$ of~$\lambda $. If $\lambda $ has
$m_1$ parts equal to~1, $m_2$ parts equal to~$2$, and so on,
we shall occasionally write $\lambda =(1^{m_1}2^{m_2}\ldots\,)$,
although strictly speaking we should write this in reverse order.

Let $\cal P$ denote the set of all partitions, and ${\cal P}_n$
the set of all partitions of~$n$ (i.e. partitions $\lambda $ such that
$\left|\lambda \right|=n$). The {\it natural} (or dominance)
{\it partial ordering} in $\cal P$ is defined as follows :
  $$\lambda \geq \mu \Longleftrightarrow\left|\lambda \right|
=\left|\mu \right|\ {\rm and}\ 
\lambda _1+\cdots+\lambda _r\geq \mu _1+\cdots
+\mu _r\ {\rm for\ all}\ r\geq 1.\leqno(1.1)$$
It is a total order on ${\cal P}_n$ for $n\leq 5$, but not for $n\geq 6$.

With each partition $\lambda $ we associate a {\it diagram}, consisting of
the points $(i,j)\in  {\bb Z}^2$ such that $1\leq j\leq \lambda _i$.
We adopt the convention (as with matrices) that the first coordinate~$i$
(the row index) increases as one goes downwards, and the second
coordinate~$j$ (the column index) increases from left to right.
Often it is more convenient to replace the lattice points
$(i,j)$ by squares, and then the diagram of $\lambda $ consists of
$\lambda _1$ boxes in the top row, $\lambda _2$ boxes in the second row,
and so on ; the whole arrangement of boxes being left-justified.

If we read the diagram of a partition $\lambda $ by columns, we obtain the
{\it conjugate partition}~$\lambda '$. 
Thus $\lambda '_j$ is the number of boxes
in the $j$-th column of~$\lambda $, and hence is equal to the number of
parts of~$\lambda $ that are $\geq j$. It is not difficult to show that
  $$\lambda \geq \mu \Longleftrightarrow \mu '\geq \lambda '.$$

%%%%%%%%%

\vskip 5pt
\noindent 
{\bf Bases of $\Lambda $.}

\nobreak
\section 1. Monomial symmetric functions|
Let $\lambda $ be a partition. It defines a monomial 
$x^\lambda =x_1^{\lambda _1}x_2^{\lambda _2}\ldots$\ The monomial
symmetric function $m_\lambda $ is the sum of all
{\it distinct} monomials obtainable from $x^\lambda $
by permutations of the $x$'s. For example,
$m_{(2,1)}=\sum x_i^2x_j$, summed over all $(i,j)$ such
that $i\not=j$.

In particular, when $\lambda =(1^r)$ we have
  $$m_{(1^r)}=e_r=\sum _{i_1<\cdots<i_r} x_{i_1}\ldots x_{i_r},$$
the $r$-th {\it elementary symmetric function}. Their generating function~is
  $$E(t)=\sum _{r\geq 0} e_rt^r=\prod (1+x_it),\leqno(1.2)$$
where $t$ is another indeterminate, and $e_0=1$.

\goodbreak
At the other extreme, when $\lambda =(r)$ we have
  $$m_{(r)}=p_r=\sum x_i^r,$$
the $r$-th {\it power sum}.

It is clear that every $f\in  \Lambda $ is uniquely expressible as a
finite linear combination of the $m_\lambda $, so that
$(m_\lambda )_{\lambda \in  {\cal P}}$ is a 
${\bb Z}$-{\it basis} of~$\Lambda $.

\section 2|
For any partition $\lambda $, let
  $$e_\lambda =e_{\lambda _1}e_{\lambda _2}\ldots$$
The $e_\lambda $ form another ${\bb Z}$-basis of $\Lambda $. Equivalently,
we have ${\bb Z}[e_1,e_2,\ldots\,]$ and the $e_r$ are algebraically
independent. Indeed, it is not difficult to show that
  $$e_{\lambda '}=m_\lambda +\sum _{\mu <\lambda }a_{\lambda \mu }m_\mu $$
for suitable coefficients $a_{\lambda \mu }$, from which the assertion
follows immediately.

\section 3|
For each $r\geq 0$ let
  $$h_r=\sum _{\left|\lambda \right|=r} m_\lambda ,$$
the sum of {\it all} monomials of total degree $r$ in the $x$'s.
The generating function for the $h_r$ is
  $$H(t)=\sum _{r\geq 0} h_rt^r=\prod (1-x_it)^{-1},\leqno(1.3)$$
as one sees by expanding each factor $(1-x_it)^{-1}$ in the
product on the right as a geometric series, and then multiplying
these series together. From (1.2) and (1.3) it follows that
  $$\displaylines{
H(t)E(-t)=1\cr
\noalign{\hbox{so that}}
\rlap{(1.4)}\hfill 
\sum _{r=0}^n (-1)^r e_r h_{n-r} =0\hfill\cr}$$
for each $n\geq 1$.

Since the $e_r$ are algebraically independent, we may define
a ring homomorphism $\omega :\Lambda \rightarrow \Lambda $ by
  $$\omega (e_r)=h_r$$
for all $r\geq 1$. The symmetry of the relations (1.4) as between
the $e$'s and the $h$'s then shows that 
$\omega (h_r)=e_r$, i.e., $\omega ^2=1$. Thus $\omega $ is an
{\it automorphism} (of period~2) of~$\Lambda $, and therefore
we have $\Lambda ={\bb Z}[h_1,h_2,\ldots\,]$. Equivalently,
the products
  $$h_\lambda =h_{\lambda _1}h_{\lambda _2}\ldots\,=\omega (e_\lambda )$$
form another ${\bb Z}$-basis of $\Lambda $.

\section 4|
The generating function for the power-sums
$p_r=\sum x_i^r$ is
  $$\leqalignno{P(t)&=\sum _{r\geq 1} p_r t^{r-1}\cr
&=\sum _i\sum _{r\geq 1} x_i^r t^{r-1}\cr
&=\sum _i {x_i \over  1-x_it}\cr
\noalign{\hbox{and therefore}}
P(t)&={d\over dt}\log H(t)={H'(t) \over  H(t)}.&(1.5)\cr}$$
Hence we have $H'(t)=H(t)P(t)$, so that
  $$nh_n=\sum _{r=1}^n p_rh_{n-r}$$
for all $n\geq 1$. These relations enable us to express the
$h$'s in terms of the $p$'s, and vice versa, and show that
  $$\displaylines{h_n\in  {\bb Q}[p_1,\ldots,p_n],\cr
p_n\in  {\bb Z}[h_1,\ldots,h_n]\cr
\noalign{\hbox{so that}}
{\bb Q}[p_1,\ldots,p_n]={\bb Q}[h_1,\ldots,h_n]\cr}$$
for all $n\geq 1$. Letting $n\rightarrow \infty $, we see that
  $$\Lambda _{\bb Q}={\bb Q}[h_1,h_2,\ldots\,]=
{\bb Q}[p_1,p_2,\ldots\,].$$

For each partition $\lambda $ let
  $$p_\lambda =p_{\lambda _1}p_{\lambda _2}\ldots$$
(with the understanding that $p_0=1$). The
$p_\lambda $ form a ${\bb Q}$-basis of 
$\Lambda _{\bb Q}$, but do {\it not} form a ${\bb Z}$-basis of~$\Lambda $.

Analogously to (1.5) we have
  $$P(-t)={d\over dt}\log E(t).\leqno(1.6)$$
Since the involution $\omega $ interchanges $E(t)$ and $H(t)$,
it follows from (1.5) and (1.6) that it interchanges
$P(t)$ and $P(-t)$, so that
  $$\omega (p_r)=(-1)^{r-1} p_r\leqno(1.7)$$
for all $r\geq 1$.

Finally, we may compute $h_n$ as a polynomial in the power
sums, as follows : from (1.5) we have
  $$\eqalign{H(t)&=\exp\biggl(\sum _{r\geq 1} {p_rt^r\over r}\biggr)\cr
&=\prod _{r\geq 1}\exp{p_rt^r \over  r}\cr
&=\prod _{r\geq 1}\sum _{m_r\geq 0}{1\over m_r!}
\biggl({p_rt^r \over  r}\biggr)^{\kern-1pt m_r}.\cr}$$
Let us pick out the coefficient of $p_\lambda $ in the  product.
If $\lambda =(1^{m_1}2^{m_2}\ldots\,)$, it is $z_\lambda ^{-1}$, where
  $$\leqalignno{z_\lambda &=\prod _{r\geq 1} \bigl(r^{m_r}.m_r!\bigr)&(1.8)\cr
\noalign{\hbox{and therefore}}
h_n&=\sum _{\left|\lambda \right|=n} z_\lambda ^{-1}p_\lambda .\cr}$$

This numerical function $z_\lambda $ (which will occur frequently in the
sequel) has the following interpretation. Let $\left|\lambda \right|=n$,
and let $w\in  {\goth S}_n$ be a permutation of cycle-type~$\lambda $.
Then $z_\lambda $ is the order of the centralizer of $w$ in~${\goth S}_n$.

\section 5. Schur functions|
Let $\lambda $ be a partition of length $\leq n$, and form the determinant
  $$D_\lambda =\det\Bigl(x_i^{\lambda _j+n-j}\Bigr)_{1\leq i,j\leq n}.$$
This vanishes whenever any two of the $x$'s are equal, hence is divisible
by the Vandermonde determinant
  $$D_0=\smash{\prod _{i<j}(x_i-x_j).}$$
The quotient
  $$s_\lambda (x_1,\ldots,x_n)=D_\lambda /D_0$$
is a homogeneous symmetric polynomial of degree
$|\lambda |$ in $x_1$, \dots, $x_n$. Moreover we have
  $$s_\lambda (x_1,\ldots,x_n,0)=s_\lambda (x_1,\ldots,x_n)$$
and hence for each partition $\lambda $ a well-defined
element $s_\lambda \in  \Lambda $, homogeneous of degree~$|\lambda |$. These
are the {\it Schur functions}.

Let us define a scalar product 
$\langle \kern4pt ,\kern3pt \rangle $ on $\Lambda $ as follows :
  $$\langle p_\lambda ,p_\mu \rangle =\delta _{\lambda \mu }z_\lambda \leqno(1.9)$$
where $\delta _{\lambda \mu }=1$ if $\lambda =\mu $, and $\delta _{\lambda \mu }=0$
otherwise. Then one can show (see e.g. 
$[{\rm M}_1]$, ch.~I) that the Schur functions
$s_\lambda $ have the following properties, which 
characterize them uniquely :
  $$\displaylines{\kern1cm({\rm A})\quad
s_\lambda =m_\lambda +\sum _{\lambda <\mu } K_{\lambda \mu } m_\mu \hfill\cr
\noalign{\hbox{for suitable coefficients $K_{\lambda \mu }$ ;}}
\kern1cm({\rm B})\quad
\langle s_\lambda ,s_\mu \rangle =0\quad{\rm if}\ \lambda \not=\mu .\hfill\cr}$$

\section 6. Zonal symmetric functions|
These are certain symmetric functions $Z_\lambda $, at present
more familiar to statisticians than combinatorialists, which
(when restricted to a finite number of variables
$x_1$, \dots, $x_n$) arise naturally in connection with Fourier
analysis on the homogeneous space $G/K$, where
$G={\rm GL}_n({\bb R})$ and $K=O(n)$, the orthogonal
group (so that $G/K$ may be identified, via
$X\mapsto  XX^t$, with the space of positive definite real
symmetric $n\times n$~matrices). I shall not give a direct
definition here, but will only remark that the
$Z_\lambda $ (suitably normalized) are characterized by the
following two properties :
  $$\displaylines{
\noalign{\vskip -4pt }
\kern1cm({\rm A})\quad
Z_\lambda =m_\lambda +{\rm lower\ terms}\hfill\cr
\noalign{\vbox{\hbox{where by ``lower terms" 
is meant a linear combination of
the $m_\mu $ such that} 
\hbox{$\mu <\lambda $,}}}
\kern1cm({\rm B})\quad
\langle Z_\lambda ,Z_\mu \rangle _{\lower2pt\hbox{$\scriptstyle 2$}}
=0\quad{\rm if}\ \lambda \not=\mu ,\hfill\cr
\noalign{\hbox{where the scalar product 
$\langle \kern4pt ,\kern3pt \rangle _{\lower2pt\hbox{$\scriptstyle 2$}}$
on $\Lambda $ is defined by}}
\rlap{(1.10)}\hfill
\langle p_\lambda ,p_\mu \rangle _{\lower2pt\hbox{$\scriptstyle 2$}}
=\delta _{\lambda \mu }.2^{\ell(\lambda )}z(\lambda ),\hfill\cr}$$
$\ell(\lambda )$ being the length of the partition~$\lambda $.

%%%%%%%%%%%%%
\section 7. Jack's symmetric functions {\rm [S]}|
These are a common generalization of the Schur functions and the
zonal symmetric functions, and again I shall not give a direct
construction of them at this stage. Let $\alpha \in  {\bb R}$, $\alpha >0$,
and define a scalar product 
$\langle \kern4pt ,\kern3pt \rangle _{\lower2pt\hbox{$\scriptstyle \alpha $}}$ 
on $\Lambda _{\bb R}$ by
  $$\langle p_\lambda ,p_\mu \rangle _{\lower2pt\hbox{$\scriptstyle \alpha $}}
=\delta _{\lambda \mu }.\alpha ^{\ell(\lambda )}z_\lambda .\leqno(1.11)$$
Then Jack's functions $P_\lambda =P_\lambda (x;\alpha )$ are characterized by the
two properties
  $$\displaylines{\noalign{\vskip -4pt }
\kern1cm{\rm (A)}\quad
P_\lambda =m_\lambda  + {\rm lower\ terms},\hfill\cr
\kern1cm{\rm (B)}\quad
\langle P_\lambda ,P_\mu \rangle _{\lower2pt\hbox{$\scriptstyle \alpha $}}=0\ 
{\rm if}\ \lambda \not=\mu .\hfill\cr}$$
The symmetric functions $P_\lambda $ depend rationally on $\alpha $, i.e. they lie
in $\Lambda _F$ where $F$ is the field ${\bb Q}(\alpha )$, and we may if we prefer
regard the parameter~$\alpha $ as an indeterminate rather than a real
number. Clearly when $\alpha =1$ they reduce to the Schur functions $s_\lambda $,
and when $\alpha =2$ to the zonal functions~$Z_\lambda $. They also tend to 
definite limits as $\alpha \rightarrow 0$ and as $\alpha \rightarrow \infty $ (even though the scalar
product (1.11) collapses), and in fact
  $$\eqalign{P_\lambda (x;\alpha )&\rightarrow  e_{\lambda '}\quad{\rm as}\ \alpha \rightarrow  0,\cr
P_\lambda (x;\alpha )&\rightarrow  m_\lambda \quad{\rm as}\ \alpha \rightarrow  \infty .\cr}$$
Also when $\alpha ={1\over 2}$ they occur in nature, as zonal spherical functions
on the homogeneous space $G/K$, where now
$G={\rm GL}_n({\bb H})$ and $K=U(n,{\bb H})$,
the quaternionic unitary group of $n\times n$~matrices.

\section 8. Hall-Littlewood symmetric functions 
{\rm [${\rm M}_1$, ch.~III]}|
These symmetric functions arose originally in connection
with the combinatorial and enumerative lattice properties of finite 
abelian $p$-groups (where $p$ is a prime number). Let $t$ be an
indeterminate, let $F={\bb Q}(t)$ and define a scalar product
$\langle \kern4pt ,\kern3pt \rangle _{\lower2pt\hbox{$\scriptstyle (t)$}}$ 
on $\Lambda _F$ by
  $$\langle p_\lambda ,p_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (t)$}}=\delta _{\lambda \mu }z_\lambda  
\prod _{i=1}^{\ell(\lambda )} (1-t^{\lambda _i})^{-1}.\leqno(1.13)$$
Then the Hall-Littlewood symmetric functions
$P_\lambda (x;t)$ are characterized by the two properties
  $$\displaylines{
\kern1cm{\rm (A)}\quad P_\lambda =m_\lambda  + {\rm lower\ terms},\hfill\cr
\kern1cm{\rm (B)}\quad 
\langle P_\lambda ,P_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (t)$}}=0\ {\rm if}\  
\lambda \not=\mu .\hfill\cr}$$
When $t=0$, the $P_\lambda $ reduce to the Schur functions $s_\lambda $, and
when $t=1$ to the monomial symmetric functions~$m_\lambda $.

\goodbreak
\bigskip
\noindent 
{\bf Remarks.}

\nobreak
\ssection {\rm 1}|
Let $n$ be a positive integer, and arrange the partitions
of $n$ in lexicographical order, so that $(1^n)$ comes first and
$(n)$ comes last. For example,
  $$(1^4),\,(2,1^2),\,(2^2),\,(3,1),\,(4),$$
if $n=4$. This is a {\it total} ordering $L_n$ of the set
${\cal P}_n$ of partitions of~$n$, and correspondingly defines
a totally ordered basis of $\Lambda ^n$ :
  $$m_{(1^n)}(=e_n),\ldots, m_\lambda ,\ldots,m_{(n)}(=p_n).$$
Now suppose we are given a (positive definite) scalar product
$\langle \kern4pt,\kern3pt\rangle $ on the space 
$\Lambda _{\bb R}^n$. Then by the Gram-Schmidt process we can derive
a unique basis $(u_\lambda )$ of $\Lambda _{\bb R}^n$ with the following two properties :
  $$\displaylines{
\kern1cm({\rm A}')\quad
u_\lambda =m_\lambda  
+\vtop{\hbox{a linear combination of the $m_\mu $}
       \hbox{for partitions $\mu $ that precede $\lambda $ in $L_n$ ;}}\hfill\cr
\kern1cm({\rm B}')\quad
\langle u_\lambda ,u_\mu \rangle _{\lower2pt\hbox{$\scriptstyle \alpha $}}=0\ 
{\rm if}\ \lambda \not=\mu .\hfill\cr}$$
If we replace $L_n$ by some other total ordering of ${\cal P}_n$,
and apply Gram-Schmidt as before, we should expect in general to end
up with a different basis $(u_\lambda )$. What in fact happens in each
of the cases (5) --- (8) is that, for the appropriate scalar product,
the basis obtained is {\it independent} of the total ordering
chosen, provided only that it is compatible with the partial
ordering~(1.1). (The lexicographical order $L_n$ satisfies this condition :
if $\mu <\lambda $ then $\mu $ precedes $\lambda $ in $L_n$.) In other words,
the conditions (A) and~(B) (in each of the cases (5)~---~(8))
{\it overdetermine} the corresponding family of symmetric functions.

\ssection {\rm 2}|
Let $V$ be a vector space (over some field $F$) and let
$S=S(V)$ be the symmetric algebra of $V$.
(If $x_1$, $x_2$, \dots\ is a basis of $V$ then
$S=F[x_1,x_2,\ldots\,]$, the polynomial algebra over
$F$ generated by the $x_i$.) Now suppose we are given a
scalar product $\langle u,v\rangle $ on $V$, with values in $F$. This scalar
product has a natural extension to $S$, defined by
  $$\langle u_1\ldots u_m,v_1\ldots v_n\rangle 
=\cases{0,&if $m\not=n$,\cr
{\rm per}(\langle u_i,v_j\rangle ),&if $m=n$,\cr}$$
where ${\rm per}(\langle u_i,v_j\rangle )$ is the permanent of the matrix
of scalar products $\langle u_i,v_j\rangle $ $(1\leq i,j\leq n)$. Here
the $u$'s and $v$'s are arbitrary elements of~$V$.

In particular, let $V$ be the vector subspace of $\Lambda _F$
spanned by the power sums $p_r$ $(r\geq 1)$, so that
$S=F[p_1,p_2,\ldots\,]=\Lambda _F$. Suppose that
$\langle \kern4pt,\kern3pt\rangle $ is a scalar product on $V$ for which
the $p_r$ are mutually orthogonal, say
  $$\langle p_r,p_s\rangle =\delta _{rs}a_r.$$
Then the natural extension of this scalar product to
$S=\Lambda _F$ is such that
  $$\langle p_\lambda ,p_\mu \rangle =\delta _{\lambda \mu }z_\lambda a_\lambda ,$$
where $a_\lambda =a_{\lambda _1}a_{\lambda _2}\ldots$ for any partition~$\lambda $.

For the scalar products (5)~---~(8) above, the $a_r$ are
respectively 1, 2, $\alpha $, and $(1-t^r)^{-1}$.

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

\let\fled=\Longrightarrow
\titre 2. The symmetric functions $P_\lambda (q,t)$|
Let $q,t$ be independent indeterminates and let
$F={\bb Q}(q,t)$ be the field of rational functions in $q$ and $t$.
We shall now change the scalar product yet again, and define
  $$\displaylines{\rlap{(2.1)}\hfill
\langle p_\lambda ,p_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}
=\delta _{\lambda \mu }z_\lambda (q,t)\hfill\cr
\noalign{\hbox{where}}
\rlap{(2.1)}\hfill
z_\lambda (q,t)=z_\lambda \prod _{i=1}^{\ell(\lambda )} {1-q^{\lambda _i} \over  1-t^{\lambda _i}}.\hfill\cr}$$
It is perhaps better to think of the parameters $q$ and $t$ as
real variables lying in the interval $(0,1)$ of ${\bb R}$,
so that the scalar product (2.1) is positive definite. The main
result of this section is the following existence theorem.

\th \noindent (2.3) Theorem|For each partition $\lambda $ there
is a unique symmetric function $P_\lambda =P_\lambda (q,t)\in  \Lambda _F$ such that
  $$\displaylines{
\kern 1cm{\rm (A)}\quad P_\lambda =m_\lambda +{\sum _{\mu <\lambda } u_{\lambda \mu } m_\mu }\hfill\cr
\noalign{\hbox{with coefficients $u_{\lambda \mu }\in  F$ ;}}
\kern 1cm{\rm (B)}\quad \langle P_\lambda ,P_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}=0\quad{\rm if}\ \lambda \not=\mu .\hfill\cr}$$
\finth

As remarked in the previous section, these two conditions overdetermine
the $P_\lambda $, and their existence therefore requires proof. Before
embarking on the proof, let us consider some particular cases.

(1) When $q=t$, the scalar product (2.1) reduces to the `usual'
scalar product (1.9), and hence $P_\lambda (q,q)$ is the Schur function
$s_\lambda $.

(2) When $q=0$, (2.1) reduces to (1.13), and hence $P_\lambda (0,t)$ is
the Hall-Littlewood function $P_\lambda (t)$.

(3) Let $q=t^\alpha $ $(\alpha \in  {\bb R},\ \alpha >0)$ and let $t\rightarrow 1$, so that
$q\rightarrow 1$ also. Then
  $${1-q^m \over  1-t^m}={1-t^{\alpha m} \over  1-t^m} \rightarrow \alpha ,$$
as $t\rightarrow 1$, for all $m$. Hence the scalar product (2.1) tends to (1.11)
as $t\rightarrow 1$ and hence
  $$\lim_{t\rightarrow 1} P_\lambda (t^\alpha ,t)$$
is the Jack symmetric function $P_\lambda (\alpha )$.

(4) When $t=1$ (and $q$ is arbitrary) we have $P_\lambda (q,1)=m_\lambda $.

(5) When $q=1$ (and $t$ is arbitrary) we have $P_\lambda (1,t)=e_{\lambda '}$.

(6) Finally, it is clear from (2.2) that
  $$z_\lambda (q^{-1}, t^{-1})=(q^{-1}t)^n z_\lambda (q,t),$$
if $\lambda $ is a partition of $n$. Hence on each $\Lambda _F^n$, the
scalar products 
$\langle \kern4pt,\kern3pt\rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}$ 
and 
$\langle \kern4pt,\kern3pt\rangle _{\lower2pt\hbox{$\scriptstyle q^{-1},t^{-1}$}}$ 
are proportional, and therefore
  $$P_\lambda (q^{-1},t^{-1})=P_\lambda (q,t).$$

We may summarize these special cases in the following diagram,
in which the point $(q,t)$ represents the basis
$(P_\lambda (q,t))$ of $\Lambda _F$ (or of $\Lambda _{\bb R}$, since we are regarding
$q$, $t$ as real numbers).

 $$\vbox{\tabskip=0pt \offinterlineskip
\segment(12.5,-51.5)\dir(1,1)\long{36}
\newbox\stratbox
\setbox\stratbox=\hbox{\vrule height 15pt depth 5pt width 0pt}
\def\strat{\relax\ifmmode\copy\stratbox\else\unhcopy\stratbox\fi}
\newbox\strotbox
\setbox\strotbox=\hbox{\vrule height 1.8cm depth 1.8cm width 0pt}
\def\strot{\relax\ifmmode\copy\strotbox\else\unhcopy\strotbox\fi}
\def\cle#1{\hbox to 3.6cm{\hfil#1\hfil}}
%\def\trait#1{\llap{\strat#1\thinspace}\vrule }
\setbox3=\hbox{$\uparrow$} \dimen3=\wd3 \divide\dimen3 by 2
\setbox4=\hbox{$t$} \dimen4=\wd4 \divide\dimen4 by 2
\halign{#&#&#\cr
\multispan2\phantom{$\bigl(P_\lambda (t)\bigr)\;$}\kern-\dimen4\kern.2pt\box4\hfil&\cr
\noalign{\vskip 3pt }
\multispan2\phantom{$\bigl(P_\lambda (t)\bigr)\;$}\kern-\dimen3\kern.3pt\box3\hfil&\cr
\noalign{\vskip -3pt }
\hfil$(0,1)\;$&\strat\vrule\cle{$(m_\lambda )$}&\vrule$\;(1,1)$\cr
\noalign{\hrule}
$\bigl(P_\lambda (t)\bigr)\;$&\strot\vrule\hfil\raise12pt\cle{$(s_\lambda )$}\hfil&
\vrule$\;(e_{\lambda '})$\hfil\cr
\noalign{\vbox to 0pt{\vss
                      \hbox{\hbox to 6cm{\rightarrowfill}\kern5pt\smash{$q$}}
                      \vss}}
\hfil$(0,0)\;$&\strat\vrule\hfil&\vrule$\;(1,0)$\hfil\cr}}$$

\thinspace
\noindent 
At each point $(q,q)$ on the diagonal of the square
we have the Schur functions $s_\lambda $, at each point on the upper
edge $(t=1)$ the monomial symmetric functions $m_\lambda $, and so on.
In this scheme the Jack functions $P_\lambda (\alpha )$, for varying $\alpha $,
correspond to the points in the infinitesimal neighbourhood of the
point $(1,1)$, and more precisely $P_\lambda (\alpha )$ corresponds to the
direction through $(1,1)$ with slope $1/\alpha $. Notice that the
bottom edge $(t=0)$ of the square remains unmarked ; I do not
know if the $P_\lambda (q,0)$ have any reasonable interpretation
(except that they can be derived from the Hall-Littlewood
functions $P_\lambda (0,t)$ by duality (\S \kern 2pt 3)).

\medskip
Let $x=(x_1,x_2,\ldots\,)$
and $y=(y_1,y_2,\ldots\,)$
be two sequences of independent indeterminates over
$F={\bb Q}(q,t)$, and define
  $$\displaylines{\rlap{(2.4)}\hfill
\Pi =\Pi (x,y;q,t)=\prod _{i,j} {(tx_iy_j;q)_\infty  \over  (x_iy_j;q)_\infty },\hfill\cr
\noalign{\hbox{where as usual}}
(a;q)_\infty =\prod _{r=0}^\infty  (1-aq^r)\cr}$$
for any $a$ for which the product on the right makes sense.

Then we have
  $$\Pi (x,y;q,t)=\sum _\lambda  z_\lambda (q,t)^{-1} p_\lambda (x)p_\lambda (y).\leqno(2.5)$$

{\it Proof}\pointir
We compute $\exp(\log \Pi )$ ; first of all,
  $$\leqalignno{\log \Pi &=\sum _{i,j} \sum _{r=0}^\infty 
\bigl(\log(1-x_iy_jq^r)^{-1}
-\log(1-tx_iy_jq^r)^{-1}\bigr)\cr
&=\sum _{i,j} \sum _{r\geq 0} \sum _{n\geq 1} {1\over n}(x_iy_jq^r)^n(1-t^n)\cr
&=\sum _{n\geq 1} {1\over n} {1-t^n \over  1-q^n} p_n(x) p_n(y)\cr
\noalign{\hbox{and therefore}}
\Pi &=\prod _{n\geq 1} \exp\biggl({1\over n} {1-t^n \over  1-q^n}
p_n(x)p_n(y)\biggr)\cr
&=\prod _{n\geq 1} \sum _{m_n=0}^\infty  {1\over m_n!}
\biggl({1\over n} {1-t^n \over  1-q^n} p_n(x) p_n(y)\biggr)^{\kern-2pt m_n}\cr}$$
in which the coefficient of $p_\lambda (x)p_\lambda (y)$,
where $\lambda =(1^{m_1}2^{m_2}\ldots\,)$ is seen
to be $z_\lambda (q,t)^{-1}$.\cqfd

\thc (2.6)|For each integer $n\geq 0$ let $(u_\lambda )$, $(v_\lambda )$ be
$F$-bases of $\Lambda _F^n$, indexed by the partitions $\lambda $ of~$n$.
Then the following statements are equivalent :

\thinspace
\noindent 
{\rm (a)}\quad 
$\langle u_\lambda ,v_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}=\delta _{\lambda \mu }$ 
for all $\lambda $, $\mu $
$($i.e., $(u_\lambda )$, $(v_\lambda )$ are dual bases of
$\Lambda _F^n$ for the scalar product $(2.1))$ ;

\thinspace
\noindent 
{\rm (b)}\quad 
$\displaystyle \sum _\lambda  u_\lambda (x) v_\lambda (y) = \Pi (x,y;q,t)$.
\finthc

{\it Proof}\pointir
Let $p_\lambda ^*=z_\lambda (q,t)^{-1} p_\lambda $, so that
  $$
\langle p_\lambda ^*,p_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}=\delta _{\lambda \mu }.$$
Suppose that
  $$\displaylines{u_\lambda =\sum _\rho  a_{\lambda \rho }p_\rho ^*, \qquad
v_\lambda =\sum _\sigma  a_{\mu \sigma }p_\sigma .\cr
\noalign{\hbox{Then we have}}
\langle u_\lambda ,v_\mu \rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}=\sum _\rho  a_{\lambda \rho } b_{\mu \rho },\cr
\noalign{\hbox{so that (a) is equivalent to}}
\rlap{(${\rm a}'$)}\hfill
\sum _\rho  a_{\lambda \rho } b_{\mu \rho }=\delta _{\lambda \mu }.\hfill\cr}$$
On the other hand, by (2.5), (b) is equivalent to
  $$\leqalignno{
\sum _\lambda  u_\lambda (x) v_\lambda (y)&=\sum _\rho  p_\rho ^*(x) p_\rho (y)\cr
\noalign{\hbox{and hence to}}
\sum _\lambda  a_{\lambda \rho } b_{\lambda \sigma }&=\delta _{\rho \sigma }.&({\rm b}')\cr}$$
Since $({\rm a}')$ and $({\rm b}')$ are equivalent
($({\rm a}')$ says that $AB^t=1$, where $A$ is the
matrix $(a_{\lambda \mu })$
and $B$ the matrix $(b_{\lambda \mu })$, and
$({\rm b}')$ says that $A^tB=1$), it follows that
(a) and (b) are equivalent.\cqfd

After these preliminaries we can embark on the
proof of the existence theorem~(2.3). The idea of
the proof is as follows : we shall work initially with a finite set of
variables $x=(x_1,\ldots,x_n)$ and construct
an $F$-linear map (or operator)
  $$\leqalignno{D&=D_{q,t}:\Lambda _{n,F}\rightarrow \Lambda _{n,F}\cr
\noalign{\hbox{having the following properties :}}
Dm_\lambda &=\sum _{\mu \leq \lambda } c_{\lambda \mu } m_\mu &(2.7.1)\cr
\noalign{\hbox{{\it for each partition $\lambda $ of length $\leq n$} ;}}
\langle Df,g\rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}
&=\langle f,Dg\rangle _{\lower2pt\hbox{$\scriptstyle (q,t)$}}&(2.7.2)\cr
\noalign{\hbox{for all $f,g\in  \Lambda _F$ ;}}
\lambda \not=\mu &\Rightarrow  c_{\lambda \lambda }\not=c_{\mu \mu }.&(2.7.3)\cr}$$
These three properties say respectively that the matrix of $D$ relative
to the basis $(m_\lambda )$ is triangular (2.7.1) ; that $D$ is
self-adjoint (2.7.2) ; and that the eigenvalues of $D$
are distinct (2.7.3).

The $P_\lambda $ are then just the eigenfunctions (or eigenvectors)
of the operator $D$. Namely we have

\thc (2.8)|For each partition $\lambda $ 
$($of length $\leq n)$ there is a unique
symmetric polynomial $P_\lambda \in  \Lambda _{n,F}$ satisfying the two conditions
  $$\displaylines{
\kern1cm {\rm (A)}\quad
P_\lambda =\sum _{\mu \leq \lambda } u_{\lambda \mu } m_\mu \hfill\cr
\noalign{\hbox{where $u_{\lambda \mu }\in  F$ and $u_{\lambda \lambda }=1$ ;}}
\kern1cm {\rm (C)}\quad
DP_\lambda =c_{\lambda \lambda }P_\lambda .\hfill\cr}$$
\finthc

{\it Proof}\pointir
From (A) and (2.7.1) we have
  $$\eqalignno{
DP_\lambda &=\sum _{\mu \leq \lambda } u_{\lambda \mu } Dm_\mu \cr
&={\sum _{\nu \leq \mu \leq \lambda } u_{\lambda \mu }c_{\mu \nu }m_\nu }\cr
\noalign{\hbox{and}}
c_{\lambda \lambda }P_\lambda &=\sum _{\nu \leq \lambda } c_{\lambda \lambda }u_{\lambda \nu }m_{\nu },\cr
\noalign{\hbox{so that (A) and (C) are satisfied if and only if}}
c_{\lambda \lambda }u_{\lambda \nu }&=\sum _{\nu \leq \mu \leq \lambda } u_{\lambda \mu } c_{\mu \nu },\cr
\noalign{\hbox{that is to say, if and only if}}
(c_{\lambda \lambda }-c_{\nu \nu })u_{\lambda \nu }&=
\sum _{\nu <\mu \leq \lambda } u_{\lambda \mu } c_{\mu \nu }\cr}$$
whenever $\nu <\lambda $.
Since $c_{\lambda \lambda }\not=c_{\nu \nu }$ by (2.7.3), this relation determines
$u_{\lambda \mu }$ uniquely in terms of the $u_{\lambda \mu }$ such that
$\nu <\mu \leq \lambda $. Hence the coefficients $u_{\lambda \mu }$ in (A) are
uniquely determined, given that $u_{\lambda \lambda }=1$.\cqfd

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

With the $P_\lambda $ as defined in (2.8) we have, by the
self-adjointness of~$D$,
  $$\eqalign{
c_{\lambda \lambda }\langle P_\lambda ,P_\mu \rangle _{q,t}
&=\langle DP_\lambda ,P_\mu \rangle _{q,t}\cr
&=\langle P_\lambda ,DP_\mu \rangle _{q,t}
=c_{\mu \mu }\langle P_\lambda ,P_\mu \rangle _{q,t}.\cr}$$
But $c_{\lambda \lambda }\not=c_{\mu \mu }$ if $\lambda \not=\mu $, hence
$\langle P_\lambda ,P_\mu \rangle _{q,t}=0$ if $\lambda \not=\mu $.

This establishes (2.3) when the number of variables is finite
(i.e. in $\Lambda _{n,F}$ rather $\Lambda _F$). 
As explained in \S \kern2pt 1, we
may then compute the coefficients $u_{\lambda \mu }$ in (A) by
Gram-Schmidt ; they will involve only the scalar products
$\langle m_\mu ,m_\nu \rangle _{q,t}$, which are independent of~$n$.
Hence the $P_\lambda $ are well-defined as elements of~$\Lambda _F$.

The proof of (2.3) therefore reduces to the construction of an
operator $D$ satisfying (2.7.1) --- (2.7.3). Let
  $$\leqalignno{
\Delta &=\prod _{1\leq i<j\leq n} (x_i-x_j)\cr
&=\sum _{w\in  {\goth S}_n} \epsilon (w) x^{w\delta }&(2.9)\cr}$$
be the Vandermonde determinant in $x_1$, \dots, $x_n$, where
$\epsilon (w)$ is the sign of the permutation $w$, and
$\delta =(n-1,n-2,\ldots,1,0)$.
Next, for any polynomial $f(x_1,\ldots,x_n)$, symmetric or not,
we define
  $$\displaylines{
(T_{q,x_i}f)(x_1,\ldots,x_n)=f(x_1,\ldots,qx_i,\ldots,x_n),\cr
(T_{t,x_i}f)(x_1,\ldots,x_n)=f(x_1,\ldots,tx_i,\ldots,x_n)\cr}$$
for $1\leq i\leq n$. Then $D$ is defined as follows :
  $$\leqalignno{D&=\Delta ^{-1} \sum _{i=1}^n \bigl(T_{t,x_i}\Delta \bigr)T_{q,x_i}\cr
&=\sum _{i=1}^n \biggl(\prod _{j\not=i} {tx_i-x_j \over  x_i-x_j}\biggr)
T_{q,x_i}.&(2.10)\cr}$$
A more useful expression for $D$ is,
from (2.9) above,
  $$D=\Delta ^{-1} \sum _{w\in  {\goth S}_n} \epsilon (w)
\sum _{i=1}^n t^{(w\delta )_i} x^{w\delta } T_{q,x_i}.\leqno(2.10')$$
 
We must now verify that $D$ satisfies (2.7.1) --- (2.7.3). Let
$\lambda $ be a partition of length $\leq n$, and let
${\goth S}_n^\lambda $ be the subgroup of ${\goth S}_n$ that
fixes $\lambda $, so that
  $$m_\lambda =\left|{\goth S}_n^\lambda \right|^{-1}
\sum _{w_1\in  {\goth S}_n} x^{w_1\lambda }.$$
From $(2.10')$ we have
  $$\left|{\goth S}_n^\lambda \right|Dm_\lambda 
=\Delta ^{-1} \sum _{w,w_1} \epsilon (w)
\sum _{i=1}^n t^{(w\delta )_i} q^{(w_1\lambda )_i}
x^{w\delta +w_1\lambda }.$$
In this sum $w$ and $w_1$ run independently through ${\goth S}_n$.
Put $w_1=ww_2$, and we obtain :
  $$\leqalignno{
\left|{\goth S}_n^\lambda \right|Dm_\lambda 
&=\Delta ^{-1} \sum _{w,w_2} \epsilon (w)
\biggl(\sum _{i=1}^n q^{(w_2\lambda )_i} t^{\delta _i}\biggr)
x^{w(w_2\lambda +\delta )}\cr
&=\sum _{w_2\in  {\goth S}_n}
\biggl(\sum _{i=1}^n 
q^{(w_2\lambda )_i} t^{n-i}\biggr)
s_{w_2\lambda }.\cr}$$
Hence if we define 
  $$\displaylines{u(\mu )=\sum _{i=1}^n q^{\mu _i} t^{n-i}\cr
\noalign{\hbox{for any $\mu =(\mu _1,\ldots,\mu _n)\in  {\bb N}^n$, we have}}
Dm_\lambda =\sum _\mu  u(\mu )s_\mu \cr
}$$
summed over all distinct derangements $\mu $ of
$\lambda =(\lambda _1,\ldots,\lambda _n)$. 
In this sum $\mu $ is not a partition
(unless $\mu =\lambda $), but $s_\mu $ is defined for all
$\mu \in  {\bb N}^n$ and is either zero or equal to
$\pm s_\nu $ for some partition $\nu <\lambda $. Hence
  $$Dm_\lambda =u(\lambda )s_\lambda +\cdots$$
where the terms not written are a linear combination
of the Schur functions $s_\nu $ such that $\nu <\lambda $, and
therefore finally
  $$\displaylines{
Dm_\lambda =\smash{\sum _{\mu \leq \lambda } c_{\lambda \mu } m_\mu }\cr
\noalign{\hbox{with}}
\rlap{(2.11)}\hfill
c_{\lambda \lambda }=u(\lambda )=\sum _{i=1}^n q^{\lambda _i} t^{n-i}.\hfill\cr}$$
This establishes (2.7.1) and also (2.7.3), since
the eigenvalues $c_{\lambda \lambda }$ of $D$ given by (2.11) are
visibly all distinct.

It remains to show that $D$ is self-adjoint (2.7.2). The proof
is in several stages :

\goodbreak
\thc(2.12)|$D$ is self-adjoint when $q=t$.
\finthc

{\it Proof}\pointir
We have, when $q=t$,
  $$\leqalignno{
D&=\Delta ^{-1}\sum _{i=1}^n \bigl(T_{t,x_i} \Delta \bigr) T_{t,x_i}\cr
\noalign{\hbox{so that}}
Ds_\lambda &=\Delta ^{-1} \sum _{i=1}^n T_{t,x_i} \bigl(\Delta  s_\lambda \bigr).\cr
\noalign{\hbox{Since}}
\Delta s_\lambda &=\sum _{w\in  {\goth S}_n} 
\epsilon (w) x^{w(\lambda +\delta )}\cr
\noalign{\hbox{it follows that}}
Ds_\lambda &=\biggl(\sum _{i=1}^n t^{\lambda _i+n-i}\biggr) s_\lambda \cr
\noalign{\hbox{and hence that
$\langle Ds_\lambda ,s_\mu \rangle =0$ if $\lambda \not=\mu $. So}}
\langle Ds_\lambda ,s_\mu \rangle &
=\langle s_\lambda ,Ds_\mu \rangle \cr}$$
for all $\lambda ,\mu $, and therefore $D$ is self-adjont.\cqfd

\thc(2.13)|$D=D_{q,t}$ is self-adjoint if and only if
  $$D_x\Pi =D_y\Pi ,$$
where $\Pi =\Pi (x,y;q,t)$ $(2.4)$ and $D_x$ $($resp. $D_y)$
means $D$ operating on symmetric functions in the
$x$ $($resp. $y)$ variables.
\finthc

\medskip
{\it Proof}\pointir
Let $(u_\lambda )$, $(v_\lambda )$ be dual bases of $\Lambda _{n,F}$ as in
(2.6), and let
  $$a_{\lambda \mu }=\langle Du_\lambda ,u_\mu \rangle _{q,t}.\leqno(2.14)$$
Clearly, $D$ is self-adjoint if and only if
$a_{\lambda \mu }=a_{\mu \lambda }$ for all $\lambda ,\mu $.

From (2.14) we have
  $$\leqalignno{
Du_\lambda &=\sum _\mu  a_{\lambda \mu } v_\mu \cr
\noalign{\hbox{and therefore, since
$\Pi =\sum  u_{\lambda }(x)v_\lambda (y)$,}}
D_x\Pi &=\sum _{\lambda ,\mu } a_{\lambda \mu } v_\mu (x) v_\lambda (y).\cr
\noalign{\hbox{Likewise}}
D_y\Pi &=\sum _{\lambda ,\mu } a_{\lambda \mu } v_\mu (y) v_\lambda (x)\cr}$$
and therefore $D_x\Pi =D_y\Pi $ if and only if
$a_{\lambda \mu }=a_{\mu \lambda }$ for all $\lambda ,\mu $.\cqfd

From the definition (2.4) of $\Pi $ we have
  $$\Pi ^{-1} T_{q,x_i} \Pi = \prod _{j=1}^n
{1-x_iy_j \over 1 -tx_iy_j}$$
which is independent of $q$. Hence
$\Pi ^{-1}D_x\Pi $ is independent of~$q$.
Hence we may assume that $q=t$. But then by (2.13) and
(2.14) we have
$\Pi ^{-1}D_x\Pi =\Pi ^{-1}D_y\Pi $,
and so by (2.14) again $D_{q,t}$ is self-adjoint
for all $q,t$. This completes the proof of (2.3).\cqfd

We shall see in the subsequent sections that the formal
properties of the symmetric functions $P_\lambda (x;q,t)$ generalize
to a remarkable extent familiar properties of Schur functions.
What is lacking (at any rate at present) is any sort of
usable ``closed formula" for $P_\lambda $. This has the effect
that the proofs we shall give are usually indirect and
quite complicated in detail, compared to the usual proofs
of the corresponding properties of Schur functions.

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

\titre 3. Duality|
It is well known (see, e.g., $[{\rm M}_1]$, ch.~I, (3.8)) that
the involution $\omega :\Lambda \rightarrow \Lambda $ of \S \kern 2pt 1 permutes the Schur
functions, namely that
  $$\omega (s_\lambda )=s_{\lambda '}.$$
The duality theorem to be stated below generalizes this fact.
Let
  $$b_\lambda =b_\lambda (q,t)=\scal P_\lambda ,P_\mu |q,t|^{-1}\in  F.\leqno(3.1)$$
(We shall later (\S \kern2pt 5) obtain an explicit formula
for $b_\lambda (q,t)$ in terms of $\lambda $, $q$ and $t$.) Now define
  $$\displaylines{\rlap{(3.2)}\hfill
Q_\lambda =b_\lambda P_\lambda \hfill\cr
\noalign{\hbox{so that}}
\scal P_\lambda ,P_\mu |q,t|=\delta _{\lambda \mu },\cr}$$
i.e., $(P_\lambda )$, $(Q_\lambda )$ are dual bases of $\Lambda _F$ for the
scalar product
$\scal \kern4.5pt,\kern3pt|q,t|$.

We also define an automorphism
  $$\displaylines{\omega _{q,t}:\Lambda _F\rightarrow \Lambda _F\cr
\noalign{\hbox{by}}
\rlap{(3.3)}\hfill
\omega _{q,t}(p_r)=(-1)^{r-1} {1-q^r \over  1-t^r} p_r.\hfill\cr}$$
Clearly $\omega _{q,t}^{-1}=\omega _{t,q}$, and $\omega _{t,t}=\omega $. Also
we have
  $$\scal \omega _{q,t}f,g|t,q|=\langle \omega f,g\rangle \leqno(3.4)$$
for all $f,g\in  \Lambda _F$, where the scalar product on the
right is that defined by~(1.9).

It is enough to check (3.4) when $f=p_\lambda $ and $g=p_\mu $,
and then it is immediate from the definitions.

We can now state the duality theorem :

\th \noindent (3.5) Theorem|For all partitions $\lambda $ we have
  $$\eqalignno{\omega _{q,t}P_\lambda (q,t)&=Q_{\lambda '}(t,q)\cr
\noalign{\hbox{or equivalently}}
\omega _{q,t}Q_\lambda (q,t)&=P_{\lambda '}(t,q).\cr}$$
{\rm (The equivalence of these two statements follows from the
fact that $\omega _{q,t}^{-1}=\omega _{t,q}$.)}
\finth

For the proof of (3.5) we require the following lemma, whose
proof we leave as an exercise.

\thc (3.6)|Let $\lambda $ be a partition, thought of as an infinite
sequence, and let
  $$f_\lambda (q,t)=(1-t)\sum _{i=1}^\infty  q^{\lambda _i} t^{i-1}.$$
Then $f_\lambda (q,t)=f_{\lambda '}(t,q)$.
\finthc

Let $D=D_{q,t}$ be the operator defined in \S \kern2pt 2, acting
on symmetric polynomials in $n$ variables $x_1$, \dots, $x_n$.
We need to modify it slightly : we define
  $$E=E_{q,t}=t^{-n}\bigl(1+(t-1)D_{q,t}\bigr).\leqno(3.7)$$
From (2.11) we have, for any partition $\lambda $ of length $\leq n$,
  $$\leqalignno{EP_\lambda (q,t)&=t^{-n}\bigl(1+(t-1)\sum _{i=1}^n
q^{\lambda _i} t^{n-i}\bigr)P_\lambda (q,t)\cr
&=f_\lambda (q,t^{-1})P_\lambda (q,t).&(3.8)\cr}$$
\indent 
Next we have
  $$\omega _{q,t} E_{q,t} \omega _{q,t}^{-1} = E_{t^{-1},q^{-1}}.\leqno(3.9)$$
The only proof I have of (3.9) at present is rather messy, and
I shall not give it here. Assuming (3.9), the proof of (3.5)
proceeds as follows : we have
  $$\eqalign{E_{t^{-1},q^{-1}} \omega _{q,t} P_\lambda (q,t)
&=\omega _{q,t} E_{q,t} P_\lambda (q,t)\cr
&=f_\lambda (q,t^{-1}) \omega _{q,t} P_\lambda (q,t)\cr
&=f_{\lambda '}(t^{-1},q) \omega _{q,t} P_\lambda (q,t)\cr}$$
by (3.9), (3.8) and (3.6). Hence $\omega _{q,t} P_\lambda (q,t)$ is
an eigenfunction of $E_{t^{-1},q^{-1}}$ with eigenvalue
$f_{\lambda '}(t^{-1},q)$. Hence it must be a scalar multiple of
$P_{\lambda '}(t^{-1},q^{-1})=P_{\lambda '}(t,q)$.
We want to show that it is actually $Q_{\lambda '}(t,q)$, so we
must show that
  $$\scal \omega _{q,t}P_\lambda (q,t),P_{\lambda '}(t,q)|t,q|=1.$$
By (3.4) this is equivalent to showing that
  $$\langle \omega P_\lambda (q,t),P_{\lambda '}(t,q)\rangle =1\leqno(3.10)$$
for the ``usual" scalar product (1.9).

To prove (3.10), we shall express $P_\lambda $ and $P_{\lambda '}$ as
linear combinations of Schur functions : say
  $$\eqalign{P_\lambda (q,t)&=s_\lambda +\sum _{\mu <\lambda } a_{\lambda \mu } s_\mu ,\cr
P_{\lambda '}(t,q)&=s_{\lambda '}+\sum _{\nu '<\lambda '} b_{\lambda \nu } s_{\nu '}.\cr}$$
Since $\omega s_\mu =s_{\mu '}$ it follows that
  $$\omega P_\lambda (q,t)=s_{\lambda '}+\sum _{\mu '>\lambda '} a_{\lambda \mu } s_{\mu '}$$
and therefore $P_{\lambda '}(t,q)$ and $\omega P_\lambda (q,t)$ have only
$s_{\lambda '}$ in common, so that
  $$\langle \omega P_\lambda (q,t),P_{\lambda '}(t,q)\rangle =\langle s_{\lambda '},s_{\lambda '}\rangle =1$$
as required. This completes the proof of (3.5).\cqfd

Since $(P_\lambda )$, $(Q_\lambda )$ are dual bases of $\Lambda _F$, we have from (2.6)
  $$\sum _\lambda  P_\lambda (x;q,t)Q_\lambda (y;q,t)=\Pi (x,y;q,t).\leqno(3.11)$$
Apply $\omega _{q,t}$ to the $y$-variables ; from (2.5) it is easily
computed that
  $$\omega _{q,t}\Pi (x,y;q,t)=\prod _{i,j} (1+x_iy_j).$$
Hence it follows from (3.5) that
  $$\sum _\lambda  P_\lambda (x;q,t)P_{\lambda '}(y;t,q)=\prod _{i,j} (1+x_iy_j).\leqno(3.12)$$
When $q=t$, (3.11) and (3.12) reduce to the familiar identities
  $$\displaylines{
\sum _\lambda  s_\lambda (x)s_\lambda (y)=\prod _{i,j} (1-x_iy_j)^{-1}\cr
\noalign{\hbox{and}}
\sum _\lambda  s_\lambda (x)s_{\lambda '}(y)=\prod _{i,j} (1+x_iy_j)\cr}$$
respectively.

Finally, we can now identify the symmetric functions
$P_\lambda (q,t)$ when either $q=1$ or $t=1$. Suppose first that
$t=1$. Then from (2.10) we have
  $$D_{q,1}=\sum _{i=1}^n T_{q,x_i}$$
so that (for any partition $\lambda $ of length $\leq n$)
  $$D_{q,1}m_\lambda =\biggl(\sum _{i=1}^n q^{\lambda _i}\biggr) m_\lambda .$$
Hence the $m_\lambda $ are the eigenfunctions of the operator
$D_{q,1}$, and therefore
  $$P_\lambda (q,1)=m_\lambda \leqno(3.13)$$
for all partitions $\lambda $.

Next, it follows from (3.5) that
  $$\scal \omega _{q,t}P_\lambda (q,t),P_{\mu '}(t,q)|t,q|=\delta _{\lambda \mu }$$
and hence by (3.4) that
  $$\langle P_\lambda (q,t),\omega P_{\mu '}(t,q)\rangle =\delta _{\lambda \mu }.$$
By (3.13) this gives
  $$\langle m_\lambda ,\omega P_{\mu '}(1,q)\rangle =\delta _{\lambda \mu }$$
when $t=1$. But the basis dual to $(m_\lambda )$ for the usual
scalar product is $(h_\lambda )$ ($[{\rm M}_1]$, ch.~I). Hence
$\omega P_{\mu '}(1,q)=h_\mu $, i.e., $P_{\mu '}(1,q)=\omega h_\mu =e_\mu $.
Hence (replacing $q$, $\mu '$ by $t$, $\lambda $)
  $$P_\lambda (1,t)=e_{\lambda '}.\leqno(3.14)$$

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

\titre 4. Skew $P$ and $Q$ functions|
Let $\mu $ and $\nu $ be two partitions. Then the product
$P_\mu P_\nu $ is a linear combination of the $P_\lambda $, say
  $$\displaylines{\rlap{(4.1)}\hfill
P_\mu P_\nu =\sum _\lambda  f_{\mu \nu }^\lambda  P_\mu \hfill\cr
\noalign{\hbox{where}}
f_{\mu \nu }^\lambda =f_{\mu \nu }^\lambda (q,t)
=\scal Q_\lambda ,P_\mu P_\nu |q,t|\in  F.\cr}$$
In particular

(1) $f_{\mu \nu }^\lambda (t,t)$ is the coefficient $c_{\mu \nu }^\lambda $ of
$s_\lambda $ in $s_\mu s_\nu $, which may be calculated by the 
Littlewood-Richardson rule.

\thinspace 
(2) $f_{\mu \nu }^\lambda (0,t)$ is the {\it Hall polynomial}
$f_{\mu \nu }^\lambda (t)$ ($[{\rm M}_1]$, ch.~II). It may be written
as a sum
  $$f_{\mu \nu }^\lambda (t)=\sum _T f_{\lower 2pt\hbox{$\scriptstyle T$}}(t)$$
of monic polynomials, where $T$ runs through the set of LR-tableaux
of shape $\lambda '-\mu '$ and weight $\nu '$ ({\it loc. cit.\/})

(3) $f_{\mu \nu }^\lambda (q,1)$ is the coefficient of $m_\lambda $ in $m_\mu m_\nu $, and
so is independent of~$q$.

(4) $f_{\mu \nu }^\lambda (1,t)=1$ if $\lambda =\mu +\nu $, and is zero otherwise. For
$P_\mu (1,t)=e_{\mu '}$, and $e_{\mu '}e_{\nu '}=e_{\mu '\cup \nu '}=e_{(\mu +\nu )'}$.


(5) $f_{\mu \nu }^\lambda (q^{-1},t^{-1})=f_{\mu \nu }^\lambda (q,t)$, since
$P_\lambda (q^{-1},t^{-1})=P_\lambda (q,t)$
(\S \kern2pt 2).

(6) By duality (3.5) we have
  $$f_{\mu \nu }^\lambda (q,t)
=f_{\mu '\nu '}^{\lambda '}(t,q)b_{\mu '}(t,q)b_{\nu '}(t,q)/b_{\lambda '}(t,q).$$
In view of (1) and (2) above, it is natural to ask whether
it is possible to attach to each LR-tableau $T$ a non zero
rational function $f_{\lower 2pt\hbox{$\scriptstyle T$}}(q,t)$ so that
  $$f_{\mu \nu }^\lambda (q,t)=\sum _T f_{\lower 2pt\hbox{$\scriptstyle T$}}(q,t),$$
where $T$ runs through the set of LR-tableaux of shape
$\lambda -\mu $ and weight~$\nu $. (If so, then by duality ((6) above)
there will be likewise a decomposition 
$f_{\mu \nu }^\lambda $ over the LR-tableaux of shape
$\lambda '-\mu '$ and weight $\nu '$.) Again, is it true that
$f_{\mu \nu }^\lambda (q,t)\not=0$ if and only if $c_{\mu \nu }^\lambda \not=0$ ?
The answers to these questions are not known, at 
any rate to the author.

Clearly we shall have $f_{\mu \nu }^\lambda =0$ unless
$\left|\lambda \right|=\left|\mu \right|+\left|\nu \right|$.
In fact, more is true :

\thc (4.2)|$f_{\mu \nu }^\lambda =0$ unless $\lambda \supset \mu $ and $\lambda \supset \nu $ $($i.e.,
the diagram of $\lambda $ contains those of $\mu $ and $\nu $$)$.
\finthc

\pc STANLEY| [S] gives a proof of (4.2) in the context
of Jack's symmetric functions. His proof can be
transposed to the present context without difficulty.

We now define skew $Q$-functions as follows. If 
$\lambda $, $\mu $ are partitions, define
  $$\displaylines{\rlap{(4.3)}\hfill
Q_{\lambda /\mu }=\sum _\nu  f_{\mu \nu }^\lambda  Q_\nu \hfill\cr
\noalign{\hbox{so that}}
\scal Q_{\lambda /\mu },P_\nu |q,t|=\scal Q_\lambda ,P_\mu P_\nu |q,t|.\cr}$$
From (4.2) if follows that $Q_{\lambda /\mu }=0$ unless
$\lambda \supset \mu $.

Let $x=(x_1,x_2,\ldots\,)$ and
$y=(y_1,y_2,\ldots\,)$ be two
sequences of independent indeterminates. If $f$ is 
any symmetric function, $f(x,y)$ shall mean
$f(x_1,y_1,x_2,y_2,\ldots\,)$, and
likewise we define $f(x,y,z,\ldots\,)$ for three or more
sequences $x,y,z,\ldots\,$ Then we have
   $$Q_{\lambda /\mu }(x,y)=\sum _\nu  Q_{\lambda /\nu }(x) Q_{\nu /\mu }(y)\leqno(4.4)$$
{\it summed over partitions $\nu $ such that $\lambda \supset \nu \supset \mu $.}

\medskip
{\it Proof}\pointir
We have
  $$\leqalignno{\sum _\lambda  Q_{\lambda /\mu }(x)P_\lambda (y)
&=\sum _{\lambda ,\nu } f_{\mu \nu }^\lambda  Q_\nu (x) P_\lambda (y)\cr
&=\sum _\nu  Q_\nu (x) P_\mu (y) P_\nu (y)\cr
&=P_\mu (y) \Pi (x,y)\cr
\noalign{\hbox{by (3.11). Hence}}
\sum _{\lambda ,\mu } Q_{\lambda /\mu }(x) P_\lambda (y) Q_\mu (z)
&=\Pi (x,y)\Pi (z,y)\cr
&=\sum _\lambda  P_\lambda (y) Q_\lambda (x,z)\cr}$$
by (3.11) again. By comparing the coefficients of 
$P_\lambda (y)$ on either side, we obtain
  $$Q_\lambda (x,z)=\sum _\mu  Q_{\lambda /\mu }(x) Q_\mu (z).\leqno(4.5)$$
If we now replace $x$ by $x,y$ in (4.5), we have
  $$\eqalign{\sum _\mu  Q_{\lambda /\mu }(x,y) Q_\mu (z)
&=Q_\lambda (x,y,z)\cr
&=\sum _\nu  Q_{\lambda /\nu }(x) Q_\nu (y,z)\cr
&=\sum _{\mu ,\nu } Q_{\lambda /\nu }(x) Q_{\nu /\mu }(y)Q_\mu (z)\cr}$$
by two applications of (4.5). If we now equate the coefficients of
$Q_\mu (s)$ at either end of this string of equalities, we shall
obtain~(4.4).\cqfd

The identity (4.4) clearly generalizes to $n$ sets of
variables $x^{(1)},\ldots, x^{(n)}$ : if $\lambda $, $\mu $ are
partitions, then
  $$Q_{\lambda /\mu }(x^{(1)},\ldots,x^{(n)})
=\sum _{(\nu )}\prod _{i=1}^n Q_{\nu ^i/\nu ^{i-1}}(x^{(i)})\leqno(4.6)$$
{\it summed over all sequences 
$(\nu )=(\nu ^0,\nu ^1,\ldots,\nu ^n)$
of partitions such that
$\mu =\nu ^0\subset \nu ^1\subset \cdots\subset \nu ^n=\lambda $.}

\medskip
Let us apply (4.6) in the case where each set of variables
$x^{(i)}$ consists of a single element~$x_i$. For a single~$x$,
we have
  $$Q_{\lambda /\mu }(x)=\varphi _{\lambda /\mu }\,x^{|\lambda -\mu |}\leqno(4.7)$$
say, where $\varphi _{\lambda /\mu }=\varphi _{\lambda /\mu }(q,t)\in  F$, and in fact

\thc (4.8)|$\varphi _{\lambda /\mu }=0$ unless $\lambda \supset \mu $ and $\lambda -\mu $ is
a horizontal strip, i.e., unless the partitions $\lambda $, $\mu $
are interlaced :
  $$\lambda _1\geq \mu _1\geq \lambda _2\geq \mu _2\geq \cdots$$
\finthc

Again we refer to [S] for a proof of this. As in the
case of (4.2), \pc STANLEY|'s proof (for Jack's functions)
can be transposed without difficulty (and indeed (4.2) is a
a consequence of (4.8)).

From (4.6) and (4.7) we obtain
  $$Q_{\lambda /\mu }(x_1,\ldots,x_n)=\sum _{(\nu )} \prod _{i=1}^n \varphi _{\nu ^i/\nu ^{i-1}}
\,x_i^{|\nu ^i-\nu ^{i-1}|}\leqno(4.9)$$
with $(\nu )$ as in (4.6) and each skew diagram 
$\nu ^i-\nu ^{i-1}$  a horizontal strip. The sequence
$(\nu )$ of partitions determines a 
(column strict) {\it tableau} $T$ of shape $\lambda -\mu $, in which
the symbol $i$ occurs in each square of $\nu ^i-\nu ^{i-1}$, for
$1\leq i\leq n$. If we now define
  $$\displaylines{
\rlap{(4.10)}\hfill
\varphi _{\lower2pt\hbox{$\scriptstyle T$}}=\prod _{i\geq 1} \varphi _{\nu ^i/\nu ^{i-1}}\hfill\cr
\noalign{\hbox{then we shall have}}
\rlap{(4.11)}\hfill
Q_{\lambda /\mu }=\sum _T 
\varphi _{\lower2pt\hbox{$\scriptstyle T$}}\, x^T\hfill\cr}$$
summed over all tableaux $T$ of shape $\lambda -\mu $, where
$x^T$ is the monomial determined by the tableau $T$, i.e.,
$x^T=x^\alpha $, where $\alpha $ is the weight of~$T$.

Later (\S \kern2pt 5) we shall derive an explicit formula
for $\varphi _{\lambda /\mu }$ and hence also for $\varphi _{\lower2pt\hbox{$\scriptstyle T$}}$, 
and then (4.11) will provide
an explicit (if complicated) expression for $Q_{\lambda /\mu }$ as a sum
of monomials.

Finally, one can define skew $P$-functions
$P_{\lambda /\mu }$ by interchanging the roles of the $P$'s and
the $Q$'s throughout. The relation between the two is
  $$P_{\lambda /\mu }=b_\lambda ^{-1} b_\mu  Q_{\lambda /\mu }\leqno(4.12)$$
with $b_\lambda $ as defined by (3.1).

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

\titre 5. Explicit formulas|
We have introduced various scalars
$b_\lambda $, $\varphi _{\lambda /\mu }$, $\varphi _{\lower 2pt\hbox{$\scriptstyle T$}}$ in the
preceding sections, but so far we have no way of computing
them explicitly. The key to this is a specialization
theorem, which will be stated in a moment.

Let $u$ be a new indeterminate, and define a homomorphism
(or specialization)
  $$\displaylines{
\epsilon _{u,t}:\Lambda _F\rightarrow F[u]\cr
\noalign{\hbox{by}}
\rlap{(5.1)}\hfill
\epsilon _{u,t}(p_r)={1-u^r \over  1-t^r}\hfill\cr}$$
for each $r\geq 1$.

To motivate this definition, suppose that
$u=t^n$, $n$ a positive
integer. Then
  $$\leqalignno{
\epsilon _{t^n,t}(p_r)&={1-t^{nr} \over  1-t^r}\cr
&=1+t^r+\cdots+t^{(n-1)r}\cr
&=p_r(1,t,\ldots,t^{n-1})\cr}$$
so that
  $$\epsilon _{t^n,t}(f)=f(1,t,\ldots,t^{n-1})$$
for any symmetric function $f$ : i.e., the effect of
$\epsilon _{t^n,t}$ is to evaluate at
$(x_1,\ldots,x_n,x_{n+1},\ldots\,)
=(1,t,\ldots,t^{n-1},0,0,\ldots\,)$.

There is a nice formula for 
$\epsilon _{u,t}(P_\lambda (q,t))$. In order to state it in a convenient
form, let us introduce the following notation. For each
square $s=(i,j)$ in the diagram of a partition~$\lambda $, let
  $$\left\{ \vcenter{\halign{$#\hfil$&\quad$#$\hfil\cr
a(s)=\lambda _i-j,&a'(s)=j-1,\cr
l(s)=\lambda '_j-i,&l'(s)=i-1,\cr}}\right.\leqno(5.2)$$
so that $l'(s)$, $l(s)$, $a(s)$ and $a'(s)$ are
respectively the numbers of squares in the diagram of $\lambda $ to the 
north, south, east and west of the square~$s$. The numbers
$a(s)$ and $a'(s)$ may be called respectively the
{\it arm-length} and the {\it arm-colength} of~$s$, and
$l(s)$, $l'(s)$ the {\it leg-length} and {\it leg-colength}. 
The {\it hook-length} at~$s$ is $a(s)+l(s)+1$.

\th \noindent (5.3) Theorem|We have
  $$\epsilon _{u,t}\bigl(P_\lambda (q,t)\bigr)=
\prod _{s\in  \lambda } {q^{a'(s)}u-t^{l'(s)} \over  q^{a(s)} t^{l(s)+1} - 1}.$$
\finth

{\it Proof}\pointir
Since by (5.1) $\epsilon _{u,t}(p_r)$ is a polynomial of
degree~$r$ in~$u$ with coefficients in~$F$, it follows
that $\epsilon _{u,t}(P_\lambda )$ is a polynomial of degree $\leq \left|\lambda \right|$, say
  $$\epsilon _{u,t}\bigl(P_\lambda (q,t)\bigr)=\Phi _\lambda (u;q,t)\in  F[u].$$
 The idea of the proof is first to locate the zeros of this
polynomial, which will give the numerator of the expression in (5.3). For
this we require two relations. The first comes from the formula (4.5)
(with $Q$'s replaced by $P$'s), which gives
  $$P_\lambda (x_1,\ldots,x_n)=\sum _\mu  P_{\lambda /\mu }(x_1)P_\mu (x_2,\ldots,x_n).\leqno(5.4)$$
Set $(x_1,\ldots,x_n)=(1,t,\ldots,t^{n-1})$
and let
  $$\psi _{\lambda /\mu }=P_{\lambda /\mu }(1;q,t).\leqno(5.5)$$
Then (5.4) becomes
  $$\Phi _\lambda (t^n;q,t)=\sum _\mu  \psi _{\lambda /\mu }(q,t) t^{|\mu |} \Phi _\mu (t^{n-1};q,t)$$
for all $n\geq 1$, and hence we have
  $$\Phi _\lambda (u;q,t)=\sum _\mu  \psi _{\lambda /\mu }(q,t) t^{|\mu |} \Phi _\mu (ut^{-1};q,t).\leqno({\rm A})$$
Next we have
  $$\epsilon _{u,t} \omega _{t,q}(f)=(-q)^{-n} \epsilon _{u,q^{-1}}(f)\leqno(5.6)$$
if $f\in  \Lambda _F^n$. (Since both $\epsilon $ and $\omega $ are ring homomorphisms,
it is enough to check (5.6) when $f=p_n$.) Hence by duality
(3.5) we have
  $$\eqalign{\epsilon _{u,t}P_\lambda (q,t)&=\epsilon _{u,t} \omega _{t,q} Q_{\lambda '}(t,q)\cr
&=(-q)^{-|\lambda |} \epsilon _{u,q^{-1}} Q_\lambda (t,q)\cr
&=(-q)^{-|\lambda |} b_{\lambda '}(t,q) \epsilon _{u,q^{-1}} P_{\lambda '}(t^{-1},q^{-1})\cr}$$
(since $P_{\lambda '}(t,q)=P_{\lambda '}(t^{-1},q^{-1})$) 
and therefore
  $$\Phi _\lambda (u;q,t)=(-q)^{-|\lambda |} b_{\lambda '}(t,q) 
\Phi _{\lambda '}(u;t^{-1},q^{-1}).\leqno({\rm B})$$

We observe next that 

\thc (5.7)|$P_\lambda (x_1,\ldots,x_n)=0$
if $n<\ell(\lambda )$.
\finthc

\noindent 
For $P_\lambda $ is a linear combination of the
$m_\mu $ such that $\mu \leq \lambda $, and

\smallskip
\centerline{$\mu \leq \lambda \Rightarrow \mu '\geq \lambda '\Rightarrow \ell(\mu )=\mu '_1\geq \lambda '_1=\ell(\lambda )>n$}
\centerline{${}\Rightarrow \ell(\mu )>n\Rightarrow m_\mu (x_1,\ldots,x_n)=0$.\qed}
\medskip
It follows from (5.7) that
  $$\Phi _\lambda (u;q,t)=0\quad
{\rm for}\ u=1,t,\ldots,t^{\ell(\lambda )-1}.$$
Hence by (B) the polynomial
$\Phi _{\lambda '}(u;t^{-1},q^{-1})$
vanishes also for these values of~$u$. By replacing
$(\lambda ,q,t)$ by $(\lambda ',t^{-1},q^{-1})$, we see that
  $$\Phi _\lambda (u;q,t)=0\quad
{\rm for}\ u=1,q^{-1},\ldots,q^{1-\lambda _1}$$
and therefore
  $$\Phi _\lambda (u;q,t)\ {\it is\ divisible\ in}\ 
F[u]\ {\it by}\ 
\prod _{j=1}^{\lambda _1} (q^{j-1}u -1).\leqno({\rm C})$$

Now we consider the relation (A) above. By (C), each term on
the right of (A) is divisible by
  $$\prod _{j=1}^{\lambda _2} (q^{j-1} u - t)$$
(because the sum in (A) is over partitions $\mu $ that interlace~$\lambda $).
Hence $\Phi _\lambda (u;q,t)$ is divisible by this product. We now repeat
the argument : each term on the right-hand side of~(A) is
divisible by
  $$\prod _{j=1}^{\lambda _3} (q^{j-1} u - t^2)$$
(since $\mu _2\geq \lambda _3$) and therefore
$\Phi _\lambda (u;q,t)$ is also divisible by this product.
Thus finally $\Phi _\lambda (u;q,t)$ is divisible in $F[u]$ by
  $$\prod _{i=1}^{\ell(\lambda )} \prod _{j=1}^{\lambda _i} (q^{j-1} u - t^{i-1})
=\prod _{s\in  \lambda } (q^{a'(s)} u - t^{l'(s)}).$$
(Observe that all these linear factors are distinct.) But
we know that $\Phi _\lambda $ has degree at most $|\lambda |$ in~$u$. Hence
we have
  $$\Phi _\lambda (u;q,t)=v_\lambda (q,t) 
\prod _{s\in  \lambda } (q^{a'(s)} u - t^{l'(s)}),\leqno({\rm D})$$
and it remains to identify the scalar factor $v_\lambda (q,t)$.
For this purpose we require

\thc (5.8)|Let $\lambda $ be a partition of length $n$, and let
$\lambda =(\lambda _1-1,\ldots,\lambda _n-1)$. Then
  $$P_\lambda (x_1,\ldots,x_n)=x_1\ldots x_nP_\mu (x_1,\ldots,x_n).$$
\finthc

\noindent 
(Compute $D_{q,t}(x_1\ldots x_nP_\mu )$ and show that
$x_1\ldots x_n P_\mu $ is an eigenfunction of $D_{q,t}$
with eigenvalue
$\sum q^{\lambda _i} t^{n-i}$.)

\medskip
From (5.8) it follows that
  $$\Phi _\lambda (t^n;q,t)= t^{n(n-1)/2} \Phi _\mu (t^n;q,t)$$
if $\ell(\lambda )=n$, and hence from (D) that
  $$\eqalign{
{v_\lambda (q,t) \over  v_\mu (q,t)} &=t^{n(n-1)/2}
\prod _{s\in  \lambda -\mu } \bigl(q^{a'(s)} t^n - t^{l'(s)}\bigr)^{-1}\cr
&=\prod _{i=1}^n \bigl(q^{\lambda _i-1} t^{n-i+1} -1\bigr)^{-1}.\cr}$$
Now the factors in this product are precisely
$\bigl(q^{a(s)} t^{l(s)+1}-1\bigr)^{-1}$
for $s$ in the first column of~$\lambda $. Hence by induction on
the number of columns of~$\lambda $ we conclude that
  $$v_\lambda (q,t)=\prod _{s\in  \lambda }\bigl(q^{a(s)} t^{l(s)+1}-1\bigr)^{-1}$$
and the proof of (5.3) is complete.\cqfd

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

From (B) we have
  $$b_{\lambda '}(t,q)=(-q)^{|\lambda |} 
{\Phi _\lambda (u;q,t) \over  \Phi _{\lambda '}(u;t^{-1},q^{-1})}$$
which together with (5.3) leads to the formula
  $$b_\lambda (q,t)= \prod _{s\in  \lambda } {1-q^{a(s)}t^{l(s)+1} \over  1-q^{a(s)+1} t^{l(s)}}.$$
So we know now the value of
$\scal P_\lambda ,P_\mu |q,t|=b_\lambda (q,t)^{-1}$.

Knowing the $b_\lambda $, it turns out that it is now quite straightforward
to calculate the scalars
  $$\varphi _{\lambda /\mu }=Q_{\lambda /\mu }(1)$$
defined in \S \kern 2pt 4. I will omit the details and merely state
the result. For each square $s$ and each partition~$\lambda $, define
  $$b_\lambda (s)=b_\lambda (s;q,t)={1-q^{a(s)} t^{l(s)+1} \over  1-q^{a(s)+1}t^{l(s)}}
\leqno(5.10)$$
if $s\in  \lambda $, and $b_\lambda (s)=1$ if $s\notin \lambda $.

Next, if $S$ is any set of squares (contained in the diagram of
$\lambda $ or not), let
  $$b_\lambda (S)=\prod _{s\in  S} b_\lambda (s).\leqno(5.11)$$
Now let $\lambda $, $\mu $ be partitions such that
$\lambda _1\geq \mu _1\geq \lambda _2\geq \mu _2\geq \cdots$ or equivalently such that
$\lambda \supset \mu $ and $\lambda -\mu $ is a horizontal strip. Let 
$C_{\lambda /\mu }$ denote the union of the columns that
contain squares of $\lambda -\mu $. Then the formula for $\varphi _{\lambda /\mu }$ is
  $$\leqalignno{\varphi _{\lambda /\mu }&=b_\lambda (C_{\lambda /\mu })/b_\mu (C_{\lambda /\mu })&(5.12)\cr
\noalign{\hbox{and for a tableau $T$}}
\varphi _{\lower2pt\hbox{$\scriptstyle T$}}
&=\prod _{i\geq 0} b_{\lambda ^i}(C_i)/b_{\lambda ^i}(C_{i+1}),&(5.13)\cr}$$
where $\lambda ^0\subset \lambda ^1\subset \cdots$ is the sequence of partitions defined
by the tableau, and $C_i$ is the union of the columns
that contain a symbol~$i$ (so that $C_0$ is empty).

In the case of Jack's symmetric functions, the results of this section
are all due to Richard \pc STANLEY|~[S], and the transposition
of his proofs into the present context is quite straightforward. Also
when $q=0$ the formulas (5.9), (5.12) and (5.13) reduce
respectively to $[{\rm M}_1]$, ch.~III, (2.12), (5.8) and~(5.9).

Finally we may remark that the identities of Schur and Littlewood
  $$\eqalign{\sum _\lambda  s_\lambda &=\prod (1-x_i)^{-1} \prod _{i<j} (1-x_ix_j)^{-1},\cr
\sum _\mu s_\mu &=\prod _{i\leq j} (1-x_ix_j)^{-1},\cr
\sum _\nu  s_\nu &=\prod _{i<j}(1-x_ix_j)^{-1}\cr}$$
in which $\lambda $ runs through all partitions, $\mu $ through all even
partitions (i.e. with all parts even) and $\nu $ through all partitions
such that $\nu '$ is even, generalize to identities for the
$P_\lambda (q,t)$ as follows. For any partition~$\lambda $, let
  $$b_\lambda ^{{\rm el}}=\prod _{\scriptstyle s\in  \lambda  \atop
\scriptstyle l(s)\ {\rm even}} b_\lambda (s),\qquad
b_\lambda ^{{\rm oa}}=\prod _{\scriptstyle s\in  \lambda  \atop
\scriptstyle a(s)\ {\rm odd}} b_\lambda (s),$$
(so the superscripts el and oa stand for ``even legs" and
``odd arms," respectively. Then there are product expressions for the
four series
  $$\displaylines{
({\rm a})\quad \sum _\lambda  b_\lambda ^{{\rm el}}(q,t) P_\lambda (q,t),\qquad
({\rm b})\quad \sum _\lambda  b_\lambda ^{{\rm oa}}(q,t) P_\lambda (q,t),\cr
({\rm c})\quad \sum _\mu  b_\mu ^{{\rm oa}}(q,t) P_\mu (q,t),\qquad
({\rm d})\quad \sum _\nu  b_\nu ^{{\rm el}}(q,t) P_\nu (q,t),\cr}$$
where as before $\lambda $ runs through all partitions, $\mu $ through
all even partitions and $\nu $ through partitions such that
$\nu '$ is even. For example, the sum (a) is equal to the product
  $$\displaylines{
\prod _i {(tx_i;q)_\infty  \over  (x_i;q)_\infty }
\prod _{i<j} {(tx_ix_j;q)_\infty  \over  (x_ix_j;q)_\infty }\cr
\noalign{\hbox{and (d) is equal to}}
\prod _{i<j} {(tx_ix_j;q)_\infty  \over  (x_ix_j;q)_\infty }.\cr}$$
The products for (b) and (c) are a little more complicated :
they may be derived from (a) and (b) by duality (3.5). In the
case of Jack's symmetric functions, (a) is due to
K.~\pc KADELL|.


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

\titre 6. The Kostka matrix|
Let
  $$\leqalignno{h_\lambda (q,t)&=\prod _{s\in  \lambda }(1-q^{a(s)}t^{l(s)+1}),&(6.1)\cr
h'_\lambda (q,t)&=\prod _{s\in  \lambda }(1-q^{a(s)+1}t^{l(s)}),&(6.2)\cr
&=h_{\lambda '}(t,q),\cr}$$
so that by (5.9) we have
  $$b_\lambda (q,t)=h_\lambda (q,t)/h'_\lambda (q,t).$$
Now define
  $$\leqalignno{J_\lambda (q,t)&=h_\lambda (q,t) P_\lambda (q,t)&(6.3)\cr
&=h'_\lambda (q,t) Q_\lambda (q,t).\cr}$$
It seems likely that when the $J_\lambda (q,t)$ are expressed in terms
of the monomial symmetric functions, the coefficients are
{\it polynomials}, i.e. elements of ${\bb Z}[q,t]$. I shall
make a more precise conjecture later. 

When $q=t$, we have
  $$\eqalignno{J_\lambda (t,t)&=H_\lambda (t)s_\lambda ,\cr
\noalign{\hbox{where}}
H_\lambda (t)&=\prod _{s\in  \lambda } (1-t^{h(s)})\cr}$$
is the hook-length polynomial
$(h(s)=a(s)+l(s)+1)$.

When $q=0$,
  $$J_\lambda (0,t)=Q_\lambda (0,t)\leqno(6.4)$$
because $h'_\lambda (0,t)=1$.

Duality (3.5) now takes the form
  $$\omega _{q,t}J_\lambda (q,t)=J_{\lambda '}(t,q)\leqno(6.5)$$
and the specialization theorem (5.3) takes the form
  $$\epsilon _{u,t} J_\lambda (q,t) = \prod _{s\in  \lambda } \bigl(t^{l'(s)} - q^{a'(s)}u\bigr).$$
From the fact (\S \kern 2pt2) that $P_\lambda (q,t)=P_\lambda (q^{-1}, t^{-1})$
we deduce that
  $$\displaylines{\rlap{(6.6)}\hfill
J_\lambda (q^{-1},t^{-1})=(-1)^{|\lambda |} q^{-n(\lambda ')} t^{-n(\lambda )-\left|\lambda \right|}
J_\lambda (q,t),\hfill\cr
\noalign{\hbox{where}}
n(\lambda )=\sum _{i\geq 1} (i-1) \lambda _i=\sum _{i\geq 1} {\lambda '_i\choose 2}.\cr}$$
\indent Recall ($[{\rm M}_1]$, ch.~I, \S \kern2pt 6) that the
{\it Kostka numbers} $K_{\lambda \mu }$ are defined by
  $$s_\lambda =\sum _\mu  K_{\lambda \mu } m_\mu .\leqno(6.7)$$
We have $K_{\lambda \mu }=0$ unless $\mu \leq \lambda $, and $K_{\lambda \lambda }=1$.
They generalize to the {\it Kostka-Foulkes polynomials}
$K_{\lambda \mu }(t)$ ($[{\rm M}_1]$, ch.~III) defined as follows :
  $$s_\lambda (x)=\sum _\mu  K_{\lambda \mu }(t) P_\mu (x;t),\leqno(6.8)$$
where the $P_\mu (x;t)=P_\mu (0;x,t)$ are the Hall-Littlewood functions.
Let $S_\lambda (x;t)$ denote the Schur functions associated with the
product 
  $$\prod (1-tx_i)/(1-x_i)\,;$$
they form a basis of 
$\Lambda _{{\bb Q}(t)}$ dual to the basis $(s_\lambda (x))$, relative to
the scalar product (1.13), and therefore (6.8) is equivalent to
  $$Q_\mu (x;t)=\sum _\lambda  K_{\lambda \mu }(t) S_\lambda (x;t).\leqno(6.8')$$
\indent Since $P_\mu (x;1)=m_\mu $ it follows from (6.7) and
(6.8) that $K_{\lambda \mu }(1)=K_{\lambda \mu }$. Now $K_{\lambda \mu }$ is the number of
tableaux of shape~$\lambda $ and weight~$\mu $. \pd FOULKES conjectured,
and \pd LASCOUX and \pd SCH\"UTZENBERGER proved, that 
$K_{\lambda \mu }(t)$ is a polynomial in~$t$ with positive integral
coefficients, and more precisely that
  $$K_{\lambda \mu }(t)=\sum _T t^{c(T)}$$
summed over all tablelaux $T$ of shape $\lambda $ and weight~$\mu $, where
$c(T)$ (the {\it charge} of $T$) is a well-defined
$\bb N$-valued function of the tableau~$T$.

In the present context we now define $K_{\lambda \mu }(q,t)\in  F$ by
  $$J_\mu (x;q,t)=\sum _\lambda  K_{\lambda \mu }(q,t) S_\lambda (x;t).\leqno(6.9)$$
Computations of the $K_{\lambda \mu }(q,t)$ suggest the

\th Conjecture|$K_{\lambda \mu }(q,t)$ is a polynomial in $q$ and $t$ with
positive integral coefficients.
\finth

Here are some partial results which tend to confirm this conjecture.

(1) We have $K_{\lambda \mu }(0,t)=K_{\lambda \mu }(t)$ (by (6.4) and $(6.8')$). In particular,
$K_{\lambda \mu }(0,0)=\delta _{\lambda \mu }$ and $K_{\lambda \mu }(0,1)=K_{\lambda \mu }$.

(2) From (6.5) and (6.6) we deduce that
  $$\eqalign{K_{\lambda \mu }(q,t)&=K_{\lambda '\mu '}(t,q),\cr
K_{\lambda \mu }(q,t)&=q^{n(\mu ')} t^{n(\mu )} K_{\lambda '\mu }(q^{-1},t^{-1}).\cr}$$

(3) Another special case is ($\lambda $, $\mu $ partitions of $n$)
  $$K_{\lambda \mu }(1,1)={n! \over  h(\lambda )},$$
where $h(\lambda )$ is the product of the hook lengths of~$\lambda $. In other words,
$K_{\lambda \mu }(1,1)$ is the number of standard tableaux of shape~$\lambda $
(and in particular does not depend on~$\mu $). This prompts the following
question, which refines the conjecture above : can one find
${\bb N}$-valued functions $a_\mu (T)$, $b_\mu (T)$, defined for standard
tableaux~$T$, such that
  $$K_{\lambda \mu }(q,t)=\sum _T q^{a_\mu (T)} t^{b_\mu (T)}$$
summed over the standard tableaux $T$ of shape $\lambda $ ? Of course
this question, as I have stated it, is not well-posed. The point
is to find some ``natural" bijection between the monomials
$q^at^b$ that occur in $K_{\lambda \mu }(q,t)$ (assuming the truth of the
conjecture) and the standard tableaux of shape~$\lambda $.

(4) When $q=t$ we have 
$K_{\lambda \mu }(t,t)\in  {\bb Z}[t]$, by a result of \pc STANLEY|.

(5) $K_{\lambda \mu }(1,t)$ is a polynomial in $t$ with positive integer
coefficients. (Recall that $P_\lambda (1,t)=e_{\lambda '}$ ; this makes it feasible
to compute $K_{\lambda \mu }(1,t)$.) By duality ((2) above) $K_{\lambda \mu }(q,1)$ is
a polynomial in $q$ with positive integral coefficients.

(6) Let $\lambda =(r,1^s)$. Then for each partition $\mu $ of $r+s$, 
$K_{\lambda \mu }(q,t)$ is the coefficient of $u^s$ in the product
  $$\prod (t^{i-1} + q^{j-1}u)$$
over all $(i,j)\in  \mu $ with the exception of (1,1).

\medskip
The symmetric functions $S_\lambda (x,t)$ are linear combinations
of the $m_\mu (x)$ with coefficients in ${\bb Z}[t]$,
from the table on
p.~128 of~$[{\rm M}_1]$. Hence the conjecture would imply that
  $$J_\lambda (q,t)=\sum _{\mu \leq \lambda } v_{\lambda \mu }(q,t) m_\mu $$
with coefficients $v_{\lambda \mu }\in  {\bb Z}[q,t]$, divisible by
$(1-t)^{l(\mu )}$. Another consequence of the conjecture is that
  $$\scal J_\lambda ,J_\mu J_\nu |q,t|\in  {\bb Z}[q,t]$$
for any three partitions $\lambda $, $\mu $, $\nu $.

Let $K_n(q,t)$ denote the matrix
$\bigl(K_{\lambda \mu }(q,t)\bigr)_{\lambda ,\mu \in  {\cal P}_n}$. For $n=1,\ldots,6$
the matrices $K_n$ (or rather their transposes $K'_n$) are
shown in the appendix.

\vskip -12pt

\titre 7. Another scalar product|
We have seen that the symmetric functions
$P_\lambda (q,t)$ are pairwise orthogonal with
respect to the scalar product
$\scal \kern4pt,\kern3pt|q,t|$.
It will appear that they are also pairwise orthogonal with respect to 
{\it another} scalar product
$\scal \kern4pt,\kern3pt|q,t|'$, which I shall now define.

We shall work throughout with a finite number of indeterminates
$x=(x_1,\ldots,x_n)$ (i.e., in $\Lambda _{n,F}$ rather $\Lambda _F$). Moreover
we shall assume (although it is not strictly necessary) that
$t=q^k$ where $k\in  {\bb N}$.

Let
  $$L_n=F[x_1^{\pm 1},\ldots,x_n^{\pm 1}]$$
be the $F$-algebra of Laurent polynomials in
$x_1$, \dots, $x_n$, i.e. of polynomials
in the $x_i$ and $x_i^{-1}$. If $f\in  L_n$ let
$\overline f=f(x_1^{-1},\ldots,x_n^{-1})$ and let
$[f]_1$ denote the constant term in~$f$. Moreover let
  $$\leqalignno{\Delta =\Delta (x;q,t)&=
\prod ^n_{\scriptstyle i,j=1\atop \scriptstyle i\not=j}
{(x_ix_j^{-1};q)_\infty  \over  (tx_ix_j^{-1};q)_\infty }&(7.1)\cr
&=\prod _{i\not=j} \prod _{r=0}^{k-1} (1-q^rx_ix_j^{-1})\cr}$$
so that $\Delta \in  L_n$.

For $f,g\in  \Lambda _{n,F}$ we now define
  $$\scal f,g|q,t|'={1\over n!} [f\overline g\Delta ]_1.\leqno(7.2)$$
This is a symmetric, positive definite scalar product on
$\Lambda _{n,F}$. It is not difficult (in fact, it is a good deal
easier than it was in \S \kern 2pt 2) to show that

\thc (7.3)|The operator $D_{q,t}$ $(2.10)$ is 
self-adjoint for this scalar product :
  $$\scal Df,g|q,t|'=\scal f,Dg|q,t|'$$
for all $f,g\in  \Lambda _{n,F}$.
\finthc

\goodbreak
From (7.3) it follows, just as in \S \kern 2pt 2, that
  $$\scal P_\lambda ,P_\mu |q,t|'=0$$
if $\lambda \not=\mu $.

\rem Remark|
When $q=t$ (i.e., when $k=1$) the two scalar products are the
same. Otherwise they are different.

\medskip
It remains to compute $\scal P_\lambda ,P_\lambda |q,t|'$, and the answer
is as follows :
  $$\displaylines{\rlap{(7.4)}\hfill
\eqalign{\scal P_\lambda ,P_\lambda |q,t|'&=\prod _{1\leq i<j\leq n}
\prod _{r=1}^{k-1} {1-q^{\lambda _i-\lambda _j+r}\, t^{j-i} \over 
1-q^{\lambda _i-\lambda _j-r} \,t^{j-i} }\cr
&=c_n \prod _{s\in  \lambda } {1-q^{a'(s)} \,t^{n-l'(s)} \over 
1-q^{a'(s)+1} \,t^{n-l'(s)-1}},\cr}\hfill\cr
\noalign{\hbox{where}}
\rlap{(7.5)}\hfill
c_n=\scal 1,1|q,t|'={1\over n!} [\Delta ]_1=\prod _{i=2}^n 
{ik-1\brack k-1},\hfill\cr}$$
a product of $q$-binomial coefficients.
(Notice that when $\lambda =0$,
(7.4) reduces to a special case of the $q$-Dyson
conjecture.)


\titre 8. Conclusion|
In the definition (7.1) of $\Delta $, and in the first of the two
products (7.4) the structure of the root system of type $A_{n-1}$ is 
clearly visible. In fact this aspect of the theory generalizes to other
root systems, and I shall conclude these lectures with a brief and
simplified account of this generalization. For full details and
proofs, see $[{\rm M}_3]$.

So let $R$ be a reduced root system, $R^+$ a system of positive
roots in~$R$ ; let $Q$ be the root lattice of $R$, and $Q^+$ the
positive cone in~$Q$, spanned by~$R^+$ ; let $P$ be the
weight lattice, and $P^{++}$ the cone of dominant weights ; and
let $W$ be the Weyl group of~$R$. Define a partial order on~$P$~by
  $$\lambda \geq \mu \Longleftrightarrow \lambda -\mu \in  Q^+.$$
\indent 
Now let $q$ and $t$ be indeterminates, and $F$ as before the field
${\bb Q}(q,t)$. Let $A=F[P]$ be the group algebra of the lattice
(or free abelian group) $P$ over $F$. To each $\lambda \in  P$ there corresponds
an element $e^\lambda $ of $A$, such that $e^\lambda .e^\mu =e^{\lambda +\mu }$, and
$e^0=1$ is the identity element of $A$. The Weyl group $W$ acts
on $P$, hence on $A$ : $w(e^\lambda )=e^{w\lambda }$ ($w\in  W,\;\lambda \in  P$).
Let $A^W$ be the subalgebra of $W$-invariants in~$A$.

We can easily describe two $F$-bases of $A^W$. One consists of
the {\it orbit-sums}
  $$m_\lambda =\sum _{\mu \in  W\lambda } e^\mu \qquad (\lambda \in  P^{++})$$
and is the counterpart of the monomial symmetric functions denoted
by the same symbols. The other basis consists of the
{\it Weyl characters} : let
  $$\displaylines{\rho ={1\over 2}\sum _{\alpha \in  R^+} \alpha \cr
\noalign{\hbox{and let}}
\delta =\prod _{\alpha \in  R^+} \bigl(e^{\alpha /2}-e^{-\alpha /2}\bigr)
=\sum _{w\in  W} \epsilon (w) e^{w\rho }\cr}$$
(where $\epsilon (w)=\det(w)=\pm 1$). 
For each $\lambda \in  P^{++}$ we define the
{\it Weyl character}
  $$\chi _{\lower 2pt\hbox{$\scriptstyle \lambda $}}
=\delta ^{-1} \sum _{w\in  W} \epsilon (w) e^{w(\lambda +\rho )}$$
which is the counterpart of the Schur function $s_\lambda $. We have
  $$\chi _{\lower 2pt\hbox{$\scriptstyle \lambda $}}
=m_\lambda +\sum _{\scriptstyle \mu <\lambda \atop \scriptstyle \mu \in  P^{++}}
K_{\lambda \mu }m_\mu $$
with coefficients $K_{\lambda \mu }\in  {\bb N}$.

If $f\in  A$, say $f=\sum _{\lambda \in  P} f_\lambda  e^\lambda $, let
  $$\overline f=\sum  f_\lambda  e^{-\lambda }$$
and let $[f]_1$ denote the constant term $f_0$ of~$f$.

For simplicity we shall assume as in \S \kern2pt 7 that
$t=q^k$, $k\in  {\bb N}$, and define, in analogy with~(7.1),
  $$\eqalign{\Delta =\Delta (q,t)&=\prod _{\alpha \in  R} {(e^a;q)_\infty  \over  (te^\alpha ;q)_\infty }\cr
&=\prod _{\alpha \in  R} (e^\alpha ;q)_k\cr}$$
so that $\Delta \in  A^W$. Now define a scalar product on $A^W$ by
  $$\langle f,g\rangle =\left|W\right|^{-1} [f\overline g\Delta ]_1$$
for $f,g\in  A^W$. Then we have

\th Theorem|There exists a unique basis
$(P_\lambda )_{\lambda \in  P^{++}}$ of $A^W$ such that
  $$\displaylines{
\kern 1cm{\rm (A)}\quad
P_\lambda =m_\lambda +\sum _{\scriptstyle \mu <\lambda \atop \scriptstyle \mu \in  P^{++}}
u_{\lambda \mu }m_\mu \hfill\cr
\noalign{\hbox{with coefficients $u_{\lambda \mu }\in  F$ ;}}
\kern 1cm{\rm (B)}\quad
\langle P_\lambda ,P_\mu \rangle =0\quad{\rm if}\ \lambda \not=\mu .\hfill\cr}$$
\finth

When $R$ is of type $A_{n-1}$, these $P_\lambda $ are essentially
the symmetric functions $P_\lambda (q,t)$, restricted to
$n$ variables $x_1$, \dots, $x_n$. For
arbitrary $R$, when $k=0$ (i.e., $t=1$) we have
$P_\lambda =m_\lambda $, and when $k=1$ (i.e., $q=t$) we have
$P_\lambda =\chi _{\lower 2pt\hbox{$\scriptstyle \lambda $}}$.

I will conclude with two conjectures which generalize
(7.4) and the specialization theorem (5.3) respectively.
For each root $\alpha \in  R$ let $\alpha ^{\scriptstyle \vee}$ be the corresponding
co-root, and let
  $$\sigma ={1\over 2} \sum _{\alpha \in  R^{+}} \alpha ^{\scriptstyle \vee}.$$

\th Conjecture 1|For all $\lambda \in  P^{++}$,
  $$\langle P_\lambda ,P_\mu \rangle =\prod _{\alpha \in  R^+}
\prod _{i=1}^{k-1} {1-q^{\alpha ^{\scriptscriptstyle \vee}(\lambda +k\rho )+i} \over 
1-q^{\alpha ^{\scriptscriptstyle \vee}(\lambda +k\rho )-i}}.$$
\finth

This is non trivial even when $\lambda =0$ (so that $P_\lambda =1$) ; in that
case it reduces to the constant term conjectures of~$[{\rm M}_2]$.

\th Conjecture 2|Let $P_\lambda (k\sigma )$ denote the image of $P_\lambda $ under
the mapping $e^\mu \mapsto  q^{k\sigma (\mu )}=t^{\sigma (\mu )}$ $(\mu \in  P$). 
Then for all $\lambda \in  P^{++}$,
  $$P_\lambda (k\sigma )=q^{-k\sigma (\lambda )} \prod _{\alpha \in  R^+}
{(q^{\alpha ^{\scriptscriptstyle \vee}(\lambda +k\rho )};q)_k \over 
(q^{\alpha ^{\scriptscriptstyle \vee}(k\rho )};q)_k }.$$
\finth

\vfill
\eject

\vglue 60pt
\eightpoint
\centerline{REFERENCES}
\nobreak
\vskip 10pt
\livre ${\rm M}_1$|\pd MACDONALD (I.G.)|Symmetric functions and
Hall polynomials|Oxford University Press, $\oldstyle 1979$|
\article ${\rm M}_2$|\pd MACDONALD (I.G.)|Some conjectures for
root systems|SIAM J. Math. Anal.|13|1982|988--1007|
\divers ${\rm M}_3$|\pd MACDONALD (I.G.)|Orthogonal polynomials
associated with root systems, preprint, $\oldstyle 1988$|
\divers S|\pd STANLEY (Richard)|Some combinatorial properties
of Jack symmetric functions, preprint, $\oldstyle 1988$|

\tenpoint
\vskip 1.5cm
\rightline{\vbox{\halign{#\hfil\cr
I. G. {\petcap Macdonald},\cr
School of Mathematical Sciences,\cr
Queen Mary College,\cr
University of London,\cr
Mile End Road,\cr
London E1 4NS,\cr
United Kingdom.\cr}}\quad}

\vfill
\eject
 

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\omit&\multispan{4}\hrulefill&\omit\cr
1^5&t^3+t^4+2t^5+t^6+t^7&t^2+t^3+t^4+t^5+t^6&
t+t^2+t^3+t^4&1&\cr
\omit&\multispan{4}\hrulefill&\omit\cr
}}

\titre 9. Appendix|
\centerline{{\bf The matrices $K(q,t)'$, $n\leq 6$}}
\vfill
\eightpoint
\line{\hfil\box\koun\hfil\box\kodeux\hfil\box\kotrois\hfil}
\vfill
\centerline{\box\koquatre}
\vfill
\centerline{\box\kocinq}
\vfill
\centerline{\box\kosix}

\vfill\eject


\newbox\kosept
\newbox\kohuit
\newbox\koneuf 
\newbox\kodix   
\newbox\koonze

\setbox\kosept=\vbox{\eightpoint \offinterlineskip
\halign{\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv#\cr
\omit\strat&\omit\hfil 6\hfil&\omit\hfil 51\hfil&
\omit\hfil 42\hfil&\omit\cr
\omit&\multispan{3}\hrulefill&\omit\cr
6&1&q+q^2+q^3+q^4+q^5&q^2+q^3+2q^4+q^5+2q^6+q^7+q^8&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
51&t&1+qt+q^2t+q^3t+q^4t&q+q^2+q^3+q^2t+q^3t+2q^4t+q^5t+q^6t&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
42&t^2&t+qt+qt^2+q^2t^2+q^3t^2&1+qt+2q^2t+q^3t+q^2t^2+q^3t^2+2q^4t^2&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
41^2&t^3&t+t^2+qt^3+q^2t^3+q^3t^3&t+qt+q^2t+qt^2+q^2t^2+q^3t^2+q^2t^3
+q^3t^3+q^4t^3&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
3^2&t^3&t^2+qt^2+q^2t^2+qt^3+q^2t^3&t+qt+q^2t+qt^2+q^2t^2+q^3t^2+q^2t^3+q^3t^3
+q^4t^3&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
321&t^4&t^2+t^3+qt^3+qt^4+q^2t^4&t+t^2+2qt^2+qt^3+2q^2t^3+q^2t^4+q^3t^4&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
31^3&t^6&t^3+t^4+t^5+qt^6+q^2t^6&t^2+t^3+qt^3+t^4+qt^4+q^2t^4+qt^5+q^2t^5+q^2t^6&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
2^3&t^6&t^4+qt^4+t^5+qt^5+qt^6&t^2+t^3+qt^3+t^4+qt^4+q^2t^4+qt^5+q^2t^5+q^2t^6&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
2^21^2&t^7&t^4+t^5+t^6+qt^6+qt^7&2t^3+t^4+qt^4+t^5+2qt^5+qt^6+q^2t^7&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
21^4&t^{10}&t^6+t^7+t^8+t^9+qt^{10}&t^4+t^5+2t^6+t^7+qt^7+t^8+qt^8+qt^9&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
1^6&t^{15}&t^{10}+t^{11}+t^{12}+t^{13}+t^{14}&t^7+t^8+2t^9+t^{10}+2t^{11}
+t^{12}+t^{13}&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
}}



\setbox\kohuit=\vbox{\eightpoint \offinterlineskip
\halign{\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv#\cr
\omit\strat&\omit\hfil $41^2$\hfil&
\omit\hfil $3^2$\hfil&\omit\cr
\omit&\multispan2\hrulefill&\omit\cr
6&q^3+q^4+2q^5+2q^6+2q^7+q^8+q^9&q^3+q^5+q^6+q^7+q^9&\cr
\omit&\multispan2\hrulefill&\omit\cr
51&q+q^2+q^3+q^4+q^3t+q^4t+2q^5t+q^6t+q^7t&q^2+q^4+q^3t+q^5t+q^6t&\cr
\omit&\multispan2\hrulefill&\omit\cr
42&q+qt+2q^2t+2q^3t+q^4t+q^3t^2+q^4t^2+q^5t^2&q+q^2t+q^3t+q^3t^2+q^5t^2&\cr
\omit&\multispan2\hrulefill&\omit\cr
41^2&1+qt+q^2t+q^3t+qt^2+q^2t^2+q^3t^2+q^3t^3+q^4t^3+q^5t^3&
qt+q^2t+q^2t^2+q^4t^2+q^3t^3&\cr
\omit&\multispan2\hrulefill&\omit\cr
3^2&qt+q^2t+q^3t+qt^2+2q^2t^2+2q^3t^2+q^4t^2+q^3t^3&
1+q^2t^2+q^3t^2+q^4t^2+q^3t^3&\cr
\omit&\multispan2\hrulefill&\omit\cr
321&t+qt+2qt^2+q^2t^2+qt^3+2q^2t^3+q^3t^3+q^3t^4&
t+qt^2+q^2t^2+q^2t^3+q^3t^4&\cr
\omit&\multispan2\hrulefill&\omit\cr
31^3&t+t^2+t^3+qt^3+q^2t^3+qt^4+q^2t^4+qt^5+q^2t^5+q^3t^6&
qt^2+t^3+qt^4+q^2t^4+q^2t^5&\cr
\omit&\multispan2\hrulefill&\omit\cr
2^3&qt^2+t^3+2qt^3+q^2t^3+2qt^4+q^2t^4+qt^5+q^2t^5&
qt^2+t^3+qt^3+qt^4+q^3t^6&\cr
\omit&\multispan2\hrulefill&\omit\cr
2^21^2&t^2+t^3+qt^3+t^4+2qt^4+2qt^5+qt^6+q^2t^6&
t^2+t^4+qt^4+qt^5+q^2t^6&\cr
\omit&\multispan2\hrulefill&\omit\cr
21^4&t^3+t^4+2t^5+t^6+qt^6+t^7+qt^7+qt^8+qt^9&
t^4+t^5+qt^6+t^7+qt^8&\cr
\omit&\multispan2\hrulefill&\omit\cr
1^6&t^6+t^7+2t^8+2t^9+2t^{10}+t^{11}+t^{12}&t^6+t^8+t^9+t^{10}+t^{12}&\cr
\omit&\multispan2\hrulefill&\omit\cr
}}



\setbox\koneuf=\vbox{\eightpoint \offinterlineskip
\halign{\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv#\cr
\omit\strat&\omit\hfil 321\hfil&\omit\cr
\omit&\omit\hrulefill&\omit\cr
6&q^4+2q^5+2q^6+3q^7+3q^8+2q^9+2q^{10}+q^{11}&\cr
\omit&\omit\hrulefill&\omit\cr
51&q^2+2q^3+2q^4+2q^5+q^6+q^4t+2q^5t+2q^6t+2q^7t+q^8t&\cr
\omit&\omit\hrulefill&\omit\cr
42&q+2q^2+q^3+q^2t+3q^3t+3q^4t+q^5t+q^4t^2+2q^5t^2+q^6t^2&\cr
\omit&\omit\hrulefill&\omit\cr
41^2&q+q^2+qt+2q^2t+2q^3t+q^4t+q^2t^2+2q^3t^2+2q^4t^2+q^5t^2+q^4t^3+q^5t^3&\cr
\omit&\omit\hrulefill&\omit\cr
3^2&q+q^2+qt+2q^2t+2q^3t+q^4t+q^2t^2+2q^3t^2+2q^4t^2+q^5t^2+q^4t^3+q^5t^3&\cr
\omit&\omit\hrulefill&\omit\cr
321&1+3qt+q^2t+qt^2+4q^2t^2+q^3t^2+q^2t^3+3q^3t^3+q^4t^4&\cr
\omit&\omit\hrulefill&\omit\cr
31^3&t+qt+t^2+2qt^2+q^2t^2+2qt^3+2q^2t^3+qt^4+2q^2t^4+q^3t^4+q^2t^5+q^3t^5&\cr
\omit&\omit\hrulefill&\omit\cr
2^3&t+qt+t^2+2qt^2+q^2t^2+2qt^3+2q^2t^3+qt^4+2q^2t^4+q^3t^4+q^2t^5+q^3t^5&\cr
\omit&\omit\hrulefill&\omit\cr
2^21^2&t+2t^2+qt^2+t^3+3qt^3+3qt^4+q^2t^4+qt^5+2q^2t^5+q^2t^6&\cr
\omit&\omit\hrulefill&\omit\cr
21^4&t^2+2t^3+2t^4+qt^4+2t^5+2qt^5+t^6+2qt^6+2qt^7+qt^8&\cr
\omit&\omit\hrulefill&\omit\cr
1^6&t^4+2t^5+2t^6+3t^7+3t^8+2t^9+2t^{10}+t^{11}&\cr
\omit&\omit\hrulefill&\omit\cr
}}


\vfill
\eightpoint
\centerline{\box\kosept}
\vfill
\centerline{\box\kohuit}
\vfill
\centerline{\box\koneuf}

\vfill\eject


\setbox\kodix=\vbox{\eightpoint \offinterlineskip
\halign{\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv#\cr
\omit\strat&\omit\hfil $31^3$\hfil&
\omit\hfil $2^3$\hfil&\omit\cr
\omit&\multispan2\hrulefill&\omit\cr
6&q^6+q^7+2q^8+2q^9+2q^{10}+q^{11}+q^{12}&
q^6+q^8+q^9+q^{10}+q^{12}&\cr
\omit&\multispan2\hrulefill&\omit\cr
51&q^3+q^4+2q^5+q^6+q^7+q^6t+q^7t+q^8t+q^9t&
q^4+q^5+q^7+q^6t+q^8t&\cr
\omit&\multispan2\hrulefill&\omit\cr
42&q^2+q^3+q^4+q^3t+2q^4t+2q^5t+q^6t+q^6t^2&
q^2+q^4+q^4t+q^5t+q^6t^2&\cr
\omit&\multispan2\hrulefill&\omit\cr
41^2&q+q^2+q^3+q^3t+q^4t+q^5t+q^3t^2+q^4t^2+q^5t^2+q^6t^3&
q^3+q^2t+q^4t+q^4t^2+q^5t^2&\cr
\omit&\multispan2\hrulefill&\omit\cr
3^2&q^3+q^2t+2q^3t+2q^4t+q^5t+q^3t^2+q^4t^2+q^5t^2&
q^3+q^2t+q^3t+q^4t+q^6t^3&\cr
\omit&\multispan2\hrulefill&\omit\cr
321&q+qt+2q^2t+q^3t+q^2t^2+2q^3t^2+q^3t^3+q^4t^3&
q+q^2t+q^2t^2+q^3t^2+q^4t^3&\cr
\omit&\multispan2\hrulefill&\omit\cr
31^3&1+qt+q^2t+qt^2+q^2t^2+qt^3+q^2t^3+q^3t^3+q^3t^4+q^3t^5&
qt+qt^2+q^2t^2+q^3t^3+q^2t^4&\cr
\omit&\multispan2\hrulefill&\omit\cr
2^3&qt+q^2t+qt^2+2q^2t^2+qt^3+2q^2t^3+q^3t^3+q^2t^4&
1+q^2t^2+q^2t^3+q^3t^3+q^2t^4&\cr
\omit&\multispan2\hrulefill&\omit\cr
2^21^2&t+qt+2qt^2+2qt^3+q^2t^3+qt^4+q^2t^4+q^2t^5&
t+qt^2+qt^3+q^2t^3+q^2t^5&\cr
\omit&\multispan2\hrulefill&\omit\cr
21^4&t+t^2+t^3+qt^3+t^4+qt^4+2qt^5+qt^6+qt^7&
t^2+qt^3+t^4+qt^5+qt^6&\cr
\omit&\multispan2\hrulefill&\omit\cr
1^6&t^3+t^4+2t^5+2t^6+2t^7+t^8+t^9&
t^3+t^5+t^6+t^7+t^9&\cr
\omit&\multispan2\hrulefill&\omit\cr
}}


\setbox\koonze=\vbox{\eightpoint \offinterlineskip
\halign{\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv\thinspace\hfil$#$\hfil\thinspace&
\tv#\cr
\omit\strat&\omit\hfil $2^21^2$\hfil&\omit\hfil $21^4$\hfil&
\omit\hfil $1^6$\hfil&\omit\cr
\omit&\multispan{3}\hrulefill&\omit\cr
6&q^7+q^8+2q^9+q^{10}+2q^{11}+q^{12}+q^{13}&
q^{10}+q^{11}+q^{12}+q^{13}+q^{14}&q^{15}&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
51&q^4+q^5+2q^6+q^7+q^8+q^7t+q^8t+q^9t&
q^6+q^7+q^8+q^9+q^{10}t&q^{10}&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
42&2q^3+q^4+q^5+q^4t+2q^5t+q^6t+q^7t^2&
q^4+q^5+q^6+q^6t+q^7t&q^7&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
41^2&q^2+q^3+q^4+q^3t+q^4t+q^5t+q^4t^2+q^5t^2+q^6t^2&
q^3+q^4+q^5+q^6t+q^6t^2&q^6&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
3^2&q^2+q^3+q^4+q^3t+q^4t+q^5t+q^4t^2+q^5t^2+q^6t^2&
q^4+q^5+q^4t+q^5t+q^6t&q^6&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
321&q+q^2+2q^2t+q^3t+2q^3t^2+q^4t^2+q^4t^3&
q^2+q^3+q^3t+q^4t+q^4t^2&q^4&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
31^3&q+qt+q^2t+qt^2+q^2t^2+q^3t^2+q^2t^3+q^3t^3+q^3t^4&
q+q^2+q^3t+q^3t^2+q^3t^3&q^3&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
2^3&q+qt+q^2t+qt^2+q^2t^2+q^3t^2+q^2t^3+q^3t^3+q^3t^4&
q^2+q^2t+q^3t+q^2t^2+q^3t^2&q^3&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
2^21^2&1+qt+2qt^2+q^2t^2+qt^3+q^2t^3+2q^2t^4&
q+qt+q^2t+q^2t^2+q^2t^3&q^2&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
21^4&t+t^2+qt^2+t^3+qt^3+2qt^4+qt^5+qt^6&1+qt+qt^2+qt^3+qt^4&q&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
1^6&t^2+t^3+2t^4+t^5+2t^6+t^7+t^8&
t+t^2+t^3+t^4+t^5&1&\cr
\omit&\multispan{3}\hrulefill&\omit\cr
}}


\centerline{\box\kodix}
\vskip 1cm
\centerline{\box\koonze}
\vfill
\eject



\bye

