%%%@@@ input from procsund
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% PAPER.TEX: Started May 16, 1994														%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\documentstyle[amssymb,12pt]{amsart}


%%%@@@ input from pmacros
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Makros fuer Proctor-Sundaram-Bijektion.                                   %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%@@@ input from mmacros
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                      Some LaTeX declarations:                            %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
 
\theoremstyle{definition}
\newtheorem{dfn}{Definition}[section]
 
\theoremstyle{remark}
\newtheorem{rem}{Remark}[section]
\newtheorem{ex}[rem]{Example}
 
\numberwithin{equation}{section}
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                      Some abbreviations:                                 %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\catcode`\@=11
\def\section{\@startsection{\@string\section}%
        1\z@{6\p@\@plus4\p@}{1\p@}%
        {\defaultfont\bf}}
\def\subsection{\@startsection{\@string\subsection}%
        2\z@{1\p@\@plus4\p@}{-.5em}%
        {\defaultfont\sc}}
\catcode`\@=12
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Christians Original-Figure-environment: Keine LaTeX-figures!             %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \newcounter{Figure} % Brauch' ma des?
 \newenvironment{Figure}%
    {\refstepcounter{Figurenum}%
         \vskip10pt minus 5pt \vbox\bgroup\hsize\textwidth\begin{center}}%
    {\vskip6pt\centerline{\hfil\small Figure~%
         \arabic{Figurenum}\hfil}\end{center}\egroup\vskip10pt minus 5pt}
 \newcounter{Figurenum}

\def\la{\lambda}
\def\ep{\varepsilon}
\def\si{\sigma}
 
\newcommand{\esf}{{e}}                  % elementary symmetric function
\newcommand{\hsf}{{h}}                  % homogeneous symmetric function
\newcommand{\ksf}{{k}}                  % Delta^2 of homogeneous symmetric
\newcommand{\schf}[2]{
   {s}_{#1}\of{#2,\bold{x}}
	}                                    % Schur function
\newcommand{\defeq}{:=}             % define equal...
\newcommand{\tableau}{T}                % semistandard tableau
\newcommand{\colT}[1]{\tableau_{.,#1}}  % column of tableau
\newcommand{\rowT}[1]{\tableau_{#1,.}}  % row of tableau
\newcommand{\Ferrer}{B}                 % Ferrers board
\newcommand{\digraph}{D}                % directed graph
\newcommand{\sdigraph}{S}               % Stembridge's directed
\newcommand{\horedge}{a}                % horizontal edge
\newcommand{\veredge}{\hat{a}}          % vertical edge
\newcommand{\path}{P}                   % lattice path
\newcommand{\dpath}{\overline{P}}       % dual lattice path
\newcommand{\pathset}{M}                % lattice path's
\newcommand{\fpath}{{\cal P}}           % family of lattice paths
\newcommand{\nfpath}{{\cal N}}          % number of lattice paths
\newcommand{\nifpath}{{\cal P}_o}       % nonintersecting family of lattice
% \newcommand{\SG}[1]{\boldsymbol{S}_{#1}}% Symmetric group
\newcommand{\SG}[1]{{\frak S}_{#1}}     % Symmetric group
\newcommand{\idty}{\operatorname{id}}   % Identity (in some group)
\newcommand{\GF}[1]{\bold{GF}\of{#1}}   % Generating function
\newcommand{\sgn}{\operatorname{sgn}}   % signum
\newcommand{\inv}{\operatorname{inv}}   % inversions
\newcommand{\cl}{\operatorname{cl}}   	% column-length

\newcommand{\sschf}[2]{
	{sp}_{#1}\of{#2,\bold{x}}
	}                                    % symplectic Schur function
\newcommand{\isschf}[2]{
   {sp}_{n,m}\of{#1,#2;\bold{x},\bold z}
	}                                    % intermediate sympl. Schur function
\newcommand{\osschf}[2]{
   {sp}_{n,1}\of{#1,#2,\bold{x},z}
   }                                    % intermediate sympl. Schur function
\newcommand{\reflect}{\bold{R}}         % reflection
\newcommand{\mreflect}{\bar{\bold{R}}}  % modified reflection
\newcommand{\oschf}[2]{
   {so}_{#1}\of{#2,\bold{x}}
   }                                    % orthogonal Schur function
\newcommand{\ooschf}[2]{
   {so}_{#1}\of{x_1,x_1^{-1},\dots,x_{#2},x_{#2}^{-1},1}
   }                                    % odd orthogonal Schur function
\newcommand{\oschftriv}[2]{
   {so}_{#1}\of{#2,\seqof{1,1,\cdots}}
   }                                    % "coarse" orthogonal Schur function
\newcommand{\liff}{\text{ iff }}        % logical: if and only if
\newcommand{\limp}{\text{ implies }}    % logical: implies
\renewcommand{\land}{\text{ and }}      % logical: and
 
\newcommand{\thmref}[1]{Theorem~\ref{#1}}
\newcommand{\secref}[1]{\S\ref{#1}}
\newcommand{\lemref}[1]{Lemma~\ref{#1}}
\newcommand{\figref}[1]{Figure~\ref{#1}}
 
\newcommand{\proof}{{\par\noindent\bf Proof: }}
\newcommand{\QED}{\nobreak\strut\hfill\qedsymbol\par\vskip3mm}
\newcommand{\noproof}{\nobreak\hfill\qedsymbol\par\vskip3mm}
 
\newcommand{\seqof}[1]{\left\langle#1\right\rangle}
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\newcommand{\numof}[1]{\big|#1\big|}
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\newcommand{\brkof}[1]{\left[#1\right]}
\newcommand{\intof}[1]{\lfloor #1\rfloor}
\newcommand{\absof}[1]{|#1|}
\newcommand{\invof}[1]{\inv\of{#1}}
\newcommand{\sgnof}[1]{\sgn\of{#1}}
\newcommand{\ceilof}[1]{\left\lceil #1\right\rceil}
\newcommand{\floorof}[1]{\left\lfloor #1\right\rfloor}

\newcommand{\singlecolumn}[1]{#1\quad \vdots\quad }

\newcommand{\truthfunc}[1]{\chi_{[#1]}}

\newcommand{\firstrow}[1]{#1\quad \vdots\quad }
 

%%%@@@ end input from mmacros

%%%@@@ input from paths
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%               PATHS.TEX: Begonnen am 16. Juli 1993                       %
% Dieses kleine Macropaket soll einfache Befehle zum Zeichnen von Gittern  %
% und Gitterpunktwegen einfuehren. (Fuer AMSLaTeX!)                        %
% Wesentlich ueberarbeitet und gestrafft am 1. Mai 1994.                   %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Fonts:                                                                   %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\font\scalefont = cmti8
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Basic parameters:                                                        %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\unitlength = 5mm					% instead of 4 mm
\thicklines

\newcount\xmin                % coordinates of extremal points of the grid
\newcount\xmax
\newcount\ymin
\newcount\ymax

\newcount\gridwidth
\newcount\gridheight

\newcount\nofxpoints          % number of points in horizontal direction
\newcount\nofypoints          % number of points in vertical direction

\newcount\dcnta               % dummy-counter A
\newcount\dcntb               % dummy-counter B

% Special macros for drawing a grid with lower left corner (xmin, ymin)
% and upper right corner (xmax, ymax):
\def\begingrid#1#2#3#4{			% arguments: xmin, ymin, xmax, ymax
	\global\xmin = #1
	\global\ymin = #2
	\global\xmax = #3
	\global\ymax = #4
	% simple test:
	\ifnum\xmin > \xmax\errmessage{PATHS: \xmin > \xmax|}\fi
	\ifnum\ymin > \ymax\errmessage{PATHS: \ymin > \ymax|}\fi
	% computation of heigth and width
	\global\gridwidth = \xmax
	\global\gridheight = \ymax
	\global\advance\gridwidth by -\xmin
	\global\advance\gridheight by -\ymin
	\nofxpoints = \gridwidth
	\advance\nofxpoints by 1
	\nofypoints = \gridheight
	\advance\nofypoints by 1
	% adjust some space around the picture
	\advance\gridwidth by 4
	\advance\gridheight by 4
	\advance\xmin by -2
	\advance\ymin by -2
	% LaTeX's picture command
	\begin{picture}(\gridwidth, \gridheight)(\xmin, \ymin)
	% restore correct the lengths
	\advance\gridwidth by -4
	\advance\gridheight by -4
	\advance\xmin by 2
	\advance\ymin by 2
	% for debugging
	% \message{PATHS: width \number\gridwidth, height \number\gridheight; %
	% at (\number\xmin, \number\ymin)}
	% draw gridpoints:
	\global\dcnta = \ymin
	\loop\ifnum\dcnta<\ymax
		\makegridline\dcnta	% do inner loop in a separate macro
		\global\advance\dcnta by 1
	\repeat
	\makegridline\ymax
}
 
% inner loop of grid-drawing procedure
\def\makegridline#1{
   \begingroup
   \global\dcntb = \xmin
   \loop\ifnum\dcntb<\xmax
      \put(\dcntb, #1){\circle*{0.1}}
      \global\advance\dcntb by 1
   \repeat
   \put(\xmax, #1){\circle*{0.1}}
   \endgroup
}
 
% draw axes
\def\makexaxis{\thinlines
        \dcnta = \gridwidth
        \advance\dcnta by 1
        \put(\xmin, 0){\vector(1,0){\dcnta}}
        \thicklines
}
\def\makeyaxis{\thinlines
        \dcntb = \gridheight
        \advance\dcntb by 1
        \put(0, \ymin){\vector(0,1){\dcntb}}
        \thicklines
}
\def\makeaxes{\thinlines
        \dcnta = \gridwidth
        \advance\dcnta by 1
        \put(\xmin, 0){\vector(1,0){\dcnta}}
        \dcntb = \gridheight
        \advance\dcntb by 1
        \put(0, \ymin){\vector(0,1){\dcntb}}
        \thicklines
}
 
% scaling on axes
\def\makexscale{
   \dcnta = \xmin
	\ifnum\dcnta<-9\dcnta=-9\fi
   \loop\ifnum\dcnta<0
		\put(\dcnta,0){\line(0,-1){0.2}}
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		{\scalefont\number\dcnta}}}
      \advance\dcnta by 1
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  \dcntb=\xmax
  \ifnum\dcntb>9\dcntb=9\fi
  \loop\ifnum\dcnta<\dcntb
      \put(\dcnta,0){\line(0,-1){0.2}}
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  \put(\dcntb,0){\line(0,-1){0.2}}
  \put(\dcntb, -0.4){\makebox(0,0){{\scalefont\number\dcntb}}}
}
\def\makeyscale{
   \dcnta = \ymin
   \loop\ifnum\dcnta<0
      \put(0,\dcnta){\line(-1,0){0.2}}
      \put(-0.4, \dcnta){\makebox(0,0){
		{\scalefont\number\dcnta}}}
      \advance\dcnta by 1
   \repeat
   \dcnta = 1
   \loop\ifnum\dcnta<\ymax
      \put(0,\dcnta){\line(-1,0){0.2}}
      \put(-0.4, \dcnta){\makebox(0,0){
		{\scalefont\number\dcnta}}}
      \advance\dcnta by 1
   \repeat
   \put(-0.4, \ymax){\makebox(0,0){{\scalefont\number\ymax}}}
}

% macro for captions:
\def\gridcaption#1{
	\dcnta=\xmin
	\dcntb=\gridwidth
	\divide\dcntb by 2
	\advance\dcnta by \dcntb
	\dcntb=\ymax
	\advance\dcntb by 1
	\put(\dcnta,\dcntb){\makebox(0,0){#1}}
}

% macros for drawing skew dotted lines
\newsavebox{\skewdowndottedbox}
\newsavebox{\skewupdottedbox}
\savebox{\skewdowndottedbox}(1,1)[lb]{
	\put(0.2,-0.2){\circle*{0.01}}
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 	\put(0.6,-0.6){\circle*{0.01}}
	\put(0.8,-0.8){\circle*{0.01}}
}
\savebox{\skewupdottedbox}(1,1)[lb]{
	\put(0.2,0.2){\circle*{0.01}}
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 	\put(0.6,0.6){\circle*{0.01}}
	\put(0.8,0.8){\circle*{0.01}}
}
\def\downskewline#1#2#3{
	\multiput(#1,#2)(1,-1){#3}{\usebox{\skewdowndottedbox}}
}
\def\upskewline#1#2#3{
	\multiput(#1,#2)(1,1){#3}{\usebox{\skewupdottedbox}}
}

% Special macros for drawing lattice paths:
\newsavebox{\hplainbox}
\savebox{\hplainbox}{\put(0,0){\line(1,0){1}}}
\newsavebox{\hdottedbox}
\savebox{\hdottedbox}(1,0)[bl]{
	\put(0.2,0){\circle*{0.01}}
	\put(0.4,0){\circle*{0.01}}
	\put(0.6,0){\circle*{0.01}}
	\put(0.8,0){\circle*{0.01}}
}
\newsavebox{\vplainbox}
\savebox{\vplainbox}{\put(0,0){\line(0,1){1}}}
\newsavebox{\vdottedbox}
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	\put(0,0.2){\circle*{0.01}}
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	\put(0,0.6){\circle*{0.01}}
	\put(0,0.8){\circle*{0.01}}
}
\newsavebox{\vplaindownbox}
\savebox{\vplaindownbox}{\put(0,0){\line(0,-1){1}}}
\newsavebox{\vdotteddownbox}
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	\put(0,-0.8){\circle*{0.01}}
}
\newsavebox{\skewupbox}
\savebox{\skewupbox}{\put(0,0){\line(1,1){1}}}
\newsavebox{\skewdownbox}
\savebox{\skewdownbox}{\put(0,0){\line(1,-1){1}}}
\newsavebox{\hstep}		% horizontal step
\newsavebox{\vstep}		% vertical step
\newsavebox{\sstep}		% skew step
\newcount\updownincrement
\def\uppath#1#2#3{
	\savebox{\hstep}{\usebox{\hplainbox}}
	\savebox{\vstep}{\usebox{\vplainbox}}
	\savebox{\sstep}{\usebox{\skewupbox}}
	\updownincrement=1
	\dcnta=#1
	\dcntb=#2
	\thickpoint{\dcnta}{\dcntb}
	\dosteplist#3\endlist
	\thickpoint{\dcnta}{\dcntb}
}
\def\updottedpath#1#2#3{
	\savebox{\hstep}{\usebox{\hdottedbox}}
	\savebox{\vstep}{\usebox{\vdottedbox}}
	\savebox{\sstep}{\usebox{\skewupdottedbox}}
	\updownincrement=1
	\dcnta=#1
	\dcntb=#2
	\thickcircle{\dcnta}{\dcntb}
	\dosteplist#3\endlist
	\thickcircle{\dcnta}{\dcntb}
}
\def\downpath#1#2#3{
	\savebox{\hstep}{\usebox{\hplainbox}}
	\savebox{\vstep}{\usebox{\vplaindownbox}}
	\savebox{\sstep}{\usebox{\skewdownbox}}
	\updownincrement=-1
	\dcnta=#1
	\dcntb=#2
	\thickpoint{\dcnta}{\dcntb}
	\dosteplist#3\endlist
	\thickpoint{\dcnta}{\dcntb}
}
\def\downdottedpath#1#2#3{
	\savebox{\hstep}{\usebox{\hdottedbox}}
	\savebox{\vstep}{\usebox{\vdotteddownbox}}
	\savebox{\sstep}{\usebox{\skewdowndottedbox}}
	\updownincrement=-1
	\dcnta=#1
	\dcntb=#2
	\thickcircle{\dcnta}{\dcntb}
	\dosteplist#3\endlist
	\thickcircle{\dcnta}{\dcntb}
}
\def\endlist{\endlist}
\def\dosteplist{\afterassignment\handlenextstep\let\next=}
\def\handlenextstep{
	\ifx\next\endlist
   	\let\next=\relax
	\else
		\ifx\next-
			\put(\dcnta,\dcntb){\usebox{\hstep}}
			\advance\dcnta by 1
		\else
			\ifx\next|
				\put(\dcnta,\dcntb){\usebox{\vstep}}
				\advance\dcntb by\updownincrement
			\else
				\ifx\next/
					\put(\dcnta,\dcntb){\usebox{\sstep}}
					\advance\dcnta by 1
					\advance\dcntb by\updownincrement
				\else
					\errmessage{PATHS: Wrong symbol path argument.}
				\fi
			\fi
		\fi
		\let\next=\dosteplist
	\fi
	\next
}
 
% draw thick point:
\def\thickpoint#1#2{\put(#1, #2){\circle*{0.4}}}

% draw thick circle
\def\thickcircle#1#2{\put(#1, #2){\circle{0.4}}}

% label thick point above or below:
\def\labelpointal#1#2#3{							% above, to the left
	\put(#1,#2){
		\begin{picture}(1,1)(0,0)
			\put(-0.85,0.35){$\scriptstyle #3$}
		\end{picture}
	}
}

\def\labelpointar#1#2#3{							% above, to the right
	\put(#1,#2){
		\begin{picture}(1,1)(0,0)
			\put(-0.15,0.35){$\scriptstyle #3$}
		\end{picture}
	}
}

\def\labelpointbl#1#2#3{							% below, to the left
	\put(#1,#2){
		\begin{picture}(1,1)(0,0)
			\put(-0.85,-0.8){$\scriptstyle #3$}
		\end{picture}
	}
}

\def\labelpointbr#1#2#3{							% below, to the right
	\put(#1,#2){
		\begin{picture}(1,1)(0,0)
			\put(-0.15,-0.8){$\scriptstyle #3$}
		\end{picture}
	}
}

% label horizontal steps
\def\hlabel#1#2#3{\put(#1,#2){\makebox(1,0.6){\scalefont #3}}}

\def\endgrid{\end{picture}}

\def\drawrect#1#2#3#4{
	\put(#1,#2){%
		\begin{picture}(#3,#4)(0,0)\thicklines%
			\put(0,0){\line(0,1){#4}}
			\put(0,0){\line(1,0){#3}}
			\put(#3,#4){\line(0,-1){#4}}
			\put(#3,#4){\line(-1,0){#3}}
		\end{picture}
	}
}


%%%@@@ end input from paths

%%%@@@ input from tableaux
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%  TABLEAUX.TEX: Begonnen am 20. Juli 1993                                 %
% Dieses kleine Macropaket soll einfache Befehle zur Darstellung von       %
% Tableaux  einfuehren. (Nur fuer AMSLaTeX|)                               %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcount\rowcount
\newcount\columncount

\def\begintableau#1#2{									% rows, columns
\global\rowcount=#1
\begin{picture}(#2,#1)(0,0)
\linethickness{0.1pt}
% \put(0,0){\line(0,1){#1}}
% \put(0,#1){\line(1,0){#2}}
}

\def\endtableau{\end{picture}}

\def\row#1{
	\global\advance\rowcount by -1
	\global\columncount=0
	\dorowlist#1\endlist
}

\def\endlist{\endlist}

\def\dorowlist{\afterassignment\handlenextentry\let\next=}

\def\handlenextentry{
	\ifx\next\endlist
		\let\next=\relax
	\else
		\ifx\next o
			\put(\columncount,\rowcount){\framebox(1,1){\space}}
			\global\advance\columncount by 1
		\else
			\ifx\next-
				% \put(\columncount,\rowcount){\framebox(1,1){\space}}
				\global\advance\columncount by 1
			\else
				\put(\columncount,\rowcount){\framebox(1,1){$\next$}}
				\global\advance\columncount by 1
			\fi
		\fi
		\let\next=\dorowlist
	\fi
	\next
}

\newcount\ca
\newcount\cb
\def\printrow#1#2{
	\global\ca=0
	\global\cb=#1
	\doprintrowlist#2\endlist
}

\def\doprintrowlist{\afterassignment\handlenextprintentry\let\next=}

\def\handlenextprintentry{
	\ifx\next\endlist
		\let\next=\relax
	\else
		\put(\ca,\cb){\makebox(1,1){$\scriptstyle\next$}}
		\global\advance\ca by 1
		\let\next=\doprintrowlist
	\fi
	\next
}



%%%@@@ end input from tableaux

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Spezielle Makros fuer Proctor-Sundaram-Bijektion.                         %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcommand{\proctor}{{\Bbb P}}		% All Proctor objects
\newcommand{\varproctor}{{\phull^3}}% Variant of (restricted) Proctor objects
\newcommand{\pcore}{G_\proctor}% Core of Proctor objects:
\newcommand{\phull}{B_\proctor}	% All \setminus Core
\newcommand{\mphull}{\overline {B_\proctor^3}}	% modified third set
\newcommand{\varphull}{\varproctor}	% All \setminus Core
\newcommand{\sundaram}{{\Bbb S}}		% All Sundaram objects
\newcommand{\score}{G_\sundaram}% Core of Sundaram objects:
\newcommand{\shull}{B_\sundaram}	% All \setminus Core
\newcommand{\esfone}{\esf^{\left[1\right]}}
\newcommand{\esfinfty}{\esf^{\left[\infty\right]}}
\newcommand{\inftiesincolumn}[2]{\numof{\setof{#1_{.,#2}=\infty}}}
\newcommand{\numofentries}[2]{\#(#1\in#2)}
\newcommand{\leqincol}[3]{\setof{#3_{.,#2}\leq #1}}
\newcommand{\leqincolT}[2]{\setof{\colT{#2}\leq #1}}
\newcommand{\morethanincol}[3]{\setof{#3_{.,#2}> #1}}
% \newcommand{\pairsincol}[3]{\left[\left[\boldsymbol{2}\right]\right]\of{
%	\leqincol{#1}{#2}{#3}}}
% \newcommand{\singsincol}[3]{\left[\left[\boldsymbol{1}\right]\right]\of{
%	\leqincol{#1}{#2}{#3}}}
\newcommand{\pairsincol}[3]{\boldsymbol{\left[2\right]}_{
	\leqincol{#1}{#2}{#3}}}
\newcommand{\singsincol}[3]{\boldsymbol{\left[1\right]}_{
	\leqincol{#1}{#2}{#3}}}
\newcommand{\inverted}[1]{\overline{#1}}
\newcommand{\nivsecpath}[2]{\brkof{N_{#1}\cap P_{#2}}}
\newcommand{\genlin}[1]{{\bold{GL}}\of{#1}}	% general linear group
\newcommand{\orthog}[1]{SO\of{#1}}		% orthogonal group
\newcommand{\abij}{{\frak a}} 			% Bijection a in involution principle
\newcommand{\ione}{{\frak i}_\sundaram}			% Involution i1 in -"-
\newcommand{\itwo}{{\frak i}_\proctor}			% Involution i2 in -"-

%%%@@@ end input from pmacros

% \family{ccr}\selectfont

\begin{document}

\title[Bijection between odd orthogonal tableaux]
{A bijection between Proctor's and Sundaram's odd orthogonal tableaux}

\author[Markus Fulmek and Christian Krattenthaler]{Markus Fulmek
and Christian Krattenthaler$^\dagger$\\
Institut f\"ur Mathematik\\Universit\"at Wien\\
Strudlhofgasse~4\\A-1090 Vienna, Austria}
\thanks{$^\dagger$ Supported in
part by EC's Human Capital and Mobility Program, grant
CHRX-CT93-0400, the second author was also supported by the
Austrian Science Foundation FWF, grant P10191-PHY}
%\date{\today}


\maketitle % Unverzichtbar!!!
\begin{abstract}
An explicit bijection between Proctor's odd orthogonal tableaux and
Sundaram's odd 
orthogonal tableaux is given.
\end{abstract}

% \tableofcontents
\section{Introduction}

%%%@@@ input from ps_intro
\label{sec:introduction}
Over the last twenty years many enumerative combinatorialists have
been interested in identities which count the number of tableaux 
or plane partitions of a
given type. Sometimes two families of tableaux or plane partitions 
have the same counts. But it seems to be very rare to find bijections
between two such families. In this paper we use the Garsia--Milne
technique to generate a bijection in such a situation.

This situation is the following.
Proctor \cite{Proctor:A-Schensted,Proctor:Young-tableaux} and
Sundaram \cite{Sundaram:Orthogonal} defined {\em different\/}
tableaux that both enumerate the odd orthogonal characters. To be more
precise, the generating function with respect to a certain weight for
either Proctor's tableaux of shape $\la$ or Sundaram's tableaux of
shape $\la$ equals the irreducible
character of $\orthog{2n+1}$ indexed by $\la$. Since the generating
functions are equal, there must exist a weight-preserving
bijection between the two sets of tableaux. The purpose of this paper is
to construct an {\em explicit\/} such bijection.

Our bijection bases on the fact that both types of tableaux
``generate" Jacobi--Trudi-type identities for their generating
functions, as shown in \cite{Fulmek-Krattenthaler:Lattice-path}. In
fact, the bijection combines mappings from
\cite{Fulmek-Krattenthaler:Lattice-path} with the involution principle
of Garsia and Milne \cite{Garsia-Milne:Method-for}.

We remark that lately there has been considerable interest in
combinatorial descriptions of classical group characters. Such
descriptions evolved mainly from either algebraic combinatorics or
algebraic geometry, culminating in Proctor's
\cite{Proctor:Young-tableaux} and Littelmann's \cite{Littelmann:Kac-Moody}
papers, respectively, which can also serve as reference papers.
However, the relations between 
the
various combinatorial
descriptions that have been developed 
are in most cases unclear, except
for the general linear characters where all descriptions are more or
less the same. In this paper we relate two combinatorial descriptions
for orthogonal characters which come from the algebraic combinatorics
side. Recently, Sheats \cite{Sheats:jeu} related the
symplectic tableaux of King and El-Sharkaway
\cite{King-El-Sharkaway:Standard} and the symplectic tableaux of
DeConcini \cite{DeConcini:Symplectic}, the first coming from the
algebraic combinatorics side, the second coming from the algebraic
geometry side. But there is still much more work to be done to
understand the relations, in particular between the descriptions that
come from different sides.

Our paper is organized as follows. In section~\ref{sec:definitions} we
define the odd orthogonal tableaux of Proctor and
Sundaram and their weights. Since we use the involution principle we
briefly review it in section~\ref{sec:principle}. Then, in
section~\ref{sec:outline} we give a rough outline of our bijection.
The details of the underlying involutions are provided in
section~\ref{sec:involutions}. An elaborate example is the subject of
section~\ref{sec:example}. As indicated at the beginning, the
ideas behind the involutions in section~\ref{sec:involutions} stem
from lattice path proofs for Jacobi--Trudi-type identities for the
irreducible characters of $\orthog{2n+1}$ given in
\cite{Fulmek-Krattenthaler:Lattice-path}. This background is explained
in section~\ref{sec:explanation}. It may be helpful to read this
section immediately after having digested what Proctor tableaux and
Sundaram tableaux are and how the Garsia--Milne involution principle
works, since it gives an idea of the ``program" for our
bijection,
% markus: Klingt das nicht ein bisschen praetentioes?
% - es duerfte doch der Normalfall sein, mit [2] nicht vertraut zu
% sein!
%Christian:
% Hmm, was ich eigentlich sagen will, ist: Dieser Abschnitt erklaert den
% Hintergrund unserer Bijektion. Und diese Erklaerungen koennen
% verstanden werden, SELBST wenn man unsere erste Arbeit nicht gelesen
% hat. So wie Du es formulierst, kann man folgern: Wenn man unsere erste
% Arbeit gelesen hat, dann kennt man das Programm unserer Bijektion
% sowieso schon, was natuerlich ueberhaupt nicht zutreffen kann. Daher
% vielleicht so:
not requiring that the reader is familiar with
\cite{Fulmek-Krattenthaler:Lattice-path}. Finally, in
section~\ref{sec:concluding} we comment on bijections between
some more odd orthogonal tableaux, due to
Proctor \cite{Proctor:Young-tableaux} and King and Welsh
\cite{King-Welsh:Construction}.


%%%@@@ end input from ps_intro
\section{Proctor's and Sundaram's odd orthogonal tableaux}

%%%@@@ input from ps_def
\label{sec:definitions}
% Definition der Proctor und Sundaram tableaux
% \begin{enumerate}
% \item Tableaux of shape $\lambda$
% \item Example of tableau.
% \item Proctor's orthogonal tableaux of shape $\lambda$:
%	The orthogonal constraints. Weights.
% \item Example.
% \item Sundaram's odd orthogonal tableaux of shape $\lambda$ and
%	$\ell\of\la\leq n$. The symplectic constraints. Weights.
% \item Example.
% end{enumerate}

For readers convenience, we recall some basic concepts.
An $r$-tuple $\lambda = \parof{\lambda_1,
% \lambda_2, % wegen Zeilenumbruch!
\dots, \lambda_r}$ with $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_r> 0$
is called a {\em partition of length $r$\/}. We denote the length of
the partition $\lambda $ by $\ell(\lambda)$. The {\em Ferrers
board\/} of $\lambda$ is an array of cells with $\ell(\lambda)$
left-justified rows and $\lambda_i$ cells in row $i$. The
{\em conjugate of\/} $\lambda$ is the partition 
$(\lambda^\prime_1, \dots, \lambda^\prime_{\lambda_1})$ where
$\lambda_j'$ is the length of the $j$-th column in the Ferrers board of
$\lambda$.

We shall consider fillings of the cells of a Ferrers board with
positive integers. Given a filling $T$ we write $T_{i,j}$ for the
entry in cell $(i,j)$. Also, we write $T_{.,j}$ for the entries in the $j$-th
column of $T$. The notation $T_{.,j}$ has to be understood in a
suggestive sense, for instance, $\{\colT{j}\le p\}$ means the multiset of all
entries in the $j$-th column of $T$ that are at most $p$.

\subsection{Sundaram tableaux}
Now let $\la$ be a partition with $\ell\of{\la}\leq n+1$.
A {\em $(2n+1)$-Sundaram tableau of shape\/} $\la$ is a filling $\tableau$
of the cells of the Ferrers board of $\la$ with elements from
$\setof{1,2,\dots,2n,\infty}$ such that
\begin{enumerate}
\item entries along rows are weakly increasing, each row is
allowed to contain only at most one entry $\infty$,
\item entries along columns are weakly increasing, and strictly
increasing with respect to the finite entries,
% markus: Hier koennte ich mir auch vorstellen:
% entries along columns are strictly increasing except for the entries
% $\infty$, which may appear more than once at the bottom of any
% column,
%Christian:
% Das ist mir nicht praezise genug, weil Du sagst, dass $\infty$ ``may
% appear at the bottom", was nicht ausschliesst, dass es woanders auch
% vorkommt. Ich bin mit meiner Formulierung auch nicht besonders
% gluecklich. Es faellt mir bloss nichts Besseres ein.
\item all entries in row $i$ are at least $2i-1$.
\end{enumerate}
Obviously it 
suffices 
to require (3) to hold for the first
element in each row only, since rows are weakly increasing. So (3) is
equivalent with
\begin{equation}
\label{eq:sundaram-constraint}
\tableau_{i,1}\geq 2i-1\;\text{ for } 1\leq i\leq\ell\of{\la}.
\end{equation}

\figref{fig:ex-Sundaram-tableau} shows an example of a
$9$-Sundaram tableau of shape $(5,4,3,1,1)$.
\begin{Figure}
\label{fig:ex-Sundaram-tableau}
\centerline{\small Example of a 9-Sundaram tableau}
\begin{picture}(15,6)(0,0)
\put(6,0.5){
	\begintableau{5}{5}
		\row{14467}
		\row{4556}
		\row{57\infty}
		\row{\infty}
		\row{\infty}
	\endtableau
}
\end{picture}
\end{Figure}


\subsection{Proctor tableaux}
Let $\lambda$ be a partition. 
A {\em $(2n+1)$-Proctor tableau of shape\/} $\la$ is a filling of the
cells of the Ferrers
board of $\la$ with elements from $\setof{1,2,\dots,2n,2n+1}$ such
that
\begin{enumerate}
\item entries along rows are weakly increasing,
\item entries along columns are strictly increasing,
\item the $(2c)$-th orthogonal condition is satisfied for all $c$ (see
below), and 
the $(2n+1)$-st orthogonal condition is satisfied,
\item the $(2c)$-th protection condition is satisfied for all $c$ (see below).
\end{enumerate} 

A filling $\tableau$ {\em satisfies the $p$-th orthogonal condition\/}
if the number 
of entries less or equal than $p$ in the {\em first two\/} columns of
$\tableau$ is less or equal to $p$. Stated more formally, this is
\begin{equation}
\label{eq:orthogonal-condition}
\numof{\leqincolT{p}{1}} + \numof{\leqincolT{p}{2}} \leq p,
\end{equation}
where $|S|$ denotes the cardinality of a multiset $S$, as usual.

A filling $\tableau$ {\em satisfies the $(2c)$-th protection condition\/} if
whenever $i+j=2c$ and $\tableau_{i,1} = 2c-1$ and 
$\tableau_{j,2}=\tableau_{j,3}=\dots=\tableau_{j,h-1} = 2c-1$ and
$\tableau_{j,h}=2c$, then we have $\tableau_{j-1, h} = 2c-1$.
Stated in a more formal manner, this is
\begin{align}
i+j=2c &\land \tableau_{i,1}=2c-1 \notag\\
		&\land
\tableau_{j,2}=\dots=\tableau_{j,h-1} = 2c-1 \hbox{ (for $h\geq2$)} \notag\\
	&\land
\tableau_{j,h} = 2c \notag\\
	\limp&
\tableau_{j-1,h} = 2c-1.\label{eq:protection-condition}
\end{align}

The filling in \figref{fig:ex-Proctor-tableau} satisfies the
$6$-th protection condition if and only if $5$ is inserted instead of
$*$ (choose $i=4$ and $j=2$). In fact, if $*=5$
then it is a $9$-Proctor
tableau of shape $(5,4,3,1)$.

%%%@@@ input from psf_ep
\begin{Figure}
\label{fig:ex-Proctor-tableau}
\centerline{\small Example of a 9-Proctor tableau, provided $*=5$}
\begin{picture}(15,5)(0,0)
\put(6,0.5){
	\begintableau{4}{5}
		\row{144*6}
		\row{2556}
		\row{378}
		\row{5}
	\endtableau
}
\end{picture}
\end{Figure}

%%%@@@ end input from psf_ep

Note that by the $(2n+1)$-st orthogonal condition for every
$(2n+1)$-Proctor tableau 
of shape $\la$ we must have $\la^\prime_1 + \la^\prime_2\leq 2n+1$.

\subsection{Odd orthogonal characters}
For both types of $(2n+1)$-orthogonal tabl\-eaux just introduced we define
weights in the following way (the definition is the same for both types):
\begin{equation}
\label{eq:general-weight-definition}
w\of{\tableau}\defeq
	\prod_{i=1}^n x_i^{\numofentries{2i-1}{\tableau}-
		\numofentries{2i}{\tableau}}.
\end{equation}

With this definition of weights, the irreducible character of the
orthogonal group $\orthog{2n+1}$ indexed by $\la$ is given as
\begin{equation}\label{eq:odd-character}
\oschf{2n+1}{\lambda} =
\sum_{\tableau}w\of{\tableau},
\end{equation}
where the sum is over all $(2n+1)$-Proctor tableaux $\tableau$ of
shape $\lambda$,
or {\em equivalently\/} over all $(2n+1)$-Sundaram tableaux $\tableau$
of shape $\la$. 
Hence there exists a weight-preserving bijection between
odd orthogonal Proctor tableaux and odd orthogonal Sundaram tableaux
of the same shape.
The purpose of this work is to construct such a bijection explicitly. 

%%%@@@ end input from ps_def

\section{The involution principle of Garsia and Milne}

%%%@@@ input from ps_princ
\label{sec:principle}

Since we are going to rely on the involution principle of Garsia and
Milne \cite{Garsia-Milne:Method-for} (see also 
\cite[sec.~4.6]{Stanton-White:Constructive}), we recall it first,
for convenience.

Suppose we are given two finite sets $\sundaram$ and $\proctor$ with
{\em signed\/} weight 
functions 
$w_\sundaram$, $w_\proctor$, respectively. Denote the subsets of
positively/negatively signed elements of $\sundaram$ and $\proctor$
by $\sundaram^+$/$\sundaram^-$ and $\proctor^+$/$\proctor^-$,
respectively. Let there be sets $\score$ and $\pcore$ of
``good" elements contained in the respective positive sets, i.e\@.
$\score\subseteq \sundaram^+$ and $\pcore\subseteq
\proctor^+$. Denote the complementary sets of ``bad" elements by
$\shull\defeq \sundaram\setminus \score$ and $\phull\defeq
\proctor\setminus \pcore$. 
Suppose there is a 
{\em weight- and sign-preserving\/} bijection $\abij$ between
$\sundaram$ and $\proctor$, $\abij:\,\sundaram\mapsto \proctor$. 
Furthermore,
suppose that there are {\em weight-preserving but
sign-reversing 
involutions\/} $\ione$ and $\itwo$ on the respective sets of bad
elements, $\ione:\;\shull\mapsto \shull$ and
$\itwo:\;\phull\mapsto \phull$.
(One can think of these involutions as ``cancelling'' all the bad elements.)
Then there must be a weight-preserving bijection
between the sets
$\score$ and $\pcore$ of respective good elements. 
And an explicit such bijection, $g$ say, can be constructed in the following
algorithmic way.
Start with some element $x\in \score$. Then $g(x)$ is given by 
$(\abij\circ \ione\circ \abij^{-1}\circ \itwo)^n\circ \abij\of{x}$,
for some $n\ge0$ (depending on $x$). To be precise, $n$ is the
minimal
nonnegative integer such that 
$(\abij\circ \ione\circ \abij^{-1}\circ \itwo)^n\circ \abij\of{x}$ is
in $\pcore$. (In fact, $n$ is unique, since $\itwo$ is not defined
for an element of $\pcore$.) For a proof see \cite{Garsia-Milne:Method-for} or
\cite[sec.~4.6]{Stanton-White:Constructive}.
 

%%%@@@ input from psf_invp
\begin{Figure}
\label{fig:involution-principle}
\centerline{\small Illustration of involution principle}

\begin{picture}(30,7)(0,0)
\put(8,3){\oval(8,6)}
\put(8,3){\oval(4,3)}
\put(4,3){\line(1,0){2}}
\put(10,3){\line(1,0){2}}
\put(5.5,5){\makebox(0,0){$\bold +$}}
\put(8,5.2){\makebox(0,0){$\shull^+$}}
\put(10.5,5){\makebox(0,0){$\bold +$}}
\put(5,1.5){\makebox(0,0){$\bold -$}}
\put(8,0.6){\makebox(0,0){$\shull^-$}}
\put(11,1.5){\makebox(0,0){$\bold -$}}
\put(8,3.8){\makebox(0,0){{\small }}}
\put(8,3){\makebox(0,0){$\bold +\;{\score}\;\bold +$}}
\put(8,2.2){\makebox(0,0){{\small }}}
\thicklines
\put(5.5,3){\vector(0,1){1}}
\put(4.5,3.5){$\ione$}
\put(5,3){\vector(0,-1){1}}
\put(10.5,3){\vector(0,1){1}}
\put(11,3.5){$\ione$}
\put(11,3){\vector(0,-1){1}}

\put(22,3){\oval(8,6)}
\put(22,3){\oval(4,3)}
\put(18,3){\line(1,0){2}}
\put(24,3){\line(1,0){2}}
\put(19.5,5){\makebox(0,0){$\bold +$}}
\put(22,5.2){\makebox(0,0){$\phull^+$}}
\put(24.5,5){\makebox(0,0){$\bold +$}}
\put(19,1.5){\makebox(0,0){$\bold -$}}
\put(22,0.6){\makebox(0,0){$\phull^-$}}
\put(25,1.5){\makebox(0,0){$\bold -$}}
\put(22,3.8){\makebox(0,0){{\small }}}
\put(22,3){\makebox(0,0){$\bold +\;{\pcore}\;\bold +$}}
\put(22,2.2){\makebox(0,0){{\small }}}
\thicklines
\put(19.5,3){\vector(0,1){1}}
\put(18.5,3.5){$\itwo$}
\put(19,3){\vector(0,-1){1}}
\put(24.5,3){\vector(0,1){1}}
\put(25,3.5){$\itwo$}
\put(25,3){\vector(0,-1){1}}

\put(13,4.5){\vector(1,0){4}}
\put(15,3){\makebox(0,0){$\abij$}}
\put(17,1.5){\vector(-1,0){4}}

\end{picture}
\end{Figure}

%%%@@@ end input from psf_invp

%%%@@@ end input from ps_princ

\section{Outline of the bijection}

%%%@@@ input from ps_obj
\label{sec:outline}
% As mentioned before, in this paper we want to give an explicit
% weight-preserving bijection between $(2n+1)$-Proctor tableau and 
% $(2n+1)$-Sundaram tableau of the same shape $\la$.
% In the
% following, we take this two conditions for granted without further notice.

The reason that our bijection is not
straight-forward but 
employs
the involution principle, 
is that $(2n+1)$-Proctor tableaux and $(2n+1)$-Sundaram tableaux are
rather ``far away from one another''. The first obvious difference is the set
of entries. 
This is, of course, very easy to remove. Just replace $2n+1$ by $\infty$
in every $(2n+1)$-Proctor tableau. Since neither entries $2n+1$ nor
entries $\infty$ do contribute to the tableau's weight, this certainly is
a weight-preserving conversion. 
From now on
we shall consider these
converted tableaux and, by abuse of terminology, call them also
$(2n+1)$-Proctor tableaux. 

In the following, let $\la$ be a fixed partition with
$\la'_1+\la'_2\le 2n+1$ and 
$\ell(\la)\le n+1$. (Recall that $\la_1'+\la_2'\le 2n+1$ is needed to
have well-defined $(2n+1)$-Proctor tableaux, and $\ell(\la)\le n+1$ is
needed to have well-defined $(2n+1)$-Sundaram tableaux.) Basically, in
view of section~\ref{sec:principle}, 
the only thing we have to do is to specify the sets $\sundaram$,
$\proctor$, their weights $w_\sundaram$, $w_\proctor$, the sets
$\score$, $\pcore$ of good elements, the bijection $\abij$,
and the involutions $\ione$, $\itwo$ on the sets of bad elements. 

Of course, the sets $\score$ and $\pcore$ of good elements are
the set of $(2n+1)$-Sundaram tableaux of shape $\la$ and the set of
$(2n+1)$-Proctor tableaux of shape $\la$, respectively.
Next we have to embed $\score$ and $\pcore$ into the bigger sets
$\sundaram$ and $\proctor$. To describe these sets we have to set up
some more terminology.

Consider our partition $\la$ and
think of the Ferrers board of $\la$ in the following way: The lower
edge forms a sequence of $\la_1$ ``steps''. Columns may be viewed as
``heaps of cells'' stacked on top of each ``step''. The number of
cells in the heap stacked on the $i$-th step is precisely $\la_i^\prime$ (by
definition the length of column $i$).
Now define a {\em generalized Ferrers board of $\la$\/} to consist of $\la_1$
columns {\em of arbitrary non-negative length\/}, stacked on the
``steps'' of the Ferrers board of $\la$. \figref{fig:generalized-Ferrer}
gives an example of this concept, where $\la=(4,3,2)$.
\figref{fig:generalized-Ferrer}.a shows the Ferrers board of
$(4,3,2)$, while \figref{fig:generalized-Ferrer}.b shows a
generalized Ferrers board of $(4,3,2)$ with column-lengths $1,5,1,0$.
By the rows of such a generalized
Ferrers board of $\la$ we mean the rows (in the usual sense) that contain
at least one cell of the board (but may contain ``holes'', too). The
bottom-most row gets number $\la_1^\prime$, the other rows are numbered
``from bottom to top'', as indicated in
\figref{fig:generalized-Ferrer}.

%%%@@@ input from psf_gf
\begin{Figure}
\label{fig:generalized-Ferrer}
\centerline{\small a. Ferrers board of $(4,3,2)$
\hskip1cm
b. A generalized Ferrers board of $(4,3,2)$}
\begin{picture}(20,8)(0,-1)

\put(1,0.5){
	\begintableau{3}{4}
		\row{oooo}
		\row{ooo}
		\row{oo}
	\endtableau
	}

\put(13,0.5){
	\begintableau{5}{4}
		\row{-o--}\put(-4,4.2){\makebox(0,0)[br]{$-1$}}
		\row{-o--}\put(-4,3.2){\makebox(0,0)[br]{$0$}}
		\row{-o--}\put(-4,2.2){\makebox(0,0)[br]{$1$}}
		\row{-oo} \put(-4,1.2){\makebox(0,0)[br]{$2$}}
		\row{oo}  \put(-4,0.2){\makebox(0,0)[br]{$3$}}
		\linethickness{1mm}
		\put(0,0){\line(1,0){2}}\put(0.3,-0.8){$1$}\put(1.3,-0.8){$2$}
		\put(2,1){\line(1,0){1}}\put(2.3,0.2){$3$}
		\put(3,2){\line(1,0){1}}\put(3.3,1.2){$4$}
	\endtableau
	}

\end{picture}
\end{Figure}


%%%@@@ end input from psf_gf

Let $\si\in\SG{\la_1}$ ($\SG{\la_1}$ denotes the symmetric group of order
$\la_1$) be an arbitrary
permutation and $\ep\in\setof{-1,+1}^{\la_1}$ be an arbitrary vector
of $+1$'s and $-1$'s.
The {\em generalized Ferrers board with parameters $\parof{\la,\si,\ep}$\/}
is the generalized Ferrers board $F$ of $\la$ where the length
of the $j$-th column of $F$, denoted by $\cl_j(F)$, is given by
\begin{equation}
\label{eq:column-length}
\cl_j(F) = \la^\prime_j-j+\si\of j\cdot\ep_{\si\of j}.
\end{equation}
An example with $\la=(4,3,2)$, $\si=1423$, and $\ep=(-1,1,1,1)$ is
displayed in \figref{fig:generalized-Ferrer}.b.
Note that for
$\si=\idty$ and $\ep=(1,1\dots,1)$ the generalized Ferrers board with
parameters $\parof{\la,\si,\ep}$ is precisely the (ordinary) Ferrers board
of $\la$. 

Our next definition concerns fillings of the cells of generalized
Ferrers boards.
A {\em  $(2n+1)$-tabloid $\tableau$
with parameters
$\parof{\la,\si,\ep}$\/} (we shall frequently write
$\tableau\of{\la,\si,\ep}$ for short) is a filling of the cells of the
generalized 
Ferrers board with parameters $\parof{\la,\si,\ep}$ with
elements from
$\setof{1,2\dots,2n,\infty}$ such that columns are weakly increasing,
and {\em strictly\/} increasing 
with respect to the finite entries. So, entry $\infty$ may occur more than
once in any column, occupying the bottom entries. 
When we 
display
a tabloid we indicate the corresponding
permutation $\si$ by writing $\si(j)$ beneath the $j$-th
column, and the corresponding
sign vector $\ep$ by writing the sign of
$\ep_{\si(j)}$ beneath the $j$-th column.
For example, the second tabloid in the second row of
\figref{fig:example-bijection} is a tabloid with parameters
$\big((3,3,2,2,1,1), 132, (-1,+1,+1)\big)$.

 The {\em signed weight\/} of
a tabloid $\tableau\of{\la,\si,\ep}$ is defined by
\begin{equation}
\label{eq:tabloid-weight}
w\of{\tableau\of{\la,\si,\ep}} \defeq\sgnof\si\cdot\sgnof\ep
	\cdot\prod_{i=1}^n x_i^{\numofentries{2i-1}{\tableau}-
		\numofentries{2i}{\tableau}},
\end{equation}
where $\sgnof\ep\defeq\prod_{k=1}^{\la_1}\ep_k$.

Now we are in the position to say what the sets $\sundaram$ and
$\proctor$ are.
We define the set $\sundaram$, and also the set $\proctor$,
to be the set of all $(2n+1)$-tabloids
$\tableau\of{\la,\si,\ep}$, where $\si$ varies over $\SG{\la_1}$ and
$\ep$ varies over $\setof{-1,+1}^{\la_1}$ arbitrarily.
For notational reasons we consider $\sundaram$ and $\proctor$ as
two different sets. 
Observe that the set $\score$ of all $(2n+1)$-Sundaram 
tableaux of shape $\la$, 
viewed as tabloids with parameters $\si=\idty$ and
$\ep=(1,1,\dots,1)$, 
is indeed contained in $\sundaram$, as well as the set $\pcore$ of
$(2n+1)$-Proctor tableaux of shape $\la$, 
viewed as tabloids with parameters $\si=\idty$ and
$\ep=(1,1,\dots,1)$, 
is contained in $\proctor$.
Of course, the weights $w_\sundaram$ and $w_\proctor$ 
are given by $w$
in \eqref{eq:tabloid-weight}.

Since $\sundaram$ and $\proctor$ are the same set,
we do not have to say much about the bijection $\abij$. It is simply
the identity mapping.
Note that this reduces the bijection $g$ given by the involution
principle to 
$x\mapsto (\ione\circ\itwo)^n(x)$,
for some $n$
depending on $x$. In plain words, $\itwo$ and $\ione$ are applied
alternatively to $x$, as long as possible.

Finally we have to specify the involutions $\ione$ and $\itwo$ on the sets
$\shull=\sundaram\setminus \score$ and 
$\phull=\proctor\setminus \pcore$ of bad elements,
respectively. This is done in full detail in the next section.
Actually, we partition $\shull$ into two disjoint subsets
$\shull^1$ and $\shull^2$ and construct involutions
$\ione^1$ and $\ione^2$ for each of these. The involution $\ione$ then
is simply the union of these two ``subinvolutions". Similarly, 
we partition $\phull$ into three disjoint subsets
$\phull^1$, $\phull^2$, $\phull^3$ and construct involutions
$\itwo^1$, $\itwo^2$, $\itwo^3$ for each of these. The involution $\itwo$ then
is the union of these three ``subinvolutions".

%%%@@@ end input from ps_obj
\section{The involutions}

%%%@@@ input from ps_invol
\label{sec:involutions}
We start with the involution $\ione:\;\shull\mapsto\shull$, since it is
much easier to describe.

\subsection{The involution $\ione$}\label{ione}
As promised in the previous section, we partition $\shull$ into two
disjoint subsets $\shull^1$ and $\shull^2$, and construct involutions
$\ione^1$ and $\ione^2$ on each of these. To define $\shull^1$ and
$\shull^2$ we introduce two types of {\em violations\/}. Then, given
some tabloid in $\shull$, we look at all the violations in the tabloid
(there will be at least one), and pick up the ``minimal" one (in a
certain sense). If this violation is of the first type then the
tabloid belongs to $\shull^1$, otherwise it belongs to $\shull^2$.

The first type of violation stems from \eqref{eq:sundaram-constraint}.
We define a {\em Sundaram violation in row $i$\/} of a
$(2n+1)$-tabloid $T$ to be
\begin{equation}
\label{eq:general-sundaram-constraint}
T_{i,1} < 2i-1 \text{ or cell $(i,1)$ does not belong to $T$ }.
\end{equation}

To motivate the second type of violation,
recall that for Sundaram tabloids we required that columns are weakly
increasing, and {\em strictly\/}
increasing with respect to the finite entries (so that entry $\infty$
is allowed to occur more than once in any column). However, we did not
say anything about the rows. So, tabloids might very well
have nonincreasing rows or more than one entry $\infty$ in some row,
both of which is forbidden for Sundaram tableaux.
This leads us to define a
{\em violation of row monotonicity in row $i$ and column $j$\/} of a
$(2n+1)$-tabloid $T$ to be 
the following:
\begin{tabular}{rll}
\hskip10pt (1)&cell $(i,j)$ belongs to $T$, $j<\la_1$ and $i\le
\la_{j+1}'$, \hbox{\hskip-7cm}\\
\hskip10pt (2)&either&$T_{i,j} > T_{i,j+1}$, \\
&or&$T_{i,j} = T_{i,j+1} = \infty$,\\
&or&cell $(i,j+1)$ does not belong to $T$.
\end{tabular}


Now, given a $(2n+1)$-tabloid $T$ in $\shull$, choose the maximal row, row
$I$ say, containing either a Sundaram or a row monotonicity violation,
and choose the minimal column, column $J$ say, which contains a
violation in row $I$. (Of course, Sundaram violations always occur in
column $1$.) If in this row and column we have a Sundaram violation,
then $T$ belongs to $\shull^1$, otherwise $T$ belongs to $\shull^2$.

There is one thing to check: We must show that {\em all\/} 
tabloids in $\shull$ do in fact contain either a Sundaram or a row
monotonicity violation. Note that the tabloids in
$\score=\sundaram\setminus\shull$ 
% markus: basically ist doch hier nicht noetig?
%Christian: Eigentlich doch, da sich die tabloids durch ihre
% zusaetzlichen Parameter von den Sundaram tableaux unterscheiden. Ich
% weiss allerdings nicht, ob diese Genauigkeit nicht eher die
% Verwirrung erhoeht.
(basically the Sundaram tableaux) 
all have $\la=\idty$ and
$\ep=\parof{1,1,\dots,1}$ and no
monotonicity violations and no Sundaram violations. Therefore it suffices
to show that there are {\em no\/} tabloids in $\sundaram$ with $\la\neq\idty$
{\em or\/} $\ep\neq\parof{1,1,\dots, 1}$ that {\em do not\/} exhibit any of
the above violations.

Suppose there is a tabloid $T\of{\la,\si,\ep}$ with {\em no\/}
monotonicity violation. In particular, there is {\em no\/} entry
$T_{i,j}$, with $j<\la_1$ and $i\le \la'_{j+1}$, for which cell
$(i,j+1)$ does not belong 
to $T$. This implies that the
{\em upper edge\/} of the generalized Ferrers board of $\la$ (forming
the shape of $T$) consists {\em only of ascending steps\/}. 
In terms of $\ep$ and $\si$ this means the following:
Let $\ep_{i_1} = \dots =\ep_{i_k}=-1$,
$\ep_{j_1} = \dots=\ep_{j_l}=1$,  $i_1<i_2<\dots i_k$,
$j_1<j_2\dots<j_l$, $k+l=\la_1$. Then because of
\eqref{eq:column-length} we must have
\begin{align}
\si\of 1 = i_k,\;    \si\of 2 = i_{k-1}, &\dots,\;\si\of k = i_1, \notag\\
\si\of{k+1} = j_1,\; \si\of{k+2} = j_2,  &\dots,\;\si\of{\la_1} = j_l.
\label{eq:ascending-permutation}
\end{align}
Hence, if {\em some\/} $\ep_j=-1$, then also
$\ep_{\si\of 1}=-1$. But then cell $(1,1)$ does not belong to $T$, which is a
Sundaram violation. On the other hand, if all $\ep_j=1$, then
we must have $\si=\idty$ by \eqref{eq:ascending-permutation}.

Finally we define the involutions $\ione^1$ on $\shull^1$ and
$\ione^2$ on $\shull^2$. We start with the definition of $\ione^1$. This
mapping relies on an operation which we call {\em inversion up to
$2c$ of a column $j$\/} of a tabloid $T$, where $c$ is some positive
integer $\le n$. (Also $\itwo^3$ will rely on this operation.) 
This is defined as follows. If for some $p$, $1\leq p\leq c$, {\em both\/}
entries $2p-1$ and $2p$ occur in column $j$ then {\em remove\/} both of them.
If {\em neither\/} $2p-1$ {\em nor\/} $2p$ is in column $j$ then
{\em insert\/} both of them. 
Apart from that,
let the remaining entries remain
unchanged (these are 
the ones where either $2p-1$ or $2p$ are in column $j$, but not both).
Of course, put all the obtained entries, old and new, into increasing
order to form the new column $j$. \figref{fig:column-inversion}
shows an example of an inversion up to $2c$ with $c=4$.
Observe that inversion up to $2c$ is a weight-preserving operation,
since a pair of entries $2k-1,2k$ contributes $x_kx_k^{-1}=1$ to the
weight of the tableau, hence it can be removed or inserted without
change of the weight. Observe further that it is an involution. Another
application to the same column $j$ restores the original situation.


%%%@@@ input from psf_ci
\begin{Figure}
\label{fig:column-inversion}
\centerline{\small Example of the ``inversion up to $2c$''}
\begin{picture}(27,8)(0,-1)

\put(9,0){
	\begintableau{4}{1}
		\row{1}
		\row{2}
		\row{6}
		\row{9}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
	\endtableau
	}

\put(13,3){$\longrightarrow$}

\put(17,0){
	\begintableau{6}{2}
		\row{3}
		\row{4}
		\row{6}
		\row{7}
		\row{8}
		\row{9}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
	\endtableau
	}

\end{picture}
\end{Figure}


%%%@@@ end input from psf_ci

For later use, we have to compute the effect of inversion up to $2c$
on the length of a column. Denote the tabloid that is obtained from
tabloid $T$ by inversion up to $2c$ of column $j$ by $\inverted{T}$.
Also denote the lengths of the $j$-th column of $T$ and $\inverted{T}$
by $\cl_j(T)$ and $\cl_j(\inverted{T})$, respectively.
Then it is obvious that 
\begin{equation}\label{eq:inversions-effect}
\numof{\setof{T_{.,j}\le 2c}}+\numof{\setof{\inverted{T}_{.,j}\le 2c}}=2c.
\end{equation}
Hence the difference of column lengths is
\begin{equation}
\label{eq:sundaram-length-difference}
\cl_j(\inverted{T}) - \cl_j(T) =
	2c - 2\cdot\numof{\leqincol{2c}{j}{T}}.
\end{equation}


Now we construct the involution $\ione^1$ on $\shull^1$. Let
$T(\la,\si,\ep)$ be a tabloid in $\shull^1$, with the Sundaram
violation in row $I$ that was selected before. Invert column $1$ up to
$2I-2$, leave $\si$ unchanged, and replace $\ep_{\si(1)}$ by
$-\ep_{\si(1)}$. Let $\inverted{T}$ be the new tabloid.
For examples, see operations 2, 6, 10, 14 and 18 in
\figref{fig:example-bijection}.

We have to check that $\ione^1$ is well-defined, and a weight-preserving
and sign-reversing involution. For being well-defined, we have
to check that $\inverted{T}$ lies in $\shull^1$, i.e, that it lies in
$\sundaram$ and contains a Sundaram violation that is minimal with
respect to our selection scheme. To show that
$\inverted{T}$ lies in $\sundaram$, we have to show that
column $j$ has the correct length with respect to
\eqref{eq:column-length}, for all $j$. Since only the first column was
changed it suffices to check the length of the first column of $\inverted{T}$.
The length of the first column of $T$ is
$\la^\prime_1-1+\si\of 1\cdot\ep_{\si\of 1}$, hence 
\begin{equation*}
\numof{\leqincol{2I - 2}{1}{T}} = I-1+\si\of{1}\ep_{\si\of{1}}.
\end{equation*}
According to \eqref{eq:sundaram-length-difference}, we therefore
have
\begin{equation*}
\cl_1(\inverted{T}) = \cl_1(T) - 2\si\of{1}\ep_{\si\of{1}}
= \la^\prime_1-1+\si\of 1(-\ep_{\si\of 1}),
\end{equation*}
which is exactly what we need.
That $\ione^1$ reverses sign is obvious since it changes
$\sgnof{\ep}$ and does not change $\sgnof{\si}$. 
It is weight-preserving since inversion
up to $2c$ has these
properties. Moreover, it is clear that 
$\inverted{T}$ again satisfies
$\inverted{T}_{I,1}< 2I - 1$ or that cell $(I,1)$ does not belong to
$\inverted{T}$. And, $I$ is maximal with respect to this property.
Therefore, if $\ione^1$ is applied to $\inverted{T}$ 
then we recover
$T$, since it means applying inversion up to $2I-2$ to the first
column again, and this restores the original object, as we already
observed. Hence, $\ione^1$ is an involution.

Now we define $\ione^2$. Let $T(\la,\si,\ep)$ be a $(2n+1)$-tabloid in
$\shull^2$ with the violation of row monotonicity in row $I$ and
column $J$ that 
were selected above. 
Again, think of a tabloid $\tableau$ as heaps of
cells (making up the columns) side by side. Now, interchange the heap
``above and including
$\tableau_{I-1,J}$'' and the heap ``above and including $\tableau_{I, J+1}$'',
replace $\si$ by
$\si^\prime\defeq\si\circ(J,J+1)$,
and leave $\ep$ unchanged. This
yields some new object $\ione^2\of T=T^\prime\of{\la,\si^\prime,\ep}$.
This
operation is shown with some examples in \figref{fig:crossing-over},
only displaying the involved columns. See operations 4, 8 and 16
in \figref{fig:example-bijection} for more elaborate examples. Note
that the effect on the numbers
beneath the displayed tabloid that encode the permutation $\si$
is that those beneath the $J$-th and the $(J+1)$-st
column are simply interchanged, the same being true for the 
the $+$'s and $-$'s 
encoding $\ep$.


%%%@@@ input from psf_co
\begin{Figure}
\label{fig:crossing-over}
\centerline{\small Example of the operation defined for
violations of row monotonicity}
\begin{picture}(27,7.5)(0,-1)

\put(0,1){
	\begintableau{3}{2}
		\row{23}
		\row{54}
		\row{99}
		\linethickness{1mm}
		\put(0,2){\line(1,0){1}}
		\put(1,2){\line(0,-1){1}}
		\put(1,1){\line(1,0){1}}
		\printrow{-1}{++}
		\printrow{-2}{21}
	\endtableau
	}

\put(2.8,2.5){$\leftrightarrow$}

\put(4,1){
	\begintableau{4}{2}
		\row{3}
		\row{4}
		\row{52}
		\row{99}
		\printrow{-1}{++}
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		\linethickness{1mm}
		\put(0,2){\line(1,0){1}}
		\put(1,2){\line(0,-1){1}}
		\put(1,1){\line(1,0){1}}
	\endtableau
	}

\put(10,1){
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		\row{-6}
		\row{77}
		\row{\infty\infty}
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		\put(0,1){\line(1,0){1}}
		\put(1,1){\line(0,-1){1}}
		\put(1,0){\line(1,0){1}}
		\printrow{-1}{+-}
		\printrow{-2}{32}
	\endtableau
	}

\put(12.8,2.5){$\leftrightarrow$}

\put(14,1){
	\begintableau{5}{2}
		\row{3-}
		\row{6-}
		\row{7-}
		\row{\infty-}
		\row{\infty7}
		\printrow{-1}{-+}
		\printrow{-2}{23}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
		\put(1,1){\line(0,-1){1}}
		\put(1,0){\line(1,0){1}}
	\endtableau
	}

\put(20,1){
	\begintableau{4}{2}
		\row{3}
		\row{6}
		\row{7}
		\row{8}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
		\put(1,1){\line(0,-1){1}}
		\put(1,0){\line(1,0){1}}
		\printrow{-1}{-+}
		\printrow{-2}{13}
	\endtableau
	}

\put(22.8,2.5){$\leftrightarrow$}

\put(24,1){
	\begintableau{3}{2}
		\row{-3}
		\row{-6}
		\row{87}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
		\put(1,1){\line(0,-1){1}}
		\put(1,0){\line(1,0){1}}
		\printrow{-1}{+-}
		\printrow{-2}{31}
	\endtableau
	}


\end{picture}
\end{Figure}


%%%@@@ end input from psf_co

We have to check that our new object $T'(\la,\si',\ep)$ is again a
$(2n+1)$-tabloid in $\shull^2$, and that $\ione^2$ is a
weight-preserving and sign-reversing involution. First, $T'$ has the
correct column length with respect to \eqref{eq:column-length} and is
therefore an element of $\sundaram$. For, by construction we have
\begin{align*}
\cl_J(T^\prime) &= \parof{\la_J^\prime - I + 1} +
	\parof{\cl_{J+1}(T) - \parof{\la_{J+1}^\prime - I}}  \\
	& = \la^\prime_J-J+\si\of{J+1}\cdot\ep_{\si\of{J+1}} \\
	& =	\la^\prime_J-J+\si^\prime\of J\cdot\ep_{\si^\prime\of J},
\end{align*}
and an analogous computation yields that column $J+1$ has the correct
length, too. Second, $T'$ again shows a violation of row monotonicity
in the same row $I$ and same column $J$ (just look at the pictures of
\figref{fig:crossing-over}). Besides, since the entry in cell $(I,J)$
and entries to the left or in lower rows were not changed, this
violation is again the ``minimal" with respect to our selection
scheme. Therefore $T'(\la,\si',\ep)$ is an element of $\shull^2$.

The observation that $T'(\la,\si',\ep)$ shows a violation of row
monotonicity at the same place as $T(\la,\si,\ep)$, implies that
if $\ione^2$ is applied to $T'(\la,\si,\ep)$ 
then one recovers
$T(\la,\si,\ep)$. Therefore $\ione^2$ is an involution on $\shull^2$.
That $\ione^2$ is weight-preserving is obvious. It reverses sign since
$\ep$ is not changed and $\si$ and $\si'$ are related by a
transposition and hence have different sign.



\subsection{The involution $\itwo$}\label{itwo}
Proctor tableaux are somewhat more complicated than Sundaram tableaux,
and so is the corresponding involution $\itwo$ on $\phull$.
As promised, we shall partition
the set $\phull$ 
into
three disjoint subsets $\phull^1$, $\phull^2$,
and $\phull^3$ and construct involutions $\itwo^1$, $\itwo^2$, and $\itwo^3$
for each of these, respectively. The involution $\itwo$ then is the
union of these three subinvolutions. 

First, we define the set $\phull^1$. 
$\phull^1$ consists of all $(2n+1)$-tabloids with more
than $2\si\of j - 1$ entries $\infty$ in column $j$ or with $\ep_{\si\of j} = -1$
for some $j$. (In terms of the preceding subsection these properties are
considered as certain {\em violations\/} of
``Proctor-tableau-ness''.)
In symbols,
\begin{equation}
\label{eq:phull1}
\phull^1\defeq\setof{T\of{\la,\si,\ep}\in\phull:\;
	\parof{\exists j:\;\inftiesincolumn{T}{j}> 2\si\of j -1}
		\text{ or }
	\parof{\exists j:\; \ep_{\si\of j}=-1}
}.
\end{equation}


We define operation $\itwo^1:\;\phull^1\rightarrow\phull^1$ as follows:
Given some $T\in\phull^1$, search for $J$ {\em minimal\/} with
$\inftiesincolumn{T}{J}> 2\si\of J - 1$ or $\ep_{\si\of J} = -1$. If $\ep_{\si\of J} = -1$,
then insert $2\si\of J$ entries $\infty$ at the bottom of column $J$, else
remove $2\si\of J$ entries $\infty$ from column $J$. In both cases,
leave $\si$ unchanged, and
replace $\ep_{\si\of J}$ by $-\ep_{\si\of J}$. For elaborate examples, see operations 1, 7,
9, 15 and 
17 in \figref{fig:example-bijection}.

We have to check that this operation is a
weight-preserving and sign-reversing involution on $\phull^1$.
$\itwo^1$ is clearly an involution, and it is easy to see that the
combined effect of insertion/removal of entries in column $J$
together with the change of $\ep_{\si\of J}$ yields a new column of the correct
length (with respect to \eqref{eq:column-length}), which either
contains more than $2\si\of J - 1 $ entries $\infty$
or has  $\ep_{\si\of J}=-1$. So for every $T\in\phull^1$ we have
$\itwo^1\of T\in \phull^1$.
$\itwo^1$ clearly is weight-preserving since entries $\infty$ do not contribute
to the weight at all. It is sign-reversing since $\sgnof{\ep}$ is
changed and $\si$ is left 
unchanged.

Next we define set $\phull^2$.
$\phull^2$ consists of all $(2n+1)$-tabloids {\em not in\/}
$\phull^1$ where the number of entries $\infty$ is greater than $1$ and less
than $2\si\of J - 2$ in some column $J$. (In terms of the preceding
subsection, this property is another
{\em violation\/} of ``Proctor-tableau-ness''.)
In symbols,
\begin{equation}
\label{eq:phull2}
\phull^2\defeq\setof{T\of{\la,\si,\ep}\in\phull\setminus\phull^1:\;
	\parof{\exists J:\;\inftiesincolumn{T}{J}\in\brkof{2, 2\si\of J -3}}
}.
\end{equation}
(Observe, that 
$\ep$ actually equals $(1,1,\dots,1)$ for all elements
of $\phull^2$.)

For this set we define mapping $\itwo^2$ as follows:
Given some $T\in\phull^2$, search for column $J$
{\em with $\si\of J$ minimal\/}, where
$\inftiesincolumn{T}{J}\in\brkof{2, 2\si\of J -3}$. Write $l$ for
$\numof{\setof{T_{.,J}=\infty}}$. Suppose, $l$
lies in the set
\begin{equation*}
S_k\defeq\setof{2k, 2k+1}\cup\setof{2\si\of J - 2k - 2, 2\si\of J - 2k - 1}
\end{equation*}
for some $k$,
$1\leq k\leq\ceilof{\frac{\si\of J}{2}} - 1$. (It is obvious that the
sets $S_k$, $1\leq k\leq\ceilof{\frac{\si\of J}{2}} - 1$, form a partition
of the 
set $\setof{2, 3, \dots, 2\si\of J -3}$,
hence $k$ is uniquely determined.)
We can deduce immediately that
\begin{equation*}
\si\of J - 2k \geq \si\of J - 2\ceilof{\frac{\si\of J}{2}} + 2 \geq 1.
\end{equation*}
Therefore, there must exist an index $S$ with $\si\of S = \si\of J - 2k$.
Now remove $2k$ entries $\infty$ from column $J$, insert them in
column $S$, replace $\si$ by $\si'\defeq\si\circ(J,S)$, and
leave $\ep$ 
unchanged. This yields the new object $T'(\la,\si',\ep)$.
For elaborate examples see operations 3 and 5 in
\figref{fig:example-bijection}.

We have to check that this operation is a weight-preserving and
sign-reversing involution on $\phull^2$. 
It is easy to check that
$\itwo^2(T)=T'(\la,\si',\ep)$ has the correct column lengths with
respect to \eqref{eq:column-length} by using the fact that
$\ep_i=1$ for all $i$.
The number $\inftiesincolumn{T}{S}$ of entries $\infty$ in column $S$
(before application of mapping $\itwo^2$) lies in
$\brkof{0,1}\cup\brkof{2\si\of S - 2, 2\si\of S - 1}$, since otherwise
$J$ were not the column with $\si\of J$ minimal (or $T$ would not belong
to $\phull^2$ at all). Since $\si\of S = \si\of J - 2k$, we have
\begin{align*}
\inftiesincolumn{T'}{S}\in&
	\brkof{2k, 2k+1}\cup\brkof{2\si\of S + 2k - 2, 2\si\of S + 2k - 1} \\
	&=
	\brkof{2k, 2k+1}\cup\brkof{2\si\of J -2k - 2, 2\si\of J -2k - 1}\\
	&=
	\brkof{2k, 2k+1}\cup\brkof{2\si'\of S -2k - 2, 2\si'\of S -2k - 1},
\end{align*}
and
\begin{align*}
\inftiesincolumn{T'}{J} \in&
	\brkof{0, 1}\cup\brkof{2\si\of J - 4k - 2, 2\si\of J - 4k - 1} \\
	&=
	\brkof{0, 1}\cup\brkof{2\si'\of J - 2, 2\si'\of J - 1}.
\end{align*}
% markus: Schlage vor
%All the other columns except
% anstatt
%Christian: Da gefaellt mir meine Version besser. Insbesondere bezieht
% sich Dein ``other" auf nichts, das vorher erwaehnt worden waere.
All the columns other than columns
$J$ and $S$ were not changed at all.
Therefore $T'$ is again an element of $\phull^2$. Besides, in the set
of all columns of $T'$ having the defining property of
\eqref{eq:phull2}, column $S$ is obviously the one with $\si'(S)$
minimal. So, by applying $\itwo^2$ to $T'(\la,\si',\ep)$ we would
recover $T(\la,\si,\ep)$. Hence, $\itwo^2$ is an involution.
$\itwo^2$ is clearly weight-preserving, since moving around entries
$\infty$ has no influence on the weight. It is sign-reversing, since
the the sign of $\si$ and $\si'$ is different and $\ep$ was left 
unchanged.

The ``remaining'' tabloids
$\phull\setminus\parof{\phull^1\cup\phull^2}$ constitute our set
$\phull^3$. They are
$(2n+1)$-tabloids $T\of{\la,\si,\ep}$ with $\ep=\parof{1,1,\dots,1}$
(since otherwise $T$ were in $\phull^1$) and $\inftiesincolumn{T}{j}\in
\brkof{0,1}\cup\brkof{2\si\of j - 2, 2\si\of j - 1}$ (since otherwise
$T$ were in $\phull^1$ or in $\phull^2$) that are no $(2n+1)$-Proctor
tableaux. 
Instead of directly working with these tabloids, it is more convenient
to map them by a weight- and sign-preserving bijection, $P$ say, to
another set, $\mphull$ say. 
$\mphull$ consists of fillings of slightly different
generalized Ferrers boards of $\la$. If $\si$ is a permutation in
$\SG{\la_1}$ and $\ep$ is a sign-vector in
$\setof{1}\times\setof{1,-1}^{\la_1-1}$ (note the difference to the
previous definition of vector $\ep$) then for $\mphull$ we
consider generalized Ferrers boards $F$ of $\la$ where the length of
the $j$-th column of $F$ is given by
\begin{equation}
\label{eq:proctor-column-length}
\cl_j(F) = \lambda^\prime_j-j+\parof{\si\of j - 1}\ep_{\si\of j} + 1.
\end{equation}
(Note the difference to
\eqref{eq:column-length}). By definition,
$\mphull$ consists of all fillings $T$ of a generalized
Ferrers board $F$ with parameters $(\la,\si,\ep)$, but with
column-lengths according to 
\eqref{eq:proctor-column-length}, with elements from
$\setof{1,2,\dots,2n,\infty}$, that are strictly increasing along
columns 
% markus: Hier gefiele mir
%which 
% besser als
%Christian: Das waere aber nicht so gut, da sich das ``which", so wie
% es dasteht, eigentlich auf 
% auf die ``columns" beziehen wuerde. Vielleicht also so:
and that
are no Proctor tableaux. (Note that column-strictness
implies that in particular there is at most {\em one\/} entry $\infty$ in any
column.) To differentiate from our previous tabloids, we call objects
in $\mphull$ {\em new $(2n+1)$-tabloids\/}. 
The definition of signed weight for a tabloid in $\mphull$ is
quite similar to \eqref{eq:tabloid-weight}, 
\begin{equation}
\label{eq:modified-tabloid-weight}
w\of{\tableau\of{\la,\si,\ep}} \defeq\sgnof\si
	\prod_{i=1}^n x_i^{\numofentries{2i-1}{\tableau}-
		\numofentries{2i}{\tableau}}.
\end{equation}
Note, however, that in contrast to \eqref{eq:tabloid-weight} vector
$\ep$ does not contribute to the sign here.

The bijection $P$ between $\varproctor$ and $\mphull$ is
defined as follows.
Consider some tabloid $T\of{\la,\si,\ep}\in\varproctor$. As mentioned
above, we have $\ep=({1,1,\dots,1})$ for all
tabloids in $\varproctor$. Now, for {\em every\/} column $j$ with
$\si\of j > 1$ and $\inftiesincolumn{T}{j}\geq 2\si\of j - 2$, remove
$2\si\of j - 2$ entries $\infty$ and set $\ep_{\si\of j}=-1$.
For an example, see operation 11 in \figref{fig:example-bijection}.
This is indeed a bijection. The inverse mapping is easily described.
Given a tabloid $T\of{\la,\si,\ep}\in\mphull$,
insert $2\si\of j - 2$ entries $\infty$ in every column $j$ with
$\ep_{\si\of j} = -1$.

Recalling the definitions of signed weights \eqref{eq:tabloid-weight}
and \eqref{eq:modified-tabloid-weight} as well as the fact that entries
$\infty$ do not contribute to the weight, it is easy to see that $P$ is
a weight- and sign-preserving bijection between
$\varproctor$ and $\mphull$.

So the remaining task is to give a weight-preserving and sign-reversing
involution on $\mphull$, which is actually the
most intricate part. 

Similar to the definition of $\ione$, we first define a list of {\em
violations\/} for new tabloids. Then, among all violations we pick up
the ``minimal" (in some sense), and finally we
apply a certain operation which depends on the type of violation that
we are considering.

To motivate our 
definitions of
violations, observe that $\mphull$
consists precisely
of those new $(2n+1)$-tabloids $\parof{T,\si,\ep}$ where either
$\si\neq\idty$ or $\ep\neq(1,1,\dots, 1)$ or
there is some violation of row monotonicity or some violation of the orthogonal
or protection conditions. 

We define a {\em violation of row monotonicity at $p$\/}
of a new $(2n+1)$-tabloid $T$ to be:
\begin{tabular}{rll}
\hskip10pt (1)&cell $(i,j)$ belongs to $T$, $j<\la_1$, $i\le
\la_{j+1}'$, and $T_{i,j}=p$, \hbox{\hskip-8.5cm}\\
\hskip10pt (2)&either&$T_{i,j} > T_{i,j+1}$, \\
&or&cell $(i,j+1)$ does not belong to $T$.
\end{tabular}

\noindent
(Note that here we {\em do not\/} consider two entries $\infty$ in the
same row as a violation since this is not forbidden for Proctor
tableaux.) 

The other types of violations (and their namings) stem from
\cite{Fulmek-Krattenthaler:Lattice-path} and come from the underlying
lattice path interpretations given there. This is commented in
section~\ref{sec:explanation}. 

We say that a new $(2n+1)$-tabloid $T(\la,\si,\ep)$ contains an
{\em even reflected intersection at $2c$}, for some $c$ with $1\le c\le n$, if
\begin{equation}\label{eq:even-reflected-intersection}
\numof{\leqincolT{2c}{1}} + \numof{\leqincolT{2c}{2}} =
2c + \parof{\si\of 1 - 1}\ep_{\si\of 1} + \parof{\si\of 2 - 1}\ep_{\si\of 2}.
\end{equation}
For an example consider the second tabloid in the third row of
\figref{fig:example-bijection}. It contains an even reflected intersection
at $6$.

We say that a new $(2n+1)$-tabloid $T(\la,\si,\ep)$ contains an {\em
odd reflected 
crossing at $2c-1$}, for some $c$ with $1\le c\le n$, if the first two
columns of $T$ contain neither $2c-1$ nor $2c$ and
\begin{equation*}
\numof{\leqincolT{2c-2}{1}} + \numof{\leqincolT{2c-2}{2}} =
	2c + \parof{\si\of 1 - 1}\ep_{\si\of 1} +
		\parof{\si\of 2 - 1}\ep_{\si\of 2} - 1.
\end{equation*}
The right-most tabloid in
\figref{fig:odd-meeting-odd-crossing} contains an odd reflected
crossing at $6$.

And, we say that a new tabloid $T(\la,\si,\ep)$ contains an {\em odd
reflected 
violation at $2c-1$}, for some $c$ with $1\le c\le n$, if the following 
specialization of \eqref{eq:protection-condition} holds 
(the additions are underlined in the text; note that they imply that
there is {\em no\/} even reflected intersection
at $2c$ or $2c-2$):
\begin{align*}
i+j=2c&\land \tableau_{i,1}=2c-1 \underline{\land
\tableau_{i+1,1}>2c}\\ 
		&\land
\tableau_{j,2}=\dots=\tableau_{j,h-1} = 2c-1 \hbox{ (for $h\geq2$)}\\
	&\underline{\land
\tableau_{j+1,2}=\dots=\tableau_{j+1,h-1} > 2c \hbox{ (for $h\geq2$)}}\\
	&\land
\tableau_{j,h} = 2c	\underline{\land \tableau_{j-1,h} < 2c-1}.
\end{align*}
The left-most tabloid in
\figref{fig:odd-meeting-odd-crossing} contains an odd reflected
violation at $6$ (choose $i=4$, $j=2$, $h=3$).

The important fact is that every element of $\mphull$
contains either a row monotonicity violation, or an even reflected
intersection, or an odd reflected crossing, or an odd reflected
violation. For, suppose that there is $T(\la,\si,\ep)$ in
$\mphull$  containing 
none 
of these violations. Then we claim that
there holds
\begin{equation}
\label{eq:generalized-orthogonal-condition}
\numof{\leqincolT{2c}{1}} + \numof{\leqincolT{2c}{2}} <
2c + \parof{\si\of 1 - 1}\ep_{\si\of 1} + \parof{\si\of 2 - 1}\ep_{\si\of 2},
\end{equation}
for $c=0,1,\dots,n$. (Note that this inequality generalizes the
orthogonal condition \eqref{eq:orthogonal-condition} for $p=2c$, to
which it reduces for $\si=\idty$ and $\ep=(1,1,\dots,1)$.) In fact, these
inequalities follow from the observation that 
\begin{align*}
\numof{\leqincolT{\infty}{1}}+&\numof{\leqincolT{\infty}{2}}=
\cl_1(T)+\cl_2(T)\\
&\hskip-1.5cm=\la_1'-1+(\si(1)-1)\ep_{\si(1)}+1+\la_2'-2+(\si(2)-1)\ep_{\si(2)}+1
\quad \quad \text{(by \eqref{eq:proctor-column-length})}\\
&\hskip-1.5cm<2n+1+(\si(1)-1)\ep_{\si(1)}+(\si(2)-1)\ep_{\si(2)}\quad \quad
\text{(since $\la_1'+\la_2'\le 2n+1$),}
\end{align*}
and a downward induction on $c$ which uses that $T$ contains neither
an even intersection nor an odd reflected crossing. Since $T$ does not
contain any row monotonicity violation
by assumption, 
the arguments leading to
\eqref{eq:ascending-permutation} also apply here. Therefore, if there is
some $\ep_j=-1$ then we must have $\ep_{\si(1)}=-1$ by 
\eqref{eq:ascending-permutation}. The generalized orthogonal
condition \eqref{eq:generalized-orthogonal-condition} with $c=0$ then
enforces $\ep_{\si(2)}=1$. Therefore $\si(2)=1$ by 
\eqref{eq:ascending-permutation}, and so
\eqref{eq:generalized-orthogonal-condition} does not hold for $c=0$,
which is a contradiction. Hence we must have $\ep=(1,1,\dots,1)$
which in turn implies $\si=\idty$, again by 
\eqref{eq:ascending-permutation}. Summarizing, $T$ is a new
$(2n+1)$-tabloid with parameters $\si=\idty$ and $\ep=(1,1,\dots,1)$
(which by \eqref{eq:proctor-column-length} means that it is a filling
of the Ferrers board of $\la$), that satisfies the $(2c)$-th
orthogonal condition (since it satisfies
\eqref{eq:generalized-orthogonal-condition}) and the $(2c)$-th
protection condition (since it does not contain any odd reflected
violation), for all $c$. In short, $T$ is a $(2n+1)$-Proctor tableau,
which is a contradiction since Proctor tableaux are excluded from
$\mphull$. 

Finally we turn to the definition of $\itwo^3$ on $\mphull$.
Let $T(\la,\si,\ep)$ be in $\mphull$. Among all the
row monotonicity violations, even reflected
intersections, odd reflected crossings, odd reflected
violations, pick out these which are at $p$ with $p$ maximal, and
among all these pick the ones in the minimal column.
If we encounter a row monotonicity violation among the latter, then
$\itwo^3$ acts in the same way as $\ione^2$ did. In the same way as it
was done for $\ione^2$ and the column-lengths
\eqref{eq:column-length}, it is checked that this operation takes care
of the (new) column lengths \eqref{eq:proctor-column-length}. Besides,
the image under this operation will again contain a row monotonicity
violation, at the same $p$ and in the same column. Hence, by this
operation we obtain an element of $\mphull$, and by applying
$\itwo^3$ again we would 
recover our original tabloid.


%%%@@@ input from psf_oc
\begin{Figure}
\label{fig:orthogonal-crossing}
\centerline{\small Example for the operation in case of an even
reflected intersection}

\begin{picture}(18,6)(0,-1)

\put(2,0){
	\begintableau{4}{2}
		\row{13}
		\row{34}
		\row{45}
		\row{6}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
		\put(1,2){\line(1,0){1}}
	\endtableau
	}

\put(5.5,3){$\longrightarrow$}

\put(8,0){
	\begintableau{4}{2}
		\row{-1}
		\row{-2}
		\row{15}
		\row{6}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
		\put(1,2){\line(1,0){1}}
	\endtableau
	}

\put(11.5,3){$\longrightarrow$}

\put(14,0){
	\begintableau{3}{2}
		\row{11}
		\row{25}
		\row{6}
		\linethickness{1mm}
		\put(0,1){\line(1,0){1}}
		\put(1,2){\line(1,0){1}}
	\endtableau
	}

\end{picture}
\end{Figure}


%%%@@@ end input from psf_oc

If we do not encounter a row monotonicity violation but an even
intersection at $2c$, or one of the other two violations at $2c-1$,
then the {\em basic\/} idea for the definition of $\itwo^3$ is as follows:
Apply inversion up to $2c$ to both columns
$1$ and $2$ and {\em interchange\/} the two
``inverted heaps'', replace
$\si$ by $\inverted\si\defeq\si\circ(1,2)$, and replace
$\ep_{\si\of 1}$ and $\ep_{\si\of 2}$ by their negatives, except that
$\ep_1$ is not changed. ($\ep_1$ has to be $1$ by definition of
$\mphull$.) 
This operation is weight-preserving since inversion up to $2c$ is, and
it is sign-reversing since $\si$ and $\inverted{\si}$ differ in sign
and the weight \eqref{eq:modified-tabloid-weight} 
does not depend on vector $\ep$.

\figref{fig:orthogonal-crossing} gives an example for this operation
with $2c=4$.

In the case of an even reflected intersection we apply this idea
without further 
modification, thus obtaining
$\inverted{T}(\la,\inverted{\si},\inverted{\ep})$. For an elaborate
example, see operation 12 in \figref{fig:example-bijection}.

We have to check that
$\inverted{T}(\la,\inverted{\si},\inverted{\ep})$ is in
$\mphull$ and that application of $\itwo^3$ to
$\inverted{T}(\la,\inverted{\si},\inverted{\ep})$ would give back
$T(\la,\si,\ep)$. First, $\inverted{T}$ has the correct column lengths
with respect to \eqref{eq:proctor-column-length} since for
$\setof{i,j}=\setof{1,2}$ we have
\begin{align*}
\cl_i(\inverted{T}) &= 
\la'_i - i + \parof{\si\of i - 1}\ep_{\si\of i}  + 1 -
	\numof{\leqincol{2c}{i}{T}} + 2c - \numof{\leqincol{2c}{j}{T}}\\
&= \la'_i - i + \parof{\si\of i - 1}\ep_{\si\of i}  + 1 -
\parof{\parof{\si\of i - 1}\ep_{\si\of i} +
	\parof{\si\of j - 1}\ep_{\si\of j}}\\
&= \la'_i - i + \parof{\si\of j - 1}(-\ep_{\si\of j})  + 1\\
&= \la'_i - i + \parof{\inverted\si\of i -
	1}\inverted\ep_{\inverted\si\of i}  + 1.
\end{align*}
Note that the exceptional
case of $\ep_1$ (which must always equal $1$) does not matter in this
(and the following)
computation, since $\ep_1$ is always multiplied by $(1-1)=0$.
Moreover, 
\begin{align*}
\numof{\leqincol{2c}{1}{\inverted{\tableau}}} +
	\numof{\leqincol{2c}{2}{\inverted{\tableau}}} &=
4c - (2c + \parof{\si\of 1 - 1}\ep_{\si\of 1} +
	\parof{\si\of 2 - 1}\ep_{\si\of 2}) \\
&= 2c + \parof{\inverted{\si}\of 1 - 1}\inverted{\ep}_{\inverted\si\of 1} +
	\parof{\inverted{\si}\of 2 - 1}\inverted\ep_{\inverted{\si}\of 2},
\end{align*}
which says that in $\inverted T$
we encounter an even
reflected intersection again, for the same
value $2c$. Hence, $\inverted{T}(\la,\inverted{\si},\inverted{\ep})$ is in
$\mphull$. Besides, this operation did not introduce a row
monotonicity violation at $2c$ in the first column. For, suppose that
$l$ is minimal such that $T_{l,1}>2c$. Then we must have $T_{l,2}\ge
T_{l,1}>2c$ because $T$ does not contain a row monotonicity violation
at some integer $>2c$. Now, if we assume that $T_{l-1,2}\le 2c$, then
by using \eqref{eq:proctor-column-length} we obtain for the number of
entries $\le 2c$ in the first two columns of $T$
$$\numof{\leqincolT{2c}{1}} + \numof{\leqincolT{2c}{2}} =
l-1+(\si(1)-1)\ep_{\si(1)}+l-2+(\si(2)-1)\ep_{\si(2)},$$
which contradicts \eqref{eq:even-reflected-intersection} in parity.
Hence we have $T_{l-1,2}>2c$. So our operation cannot introduce a row
monotonicity violation at $2c$ in the first column of $\inverted T$. 
 Therefore, by
applying $\itwo^3$ to
$\inverted{T}(\la,\inverted{\si},\inverted{\ep})$ we 
recover
$T(\la,\si,\ep)$. 



In case of odd reflected violations and odd reflected crossings, the
basic idea of the 
operation as explained above is modified in the following way. Suppose
we encounter 
an odd reflected violation at $2c$. First, set $\tableau_{i,1}=2c$
and, to compensate for the according change in weight (with respect to
\eqref{eq:modified-tabloid-weight}), set
$\tableau_{j,h}=2c-1$. This amounts to an interchange
of entries $2c$ and $2c-1$ in $\tableau$ which {\em by assumption\/}
does not violate the
strict monotonicity in {\em columns\/}. Moreover, this interchange clearly
preserves weight. Next, {\em insert\/} $2c-1$ in column $1$ and $2c$ in
column $2$
(in the appropriate rows). This is possible {\em by assumption\/} without
violating column-strictness. Again, this insertion does not
change weight. Now apply the operation from above.
\figref{fig:odd-meeting-odd-crossing} gives an example for this
operation with $2c=6$, $i=4$, $j=2$, $h=3$.




%%%@@@ input from psf_oo
\begin{Figure}
\label{fig:odd-meeting-odd-crossing}
\centerline{\small Conversion of odd reflected violation into odd
reflected crossing}
\begin{picture}(30,9)(0,0)
\put(2,3){
	\begintableau{4}{3}
	\row{234}
	\row{356}
	\row{477}
	\row{5}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(0,0){\line(1,0){1}}
	\put(0,0){\line(0,1){1}}
	\put(1,1){\line(-1,0){1}}
	\put(1,1){\line(0,-1){1}}
	\put(1,2){\line(1,0){2}}
	\put(1,2){\line(0,1){1}}
	\put(3,3){\line(-1,0){2}}
	\put(3,3){\line(0,-1){1}}
	\endtableau
}

\put(7,5){$\rightarrow$}

\put(9,3){
	\begintableau{4}{3}
	\row{234}
	\row{355}
	\row{477}
	\row{6}
	\endtableau
}

\put(14,5){$\rightarrow$}

\put(16,3){
	\begintableau{5}{3}
	\row{23}
	\row{354}
	\row{465}
	\row{577}
	\row{6}
	\endtableau
}

\put(21,5){$\rightarrow$}

\put(23,3){
	\begintableau{4}{3}
	\row{--4}
	\row{125}
	\row{277}
	\row{3}
	\printrow{-1}{-++}
	\printrow{-2}{213}
	\endtableau
}

\end{picture}
\end{Figure} 

%%%@@@ end input from psf_oo

Again, we have to check that this operation gives us an element of
$\mphull$. 
To see this, we compute the resulting column
lengths. For $\setof{i,j}=\setof{1,2}$ we have
\begin{align*}
\cl_i(\inverted T) &= \la^\prime_i-i+\parof{\si\of i - 1}\ep_{\si\of i} + 1
-{\numof{\leqincolT{2c}{i}}} + 2c -
{\parof{\numof{\leqincol{2c}{j}{T}}+1}}\\
&= \lambda^\prime_i-i+\parof{\si\of i - 1}\ep_{\si\of i} + 1  -
	\parof{ \parof{\si\of i - 1}\ep_{\si\of i} +
					\parof{\si\of j - 1}\ep_{\si\of j} - 1} -1 \\
&= \lambda^\prime_i-i+\parof{\si\of j - 1}(-\ep_{\si\of j}) + 1.	\\
&= \lambda^\prime_i-i+\parof{\inverted\si\of i -
	1}\inverted\ep_{\inverted\si\of i} + 1.	
\end{align*}
Observe that neither column $1$ nor
column $2$ of the resulting tableau $\inverted{\tableau}$ contains
$2c-1$ or $2c$. Hence $\numof{\leqincol{2c}{1}{\inverted T}} +
\numof{\leqincol{2c}{2}{\inverted T}} = 
\numof{\leqincol{2c-2}{1}{\inverted T}} +
\numof{\leqincol{2c-2}{2}{\inverted T}}$ and so 
\begin{align*}
\numof{\leqincol{2c-2}{1}{\inverted T}} +&
\numof{\leqincol{2c-2}{2}{\inverted T}}\\
&={2c-\numof{\leqincolT{2c}{1}} - 1 + 
	2c-\numof{\leqincolT{2c}{2}} - 1}\\
&=2c - \parof{\si\of 1 - 1}\ep_{\si\of 1} -
	\parof{\si\of 2 - 1}\ep_{\si\of 2} - 1 \\
&=2c+ \parof{\inverted\si\of 1 - 1}\inverted\ep_{\inverted\si\of 1} +
	\parof{\inverted\si\of 2 - 1}\inverted\ep_{\inverted\si\of 2} - 1.
\end{align*}
So this operation converted an odd reflected violation at $2c$ into an
odd reflected crossing 
at $2c$. Hence, $\inverted{T}(\la,\inverted{\si},\inverted{\ep})$ is
indeed in $\mphull$. Also, this operation did not introduce
a row monotonicity violation at $2c-1$ in the first column (there is
no entry $2c-1$ in 
the first column of $\inverted T$), and, of course, no even intersection
at $2c-1$ (which simply does not exist). Therefore this odd reflected crossing
in $\inverted T$
is the ``minimal" violation according to our selection scheme. Since
$\itwo^3$ should become an involution, the definition of $\itwo^3$ for
odd reflected crossings has to be the inverse of the above described
conversion of an odd reflected violation into an odd reflected crossing.
A close look shows that this operation can indeed be reversed.
This completes the definition of the involution $\itwo^3$ and hence of
$\itwo$. 


%%%@@@ end input from ps_invol
\section{The example}

%%%@@@ input from ps_xampl
\label{sec:example}


%%%@@@ input from psf_ex
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Some adhoc macro-definitions: %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\newcount\localdummy
\localdummy=0
\def\printnextarrow#1#2#3{\global\advance\localdummy by 1
\put(#1,#2){
	\begin{picture}(1,1)(0,0)
	\put(0,0){$\overrightarrow{#3}$}
	\put(.05,1.2){$\scriptstyle\bold{(\number\localdummy)}$}
	\end{picture}
	}
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%     End of adhoc macros:      %
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}
\begin{Figure}
\label{fig:example-bijection}
\centerline{\small Example for the algorithmic bijection}
\def\ten{10}
\def\elv{11}
\begin{picture}(30,38.5)(0,0) % vertical spacing 9.5, horizontal 6
\put(0,29.5){
	\begintableau{6}{3}
	\row{14\infty}
	\row{67\infty}
	\row{9\ten}
	\row{\elv\elv}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(0,2){\line(1,0){1}}
	\put(0,2){\line(0,-1){2}}
	\put(0,0){\line(1,0){1}}
	\put(1,0){\line(0,1){2}}
	\endtableau
}

%\put(4.5,34.5){$\overrightarrow{\itwo^1}$}
\printnextarrow{4.5}{34.5}{\itwo^1}

\put(6,29.5){
	\begintableau{6}{3}
	\row{-4\infty}
	\row{-7\infty}
	\row{1\ten}
	\row{6\elv}
	\row{9}
	\row{\elv}
	\printrow{-1}{-++}
	\printrow{-2}{123}
	\put(-1.1,2.35){$\scriptstyle\bold{7>}$}
	\put(-0.8,2.5){\circle{1}}
	\linethickness{1mm}
	\put(0,3){\line(1,0){1}}
	\put(0,3){\line(0,-1){1}}
	\put(0,2){\line(1,0){1}}
	\put(1,2){\line(0,1){1}}
	\endtableau
}

%\put(10.5,34.5){$\overrightarrow{\ione^1}$}
\printnextarrow{10.5}{34.5}{\ione^1}

\put(12,29.5){
	\begintableau{6}{3}
	\row{14\infty}
	\row{37\infty}
	\row{4\ten}
	\row{6\elv}
	\row{9}
	\row{\elv}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(2,6){\line(1,0){1}}
	\put(2,6){\line(0,-1){2}}
	\put(2,4){\line(1,0){1}}
	\put(3,4){\line(0,1){2}}
	\endtableau
}

\put(15.,32.5){$\scriptstyle k=1,$}
\put(14.6,31.7){$\scriptstyle \si(S)=3-2k$}
\put(14.6,31.1){$\scriptstyle \hphantom{\si(S)}{}=1.$}
%\put(16.5,34.5){$\overrightarrow{\itwo^2}$}
\printnextarrow{16.5}{34.5}{\itwo^2}

\put(18,29.5){
	\begintableau{8}{3}
	\row{1}
	\row{3}
	\row{44}
	\row{67}
	\row{9\ten}
	\row{\elv\elv}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{321}
	\put(2.5,4.4){$\bold{!}$}
	\put(2.65,4.65){\circle{1}}
	\linethickness{1mm}
	\put(1,5){\line(1,0){1}}
	\put(2,5){\line(0,-1){1}}
	\put(2,4){\line(1,0){1}}
	\endtableau
}

%\put(22.5,34,5){$\overrightarrow{\ione^2}$}
\printnextarrow{22.5}{34,5}{\ione^2}

\put(24,29.5){
	\begintableau{8}{3}
	\row{1}
	\row{3}
	\row{4}
	\row{674}
	\row{9\ten}
	\row{\elv\elv}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{312}
	\linethickness{1mm}
	\put(0,2){\line(1,0){1}}
	\put(0,2){\line(0,-1){2}}
	\put(0,0){\line(1,0){1}}
	\put(1,0){\line(0,1){2}}
	\endtableau
}

\put(27.7,32.5){$\scriptstyle k=1,$}
\put(27.3,31.7){$\scriptstyle \si(S)=3-2k$}
\put(27.3,31.1){$\scriptstyle \hphantom{\si(S)}{}=1.$}
\printnextarrow{28.5}{34.5}{\itwo^2}

\put(0,20){
	\begintableau{7}{3}
	\row{-7}
	\row{1\ten}
	\row{3\elv4}
	\row{4\infty}
	\row{6\infty}
	\row{9}
	\row{\elv}
	\printrow{-1}{+++}
	\printrow{-2}{132}
	\put(-1.1,2.35){$\scriptstyle\bold{7>}$}
	\put(-0.8,2.5){\circle{1}}
	\linethickness{1mm}
	\put(0,3){\line(1,0){1}}
	\put(0,3){\line(0,-1){1}}
	\put(0,2){\line(1,0){1}}
	\put(1,2){\line(0,1){1}}
	\endtableau
}

%\put(4.5,25){$\overrightarrow{\ione^1}$}
\printnextarrow{4.5}{25}{\ione^1}

\put(6,20){
	\begintableau{7}{3}
	\row{-7}
	\row{-\ten}
	\row{-\elv4}
	\row{1\infty}
	\row{6\infty}
	\row{9}
	\row{\elv}
	\printrow{-1}{-++}
	\printrow{-2}{132}
	\linethickness{1mm}
	\put(0,-1){\line(1,0){1}}
	\put(0,-1){\line(0,1){1}}
	\put(1,0){\line(-1,0){1}}
	\put(1,0){\line(0,-1){1}}
	\endtableau
}

%\put(10.5,25){$\overrightarrow{\itwo^1}$}
\printnextarrow{10.5}{25}{\itwo^1}

\put(12,20){
	\begintableau{7}{3}
	\row{-7}
	\row{1\ten}
	\row{6\elv4}
	\row{9\infty}
	\row{\elv\infty}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{132}
	\linethickness{1mm}
	\put(1,5){\line(1,0){1}}
	\put(2,5){\line(0,-1){1}}
	\put(2,4){\line(1,0){1}}
	\endtableau
}

%\put(16.5,25){$\overrightarrow{\ione^2}$}
\printnextarrow{16.5}{25}{\ione^2}

\put(18,20){
	\begintableau{6}{3}
	\row{147}
	\row{6\elv\ten}
	\row{9\infty}
	\row{\elv\infty}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(0,2){\line(1,0){1}}
	\put(0,2){\line(0,-1){2}}
	\put(0,0){\line(1,0){1}}
	\put(1,0){\line(0,1){2}}
	\endtableau
}

%\put(22.5,25){$\overrightarrow{\itwo^1}$}
\printnextarrow{22.5}{25}{\itwo^1}

\put(24,20){
	\begintableau{6}{3}
	\row{-47}
	\row{-\elv\ten}
	\row{1\infty}
	\row{6\infty}
	\row{9}
	\row{\elv}
	\printrow{-1}{-++}
	\printrow{-2}{123}
	\put(-1.1,2.35){$\scriptstyle\bold{7>}$}
	\put(-0.8,2.5){\circle{1}}
	\linethickness{1mm}
	\put(0,3){\line(1,0){1}}
	\put(0,3){\line(0,-1){1}}
	\put(0,2){\line(1,0){1}}
	\put(1,2){\line(0,1){1}}
	\endtableau
}

%\put(28.5,25){$\overrightarrow{\ione^1}$}
\printnextarrow{28.5}{25}{\ione^1}

\put(0,10.5){
	\begintableau{6}{3}
	\row{147}
	\row{3\elv\ten}
	\row{4\infty}
	\row{6\infty}
	\row{9}
	\row{\elv}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(1,2){\line(1,0){1}}
	\put(1,2){\line(0,1){2}}
	\put(1,4){\line(1,0){1}}
	\put(2,4){\line(0,-1){2}}
	\endtableau
}

%\put(4.5,15.5){$\overrightarrow{P}$}
\printnextarrow{4.5}{15.5}{P}

\put(6,10.5){
	\begintableau{6}{3}
	\row{1-7}
	\row{3-\ten}
	\row{44}
	\row{6\elv}
	\row{9}
	\row{\elv}
	\printrow{-1}{+-+}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(0,2){\line(1,0){1}}
	\put(1,2){\line(0,1){1}}
	\put(1,3){\line(1,0){1}}
	\endtableau
}

\put(9.2,13.3){{\tiny even}}
\put(8.6,12.5){{\tiny reflected}}
\put(8.2,11.7){{\tiny intersection}}
\put(9.2,10.9){{\tiny at $6$}}
\printnextarrow{10.5}{15.5}{\itwo^3}

\put(12,10.5){
	\begintableau{7}{3}
	\row{1}
	\row{2-7}
	\row{41\ten}
	\row{56}
	\row{6\elv}
	\row{9}
	\row{\elv}
	\printrow{-1}{+++}
	\printrow{-2}{213}
	\endtableau
}

%\put(16.5,15.5){$\overrightarrow{\ione^1}$}
\printnextarrow{16}{15.5}{P^{-1}}

\put(18,10.5){
	\begintableau{7}{3}
	\row{1}
	\row{2-7}
	\row{41\ten}
	\row{56}
	\row{6\elv}
	\row{9}
	\row{\elv}
	\printrow{-1}{+++}
	\printrow{-2}{213}
	\put(-1.1,2.35){$\scriptstyle\bold{7>}$}
	\put(-0.8,2.5){\circle{1}}
	\linethickness{1mm}
	\put(0,3){\line(1,0){1}}
	\put(0,3){\line(0,-1){1}}
	\put(0,2){\line(1,0){1}}
	\put(1,2){\line(0,1){1}}
	\endtableau
}

%\put(16.5,15.5){$\overrightarrow{\ione^1}$}
\printnextarrow{22.5}{15.5}{\ione^1}

\put(24,10.5){
	\begintableau{6}{3}
	\row{--7}
	\row{-1\ten}
	\row{-6}
	\row{4\elv}
	\row{9}
	\row{\elv}
	\printrow{-1}{-++}
	\printrow{-2}{213}
	\linethickness{1mm}
	\put(0,-1){\line(1,0){1}}
	\put(0,-1){\line(0,1){1}}
	\put(1,0){\line(-1,0){1}}
	\put(1,0){\line(0,-1){1}}
	\endtableau
}

%\put(22.5,15.5){$\overrightarrow{\itwo^1}$}
\printnextarrow{28.5}{15.5}{\itwo^1}

\put(0,1){
	\begintableau{7}{3}
	\row{4}
	\row{9-7}
	\row{\elv1\ten}
	\row{\infty6}
	\row{\infty\elv}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{213}
	\linethickness{1mm}
	\put(0,3){\line(1,0){1}}
	\put(1,3){\line(0,-1){1}}
	\put(1,2){\line(1,0){1}}
	\endtableau
}

%\put(28.5,15.5){$\overrightarrow{\ione^2}$}
\printnextarrow{4.5}{6}{\ione^2}

\put(6,1){
	\begintableau{6}{3}
	\row{147}
	\row{69\ten}
	\row{\elv\elv}
	\row{\infty\infty}
	\row{\infty}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\linethickness{1mm}
	\put(0,0){\line(1,0){1}}
	\put(0,0){\line(0,1){2}}
	\put(0,2){\line(1,0){1}}
	\put(1,2){\line(0,-1){2}}
	\endtableau
}

%\put(10.5,6){$\overrightarrow{\itwo^1}$}
\printnextarrow{10.5}{6}{\itwo^1}

\put(12,1){
	\begintableau{6}{3}
	\row{-47}
	\row{-9\ten}
	\row{1\elv}
	\row{6\infty}
	\row{\elv}
	\row{\infty}
	\printrow{-1}{-++}
	\printrow{-2}{123}
	\put(-1.1,2.35){$\scriptstyle\bold{7>}$}
	\put(-0.8,2.5){\circle{1}}
	\linethickness{1mm}
	\put(0,2){\line(1,0){1}}
	\put(0,2){\line(0,1){1}}
	\put(0,3){\line(1,0){1}}
	\put(1,3){\line(0,-1){1}}
	\endtableau
}

%\put(16.5,6){$\overrightarrow{\ione^1}$}
\printnextarrow{16.5}{6}{\ione^1}

\put(18,1){
	\begintableau{6}{3}
	\row{147}
	\row{39\ten}
	\row{4\elv}
	\row{6\infty}
	\row{\elv}
	\row{\infty}
	\printrow{-1}{+++}
	\printrow{-2}{123}
	\endtableau
}

\end{picture}
\end{Figure} 
\end{figure}

%%%@@@ end input from psf_ex

After these sometimes tricky constructions time has come to have a look
at a concrete example, worked out in detail. In
\figref{fig:example-bijection}
we start with a 13-Sundaram tableau of shape $\la=\parof{3,3,2,2,1,1}$
($\la^\prime=\parof{6,4,2}$), and apply the
algorithmic bijection until we finally reach the corresponding
13-Proctor tableau.
The algorithm consists of alternatively applying $\itwo$ and $\ione$,
as long as possible. Each step is labelled by the corresponding
subinvolution. 
All the time, we keep track of permutation $\si$ and vector $\ep$ by
writing $\si\of j$ as well as the sign of $\ep_{\si\of j}$ below column $j$.
Moreover, the violation that is responsible for the respective
involution is graphically marked with thick lines.


%%%@@@ end input from ps_xampl
\section{The background}

%%%@@@ input from ps_expln
\label{sec:explanation}
In this section we explain the background behind our bijection. As
indicated in the introduction, the basic source for the operations of
the previous section is the paper 
\cite{Fulmek-Krattenthaler:Lattice-path}. In
\cite[sec.~8]{Fulmek-Krattenthaler:Lattice-path} it was shown how to
prove combinatorially that the generating function $\sum _{} ^{}w(T)$
for all $(2n+1)$-Sundaram tableaux $T$ of shape $\la$ equals the determinant
\begin{equation}
\Big|{
	\sum _{k\ge0} ^{}e_{\lambda_i'-i+j-k}\of{\bold{x}} -
	\sum _{k\ge0} ^{}e_{\lambda_i'-i-j-k}\of{\bold{x}} 
}\Big|_{\la_1\times\la_1},\label{eq:sundaram-determinant}
\end{equation}
where $\bold x=(x_1,x_1^{-1},x_2,x_2^{-1},\dots,x_n,x_n^{-1})$ and
$e_m(z_1,z_2,\dots,z_s)$ denotes the $m$-th elementary symmetric
function in $z_1,z_2,\dots,z_s$ (cf\@.
\cite[p.~12]{Macdonald:Symmetric-Functions}). Also in
\cite{Fulmek-Krattenthaler:Lattice-path} it was shown how to
prove combinatorially that the generating function $\sum _{} ^{}w(T)$
for all $(2n+1)$-Proctor tableaux $T$ of shape $\la$ equals the determinant
\begin{equation}
\big|{
	\firstrow{e_{\la^\prime_i-i+1}\of{\bold{x},1}}
	e_{\la_i^\prime-i+j}\of{\bold{x},1} +
	e_{\la_i^\prime-i-j+2}\of{\bold{x},1}
}\big|_{\la_1\times\la_1}, \label{eq:proctor-determinant}
\end{equation}
where $e_m(\bold x,1)=e_m(x_1,x_1^{-1},\dots,x_n,x_n^{-1},1)=e_m(\bold
x)+e_{m-1}(\bold x)$. 
The notation of the determinant means that the first expression
gives the entries of the first column and the second the entries for
the remaining columns, $j\geq 2$.
Actually, by \eqref{eq:odd-character} both determinants equal the odd
orthogonal character $\oschf{2n+1}{\la}$. That they are the same can
also be shown directly by first subtracting the sums in
\eqref{eq:sundaram-determinant}, thereby writing
\eqref{eq:sundaram-determinant} as
$$\Big|{\sum _{k=1} ^{2j}e_{\la_i'-i-j+k}(\bold x)}\Big|_{\la_1\times\la_1}=
\Big|{\sum _{k=1} ^{j}e_{\la_i'-i-j+2k}(\bold x,1)}\Big|_{\la_1\times\la_1},$$
and then subtracting column $1$ from columns $2$ through $\la_1$,
column $2$ from columns $3$ through $\la_1$, etc. This results
precisely in \eqref{eq:proctor-determinant}.

Now, here is the idea for our bijection between $(2n+1)$-Sundaram
tableaux and $(2n+1)$-Proctor tableaux of the same shape $\la$: Expand
the determinant \eqref{eq:sundaram-determinant} and interpret the
resulting sum of (signed) monomials as a generating function for
certain objects, $O_\sundaram$ say. The $(2n+1)$-Sundaram tableaux of
shape $\la$ will appear in $O_\sundaram$ since
\eqref{eq:sundaram-determinant} equals their generating function.
Subsequently, find a weight-preserving and sign-reversing involution
that cancels all the remaining objects in $O_\sundaram$. Likewise,
expand the determinant \eqref{eq:proctor-determinant} and interpret
the resulting sum of (signed) monomials as a generating function for
certain objects. In fact, as we saw by the reduction of
\eqref{eq:sundaram-determinant} to \eqref{eq:proctor-determinant}, the
determinant \eqref{eq:proctor-determinant}, when expanded, is a
generating function for a {\em subset\/} of $O_\sundaram$. The
$(2n+1)$-Proctor tableaux of shape $\la$ will appear in this subset
since \eqref{eq:proctor-determinant} equals their generating function.
Subsequently, find a weight-preserving and sign-reversing involution
that cancels all the remaining objects in this subset. And also, find
a weight-preserving and sign-reversing involution that cancels all the
objects in $O_\sundaram$ that are not in this subset. (I.e\@.,
translate the algebraic steps that lead from
\eqref{eq:sundaram-determinant} to \eqref{eq:proctor-determinant} into
combinatorics.) Provided these involutions are found, the involution
principle (see section~\ref{sec:principle}) applies and gives us an
explicit bijection between $(2n+1)$-Sundaram tableaux and
$(2n+1)$-Proctor tableaux of the same shape $\la$.

The proof for \eqref{eq:sundaram-determinant} in
\cite{Fulmek-Krattenthaler:Lattice-path} consisted exactly of
expanding the determinant, interpreting the resulting sum as a
generating function for certain families of lattice paths, and finally
cancelling those families of lattice paths that did not correspond to
Sundaram tableaux by a weight-preserving and sign-reversing involution.
The set $\sundaram$ ($=\proctor$) defined in section~\ref{sec:outline}
(paragraph after (\ref{eq:tabloid-weight}))
is the tabloid translation of the families of lattice paths.
(Each column of a tabloid corresponds to a lattice path. The entries
of the columns correspond to the labels of the horizontal steps in the
lattice path.) The involution in
\cite{Fulmek-Krattenthaler:Lattice-path} consisted of two different
operations, one being the usual Gessel--Viennot mapping
\cite{Gessel-Viennot:Binomial,Gessel-Viennot:Determinants-paths},
corresponding to our $\ione^2$ (see section~\ref{ione}), and one being a modified reflection of
paths, corresponding to our $\ione^1$ (see section~\ref{ione}).

Likewise, the proof for \eqref{eq:proctor-determinant} in
\cite{Fulmek-Krattenthaler:Lattice-path} consisted exactly of
expanding the determinant, interpreting the resulting sum as a
generating function for certain families of lattice paths, and finally
cancelling those families of lattice paths that did not correspond to
Proctor tableaux by a weight-preserving and sign-reversing involution.
The families of lattice paths correspond to the union
$\mphull\cup\setof{(2n+1)\text{-Proctor tableaux of shape }\la}$. 
The involution in
\cite{Fulmek-Krattenthaler:Lattice-path} consisted 
of two different
operations, too.
One was the usual Gessel--Viennot mapping,
corresponding to our $\itwo^3$ (see section~\ref{itwo}) in the case of a row monotonicity
% markus: Wuerde die folgenden 2 Zeile ersetzen,
%Christian: Dann sollten wir den Bandwurm ueberhaupt aufloesen.
violation. The other one was called ``modified
reflection-Gessel--Viennot mapping" and mapped even intersections
to even intersections and odd reflected violations to odd reflected
crossings. This operation corresponds to our $\itwo^3$ (see
section~\ref{itwo}) in the case of
the other violations. (Actually, in
\cite{Fulmek-Krattenthaler:Lattice-path} we even had to modify this
mapping because we did not want it 
% markus: Das gefaellt mir nicht:
% to preserve nonintersectingness.
% schlage vor:
%Christian: Vielleicht ist diese Version noch besser.
to introduce points of intersections.
Fortunately, this is not necessary here.) Finally, the subtraction of
the sums in the entries in \eqref{eq:sundaram-determinant} and the
subsequent column operations that transformed
\eqref{eq:sundaram-determinant} into \eqref{eq:proctor-determinant} are
combinatorially modelled by $\itwo^1$ and $\itwo^2$ (see
section~\ref{itwo}), respectively.

%%%@@@ end input from ps_expln

\section{Concluding remarks}
\label{sec:concluding}
In \cite[sec.~6]{Proctor:Young-tableaux} Proctor defined 4 different
types of odd orthogonal tableaux, one type being due to King and Welsh
\cite{King-Welsh:Construction}. Let us call them odd orthogonal
tableaux of the first, second, third, and fourth kind (as was done in
\cite{Fulmek-Krattenthaler:Lattice-path}). The first kind is the one
considered in the previous sections. King and Welsh \cite[App.~A,
p.~276f]{King-Welsh:Construction} gave a bijection between the
tableaux of first and second kind, and an analogous argument provides a
bijection between the tableaux of third and fourth kind (see
\cite[sec.~7, next-to-last
paragraph]{Fulmek-Krattenthaler:Lattice-path}). We remark that it is
not difficult to construct an explicit bijection between the tableaux
of first and third kind using the involution principle again, by
embedding both types of tableaux in a set 
of tabloids with new column lengths \eqref{eq:proctor-column-length}
and using the involution $\itwo^3$ and a variation of it. Thus all
five types of odd orthogonal tableaux (including Sundaram tableaux)
are related by an explicit bijection.


\bibliographystyle{amsplain}
\bibliography{procsund}

\end{document}

%%%@@@ end input from procsund
