% LaTeX file for a 46 page document requires 3 eps files:
% c4g.eps, c5g.eps, c6g.eps
%   Color versions of these eps-figures are in the files
% c4c.eps, c5c.eps, c6c.eps
\documentclass[dvips,12pt]{article}
\usepackage{graphics}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{latexsym}

%%%%%*****-----^^^^^

% Tableau macros -- tabmac.sty, by Glenn Tesler
% Do not further distribute until further notice; it's incomplete now.
%
% \documentstyle[...,tabmac,...]{article}
% or to use arrows through pb-diagram, things must go in this order:
%    \documentstyle[...,pb-diagram,tabmac,...]{article}
% or \documentstyle[...,pb-diagram,lamsarrow,pb-lams,tabmac,...]{article}
%
% \tableau[options]{row1|row2|...|rowk}       (can use \\ instead of |)
% \tableau*[options]{row1\\row2\\...\\rowk}
% rowk may be followed by optional | or \\, so to actually produce
% a final row with only one entry, terminate with || or \\ \\
%
% options:
%       alignment:
%               c: center,  b: bottom,  t: top
%       diagram type:
%                          F: Ferrer's diagram            Y: Young diagram
%             dflt cell frame  none                             thin frame
%             null cells       \bullet                          thin frame
%             ~ (for skew)     \bullet                          no frame
%          a: use arrow font dimension as base instead of \strut
%       cell size in multiples of current font's \strut's dimensions
%              s: small (x1), m: medium (x2 -- default), l: large (x3)
%              p: small partition sized
% can set variable \Tscale for scaling as well.
% If you do this, don't use s,m,l as they will override this setting.
% can use font size declarations -- \tiny, \Large, etc. -- to change
% font size and cell size, too.
%
% rowi has the form
%       c1,c2,c3,c4,...,cn
% where these are the entries to go on the row.  Can use & instead of ,
% in * form, must use &
% If no entry is given, the cell is a frame (Young) or a bullet (Ferrers)
% special entries (only \ forms available with \tableau*)
%  \bullet, @ = bullet (unframed in Ferrers, framed in Young)
%  \bl        = blank (no frame in either)
%  \sk, ~     = skew cell (use in upper left corner of tableau;
%                      no frame in Young, bullet in Ferrer)
%  \tf, *     = thick frame
%  \fl        = filled square
%  \overlay ... = superimpose ... on this cell (terminate with another
%                 \overlay, or with the end of the tab field)
%  \overlay[??] ... = superimpose ... on this cell with alignment given by
%                      combination of the characters in ??
%                      t, c, b -- top, center, bottom
%                      l, m, r -- left, middle, right
% To get a blank row, use the one entry \bl on that row, i.e., |\bl|
% because doing || will yield a one-square row (Young) or one bullet row
% (Ferrers)
%
% The \arrow command from pb-diagram is implemented.
% Use the exact same syntax for \arrow as in diagram.
% Additional parameters which dgo@???? routines may set are two routines
%    \def\tb@getnodesizehead#1#2{...}
%    \def\tb@getnodesizetail#1#2{...}
% which put the width & height of the text at the head or tail nodes
% into number registers #1, #2, in units of scaled points.
%
% Arrow bugs:
%    The tableau dimensions don't correspond to the restricted
%    discrete arrow lengths, so when real small arrow wanted,
%    only a longer one might be available; option ``a'' may help.
%
%    Dashed arrows may not be uniformly dashed.
%
% More to do
%    Add in LaTeX picture mode; put in alternate frames (dashed, oval, circle)
%    from LaTeX picture mode instead of just the thin and bold frames,
%    and use the picture mode \linethickness declaration instead of
%    \tb@fframe,...
%
%
%
% w/in tableau,   \dimen0=cell size, \dimen1=frame size, \dimen2=strut size
% with option "a", \dimen2 is replaced by value better suited to arrows.
%
% Be very careful that all lines end in %
% so that no spurious spaces are introduced anywhere.

\edef\savecatcodeat{\the\catcode`@}
\catcode`\@=11

\def\tb@ifSpecChars#1#2{#1}
\def\tb@ifNoSpecChars#1#2{#2}

% Determine if * form used
\def\tableau{%
  \bgroup% matched in \tb@tableauD
  \@ifstar{\let\Tif\tb@ifNoSpecChars\tb@tableauB}% *, don't use special chars
          {\let\Tif\tb@ifSpecChars\tb@tableauB}}% no *, use special chars

\def\tb@tableauB{% add [] if no [options]
  \@ifnextchar[{\tb@tableauC}{\tb@tableauC[]}}

\def\tb@tableauC[#1]{\hbox\bgroup%
    \let\\=\cr% end line
    \def\bl{\global\let\tbcellF\tb@cellNF}%
    \def\tf{\global\let\tbcellF\tb@cellH}% highlighted cell
%
%   \dimen0=cell size, \dimen1=frame size, \dimen2=strut size
    \dimen2=\ht\strutbox \advance\dimen2 by\dp\strutbox%
    \ifx\baselinestretch\undefined\relax%
    \else%
% want \dimen2 = \dimen2 / \baselinestretch, but can't do that if
% \baselinestretch has a decimal, so fake it to 2 decimal places.
       \dimen0=100sp \dimen0=\baselinestretch\dimen0%
       \dimen2=100\dimen2 \divide\dimen2 by\dimen0%
    \fi%
%
    \let\tpos\tb@vcenter% default position
    \tb@initYoung% default tableau type
    \tb@options#1\eoo% parse options
%
    \let\arrow\tb@arrow%
%
    \dimen0=\Tscale\dimen2%
    \dimen1=\dimen0 \advance\dimen1 by \tb@fframe%
%
    \lineskip=0pt\baselineskip=0pt% line spacing will be from \vbox to \dimen0
%
    % \mkcell{#1} --  format individual cell:
    %   set cellF to default frame.
    %   supply argument for cell if none given
    %   format argument in sized box, then superimpose cell frame.
    \def\tb@nothing{}%
    \def\endcellno{$\rss\egroup\bss\egroup}% end cell w/o overlap
    \def\endcell{\endcellno\kern-\dimen0}% end cell & prepare to overlap it
    \def\begincell{\vbox to\dimen0\bgroup\vss\hbox to\dimen0\bgroup\hss$}%
%    \def\overlay{\endcell\begincell}%
    \let\overlay\tb@overlay%
    \let\fl\tb@fl%
    \let\lss\hss\let\rss\hss\let\tss\vss\let\bss\vss% cell alignment
    \def\mkcell##1{% format individual cell
        \let\tbcellF\tb@cellD% default cell frame
        \def\tb@cellarg{##1}% store cell contents
        % provide default contents if cell empty
        \ifx\tb@cellarg\tb@nothing\let\tb@cellarg\tb@cellE\fi%
        \begincell\tb@cellarg\endcellno% the actual cell content
        \tbcellF}% draw cell frame
%
    \let\savecellF\tbcellF% save global value of cellF in case of nested tableau
     \Tif{\catcode`,=4\catcode`|=\active}{}\tb@tableauD}%




\let\tb@savetableauD\tableauD% save any current definition
{% set up characters which will be interpreted as command characters
% when the definition of \tableauD is read in
    \catcode`|=\active \catcode`*=\active \catcode`~=\active%
    \catcode`@=\active% command characters
\gdef\tableauD#1{%
  \Tif{% make all the command characters active in math mode when #1 parsed
    \mathcode`|="8000 \mathcode`*="8000%
    \mathcode`~="8000 \mathcode`@="8000%
    \def@{\bullet}%
    \let|\cr% end line
    \let*\tf% highlighted cell
    \let~\sk% skew cell
  }{}%
  \tpos{\tabskip=0pt\halign{&\mkcell{##}\cr#1\crcr}}%
  \global\let\tbcellF\savecellF% restore global value
  \egroup% match \hbox\bgroup at start of \tableauC
  \egroup}% match \bgroup at start of \tableau
}
\let\tb@tableauD\tableauD% rename the command
\let\tableauD\tb@savetableauD% restore old command with this name
\let\tb@savetableauD\undefined

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                      Parse options in \tableau[options]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% pick out one option at a time until \eoo reached.
\def\tb@options#1{\ifx#1\eoo\relax\else\tb@option#1\expandafter\tb@options\fi}

% dispatch handler for each option
\def\tb@option#1{%
  \if#1t\let\tpos\tb@vtop\fi%        t = align at top
  \if#1c\let\tpos\tb@vcenter\fi%     c = align at center
  \if#1b\let\tpos\vbox\fi%           b = align at bottom
  \if#1F\tb@initFerrers\fi%          F = Ferrers diagram
  \if#1Y\tb@initYoung\fi%            Y = Young diagram
  \if#1s\tb@initSmall\fi%            s = small boxes
  \if#1m\tb@initMedium\fi%           m = medium boxes
  \if#1l\tb@initLarge\fi%            l = large boxes
  \if#1p\tb@initPartition\fi%            p = small partition sized boxes
  \if#1a\tb@initArrow\fi%            a = use arrow font as base dimension
}

% alternate to \vcenter which starts math mode if necessary
\def\tb@vcenter#1{\ifmmode\vcenter{#1}\else$\vcenter{#1}$\fi}

% alternate to \vtop; the array produced by \halign has as its baseline
% the lower border of the first row of the tableau, so move down by this
% amount, and move up \ht\strutbox to align the top border with the
% top of the text on this line.
\def\tb@vtop#1{\hbox{\raise\ht\strutbox\hbox{\lower\dimen0\vtop{#1}}}}

\def\tb@initPartition{\def\Tscale{.3}}
\def\tb@initSmall{\def\Tscale{1}}
\def\tb@initMedium{\def\Tscale{2}}
\def\tb@initLarge{\def\Tscale{3}}

% diagram has dimens hardcoded.  Arrow is .625em
\def\tb@initArrow{\dimen2=1.25em}


\def\tb@initYoung{%
  \def\tb@cellE{}% empty cell stays empty
  \let\tb@cellD\tb@cellN% default frame is normal frame
  \def\sk{\global\let\tbcellF\tb@cellNF}}% skew cells are empty
\def\tb@initFerrers{%
  \def\tb@cellE{\bullet}% empty cell gets bullet
  \let\tb@cellD\tb@cellNF% default frame is no frame
  \def\sk{\bullet}}% skew cell gets bullet

\tb@initMedium% default scale


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%               draw square frame with edges of given thickness
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% initial refrence pt = upper right corner of cell
\def\tb@sframe#1{%
  \vbox to0pt{%            Embed frame in a box of no vert or hor extent
    \vss%                            pull box above reference point
    \hbox to0pt{%
      \hss%                          pull box left of reference point
      \vbox to\dimen1{%              Actual width of frame
        \hrule depth #1 height0pt% draw top edge of frame
        \vss%                     begin vcenter sides
        \hbox to\dimen1{%           horiz box with side edges just inside
          \vrule width #1 height\dimen1% left edge
          \hss%                     stretch center
          \vrule width #1%         right edge
          }%
        \vss%                     end vcenter sides
        \hrule height #1 depth 0in% bottom edge
        }%
      \kern-\tb@hframe%           horiz alignment off by half line width
      }%
    \kern-\tb@hframe}}%           vert alignment off by half line width
% alignment off by \dimen1-\dimen0=2\tb@hframe


% should this be combined with picture modes' linethickness?
% \tb@fframe = width of ordinary divider line
% \tb@hframe = \tb@fframe/2
% \tb@bframe = width of bold divider line;
%     ought to have \tb@bframe >= \tb@fframe
\def\tb@hframe{.2pt}\def\tb@fframe{.4pt}\def\tb@bframe{2pt}
\def\tb@cellH{\tb@sframe{\tb@bframe}}       % bold frame
\def\tb@cellNF{}                            % no frame
\def\tb@cellN{\tb@sframe{\tb@fframe}}       % normal frame
\let\tbcellF\tb@cellN                       % default is normal


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                             overlay cell contents
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\def\tb@rpad{1pt}
\def\tb@lpad{1pt}
\def\tb@tpad{1.8pt}
\def\tb@bpad{1.8pt}

\def\tb@overlay{\endcell\@ifnextchar[{\tb@overlaya}{\begincell}}
%\def\tb@overlay{\endcell\@ifnextchar[{\begincell}{\begincell}}
\def\tb@overlaya[#1]{\vbox to\dimen0\bgroup%
  \tb@overlayoptions#1\eoo%
  \tss\hbox to\dimen0\bgroup\lss$}
% pick out one option at a time until \eoo reached.
\def\tb@overlayoptions#1{\ifx#1\eoo\relax\else\tb@overlayoption#1\expandafter\tb@overlayoptions\fi}

% dispatch handler for each option
\def\tb@overlayoption#1{
  \if#1t\def\tss{\vskip\tb@tpad}\let\bss\vss\fi% t = align at top
  \if#1c\let\tss\vss\let\bss\vss\fi%             c = align at center
  \if#1b\def\bss{\vskip\tb@bpad}\let\tss\vss\fi% b = align at bottom
  \if#1l\def\lss{\hskip\tb@lpad}\let\rss\hss\fi% l = align at left
  \if#1m\let\lss\hss\let\rss\hss\fi%             m = align at middle
  \if#1r\def\rss{\hskip\tb@rpad}\let\lss\hss\fi% r = align at right
}

\def\tb@fl{\endcell\begincell\vrule depth 0pt width \dimen0 height \dimen0 \endcell\begincell}


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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%                            arrows, if diagram loaded
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\@ifundefined{diagram}{}{
% should \tb@arrowpad should be a multiple of .5?
\def\tb@arrowpad{.5}

% stripped down/altered from \dg@arrow
% Usage: \tb@arrow[USERSIZE]{ARROW_SPECS}LABELS...
\newoptcommand{\tb@arrow}{\@ne}[2]{%
  \endcell% end previous cell contents
   % Keep all arrow geometry parameters local.
   % The \endgroup is in \tb@draw
   \begingroup%
   \let\dg@getnodesize\tb@getnodesize% substitute routine to get nodesize
   % Get optional arrow USERSIZE.
   \dg@USERSIZE=#1\relax%
   \ifnum\dg@USERSIZE<\@ne \dg@USERSIZE=\@ne \fi%
   % Parse arrow specification.
   \dg@parse{#2}%
   \dg@label{\tb@draw{#1}{#2}}}% draw arrow

% substitute alternate routine for getting node sizes
% Usage: \tb@getnodesize{SLIST}{XCOORD}{YCOORD}{\WIDTH}{\HEIGHT}
\def\tb@getnodesize#1#2#3#4#5{\dimen3=\tb@arrowpad\dimen2 #4=\dimen3 #5=\dimen3\relax}
\def\tb@getnodesize#1#2#3#4#5{\ifnum#2=0\ifnum#3=0\tb@getnodesizetail{#4}{#5}\else\tb@getnodesizehead{#4}{#5}\fi\else\tb@getnodesizehead{#4}{#5}\fi}
\def\tb@getnodesizetail#1#2{\dimen3=.5\dimen2 #1=\dimen3 #2=\dimen3}
\def\tb@getnodesizehead#1#2{\dimen3=.5\dimen2 #1=\dimen3 #2=\dimen3}

% like \dg@drawcalc & \dg@draw; do the actual drawing
\def\tb@draw#1#2#3#4{%
  % grid geometry is determined by the tableau grid
        \dg@X=0\dg@Y=0\dg@XGRID=1\dg@YGRID=1\unitlength=.001\dimen0%
        \dg@LBLOFF=\dgLABELOFFSET \divide\dg@LBLOFF\unitlength%
%
        \dg@drawcalc% compute arrow geometry
        \begincell% start tableau cell
        \let\lams@arrow\tb@lams@arrow% substitute routine
  % draw arrow              VVVVV matches endgroup in \dg@draw
        \begin{picture}(0,0)\begingroup\dg@draw{#1}{#2}{#3}{#4}\end{picture}%
        \endcell% end tableau cell
        \endgroup% match \begingroup in \tb@arrow
        \begincell}% start new entry in this cell
}



% \lamssource{V} (= \arrow{...,V}), \lamstarget{V}, \lamstarget{^}
% cause placement of arrow to be ``adjusted'', but then the
% 45 deg. arrows don't go through grid points on the tableau.
%
% So this code from lamsarrow.sty has the adjustment section disabled.
%
% Perhaps \dgo@V should be modified to provide extra padding around cell
% contents to compensate for the removed adjustment.
%
%\def\lams@arrow#1#2{%
\def\tb@lams@arrow#1#2{%
 \lams@firstx\z@\lams@firsty\z@
 \lams@lastx#1\relax\lams@lasty#2\relax
 \lams@center\z@
 %
 % compute direction flags
 \N@false\E@false\H@false\V@false
 \ifdim\lams@lastx>\z@\E@true\fi
 \ifdim\lams@lastx=\z@\V@true\fi
 \ifdim\lams@lasty>\z@\N@true\fi
 \ifdim\lams@lasty=\z@\H@true\fi
 \NESW@false
 \ifN@\ifE@\NESW@true\fi\else\ifE@\else\NESW@true\fi\fi
 %
 % compute shaft char info with help of \lams@slope macro
 \ifH@\else\ifV@\else
  \lams@slope
  \ifnum\lams@tani>\lams@tanii
   \lams@ht\ten@\p@\lams@wd\ten@\p@
   \multiply\lams@wd\lams@tanii\divide\lams@wd\lams@tani
  \else
   \lams@wd\ten@\p@\lams@ht\ten@\p@
   \divide\lams@ht\lams@tanii\multiply\lams@ht\lams@tani
  \fi
 \fi\fi
%%%%%%%%%%%%%%%%% BEGIN disable
% %
% % adjust arrow placement for source and target style
% \ifH@\else\ifV@\else
%  \ifnum\lams@target=\thr@@
%   \ifN@\advance\lams@lasty-.3\lams@ht
%   \else\advance\lams@lasty.3\lams@ht\fi
%  \fi
%  \ifnum\lams@source=\tw@
%   \ifE@\advance\lams@firstx.3\lams@ht
%   \else\advance\lams@firstx-.3\lams@ht\fi
%  \fi
%  \ifnum\lams@target=12
%   \ifN@\advance\lams@lasty-\lams@ht
%   \else\advance\lams@lasty\lams@ht\fi
%  \fi
% \fi\fi
%%%%%%%%%%%%%%%%% END disable
 %
 % draw arrow, special-casing horizontal and vertical ones
 \ifH@  \lams@harrow
 \else\ifV@ \lams@varrow
 \else \lams@darrow
 \fi\fi
}




\catcode`\@=\savecatcodeat
\let\savecatcodeat\undefined

%%%%%*****-----^^^^^

\overfullrule=0pt
\baselineskip 14pt
%\settabs 8 \columns
\parindent=.5truein
\hfuzz=3.44182pt
\def\part {\vdash}
\def\lb {\left[}
\def\rb {\right]}
\def\X {X}
\def\coef {\Big|}
\def\Hm {{\cal H}_m}
\def\snake {\rfloor}
\def\attic {\rceil}
\def\la {\lambda}
\def\vsp {\vskip 8pt}
\def\sp {\hskip .05in}
\def\bsp {\hskip .5in}
\def\bspp {\hskip .5in}
\def\coeff {{\Big|}}
\def\sab {\vskip .1in}
\def\sabm {\vskip .25in}
\def \ess {\enskip}
\def \sab {\vskip .5in}
\def \sa {\vskip .2in}
\def\ll {{l}}
\def\contained {\subseteq}
\def\V {{\mathcal V}}
\def\H {{\mathcal H}}
\def\HH {{\bf H}}
\def\HHH {{\bf H}}
\def\Hmu {{H_\mu[X;t]}}
\def\wH {{\tilde H}}
\def\wiggle {\tilde}

\def\Hto {H_2^\tto}
\def\Hot {H_2^\tot}
\def\ul {\underline}
\def\ol {\bar}
\def\XST {{XST}}
\def\ih {{i}}
\def\iv {{i'}}
\def\to {{\def\Tscale{.65} \tableau[Y]{1}}}
\def\tot {{\def\Tscale{.65} \tableau[Y]{1& 2}}}
\def\tto {{\def\Tscale{.65} \tableau[Y]{2\\1}}}
\def\TSY {\tableau[sY]}
\def\Hti {{\bf H}_2^{-1}}
\def\AH {{\bf H}_2}
\def\AV {{\bar{\bf H}}_2}
\def\AHr {\AH^{\rho}}
\def\AVr {\AV^{\rho}}

\def\endofproof {\hskip .1in $\Box$}

\newcommand{\ontop}[2]{\substack{#1\\#2}}
\def\today{\ifcase\month\or
January\or February\or March\or April\or may\or June\or
July\or August\or September\or October\or November\or
December\fi
\space\number\day, \number\year}
%\headline={\small  \hfill Macdonald Vertex Operator \& Two-Column (q,t)-Kostka Statistics $\ess$
%\hfill \today $\ess$$\ess$ Page \folio} \footline={\hfil}
%\font\normal=cmr10
%\font\small=cmr6
%\font\ninerm=cmr9
%\font\ita=cmti10
%\font\bol=cmbx10
%\font\abs=cmr9
%\nopagenumbers
\vskip 24pt

\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{definition}[thm]{Definition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{conjecture}[thm]{Conjecture}
\newtheorem{example}[thm]{Example}

\textwidth 6.5 in
\hoffset -.5in
\vsize 9in
\begin{document}

\pagestyle{myheadings} \markright{\sc the electronic journal of
combinatorics 5 (1998), \#R45\hfill} \thispagestyle{empty}
\normalbaselineskip .3in \baselineskip \normalbaselineskip

\title{A Macdonald Vertex Operator and
Standard Tableaux Statistics
for the Two-Column $(q,t)$-Kostka Coefficients}
\author{Mike Zabrocki\\ Centre de Recherche Math\'ematiques, Universit\'e \\ de
Montr\'eal/LaCIM, Universit\'e de Qu\'ebec \`a Montr\'eal \\ email: {\small\tt
zabrocki@math.ucsd.edu}}
\date{}
\maketitle
\smallskip
\centerline{Submitted: September 30, 1998; Accepted: November 2, 1998}
\smallskip
{\small MR Subject Number: {05E10}\hfill}

\smallskip
{\small Keywords: Macdonald polynomials, tableaux, symmetric functions,
q,t-Kostka coefficients\hfill}
\bigskip

\begin{abstract}
The two parameter family of coefficients $K_{\la\mu}(q,t)$
introduced by Macdonald are conjectured to $(q,t)$ count the standard
tableaux of shape $\la$.  If this conjecture is correct, then there
exist statistics $a_\mu(T)$ and $b_\mu(T)$ such that the family of
symmetric functions $H_\mu[X;q,t] = \sum_\la K_{\la \mu}(q,t) s_\la[X]$
are generating functions for the
standard tableaux of size $|\mu|$ in the sense that
$H_\mu[X;q,t] = \sum_{T} q^{a_\mu(T)} t^{b_\mu(T)} s_{\la(T)}[X]$
where the sum is over standard
tableau of of size $|\mu|$.  We give a formula for a symmetric function
operator $H_2^{qt}$ with the property that $H_2^{qt} H_{(2^a1^b)}[X;q,t]=
H_{(2^{a+1}1^b)}[X;q,t]$.  This operator has a combinatorial action
on the Schur function basis.  We use this Schur function
action to show by induction that
$H_{(2^a1^b)}[X;q,t]$ is the generating function for standard tableaux
of size $2a+b$ (and hence that $K_{\la(2^a1^b)}(q,t)$ is a
polynomial with non-negative integer coefficients).
The inductive proof gives an algorithm for 'building' 
the standard tableaux of size $n+2$  from the
standard tableaux of size $n$ and divides the standard tableaux into
classes that are generalizations of the catabolism type.
We show that reversing this construction gives the
statistics $a_\mu(T)$ and $b_\mu(T)$ when $\mu$ is of the form
$(2^a1^b)$ and that these statistics prove conjectures about
the relationship between adjacent rows of the $(q,t)$-Kostka
matrix that were suggested by Lynne Butler.
\end{abstract}


\section{Introduction}

The Macdonald basis for the symmetric functions
generalizes many other bases by specializing the values of
$t$ and $q$.  The symmetric function basis $\{ P_\mu[X;q,t] \}_\mu$
is defined (\cite{M} p. 321) as being self-orthogonal and having
an upper triangularity condition with the monomial
symmetric functions and the integral form of the basis is
defined by setting $J_\mu[X;q,t] = P_\mu[X;q,t] h_\mu(q,t)$ for
some $q,t$-polynomial coefficients $h_\mu(q,t)$.  The
$\{ J_\mu[X;q,t] \}_\mu$ have the expansion
$$J_\mu[X;q,t] = \sum_{\la} K_{\la\mu}(q,t) S_\la[X;t]$$
where $S_\la[X;t]$ is the dual Schur basis. %CHECK THIS NAME/DEF
The coefficients $K_{\la\mu}(q,t)$ are referred to as the
Macdonald $(q,t)$-Kostka coefficients.  These coefficients
are known to be polynomials and conjectured to have non-negative
integer coefficients.  It is known that $K_{\la\mu}(1,1) = K_\la$
and so it is conjectured that these coefficients $(q,t)$ count
the standard tableau of shape $\la$.

We are interested here in the basis
$$H_\mu[X;q,t] = \sum_\la K_{\la\mu}(q,t) s_\la[X]$$
It has the specializations that $H_\mu[X;0,t] = H_\mu[X;t]$
(the Hall-Littlewood basis of symmetric functions), $H_\mu[X;0,0] =
s_\mu[X]$, $H_\mu[X;0,1] = h_\mu[X]$, and the property that
$H_\mu[X; q, t] = q^{n(\mu')} t^{n(\mu)} \omega H_\mu[X;1/q,1/t]$ and
$H_\mu[X; q, t] = \omega H_{\mu'}[X; t,q]$.

For each of the homogeneous, Schur, and Hall-Littlewood symmetric
functions there are vertex operators with the property that for
$m \geq \mu_1$
$h_m h_{\mu}[X] = h_{(m,\mu)}[X]$, $S_m s_\mu[X] = s_{(m,\mu)}[X]$,
and $H_m^t H_{\mu}[X;t] = H_{(m,\mu)}[X;t]$ where
$(m, \mu)$ represents the partition $(m, \mu_1, \mu_2, \\ \ldots,
 \mu_k)$.
These are each given by the following formulas:

\begin{align}
i)& \hskip .2in h_m = h_m[X] \label{hvertex} \\
ii)& \hskip .2in S_m = \sum_{i\geq 0} (-1)^i h_{m+i}[X] e_i^{\perp}
\label{svertex} \\
iii)& \hskip .2in H_m^t = \sum_{j\geq 0} t^j S_{m+j} h_j^{\perp}
\label{Htvertex}
\end{align}

The action of each of these operators on the
Schur basis is known (\cite{Za}).
It is hopeful that a similar vertex operator
can be found for the $H_m[X;q,t]$
symmetric functions and the action on the Schur basis can be expressed
easily.

Define $H_m^{qt}$ to be ''the'' operator that has the property
that $H_m^{qt} H_\mu[X;q,t] = \\ H_{(m,\mu)}[X;q,t]$.  This condition alone
is not sufficient to define this operator uniquely, but it is sufficient
to calculate the action on the Schur basis for certain partitions.
Since the $\{H_\mu[X;q,t]\}_\mu$ is a basis for the symmetric functions,
$s_\la[X] = \sum_\mu d_{\la\mu}(q,t) H_\mu[X;q,t]$, and for
$m \geq |\la|$, $H_m^{qt}$ may be calculated by the expression
$$H_m^{qt} s_\la[X] = \sum_\mu d_{\la\mu}(q,t) H_{(m,\mu)}[X;q,t]$$

These calculations are enough to inspire the following conjecture

\begin{conjecture}
$$H_m^{qt} = \sum_{T \in ST^m} q^{co(T)} H_m^T(t)$$
for some polynomial symmetric functions operators $H_m^T(t)$ that are only
dependent on $t$ with the following properties:
\def\Tscale{.7}
$$\hskip .15in i) \hskip .2in H_m^{T}(1) = s_{\la(T)}[X]$$
$$\hskip .5in ii) \hskip .2in H_m^{\omega T}(t) = \omega H_m^T(1/t) \omega R^t$$
$$iii) \hskip .2in H_m^{{{\tableau[Y]{1,2, ,m}}}} = H_m^t$$
where $T$ is a standard tableau of size $m$, $co(T)$ is the cocharge
statistic on the tableau, $\la(T)$ is the shape
of the tableau, $H_m^t$ is the Hall-Littlewood
vertex operator, $\omega T$ is the tableau flipped about the diagonal
and $R^t$ is a linear operator that acts on homogeneous symmetric functions
$P[X]$ of degree $n$ with the action $R^t P[X] = t^n P[X]$.
\end{conjecture}

These vertex operators do not seem to be transformed versions of the
vertex operators known for the $\{P_\mu[X;q,t]\}_\mu$ (\cite{LV}, \cite{KN}).

In the case that $m=2$, this conjecture completely determines the
operator $H_2^{qt}$ and the main result presented in the first
section of this paper will be

\begin{thm} The operator
$$H_2^{qt} = H_2^t + q \omega H_2^{\frac{1}{t}} \omega R^t$$
has the property that
$H_2^{qt} H_{(2^a1^b)}[X;q,t] = H_{(2^{a+1}1^b)}[X;q,t]$.
\end{thm}

This theorem will follow from a formula by John Stembridge \cite{St}
that gives an expression for the Macdonald polynomial indexed by a
shape with two columns in terms of Hall-Littlewood polynomials.
Susanna Fischel \cite{Fi} has already used this result to find statistics
on rigged configurations that are known to be isomorphic to standard
tableaux.  It would be better to have these statistics directly for
standard tableau since the bijection between standard tableau and
rigged configurations is not trivial (\cite{KR1}, \cite{KR2}, \cite{H}).

\vskip .2in
Our main purpose for finding the vertex operator $H_m^{qt}$
and its action on the
Schur function basis is to use it to discover statistics $a_\mu(T)$
and $b_\mu(T)$ on standard
tableau so that $K_{\la\mu}(q,t) = \sum_{T \in ST^\la} q^{a_\mu(T)}t^{b_\mu(T)}$.
If these statistics exist, then the family of symmetric functions
$\{H_\mu[X; q,t]\}_\mu$ can be thought of as generating functions for
the standard tableaux in the sense that $H_\mu[X;q,t] = \sum_{T \in
ST^{|\mu|}} q^{a_\mu(T)}t^{b_\mu(T)} s_{\la(T)}[X]$.

The vertex operator property has the interpretation that $H_m^{qt}$ changes
the generating function for the standard tableaux of size $n$ to the
generating function for the standard tableaux of size $n+m$.  Knowing
the action of $H_m^{qt}$ on the Schur function basis gives a description
of how the shape of the tableau changes when a block of size $m$ is
added.

In the case of $m=2$, the action of $H_2^t$ (and $\omega
H_2^{\frac{1}{t}} \omega R^t$ and hence $H_2^{qt}$)
on the Schur function basis
is well understood.  The operator $H_2^{qt}$ can be
interpreted as instructions for building the standard tableaux of
size $n+2$ from the standard tableaux of size $n$. The second
section of this paper  will define a tableaux operator and show how it
can be used to build tableaux of larger content from smaller and
state explicitly how cancellation of any negative terms in the
expression $H_2^{qt} H_{(2^a1^b)}[X;q,t] = H_{(2^{a+1}1^b)}[X;q,t]$
occurs.  This operator suggests that the standard tableaux are divided
into subclasses of tableaux and that each subclass is represented by
a piece of the expression for $H_{(2^a1^b)}[X;q,t]$.
  The last section will be exposition of the statistics
$a_\mu(T)$ and $b_\mu(T)$ and on the subclasses of tableaux.

\sabm
\subsection{Notation}

A partition $\la$ is a
weakly decreasing sequence of non-negative integers with
$\la_1 \geq \la_2 \geq \ldots \geq \la_k \geq 0$.  The length $\ll(\la)$
of the partition is the largest $i$ such that $\la_i >0$.  The partition
$\la$ is a partition of $n$ if $\la_1 + \la_2 + \cdots + \la_{\ll(\la)} = n $.
We associate a partition with its diagram and often use the two
interchangeably.  We use the French convention and draw the largest
part on the bottom of the diagram.
One partition is contained in another, $\la \contained \mu$ if $\la_i
\leq \mu_i$ for all $i$ (the notation is to suggest that if the diagram
for $\la$ were placed over the diagram for $\mu$ that one would
be contained in the other).

For every partition $\la$ there is a corresponding conjugate partition denoted
by $\la'$ where  $\la_i' = $ the number of cells in the $i^{th}$ column of
$\la$.

A skew partition is denoted by $\la \slash \mu$, where it is assumed
that $\mu \contained \la$, and represents the
cells that are in $\la$ but are not in $\mu$.
A skew partition $\la \slash \mu$
is said to be a horizontal strip if there is at most
one cell in each column.  Denote the class of horizontal
strips of size $k$ by $\H_k$ so that the notation $\la \slash \mu
\in \H_k$ means that $\la \slash \mu$ is a horizontal strip with $k$
cells.  Similarly, the class of vertical strips (skew partitions
with only one cell in each row) will be denoted by $\V_k$.


A useful statistic defined on compositions, $\mu$, is
$n(\mu) = \sum_i \mu_i (i-1)$.

If $\la$ is a partition, then let $\la^r$ denote the partition with
the first row removed, that is $\la^r = (\la_2, \la_3,
\ldots, \la_{\ll(\la)})$.
Let $\la^c$ denote the partition with the first column removed, so that
$\la^c = (\la_1 -1, \la_2 -1, \ldots, \la_{\ll(\la)} -1 )$.
This allows us to define the border of a partition $\mu$
to be the skew partition $\mu \slash
\mu^{rc}$.


Define the $k$-snake of a partition $\mu$ to be the
$k$ bottom most right hand cells of the border of $\mu$ (the choice of
the word "snake" is supposed to suggest the cells that slink with its
belly on the ground from
the bottom of the partition up along the right hand edge).  We use the
symbol $ht_k(\mu)$ to denote the height of the $k$-snake.  The symbol
$\mu \snake_k = (\mu_2 -1, \mu_3-1, \ldots, \mu_h-1, \mu_1 +h -k-1,
\mu_{h+1}, \ldots, \mu_{\ll(\la)})$
will be used to represent a partition with the $k$-snake removed
with the understanding that if removing the $k$-snake does not leave
a partition that this symbol is undefined.

Define the $k$-attic of a partition $\mu$ to be the top most
left hand cells of the border of $\mu$.  The symbol
${\bar{ht}}_k(\mu)$ will represent the width of the $k$-attic
(${\bar{ht}}_k(\mu) = ht_k(\mu')$), and
$\mu \attic^k = \mu' \snake_k'$ will  represent a
partition with the $k$-attic removed with the understanding that
if removing the $k$-attic does not leave a partition that this
symbol is undefined.

Assume the convention that a Schur symmetric function indexed
by a partition $\rho \snake_n$ or $\rho \attic^n$ that does not
exist is $0$.

\sabm

\begin{example}
\def\Tscale{.7}
\def\ts{{\hskip .1in}}
$$\begin{array}{cccccc}
\ts {\tableau[Y]{|,|,|,,,|,,,,}}\ts  &
\ts {\tableau[Y]{~||,|,|,,,}}\ts  &
\ts {\tableau[Y]{~|||,,|,,,}} \ts &
\ts {\tableau[Y]{~|~||,,,|,,,,}}\ts  &
\ts {\tableau[Y]{|,|,|,|,,}}\ts  &
\ts {\tableau[Y]{|,|,||,,}}\ts  \\
\la & \la^r & \la^c & \la \attic^4 & \la \snake_4 & \la \snake_5 (DNE)
\end{array}$$

If $\la = (5,4,2,2,1)$ is the partition, then the $\la^r = (4,2,2,1)$,
$\la^c = (4,3,1,1)$,  $\la \attic^4 = (5,4,1)$,
$\la \snake_4 = (3,2,2,2,1)$ can all be calculated by drawing the diagram
for $\la$ and crossing off the appropriate cells.
Note that in this example that $\la \snake_5$ does not exist.
\end{example}

If the shape of $\rho = \la \snake_k$ is given and the height of the
$k$-snake is specified then $\la$ can be recovered
($\la$ is determined from $\rho$ by adding
a $k$-snake of height $h$).  This is because
\begin{equation}\label{addksnake}
\la = (\rho_h + k - h +1,
\rho_1+1, \rho_2+1, \ldots, \rho_{h-1}+1, \rho_{h+1}, \rho_{h+2}, \ldots,
\rho_{\ll(\rho)})
\end{equation}
and so $\la$ will be a partition as long as $k$ is
sufficiently large.

A standard tableau
is a diagram of a partition of $n$ filled with the numbers
$1$ to $n$ such that the labels increase moving from left to right
in the rows and from bottom to top in the columns.  The set of standard
tableaux of size $n$ will be denoted by $ST^n$.

We will consider the ring of symmetric functions in an
infinite number of variables as a subring of
$\mathbb{Q}[x_1, x_2, \ldots]$.  A more precise construction of this
ring can be found in \cite{M} section I.2.

We make use of plethystic notation for symmetric functions here.
This is a notational device for expressing the substitution of
the monomials of one expression, $E = E(t_1, t_2, t_3, \ldots)$
for the variables of a
symmetric function, $P$.  The result will be denoted by $P[E]$
and represents the expression found by expanding $P$ in terms
of the power symmetric functions and then substituting for
$p_k$ the expression $E(t_1^k, t_2^k, t_3^k, \ldots)$.

More precisely, if the power sum expansion of the symmetric
function $P$ is given by
$$P = \sum_\la c_\la p_\la$$ then the $P[E]$ is given by the formula
$$P[E] = \sum_\la c_\la p_\la
\coeff_{p_k \rightarrow E(t_1^k, t_2^k, t_3^k, \ldots)}$$.

To express a symmetric function in a single set of variables
$x_1, x_2, \ldots, x_n$, let $X_n = x_1 + x_2 + \cdots + x_n$.
The expression $P[X_n]$ represents the symmetric
function $P$ evaluated at the variables $x_1, x_2, \ldots, x_n$
since

$$P(x_1, x_2, \ldots, x_n) = \sum_\la c_\la p_\la \coeff_{p_k \rightarrow
x_1^k + x_2^k + \cdots + x_n^k} = P[X_n]$$

The Cauchy kernel is a ubiquitous formula in the theory
of symmetric functions (especially when working with
plethystic notation).
\begin{definition} The Cauchy kernel
$$\Omega[X] = \prod_i \frac{1}{1-x_i}$$
\end{definition}

It follows using plethystic notation that
$\Omega[X] \Omega[Y] = \Omega[X+Y]$ and
$\Omega[-X] = \prod_i (1 - x_i)$.

The Cauchy kernel
evaluated at the product of two sets of variables
has the formula (\cite{M} p 63)
$$\Omega[XY] = \prod_{i,j} \frac{1}{1-x_i y_j}
= \sum_\la s_\la[X] s_\la[Y] = \sum_\la h_\la[X] m_\la[Y]$$

We will use the notation that $f^\perp$ to denote
the adjoint to multiplication for a symmetric function $f$
with respect to the standard inner product.
Therefore $\left< f^\perp g, h \right> = \left< g, f h \right>$.
Note that $h_k^\perp$ and $e_k^\perp$ act on the Schur function
basis with the formulas
$$e_k^\perp s_\mu = \sum_{\mu\slash\la \in \V_k} s_\la$$
$$h_k^\perp s_\mu = \sum_{\mu\slash\la \in \H_k} s_\la$$

The Macdonald basis \cite{M} for the symmetric functions are defined by
the following two conditions

$$a) \hskip .2in P_\la = s_\la + \sum_{\mu < \la} s_\mu c_{\mu\la}(q,t)$$
$$b) \left< P_\la, P_\mu \right>_{qt} = 0 \hskip .1in for \hskip .1in \la \neq \mu$$

where $\left< , \right>_{qt}$ denotes the scalar product of symmetric functions defined
on the power symmetric functions by
$\left< p_\la, p_\mu \right>_{qt} = \delta_{\la\mu} z_\la p_\la\left[ \frac{1-q}{1-t} \right]$
($z_\la$ is the size of the stablizer of the permuations of cycle structure $\la$ and
$\delta_{xy}=1$ if $x=y$ and $0$ otherwise).  We will also refer to the basis $H_\mu[X;q,t] =
\prod_{s \in \mu} (1-q^{a_\mu(s)} t^{l_\mu(s)+1}) P_\mu\left[ \frac{X}{1-t}; q,t \right]
= \sum_\la K_{\la\mu}(q,t) s_\la[X]$ that is of interest in this paper
as Macdonald symmetric functions ($s \in \mu$ means run over all cells $s$ in $\mu$ and
$a_\mu(s)$ and $l_\mu(s)$ are the arm and leg of $s$ in $\mu$ respectively).

The Hall-Littlewood symmetric functions $H_\mu[ X; t]$
can be defined by the following formula.

\begin{definition}The Hall-Littlewood symmetric function
\label{def:Hmu}
$$\Hmu = \prod_{i\geq 0, 1\leq j \leq k} \frac{1}{1-z_j x_i}
\prod_{1 \leq i \leq j \leq k}
\frac{1 - z_j / z_i}{1 - t z_j / z_i}
\coeff_{Z^\mu} $$
where $\mu$ is a partition with $k$ parts and $\coeff_{Z^\mu}$ represents
taking the coefficient of the monomial $z_1^{\mu_1} z_2^{\mu_2} \cdots z_k^{\mu_k}$.
\end{definition}

These symmetric functions are not the same, but are related to
the symmetric functions referred to as Hall-Littlewood polynomials
in \cite{M} p. 208.
The Hall-Littlewood functions are
related to the Schur symmetric functions by letting $t \rightarrow 0$
and to the homogeneous symmetric functions by letting $t \rightarrow 1$.

The Hall-Littlewood functions can be expanded in terms of the
Schur symmetric function basis with coefficients $K_{\la\mu}(t)$, that
is, $H_\mu[X;t] =$ $ \sum_\la K_{\la\mu}(t) s_\la[X]$.  The
$K_{\la\mu}(t)$ are well studied and referred to as the Kostka-Foulkes
polynomials.
The vertex operator, $H_m^t$ in formula (\ref{Htvertex}),
that has $H_m^t H_\mu[X; t] = H_{(m,\mu)}[X;t]$ is due
to Jing (\cite{J}, \cite{G}).  The Schur function
vertex operator of equation (\ref{svertex})
is due to Bernstein \cite{Ze} (p. 69).


\section{The Vertex Operator}

Define the following symmetric function operator by the following
equivalent formulas

\begin{definition}
Let $P[X]$ be a homogeneous symmetric function of degree $n$.
\begin{align}
H_2^{qt} P[X] &= (H_2^t + q \omega H_2^{\frac{1}{t}} \omega R^t) P[X] \\
&= P \left[ X - \frac{1-t}{z} \right] \Omega[zX] +
q P \left[ tX - \frac{1-t}{z} \right] \Omega[-zX] \coeff_{z^2} \\
&= \sum_{i \geq 0}
(t^i S_{2+i} h_i^\perp + q t^{n-i} \omega S_{2+i} \omega e_i^\perp) P[X] \\
&= \sum_{i,j \geq 0} (t^j (-1)^i h_{2+i+j}[X] +
q t^{n-i} (-1)^j e_{2+i+j}[X]) e_i^{\perp} h_j^{\perp} P[X]
\end{align}
where the symbol $\coeff_{z^2}$ means take the coefficient of $z^2$ in the expression
and $R^t$ is an operator that has the property $R^t P[X] = t^n P[X]$.
\end{definition}

For the remainder of this paper the symbol
$\Hto$ will represent the expression
$\omega H_2^{\frac{1}{t}} \omega R^t$ and the symbol $\Hot$ will
represent the operator $H_2^t$
so that $H_2^{qt} = \Hot + q \Hto$.

A formula for the $(q,t)$ Kostka coefficients $K_{\la \mu}(q,t)$ when
$\mu$ is a two column partition was given in \cite{St}.  That
result will be used to prove that the $H_2^{qt}$ operator has the
vertex operator property.  The proof first requires the following four lemmas:

\begin{lemma}
$$H_{(1^{b+2})}[X;t] = t^{b+1} H_{(21^b)}[X;t]
+ t^{\left( {\ontop{b+1}{2}} \right)} \omega
H_{(21^b)}[X;t^{-1}]$$
\end{lemma}

\noindent{\bf Proof} \hskip .1in
There are combinatorial interpretations of each term of this
equation and a bijective proof is easy enough to state.
The left hand side of this equation is given by
$$H_{(1^{b+2})}[X;t] = \sum_{T \in ST^{b+2}} t^{c(T)} s_{\la(T)}[X]$$
Each term on the right hand side of the equation is given by
the sums
$$t^{b+1} H_{(21^b)}[X;t] = \sum_{T \in CST^{(21^b)}}
t^{c(T) +b + 1} s_{\la(T)}[X]$$
$$t^{\left( {\ontop{b+1}{2}} \right)} \omega
H_{(21^b)}[X;t^{-1}] = \sum_{T \in CST^{(21^b)}}
t^{ \left( {\ontop{b+1}{2}} \right)-c(T) } s_{\la(\omega T)}[X]$$
where $\omega T$ is the tableau that is flipped about the diagonal.

Each standard tableau has either the label of $2$ lying to the
immediate right of $1$ or above it.

A tableau
that has a $2$ that lies immediately to the right of the $1$
is isomorphic to a tableau that has content $(21^b)$ and charge that
is $b+1$ higher.  The isomorphism simply decreases the label
any cell with a label higher than $2$ by $1$ and the inverse is
to increase the label of every cell except the $1$ in the corner.
The charge of the standard tableau is $b+1$ more than the charge
of the corresponding tableau of content $(21^b)$ because in the
word definition of charge, the index of every letter (except the $1$)
of the word of the tableau decreases by $1$ when the labels are
decreased.

A tableau that has a label of $2$ lying above the $1$ can be
transposed about the diagonal and this tableau is isomorphic to
a tableau of content $(21^b)$ by the same map.  The charge of
standard tableau is the cocharge of the transposed tableau so
$c(T) = \left( {\ontop{b+2}{2}} \right) - c(\omega T)$.  The
transformation that decreases the label in each cell by $1$
(except the first cell) decreases the charge of the tableau by $b+1$
and so the charge of the tableau of content $(21^b)$ is
$\left( {\ontop{b+2}{2}} \right) - (b+1) - c(T)$.
{\endofproof}

\sa


\begin{lemma}\label{bHonHt}
$$\Hto H_{(1^b)}[X;t] =
H_{(1^{b+2})}[X;t] - t^{b+1}H_{(21^b)}[X;t]$$
\end{lemma}

\noindent{\bf Proof} \hskip .1in
Note that for the Hall-Littlewood symmetric function indexed by
the partition $(1^b)$ we know from \cite{M} p. 364 that $H_{(1^b)}[X;t] =
(t;t)_n e_n \left[ {\frac{X}{1-t}} \right]$.
 From this we derive
\begin{align*}
H_{(1^b)}[X;t] &= (t;t)_b e_b \left[ {\frac{X}{1-t}} \right]\cr
&= (-1)^b t^{\left( {\ontop{b+1}{2}}\right)} (t^{-1};t^{-1})_b e_b
\left[ -{\frac{X}{(1 - 1/t)t}} \right] \cr
&= (-1)^{2b} t^{\left( {\ontop{b+1}{ 2}}\right) - b} (t^{-1};t^{-1})_b
h_b \left[ {\frac{X}{(1 - 1/t)}} \right] \cr
&= t^{\left( {\ontop{b}{ 2}} \right) }\omega
H_{(1^b)}[X;t^{-1}]
\end{align*}

So that using the last lemma and the vertex operator property gives
that
\begin{align*}
\omega H_2^{\frac{1}{t}} \omega t^b H_{(1^b)}[X;t] &=
t^{\left( {\ontop{b}{ 2}} \right) +b}
\omega H_2^{\frac{1}{t}} H_{(1^b)}[X;t^{-1}] \cr
&= t^{\left( {\ontop{b+1}{ 2}} \right)} \omega H_{(21^b)}[X;t^{-1}] \cr
&= H_{(1^{b+2})}[X;t] - t^{b+1} H_{(21^b)}[X;t]
\end{align*}
{\endofproof}

\begin{lemma}\label{commHbH}
$$\Hto \Hot = t \Hot \Hto$$
\end{lemma}

\noindent{\bf Proof} \hskip .1in
Let $H(z) = P \left[ X - {\frac{1-t}{z}} \right] \Omega [ z X ]$
so that $H_2 = H(z) \coeff_{z^2} = t^2 H(z/t) \coeff_{z^2}$.

\begin{align*}
\Hto \Hot P[X] &= t^2 \Hto P \left[ X - {\frac{1-t}{(z/t)}} \right]
\Omega[(z/t) X] \coeff_{z^2} \cr
&= t^2 P \left[ tX - {\frac{1-t}{ u}} - {\frac{t-t^2}{ z}} \right]
\Omega \left[ (z/t) \left(tX - {\frac{1-t}{ u}} \right) \right]
\Omega \left[ -uX \right] \coeff_{z^2 u^2} \cr
&= t^2 P \left[ tX - {\frac{1-t}{ u}} - {\frac{t - t^2}{ z}} \right]
\Omega[ zX ] \Omega[ -uX] \Omega \left[ {\frac{z}{ u}} - {\frac{z}{ut}}
\right] \coeff_{z^2 u^2} \cr
&= t^2 P \left[ t \left( X - {\frac{1-t}{ z}} \right) - {\frac{1-t}{ u}} \right]
\Omega[-uX] \Omega[zX] {\frac{1 - {\frac{z}{ut}}}{1 - {\frac{z}{u}}}}
\coeff_{z^2 u^2} \cr
&= t P \left[ t \left( X - {\frac{1-t}{ z}} \right) - {\frac{1-t}{ u}} \right]
\Omega[-uX] \Omega[zX] {\frac{ 1 - {\frac{t u}{z}}}{ 1- {\frac{u}{ z}}}}
\coeff_{z^2 u^2} \cr
&= t P \left[ t \left( X - {\frac{1-t}{ z}} \right) - {\frac{1-t}{ u}} \right]
\Omega[-uX] \Omega[zX] \Omega \left[ u {\frac{1-t}{ z}} \right]
\coeff_{z^2 u^2} \cr
&= t \Hot P \left[ t X - {\frac{1-t}{ u}} \right]
\Omega[-uX] \coeff_{ u^2} \cr
&= t \Hot \Hto P[X]
\end{align*}
{\endofproof}

\begin{lemma}
$$\Hto H_{(2^a1^b)}[X;t] = t^a H_{(2^a1^{b+2})}[X;t]
- t^{a+b+1} H_{(2^{a+1}1^{b})}[X;t]$$
\end{lemma}


\noindent{\bf Proof} \hskip .1in
We show by induction on $a$ that this is true.
By the Lemma \ref{bHonHt}, the statement holds for $a=0$
and by using the previous lemma we have that

\begin{align*}
\Hto H_{(2^a1^{b})}[ X; t]
&= t \Hot \Hto H_{(2^{a-1}1^{b})}[ X; t] \cr
&= t \Hot( t^{a-1} H_{(2^{a-1}1^{b+2})}[X;t]
- t^{a+b} H_{(2^{a}1^{b})}[X;t]) \cr
&= t^a H_{(2^a1^{b+2})}[X;t] - t^{a+b+1} H_{(2^{a+1}, 1^{b})}[X; t]
\end{align*}
{\endofproof}

\sab
Note that there is a bijective proof of this identity that
follows by rewriting the equation as
$$\Hto H_{(2^a1^{b})}[X;t] + t^{a+b+1} H_{(2^{a+1}1^{b})}[X;t]
= t^a H_{(2^a1^{b+2})}[X;t]
$$
and realizing the combinatorial interpretation of each
piece of this equation as the sum over tableaux.
The combinatorial interpretation of the operator
$\Hto$ will be explained later and so
an algebraic proof (except for the first lemma) is provided
here instead.

\sa

\begin{thm}
$$H_2^{qt} H_{(2^a1^{b})}[X;q,t] = H_{(2^{a+1}1^{b})}[X;q,t]$$
\end{thm}

\noindent{\bf Proof} \hskip .1in
For integers $n\geq 0$, define
$$(a;t)_n = (1-a)(1-at) \cdots (1-at^{n-1})$$
In Theorem 1.1 of \cite{St}, an expansion of the $2$-column
Macdonald polynomials in terms of the Hall-Littlewood
polynomials is given as
$$H_{(2^a1^{b})}[X;q,t] = \sum_{i=0}^a q^{a-i} (q t^{a+b}; t^{-1})_i
{\frac{(t^a; t^{-1})_i}{(t^i; t^{-1})_i}} H_{(2^i1^{b+2a-2i})}[X;t]$$
By Lemma \ref{bHonHt}, we have that
\begin{align*}
H_2^{qt} H_{(2^i1^{b+2a-2i})}[X;t] &=
(1-q t^{b+2a+1-i}) H_{(2^{i+1}1^{b+2a-2i})}[X;t]\\ &\hskip .2in
+ q t^i H_{(2^i1^{b+2a-2i+2})}[X;t]
\end{align*}
So then using these two expressions we have that

\begin{align*}
H_2^{qt} H_{(2^a1^{b})}[X;q,t] &=
\sum_{i=0}^a q^{a-i} (q t^{a+b}; t^{-1})_i
{\frac{(t^a; t^{-1})_i}{(t^i; t^{-1})_i}}
H_2^{qt} H_{(2^i1^{b+2a-2i})}[X;t] \cr
&= \sum_{i=0}^a q^{a-i} (q t^{a+b}; t^{-1})_i
{\frac{(t^a; t^{-1})_i}{(t^i; t^{-1})_i}} \cr
& \hskip 1in
( (1-q t^{b+2a+1-i}) H_{(2^{i+1}1^{b+2a-2i})}[X;t] \\
&\hskip 1in + q t^i H_{(2^i1^{b+2a-2i+2})}[X;t] )
\end{align*}


Algebraic manipulation and changing the index of the sums
reduces this expression to one for the symmetric function
$H_{(2^{a+1}1^{b})}[X;q,t]$.

$$
= \sum_{i=0}^{a+1} q^{a-i+1} (q t^{a+b+1}; t^{-1})_{i}
{\frac{(t^{a+1}; t^{-1})_{i}}{(t^i; t^{-1})_i}}
H_{(2^i1^{b+2a-2i+2})}[X;t] = H_{(2^{a+1}1^{b})}[X;q,t]
$$
{\endofproof}

One result that follows from this theorem is that the $H_\mu[X;q,t]$
when $\mu=(2^a1^b)$ has an unusual breakdown into 'atoms' as in
the following formula.

\begin{cor}\label{atoms}
\def\Tscale{.6}
$$H_{(2^a1^b)}[X;q,t] =
\sum_{s \in \left\{ \tableau[Y]{1,2},\tableau[Y]{2|1} \right\}^a}
H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t] q^{\sum_i co(s_i)}$$
where  $co(\tot) = 0$
and $co\left( \tto \right) = 1$.
\end{cor}

The interesting thing about this corollary is that the
symmetric functions $H_2^{s_1} H_2^{s_2} \\ \cdots H_2^{s_a} H_{(1^b)}[X;t]$
are each generating functions for a subset of the standard tableaux
and are all Schur positive.  This will be the main result of the
next section and in the third section we will consider these as the
atoms of the symmetric functions $H_{(2^a1^b)}[X;q,t]$.

Because of the relation from Lemma \ref{commHbH}, for
$\sum_i co(s_i) = k$ we have that
\begin{equation}\label{alliso}
H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t] =
t^x \Hot \cdots \Hot \Hto \cdots \Hto H_{(1^b)}[X;t]
\end{equation}
for some $x \geq 0$
where the $\Hot$ occurs $a-k$ times and $\Hto$ occurs $k$ times.
In fact we may derive the following identity.

\begin{cor}
$$H_{(2^a1^b)}[X;q,t] = \sum_{i=0}^a
\left[ \begin{array}{l}a \\ i\end{array} \right]_t
(\Hot)^{a-i} (\Hto)^{i} H_{(1^b)}[X;t] q^i$$
where
$$\left[ \begin{array}{c}n \\ k\end{array} \right]_t =
\frac{(t^n;t^{-1})_k}{(t^k;t^{-1})_k}$$
\end{cor}

\noindent{\bf Proof} \hskip .1in
\def\Tscale{.6}
Let $T(a) = \left\{ \tableau[Y]{1,2},
\tableau[Y]{2|1} \right\}^a$, the set of tuples of length $a$ with
entries that are standard tableaux of size $2$.
For $s \in T(a)$, let
$co(s) = \sum_i co(s_i)$ and let \\ $inv(s) = \sum_{1\leq j<i\leq a}
\chi\left(co\left(s_i\right)  <  co\left(s_j\right)\right)$.

The expression for $H_{(2^a1^b)}[X;q,t]$ from the previous
corollary and relation in Lemma \ref{commHbH} gives
that
$$H_{(2^a1^b)}[X;q,t] =
\sum_{s \in T(a)}
H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t] q^{co(s)}$$
$$\hskip .95in =\sum_{l=0}^{a} q^l \sum_{\ontop{s \in T(a)}{co(s)=l}}
H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]$$
$$\hskip 1.45in =\sum_{l=0}^{a} q^l \sum_{\ontop{s \in T(a)}{co(s)=l}}
t^{inv(s)} \left( \Hot \right)^{b+2a - l} \left( \Hto \right)^l
H_{(1^b)}[X;t]$$

Note that $\sum_{\ontop{s \in T(n)}{co(s)=k}} t^{inv(s)}$ satisfies the
relations
$$\sum_{\ontop{s \in T(n)}{co(s)=k}} t^{inv(s)}
= t^k \sum_{\ontop{s \in T(n)}{co(s)=k}} t^{inv(s)}
+ \sum_{\ontop{s \in T(n)}{co(s)=k-1}} t^{inv(s)}$$
and $\sum_{\ontop{s \in T(n)}{co(s)=n}} t^{inv(s)} =
\sum_{\ontop{s \in T(n)}{co(s)=0}} t^{inv(s)} = 1$.  The $t$ binomial
coefficient also satisfies the same recursion
$\left[ \begin{array}{c}n \\ k\end{array} \right]_t = t^k
\left[ \begin{array}{c}n-1 \\ k\end{array} \right]_t +
\left[ \begin{array}{c}n-1 \\ k-1\end{array} \right]_t$
and $\left[ \begin{array}{c}n \\ n\end{array} \right]_t =
\left[ \begin{array}{c}n \\ 0\end{array} \right]_t = 1$ therefore
they have the same values.
{\endofproof}

In the next section we will give a combinatorial interpretation to these
polynomials and show that when expanded in terms of Schur functions
that the coefficients are polynomials with non-negative integer coefficients.
The 'atoms' that the Macdonald polynomials break down into are
related to the Butler conjectures of the $K_{\la\mu}(q,t)$.  This relation
will be made more precise in the last section with the exposition of the
tableaux statistics.


The $H_2^{qt}$ operator can be expressed in terms of
the Hall-Littlewood vertex operator and the action of $\Hot$
on the Schur functions is known and given by the formula,
\begin{prop}
Let $\la$ be a partition of $n$, for $k\geq 0$ then
$$H_k^{t} s_\la[X] = \sum_{\rho \in \H_{n+k}} (-1)^{ht_n(\rho)-1}
t^{|\la \slash \rho^r|} s_{\rho \snake_n}[X]$$
\end{prop}

Because of this last proposition, the action of the $H_2^{qt}$
on the Schur functions follows and can be stated as

\begin{prop}
Let $\la$ be a partition of $n$ then
$$H_2^{qt} s_\la[X] = \sum_{\rho \in \H_{n+2}} (-1)^{ht_n(\rho)-1}
t^{|\la \slash \rho^r|} s_{\rho \snake_n}[X]
+ q \sum_{\rho \in \V_{n+2}} (-1)^{{\bar {ht}}_n(\rho)-1} t^{|\rho^c|}
s_{\rho \attic^n}[X]$$
\end{prop}


\section{The Tableaux Operators}
Define the class of {\it x-strict tableaux with n cells}
(denoted by $\XST^n$)
to be the tableaux in the alphabet $\{  i, \iv \}_{i \geq 1}$
with the following restrictions:
\begin{itemize}
\item
For each $i$, the tableau contains either no cells labeled by
$i$ or $\iv$, one cell labeled by $i$ and none by $\iv$,
two cells labeled by $i$ and none by $\iv$,
or two cells labeled by $\iv$ and none by $i$.  No other combinations
are allowed.
\item The cell to the right of $i$ can be labeled with an $i$ or higher.  The
cell above $i$ must be label larger than $i$.
\item The cell to the right of $\iv$ must have a label strictly higher
than $i$.  The cell just above $\iv$ must have a label of $\iv$ or
higher.\end{itemize}

Define the content of $T \in \XST^n$ to be the tuple $s \in
\left\{ \tot, \tto, \to, \cdot\right\}^k$
such that $s_i = \tot$ if
$T$ contains two $\ih$, $s_i = \tto$ if $T$ contains two $\iv$,
$s_i = \to$ if $T$ contains just one cell labeled by $i$
and finally $s_i = \cdot$ if it does not contain $i$ or $\iv$.
Denote the content of the tableau $T$ by the symbol $\mu(T)$.
The standard tableaux of size $n$ are the set of tableaux $T \in XST^n$
such that $\mu(T) = \left( \to^n \right)$.

\vskip .3in

\noindent
{\bf Example }

$$T = \tableau[sY]{6'|4,6'|2',3,3|1,1,2',5}$$

\vskip .1in
$T \in \XST^{10}$ and $\mu(T) = \left(\tot, \tto, \tot,\to,\to,\tto \right)$

\vskip .3in
Let the operation $V$ act on $T \in \XST^n$ such that $\mu(T) = (\to, \to,
s)$.  The operator $V$ changes $T$ to a tableau of either type $(\tot,s)$
or of type $\left( \tto,s \right)$ depending on if the label $2$ lies to the right or
above the $1$ respectively by changing the cells labeled by $1$ and $2$
to either $1$s or $1'$s and decreasing the labels of each of
the cells labeled with a $3$ or higher by $1$.   $V^{-1}$ will be the
operator that acts on $T \in XST^n$ with $\mu(T)_1 = \tot$ or $\tto$
that is the reverse of the operator $V$.

The game of Jeu-de-Taquin may be played on these tableau with the
consideration that the cells have the ordering that
the cell with a label $\iv$ that lies above the other $\iv$ has a value
of $i + \frac{1}{2}$.  This same consideration on the ordering of the cells
allows us to define row and
column insertion and deletion using the usual Robinson-Schenstead
correspondence.
%I think that this is correct the question is does the language make
%sense and is it detailed enough for others to draw the same
%conclusion that I do from it.

Define a symmetric group action on the type of the tableau.  For $T \in \XST$,
let $s = \mu(T)$.  The operator $(i, i+1)$ will have the property
that $\mu((i, i+1)T) = (i,i+1)s = (s_1, \ldots, s_{i+1}, s_i, \ldots, s_k)$.
The operation $(i,i+1)T$ has
the following definition:

\begin{itemize}
\item If $s_i = s_{i+1}$ then $(i, i+1) T = T$.
\item If $s_i \neq s_{i+1}$ then ignore all the cells in
$T$ except those with a label in the set $\{ i,\iv,
i+1,i+1'\}$
and bring them to straight shape.  The possible configurations of these cells
are listed below in pairs.  The action of $(i,i+1)$ is to replace
the configuration by the corresponding one in the same row and then play
Jeu-de-Taquin in reverse to restore the cells to their original position (see
\cite{Fu} to justify that this is a well defined operation).

\vskip .3in
Let $k=i+1$ (just so that in the following diagrams, $i+1$ fits in the cell)
$$\TSY{i} \hskip .5in \longleftrightarrow \hskip .5in \TSY{k}$$
$$\TSY{i'|i'} \hskip .5in \longleftrightarrow \hskip .5in \TSY{k'|k'}$$
$$\TSY{i,i} \hskip .5in \longleftrightarrow \hskip .5in \TSY{k,k}$$
$$\TSY{k|i,k} \hskip .5in \longleftrightarrow \hskip .5in \TSY{k|i,i}$$
$$\TSY{i,k,k} \hskip .5in \longleftrightarrow \hskip .5in \TSY{i,i,k}$$
$$\TSY{k'|k'|i} \hskip .5in \longleftrightarrow \hskip .5in \TSY{k|i'|i'}$$
$$\TSY{k'|i,k'} \hskip .5in \longleftrightarrow \hskip .5in \TSY{i'|i',k}$$
$$\TSY{k'|k'|i,i} \hskip .5in \longleftrightarrow \hskip .5in \TSY{k|i'|i',k}$$
$$\TSY{k'|i,i,k'} \hskip .5in \longleftrightarrow \hskip .5in \TSY{i'|i',k,k}$$
\vskip .3in

\end{itemize}



Define an operator $\Hti$ on x-standard tableaux of content $\mu(T) = (\tot,s)$
or  $\mu(T)=\left( \tto,s \right)$ and transforms it into an x-standard tableau
of content $(s)$ by the following procedure:

1.  If $\mu(T) = (\tot,s)$ then let $R_1$ be the first row of $T$ and
${\wiggle T}$ be $T$ with the first row removed.
Row insert the cells of $R_1$
that are not $1$ into ${\wiggle T}$ from largest to smallest
and decrease each label by $1$
in this new tableau.  The result will be $\Hti T$.

2.  If $\mu(T) = \left( \tto, s\right)$ then let $C_1$ be the first column of $T$ and
let ${\wiggle T}$ be $T$ with the first column removed.
Column insert the cells of $C_1$ that are not
$1$ or $1'$ into ${\wiggle T}$ from largest to smallest
and decrease by $1$ each of the labels of the
cells in this new tableau.  The result will be $\Hti T$.

Clearly, if $\mu(T) = (\tot, s)$ or $\left( \tto, s\right)$ then $\mu(\Hti T) = (s)$.

This operator will be used to define the
{\it type} of a standard tableau.
Let $\mu = (2^a1^b)$.  Let $T$ be a standard tableau
of size $2a+b$.   The $\mu-type$ will represent the orientation of
the ''building blocks'' of the standard tableau.  It will be represented by
the symbol $type_\mu(T)$ and be defined as the tuple of standard tableaux
of size $1$ or $2$ with the following properties:
\begin{itemize}
\item If $a=0$ and $\mu=(1^b)$ then $type_\mu(T) = (\to^b)$.
\item If $a=1$ then $\mu(V T)_1 = \tto$ or $\tot$ and
$type_{(21^b)}(T) = (\mu(V T)_1, \to^b)$.
\item If $a>1$ then $\mu(V T)_1 = \tto$ or $\tot$ and \\
$type_{(2^a1^b)}(T) = (\mu(V T)_1, type_{(2^{a-1}1^b)}(\Hti V T))$
\end{itemize}

We wish to show that there is a relation between the $\mu-type$
of a standard tableau and a method for unstandardization of the tableau
so that the content matches the $\mu-type$.

\begin{lemma}
For a $T \in \XST^n$ ($n \geq 4$) and $\mu(T) = \left(\tto, \to, \to, s\right)$
or $\mu(T) = (\tot, \to, \to, s)$ (where $s$ is the remainder
of the type-list) the tableaux operators have the
following relationship
$$V \Hti T = \Hti (1,2) V (2,3) (1,2) T$$
\end{lemma}

\noindent{\bf Proof} \hskip .1in
The $V$ and $(1,2) V (2,3) (1,2)$ operators
only change the values of the cells that are
labeled with $1,1',2,$ or $3$.  The relative values
of the cells of $T$ do not change so it should be
clear that if we verify this is true for the 10 tableaux
below that it will be true for all x-standard tableaux that contain
these as sub-tableaux.

$$\TSY{~|~|1,1,2,3} \hskip .2in
\TSY{~|2,3|1,1,~} \hskip .2in
\TSY{~|3|1,1,2} \hskip .2in
\TSY{~|1'|1',2,3} \hskip .2in
\TSY{3|1'|1',2}$$
$$\TSY{~|3|2|1,1,~,~} \hskip .2in
\TSY{~|~|2|1,1,3} \hskip .2in
\TSY{~|~|1',3|1',2,~} \hskip .2in
\TSY{~|2|1'|1',3,~} \hskip .2in
\TSY{3|2|1'|1',~}$$
If $T$ is any of the first $5$ tableaux then
$V \Hti T = \Hti (1,2) V (2,3) (1,2) T = \TSY{1,1}$ and
if $T$ is any of the second $5$ tableaux then
$V \Hti T =$ $\Hti (1,2) V (2,3) (1,2) T = \TSY{1'|1'}$.
{\endofproof}

\begin{lemma}
For a $T \in \XST^n$ and for $i>1$
the tableaux operators have the
following relationship
$$(i-1, i) \Hti T = \Hti (i,i+1) T$$
\end{lemma}

\noindent{\bf Proof} \hskip .1in
As in the previous lemma, it is only necessary to check what
these two operators do to the cells that they change.
This means that there is nothing to check if $\mu(T)_i =
\mu(T)_{i+1}$.
A brute force proof this time however has MANY more cases
to check.  For each of the possible 18 configurations
in $(1)$ an application of $\Hti$ will rearrange
the positions of cells labeled by $i$ and $i+1$ if $\mu(T)_1
= \tot$ and there are cells labeled by $i$ or $i+1$ in the
first row of $T$, or $\mu(T)_1 = \tto$ and there are cells
labeled by $i$,$i'$,$i+1$, or $i+1'$  in the first column of $T$.

Take one of the $18$ configurations of the cells
$i$,$i'$,$i+1$, or $i+1'$ in $T$ from $(1)$, we
we will write the possible configurations of the cells
that are labeled by $i-1$,$i-1'$, $i$ and $i'$ after
an application of $\Hti$ to $T$.  We need only verify that the
images of these possible configurations under $(i-1,i)$ are
the same as the possible configurations
of the cells
that are labeled by $i-1$,$i-1'$, $i$ and $i'$ after
an application of $\Hti$ to $(i,i+1) T$.

\vskip .2in
For notational purposes, let $k=i+1$ and $h=i-1$.  Cells that were
in the first row of $T$ and change position because $\mu(T)_1 = \tot$
will have an underline under the label, cells that were in the first
column of $T$ and change position because $\mu(T)_1 = \tto$
will have a bar over the label.

\noindent
$\TSY{i,~,~} \hskip .5in \TSY{h}, \TSY{{\ul h}}, \TSY{{\ol h}}$\\ \\
$\TSY{k,~,~} \hskip .5in \TSY{i}, \TSY{{\ul i}}, \TSY{{\ol i}}$\\ \\
$\TSY{i'|i',~,~} \hskip .5in \TSY{h'|h'}, \TSY{{\ul h'}|h'}, \TSY{h'|{\ol h'}},
\TSY{{\ol h'}|{\ol h'}}$\\ \\
$\TSY{k'|k',~,~} \hskip .5in \TSY{i'|i'}, \TSY{{\ul i'}|i'}, \TSY{i'|{\ol i'}},
\TSY{{\ol i'}|{\ol i'}}$\\ \\
$\TSY{i,i,~} \hskip .5in \TSY{h,h}, \TSY{{\ul h},h}, \TSY{{\ul h},{\ul h}},
\TSY{h,{\ol h}}$\\ \\
$\TSY{k,k,~} \hskip .5in \TSY{i,i}, \TSY{{\ul i},i}, \TSY{{\ul i},{\ul i}},
\TSY{i,{\ol i}}$\\ \\
$\TSY{k|i,k,~} \hskip .5in \TSY{i|h,i}, \TSY{{\ul i}|h,i},
\TSY{~|{\ul h},{\ul i},i}, \TSY{~|h,i,{\ol i}}, \TSY{i|{\ol h},{\ol i}} $\\ \\
$\TSY{k|i,i,~} \hskip .5in \TSY{i|h,h}, \TSY{i|{\ul h},h},
\TSY{~|{\ul h},{\ul h},i}, \TSY{~|h,h,{\ol i}}, \TSY{{\ol i}|h,{\ol h}} $\\ \\
$\TSY{~|i,k,k} \hskip .5in \TSY{~|h,i,i}, \TSY{{\ul i}|h,i}, \TSY{{\ul i}|h,{\ul i}},
\TSY{~|{\ul h}, {\ul i}, {\ul i}}, \TSY{k|{\ol i},k}$ \\ \\
$\TSY{~|i,i,k} \hskip .5in \TSY{~|h,h,i}, \TSY{k|i,{\ul i}}, \TSY{{\ul k}|{\ul i},i},
\TSY{~|{\ul i},{\ul i},{\ul k}}, \TSY{k|i,{\ol i}}$ \\ \\
$\TSY{k'|k'|i,~,~} \hskip .5in \TSY{i'|i'|h}, \TSY{~|i'|{\ul h},i'}, \TSY{~|i'|h,{\ol i'}},
\TSY{~|{\ol i'}|h,{\ol i'}},\TSY{{\ol i'}|{\ol i'}|{\ol h}}$ \\ \\
$\TSY{k|i'|i',~,~} \hskip .5in \TSY{i|h'|h'}, \TSY{~|{\ul h'}|h',i}, \TSY{~|h'|h',{\ol i}},
\TSY{~| h'|{\ol h'},{\ol i}}, \TSY{{\ol i}|{\ol h'}|{\ol h'}}$ \\ \\
$\TSY{~|k'|i,k',~} \hskip .5in \TSY{~|i'|h,i'}, \TSY{{\ul i'}|i'|h},
\TSY{~|{\ul i'}|{\ul h},i'}, \TSY{~|i'|h,{\ol i'}}, \TSY{i'|{\ol i'}|{\ol h}}$ \\ \\
$\TSY{~|i'|i',k,~} \hskip .5in \TSY{~|h'|h',i}, \TSY{{\ul i}|h'|h'},
\TSY{~|{\ul h'}|h',i}, \TSY{~|h'|{\ol h'},i}, \TSY{i|{\ol h'}|{\ol h'}}$ \\ \\
$\TSY{k'|k'|i,i,~} \hskip .5in \TSY{i'|i'|h,h}, \TSY{i'|i'|{\ul h},h},
\TSY{i'|i'|{\ul h}, {\ul h}}, \TSY{~|i'|h,h,{\ol i'}},
\TSY{~|{\ol i'}|h,h,{\ol i'}}, \TSY{{\ol i'}|{\ol i'}|h,{\ol h}}$ \\ \\
$\TSY{k|i'|i',k,~} \hskip .5in \TSY{i|h'|h',i}, \TSY{{\ul i}|h'|h',i},
\TSY{{\ul i}|h'|{\ul h'},i }, \TSY{~|h'|h',i,{\ol i}},
\TSY{~|h'|{\ol h'},i,{\ol i}},
\TSY{i|{\ol h'}|{\ol h'},{\ol i}} $\\ \\
$\TSY{~|k'|i,i,k'} \hskip .5in
\TSY{~|i'|h,h,i'}, \TSY{{\ul i'}|i'|h,h}, \TSY{{\ul i'}|i'|{\ul h},h},
\TSY{~|{\ul i'}|{\ul h},{\ul h},i'}, \TSY{~|i'|h,h,{\ol i'}},
\TSY{i'|{\ol i'}|h,{\ol h}}$\\ \\
$\TSY{~|i'|i',k,k} \hskip .5in
 \TSY{~|h'|h',i,i}, \TSY{{\ul i}|h'|h',i}, \TSY{{\ul i}|h'|h',{\ul i}},
 \TSY{~|{\ul h'}|h',{\ul i},{\ul i}}, \TSY{~|h'|{\ol h'},i,i},
\TSY{i|{\ol h'}|{\ol h'},i}$
{\endofproof}

\vskip .3in
Define the tableau operator $M_1=V$ and $M_i$ for $i\geq 2$
by the composition of the
$\sigma_i = (i,i+1)$ operators and the $V$ operators \\
$M_i = \sigma_{i-1} \sigma_{i-2} \cdots \sigma_1 V
\sigma_2 \sigma_3 \cdots \sigma_{i} \sigma_1 \sigma_2 \cdots \sigma_{i-1}$
Notice that $M_i$ is simply defined so that it has the property
$$M_i \Hti T = \Hti M_{i+1} T$$
These $M_i$ operators are ''unstandardization'' operators in the sense of
the following proposition.

\begin{prop}
Let $\mu=(2^a1^b)$ and $T$ is a standard tableau of size $2a+b$ then
$$\mu(M_{a} M_{a-1} \cdots M_1 T) = type_{(2^a1^b)}(T)$$
\end{prop}

\noindent{\bf Proof} \hskip .1in
Note that the $type_{(2^a1^b)}(T)_i = \mu(V (\Hti V)^i T)_1$.
We observe that $\mu(\Hti T)_j = \mu( T)_{j+1}$ and hence
$type_{(2^a1^b)}(T)_i =
\mu( {(\Hti)}^i M_i M_{i-1} \cdots M_1 T)_1$ $ = \mu(M_i M_{i-1} \cdots M_1 T)_i
= \mu(M_a M_{a-1} \cdots M_1 T)_i$ (and the last equality follows since
the $M_j$ for $j > i$ does not change the $i^{th}$ entry in the content).

{\endofproof}


Define the operator $N_a$ to be the sequence of operators
$M_a M_{a-1} \cdots M_1$.  When $N_a$ acts on a standard tableau,
it maps it to an x-standard tableaux with the relation
$\mu(N_a T) = type_{(2^a1^b)}(T)$ for $T \in ST^{2a+b}$.
This operator is a bijection between standard tableaux and
x-standard tableaux with content that is a tuple in $\left\{ \tot,
\tto \right\}^a \times \{\to\}^b$.

\begin{example}
$$\TSY{8|5,7|4,6|1,2,3} \hskip .3in
\TSY{7|4,6|3,5|1,1,2} \hskip .3in
\TSY{6|3,5|2,4|1,1,2} \hskip .3in
\TSY{5|3',4|2,3'|1,1,2}\hskip .3in
\TSY{4'|3',4'|2,3'|1,1,2}$$
$$\hskip .1in T \hskip .6in N_1 T \hskip .5in N_2 T
\hskip .6in N_3 T \hskip .6in
N_4 T$$
\vskip .2in

$$\TSY{8|5,7|4,6|1,2,3} \hskip .5in
\TSY{~|6|3,5|1,2,4} \hskip .7in
\TSY{~|~|2,4|1,3} \hskip 1in
\TSY{~|~|2|1}$$
$$\hskip .6in T \hskip .7in \Hti V T \hskip .4in
\Hti V \Hti V T \hskip .1in
\Hti V \Hti V \Hti V T$$
\end{example}

Note that the operators $M_i$ are completely reversible so that they
describe a procedure for mapping the standard tableaux of size $2a+b$
bijectively to the x-standard tableaux with
content in the set
$\left\{ \tot, \tto \right\}^a \times \left\{ \to \right\}^b$.


Let $T \in ST^{2a+b}$ and let $\mu$ be a partition with two columns with
$\mu=(2^a1^b)$.
We will let the statistic $b_\mu(T)$ on standard tableaux be
the number of occurrences of $\tto$ in the $type_\mu(T)$.
Let the statistic $a_\mu(T)$ be defined recursively with a base case
of $a=0$ so that $a_{(1^b)}(T) = c(T)$.  For $a>0$ let
$a_\mu(T) = a_{\mu^r}(\Hti V T) + (\la(T)_1 - 2)$ if $type_\mu(T)_1 = \tot$
and $a_\mu(T) = a_{\mu^r}(\Hti V T) + |\la(T)^c|$ if $type_\mu(T)_1 = \tto$.

\begin{prop}  \label{amutype}
Let $n=2a+b$.
The statistic $a_\mu(T)$ where $\mu = (2^a1^b)$
satisfies the formula
$$a_\mu(T) = c(T) - \sum_{i=1}^a ((n+1) - 2i) \chi(type_\mu(T)_i = \tot)$$
\end{prop}

\noindent{\bf Proof} \hskip .1in
By induction on $a$
we assume that it is true for partitions with fewer than $a$
parts equal to $2$ (with the base case of $a=0$ true by definition),
then when $type_\mu(T)_1 = \tot$ we have that
\begin{align*}
a_\mu(T) &= a_{\mu^r}(\Hti V T) + (\la(T)_1 - 2)\\
&= c( \Hti V T) -
\sum_{i=1}^{a-1} ((n-1)-2i) \chi(type_{\mu^r}(\Hti V T)_i = \tot)
+ (\la(T)_1 - 2)\\
&= c( \Hti V T ) - \sum_{i=2}^a ((n+1)-2i) \chi(type_\mu(T)_i = \tot)
+ (\la(T)_1 - 2)
\end{align*}



The charge statistic is well understood and several methods for computing
the charge exist.  For  $type_\mu(T)_1 = \tot$ we have that
$c(\Hti V T) = c(V T) - (\la_1 -2)$ because $\Hti$ is the operation of cyclage
of the $(\la_1 -2)$ cells in the first row and then reducing the content.
We also know that
$c(V T) = c(T) - (n-1)$ by using the word definition of charge.
This implies that $c(\Hti V T) = c(T) - (n-1) - (\la_1 - 2)$ and hence
that
$$a_\mu(T) = c(T) - (n-1) -
\sum_{i=2}^a ((n+1)-2i) \chi(type_\mu(T)_i = \tot) + (\la(T)_1 - 2)
$$

When $type_\mu(T)_1 = \tto$ then a similar calculation gives that
$$a_\mu(T) = c( \Hti V T ) -
\sum_{i=2}^a ((n+1)-2i) \chi(type_\mu(T)_i = \tot) + |\la(T)^c|$$
The charge of
the tableau $\Hti V T$ can be calculated by noting that the $\Hti$
operator is one cyclage operation for every cell in $|\la(T)^c|$
and then reducing the content so that $c(\Hti V T) = c(T) - |\la(T)^c|$.
In both cases $a_\mu(T) =  c(T) - \sum_{i=1}^a ((n+1) - 2i)
\chi(type_\mu(T)_i = \tot)$.
{\endofproof}

We are now ready to state the main theorem of this section.

\begin{thm}\label{atomgf}
Let $\mu=(2^a1^b)$.
For $s \in \left\{ \tot, \tto \right\}^a$
the symmetric functions
$H_2^{s_1} H_2^{s_2} \\ \cdots H_2^{s_a} H_{(1^b)}[X;t]$
are generating functions for the standard tableaux of $\mu-type =
(s,\to^b)$ in the sense that
$$H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]
= \sum_{\ontop{T \in ST^{2a+b}}{type_\mu(T) = (s,\to^b)}}
t^{a_\mu(T)} s_{\la(T)}[X]$$
\end{thm}


This gives the following logical corollary that follows from the
theorem and Corollary {\ref{atoms}}
\begin{cor}
Let $\mu = (2^a1^b)$.  The $H_\mu[X;q,t]$ are generating functions
for the standard tableaux in the sense that
$$H_\mu[X;q,t] = \sum_{T \in ST^{2a+b}} t^{a_\mu(T)} q^{b_\mu(T)} s_{\la(T)}[X]$$
\end{cor}

We will show that this this is true by describing two procedures,
one, $\AH$, that takes as input a standard tableau of
shape $\la \vdash n$ and $\mu-type = s$,
and returns a tableau for every term in the expression
\begin{equation}\label{snakerule}
\Hot s_\la[X] =
\sum_{\rho \slash \la \in \H_{n+2}} (-1)^{ht_n(\rho)-1}
t^{|\la \slash \rho^r|} s_{\rho \snake_n}[X]
\end{equation}
The other
procedure, $\AV$, will take as input a standard tableau of shape $\la$, and
return a tableau for every term in the expression
\begin{equation}\label{atticrule}
\Hto s_\la[X] =
\sum_{\rho \slash \la \in \V_{n+2}} (-1)^{{\bar {ht}}_n(\rho)-1}
t^{|\rho^c|} s_{\rho \attic^n}[X]
\end{equation}

Let $n \geq 0$ and $T \in XST^n$ and then let $\rho$ be any partition
of $2n+2$ such that $\rho \slash \la(T)  \in \H_{n+2}$.  Note that
$\la(T) \slash \rho^r$ is also a horizontal strip.  Create a tableau
$\wiggle T$ such that $\la(\wiggle T) = \rho^r$ by performing one
column evacuation for the each cell in $\la(T) \slash \rho^r$ from
right to left.  Because the bumping paths of the cells do not
cross, the cells will be evacuated in weakly increasing order (that is,
it may contain two cells of label $i$).  Let $R$ be the row of cells
that are evacuated from $T$.  Increase all of the labels of the cells in
$R$ and $\wiggle T$ by one.  Create a new tableau by row inserting
the labels $1$, $1$, and all of the labels in $R$ in increasing
order into $\wiggle T$.
Call this new tableau $\AHr T$.

The purpose of the definition of $\AHr$
is to create a tableau for every term
in the expression
$$\Hot \left(t^{a_\mu(T)} s_{\la(T)}[X]\right) =
\sum_{\rho \slash \la(T) \in \H_{n+2}}
t^{a_\mu(T)+|\la(T) \slash \rho^r|} (-1)^{ht_n(\rho)-1} s_{\rho \snake_n}[X]$$
where we have used the convention that $s_{\rho \snake_n}[X] = 0$
if $\rho \snake_n$ is not a partition.  The tableaux such that
$\la( \AHr T) \neq \rho \snake_n$ will correspond to terms that either cancel or
have weight zero when $\Hot$ acts on the symmetric functions
$H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]$ (where
$s_i = \tot$ or $\tto$).


\begin{example}
$n=6$ and $T = \TSY{1',4|1',2,2,3} \in XST^{6}$.
Consider the three following choices for the partition $\rho$
corresponding to the three types of resulting tableau that will
be created.
\begin{itemize}
\item $\rho = (9,3,2)$ \\
$R = \TSY{1'}$ and ${\wiggle T} = \TSY{2,4|1',2,3}$. \\
$$\AHr T = \TSY{3,5|2',3,4|1,1,2'}$$
$\Hti \AH^{(9,3,2)} T = T$ and $\rho \snake_6 = \la(\AHr T)$.

\item $\rho = (9,4,1)$ \\
$R = \TSY{1'}$ and $\wiggle T = \TSY{4|1',2,2,3}$. \\
$$\AHr T = \TSY{5|2',3,3|1,1,2',4}$$
$\rho \snake_6$ is not a partition.

\item $\rho=(10,4)$\\
$R = \TSY{1',4}$ and ${\wiggle T} = \TSY{1',2,2,3}$. \\
$$\AHr T = \TSY{2',3,3|1,1,2',4,5}$$
$\la(\AH^{(10,4)} T) \neq \rho \snake_6$ but $\rho \snake_6$ is
a partition.
\end{itemize}
\end{example}



Notice that $\AHr T$ falls into one of the three following categories:
\begin{enumerate}

\item $\la(\AHr T) = \rho \snake_n$.  We make
the following observations in this case:
\begin{itemize}
\item $\la(\AHr T) = \rho \snake_n = (\rho_1 -n, \rho^r)$
\item $ht_n(\rho) = 1$
\item $\mu(\AHr T) = (\tot, \mu(T))$
\item For $T \in ST^n$,
$type_{(2^{a+1}1^b)}(V^{-1} \AHr T) = (\tot, type_{(2^a1^b)}(T))$
\item $\Hti \AHr T = T$
\item If $T \in ST^n$ then $a_{(2^{a+1}1^b)}(V^{-1} \AHr T)
= a_{(2^a1^b)}(T) + (\rho_1 - n -2) =
a_{(2^a1^b)}(T) + |\la(T)/\rho^r|$
\end{itemize}

\item $\rho \snake_n$ does not exist.

\item $\rho \snake_n$ exists but $\la(\AHr T) \neq \rho \snake_n$.
We still have $\mu(\AHr T) = (\tot, \mu(T))$, but we no longer have
the relationship $\Hti \AHr T = T$.  About $\la(\AHr T)$ we
can really only say that $\rho \slash \la(\AHr T)$ is a horizontal strip
of size $n$.
\end{enumerate}

Similarly, there is a procedure that adds a
column block of size $2$.
Let $n \geq 0$ and $T \in XST^n$ and then let $\rho$ be any partition
of $2n+2$ such that $\rho \slash \la(T)  \in \V_{n+2}$.  Note that
$\la(T) \slash \rho^c$ is also a vertical strip.  Create a tableau
$\wiggle T$ such that $\la(\wiggle T) = \rho^c$ by performing one
row evacuation for the each cell in $\la(T) \slash \rho^c$ from
top to bottom.  Because the bumping paths of the cells do not
cross, the cells will be evacuated in strictly increasing order (
$C$ may contain two cells of label $i'$).  Let $C$ be the column of cells
that are evacuated from $T$.  Increase all of the labels of the cells in
$C$ and $\wiggle T$ by one.  Create a new tableau by column inserting
the labels $1'$, $1'$, and all of the labels in $C$ in order into $\wiggle T$.
Call this new tableau $\AVr T$.

Again, we observe the following three categories for $\AVr$:
\begin{enumerate}

\item $\la(\AVr T) = \rho \attic^n$.  We make
the following observations in this case:
\begin{itemize}
\item $\la(\AVr T) = \rho \attic^n = (\rho_1, \rho_2, \ldots, \rho_{l(\rho)-n})$
\item ${\bar{ht}}_n(\rho) = 1$
\item $\mu(\AVr T) = \left(\tto, \mu(T)\right)$
\item For $T \in ST^n$,
$type_{(2^{a+1}1^b)}(V^{-1} \AVr T) = \left(\tto, type_{(2^a1^b)}(T)\right)$
\item $\Hti \AVr T = T$
\item If $T \in ST^n$ then $a_{(2^{a+1}1^b)}(V^{-1} \AVr T)
= a_{(2^a1^b)}(T) + (2n +2 - \rho_1') =
a_{(2^a1^b)}(T) + |\rho^c|$
\end{itemize}

\item $\rho \attic^n$ does not exist.

\item $\rho \attic^n$ exists but $\la(\AVr T) \neq \rho \attic^n$.
We still have $\mu(\AVr T) = \left(\tto, \mu(T)\right)$, but we no longer have
the relationship $\Hti \AVr T = T$.  We also have that
$\rho \slash \la(\AVr T)$ is a vertical strip of size $n$.
\end{enumerate}


The operators $\Hti$ and $\AHr$/$\AVr$ are not true inverses of each other,
but $\AHr$ and $\AVr$ are both
invertible and there is no problem reversing the steps
to find $T$ from $\AHr T$ or $\AVr T$ as long as $\rho$ is known.
The $\AVr$ and $\AHr$ operators do have a similar relationship
with the $(i, i+1)$ and $V$ operators.


\begin{lemma}
For a $T \in \XST^n$ ($n \geq 2$) and $\mu(T) = (\to, \to, s)$
(where $s$ is the remainder of the type-list)
the tableaux operators have the
following relationship
$$(1,2) V (2,3) (1,2) \AHr T = \AHr V T$$
$$(1,2) V (2,3) (1,2) \AVr T = \AVr V T$$

\end{lemma}

\begin{lemma}
For a $T \in \XST^n$ and for $1 <  i \leq n$
the tableaux operators have the
following relationship
$$(i, i+1) \AHr T = \AHr (i-1,i) T$$
$$(i, i+1) \AVr T = \AVr (i-1,i) T$$
\end{lemma}

The proofs of these lemmas are nearly the same as in the
corresponding lemmas for the $\Hti$ operator.

The main result that we need from these relationships
can be stated as follows:

\begin{cor}
Let $n=2a + b$ and $T \in ST^{n}$ and let $\rho$ be a partition of
$2n+2$ such that $\rho \slash \la(T) \in \H_{n+2}$ then
$$type_{(2^{a+1}1^b)}(V^{-1} \AHr T) = (\tot, type_{(2^a1^b)}(T))$$
Similarly, if $\rho \slash \la(T) \in \V_{n+2}$ then
$$type_{(2^{a+1}1^b)}(V^{-1} \AVr T) = \left(\tto, type_{(2^a1^b)}(T)\right)$$
\end{cor}

\noindent{\bf Proof} \hskip .1in
 From the previous two lemmas we may derive that $M_{i+1} \AHr
= \AHr M_i$ and $M_{i+1} \AVr = \AVr M_i$
for $i>1$.  Therefore since the relation between the
type of $T \in ST^n$ and the content of $N_a T \in XST^n$ is known then
\begin{align*}
type_{(2^{a+1}1^b)}(V^{-1} \AHr T) &= \mu(N_{a+1} V^{-1} \AHr T) \\
&= \mu( M_{a+1} M_{a} \cdots M_1 V^{-1} \AHr T) \\
&= \mu( M_{a+1} M_{a} \cdots M_2 \AHr T) \\
&= \mu( \AHr M_{a} M_{a-1} \cdots M_1 T) \\
&= ( \tot, \mu(N_a T)) = (\tot, type_{(2^a1^b)}(T))
\end{align*}
The proof is exactly analogous for the statement for the
$\AVr$ operator.
{\endofproof}

This also gives the following result about the $a_\mu$ statistic

\begin{cor}
Let $n=2a + b$ and $T \in ST^{n}$ and let $\rho$ be a partition of
$2n+2$ such that $\rho \slash \la(T) \in \H_{n+2}$ then
$$a_{(2^{a+1}1^b)}(V^{-1} \AHr T) = a_{(2^{a}1^b)}(T) +
|\la(T) \slash \rho^r|$$
and if $\rho \slash \la(T) \in \V_{n+2}$ then
$$a_{(2^{a+1}1^b)}(V^{-1} \AVr T) = a_{(2^{a}1^b)}(T) +
|\rho^c|$$
\end{cor}

\noindent{\bf Proof} \hskip .1in
Using the previous lemma and Proposition \ref{amutype}
we know that
\begin{align*}
a_{(2^{a+1}1^b)}(V^{-1} \AHr T) &= c(V^{-1} \AHr T) - (n+1) \\
& \hskip .1in- \sum_{i=2}^{a+1} ((n+3) - 2i) \chi( type_{(2^a1^b)}(T)_{i-1} = \tot ) \\
&= c(T) + |\la(T) \slash \rho^r| + (n+1)- (n+1)\\
& \hskip .1in - \sum_{i=1}^a ((n+1) - 2i) \chi(type_{(2^a1^b)}(T)_i = \tot) \\
&= a_{(2^{a}1^b)}(T) + |\la(T) \slash \rho^r|
\end{align*}
using the methods that we have to calculate the charge.

Similarly for the $\AVr$ operator, we have that
\begin{align*}
a_{(2^{a+1}1^b)}(V^{-1} \AVr T) &= c(V^{-1} \AVr T) -
\sum_{i=2}^{a+1} ((n+3) - 2i) \chi( type_{(2^a1^b)}(T)_{i-1} = \tot ) \\
&= c(T) + |\rho^c| -
\sum_{i=1}^a ((n+1) - 2i) \chi(type_{(2^a1^b)}(T)_i = \tot) \\
&= a_{(2^a1^b)}(T) + |\rho^c|
\end{align*}
{\endofproof}

The result of this is that we have a formula for the action of
$\Hot$ and $\Hto$ on $t^{a_\mu(T)} s_{\la(T)}[X]$ for
a standard tableau $T$ in terms of pictures that follows directly from
equations (\ref{snakerule}), (\ref{atticrule}) and the previous
corollaries.

\begin{prop}\label{Hotaction}
Let $T \in ST^n$ and $\mu = (2^a1^b)$ where $n=2a+b$ then
$$\Hot \left( t^{a_\mu(T)} s_{\la(T)}[X]\right) = \sum_{\rho \slash \la(T) \in \H_{n+2}}
(-1)^{ht_n(\rho)-1} t^{a_\mu(V^{-1} \AHr T)} s_{\rho \snake_n}[X]$$
\end{prop}

and
\begin{prop}\label{Htoaction}
Let $T \in ST^n$ and $\mu = (2^a1^b)$ where $n=2a+b$ then
$$\Hto t^{a_\mu(T)} s_{\la(T)}[X] = \sum_{\rho \slash \la(T) \in \V_{n+2}}
(-1)^{{\bar {ht}}_n(\rho)-1} t^{a_\mu(V^{-1} \AVr T)} s_{\rho \attic^n}[X]$$
\end{prop}

We are now ready to prove Theorem \ref{atomgf}.

\noindent{\bf Proof} \hskip .1in (of Theorem \ref{atomgf})
Let $\mu=(2^a1^b)$ and $n=2a+b$ and $s \in \left\{ \tot, \tto \right\}$.
If $a=0$ then it is a well known result that $H_{(1^b)}[X;t] = \sum_{T \in
ST^b} t^{c(T)} s_{\la(T)}[X]$ so the base case is true.

Assume that
$$H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]
=  \sum_{\ontop{T \in ST^{2a+b}}{type_\mu(T) = (s,\to^b)}}
t^{a_\mu(T)} s_{\la(T)}[X]$$
then by Proposition \ref{Hotaction} we have that

\begin{align}
\Hot H_2^{s_1} &H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]
=  \label{} \\
&\sum_{\ontop{T \in ST^{2a+b}}{type_\mu(T) = (s,\to^b)}}
\sum_{\rho \slash \la(T) \in \H_{n+2}} (-1)^{ht_n(\rho)-1}
t^{a_{(2,\mu)}(\AHr T)} s_{\rho \snake_n}[X] \label{H2ongf}
\end{align}

As was mentioned in the definition of $\AHr$, $\AHr T$ falls into
three categories.  If $\la(\AHr T) = \rho \snake_n$ then
we have that $ht_n(\rho)-1 = 0$ and so
$$ (-1)^{ht_n(\rho)-1} t^{a_{(2,\mu)}(\AHr T)} s_{\rho \snake_n}[X]
= t^{a_{(2,\mu)}(\AHr T)} s_{\la(\AHr T)}[X]$$

If $\rho \snake_n$ does not exist then $s_{\rho\snake_n}[X] = 0$.
These terms may be ignored since they do not change the sum (\ref{H2ongf}).

If $\rho \snake_n$ does exist but $\la(\AHr T) \neq \rho \snake_n$ then
the corresponding term
$(-1)^{ht_n(\rho)-1} \\
t^{a_{(2,\mu)}(\AHr T)} s_{\rho \snake_n}[X]$ will cancel, but it is
necessary to demonstrate a sign reversing
involution on this set.  We require the following lemma

\begin{lemma}
There exists an involution $I_\la^n$ on partitions $\rho$ such
that $\rho \slash \la \in \H_n$, $\rho \snake_n$ exists and
$\la \neq \rho \snake_n$ with the property that $ht_n(I_\la^n(\rho)) =
ht_n(\rho) \pm 1$ and $\rho \snake_n = I_\la^n(\rho) \snake_n$.
\end{lemma}

\noindent{\bf Proof} \hskip .1in  (of Lemma)  Let $h=ht_n(\rho)$.  Let $\gamma = \rho
\snake_n$.
$I_\la^n$ maps the set of $\rho$ that satisfy the conditions of the lemma with
$\la_h > \gamma_h$ to the set of $\rho$ that satisfy the conditions of the
lemma with $\la_h \leq \gamma_h$.

If $\la_h > \gamma_h$ then $I_\la^n(\rho) = \gamma$ with an $n-snake$ of
height $h+1$ added.  If $\la_h \leq \gamma_h$ then $I_\la^n(\rho) = \gamma$
with an $n-snake$ of height $h-1$ added (note that if $h=1$ and
$\la_1 \leq \gamma_1$ then $\la = \gamma$).

If $\la_h > \gamma_h$ then $\la_{h+1}
\leq \rho_{h+1} = \gamma_{h+1}$ so that $\la_{h+1} \leq \gamma_{h+1}$.
Also if $\la_h \leq \gamma_h$ then $\la_{h-1} \geq
\rho_{h} > \rho_{h} - 1 = \gamma_{h-1}$ so that $\la_{h-1} > \gamma_{h-1}$.
These two statements together show that $I_\la^n$ is an involution.
{\endofproof}

With this lemma we have a sign reversing involution on the tableaux
$\AHr T$ such that $\la(\AHr T) \neq \rho \snake_n$.  For if there is
a tableau $\AHr T$ with the property that $\la(\AHr T) \neq \rho$
and $\rho \snake_n$ exists then if we let
$\wiggle \rho = I_\la^n(\rho)$, the involution says
that $S = (\AH^{\wiggle \rho})^{-1} \AHr T$ also has this property
and $  (-1)^{ht_n(\rho)-1} t^{a_{(2,\mu)}(\AHr T)} s_{\rho \snake_n}[X] =
-  (-1)^{ht_n(\wiggle \rho)-1}
t^{a_{(2,\mu)}(\AH^{\wiggle \rho} S)} s_{\wiggle \rho \snake_n}[X]$.

Therefore equation (\ref{H2ongf}) has only positive terms and
the negative ones cancel.  The terms that survive all have the
property that $ \\ \Hti \AHr T = T$.  Every tableau $T$ such
that $type_\mu(T) = (\tot, s)$ will correspond to exactly one
of the terms in this sum since $\AH^{(n+\la(T)_1,\la(T)^r)} \Hti T = T$.
Therefore equation (\ref{H2ongf}) becomes
$$ \Hot H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]
= \sum_{\ontop{T \in ST^{2a+b+2}}{type_{(2,\mu)}(T) = \left(\tot,s,\to^b\right)}}
t^{a_{(2,\mu)}(T)} s_{\la(T)}[X]
$$

The proof that
$$ \Hto H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^b)}[X;t]
= \sum_{\ontop{T \in ST^{2a+b+2}}{type_{(2,\mu)}(T) = \left(\tto,s,\to^b\right)}}
t^{a_{(2,\mu)}(T)} s_{\la(T)}[X]
$$ is exactly analogous.
{\endofproof}


\section{More About Statistics on Tableaux}

Some very interesting properties about the standard tableaux
follow from the definitions in the previous section.  The 'atoms'
of the Macdonald polynomials and the $\mu-type$ of the standard
tableaux suggest that the tableaux naturally fall into standard
tableaux classes.  For a sequence $s \in \left\{ \tot, \tto \right\}^a
\times \left\{ \to \right\}^b$ set $STC^s = \{ T \in ST^{2a+b} |
type_{(2^a1^b)}(T) = s\}$.

It will never be clear how beautiful this breakdown of the standard tableaux
into classes is until the picture of where the 'atoms' lie in
the standard tableaux
when they are ranked by the charge is clear.  The figures at the end of this
paper are the posets of the standard tableaux of size $4, 5$ and $6$ when
they are ranked by  the charge.  The standard tableau classes are grouped
together in this poset and shaded so that each class is separated.  The
horizontal position of each tableau is slightly related to cyclage, but
not as much as it was in the case of the column strict tableaux.
Many of the properties of the Macdonald polynomials can be observed in
these diagrams (especially my favorite: $\omega H_{(2^a1^b)}[X;q,t] =
q^a t^{n(2^a1^b)} H_{(2^a1^b)}[X;1/q,1/t]$) and expansions for
$H_{(2^a1^b)}[X;q,t]$ in terms of Schur functions can be
immediately written down.

\begin{example}
The tableaux class $STC^{\left(\tot, \tto, \tto \right)}$ is given by
$$\begin{array}{ccc}
&{\tableau[sbY]{\bl|6|5|4|3|1,2|\bl}}& \\
&{\tableau[sbY]{\bl|5|4|3|1,2,6|\bl}} &
{\tableau[sbY]{\bl|5|4|3,6|1,2|\bl}} \\
{\tableau[sbY]{\bl|4,6|3,5|1,2|\bl}} &
{\tableau[sbY]{\bl|6|4|3|1,2,5|\bl}} &
{\tableau[sbY]{\bl|4|3,6|1,2,5|\bl}} \\
{\tableau[sbY]{\bl|6|3,5|1,2,4|\bl}} &
{\tableau[sbY]{\bl|4|3|1,2,5,6|\bl}} & \\
&{\tableau[sbY]{\bl|3,5|1,2,4,6|\bl}}&
\end{array}
$$

The symmetric function that this corresponds to is
$$\Hot \Hto \Hto 1 = s_{(21111)}[X] + t s_{(3111)}[X] + t s_{(2211)}[X]
+t^2 s_{(222)}[X] + t^2 s_{(3111)}[X]$$
$$\hskip .3in + t^2 s_{(321)}[X] +t^3 s_{(321)}[X]
+t^3 s_{(411)}[X] + t^4 s_{(42)}[X]$$
\end{example}


These classes have the property that $STC^{\left(s,\tot,\to^b\right)}
\cup STC^{\left(s,\tto,\to^b\right)} = STC^{(s,\to^{b+2})}$ simply by
definition of the type.  There is also a relation between the $a_\mu$ and
$b_\mu$ statistics over this set of tableaux.

\begin{prop} \label{butler}
If $type_{(2^{a+1}1^b)}(T)_{a+1} = \tto$ then
$a_{(2^{a+1}1^b)}(T) = a_{(2^a1^{b+2})}(T)$ and $b_{(2^{a+1}1^b)}(T) =
b_{(2^{a+1}1^b)}(T) +1$.

If $type_{(2^{a+1}1^b)}(T)_{a+1} = \tot$ then
$a_{(2^{a+1}1^b)}(T) = a_{(2^a1^{b+2})}(T)+ (b+1)$ and $b_{(2^{a+1}1^b)}(T) =
b_{(2^{a+1}1^b)}(T)$.
\end{prop}

\begin{example}
$$T = \TSY{4,6|2,5|1,3}$$
$type_{(2^3)}(T) = {\left(\tto,\tot,\tto\right)}$

$a_{(2^3)}(T) = 3$, $a_{(2^21^2)}(T) = 3$,
$a_{(21^4)}(T) = 6$, $a_{(1^6)}(T) = 6$

$b_{(2^3)}(T) = 2$, $b_{(2^21^2)}(T) = 1$,
$b_{(21^4)}(T) = 1$, $b_{(1^6)}(T) = 0$

\end{example}

This relationship is consistent with observations made by Lynne Butler
\cite{Bu} about adjacent rows of the $q,t$-Kostka
matrix.  Comparing $K_{\la(2^a1^{b+2})}(q,t)$ to $K_{\la(2^{a+1}2^b)}(q,t)$,
one notices that every term either changes by a factor of $q$ or a factor
of $t^{b+1}$.

In fact we derive the following Corollary from the proposition.
\begin{cor}
For $s \in \left\{ \tot, \tto \right\}^a$ we have the following
relationship

\begin{align*}
H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_2^\tto &H_{(1^b)}[X;t]+
t^{b+1} H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} \Hot H_{(1^b)}[X;t]\\
&= H_2^{s_1} H_2^{s_2} \cdots H_2^{s_a} H_{(1^{b+2})}[X;t]
\end{align*}
\end{cor}

\vskip .2in
Notice that equation (\ref{alliso}) suggests that the standard
tableaux classes are isomorphic if the type has the same number
of occurrences of $\tto$.  The isomorphism between these classes
is given by the composition of the operators that were introduced
in the last section.  The $N_a$ (invertible) operator changes
a standard tableau in the class $STC^s$ to an x-standard tableau
such that the content is $s$.  The $(i, i+1)$ operators change the
tableau to define a symmetric group action on the content.  To
make this more precise we define the bijection in the following
proposition:

\begin{prop}
For $1 \leq i < a$ and $s \in \left\{ \tot, \tto \right\}^a \times
\left\{ \to^b \right\}$
the operator $N_{a}^{-1} (i, i+1) N_{a}$ is a bijection between
$STC^s$ and $STC^{(i,i+1)s}$.  Furthermore this operator has the
property that if $s_i = \tot$ and $s_{i+1} = \tto$ then
$a_{(2^a1^b)}(N_{a}^{-1} (i, i+1) N_{a} T) = a_{(2^a1^b)}(T) +1$
and $b_{(2^a1^b)}(N_{a}^{-1} (i, i+1) N_{a} T) = b_{(2^a1^b)}(T)$.
\end{prop}

\begin{example}
$$\TSY{4,6|2,5|1,3} \hskip .5in \TSY{4,6|3,5|1,2} \hskip .5in
\TSY{5,6|2,4|1,3}$$
$$\hskip .35in T \hskip .3in N_3^{-1}(1,2) N_3 T\hskip .15in
N_3^{-1} (2,3) N_3 T$$
$T$ has $type_{(2^3)}(T) = \left(\tto, \tot, \tto \right)$, so
$type_{(2^3)}(N_3^{-1}(1,2) N_3 T) = \left(\tot, \tto, \tto \right)$
and \\ $type_{(2^3)}(N_3^{-1}(2,3) N_3 T) = \left(\tot, \tto, \tto \right)$.

$a_{(2^3)}(T) = 3$, $a_{(2^3)}(N_3^{-1}(1,2) N_3 T) = 2$,
$a_{(2^3)}(N_3^{-1} (2,3) N_3 T) = 4$.
\end{example}

\vskip .2in

Taking the transpose (flipping the shape and entries of the diagram about
the $x=y$ line) of a standard tableau $T$ will be represented by the operator
$\omega T$.  It has the property
that if the $type_\mu(T) = s$ then the $type_\mu(\omega T) = (\omega s_1,
\omega s_2, \ldots, \omega s_k)$.  This follows directly from the
definition of the $\mu-type$ since $\Hti V \omega T = \omega \Hti V T$.
This gives a simple method for computing the $a_\mu$ and $b_\mu$ statistics
of $\omega T$ from the $a_\mu$ and $b_\mu$ statistics of $T$.

\begin{prop}
$a_{(2^a1^b)}( \omega T) = \left(  \ontop{a+b}{2} \right) +
\left(  \ontop{a}{2} \right) -
a_{(2^a1^b)}(T)$ and
$b_{(2^a1^b)}(\omega T) = a - b_{(2^a1^b)}(T)$.
\end{prop}

This result follows from the fact that
$c(\omega T) = \left(  \ontop{n}{2} \right) - c(T)$ and
Proposition \ref{amutype} and the definition of $b_\mu$.
This is consistent with the symmetric function identity
$\omega H_\mu[X;q,t] = q^{n(\mu')} t^{n(\mu)} H_\mu[X;1/q,1/t]$
since $n((2^a1^b)) = \left(  \ontop{a+b}{2} \right) +
\left(  \ontop{a}{2} \right)$ and $n((2^a1^b)') = a$.

Before the end of this paper, we would like to point out
a less obvious observation about the standard tableaux classes.
After a conversation with Will Brockman on about the Hall-Littlewood
polynomials, he showed me several conjectures about the
number of standard tableaux that fall in a catabolism type when ranked by
charge.  Since
the standard tableaux classes that we have defined here are
generalizations for the catabolism type, it seems likely that
the same conjectures will hold true for these classes.
Again we let $s \in \left\{ \tot, \tto \right\}^a \times \left\{
\to \right\}^b$ then define the symbol $A_s^i = \# \{ T | T \in STC^s,
a_{(2^a1^b)}(T) = i \}$.

\begin{conjecture}
The sequence $A_s^* = (A_s^0, A_s^1, A_s^2, \ldots )$ is a unimodal sequence
(that is, it increase and then decreases).
\end{conjecture}

\begin{example}
\begin{align*}
\mu = (2^3) \hskip .3in s &= \left( \tot, \tot, \tot \right) \hskip .3in
A_s^* = (1,1,2,3,2,1,1) \\
s&= \left( \tot, \tot, \tto \right) \hskip .3in A_s^* = (1,2,3,2,1) \\
s&= \left( \tot, \tto, \tto \right) \hskip .3in A_s^* = (1,2,3,2,1) \\
s&= \left( \tto, \tto, \tto \right) \hskip .3in A_s^* = (1,1,2,3,2,1,1)
\end{align*}
\begin{align*}
\mu = (2^21^2) \hskip .3in s &= \left( \tot, \tot, \to^2 \right) \hskip .3in
A_s^* = (1,3,4,4,4,2,1,1) \\
s&= \left( \tot, \tto, \to^2 \right) \hskip .3in A_s^* = (1,2,4,4,4,2,1) \\
s&= \left( \tto, \tto, \to^2 \right) \hskip .3in A_s^* = (1,1,2,4,4,4,3,1)
\end{align*}
\begin{align*}
\mu = (2^11^4) \hskip .3in s &= \left( \tot, \to^4 \right) \hskip .3in
A_s^* = (1,2,4,5,7,6,5,4,2,1,1) \\
s&= \left( \tto, \to^4 \right) \hskip .3in A_s^* = (1,1,2,4,5,6,7,5,4,2,1)
\end{align*}
$$\mu = (1^6) \hskip .3in s = (\to^6) \hskip .3in
A_s^* = (1,1,2,4,5,7,9,9,9,9,7,5,4,2,1,1)$$

We list here the $A_s^*$ sequences for only the classes $\left( \tot^l,
\tto^{a-l}, \to^b \right)$ since the other classes are isomorphic to these.
By the observations from Proposition \ref{butler} we know that
for $s \in \left\{ \tot, \tto \right\}^a$ we have that
the sequence $A_{\left(s, \to^{b+2} \right)}^*$ can be calculated from the sequences
$A_{\left(s, \tot, \to^b\right)}^*$ and $A_{\left(s, \tto, \to^b\right)}^*$ since
$A_{\left(s, \to^{b+2} \right)}^i = A_{\left(s, \tot, \to^b\right)}^{i-b-1}
+ A_{\left(s, \tto, \to^b\right)}^i$.
\end{example}

\vfill

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\vspace{.5in}

\noindent\texttt{zabrocki@math.ucsd.edu}

\newpage

\begin{figure}[htbp]
\vskip -.25in
\hskip -.75in
\includegraphics{c4c.eps} % for color figs
%\includegraphics{c4g.eps}
\vskip 6.75in
\caption{Charge Poset for $n=4$}
\end{figure}
\newpage

\begin{figure}[htbp]
\hskip .5in
\includegraphics{c5c.eps} % for color figs
%\includegraphics{c5g.eps}
\vskip 6.75in
\caption{Charge Poset for $n=5$}
\end{figure}
\newpage

\begin{figure}[htbp]
\hskip .75in
\includegraphics{c6c.eps} % for color figs
%\includegraphics{c6g.eps}
\vskip 7in
\caption{Charge Poset for $n=6$}
\end{figure}



\end{document}