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\topmatter 
\title A bijective proof of the
hook-content formula for super Schur functions\\
and a modified jeu de taquin
\endtitle 
\author C.~Krattenthaler\footnote"$^\dagger$"{Supported in part by EC's Human
Capital and Mobility Program, grant CHRX-CT93-0400 and the\linebreak 
\hbox{Austrian Science Foundation FWF, grant P10191-MAT}}
\endauthor 
\affil 
Institut f\"ur Mathematik der Universit\"at Wien,\\
Strudlhofgasse 4, A-1090 Wien, Austria.\\
e-mail: KRATT\@Pap.Univie.Ac.At
\endaffil
\address Institut f\"ur Mathematik der Universit\"at Wien,
Strudlhofgasse 4, A-1090 Wien, Austria.
\endaddress
%\email KRATT\@Pap.Univie.Ac.At\endemail
\dedicatory Dedicated to Dominique Foata\enddedicatory
%\date \enddate
%\thanks \endthanks
\subjclass Primary 05A15;
 Secondary 05A17, 05A30, 05E10, 05E15, 11P81
\endsubjclass
\keywords Plane partitions, tableaux, hook-content formula,
Hillman--Grassl algorithm, jeu de taquin, super Schur function
\endkeywords
\abstract 
A bijective proof of the product formula for the principal
specialization of super Schur functions (also called hook Schur functions)
is given using the combinatorial description of super Schur functions
in terms of certain tableaux due to Berele and Regev. Our
bijective proof is based on the Hillman--Grassl algorithm and
a modified version of Sch\"utzenberger's jeu de taquin. We then
explore the relationship between our modified jeu de taquin and a
modified jeu de taquin by Goulden and Greene. We define a common
extension and prove an invariance property for it, thus discovering
that both modified jeu de taquins are, though different, equivalent.


\endabstract
\endtopmatter

\font\smcp=cmcsc8
\headline={\ifnum\pageno>1 {\smcp the electronic journal of  
combinatorics 3 (1996), \#R14\hfill\folio} \fi}

\document

\subhead 1. Introduction\endsubhead
Let $\la$ be a partition, i.e\. $\la=(\la_1,\la_2,\dots,\la_r)$ is a
sequence of integers with
$\la_1\ge\la_2\ge\dots\ge\la_r>0$. The {\it super Schur function\/}
$\S_\la(\bold x,\bold y)$ (or {\it hook Schur function}), where $\bold
x=(x_1,x_2,\dots)$ and $\bold y=(y_1,y_2,\dots)$ are two sequences of
indeterminates, is defined by
$$\S_\la(\bold x,\bold y)=\det_{1\le i,j\le r}(\S_{\la_i-i+j}(\bold
x,\bold y)),\tag1.1$$
where the series $\S_m(\bold x,\bold y)$ are determined by their
generating function
$$\sum _{m\ge 0} ^{}\S_m(\bold x,\bold y)z^m=\prod _{i\ge1} ^{}\frac
{1+y_iz} {1-x_iz}.\tag1.2$$
Super Schur functions appear naturally in the representation theory of
Lie superalgebras (cf\. \cite{\BeReAA, \BeRJAA, \KacVAC, \KacVAD,
\KacVAE, \ScheAA}), and were also discovered independently under the
name of supersymmetric polynomials \cite{\MeNRAA}.

In order to be able to state the formula alluded to in the title, we
have to recall some basic notions
from partition theory. The {\it Ferrers
diagram\/} of a partition $\lambda=(\la_1,\la_2,\dots,\la_r)$
is an array of cells with $r$ left-justified rows and $\lambda_i$
cells in row $i$. Figure~1.a shows the Ferrers
diagram corresponding to $(4,3,3,1)$.
The {\it 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 diagram of
$\lambda$. We label the cell in the $i$-th row and $j$-th column of
(the Ferrers diagram of) $\la$ by the pair $(i,j)$. Also, if we write
$\rh\in\la$ we mean `$\rh$ is a cell of $\la$'. The {\it hook length\/}
$h_\rh$ of a cell $\rh=(i,j)$ of $\la$ is $(\la_i-j)+(\la'_j-i)+1$, \
the number of cells in the
{\it hook\/} of $\rh$, which is the set of cells that are either in
the same row as $\rh$ and to the right of $\rh$, or in the same column
as $\rh$ and below $\rh$, $\rh$ included. The {\it content\/} $c_\rh$
of a cell $\rh=(i,j)$ of $\la$ is $j-i$.
\vskip10pt
\vbox{
$$
\gather
\smatrix \format \sa\c\quad \s\c\quad \s\c\quad \s\c\quad \se\\
\hlinefor9\\
&\vphantom{f}&& && && &\\
\hlinefor9\\
&\vphantom{f}&& && &\\
\hlinefor7\\
&\vphantom{f}&& && &\\
\hlinefor7\\
&\vphantom{f} &\\
\hlinefor3
\endsmatrix
\hskip1cm
\smatrix \format\sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&\hbox to10pt{\hss 1\hss}&&\hbox to10pt{\hss 3\hss}&&\hbox to10pt{\hss
3\hss}&&\hbox to10pt{\hss $\3$\hss}&\\
\hlinefor9\\
&2&&\1&&\3&\\
\hlinefor7\\
&4&&\2&&\3&\\
\hlinefor7\\
&\2&\\
\hlinefor3
\endsmatrix\\
 \text{\eightpoint \hskip1cm a. Ferrers diagram\quad \quad 
b. super semistandard tableau}
\endgather
$$
\centerline{\eightpoint Figure 1}
}
\vskip10pt
Now we are in the position to state the hook-content formula for the
principal specialization of super Schur functions.

\proclaim{Theorem 1}Let $\la=(\la_1,\la_2,\dots,\la_r)$ be a
partition. For the specialization $x_i=aq^i$, $y_i=bq^i$,
$i=1,2,\dots$, in the super Schur function $\S_\la(\bold x,\bold y)$
there holds
$$\S_\la((aq,aq^2,aq^3,\dots),(bq,bq^2,bq^3,\dots))
=q^{\sum _{i=1} ^{r}i\la_i}\prod _{\rh\in\la} ^{}\frac
{a+bq^{c_\rh}} {1-q^{h_\rh}}.\tag1.3$$ 
\endproclaim
Theorem~1 can be found, as everything about symmetric functions, in a
slightly different, but equivalent form, in Macdonald's book
\cite{\MacdAC, p.~27, Ex.~5, p.~45, Ex.~3}.

For our bijective proof of this hook-content formula we use the
combinatorial description of super Schur functions in terms of what
I call {\it super semistandard tableaux\/} (see section~2 for the
definition), originally introduced 
by Berele and Regev \cite{\BeReAA} under the name of
{\it $(k,l)$-semistandard tableaux\/}, a name however which does not
make sense in 
our context. Our proof is based on the Hillman--Grassl algorithm
\cite{\HiGrAA} and on Sch\"utzenberger's \cite{\SchuAA} jeu de taquin.
It is inspired by our bijective proof \cite{\KratAY} of Stanley's
hook-content formula for the generating function for column-strict
reverse plane partitions of a given shape with bounded entries. Though there
are quite a few similarities with \cite{\KratAY}, it turns out that
the algorithms and arguments in \cite{\KratAY} have to be adapted
quite a bit in order to provide a bijective proof of (1.3).

We recall all the needed tableau and plane partition definitions in
section~2. Then, in section~3, we present our bijective proof of the
hook-content formula (1.3) for super Schur functions. In the appendix
we carry out a complete example for our bijection. After a first
version of this article was written, I discovered that a jeu de
taquin procedure similar to ours already appeared in a paper by
Goulden and Greene \cite{\GoGrAA}. In fact, they are different
{\it algorithms\/}, but it turns out that as {\it mappings\/} they
are the same. We explore the relationship between the two algorithms
in section~4. In order to prove the equivalence of the two algorithms, we
define a joint extension of the algorithms, and establish an
invariance property of the underlying modified jeu de taquin.

\subhead 2. Super semistandard tableaux\endsubhead
Super semistandard tableaux are hybrids of column-strict reverse plane
partitions and row-strict reverse plane partitions. We are going
to define the latter first. Let $\la=(\la_1,\la_2,\dots,\la_r)$ be a
partition. A {\it reverse plane
partition of shape\/} $\la$ is a filling of the cells of (the Ferrers
diagram of) $\la$
with entries from an ordered alphabet such that the entries along rows
and along columns are 
weakly increasing. A reverse plane partition is called {\it
column-strict\/} if in addition columns are {\it strictly\/}
increasing. Likewise, a reverse plane partition is called {\it
row-strict\/} if rows are {\it strictly\/}
increasing. We will also consider {\it skew shapes\/} for row-strict
reverse plane partitions and column-strict
reverse plane partitions. If $\la$ and $\mu$ are two partitions with
$\la\ge\mu$ (i.e\. the Ferrers diagram of $\mu$ is contained in the
Ferrers diagram of $\la$), then the {\it skew shape\/} $\la/\mu$
consists of all cells that are contained in $\la$ but not in $\mu$. As
expected, by a {\it row-strict \rm(\it respectively column-strict\rm) \it 
reverse plane partition of shape
$\la/\mu$\/} we mean a filling of the cells of $\la/\mu$ with entries
from some ordered alphabet such that columns (respectively rows) 
are weakly increasing and
rows (respectively columns) are strictly increasing.

Now, a {\it super semistandard tableau of shape\/} $\la$ is a filling $S$
of the cells of $\la$ with entries from the ordered alphabet
$1<2<3<\dots<\1<\2<\3<\cdots$ such that
\roster 
\item "(SST1)"the unbarred entries form a column-strict reverse plane
partition of some shape $\nu$, where $\nu$ is a partition,
\item "(SST2)"the barred entries form a row-strict reverse plane
partition of shape $\la/\nu$.
\endroster
Figure~1.b shows a super semistandard tableau of shape $(4,3,3,1)$. We
call the part of a super semistandard tableau $S$ that is occupied by
the unbarred entries the {\it inner part\/} of $S$ and denote it by
$S^I$. Likewise, we
call the part of $S$ that is occupied by
the barred entries the {\it outer part\/} of $S$ and denote it by
$S^O$.

The combinatorial interpretation of the super Schur function
$\S_\la(\bold x,\bold y)$ is
(here SST is short for `super semistandard tableau')
$$\S_\la(\bold x,\bold y)=\underset \text{of shape }\la\to{\sum
_{S\text{ a SST}} ^{}} \prod _{i\ge1} ^{}x_i^{\text{\#$i$'s in
$S$}}y_i^{\text{\#$\bar i$'s in $S$}},\tag2.1$$
which can be proven combinatorially by the Gessel--Viennot method
\cite{\GeViAA, \GeViAB} of nonintersecting lattice paths as shown by
Remmel \cite{\RemmAC, sec.~5} (see also Brenti \cite{\BretAE, Theorem~3.4}).

For convenience, we introduce the following two terms. Given any
filling $P$ (a super semistandard tableau, or a reverse plane
partition, or \dots) of some (possibly skew) shape,
we write $P_\rh$, respectively $\overline{ P_\rh}$, for the entry in cell
$\rh$ of $P$, depending on whether we mean the unbarred or barred
integer, respectively. 
Also, we
call the sum of all the entries of $P$ (forgetting about all the bars
when taking the sum) the 
{\it norm\/} of $P$, and denote it by $n(P)$. For example, the norm of
the super semistandard tableau in Figure~1.b is $27$.

\subhead 3. Bijective proof of the hook-content formula for super
Schur functions\endsubhead
In view of (2.1), our bijective proof of the hook-content
formula (1.3) will consist of proving
$$\underset \text{of shape} \la\to{\sum _{S \text{ a SST}} ^{}}
a^{\vert S^I\vert} b^{\vert S^O\vert}q^{n(S)}=
 q^{\sum _{i=1} ^{r}i\la_i}\prod _{\rh\in\la} ^{}\frac {1}
{1-q^{h_\rh}}\prod _{\rh\in\la} ^{}(a+bq^{c_\rh})\tag3.1$$
bijectively, where, as expected, $\vert S^I\vert$ and $\vert S^O\vert$
denote the numbers of entries in the inner part and outer part of $S$,
respectively. 

While it is obvious what the combinatorial interpretation of the
left-hand side of (3.1) is, we need to introduce yet a few more terms
for being able to conveniently describe the right-hand side of (3.1)
in combinatorial terms. 
We call an {\it arbitrary\/} filling of the cells of $\la$ with
{\it nonnegative\/} integers a {\it tabloid of shape\/} $\la$. 
We define the {\it hook weight\/} $w_h(T)$ of a tabloid $T$ of
shape $\la$ by $\sum _{\rh\in\la} ^{}T_\rh\cdot h_\rh$. Furthermore,
we call a 
filling of the cells of $\la$ with only $0$'s and $1$'s a {\it
$(0\text{--}1)$-tabloid\/}. We define the
{\it content weight\/} $w_c(T)$ of a $(0\text{--}1)$-tabloid $T$ by
$\sum _{\rh\in\la} 
^{}T_\rh\cdot c_\rh$. Then the right-hand side of (3.1) is the
generating function 
$$\sum _{}
^{}q^{n(P_0)}q^{w_h(T_R)}q^{w_c(U_R)}a^{\text{\#$0$'s in
$U_R$}}b^{\text{\#$1$'s in $U_R$}},$$ 
where the sum
is over all triples $(P_0,T_R,U_R)$, with $P_0$ being the ``minimal"
column-strict reverse plane partition of shape $\la$ with entries
from $1<2<3<\cdots$, i.e\. the 
column-strict reverse plane partition with all entries in row $i$
equal to $i$ for all $i$, with $T_R$ varying over all tabloids of
shape $\la$, and with $U_R$ varying over all $(0\text{--}1)$-tabloids of
shape $\la$. So the task is to set
up a bijection that maps a super semistandard tableau $S_L$ of shape
$\la$ to such a triple $(P_0,T_R,U_R)$, such that 
$$n(S_L)=n(P_0)+w_h(T_R)+w_c(U_R)\tag3.2$$
and
$$\vert (S_L)^I\vert=\text{number of $0$'s in $U_R$},\quad \quad
\text{and}\quad \quad 
\vert (S_L)^O\vert=\text{number of $1$'s in $U_R$}.\tag3.3$$

One step in our bijection was already done much earlier. In their
celebrated paper \cite{\HiGrAA}, Hillman and Grassl constructed an
algorithmic bijection between tabloids $T_R$ of shape $\la$ and reverse
plane partitions $\bP_R$ of shape $\la$ with entries from $0<1<2<\cdots$
such that $n(\bP_R)=w_h(T_R)$. If we add such a reverse plane
partition $\bP_R$ to $P_0$ cell-wise, then we obtain a
{\it column-strict\/} reverse plane partition $P_R$ of shape $\la$ with
entries from $1<2<3<\cdots$, and we have $n(P_R)=n(P_0)+n(
\bP_R)=n(P_0)+w_h(T_R)$. Therefore the new task is to set up a bijection
that maps a super semistandard tableau $S_L$ of shape $\la$ 
to a pair $(P_R,U_R)$, where $P_R$ is a column-strict
reverse plane partition of shape $\la$ with entries from
$1<2<3\cdots$, and where $U_R$ is a $(0\text{--}1)$-tabloid of shape
$\la$, such that 
$$n(S_L)=n(P_R)+w_c(U_R)\tag3.4$$
and (3.3) hold.

We claim that the following algorithm performs this task.

\bigskip
\noindent
{\bf Algorithm S}. The input for the algorithm is a super semistandard
tableau $S_L$ of shape $\la$.

(S0) Set $(S,U):=(S_L,\bold 0)$, where $\bold 0$ denotes the
$(0\text{--}1)$-tabloid
of shape $\la$ with $0$ in each cell.

\smallskip
(S1) If $S$ does not contain any barred entry then stop. The output of
the algorithm is $(S,U)$.

Otherwise, consider all {\it inward corner cells\/} of the
outer part $S^O$ of $S$
(which are the 
cells in $S^O$ with left and upper neighbour cells in $S^I$). Choose all such
inward corner cells that 
contain the minimal barred entry of $S$, and among all these pick the
top-most, cell $\om$ say. (Note that the minimal barred entry of $S$ {\it
must\/} appear in an inward corner cell of $S^O$ since $S^O$ is a
row-strict reverse plane partition.)
Replace the entry $\overline {S_\om}$ in cell $\om$ by
$S_\om-c_\om$. (Note that we ``unbarred" the entry in this cell. Also
note that $S_\om-c_\om$ might be nonpositive, which does not bother us
at this point.) Call
this entry {\it special\/}. Continue with (S2). 

\smallskip
(S2) If the special entry, $s$ say, does not violate (SST1), i.e\.
if $s$ does not violate weak increase along rows and strict increase along
columns (in the unbarred entries, of course),
then continue with (S3).

If it does, then we have the following
situation,
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&y&\\
\hlinefor5\\
&x&&s&\\
\hlinefor5
\endsmatrix\ ,\tag3.5
$$
where at least one of $x>s$ and $y\ge s$ holds. (One of $x$ or $y$ is
also allowed to 
be actually not there.) If $x>y$ then do the move
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&y&\\
\hlinefor5\\
&s+1&&x&\\
\hlinefor5
\endsmatrix\ .\tag3.6
$$
If $x\le y$ then do the move
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&s-1&\\
\hlinefor5\\
&x&&y&\\
\hlinefor5
\endsmatrix\ .\tag3.7
$$
The new special entry in (3.6) is $s+1$, the new special entry in
(3.7) is $s-1$. Repeat (S2). (Note that always after either type of
move the only possible violations of increase along rows or strict
increase along columns involve the new special entry and the entry to
the left or/and above.)

\smallskip
(S3) Let $S$ be the super semistandard tableau just
obtained. (The fact that indeed a super semistandard tableau is
obtained will be proved right after Example~S.) If we ended up with
the special entry in cell $\rh$ then add 
$1$ to the entry in cell $\rh$ of $U$. The tabloid
thus obtained is 
the new $U$. (It will be proved in Lemma~S below that $U$ is in fact a
$(0\text{--}1)$-tabloid.) Continue with (S1).\quad \quad \qed

\bigskip
\example{Example S}A complete example for Algorithm~S can be found in
the appendix. There we map the super semistandard tableau
of shape $(4,3,3,1)$ on the left of Figure~2 to the pair on
the right of Figure~2, consisting of a column-strict reverse plane partition
of shape $(4,3,3,1)$ with entries from $1<2<3<\cdots$ and a
$(0\text{--}1)$-tabloid of shape $(4,3,3,1)$, such that the
weight properties (3.3) and (3.4) hold. In fact, the norm of the
semistandard tableau
on the left of Figure~2 is $27$, while the norm of the
column-strict reverse plane partition on the right is $31$ and the
content weight of the $(0\text{--}1)$-tabloid is $-4$, which verifies
(3.4) in this case. That (3.3) holds in our example is obvious.
\vskip10pt
\vbox{
$$
\matrix \format \c&\quad \c\quad &\c\\
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&3&&3&&\3&\\
\hlinefor9\\
&2&&\1&&\3&\\
\hlinefor7\\
&4&&\2&&\3&\\
\hlinefor7\\
&\2&\\
\hlinefor3
\endsmatrix
&\longleftrightarrow&
\(
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&1&&1&&3&\\
\hlinefor9\\
&2&&3&&3&\\
\hlinefor7\\
&4&&4&&4&\\
\hlinefor7\\
&5&\\
\hlinefor3
\endsmatrix
, 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
&1&&0&&1&&0&\\
\hlinefor9\\
&0&&1&&0&\\
\hlinefor7\\
&1&&1&&0&\\
\hlinefor7\\
&1&\\
\hlinefor3
\endsmatrix
\)\\
\matrix \format\r&\l\\
 n(.)&{}=27\\
\vert \text{inner part}\vert&{}=5\\
\vert \text{outer part}\vert&{}=6
\endmatrix& &
\matrix \format \r&\l\\
n(.)=31\ ,\ w_c(.)&{}=-4\\
\text{\#0's}&{}=5\\
\text{\#1's}&{}=6
\endmatrix
\endmatrix
$$
\centerline{\eightpoint Figure 2}
}
\vskip10pt
The appendix has to be read in the following way. First of all, ignore
all double circles, and all even rows in the right columns. What the
left columns show is the pair $(S,U)$ that is obtained after each loop
(S1)-(S2)-(S3). Together with the pair $(S,U)$ a filling of the shape
$(4,3,3,1)$ is displayed that 
shows all values $S_\rh+c_\rh$ for all cells $\rh$ with $U_\rh\ne0$.
This will be important for understanding Lemma~S but can be ignored for
the moment.
At each stage, the entry that is chosen by (S1) is circled. Then each
intermediate step during the loop (S1)-(S2)-(S3) is displayed in the
odd rows of the right columns. The special entry is always
underlined. When a super semistandard tableau is reached,
the special entry is boxed. The entry in the
corresponding cell of the $(0\text{--}1)$-tabloid is subsequently
increased by $1$ in step (S3).\quad \quad \qed
\endexample

It should be noticed that, aside from adding/subtracting $1$ to/from
the special entry, what happens from (3.5) to (3.6), respectively
(3.7), is a {\it jeu de taquin forward move\/} (cf\. \cite{\SchuAA,
sec.~2; \SagaAL, pp.~120/169}).

We have to justify the claim in (S3) that when we arrive at (S3) we
obtained a super semistandard tableau. We do this by induction on the
number of loops (S1)-(S2)-(S3). For the induction hypothesis assume
that we enter (S1) with a pair $(S,U)$, where $S$ is a super
semistandard tableau. (This is true for our initial pair $(S_L,\bold
0)$). It is obvious that when we arrive at (S3) that $S$ was
transformed into a filling that satisfies (SST1) (otherwise the step
(S2) was not finished) and (SST2) (because the only change in the
outer part, the part containing the barred entries, was to remove an
entry in an inward corner cell from it and make the entry into an
unbarred entry). Therefore the only 
problem that could arise is when the special entry in the end is $\le0$.
However, when we arrive at (S3), a special
entry $\le0$ can only occur in cell $(1,1)$, because otherwise step
(S2) was not finished. Each loop (S1)-(S2)-(S3) of the algorithm
starts with some entry $\overline {S_{\om}}\ge\1$ in an inward corner
cell $\om$ of 
the outer part $S^O$ of $S$. It is replaced
by $S_{\om}-c_{\om}$. Then it is (possibly) moved according to (3.6) and
(3.7). It is easy to check that at each stage
during performing the steps (S2), the special entry, if located in
cell $\rh$, will equal $S_\om-c_\rh$. This is a property so
important that it has to be recorded for later use,
$$\text{(special entry in $\rh$)}=S_\om-c_\rh.\tag3.8$$
Suppose we reach cell $(1,1)$. 
When we arrive at $(1,1)$, by
(3.8) and since $c_{(1,1)}=0$, our special entry has become
$S_{\om}$. But this is $\ge1$ since $\overline{S_{\om}}\ge\1$.

So we have an algorithm that maps a
super semistandard tableau $S_L$ 
to a pair $(P_R,U_R)$, where $P_R$ is a super semistandard tableau
without any barred entry, or in plain words, a column-strict reverse plane
partition, and where $U_R$, thanks to Lemma~S (cf\. the Remark after
Lemma~S), is a $(0\text{--}1)$-tabloid. Stated in different terms, the
algorithm maps left-hand side objects of (3.1) to right-hand side
objects of (3.1). Besides, this
mapping satisfies the weight properties (3.3) and (3.4). While (3.3)
is trivially satisfied, (3.4) follows from (3.8).

What remains is to establish that our algorithm is actually a
{\it bijection\/} between left-hand side and right-hand side objects. This
will be accomplished by constructing an algorithm, Algorithm~S* below,
that will turn out to be the inverse of Algorithm~S. To motivate the
definition of Algorithm~S*, we note the following lemma.

\proclaim{Lemma S}Let $(S,U)$ be obtained after some loop
{\rm (S1)-(S2)-(S3)} during Algorithm~S. 
Suppose that the loop terminated in
cell $\ga$ when reaching (S3). Then $\ga$ cannot be reached by a
special entry during succeeding loops (S1)-(S2)-(S3). Also, among all
cells $\rh$ with 
$U_\rh\ne0$, $\ga$ is a 
cell for which the value $S_\rh+c_\rh$ attains its maximum, and if
there are several cells $\rh$ with $U_\rh\ne0$ where the maximum is
attained, then $\ga$ is the left-most and bottom-most of those. Besides,
there holds
$$\overline{S_\ga+c_\ga}\le \min\{\text{\rm entries in }S^O\}.\tag3.9$$
\endproclaim
\remark{Remark}Note that the property that $\ga$ cannot be reached by
``later" special entries inductively implies that after each loop
(S1)-(S2)-(S3) the tabloid $U$ is in fact a $(0\text{--}1)$-tabloid.
\endremark
\demo{Proof of the Lemma}We prove the assertions by induction on the
number of loops (S1)-(S2)-(S3).

For the start of the induction it will suffice to recall that we start
Algorithm~S with the pair $(S_L,\bold0)$, where $S_L$ is a super
semistandard tableau.

As induction hypothesis we assume that $(S,U)$ is either the initial
pair $(S_L,\bold0)$, or obtained after some loop (S1)-(S2)-(S3) in
which case all the assertions of the Lemma are assumed to be true.
Let $\om$ be the cell where the next loop starts, i.e\. the
inward corner cell of $S^O$ chosen by (S1), and let $\ze$ be the cell
where this 
next loop terminates at (S3). Let $(\bS, \bU)$ be the outcome after
this loop. Note that by construction the cells $\rh$ with $\bU_\rh\ne0$ are
$\ze$ and the cells with $U_\rh\ne0$. In particular, $U_\rh\ne0$
implies $\bU_\rh\ne0$.

\smallskip
First we prove (3.9) for $\bS$ and $\ze$. By (3.8),
the entry of $\bS$ in cell $\ze$ is 
$$\bS_\ze=S_\om-c_\ze.\tag3.10$$
Since by definition of (S1) we have
$\overline{S_\om}=\min\{\text{entries in }S^O\}$, this immediately implies
$$\overline{\bS_\ze+c_\ze}=\overline{S_\om}=\min\{\text{entries in }S^O\}\le
\min\{\text{entries in }\bS^O\}.\tag3.11$$
The last inequality is due to the fact that the multiset of entries
in $\bS^O$ comes out of the multiset of entries
in $S^O$ by removing
$\overline {S_\om}$. 
Obviously, (3.11) proves (3.9) for $\bS$ and $\ze$, as desired.

\smallskip
Next we prove that the cell $\ze$ cannot be reached during a ``later"
loop (S1)-(S2)-(S3). We distinguish between two
cases. Namely, by 
definition of (S1), a ``later"
loop (S1)-(S2)-(S3) starts either at a cell, $\ka$ say,
strictly to the right of $\om$ or weakly to the
left and strictly below of $\om$. 

In the former
case, by definition of (S1) and the fact that the outer part $S^O$ of
$S$ is a row-strict reverse plane partition, we have 
$$\bS_\ka=S_\ka>S_\om.\tag3.12$$ 
Then, when
performing the loop (S1)-(S2)-(S3), first $\overline{\bS_\ka}$ is replaced by
$\bS_\ka-c_\ka$ and then (possibly) moved inwards according to (3.6)
and (3.7). If we want to reach $\ze$, we have to reach the neighbour
cell to the 
right or the neighbour cell below first. By (3.8) we would reach the
neighbour cell to the right of $\ze$ with the special entry being equal to
$\bS_\ka-(c_\ze+1)$. By (3.12) this is $\ge S_\om-c_\ze=\bS_\ze$, the
latter equality holding because of (3.10). Therefore, by definition
of (S2), the loop either already stops here or continues with an
upward move (3.7), and so does not reach $\ze$ in any case. On
the other hand, if we reach the neighbour cell below $\ze$, by (3.8)
the special entry would equal $\bS_\ka-(c_\ze-1)$. By (3.12) this is $>
S_\om-c_\ze=\bS_\ze$. Again, by definition of (S2), this says that the
loop either already stops here or continues with a left move (3.6), so
it cannot reach $\ze$ in any case. 

The second case is that we start our ``later" loop in a cell $\ka$
that is located weakly to the left and strictly below of $\om$. By
definition of (S1) we have  
$$\bS_\ka=S_\ka\ge S_\om.\tag3.13$$ 
Again, if we want to reach $\ze$, we have to reach
either the neighbour cell to the right or the neighbour cell below of
$\ze$ first. We claim 
that it cannot happen that the cell to the right is reached. This
would follow immediately from the following easy-to-check property
of our modified forward jeu de taquin (S2): If the second ``jeu de
taquin path" is below the first ``jeu de taquin path" somewhere,
then it has to stay below from thereon. To make this precise,
suppose that during the first loop the special entry, $s_1$ say, went
to the left by (3.7), see the left half of Figure~3. (The arrows mark the
direction of move of the special entry.)
\vskip10pt
\vbox{
$$
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&y&&s_1&\\
\hlinefor5\\
&z&&*&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to{\overset \text{first loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&\hbox to5pt{$*\kern-4pt\leftarrow$\hss}&&y&\\
\hlinefor5\\
&z&&*&\\
\hlinefor5
\endsmatrix
\hskip2cm
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&y&\\
\hlinefor5\\
&z&&s_2&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to {\overset \text{second loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&y&\\
\hlinefor5\\
&\hbox to5pt{$*\kern-4pt\leftarrow$\hss}&&z&\\
\hlinefor5
\endsmatrix
$$
\centerline{\eightpoint Figure 3}
}
\vskip10pt
\noindent
Since columns are strictly increasing in the inner part, we have $y<
z$. Suppose that 
during the second loop we reach the cell neighbouring $y$ and $z$ with
a special entry $s_2$, see the right half of Figure~3. Then by
definition of the algorithm we have to stop here or, if not,
we are forced to move left in the next step
(S2). It is our assumption that the second ``jeu de taquin path" starts
below the first, therefore it has to stay below
always. 

Now we know that the only possibility to reach cell $\ze$ is by
reaching the neighbour cell below first. However, when we reach this
neighbour cell below of $\ze$, by (3.8) the special entry
equals $\bS_\ka-(c_\ze-1)$. By (3.13) this is $>
S_\om-c_\ze=\bS_\ze$. Once more, by definition of (S2), this says that the
loop either already stops here or continues with a left move (3.6), so
it cannot reach $\ze$ in any case.

\smallskip
Next we show that $\bS_\rh+c_\rh$, evaluated at cells $\rh$ with
$\bU_\rh\ne0$, attains its maximal value at $\ze$. 
This is evident if $(S,U)$ is the initial pair $(S_L,\bold0)$, i.e\.
if $(\bS,\bU)$ is the pair obtained by applying one loop
(S1)-(S2)-(S3) to $(S_L,\bold0)$, because then $\ze$ is the only cell
with $\bU_\ze\ne0$. So, let us assume that $(S,U)$ is
obtained after some loop (S1)-(S2)-(S3). Let $\ga$ be the
cell where the next-to-last loop, which gave rise to $(S,U)$, terminated
at (S3). 

Let $\rh$ be any cell with $\bU_\rh\ne0$. Recall that this means $\rh=\ze$ or
$U_\rh\ne0$. We want to show
$$\bS_\rh+c_\rh\le \bS_\ze+c_\ze.\tag3.14$$
If $\rh=\ze$ then there is nothing to show. So let $\rh\ne \ze$. Then we
have $U_\rh\ne0$. By induction hypothesis for $\ga$, we have
$S_\ga+c_\ga\ge S_\rh+c_\rh$ for all cells with $U_\rh\ne0$. Also by
induction hypothesis, we know that $\rh$ cannot be reached during a
``later" loop (S1)-(S2)-(S3), therefore we have
$\bS_\rh=S_\rh$. And, again by induction
hypothesis, (3.9) holds for $S$ and $\ga$. Hence we have
$$\overline{S_\ga+c_\ga}\le\min\{\text{entries in
}S^O\}=\overline{S_\om}=\overline{\bS_\ze+c_\ze},
$$
the equalities holding because of (3.11).
Combining everything, we conclude
$$\bS_\rh+c_\rh=S_\rh+c_\rh\le S_\ga+c_\ga\le \bS_\ze+c_\ze,\tag3.15$$
which verifies (3.14).

\smallskip
Finally we show that $\ze$ is the left-most and bottom-most among all
cells $\rh$ with $\bU_\rh\ne0$ where $\bS_\rh+c_\rh$ attains its maximal
value. Again, this is trivially satisfied if $(S,U)$ is the initial
pair $(S_L,\bold0)$ since then $\ze$ is the only cell with
$\bU_\ze\ne0$. So, let us assume that $(S,U)$ is
obtained after some loop (S1)-(S2)-(S3). Also here, let $\ga$ be the
cell where the next-to-last loop, which gave rise to $(S,U)$, terminated
at (S3).  
Let $\rh$ be a cell with $\bU_\rh\ne0$ where $\bS_\rh+c_\rh$
attains its maximal value. In particular, we have
$$\bS_\rh+c_\rh=\bS_\ze+c_\ze.\tag3.16$$

Of course, nothing is to show for $\rh=\ze$, so we
may assume $\rh\ne\ze$. This implies $U_\rh\ne0$. We already observed
that then the relations (3.15) hold.
Combining (3.15) and (3.16) we are forced to conclude
$$S_\rh+c_\rh=S_\ga+c_\ga=\bS_\ze+c_\ze.\tag3.17$$

We shall show that $\ze$ has to lie in the region (weakly) to the left
and (weakly) below of $\ga$. Since $\ga$ was the left-most and
bottom-most of all the cells $\rh$ with $U_\rh\ne0$ where the maximal
value of $S_\rh+c_\rh$ is attained, and since
$S_\rh+c_\rh=\bS_\rh+c_\rh$ for all cells with $U_\rh\ne0$, 
this would establish that $\ze$
is the left-most and bottom-most of all the cells $\rh$ with
$\bU_\rh\ne0$ where the maximal value of $\bS_\rh+c_\rh$ is
attained, as desired.

We prove the claim of the previous paragraph by excluding the other
three quarter regions that are determined by the horizontal line and the
vertical line running through $\ga$. 

First, suppose that $\ze$ lies in the region strictly to the right and
weakly below of $\ga$. 
For these cells there holds
the following basic computation. 
For convenience, let $\ga=(i_1,j_1)$ and 
$\ze=(i_2,j_2)$, where $\ze$ is located in this region strictly to the
right and weakly below of $\ga$, i.e\. 
$i_1\le i_2$ and $j_1< j_2$. Then, since the inner part $\bS^I$ of
$\bS$ is a column-strict
reverse plane partition, we have
$$\align \bS_\ze+c_\ze&=\bS_\ze+(j_2-i_2)\\
&\ge (\bS_\ga+i_2-i_1)+(j_2-i_2)=\bS_\ga+j_2-i_1\\
&> \bS_\ga+j_1-i_1=\bS_\ga+c_\ga.
\tag3.18
\endalign$$
This contradicts (3.17) because of
$S_\ga=\bS_\ga$, which again follows from the induction hypothesis that
$\ga$ cannot be reached during ``later" loops (S1)-(S2)-(S3). Thus,
this region is excluded. 

Next we show that $\ze$ cannot lie in the region (weakly) to the right
and (weakly) above $\ga$, $\ga$ excluded. This would follow
immediately from the claim that if two successive loops (S1)-(S2)-(S3)
start with the same size of entry in the inward corner cells chosen by (S1)
(which applies in our case since the loop that lead to $S$ started
with an entry $\overline{S_\ga+c_\ga}$ in some inward corner cell, and
the loop that 
lead from $S$ to $\bS$ started with $\overline{\bS_\ze+c_\ze}$, both
quantities being the same by (3.17)) then the second path of moves has
to stay below the first path of moves.

To check this claim, once again note that both loops started with the
same size of entries in the cells chosen by (S1). By the rules in
(S1), and because the outer part $S^O$ of $S$ is a row-strict reverse
plane partition, this means that the 
second loop started strictly to the left and strictly below of the first.
We already proved that
our modified forward jeu de taquin (S2) has the property that if the
second ``jeu de 
taquin path" is below of the first ``jeu de taquin path" somewhere,
then it has to stay below from thereon.
So the region (weakly) to the right and (weakly) above $\ga$ can never
be reached. 

Finally, we examine if $\ze$ could be located in the region strictly
to the left and weakly above of $\ga$. Since $S_\ga=\bS_\ga$, once
more the computation (3.18), with $\ga$ and $\ze$ interchanged
applies,
implying $\bS_\ga+c_\ga>\bS_\ze+c_\ze$, which contradicts (3.17)
because of $S_\ga=\bS_\ga$. 
Therefore we cannot reach the region under consideration.

This completes the proof of the Lemma.\quad \quad \qed
\enddemo

From Lemma~S it is pretty obvious what the inverse algorithm of
Algorithm~S could be.

\bigskip
\noindent
{\bf Algorithm S*}. The input for the algorithm is a pair $(P_R,U_R)$,
where $P_R$ is a column-strict
reverse plane partition of shape $\la$ with entries from
$1<2<3<\cdots$, and where $U_R$ is a $(0\text{--}1)$-tabloid of shape $\la$.

(S*0) Set $(S,U):=(P_R,U_R)$.

\smallskip
(S*1) If $U=\bold 0$ then stop. The output of
the algorithm is $S$.

Otherwise, consider all cells $\rh$ with $U_\rh\ne0$. Among these
choose the cells for which $S_\rh+c_\rh$ is maximal, and among all
these pick the left-most and bottom-most, cell $\ze$ say. (Observe that
among two different cells
attaining the same value of $S_\rh+c_\rh$ one is always weakly to
the left and strictly below of the other, again because of the
computation (3.18), with $\bS$ replaced by $S$.
So the left-most and bottom-most of these does exist.) 
Replace the entry $S_\ze$ in cell $\ze$ by
$\overline{S_\ze+c_\ze}$. (Note that we ``barred" the entry.) Call
this entry {\it special\/}. Continue with (S*2). 

\smallskip
(S*2) If the special entry, $s$ say, does not have a right or bottom
neighbour entry that is unbarred,
then continue with (S*3).

If the special entry does have a right or bottom
neighbour entry that is unbarred, then we have the following
situation,
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&s&&x&\\
\hlinefor5\\
&y&\\
\hlinefor3
\endsmatrix\ ,\tag3.19
$$
where at least one of $x$ or $y$ is an unbarred entry.
(One of $x$ or $y$ is
also allowed to be  
actually not there.) If $x<y$ (in particular, this is the case if
$y$ is a barred entry) then do the move
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&x&&s&\\
\hlinefor5\\
&y&\\
\hlinefor3
\endsmatrix\ .\tag3.20
$$
If $x\ge y$ (in particular, this is the case if
$x$ is a barred entry) then do the move
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&y&&x&\\
\hlinefor5\\
&s&\\
\hlinefor3
\endsmatrix\ .\tag3.21
$$
The new special entry in either case is $s$.
Repeat (S*2). 

\smallskip
(S*3) Let $S$ be the super semistandard tableau just
obtained. (The fact that indeed a super semistandard tableau
is obtained will be proved in the subsequent
Lemma~S*.) Replace the entry $1$ in cell $\ze$ of $U$ by $0$.
The $(0\text{--}1)$-tabloid thus obtained is
the new $U$. Continue with (S*1).\quad \quad \qed

\bigskip
\example{Example S*}A complete example for Algorithm~S* can be found in
the appendix. There we map the pair on
the right of Figure~2, consisting of a column-strict reverse plane partition
of shape $(4,3,3,1)$ with entries from $1<2<3<\cdots$ and a
$(0\text{--}1)$-tabloid of shape $(4,3,3,1)$, to 
the super semistandard tableau
of shape $(4,3,3,1)$ on the left of Figure~2, such that the
weight properties (3.3) and (3.4) hold. It is simply the inverse of
the example 
for Algorithm~S given in Example~S. Therefore, here the appendix has to
be read in the reverse direction,
and in the following way. First of all, ignore
all single circles, and all odd rows in the right columns. What the
left columns show is the pair $(S,U)$ that is obtained after each loop
(S*1)-(S*2)-(S*3) together with a filling of the shape $(4,3,3,1)$ that
shows all values $S_\rh+c_\rh$ for all cells $\rh$ with $U_\rh\ne0$.
At each stage, the entry that is chosen by (S*1) is doubly circled. Then each
intermediate step during the loop (S*1)-(S*2)-(S*3) is displayed in the
even rows of the right columns. The special entry is always
doubly underlined. When a super semistandard tableau is reached,
the special entry is doubly boxed. The entry in the
corresponding cell of the $(0\text{--}1)$-tabloid is subsequently
decreased by $1$ in step (S*3).\quad \quad \qed
\endexample

Again, it should be noticed that (3.20) and (3.21) are exactly {\it jeu de
taquin backward moves\/} (cf\. \cite{\SchuAA, sec.~2; \SagaAL,
pp.~120/169}),
which reverse the forward moves (3.6) 
and (3.7), respectively, except for the subtraction/addition of $1$
in (3.6) and (3.7).

In order to show that Algorithm~S* is always well-defined, we have to
confirm that when arriving at (S*3) we always obtained a super semistandard
tableau. This is established in the following lemma.
Besides, this lemma contains the facts about Algorithm~S* that are
needed to prove that the Algorithms~S and S* are inverses of each other.

\proclaim{Lemma S*}Let $(S,U)$ be obtained after some loop
{\rm (S*1)-(S*2)-(S*3)} during Algorithm~S*. Then for all cells $\rh$ with
$U_\rh\ne0$ there holds
$$\overline{S_\rh+c_\rh}\le\min\{\text{\rm entries in }S^O\}.\tag3.22$$
Also, $S$ is a super semistandard tableau. Besides, if
$\om$ is the inward corner cell of $S^O$ that contained the special
entry at the end 
of the loop {\rm (S*1)-(S*2)-(S*3)} that lead to $S$, then $\om$ is the
top-most inward corner cell of $S^O$ that contains the minimal entry of $S^O$.
\endproclaim
\demo{Proof}We prove the assertions by induction on the number of
loops (S*1)-(S*2)-(S*3). To be precise, we prove the following
(slightly stronger) statement inductively: All the assertions of the
Lemma hold. Besides, during the loop (S*1)-(S*2)-(S*3) that leads to
$(S,U)$ the special entry never occupies a cell $\rh$ with $U_\rh\ne0$.

To begin with,
we know that when we start with Algorithm~S* we have
a pair $(S,U)$, where $S$ is a super semistandard tableau (since
$S=P_R$ is a column-strict reverse plane partition in $1<2<3<\cdots$).
Also, (3.22) is vacuously satisfied (there is no barred entry in $S=P_R$).
This will suffice for the
start of the induction.

As induction hypothesis let us assume that the assertions of the Lemma
and the assertion that the special entry never occupies a cell $\rh$
with $U_\rh\ne0$
are true for $(S,U)$ and all preceding pairs occuring in step (S*3)
during the process of the algorithm, except of course that the
assertion about the inward corner cell $\om$ and the assertion about
the special entry do not hold for the initial
pair (because they do not make sense for the initial pair).

Let $\ze$ be the cell where the loop (S*1)-(S*2)-(S*3) starts from
$(S,U)$, i.e\. the cell that is chosen by applying (S*1) to $(S,U)$,
and let $\ka$ be the cell where the loop stops at (S*3).
Furthermore, let $(\bS,\bU)$ be the outcome after this loop. Then, by
definition of the algorithm we have
$$\overline{\bS_\ka}=\overline{S_\ze+c_\ze}.\tag3.23$$
Note that, also by definition of the algorithm,
the cells $\rh$ with $\bU_\rh\ne0$ are those with
$U_\rh\ne0$, excluding $\ze$. In particular, $\bU_\rh\ne0$
implies $U_\rh\ne0$.

\smallskip
First we prove that during the loop (S*1)-(S*2)-(S*3) leading from
$(S,U)$ to $(\bS,\bU)$ the special entry never occupies a cell $\rh$ with
$\bU_\rh\ne0$. Recall that, by definition of (S*1), the cell $\ze$,
the cell where the loop starts, is the
left-most and bottom-most cell among the cells with $U_\ze\ne0$ 
where the value $S_\ze+c_\ze$ is
maximal. Let the special entry be located in some cell $\rh$ with 
$\bU_\rh\ne0$, at some
stage during the performance of the loop. 
Clearly, $\rh$ is located 
(weakly) to the right and (weakly) below of $\ze$. Since, by definition of
(S*3), we have $\bU_\ze=0$, cell $\rh$ is different from $\ze$. If
$\rh$ were located {\it strictly\/} to the right of $\ze$, then the
computation (3.18), with $\bS$ replaced by $S$, $\ze$ replaced by
$\rh$, and $\ga$ replaced by $\ze$, would imply that
$S_\rh+c_\rh>S_\ze+c_\ze$. But this contradicts the definition of
$\ze$. So the only other possibility is that $\rh$ is located
in the same column as $\ze$ and strictly below $\ze$. Then a
computation similar to (3.18) would show that
$S_\rh+c_\rh=S_\ze+c_\ze$, which again contradicts the definition of
$\ze$. Therefore the special entry can only meet cells $\rh$ with
$\bU_\rh=0$. In particular, this implies that the entries in cells $\rh$
with $\bU_\rh\ne0$ do not change during the loop. Therefore we have
$$\bS_\rh=S_\rh\tag3.24$$
for all cells $\rh$ with $\bU_\rh\ne0$.

\smallskip
Now we prove (3.22) for $\bS$. Let $\rh$ be any cell
with $\bU_\rh\ne0$. This implies
$U_\rh\ne0$. Therefore, by (3.24) and by construction of $\ze$ in
(S*1), we have 
$$\bS_\rh+c_\rh= S_\rh+c_\rh\le S_\ze+c_\ze.\tag3.25$$
Also by construction of $\ze$, we have $U_\ze\ne0$, and hence by
induction hypothesis (3.22) that
$\overline{S_\ze+c_\ze}\le\min\{\text{entries 
in }S^O\}$. This implies that
$\overline{S_\ze+c_\ze}=\min\{\text{entries in 
}\bS^O\}$, since the multiset of entries of $\bS^O$ equals the
multiset of 
entries in $S^O$, with the special entry $\overline{S_\ze+c_\ze}$,
created in 
(S*1) and finally located in cell $\ka$ in $\bS$, added. Hence, (3.25) proves
(3.22) with $S$ replaced by $\bS$, as desired. 

\smallskip
Now we prove that $\bS$ is a super semistandard tableau. By
construction of Algorithm~S*, the inner part $\bS^I$ of $\bS$
automatically is a column-strict reverse plane partition. Therefore,
it is only to prove that the outer part $\bS^O$ of $\bS$ is a
row-strict reverse plane partition. If
$\overline{S_\ze+c_\ze}<\min\{\text{entries in }S^O\}$ then this assertion
certainly holds, since $\overline{\bS_\ka}=\overline{S_\ze+c_\ze}$ is
the only new entry in 
$\bS$, and it is located in an inward corner cell of $\bS^O$. Note in
particular, that we are in this case at the 
very beginning.

By induction hypothesis, (3.22) holds for $\ze$, so the only other case
is 
$$\overline{S_\ze+c_\ze}=\min\{\text{entries in }S^O\}.\tag3.26$$
Observe that the only difficulty arises when we reach the cell
$\ka$ at the end of a loop (S*1)-(S*2)-(S*3), and the neighbour cell
to the right of $\ka$ contains the entry $\overline{S_\ze+c_\ze}$.
In this case row-strictness of the outer part $\bS^O$ of $\bS$ would
be violated. We have to show that this case cannot occur.

Let $(S',U')$ be the pair preceding $(S,U)$, i.e\. $(S,U)$ is obtained
by applying one loop (S*1)-(S*2)-(S*3) to $(S',U')$. As we just noted,
$(S',U')$ exists, since if $(S,U)$ were the initial pair we would not
be in this case.
Furthermore, let $\ga$ be the cell where this loop
starts, and let $\om$ be the corner cell where it stops. By definition
of the algorithm we have
$$\overline{S_\om}=\overline{S'_\ga+c_\ga}.\tag3.27$$ 
Now, by
induction hypothesis and the definition of the algorithm, there holds
$$\overline{S'_\ga+c_\ga}=\min\{\text{entries in }S^O\}.\tag3.28$$
Furthermore, there holds $S'_\ze= S_\ze$, since by induction
hypothesis the cell $\ze$ (which is a cell with $U_\ze\ne0$) was not met by the
special entry during the loop (S*1)-(S*2)-(S*3) that lead from
$(S',U')$ to $(S,U)$.
Hence, by definition of (S*1), $S'_\ga+c_\ga\ge
S'_\ze+c_\ze= S_\ze+c_\ze$.
Combining this with (3.26) and (3.28), we are forced to conclude
$$S'_\ga+c_\ga=S'_\ze+c_\ze\,(=S_\ze+c_\ze).\tag3.29$$
Therefore, again by definition of (S*1), $\ze$ lies weakly to the
right and strictly above $\ga$.

It is an easy-to-check property of backward jeu de taquin (S*2) that
if the second ``jeu de
taquin path" is above the first ``jeu de taquin path" somewhere,
then it has to stay above from thereon. To be precise,
suppose that during the first loop (S*1)-(S*2)-(S*3) the special
entry, $s_1$ say, went 
to the right by (3.20), see the left half of Figure~4. (Again, the
arrows mark the 
direction of move of the special entry.)
\vskip10pt
\vbox{
$$
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&y&\\
\hlinefor5\\
&s_1&&z&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to{\overset \text{first loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&*&&y&\\
\hlinefor5\\
&\hbox to5pt{$z\kern-4pt\rightarrow$\hss}&&*&\\
\hlinefor5
\endsmatrix
\hskip2cm
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&s_2&&y&\\
\hlinefor5\\
&z&&*&\\
\hlinefor5
\endsmatrix
\overset \text{during}\to {\overset \text{second loop}\to\longrightarrow}
\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&\hbox to5pt{$y\kern-4pt\rightarrow$\hss}&&*&\\
\hlinefor5\\
&z&&*&\\
\hlinefor5
\endsmatrix
$$
\centerline{\eightpoint Figure 4}
}
\vskip10pt
\noindent
Since columns are strictly increasing in the inner part, we have
$y<z$. Suppose that 
during the second loop we reach the cell neighbouring $y$ and $z$ with
a special entry $s_2$, see the right half of Figure~4. Then by
definition of the algorithm we have to stop here or, if not,
we are forced to move right in the next step
(S*2). 

We know that the second ``jeu de taquin path" starts at $\ze$,
which is weakly to the right and strictly above of $\ga$, the
starting cell of the first ``jeu de taquin path". Therefore the second
path has to stay above the first path always. Hence $\ka$, the cell
where the second path terminates, cannot be
the left neighbour cell of $\om$. Therefore, $\ka$ is in a row above
$\om$. By induction hypothesis, $\om$ is the
top-most inward corner 
cell of $S^O$ that contains the minimal entry of $S^O$. This implies
that the cell to the right of $\ka$ contains and entry that is
$>\overline{S_\om}=\overline{\bS_\ka}$, the equality following from a
combination of (3.27), (3.29), and (3.23). As noted above, this
guarantees that $\bS$ is a super semistandard tableau.

\smallskip
Finally, we prove that $\ka$ is the top-most inward corner cell in $\bS^O$
that contains the minimal entry in $\bS^O$. This is trivially true if
$\overline{S_\ze+c_\ze}<\min\{\text{entries in }S^O\}$, again by remembering
(3.23). Note that this inequality 
is in particular true at the very beginning of Algorithm~S*.
Because of the induction hypothesis (3.22), the only other case is
$\overline{S_\ze+c_\ze}=\min\{\text{entries in }S^O\}$. Since we are
{\it not\/} 
at the very beginning, we are allowed to assume that this last
assertion of Lemma~S* holds for $S$ and $\om$. However, we already
considered the case $\overline{S_\ze+c_\ze}=\min\{\text{entries in }S^O\}$
before (see (3.26)) and showed that the ``jeu de taquin 
path" leading from $\ze$ to $\ka$ has to stay above of the ``jeu
de taquin path" leading from $\ga$ to $\om$. Hence $\ka$ is strictly
above $\om$. By induction hypothesis, $\om$ was the top-most
corner cell containing the minimal entry of $S^O$. So $\ka$, which by (3.23)
contains $\overline{S_\ze+c_\ze}=\min\{\text{entries in }S^O\}$ in
$\bS$, is the 
top-most corner cell of $\bS$ containing $\min\{\text{entries in
}\bS^O\}$ ($=\min\{\text{entries in }S^O\}$), as desired.

This completes the proof of the Lemma.\quad \quad \qed
\enddemo

From Lemmas~1 and 2 it is abundantly clear that the Algorithms~S and
S* are inverses of each other. This finishes the bijective proof of
(3.1).


\subhead 4. More on modified jeu de taquin\endsubhead
In this section we explore the relations between our modified jeu de
taquin in Algorithm~S and a modified jeu de taquin in a paper of
Goulden and Greene \cite{\GoGrAA, sec.~3}. The main result of this
section is an invariance property (Theorem~SG) of 
modified jeu de taquin, which
somehow is the analogue of the invariance of standard jeu de taquin 
(see \cite{\SchuAA%, ??
; \SagaAL, Theorem~3.9.7}).

Super Schur functions are also defined for skew shapes $\la/\mu$. For
the purposes of this section, we assume that $\bold x$ and $\bold y$
are {\it doubly\/} infinite sequences of variables, $\bold
x=(\dots,x_{-1},x_0,x_1,\dots)$ and $\bold
y=(\dots,y_{-1},y_0,y_1,\dots)$. Then, the
skew super Schur function $S_{\la/\mu}(\bold x,\bold y)$ is defined
by
$$\S_{\la/\mu}(\bold x,\bold y)=\det_{1\le i,j\le
r}(\S_{\la_i-i-\mu_j+j}(\bold
x,\bold y)),\tag4.1$$
where the series $\S_m(\bold x,\bold y)$ are determined by their
generating function 
$$\sum _{m\ge 0} ^{}\S_m(\bold x,\bold y)z^m=\prod _{i=-\infty} ^{\infty}\frac
{1+y_iz} {1-x_iz},$$
(note the difference to (1.2) in the range of the product).
The combinatorial description (2.1)
extends to skew super Schur functions. Namely, if we extend the
definition of super semistandard tableau by saying that it is a
filling of some skew shape with entries from the ordered alphabet
$\cdots<-1<0<1<\cdots<\overline{-1}<\overline{0}<\1<\cdots$ satisfying
(SST1) (but the shape $\nu$ replaced by $\nu/\mu$) and (SST2), then
(again SST is short for `super semistandard tableau')
$$\S_{\la/\mu}(\bold x,\bold y)=\underset \text{of shape }\la/\mu\to{\sum
_{S\text{ a SST}} ^{}} \prod _{i=-\infty} ^{\infty}x_i^{\text{\#$i$'s in
$S$}}y_i^{\text{\#$\bar i$'s in $S$}}.\tag4.2$$
As in the nonskew case
(4.2) can be proven combinatorially by the Gessel--Viennot method
of nonintersecting lattice paths, see
\cite{\RemmAC, sec.~5; \BretAE, Theorem~3.4}.


Algorithm~S was defined for super semistandard tableaux of a nonskew
shape $\la$. 
However, everything can be defined and works fine also for a skew
shape $\la/\mu$. To be precise, it is readily checked that Algorithm~S for
a skew shape $\la/\mu$ proves the following alternative combinatorial
desription for (skew) super Schur functions due to Goulden and Greene
\cite{\GoGrAA, Theorem~1.1} (here CRPP is short for column-strict
reverse plane partition with integer entries)
$$\S_{\la/\mu}(\bold x,\bold y)=\underset \text{of shape }\la/\mu\to{\sum
_{P\text{ a CRPP}} ^{}} \prod _{\rh\in\la/\mu}
^{}(x_{P_\rh}+y_{P_\rh+c_\rh}).\tag4.3$$
(In fact, we needed the restriction of a nonskew shape only at one place,
namely, in the second paragraph after Example~S, when showing that
all the entries are at least 1 when arriving at (S3). But now we are
considering alphabets that are unbounded above {\it and\/} below, so
this step is superfluous in this context.)
Goulden and Greene also construct a bijection for proving that the
representations (4.2) and (4.3) for $\S_{\la/\mu}(\bold x,\bold y)$
agree. This bijection is based on a generalization of Bender and
Knuth's involution \cite{\BeKnAA, pp.~46/47; \SagaAL, pp.~152/153} 
that proves
the symmetry of Schur functions combinatorially. In passing, they
provide (without proof) an alternative description of this bijection,
in terms of a jeu de taquin-like procedure. Though similar, this
procedure is different from ours. To understand the relationship
between the two procedures, we introduce a {\it
nondeterministic\/} jeu de taquin procedure which contains both
Algorithm~S and Goulden and Greene's algorithm as special cases.

The following (nondeterministic) algorithm maps a super semistandard
tableau $S$ of shape $\la/\mu$ to an array $A$ of shape $\la/\mu$
containing unbarred and barred positive integers such that
\roster
\item"(A1)" the multiset of entries in $S$ and $A$ agree,
\item"(A2)" if in $A$ each barred entry, $\bar y$ say, contained in cell
$\rh$, is replaced by $y-c_\rh$, then $A$ becomes a column-strict
reverse plane partition.
\endroster
It should be observed that such an array $A$ is exactly equivalent to
a pair $(P,U)$ of a column-strict reverse plane partition $P$ and a
$(0\text {--}1)$-tabloid $U$, the latter pairs occuring as output of
Algorithm~S. Given such an array $A$ the corresponding pair $(P,U)$
is obtained by defining $P$ to be the column-strict reverse plane
partition constructed in (A2) above, and by defining $U$ to be the $(0\text
{--}1)$-tabloid with $1$ in each cell that contains a barred entry in
$A$ and with $0$ in each cell that contains an unbarred entry in $A$.
Clearly, this correspondence is a bijection.

\bigskip
\noindent
{\bf Algorithm SG}. The input for the algorithm is a super semistandard
tableau $S$ of shape $\la/\mu$.

(SG0) Set $A:=S$.

\smallskip
(SG1) Consider all pairs $(x_1,\bar y_1)$, where $x_1$ is an unbarred
entry and $\bar y_1$ is a barred entry located in cell $\rh_1$, 
$x_1$ being the left neighbour of $\bar y_1$, and all pairs 
$(x_2,\bar y_2)$, where $x_2$ is an unbarred
entry and $\bar y_2$ is a barred entry located in cell $\rh_2$, 
$x_2$ being the top neighbour of $\bar y_2$. If for all such pairs
there holds
$$x_1\le y_1-c_{\rh_1}\tag4.4$$
and
$$x_2< y_2-c_{\rh_2},\tag4.5$$
then stop. The output of the algorithm is $A$. (Observe that (4.4)
and (4.5) are required for $A$ to satisfy property (A2) above.)

Otherwise, do one of the following two steps B or U.

B. Consider all violations of (4.4) or (4.5). Let $\bar y$ be the
{\it top-most\/} among the {\it minimal barred\/} entries involved.
(From Lemma~1, (2), it will follow that $\bar y$ is unique.)
Let $\bar y$ be located in cell $\rh$. 
Then we have the following
situation,
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&x_t&\\
\hlinefor5\\
&x_l&&\bar y&\\
\hlinefor5
\endsmatrix\ ,\tag4.6
$$
where at least one of $x_l>y-c_\rh$ and $x_t\ge y-c_\rh$ holds. 
(One of $x_l$ or $x_t$ is
also allowed to 
be actually not there, or, in abuse of notation, to be a barred entry.
In the latter case, by convention the corresponding inequality does
not hold.) If $x_l>x_t$ (or if $x_t$ is a barred entry) then do the move
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&x_t&\\
\hlinefor5\\
&\bar y&&x_l&\\
\hlinefor5
\endsmatrix\ .\tag4.7
$$
If $x_l\le x_t$ (or if $x_l$ is a barred entry) then do the move
$$\smatrix \format \sa\c\s\c\se\\
\omit&\omit&\hlinefor3\\
\omit&&&\bar y&\\
\hlinefor5\\
&x_l&&x_t&\\
\hlinefor5
\endsmatrix\ .\tag4.8
$$

U. Consider all violations of (4.4) or (4.5). Let $x$ be the
{\it right-most\/} among the {\it maximal unbarred\/} entries involved.
(From Lemma~1, (1), it will follow that $x$ is unique.)
Let the cell to the right of $x$ be $\rh_r$ and the cell below of $x$
be $\rh_b$. 
Then we have the following
situation,
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&x&&\bar y_r&\\
\hlinefor5\\
&\bar y_b&\\
\hlinefor3
\endsmatrix\ ,\tag4.9
$$
where at least one of $x>y_r-c_{\rh_r}$ and $x\ge y_b-c_{\rh_b}$ holds. 
(One of $\bar y_r$ or $\bar y_b$ is
also allowed to 
be actually not there, or, in abuse of notation, to be an unbarred entry.
In the latter case, by convention the corresponding inequality does
not hold.) If $\bar y_r\le \bar y_b$ (or if $\bar y_b$ is an 
unbarred entry) then do the move
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&\bar y_r&&x&\\
\hlinefor5\\
&\bar y_b&\\
\hlinefor3
\endsmatrix\ .\tag4.10
$$
If $\bar y_r>\bar y_b$ (or if $\bar y_r$ is an unbarred entry) then do the move
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&\bar y_b&&\bar y_r&\\
\hlinefor5\\
&x&\\
\hlinefor3
\endsmatrix\ .\tag4.11
$$

Let the new $A$ be the array of shape $\la/\mu$ just obtained. Repeat
(SG1).\quad \quad \qed

\bigskip
The output of the algorithm trivially satisfies condition (A1) above.
It will follow from Lemma~1 that the output also satisfies (A2), see
the Remark after Lemma~1.

It is an easy observation that when always choosing step B in
Algorithm~SG we basically do the same as in Algorithm~S. The only
difference is that in Algorithm~SG we do not subtract the content
from the barred entries and keep the barred entries as such, unlike
in Algorithm~S. The translation of an output of Algorithm~SG into the
corresponding output of Algorithm~S was described in the paragraph
preceding the description of Algorithm~SG. Similarly, when always
choosing step U in Algorithm~SG we (basically) do the same as in
Goulden and Greene's algorithm. So, roughly speaking, what we do in
Algorithm~S is to move the barred entries, one after the other,
``into" the unbarred entries, while Goulden and Greene move the
unbarred entries, one after the other, ``into" the barred entries.
Nevertheless, both procedures turn out to be equivalent, i.e\. the
final result is the same in both cases. In fact, the main result of
this section (Theorem~SG below) is that, regardless which choices we
take in step (SG1), the final result will be the same. 
Therefore, though Goulden and Greene write that,
unlike the standard jeu de taquin, their algorithm is deterministic
and different sequences of moves may give different results, there is
{\it some\/} freedom in choosing the order of moves. But there is not
as much freedom as in standard jeu de taquin where there is complete
freedom.
\proclaim{Theorem SG}The output of Algorithm~SG applied to some given
super semistandard tableau $S$ is independent of the choices between
$B$ and $U$ that are taken in step (SG1).
\endproclaim
The proof of Theorem~SG is based on the following two Lemmas.
\proclaim{Lemma 1} After each application of {\rm(SG1)} during
Algorithm~SG, the obtained array $A$ has the following properties.
\roster
\item Let $x$ be an unbarred entry located (weakly) to the left and
(weakly) above the unbarred entry $y$. Then $x\le y$. Also, each
column is {\it strictly\/} increasing in the unbarred entries.
\item Let $\bar x$ be a barred entry located (weakly) to the left and
(weakly) above the barred entry $\bar y$. Then $\bar x\le \bar y$. Also, each
row is {\it strictly\/} increasing in the barred entries.
\item For all pairs $(\bar x_1,y_1)$, where $\bar x_1$ is a barred
entry located in cell $\rh_1$ and $y_1$ is an unbarred entry,
$\bar x_1$ being the left neighbour of $y_1$, there holds
$$x_1-c_{\rh_1}\le y_1.\tag4.12$$
For all pairs 
$(\bar x_2,y_2)$, where $\bar x_2$ is a barred
entry located in cell $\rh_2$ and $y_2$ is an unbarred entry,
$\bar x_2$ being the top neighbour of $y_2$, there holds
$$x_2-c_{\rh_2}< y_2.\tag4.13$$
\endroster
\endproclaim
\remark{Remark} Clearly, this implies that the output of Algorithm~SG
satisfies (A2). For, in addition to the properties (1)--(3) of the Lemma, the
output satisfies also (4.4) and (4.5) since otherwise the step (SG1)
was not finished.
\endremark

\demo{Proof}We prove the Lemma by induction on the number of
applications of (SG1). The claim of the Lemma is certainly true at
the very beginning since we start with a super semistandard tableau.

For the induction step a careful case-by-case analysis has to be
performed. We consider just one case out of four (two cases each for
steps B and U, depending on whether moving horizontally or
vertically), the others are similar.

Suppose that by performing step U, (4.11), an unbarred entry, $x$
say, is moved down one row. Figure~5 displays the generic situation before
and after the move. Here, $m,n,s,t$ can be any nonnegative integers.
\vskip10pt
\vbox{
$$\gather
\raise1cm\hbox to0pt{\hskip3.8cm \eightpoint cell $\rh$\hss}
\raise0.7cm\hbox to0pt{\hskip3.6cm /\hss}
\raise-1cm\hbox to0pt{\hskip3.8cm \eightpoint cell $\om$\hss}
\raise-0.7cm\hbox to0pt{\hskip3.6cm $\backslash$\hss}
\smatrix \format\r\s\c\s\l\\
\hlinefor5\\
\dots \bar b\,x_{-m}\dots x_{-2}\,x_{-1}&&x&&x_1\,x_2\dots x_s\,\bar
z\dots\\
\hlinefor5\\
\dots a\,\bar y_{-n}\dots \bar y_{-2}\,\bar y_{-1}&&\bar y&&\bar
y_1\,\bar y_2\dots \bar y_t\,u\dots\\
\hlinefor5\\
\endsmatrix
\quad 
\raise1cm\hbox to0pt{\hskip3.8cm \eightpoint cell $\rh$\hss}
\raise0.7cm\hbox to0pt{\hskip3.6cm /\hss}
\raise-1cm\hbox to0pt{\hskip3.8cm \eightpoint cell $\om$\hss}
\raise-0.7cm\hbox to0pt{\hskip3.6cm $\backslash$\hss}
\smatrix \format\r\s\c\s\l\\
\hlinefor5\\
\dots \bar b\,x_{-m}\dots x_{-2}\,x_{-1}&&\bar y&&x_1\,x_2\dots x_s\,\bar
z\dots\\
\hlinefor5\\
\dots a\,\bar y_{-n}\dots \bar y_{-2}\,\bar y_{-1}&&x&&\bar
y_1\,\bar y_2\dots \bar y_t\,u\dots\\
\hlinefor5\\
\endsmatrix\\
\text {\eightpoint a. before the move\hskip4cm b. after the move}
\endgather$$
\centerline{\eightpoint Figure 5}
}
\vskip10pt
\noindent
The pictures have to be taken symbolically in the sense that all
combinations of $n>m$ or
$n\le m$, and $s>t$ or $s\le t$ are possible.

By definition of step U, $x$ is the right-most among the maximal
unbarred entries that violate (4.4) or (4.5) in Figure~5.a. Moreover,
since we moved down by (4.11),
we have (for the definition of $\om$ see Figure~5)
$$x\ge y-c_\om.\tag4.14$$
Of course, by induction hypothesis, nothing has to be shown for
entries that did not move. In fact, it suffices to prove
$$\align \text {ad (1):}\quad &a\le x\le u\tag4.15\\
\text {ad (2):}\quad &\bar b<\bar y<\bar z\tag4.16\\
\text {ad (3):}\quad &y-c_\rh<x,\ y-c_\rh\le x_1 \text { and
}y_{-1}-c_{\om_{-1}}\le x,\tag4.17
\endalign$$
where we assume that $\bar y_i$ is located in cell $\om_i$,
$i=-n,\dots,-1,1,\dots,t$.

In order to prove $a\le x$, we distinguish between two cases. First
suppose $a>y_{-n}-c_{\om_{-n}}$. This is a violation of (4.4). Since $x$
is the right-most among the maximal
unbarred entries that violate (4.4) or (4.5) in Figure~5.a, we must
have $x\ge a$. On the other hand, suppose $a\le y_{-n}-c_{\om_{-n}}$.
Then, by induction hypothesis (2) (applied to Figure~5.a of course),
and by (4.14), we have
$$a\le y_{-n}-c_{\om_{-n}}\le y_{-n+1}-c_{\om_{-n+1}}\le\dots\le
y_{-1}-c_{\om_{-1}}\le y-c_\om\le x.$$

The inequality $x\le u$ follows from the induction hypothesis (1).

For proving $\bar b<\bar y$, we first note that by induction
hypothesis (2) we have $\bar b\le \bar y$. So, we only have to show
that $\bar b\ne\bar y$. By way of contradiction, let us assume $\bar
b=\bar y$. Then, by induction hypothesis (2), it is impossible that
there are any barred entries to the left of $\bar y$ in Figure~5.a.
This means that $n=0$, and that $a$ is the left neighbour of $\bar y$
in Figure~5.a. At the very beginning the entry $\bar b$ was strictly
above and weakly to the right of the entry $\bar y$, because
otherwise $\bar b$ and $\bar y$ would have been in the same row at
some stage, contradicting induction hypothesis (2). For similar
reasons it is impossible that a rectangle determined by $\bar b$ as
its top-left corner and $\bar y$ as its bottom-right corner contained
another barred entry at any time. Finally, because of the argument
visualized in Figure~3, the path of entry $\bar y$ always stays below
the path of entry $\bar b$. Therefore $\bar b$ must have passed all
the way through from cell $\rh$ to the cell that it is now occupying
in Figure~5. Moreover, $x_{-m},\dots,x_{-1},x$ and $a$ did not move
after $\bar b$ passed through. But then, at some point, we must have
considered the situation
$$\smatrix \format \sa\c\s\c\se\\
\hlinefor5\\
&x&&\bar b&\\
\hlinefor5\\
&a&\\
\hlinefor3
\endsmatrix
$$
before. By induction hypothesis (1), this implies $x<a$. This
contradicts $x\ge a$ which was already established.

Next we show $\bar y<\bar z$. Let $\bar z$ be located in cell $\ze$.
If $s=0$, i.e\. if $\bar z$ is the right neighbour of $x$ in
Figure~5.a, then $\bar y<\bar z$ holds by definition of step U,
(4.11). If $s>0$, then we must have $x_s\le z-c_\ze$. For, otherwise
(4.4) would be violated. Besides, by induction hypothesis (1) we
have $x\le x_s$, and so $x$ would not be the 
right-most among the maximal
unbarred entries that violate (4.4) or (4.5) in Figure~5.a. Combining
this with (4.14), we conclude $y-c_\om\le x\le x_s\le z-c_\ze$ and
hence $y\le z-s-2<z$.

All three inequalities in (4.17) follow directly from (4.14), $x\le
x_1$, and $\bar y_{-1}<\bar y$, 
the latter two being due to the induction hypothesis (1), (2).
\quad \quad \qed
\enddemo
\proclaim{Lemma 2}Suppose that $A$ is an array satisfying
{\rm(1)--(3)} in
Lemma~1. Furthermore suppose that the right-most among the maximal
unbarred entries that violate {\rm(4.4)} or {\rm(4.5)} is not the left or top
neighbour of the top-most among the minimal barred entries 
that violate {\rm(4.4)} or {\rm(4.5)}. Then the steps {\rm U} and\/
{\rm B} commute,
$$\text {\rm U}(\text {\rm B}(A))=\text {\rm B}(\text {\rm U}(A)).$$
\endproclaim

\demo{Proof}This is established by a straight-forward case-by-case
analysis, utilizing Lem\-ma~1.\quad \quad \qed
\enddemo
Finally we are in the position to prove Theorem~SG.
\demo{Proof of Theorem SG} Suppose that during Algorithm~SG we
applied the sequence $W(\text {U},\text {B})$ of steps U and B to the
initial semistandard tableau $S$. One should think of $W(\text {U},\text
{B})$ as a word in U's and B's. Let the length of $W(\text {U},\text {B})$
be $L$. We shall show that 
$$W(\text {U},\text {B})(S)=\text{B}^L(S),\tag4.18$$
i.e\. that the result is the same when we apply step B all the time
(which is basically the same as Algorithm~S, as we already noted).
Clearly this would suffice to prove the Theorem.

We establish (4.18) by successively changing U's in $W(\text {U},\text {B})$
into B's and by applying the commutation rule in Lemma~2. First we claim
that there is a word $W_1(\text {U},\text {B})$ such that 
$$W(\text {U},\text {B})(S)=\text{B}\,W_1(\text {U},\text {B})(S).\tag4.19$$
This is trivially true if $W(\text {U},\text {B})$ already has the
form $\text{B}\,W_1(\text {U},\text {B})$. So let us assume that $W(\text {U},\text {B})$
has the form $\text{U}\,W_1(\text {U},\text {B})$. Consider the array 
$W_1(\text {U},\text {B})(S)$, the array obtained before the final
step. Since $W(\text {U},\text {B})(S)=\text {U}\,
W_1(\text {U},\text {B})(S)$ satisfies (4.4) and (4.5), it is obvious
that the right-most among the maximal
unbarred entries that violate (4.4) or (4.5) in $W_1(\text {U},\text {B})(S)$
is the left or top
neighbour of the top-most among the minimal barred entries 
that violate (4.4) or (4.5) in $W_1(\text {U},\text {B})(S)$. Hence,
it does not make any difference whether we apply U or B to $W_1(\text {U},\text
{B})(S)$. This settles (4.19) also in this case.

Now we use the commutation rule in Lemma~2 to move the left B in the
right-hand side of (4.19) inside the word $W_1(\text {U},\text
{B})$. The only problem arises, when we arrive at $
W_2(\text {U},\text {B})\,\text{B}\,\text{U}\,W_3(\text {U},\text {B})(S)$,
where $W_3(\text {U},\text {B})(S)$ is an array in which the
right-most among the maximal
unbarred entries that violate (4.4) or (4.5) is the left or top
neighbour of the top-most among the minimal barred entries 
that violate (4.4) or (4.5). In this case Lemma~2 does not apply. But
then we have $\text{U}\,W_3(\text {U},\text {B})(S)=\text{B}\,W_3(\text {U},\text
{B})(S)$, and therefore
$$W(\text {U},\text {B})(S)=W_2(\text {U},\text {B})\,\text{B}\,\text{B}\,
W_3(\text {U},\text {B}).\tag4.20$$
We continue by moving this ``new" B inside the word $W_3(\text {U},\text
{B})$, until we arrive at
$$W(\text {U},\text {B})(S)=\big(W_4(\text {U},\text
{B})\,\text{B}\big)(S),\tag4.21$$
for some word $W_4(\text {U},\text {B})$. By iterating the process
(4.19)-(4.20)-(4.21) we finally obtain (4.18).\quad \quad \qed
\enddemo



\remark{Acknowledgement}This work was carried out while the author
visited the University of California at San Diego. He thanks the
University of California and in particular Adriano Garsia for making
this visit possible. Besides, he is indebted to Adriano Garsia for
drawing his attention to the problem of finding ``nice" combinatorial
proofs of hook formulas.
\endremark



\subhead Appendix\endsubhead
The appendix contains a complete example for Algorithms~S and S* for
$\la=(4,3,3,1)$, setting up a mapping between the two sides
of Figure~2. See the specific descriptions given in Examples~1 and 2
of how to read the following
tables.

\newpage
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%1. Zeile, 2. Spalte
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%2. Zeile, 1. Spalte 
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%3. Zeile, 2. Spalte
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%4. Zeile, 2. Spalte
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%5. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(S1)\kern2.5pt}\to\longrightarrow
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\overset\hbox{\kern2.5pt(S3)\kern2.5pt}\to\longrightarrow
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%6. Zeile, 1. Spalte
\hskip46pt 
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
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&&
%6. Zeile, 2. Spalte
\overset\hbox{(S*3)}\to\longleftarrow
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&1&&1&&3&&\3&\\
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\overset\hbox{(S*1)}\to\longleftarrow
%\\&&\\\hlinefor3\\&&\\
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\newpage
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%7. Zeile, 2. Spalte
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\overset\hbox{\kern2.5pt(S3)\kern2.5pt}\to\longrightarrow
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%8. Zeile, 1. Spalte
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%8. Zeile, 2. Spalte
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%9. Zeile, 2. Spalte
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%10. Zeile, 2. Spalte
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%11. Zeile, 1. Spalte
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%11. Zeile, 2. Spalte
\overset\hbox{\kern2.5pt(S1)\kern2.5pt}\to\longrightarrow
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%14. Zeile, 1. Spalte
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%14. Zeile, 2. Spalte
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\overset\hbox{(S*1)}\to\longleftarrow
%\\&&\\\hlinefor3\\&&\\
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\newpage
$$
\smatrix \format \l\s\l\\
%13. Zeile, 1. Spalte
\smatrix \format \sa\c\s\c\s\c\s\c\se\\
\hlinefor9\\
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&&
\\&&\\
%22. Zeile, 1. Spalte
\hskip46pt 
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&1&&-&&3&&-&\\
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\Refs

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Young diagrams with applications to combinatorics and representations
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\ref\no \BeRJAA\by A. Berele and J. B. Remmel \yr 1985 \paper Hook
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\ref\no \BretAE\by F.    Brenti \yr 1993 \paper Determinants of
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\ref\no \GeViAA\by I. M. Gessel and X. Viennot \yr 1985 \paper
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\ref\no \GeViAB\by I. M. Gessel and X. Viennot \yr 1989 \paper
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\ref\no \KacVAD\by V. G. Kac \yr 1977 \paper Characters of typical
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\ref\no \KacVAE\by V. G. Kac \yr 1978 \paper Representations of
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\ref\no \KratAY\by C.    Krattenthaler \paper An involution
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\ref\no \MacdAC\by I. G. Macdonald \yr 1995 \book Symmetric Functions 
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\ref\no \MeNRAA\by N.    Metropolis, G. Nicoletti and G.-C. Rota \yr
1981 \paper A new class of symmetric functions\jour Math\. Anal\.
Appl\. Part~B, Adv\. in Math\. Suppl\. Stud\.\vol 7\pages 563--575\endref  

\ref\no \RemmAC\by J. B. Remmel \yr 1984 \paper The combinatorics of
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\ref\no \SagaAL\by B. E. Sagan \yr 1991 \book The symmetric group\publ
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\ref\no \ScheAA\by M.    Scheunert \yr 1979 \book The theory of Lie
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\ref\no \SchuAA\by M.-P. Sch\"utzenberger \yr 1977 \paper La
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\endRefs



\enddocument
