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\ihead{Invariant Theory, Young Bitableaux, and Combinatorics}
\ohead{page \thepage}
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\begin{document}

\author{J. D\'{e}sarm\'{e}nien, Joseph P. S. Kung, Gian-Carlo Rota\thanks{dedicated to
Garrett Birkhoff}}
\title{Invariant Theory, Young Bitableaux, and Combinatorics}
\date{unofficial re-edition, \today}
\maketitle
\tableofcontents

\subsection*{***}

The present document is an unofficial re-edition of the paper
\href{https://doi.org/10.1016/0001-8708(78)90077-4}{\textquotedblleft
Invariant Theory, Young Bitableaux, and Combinatorics\textquotedblright\ by J.
D\'{e}sarm\'{e}nien, Joseph P. S. Kung, Gian-Carlo Rota (originally published
in: Advances in Mathematics, Volume 27, Issue 1, January 1978, pp. 63--92,
doi:10.1016/0001-8708(78)90077-4)}.\footnote{See
\url{http://www.sciencedirect.com/science/article/pii/0001870878900774} for
the publisher's webpage for the original article.} This re-edition attempts to
correct misprints\footnote{All corrections I have made are identified in
footnotes, marked with \textquotedblleft\textit{Correction:}\textquotedblright%
.} and add some comments (identifiable as footnotes, marked with
\textquotedblleft\textit{Comment:}\textquotedblright).

This re-edition was prepared by Darij Grinberg. It is unofficial and has been
endorsed neither by the authors of the original article nor by Elsevier (its
publishers); its legality derives from
\href{https://tqft.net/misc/elsevier-oa/}{the Elsevier Open Archives license
as it was formulated in 2013}\footnote{See
\url{https://web-beta.archive.org/web/20131022235944/http://www.elsevier.com/about/open-access/oa-and-elsevier/oa-license-policy}
for an archived version of this license on Elsevier's website.}. The article
is in copyright \copyright \ 1978 Published by Elsevier Inc.

I have tried to reproduce the original paper as faithfully as possibly while
correcting errors and making use of the possibilities of contemporary LaTeX
typography (e.g., numbering theorems, finishing proofs with $\square$ signs).
Nevertheless, errors and unintended discrepancies are likely to have happened.
Please let me know if you find anything wrong!\footnote{My email address is
\url{darijgrinberg@gmail.com} .}

\section{Introduction}

Since its emergence in the middle of the last century, invariant theory has
oscillated between two clearly distinguishable poles. The first, and the one
that was later to survive the temporary \textquotedblleft
death\textquotedblright\ of the field, is geometry. Invariants were identified
with the invariants of surfaces. Their study, the aim of which was to give
information about the solution of systems of polynomial equations, was to lead
to the rise of commutative algebra. From this standpoint, projective
invariants were eventually seen as poor relations of the richer algebraic invariants.

A casualty of this trend was the study of the projective generation of
surfaces, a problem which was condemned by Cremona as \textquotedblleft too
difficult,\textquotedblright\ and which has never quite recovered from the
blow, despite the recent excitement over finite fields. In contrast, other
heretical schools survived the \textit{Fata Morgana} of algebra with the
promise, not always fulfilled, that sooner or later they would be brought back
into the commutative fold. Thus, the genial computations of the high school
teacher Hermann Schubert were proclaimed a \textquotedblleft
problem\textquotedblright\ by Hilbert, who was articulating the general
feeling at the time that enumerative geometry required a justification in
terms of the dominant concepts of the day, namely, rings and fields.

Similarly, the mystical vision of Hermann Grassmann, another high school
teacher, was only appreciated by other oddballs like Peano, Study, and several
inevitable English gentleman-mathematicians. It took the advocacy of someone
of the stature of Elie Cartan to get Grassmann's techniques accepted by a
public by then avid for simplications, but reluctant to acknowledge
embarrassing oversights; and then, only at the cost of putting them to a use
for which they were not intended, though magically suited. The recognition
that anti-commutativity is a sibling, with an equally noble genealogy, of
commutativity is only now beginning, under the prodding of the particle
physicists, who with exquisite salesmanship have proclaimed it a law of nature.

The second pole of invariant theory was algorithmic. To be sure, all invariant
theory is ultimately concerned with one problem. In crude, oversimplified,
off-putting language, this problem is to generalize to tensors the eigenvalue
theory of matrices, and all invariant theorists from Boole to Mumford have
been, tacitly or otherwise, concerned with it. The algorithmic school,
however, saw this problem as one of \textquotedblleft explicit
computation,\textquotedblright\ an expression which was to smack of
mathematical bad taste in the 1930s. In a century which prefers existence to
construction, structure to algorithm, algebra to combinatorics, such a school
could not thrive, and it did not, supported as it was more by the English and
Italians than by the Germans and French. There were, however, weightier
reasons for this defeat of the algorithmic school. Their most striking
productions, the expansions that go under the names of Capelli, Clebsch,
Gordan, and Young, were hopelessly tethered to characteristic zero, and seemed
to belie the avowed combinatorial ideal of doing away with all numbers that
are not integers, and preferably positive ones at that. To top it all, Igusa
showed that, with the massive machinery of algebraic geometry, some of the
results of classical invariant theory could be extended to fields of positive characteristic.

In this environment, the 1974 paper of Doubilet, Rota, and Stein \cite{3},
which for the first time succeeded in extending to arbitrary infinite fields,
by constructive algorithmic methods, the two \textquotedblleft fundamental
theorems\textquotedblright\ of invariant theory, could only appear as an
intrusion. To make things worse, the authors' sympathy for the nineteenth
century went as far as to embrace matters of style, thus alienating many
readers in a less romantic century. In 1976, de Concini and Procesi \cite{1}
charitably rewrote parts of that paper and developed some of the suggestions
made therein, thus showing that the authors' claims were indeed well-founded.

In this paper we give a self-contained combinatorial presentation -- the first
one, to be sure -- of vector invariant theory over an arbitrary infinite
field. We begin by proving the \textit{Straightening Formula}, which is
probably one of the fundamental algorithms of multilinear algebra. This
formula is the culmination of a trend of thought that can be traced back to
Capelli, and was developed most notably by Alfred Young and the Scottish
invariant theorists. Had it not been for the disrepute into which algorithmic
methods had fallen in the thirties, the full proof of this formula would have
appeared earlier than in 1974, and might have anticipated the current revival
of classical invariant theory.

In comparison with other classical expansions, the straightening formula
offers two advantages. First, it holds over the ring of integers. Second, it
recognizes the crucial role played by the notion of a bitableau in obtaining a
characteristic-free proof of the first fundamental theorem. In fact, we give
two proofs of this result. Both of these proofs are based on new ideas, first
presented in 1974. Even in characteristic zero, either of these proofs differs
from any previously given, and is, in addition, much simpler as it only relies
on elementary linear algebra and some combinatorics. The success of the notion
of a bitableau also shows why previous attempts to prove the first fundamental
theorem by expansion into \textit{single} Young tableaux were bound to fail.
Strangely, Alfred Young himself was the first to consider bitableaux in his
study of the representations of the octahedral group, but it did not occur to
him that they would be useful in the study of the projective group.

Less surprisingly, the straightening formula is also used to give a simple
proof of the second fundamental theorem, in a version that has been proved by
van der Waerden in characteristic zero. The present proof shows that the
straightening formula is indeed the characteristic-free replacement of the
Gordan-Capelli expansion.

The second fundamental theorem has lived in a limbo ever since Weyl's fumbling
justification in \textquotedblleft The Classical Groups\textquotedblright%
\ \cite{4}. Some invariant theorists have taken the easy way out and claimed
it as a result in algebraic geometry, stating certain facts about the
coordinate rings of Grassmannians or flag manifolds. We believe on the
contrary that the second fundamental theorem plays a crucial role in invariant
theory which can perhaps be best understood by analogy with the predicate
calculus. Here, two aspects have long been recognized as complementary: a
syntactical aspect, where the subject is presented as a purely algebraic
system subject to formal rules; and a semantical aspect, where the possible
set-theoretic interpretations, or models, are classified. These two aspects
are connected by the G\"{o}del completeness theorem.

A corresponding situation obtains in invariant theory. Here, what we call the
letter place algebra is the syntactic counterpart to the semantics of
representing abstract brackets by actual inner products of vectors and
covectors in a vector space. The second fundamental theorem is the
invariant-theoretic analog of the G\"{o}del completeness theorem. This
suggests a host of questions on invariants which can be gleaned from analogous
questions in the predicate calculus.

Other applications of the straightening formula, some of which were adumbrated
in 1974, will be given elsewhere. We mention, as examples, a
characteristic-free theory of symmetric functions, the study of polynomial
identities in an associative algebra, the classification of transvectants, and
connections with the algebra of second quantization. The present work is
merely the first in what is hoped to be a far-reaching extension of the
research program of projective invariant theory.

\section{\label{sect.young}Young Tableaux}

The fundamental combinatorial notion in this study is that of a \textit{Young
tableau}. Let $\left(  \lambda\right)  =\left(  \lambda_{1},\ldots,\lambda
_{p}\right)  $ be a \textit{partition} of the integer $n$: that is, $\left(
\lambda\right)  $ is a finite sequence of positive integers such that%
\[
\lambda_{1}+\cdots+\lambda_{p}=n
\]
and%
\[
\lambda_{1}\geq\cdots\geq\lambda_{p}>0.
\]
If $\left(  \lambda\right)  $ is a partition of $n$,
then\footnote{\textit{Correction:} Added \textquotedblleft
then\textquotedblright\ to make the sentence unambiguous.} its \textit{shape},
also denoted by $\left(  \lambda\right)  $, is the set of integer points
$\left(  i,-j\right)  $ in the plane, with $1\leq j\leq p$ and $1\leq
i\leq\lambda_{j}$. The shape $\left(  \lambda\right)  =\left(  \lambda
_{1},\ldots,\lambda_{s}\right)  $ is said to be \textit{longer} than the shape
$\left(  \mu\right)  =\left(  \mu_{1},\ldots,\mu_{t}\right)  $ if, considered
as a finite sequence, $\left(  \lambda\right)  $ is greater than $\left(
\mu\right)  $ in the lexicographic order from left to right. Here, the
definition of the lexicographic order is subject to the following
caveat:\footnote{\textit{Correction:} Added this caveat. Without it, for
example, the Second Fundamental Theorem of Invariant Theory (as stated in this
paper) would be false, because the partition $\left(  d,1\right)  $ would be
strictly longer than $\left(  d\right)  $.} If a finite sequence $\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{s}\right)  $ is a proper prefix of a
finite sequence $\left(  \mu_{1},\mu_{2},\ldots,\mu_{t}\right)  $ (that is, we
have $t>s$ and $\left(  \mu_{1},\mu_{2},\ldots,\mu_{s}\right)  =\left(
\lambda_{1},\lambda_{2},\ldots,\lambda_{s}\right)  $), then $\left(
\lambda\right)  $ is understood to be greater than $\left(  \mu\right)  $ (not
smaller than $\left(  \mu\right)  $ as with the usual definition of
lexicographic order).\footnote{Thus, the lexicographic order is defined as
follows:
\par
A finite sequence $\left(  \lambda_{1},\lambda_{2},\ldots,\lambda_{s}\right)
$ is greater than a finite sequence $\left(  \mu_{1},\mu_{2},\ldots,\mu
_{t}\right)  $ in the lexicographic order if and only if one of the following
two statements holds:
\par
\begin{itemize}
\item There exists a $k\in\left\{  1,2,\ldots,\min\left\{  s,t\right\}
\right\}  $ such that $\lambda_{k}>\mu_{k}$ but each $i\in\left\{
1,2,\ldots,k-1\right\}  $ satisfies $\lambda_{i}=\mu_{i}$.
\par
\item We have $t>s$ and each $i\in\left\{  1,2,\ldots,s\right\}  $ satisfies
$\lambda_{i}=\mu_{i}$.
\end{itemize}
} Thus, for example, $\left(  3,1\right)  $ is longer (i.e., greater) than
$\left(  3,1,2\right)  $.\ \ \ \ \footnote{\textit{Comment:} If this sounds
counterintuitive to you, think of the word \textquotedblleft
longer\textquotedblright\ as referring to the \textit{horizontal} length of
the partition's Young diagram (not the length of the partition as a list of
numbers).}

A \textit{Young tableau} of\footnote{\textit{Correction:} Replaced
\textquotedblleft on\textquotedblright\ by \textquotedblleft
of\textquotedblright.} the shape $\left(  \lambda\right)  $ with values in the
set $E$ is an assignment of an element of $E$ to each point in the shape
$\left(  \lambda\right)  $. For example, $T_{1}$ and $T_{2}$ are Young
tableaux of shape $\left(  \lambda\right)  =\left(  5,4,2,2,1,1\right)  $ with
values in the integers:\footnote{\textit{Correction:} I have corrected these
two tableaux. The original versions were their transposes, and they would fail
the claim that $T_{2}$ is standard.}%
\[
T_{1}=%
\begin{array}
[c]{cccccc}%
3 & 2 & 4 & 4 & 7 & 8\\
1 & 2 & 3 & 5 &  & \\
2 & 1 &  &  &  & \\
6 & 2 &  &  &  & \\
4 &  &  &  &  &
\end{array}
,\ \ \ \ \ \ \ \ \ \ T_{2}=%
\begin{array}
[c]{cccccc}%
1 & 2 & 4 & 5 & 7 & 8\\
1 & 2 & 4 & 6 &  & \\
2 & 3 &  &  &  & \\
2 & 4 &  &  &  & \\
3 &  &  &  &  &
\end{array}
.
\]


\footnote{\textit{Correction:} This paragraph and the next two have been added
by me.} If $p$ and $q$ are two integers, then the \textit{cell }$\left(
p,q\right)  $ shall mean the point $\left(  q,-p\right)  $ in the plane. Thus,
if $\left(  \lambda\right)  =\left(  \lambda_{1},\lambda_{2},\ldots
,\lambda_{\ell}\right)  $ is a partition, then the shape of $\left(
\lambda\right)  $ is the set of all cells $\left(  p,q\right)  $ with $1\leq
p\leq\ell$ and $1\leq q\leq\lambda_{p}$. These cells are called the
\textit{cells of }$\left(  \lambda\right)  $.

Recall that a Young tableau of shape $\left(  \lambda\right)  $ with values in
$E$ is an assignment of an element of $E$ to each point in the shape $\left(
\lambda\right)  $; in other words, a Young tableau of shape $\left(
\lambda\right)  $ with values in $E$ is an assignment of an element of $E$ to
each cell of $\left(  \lambda\right)  $.

If $R$ is any tableau, and if $p$ and $q$ are two positive integers such that
the cell $\left(  p,q\right)  $ (that is, the point $\left(  q,-p\right)  $)
belongs to $\left(  \lambda\right)  $, then we let $R\left(  p,q\right)  $
denote the element of $\mathcal{U}$ assigned to this cell $\left(  p,q\right)
$. This element $R\left(  p,q\right)  $ is called the \textit{entry} of $R$ in
cell $\left(  p,q\right)  $. For a given $p\geq1$, the $p$\textit{-th row} of
the tableau $R$ consists of the entries of $R$ in all cells of the form
$\left(  p,q\right)  $ with $q\geq1$ (arranged in the order of increasing
$q$). For a given $q\geq1$, the $q$\textit{-th column} of the tableau $R$
consists of the entries of $R$ in all cells of the form $\left(  p,q\right)  $
with $p\geq1$ (arranged in the order of increasing $p$). For example, the
$2$-nd row of the above tableau $T_{1}$ is $1235$, whereas its $3$-rd column
is $43$.

In this paper, $E$ is always a totally ordered set. A Young tableau is said to
be \textit{standard} if the entries in each row are
increasing\footnote{\textit{Comment:} The word \textquotedblleft
increasing\textquotedblright\ means \textquotedblleft strictly
increasing\textquotedblright\ throughout this paper.} from left to right, and
the entries in each column are nondecreasing
downward\footnote{\textit{Comment:} This notion of \textquotedblleft
standard\textquotedblright\ is not the one commonly used nowadays in
combinatorics (although it seems to have had some popularity in invariant
theory). What is called a \textquotedblleft standard Young
tableau\textquotedblright\ in this paper would probably be called
\textquotedblleft cosemistandard Young tableau\textquotedblright\ or
\textquotedblleft transpose semistandard Young tableau\textquotedblright%
\ (indeed, it is a Young tableau whose transpose is semistandard, in today's
language). The currently popular use of the word \textquotedblleft
standard\textquotedblright\ is different: It stands for a Young tableau $T$
with the following properties:
\par
\begin{itemize}
\item The entries of $T$ are $1,2,\ldots,n$ (for a fixed $n\in\mathbb{N}$).
\par
\item The entries of $T$ are pairwise distinct.
\par
\item The entries in each row of $T$ are increasing from left to right.
\par
\item The entries in each column of $T$ are increasing downward.
\end{itemize}
}. In our previous example, $T_{2}$ is standard but $T_{1}$ is not. This
definition, though unconventional, is the natural one for dealing with
bitableaux (which are introduced in the sequel).

A word on notation: $\mathfrak{S}_{p}$ denotes the symmetric group on $p$
symbols, and for a permutation $\sigma\in\mathfrak{S}_{p}$, its signature is
denoted $\operatorname*{sgn}\left(  \sigma\right)  $.

\section{\label{sect.straight}The Straightening Formula}

Let $\mathcal{X}=\left\{  x_{1},\ldots,x_{n}\right\}  $ and $\mathcal{U}%
=\left\{  u_{1},\ldots,u_{k}\right\}  $ be two alphabets, and let $P$ be the
algebra of polynomials over the field $K$ in the indeterminates $\left(
x_{i}\mid u_{j}\right)  $; this $K$-algebra\footnote{\textit{Correction:}
Added the words \textquotedblleft this $K$-algebra\textquotedblright.} $P$ is
called the \textit{letter place algebra}. Suppose $\left(  x_{i_{1}}%
,\ldots,x_{i_{p}}\right)  $ and $\left(  u_{i_{1}},\ldots,u_{i_{p}}\right)  $
are two finite sequences with the same length of letters from $\mathcal{X}$
and $\mathcal{U}$. Their inner product\footnote{\textit{Correction:} The
original wrote \textquotedblleft$\left(  x_{i_{1}}\cdots x_{i_{p}}\mid
u_{j_{1}},\cdots,u_{j_{p}}\right)  $\textquotedblright\ instead of
\textquotedblleft$\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots
u_{j_{p}}\right)  $\textquotedblright. I have removed the commas, since they
seem to be unintentional.} $\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}%
}\cdots u_{j_{p}}\right)  $ is the polynomial in $P$ defined by%
\[
\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\sum_{\sigma\in\mathfrak{S}_{p}}\operatorname*{sgn}\left(  \sigma\right)
\left(  x_{i_{\sigma1}}\mid u_{j_{1}}\right)  \cdots\left(  x_{i_{\sigma p}%
}\mid u_{j_{p}}\right)  .
\]


The inner product is an antisymmetric function in $x_{i}$ and
in\footnote{\textit{Comment:} I have added this \textquotedblleft
in\textquotedblright\ in order to stress that it is antisymmetric in the $p$
variables $i_{1},i_{2},\ldots,i_{p}$ and in the $p$ variables $j_{1}%
,j_{2},\ldots,j_{p}$ separately, but not in all the $2p$ variables taken
together.} $u_{j}$\ \ \ \ \footnote{\textit{Comment:} Explicitly, this is
saying the following:
\par
\begin{itemize}
\item If two of the numbers $i_{1},i_{2},\ldots,i_{p}$ are equal, then
$\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  =0$.
The same holds if two of the numbers $j_{1},j_{2},\ldots,j_{p}$ are equal.
\par
\item If we interchange two of the numbers $i_{1},i_{2},\ldots,i_{p}$, then
the inner product $\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots
u_{j_{p}}\right)  $ gets multiplied by $-1$. The same happens if we
interchange two of the numbers $j_{1},j_{2},\ldots,j_{p}$.
\end{itemize}
\par
Both of these properties follow from the fact that%
\begin{align*}
\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)   &
=\sum_{\sigma\in\mathfrak{S}_{p}}\operatorname*{sgn}\left(  \sigma\right)
\left(  x_{i_{\sigma1}}\mid u_{j_{1}}\right)  \cdots\left(  x_{i_{\sigma p}%
}\mid u_{j_{p}}\right) \\
&  =\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert
\end{align*}
(and from the antisymmetry of the determinant).}. Thus, we may suppose, up to
a change in sign, that in any inner product, the indices of $x$ and $u$ are
increasing. Moreover, an inner product is nonzero if and only if no letter is repeated.

We define a total order on the set $\mathcal{U}$ by setting $u_{1}%
<u_{2}<\cdots<u_{k}$. Similarly, we define a total order on the set
$\mathcal{X}$ by setting $x_{1}<x_{2}<\cdots<x_{n}$. Thus, a tableau with
entries from $\mathcal{X}$ or from $\mathcal{U}$ may be
standard.\footnote{\textit{Correction:} I added these three sentences to
clarify the definition of a \textquotedblleft standard\textquotedblright%
\ tableau with entries in $\mathcal{X}$ or $\mathcal{U}$.}

The \textit{content} of a monomial in $P$ is the pair of vectors%
\[
\left(  \alpha,\beta\right)  =\left(  \left(  \alpha_{1},\ldots,\alpha
_{n}\right)  ,\left(  \beta_{1},\ldots,\beta_{k}\right)  \right)  ,
\]
where $\alpha_{s}$ (resp. $\beta_{t}$) is the total degree of the factors in
the monomial of the form $\left(  x_{s}\mid u_{j}\right)  $, $1\leq j\leq k$
(resp. $\left(  x_{i}\mid u_{t}\right)  $, $1\leq i\leq n$). The
monomials\footnote{\textit{Correction:} Replaced \textquotedblleft
monomial\textquotedblright\ by \textquotedblleft monomials\textquotedblright%
\ here.} of content $\left(  \alpha,\beta\right)  $
generate\footnote{\textit{Correction:} Replaced \textquotedblleft
generates\textquotedblright\ by \textquotedblleft generate\textquotedblright%
\ here.} a subspace of $P$, denoted by $P\left(  \alpha,\beta\right)  $. The
elements of $P\left(  \alpha,\beta\right)  $ are homogeneous polynomials, in
which each monomial has the same content; we say that a polynomial in
$P\left(  \alpha,\beta\right)  $ has content $\left(  \alpha,\beta\right)  $.
It is clear that the product of a polynomial of content $\left(  \alpha
,\beta\right)  $ and\footnote{\textit{Correction:} Removed a misleading comma
before this \textquotedblleft and\textquotedblright.} a polynomial of content
$\left(  \alpha^{\prime},\beta^{\prime}\right)  $ is a polynomial of content
$\left(  \alpha+\alpha^{\prime},\beta+\beta^{\prime}\right)  $. For example,
the inner product $\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots
u_{j_{p}}\right)  $ has content $\left(  \alpha,\beta\right)  $ where
$\alpha_{i}$ (resp. $\beta_{i}$) is $1$ if $x_{i}$ is in the sequence
$x_{i_{1}},\ldots,x_{i_{p}}$ (resp. $u_{i}$ is in the sequence $u_{j_{1}%
},\ldots,u_{j_{p}}$) and $0$ otherwise.

A \textit{bitableau} is a pair $\left[  T,T^{\prime}\right]  $ of Young
tableaux of the same shape $\left(  \lambda\right)  $, where the tableau $T$
has entries from $\mathcal{X}$ and the tableau $T^{\prime}$ has entries from
$\mathcal{U}$. The \textit{content} of the bitableau $\left[  T,T^{\prime
}\right]  $ is the pair of vectors $\left(  \alpha,\beta\right)  $ where
$\alpha_{i}$ (resp. $\beta_{j}$) is the number of occurrences of $x_{i}$ in
$T$ (resp. $u_{j}$ in $T^{\prime}$). With a bitableau $\left[  T,T^{\prime
}\right]  $ of content $\left(  \alpha,\beta\right)  $, we associate the
polynomial, denoted by $\left(  T\mid T^{\prime}\right)  $, obtained by taking
the product of the inner products of each row of $T$ with the corresponding
row in $T^{\prime}$. The polynomial $\left(  T\mid T^{\prime}\right)  $, which
is in $P\left(  \alpha,\beta\right)  $, is called the \textit{bideterminant}
of the bitableau $\left[  T,T^{\prime}\right]  $, or simply, the bideterminant
$\left(  T\mid T^{\prime}\right)  $.

\begin{example}%
\[
\left(
\begin{array}
[c]{ccc}%
x_{1} & x_{2} & x_{3}\\
x_{2} & x_{3} & \\
x_{1} &  &
\end{array}
\ \middle\vert\
\begin{array}
[c]{ccc}%
u_{1} & u_{3} & u_{4}\\
u_{1} & u_{2} & \\
u_{3} &  &
\end{array}
\right)  =\left(  x_{1}x_{2}x_{3}\mid u_{1}u_{3}u_{4}\right)  \left(
x_{2}x_{3}\mid u_{1}u_{2}\right)  \left(  x_{1}\mid u_{3}\right)  .
\]

\end{example}

As for inner products, the bideterminant $\left(  T\mid T^{\prime}\right)  $
is nonzero if and only if no letter is repeated in any row of $T$ or
$T^{\prime}$. Moreover, we can suppose, up to a change of sign, that the
entries in each row of $T$ and $T^{\prime}$ in the bideterminant are increasing.

A bitableau $\left[  T,T^{\prime}\right]  $ is \textit{standard} if both $T$
and $T^{\prime}$ are standard. For example, the bitableau%
\[
\left[
\begin{array}
[c]{ccc}%
x_{1} & x_{2} & x_{3}\\
x_{1} & x_{3} & \\
x_{2} &  &
\end{array}
,\
\begin{array}
[c]{ccc}%
u_{1} & u_{2} & u_{4}\\
u_{1} & u_{3} & \\
u_{3} &  &
\end{array}
\right]
\]
is standard.

We can now state the main result of this
section.\footnote{\textit{Correction:} In the following theorem and also in
the Corollary that follows it, I have replaced \textquotedblleft standard
tableaux\textquotedblright\ by \textquotedblleft standard
bitableaux\textquotedblright.}

\begin{theorem}
[the straightening formula]\label{thm.1}Suppose $\left[  T,T^{\prime}\right]
$ is a bitableau of shape $\left(  \lambda\right)  $ and content $\left(
\alpha,\beta\right)  $. Then, its bideterminant $\left(  T\mid T^{\prime
}\right)  $ is a linear combination, with integer coefficients, of
bideterminants of standard bitableaux of the same content and of the same or
longer shape\footnotemark.
\end{theorem}

\footnotetext{\textit{Comment:} Here is a less ambiguous way to state this:
Its bideterminant $\left(  T\mid T^{\prime}\right)  $ can be written in the
form $\left(  T\mid T^{\prime}\right)  =\sum_{j\in J}\alpha_{j}\left(
S_{j}\mid S_{j}^{\prime}\right)  $, where $J$ is a finite set, each
$\alpha_{j}$ is an integer, and each $\left[  S_{j},S_{j}^{\prime}\right]  $
is a bitableau having the same content as $\left[  T,T^{\prime}\right]  $ and
with the following property: The shape of $\left[  S_{j},S_{j}^{\prime
}\right]  $ is longer or equal to the shape of $\left[  T,T^{\prime}\right]
$. (The \textit{shape} of a bitableau $\left[  T,T^{\prime}\right]  $ is
defined to be the shape of the tableau $T$ (or, equivalently, the shape of the
tableau $T^{\prime}$).)
\par
In general, we often speak of the \textquotedblleft shape\textquotedblright%
\ or the \textquotedblleft content\textquotedblright\ of a bideterminant, when
meaning the shape or the content of the underlying bitableau. For example, a
bideterminant of shape $\left(  \lambda\right)  $ and content $\left(
\alpha,\beta\right)  $ means a bideterminant of a bitableau of shape $\left(
\lambda\right)  $ and content $\left(  \alpha,\beta\right)  $.}

\begin{example}%
\[
\left(
\begin{array}
[c]{c}%
x_{2}\\
x_{1}%
\end{array}
\ \middle|\
\begin{array}
[c]{c}%
u_{1}\\
u_{2}%
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
x_{1}\\
x_{2}%
\end{array}
\ \middle|\
\begin{array}
[c]{c}%
u_{1}\\
u_{2}%
\end{array}
\right)  -\left(
\begin{array}
[c]{cc}%
x_{1} & x_{2}%
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
u_{1} & u_{2}%
\end{array}
\right)  .
\]

\end{example}

\begin{corollary}
\label{cor.strf.gen}The vector space $P\left(  \alpha,\beta\right)  $ is
generated by the bideterminants of standard bitableaux of content $\left(
\alpha,\beta\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.strf.gen}.]We only need to observe that the
monomial $\left(  x_{i_{1}}\mid u_{j_{1}}\right)  \cdots\left(  x_{i_{p}}\mid
u_{j_{p}}\right)  $ is the bideterminant of the bitableau:%
\[
\left[
\begin{array}
[c]{c}%
x_{i_{1}}\\
\vdots\\
x_{i_{p}}%
\end{array}
,\
\begin{array}
[c]{c}%
u_{j_{1}}\\
\vdots\\
u_{j_{p}}%
\end{array}
\right]  .
\]

\end{proof}

To facilitate the proof of Theorem \ref{thm.1}, we introduce the notion of a
\textit{shuffle product}. Let%
\[
\left(  i_{1},\ldots,i_{p},l_{1},\ldots,l_{q}\right)
\]
be an increasing sequence of integers, and%
\begin{align*}
A  &  =\left(  x_{i_{1}}\cdots x_{i_{p}}x_{i_{p+1}}\cdots x_{i_{s}}\mid
u_{j_{1}}\cdots u_{j_{s}}\right)  ,\\
B  &  =\left(  x_{l_{1}}\cdots x_{l_{q}}x_{l_{q+1}}\cdots x_{l_{t}}\mid
u_{m_{1}}\cdots u_{m_{t}}\right)
\end{align*}
be two inner products. The \textit{shuffle product }$AB$\textit{ supported by
the variables }$x_{i_{1}},\ldots,x_{i_{p}},x_{l_{1}},\ldots,x_{l_{q}}$ is
defined by%
\begin{align*}
&  \left(  \dot{x}_{i_{1}}\cdots\dot{x}_{i_{p}}x_{i_{p+1}}\cdots x_{i_{s}}\mid
u_{j_{1}}\cdots u_{j_{s}}\right)  \left(  \dot{x}_{l_{1}}\cdots\dot{x}_{l_{q}%
}x_{l_{q+1}}\cdots x_{l_{t}}\mid u_{m_{1}}\cdots u_{m_{t}}\right) \\
&  =\left.  \sum_{\sigma}\right.  ^{\prime}\operatorname*{sgn}\left(
\sigma\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i_{1}}\cdots x_{\sigma i_{p}%
}x_{i_{p+1}}\cdots x_{i_{s}}\mid u_{j_{1}}\cdots u_{j_{s}}\right)  \left(
x_{\sigma l_{1}}\cdots x_{\sigma l_{q}}x_{l_{q+1}}\cdots x_{l_{t}}\mid
u_{m_{1}}\cdots u_{m_{t}}\right)  ,
\end{align*}
where the summation is over all permutations $\sigma$ of the set $\left\{
i_{1},\ldots,i_{p},l_{1},\ldots,l_{q}\right\}  $ for which $\sigma
i_{1}<\cdots<\sigma i_{p}$ and $\sigma l_{1}<\cdots<\sigma l_{q}$. This
restricted summation is indicated by the notation $\left.  \sum\right.
^{\prime}$. Another notational device is: A \textit{dot} over a letter
indicates that the letter is in the support of the shuffle
product\footnote{\textit{Comment:} We say that a letter is \textquotedblleft
in the support\textquotedblright\ of the shuffle product if this letter is one
of the variables on which the shuffle product is supported.}. The notion of a
shuffle product supported by letters in $\mathcal{U}$ is similar.

\begin{example}
The shuffle product $\left(  x_{1}x_{2}x_{3}\mid\mathbf{u}\right)  \left(
x_{3}x_{4}x_{1}\mid\mathbf{u}^{\prime}\right)  $ supported by $x_{1},x_{2}$ in
the first term and $x_{3},x_{4}$ in the second is given by%
\begin{align*}
&  \left(  \dot{x}_{1}\dot{x}_{2}x_{3}\mid\mathbf{u}\right)  \left(  \dot
{x}_{3}\dot{x}_{4}x_{1}\mid\mathbf{u}^{\prime}\right) \\
&  =\left(  x_{1}x_{2}x_{3}\mid\mathbf{u}\right)  \left(  x_{3}x_{4}x_{1}%
\mid\mathbf{u}^{\prime}\right)  -\left(  x_{1}x_{3}x_{3}\mid\mathbf{u}\right)
\left(  x_{2}x_{4}x_{1}\mid\mathbf{u}^{\prime}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(  x_{1}x_{4}x_{3}\mid\mathbf{u}\right)  \left(
x_{2}x_{3}x_{1}\mid\mathbf{u}^{\prime}\right)  +\left(  x_{2}x_{3}x_{3}%
\mid\mathbf{u}\right)  \left(  x_{1}x_{4}x_{1}\mid\mathbf{u}^{\prime}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ -\left(  x_{2}x_{4}x_{3}\mid\mathbf{u}\right)  \left(
x_{1}x_{3}x_{1}\mid\mathbf{u}^{\prime}\right)  +\left(  x_{3}x_{4}x_{3}%
\mid\mathbf{u}\right)  \left(  x_{1}x_{2}x_{1}\mid\mathbf{u}^{\prime}\right)
.
\end{align*}
Only two of the terms in the expansion are nonzero, and after an appropriate
reordering, we have%
\begin{align*}
&  \left(  \dot{x}_{1}\dot{x}_{2}x_{3}\mid\mathbf{u}\right)  \left(  x_{1}%
\dot{x}_{3}\dot{x}_{4}\mid\mathbf{u}^{\prime}\right) \\
&  =\left(  x_{1}x_{2}x_{3}\mid\mathbf{u}\right)  \left(  x_{1}x_{3}x_{4}%
\mid\mathbf{u}^{\prime}\right)  -\left(  x_{1}x_{3}x_{4}\mid\mathbf{u}\right)
\left(  x_{1}x_{2}x_{3}\mid\mathbf{u}^{\prime}\right)  .
\end{align*}

\end{example}

Now, observe that, by definition,%
\begin{equation}
\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert . \label{eq.inner-prod-as-det}%
\end{equation}
We can expand the determinant by the first column to obtain the identity%
\begin{equation}
\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\left(  \dot{x}_{i_{1}}\mid u_{j_{1}}\right)  \left(  \dot{x}_{i_{2}}%
\cdots\dot{x}_{i_{p}}\mid u_{j_{2}}\cdots u_{j_{p}}\right)  .
\label{eq.laplace.1}%
\end{equation}
Similarly, using Laplace's expansion, we see that each $s\in\left\{
0,1,\ldots,p\right\}  $ satisfies
\[
\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\left(  \dot{x}_{i_{1}}\cdots\dot{x}_{i_{s}}\mid u_{j_{1}}\cdots u_{j_{s}%
}\right)  \left(  \dot{x}_{i_{s+1}}\cdots\dot{x}_{i_{p}}\mid u_{j_{s+1}}\cdots
u_{j_{p}}\right)
\]
\footnote{\textit{Comment:} This latter formula will be referred to as
\textquotedblleft\textit{Laplace's identity}\textquotedblright\ further on in
this paper. It follows from the following apocryphal property of determinants
(sometimes known as \textquotedblleft Laplace expansion in the first $s$
columns\textquotedblright):
\par
\textbf{Laplace expansion in the first }$s$ \textbf{columns.} Given
$p\in\mathbb{N}$ and $s\in\left\{  0,1,\ldots,p\right\}  $ and a $p\times
p$-matrix $A=\left(  a_{i,j}\right)  _{1\leq i\leq p,\ 1\leq j\leq p}$, we
have%
\begin{equation}
\left\vert A\right\vert =\sum_{\substack{\sigma\in\mathfrak{S}_{p}%
;\\\sigma1<\sigma2<\cdots<\sigma s;\\\sigma\left(  s+1\right)  <\sigma\left(
s+2\right)  <\cdots<\sigma p}}\operatorname*{sgn}\left(  \sigma\right)
\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma1,1} & \cdots & a_{\sigma1,s}\\
\vdots & \ddots & \vdots\\
a_{\sigma s,1} & \cdots & a_{\sigma s,s}%
\end{array}
\right\vert \cdot\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma\left(  s+1\right)  ,s+1} & \cdots & a_{\sigma\left(  s+1\right)
,p}\\
\vdots & \ddots & \vdots\\
a_{\sigma p,s+1} & \cdots & a_{\sigma p,p}%
\end{array}
\right\vert . \label{eq.laplace-expansion}%
\end{equation}
\par
Let me outline two proofs of (\ref{eq.laplace-expansion}):
\par
\begin{itemize}
\item \textit{First proof:} Argue that the right hand side of
(\ref{eq.laplace-expansion}) is an alternating multilinear function in the
rows of $A$ (the alternating property follows from showing that the right hand
side vanishes whenever two consecutive rows of $A$ are equal) that takes the
value $1$ when $A=I_{p}$. But it is well-known that the only such function is
$\left\vert A\right\vert $.
\par
\item \textit{Second proof:} For any permutations $\alpha\in\mathfrak{S}_{s}$
and $\beta\in\mathfrak{S}_{p-s}$, we define a permutation $\alpha\oplus
\beta\in\mathfrak{S}_{p}$ as the map%
\[
\left\{  1,2,\ldots,p\right\}  \rightarrow\left\{  1,2,\ldots,p\right\}
,\ \ \ \ \ \ \ \ \ \ i\mapsto%
\begin{cases}
\alpha\left(  i\right)  , & \text{if }i\leq s;\\
\beta\left(  i-s\right)  +s, & \text{if }i>s
\end{cases}
.
\]
Notice that the map $\mathfrak{S}_{s}\times\mathfrak{S}_{p-s}\rightarrow
\mathfrak{S}_{p},\ \left(  \alpha,\beta\right)  \mapsto\alpha\oplus\beta$ is a
group homomorphism. Now, it is not hard to show that each permutation $\tau
\in\mathfrak{S}_{p}$ can be \textbf{uniquely} written in the form $\tau
=\sigma\circ\left(  \alpha\oplus\beta\right)  $, where $\sigma$ is a
permutation in $\mathfrak{S}_{p}$ satisfying $\sigma1<\sigma2<\cdots<\sigma s$
and $\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma p$,
and where $\alpha\in\mathfrak{S}_{s}$ and $\beta\in\mathfrak{S}_{p-s}$. Thus,%
\begin{align*}
&  \sum_{\tau\in\mathfrak{S}_{p}}\operatorname*{sgn}\left(  \tau\right)
\prod_{i=1}^{p}a_{\tau i,i}\\
&  =\sum_{\substack{\sigma\in\mathfrak{S}_{p};\\\sigma1<\sigma2<\cdots<\sigma
s;\\\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma
p}}\sum_{\alpha\in\mathfrak{S}_{s}}\sum_{\beta\in\mathfrak{S}_{p-s}%
}\underbrace{\operatorname*{sgn}\left(  \sigma\circ\left(  \alpha\oplus
\beta\right)  \right)  }_{=\operatorname*{sgn}\left(  \sigma\right)
\operatorname*{sgn}\left(  \alpha\oplus\beta\right)  }\underbrace{\prod
_{i=1}^{p}a_{\left(  \sigma\circ\left(  \alpha\oplus\beta\right)  \right)
i,i}}_{=\left(  \prod_{i=1}^{s}a_{\left(  \sigma\circ\left(  \alpha\oplus
\beta\right)  \right)  i,i}\right)  \left(  \prod_{i=s+1}^{p}a_{\left(
\sigma\circ\left(  \alpha\oplus\beta\right)  \right)  i,i}\right)  }\\
&  =\sum_{\substack{\sigma\in\mathfrak{S}_{p};\\\sigma1<\sigma2<\cdots<\sigma
s;\\\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma
p}}\sum_{\alpha\in\mathfrak{S}_{s}}\sum_{\beta\in\mathfrak{S}_{p-s}%
}\operatorname*{sgn}\left(  \sigma\right)  \underbrace{\operatorname*{sgn}%
\left(  \alpha\oplus\beta\right)  }_{=\operatorname*{sgn}\left(
\alpha\right)  \operatorname*{sgn}\left(  \beta\right)  }\left(  \prod
_{i=1}^{s}\underbrace{a_{\left(  \sigma\circ\left(  \alpha\oplus\beta\right)
\right)  i,i}}_{\substack{=a_{\sigma\left(  \alpha i\right)  ,i}\\\text{(since
}\left(  \alpha\oplus\beta\right)  i=\alpha i\text{)}}}\right)  \left(
\prod_{i=s+1}^{p}\underbrace{a_{\left(  \sigma\circ\left(  \alpha\oplus
\beta\right)  \right)  i,i}}_{\substack{=a_{\sigma\left(  \beta\left(
i-s\right)  +s\right)  ,i}\\\text{(since }\left(  \alpha\oplus\beta\right)
i=\beta\left(  i-s\right)  +s\text{)}}}\right) \\
&  =\sum_{\substack{\sigma\in\mathfrak{S}_{p};\\\sigma1<\sigma2<\cdots<\sigma
s;\\\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma
p}}\sum_{\alpha\in\mathfrak{S}_{s}}\sum_{\beta\in\mathfrak{S}_{p-s}%
}\operatorname*{sgn}\left(  \sigma\right)  \operatorname*{sgn}\left(
\alpha\right)  \operatorname*{sgn}\left(  \beta\right)  \left(  \prod
_{i=1}^{s}a_{\sigma\left(  \alpha i\right)  ,i}\right)  \underbrace{\left(
\prod_{i=s+1}^{p}a_{\sigma\left(  \beta\left(  i-s\right)  +s\right)
,i}\right)  }_{\substack{=\prod_{i=1}^{p-s}a_{\sigma\left(  \beta\left(
i\right)  +s\right)  ,i+s}\\\text{(here, we have substituted }i+s\\\text{for
}i\text{ in the sum)}}}\\
&  =\sum_{\substack{\sigma\in\mathfrak{S}_{p};\\\sigma1<\sigma2<\cdots<\sigma
s;\\\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma
p}}\sum_{\alpha\in\mathfrak{S}_{s}}\sum_{\beta\in\mathfrak{S}_{p-s}%
}\operatorname*{sgn}\left(  \sigma\right)  \operatorname*{sgn}\left(
\alpha\right)  \operatorname*{sgn}\left(  \beta\right)  \left(  \prod
_{i=1}^{s}a_{\sigma\left(  \alpha i\right)  ,i}\right)  \left(  \prod
_{i=1}^{p-s}a_{\sigma\left(  \beta\left(  i\right)  +s\right)  ,i+s}\right) \\
&  =\sum_{\substack{\sigma\in\mathfrak{S}_{p};\\\sigma1<\sigma2<\cdots<\sigma
s;\\\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma
p}}\operatorname*{sgn}\left(  \sigma\right)  \underbrace{\left(  \sum
_{\alpha\in\mathfrak{S}_{s}}\operatorname*{sgn}\left(  \alpha\right)
\prod_{i=1}^{s}a_{\sigma\left(  \alpha i\right)  ,i}\right)  }_{=\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma1,1} & \cdots & a_{\sigma1,s}\\
\vdots & \ddots & \vdots\\
a_{\sigma s,1} & \cdots & a_{\sigma s,s}%
\end{array}
\right\vert }\underbrace{\left(  \sum_{\beta\in\mathfrak{S}_{p-s}%
}\operatorname*{sgn}\left(  \beta\right)  \prod_{i=1}^{p-s}a_{\sigma\left(
\beta\left(  i\right)  +s\right)  ,i+s}\right)  }_{=\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma\left(  s+1\right)  ,s+1} & \cdots & a_{\sigma\left(  s+1\right)
,p}\\
\vdots & \ddots & \vdots\\
a_{\sigma p,s+1} & \cdots & a_{\sigma p,p}%
\end{array}
\right\vert }\\
&  =\sum_{\substack{\sigma\in\mathfrak{S}_{p};\\\sigma1<\sigma2<\cdots<\sigma
s;\\\sigma\left(  s+1\right)  <\sigma\left(  s+2\right)  <\cdots<\sigma
p}}\operatorname*{sgn}\left(  \sigma\right)  \left\vert
\begin{array}
[c]{ccc}%
a_{\sigma1,1} & \cdots & a_{\sigma1,s}\\
\vdots & \ddots & \vdots\\
a_{\sigma s,1} & \cdots & a_{\sigma s,s}%
\end{array}
\right\vert \cdot\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma\left(  s+1\right)  ,s+1} & \cdots & a_{\sigma\left(  s+1\right)
,p}\\
\vdots & \ddots & \vdots\\
a_{\sigma p,s+1} & \cdots & a_{\sigma p,p}%
\end{array}
\right\vert .
\end{align*}
Since the left hand side of this equality is precisely $\left\vert
A\right\vert $ (by the definition of $\left\vert A\right\vert $), this
rewrites as%
\[
\left\vert A\right\vert =\sum_{\substack{\sigma\in\mathfrak{S}_{p}%
;\\\sigma1<\sigma2<\cdots<\sigma s;\\\sigma\left(  s+1\right)  <\sigma\left(
s+2\right)  <\cdots<\sigma p}}\operatorname*{sgn}\left(  \sigma\right)
\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma1,1} & \cdots & a_{\sigma1,s}\\
\vdots & \ddots & \vdots\\
a_{\sigma s,1} & \cdots & a_{\sigma s,s}%
\end{array}
\right\vert \cdot\left\vert
\begin{array}
[c]{ccc}%
a_{\sigma\left(  s+1\right)  ,s+1} & \cdots & a_{\sigma\left(  s+1\right)
,p}\\
\vdots & \ddots & \vdots\\
a_{\sigma p,s+1} & \cdots & a_{\sigma p,p}%
\end{array}
\right\vert .
\]
This proves (\ref{eq.laplace-expansion}).
\end{itemize}
}. These two identities are examples of the fact that, under certain
assumptions, the shuffle product of two inner products is equal to an inner
product of length greater than that of each of the two original inner
products.\footnote{\textit{Correction:} Replaced \textquotedblleft$\left(
x_{i_{1}}\cdots x_{i_{p}}x_{i_{p+1}}\cdots x_{i_{1}}\mid u_{j_{1}}\cdots
u_{j_{s}}\right)  $\textquotedblright\ by \textquotedblleft$\left(  x_{i_{1}%
}\cdots x_{i_{p}}x_{i_{p+1}}\cdots x_{i_{s}}\mid u_{j_{1}}\cdots u_{j_{s}%
}\right)  $\textquotedblright\ in the following lemma.}

\begin{lemma}
\label{lem.2}Let $\left(  x_{i_{1}}\cdots x_{i_{p}}x_{i_{p+1}}\cdots x_{i_{s}%
}\mid u_{j_{1}}\cdots u_{j_{s}}\right)  $ and $\left(  x_{l_{1}}\cdots
x_{l_{q}}x_{l_{q+1}}\cdots x_{l_{t}}\mid u_{m_{1}}\cdots u_{m_{t}}\right)  $
be two inner products satisfying:%
\begin{align*}
i_{1}  &  <\cdots<i_{p}<l_{1}<\cdots<l_{q},\\
j_{1}  &  <\cdots<j_{s},\\
m_{1}  &  <\cdots<m_{t},\\
s  &  <p+q,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ t<p+q.
\end{align*}
Then the shuffle product%
\[
C=\left(  \dot{x}_{i_{1}}\cdots\dot{x}_{i_{p}}x_{i_{p+1}}\cdots x_{i_{s}}\mid
u_{j_{1}}\cdots u_{j_{s}}\right)  \left(  \dot{x}_{l_{1}}\cdots\dot{x}_{l_{q}%
}x_{l_{q+1}}\cdots x_{l_{t}}\mid u_{m_{1}}\cdots u_{m_{t}}\right)
\]
is a linear combination, with integer coefficients, of bideterminants of
bitableaux of shape strictly longer than each of the partitions $\left(
s\right)  $ and $\left(  t\right)  $ of the integers $s$ and $t$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.2}.]The proof is a computation with four steps.
First, expand the shuffle product $C$:%
\begin{align*}
C  &  =\left.  \sum_{\sigma}\right.  ^{\prime}\operatorname*{sgn}\left(
\sigma\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i_{1}}\cdots x_{\sigma i_{p}%
}x_{i_{p+1}}\cdots x_{i_{s}}\mid u_{j_{1}}\cdots u_{j_{s}}\right)  \left(
x_{\sigma l_{1}}\cdots x_{\sigma l_{q}}x_{l_{q+1}}\cdots x_{l_{t}}\mid
u_{m_{1}}\cdots u_{m_{t}}\right)  .
\end{align*}
Now apply Laplace's identity to the letters in $U$;%
\begin{align*}
C  &  =\left.  \sum_{\sigma}\right.  ^{\prime}\operatorname*{sgn}\left(
\sigma\right)  \left(  x_{\sigma i_{1}}\cdots x_{\sigma i_{p}}\mid\dot
{u}_{j_{1}}\cdots\dot{u}_{j_{p}}\right)  \left(  x_{i_{p+1}}\cdots x_{i_{s}%
}\mid\dot{u}_{j_{p+1}}\cdots\dot{u}_{j_{s}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  x_{\sigma l_{1}}\cdots x_{\sigma l_{q}%
}\mid\overline{u}_{m_{1}}\cdots\overline{u}_{m_{q}}\right)  \left(
x_{l_{q+1}}\cdots x_{l_{t}}\mid\overline{u}_{m_{q+1}}\cdots\overline{u}%
_{m_{t}}\right)  .
\end{align*}
To distinguish between the two shuffle products, a bar instead of a dot is
used in the second. We next group together the first and third factor:%
\begin{align*}
C  &  =\left.  \sum_{\sigma,\tau,\mu}\right.  ^{\prime}\operatorname*{sgn}%
\left(  \sigma\right)  \operatorname*{sgn}\left(  \tau\right)
\operatorname*{sgn}\left(  \mu\right)  \left[  \left(  x_{\sigma i_{1}}\cdots
x_{\sigma i_{p}}\mid u_{\tau j_{1}}\cdots u_{\tau j_{p}}\right)  \left(
x_{\sigma l_{1}}\cdots x_{\sigma l_{q}}\mid u_{\mu m_{1}}\cdots u_{\mu m_{q}%
}\right)  \right] \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  x_{i_{p+1}}\cdots x_{i_{s}}\mid u_{\tau
j_{p+1}}\cdots u_{\tau j_{s}}\right)  \left(  x_{l_{q+1}}\cdots x_{l_{t}}\mid
u_{\mu m_{q+1}}\cdots u_{\mu m_{t}}\right)  .
\end{align*}
Finally, we apply Laplace's identity, this time on the letters $x_{i_{1}%
},\ldots,x_{i_{p}},x_{l_{1}},\ldots,x_{l_{q}}$:%
\begin{align*}
C  &  =\left.  \sum_{\tau,\mu}\right.  ^{\prime}\operatorname*{sgn}\left(
\tau\right)  \operatorname*{sgn}\left(  \mu\right)  \left(  x_{i_{1}}\cdots
x_{i_{p}}x_{l_{1}}\cdots x_{l_{q}}\mid u_{\tau j_{1}}\cdots u_{\tau j_{p}%
}u_{\mu m_{1}}\cdots u_{\mu m_{q}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  x_{i_{p+1}}\cdots x_{i_{s}}\mid u_{\tau
j_{p+1}}\cdots u_{\tau j_{s}}\right)  \left(  x_{l_{q+1}}\cdots x_{l_{t}}\mid
u_{\mu m_{q+1}}\cdots u_{\mu m_{t}}\right)  .
\end{align*}
Each term in this last expansion for $C$ is a bideterminant with three
rows\footnote{\textit{Comment:} Some of these rows can be empty. (This happens
when $p=s$ or $q=t$.)}, with the first row of length $p+q>s,t$. This concludes
the proof of the lemma.
\end{proof}

Since the summation in the shuffle product always includes the identity
permutation, we can restate the previous lemma in the following equivalent
form:\footnote{\textit{Correction:} Added \textquotedblleft of $\left\{
i_{1},\ldots,i_{p},l_{1},\ldots,l_{q}\right\}  $\textquotedblright\ in the
following lemma.}

\begin{lemma}
\label{lem.3}Let $\left(  x_{i_{1}}\cdots x_{i_{p}}x_{i_{p+1}}\cdots x_{i_{s}%
}\mid u_{j_{1}}\cdots u_{j_{s}}\right)  $ and $\left(  x_{l_{1}}\cdots
x_{l_{q}}x_{l_{q+1}}\cdots x_{l_{t}}\mid u_{m_{1}}\cdots u_{m_{t}}\right)  $
be two inner products satisfying%
\begin{align*}
i_{1}  &  <\cdots<i_{p}<l_{1}<\cdots<l_{q},\\
j_{1}  &  <\cdots<j_{s},\\
m_{1}  &  <\cdots<m_{t},\\
s  &  <p+q,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ t<p+q.
\end{align*}
Then,%
\begin{align*}
&  \left(  x_{i_{1}}\cdots x_{i_{s}}\mid u_{j_{1}}\cdots u_{j_{s}}\right)
\left(  x_{l_{1}}\cdots x_{l_{t}}\mid u_{m_{1}}\cdots u_{m_{t}}\right) \\
&  =-\left.  \sum_{\substack{\sigma,\\\sigma\neq\operatorname*{id}}}\right.
^{\prime}\operatorname*{sgn}\left(  \sigma\right)  \left(  x_{\sigma i_{1}%
}\cdots x_{\sigma i_{p}}x_{i_{p+1}}\cdots x_{i_{s}}\mid u_{j_{1}}\cdots
u_{j_{s}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma l_{1}}\cdots x_{\sigma l_{q}%
}x_{l_{q+1}}\cdots x_{l_{t}}\mid u_{m_{1}}\cdots u_{m_{t}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +D,
\end{align*}
where the summation is over all the nonidentical permutations $\sigma$ of
$\left\{  i_{1},\ldots,i_{p},l_{1},\ldots,l_{q}\right\}  $ satisfying $\sigma
i_{1}<\cdots<\sigma i_{p}$ and $\sigma l_{1}<\cdots<\sigma l_{q}$, and the
term $D$ is a linear combination with integer coefficients of bideterminants
of bitableaux of shape strictly longer than $\left(  s\right)  $ and $\left(
t\right)  $.
\end{lemma}

Remarking that all we have done remains valid if we exchange the roles of the
alphabets $\mathcal{X}$ and $\mathcal{U}$, we are now ready to prove Theorem
\ref{thm.1}.

\begin{proof}
[Proof of Theorem \ref{thm.1}.]We begin by defining a total order on
bitableaux of the same shape. Let $\left[  T,T^{\prime}\right]  $ be a
bitableau of shape\footnote{\textit{Correction:} Replaced \textquotedblleft
form\textquotedblright\ by \textquotedblleft shape\textquotedblright.}
$\left(  \lambda\right)  =\left(  \lambda_{1},\ldots,\lambda_{p}\right)  $.
The entry in $T$ (resp. $T^{\prime}$) on row $s$ and column $t$ is denoted by
$x_{i\left(  s,t\right)  }$ (resp. $u_{j\left(  s,t\right)  }$). With the
bitableau $\left[  T,T^{\prime}\right]  $, we associate the sequence%
\begin{align*}
&  \left(  i\left(  1,1\right)  ,\ldots,i\left(  1,\lambda_{1}\right)
,\ i\left(  2,1\right)  ,\ldots,i\left(  2,\lambda_{2}\right)  ,\ \ldots
,\ i\left(  p,1\right)  ,\ldots,i\left(  p,\lambda_{p}\right)  ,\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  j\left(  1,1\right)  ,\ldots,j\left(
1,\lambda_{1}\right)  ,\ j\left(  2,1\right)  ,\ldots,j\left(  2,\lambda
_{2}\right)  ,\ \ldots,\ j\left(  p,1\right)  ,\ldots,j\left(  p,\lambda
_{p}\right)  \right)  ,
\end{align*}
which is the sequence obtained by reading off the bitableau row by
row\footnote{\textit{Comment:} ... where we first read the tableau $T$ and
then the tableau $T^{\prime}$.}. The bitableaux are now ordered according to
the lexicographic order on their associated sequences.

Fix a pair of vectors $\left(  \alpha,\beta\right)  $, and consider from now
on solely the bitableaux of content $\left(  \alpha,\beta\right)  $. Now,
suppose the theorem is false for at least one of these
bitableaux.\footnote{\textit{Correction:} I have added the preceding two
sentences (replacing \textquotedblleft Now, suppose the theorem is
false.\textquotedblright) and also added the words \textquotedblleft of
content $\left(  \alpha,\beta\right)  $\textquotedblright\ in the next two
sentences. Otherwise, the following sentence would not make sense (there might
not be a \textquotedblleft longest shape\textquotedblright).} Let $\left(
\lambda\right)  $ be the longest shape with a bitableau of content $\left(
\alpha,\beta\right)  $ not satisfying the theorem\footnote{\textit{Comment:}
This is well-defined, because there are only finitely many shapes that afford
bitableaux of content $\left(  \alpha,\beta\right)  $, and at least one of
these bitableaux fails to satisfy the theorem.}. Among the bitableaux of shape
$\left(  \lambda\right)  $ and content $\left(  \alpha,\beta\right)  $, let
$\left[  T,T^{\prime}\right]  $ be the smallest bitableau not satisfying the
theorem:%
\[
\left[  T,T^{\prime}\right]  =\left[
\begin{array}
[c]{cccc}%
x_{i\left(  1,1\right)  } & \cdots & \cdots & x_{i\left(  1,\lambda
_{1}\right)  }\\
\vdots &  & \vdots & \\
x_{i\left(  p,1\right)  } & \cdots & x_{i\left(  p,\lambda_{p}\right)  } &
\end{array}
,%
\begin{array}
[c]{cccc}%
u_{j\left(  1,1\right)  } & \cdots & \cdots & u_{j\left(  1,\lambda
_{1}\right)  }\\
\vdots &  & \vdots & \\
u_{j\left(  p,1\right)  } & \cdots & u_{j\left(  p,\lambda_{p}\right)  } &
\end{array}
\right]  .
\]
Suppose that $\left[  T_{1},T_{1}^{\prime}\right]  $ is obtained from $\left[
T,T^{\prime}\right]  $ by putting each row in increasing order. Then, $\left[
T_{1},T_{1}^{\prime}\right]  $ has a lexicographically smaller associated
sequence than $\left[  T,T^{\prime}\right]  $ (unless $\left[  T_{1}%
,T_{1}^{\prime}\right]  =\left[  T,T^{\prime}\right]  $%
)\ \ \ \ \footnote{\textit{Correction:} I have added the preceding
parenthetical.}. But $\left(  T_{1}\mid T_{1}^{\prime}\right)  =\pm\left(
T\mid T^{\prime}\right)  $, and hence, if $\left[  T,T^{\prime}\right]  $ is a
counterexample, so is $\left[  T_{1},T_{1}^{\prime}\right]  $. We conclude
that all the rows in $\left[  T,T^{\prime}\right]  $ are in increasing order.

Clearly, $\left[  T,T^{\prime}\right]  $ is not standard; let us suppose that
$T$ is nonstandard. Then, there exist integers $l$ and $m$,
with\footnote{\textit{Correction:} Added the word \textquotedblleft
with\textquotedblright. Also, replaced \textquotedblleft$1\leq m\leq
\lambda_{l}$\textquotedblright\ by \textquotedblleft$1\leq m\leq\lambda_{l+1}%
$\textquotedblright.} $1\leq l\leq p$ and\footnote{\textit{Correction:}
Replaced the ambiguous comma by an \textquotedblleft and\textquotedblright.}
$1\leq m\leq\lambda_{l+1}$, such that $i\left(  l,m\right)  >i\left(
l+1,m\right)  $. That is, we have the following
situation:\footnote{\textit{Correction:} Fixed the size of the parentheses in
the table below, and removed an unnecessary comma.}%
\[%
\begin{array}
[c]{cccccccc}%
x_{i\left(  l,1\right)  } & \cdots & x_{i\left(  l,m-1\right)  } & x_{i\left(
l,m\right)  } & x_{i\left(  l,m+1\right)  } & \cdots & \cdots & x_{i\left(
l,\lambda_{l}\right)  }\\
x_{i\left(  l+1,1\right)  } & \cdots & x_{i\left(  l+1,m-1\right)  } &
x_{i\left(  l+1,m\right)  } & x_{i\left(  l+1,m+1\right)  } & \cdots &
x_{i\left(  l+1,\lambda_{l+1}\right)  } &
\end{array}
\ \ .
\]
We call such a situation a \textit{violation}.

In the bideterminant $\left(  T\mid T^{\prime}\right)  $, consider the shuffle
product of the two inner products corresponding to rows $l$ and $l+1$, which
support the letters \newline$x_{i\left(  l+1,1\right)  },\ldots,x_{i\left(
l+1,m\right)  },x_{i\left(  l,m\right)  },\ldots,x_{i\left(  l,\lambda
_{l}\right)  }$. Since the support contains $\lambda_{l}+1$ letters, and the
length of each of the inner products is at most $\lambda_{l}$, we can apply
Lemma \ref{lem.3}\footnote{\textit{Correction:} Replaced \textquotedblleft the
lemma\textquotedblright\ by \textquotedblleft Lemma \ref{lem.3}%
\textquotedblright.} to obtain%
\begin{equation}
\left(  T\mid T^{\prime}\right)  =-\left.  \sum_{\substack{\sigma,\\\sigma
\neq\operatorname*{id}}}\right.  ^{\prime}\pm\left(  T_{\sigma}\mid T^{\prime
}\right)  +D, \label{pf.thm.1.3}%
\end{equation}
where $D$ is a linear combination of bideterminants of shape
greater\footnote{\textit{Comment:} Here and in the following,
\textquotedblleft greater\textquotedblright\ means \textquotedblleft strictly
longer\textquotedblright.} than $\left(  \lambda\right)  $%
.\ \ \ \ \footnote{\textit{Comment:} Let us explain why this holds.
\par
For each $k\in\left\{  1,2,\ldots,p\right\}  $, let
\[
r_{k}=\left(  x_{i\left(  k,1\right)  }\cdots x_{i\left(  k,\lambda
_{k}\right)  }\mid u_{j\left(  k,1\right)  }\cdots u_{j\left(  k,\lambda
_{k}\right)  }\right)  .
\]
Then, $\left(  T\mid T^{\prime}\right)  =r_{1}r_{2}\cdots r_{p}$ (by the
definition of the bideterminant $\left(  T\mid T^{\prime}\right)  $).
\par
Now, consider the inner products%
\[
r_{l+1}=\left(  x_{i\left(  l+1,1\right)  }\cdots x_{i\left(  l+1,m-1\right)
}x_{i\left(  l+1,m\right)  }x_{i\left(  l+1,m+1\right)  }\cdots x_{i\left(
l+1,\lambda_{l+1}\right)  }\mid u_{j\left(  l+1,1\right)  }\cdots u_{j\left(
l+1,\lambda_{l+1}\right)  }\right)
\]
and
\begin{align*}
r_{l}  &  =\left(  x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)
}x_{i\left(  l,m\right)  }x_{i\left(  l,m+1\right)  }\cdots x_{i\left(
l,\lambda_{l}\right)  }\mid u_{j\left(  l,1\right)  }\cdots u_{j\left(
l,\lambda_{l}\right)  }\right) \\
&  =\left(  x_{i\left(  l,m\right)  }x_{i\left(  l,m+1\right)  }\cdots
x_{i\left(  l,\lambda_{l}\right)  }x_{i\left(  l,1\right)  }\cdots x_{i\left(
l,m-1\right)  }\mid u_{j\left(  l,1\right)  }\cdots u_{j\left(  l,\lambda
_{l}\right)  }\right)  .
\end{align*}
These two inner products satisfy%
\[
i\left(  l+1,1\right)  <\cdots<i\left(  l+1,m\right)  <i\left(  l,m\right)
<\cdots<i\left(  l,\lambda_{l}\right)
\]
(since $i\left(  l,m\right)  >i\left(  l+1,m\right)  $ and since the rows of
$T$ are increasing) and $j\left(  l+1,1\right)  <\cdots<j\left(
l+1,\lambda_{l+1}\right)  $ and $j\left(  l,1\right)  <\cdots<j\left(
l,\lambda_{l}\right)  $ and $\lambda_{l+1}<\lambda_{l}+1$ and $\lambda
_{l}<\lambda_{l}+1$. Hence, Lemma \ref{lem.3} can be applied to these two
inner products, showing that%
\begin{align}
&  r_{l+1}r_{l}\nonumber\\
&  =-\left.  \sum_{\substack{\sigma,\\\sigma\neq\operatorname*{id}}}\right.
^{\prime}\operatorname*{sgn}\left(  \sigma\right)  \left(  x_{\sigma i\left(
l+1,1\right)  }\cdots x_{\sigma i\left(  l+1,m\right)  }x_{i\left(
l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda_{l+1}\right)  }\mid
u_{j\left(  l+1,1\right)  }\cdots u_{j\left(  l+1,\lambda_{l+1}\right)
}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma
i\left(  l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)
}x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(
l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +D^{\prime}, \label{pf.thm.1.fn5.rl+1rl}%
\end{align}
where the summation is over all the nonidentical permutations $\sigma$ of
\newline$\left\{  i\left(  l+1,1\right)  ,\ldots,i\left(  l+1,m\right)
,i\left(  l,m\right)  ,\ldots,i\left(  l,\lambda_{l}\right)  \right\}  $
satisfying $\sigma i\left(  l+1,1\right)  <\cdots<\sigma i\left(
l+1,m\right)  $ and $\sigma i\left(  l,m\right)  <\cdots<\sigma i\left(
l,\lambda_{l}\right)  $, and the term $D^{\prime}$ is a linear combination
with integer coefficients of bideterminants of bitableaux of shape strictly
longer than $\left(  \lambda_{l}\right)  $ and $\left(  \lambda_{l+1}\right)
$.
\par
Now, multiplying the equality (\ref{pf.thm.1.fn5.rl+1rl}) by $\left(
r_{1}r_{2}\cdots r_{l-1}\right)  \left(  r_{l+2}r_{l+3}\cdots r_{p}\right)  $,
we obtain%
\begin{align*}
&  r_{l+1}r_{l}\left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(  r_{l+2}%
r_{l+3}\cdots r_{p}\right) \\
&  =-\left.  \sum_{\substack{\sigma,\\\sigma\neq\operatorname*{id}}}\right.
^{\prime}\operatorname*{sgn}\left(  \sigma\right)  \left(  x_{\sigma i\left(
l+1,1\right)  }\cdots x_{\sigma i\left(  l+1,m\right)  }x_{i\left(
l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda_{l+1}\right)  }\mid
u_{j\left(  l+1,1\right)  }\cdots u_{j\left(  l+1,\lambda_{l+1}\right)
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma
i\left(  l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)
}x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(
l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +D^{\prime}\left(  r_{1}r_{2}\cdots r_{l-1}\right)
\left(  r_{l+2}r_{l+3}\cdots r_{p}\right)  .
\end{align*}
Since $r_{l+1}r_{l}\left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right)  =r_{1}r_{2}\cdots r_{p}=\left(  T\mid
T^{\prime}\right)  $, this rewrites as%
\begin{align*}
&  \left(  T\mid T^{\prime}\right) \\
&  =-\left.  \sum_{\substack{\sigma,\\\sigma\neq\operatorname*{id}}}\right.
^{\prime}\operatorname*{sgn}\left(  \sigma\right)  \left(  x_{\sigma i\left(
l+1,1\right)  }\cdots x_{\sigma i\left(  l+1,m\right)  }x_{i\left(
l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda_{l+1}\right)  }\mid
u_{j\left(  l+1,1\right)  }\cdots u_{j\left(  l+1,\lambda_{l+1}\right)
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma
i\left(  l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)
}x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(
l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right) \\
&  \ \ \ \ \ \ \ \ \ \ +D^{\prime}\left(  r_{1}r_{2}\cdots r_{l-1}\right)
\left(  r_{l+2}r_{l+3}\cdots r_{p}\right)  .
\end{align*}
Thus, in order to prove (\ref{pf.thm.1.3}), we need to verify the following
two claims:
\par
\begin{quote}
\textit{Claim 1:} Let $\sigma$ be a nonidentical permutation of \newline%
$\left\{  i\left(  l+1,1\right)  ,\ldots,i\left(  l+1,m\right)  ,i\left(
l,m\right)  ,\ldots,i\left(  l,\lambda_{l}\right)  \right\}  $ satisfying
$\sigma i\left(  l+1,1\right)  <\cdots<\sigma i\left(  l+1,m\right)  $ and
$\sigma i\left(  l,m\right)  <\cdots<\sigma i\left(  l,\lambda_{l}\right)  $.
Then,%
\begin{align*}
&  \left(  x_{\sigma i\left(  l+1,1\right)  }\cdots x_{\sigma i\left(
l+1,m\right)  }x_{i\left(  l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda
_{l+1}\right)  }\mid u_{j\left(  l+1,1\right)  }\cdots u_{j\left(
l+1,\lambda_{l+1}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma
i\left(  l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)
}x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(
l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right) \\
&  =\pm\left(  T_{\sigma}\mid T^{\prime}\right)
\end{align*}
for some tableau $T_{\sigma}$ which differs from $T$ only in rows $l$ and
$l+1$, and whose rows $l$ and $l+1$ are%
\[%
\begin{array}
[c]{cccccccc}%
x_{i\left(  l,1\right)  } & \cdots & x_{i\left(  l,m-1\right)  } & x_{\sigma
i\left(  l,m\right)  } & x_{\sigma i\left(  l,m+1\right)  } & \cdots & \cdots
& x_{\sigma i\left(  l,\lambda_{l}\right)  }\\
x_{\sigma i\left(  l+1,1\right)  } & \cdots & x_{\sigma i\left(
l+1,m-1\right)  } & x_{\sigma i\left(  l+1,m\right)  } & x_{i\left(
l+1,m+1\right)  } & \cdots & x_{i\left(  l+1,\lambda_{l+1}\right)  } &
\end{array}
\ \ \ .
\]
\par
\textit{Claim 2:} Let $\left[  S,S^{\prime}\right]  $ be a bitableau of shape
strictly longer than $\left(  \lambda_{l}\right)  $ and $\left(  \lambda
_{l+1}\right)  $. Then,%
\[
\left(  S\mid S^{\prime}\right)  \left(  r_{1}r_{2}\cdots r_{l-1}\right)
\left(  r_{l+2}r_{l+3}\cdots r_{p}\right)
\]
is a bideterminant of shape greater than $\left(  \lambda\right)  $.
\end{quote}
\par
[\textit{Proof of Claim 1:} Let $T_{\sigma}$ be the tableau which differs from
$T$ only in rows $l$ and $l+1$, and whose rows $l$ and $l+1$ are%
\[%
\begin{array}
[c]{cccccccc}%
x_{i\left(  l,1\right)  } & \cdots & x_{i\left(  l,m-1\right)  } & x_{\sigma
i\left(  l,m\right)  } & x_{\sigma i\left(  l,m+1\right)  } & \cdots & \cdots
& x_{\sigma i\left(  l,\lambda_{l}\right)  }\\
x_{\sigma i\left(  l+1,1\right)  } & \cdots & x_{\sigma i\left(
l+1,m-1\right)  } & x_{\sigma i\left(  l+1,m\right)  } & x_{i\left(
l+1,m+1\right)  } & \cdots & x_{i\left(  l+1,\lambda_{l+1}\right)  } &
\end{array}
\ \ \ .
\]
Then,%
\begin{align*}
&  \left(  T_{\sigma}\mid T^{\prime}\right) \\
&  =\left(  r_{1}r_{2}\cdots r_{l-1}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  x_{i\left(  l,1\right)  }\cdots
x_{i\left(  l,m-1\right)  }x_{\sigma i\left(  l,m\right)  }x_{\sigma i\left(
l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)  }\mid
u_{j\left(  l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  x_{\sigma i\left(  l+1,1\right)  }\cdots
x_{\sigma i\left(  l+1,m\right)  }x_{i\left(  l+1,m+1\right)  }\cdots
x_{i\left(  l+1,\lambda_{l+1}\right)  }\mid u_{j\left(  l+1,1\right)  }\cdots
u_{j\left(  l+1,\lambda_{l+1}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot\left(  r_{l+2}r_{l+3}\cdots r_{p}\right) \\
&  =\left(  x_{\sigma i\left(  l+1,1\right)  }\cdots x_{\sigma i\left(
l+1,m\right)  }x_{i\left(  l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda
_{l+1}\right)  }\mid u_{j\left(  l+1,1\right)  }\cdots u_{j\left(
l+1,\lambda_{l+1}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \underbrace{\left(  x_{i\left(  l,1\right)  }\cdots
x_{i\left(  l,m-1\right)  }x_{\sigma i\left(  l,m\right)  }x_{\sigma i\left(
l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)  }\mid
u_{j\left(  l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right)
}_{\substack{=\pm\left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma i\left(
l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)  }x_{i\left(
l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(  l,1\right)
}\cdots u_{j\left(  l,\lambda_{l}\right)  }\right)  \\\text{(by the
antisymmetry of the inner product)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right) \\
&  =\pm\left(  x_{\sigma i\left(  l+1,1\right)  }\cdots x_{\sigma i\left(
l+1,m\right)  }x_{i\left(  l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda
_{l+1}\right)  }\mid u_{j\left(  l+1,1\right)  }\cdots u_{j\left(
l+1,\lambda_{l+1}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma
i\left(  l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)
}x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(
l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right)  .
\end{align*}
Therefore,
\begin{align*}
&  \left(  x_{\sigma i\left(  l+1,1\right)  }\cdots x_{\sigma i\left(
l+1,m\right)  }x_{i\left(  l+1,m+1\right)  }\cdots x_{i\left(  l+1,\lambda
_{l+1}\right)  }\mid u_{j\left(  l+1,1\right)  }\cdots u_{j\left(
l+1,\lambda_{l+1}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  x_{\sigma i\left(  l,m\right)  }x_{\sigma
i\left(  l,m+1\right)  }\cdots x_{\sigma i\left(  l,\lambda_{l}\right)
}x_{i\left(  l,1\right)  }\cdots x_{i\left(  l,m-1\right)  }\mid u_{j\left(
l,1\right)  }\cdots u_{j\left(  l,\lambda_{l}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right) \\
&  =\pm\left(  T_{\sigma}\mid T^{\prime}\right)  .
\end{align*}
This proves Claim 1.]
\par
[\textit{Proof of Claim 2:} A \textit{composition} shall mean a finite
sequence of positive integers. Note that each partition is a composition, but
not each composition is a partition. If $\left(  \mu\right)  $ is a
composition, then the notion of the \textquotedblleft shape of $\left(
\mu\right)  $\textquotedblright\ is defined in the same way as it has been
defined when $\mu$ is a partition. If $\left(  \mu\right)  $ is a composition,
then an \textit{almost-tableau} of shape $\left(  \mu\right)  $ with values in
the set $E$ means an assignment of an element of $E$ to each point in the
shape $\left(  \mu\right)  $.
\par
Define a Young tableau $\widetilde{T}$ by the following procedure:
\par
\begin{itemize}
\item \textit{Step 1:} Let $T_{\circ}$ be the Young tableau $T$.
\par
\item \textit{Step 2:} Replace the $l$-th and the $\left(  l+1\right)  $-st
rows of the tableau $T_{\circ}$ by the rows of $S$. (Notice that $S$ may have
fewer or more than $2$ rows; thus, the number of rows of $T_{\circ}$ is not
necessarily preserved in this step. Also, the resulting array $T_{\circ}$ is
an almost-tableau, but not necessarily a Young tableau, because the lengths of
its rows might no longer be in weakly decreasing order.)
\par
\item \textit{Step 3:} Permute the rows of the almost-tableau $T_{\circ}$ in
such a way that their lengths become weakly decreasing (i.e., it becomes a
Young tableau once again). Denote the resulting Young tableau $T_{\circ}$ by
$\widetilde{T}$.
\end{itemize}
\par
Similarly, define a Young tableau $\widetilde{T^{\prime}}$ (using the tableaux
$T^{\prime}$ and $S^{\prime}$ instead of $T$ and $S$); in doing so, make sure
to permute the rows in the same way as during the construction of
$\widetilde{T}$.
\par
The resulting bitableau $\left[  \widetilde{T},\widetilde{T^{\prime}}\right]
$ has the bideterminant%
\[
\left(  \widetilde{T}\mid\widetilde{T^{\prime}}\right)  =\left(  S\mid
S^{\prime}\right)  \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right)
\]
(because its rows are precisely the rows of the bitableau $\left[
T,T^{\prime}\right]  $ except for its $l$-th and $\left(  l+1\right)  $-st
rows, which have been replaced by the rows of the bitableau $\left[
S,S^{\prime}\right]  $).
\par
Now, we shall show that the shape of $\widetilde{T}$ is greater than $\left(
\lambda\right)  $.
\par
Indeed, the bitableau $\left[  S,S^{\prime}\right]  $ has shape strictly
longer than $\left(  \lambda_{l}\right)  $. In other words, the tableau $S$
has shape strictly longer than $\left(  \lambda_{l}\right)  $. Thus, the first
row of $S$ has length $>\lambda_{l}$. Thus, in Step 2 of our procedure by
which we defined $\widetilde{T}$, the shape of the almost-tableau $T_{\circ}$
has increased in lexicographic order (because the $l$-th row, which used to
have length $\lambda_{l}$, now has length $>\lambda_{l}$, while the rows above
it have preserved their lengths). Furthermore, in Step 3 of the procedure, the
shape of $T_{\circ}$ has either stayed unchanged or increased in lexicographic
order (because when we sort a list of numbers in weakly decreasing order, this
list either stays unchanged or increases in lexicographic order). Thus,
throughout the procedure, the shape of $T_{\circ}$ has increased in
lexicographic order. Therefore, the shape of the tableau obtained at the end
of the procedure is greater than the shape of the tableau at its beginning. In
other words, the shape of $\widetilde{T}$ is greater than the shape of $T$
(because the tableau obtained at the end of the procedure is $\widetilde{T}$,
while the tableau at its beginning is $T$). In other words, the shape of
$\widetilde{T}$ is greater than $\left(  \lambda\right)  $ (since the shape of
$T$ is $\left(  \lambda\right)  $). Hence, $\left(  \widetilde{T}%
\mid\widetilde{T^{\prime}}\right)  $ is a bideterminant of shape greater than
$\left(  \lambda\right)  $.
\par
Because of $\left(  \widetilde{T}\mid\widetilde{T^{\prime}}\right)  =\left(
S\mid S^{\prime}\right)  \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right)  $, this rewrites as follows: $\left(  S\mid
S^{\prime}\right)  \left(  r_{1}r_{2}\cdots r_{l-1}\right)  \left(
r_{l+2}r_{l+3}\cdots r_{p}\right)  $ is a bideterminant of shape greater than
$\left(  \lambda\right)  $. This proves Claim 2.]} By our choice of $\left(
\lambda\right)  $, the polynomial\footnote{\textit{Correction:} Added the
words \textquotedblleft the polynomial\textquotedblright\ in order to separate
expressions.} $D$ is also a linear combination of bideterminants of standard
tableaux of shape greater than $\left(  \lambda\right)  $.

Now, each tableau $T_{\sigma}$ differs from $T$ only in rows $l$ and $l+1$,
which in $T_{\sigma}$ are:\footnote{\textit{Correction:} Added the words
\textquotedblleft which in $T_{\sigma}$ are\textquotedblright.}%
\[%
\begin{array}
[c]{cccccccc}%
x_{i\left(  l,1\right)  } & \cdots & x_{i\left(  l,m-1\right)  } & x_{\sigma
i\left(  l,m\right)  } & x_{\sigma i\left(  l,m+1\right)  } & \cdots & \cdots
& x_{\sigma i\left(  l,\lambda_{l}\right)  }\\
x_{\sigma i\left(  l+1,1\right)  } & \cdots & x_{\sigma i\left(
l+1,m-1\right)  } & x_{\sigma i\left(  l+1,m\right)  } & x_{i\left(
l+1,m+1\right)  } & \cdots & x_{i\left(  l+1,\lambda_{l+1}\right)  } &
\end{array}
\ \ \ .
\]
In the tableau $T$, however, we have the inequalities%
\[%
%TCIMACRO{\TeXButton{arraycolsep}{\arraycolsep=2pt}}%
%BeginExpansion
\arraycolsep=2pt%
%EndExpansion%
\begin{array}
[c]{ccccccccccccc}
&  &  &  &  &  & i\left(  l,m\right)  & < & i\left(  l,m+1\right)  & < &
\cdots & < & i\left(  l,\lambda_{l}\right) \\
&  &  &  &  &  & \vee &  &  &  &  &  & \\
i\left(  l+1,1\right)  & < & \cdots & < & i\left(  l+1,m-1\right)  & < &
i\left(  l+1,m\right)  &  &  &  &  &  &
\end{array}
\ \ \ .
\]
For any nonidentical permutation $\sigma$ in the shuffle product, the index
$\sigma i\left(  l,m\right)  $ must equal one of the indices $i\left(
l+1,1\right)  ,\ldots,i\left(  l+1,m\right)  $; in particular, we
have\footnote{\textit{Correction:} Added \textquotedblleft we
have\textquotedblright\ and replaced comma by semicolon for disambiguation.}
$\sigma i\left(  l,m\right)  <i\left(  l,m\right)  $. Thus, the tableau
$T_{\sigma}$ has a lexicographically smaller associated sequence than $T$. By
our choice of $\left[  T,T^{\prime}\right]  $, however, this shows
that\footnote{\textit{Correction:} Added the words \textquotedblleft this
shows that\textquotedblright\ for clarity.} each of the
bitableaux\footnote{\textit{Correction:} Replaced \textquotedblleft
tableaux\textquotedblright\ by \textquotedblleft bitableaux\textquotedblright%
.} $\left[  T_{\sigma},T^{\prime}\right]  $ satisfies the theorem, and hence,
by substitution, we can write $\left(  T\mid T^{\prime}\right)  $ as a sum of
bideterminants of standard bitableaux\footnote{\textit{Correction:} Replaced
\textquotedblleft tableaux\textquotedblright\ by \textquotedblleft
bitableaux\textquotedblright.} of shape equal to or longer than $\left(
\lambda\right)  $. This contradicts our initial assumption.

It remains to observe that if $T$ were standard, then $T^{\prime}$ would have
to be nonstandard; the same reasoning can then be applied to $T^{\prime}$ to
yield a contradiction. This concludes the proof of the theorem.
\end{proof}

The proof contains implicitly an algorithm for expressing any bitableau as a
linear combination of standard bitableaux\footnote{\textit{Comment:} More
precisely: \textquotedblleft... for expressing the bideterminant of any
bitableau as a linear combination of bideterminants of standard
bitableaux\textquotedblright.} by successive corrections of violations. This
is inefficient for practical computations, as the number of bitableaux
introduced during a correction is, in general, very large.

As an exercise, apply the algorithm to obtain the following identity (only the
subscripts are shown):%
\begin{align*}
\left(
\begin{array}
[c]{cc}%
2 & 3\\
1 & 4\\
2 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
1 & 3\\
2 & 4\\
2 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)  -\left(
\begin{array}
[c]{cc}%
1 & 2\\
2 & 4\\
3 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(
\begin{array}
[c]{cc}%
1 & 2\\
2 & 3\\
4 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)  -\left(
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
2 &  & \\
4 &  &
\end{array}
\ \middle|\
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
1 &  & \\
1 &  &
\end{array}
\right)  .
\end{align*}


\section{\label{sect.basis}The Basis Theorem}

As we have seen, the standard bideterminants (i.e., the bideterminants of
standard bitableaux\footnote{\textit{Correction:} Replaced \textquotedblleft
tableaux\textquotedblright\ by \textquotedblleft bitableaux\textquotedblright%
.}) of content $\left(  \alpha,\beta\right)  $
span\footnote{\textit{Correction:} Replaced \textquotedblleft
spans\textquotedblright\ by \textquotedblleft span\textquotedblright.} the
vector space $P\left(  \alpha,\beta\right)  $. In this section, using the
technique of set polarization operators, we show that, in fact, they form a basis.

We augment the alphabets $\mathcal{X}$ and $\mathcal{U}$ by adding new letters
from the sets $\mathcal{S}$ and $\mathcal{T}$, respectively. The sets
$\mathcal{S}$ and $\mathcal{T}$ are supposed finite, but large enough that the
ensuing constructions can be performed. This enlarges the algebra $P$, even
though the vector space $P\left(  \alpha,\beta\right)  $ remains
unchanged.\footnote{\textit{Comment:} Here, we identify each pair of vectors
$\left(  \alpha,\beta\right)  =\left(  \left(  \alpha_{1},\alpha_{2}%
,\ldots,\alpha_{n}\right)  ,\left(  \beta_{1},\beta_{2},\ldots,\beta
_{k}\right)  \right)  $ with the pair of vectors%
\[
\left(  \left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{n},\underbrace{0,0,\ldots
,0}_{\left\vert \mathcal{S}\right\vert \text{ many zeroes}}\right)  ,\left(
\beta_{1},\beta_{2},\ldots,\beta_{k},\underbrace{0,0,\ldots,0}_{\left\vert
\mathcal{T}\right\vert \text{ many zeroes}}\right)  \right)  ;
\]
this allows us to speak of $P\left(  \alpha,\beta\right)  $ even after the
alphabets $\mathcal{X}$ and $\mathcal{U}$ have been enlarged.}

Thus, a bitableau can have entries from all of the four alphabets
$\mathcal{X}$, $\mathcal{U}$, $\mathcal{S}$ and $\mathcal{T}$ (more precisely,
it consists of a tableau with entries from $\mathcal{X}\cup\mathcal{S}$ and a
tableau with entries from $\mathcal{U}\cup\mathcal{T}$). A bitableau with
entries from the alphabets $\mathcal{X}$ and $\mathcal{U}$ only will be called
an $\left(  \mathcal{X},\mathcal{U}\right)  $\textit{-bitableau}. The $\left(
\mathcal{X},\mathcal{U}\right)  $\textit{-content} of a bitableau $\left[
T,T^{\prime}\right]  $ is defined as the pair of vectors $\left(  \alpha
,\beta\right)  $ where $\alpha_{i}$ (resp. $\beta_{j}$) is the number of
occurrences of $x_{i}$ in $T$ (resp. $u_{j}$ in $T^{\prime}$). (Thus, the
$\left(  \mathcal{X},\mathcal{U}\right)  $-content differs from the usual
content only in that it disregards the new letters from $\mathcal{S}$ and
$\mathcal{T}$.) An $\left(  \mathcal{X},\mathcal{U}\right)  $%
\textit{-bideterminant} means a bideterminant of an $\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bitableau.\ \ \ \ \footnote{\textit{Correction:} I have
added the preceding paragraph, since the original paper was cavalier about
where these issues (i.e., in which contexts to disregard the new letters from
$\mathcal{S}$ and $\mathcal{T}$, and in which context to count them in). I
hope my correction is correct!}

The $\left(  \mathcal{X},\mathcal{U}\right)  $\textit{-content} of a monomial
in $P$ is the pair of vectors%
\[
\left(  \alpha,\beta\right)  =\left(  \left(  \alpha_{1},\ldots,\alpha
_{n}\right)  ,\left(  \beta_{1},\ldots,\beta_{k}\right)  \right)  ,
\]
where $\alpha_{s}$ (resp. $\beta_{t}$) is the total degree of the factors in
the monomial of the form $\left(  x_{s}\mid u\right)  $ with $u\in
\mathcal{U}\cup\mathcal{T}$ (resp. $\left(  x\mid u_{t}\right)  $ with
$x\in\mathcal{X}\cup\mathcal{S}$).\ \ \ \ \footnote{\textit{Correction:} I
have added the preceding paragraph, since the original paper was cavalier
about where these issues (i.e., in which contexts to disregard the new letters
from $\mathcal{S}$ and $\mathcal{T}$, and in which context to count them in).
I hope my correction is correct!}

Let $x_{i}$, $u_{i}$, $s_{j}$, and $t_{j}$ be letters from the alphabets
$\mathcal{X}$, $\mathcal{U}$, $\mathcal{S}$, and $\mathcal{T}$. The
\textit{set polarization operators }$D^{l}\left(  s_{j},x_{i}\right)  $ (for
integers $l\geq0$)\ \ \ \ \footnote{\textit{Correction:} Added
\textquotedblleft for integers $l\geq0$\textquotedblright.} are defined as
follows: Let
\[
M=\left(  x_{i_{1}}\mid u_{1}^{\prime}\right)  \cdots\left(  x_{i_{p}}\mid
u_{p}^{\prime}\right)  \cdot\left(  s_{j_{1}}\mid v_{1}^{\prime}\right)
\cdots\left(  s_{j_{q}}\mid v_{q}^{\prime}\right)
\]
be a monomial of $\left(  \mathcal{X},\mathcal{U}\right)  $%
-content\footnote{\textit{Correction:} Replaced \textquotedblleft
content\textquotedblright\ by \textquotedblleft$\left(  \mathcal{X}%
,\mathcal{U}\right)  $-content\textquotedblright.} $\left(  \alpha
,\beta\right)  $, where $u_{1}^{\prime},\ldots,u_{p}^{\prime},v_{1}^{\prime
},\ldots,v_{q}^{\prime}$ are elements of $\mathcal{U}\cup\mathcal{T}$. Then,

\begin{enumerate}
\item[(i)] if $\alpha_{i}<l$, we set $D^{l}\left(  s_{j},x_{i}\right)  M=0$,

\item[(ii)] if $\alpha_{i}\geq l$, we set $D^{l}\left(  s_{j},x_{i}\right)
M=\sum_{r}M_{r}$, where $M_{1},\ldots,M_{r},\ldots,M_{\dbinom{\alpha_{i}}{l}}$
are all the $\dbinom{\alpha_{i}}{l}$ distinct monomials obtained from $M$ by
replacing each subset of $l$ letters $x_{i}$ by $l$ letters $s_{j}$. (In
particular, each of the monomials $M_{r}$ contains the letter $x_{i}$
exactly\footnote{\textit{Correction:} Added the word \textquotedblleft
exactly\textquotedblright\ in order to separate two unrelated expressions.}
$\left(  \alpha_{i}-l\right)  $ times and the letter $s_{j}$
exactly\footnote{\textit{Correction:} Added the word \textquotedblleft
exactly\textquotedblright\ in order to separate two unrelated expressions.}
$l$ times.)\footnote{\textit{Comment:} Notice that case (i) should be regarded
as a particular case of case (ii). Indeed, if $\alpha_{i}<l$, then
$\dbinom{\alpha_{i}}{l}=0$, and there is no subset of $l$ letters $x_{i}$ in
$M$ (because $M$ has fewer than $l$ letters $x_{i}$ in total); therefore, if
we apply the rule for defining $D^{l}\left(  s_{j},x_{i}\right)  M$ in case
(ii) to case (i), then we obtain $D^{l}\left(  s_{j},x_{i}\right)  M=\sum
_{r}M_{r}=\left(  \text{empty sum}\right)  =0$, which is precisely how we
defined $D^{l}\left(  s_{j},x_{i}\right)  M$ in case (i). Therefore, strictly
speaking, there was no need to treat case (i) separately.}
\end{enumerate}

The operator $D^{l}\left(  s_{j},x_{i}\right)  $ is now extended to all of
$P\left(  \alpha,\beta\right)  $ by linearity.

The operator $D^{0}\left(  s_{j},x_{i}\right)  $%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$D^{o}%
$\textquotedblright\ by \textquotedblleft$D^{0}$\textquotedblright.} is the
identity operator, and for $1\leq l\leq\alpha_{i}$, the set polarization
operator $D^{l}\left(  s_{j},x_{i}\right)  $ maps a polynomial in $P\left(
\alpha,\beta\right)  $ to a polynomial lying outside $P\left(  \alpha
,\beta\right)  $.\ \ \ \ \footnote{\textit{Comment:} It is not hard to prove
that for a given choice of $x_{i}$ and $s_{j}$, the sequence $\left(
D^{0}\left(  s_{j},x_{i}\right)  ,D^{1}\left(  s_{j},x_{i}\right)
,D^{2}\left(  s_{j},x_{i}\right)  ,\ldots\right)  $ is a
\textit{divided-powers Hasse-Schmidt derivation} in the sense of
\cite[Definition 2.6]{KacGri}. Explicitly, this means that the following
holds:
\par
\begin{itemize}
\item We have $D^{0}\left(  s_{j},x_{i}\right)  =\operatorname*{id}$.
\par
\item We have $\left(  D^{l}\left(  s_{j},x_{i}\right)  \right)  \left(
ab\right)  =\sum_{g=0}^{l}\left(  D^{g}\left(  s_{j},x_{i}\right)  a\right)
\left(  D^{l-g}\left(  s_{j},x_{i}\right)  b\right)  $ for all $a\in P$ and
$b\in P$.
\par
\item We have $D^{p}\left(  s_{j},x_{i}\right)  \cdot D^{q}\left(  s_{j}%
,x_{i}\right)  =\dbinom{p+q}{p}D^{p+q}\left(  s_{j},x_{i}\right)  $ for all
$p\in\mathbb{N}$ and $q\in\mathbb{N}$.
\end{itemize}
\par
As a consequence, the operator $D^{1}\left(  s_{j},x_{i}\right)  $ is a
derivation on $P$, and each $l\in\mathbb{N}$ satisfies $l!\cdot D^{l}\left(
s_{j},x_{i}\right)  =\left(  D^{1}\left(  s_{j},x_{i}\right)  \right)  ^{l}$.
Thus, if the ground field $K$ has characteristic $0$, then the operators
$D^{l}\left(  s_{j},x_{i}\right)  $ can all be computed from $D^{1}\left(
s_{j},x_{i}\right)  $.
\par
Operators such as $D^{l}\left(  s_{j},x_{i}\right)  $ are the hallmark of
characteristic-free invariant theory (and, generally, tend to occur when
studying polynomials over fields of arbitrary characteristic).}

For bideterminants, the set polarization operators act in the following simple
way.\footnote{\textit{Correction:} In the following lemma, I replaced
\textquotedblleft content\textquotedblright\ by \textquotedblleft$\left(
\mathcal{X},\mathcal{U}\right)  $-content\textquotedblright. Also, added the
words \textquotedblleft then\textquotedblright\ to parts (i) and (ii) in order
to separate formulas, and replaced run-on-sentence by three separate
sentences.}

\begin{lemma}
\label{lem.basis.1}Let $\left(  T\mid T^{\prime}\right)  $ be a bideterminant
of $\left(  \mathcal{X},\mathcal{U}\right)  $-content $\left(  \alpha
,\beta\right)  $. Then:

\begin{enumerate}
\item[(i)] If $\alpha_{i}<l$, then $D^{l}\left(  s_{j},x_{i}\right)  \left(
T\mid T^{\prime}\right)  =0$.

\item[(ii)] If $\alpha_{i}\geq l$, then $D^{l}\left(  s_{j},x_{i}\right)
\left(  T\mid T^{\prime}\right)  =\sum_{r}\left(  T_{r}\mid T^{\prime}\right)
$, where $T_{1},\ldots,T_{r},\ldots,T_{\dbinom{\alpha_{i}}{l}}$ are all the
distinct $\dbinom{\alpha_{i}}{l}$ tableaux obtained from $T$ by replacing each
subset of $l$ letters $x_{i}$ by $l$ letters $s_{j}$.
\end{enumerate}
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.basis.1}.]Expand $\left(  T\mid T^{\prime}\right)  $
into a sum of monomials $M^{t}$ of $\left(  \mathcal{X},\mathcal{U}\right)
$-content\footnote{\textit{Correction:} Replaced \textquotedblleft
content\textquotedblright\ by \textquotedblleft$\left(  \mathcal{X}%
,\mathcal{U}\right)  $-content\textquotedblright.} $\left(  \alpha
,\beta\right)  $:%
\[
\left(  T\mid T^{\prime}\right)  =\sum_{t}M^{t}.
\]
Now, if $\alpha_{i}<l$, then\footnote{\textit{Correction:} Added the word
\textquotedblleft then\textquotedblright.} $D^{l}\left(  s_{j},x_{i}\right)
M^{t}=0$ for all the monomials $M^{t}$. Hence (i).

Now, suppose that $\alpha_{i}\geq l$. Then,%
\[
D^{l}\left(  s_{j},x_{i}\right)  \left(  T\mid T^{\prime}\right)  =\sum
_{t}D^{l}\left(  s_{j},x_{i}\right)  M^{t}=\sum_{t}\sum_{r}M_{r}^{t},
\]
where $M_{r}^{t}$ (for a fixed $t$)\ \ \ \ \footnote{\textit{Correction:}
Added \textquotedblleft(for a fixed $t$)\textquotedblright.} are the
$\dbinom{\alpha_{i}}{l}$ monomials obtained from $M^{t}$ according to rule
(ii) of the definition of $D^{l}\left(  s_{j},x_{i}\right)  $%
\ \ \ \ \footnote{\textit{Correction:} Added the words \textquotedblleft of
the definition of $D^{l}\left(  s_{j},x_{i}\right)  $\textquotedblright.}.
Interchanging the order of summation, we have%
\[
D^{l}\left(  s_{j},x_{i}\right)  \left(  T\mid T^{\prime}\right)  =\sum
_{r}\sum_{t}M_{r}^{t},
\]
But\footnote{\textit{Correction:} Removed a useless comma at this point.}%
\[
\sum_{t}M_{r}^{t}=\left(  T_{r}\mid T^{\prime}\right)  ,
\]
where the same set of $l$ letters $x_{i}$ are replaced by $l$ letters $s_{j}$
on both sides of the equation. Hence,%
\[
D^{l}\left(  s_{j},x_{i}\right)  \left(  T\mid T^{\prime}\right)  =\sum
_{r}\left(  T_{r}\mid T^{\prime}\right)  .
\]
This proves the lemma.
\end{proof}

The set polarization operators $D^{l}\left(  t_{j},u_{i}\right)  $ are defined
in an analogous manner, and the analog of the previous lemma is true for these operators.

\begin{example}%
\begin{align*}
&  D^{2}\left(  s_{1},x_{2}\right)  \left(
\begin{array}
[c]{ccc}%
x_{1} & x_{2} & x_{3}\\
x_{1} & x_{2} & \\
x_{2} &  &
\end{array}
\ \middle|\ T^{\prime}\right) \\
&  =\left(
\begin{array}
[c]{ccc}%
x_{1} & s_{1} & x_{3}\\
x_{1} & s_{1} & \\
x_{2} &  &
\end{array}
\ \middle|\ T^{\prime}\right)  +\left(
\begin{array}
[c]{ccc}%
x_{1} & s_{1} & x_{3}\\
x_{1} & x_{2} & \\
s_{1} &  &
\end{array}
\ \middle|\ T^{\prime}\right)  +\left(
\begin{array}
[c]{ccc}%
x_{1} & x_{2} & x_{3}\\
x_{1} & s_{1} & \\
s_{1} &  &
\end{array}
\ \middle|\ T^{\prime}\right)  ,
\end{align*}
while%
\[
D^{2}\left(  s_{1},x_{3}\right)  \left(
\begin{array}
[c]{ccc}%
x_{1} & x_{2} & x_{3}\\
x_{1} & x_{2} & \\
x_{2} &  &
\end{array}
\ \middle|\ T^{\prime}\right)  =0.
\]

\end{example}

Note that since the alphabets $\mathcal{X}$, $\mathcal{U}$, $\mathcal{S}$, and
$\mathcal{T}$ are disjoint, the set polarization operators commute.

The set polarization operators are the building blocks of the \textit{Capelli
operator} $C\left(  T,T^{\prime}\right)  $%
\ \ \ \ \footnote{\textit{Correction:} Added \textquotedblleft$C\left(
T,T^{\prime}\right)  $\textquotedblright.}, which is defined for each
bitableau $\left[  T,T^{\prime}\right]  $ of shape $\left(  \lambda\right)  $
as follows: Let $\alpha_{i}\left(  q\right)  $ (resp. $\beta_{j}\left(
q\right)  $) be the number of occurrences of $x_{i}$%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$x_{j}%
$\textquotedblright\ by \textquotedblleft$x_{i}$\textquotedblright.} (resp.
$u_{j}$) in the $q$th column of $T$ (resp. $T^{\prime}$). The Capelli operator
$C\left(  T,T^{\prime}\right)  $\ \ \ \ \footnote{\textit{Correction:} Added
\textquotedblleft$C\left(  T,T^{\prime}\right)  $\textquotedblright.} is
defined by the following formula:\footnote{\textit{Correction:} In the
following formula, I replaced \textquotedblleft$D^{\beta_{l}\left(  q\right)
}\left(  t_{q},u_{j}\right)  $\textquotedblright\ by \textquotedblleft%
$D^{\beta_{j}\left(  q\right)  }\left(  t_{q},u_{j}\right)  $%
\textquotedblright.}%
\[
C\left(  T,T^{\prime}\right)  =\prod_{1\leq q\leq\lambda_{1}}\left(
\prod_{1\leq i\leq n}D^{\alpha_{i}\left(  q\right)  }\left(  s_{q}%
,x_{i}\right)  \right)  \left(  \prod_{1\leq j\leq k}D^{\beta_{j}\left(
q\right)  }\left(  t_{q},u_{j}\right)  \right)  .
\]
\footnote{\textit{Comment:} Note that this is a product of commuting operators
(since the set polarization operators commute). Note also that $\lambda_{1}$
is the number of columns of the shape $\left(  \lambda\right)  $.}

\begin{example}
Suppose
\[
\left[  T,T^{\prime}\right]  =\left[
\begin{array}
[c]{ccc}%
x_{1} & x_{2} & x_{3}\\
x_{1} & x_{2} & \\
x_{2} &  &
\end{array}
,\
\begin{array}
[c]{ccc}%
u_{1} & u_{2} & u_{3}\\
u_{1} & u_{3} & \\
u_{1} &  &
\end{array}
\right]  .
\]
Then%
\begin{align*}
C\left(  T,T^{\prime}\right)   &  =D^{2}\left(  s_{1},x_{1}\right)
D^{1}\left(  s_{1},x_{2}\right)  D^{2}\left(  s_{2},x_{2}\right)  D^{1}\left(
s_{3},x_{3}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \cdot D^{3}\left(  t_{1},u_{1}\right)  D^{1}\left(
t_{2},u_{2}\right)  D^{1}\left(  t_{2},u_{3}\right)  D^{1}\left(  t_{3}%
,u_{3}\right)  .
\end{align*}

\end{example}

We now impose a new total order on $\left(  \mathcal{X},\mathcal{U}\right)
$-bitableaux\footnote{\textit{Correction:} Replaced \textquotedblleft
bitableaux\textquotedblright\ by \textquotedblleft$\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bitableaux\textquotedblright.} of the same shape.
Associate with each $\left(  \mathcal{X},\mathcal{U}\right)  $%
-bitableau\footnote{\textit{Correction:} Replaced \textquotedblleft
bitableau\textquotedblright\ by \textquotedblleft$\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bitableau\textquotedblright.} the sequence formed by
reading off the indices down each column, successively, first in $T$ and then
in $T^{\prime}$. The bitableaux are then ordered according to the
lexicographic order of their associated column
sequences.\footnote{\textit{Comment:} This is indeed a total order on the set
of all bitableaux of a given shape $\left(  \lambda\right)  $. This is because
a $\left(  \mathcal{X},\mathcal{U}\right)  $-bitableau \textbf{of a given
shape} $\left(  \lambda\right)  $ is uniquely determined by its associated
column sequence.}

\begin{example}
For the bitableau in the preceding example, the associated column sequence is%
\[
\left(  1,1,2,2,2,3,1,1,1,2,3,3\right)  .
\]

\end{example}

If the $\left(  \mathcal{X},\mathcal{U}\right)  $%
-bitableau\footnote{\textit{Correction:} Added the words \textquotedblleft the
$\left(  \mathcal{X},\mathcal{U}\right)  $-bitableau\textquotedblright.}
$\left[  T,T^{\prime}\right]  $ is standard, the associated column sequence
can be written%
\[
\left(  1^{\alpha_{1}\left(  1\right)  }\cdots n^{\alpha_{n}\left(  1\right)
}2^{\alpha_{2}\left(  2\right)  }\cdots n^{\alpha_{n}\left(  2\right)  }\cdots
n^{\alpha_{n}\left(  \lambda_{1}\right)  }1^{\beta_{1}\left(  1\right)
}\cdots k^{\beta_{k}\left(  1\right)  }\cdots k^{\beta_{k}\left(  \lambda
_{1}\right)  }\right)  .
\]
We have used the fact that for the $\left(  \mathcal{X},\mathcal{U}\right)
$-bitableau $\left[  T,T^{\prime}\right]  $ to be
standard\footnote{\textit{Correction:} Replaced \textquotedblleft for the
bideterminant $\left(  T\mid T^{\prime}\right)  $ to be
nonzero\textquotedblright\ by \textquotedblleft for the $\left(
\mathcal{X},\mathcal{U}\right)  $-bitableau $\left[  T,T^{\prime}\right]  $ to
be standard\textquotedblright. (Indeed, the bideterminant $\left(  T\mid
T^{\prime}\right)  $ being nonzero was not a requirement.)}, we must have
$\alpha_{i}\left(  q\right)  =\beta_{i}\left(  q\right)  =0$ for
$i<q$\ \ \ \ \footnote{\textit{Comment:} This is because the entries in each
row must be strictly increasing.}.

\begin{theorem}
\label{thm.2.1}Let $\left[  T,T^{\prime}\right]  $ and $\left[  V,V^{\prime
}\right]  $ be two standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bitableaux of shape $\left(  \lambda\right)  $ and $\left(  \mu\right)  $
with the same content. Then:

\begin{enumerate}
\item[(i)] We have $C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  \neq0$.

\item[(ii)] If $\left(  \mu\right)  $ is longer than $\left(  \lambda\right)
$, then $C\left(  T,T^{\prime}\right)  \left(  V\mid V^{\prime}\right)  =0$.

\item[(iii)] If $\left(  \lambda\right)  =\left(  \mu\right)  $, and if
$\left[  V,V^{\prime}\right]  $ is greater than $\left[  T,T^{\prime}\right]
$ in the lexicographic order of their associated column sequences, then
$C\left(  T,T^{\prime}\right)  \left(  V\mid V^{\prime}\right)  =0$.
\end{enumerate}
\end{theorem}

\footnote{\textit{Correction:} In this theorem, I have replaced
\textquotedblleft Let $\left(  T\mid T^{\prime}\right)  $ and $\left(  V\mid
V^{\prime}\right)  $ be two standard bideterminants\textquotedblright\ by
\textquotedblleft Let $\left[  T,T^{\prime}\right]  $ and $\left[
V,V^{\prime}\right]  $ be two standard $\left(  \mathcal{X},\mathcal{U}%
\right)  $-bitableaux\textquotedblright\ for precision. Also, I have made each
of (i), (ii) and (iii) a separate sentence. Also, I have replaced
\textquotedblleft smaller than\textquotedblright\ by \textquotedblleft smaller
or equal to\textquotedblright\ in part (iii). Finally, I have rewritten part
(iii) in contrapositive form, since this is how this part is actually being
proved and used (and it is more natural to state it like this from a
constructive point of view).}

\begin{proof}
[Proof of Theorem \ref{thm.2.1}.](i) We calculate $C\left(  T,T^{\prime
}\right)  \left(  T\mid T^{\prime}\right)  $ explicitly; we
have\footnote{\textit{Correction:} In the formula below, replaced
\textquotedblleft$\alpha_{n}\left(  a\right)  $\textquotedblright\ by
\textquotedblleft$\alpha_{n}\left(  1\right)  $\textquotedblright; added more
\textquotedblleft$\vdots$\textquotedblright s and \textquotedblleft$\cdots
$\textquotedblright s; added the words \textquotedblleft
rows\textquotedblright\ for clarity.}%
\[
\left(  T\mid T^{\prime}\right)  =\left(
\begin{array}
[c]{@{}r@{\,}l}%
\alpha\text{$_{1}\left(  1\right)  $ rows} & \left\{
\begin{array}
[c]{cc}%
x_{1} & \cdots\\
\vdots & \vdots\\
x_{1} & \cdots
\end{array}
\right. \\
& \ \ \ \ \ \ \vdots\\
\alpha_{l}\left(  1\right)  \text{ rows} & \left\{
\begin{array}
[c]{cc}%
x_{l} & \cdots\\
\vdots & \vdots\\
x_{l} & \cdots
\end{array}
\right. \\
& \ \ \ \ \ \ \vdots\\
\alpha_{n}\left(  1\right)  \text{ rows} & \left\{
\begin{array}
[c]{cc}%
x_{n} & \cdots\\
\vdots & \vdots\\
x_{n} & \cdots
\end{array}
\right.
\end{array}
\ \middle|\ T^{\prime}\right)  .
\]
Now, the letter $x_{1}$ can only be found in the first column of $T$. Hence,
$D^{\alpha_{1}\left(  1\right)  }\left(  s_{1},x_{1}\right)  \left(  T\mid
T^{\prime}\right)  $\ \ \ \ \footnote{\textit{Correction:} Replaced
\textquotedblleft$D^{\alpha_{1}\left(  1\right)  }\left(  s_{1},x_{i}\right)
$\textquotedblright\ by \textquotedblleft$D^{\alpha_{1}\left(  1\right)
}\left(  s_{1},x_{1}\right)  $\textquotedblright.} consists of a single term,
obtained by substituting $s_{1}$ for all the $x_{1}$ in $T$.

Assume that $\left(  \prod_{1\leq i\leq l-1}D^{\alpha_{i}\left(  1\right)
}\left(  s_{1},x_{i}\right)  \right)  \left(  T\mid T^{\prime}\right)
$\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\alpha
_{1}\left(  1\right)  $\textquotedblright\ by \textquotedblleft$\alpha
_{i}\left(  1\right)  $\textquotedblright. Added parentheses around the
product. Similarly in the equalities below.} consists of a single term,
obtained by substituting all the letters $x_{i}$, $1\leq i\leq l-1$, in the
first column of $T$ by the letter $s_{1}$. Then,%
\begin{align*}
&  \left(  \prod_{1\leq i\leq l}D^{\alpha_{i}\left(  1\right)  }\left(
s_{1},x_{i}\right)  \right)  \left(  T\mid T^{\prime}\right) \\
&  =D^{\alpha_{l}\left(  1\right)  }\left(  s_{1},x_{l}\right)  \left(
\prod_{1\leq i\leq l-1}D^{\alpha_{i}\left(  1\right)  }\left(  s_{1}%
,x_{i}\right)  \right)  \left(  T\mid T^{\prime}\right) \\
&  =D^{\alpha_{1}\left(  1\right)  }\left(  s_{1},x_{l}\right)  \left(
\begin{array}
[c]{@{}r@{\,}l}%
\sum_{1\leq i\leq l-1}\alpha\text{$_{i}\left(  1\right)  $ rows} & \left\{
\begin{array}
[c]{cc}%
s_{1} & \cdots\\
\vdots & \vdots\\
s_{1} & \cdots
\end{array}
\right. \\
\alpha_{l}\left(  1\right)  \text{ rows} & \left\{
\begin{array}
[c]{cc}%
x_{l} & \cdots\\
\vdots & \vdots\\
x_{l} & \cdots
\end{array}
\right. \\
& \ \ \ \ \ \ \vdots\\
\alpha_{n}\left(  1\right)  \text{ rows} & \left\{
\begin{array}
[c]{cc}%
x_{n} & \cdots\\
\vdots & \vdots\\
x_{n} & \cdots
\end{array}
\right.
\end{array}
\ \middle|\ T^{\prime}\right)  .
\end{align*}
Since tableau $T$ is standard, any occurrence of $x_{l}$ in other than the
first column must be in the first $\sum_{1\leq i\leq l-1}\alpha_{i}\left(
1\right)  $ rows. If any of these $x_{l}$ are chosen for substitution during
the polarization\footnote{\textit{Comment:} \textquotedblleft
Polarization\textquotedblright\ here means the application of the operator
$D^{\alpha_{1}\left(  1\right)  }\left(  s_{1},x_{l}\right)  $.}, the letter
$s_{1}$ would be repeated within a row, and the resulting bideterminant would
be zero. Hence, the only nonzero term in the above expression is the term
obtained by substituting $s_{1}$ for all the $\alpha_{l}\left(  1\right)  $
letters $x_{l}$ in the first column of $T$. By induction, we have shown that
the expression $\left(  \prod_{1\leq i\leq n}D^{\alpha_{1}\left(  1\right)
}\left(  s_{1},x_{i}\right)  \right)  \left(  T\mid T^{\prime}\right)  $
consists of a single nonzero term, obtained by substituting $s_{1}$ for all
the letters in the first column of $T$.

Repeating this argument for the other columns, we can easily see that
$C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  $ is obtained
by substituting $s_{q}$ (resp. $t_{q}$) for all the letters in the $q$th
column of $T$ (resp. $T^{\prime}$); that is,\footnote{\textit{Correction:}
Added a few more dots and an extra $s_{1}$ to the bitableau below to hopefully
make it clearer.}%
\[
C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  =\left(
\begin{array}
[c]{cccc}%
s_{1} & s_{2} & \cdots & s_{\lambda_{1}}\\
\vdots & \vdots & \ddots & \vdots\\
s_{1} & s_{2} & \cdots & s_{\lambda_{1}}\\
\vdots & \vdots & \iddots & \\
s_{1} & s_{2} &  & \\
\vdots &  &  & \\
s_{1} &  &  &
\end{array}
\ \middle|\
\begin{array}
[c]{cccc}%
t_{1} & t_{2} & \cdots & t_{\lambda_{1}}\\
\vdots & \vdots & \ddots & \vdots\\
t_{1} & t_{2} & \cdots & t_{\lambda_{1}}\\
\vdots & \vdots & \iddots & \\
t_{1} & t_{2} &  & \\
\vdots &  &  & \\
t_{1} &  &  &
\end{array}
\right)  .
\]
The two tableaux appearing on the right hand side will be denoted by $U$ and
$U^{\prime}$. Thus, $C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  =\left(  U\mid U^{\prime}\right)  \neq0$%
.\ \ \ \ \footnote{\textit{Correction:} Added the previous two sentences.}

(ii) The expression $C\left(  T,T^{\prime}\right)  \left(  V\mid V^{\prime
}\right)  $ consists of a sum of bideterminants of the same shape $\left(
\mu\right)  $. If it is nonzero, then one of the bideterminants, say $\left(
W\mid W^{\prime}\right)  $, is nonzero. The content of $\left(  W\mid
W^{\prime}\right)  $ is the same as that of $\left(  U\mid U^{\prime}\right)
=C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  $ (since the
set polarization operators transform the content of a polynomial in a
predictable way (i.e., the content of the output is determined by the content
of the input), and therefore so does the operator $C\left(  T,T^{\prime
}\right)  $)\ \ \ \ \footnote{\textit{Correction:} Inserted the preceding
parenthetical sentence.}; that is, for $1\leq l\leq\lambda_{1}$, the letters
$s_{l}$ and $t_{l}$ occur in $\left(  W\mid W^{\prime}\right)  $
exactly\footnote{\textit{Correction:} Added the word \textquotedblleft
exactly\textquotedblright.} $\widetilde{\lambda}_{l}$ times, where%
\begin{align*}
\widetilde{\lambda}_{l}  &  =\sum_{1\leq i\leq n}\alpha_{i}\left(  l\right) \\
&  =\text{the height of the }l\text{th column.}%
\end{align*}


We are required to show that $\left(  \mu\right)  $ is shorter or equal
to\footnote{\textit{Correction:} Replaced \textquotedblleft shorter
than\textquotedblright\ by \textquotedblleft shorter or equal
to\textquotedblright.} $\left(  \lambda\right)  $. If $\left(  \mu\right)
\neq\left(  \lambda\right)  $, let $m$ be the smallest integer such that
$\lambda_{m}\neq\mu_{m}$\ \ \ \ \footnote{\textit{Comment:} Such an $m$
exists, for the following reason: The tableaux $T$ and $V$ have the same
content. Thus, they have the same number of boxes. In other words, $\left(
\lambda\right)  $ and $\left(  \mu\right)  $ are partitions of one and the
same integer. Hence, $\left(  \lambda\right)  $ cannot be a proper subsequence
of $\left(  \mu\right)  $, and $\left(  \mu\right)  $ cannot be a proper
subsequence of $\left(  \lambda\right)  $. Therefore, if $\left(  \mu\right)
\neq\left(  \lambda\right)  $, then there exists an integer $m$ satisfying
$\lambda_{m}\neq\mu_{m}$.}.\ \ \ \ \footnote{\textit{Correction:} At this
spot, I have removed the following two sentences:
\par
\textquotedblleft If $m=1$, it must be the case that $\mu_{1}<\lambda_{1}$,
for the first row of $W$ contains $\mu_{1}$ \textit{distinct} letters chosen
from the set $\left\{  s_{1},\ldots,s_{\lambda_{1}}\right\}  $.
\par
Now, suppose that $m\geq2$.\textquotedblright.
\par
In fact, these sentences are not wrong, but they are unnecessary, since the
argument that follows does not require $m\geq2$, and thus the $m=1$ case needs
not be treated separately.} We claim that: For $1\leq i\leq m-1$, the contents
of the $i$th row in $U$ and in $W$ are identical.

The proof is by strong induction\footnote{\textit{Correction:} Replaced
\textquotedblleft induction\textquotedblright\ by \textquotedblleft strong
induction\textquotedblright.} on $i$.\ \ \ \ \footnote{\textit{Correction:} At
this spot, I have removed the following sentence:
\par
\textquotedblleft If $i=1$, then $\mu_{1}=\lambda_{1}$; by our preceding
observations, the first row in both $W$ and $U$ consists of the set $\left\{
s_{1},\ldots,s_{\lambda_{1}}\right\}  $ arranged in some
order.\textquotedblright.
\par
In fact, this sentence is useless here, since a strong induction does not
contain an induction base.} Namely,\footnote{\textit{Correction:} Replaced
\textquotedblleft Now\textquotedblright\ by \textquotedblleft
Namely,\textquotedblright.} assume that the proposition is true up to the
$i$th row. The letters $s_{l}$, for $l\geq\lambda_{i}+1$, have all been used
in the first $i-1$ rows in $U$, hence in $W$ (since $U$ and $W$ have the same
content)\footnote{\textit{Correction:} Inserted the preceding parenthetical
sentence.}. For the $i$th row in $W$, we thus\footnote{\textit{Correction:}
Added the word \textquotedblleft thus\textquotedblright\ for clarity of the
argument.} have to choose $\mu_{i}$ distinct letters from $\left\{
s_{1},\ldots,s_{\lambda_{i}}\right\}  $. But $\lambda_{i}=\mu_{i}$; hence the
contents of the $i$th row in $U$ and $W$ are identical. This proves the
claim.\footnote{\textit{Correction:} Added the preceding sentence.}

Now, consider the $m$th row. As the contents of
the\footnote{\textit{Correction:} Added \textquotedblleft contents of
the\textquotedblright.} first $m-1$ rows are identical, the $m$th row in $W$
contains $\mu_{m}$ distinct letters from the set $\left\{  s_{1}%
,\ldots,s_{\lambda_{m}}\right\}  $\ \ \ \ \footnote{\textit{Comment:} This
follows from noticing that the letters $s_{l}$, for $l\geq\lambda_{m}+1$, have
all been used in the first $m-1$ rows in $U$, hence in $W$ (since $U$ and $W$
have the same content), and thus are not available for the $m$th row any
more.}. Since $\mu_{m}\neq\lambda_{m}$, we must have $\mu_{m}<\lambda_{m}$.

(iii) We can now suppose that $\left(  \mu\right)  =\left(  \lambda\right)
$.\ \ \ \ \footnote{\textit{Correction:} Removed \textquotedblleft and
$C\left(  T,T^{\prime}\right)  \left(  V\mid V^{\prime}\right)  \neq
0$\textquotedblright\ from this sentence, as it was an instance of useless
proof by contradiction.} Recall that the associated column sequence of
$\left[  T,T^{\prime}\right]  $\ \ \ \ \footnote{\textit{Correction:} Replaced
\textquotedblleft of $\left(  T\mid T^{\prime}\right)  $\textquotedblright\ by
\textquotedblleft of $\left[  T,T^{\prime}\right]  $\textquotedblright.} is%
\[
\left(  1^{\alpha_{1}\left(  1\right)  }\cdots n^{\alpha_{n}\left(  1\right)
}\cdots n^{\alpha_{n}\left(  \lambda_{1}\right)  }1^{\beta_{1}\left(
1\right)  }\cdots k^{\beta_{k}\left(  1\right)  }\cdots k^{\beta_{k}\left(
\lambda_{1}\right)  }\right)  .
\]
We shall denote by $\gamma_{i}\left(  q\right)  $ (resp. $\delta_{j}\left(
q\right)  $) the number of occurrences of $x_{i}$ (resp. $u_{j}$%
)\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft(resp.
$u_{i}$)\textquotedblright\ by \textquotedblleft(resp. $u_{j}$%
)\textquotedblright.} in the $q$th column of $V$ (resp. $V^{\prime}$). The
associated column sequence of $\left[  V,V^{\prime}\right]  $%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft of $\left(
V\mid V^{\prime}\right)  $\textquotedblright\ by \textquotedblleft of $\left[
V,V^{\prime}\right]  $\textquotedblright.} is\footnote{\textit{Correction:} In
the column sequence below, I replaced \textquotedblleft$\delta_{k}\left(
3\right)  $\textquotedblright\ by \textquotedblleft$\delta_{k}\left(
1\right)  $\textquotedblright.}%
\[
\left(  1^{\gamma_{1}\left(  1\right)  }\cdots n^{\gamma_{n}\left(  1\right)
}\cdots n^{\gamma_{n}\left(  \lambda_{1}\right)  }1^{\delta_{1}\left(
1\right)  }\cdots k^{\delta_{k}\left(  1\right)  }\cdots k^{\delta_{k}\left(
\lambda_{1}\right)  }\right)  .
\]


Suppose now that $\left[  T,T^{\prime}\right]  $ and $\left[  V,V^{\prime
}\right]  $ differ in the left tableau; the reasoning is similar if the
only\footnote{\textit{Correction:} Added the word \textquotedblleft
only\textquotedblright\ for clarity.} difference lies in the right tableau.

Let $p$ be the first column where $T$ and $V$ differ, and in the $p$th column,
let $x_{l}$ be the smallest index that is different. That is, we
have\footnote{\textit{Correction:} Added the words \textquotedblleft we
have\textquotedblright\ in the following statements, and replaced
\textquotedblleft$\delta_{i}\left(  p\right)  $\textquotedblright\ by
\textquotedblleft$\gamma_{i}\left(  p\right)  $\textquotedblright.}%
\begin{align*}
&  \text{for }1\leq i\leq n\text{ and }1\leq q\leq p-1\text{, we have }%
\alpha_{i}\left(  q\right)  =\gamma_{i}\left(  q\right)  ;\\
&  \text{for }1\leq i\leq l-1\text{, we have }\alpha_{i}\left(  p\right)
=\gamma_{i}\left(  p\right)  ;\\
&  \text{but }\alpha_{l}\left(  p\right)  \neq\gamma_{l}\left(  p\right)
\text{.}%
\end{align*}


Now, assume that $\left[  V,V^{\prime}\right]  $%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\left(
V\mid V^{\prime}\right)  $\textquotedblright\ by \textquotedblleft$\left[
V,V^{\prime}\right]  $\textquotedblright.} is lexicographically
\textit{greater} than $\left[  T,T^{\prime}\right]  $%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\left(
T\mid T^{\prime}\right)  $\textquotedblright\ by \textquotedblleft$\left[
T,T^{\prime}\right]  $\textquotedblright.}; that is, $\alpha_{l}\left(
p\right)  >\gamma_{l}\left(  p\right)  $\ \ \ \ \footnote{\textit{Comment:}
Let me explain why the assumption (that $\left[  V,V^{\prime}\right]  $ is
lexicographically \textit{greater} than $\left[  T,T^{\prime}\right]  $)
yields $\alpha_{l}\left(  p\right)  >\gamma_{l}\left(  p\right)  $.
\par
Indeed, assume the contrary. Thus, $\alpha_{l}\left(  p\right)  \leq\gamma
_{l}\left(  p\right)  $, so that $\alpha_{l}\left(  p\right)  <\gamma
_{l}\left(  p\right)  $ (since $\alpha_{l}\left(  p\right)  \neq\gamma
_{l}\left(  p\right)  $). Hence, the topmost $\sum_{1\leq i\leq l}\alpha
_{i}\left(  p\right)  $ indices in the $p$th column of $T$ equal the
corresponding indices in the $p$th column of $V$ (because for $1\leq i\leq
l-1$, we have $\alpha_{i}\left(  p\right)  =\gamma_{i}\left(  p\right)  $),
but the $\left(  \sum_{1\leq i\leq l}\alpha_{i}\left(  p\right)  \right)
+1$-st index from the top is greater in $T$ than in $V$ (in fact, the index in
$T$ is $>p$ whereas the index in $V$ is $=p$). Therefore, the associated
column sequence of $\left[  T,T^{\prime}\right]  $ is greater than the
associated column sequence of $\left[  V,V^{\prime}\right]  $ (because the
first $p-1$ columns of $T$ and of $V$ are equal). This contradicts the
assumption that $\left[  V,V^{\prime}\right]  $ is lexicographically
\textit{greater} than $\left[  T,T^{\prime}\right]  $. This contradiction
completes the proof.}. Consider the action of the Capelli operator $C\left(
T,T^{\prime}\right)  $ on $\left(  V\mid V^{\prime}\right)  $. The
polarizations $D^{\alpha_{1}\left(  1\right)  }\left(  s_{1},x_{1}\right)
,D^{\alpha_{2}\left(  1\right)  }\left(  s_{1},x_{2}\right)  ,\ldots
,D^{\alpha_{l-1}\left(  p\right)  }\left(  s_{p},x_{l-1}\right)  $ act on
$\left(  V\mid V^{\prime}\right)  $ exactly as they do on $\left(  T\mid
T^{\prime}\right)  $. At this instant\footnote{\textit{Comment:} i.e., after
these polarizations have been applied to $\left(  V\mid V^{\prime}\right)  $},
the expression\footnote{\textit{Correction:} Replaced \textquotedblleft%
$1\leq\alpha\leq p-1$\textquotedblright\ by \textquotedblleft$1\leq q\leq
p-1$\textquotedblright\ in the following equality. Also, added some
parenthesis to disambiguate the products.}%
\[
\left(  \prod_{1\leq i\leq l-1}D^{\alpha_{i}\left(  p\right)  }\left(
s_{p},x_{i}\right)  \right)  \cdot\left(  \prod_{1\leq q\leq p-1}\left(
\prod_{1\leq i\leq n}D^{\alpha_{i}\left(  q\right)  }\left(  s_{q}%
,x_{i}\right)  \right)  \right)  \left(  V\mid V^{\prime}\right)
\]
is a bideterminant of the form\footnote{\textit{Correction:} Added
\textquotedblleft$\cdots$\textquotedblright s at the end of the rows of the
left tableau.}%
\[
\left(
\begin{array}
[c]{@{}r@{\,}l}%
\sum_{1\leq i\leq l-1}\alpha\text{$_{i}\left(  p\right)  $ rows} & \left\{
\begin{array}
[c]{cccccc}%
s_{1} & s_{2} & \cdots & s_{p-1} & s_{p} & \cdots\\
\vdots & \vdots & \ddots & \vdots & \vdots & \cdots\\
s_{1} & s_{2} & \cdots & s_{p-1} & s_{p} & \cdots
\end{array}
\right. \\
\gamma_{l}\left(  p\right)  \text{ rows} & \left\{
\begin{array}
[c]{cccccc}%
s_{1} & s_{2} & \cdots & s_{p-1} & x_{l} & \cdots\\
\vdots & \vdots & \ddots & \vdots & \vdots & \cdots\\
s_{1} & s_{2} & \cdots & s_{p-1} & x_{l} & \cdots
\end{array}
\right. \\
& \ \ \ \
\begin{array}
[c]{cccccc}%
\vdots\ \  & \ \vdots\ \  & \ddots & \ \ \ \vdots\ \ \  & \ \ \vdots\  &
\cdots
\end{array}
\\
& \ \ \ \left.
\begin{array}
[c]{cccccc}%
s_{1} & s_{2} & \cdots & s_{p-1} &  & \\
\vdots & \vdots & \iddots &  &  & \\
s_{1} & s_{2} &  &  &  & \\
\vdots &  &  &  &  & \\
s_{1} &  &  &  &  &
\end{array}
\right.
\end{array}
\ \middle|\ V^{\prime}\right)  .
\]
That is, the first $p-1$ columns are replaced by the appropriate letters
$s_{q}$; in the $p$th column, the first $\sum_{1\leq i\leq l-1}\alpha
_{i}\left(  p\right)  $ letters are replaced by $s_{p}$, and the remainder of
the tableau is unchanged.

Since $\alpha_{l}\left(  p\right)  >\gamma_{l}\left(  p\right)  $, and $V$ is
standard, any choice of $\alpha_{l}\left(  p\right)  $%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\gamma
_{l}\left(  p\right)  $\textquotedblright\ by \textquotedblleft$\alpha
_{l}\left(  p\right)  $\textquotedblright.} letters $x_{l}$ must involve a
letter $x_{l}$ lying in the first $\sum_{1\leq i\leq l-1}\alpha_{i}\left(
p\right)  $ rows of $V$. The set polarization operator $D^{\alpha_{l}\left(
p\right)  }\left(  s_{p},x_{l}\right)  $\ \ \ \ \footnote{\textit{Correction:}
Replaced \textquotedblleft$D^{\alpha_{i}\left(  p\right)  }\left(  s_{p}%
,x_{l}\right)  $\textquotedblright\ by \textquotedblleft$D^{\alpha_{l}\left(
p\right)  }\left(  s_{p},x_{l}\right)  $\textquotedblright.} substitutes
$s_{p}$ for this particular $x_{l}$. The resulting bideterminant is zero,
since there are two letters $s_{p}$ in a single row. This proves $C\left(
T,T^{\prime}\right)  \left(  V\mid V^{\prime}\right)  =0$%
.\ \ \ \ \footnote{\textit{Correction:} Replaced the original sentence
\textquotedblleft This contradicts the assumption; we have therefore proved
that $\left(  V\mid V^{\prime}\right)  $ is lexicographically smaller than
$\left(  T\mid T^{\prime}\right)  $\textquotedblright\ (which was imprecise
and also a needless proof by contradiction) by \textquotedblleft This proves
$C\left(  T,T^{\prime}\right)  \left(  V\mid V^{\prime}\right)  =0$%
\textquotedblright.}
\end{proof}

With Theorem \ref{thm.2.1} proved, we can now proceed to the main result of
this section.\footnote{\textit{Correction:} In the following theorem, I have
replaced \textquotedblleft bideterminant\textquotedblright\ by
\textquotedblleft$\left(  \mathcal{X},\mathcal{U}\right)  $%
-bideterminants\textquotedblright.}

\begin{theorem}
\label{thm.2.2}The standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminants of content $\left(  \alpha,\beta\right)  $ form a basis of
the vector space $P\left(  \alpha,\beta\right)  $.
\end{theorem}

\begin{proof}
[Proof of Theorem \ref{thm.2.2}.]By Corollary \ref{cor.strf.gen}%
\footnote{\textit{Correction:} Replaced \textquotedblleft By Theorem
\ref{thm.1}\textquotedblright\ by \textquotedblleft By Corollary
\ref{cor.strf.gen}\textquotedblright.}, the standard $\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bideterminants\footnote{\textit{Correction:} Replaced
\textquotedblleft bideterminants\textquotedblright\ by \textquotedblleft%
$\left(  \mathcal{X},\mathcal{U}\right)  $-bideterminants\textquotedblright.}
span the vector space $P\left(  \alpha,\beta\right)  $. Suppose we have a
nontrivial linear relation between these bideterminants. We can write this
linear relation as follows:%
\[
a\left(  T\mid T^{\prime}\right)  +A+B=0,
\]
where $a$ is a nonzero scalar in $K$, $\left(  \lambda\right)  $ is the
shortest shape occurring in the linear relation, $\left[  T,T^{\prime}\right]
$\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\left(
T\mid T^{\prime}\right)  $\textquotedblright\ by \textquotedblleft$\left[
T,T^{\prime}\right]  $\textquotedblright.} is the bitableau of shape $\left(
\lambda\right)  $ with the lexicographically smallest associated
\textit{column} sequence, $A$ is the linear combination of the remaining
bitableaux\footnote{\textit{Correction:} Replaced \textquotedblleft
tableaux\textquotedblright\ by \textquotedblleft bitableaux\textquotedblright%
.} of shape $\left(  \lambda\right)  $, and $B$ is the linear combination of
the remainder of the bitableaux, which are necessarily of shape longer than
$\left(  \lambda\right)  $.

Applying the Capelli operator $C\left(  T,T^{\prime}\right)  $ to the relation
we have, by Theorem \ref{thm.2.1},\footnote{\textit{Correction:} Here, I have
added the comments \textquotedblleft by Theorem \ref{thm.2.1}
(iii)\textquotedblright, \textquotedblleft by Theorem \ref{thm.2.1}
(ii)\textquotedblright\ and \textquotedblleft by Theorem \ref{thm.2.1}
(i)\textquotedblright.}
\begin{align*}
C\left(  T,T^{\prime}\right)  A  &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem \ref{thm.2.1} (iii)}\right)  ,\\
C\left(  T,T^{\prime}\right)  B  &  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem \ref{thm.2.1} (ii)}\right)  ,
\end{align*}
but%
\[
C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  \neq
0\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.2.1} (i)}\right)  .
\]
This implies $aC\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)
=0$, which is a contradiction.\footnote{\textit{Comment:} Here is (what I
believe to be) a clearer way to write up this argument:
\par
Applying the map $C\left(  T,T^{\prime}\right)  $ to the equality $a\left(
T\mid T^{\prime}\right)  +A+B=0$, we obtain%
\[
C\left(  T,T^{\prime}\right)  \left(  a\left(  T\mid T^{\prime}\right)
+A+B\right)  =C\left(  T,T^{\prime}\right)  0=0.
\]
Thus,
\begin{align*}
0  &  =C\left(  T,T^{\prime}\right)  \left(  a\left(  T\mid T^{\prime}\right)
+A+B\right) \\
&  =aC\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)
+\underbrace{C\left(  T,T^{\prime}\right)  A}_{=0}+\underbrace{C\left(
T,T^{\prime}\right)  B}_{=0}\\
&  =aC\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  .
\end{align*}
In other words, $aC\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  =0$. Since $C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  \neq0$, this entails that $a=0$. This contradicts $a\neq0$.}
\end{proof}

The proof of this theorem contains another algorithm for expressing any
bideterminant as a linear combination of standard
bideterminants\footnote{\textit{Correction:} Replaced \textquotedblleft any
bitableau as a linear combination of standard bitableaux\textquotedblright\ by
\textquotedblleft any bideterminant as a linear combination of standard
bideterminants\textquotedblright.}. Suppose%
\[
\left(  T\mid T^{\prime}\right)  =\sum_{i}a_{i}\left(  T_{i}\mid T_{i}%
^{\prime}\right)
\]
is the unique decomposition of $\left(  T\mid T^{\prime}\right)  $ into
standard bitableaux, written so that if $i<j$, then either $T_{i}$ is of shape
shorter than $T_{j}$, or $T_{i}$ and $T_{j}$%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft or $T_{j}%
$\textquotedblright\ by \textquotedblleft or $T_{i}$ and $T_{j}$%
\textquotedblright.} have the same shape, and $\left[  T_{i},T_{i}^{\prime
}\right]  $ has a lexicographically smaller associated column sequence than
$\left[  T_{j},T_{j}^{\prime}\right]  $\ \ \ \ \footnote{\textit{Correction:}
Replaced \textquotedblleft and $T_{i}$ has a lexicographically smaller
associated column sequence than $T_{j}$\textquotedblright\ by
\textquotedblleft and $\left[  T_{i},T_{i}^{\prime}\right]  $ has a
lexicographically smaller associated column sequence than $\left[  T_{j}%
,T_{j}^{\prime}\right]  $\textquotedblright.}. The coefficients $a_{i}$ are
called the \textit{straightening coefficients}. They can be computed by
applying the Capelli operators $C\left(  T_{j},T_{j}^{\prime}\right)  $ to
both sides of the linear relation; by Theorem \ref{thm.2.1}, we obtain a
triangular array of equations between bideterminants with entries from the
alphabets $\mathcal{S}$ and $\mathcal{T}$. From this, we can extract a
triangular system of linear equations for the coefficients $a_{i}$, which can
then be solved.\footnote{\textit{Correction:} In the following example, I have
replaced \textquotedblleft bideterminant\textquotedblright\ by
\textquotedblleft$\left(  \mathcal{X},\mathcal{U}\right)  $%
-bideterminant\textquotedblright.}

\begin{example}
Consider the $\left(  \mathcal{X},\mathcal{U}\right)  $-bideterminant (where,
for simplicity, all but the subscripts are suppressed):%
\[
\left(
\begin{array}
[c]{cc}%
2 & 3\\
1 & 4\\
2 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)  .
\]
The standard bitableaux of the same or longer shape of the same content are%
\begin{align*}
&  \left[
\begin{array}
[c]{cc}%
1 & 3\\
2 & 4\\
2 &
\end{array}
,\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right]  ,\ \ \ \ \ \ \ \ \ \ \left[
\begin{array}
[c]{cc}%
1 & 2\\
2 & 4\\
3 &
\end{array}
,\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right]  ,\ \ \ \ \ \ \ \ \ \ \left[
\begin{array}
[c]{cc}%
1 & 2\\
2 & 3\\
4 &
\end{array}
,\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right]  ,\\
&  \left[
\begin{array}
[c]{ccc}%
1 & 3 & 4\\
2 &  & \\
2 &  &
\end{array}
,\
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
1 &  & \\
1 &  &
\end{array}
\right]  ,\ \ \ \ \ \ \ \ \ \ \left[
\begin{array}
[c]{ccc}%
1 & 2 & 4\\
2 &  & \\
3 &  &
\end{array}
,\
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
1 &  & \\
1 &  &
\end{array}
\right]  ,\\
&  \left[
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
2 &  & \\
4 &  &
\end{array}
,\
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
1 &  & \\
1 &  &
\end{array}
\right]  .
\end{align*}
Let $a_{1},\ldots,a_{6}$ be the corresponding straightening coefficients. We
obtain, through the Capelli operators, the equations:%
\[%
\begin{array}
[c]{cccccccc}%
a_{1} &  &  &  &  &  & = & 1\\
& a_{2} &  &  &  &  & = & -1\\
-a_{1} &  & +a_{3} &  &  &  & = & 0\\
&  &  & a_{4} &  &  & = & 0\\
&  &  &  & a_{5} &  & = & 0\\
&  & a_{3} &  &  & +a_{6} & = & 0
\end{array}
\]
Therefore, we obtain%
\begin{align*}
\left(
\begin{array}
[c]{cc}%
2 & 3\\
1 & 4\\
2 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)   &  =\left(
\begin{array}
[c]{cc}%
1 & 3\\
2 & 4\\
2 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)  -\left(
\begin{array}
[c]{cc}%
1 & 2\\
2 & 4\\
3 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ +\left(
\begin{array}
[c]{cc}%
1 & 2\\
2 & 3\\
4 &
\end{array}
\ \middle|\
\begin{array}
[c]{cc}%
1 & 2\\
1 & 3\\
1 &
\end{array}
\right)  -\left(
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
2 &  & \\
4 &  &
\end{array}
\ \middle|\
\begin{array}
[c]{ccc}%
1 & 2 & 3\\
1 &  & \\
1 &  &
\end{array}
\right)  .
\end{align*}

\end{example}

From Theorem \ref{thm.2.2}, we obtain a basis of the whole polynomial ring in
the $\left(  x_{i}\mid u_{j}\right)  $:\ \ \ \ \footnote{\textit{Correction:}
Added this sentence and the next corollary.}

\begin{corollary}
\label{cor.2.2.basis}The standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminants form a basis of the algebra of polynomials over the field $K$
in the indeterminates $\left(  x_{i}\mid u_{j}\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.2.2.basis}.]This algebra (as a vector space) is
the direct sum $\bigoplus_{\left(  \alpha,\beta\right)  }P\left(  \alpha
,\beta\right)  $. Each of the addends has a basis consisting of the standard
$\left(  \mathcal{X},\mathcal{U}\right)  $-bideterminants of content $\left(
\alpha,\beta\right)  $ (by Theorem \ref{thm.2.2}). Thus, the whole algebra has
a basis consisting of all the standard $\left(  \mathcal{X},\mathcal{U}%
\right)  $-bideterminants.
\end{proof}

\section{\label{sect.invar}Invariant Theory}

\subsection{\label{subsect.invar.sft}The Second Fundamental Theorem}

Classical invariant theory is concerned with the behavior of forms under the
action of linear transformations. Let $\left\{  \mathbf{u}_{1},\ldots
,\mathbf{u}_{d}\right\}  $ be a dual basis for the vector space $V_{d}$ (of
dimension $d$).\ \ \ \ \footnote{\textit{Comment:} I am not sure what
\textquotedblleft dual basis\textquotedblright\ means here; I suspect that it
means \textquotedblleft basis for the dual space\textquotedblright\ (i.e., a
\textit{dual basis} of a vector space $V$ means a basis of the dual space
$V^{\ast}$ of $V$). However, this is not very important, because the vector
space $V_{d}$ and its dual basis $\left\{  \mathbf{u}_{1},\ldots
,\mathbf{u}_{d}\right\}  $ are used only for motivation. A proper definition
of the notion of forms shall be given two sentences later.} A \textit{form}
$F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  $ on $m$ vectors in
$V_{d}$ is a polynomial in the $md$ scalar products of the $m$ vectors
$\mathbf{x}_{i}$ with the $d$ covectors $\mathbf{u}_{j}$%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft%
$\mathbf{u}_{i}$\textquotedblright\ by \textquotedblleft$\mathbf{u}_{j}%
$\textquotedblright.} in the dual basis. More pedantically, consider the
polynomial algebra $K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}%
\right]  $ in the indeterminates $x_{ir},u_{jr},s_{pr},t_{qr}$, where $1\leq
i\leq n$, $1\leq j\leq k$, $1\leq p$, $1\leq q$, and $1\leq r\leq d$. In this
algebra, we distinguish the following polynomials:%
\begin{align*}
\left\langle x_{i}\mid u_{j}\right\rangle  &  =\sum_{1\leq r\leq d}%
x_{ir}u_{jr},\\
\left\langle s_{p}\mid u_{j}\right\rangle  &  =\sum_{1\leq r\leq d}%
s_{pr}u_{jr},\\
\left\langle x_{i}\mid t_{q}\right\rangle  &  =\sum_{1\leq r\leq d}%
x_{ir}t_{qr},\\
\left\langle s_{p}\mid t_{q}\right\rangle  &  =\sum_{1\leq r\leq d}%
s_{pr}t_{qr}.
\end{align*}
Let $\widehat{P}$ be the subalgebra of $K\left[  \mathbf{x},\mathbf{u}%
,\mathbf{s},\mathbf{t}\right]  $ generated by these polynomials; $\widehat{P}$
is called the \textit{algebra of forms}. Rigorously speaking, we define a
\textit{form} to be an element of $\widehat{P}$%
.\ \ \ \ \footnote{\textit{Correction:} Added the preceding sentence.}

There exists an algebra homomorphism\footnote{\textit{Correction:} Replaced
\textquotedblleft a homomorphism\textquotedblright\ by \textquotedblleft an
algebra homomorphism\textquotedblright.} $\phi$ from $P$ (constructed as in
the previous section from the alphabets $\mathcal{X}$, $\mathcal{U}$,
$\mathcal{S}$, and $\mathcal{T}$) to $\widehat{P}$ defined by%
\[
\phi:%
\begin{array}
[c]{ccc}%
\left(  x_{i}\mid u_{j}\right)  & \mapsto & \left\langle x_{i}\mid
u_{j}\right\rangle ,\\
\left(  s_{p}\mid u_{j}\right)  & \mapsto & \left\langle s_{p}\mid
u_{j}\right\rangle ,\\
\left(  x_{i}\mid t_{q}\right)  & \mapsto & \left\langle x_{i}\mid
t_{q}\right\rangle ,\\
\left(  s_{p}\mid t_{q}\right)  & \mapsto & \left\langle s_{p}\mid
t_{q}\right\rangle .
\end{array}
\]
Consider a monomial $m=\left(  x_{i_{1}}\mid u_{j_{1}}\right)  \cdots\left(
x_{i_{a}}\mid u_{j_{a}}\right)  $ in $P$. Its image in $\widehat{P}$ under
$\phi$ is given by
\[
\phi m=\sum_{f}x_{i_{1}f_{1}}u_{j_{1}f_{1}}\cdots x_{i_{a}f_{a}}u_{j_{a}f_{a}%
},
\]
where the summation is over the set of all functions $f:i\mapsto f_{i}$ from
$\left\{  1,\ldots,a\right\}  $ to $\left\{  1,\ldots,d\right\}  $. We shall
use the simpler notation%
\[
\phi m=\sum_{f}m_{f}.
\]
\footnote{\textit{Comment:} Thus, $m_{f}$ denotes the monomial $x_{i_{1}f_{1}%
}u_{j_{1}f_{1}}\cdots x_{i_{a}f_{a}}u_{j_{a}f_{a}}$ whenever $f:i\mapsto
f_{i}$ is a function from $\left\{  1,\ldots,a\right\}  $ to $\left\{
1,\ldots,d\right\}  $. Of course, this definition depends not just on the
monomial $m$, but also on the order of the factors in the product $\left(
x_{i_{1}}\mid u_{j_{1}}\right)  \cdots\left(  x_{i_{a}}\mid u_{j_{a}}\right)
$; we can thus only use it when this order is fixed.} The restriction of the
homomorphism $\phi$ to $P\left(  \alpha,\beta\right)  $ is called the
\textit{Pascal homomorphism}.\footnote{\textit{Correction:} In the following
theorem, I have replaced \textquotedblleft bideterminants\textquotedblright%
\ by \textquotedblleft$\left(  \mathcal{X},\mathcal{U}\right)  $%
-bideterminants\textquotedblright.}

\begin{theorem}
[the second fundamental theorem of invariant theory]\label{thm.3.1}The kernel
of the Pascal homomorphism is the subspace of $P\left(  \alpha,\beta\right)  $
spanned by the standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminants of shape strictly longer than $\left(  d\right)  $.
\end{theorem}

Some preliminary observations are in order.

In the same fashion as for $P$, we define set polarization operators for
$K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  $. Let
$v=x_{ir_{1}}\cdots x_{ir_{a}}$ be a monomial containing only variables of the
form $x_{ir}$, where $i$ is fixed and $r$ is arbitrary. Suppose $p$ is a
positive integer, and $E$ a subset of $\left\{  1,\ldots,a\right\}  $. Then,
$v^{E,p}$\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft%
$V^{E,p}$\textquotedblright\ by \textquotedblleft$v^{E,p}$\textquotedblright.}
is the monomial obtained from $v$ by replacing the variable $x_{ir_{b}}$ by
the variable $s_{pr_{b}}$ whenever $b\in E$%
.\ \ \ \ \footnote{\textit{Comment:} This monomial $v^{E,p}$ depends not just
on the monomial $v$, but also on the order of the factors in the product
$x_{ir_{1}}\cdots x_{ir_{a}}$; we can thus only use this notation when this
order is fixed.} Now, for a given nonnegative\footnote{\textit{Correction:}
Replaced \textquotedblleft positive\textquotedblright\ by \textquotedblleft
nonnegative\textquotedblright.} integer $l$, the \textit{set polarization
operator} $\widehat{D}^{l}\left(  s_{p},x_{i}\right)  $ acts on the monomial
$v$ as follows:%
\[
\widehat{D}^{l}\left(  s_{p},x_{i}\right)  v=\sum_{E}v^{E,p},
\]
where the summation is over all the $l$-subsets\footnote{\textit{Comment:}
\textquotedblleft$l$-subset\textquotedblright\ means \textquotedblleft%
$l$-element subset\textquotedblright.} of $\left\{  1,\ldots,a\right\}
$\ \ \ \ \footnote{\textit{Comment:} We can rewrite this equality (without
using the slippery notation $v^{E,p}$) as follows:%
\begin{align}
\widehat{D}^{l}\left(  s_{p},x_{i}\right)  v  &  =\sum_{\substack{E\subseteq
\left\{  1,2,\ldots,a\right\}  ;\\\left\vert E\right\vert =l}}\left(
\prod_{b\in E}s_{pr_{b}}\right)  \left(  \prod_{\substack{b\in\left\{
1,2,\ldots,a\right\}  ;\\b\notin E}}x_{ir_{b}}\right) \label{eq.Dhat.def2}\\
&  =\sum_{\substack{E\subseteq\left\{  1,2,\ldots,a\right\}  ;\\\left\vert
E\right\vert =l}}\left(  \prod_{b\in\left\{  1,2,\ldots,a\right\}  }%
\begin{cases}
s_{pr_{b}}, & \text{if }b\in E\\
x_{ir_{b}}, & \text{if }b\notin E
\end{cases}
\right)  . \label{eq.Dhat.def3}%
\end{align}
(It is easy to see that this does not depend on the order of the factors in
$x_{ir_{1}}\cdots x_{i_{r_{a}}}$, and thus the image $\widehat{D}^{l}\left(
s_{p},x_{i}\right)  v$ is really well-defined.)}. Consider now an arbitrary
monomial $w$. We can write $w$ as the product of two monomials $w^{\prime}$
and $w^{\prime\prime}$, where $w^{\prime}$ is the product of all the variables
in $w$ of the form $x_{ir}$, and $w^{\prime\prime}$ is the product of the
remaining variables in $w$\ \ \ \ \footnote{\textit{Correction:} Added
\textquotedblleft in $w$\textquotedblright.}. Then, we set%
\begin{equation}
\widehat{D}^{l}\left(  s_{p},x_{i}\right)  w=\left(  \widehat{D}^{l}\left(
s_{p},x_{i}\right)  w^{\prime}\right)  w^{\prime\prime}.
\label{eq.Dhat.def-full}%
\end{equation}
The operator $\widehat{D}^{l}\left(  s_{p},x_{i}\right)  $ is extended to all
of $K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  $ by linearity.

The operators $\widehat{D}^{l}\left(  t_{q},u_{j}\right)  $ are defined
analogously. It is clear that, as in the case of the operators $D$, the
operators $\widehat{D}$ commute\footnote{\textit{Comment:} I.e., the operators
$\widehat{D}^{l}\left(  s_{p},x_{i}\right)  $ for all $p$ and $i$ and the
operators $\widehat{D}^{l}\left(  t_{q},u_{j}\right)  $ for all $q$ and $j$
all commute.}.

We have the following identity:

\begin{lemma}
\label{lem.Dhat-D}%
\[
\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \phi=\phi D^{l}\left(  s_{p}%
,x_{i}\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.Dhat-D}.]\footnote{\textit{Comment:} I don't like
this proof, as it somewhat lacks rigor (in fact, it uses the ambiguous
notation $\left(  m_{f}\right)  ^{E,p}$ without properly explaining how
precisely it is defined). Thus, I have added an alternative proof of Lemma
\ref{lem.Dhat-D} further below.} It suffices to verify the identity for
monomials of the form\footnote{\textit{Comment:} This is somewhat imprecise,
since $m$ should also be allowed to contain factors of the form $\left(
x_{i}\mid t_{q}\right)  $.}%
\[
m=\left(  x_{i}\mid u_{j_{1}}\right)  \cdots\left(  x_{i}\mid u_{j_{a}%
}\right)  ,
\]
where all the letters $x$ have index $i$%
.\ \ \ \ \footnote{\textit{Correction:} Replaced a comma by a period here.}
Let $m^{E,p}$ denote the monomial obtained from $m$ by replacing each variable
$\left(  x_{i}\mid u_{j_{b}}\right)  $ by $\left(  s_{p}\mid u_{j_{b}}\right)
$ whenever $b\in E$. Then, we observe (notation as earlier):%
\[
\left(  m^{E,p}\right)  _{f}=\left(  m_{f}\right)  ^{E,p}.
\]
We can now finish the proof through the following computation:%
\begin{align*}
\phi D^{l}\left(  s_{p},x_{i}\right)  m  &  =\phi\sum_{E}m^{E,p}=\sum_{E}\phi
m^{E,p}\\
&  =\sum_{E}\sum_{f}\left(  m^{E,p}\right)  _{f}=\sum_{E}\sum_{f}\left(
m_{f}\right)  ^{E,p}=\sum_{f}\widehat{D}^{l}\left(  s_{p},x_{i}\right)
m_{f}\\
&  =\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \left(  \sum_{f}m_{f}\right)
=\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \phi m.
\end{align*}

\end{proof}

\begin{proof}
[Alternative proof of Lemma \ref{lem.Dhat-D}.]\footnote{\textit{Correction:} I
have added this proof, since I do not fully trust the original proof given
above.} Introduce a new indeterminate $T$, and consider the polynomial rings
$P\left[  T\right]  $, $\left(  K\left[  \mathbf{x},\mathbf{u},\mathbf{s}%
,\mathbf{t}\right]  \right)  \left[  T\right]  $ and $\widehat{P}\left[
T\right]  $. The $K$-algebra homomorphism $\phi$ gives rise to a $K$-algebra
homomorphism $P\left[  T\right]  \rightarrow\widehat{P}\left[  T\right]  $
which sends $T$ to $T$ while acting on $P$ as $\phi$. This latter $K$-algebra
homomorphism $P\left[  T\right]  \rightarrow\widehat{P}\left[  T\right]  $
shall be denoted by $\phi\left[  T\right]  $.

Let $\mathcal{V}_{i}$ be the set of all indeterminates in the polynomial ring
$K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  $ that are
\textbf{not} of the form $x_{ir}$ with $1\leq r\leq d$. In other words,%
\begin{align*}
\mathcal{V}_{i}  &  =\left\{  x_{i^{\prime}r}\ \mid\ 1\leq i^{\prime}\leq
n\text{ and }1\leq r\leq d\text{ satisfying }i^{\prime}\neq i\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \cup\left\{  u_{jr}\ \mid\ 1\leq j\leq k\text{ and
}1\leq r\leq d\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \cup\left\{  s_{p^{\prime}r}\ \mid\ 1\leq p^{\prime
}\text{ and }1\leq r\leq d\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \cup\left\{  t_{qr}\ \mid\ 1\leq q\text{ and }1\leq
r\leq d\right\}  .
\end{align*}


Let $\widehat{\mathbf{D}}$ be the $K$-algebra homomorphism $K\left[
\mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  \rightarrow\left(
K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  \right)  \left[
T\right]  $ that sends%
\begin{align*}
x_{ir}  &  \mapsto x_{ir}+Ts_{pr}\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq
r\leq d;\\
\eta &  \mapsto\eta\ \ \ \ \ \ \ \ \ \ \text{for each }\eta\in\mathcal{V}_{i}.
\end{align*}
Then, each $w\in K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]
$ satisfies%
\begin{equation}
\widehat{\mathbf{D}}w=\sum_{l\geq0}\widehat{D}^{l}\left(  s_{p},x_{i}\right)
w\cdot T^{l} \label{pf.lem.Dhat-D.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.Dhat-D.1}):} It is clearly enough to
prove the equality (\ref{pf.lem.Dhat-D.1}) in the case when $w$ is a monomial.
Thus, assume that $w$ is a monomial. Write $w$ in the form $w=w^{\prime
}w^{\prime\prime}$, where $w^{\prime}$ is the product of all the variables in
$w$ of the form $x_{ir}$, and $w^{\prime\prime}$ is the product of the
remaining variables in $w$.
\par
The monomial $w^{\prime\prime}$ contains no variables of the form $x_{ir}$
(because of how $w^{\prime\prime}$ was defined). Thus, $w^{\prime\prime}$ is a
product of variables from the set $\mathcal{V}_{i}$. But the $K$-algebra
homomorphism $\widehat{\mathbf{D}}$ leaves all variables from the set
$\mathcal{V}_{i}$ fixed (by the definition of $\widehat{\mathbf{D}}$). Thus,
$\widehat{\mathbf{D}}$ also leaves any product of such variables fixed. Hence,
$\widehat{\mathbf{D}}$ leaves $w^{\prime\prime}$ fixed (since $w^{\prime
\prime}$ is a product of variables from the set $\mathcal{V}_{i}$). In other
words, $\widehat{\mathbf{D}}w^{\prime\prime}=w^{\prime\prime}$.
\par
Now, write the monomial $w^{\prime}$ in the form $w^{\prime}=\prod_{b=1}%
^{a}x_{ir_{b}}$. (This is clearly possible due to how $w^{\prime}$ was
defined.) Thus, $w^{\prime}=\prod_{b=1}^{a}x_{ir_{b}}=x_{ir_{1}}x_{ir_{2}%
}\cdots x_{ir_{a}}$. Hence, (\ref{eq.Dhat.def2}) (applied to $v=w^{\prime}$)
yields%
\begin{equation}
\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \left(  w^{\prime}\right)
=\sum_{\substack{E\subseteq\left\{  1,2,\ldots,a\right\}  ;\\\left\vert
E\right\vert =l}}\left(  \prod_{b\in E}s_{pr_{b}}\right)  \left(
\prod_{\substack{b\in\left\{  1,2,\ldots,a\right\}  ;\\b\notin E}}x_{ir_{b}%
}\right)  \label{pf.lem.Dhat-D.1.pf.1}%
\end{equation}
for each $l\geq0$.
\par
Applying the map $\widehat{\mathbf{D}}$ to the equality $w^{\prime}%
=\prod_{b=1}^{a}x_{ir_{b}}$, we obtain%
\begin{align*}
\widehat{\mathbf{D}}w^{\prime}  &  =\widehat{\mathbf{D}}\left(  \prod
_{b=1}^{a}x_{ir_{b}}\right)  =\prod_{b=1}^{a}\underbrace{\widehat{\mathbf{D}%
}\left(  x_{ir_{b}}\right)  }_{=x_{ir_{b}}+Ts_{pr_{b}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widehat{\mathbf{D}}\text{ is a
}K\text{-algebra homomorphism}\right) \\
&  =\prod_{b=1}^{a}\left(  x_{ir_{b}}+Ts_{pr_{b}}\right)  =\underbrace{\sum
_{E\subseteq\left\{  1,2,\ldots,a\right\}  }}_{=\sum_{l\geq0}\sum
_{\substack{E\subseteq\left\{  1,2,\ldots,a\right\}  ;\\\left\vert
E\right\vert =l}}}\underbrace{\left(  \prod_{b\in E}\left(  Ts_{pr_{b}%
}\right)  \right)  }_{=T^{\left\vert E\right\vert }\prod_{b\in E}s_{pr_{b}}%
}\left(  \prod_{\substack{b\in\left\{  1,2,\ldots,a\right\}  ;\\b\notin
E}}x_{ir_{b}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have expanded the product}%
\right) \\
&  =\sum_{l\geq0}\sum_{\substack{E\subseteq\left\{  1,2,\ldots,a\right\}
;\\\left\vert E\right\vert =l}}\underbrace{T^{\left\vert E\right\vert }%
}_{\substack{=T^{l}\\\text{(since }\left\vert E\right\vert =l\text{)}}}\left(
\prod_{b\in E}s_{pr_{b}}\right)  \left(  \prod_{\substack{b\in\left\{
1,2,\ldots,a\right\}  ;\\b\notin E}}x_{ir_{b}}\right) \\
&  =\sum_{l\geq0}\sum_{\substack{E\subseteq\left\{  1,2,\ldots,a\right\}
;\\\left\vert E\right\vert =l}}T^{l}\left(  \prod_{b\in E}s_{pr_{b}}\right)
\left(  \prod_{\substack{b\in\left\{  1,2,\ldots,a\right\}  ;\\b\notin
E}}x_{ir_{b}}\right) \\
&  =\sum_{l\geq0}T^{l}\underbrace{\sum_{\substack{E\subseteq\left\{
1,2,\ldots,a\right\}  ;\\\left\vert E\right\vert =l}}\left(  \prod_{b\in
E}s_{pr_{b}}\right)  \left(  \prod_{\substack{b\in\left\{  1,2,\ldots
,a\right\}  ;\\b\notin E}}x_{ir_{b}}\right)  }_{\substack{=\widehat{D}%
^{l}\left(  s_{p},x_{i}\right)  \left(  w^{\prime}\right)  \\\text{(by
(\ref{pf.lem.Dhat-D.1.pf.1}))}}}\\
&  =\sum_{l\geq0}T^{l}\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \left(
w^{\prime}\right)  .
\end{align*}
\par
Applying the map $\widehat{\mathbf{D}}$ to the equality $w=w^{\prime}%
w^{\prime\prime}$, we obtain%
\begin{align*}
\widehat{\mathbf{D}}w  &  =\widehat{\mathbf{D}}\left(  w^{\prime}%
w^{\prime\prime}\right)  =\underbrace{\left(  \widehat{\mathbf{D}}w^{\prime
}\right)  }_{=\sum_{l\geq0}T^{l}\widehat{D}^{l}\left(  s_{p},x_{i}\right)
\left(  w^{\prime}\right)  }\underbrace{\left(  \widehat{\mathbf{D}}%
w^{\prime\prime}\right)  }_{=w^{\prime\prime}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\widehat{D}_{T}\left(  s_{p}%
,x_{i}\right)  \text{ is a }K\text{-algebra homomorphism}\right) \\
&  =\left(  \sum_{l\geq0}T^{l}\widehat{D}^{l}\left(  s_{p},x_{i}\right)
\left(  w^{\prime}\right)  \right)  w^{\prime\prime}=\sum_{l\geq0}%
T^{l}\underbrace{\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \left(  w^{\prime
}\right)  w^{\prime\prime}}_{\substack{=\widehat{D}^{l}\left(  s_{p}%
,x_{i}\right)  w\\\text{(by (\ref{eq.Dhat.def-full}))}}}\\
&  =\sum_{l\geq0}T^{l}\widehat{D}^{l}\left(  s_{p},x_{i}\right)  w=\sum
_{l\geq0}\widehat{D}^{l}\left(  s_{p},x_{i}\right)  w\cdot T^{l}.
\end{align*}
This proves (\ref{pf.lem.Dhat-D.1}).}.

Let $\mathbf{D}$ be the $K$-algebra homomorphism $P\rightarrow P\left[
T\right]  $ that sends%
\begin{align*}
\left(  x_{i}\mid\eta\right)   &  \mapsto\left(  x_{i}\mid\eta\right)
+T\left(  s_{p}\mid\eta\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }\eta
\in\mathcal{U}\cup\mathcal{T};\\
\left(  \xi\mid\eta\right)   &  \mapsto\left(  \xi\mid\eta\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }\xi\in\left(  \mathcal{X}\cup
\mathcal{S}\right)  \setminus\left\{  x_{i}\right\}  \text{ and }\eta
\in\mathcal{U}\cup\mathcal{T}.
\end{align*}
Then, each $w\in P$ satisfies%
\begin{equation}
\mathbf{D}w=\sum_{l\geq0}D^{l}\left(  s_{p},x_{i}\right)  w\cdot T^{l}.
\label{pf.lem.Dhat-D.2}%
\end{equation}
\footnote{The proof of this is similar to the proof of (\ref{pf.lem.Dhat-D.1}%
).}

Now, it is easy to see that
\[
\widehat{\mathbf{D}}\circ\phi=\phi\left[  T\right]  \circ\mathbf{D}.
\]
Indeed, both sides of this equality are $K$-algebra homomorphisms from $P$ to
$\left(  K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  \right)
\left[  T\right]  $, and thus their equality can be comfortably proven by
verifying it only on the generators $\left(  \xi\mid\eta\right)  $ of the
$K$-algebra $P$ (which verification is straightforward and easy, requiring
only two cases depending on whether $\xi=x_{i}$ or not).

Now, let $w\in P$. Then,
\[
\left(  \widehat{\mathbf{D}}\circ\phi\right)  \left(  w\right)
=\widehat{\mathbf{D}}\left(  \phi w\right)  =\sum_{l\geq0}\widehat{D}%
^{l}\left(  s_{p},x_{i}\right)  \phi w\cdot T^{l}%
\]
(by (\ref{pf.lem.Dhat-D.1}), applied to $\phi w$ instead of $w$). Comparing
this with%
\begin{align*}
\underbrace{\left(  \widehat{\mathbf{D}}\circ\phi\right)  }_{=\phi\left[
T\right]  \circ\mathbf{D}}\left(  w\right)   &  =\left(  \phi\left[  T\right]
\circ\mathbf{D}\right)  \left(  w\right)  =\left(  \phi\left[  T\right]
\right)  \underbrace{\left(  \mathbf{D}w\right)  }_{\substack{=\sum_{l\geq
0}D^{l}\left(  s_{p},x_{i}\right)  w\cdot T^{l}\\\text{(by
(\ref{pf.lem.Dhat-D.2}))}}}\\
&  =\left(  \phi\left[  T\right]  \right)  \left(  \sum_{l\geq0}D^{l}\left(
s_{p},x_{i}\right)  w\cdot T^{l}\right) \\
&  =\sum_{l\geq0}\phi D^{l}\left(  s_{p},x_{i}\right)  w\cdot T^{l},
\end{align*}
we obtain%
\[
\sum_{l\geq0}\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \phi w\cdot T^{l}%
=\sum_{l\geq0}\phi D^{l}\left(  s_{p},x_{i}\right)  w\cdot T^{l}.
\]
Comparing coefficients in front of $T^{l}$ on both sides of this equality, we
obtain $\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \phi w=\phi D^{l}\left(
s_{p},x_{i}\right)  w$. Since this holds for all $w\in P$, we thus find
$\widehat{D}^{l}\left(  s_{p},x_{i}\right)  \phi=\phi D^{l}\left(  s_{p}%
,x_{i}\right)  $. This proves Lemma \ref{lem.Dhat-D}.
\end{proof}

An analogue of Lemma \ref{lem.Dhat-D} for the operators $\widehat{D}%
^{l}\left(  t_{q},u_{j}\right)  $ and $D^{l}\left(  t_{q},u_{j}\right)  $ also
holds.\footnote{\textit{Correction:} Added this sentence.}

Consider now a bitableau $\left[  T,T^{\prime}\right]  $. The \textit{Capelli
operator} $\widehat{C}\left(  T,T^{\prime}\right)  $
is\footnote{\textit{Correction:} Replaced \textquotedblleft
are\textquotedblright\ by \textquotedblleft is\textquotedblright.} defined on
$K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  $ by mimicking
the definition of $C\left(  T,T^{\prime}\right)  $ with $\widehat{D}$ instead
of $D$.\ \ \ \ \footnote{\textit{Comment:} Explicitly, this means that
$\widehat{C}\left(  T,T^{\prime}\right)  $ is defined as follows: Let $\left(
\lambda\right)  $ be the shape of $T$. Let $\alpha_{i}\left(  q\right)  $
(resp. $\beta_{j}\left(  q\right)  $) be the number of occurrences of $x_{i}$
(resp. $u_{j}$) in the $q$th column of $T$ (resp. $T^{\prime}$). The Capelli
operator $\widehat{C}\left(  T,T^{\prime}\right)  $ is defined by the
following formula:%
\[
\widehat{C}\left(  T,T^{\prime}\right)  =\prod_{1\leq q\leq\lambda_{1}}\left(
\prod_{1\leq i\leq n}\widehat{D}^{\alpha_{i}\left(  q\right)  }\left(
s_{q},x_{i}\right)  \right)  \left(  \prod_{1\leq j\leq k}\widehat{D}%
^{\beta_{j}\left(  q\right)  }\left(  t_{q},u_{j}\right)  \right)  .
\]
The right hand side of this equality is a product of commuting operators
(since the operators $\widehat{D}$ all commute). Notice that $\lambda_{1}$ is
the number of columns of $T$.} Lemma \ref{lem.Dhat-D} (and its analogue for
the operators $\widehat{D}^{l}\left(  t_{q},u_{j}\right)  $ and $D^{l}\left(
t_{q},u_{j}\right)  $) yields as a corollary the following
identity:\footnote{\textit{Correction:} Slightly rewrote this sentence in
order to make the identity an actual corollary.}

\begin{corollary}
\label{cor.Chat-C}%
\[
\widehat{C}\left(  T,T^{\prime}\right)  \phi=\phi C\left(  T,T^{\prime
}\right)  .
\]

\end{corollary}

Let us next prove some further observations.\footnote{\textit{Correction:}
Everything between this sentence and the proof of Theorem \ref{thm.3.1} has
been added by me.}

\begin{lemma}
\label{lem.3.1.o1}Each inner product $\left(  x_{i_{1}}\cdots x_{i_{l}}\mid
u_{j_{1}}\cdots u_{j_{l}}\right)  $ with $l>d$ satisfies $\phi\left(
x_{i_{1}}\cdots x_{i_{l}}\mid u_{j_{1}}\cdots u_{j_{l}}\right)  =0$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.3.1.o1}.]Let $\left(  x_{i_{1}}\cdots x_{i_{l}}\mid
u_{j_{1}}\cdots u_{j_{l}}\right)  $ be an inner product with $l>d$. It is
straightforward to see that%
\[
\left(
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right)  =\left(
\begin{array}
[c]{ccc}%
x_{i_{1},1} & \cdots & x_{i_{1},d}\\
\vdots & \ddots & \vdots\\
x_{i_{l},1} & \cdots & x_{i_{l},d}%
\end{array}
\right)  \cdot\left(
\begin{array}
[c]{ccc}%
u_{j_{1},1} & \cdots & u_{j_{l},1}\\
\vdots & \ddots & \vdots\\
u_{j_{1},d} & \cdots & u_{j_{l},d}%
\end{array}
\right)  .
\]
Hence, the matrix $\left(
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right)  $ is the product of an $l\times d$-matrix with a $d\times l$-matrix.
Thus, the matrix $\left(
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right)  $ has determinant $0$ (since $d<l$). In other words, $\left\vert
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right\vert =0$.

But%
\[
\left(  x_{i_{1}}\cdots x_{i_{l}}\mid u_{j_{1}}\cdots u_{j_{l}}\right)
=\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{l}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{l}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{l}}\mid
u_{j_{l}}\right)
\end{array}
\right\vert .
\]
Applying the map $\phi$ to this identity, we obtain%
\begin{align*}
\phi\left(  x_{i_{1}}\cdots x_{i_{l}}\mid u_{j_{1}}\cdots u_{j_{l}}\right)
&  =\phi\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{l}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{l}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{l}}\mid
u_{j_{l}}\right)
\end{array}
\right\vert \\
&  =\left\vert
\begin{array}
[c]{ccc}%
\phi\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \phi\left(  x_{i_{1}%
}\mid u_{j_{l}}\right) \\
\vdots & \ddots & \vdots\\
\phi\left(  x_{i_{l}}\mid u_{j_{1}}\right)  & \cdots & \phi\left(  x_{i_{l}%
}\mid u_{j_{l}}\right)
\end{array}
\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\text{ is a }K\text{-algebra
homomorphism}\right) \\
&  =\left\vert
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\left(  x_{i}\mid
u_{j}\right)  =\left\langle x_{i}\mid u_{j}\right\rangle \text{ for all
}i\text{ and }j\right) \\
&  =0.
\end{align*}
This proves Lemma \ref{lem.3.1.o1}.
\end{proof}

\begin{lemma}
\label{lem.3.1.o2}Let $\omega:K\left[  \mathbf{x},\mathbf{u},\mathbf{s}%
,\mathbf{t}\right]  \rightarrow K$ be the $K$-algebra homomorphism given by%
\begin{align*}
x_{ir}  &  \mapsto0\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq i\leq n\text{ and
}1\leq r\leq d;\\
u_{jr}  &  \mapsto0\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq j\leq k\text{ and
}1\leq r\leq d;\\
s_{pr}  &  \mapsto\delta_{p,r}\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq p\text{
and }1\leq r\leq d;\\
t_{qr}  &  \mapsto\delta_{q,r}\ \ \ \ \ \ \ \ \ \ \text{for all }1\leq q\text{
and }1\leq r\leq d
\end{align*}
(where $\delta_{u,v}$ stands for $%
\begin{cases}
1, & \text{if }u=v;\\
0, & \text{if }u\neq v
\end{cases}
$). Then, each $k\in\left\{  0,1,\ldots,d\right\}  $ satisfies $\omega\left(
\phi\left(  s_{1}s_{2}\cdots s_{k}\mid t_{1}t_{2}\cdots t_{k}\right)  \right)
=1$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.3.1.o2}.]Let $k\in\left\{  0,1,\ldots,d\right\}  $.
Then, for each $p,q\in\left\{  1,2,\ldots,k\right\}  $, we have%
\begin{align}
\omega\left\langle s_{p}\mid t_{q}\right\rangle  &  =\omega\left(  \sum_{1\leq
r\leq d}s_{pr}t_{qr}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\left\langle s_{p}\mid t_{q}\right\rangle =\sum_{1\leq r\leq d}s_{pr}%
t_{qr}\right) \nonumber\\
&  =\sum_{1\leq r\leq d}\underbrace{\omega\left(  s_{pr}\right)
}_{\substack{=\delta_{p,r}\\\text{(by the definition of }\omega\text{)}%
}}\underbrace{\omega\left(  t_{qr}\right)  }_{\substack{=\delta_{q,r}%
\\\text{(by the definition of }\omega\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\text{ is a }K\text{-algebra
homomorphism}\right) \nonumber\\
&  =\sum_{1\leq r\leq d}\delta_{p,r}\delta_{q,r}=%
\begin{cases}
1, & \text{if }p=q\leq d;\\
0, & \text{otherwise}%
\end{cases}
=%
\begin{cases}
1, & \text{if }p=q;\\
0, & \text{otherwise}%
\end{cases}
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }p=q\leq d\text{ holds if and only if }p=q\\
\text{(because }q\leq d\text{ always holds (since }q\leq k\leq d\text{))}%
\end{array}
\right) \nonumber\\
&  =\delta_{p,q}. \label{pf.thm.3.1.fn1.obs2.pf.2}%
\end{align}
Now,
\[
\left(  s_{1}s_{2}\cdots s_{k}\mid t_{1}t_{2}\cdots t_{k}\right)  =\left\vert
\begin{array}
[c]{ccc}%
\left(  s_{1}\mid t_{1}\right)  & \cdots & \left(  s_{1}\mid t_{k}\right) \\
\vdots & \ddots & \vdots\\
\left(  s_{k}\mid t_{1}\right)  & \cdots & \left(  s_{k}\mid t_{k}\right)
\end{array}
\right\vert .
\]
Applying the map $\phi$ to this identity, we obtain%
\begin{align*}
\phi\left(  s_{1}s_{2}\cdots s_{k}\mid t_{1}t_{2}\cdots t_{k}\right)   &
=\phi\left\vert
\begin{array}
[c]{ccc}%
\left(  s_{1}\mid t_{1}\right)  & \cdots & \left(  s_{1}\mid t_{k}\right) \\
\vdots & \ddots & \vdots\\
\left(  s_{k}\mid t_{1}\right)  & \cdots & \left(  s_{k}\mid t_{k}\right)
\end{array}
\right\vert =\left\vert
\begin{array}
[c]{ccc}%
\phi\left(  s_{1}\mid t_{1}\right)  & \cdots & \phi\left(  s_{1}\mid
t_{k}\right) \\
\vdots & \ddots & \vdots\\
\phi\left(  s_{k}\mid t_{1}\right)  & \cdots & \phi\left(  s_{k}\mid
t_{k}\right)
\end{array}
\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\text{ is a }K\text{-algebra
homomorphism}\right) \\
&  =\left\vert
\begin{array}
[c]{ccc}%
\left\langle s_{1}\mid t_{1}\right\rangle  & \cdots & \left\langle s_{1}\mid
t_{k}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle s_{k}\mid t_{1}\right\rangle  & \cdots & \left\langle s_{k}\mid
t_{k}\right\rangle
\end{array}
\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\left(  s_{p}\mid
t_{q}\right)  =\left\langle s_{p}\mid t_{q}\right\rangle \text{ for all
}p\text{ and }q\right)  .
\end{align*}
Applying the map $\omega$ to this identity, we find%
\begin{align*}
\omega\left(  \phi\left(  s_{1}s_{2}\cdots s_{k}\mid t_{1}t_{2}\cdots
t_{k}\right)  \right)   &  =\omega\left\vert
\begin{array}
[c]{ccc}%
\left\langle s_{1}\mid t_{1}\right\rangle  & \cdots & \left\langle s_{1}\mid
t_{k}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle s_{k}\mid t_{1}\right\rangle  & \cdots & \left\langle s_{k}\mid
t_{k}\right\rangle
\end{array}
\right\vert \\
&  =\left\vert
\begin{array}
[c]{ccc}%
\omega\left\langle s_{1}\mid t_{1}\right\rangle  & \cdots & \omega\left\langle
s_{1}\mid t_{k}\right\rangle \\
\vdots & \ddots & \vdots\\
\omega\left\langle s_{k}\mid t_{1}\right\rangle  & \cdots & \omega\left\langle
s_{k}\mid t_{k}\right\rangle
\end{array}
\right\vert \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\text{ is a }K\text{-algebra
homomorphism}\right) \\
&  =\left\vert \underbrace{%
\begin{array}
[c]{ccc}%
\delta_{1,1} & \cdots & \delta_{1,k}\\
\vdots & \ddots & \vdots\\
\delta_{k,1} & \cdots & \delta_{k,k}%
\end{array}
}_{=I_{k}}\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.3.1.fn1.obs2.pf.2})}\right) \\
&  =\left\vert I_{k}\right\vert =1.
\end{align*}
This proves Lemma \ref{lem.3.1.o2}.
\end{proof}

\begin{lemma}
\label{lem.3.1.o3}Let $\left[  T,T^{\prime}\right]  $ be a standard $\left(
\mathcal{X},\mathcal{U}\right)  $-bitableau of shape $\left(  \lambda\right)
$, where $\left(  \lambda\right)  $ is not longer than $\left(  d\right)  $.
Then,
\begin{equation}
\phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  \neq0.
\label{pf.thm.3.1.fn1.obs3.eq}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.3.1.o3}.]Consider the $K$-algebra homomorphism
$\omega:K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]
\rightarrow K$ defined in Lemma \ref{lem.3.1.o2}.

We have assumed that $\left(  \lambda\right)  $ is not longer than $\left(
d\right)  $. Thus, $\lambda_{1}\leq d$. Hence, $\lambda_{r}\leq d$ for all
$r\in\left\{  1,2,3,\ldots\right\}  $. Therefore, for each $r\in\left\{
1,2,3,\ldots\right\}  $, we have $\lambda_{r}\in\left\{  0,1,\ldots,d\right\}
$ and thus%
\begin{equation}
\omega\left(  \phi\left(  s_{1}s_{2}\cdots s_{\lambda_{r}}\mid t_{1}%
t_{2}\cdots t_{\lambda_{r}}\right)  \right)  =1 \label{pf.lem.3.1.o3.1}%
\end{equation}
(by Lemma \ref{lem.3.1.o2}, applied to $k=\lambda_{r}$).

Define two tableaux $U$ and $U^{\prime}$ as in the proof of Theorem
\ref{thm.2.1} (i). Then, $C\left(  T,T^{\prime}\right)  \left(  T\mid
T^{\prime}\right)  =\left(  U\mid U^{\prime}\right)  $ (this was shown in the
proof of Theorem \ref{thm.2.1} (i)). But from the definition of $U$ and
$U^{\prime}$, we obtain%
\[
\left(  U\mid U^{\prime}\right)  =\prod_{r\geq1}\left(  s_{1}s_{2}\cdots
s_{\lambda_{r}}\mid t_{1}t_{2}\cdots t_{\lambda_{r}}\right)  .
\]
Applying the map $\phi$ to both sides of this equality, we obtain%
\[
\phi\left(  U\mid U^{\prime}\right)  =\phi\left(  \prod_{r\geq1}\left(
s_{1}s_{2}\cdots s_{\lambda_{r}}\mid t_{1}t_{2}\cdots t_{\lambda_{r}}\right)
\right)  =\prod_{r\geq1}\phi\left(  s_{1}s_{2}\cdots s_{\lambda_{r}}\mid
t_{1}t_{2}\cdots t_{\lambda_{r}}\right)
\]
(since $\phi$ is a $K$-algebra homomorphism). Applying the map $\phi$ to the
equality $C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)
=\left(  U\mid U^{\prime}\right)  $, we now find%
\[
\phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)
=\phi\left(  U\mid U^{\prime}\right)  =\prod_{r\geq1}\phi\left(  s_{1}%
s_{2}\cdots s_{\lambda_{r}}\mid t_{1}t_{2}\cdots t_{\lambda_{r}}\right)  .
\]
Applying the map $\omega$ to both sides of this equality, we find%
\begin{align*}
\omega\left(  \phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  \right)   &  =\omega\left(  \prod_{r\geq1}\phi\left(  s_{1}%
s_{2}\cdots s_{\lambda_{r}}\mid t_{1}t_{2}\cdots t_{\lambda_{r}}\right)
\right) \\
&  =\prod_{r\geq1}\underbrace{\omega\left(  \phi\left(  s_{1}s_{2}\cdots
s_{\lambda_{r}}\mid t_{1}t_{2}\cdots t_{\lambda_{r}}\right)  \right)
}_{\substack{=1\\\text{(by (\ref{pf.lem.3.1.o3.1}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\omega\text{ is a }K\text{-algebra
homomorphism}\right) \\
&  =\prod_{r\geq1}1=1\neq0.
\end{align*}
Therefore, $\phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  \neq0$. This proves Lemma \ref{lem.3.1.o3}.
\end{proof}

\begin{lemma}
\label{lem.3.1.o4}Let $\left[  T,T^{\prime}\right]  $ be a standard $\left(
\mathcal{X},\mathcal{U}\right)  $-bitableau of shape $\left(  \lambda\right)
$, where $\left(  \lambda\right)  $ is longer than $\left(  d\right)  $.
Then,
\begin{equation}
\phi\left(  T\mid T^{\prime}\right)  =0. \label{pf.thm.3.1.fn1.obs4.eq}%
\end{equation}

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.3.1.o4}.]We have assumed that $\left(  \lambda
\right)  $ is longer than $\left(  d\right)  $. Thus, $\lambda_{1}>d$. Now,
for each $p\geq1$ and $q\in\left\{  1,2,\ldots,\lambda_{p}\right\}  $, let
$x_{i\left(  p,q\right)  }$ be the $\left(  p,q\right)  $-th entry of the
tableau $T$, and let $u_{j\left(  p,q\right)  }$ be the $\left(  p,q\right)
$-th entry of the tableau $T^{\prime}$. Then,%
\begin{equation}
\left(  T\mid T^{\prime}\right)  =\prod_{r\geq1}\left(  x_{i\left(
r,1\right)  }x_{i\left(  r,2\right)  }\cdots x_{i\left(  r,\lambda_{r}\right)
}\mid u_{j\left(  r,1\right)  }u_{j\left(  r,2\right)  }\cdots u_{j\left(
r,\lambda_{r}\right)  }\right)  . \label{pf.thm.3.1.fn1.obs4.pf.1}%
\end{equation}
But $\lambda_{1}>d$ and thus $\phi\left(  x_{i\left(  1,1\right)  }x_{i\left(
1,2\right)  }\cdots x_{i\left(  1,\lambda_{1}\right)  }\mid u_{j\left(
1,1\right)  }u_{j\left(  1,2\right)  }\cdots u_{j\left(  1,\lambda_{1}\right)
}\right)  =0$ (by Lemma \ref{lem.3.1.o1}, applied to $l=\lambda_{1}$,
$i_{k}=i\left(  1,k\right)  $ and $j_{k}=j\left(  1,k\right)  $). Hence, the
product $\prod_{r\geq1}\phi\left(  x_{i\left(  r,1\right)  }x_{i\left(
r,2\right)  }\cdots x_{i\left(  r,\lambda_{r}\right)  }\mid u_{j\left(
r,1\right)  }u_{j\left(  r,2\right)  }\cdots u_{j\left(  r,\lambda_{r}\right)
}\right)  $ has at least one factor equal to $0$ (namely, the factor for
$r=1$), and thus equals $0$ itself. In other words,%
\[
\prod_{r\geq1}\phi\left(  x_{i\left(  r,1\right)  }x_{i\left(  r,2\right)
}\cdots x_{i\left(  r,\lambda_{r}\right)  }\mid u_{j\left(  r,1\right)
}u_{j\left(  r,2\right)  }\cdots u_{j\left(  r,\lambda_{r}\right)  }\right)
=0.
\]


But applying the map $\phi$ to the equality (\ref{pf.thm.3.1.fn1.obs4.pf.1}),
we obtain%
\begin{align*}
\phi\left(  T\mid T^{\prime}\right)   &  =\phi\left(  \prod_{r\geq1}\left(
x_{i\left(  r,1\right)  }x_{i\left(  r,2\right)  }\cdots x_{i\left(
r,\lambda_{r}\right)  }\mid u_{j\left(  r,1\right)  }u_{j\left(  r,2\right)
}\cdots u_{j\left(  r,\lambda_{r}\right)  }\right)  \right) \\
&  =\prod_{r\geq1}\phi\left(  x_{i\left(  r,1\right)  }x_{i\left(  r,2\right)
}\cdots x_{i\left(  r,\lambda_{r}\right)  }\mid u_{j\left(  r,1\right)
}u_{j\left(  r,2\right)  }\cdots u_{j\left(  r,\lambda_{r}\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\phi\text{ is a }K\text{-algebra
homomorphism}\right) \\
&  =0.
\end{align*}
This proves Lemma \ref{lem.3.1.o4}.
\end{proof}

With these tools in hand, we can begin the proof of the theorem.

\begin{proof}
[Proof of Theorem \ref{thm.3.1}.]\footnote{\textit{Correction:} I have removed
the first two paragraphs of this proof, since they were a distraction from the
proof. (To be more precise, the useful parts of these two paragraphs have been
preserved in Lemmas \ref{lem.3.1.o1}, \ref{lem.3.1.o2}, \ref{lem.3.1.o3}, and
\ref{lem.3.1.o4}. The two paragraphs I removed can be found in Remark
\ref{rmk.3.1.old}.)} Indeed\footnote{\textit{Correction:} Replaced
\textquotedblleft Finally\textquotedblright\ by \textquotedblleft
Indeed\textquotedblright.}, consider an element $M\in P\left(  \alpha
,\beta\right)  $\ \ \ \ \footnote{\textit{Correction:} Replaced
\textquotedblleft$M$\textquotedblright\ by \textquotedblleft$M\in P\left(
\alpha,\beta\right)  $\textquotedblright.} in the kernel of $\phi$. Using
Corollary \ref{cor.strf.gen}\footnote{\textit{Correction:} Replaced
\textquotedblleft the straightening formula\textquotedblright\ by
\textquotedblleft Corollary \ref{cor.strf.gen}\textquotedblright.}, write $M$
as a linear combination of standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminants\footnote{\textit{Correction:} Replaced \textquotedblleft
bideterminants\textquotedblright\ by \textquotedblleft$\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bideterminants\textquotedblright.}:%
\[
M=a\left(  T\mid T^{\prime}\right)  +N,
\]
where $\left(  \lambda\right)  $ is the shortest shape occurring in the
expansion, and, of all the bideterminants of shape $\left(  \lambda\right)  $
occurring in the expansion, $\left(  T\mid T^{\prime}\right)  $ is the one
with the lexicographically smallest column sequence\footnote{\textit{Comment:}
... and, of course, $a$ is a nonzero element of $K$.}. Applying the Capelli
operator $C\left(  T,T^{\prime}\right)  $, and observing that, by Theorem
\ref{thm.2.1}, $C\left(  T,T^{\prime}\right)  N=0$, we obtain%
\[
C\left(  T,T^{\prime}\right)  M=aC\left(  T,T^{\prime}\right)  \left(  T\mid
T^{\prime}\right)  .
\]
\footnote{\textit{Comment:} Here is this argument in slightly more detail:
\par
By the definition of $\left(  \lambda\right)  $ and $\left(  T\mid T^{\prime
}\right)  $, we know that $N$ is a $K$-linear combination of bideterminants
$\left(  V\mid V^{\prime}\right)  $ for standard $\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bitableaux $\left[  V,V^{\prime}\right]  $ which either
have shape longer than $\left(  \lambda\right)  $ or have the same shape as
$\left(  \lambda\right)  $ but have a greater column sequence than $\left[
T,T^{\prime}\right]  $. All such bideterminants $\left(  V\mid V^{\prime
}\right)  $ satisfy $C\left(  T,T^{\prime}\right)  \left(  V\mid V^{\prime
}\right)  =0$ (indeed, in the former case, this follows from Theorem
\ref{thm.2.1} (ii), whereas in the latter case it follows from Theorem
\ref{thm.2.1} (iii)). Hence, $N$ (being a $K$-linear combination of such
bideterminants) must also satisfy $C\left(  T,T^{\prime}\right)  N=0$. Now,
applying the operator $C\left(  T,T^{\prime}\right)  $ to the equality
$M=a\left(  T\mid T^{\prime}\right)  +N$, we obtain%
\[
C\left(  T,T^{\prime}\right)  M=C\left(  T,T^{\prime}\right)  \left(  a\left(
T\mid T^{\prime}\right)  +N\right)  =aC\left(  T,T^{\prime}\right)  \left(
T\mid T^{\prime}\right)  +\underbrace{C\left(  T,T^{\prime}\right)  N}%
_{=0}=aC\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  ,
\]
qed.} Applying $\phi$ and using the identity in Corollary \ref{cor.Chat-C}%
\footnote{\textit{Correction:} Replaced \textquotedblleft the previous
corollary\textquotedblright\ by \textquotedblleft Corollary \ref{cor.Chat-C}%
\textquotedblright.}, we have%
\[
a\phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  =\phi
C\left(  T,T^{\prime}\right)  M=\widehat{C}\left(  T,T^{\prime}\right)  \phi
M.
\]
But $\phi M=0$; hence, $C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  $ must be a bideterminant of shape strictly longer than $\left(
d\right)  $\ \ \ \ \footnote{\textit{Comment:} Here is why this holds:
\par
Define two tableaux $U$ and $U^{\prime}$ as in the proof of Theorem
\ref{thm.2.1} (i). Then, $C\left(  T,T^{\prime}\right)  \left(  T\mid
T^{\prime}\right)  =\left(  U\mid U^{\prime}\right)  $ (this was shown in the
proof of Theorem \ref{thm.2.1} (i)). Thus, $C\left(  T,T^{\prime}\right)
\left(  T\mid T^{\prime}\right)  $ is a bideterminant. It only remains to show
that this bideterminant $C\left(  T,T^{\prime}\right)  \left(  T\mid
T^{\prime}\right)  $ has shape longer than $\left(  d\right)  $. In other
words, it remains to prove that $\left(  \lambda\right)  $ is longer than
$\left(  d\right)  $ (since this bideterminant has shape $\left(
\lambda\right)  $).
\par
Assume the contrary. Thus, $\left(  \lambda\right)  $ is not longer than
$\left(  d\right)  $. Hence, (\ref{pf.thm.3.1.fn1.obs3.eq}) yields $\phi
C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  \neq0$.
\par
But $a\phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)
=\widehat{C}\left(  T,T^{\prime}\right)  \underbrace{\phi M}_{=0}=0$, so that
$a=0$ (since $\phi C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime
}\right)  \neq0$). This contradicts $a\neq0$. This contradiction shows that
our assumption was false. Hence, we have shown that $\left(  \lambda\right)  $
is longer than $\left(  d\right)  $.}. As $C\left(  T,T^{\prime}\right)
\left(  T\mid T^{\prime}\right)  $ and $\left(  T\mid T^{\prime}\right)  $
have the same shape, we conclude that $\left(  \lambda\right)  $ is strictly
longer than $\left(  d\right)  $. But we have chosen $\left(  \lambda\right)
$ to be the shortest shape occurring in the expansion of $M$. Therefore, $M$
is a linear combination of standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminants\footnote{\textit{Correction:} Replaced \textquotedblleft
bideterminants\textquotedblright\ by \textquotedblleft$\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bideterminants\textquotedblright.} of shape strictly
longer than $\left(  d\right)  $. This concludes the proof of the
theorem\footnote{\textit{Comment:} Let me explain why this concludes the proof
of Theorem \ref{thm.3.1}:
\par
We have just shown that each $M\in P\left(  \alpha,\beta\right)  $ lying in
the kernel of $\phi$ must be a linear combination of standard $\left(
\mathcal{X},\mathcal{U}\right)  $-bideterminants of shape strictly longer than
$\left(  d\right)  $. In other words, each $M$ in the kernel of the Pascal
homomorphism must be a linear combination of standard $\left(  \mathcal{X}%
,\mathcal{U}\right)  $-bideterminants of shape strictly longer than $\left(
d\right)  $.
\par
On the other hand, each standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminant of shape strictly longer than $\left(  d\right)  $ lies in the
kernel of $\phi$ (by (\ref{pf.thm.3.1.fn1.obs4.eq})), thus in the kernel of
the Pascal homomorphism. Hence, each linear combination of standard $\left(
\mathcal{X},\mathcal{U}\right)  $-bideterminants of shape strictly longer than
$\left(  d\right)  $ lies in the kernel of the Pascal homomorphism.
\par
Thus, we have shown the following two facts:
\par
\begin{itemize}
\item Each $M$ in the kernel of the Pascal homomorphism must be a linear
combination of standard $\left(  \mathcal{X},\mathcal{U}\right)
$-bideterminants of shape strictly longer than $\left(  d\right)  $;
\par
\item Each linear combination of standard $\left(  \mathcal{X},\mathcal{U}%
\right)  $-bideterminants of shape strictly longer than $\left(  d\right)  $
lies in the kernel of the Pascal homomorphism.
\end{itemize}
\par
Combining these two facts, we conclude that the elements of the kernel of the
Pascal homomorphism are precisely the linear combinations of standard $\left(
\mathcal{X},\mathcal{U}\right)  $-bideterminants of shape strictly longer than
$\left(  d\right)  $. This proves Theorem \ref{thm.3.1}.}.
\end{proof}

\begin{remarkfns}
\label{rmk.3.1.old}For the sake of completeness, here are the first two
paragraphs from the original proof of Theorem \ref{thm.3.1} (which have been
omitted from our above version of this proof)\footnote{As already mentioned,
they are not necessary to the proof.}:

The image under $\phi$ of an inner product $\left(  x_{i_{1}}\cdots x_{i_{l}%
}\mid u_{j_{1}}\cdots u_{j_{l}}\right)  $ is the
determinant\footnote{\textit{Correction:} Replaced \textquotedblleft$x_{j_{1}%
}$\textquotedblright\ by \textquotedblleft$u_{j_{1}}$\textquotedblright\ in
this determinant.}%
\[
\left\vert
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right\vert .
\]
This is the determinant of the matrix%
\[
\left[
\begin{array}
[c]{ccc}%
x_{i_{1},1} & \cdots & x_{i_{1},d}\\
\vdots & \ddots & \vdots\\
x_{i_{l},1} & \cdots & x_{i_{l},d}%
\end{array}
\right]  \cdot\left[
\begin{array}
[c]{ccc}%
u_{j_{1},1} & \cdots & u_{j_{l},1}\\
\vdots & \ddots & \vdots\\
u_{j_{1},d} & \cdots & u_{j_{l},d}%
\end{array}
\right]  .
\]
\footnote{\textit{Comment:} This is because%
\[
\left[
\begin{array}
[c]{ccc}%
\left\langle x_{i_{1}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{1}}\mid u_{j_{l}}\right\rangle \\
\vdots & \ddots & \vdots\\
\left\langle x_{i_{l}}\mid u_{j_{1}}\right\rangle  & \cdots & \left\langle
x_{i_{l}}\mid u_{j_{l}}\right\rangle
\end{array}
\right]  =\left[
\begin{array}
[c]{ccc}%
x_{i_{1},1} & \cdots & x_{i_{1},d}\\
\vdots & \ddots & \vdots\\
x_{i_{l},1} & \cdots & x_{i_{l},d}%
\end{array}
\right]  \cdot\left[
\begin{array}
[c]{ccc}%
u_{j_{1},1} & \cdots & u_{j_{l},1}\\
\vdots & \ddots & \vdots\\
u_{j_{1},d} & \cdots & u_{j_{l},d}%
\end{array}
\right]  .
\]
} As the variables $x_{ir},u_{jr}$ are algebraically independent, the above
matrix is of maximum possible rank\footnote{\textit{Comment:}
\textquotedblleft Maximum possible rank\textquotedblright\ here means
\textquotedblleft rank $\min\left\{  l,d\right\}  $\textquotedblright\ (where
\textquotedblleft rank\textquotedblright\ means the rank over the quotient
field of $K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}\right]  $). Of
course, the claim that this matrix has rank $\min\left\{  l,d\right\}  $ is
only made under the tacit assumption that the indices $i_{1},\ldots,i_{l}$ are
distinct and the indices $j_{1},\ldots,j_{l}$ are distinct.
\par
Let me explain why this claim holds. We must prove that the matrix%
\[
\left[
\begin{array}
[c]{ccc}%
x_{i_{1},1} & \cdots & x_{i_{1},d}\\
\vdots & \ddots & \vdots\\
x_{i_{l},1} & \cdots & x_{i_{l},d}%
\end{array}
\right]  \cdot\left[
\begin{array}
[c]{ccc}%
u_{j_{1},1} & \cdots & u_{j_{l},1}\\
\vdots & \ddots & \vdots\\
u_{j_{1},d} & \cdots & u_{j_{l},d}%
\end{array}
\right]
\]
has rank $\min\left\{  l,d\right\}  $. Indeed, it clearly has rank $\leq
\min\left\{  l,d\right\}  $ (since it is the product of an $l\times d$-matrix
with a $d\times l$-matrix). Thus, it remains to show that it has rank
$\geq\min\left\{  l,d\right\}  $. To achieve this, it will suffice to prove
that it has a nonvanishing minor of size $\min\left\{  l,d\right\}  $. We
shall distinguish between the cases $l\geq d$ and $l<d$:
\par
\begin{itemize}
\item If $l\geq d$, then the northwesternmost $d\times d$-minor of our matrix
is%
\[
\det\left(  \left[
\begin{array}
[c]{ccc}%
x_{i_{1},1} & \cdots & x_{i_{1},d}\\
\vdots & \ddots & \vdots\\
x_{i_{d},1} & \cdots & x_{i_{d},d}%
\end{array}
\right]  \cdot\left[
\begin{array}
[c]{ccc}%
u_{j_{1},1} & \cdots & u_{j_{d},1}\\
\vdots & \ddots & \vdots\\
u_{j_{1},d} & \cdots & u_{j_{d},d}%
\end{array}
\right]  \right)  ,
\]
which is nonzero (because if we substitute $\delta_{s,t}$ for each variable
$x_{i_{s},t}$ and for each variable $u_{j_{s},t}$, then this $d\times d$-minor
becomes $\det\left(  I_{d}\right)  =1$); this is thus a nonvanishing minor of
size $d=\min\left\{  l,d\right\}  $ (since $l\geq d$).
\par
\item If $l<d$, then the $l\times l$-minor of our matrix is%
\[
\det\left(  \left[
\begin{array}
[c]{ccc}%
x_{i_{1},1} & \cdots & x_{i_{1},d}\\
\vdots & \ddots & \vdots\\
x_{i_{l},1} & \cdots & x_{i_{l},d}%
\end{array}
\right]  \cdot\left[
\begin{array}
[c]{ccc}%
u_{j_{1},1} & \cdots & u_{j_{l},1}\\
\vdots & \ddots & \vdots\\
u_{j_{1},d} & \cdots & u_{j_{l},d}%
\end{array}
\right]  \right)  ,
\]
which is nonzero (because if we substitute $\delta_{s,t}$ for each variable
$x_{i_{s},t}$ and for each variable $u_{j_{s},t}$, then this $l\times l$-minor
becomes $\det\left(  I_{l}\right)  =1$); this is thus a nonvanishing minor of
size $l=\min\left\{  l,d\right\}  $ (since $l<d$).
\end{itemize}
}; that is to say, its determinant is zero iff $l>d$.

Now, consider a bideterminant $\left(  T\mid T^{\prime}\right)  $. It is a
product of inner products of lengths $\lambda_{1},\ldots,\lambda_{l}$. As
$\widehat{P}$\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft%
$P$\textquotedblright\ by \textquotedblleft$\widehat{P}$\textquotedblright.}
(being a subring of $K\left[  \mathbf{x},\mathbf{u},\mathbf{s},\mathbf{t}%
\right]  $) is an integral domain, $\phi\left(  T\mid T^{\prime}\right)  $ is
zero iff one of its constituent inner products has zero image. This happens
iff $\lambda_{1}>d$, or $\left(  \lambda\right)  $ is strictly longer than
$\left(  d\right)  $.
\end{remarkfns}

\subsection{\label{subsect.invar.fft}The First Fundamental Theorem: statement}

Consider the algebra $P$, defined on the alphabets $\mathcal{X}=\left\{
x_{1},\ldots,x_{n}\right\}  $ and $\mathcal{U}=\left\{  u_{1},\ldots
,u_{d}\right\}  $.\ \ \ \ \footnote{\textit{Comment:} In other words, from
here on, we return to the original version of the $K$-algebra $P$, which does
not include the indeterminates from the alphabets $\mathcal{S}$ and
$\mathcal{T}$; we furthermore set $d=k$.
\par
From here on, the letter $k$ shall no longer stand for the size of the
alphabet $\mathcal{U}$ (as it did in \textquotedblleft$\mathcal{U}=\left\{
u_{1},\ldots,u_{k}\right\}  $\textquotedblright). Indeed, we do not need it in
this role any more, since the letter $d$ serves the same purpose (because
$\mathcal{U}=\left\{  u_{1},\ldots,u_{d}\right\}  $). Thus, the letter $k$ is
now free for any other use.} In $P$, all the bideterminants of shape strictly
longer than $\left(  d\right)  $ are identically $0$%
\ \ \ \ \footnote{\textit{Comment:} This is because if $\left(  T\mid
T^{\prime}\right)  $ is a bideterminant of shape strictly longer than $\left(
d\right)  $, then the first row of $T^{\prime}$ has length $>d$ and thus
contains two equal letters.}; hence, by Theorem \ref{thm.3.1}, the
map\footnote{\textit{Correction:} Added the words \textquotedblleft the
map\textquotedblright.} $\phi$ is an injection. This allows us to transfer
questions about forms to questions about elements of $P$.

(Recall that $P$ is the $K$-algebra of polynomials in the indeterminates
$\left(  x_{i}\mid u_{j}\right)  $ with $1\leq i\leq n$ and $1\leq j\leq d$
over $K$.)\ \ \ \ \footnote{\textit{Correction:} Added the preceding
sentence.}

Let $L$ be a linear transformation from the vector space $V_{d}$ to
itself.\footnote{\textit{Comment:} This paragraph, again, just serves the
purpose of providing motivation.} If $F$ is a form on $V_{d}$,
then\footnote{\textit{Correction:} Added the word \textquotedblleft
then\textquotedblright.} $L$ acts on $F$ by%
\[
LF\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  =F\left(
L\mathbf{x}_{1},\ldots,L\mathbf{x}_{m}\right)  .
\]
A form $F$\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft
form\textquotedblright\ by \textquotedblleft form $F$\textquotedblright.} is
\textit{invariant} if, for all invertible linear transformations $L$, there
exists a scalar $a\left(  L\right)  $ such that $LF=a\left(  L\right)  F$.

Transferring to the algebra $P$, a \textit{form} $F\left(  \mathbf{x}%
_{1},\ldots,\mathbf{x}_{n}\right)  $ is an element of $P$, that is, a
polynomial in the variables $\left(  x_{i}\mid u_{j}\right)  $ with $1\leq
i\leq n$ and $1\leq j\leq d$.\ \ \ \ \footnote{\textit{Correction:} I have
simplified the preceding sentence, removing some confusing generality. I
re-introduce said generality later on where it becomes necessary (in the
second proof of Theorem \ref{thm.3.2}).} A $d \times d$%
-matrix\footnote{\textit{Correction:} Here and in the following, I have
replaced the word \textquotedblleft transformation\textquotedblright\ (or
``linear transformation'') by ``$d \times d$-matrix $L \in K^{d \times d}$''.
Indeed, the algebra $P$ is not defined in a basis-free fashion, so the use of
transformations instead of matrices does not serve any useful purpose (but it
does create an extra level of indirection).
\par
Also, in this specific place, I have replaced \textquotedblleft An invertible
linear transformation $L$, given by an invertible $d\times d$ square matrix
$\left(  l_{jk}\right)  $\textquotedblright\ by \textquotedblleft A a $d\times
d$-matrix $L = \left(  l_{kj}\right)  _{1\leq j,k\leq d} \in K^{d \times d}%
$\textquotedblright. (In particular, I have replaced the matrix $\left(
l_{jk}\right)  $ by $\left(  l_{kj}\right)  $ in order for the action to be
left-associative.)} $L = \left(  l_{kj}\right)  _{1\leq j,k\leq d} \in K^{d
\times d}$ acts upon $P$ as an algebra homomorphism as follows:%
\[
L\left(  x_{i}\mid u_{j}\right)  =\sum_{1\leq k\leq d}l_{jk}\left(  x_{i}\mid
u_{k}\right)  ,\ \ \ \ \ \ \ \ \ \ 1\leq i\leq n,\ \ \ \ \ \ \ \ \ \ 1\leq
j\leq d.
\]
A form $F$ is \textit{invariant} if, for all invertible $d \times d$-matrices
$L \in K^{d \times d}$, there exists a scalar $a\left(  L\right)  $ such that
$LF=a\left(  L\right)  F$.

An example of an invariant form is the inner product $\left(  x_{i_{1}}\cdots
x_{i_{d}}\mid u_{1}\cdots u_{d}\right)  $%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\left(
x_{i_{1}}\cdots d_{i_{d}}\mid u_{1}\cdots u_{d}\right)  $\textquotedblright%
\ by \textquotedblleft$\left(  x_{i_{1}}\cdots x_{i_{d}}\mid u_{1}\cdots
u_{d}\right)  $\textquotedblright.}; it is the determinant $\left\vert \left(
x_{i_{j}}\mid u_{k}\right)  \right\vert _{1\leq j,k\leq d}$, and in this case
$a\left(  L\right)  =\det\left(  l_{jk}\right)  $%
\ \ \ \ \footnote{\textit{Comment:} Let me show why this is true:
\par
Let $F=\left(  x_{i_{1}}\cdots x_{i_{d}}\mid u_{1}\cdots u_{d}\right)  $ for
some choice of indices $i_{1},\ldots,i_{d}\in\left\{  1,2,\ldots,n\right\}  $.
Let $L=\left(  l_{kj}\right)  _{1\leq j,k\leq d}$ be any $d\times d$-matrix.
(It needs not be invertible.) Set $a\left(  L\right)  =\det\left(
l_{jk}\right)  $. We must then prove that $LF=a\left(  L\right)  F$.
\par
From $L=\left(  l_{kj}\right)  _{1\leq j,k\leq d}$, we obtain $L^{T}=\left(
l_{jk}\right)  _{1\leq j,k\leq d}$, so that $\det\left(  L^{T}\right)
=\det\left(  l_{jk}\right)  _{1\leq j,k\leq d}=\det\left(  l_{jk}\right)  $.
Hence, $\det\left(  l_{jk}\right)  =\det\left(  L^{T}\right)  =\det
L=\det\left(  \left(  l_{kj}\right)  _{1\leq j,k\leq d}\right)  $ (since
$L=\left(  l_{kj}\right)  _{1\leq j,k\leq d}$).
\par
We have $F=\left(  x_{i_{1}}\cdots x_{i_{d}}\mid u_{1}\cdots u_{d}\right)
=\left\vert \left(  x_{i_{j}}\mid u_{k}\right)  \right\vert _{1\leq j,k\leq
d}$ and thus%
\begin{align*}
LF  &  =L\left\vert \left(  x_{i_{j}}\mid u_{k}\right)  \right\vert _{1\leq
j,k\leq d}\\
&  =\left\vert L\left(  x_{i_{j}}\mid u_{k}\right)  \right\vert _{1\leq
j,k\leq d}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }L\text{ acts on }P\text{ by
an algebra homomorphism}\right) \\
&  =\left\vert \underbrace{L\left(  x_{i_{j}}\mid u_{r}\right)  }%
_{\substack{=\sum_{1\leq k\leq d}l_{rk}\left(  x_{i_{j}}\mid u_{k}\right)
\\\text{(by the definition of the action of }L\text{)}}}\right\vert _{1\leq
j,r\leq d}=\left\vert \sum_{1\leq k\leq d}l_{rk}\left(  x_{i_{j}}\mid
u_{k}\right)  \right\vert _{1\leq j,r\leq d}\\
&  =\det\left(  \underbrace{\left(  \sum_{1\leq k\leq d}l_{rk}\left(
x_{i_{j}}\mid u_{k}\right)  \right)  _{1\leq j,r\leq d}}_{=\left(  \left(
x_{i_{j}}\mid u_{k}\right)  \right)  _{1\leq j,k\leq d}\cdot\left(
l_{kj}\right)  _{1\leq j,k\leq d}}\right)  =\det\left(  \left(  \left(
x_{i_{j}}\mid u_{k}\right)  \right)  _{1\leq j,k\leq d}\cdot\left(
l_{kj}\right)  _{1\leq j,k\leq d}\right) \\
&  =\underbrace{\det\left(  \left(  x_{i_{j}}\mid u_{k}\right)  _{1\leq
j,k\leq d}\right)  }_{=\left\vert \left(  x_{i_{j}}\mid u_{k}\right)
\right\vert _{1\leq j,k\leq d}=F}\cdot\underbrace{\det\left(  \left(
l_{kj}\right)  _{1\leq j,k\leq d}\right)  }_{=\det\left(  l_{jk}\right)
=a\left(  L\right)  }\\
&  =F\cdot a\left(  L\right)  =a\left(  L\right)  F,
\end{align*}
qed.}. Similarly, the bideterminants of shape $\left(  d,\ldots,d\right)  $
are invariant forms\footnote{\textit{Comment:} This holds for the following
reason: Let $\left[  T,T^{\prime}\right]  $ be a bitableau of shape $\left(
d,\ldots,d\right)  $. Then, we claim that the bideterminant $\left(  T\mid
T^{\prime}\right)  $ is an invariant form.
\par
\textit{Proof of the claim:} We are in one of the following two cases:
\par
\begin{statement}
\textit{Case 1:} Each row of $T^{\prime}$ consists of the $d$ entries
$u_{1},\ldots,u_{d}$ (possibly reordered).
\par
\textit{Case 2:} At least one row of $T^{\prime}$ does not consist of the $d$
entries $u_{1},\ldots,u_{d}$ (possibly reordered).
\end{statement}
\par
Let us first consider Case 1. In this case, each row of $T^{\prime}$ consists
of the $d$ entries $u_{1},\ldots,u_{d}$ (possibly reordered). Hence, upon
reordering the entries in the rows of $T^{\prime}$, we can ensure that each
row of $T^{\prime}$ has the form $u_{1}\cdots u_{d}$. Thus, the bideterminant
$\left(  T\mid T^{\prime}\right)  $ is a product of inner products of the form
of the form $\left(  x_{i_{1}}\cdots x_{i_{d}}\mid u_{1}\cdots u_{d}\right)
$. Since all inner products of the latter form are invariant forms, we
therefore conclude that the bideterminant $\left(  T\mid T^{\prime}\right)  $
is an invariant form as well (since any product of invariant forms must be an
invariant form). Thus, the claim is proven in Case 1.
\par
Let us now consider Case 2. In this case, at least one row of $T^{\prime}$
does not consist of the $d$ entries $u_{1},\ldots,u_{d}$ (possibly reordered).
Hence, this row must have two equal entries (because it has $d$ entries chosen
from the set $\mathcal{U}=\left\{  u_{1},\ldots,u_{d}\right\}  $, and thus it
either consists of the $d$ entries $u_{1},\ldots,u_{d}$ or has two equal
entries). Consequently, the inner product in $\left(  T\mid T^{\prime}\right)
$ corresponding to this row is $0$. Therefore, the bideterminant $\left(
T\mid T^{\prime}\right)  $ is $0$. Thus, of course, $\left(  T\mid T^{\prime
}\right)  $ is an invariant form (since $0$ is an invariant form). Hence, the
claim is proven in Case 2.
\par
We thus have proven the claim in both possible cases.}; these bideterminants
are called \textit{rectangular}. For a rectangular bideterminant with $g$
rows, $a\left(  L\right)  =\left(  \det\left(  l_{jk}\right)  \right)  ^{g}$.
Note that since any bideterminant of shape longer than $\left(  d\right)  $ is
zero, a rectangular bideterminant is a linear combination of standard
rectangular bideterminants (with the same number of
rows)\footnote{\textit{Comment:} This is a consequence of Theorem \ref{thm.1}.
For a more detailed proof, see Lemma \ref{lem.bracket} further below.}.

These examples are in fact paradigmatic.\footnote{\textit{Correction:}
Replaced \textquotedblleft has\textquotedblright\ by \textquotedblleft
have\textquotedblright\ in the following theorem. Also, removed the
\textquotedblleft all of which have the same shape $\left(  d,\ldots,d\right)
$\textquotedblright\ part, since it only holds for homogeneous invariants.}

\begin{theorem}
[the first fundamental theorem of invariant theory]\label{thm.3.2}Over an
infinite field $K$, a form in $P$ is invariant iff it is a linear combination
of standard rectangular bideterminants.
\end{theorem}

Note that the \textquotedblleft if\textquotedblright\ part of Theorem
\ref{thm.3.2} is clearly true (since rectangular bideterminants are always
invariant). Thus, only the \textquotedblleft only if\textquotedblright\ part
remains to be proven.\footnote{\textit{Correction:} Added the preceding two
sentences.}

\subsection{\label{subsect.invar.fft1}The First Fundamental Theorem: first
proof}

Before we prove this theorem in two different ways, let us see how certain
matrices act on inner products:\footnote{\textit{Correction:} I have added
this sentence, the next lemma and its proof, in order to make the subsequent
first proof of Theorem \ref{thm.3.2} clearer.}

\begin{lemma}
\label{lem.3.2.a}Let $j\in\left\{  2,3,\ldots,d\right\}  $. Let $L$ be the the
$d\times d$-matrix $\left(  l_{qp}\right)  _{1\leq p,q\leq d}$, where%
\begin{align*}
l_{j-1,j}  &  =1;\\
l_{pp}  &  =1\ \ \ \ \ \ \ \ \ \ \text{for all }p;\\
l_{qp}  &  =0\ \ \ \ \ \ \ \ \ \ \text{for all }p,q\text{ satisfying }p\neq
q\text{ and }\left(  p,q\right)  \neq\left(  j,j-1\right)  .
\end{align*}
Thus, $L$ acts upon $P$ as an algebra homomorphism.

Let $p\in\mathbb{N}$. Let $x_{i_{1}},\ldots,x_{i_{p}}$ be any $p$ letters in
$\mathcal{X}$, and let $u_{j_{1}},\ldots,u_{j_{p}}$ be $p$ distinct letters in
$\mathcal{U}$.

\textbf{(a)} If no $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies $j_{g}%
=j-1$, then
\[
L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  .
\]


\textbf{(b)} If some $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies
$j_{g}=j-1$, then this $g$ satisfies%
\begin{align*}
&  L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right) \\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
+\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}}%
u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  .
\end{align*}


\textbf{(c)} If some $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies
$j_{g}=j-1$, and if some $h\in\left\{  1,2,\ldots,p\right\}  $ satisfies
$j_{h}=j$, then%
\[
L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.3.2.a}.]The action of $L$ on $P$ satisfies the
equality%
\begin{equation}
L\left(  x_{i}\mid u_{j-1}\right)  =\left(  x_{i}\mid u_{j-1}\right)  +\left(
x_{i}\mid u_{j}\right)  \label{pf.lem.3.2.a.L1}%
\end{equation}
for all $i\in\left\{  1,2,\ldots,n\right\}  $, as well as%
\begin{equation}
L\left(  x_{i}\mid u_{k}\right)  =\left(  x_{i}\mid u_{k}\right)
\label{pf.lem.3.2.a.L2}%
\end{equation}
for all $i\in\left\{  1,2,\ldots,n\right\}  $ and $k\in\left\{  1,2,\ldots
,d\right\}  $ satisfying $k\neq j-1$.

Applying the action of $L$ to the equality (\ref{eq.inner-prod-as-det}), we
obtain%
\begin{align}
&  L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
\nonumber\\
&  =L\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert \nonumber\\
&  =\left\vert
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert \label{pf.lem.3.2.a.4}%
\end{align}
(since $L$ acts upon $P$ as an algebra homomorphism).

\textbf{(b)} Assume that some $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies
$j_{g}=j-1$. Consider this $g$.

The equality (\ref{pf.lem.3.2.a.L1}) rewrites as%
\begin{equation}
L\left(  x_{i}\mid u_{j_{g}}\right)  =\left(  x_{i}\mid u_{j_{g}}\right)
+\left(  x_{i}\mid u_{j}\right)  \label{pf.lem.3.2.a.b.1}%
\end{equation}
(since $j-1=j_{g}$). Furthermore, every $h\in\left\{  1,2,\ldots,p\right\}  $
satisfying $h\neq g$ satisfies%
\begin{equation}
L\left(  x_{i}\mid u_{j_{h}}\right)  =\left(  x_{i}\mid u_{j_{h}}\right)
\label{pf.lem.3.2.a.b.2}%
\end{equation}
\footnote{\textit{Proof:} Let $h\in\left\{  1,2,\ldots,p\right\}  $ be such
that $h\neq g$. The numbers $j_{1},\ldots,j_{p}$ are distinct (since
$u_{j_{1}},\ldots,u_{j_{p}}$ are distinct letters). Thus, from $h\neq g$, we
obtain $j_{h}\neq j_{g}=j-1$. Hence, (\ref{pf.lem.3.2.a.L2}) (applied to
$k=j_{h}$) shows that $L\left(  x_{i}\mid u_{j_{h}}\right)  =\left(  x_{i}\mid
u_{j_{h}}\right)  $. This proves (\ref{pf.lem.3.2.a.b.2}).}.

Applying the equality (\ref{eq.inner-prod-as-det}) to $\left(  j_{1}%
,\ldots,j_{g-1},j,j_{g+1},\ldots,j_{p}\right)  $ instead of $\left(
j_{1},\ldots,j_{p}\right)  $, we obtain%
\begin{align}
&  \left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}}%
u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right) \nonumber\\
&  =\left\vert
\begin{array}
[c]{ccccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j}\right)  & \cdots & \left(  x_{i_{1}}\mid u_{j_{p}}\right) \\
\vdots & \ddots & \vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j}\right)  & \cdots & \left(  x_{i_{p}}\mid u_{j_{p}}\right)
\end{array}
\right\vert , \label{pf.lem.3.2.a.b.4}%
\end{align}
where the matrix on the right hand side is to be understood as the $p\times
p$-matrix whose $\left(  s,t\right)  $-th entry is $%
\begin{cases}
\left(  x_{i_{s}}\mid u_{j_{t}}\right)  , & \text{if }t\neq g;\\
\left(  x_{i_{s}}\mid u_{j}\right)  , & \text{if }t=g
\end{cases}
$.

But the equalities (\ref{pf.lem.3.2.a.b.1}) and (\ref{pf.lem.3.2.a.b.2}) show
that the matrix $\left[
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right]  $ differs from the matrix $\left[
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right]  $ only in its $g$-th column, which is $\left[
\begin{array}
[c]{c}%
\left(  x_{i_{1}}\mid u_{j_{g}}\right)  +\left(  x_{i_{1}}\mid u_{j}\right) \\
\vdots\\
\left(  x_{i_{p}}\mid u_{j_{g}}\right)  +\left(  x_{i_{p}}\mid u_{j}\right)
\end{array}
\right]  $. Since the determinant of a matrix is multilinear in its columns
(in particular, linear in its $g$-th column), we thus conclude that the
determinant of the matrix $\left[
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right]  $ can be expressed as follows:%
\begin{align*}
&  \left\vert
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert \\
&  =\underbrace{\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert }_{\substack{=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}%
}\cdots u_{j_{p}}\right)  \\\text{(by (\ref{eq.inner-prod-as-det}))}%
}}+\underbrace{\left\vert
\begin{array}
[c]{ccccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j}\right)  & \cdots & \left(  x_{i_{1}}\mid u_{j_{p}}\right) \\
\vdots & \ddots & \vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j}\right)  & \cdots & \left(  x_{i_{p}}\mid u_{j_{p}}\right)
\end{array}
\right\vert }_{\substack{=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}%
}\cdots u_{j_{g-1}}u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  \\\text{(by
(\ref{pf.lem.3.2.a.b.4}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{where the second matrix is the same as the}\\
\text{matrix on the right hand side of (\ref{pf.lem.3.2.a.b.4})}%
\end{array}
\right) \\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
+\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}}%
u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  .
\end{align*}
Hence, (\ref{pf.lem.3.2.a.4}) becomes%
\begin{align*}
&  L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right) \\
&  =\left\vert
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert \\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
+\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}}%
u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  .
\end{align*}
This proves Lemma \ref{lem.3.2.a} \textbf{(b)}.

\textbf{(c)} Assume that some $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies
$j_{g}=j-1$. Consider this $g$.

Assume that some $h\in\left\{  1,2,\ldots,p\right\}  $ satisfies $j_{h}=j$.
Consider this $h$.

We have $j_{g}=j-1\neq j=j_{h}$ and thus $g\neq h$. Hence, $h\neq g$; this
shows that $j_{h}$ is one of the numbers $j_{1},\ldots,j_{g-1},j_{g+1}%
,\ldots,j_{p}$. Hence, two of the numbers $j_{1},\ldots,j_{g-1},j,j_{g+1}%
,\ldots,j_{p}$ are equal (namely, $j_{h}$ and $j$). Thus, two of the letters
$u_{j_{1}},\ldots,u_{j_{g-1}},u_{j},u_{j_{g+1}},\ldots,u_{j_{p}}$ are equal.
Therefore, the inner product
\[
\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}}%
u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  =0
\]
(since the inner product is antisymmetric in the $u$'s). Now, Lemma
\ref{lem.3.2.a} \textbf{(b)} yields%
\begin{align*}
&  L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right) \\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
+\underbrace{\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}%
}u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  }_{=0}\\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  .
\end{align*}
This proves Lemma \ref{lem.3.2.a} \textbf{(c)}.

\textbf{(a)} Assume that no $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies
$j_{g}=j-1$. Then, every $h\in\left\{  1,2,\ldots,p\right\}  $ satisfies%
\begin{equation}
L\left(  x_{i}\mid u_{j_{h}}\right)  =\left(  x_{i}\mid u_{j_{h}}\right)
\label{pf.lem.3.2.a.a.2}%
\end{equation}
\footnote{\textit{Proof:} Let $h\in\left\{  1,2,\ldots,p\right\}  $. Recall
that no $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies $j_{g}=j-1$. In other
words, each $g\in\left\{  1,2,\ldots,p\right\}  $ satisfies $j_{g}\neq j-1$.
Applying this to $g=h$, we obtain $j_{h}\neq j-1$. Hence,
(\ref{pf.lem.3.2.a.L2}) (applied to $k=j_{h}$) shows that $L\left(  x_{i}\mid
u_{j_{h}}\right)  =\left(  x_{i}\mid u_{j_{h}}\right)  $. This proves
(\ref{pf.lem.3.2.a.a.2}).}. Thus,%
\begin{equation}
\left[
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right]  =\left[
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right]  . \label{pf.lem.3.2.a.a.2m}%
\end{equation}


Now, (\ref{pf.lem.3.2.a.4}) becomes%
\begin{align*}
&  L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right) \\
&  =\left\vert
\begin{array}
[c]{ccc}%
L\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
L\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & L\left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert \\
&  =\left\vert
\begin{array}
[c]{ccc}%
\left(  x_{i_{1}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{1}}\mid
u_{j_{p}}\right) \\
\vdots & \ddots & \vdots\\
\left(  x_{i_{p}}\mid u_{j_{1}}\right)  & \cdots & \left(  x_{i_{p}}\mid
u_{j_{p}}\right)
\end{array}
\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.3.2.a.a.2m}%
)}\right) \\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.inner-prod-as-det})}\right)  .
\end{align*}


This proves Lemma \ref{lem.3.2.a} \textbf{(a)}.
\end{proof}

A further lemma will help us analyze bitableaux:\footnote{\textit{Correction:}
I have added this sentence, the next lemma and its proof, in order to make the
subsequent first proof of Theorem \ref{thm.3.2} clearer.}

\begin{lemma}
\label{lem.3.2.b}Let $T$ be a standard tableau with values in the alphabet
$\mathcal{U}$. Let $j>1$ be an integer. Let $g$ be a positive integer.

An entry $u_{j-1}$ in $T$ shall be called \textit{malleable} if the letter
$u_{j}$ does not appear in the same row of $T$ as this entry.

Let $\widetilde{T}$ be the tableau obtained from $T$ by replacing each
malleable entry $u_{j-1}$ in the $g$-th column of $T$ by $u_{j}$.

Then, the tableau $\widetilde{T}$ is again standard.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.3.2.b}.]\footnote{This proof is here purely for the
sake of completeness. It is a long-winded writeup of a fairly straightforward
combinatorial argument, which you will probably have found faster than you can
read my proof.} Recall that the entries of $T$ and $\widetilde{T}$ belong to
the totally ordered set $\mathcal{U}$.

The tableau $T$ is standard; in other words, the entries in each row of $T$
are increasing from left to right, and the entries in each column of $T$ are
nondecreasing downward. We must prove the same for the tableau $\widetilde{T}$.

Let $\lambda_{1},\lambda_{2},\lambda_{3},\ldots$ be the lengths of the rows of
$T$ (from top to bottom), and let $\lambda_{1}^{\prime},\lambda_{2}^{\prime
},\lambda_{3}^{\prime},\ldots$ be the lengths of the columns of $T$ (from left
to right). Thus, of course, $\lambda_{1},\lambda_{2},\lambda_{3},\ldots$ are
also the lengths of the rows of $\widetilde{T}$, whereas $\lambda_{1}^{\prime
},\lambda_{2}^{\prime},\lambda_{3}^{\prime},\ldots$ are also the lengths of
the columns of $\widetilde{T}$.

Recall that the tableau $\widetilde{T}$ is obtained from $T$ by replacing each
malleable entry $u_{j-1}$ in the $g$-th column of $T$ by $u_{j}$. Thus, each
entry of $\widetilde{T}$ either equals the corresponding entry of $T$, or is
larger than it (because the only entries that were replaced were entries
$u_{j-1}$, and these were replaced by the larger letter $u_{j}$). In other
words, each cell $\left(  p,q\right)  $ of $T$ satisfies%
\begin{equation}
\widetilde{T}\left(  p,q\right)  \geq T\left(  p,q\right)  .
\label{pf.lem.3.2.b.larger}%
\end{equation}


Recall again that the tableau $\widetilde{T}$ is obtained from $T$ by
replacing each malleable entry $u_{j-1}$ in the $g$-th column of $T$ by
$u_{j}$. Thus, each malleable entry $u_{j-1}$ in the $g$-th column of $T$ was
replaced by $u_{j}$. In other words, if some cell $\left(  p,q\right)  $ in
the $g$-th column of $T$ contains a malleable entry $u_{j-1}$, then the
corresponding entry of $\widetilde{T}$ is $u_{j}$. In other words, if some
cell $\left(  p,q\right)  $ in the $g$-th column of $T$ contains a malleable
entry $u_{j-1}$, then%
\begin{equation}
\widetilde{T}\left(  p,q\right)  =u_{j}. \label{pf.lem.3.2.b.malle}%
\end{equation}
(Of course, under this condition, we must have $q=g$; but we nevertheless
chose to call the cell $\left(  p,q\right)  $ and not $\left(  p,g\right)  $
just for the sake of uniform notations.)

Recall again that the tableau $\widetilde{T}$ is obtained from $T$ by
replacing each malleable entry $u_{j-1}$ in the $g$-th column of $T$ by
$u_{j}$. Thus, each entry of $T$ has been preserved unchanged in
$\widetilde{T}$ unless it was a malleable entry $u_{j-1}$ in the $g$-th column
of $T$. In other words, if $\left(  p,q\right)  $ is a cell of $T$, then%
\begin{equation}
\widetilde{T}\left(  p,q\right)  =T\left(  p,q\right)  ,
\label{pf.lem.3.2.b.nonmalle}%
\end{equation}
unless the entry of $T$ in cell $\left(  p,q\right)  $ is a malleable entry
$u_{j-1}$ in the $g$-th column of $T$.

Let us make the following simple observation:

\begin{statement}
\textit{Observation 0:} Let $\left(  p,q\right)  $ be a cell of $T$ satisfying
$T\left(  p,q\right)  \neq\widetilde{T}\left(  p,q\right)  $. Then:

\begin{itemize}
\item We have $q=g$ and $T\left(  p,q\right)  =u_{j-1}$ and $\widetilde{T}%
\left(  p,q\right)  =u_{j}$.

\item Furthermore, the letter $u_{j}$ does not appear in the $p$-th row of $T$.
\end{itemize}
\end{statement}

[\textit{Proof of Observation 0:} We have $T\left(  p,q\right)  \neq
\widetilde{T}\left(  p,q\right)  $. In other words, $\widetilde{T}\left(
p,q\right)  \neq T\left(  p,q\right)  $.

If the entry of $T$ in cell $\left(  p,q\right)  $ was not a malleable entry
$u_{j-1}$ in the $g$-th column of $T$, then we would have $\widetilde{T}%
\left(  p,q\right)  =T\left(  p,q\right)  $ (by (\ref{pf.lem.3.2.b.nonmalle}%
)); but this would contradict $\widetilde{T}\left(  p,q\right)  \neq T\left(
p,q\right)  $. Hence, the entry of $T$ in cell $\left(  p,q\right)  $ is a
malleable entry $u_{j-1}$ in the $g$-th column of $T$.

Thus, in particular, the entry of $T$ in cell $\left(  p,q\right)  $ lies in
the $g$-th column of $T$. In other words, cell $\left(  p,q\right)  $ lies in
the $g$-th column. In other words, $q=g$.

Moreover, the entry of $T$ in cell $\left(  p,q\right)  $ is a malleable entry
$u_{j-1}$ in the $g$-th column of $T$. In particular, this entry is $u_{j-1}$.
In other words, $T\left(  p,q\right)  =u_{j-1}$.

The cell $\left(  p,q\right)  $ in the $g$-th column of $T$ contains a
malleable entry $u_{j-1}$ (since the entry of $T$ in cell $\left(  p,q\right)
$ is a malleable entry $u_{j-1}$ in the $g$-th column of $T$). Thus,
(\ref{pf.lem.3.2.b.malle}) shows that $\widetilde{T}\left(  p,q\right)
=u_{j}$.

Recall that the entry of $T$ in cell $\left(  p,q\right)  $ is a malleable
entry $u_{j-1}$ in the $g$-th column of $T$. In particular, this entry is
malleable. In other words, the letter $u_{j}$ does not appear in the same row
of $T$ as this entry (by the definition of \textquotedblleft
malleable\textquotedblright). In other words, the letter $u_{j}$ does not
appear in the $p$-th row of $T$ (since clearly, the row of $T$ that contains
the entry in cell $\left(  p,q\right)  $ must be the $p$-th row of $T$). Thus,
the proof of Observation 0 is complete.]

We know that the entries of each row of $T$ are increasing from left to right.
In other words, each positive integer $p$ satisfies%
\begin{equation}
T\left(  p,1\right)  <T\left(  p,2\right)  <\cdots<T\left(  p,\lambda
_{p}\right)  . \label{pf.lem.3.2.b.Trow}%
\end{equation}


We know that the entries of each column of $T$ are nondecreasing downward. In
other words, each positive integer $q$ satisfies%
\begin{equation}
T\left(  1,q\right)  \leq T\left(  2,q\right)  \leq\cdots\leq T\left(
\lambda_{q}^{\prime},q\right)  . \label{pf.lem.3.2.b.Tcol}%
\end{equation}


We now make the following two claims:

\begin{statement}
\textit{Claim 1:} Each positive integer $p$ satisfies%
\[
\widetilde{T}\left(  p,1\right)  <\widetilde{T}\left(  p,2\right)
<\cdots<\widetilde{T}\left(  p,\lambda_{p}\right)  .
\]

\end{statement}

\begin{statement}
\textit{Claim 2:} Each positive integer $q$ satisfies%
\[
\widetilde{T}\left(  1,q\right)  \leq\widetilde{T}\left(  2,q\right)
\leq\cdots\leq\widetilde{T}\left(  \lambda_{q}^{\prime},q\right)  .
\]

\end{statement}

[\textit{Proof of Claim 1:} Let $p$ be a positive integer. Let $q\in\left\{
1,2,\ldots,\lambda_{p}-1\right\}  $. We are going to show that $\widetilde{T}%
\left(  p,q\right)  <\widetilde{T}\left(  p,q+1\right)  $.

Indeed, assume the contrary. Thus, $\widetilde{T}\left(  p,q\right)
\geq\widetilde{T}\left(  p,q+1\right)  $.

However, $\left(  p,q+1\right)  $ is also a cell of $T$ (since $q\in\left\{
1,2,\ldots,\lambda_{p}-1\right\}  $). Thus, (\ref{pf.lem.3.2.b.larger})
(applied to $\left(  p,q+1\right)  $ instead of $\left(  p,q\right)  $) yields
$\widetilde{T}\left(  p,q+1\right)  \geq T\left(  p,q+1\right)  $. Hence,
$\widetilde{T}\left(  p,q\right)  \geq\widetilde{T}\left(  p,q+1\right)  \geq
T\left(  p,q+1\right)  $.

But (\ref{pf.lem.3.2.b.Trow}) shows that $T\left(  p,q\right)  <T\left(
p,q+1\right)  \leq\widetilde{T}\left(  p,q\right)  $ (since $\widetilde{T}%
\left(  p,q\right)  \geq T\left(  p,q+1\right)  $). Hence, $T\left(
p,q\right)  \neq\widetilde{T}\left(  p,q\right)  $. Thus, Observation 0 shows that:

\begin{itemize}
\item We have $q=g$ and $T\left(  p,q\right)  =u_{j-1}$ and $\widetilde{T}%
\left(  p,q\right)  =u_{j}$.

\item Furthermore, the letter $u_{j}$ does not appear in the $p$-th row of $T$.
\end{itemize}

From $T\left(  p,q\right)  =u_{j-1}$, we obtain $u_{j-1}=T\left(  p,q\right)
<T\left(  p,q+1\right)  $. Therefore, $T\left(  p,q+1\right)  >u_{j-1}$, so
that $T\left(  p,q+1\right)  =u_{h}$ for some $h\in\left\{  j,j+1,\ldots
,d\right\}  $. Consider this $h$. We have $h\geq j$ (since $h\in\left\{
j,j+1,\ldots,d\right\}  $). But we have $\widetilde{T}\left(  p,q\right)
=u_{j}$. Hence, $u_{j}=\widetilde{T}\left(  p,q\right)  \geq T\left(
p,q+1\right)  =u_{h}$. In other words, $j\geq h$. Combined with $h\geq j$,
this yields $h=j$. Hence, $u_{h}=u_{j}$. Thus, $T\left(  p,q+1\right)
=u_{h}=u_{j}$. Thus, the letter $u_{j}$ appears in the $p$-th row of $T$
(namely, in cell $\left(  p,q+1\right)  $). This contradicts the fact that the
letter $u_{j}$ does not appear in the $p$-th row of $T$.

This contradiction shows that our assumption was false. Hence, $\widetilde{T}%
\left(  p,q\right)  <\widetilde{T}\left(  p,q+1\right)  $ is proven.

Now, forget that we fixed $q$. We thus have shown that $\widetilde{T}\left(
p,q\right)  <\widetilde{T}\left(  p,q+1\right)  $ for each $q\in\left\{
1,2,\ldots,\lambda_{p}-1\right\}  $. In other words, $\widetilde{T}\left(
p,1\right)  <\widetilde{T}\left(  p,2\right)  <\cdots<\widetilde{T}\left(
p,\lambda_{p}\right)  $. This proves Claim 1.]

[\textit{Proof of Claim 2:} Let $q$ be a positive integer. Let $p\in\left\{
1,2,\ldots,\lambda_{q}^{\prime}-1\right\}  $. We are going to show that
$\widetilde{T}\left(  p,q\right)  \leq\widetilde{T}\left(  p+1,q\right)  $.

Indeed, assume the contrary. Thus, $\widetilde{T}\left(  p,q\right)
>\widetilde{T}\left(  p+1,q\right)  $.

However, $\left(  p+1,q\right)  $ is also a cell of $T$ (since $p\in\left\{
1,2,\ldots,\lambda_{q}^{\prime}-1\right\}  $). Thus,
(\ref{pf.lem.3.2.b.larger}) (applied to $\left(  p+1,q\right)  $ instead of
$\left(  p,q\right)  $) yields $\widetilde{T}\left(  p+1,q\right)  \geq
T\left(  p+1,q\right)  $. Hence, $\widetilde{T}\left(  p,q\right)
>\widetilde{T}\left(  p+1,q\right)  \geq T\left(  p+1,q\right)  $.

But (\ref{pf.lem.3.2.b.Tcol}) shows that $T\left(  p,q\right)  \leq T\left(
p+1,q\right)  <\widetilde{T}\left(  p,q\right)  $ (since $\widetilde{T}\left(
p,q\right)  >T\left(  p+1,q\right)  $). Hence, $T\left(  p,q\right)
\neq\widetilde{T}\left(  p,q\right)  $. Thus, Observation 0 shows that:

\begin{itemize}
\item We have $q=g$ and $T\left(  p,q\right)  =u_{j-1}$ and $\widetilde{T}%
\left(  p,q\right)  =u_{j}$.

\item Furthermore, the letter $u_{j}$ does not appear in the $p$-th row of $T$.
\end{itemize}

From $T\left(  p,q\right)  =u_{j-1}$, we obtain $u_{j-1}=T\left(  p,q\right)
\leq T\left(  p+1,q\right)  $. Therefore, $T\left(  p+1,q\right)  \geq
u_{j-1}$, so that $T\left(  p+1,q\right)  =u_{h}$ for some $h\in\left\{
j-1,j,\ldots,d\right\}  $. Consider this $h$. We have $h\geq j-1$ (since
$h\in\left\{  j-1,j,\ldots,d\right\}  $). But $u_{h}=T\left(  p+1,q\right)
<\widetilde{T}\left(  p,q\right)  =u_{j}$. In other words, $h<j$. Hence,
$h\leq j-1$. Combining this with $h\geq j-1$, we obtain $h=j-1$. Now,
$T\left(  p+1,q\right)  =u_{h}=u_{j-1}$ (since $h=j-1$). Thus, there is an
entry $u_{j-1}$ in cell $\left(  p+1,q\right)  $ of $T$.

The letter $u_{j}$ does not appear in the $\left(  p+1\right)  $-st row of
$T$\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, the letter
$u_{j}$ appears in the $\left(  p+1\right)  $-st row of $T$. In other words,
there exists an $r\in\left\{  1,2,\ldots,\lambda_{p+1}\right\}  $ such that
$T\left(  p+1,r\right)  =u_{j}$. Consider this $r$.
\par
Applying (\ref{pf.lem.3.2.b.Trow}) to $p+1$ instead of $p$, we obtain
$T\left(  p+1,1\right)  <T\left(  p+1,2\right)  <\cdots<T\left(
p+1,\lambda_{p+1}\right)  $. Hence, if we had $r\leq q$, then we would have
$T\left(  p+1,r\right)  \leq T\left(  p+1,q\right)  =u_{j-1}$, which would
contradict $T\left(  p+1,r\right)  =u_{j}>u_{j-1}$. Thus, we cannot have
$r\leq q$. Hence, we have $r>q$. In other words, $q<r$.
\par
Thus, the cell $\left(  p,r\right)  $ lies \textquotedblleft
between\textquotedblright\ the cells $\left(  p,q\right)  $ and $\left(
p+1,r\right)  $ (in the sense that there is a shortest northwest-to-southeast
lattice path from $\left(  p,q\right)  $ to $\left(  p+1,r\right)  $ passing
through $\left(  p,r\right)  $). Thus, the tableau $T$ must have a cell
$\left(  p,r\right)  $ (since $T$ has a cell $\left(  p,q\right)  $ and a cell
$\left(  p+1,r\right)  $). Now, from (\ref{pf.lem.3.2.b.Trow}), we obtain
$T\left(  p,q\right)  <T\left(  p,r\right)  $ (since $q<r$). Hence, $T\left(
p,r\right)  >T\left(  p,q\right)  =u_{j-1}$.
\par
But (\ref{pf.lem.3.2.b.Tcol}) (applied to $r$ instead of $q$) yields $T\left(
1,r\right)  \leq T\left(  2,r\right)  \leq\cdots\leq T\left(  \lambda
_{r}^{\prime},r\right)  $. Hence, $T\left(  p,r\right)  \leq T\left(
p+1,r\right)  =u_{j}$. In other words, $T\left(  p,r\right)  =u_{i}$ for some
$i\in\left\{  1,2,\ldots,j\right\}  $. Consider this $i$.
\par
But $u_{i}=T\left(  p,r\right)  >u_{j-1}$, so that $i>j-1$. In other words,
$i\geq j$. Combining this with $i\leq j$ (since $i\in\left\{  1,2,\ldots
,j\right\}  $), we obtain $i=j$. Hence, $T\left(  p,r\right)  =u_{i}=u_{j}$
(since $i=j$). Hence, the letter $u_{j}$ appears in the $p$-th row of $T$
(namely, in cell $\left(  p,r\right)  $). This contradicts the fact that the
letter $u_{j}$ does not appear in the $p$-th row of $T$. This contradiction
shows that our assumption was false, qed.}. In other words, the letter $u_{j}$
does not appear in the same row of $T$ as the entry of $T$ in cell $\left(
p+1,q\right)  $. Hence, the entry $u_{j-1}$ in cell $\left(  p+1,q\right)  $
of $T$ has the property that the letter $u_{j}$ does not appear in the same
row of $T$ as this entry. In other words, this entry $u_{j-1}$ is malleable
(by the definition of \textquotedblleft malleable\textquotedblright).

Also, $\left(  p+1,q\right)  $ is a cell in the $g$-th column of $T$ (since
$q=g$). Hence, $\left(  p+1,q\right)  $ is a cell in the $g$-th column of $T$
containing a malleable entry $u_{j-1}$ (since the entry $u_{j-1}$ in cell
$\left(  p+1,q\right)  $ of $T$ is malleable). Thus, (\ref{pf.lem.3.2.b.malle}%
) (applied to $\left(  p+1,q\right)  $ instead of $\left(  p,q\right)  $)
yields $\widetilde{T}\left(  p+1,q\right)  =u_{j}$. Now, recall that
$\widetilde{T}\left(  p,q\right)  =u_{j}$; hence, $u_{j}=\widetilde{T}\left(
p,q\right)  >\widetilde{T}\left(  p+1,q\right)  =u_{j}$. This is clearly absurd.

This contradiction shows that our assumption was false. Hence, $\widetilde{T}%
\left(  p,q\right)  \leq\widetilde{T}\left(  p+1,q\right)  $ is proven.

Now, forget that we fixed $p$. We thus have shown that $\widetilde{T}\left(
p,q\right)  \leq\widetilde{T}\left(  p+1,q\right)  $ for each $p\in\left\{
1,2,\ldots,\lambda_{q}^{\prime}-1\right\}  $. In other words, $\widetilde{T}%
\left(  1,q\right)  \leq\widetilde{T}\left(  2,q\right)  \leq\cdots
\leq\widetilde{T}\left(  \lambda_{q}^{\prime},q\right)  $. This proves Claim 2.]

Now, consider the tableau $\widetilde{T}$. The entries in each row of
$\widetilde{T}$ are increasing from left to right (by Claim 1), and the
entries in each column of $\widetilde{T}$ are nondecreasing downward (by Claim
2). In other words, the tableau $\widetilde{T}$ is standard. This proves Lemma
\ref{lem.3.2.b}.
\end{proof}

We shall say that a standard tableau with entries from the alphabet
$\mathcal{U}$ is \textit{rectangular} if and only if it has shape $\left(
d,\ldots,d\right)  $ (for some number of entries $d$, possibly zero). Notice
that such a rectangular tableau must necessarily look like this:%
\[%
\begin{array}
[c]{cccc}%
u_{1} & u_{2} & \cdots & u_{d}\\
\vdots & \vdots & \ddots & \vdots\\
u_{1} & u_{2} & \cdots & u_{d}%
\end{array}
\]
(because each of its rows has length $d$ and is strictly increasing, whence it
has the form $u_{1}u_{2}\cdots u_{d}$). Notice also that a bideterminant
$\left(  T\mid T^{\prime}\right)  $ is rectangular if and only if the tableau
$T^{\prime}$ is rectangular.\footnote{\textit{Correction:} I have added this
paragraph in order to clarify what the word \textquotedblleft
rectangular\textquotedblright\ means when it is applied to a single tableau
(as it is in the following proof).}

\begin{proof}
[First proof of Theorem \ref{thm.3.2}.]Suppose $F$ is invariant. WLOG assume
that $F\neq0$ (otherwise, everything is clear).\footnote{\textit{Correction:}
Added this sentence.} Using the straightening formula, we can express $F$ as a
linear combination of standard bideterminants:%
\begin{equation}
F=\sum_{s}\alpha_{s}\left(  T_{s}\mid T_{s}^{\prime}\right)  ,
\label{pf.thm.3.2.1st.expansion}%
\end{equation}
where the $\left[  T_{s},T_{s}^{\prime}\right]  $ are distinct standard
bitableaux and where the $\alpha_{s}$ are nonzero scalars for all
$s$\ \ \ \ \footnote{\textit{Correction:} Added \textquotedblleft where the
$\left[  T_{s},T_{s}^{\prime}\right]  $ are distinct standard bitableaux and
where the $\alpha_{s}$ are nonzero scalars for all $s$\textquotedblright%
\ (this later becomes important).}. Our goal is to prove that all tableaux
$T_{s}^{\prime}$ appearing in (\ref{pf.thm.3.2.1st.expansion}) are
rectangular.\footnote{\textit{Correction:} Added this sequence.}

Assume the contrary.\footnote{\textit{Correction:} Added this sequence.} We
shall probe $F$ with appropriate $d \times d$-matrices.

Consider first the $d\times d$-matrix $L\in K^{d\times d}$ defined by%
\begin{align}
L\left(  x_{i}\mid u_{j}\right)   &  =c\left(  x_{i}\mid u_{j}\right)
,\label{pf.thm.3.2.L1.def1}\\
L\left(  x_{i}\mid u_{k}\right)   &  =\left(  x_{i}\mid u_{k}\right)
,\ \ \ \ \ \ \ \ \ \ \text{for }k\neq j, \label{pf.thm.3.2.L1.def2}%
\end{align}
where $j$ is a fixed element of $\left\{  1,2,\ldots,d\right\}  $
and\footnote{\textit{Correction:} Added \textquotedblleft where $j$ is a fixed
element of $\left\{  1,2,\ldots,d\right\}  $ and\textquotedblright.} where $c$
is a nonzero scalar\footnote{\textit{Comment:} Here is a more precise way to
say this: Fix a $j\in\left\{  1,2,\ldots,d\right\}  $ and a nonzero scalar
$c$. Let $L$ be the $d\times d$-matrix $\left(  l_{qp}\right)  _{1\leq p,q\leq
d}\in K^{d\times d}$, where%
\begin{align*}
l_{jj}  &  =c;\\
l_{kk}  &  =1\ \ \ \ \ \ \ \ \ \ \text{for }k\neq j;\\
l_{qp}  &  =0\ \ \ \ \ \ \ \ \ \ \text{for }p\neq q.
\end{align*}
Clearly, this $d\times d$-matrix $L$ is invertible (since it is a diagonal
matrix with nonzero entries on the diagonal). It is now straightforward to see
that this $L$ satisfies (\ref{pf.thm.3.2.L1.def1}) and
(\ref{pf.thm.3.2.L1.def2}).}. If $b_{j}^{\left(  s\right)  }$ is the number of
occurrences of $u_{j}$ in $T_{s}^{\prime}$, then%
\[
L\left(  T_{s}\mid T_{s}^{\prime}\right)  =c^{b_{j}^{\left(  s\right)  }%
}\left(  T_{s}\mid T_{s}^{\prime}\right)
\]
and therefore\footnote{\textit{Correction:} Added the word \textquotedblleft
therefore\textquotedblright.}%
\begin{equation}
LF=\sum_{s}\alpha_{s}c^{b_{j}^{\left(  s\right)  }}\left(  T_{s}\mid
T_{s}^{\prime}\right)  . \label{pf.thm.3.2.L1.2}%
\end{equation}
As $F$ is invariant, we also have%
\begin{equation}
LF=a\left(  L\right)  F=\sum_{s}\alpha_{s}a\left(  L\right)  \left(  T_{s}\mid
T_{s}^{\prime}\right)  . \label{pf.thm.3.2.L1.3}%
\end{equation}
But the expansion into standard bideterminants is unique. Hence, comparing
(\ref{pf.thm.3.2.L1.2}) with (\ref{pf.thm.3.2.L1.3}), we
obtain\footnote{\textit{Correction:} Added this sequence for clarity.}%
\begin{equation}
\alpha_{s}c^{b_{j}^{\left(  s\right)  }}=\alpha_{s}a\left(  L\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }s, \label{pf.thm.3.2.L1.4}%
\end{equation}
and therefore%
\begin{equation}
c^{b_{j}^{\left(  s\right)  }}=a\left(  L\right)
\ \ \ \ \ \ \ \ \ \ \text{for all }s \label{pf.thm.3.2.L1.5}%
\end{equation}
(since the $\alpha_{s}$ are nonzero). Hence, we must have, for all $s$ and
$t$,%
\[
c^{b_{j}^{\left(  s\right)  }}=c^{b_{j}^{\left(  t\right)  }}=a\left(
L\right)  .
\]
This equality holds for all the nonzero\footnote{\textit{Correction:} Added
the word \textquotedblleft nonzero\textquotedblright.} scalars $c$ in the
infinite field $K$\ \ \ \ \footnote{\textit{Correction:} Replaced
\textquotedblleft$k$\textquotedblright\ by \textquotedblleft$K$%
\textquotedblright.}. Therefore,%
\begin{equation}
b_{j}^{\left(  s\right)  }=b_{j}^{\left(  t\right)  }%
\ \ \ \ \ \ \ \ \ \ \text{for all }s\text{ and }t \label{pf.thm.3.2.L1.7}%
\end{equation}
\footnote{\textit{Comment:} Here is this argument in more detail:
\par
Comparing (\ref{pf.thm.3.2.L1.2}) with (\ref{pf.thm.3.2.L1.3}), we obtain%
\[
\sum_{s}\alpha_{s}c^{b_{j}^{\left(  s\right)  }}\left(  T_{s}\mid
T_{s}^{\prime}\right)  =\sum_{s}\alpha_{s}a\left(  L\right)  \left(  T_{s}\mid
T_{s}^{\prime}\right)  .
\]
Since the standard bideterminants $\left(  T_{s}\mid T_{s}^{\prime}\right)  $
are $K$-linearly independent (by Theorem \ref{thm.2.2}, since the $\left[
T_{s},T_{s}^{\prime}\right]  $ are distinct standard bitableaux), this shows
that%
\[
\alpha_{s}c^{b_{j}^{\left(  s\right)  }}=\alpha_{s}a\left(  L\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }s.
\]
We can divide this equality by $\alpha_{s}$ (since $\alpha_{s}$ is nonzero),
and thus obtain%
\[
c^{b_{j}^{\left(  s\right)  }}=a\left(  L\right)
\ \ \ \ \ \ \ \ \ \ \text{for each }s.
\]
Hence, $\underbrace{c^{b_{j}^{\left(  s\right)  }}}_{=a\left(  L\right)
}-\underbrace{c^{b_{j}^{\left(  t\right)  }}}_{=a\left(  L\right)  }=a\left(
L\right)  -a\left(  L\right)  =0$ for any $s$ and $t$.
\par
Now, let $s$ and $t$ be arbitrary indices. Then, we have just shown that
$c^{b_{j}^{\left(  s\right)  }}-c^{b_{j}^{\left(  t\right)  }}=0$. We have
proven this for every nonzero scalar $c\in K$. Thus, $c^{b_{j}^{\left(
s\right)  }}-c^{b_{j}^{\left(  t\right)  }}=0$ holds for infinitely many
values of $c\in K$ (since there are infinitely many nonzero scalars $c\in K$).
\par
Now, consider the polynomial $T^{b_{j}^{\left(  s\right)  }}-T^{b_{j}^{\left(
t\right)  }}\in K\left[  T\right]  $. This polynomial has infinitely many
roots in $K$ (since $c^{b_{j}^{\left(  s\right)  }}-c^{b_{j}^{\left(
t\right)  }}=0$ holds for infinitely many values of $c\in K$), and thus must
be identically zero. Therefore, $b_{j}^{\left(  s\right)  }=b_{j}^{\left(
t\right)  }$, qed.}. We shall write $b_{j}$ for the common value of the
integers $b_{j}^{\left(  s\right)  }$. Hence, each tableau $T_{s}^{\prime}$
contains the letter $u_{j}$ exactly $b_{j}$
times.\footnote{\textit{Correction:} Added this sentence.}

Now, fix two distinct elements $j$ and $k$ of $\left\{  1,2,\ldots,d\right\}
$, and\footnote{\textit{Correction:} Added \textquotedblleft fix two distinct
elements $j$ and $k$ of $\left\{  1,2,\ldots,d\right\}  $,
and\textquotedblright.} let $L$ be the $d\times d$-matrix in $K^{d \times d}$
defined by%
\begin{align}
L\left(  x_{i}\mid u_{j}\right)   &  =\left(  x_{i}\mid u_{k}\right)
,\label{pf.thm.3.2.L2.def1}\\
L\left(  x_{i}\mid u_{k}\right)   &  =\left(  x_{i}\mid u_{j}\right)
,\label{pf.thm.3.2.L2.def2}\\
L\left(  x_{i}\mid u_{p}\right)   &  =\left(  x_{i}\mid u_{p}\right)
,\ \ \ \ \ \ \ \ \ \ \text{for }p\neq j\text{ and }p\neq k
\label{pf.thm.3.2.L2.def3}%
\end{align}
\footnote{\textit{Comment:} Here is a more precise way to say this: Fix two
distinct elements $j$ and $k$ of $\left\{  1,2,\ldots,d\right\}  $. Let $L$ be
the $d\times d$-matrix $\left(  l_{qp}\right)  _{1\leq p,q\leq d} \in K^{d
\times d}$, where%
\begin{align*}
l_{jk}  &  =1;\\
l_{jp}  &  =0\ \ \ \ \ \ \ \ \ \ \text{for }p\neq k;\\
l_{kj}  &  =1;\\
l_{kp}  &  =0\ \ \ \ \ \ \ \ \ \ \text{for }p\neq j;\\
l_{pp}  &  =1\ \ \ \ \ \ \ \ \ \ \text{for }p\neq j\text{ and }p\neq k;\\
l_{qp}  &  =0\ \ \ \ \ \ \ \ \ \ \text{for }q\notin\left\{  j,k\right\}
\text{ and }p\neq q.
\end{align*}
Clearly, this matrix $L$ is invertible (since it is a permutation matrix). It
is now straightforward to see that this $L$ satisfies
(\ref{pf.thm.3.2.L2.def1}), (\ref{pf.thm.3.2.L2.def2}) and
(\ref{pf.thm.3.2.L2.def3}).}. Each of the bideterminants $L\left(  T_{s}\mid
T_{s}^{\prime}\right)  $ contains\footnote{\textit{Correction:} Replaced
\textquotedblleft contain\textquotedblright\ by \textquotedblleft
contains\textquotedblright.} the letter $u_{k}$
exactly\footnote{\textit{Correction:} Added the word \textquotedblleft
exactly\textquotedblright.} $b_{j}$ times. As the content of a bideterminant
is preserved under straightening\footnote{\textit{Comment:} This is just a
reference to Theorem \ref{thm.1}.}, $LF$ is
therefore\footnote{\textit{Correction:} Added the word \textquotedblleft
therefore\textquotedblright.} a linear combination of standard bideterminants
each containing the letter $u_{k}$ exactly\footnote{\textit{Correction:} Added
the word \textquotedblleft exactly\textquotedblright.} $b_{j}$ times. But $F$
is invariant, and therefore\footnote{\textit{Correction:} Added the word
\textquotedblleft therefore\textquotedblright.} $LF=\sum_{s}\alpha_{s}a\left(
L\right)  \left(  T_{s}\mid T_{s}^{\prime}\right)  $, where each of the
bideterminants in this expansion contains\footnote{\textit{Correction:}
Replaced \textquotedblleft contain\textquotedblright\ by \textquotedblleft
contains\textquotedblright.} the letter $u_{k}$
exactly\footnote{\textit{Correction:} Added the word \textquotedblleft
exactly\textquotedblright.} $b_{k}$ times. Applying the basis
theorem\footnote{\textit{Comment:} i.e., Corollary \ref{cor.2.2.basis}}, we
conclude that $b_{j}=b_{k}$\ \ \ \ \footnote{\textit{Comment:} Here, we are
using the fact that $a\left(  L\right)  \neq0$. Why is this the case?
\par
Indeed, assume the contrary. Thus, $a\left(  L\right)  =0$. But since $F$ is
invariant, we have $LF=a\left(  L\right)  F$ and $L^{-1}F=a\left(
L^{-1}\right)  F$. Hence, $L^{-1}\underbrace{LF}_{=a\left(  L\right)
F}=\underbrace{a\left(  L\right)  }_{=0}L^{-1}F=0$, which contradicts
$L^{-1}LF=F\neq0$. This contradiction shows that our assumption was false;
hence, $a\left(  L\right)  \neq0$.}. We shall denote by $b$ the common value
of the integers $b_{j}$. Thus,
\begin{equation}
\text{each tableau }T_{s}^{\prime}\text{ contains the letter }u_{j}\text{
exactly }b\text{ times} \label{pf.thm.3.2.pf1.b-times}%
\end{equation}
\footnote{\textit{Correction:} Added this sentence.}.

Since each letter $u_{j}$ is repeated $b$ times, the minimum number of rows in
$T_{s}^{\prime}$ is $b$. The number of rows is exactly $b$ if and only if
$T_{s}^{\prime}$ is rectangular\footnote{\textit{Comment:} Recall that a
standard tableau $T_{s}^{\prime}$ is rectangular if and only if each of its
rows has the form $u_{1}u_{2}\cdots u_{d}$. (Thus, if $T_{s}^{\prime}$ is
rectangular, then the bideterminant $\left(  T_{s}\mid T_{s}^{\prime}\right)
$ is rectangular.)}.

Now, suppose that $T_{s}^{\prime}$ is not rectangular; that is, the number of
rows in $T_{s}^{\prime}$ is strictly greater than $b$. In $T_{s}^{\prime}$,
all the letters $u_{1}$ are in the first column. Let $u_{l}$ be the first
letter in the first column following the run of letters $u_{1}$%
\ \ \ \ \footnote{\textit{Comment:} Such a letter exists, because the run of
letters $u_{1}$ cannot cover the whole first column (since the number of rows
is greater than $b$, but the letter $u_{1}$ appears only $b$ times).}. Then,
all the letters from $u_{1}$ to $u_{l-1}$ occur in the first $b$ rows. Hence,
each of the letters from $u_{1}$ to $u_{l-1}$ occurs in each of the first $b$
rows (since it must occur $b$ times in total, but can only occur in the first
$b$ rows, and therefore has to occur at least once in each of these $b$ rows).
Thus, each of the first $b$ rows begins with $u_{1}u_{2}\cdots u_{l-1}%
$\ \ \ \ \footnote{\textit{Correction:} Added this and the preceding
sentence.}. Moreover, if $q$ is the number of occurrences of $u_{l}$ in the
first column, then\footnote{\textit{Correction:} Added \textquotedblleft
then\textquotedblright.} the remaining $b-q$ letters $u_{l}$ must all occur in
the first $b-q$ rows\footnote{\textit{Comment:} Let me justify this:
\par
We must prove that each occurrence of $u_{l}$ in $T_{s}^{\prime}$ other than
in the first column must be in one of the first $b-q$ rows. In other words, we
must prove that if the $\left(  r,c\right)  $-th entry of $T_{s}^{\prime}$
equals $u_{l}$ for some integers $r\geq1$ and $s\geq2$, then $r\leq b-q$.
\par
So let us assume that the $\left(  r,c\right)  $-th entry of $T_{s}^{\prime}$
equals $u_{l}$ for some integers $r\geq1$ and $c>1$. We must prove that $r\leq
b-q$.
\par
We are in one of the following two cases:
\par
\begin{statement}
\textit{Case 1:} We have $r\leq b$.
\par
\textit{Case 2:} We have $r>b$.
\end{statement}
\par
Let us first consider Case 1. In this case, we have $r\leq b$. Thus, the
$r$-th row of $T_{s}^{\prime}$ begins with $u_{1}u_{2}\cdots u_{l-1}$ (since
each of the first $b$ rows begins with $u_{1}u_{2}\cdots u_{l-1}$). The next
entry of this $r$-th row after $u_{1}u_{2}\cdots u_{l-1}$ must then be $u_{l}$
(since $u_{l}$ occurs in the $r$-th row of $T_{s}^{\prime}$ (since the
$\left(  r,c\right)  $-th entry of $T_{s}^{\prime}$ equals $u_{l}$)). Hence,
the $\left(  r,l\right)  $-th entry of $T_{s}^{\prime}$ is $u_{l}$. Therefore,
the first $r$ entries of the $l$-th column of $T_{s}^{\prime}$ must be of the
form $u_{g}$ with $g\leq l$ (since $T_{s}^{\prime}$ is standard). However,
these entries must also be of the form $u_{g}$ with $g\geq l$ (since any entry
in the $l$-th column of $T_{s}^{\prime}$ has this form (again since
$T_{s}^{\prime}$ is standard)). Hence, these entries must be of the form
$u_{g}$ with $g$ satisfying both $g\leq l$ and $g\geq l$ simultaneously. In
other words, these entries must be of the form $u_{l}$. Hence, there are at
least $r$ entries $u_{l}$ in the $l$-th column of $T_{s}^{\prime}$. Since
there are also $q$ entries $u_{l}$ in the $1$-st column of $T_{s}^{\prime}$,
we conclude that there are at least $r+q$ entries $u_{l}$ in $T_{s}^{\prime}$
(since the $l$-th column of $T_{s}^{\prime}$ and the $1$-st column of
$T_{s}^{\prime}$ are two different columns). In other words, $b\geq r+q$
(since the total number of entries $u_{l}$ in $T_{s}^{\prime}$ is $b$). Hence,
$r\leq b-q$. Thus, we have proven $r\leq b-q$ in Case 1.
\par
Now, let us consider Case 2. In this case, we have $r>b$. Hence, the first
entry of the $r$-th row of $T_{s}^{\prime}$ has the form $u_{g}$ with $g\geq
l$. Consider this $g$.
\par
Recall that $c>1$. Hence, the $c$-th entry of the $r$-th row of $T_{s}%
^{\prime}$ is larger than the first entry of the $r$-th row of $T_{s}^{\prime
}$. In other words, the $c$-th entry of the $r$-th row of $T_{s}^{\prime}$ has
the form $u_{h}$ for some $h>g$ (since the first entry of the $r$-th row of
$T_{s}^{\prime}$ is $u_{g}$). Consider this $h$. Thus,%
\[
u_{h}=\left(  \text{the }c\text{-th entry of the }r\text{-th row of }%
T_{s}^{\prime}\right)  =\left(  \text{the }\left(  r,c\right)  \text{-th entry
of }T_{s}^{\prime}\right)  =u_{l}%
\]
(since the $\left(  r,c\right)  $-th entry of $T_{s}^{\prime}$ equals $u_{l}%
$). Hence, $h=l\leq g$ (since $g\geq l$). This contradicts $h>g$. Thus, we
have obtained a contradiction in Case 2. Hence, Case 2 cannot occur. Thus,
$r\leq b-q$ is proven (since we have proven $r\leq b-q$ in Case 1). This
completes the proof.}. The situation is summarized
by:\footnote{\textit{Correction:} Replaced \textquotedblleft$T^{\prime}%
$\textquotedblright\ by \textquotedblleft$T_{s}^{\prime}$\textquotedblright%
\ in the display. As usual, added some \textquotedblleft$\cdots$%
\textquotedblright s and the words \textquotedblleft rows\textquotedblright.}%
\begin{equation}
T_{s}^{\prime}=\ \ \ \
\begin{array}
[c]{@{}r@{\,}l}%
b\text{ rows}\left\{
\begin{array}
[c]{@{}r}%
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\\
q\text{ rows}\left\{
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\right.
\end{array}
\right.  \!\!\!\!\!\!\!\!\!\!\!\! &
\begin{array}
[c]{@{\,}l}%
\begin{array}
[c]{ccccc}%
u_{1} & \cdots & u_{l-1} & u_{l} & \cdots\\
\vdots & \ddots & \vdots & \vdots & \cdots\\
u_{1} & \cdots & u_{l-1} & u_{l} & \cdots
\end{array}
\\%
\begin{array}
[c]{ccccc}%
u_{1} & \cdots & u_{l-1} & u_{m} & \cdots\\
\vdots & \ddots & \vdots & \iddots & \\
u_{1} & \cdots & u_{l-1} &  &
\end{array}
\end{array}
\\
q\text{ rows}\left\{
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\right.  \!\!\!\!\!\!\!\! & \!%
\begin{array}
[c]{l}%
\begin{array}
[c]{ccc}%
u_{l} & \cdots & \cdots\\
\vdots & \ddots & \iddots\\
u_{l} & \cdots &
\end{array}
\end{array}
\\
&
\begin{array}
[c]{l}%
\begin{array}
[c]{cccc}%
\vdots &  &  &
\end{array}
\end{array}
\end{array}
\label{pf.thm.3.2.pf1.Ts'=}%
\end{equation}
where $m\geq l+1$. Such a tableau is called an $l$\textit{-critical tableau},
and its \textit{parameter} is $q$.

Let $j$ be the smallest index such that there exists a $j$-critical tableau in
the expansion (\ref{pf.thm.3.2.1st.expansion})\footnote{\textit{Correction:}
Added \textquotedblleft(\ref{pf.thm.3.2.1st.expansion})\textquotedblright.} of
$F$. We break up the expansion (\ref{pf.thm.3.2.1st.expansion}%
)\footnote{\textit{Correction:} Added \textquotedblleft%
(\ref{pf.thm.3.2.1st.expansion})\textquotedblright.} of $F$ into%
\begin{equation}
F=\sum_{s}\alpha_{s}\left(  T_{s}\mid T_{s}^{\prime}\right)  +\sum_{t}%
\alpha_{t}\left(  T_{t}\mid T_{t}^{\prime}\right)  +G,
\label{pf.thm.3.2.1st.expansion3}%
\end{equation}
where the first summation is over all the indices $s$ such that $T_{s}%
^{\prime}$ is $j$-critical, the second summation is over all $t$ such that
$T_{t}^{\prime}$ is $l$-critical\footnote{\textit{Correction:} Removed a
redundant comma here.} for some $l>j$\ \ \ \ \footnote{\textit{Correction:}
Replaced \textquotedblleft$l>q$\textquotedblright\ by \textquotedblleft%
$l>j$\textquotedblright.}, and $G$ is the linear combination of all the
$b\times d$ rectangular standard bideterminants.

Now, fix any $j\in\left\{  2,3,\ldots,d\right\}  $,
and\footnote{\textit{Correction:} Added \textquotedblleft Now, fix any
$j\in\left\{  2,3,\ldots,d\right\}  $, and\textquotedblright.} let $L$ be the
$d \times d$-matrix in $K^{d\times d}$ defined by%
\begin{align}
L\left(  x_{i}\mid u_{j-1}\right)   &  =\left(  x_{i}\mid u_{j-1}\right)
+\left(  x_{i}\mid u_{j}\right)  ,\label{pf.thm.3.2.L3.def1}\\
L\left(  x_{i}\mid u_{k}\right)   &  =\left(  x_{i}\mid u_{k}\right)
,\ \ \ \ \ \ \ \ \ \ \text{for }k\neq j-1 \label{pf.thm.3.2.L3.def2}%
\end{align}
\footnote{\textit{Comment:} Here is a more precise way to say this: Fix any
$j\in\left\{  2,3,\ldots,d\right\}  $. Let $L$ be the $d\times d$-matrix
$\left(  l_{qp}\right)  _{1\leq p,q\leq d} \in K^{d \times d}$, where%
\begin{align*}
l_{j-1,j}  &  =1;\\
l_{pp}  &  =1\ \ \ \ \ \ \ \ \ \ \text{for all }p;\\
l_{qp}  &  =0\ \ \ \ \ \ \ \ \ \ \text{for all }p,q\text{ satisfying }p\neq
q\text{ and }\left(  p,q\right)  \neq\left(  j,j-1\right)  .
\end{align*}
Clearly, this matrix $L$ is invertible (since it is lower-unitriangular). It
is now straightforward to see that this $L$ satisfies
(\ref{pf.thm.3.2.L3.def1}) and (\ref{pf.thm.3.2.L3.def2}).}. Under $L$, those
bideterminants in which all the letters $u_{j-1}$ and $u_{j}$ occur in the
first $b$ rows are unchanged\footnote{\textit{Comment:} This is because
\par
\begin{itemize}
\item any inner product containing both letters $u_{j-1}$ and $u_{j}$ is
unchanged under $L$ (by Lemma \ref{lem.3.2.a} \textbf{(c)});
\par
\item any inner product that does not contain the letter $u_{j}$ is unchanged
under $L$ (by Lemma \ref{lem.3.2.a} \textbf{(a)});
\par
\item $L$ acts upon $P$ as an algebra homomorphism.
\end{itemize}
\par
(Here, we are only talking about inner products in which the indices of the
$u$ are distinct. But all the inner products forming a bideterminant have this
property.)}; in particular, the rectangular bideterminants in $G$ and the
$l$-critical bideterminants in the second summation in
(\ref{pf.thm.3.2.1st.expansion3})\footnote{\textit{Correction:} Added
\textquotedblleft in (\ref{pf.thm.3.2.1st.expansion3})\textquotedblright.}
remain unaltered. In other words,\footnote{\textit{Correction:} Inserted this
sentence.}%
\begin{equation}
LG=G, \label{pf.thm.3.2.1st.expansion3.L3}%
\end{equation}
and%
\begin{equation}
L\left(  T_{t}\mid T_{t}^{\prime}\right)  =\left(  T_{t}\mid T_{t}^{\prime
}\right)  \ \ \ \ \ \ \ \ \ \ \text{for each }t\text{ in the second summation
in (\ref{pf.thm.3.2.1st.expansion3}).} \label{pf.thm.3.2.1st.expansion3.L2}%
\end{equation}
For the $j$-critical bideterminants,\footnote{\textit{Correction:} In the
following display, I have removed the \textquotedblleft$\left(  T_{s}\mid
T_{s}^{\prime}\right)  $\textquotedblright\ addend, because I prefer to regard
this addend as part of the sum $\sum_{r}\left(  T_{s}\mid T_{s}^{r}\right)
$.}%
\begin{equation}
L\left(  T_{s}\mid T_{s}^{\prime}\right)  =\sum_{r}\left(  T_{s}\mid T_{s}%
^{r}\right)  , \label{pf.thm.3.2.pf1.LTsTs'=}%
\end{equation}
where $T_{s}^{r}$ is a tableau (not necessarily
standard)\footnote{\textit{Correction:} Added \textquotedblleft(not
necessarily standard)\textquotedblright.} of the following
form:\footnote{\textit{Correction:} As usual, added some \textquotedblleft%
$\cdots$\textquotedblright s and the words \textquotedblleft
rows\textquotedblright\ in the display below.}%
\begin{equation}
\
\begin{array}
[c]{@{}r@{\,}l}%
b\text{ rows}\left\{
\begin{array}
[c]{@{}r}%
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\\
q\text{ rows}\left\{
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\right.
\end{array}
\right.  \!\!\!\!\!\!\!\!\!\!\!\! &
\begin{array}
[c]{@{\,}l}%
\begin{array}
[c]{ccccc}%
u_{1} & \cdots & u_{j-1} & u_{j} & \cdots\\
\vdots & \ddots & \vdots & \vdots & \cdots\\
u_{1} & \cdots & u_{j-1} & u_{j} & \cdots
\end{array}
\\%
\begin{array}
[c]{ccccc}%
u_{1} & \cdots & \ \ \ast\ \ \  & u_{m} & \cdots\\
\vdots & \ddots & \vdots & \iddots & \\
u_{1} & \cdots & \ \ \ast\ \ \  &  &
\end{array}
\end{array}
\\
q\text{ rows}\left\{
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\right.  \!\!\!\!\!\!\!\! & \!%
\begin{array}
[c]{l}%
\begin{array}
[c]{ccc}%
u_{j} & \cdots & \cdots\\
\vdots & \ddots & \iddots\\
u_{j} & \cdots &
\end{array}
\end{array}
\\
&
\begin{array}
[c]{l}%
\begin{array}
[c]{cccc}%
\vdots &  &  &
\end{array}
\end{array}
\end{array}
\label{pf.thm.3.2.pf1.Tsr=}%
\end{equation}
where any of the $\ast$ symbols in column $j-1$%
\ \ \ \ \footnote{\textit{Correction:} Replaced \textquotedblleft$\ast
$\textquotedblright\ by \textquotedblleft any of the $\ast$ symbols in column
$j-1$\textquotedblright.} may be $u_{j-1}$ or $u_{j}$, and
where\footnote{\textit{Correction:} Added \textquotedblleft
where\textquotedblright.} $m\geq j+1$.\ \ \ \ \footnote{\textit{Comment:} This
follows from the following argument:
\par
Fix any $s$.
\par
Let $w$ be the number of rows of the tableau $T_{s}$; thus, $w$ is also the
number of rows of the tableau $T_{s}^{\prime}$.
\par
The tableau $T_{s}^{\prime}$ is $j$-critical; thus, it looks as follows:%
\begin{equation}
T_{s}^{\prime}=\ \ \ \
\begin{array}
[c]{@{}r@{\,}l}%
b\text{ rows}\left\{
\begin{array}
[c]{@{}r}%
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\\
q\text{ rows}\left\{
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\right.
\end{array}
\right.  \!\!\!\!\!\!\!\!\!\!\!\! &
\begin{array}
[c]{@{\,}l}%
\begin{array}
[c]{ccccc}%
u_{1} & \cdots & u_{j-1} & u_{j} & \cdots\\
\vdots & \ddots & \vdots & \vdots & \cdots\\
u_{1} & \cdots & u_{j-1} & u_{j} & \cdots
\end{array}
\\%
\begin{array}
[c]{ccccc}%
u_{1} & \cdots & u_{j-1} & u_{m} & \cdots\\
\vdots & \ddots & \vdots & \iddots & \\
u_{1} & \cdots & u_{j-1} &  &
\end{array}
\end{array}
\\
q\text{ rows}\left\{
\begin{array}
[c]{c}%
\vphantom{s_1}\\
\vphantom{\vdots}\\
\vphantom{s_1}
\end{array}
\right.  \!\!\!\!\!\!\!\! & \!%
\begin{array}
[c]{l}%
\begin{array}
[c]{ccc}%
u_{j} & \cdots & \cdots\\
\vdots & \ddots & \iddots\\
u_{j} & \cdots &
\end{array}
\end{array}
\\
&
\begin{array}
[c]{l}%
\begin{array}
[c]{cccc}%
\vdots &  &  &
\end{array}
\end{array}
\end{array}
. \label{pf.thm.3.2.L3.fn124.Ts'=}%
\end{equation}
In particular, all entries of $T_{s}^{\prime}$ in row $b+1$ and the rows below
have the form $u_{h}$ for $h\geq j$ (because the first entry of row $b+1$ is
$u_{j}$, and the entries are nondecreasing along rows and column).
\par
For each $k\in\left\{  1,2,\ldots,w\right\}  $, we set $r_{k}=\left(
x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  $, where
$x_{i_{1}}\cdots x_{i_{p}}$ is the $k$-th row of the tableau $T_{s}$, and
where $u_{j_{1}}\cdots u_{j_{p}}$ is the $k$-th row of the tableau
$T_{s}^{\prime}$. Thus, $\left(  T_{s}\mid T_{s}^{\prime}\right)  =r_{1}%
r_{2}\cdots r_{w}$. Hence,%
\begin{equation}
L\left(  T_{s}\mid T_{s}^{\prime}\right)  =L\left(  r_{1}r_{2}\cdots
r_{w}\right)  =\left(  Lr_{1}\right)  \left(  Lr_{2}\right)  \cdots\left(
Lr_{w}\right)  \label{pf.thm.3.2.L3.fn124.ah}%
\end{equation}
(since $L$ acts upon $P$ as an algebra homomorphism).
\par
Fix $k\in\left\{  1,2,\ldots,w\right\}  $. We shall now compute the value
$Lr_{k}$. First, let $x_{i_{1}}\cdots x_{i_{p}}$ be the $k$-th row of the
tableau $T_{s}$, and let $u_{j_{1}}\cdots u_{j_{p}}$ be the $k$-th row of the
tableau $T_{s}^{\prime}$. Thus, $r_{k}=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid
u_{j_{1}}\cdots u_{j_{p}}\right)  $ (by the definition of $r_{k}$). Notice
that $j_{1}<j_{2}<\cdots<j_{p}$ (since $T_{s}^{\prime}$ is a standard
tableau); thus, the letters $u_{j_{1}},\ldots,u_{j_{p}}$ are distinct. Now,
let us compute $Lr_{k}$ depending on the value of $k$:
\par
\begin{itemize}
\item Let us first assume that $k\leq b-q$. Thus, the $k$-th row of
$T_{s}^{\prime}$ contains both $u_{j-1}$ and $u_{j}$ (because of
(\ref{pf.thm.3.2.L3.fn124.Ts'=})). In other words, $u_{j_{1}}\cdots u_{j_{p}}$
contains both $u_{j-1}$ and $u_{j}$ (since $u_{j_{1}}\cdots u_{j_{p}}$ is the
$k$-th row of $T_{s}^{\prime}$). In other words, some $g\in\left\{
1,2,\ldots,p\right\}  $ satisfies $j_{g}=j-1$, and some $h\in\left\{
1,2,\ldots,p\right\}  $ satisfies $j_{h}=j$. Hence, Lemma \ref{lem.3.2.a}
\textbf{(c)} shows that $L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}%
}\cdots u_{j_{p}}\right)  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}%
}\cdots u_{j_{p}}\right)  $. Since $r_{k}=\left(  x_{i_{1}}\cdots x_{i_{p}%
}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  $, this rewrites as $Lr_{k}=r_{k}$.
Thus, we have computed $Lr_{k}$ in the case when $k\leq b-q$.
\par
\item Let us now assume that $b-q<k\leq b$. Thus, the $k$-th row of
$T_{s}^{\prime}$ contains $u_{j-1}$ but not $u_{j}$ (because of
(\ref{pf.thm.3.2.L3.fn124.Ts'=})). In other words, $u_{j_{1}}\cdots u_{j_{p}}$
contains $u_{j-1}$ but not $u_{j}$ (since $u_{j_{1}}\cdots u_{j_{p}}$ is the
$k$-th row of $T_{s}^{\prime}$). In particular, $u_{j_{1}}\cdots u_{j_{p}}$
contains $u_{j-1}$. In other words, some $g\in\left\{  1,2,\ldots,p\right\}  $
satisfies $j_{g}=j-1$. Consider this $g$. Hence, Lemma \ref{lem.3.2.a}
\textbf{(b)} shows that
\begin{align}
&  L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
\nonumber\\
&  =\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
+\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{g-1}}%
u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  . \label{pf.thm.3.2.L3.fn124.2}%
\end{align}
\par
Set $r_{k}^{\prime}=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots
u_{j_{g-1}}u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  $. Thus, $r_{k}^{\prime}$
is the same inner product $r_{k}$, except that the letter $u_{j_{g}}$ (that
is, the $g$-th letter to the right of the vertical bar) has been replaced by
$u_{j}$. In other words, $r_{k}^{\prime}$ is the result of replacing the
letter $u_{j-1}$ by $u_{j}$ in the inner product $r_{k}$ (since $j_{g}=j-1$).
Now, (\ref{pf.thm.3.2.L3.fn124.2}) rewrites as $Lr_{k}=r_{k}+r_{k}^{\prime}$
(since $r_{k}=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}%
}\right)  $ and $r_{k}^{\prime}=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid
u_{j_{1}}\cdots u_{j_{g-1}}u_{j}u_{j_{g+1}}\cdots u_{j_{p}}\right)  $). Thus,
we have computed $Lr_{k}$ in the case when $b-q<k\leq b$.
\par
\item Let us finally assume that $k>b$. Thus, all entries in the $k$-th row of
$T_{s}^{\prime}$ have the form $u_{h}$ for $h\geq j$ (because all entries of
$T_{s}^{\prime}$ in row $b+1$ and the rows below have the form $u_{h}$ for
$h\geq j$). In particular, the $k$-th row of $T_{s}^{\prime}$ does not contain
$u_{j-1}$. In other words, $u_{j_{1}}\cdots u_{j_{p}}$ does not contain
$u_{j-1}$ (since $u_{j_{1}}\cdots u_{j_{p}}$ is the $k$-th row of
$T_{s}^{\prime}$). In other words, no $g\in\left\{  1,2,\ldots,p\right\}  $
satisfies $j_{g}=j-1$. Hence, Lemma \ref{lem.3.2.a} \textbf{(a)} shows that
$L\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)
=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}}\right)  $.
Since $r_{k}=\left(  x_{i_{1}}\cdots x_{i_{p}}\mid u_{j_{1}}\cdots u_{j_{p}%
}\right)  $, this rewrites as $Lr_{k}=r_{k}$. Thus, we have computed $Lr_{k}$
in the case when $k>q$.
\end{itemize}
\par
Now, forget that we fixed $k$. Thus, for each $k\in\left\{  1,2,\ldots
,w\right\}  $, we have obtained a formula for $Lr_{k}$, namely:
\par
\begin{itemize}
\item If $k$ satisfies either $k\leq b-q$ or $k>b$, then $Lr_{k}=r_{k}$.
\par
\item If $k$ satisfies $b-q<k\leq b$, then $Lr_{k}=r_{k}+r_{k}^{\prime}$,
where $r_{k}^{\prime}$ is the result of replacing the letter $u_{j-1}$ by
$u_{j}$ in the inner product $r_{k}$.
\end{itemize}
\par
Multiplying all of these formulas, we obtain%
\begin{align*}
&  \left(  Lr_{1}\right)  \left(  Lr_{2}\right)  \cdots\left(  Lr_{w}\right)
\\
&  =r_{1}r_{2}\cdots r_{b-q}\left(  r_{b-q+1}+r_{b-q+1}^{\prime}\right)
\left(  r_{b-q+2}+r_{b-q+2}^{\prime}\right)  \cdots\left(  r_{b}+r_{b}%
^{\prime}\right)  r_{b+1}r_{b+2}\cdots r_{w}.
\end{align*}
Hence, (\ref{pf.thm.3.2.L3.fn124.ah}) becomes%
\begin{align*}
L\left(  T_{s}\mid T_{s}^{\prime}\right)   &  =\left(  Lr_{1}\right)  \left(
Lr_{2}\right)  \cdots\left(  Lr_{w}\right) \\
&  =r_{1}r_{2}\cdots r_{b-q}\left(  r_{b-q+1}+r_{b-q+1}^{\prime}\right)
\left(  r_{b-q+2}+r_{b-q+2}^{\prime}\right)  \cdots\left(  r_{b}+r_{b}%
^{\prime}\right)  r_{b+1}r_{b+2}\cdots r_{w}.
\end{align*}
Expanding the right hand side, we thus conclude that $L\left(  T_{s}\mid
T_{s}^{\prime}\right)  $ is the sum of all products obtained from $r_{1}%
r_{2}\cdots r_{w}$ by replacing some of the factors $r_{b-q+1},r_{b-q+2}%
,\ldots,r_{b}$ by $r_{b-q+1}^{\prime},r_{b-q+2}^{\prime},\ldots,r_{b}^{\prime
}$ (respectively). But of course, these products are precisely the
bideterminants of all bitableaux that are obtained from $\left(  T_{s}\mid
T_{s}^{\prime}\right)  $ by replacing some of the letters $u_{j-1}$ in rows
$b-q+1,b-q+2,\ldots,b$ by $u_{j}$. Hence, $L\left(  T_{s}\mid T_{s}^{\prime
}\right)  $ is the sum of the bideterminants of all these bitableaux. Hence,
$L\left(  T_{s}\mid T_{s}^{\prime}\right)  =\sum_{r}\left(  T_{s}\mid
T_{s}^{r}\right)  $, where each $T_{s}^{r}$ is a tableau of the form
(\ref{pf.thm.3.2.pf1.Tsr=}), where any of the $\ast$ symbols in column $j-1$
may be $u_{j-1}$ or $u_{j}$, and where $m\geq j+1$. This proves
(\ref{pf.thm.3.2.pf1.LTsTs'=}).}

Let $T_{s}^{\prime\prime}$ be the tableau such that all the $\ast$'s are
$u_{j}$\ \ \ \ \footnote{\textit{Comment:} More precisely: Let $T_{s}%
^{\prime\prime}$ be the tableau obtained from $T_{s}^{\prime}$ by replacing
\textbf{all} the letters $u_{j-1}$ in rows $b-q+1,b-q+2,\ldots,b$ by $u_{j}$.
Thus, $T_{s}^{\prime\prime}$ is one of the tableaux $T_{s}^{r}$ on the right
hand side of (\ref{pf.thm.3.2.pf1.LTsTs'=}). Actually, among all those
tableaux $T_{s}^{r}$, the tableau $T_{s}^{\prime\prime}$ is the one with the
most occurrences of the letter $u_{j}$.}; thus, $T_{s}^{\prime\prime}$ is a
standard tableau containing $u_{j}$ exactly\footnote{\textit{Correction:}
Added the word \textquotedblleft exactly\textquotedblright.} $b+q$
times\footnote{\textit{Comment:} Here is why this is true:
\par
First, let us see why $T_{s}^{\prime\prime}$ is a standard tableau.
\par
An entry $u_{j-1}$ in $T_{s}^{\prime}$ shall be called \textit{malleable} if
the letter $u_{j}$ does not appear in the same row of $T_{s}^{\prime}$ as this
entry.
\par
Indeed, recall that the tableau $T_{s}^{\prime}$ is standard. The tableau
$T_{s}^{\prime\prime}$ is obtained from it by replacing some entries (namely,
the entries $u_{j-1}$ in rows $b-q+1,b-q+2,\ldots,b$ of the $\left(
j-1\right)  $-th column) of $T_{s}^{\prime}$ by $u_{j}$. It is easy to see
(from a look at (\ref{pf.thm.3.2.L3.fn124.Ts'=})) that these entries are
precisely the malleable entries $u_{j-1}$ in the $\left(  j-1\right)  $-th
column of $T_{s}^{\prime}$. Thus, the tableau $T_{s}^{\prime\prime}$ is
obtained from $T_{s}^{\prime}$ by replacing each malleable entry $u_{j-1}$ in
the $\left(  j-1\right)  $-th column of $T$ by $u_{j}$. Hence, Lemma
\ref{lem.3.2.b} (applied to $T_{s}^{\prime}$, $j-1$ and $T_{s}^{\prime\prime}$
instead of $T$, $g$ and $\widetilde{T}$) yields the tableau $T_{s}%
^{\prime\prime}$ is standard.
\par
It now remains to prove that the tableau $T_{s}^{\prime\prime}$ contains
$u_{j}$ exactly $b+q$ times.
\par
Indeed, we know that the tableau $T_{s}^{\prime}$ contains $u_{j}$ exactly $b$
times. The tableau $T_{s}^{\prime\prime}$ differs from $T_{s}^{\prime}$ in
having $q$ further entries equal to $u_{j}$ (in fact, $T_{s}^{\prime\prime}$
was obtained from $T_{s}^{\prime}$ by replacing a total of $q$ entries by
$u_{j}$, and none of these entries had been $u_{j}$ in $T_{s}^{\prime}$).
Thus, the tableau $T_{s}^{\prime\prime}$ contains $u_{j}$ exactly $b+q$ times.
This completes our proof.}. Applying the operator $L$ to the equality
(\ref{pf.thm.3.2.1st.expansion3}), we obtain\footnote{\textit{Correction:}
This sentence has been refactored and expanded. Also, I have again removed the
\textquotedblleft$\left(  T_{s}\mid T_{s}^{\prime}\right)  $\textquotedblright%
\ addend, because I prefer to regard this addend as part of the sum $\sum
_{r}\left(  T_{s}\mid T_{s}^{r}\right)  $.}%
\begin{align*}
LF  &  =\sum_{s}\alpha_{s}\underbrace{L\left(  T_{s}\mid T_{s}^{\prime
}\right)  }_{\substack{=\sum_{r}\left(  T_{s}\mid T_{s}^{r}\right)
\\\text{(by (\ref{pf.thm.3.2.pf1.LTsTs'=}))}}}+\sum_{t}\alpha_{t}%
\underbrace{L\left(  T_{t}\mid T_{t}^{\prime}\right)  }_{\substack{=\left(
T_{t}\mid T_{t}^{\prime}\right)  \\\text{(by
(\ref{pf.thm.3.2.1st.expansion3.L2}))}}}+\underbrace{LG}%
_{\substack{=G\\\text{(by (\ref{pf.thm.3.2.1st.expansion3.L3}))}}}\\
&  =\sum_{s}\alpha_{s}\sum_{r}\left(  T_{s}\mid T_{s}^{r}\right)  +\sum
_{t}\alpha_{t}\left(  T_{t}\mid T_{t}^{\prime}\right)  +G\\
&  =\sum_{s,r}\alpha_{s}\left(  T_{s}\mid T_{s}^{r}\right)  +\sum_{t}%
\alpha_{t}\left(  T_{t}\mid T_{t}^{\prime}\right)  +G,
\end{align*}
so that%
\begin{align}
&  \sum_{s,r}\alpha_{s}\left(  T_{s}\mid T_{s}^{r}\right)  +\sum_{t}\alpha
_{t}\left(  T_{t}\mid T_{t}^{\prime}\right)  +G\nonumber\\
&  =LF=a\left(  L\right)  F\ \ \ \ \ \ \ \ \ \ \left(  \text{since }F\text{ is
invariant}\right) \nonumber\\
&  =\sum_{s}a\left(  L\right)  \alpha_{s}\left(  T_{s}\mid T_{s}^{\prime
}\right)  +\sum_{t}a\left(  L\right)  \alpha_{t}\left(  T_{t}\mid
T_{t}^{\prime}\right)  +a\left(  L\right)  G.
\label{pf.thm.3.2.1st.expansion5}%
\end{align}
(by (\ref{pf.thm.3.2.1st.expansion3})).

\footnote{\textit{Correction:} Removed the sentence \textquotedblleft Each
bideterminant $\left(  T_{s}\mid T_{s}^{r}\right)  $ contains the letter
$u_{j}$ at least $b+1$ times.\textquotedblright. This sentence was useless and
does not hold with my definition of $\left(  T_{s}\mid T_{s}^{r}\right)  $.}
Let $q^{\prime}$ be the largest parameter for the $j$-critical tableaux
$T_{s}^{\prime}$ appearing in (\ref{pf.thm.3.2.1st.expansion5}%
)\footnote{\textit{Correction:} Added the words \textquotedblleft%
$T_{s}^{\prime}$ appearing in (\ref{pf.thm.3.2.1st.expansion5}%
)\textquotedblright.}. The only bideterminants in the equality
(\ref{pf.thm.3.2.1st.expansion5})\footnote{\textit{Correction:} Replaced
\textquotedblleft the above equality\textquotedblright\ by \textquotedblleft
the equality (\ref{pf.thm.3.2.1st.expansion5})\textquotedblright.} containing
$u_{j}$ exactly $b+q^{\prime}$ times are the bideterminants $\left(  T_{v}\mid
T_{v}^{\prime\prime}\right)  $, where $T_{v}^{\prime}$ is a $j$-critical
tableau with parameter $q^{\prime}$\ \ \ \ \footnote{\textit{Comment:} Let me
explain why this is true.
\par
First, we recall that a $j$-critical tableau must have at least one entry
$u_{j}$ in its first column; thus, its parameter is positive (since its
parameter is the number of entries $u_{j}$ in its first column). Hence,
$q^{\prime}$ is positive (since $q^{\prime}$ is the parameter of a
$j$-critical tableau). Thus, $q^{\prime}\neq0$, so that $b+q^{\prime}\neq b$.
\par
Each of the bideterminants appearing in (\ref{pf.thm.3.2.1st.expansion})
contains $u_{j}$ exactly $b$ times (because of (\ref{pf.thm.3.2.pf1.b-times}%
)). In other words, each of the bideterminants appearing in
(\ref{pf.thm.3.2.1st.expansion3}) contains $u_{j}$ exactly $b$ times (since
these bideterminants are the same as the bideterminants appearing in
(\ref{pf.thm.3.2.1st.expansion})). Hence, none of the bideterminants appearing
in (\ref{pf.thm.3.2.1st.expansion3}) contains $u_{j}$ exactly $b+q^{\prime}$
times (since $b+q^{\prime}\neq b$). Thus, none of the bideterminants $\left(
T_{s}\mid T_{s}^{\prime}\right)  $ and $\left(  T_{t}\mid T_{t}^{\prime
}\right)  $ in (\ref{pf.thm.3.2.1st.expansion5}) contains $u_{j}$ exactly
$b+q^{\prime}$ times. The same holds for the bideterminants contained in $G$
(for the same reason).
\par
Therefore, any bideterminants in (\ref{pf.thm.3.2.1st.expansion5}) that
contain $u_{j}$ exactly $b+q^{\prime}$ times must be of the form $\left(
T_{s}\mid T_{s}^{r}\right)  $. However, not every bideterminant of this form
actually contains $u_{j}$ exactly $b+q^{\prime}$ times. Let us see which ones
do:
\par
Consider a bideterminant $\left(  T_{s}\mid T_{s}^{r}\right)  $ that appears
in (\ref{pf.thm.3.2.1st.expansion5}) and contains $u_{j}$ exactly
$b+q^{\prime}$ times. The corresponding tableau $T_{s}^{\prime}$ is
$j$-critical; let $q$ be its parameter. Thus, $q^{\prime}\geq q$ (since
$q^{\prime}$ is the largest parameter for the $j$-critical tableaux appearing
in (\ref{pf.thm.3.2.1st.expansion5})). Furthermore, the tableau $T_{s}^{r}$
has the form (\ref{pf.thm.3.2.pf1.Tsr=}), where any of the $\ast$ symbols in
column $j-1$ may be $u_{j-1}$ or $u_{j}$. Let $h$ be the number of $\ast$
symbols that are $u_{j}$; thus, $h\leq q$ (since the total number of $\ast$
symbols in (\ref{pf.thm.3.2.pf1.Tsr=}) is $q$). Notice that the equality $h=q$
holds if and only if each of the $\ast$ symbols is a $u_{j}$.
\par
From (\ref{pf.thm.3.2.pf1.b-times}), we know that the tableau $T_{s}^{\prime}$
contains the letter $u_{j}$ exactly $b$ times. The tableau $T_{s}^{r}$ is
obtained from $T_{s}^{\prime}$ by replacing some of the letters $u_{j-1}$ by
$u_{j}$ (indeed, the positions of the $\ast$ symbols in
(\ref{pf.thm.3.2.pf1.Tsr=}) were occupied by $u_{j-1}$ in $T_{s}^{\prime}$).
Hence, the tableau $T_{s}^{r}$ contains the letter $u_{j}$ exactly $b+h$ times
(since there are precisely $h$ letters $u_{j-1}$ that are replaced by $u_{j}$
in $T_{s}^{r}$). Comparing this with the fact that the tableau $T_{s}^{r}$
contains the letter $u_{j}$ exactly $b+q^{\prime}$ times, we conclude that
$b+h=b+q^{\prime}$. Thus, $h=q^{\prime}$. Combining $h=q^{\prime}\geq q$ with
$h\leq q$, we obtain $h=q$. In other words, exactly $q$ of the $\ast$ symbols
in (\ref{pf.thm.3.2.pf1.Tsr=}) are $u_{j}$. This means that \textbf{all} the
$\ast$ symbols in (\ref{pf.thm.3.2.pf1.Tsr=}) are $u_{j}$ (because $q$ is the
total number of $\ast$ symbols in (\ref{pf.thm.3.2.pf1.Tsr=})). Hence,
$T_{s}^{r}$ is the tableau $T_{s}^{\prime\prime}$ (because $T_{s}%
^{\prime\prime}$ was defined as the one tableau $T_{s}^{r}$ for which all the
$\ast$ symbols in (\ref{pf.thm.3.2.pf1.Tsr=}) are $u_{j}$).
\par
Furthermore, $q^{\prime}=h=q$. Hence, the $j$-critical tableau $T_{s}^{\prime
}$ has parameter $q^{\prime}$ (since it has parameter $q$). Thus, we have
shown that $T_{s}^{r}$ is the tableau $T_{s}^{\prime\prime}$, and that the
$j$-critical tableau $T_{s}^{\prime}$ has parameter $q^{\prime}$.
\par
Now, forget that we fixed $\left(  T_{s}\mid T_{s}^{r}\right)  $. We thus have
shown that if $\left(  T_{s}\mid T_{s}^{r}\right)  $ is a bideterminant that
appears in (\ref{pf.thm.3.2.1st.expansion5}) and contains $u_{j}$ exactly
$b+q^{\prime}$ times, then $T_{s}^{r}$ is the tableau $T_{s}^{\prime\prime}$,
and the $j$-critical tableau $T_{s}^{\prime}$ has parameter $q^{\prime}$.
Since any bideterminants in (\ref{pf.thm.3.2.1st.expansion5}) that contain
$u_{j}$ exactly $b+q^{\prime}$ times must be of the form $\left(  T_{s}\mid
T_{s}^{r}\right)  $, we can therefore summarize: Any bideterminants in
(\ref{pf.thm.3.2.1st.expansion5}) that contain $u_{j}$ exactly $b+q^{\prime}$
times must be of the form $\left(  T_{s}\mid T_{s}^{r}\right)  $, where
$T_{s}^{r}$ is the tableau $T_{s}^{\prime\prime}$, and the $j$-critical
tableau $T_{s}^{\prime}$ has parameter $q^{\prime}$. In other words, any
bideterminants in (\ref{pf.thm.3.2.1st.expansion5}) that contain $u_{j}$
exactly $b+q^{\prime}$ times must be of the form $\left(  T_{s}\mid
T_{s}^{\prime\prime}\right)  $, where the $j$-critical tableau $T_{s}^{\prime
}$ has parameter $q^{\prime}$. In other words, any bideterminants in
(\ref{pf.thm.3.2.1st.expansion5}) that contain $u_{j}$ exactly $b+q^{\prime}$
times must be of the form $\left(  T_{v}\mid T_{v}^{\prime\prime}\right)  $,
where $T_{v}^{\prime}$ is a $j$-critical tableau with parameter $q^{\prime}$.
\par
Conversely, any bideterminant of this form must contain $u_{j}$ exactly
$b+q^{\prime}$ times (as one can easily show by following the above argument
backwards). Thus, the bideterminants in the equality
(\ref{pf.thm.3.2.1st.expansion5}) containing $u_{j}$ exactly $b+q^{\prime}$
times are \textbf{exactly} the bideterminants $\left(  T_{v}\mid T_{v}%
^{\prime\prime}\right)  $, where $T_{v}^{\prime}$ is a $j$-critical tableau
with parameter $q^{\prime}$. Qed.}. Therefore, the projection of the equality
(\ref{pf.thm.3.2.1st.expansion5})\footnote{\textit{Correction:} Replaced
\textquotedblleft this equality\textquotedblright\ by \textquotedblleft the
equality (\ref{pf.thm.3.2.1st.expansion5})\textquotedblright.} onto the
subspace spanned by the bideterminants containing $u_{j}$ exactly
$b+q^{\prime}$ times yields%
\begin{equation}
\sum_{v}\alpha_{v}\left(  T_{v}\mid T_{v}^{\prime\prime}\right)  =0
\label{pf.thm.3.2.1st.basis-contra}%
\end{equation}
\footnote{\textit{Correction:} I have switched the left and the right hand
side of this equality, to match my modified version of
(\ref{pf.thm.3.2.1st.expansion5}).} (where the sum is over all $v$ for which
$T_{v}^{\prime}$ is a $j$-critical tableau with parameter $q^{\prime}%
$)\ \ \ \ \footnote{\textit{Correction:} Added the preceding parenthetical.}.

All the bitableaux\footnote{\textit{Correction:} Just as many times before, I
have replaced bideterminants by bitableaux.} $\left[  T_{v},T_{v}^{\prime
}\right]  $, and hence $\left[  T_{v},T_{v}^{\prime\prime}\right]  $, are
distinct and standard\footnote{\textit{Comment:} Let me explain why this is
true.
\par
We already know that the tableaux $T_{v}^{\prime\prime}$ are standard; thus,
the bitableaux $\left[  T_{v},T_{v}^{\prime\prime}\right]  $ are standard. It
remains to show that these bitableaux $\left[  T_{v},T_{v}^{\prime\prime
}\right]  $ are distinct.
\par
Assume the contrary. Thus, there exist two distinct indices $v$ and $w$
(having the property that $T_{v}^{\prime}$ and $T_{w}^{\prime}$ are
$j$-critical tableaux with parameters $q^{\prime}$) such that $\left[
T_{v},T_{v}^{\prime\prime}\right]  =\left[  T_{w},T_{w}^{\prime\prime}\right]
$. Hence, $T_{v}=T_{w}$ and $T_{v}^{\prime\prime}=T_{w}^{\prime\prime}$.
\par
Recall that $T_{v}^{\prime}$ is a $j$-critical tableau with parameter
$q^{\prime}$. Hence, the tableau $T_{v}^{\prime\prime}$ is obtained from
$T_{v}^{\prime}$ by replacing the entries $u_{j-1}$ in rows $b-q^{\prime
}+1,b-q^{\prime}+2,\ldots,b$ of the $\left(  j-1\right)  $-th column by
$u_{j}$ (by the definition of $T_{v}^{\prime\prime}$). Therefore, the tableau
$T_{v}^{\prime}$ can be obtained from the tableau $T_{v}^{\prime\prime}$ by
replacing the entries $u_{j}$ in rows $b-q^{\prime}+1,b-q^{\prime}+2,\ldots,b$
of the $\left(  j-1\right)  $-th column by $u_{j-1}$ (since the entries in
rows $b-q^{\prime}+1,b-q^{\prime}+2,\ldots,b$ of the $\left(  j-1\right)  $-th
column of $T_{v}^{\prime}$ are $u_{j-1}$ (again, because $T_{v}^{\prime}$ is
$j$-critical)). Similarly, the tableau $T_{w}^{\prime}$ can be obtained from
the tableau $T_{w}^{\prime\prime}$ by the exact same replacements. Therefore,
$T_{v}^{\prime}=T_{w}^{\prime}$ (since $T_{v}^{\prime\prime}=T_{w}%
^{\prime\prime}$). Combined with $T_{v}=T_{w}$, this yields $\left[
T_{v},T_{v}^{\prime}\right]  =\left[  T_{w},T_{w}^{\prime}\right]  $. But this
contradicts the fact that the bitableaux $\left[  T_{s},T_{s}^{\prime}\right]
$ were chosen to be distinct. This contradiction completes our proof.}. By the
basis theorem (specifically, the part of Corollary \ref{cor.2.2.basis} that
says that the standard $\left(  \mathcal{X},\mathcal{U}\right)  $%
-bideterminants are linearly independent)\footnote{\textit{Correction:} I have
expanded this sentence to be clearer.}, we thus conclude that all the
coefficients $\alpha_{v}$ in (\ref{pf.thm.3.2.1st.basis-contra}) must be $0$.
We conclude that there cannot be a $j$-critical bideterminant in the expansion
of $F$; in particular, there cannot be a nonrectangular bideterminant in the
expansion. This contradicts our assumption that not all tableaux
$T_{s}^{\prime}$ appearing in (\ref{pf.thm.3.2.1st.expansion}) are
rectangular\footnote{\textit{Correction:} Inserted this sentence.}. This
completes the first proof of Theorem \ref{thm.3.2}.
\end{proof}

\subsection{\label{subsect.invar.fft2}The First Fundamental Theorem: second
proof}

In preparation for the second proof of Theorem \ref{thm.3.2}, let us state a
few lemmas.\footnote{\textit{Correction:} I have rewritten this second proof
in many ways. Here are the most important changes:
\par
\begin{itemize}
\item Lemma \ref{lem.A} and Lemma \ref{lem.E} have been moved out of the
proof, as they are easier to understand as separate results.
\par
\item I have made \textquotedblleft Weyl's principle of the irrelevance of
algebraic inequalities\textquotedblright\ into a separate proposition
(Proposition \ref{prop.weyl}), and fixed it by requiring $g_{1},\ldots,g_{r}$
to be nonzero. I have also added a proof outline.
\par
\item I have explicitly stated two corollaries of Weyl's principle
(Corollaries \ref{cor.weyl.1} and \ref{cor.weyl.2}) as well as two further
auxiliary facts (Propositions \ref{lem.bracket} and \ref{lem.bracket2}) which
were mostly implicit in the original proof.
\par
\item Multiple wrong indices have been fixed in the treatment of the adjugate
$L^{\ast}$.
\par
\item Arguments such as the proofs of (\ref{pf.thm.3.2.2nd.acI}) and
(\ref{pf.thm.3.2.2nd.aL1L2}) have been expanded significantly, and I have
clarified why $a\left(  L\right)  $ is a polynomial in the entries of the
matrix $L$.
\par
\item I have introduced an extra polynomial ring $K\left[  \lambda\right]  $,
which allows me to clarify the various applications of Weyl's principle to
matrices as well as make Lemma \ref{lem.F} more precise (the $\lambda_{ij}$ in
Lemma \ref{lem.F} are now polynomial indeterminates rather than elements of
$K$). This has the additional advantage that an important puzzle piece (the
invariance of the polynomials $F_{i\left(  1,1\right)  ,\ldots,i\left(
m,d\right)  }$) could be proved in higher generality and in a simpler way. As
a consequence of the latter, a part of the original proof (the argument that
the $B_{jk}$ are algebraically independent, and its consequences) has been
removed for uselessness.
\par
\item Lemma \ref{lem.D} and Lemma \ref{lem.C} have been switched, thus
allowing me to deduce the latter from the former, sidestepping the somewhat
complicated proof in the original paper.
\par
\item The definition of the evaluation $\epsilon_{L}$ and various other parts
of the proof has been rewritten in a clearer fashion.
\par
\item Generally, details have been added.
\end{itemize}
}

\begin{lemma}
\label{lem.A}Let $K$ be a field. Let $d\geq1$. Considered as a polynomial in
the $d^{2}$ indeterminates $l_{ij}$ (for $1\leq i\leq d$ and $1\leq j\leq d$),
the determinant $\Delta=\left\vert l_{ij}\right\vert $ is irreducible.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.A}.]Suppose that $\Delta=AB$. Since $\Delta$ is
linear in each variable $l_{ij}$, the variable $l_{11}$ cannot occur in both
$A$ and $B$. Suppose that $l_{11}$ occurs in $A$. In the expansion of $\Delta$
into monomials, each monomial contains exactly one variable from each row and
column. Hence, none of the variables $l_{1r},l_{s1}$ with $1\leq r\leq d$ and
$1\leq s\leq d$ can occur in $B$.\ \ \ \ \footnote{\textit{Comment:} Here is
why:
\par
Assume (for the sake of contradiction) that one of these variables does occur
in $B$. Let us assume that it is a variable of the form $l_{1r}$ with $1\leq
r\leq d$ (since the case when it is a variable of the form $l_{s1}$ with
$1\leq s\leq d$ is analogous).
\par
Let $S$ be the ring of all polynomials in the indeterminates
\begin{align*}
&  l_{1,1},l_{1,2},\ldots,l_{1,d},\\
&  l_{2,1},l_{2,2},\ldots,l_{2,d},\\
&  \ldots,\\
&  l_{d,1},l_{d,2},\ldots,l_{d,d}%
\end{align*}
over $K$. For any polynomial $g\in S$, we let $\overline{\deg}g$ denote the
total degree of $g$ with respect to the indeterminates $l_{1,1},l_{1,2}%
,\ldots,l_{1,d}$ (that is, the total degree of $g$ when $g$ is considered as a
polynomial in the indeterminates $l_{1,1},l_{1,2},\ldots,l_{1,d}$, while all
the remaining indeterminates%
\begin{align*}
&  l_{2,1},l_{2,2},\ldots,l_{2,d},\\
&  l_{3,1},l_{3,2},\ldots,l_{3,d},\\
&  \ldots,\\
&  l_{d,1},l_{d,2},\ldots,l_{d,d}%
\end{align*}
are regarded as constants). Then, any two nonzero polynomials $g$ and $h$ in
$S$ satisfy $\overline{\deg}\left(  gh\right)  =\overline{\deg}g+\overline
{\deg}h$. Applying this to $g=A$ and $h=B$, we obtain $\overline{\deg}\left(
AB\right)  =\overline{\deg}A+\overline{\deg}B$.
\par
The determinant $\Delta=\left\vert l_{ij}\right\vert $ is homogeneous of
degree $1$ in the variables $l_{1,1},l_{1,2},\ldots,l_{1,d}$ (since the
determinant of a $d\times d$-matrix is linear with respect to its first row).
Thus, $\overline{\deg}\Delta\leq1$.
\par
But the variable $l_{11}$ occurs in $A$; thus, $\overline{\deg}A\geq1$. Also,
a variable of the form $l_{1r}$ with $1\leq r\leq d$ occurs in $B$ (according
to our assumption); thus, $\overline{\deg}B\geq1$. Hence, $\overline{\deg
}\left(  AB\right)  =\underbrace{\overline{\deg}A}_{\geq1}%
+\underbrace{\overline{\deg}B}_{\geq1}\geq1+1>1$. This contradicts
$\overline{\deg}\left(  \underbrace{AB}_{=\Delta}\right)  =\overline{\deg
}\Delta\leq1$. This contradiction shows that our assumption was false. This
completes the proof.}

If $B$ is not a constant polynomial, then $B$ contains a variable $l_{pq}$,
where, a fortiori, $p,q>1$. By a similar argument, $A$ cannot contain the
variables $l_{pr},l_{sq}$ with $r\neq q$ and $s\neq p$. But this implies that
neither $A$ nor $B$ contain the variables $l_{p1}$ and $l_{1q}$, both of which
appear in the expansion of $\Delta$. Hence, $B$ must be a constant polynomial,
and our lemma is proved.
\end{proof}

A technical result we shall use time and again is \textit{Weyl's principle of
the irrelevance of algebraic inequalities}:

\begin{proposition}
[Weyl's principle of the irrelevance of algebraic inequalities]%
\label{prop.weyl}Let $K$ be an infinite field. Let $\left\{  z_{1}%
,\ldots,z_{p}\right\}  $ be a finite set of indeterminates. Let $f,g_{1}%
,\ldots,g_{r}$ be polynomials in $K\left[  z_{1},\ldots,z_{p}\right]  $ such
that $g_{1},\ldots,g_{r}$ are nonzero. Suppose that
\begin{equation}
f\left(  s_{1},\ldots,s_{p}\right)  =0 \label{eq.prop.weyl.ass}%
\end{equation}
for all $s_{1},\ldots,s_{p}\in K$ such that%
\begin{equation}
g_{i}\left(  s_{1},\ldots,s_{p}\right)  \neq0\ \ \ \ \ \ \ \ \ \ \text{for all
}1\leq i\leq r. \label{eq.prop.weyl.req}%
\end{equation}
Then, $f$ is identically zero.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.weyl}.]The proof is routine commutative
algebra (see Weyl \cite[Chapter I, Lemma (1.1.A)]{4}). For the sake of
completeness, let us nevertheless sketch it: We know that $K$ is a field;
thus, the ring $K\left[  z_{1},\ldots,z_{p}\right]  $ is an integral domain.
Let $g$ be the polynomial $g_{1}g_{2}\cdots g_{r}$. Since the polynomials
$g_{1},\ldots,g_{r}$ are nonzero, their product $g_{1}g_{2}\cdots g_{r}$ must
also be nonzero (since $K\left[  z_{1},\ldots,z_{p}\right]  $ is an integral
domain). In other words, $g$ is nonzero (since $g=g_{1}g_{2}\cdots g_{r}$).

Every $s_{1},\ldots,s_{p}\in K$ satisfy%
\begin{equation}
\left(  fg\right)  \left(  s_{1},\ldots,s_{p}\right)  =0
\label{pf.prop.weyl.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.weyl.1}):} Let $s_{1},\ldots,s_{p}\in
K$. If all $1\leq i\leq r$ satisfy $g_{i}\left(  s_{1},\ldots,s_{p}\right)
\neq0$, then
\[
\left(  fg\right)  \left(  s_{1},\ldots,s_{p}\right)  =\underbrace{f\left(
s_{1},\ldots,s_{p}\right)  }_{\substack{=0\\\text{(by (\ref{eq.prop.weyl.ass}%
))}}}g\left(  s_{1},\ldots,s_{p}\right)  =0.
\]
Thus, if all $1\leq i\leq r$ satisfy $g_{i}\left(  s_{1},\ldots,s_{p}\right)
\neq0$, then (\ref{pf.prop.weyl.1}) is proven. Hence, for the rest of the
proof of (\ref{pf.prop.weyl.1}), we WLOG assume that not all $1\leq i\leq r$
satisfy $g_{i}\left(  s_{1},\ldots,s_{p}\right)  \neq0$. Thus, at least one
$1\leq i\leq r$ satisfies $g_{i}\left(  s_{1},\ldots,s_{p}\right)  =0$.
Therefore, the product $\prod_{i=1}^{r}g_{i}\left(  s_{1},\ldots,s_{p}\right)
$ has at least one factor equal to $0$; consequently, the whole product is
$0$. In other words, $\prod_{i=1}^{r}g_{i}\left(  s_{1},\ldots,s_{p}\right)
=0$. But $g=g_{1}g_{2}\cdots g_{r}=\prod_{i=1}^{r}g_{i}$ and therefore%
\[
g\left(  s_{1},\ldots,s_{p}\right)  =\prod_{i=1}^{r}g_{i}\left(  s_{1}%
,\ldots,s_{p}\right)  =0.
\]
Hence,%
\[
\left(  fg\right)  \left(  s_{1},\ldots,s_{p}\right)  =f\left(  s_{1}%
,\ldots,s_{p}\right)  \underbrace{g\left(  s_{1},\ldots,s_{p}\right)  }%
_{=0}=0.
\]
This proves (\ref{pf.prop.weyl.1}).}.

But the field $K$ is infinite. Thus, the subset $K^{p}$ of $K^{p}$ is
Zariski-dense. In elementary terms, this says the following: If $h\in K\left[
z_{1},\ldots,z_{p}\right]  $ is a polynomial such that every $s_{1}%
,\ldots,s_{p}\in K$ satisfy $h\left(  s_{1},\ldots,s_{p}\right)  =0$, then
$h=0$. We can apply this to $h=fg$ (since every $s_{1},\ldots,s_{p}\in K$
satisfy (\ref{pf.prop.weyl.1})). Thus, we conclude that $fg=0$. Since the ring
$K\left[  z_{1},\ldots,z_{p}\right]  $ is an integral domain, we thus have
either $f=0$ or $g=0$ (or both). Since $g$ is nonzero, we thus conclude that
$f=0$. This proves Proposition \ref{prop.weyl}.
\end{proof}

For the sake of convenience, let us specifically state the particular case of
Proposition \ref{prop.weyl} for $r=1$:

\begin{corollary}
\label{cor.weyl.1}Let $K$ be an infinite field. Let $\left\{  z_{1}%
,\ldots,z_{p}\right\}  $ be a finite set of indeterminates. Let $f$ be a
polynomial in $K\left[  z_{1},\ldots,z_{p}\right]  $. Let $g$ be a nonzero
polynomial in $K\left[  z_{1},\ldots,z_{p}\right]  $. Assume that $f\left(
s_{1},\ldots,s_{p}\right)  =0$ for all $s_{1},\ldots,s_{p}\in K$ such that
$g\left(  s_{1},\ldots,s_{p}\right)  \neq0$. Then, $f$ is identically zero.
\end{corollary}

For each $d$-tuple $\left(  i_{1},i_{2},\ldots,i_{d}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{d}$, we adopt the following \textit{bracket} notation:%
\[
\left[  x_{i_{1}},\ldots,x_{i_{d}}\right]  =\left(  x_{i_{1}}\cdots x_{i_{d}%
}\mid u_{1}\cdots u_{d}\right)  \in P.
\]
Any such polynomial $\left[  x_{i_{1}},\ldots,x_{i_{d}}\right]  $ will be
called a \textit{bracket}. A \textit{bracket monomial} shall mean a product of
brackets. The following is easy to see:

\begin{lemma}
\label{lem.bracket}Any bracket monomial is a linear combination of standard
rectangular bideterminants, all of which have the same shape $\left(
d,\ldots,d\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bracket}.]Clearly, any bracket monomial is a
rectangular bideterminant. It thus remains to show that each rectangular
bideterminant is a linear combination of standard rectangular bideterminants,
all of which have the same shape $\left(  d,\ldots,d\right)  $.

In other words, it remains to show that if $\left[  T,T^{\prime}\right]  $ is
a bitableau such that the tableau $T^{\prime}$ is rectangular, then $\left(
T\mid T^{\prime}\right)  $ is a linear combination of standard rectangular
bideterminants, all of which have the same shape $\left(  d,\ldots,d\right)  $.

So let $\left[  T,T^{\prime}\right]  $ be a bitableau such that the tableau
$T^{\prime}$ is rectangular. Let $\left(  \lambda\right)  $ be the shape of
$T^{\prime}$. Thus, $\left(  \lambda\right)  =\underbrace{\left(
d,\ldots,d\right)  }_{g\text{ entries}}$ for some $g\in\mathbb{N}$ (since
$T^{\prime}$ is rectangular).

Now, we notice the following:

\begin{statement}
\textit{Claim 1:} Let $\left[  W,W^{\prime}\right]  $ be a standard bitableau
of the same content as $\left[  T,T^{\prime}\right]  $ and of the same or
longer shape. Then, the tableau $W^{\prime}$ has shape $\left(  \lambda
\right)  $.
\end{statement}

[\textit{Proof of Claim 1:} Assume the contrary. Thus, $W^{\prime}$ does not
have shape $\left(  \lambda\right)  $. In other words, the tableau $W^{\prime
}$ does not have the same shape as $T^{\prime}$ (because the shape of
$T^{\prime}$ is $\left(  \lambda\right)  $).

The tableau $T^{\prime}$ has shape $\left(  \lambda\right)
=\underbrace{\left(  d,\ldots,d\right)  }_{g\text{ entries}}$. Hence, it has
$gd$ entries.

The bitableau $\left[  W,W^{\prime}\right]  $ has the same content as $\left[
T,T^{\prime}\right]  $. Thus, the tableau $W^{\prime}$ has equally many
entries as $T^{\prime}$. Therefore, the tableau $W^{\prime}$ has $gd$ entries
(since the tableau $T^{\prime}$ has $gd$ entries).

Let $\left(  \mu\right)  =\left(  \mu_{1},\mu_{2},\ldots,\mu_{h}\right)  $ be
the shape of the tableau $W^{\prime}$. Therefore, the tableau $W^{\prime}$ has
$\mu_{1}+\mu_{2}+\cdots+\mu_{h}$ entries. Since the tableau $W^{\prime}$ has
$gd$ entries, we thus conclude that $\mu_{1}+\mu_{2}+\cdots+\mu_{h}=gd$.

The bitableau $\left[  W,W^{\prime}\right]  $ has the same or longer shape
than $\left[  T,T^{\prime}\right]  $. In other words, the tableau $W^{\prime}$
has the same or longer shape than $T^{\prime}$. Hence, the tableau $W^{\prime
}$ has longer shape than $T^{\prime}$ (because the tableau $W^{\prime}$ cannot
have the same shape as $T^{\prime}$). In other words, $\left(  \mu_{1},\mu
_{2},\ldots,\mu_{h}\right)  >\underbrace{\left(  d,\ldots,d\right)  }_{g\text{
entries}}$ in lexicographic order (since the tableau $W^{\prime}$ has shape
$\left(  \mu_{1},\mu_{2},\ldots,\mu_{h}\right)  $, while the tableau
$T^{\prime}$ has shape $\underbrace{\left(  d,\ldots,d\right)  }_{g\text{
entries}}$). By the definition of lexicographic order, this means that we must
be in one of the following two cases:

\textit{Case 1:} There exists some $i\in\left\{  1,2,\ldots,\min\left\{
h,g\right\}  \right\}  $ such that $\mu_{i}>d$ and \newline$\left(  \mu
_{j}=d\text{ for each }j<i\right)  $.

\textit{Case 2:} We have $h<g$ and $\left(  \mu_{j}=d\text{ for each }%
j\in\left\{  1,2,\ldots,h\right\}  \right)  $.

Let us first consider Case 1. In this case, there exists some $i\in\left\{
1,2,\ldots,\min\left\{  h,g\right\}  \right\}  $ such that $\mu_{i}>d$ and
$\left(  \mu_{j}=d\text{ for each }j<i\right)  $. Consider this $i$. Then,
$\mu_{i}>d$. But the $i$-th row of $W^{\prime}$ has more than $d$ entries
(since $\left(  \mu_{1},\mu_{2},\ldots,\mu_{h}\right)  $ is the shape of the
tableau $W^{\prime}$). Thus, the $i$-th row of $W^{\prime}$ has more than $d$
entries (since $\mu_{i}>d$). Consequently, two of these entries must be equal
(since the only available entries for $W^{\prime}$ are the $d$ entries
$u_{1},u_{2},\ldots,u_{d}$). This contradicts the fact that the entries in the
$i$-th row of $W^{\prime}$ are strictly increasing (since $W^{\prime}$ is a
standard tableau). Thus, we have found a contradiction in Case 1.

Let us now consider Case 2. In this case, we have $h<g$ and \newline$\left(
\mu_{j}=d\text{ for each }j\in\left\{  1,2,\ldots,h\right\}  \right)  $. Now,%
\begin{align*}
\mu_{1}+\mu_{2}+\cdots+\mu_{h}  &  =\underbrace{h}_{<g}%
d\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{j}=d\text{ for each }%
j\in\left\{  1,2,\ldots,h\right\}  \right) \\
&  <gd\ \ \ \ \ \ \ \ \ \ \left(  \text{since }d>0\right)  .
\end{align*}
This contradicts $\mu_{1}+\mu_{2}+\cdots+\mu_{h}=gd$. Thus, we have obtained a
contradiction in Case 2.

Hence, we have obtained a contradiction in both Cases 1 and 2. This
contradiction shows that our assumption was wrong; hence, Claim 1 is proven.]

Now, Theorem \ref{thm.1} shows that $\left(  T\mid T^{\prime}\right)  $ is a
linear combination, with integer coefficients, of bideterminants $\left(
W\mid W^{\prime}\right)  $ of standard bitableaux $\left[  W,W^{\prime
}\right]  $ of the same content and of the same or longer shape. All these
bideterminants $\left(  W\mid W^{\prime}\right)  $ have shape $\left(
\lambda\right)  $ (since Claim 1 yields that the respective tableaux
$W^{\prime}$ have shape $\left(  \lambda\right)  $) and thus are rectangular
(since $\left(  \lambda\right)  =\underbrace{\left(  d,\ldots,d\right)
}_{g\text{ entries}}$) and have the same shape $\left(  d,\ldots,d\right)  $
(again since $\left(  \lambda\right)  =\underbrace{\left(  d,\ldots,d\right)
}_{g\text{ entries}}$). Thus, $\left(  T\mid T^{\prime}\right)  $ is a linear
combination of standard rectangular bideterminants $\left(  W\mid W^{\prime
}\right)  $, all of which have the same shape $\left(  d,\ldots,d\right)  $.
This proves Lemma \ref{lem.bracket}.
\end{proof}

\begin{lemma}
\label{lem.bracket2}Assume that $n\geq d$. Let $F\in P$ be an invariant form,
and let $g\in\mathbb{N}$ be such that $\left[  x_{1},\ldots,x_{d}\right]
^{g}\cdot F$ is a polynomial in the brackets (i.e., a linear combination of
bracket monomials). Then, $F$ is a linear combination of standard rectangular bideterminants.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.bracket2}.]By Corollary \ref{cor.strf.gen}, we can
write $F$ as a linear combination of bideterminants of standard bitableaux:%
\begin{equation}
F=\sum_{i\in I}b_{i}\left(  U_{i}\mid U_{i}^{\prime}\right)  ,
\label{pf.lem.bracket2.1}%
\end{equation}
where the $\left[  U_{i},U_{i}^{\prime}\right]  $ are distinct standard
bitableaux, and where the $b_{i}$ are nonzero elements of $K$. For each $i\in
I$, each row of the tableau $U_{i}^{\prime}$ has length $\leq d$%
\ \ \ \ \footnote{\textit{Proof.} Let $i\in I$. The tableau $U_{i}^{\prime}$
is standard; thus, each row of $U_{i}^{\prime}$ is strictly increasing. Hence,
each row of $U_{i}^{\prime}$ must have at most $d$ entries (because if it had
more than $d$ entries, then two of these entries would be equal (since the
only available entries for $U_{i}^{\prime}$ are the $d$ letters $u_{1}%
,u_{2},\ldots,u_{d}$), and this would contradict the fact that this row is
strictly increasing). In other words, each row of $U_{i}^{\prime}$ has length
$\leq d$. Qed.}, and therefore each row of the tableau $U_{i}$ has length
$\leq d$ as well (since the length of a row of $U_{i}$ equals the length of
the corresponding row of $U_{i}^{\prime}$).

If each of the bitableaux $\left[  U_{i},U_{i}^{\prime}\right]  $ is
rectangular, then Lemma \ref{lem.bracket2} is proven (because in this case,
(\ref{pf.lem.bracket2.1}) shows that $F$ is a linear combination of standard
rectangular bideterminants). Let us thus assume that this is not the case.
Hence, not all of the bitableaux $\left[  U_{i},U_{i}^{\prime}\right]  $ are rectangular.

For each $i\in I$, we let $\widehat{U}_{i}$ be the tableau%
\[%
\begin{array}
[c]{@{}r@{\,}l}%
g\text{ rows} & \left\{
\begin{array}
[c]{cccc}%
x_{1} & x_{2} & \cdots & x_{d}\\
\vdots & \vdots & \ddots & \vdots\\
x_{1} & x_{2} & \cdots & x_{d}%
\end{array}
\right. \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{i}%
\end{array}
\]
(that is, the result of piling $g$ rows of the form $x_{1}x_{2}\cdots x_{d}$
on top of the tableau $U_{i}$). This tableau $\widehat{U}_{i}$ is standard
(since $U_{i}$ is standard, since each row of the tableau $U_{i}$ has length
$\leq d$, and since $x_{k}$ is the smallest possible entry that an entry in
the $k$-th column of a standard tableau can have).

Similarly, for each $i\in I$, we let $\widehat{U}_{i}^{\prime}$ be the tableau%
\[%
\begin{array}
[c]{@{}r@{\,}l}%
g\text{ rows} & \left\{
\begin{array}
[c]{cccc}%
u_{1} & u_{2} & \cdots & u_{d}\\
\vdots & \vdots & \ddots & \vdots\\
u_{1} & u_{2} & \cdots & u_{d}%
\end{array}
\right. \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ U_{i}^{\prime}%
\end{array}
\]
(that is, the result of piling $g$ rows of the form $u_{1}u_{2}\cdots u_{d}$
on top of the tableau $U_{i}^{\prime}$). A similar argument shows that this
tableau $\widehat{U}_{i}^{\prime}$ is standard. Thus, $\left[  \widehat{U}%
_{i},\widehat{U}_{i}^{\prime}\right]  $ is a standard bitableau for each $i\in
I$. Moreover, the bitableaux $\left[  \widehat{U}_{i},\widehat{U}_{i}^{\prime
}\right]  $ for $i\in I$ are distinct (since the bitableaux $\left[
U_{i},U_{i}^{\prime}\right]  $ are distinct, and since the bitableaux $\left[
\widehat{U}_{i},\widehat{U}_{i}^{\prime}\right]  $ are obtained from $\left[
U_{i},U_{i}^{\prime}\right]  $ by piling $g$ new rows on top). Also notice
that not all of the bitableaux $\left[  \widehat{U}_{i},\widehat{U}%
_{i}^{\prime}\right]  $ are rectangular (since not all of the bitableaux
$\left[  U_{i},U_{i}^{\prime}\right]  $ are rectangular).

Clearly, each $i\in I$ satisfies%
\begin{align*}
\left(  \widehat{U}_{i}\mid\widehat{U}_{i}^{\prime}\right)   &  =\left(
\underbrace{\left(  x_{1}x_{2}\cdots x_{d}\mid u_{1}u_{2}\cdots u_{d}\right)
}_{=\left[  x_{1},x_{2},\ldots,x_{d}\right]  }\right)  ^{g}\left(  U_{i}\mid
U_{i}^{\prime}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the construction of }\widehat{U}%
_{i}\text{ and }\widehat{U}_{i}^{\prime}\right) \\
&  =\left[  x_{1},x_{2},\ldots,x_{d}\right]  ^{g}\left(  U_{i}\mid
U_{i}^{\prime}\right)  .
\end{align*}
Thus,
\begin{align*}
\sum_{i\in I}b_{i}\underbrace{\left(  \widehat{U}_{i}\mid\widehat{U}%
_{i}^{\prime}\right)  }_{=\left[  x_{1},x_{2},\ldots,x_{d}\right]  ^{g}\left(
U_{i}\mid U_{i}^{\prime}\right)  }  &  =\left[  x_{1},x_{2},\ldots
,x_{d}\right]  ^{g}\cdot\underbrace{\sum_{i\in I}b_{i}\left(  U_{i}\mid
U_{i}^{\prime}\right)  }_{\substack{=F\\\text{(by (\ref{pf.lem.bracket2.1}))}%
}}\\
&  =\left[  x_{1},x_{2},\ldots,x_{d}\right]  ^{g}\cdot F.
\end{align*}
Thus, the decomposition of $\left[  x_{1},x_{2},\ldots,x_{d}\right]  ^{g}\cdot
F$ into a linear combination of standard bideterminants is $\sum_{i\in I}%
b_{i}\left(  \widehat{U}_{i}\mid\widehat{U}_{i}^{\prime}\right)  $. (We are
allowed to speak of \textquotedblleft the\textquotedblright\ decomposition,
because Corollary \ref{cor.2.2.basis} shows that there is a unique such decomposition.)

But $\left[  x_{1},\ldots,x_{d}\right]  ^{g}\cdot F$ is a polynomial in the
brackets, i.e., a linear combination of bracket monomials. Hence, $\left[
x_{1},\ldots,x_{d}\right]  ^{g}\cdot F$ is a linear combination of standard
rectangular bideterminants (since Lemma \ref{lem.bracket} shows that any
bracket monomial is a linear combination of standard rectangular
bideterminants). Thus, the decomposition of $\left[  x_{1},\ldots
,x_{d}\right]  ^{g}\cdot F$ into a linear combination of standard
bideterminants has the property that all bideterminants appearing in it are
rectangular. Since this decomposition is $\sum_{i\in I}b_{i}\left(
\widehat{U}_{i}\mid\widehat{U}_{i}^{\prime}\right)  $, this shows that all of
the bideterminants $\left(  \widehat{U}_{i}\mid\widehat{U}_{i}^{\prime
}\right)  $ are rectangular. In other words, all of the bitableaux $\left[
\widehat{U}_{i},\widehat{U}_{i}^{\prime}\right]  $ are rectangular. This
contradicts the fact that not all of the bitableaux $\left[  \widehat{U}%
_{i},\widehat{U}_{i}^{\prime}\right]  $ are rectangular. This contradiction
shows that our assumption was wrong. As we have seen, this completes the proof
of Lemma \ref{lem.bracket2}.
\end{proof}

\begin{lemma}
\label{lem.E}Assume that $n\geq d$. In the letter place algebra $P$, we have%
\[
\left[  x_{1},\ldots,x_{d}\right]  \left(  x_{j}\mid u_{m}\right)  =\sum
_{k=1}^{d}\left[  x_{1},\ldots,x_{k-1},x_{j},x_{k+1},\ldots,x_{d}\right]
\left(  x_{k}\mid u_{m}\right)
\]
for any $1\leq j\leq n$ and $1\leq m\leq d$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.E}.]The identity follows from expanding, by Laplace's
rule, the inner product $\left(  x_{1}\cdots x_{d}x_{j}\mid u_{1}\cdots
u_{d}u_{m}\right)  $, which is identically zero in $P$ (since it is a
determinant with two equal columns).\footnote{In more detail:
\par
Applying (\ref{eq.laplace.1}) to $p=d+1$, $\left(  i_{1},i_{2},\ldots
,i_{p}\right)  =\left(  j,1,2,\ldots,d\right)  $ and $\left(  j_{1}%
,j_{2},\ldots,j_{p}\right)  =\left(  m,1,2,\ldots,d\right)  $, we obtain%
\begin{align*}
&  \left(  x_{j}x_{1}\cdots x_{d}\mid u_{m}u_{1}\cdots u_{d}\right) \\
&  =\left(  x_{j}\mid u_{m}\right)  \underbrace{\left(  x_{1}\cdots x_{d}\mid
u_{1}\cdots u_{d}\right)  }_{=\left[  x_{1},\ldots,x_{d}\right]  }\\
&  \ \ \ \ \ \ \ \ \ \ +\sum_{k=1}^{d}\left(  -1\right)  ^{k}%
\underbrace{\left(  x_{k}\mid u_{m}\right)  \left(  x_{j}x_{1}\cdots
x_{k-1}x_{k+1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)  }_{=\left(
x_{j}x_{1}\cdots x_{k-1}x_{k+1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)
\left(  x_{k}\mid u_{m}\right)  }\\
&  =\left(  x_{j}\mid u_{m}\right)  \left[  x_{1},\ldots,x_{d}\right]
+\sum_{k=1}^{d}\left(  -1\right)  ^{k}\underbrace{\left(  x_{j}x_{1}\cdots
x_{k-1}x_{k+1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)  }_{=\left(
-1\right)  ^{k-1}\left(  x_{1}\cdots x_{k-1}x_{j}x_{k+1}\cdots x_{d}\mid
u_{1}\cdots u_{d}\right)  }\left(  x_{k}\mid u_{m}\right) \\
&  =\underbrace{\left(  x_{j}\mid u_{m}\right)  \left[  x_{1},\ldots
,x_{d}\right]  }_{=\left[  x_{1},\ldots,x_{d}\right]  \left(  x_{j}\mid
u_{m}\right)  }+\sum_{k=1}^{d}\underbrace{\left(  -1\right)  ^{k}\left(
-1\right)  ^{k-1}}_{=-1}\underbrace{\left(  x_{1}\cdots x_{k-1}x_{j}%
x_{k+1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)  }_{=\left[  x_{1}%
,\ldots,x_{k-1},x_{j},x_{k+1},\ldots,x_{d}\right]  }\left(  x_{k}\mid
u_{m}\right) \\
&  =\left[  x_{1},\ldots,x_{d}\right]  \left(  x_{j}\mid u_{m}\right)
-\sum_{k=1}^{d}\left[  x_{1},\ldots,x_{k-1},x_{j},x_{k+1},\ldots,x_{d}\right]
\left(  x_{k}\mid u_{m}\right)  .
\end{align*}
But $\left(  x_{j}x_{1}\cdots x_{d}\mid u_{m}u_{1}\cdots u_{d}\right)  =0$
(since $\left(  x_{j}x_{1}\cdots x_{d}\mid u_{m}u_{1}\cdots u_{d}\right)  $ is
a determinant having two equal columns (since $u_{m}$ also appears among
$u_{1},u_{2},\ldots,u_{d}$)). Thus,%
\begin{align*}
0  &  =\left(  x_{j}x_{1}\cdots x_{d}\mid u_{m}u_{1}\cdots u_{d}\right) \\
&  =\left[  x_{1},\ldots,x_{d}\right]  \left(  x_{j}\mid u_{m}\right)
-\sum_{k=1}^{d}\left[  x_{1},\ldots,x_{k-1},x_{j},x_{k+1},\ldots,x_{d}\right]
\left(  x_{k}\mid u_{m}\right)  .
\end{align*}
In other words,%
\[
\left[  x_{1},\ldots,x_{d}\right]  \left(  x_{j}\mid u_{m}\right)  =\sum
_{k=1}^{d}\left[  x_{1},\ldots,x_{k-1},x_{j},x_{k+1},\ldots,x_{d}\right]
\left(  x_{k}\mid u_{m}\right)  .
\]
This proves Lemma \ref{lem.E}.}
\end{proof}

\begin{proof}
[Second proof of Theorem \ref{thm.3.2}.]Recall that a form $F$ is invariant
if, for all invertible $d \times d$-matrices $L \in K^{d \times d}$, there
exists a scalar $a\left(  L\right)  $ such that $LF=a\left(  L\right)  F$. Let
$F=F_{0}+\cdots+F_{t}$ be the decomposition of the polynomial $F$ into
homogeneous components with respect to the \textit{total} degree (such that
each component $F_{i}$ has total degree $i$). Then, if $F$ is invariant,%
\[
LF=LF_{0}+\cdots+LF_{t},
\]
whence%
\[
LF_{0}+\cdots+LF_{t}=LF=a\left(  L\right)  F=a\left(  L\right)  F_{0}%
+\cdots+a\left(  L\right)  F_{t}.
\]
Since the action of $L$ on $P$ preserves the total degree, we must thus have%
\[
LF_{i}=a\left(  L\right)  F_{i}\ \ \ \ \ \ \ \ \ \ \text{for all }0\leq i\leq
t.
\]
In other words, all the homogeneous components $F_{0},F_{1},\ldots,F_{t}$ of
$F$ are themselves invariant. It suffices, therefore, to consider only
invariant forms that are homogeneous.

Now, fix a nonzero homogeneous invariant form $F$ of degree $t$. Then, for all
invertible $d \times d$-matrices $L \in K^{d \times d}$, there exists a scalar
$a\left(  L\right)  $ such that $LF=a\left(  L\right)  F$ (since $F$ is
invariant). This scalar $a\left(  L\right)  $ is uniquely determined by $L$
and $F$ (since $F$ is nonzero), and is a polynomial in the entries $l_{jk}$ of
the matrix $L$\ \ \ \ \footnote{\textit{Comment:} To be more precise: There
exists a polynomial map $\mathbf{a}:K^{d\times d}\rightarrow K$ such that
every invertible $d\times d$-matrix $L \in K^{d \times d}$ satisfies $a\left(
L\right)  =\mathbf{a}\left(  L\right)  $.
\par
This can be proven as follows:
\par
Since $F$ is nonzero, there exists some monomial $\mathfrak{m}$ such that the
coefficient of $\mathfrak{m}$ in $F$ is $\neq0$. Fix such an $\mathfrak{m}$.
Let $\lambda$ be the coefficient of $\mathfrak{m}$ in $F$; thus, $\lambda
\neq0$. Now, each invertible $d\times d$-matrix $L \in K^{d \times d}$
satisfies%
\begin{align*}
&  \left(  \text{the coefficient of }\mathfrak{m}\text{ in }\underbrace{LF}%
_{=a\left(  L\right)  F}\right) \\
&  =\left(  \text{the coefficient of }\mathfrak{m}\text{ in }a\left(
L\right)  F\right)  =a\left(  L\right)  \cdot\underbrace{\left(  \text{the
coefficient of }\mathfrak{m}\text{ in }F\right)  }_{=\lambda}=a\left(
L\right)  \cdot\lambda
\end{align*}
and therefore%
\[
a\left(  L\right)  =\dfrac{1}{\lambda}\cdot\left(  \text{the coefficient of
}\mathfrak{m}\text{ in }LF\right)  .
\]
Therefore, $a\left(  L\right)  $ is a polynomial in the entries of the matrix
$L$ (since the coefficient of $\mathfrak{m}$ in $LF$ is a polynomial in the
entries of the matrix $L$). This is what we wanted to prove.}. If $c\in K$ is
a nonzero scalar, then we have%
\begin{equation}
a\left(  cI\right)  =c^{t}. \label{pf.thm.3.2.2nd.acI}%
\end{equation}


[\textit{Proof:} Let $c\in K$ be a nonzero scalar. Consider the invertible
$d\times d$-matrix $L \in K^{d \times d}$ defined by $L=cI$, where $I$ is the
identity matrix. Under the action of this $L$ on $P$, we have $\left(
x_{i}\mid u_{j}\right)  \mapsto c\left(  x_{i}\mid u_{j}\right)  $. Thus,
$LF=c^{t}F$ (since $F$ is homogeneous of degree $t$). Thus, the formula
$LF=a\left(  L\right)  F$ becomes $c^{t}F=a\left(  L\right)  F=a\left(
cI\right)  F$ (since $L=cI$). Hence, $c^{t}=a\left(  cI\right)  $ (since $F$
is nonzero), qed.]

Moreover, the function $a\left(  L\right)  $ is multiplicative, in the sense
that for any two invertible $d \times d$-matrices $L_{1}$ and $L_{2}$ in $K^{d
\times d}$, we have%
\begin{equation}
a\left(  L_{1}L_{2}\right)  =a\left(  L_{1}\right)  a\left(  L_{2}\right)  .
\label{pf.thm.3.2.2nd.aL1L2}%
\end{equation}


[\textit{Proof:} Let $L_{1}$ and $L_{2}$ be two invertible $d\times
d$-matrices in $K^{d \times d}$. The formula $LF=a\left(  L\right)  F$ then
yields
\[
L_{1}F=a\left(  L_{1}\right)  F,\ \ \ \ \ \ \ \ \ \ L_{2}F=a\left(
L_{2}\right)  F,\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ L_{1}%
L_{2}F=a\left(  L_{1}L_{2}\right)  F.
\]
Hence, $a\left(  L_{1}L_{2}\right)  F=L_{1}\underbrace{L_{2}F}_{=a\left(
L_{2}\right)  F}=a\left(  L_{2}\right)  \underbrace{L_{1}F}_{=a\left(
L_{1}\right)  F}=a\left(  L_{1}\right)  a\left(  L_{2}\right)  F$. Since $F$
is nonzero, this results in $a\left(  L_{1}L_{2}\right)  =a\left(
L_{1}\right)  a\left(  L_{2}\right)  $, qed.]

Given a $d\times d$-matrix $L = \left(  l_{jk}\right)  \in K^{d \times d} $,
its adjugate $L^{\ast}$ is the $d \times d$-matrix $\left(  l_{jk}^{\ast
}\right)  $, where%
\[
l_{jk}^{\ast}=\left(  \text{the }kj\text{th cofactor of the matrix } L\right)
=\left(  -1\right)  ^{j+k}\left\vert l_{pq}\right\vert _{p\neq k,\ q\neq j}.
\]
The adjugate has the property%
\begin{equation}
LL^{\ast}=\left\vert l_{ij}\right\vert I. \label{pf.thm.3.2.2nd.LL*}%
\end{equation}


[\textit{Proof of (\ref{pf.thm.3.2.2nd.LL*}):} Fix $1\leq s\leq d$ and $1\leq
t\leq d$. The $st$th entry of the matrix $LL^{\ast}$ is%
\[
\sum_{m}l_{sm}l_{mt}^{\ast}=\sum_{m}\left(  -1\right)  ^{m+t}l_{sm}\left\vert
l_{pq}\right\vert _{p\neq t,\ q\neq m}.
\]
By the Laplace expansion, the right-hand side is the determinant of the matrix
$\left(  l_{jk}\right)  $ with the $t$th row replaced by the row vector
$\left(  l_{sm}\right)  _{1\leq m\leq d}$. This determinant is zero if $s\neq
t$ (since it has two equal rows in this case) and equals $\left\vert
l_{ij}\right\vert $ if $s=t$. This is, of course, precisely the $st$th entry
of the matrix $\left\vert l_{ij}\right\vert I$. Hence,
(\ref{pf.thm.3.2.2nd.LL*}) follows.]

As an immediate consequence of (\ref{pf.thm.3.2.2nd.LL*}), we obtain%
\[
a\left(  L\right)  a\left(  L^{\ast}\right)  =\left\vert l_{ij}\right\vert
^{t}%
\]
(because of (\ref{pf.thm.3.2.2nd.aL1L2}) and (\ref{pf.thm.3.2.2nd.acI})). As
the determinant is irreducible (by Lemma \ref{lem.A}%
)\footnote{\textit{Comment:} This requires $d\geq1$. In the (degenerate) case
$d=0$, everything is obvious anyway.}, each of the factors on the left must
therefore also be a power of the determinant (since the polynomial ring $P$ is
a unique factorization domain). In particular, $a\left(  L\right)  $ is a
power of the determinant. In other words, there is a nonnegative integer $g$
such that every invertible $d \times d$-matrix $L \in K^{d \times d}$
satisfies $a\left(  L\right)  =\left(  \det L\right)  ^{g}$. For this $g$,
every invertible $d \times d$-matrix $L \in K^{d \times d}$ satisfies
$LF=\underbrace{a\left(  L\right)  }_{=\left(  \det L\right)  ^{g}}F=\left(
\det L\right)  ^{g}F$. We have thus proved:

\begin{lemma}
\label{lem.B}Let $F\in P$ be a nonzero homogeneous invariant form. Then, there
is a nonnegative integer $g$ such that every invertible $d \times d$-matrix $L
\in K^{d \times d}$ satisfies $LF=\left(  \det L\right)  ^{g}F$.
\end{lemma}

Let us now introduce a polynomial ring, which we will call $K\left[
\lambda\right]  $. Namely, we let $K\left[  \lambda\right]  $ be the
polynomial ring over $K$ in the $d^{2}$ indeterminates%
\begin{align*}
&  \lambda_{1,1},\lambda_{1,2},\ldots,\lambda_{1,d},\\
&  \lambda_{2,1},\lambda_{2,2},\ldots,\lambda_{2,d},\\
&  \vdots,\\
&  \lambda_{d,1},\lambda_{d,2},\ldots,\lambda_{d,d}.
\end{align*}
More generally, if $R$ is any commutative ring, then $R\left[  \lambda\right]
$ shall mean the polynomial ring over $R$ in these $d^{2}$ indeterminates.

If $f\in K\left[  \lambda\right]  $ is a polynomial, if $A$ is a commutative
$K$-algebra, and if $M=\left(  m_{j,k}\right)  _{1\leq j\leq d,\ 1\leq k\leq
d}\in A^{d\times d}$ is a $d\times d$-matrix over $A$, then $f\left[
M\right]  $ shall mean the result of substituting $a_{j,k}$ for each
indeterminate $\lambda_{j,k}$ in $f$. For example, if $f=\lambda_{1,1}%
+\lambda_{2,2}+\cdots+\lambda_{d,d}$, then $f\left[  M\right]  =m_{1,1}%
+m_{2,2}+\cdots+m_{d,d}=\operatorname*{Tr}M$ for any $M=\left(  m_{j,k}%
\right)  _{1\leq j\leq d,\ 1\leq k\leq d}\in A^{d\times d}$.

We let $\det\lambda$ be the polynomial $\sum_{\sigma\in\mathfrak{S}_{d}%
}\operatorname*{sgn}\left(  \sigma\right)  \lambda_{1,\sigma\left(  1\right)
}\lambda_{2,\sigma\left(  2\right)  }\cdots\lambda_{d,\sigma\left(  d\right)
}\in K\left[  \lambda\right]  $. Thus, $\left(  \det\lambda\right)  \left[
M\right]  =\det M$ for any $M=\left(  m_{j,k}\right)  _{1\leq j\leq d,\ 1\leq
k\leq d}\in A^{d\times d}$.

The following particular case of Corollary \ref{cor.weyl.1} will be
particularly useful:

\begin{corollary}
\label{cor.weyl.2}Let $K$ be an infinite field. Let $f$ be a polynomial in
$K\left[  \lambda\right]  $. Assume that $f\left[  L\right]  =0$ for every
invertible $d \times d$-matrix $L \in K^{d \times d}$. Then, $f=0$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.weyl.2}.]We have assumed that $f\left[  L\right]
=0$ for every invertible $d \times d$-matrix $L \in K^{d \times d}$. In other
words, $f\left[  M\right]  =0$ for every invertible $d\times d$-matrix $M\in
K^{d\times d}$ (here, we have renamed the index $L$ as $M$). In other words,
$f\left[  M\right]  =0$ for every matrix $M\in K^{d\times d}$ such that $\det
M\neq0$ (since a $d\times d$-matrix $M\in K^{d\times d}$ is invertible if and
only if $\det M\neq0$). In other words, $f\left[  M\right]  =0$ for every
matrix $M\in K^{d\times d}$ such that $\left(  \det\lambda\right)  \left[
M\right]  \neq0$ (since $\left(  \det\lambda\right)  \left[  M\right]  =\det
M$). In other words, $f\left(  \left(  m_{j,k}\right)  _{1\leq j\leq d,\ 1\leq
k\leq d}\right)  =0$ for every family $\left(  m_{j,k}\right)  _{1\leq j\leq
d,\ 1\leq k\leq d}\in K^{d\times d}$ such that $\left(  \det\lambda\right)
\left(  \left(  m_{j,k}\right)  _{1\leq j\leq d,\ 1\leq k\leq d}\right)
\neq0$. Therefore, Corollary \ref{cor.weyl.1} (applied to $\left\{
z_{1},\ldots,z_{p}\right\}  =\left\{  \lambda_{j,k}\ \mid\ 1\leq j\leq d\text{
and }1\leq k\leq d\right\}  $ and $g=\det\lambda$) shows that $f$ is
identically zero (since $\det\lambda$ is nonzero). This proves Corollary
\ref{cor.weyl.2}.
\end{proof}

We also introduce the following notation: Let $L = \left(  l_{jk}\right)  $ be
a $d \times d$-matrix in $K^{d \times d}$. The evaluation $\epsilon_{L}$ is
the $K$-algebra homomorphism from $P$ to $K$ given by%
\[
\left(  x_{j}\mid u_{k}\right)  \mapsto%
\begin{cases}
l_{kj}, & \text{if }1\leq j\leq d;\\
0, & \text{otherwise}%
\end{cases}
.
\]
An easy computation shows that the evaluation satisfies%
\begin{equation}
\epsilon_{L}F=\epsilon_{I}LF\ \ \ \ \ \ \ \ \ \ \text{for any form }F\in P,
\label{pf.thm.3.2.2nd.eLF}%
\end{equation}
where $I$ is the identity matrix.\footnote{\textit{Proof.} We need to show
that the maps%
\begin{align*}
P  &  \rightarrow K,\ \ \ \ \ \ \ \ \ \ F\mapsto\epsilon_{L}%
F\ \ \ \ \ \ \ \ \ \ \text{and}\\
P  &  \rightarrow K,\ \ \ \ \ \ \ \ \ \ F\mapsto\epsilon_{I}LF
\end{align*}
are identical. Since both of these maps are $K$-algebra homomorphism, it
suffices to show that these maps send the form $\left(  x_{j}\mid
u_{k}\right)  $ (for any $j$ and $k$) to the same image. In other words, it
suffices to show that $\epsilon_{L}\left(  x_{j}\mid u_{k}\right)
=\epsilon_{I}L\left(  x_{j}\mid u_{k}\right)  $ for all $j$ and $k$.
\par
So let us fix $j$ and $k$. The definition of the action of $L$ on $P$ yields
$L\left(  x_{j}\mid u_{k}\right)  =\sum_{1\leq r\leq d}l_{kr}\left(  x_{j}\mid
u_{r}\right)  $. Hence,%
\begin{align*}
\epsilon_{I}L\left(  x_{j}\mid u_{k}\right)   &  =\epsilon_{I}\left(
\sum_{1\leq r\leq d}l_{kr}\left(  x_{j}\mid u_{r}\right)  \right)
=\sum_{1\leq r\leq d}l_{kr}\underbrace{\epsilon_{I}\left(  x_{j}\mid
u_{r}\right)  }_{\substack{=%
\begin{cases}
1, & \text{if }j=r;\\
0, & \text{if }j\neq r
\end{cases}
\\\text{(by the definition of }\epsilon_{I}\text{)}}}=\sum_{1\leq r\leq
d}l_{kr}%
\begin{cases}
1, & \text{if }j=r;\\
0, & \text{if }j\neq r
\end{cases}
\\
&  =%
\begin{cases}
l_{kj}, & \text{if }1\leq j\leq d;\\
0, & \text{otherwise}%
\end{cases}
.
\end{align*}
Comparing this with%
\[
\epsilon_{L}\left(  x_{j}\mid u_{k}\right)  =%
\begin{cases}
l_{kj}, & \text{if }1\leq j\leq d;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\epsilon_{L}\right)
,
\]
this yields $\epsilon_{L}\left(  x_{j}\mid u_{k}\right)  =\epsilon_{I}L\left(
x_{j}\mid u_{k}\right)  $. This completes our proof.}

For our next considerations, it will be helpful to consider the rings $P$ for
various values of $n$ simultaneously. This necessitates the following
notation: We denote the $K$-algebra $P$ by $P_{n}$ when we want to stress its
dependence on $n$. Then,
\begin{equation}
\text{the }K\text{-algebra }P_{m}\text{ canonically becomes a }%
K\text{-subalgebra of }P_{n} \label{pf.thm.3.2.2nd.embed}%
\end{equation}
whenever $m\in\mathbb{N}$ and $n\in\mathbb{N}$ satisfy $m\leq n$ (because each
of the indeterminates $\left(  x_{i}\mid u_{j}\right)  $ appearing in $P_{m}$
is one of the indeterminates of $P_{n}$ as well). Any $d \times d$-matrix $L
\in K^{d \times d}$ acts on $P_{m}$ for all $m\in\mathbb{N}$, and these
actions are compatible (i.e., the canonical inclusions $P_{m}\rightarrow
P_{n}$ commute with the actions of $L$).

Thus, for every $m\in\left\{  0,1,\ldots,n\right\}  $, the $K$-algebra $P_{m}$
is a subalgebra of $P_{n}=P$, and the action of any $d \times d$-matrix $L \in
K^{d \times d}$ on $P_{m}$ is the restriction of the corresponding action on
$P_{n}$. When $F$ is an element of the $K$-algebra $P_{m}$, we shall write
$F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  $ for $F$ in order to
stress that it belongs to $P_{m}$.

(For example, if $n=4$ and $d=3$, then the form $\left(  x_{2}\mid
u_{3}\right)  \left(  x_{3}\mid u_{3}\right)  $ belongs not only to $P=P_{4}$
but also to $P_{3}$.)

\begin{lemma}
\label{lem.D}Assume that $n\geq d$. If $F\left(  \mathbf{x}_{1},\ldots
,\mathbf{x}_{d}\right)  \in P_{d}$ is a nonzero homogeneous invariant, then
\[
F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)  =c\left(  x_{1}\cdots
x_{d}\mid u_{1}\cdots u_{d}\right)  ^{g}%
\]
for some nonnegative integer $g$, and scalar $c$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.D}.]The $K$-algebra homomorphism%
\[
\psi:K\left[  \lambda\right]  \rightarrow P_{d},\ \ \ \ \ \ \ \ \ \ f\mapsto
f\left(  \left(  \left(  x_{j}\mid u_{k}\right)  \right)  _{1\leq j\leq
d,\ 1\leq k\leq d}\right)
\]
is a $K$-algebra isomorphism\footnote{In fact, $K\left[  \lambda\right]  $ is
the polynomial ring in the $d^{2}$ indeterminates $\lambda_{j,k}$ for $1\leq
j\leq d$ and $1\leq k\leq d$, whereas $P_{d}$ is the polynomial ring in the
$d^{2}$ indeterminates $\left(  x_{j}\mid u_{k}\right)  $ for $1\leq j\leq d$
and $1\leq k\leq d$. The homomorphism $\psi$ sends each of the former
indeterminates $\lambda_{j,k}$ to the corresponding $\left(  x_{j}\mid
u_{k}\right)  $; thus, it can be viewed as just a renaming of the
indeterminates.}. Define $f\in K\left[  \lambda\right]  $ by $f=\psi
^{-1}\left(  F\right)  $. Thus,%
\begin{equation}
F=\psi\left(  f\right)  =f\left(  \left(  \left(  x_{j}\mid u_{k}\right)
\right)  _{1\leq j\leq d,\ 1\leq k\leq d}\right)  . \label{pf.lem.D.0}%
\end{equation}
Hence, every $d \times d$-matrix $L \in K^{d \times d}$ satisfies%
\begin{equation}
\epsilon_{L}F=f\left[  L\right]  \label{pf.lem.D.1}%
\end{equation}
\footnote{\textit{Proof.} Let $L$ be a $d \times d$-matrix in $K^{d \times d}%
$. Write this matrix $L$ in the form $L = \left(  l_{kj}\right)  $.
\par
Let $\omega_{L}:K\left[  \lambda\right]  \rightarrow K$ be the $K$-algebra
homomorphism sending each $g\in K\left[  \lambda\right]  $ to $g\left[
L\right]  $.
\par
Every $j$ and $k$ satisfy%
\begin{align*}
\left(  \epsilon_{L}\circ\psi\right)  \left(  \lambda_{j,k}\right)   &
=\epsilon_{L}\left(  \underbrace{\psi\left(  \lambda_{j,k}\right)
}_{\substack{=\left(  x_{j}\mid u_{k}\right)  \\\text{(by the definition of
}\psi\text{)}}}\right)  =\epsilon_{L}\left(  x_{j}\mid u_{k}\right) \\
&  =%
\begin{cases}
l_{kj}, & \text{if }1\leq j\leq d;\\
0, & \text{otherwise}%
\end{cases}
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\epsilon_{L}\right) \\
&  =l_{kj}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }1\leq j\leq d\right) \\
&  =\omega_{L}\left(  \lambda_{j,k}\right)
\end{align*}
(since the definition of $\omega_{L}$ yields $\omega_{L}\left(  \lambda
_{j,k}\right)  =\lambda_{j,k}\left[  L\right]  =l_{kj}$). In other words, the
two $K$-algebra homomorphisms $\epsilon_{L}\circ\psi$ and $\omega_{L}$ from
$K\left[  \lambda\right]  $ to $K$ are equal to each other on the generators
$\lambda_{j,k}$ of the $K$-algebra $K\left[  \lambda\right]  $. Hence, these
two homomorphisms must be identical. In other words, $\epsilon_{L}\circ
\psi=\omega_{L}$. Applying both sides of this equality to $f$, we find
$\left(  \epsilon_{L}\circ\psi\right)  \left(  f\right)  =\omega_{L}\left(
f\right)  =f\left[  L\right]  $ (by the definition of $\omega_{L}$). Hence,
$f\left[  L\right]  =\left(  \epsilon_{L}\circ\psi\right)  \left(  f\right)
=\epsilon_{L}\left(  \underbrace{\psi\left(  f\right)  }_{=F}\right)
=\epsilon_{L}\left(  F\right)  $, qed.}.

Denote the scalar $\epsilon_{I}F\in K$ by $c$.

Lemma \ref{lem.B} shows that there is a nonnegative integer $g$ such that
every invertible $d \times d$-matrix $L \in K^{d \times d}$ satisfies
$LF=\left(  \det L\right)  ^{g}F$. Consider this $g$.

Now, any invertible $d \times d$-matrix $L \in K^{d \times d}$ satisfies%
\begin{align*}
f\left[  L\right]   &  =\epsilon_{L}F\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.D.1})}\right) \\
&  =\epsilon_{I}\underbrace{LF}_{=\left(  \det L\right)  ^{g}F}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.3.2.2nd.eLF})}\right) \\
&  =\left(  \det L\right)  ^{g}\underbrace{\epsilon_{I}F}_{=c}=\left(  \det
L\right)  ^{g}c=c\left(  \underbrace{\det L}_{=\left(  \det\lambda\right)
\left[  L\right]  }\right)  ^{g}=c\left(  \left(  \det\lambda\right)  \left[
L\right]  \right)  ^{g}\\
&  =\left(  c\left(  \det\lambda\right)  ^{g}\right)  \left[  L\right]  .
\end{align*}
In other words, any invertible $d \times d$-matrix $L \in K^{d \times d}$
satisfies $\left(  f-c\left(  \det\lambda\right)  ^{g}\right)  \left[
L\right]  =0$. Therefore, Corollary \ref{cor.weyl.2} (applied to $f-c\left(
\det\lambda\right)  ^{g}$ instead of $f$) shows that $f-c\left(  \det
\lambda\right)  ^{g}=0$. Thus, $f=c\left(  \det\lambda\right)  ^{g}$. Hence,
(\ref{pf.lem.D.0}) becomes%
\begin{align*}
F  &  =\underbrace{f}_{=c\left(  \det\lambda\right)  ^{g}}\left(  \left(
\left(  x_{j}\mid u_{k}\right)  \right)  _{1\leq j\leq d,\ 1\leq k\leq
d}\right) \\
&  =\left(  c\left(  \det\lambda\right)  ^{g}\right)  \left(  \left(  \left(
x_{j}\mid u_{k}\right)  \right)  _{1\leq j\leq d,\ 1\leq k\leq d}\right) \\
&  =c\left(  \underbrace{\det\left(  \left(  x_{j}\mid u_{k}\right)  \right)
_{1\leq j\leq d,\ 1\leq k\leq d}}_{=\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots
u_{d}\right)  }\right)  ^{g}=c\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots
u_{d}\right)  ^{g}.
\end{align*}
This proves Lemma \ref{lem.D}.
\end{proof}

\begin{lemma}
\label{lem.C}If $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  \in
P_{m}$ is a nonzero homogeneous invariant form with $m<d$, then $F$ is constant.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.C}.]Let $F\left(  \mathbf{x}_{1},\ldots
,\mathbf{x}_{m}\right)  \in P_{m}$ be a nonzero homogeneous invariant form
with $m<d$.

If $n<d$, then $P_{n}$ canonically becomes a $K$-subalgebra of $P_{d}$ (by
(\ref{pf.thm.3.2.2nd.embed})). Hence, we WLOG assume that $n\geq d$ (because
if $n<d$, then we can replace $n$ by $d$, and perform the argument below in
$P_{d}$ instead of $P_{n}$).

Now, $m<d$. Hence, (\ref{pf.thm.3.2.2nd.embed}) (applied to $d$ instead of
$n$) shows that $P_{m}$ canonically becomes a $K$-subalgebra of $P_{d}$. Thus,
$P_{m}\subseteq P_{d}$.

Now, $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  \in P_{m}\subseteq
P_{d}$. Therefore, we can write $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}%
_{d}\right)  $ for $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  $.
Since this form $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)
=F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  $ is invariant, we can
thus apply Lemma \ref{lem.D}, and conclude that
\begin{equation}
F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)  =c\left(  x_{1}\cdots
x_{d}\mid u_{1}\cdots u_{d}\right)  ^{g} \label{pf.lem.C.1}%
\end{equation}
for some nonnegative integer $g$, and scalar $c$. Consider these $g$ and $c$.

Consider the $K$-algebra homomorphism%
\[
\eta:P_{d}\rightarrow P_{m},\ \ \ \ \ \ \ \ \ \ \left(  x_{j}\mid
u_{k}\right)  \mapsto%
\begin{cases}
\left(  x_{j}\mid u_{k}\right)  , & \text{if }1\leq j\leq m;\\
0, & \text{otherwise}%
\end{cases}
.
\]
This homomorphism $\eta$ sends
\[
\left(  x_{d}\mid u_{k}\right)  \mapsto%
\begin{cases}
\left(  x_{d}\mid u_{k}\right)  , & \text{if }1\leq d\leq m;\\
0, & \text{otherwise}%
\end{cases}
=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }d>m\right)
\]
for each $k$. Thus, it sends the whole $d$-th row of the determinant $\left(
x_{1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)  $ to $0$. Consequently, it
also sends this determinant to $0$. In other words,%
\[
\eta\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)  =0.
\]
But on the other hand, $\eta$ acts as the identity on the subalgebra $P_{m}$
of $P_{d}$ (since it fixes $\left(  x_{j}\mid u_{k}\right)  $ whenever $1\leq
j\leq m$); therefore, it fixes $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}%
_{d}\right)  $ (since $F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)
=F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{m}\right)  \in P_{m}$). In other
words, $\eta\left(  F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)
\right)  =F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)  $. Now,
applying the map $\eta$ to the equality (\ref{pf.lem.C.1}), we find%
\begin{align*}
\eta\left(  F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)  \right)   &
=\eta\left(  c\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots u_{d}\right)
^{g}\right)  =c\left(  \underbrace{\eta\left(  x_{1}\cdots x_{d}\mid
u_{1}\cdots u_{d}\right)  }_{=0}\right)  ^{g}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\eta\text{ is a }K\text{-algebra
homomorphism}\right) \\
&  =c0^{g}.
\end{align*}
Hence, $c0^{g}=\eta\left(  F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}%
_{d}\right)  \right)  =F\left(  \mathbf{x}_{1},\ldots,\mathbf{x}_{d}\right)
\neq0$ (since $F$ is nonzero). Consequently, $0^{g}\neq0$. Therefore, $g=0$.
Hence, (\ref{pf.lem.C.1}) rewrites as $F\left(  \mathbf{x}_{1},\ldots
,\mathbf{x}_{d}\right)  =c\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots
u_{d}\right)  ^{0}=c$. Therefore, $F$ is constant. This proves Lemma
\ref{lem.C}.
\end{proof}

The next lemma is a simple variation on the multinomial theorem.

\begin{lemma}
\label{lem.F}Let $F\in P_{n}$ be a homogeneous form of degree $t$. Let
$F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  \in P_{d}\left[  \lambda\right]  $ be
the result of substituting $\sum_{i=1}^{d}\lambda_{j,i}\left(  x_{i}\mid
u_{r}\right)  $ for each indeterminate $\left(  x_{j}\mid u_{r}\right)  $
(with $1\leq j\leq n$ and $1\leq r\leq d$) in $F$.

\textbf{(a)} Then,
\begin{align}
&  F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right) \nonumber\\
&  =\sum\left(  \prod_{j=1}^{n}\prod_{r=1}^{d}\lambda_{j,r}^{i\left(
j,r\right)  }\right)  \cdot F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\left(  \mathbf{x}_{1},\ldots,\mathbf{x}%
_{d}\right)  , \label{eq.lem.F.a.eq}%
\end{align}
where the summation is over all families $\left(  i\left(  1,1\right)
,i\left(  1,2\right)  ,\ldots,i\left(  n,d\right)  \right)  =\left(  i\left(
j,r\right)  \right)  _{1\leq j\leq n,\ 1\leq r\leq d}\in\mathbb{N}^{n\times
d}$ of nonnegative integers satisfying%
\[
\sum_{q=1}^{d}\sum_{p=1}^{n}i\left(  p,q\right)  =t,
\]
and where the polynomials $F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\in P_{d}$ are homogeneous of degree $t$.

\textbf{(b)} Assume that $F$ is invariant. Then, each of the polynomials
$F_{i\left(  1,1\right)  ,i\left(  1,2\right)  ,\ldots,i\left(  n,d\right)
}\in P_{d}$ is invariant.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.F}.]\textbf{(a)} The polynomial ring $P_{d}\left[
\lambda\right]  $ can be regarded as a polynomial ring over $K$ in the
$nd+d^{2}$ indeterminates%
\begin{align*}
\left(  x_{j}\mid u_{k}\right)  \ \ \ \ \ \ \ \ \ \ \text{for }1  &  \leq
j\leq n\text{ and }1\leq k\leq d,\text{ \textbf{as well as}}\\
\lambda_{j,k}\ \ \ \ \ \ \ \ \ \ \text{for }1  &  \leq j\leq d\text{ and
}1\leq k\leq d.
\end{align*}
Let us consider this polynomial ring $P_{d}\left[  \lambda\right]  $ as a
$\mathbb{Z}\times\mathbb{Z}$-graded $K$-algebra, where each of the variables
$\left(  x_{j}\mid u_{k}\right)  $ has degree $\left(  1,0\right)  $ whereas
the variables $\lambda_{j,k}$ have degree $\left(  0,1\right)  $. Thus, the
sums $\sum_{i=1}^{d}\lambda_{j,i}\left(  x_{i}\mid u_{r}\right)  \in
P_{d}\left[  \lambda\right]  $ (for all $1\leq j\leq n$ and $1\leq r\leq d$)
are homogeneous of degree $\left(  1,1\right)  $. Therefore, substituting
these sums for the indeterminates $\left(  x_{j}\mid u_{r}\right)  $
transforms any homogeneous polynomial $G\in P_{n}$ of degree $t$ into a
homogeneous polynomial in $P_{d}\left[  \lambda\right]  $ of degree $\left(
t,t\right)  $\ \ \ \ \footnote{If you don't believe this, just check it in the
case when $G$ is a monomial (from which the general case follows by
linearity).}. Therefore, the polynomial $F\left(  \sum_{i=1}^{d}\lambda
_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)
\in P_{d}\left[  \lambda\right]  $ is homogeneous of degree $\left(
t,t\right)  $ (since it is obtained by this substitution from the polynomial
$F\in P_{n}$, which is homogeneous of degree $t$). Hence, it can be written in
the form%
\begin{align*}
&  F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right) \\
&  =\sum\left(  \prod_{j=1}^{n}\prod_{r=1}^{d}\lambda_{j,r}^{i\left(
j,r\right)  }\right)  \cdot F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\left(  \mathbf{x}_{1},\ldots,\mathbf{x}%
_{d}\right)  ,
\end{align*}
where the summation is over all families \newline$\left(  i\left(  1,1\right)
,i\left(  1,2\right)  ,\ldots,i\left(  n,d\right)  \right)  =\left(  i\left(
j,r\right)  \right)  _{1\leq j\leq n,\ 1\leq r\leq d}\in\mathbb{N}^{n\times
d}$ of nonnegative integers satisfying%
\[
\sum_{q=1}^{d}\sum_{p=1}^{n}i\left(  p,q\right)  =t,
\]
and where the polynomials $F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\in P_{d}$ are homogeneous of degree $t$. This
proves Lemma \ref{lem.F} \textbf{(a)}.

\textbf{(b)} Every $d \times d$-matrix $L \in K^{d \times d}$ acts on $P_{d}$
as a $K$-algebra endomorphism, and thus also acts on $P_{d}\left[
\lambda\right]  $ as a $K\left[  \lambda\right]  $-algebra
endomorphism\footnote{The action of a $d \times d$-matrix $L \in K^{d \times
d}$ on $P_{d}\left[  \lambda\right]  $ is obtained by extending the action of
$L$ on $P_{d}$ as a $K\left[  \lambda\right]  $-algebra endomorphism. Thus, if
we regard an element of $P_{d}\left[  \lambda\right]  $ as a polynomial in the
$\lambda_{j,k}$ with coefficients in $P_{d}$, then $L$ acts on each of these
coefficients separately.}. This leads to a notion of invariant elements of
$P_{d}\left[  \lambda\right]  $: Namely, we say that a $G\in P_{d}\left[
\lambda\right]  $ is \textit{invariant} if and only if, for all invertible $d
\times d$-matrices $L \in K^{d \times d}$, there exists a scalar $a\left(
L\right)  \in K$ such that $LG=a\left(  L\right)  G$.

Now, let us show the following claim:

\begin{statement}
\textit{Claim 1:} Each $G\in P_{n}$ satisfies
\[
\left(  LG\right)  \left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i}%
,\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  =L\cdot G\left(
\sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda
_{n,i}\mathbf{x}_{i}\right)  .
\]

\end{statement}

[\textit{Proof of Claim 1:} We consider the two maps%
\[
P_{n}\rightarrow P_{d}\left[  \lambda\right]  ,\ \ \ \ \ \ \ \ \ \ G\mapsto
\left(  LG\right)  \left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i}%
,\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)
\]
and%
\[
P_{n}\rightarrow P_{d}\left[  \lambda\right]  ,\ \ \ \ \ \ \ \ \ \ G\mapsto
L\cdot G\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum
_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  .
\]
Our goal is to show that these two maps are equal. Since both of these maps
are $K$-algebra homomorphisms\footnote{This is because $L$ acts as $K$-algebra
endomorphisms on both $P_{d}$ and $P_{d}\left[  \lambda\right]  $, and because
the map%
\[
P_{n}\rightarrow P_{d}\left[  \lambda\right]  ,\ \ \ \ \ \ \ \ \ \ G\mapsto
G\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)
\]
is a $K$-algebra homomorphism as well.}, we only need to check that these two
maps take equal values on the generators $\left(  x_{j}\mid u_{r}\right)  $ of
$P_{n}$. In other words, we need to prove Claim 1 in the case when $G=\left(
x_{j}\mid u_{r}\right)  $ for some $1\leq j\leq n$ and $1\leq r\leq d$.

So let us fix $1\leq j\leq n$ and $1\leq r\leq d$. Let us set $G=\left(
x_{j}\mid u_{r}\right)  $. We must prove Claim 1 for this $G$.

Write the $d \times d$-matrix $L$ in the form $L = \left(  l_{kj}\right)
_{1\leq j,k\leq d}$. We have $G=\left(  x_{j}\mid u_{r}\right)  $ and thus%
\[
LG=L\left(  x_{j}\mid u_{r}\right)  =\sum_{1\leq k\leq d}l_{rk}\left(
x_{j}\mid u_{k}\right)
\]
(by the definition of the action of $L$ on $P_{n}$). Hence, the definition of
\newline$\left(  LG\right)  \left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}%
_{i},\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  $ yields%
\begin{align}
\left(  LG\right)  \left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i}%
,\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)   &  =\sum_{1\leq
k\leq d}l_{rk}\sum_{i=1}^{d}\lambda_{j,i}\left(  x_{i}\mid u_{k}\right)
\nonumber\\
&  =\sum_{1\leq k\leq d}\sum_{i=1}^{d}l_{rk}\lambda_{j,i}\left(  x_{i}\mid
u_{k}\right)  . \label{pf.lem.F.b.1}%
\end{align}


On the other hand, $G=\left(  x_{j}\mid u_{r}\right)  $. Hence, the definition
of $G\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  $ yields%
\[
G\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  =\sum_{i=1}^{d}\lambda_{j,i}\left(
x_{i}\mid u_{r}\right)  .
\]
Hence,%
\begin{align*}
L\cdot G\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum
_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)   &  =L\cdot\sum_{i=1}^{d}%
\lambda_{j,i}\left(  x_{i}\mid u_{r}\right) \\
&  =\sum_{i=1}^{d}\lambda_{j,i}\underbrace{L\left(  x_{i}\mid u_{r}\right)
}_{\substack{=\sum_{1\leq k\leq d}l_{rk}\left(  x_{i}\mid u_{k}\right)
\\\text{(by the definition of the action of }L\text{)}}}\\
&  =\sum_{i=1}^{d}\lambda_{j,i}\sum_{1\leq k\leq d}l_{rk}\left(  x_{i}\mid
u_{k}\right) \\
&  =\sum_{1\leq k\leq d}\sum_{i=1}^{d}l_{rk}\lambda_{j,i}\left(  x_{i}\mid
u_{k}\right)  .
\end{align*}
Comparing this with (\ref{pf.lem.F.b.1}), we obtain%
\[
\left(  LG\right)  \left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i}%
,\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  =L\cdot G\left(
\sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda
_{n,i}\mathbf{x}_{i}\right)  .
\]
Thus, Claim 1 is proven for $G=\left(  x_{j}\mid u_{r}\right)  $. As we have
already explained, this completes the proof of Claim 1.]

Now, consider our invariant form $F\in P_{n}$. Thus, any invertible $d \times
d$-matrix $L \in K^{d \times d}$ satisfies $LF=a\left(  L\right)  F$ for some
scalar $a\left(  L\right)  $ (since $F$ is invariant). Now, any invertible $d
\times d$-matrix $L \in K^{d \times d}$ satisfies%
\begin{align*}
L\cdot F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum
_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)   &  =\underbrace{\left(
LF\right)  }_{=a\left(  L\right)  F}\left(  \sum_{i=1}^{d}\lambda
_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Claim 1, applied to }G=F\right) \\
&  =a\left(  L\right)  F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}%
_{i},\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  .
\end{align*}
Hence, the element $F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i}%
,\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  \in P_{d}\left[
\lambda\right]  $ is invariant.

But it is easy to see that if a polynomial $G\in P_{d}\left[  \lambda\right]
$ is invariant, then each of its coefficients is invariant\footnote{Here, we
are regarding $G$ as a polynomial in the indeterminates $\lambda_{k,j}$ over
the ring $P_{d}$. Thus, the indeterminates in $P_{d}$ are regarded as
constants, not as indeterminates.} (because $d \times d$-matrices $L \in K^{d
\times d}$ act on $P_{d}\left[  \lambda\right]  $ by acting on each
coefficient separately). Applying this to $G=F\left(  \sum_{i=1}^{d}%
\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}%
_{i}\right)  $, we conclude that each coefficient of $F\left(  \sum_{i=1}%
^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda_{n,i}%
\mathbf{x}_{i}\right)  $ is invariant. But the coefficients of $F\left(
\sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda
_{n,i}\mathbf{x}_{i}\right)  $ are none other than the polynomials
$F_{i\left(  1,1\right)  ,i\left(  1,2\right)  ,\ldots,i\left(  n,d\right)
}\in P_{d}$ (by the equality (\ref{eq.lem.F.a.eq})). Hence, we have shown that
each of the polynomials $F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\in P_{d}$ is invariant. This proves Lemma
\ref{lem.F} \textbf{(b)}.
\end{proof}

The preliminary lemmas are now disposed of. Let $F\in P=P_{n}$ be a
homogeneous invariant form of degree $t$. We must show that $F$ is a linear
combination of standard rectangular bideterminants. Due to Lemma
\ref{lem.bracket}, it will suffice to show that $F$ is a linear combination of
bracket monomials. We can WLOG assume that $F$ is nonconstant (since constant
forms are clearly linear combinations of bracket monomials); thus, $t>0$. If
we had $n<d$, then Lemma \ref{lem.C} (applied to $m=n$) would yield that $F$
is constant. Thus, we cannot have $n<d$. Hence, we have $n\geq d$.

Let $P^{B}$ be the $K$-subalgebra of $P=P_{n}$ generated by all brackets.

Let $P_{d}^{I}$ be the $K$-subalgebra of $P_{d}$ consisting of all forms
polynomials in $P_{d}$. Then, $P_{d}^{I}$ is spanned (as a $K$-vector space)
by all homogeneous invariants in $P_{d}$; but all the latter are scalar
multiples of powers of $\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots
u_{d}\right)  $ (by Lemma \ref{lem.D}). Hence, $P_{d}^{I}$ is spanned (as a
$K$-vector space) by scalar multiples of powers of $\left(  x_{1}\cdots
x_{d}\mid u_{1}\cdots u_{d}\right)  $. In other words, $P_{d}^{I}$ is
generated as a $K$-algebra by the form $\left(  x_{1}\cdots x_{d}\mid
u_{1}\cdots u_{d}\right)  $. In other words, $P_{d}^{I}$ is generated as a
$K$-algebra by the bracket $\left[  x_{1},\ldots,x_{d}\right]  $ (since
$\left[  x_{1},\ldots,x_{d}\right]  =\left(  x_{1}\cdots x_{d}\mid u_{1}\cdots
u_{d}\right)  $). Hence, $P_{d}^{I}\subseteq P^{B}$ (since $P^{B}$ is the
$K$-subalgebra of $P$ generated by \textbf{all} brackets).

For each $1\leq j\leq n$ and $1\leq k\leq d$, we define a bracket $B_{j,k}\in
P$ by%
\begin{equation}
B_{j,k}=\left[  x_{1},\ldots,x_{k-1},x_{j},x_{k+1},\ldots,x_{d}\right]  .
\label{pf.thm.3.2.2nd.Bjk}%
\end{equation}
Then, every $1\leq j\leq n$ and $1\leq r\leq d$ satisfy%
\begin{align}
&  \left[  x_{1},\ldots,x_{d}\right]  \left(  x_{j}\mid u_{r}\right)
\nonumber\\
&  =\sum_{k=1}^{d}\underbrace{\left[  x_{1},\ldots,x_{k-1},x_{j}%
,x_{k+1},\ldots,x_{d}\right]  }_{\substack{=B_{j,k}\\\text{(by
(\ref{pf.thm.3.2.2nd.Bjk}))}}}\left(  x_{k}\mid u_{r}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.E}, applied to }m=r\right)
\nonumber\\
&  =\sum_{k=1}^{d}B_{j,k}\left(  x_{k}\mid u_{r}\right)  =\sum_{i=1}%
^{d}B_{j,i}\left(  x_{i}\mid u_{r}\right)  . \label{pf.thm.3.2.2nd.4}%
\end{align}


Lemma \ref{lem.F} \textbf{(a)} yields that the polynomial $F\left(  \sum
_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda
_{n,i}\mathbf{x}_{i}\right)  \in P_{d}\left[  \lambda\right]  $ (defined as in
Lemma \ref{lem.F}) can be written in the form (\ref{eq.lem.F.a.eq}), where the
summation is over all families $\left(  i\left(  1,1\right)  ,i\left(
1,2\right)  ,\ldots,i\left(  n,d\right)  \right)  =\left(  i\left(
j,r\right)  \right)  _{1\leq j\leq n,\ 1\leq r\leq d}\in\mathbb{N}^{n\times
d}$ of nonnegative integers satisfying $\sum_{q=1}^{d}\sum_{p=1}^{n}i\left(
p,q\right)  =t$, and where the polynomials $F_{i\left(  1,1\right)  ,i\left(
1,2\right)  ,\ldots,i\left(  n,d\right)  }\in P_{d}$ are homogeneous of degree
$t$. Consider these polynomials $F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\in P_{d}$. Lemma \ref{lem.F} \textbf{(b)} shows
that each of these polynomials $F_{i\left(  1,1\right)  ,i\left(  1,2\right)
,\ldots,i\left(  n,d\right)  }\in P_{d}$ is invariant, i.e., belongs to
$P_{d}^{I}$. Thus, the equality (\ref{eq.lem.F.a.eq}) shows that
\[
F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  \in P_{d}^{I}\left[  \lambda\right]
.
\]
Let $\omega:P_{d}^{I}\left[  \lambda\right]  \rightarrow P^{B}$ be the unique
$P_{d}^{I}$-algebra homomorphism sending each indeterminate $\lambda_{j,k}$ to
$B_{j,k}$. This is well-defined, since both $B_{j,k}\in P^{B}$ for all $j$ and
$k$ (since $B_{j,k}$ is a bracket) and $P_{d}^{I}\subseteq P^{B}$.

We claim that the polynomial%
\[
\left[  x_{1},\ldots,x_{d}\right]  ^{t}F
\]
equals a polynomial in the brackets $\left[  x_{i_{1}},\ldots,x_{i_{d}%
}\right]  $. Indeed, as $F$ is homogeneous of degree $t$, we have%
\begin{equation}
\left[  x_{1},\ldots,x_{d}\right]  ^{t}\cdot F=F\left(  \left[  x_{1}%
,\ldots,x_{d}\right]  \mathbf{x}_{1},\ldots,\left[  x_{1},\ldots,x_{d}\right]
\mathbf{x}_{n}\right)  \label{pf.thm.3.2.2nd.bratF}%
\end{equation}
\footnote{We are here using the following notation: If $a_{1},a_{2}%
,\ldots,a_{n}$ are $n$ elements of $P_{n}$, then $F\left(  a_{1}\mathbf{x}%
_{1},\ldots,a_{n}\mathbf{x}_{n}\right)  $ denotes the result of substituting
$a_{j}\cdot\left(  x_{j}\mid u_{r}\right)  $ for each indeterminate $\left(
x_{j}\mid u_{r}\right)  $ (with $1\leq j\leq n$ and $1\leq r\leq d$) in $F$.}.
But%
\begin{equation}
F\left(  \left[  x_{1},\ldots,x_{d}\right]  \mathbf{x}_{1},\ldots,\left[
x_{1},\ldots,x_{d}\right]  \mathbf{x}_{n}\right)  =\omega\left(  F\left(
\sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}^{d}\lambda
_{n,i}\mathbf{x}_{i}\right)  \right)  \label{pf.thm.3.2.2nd.omega-use}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.thm.3.2.2nd.omega-use}):} Recall that
$F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  $ is the result of substituting
$\sum_{i=1}^{d}\lambda_{j,i}\left(  x_{i}\mid u_{r}\right)  $ for each
indeterminate $\left(  x_{j}\mid u_{r}\right)  $ in $F$. Applying the map
$\omega$ further replaces each $\lambda_{j,i}$ by $B_{j,i}$ (by the definition
of $\omega$). Combining these two facts, we conclude that $\omega\left(
F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots,\sum_{i=1}%
^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  \right)  $ is the result of
substituting $\sum_{i=1}^{d}B_{j,i}\left(  x_{i}\mid u_{r}\right)  $ for each
indeterminate $\left(  x_{j}\mid u_{r}\right)  $ in $F$.
\par
But the definition of $F\left(  \left[  x_{1},\ldots,x_{d}\right]
\mathbf{x}_{1},\ldots,\left[  x_{1},\ldots,x_{d}\right]  \mathbf{x}%
_{n}\right)  $ shows that $F\left(  \left[  x_{1},\ldots,x_{d}\right]
\mathbf{x}_{1},\ldots,\left[  x_{1},\ldots,x_{d}\right]  \mathbf{x}%
_{n}\right)  $ is the result of substituting $\left[  x_{1},\ldots
,x_{d}\right]  \left(  x_{j}\mid u_{r}\right)  $ for each indeterminate
$\left(  x_{j}\mid u_{r}\right)  $ in $F$. In light of (\ref{pf.thm.3.2.2nd.4}%
), this rewrites as follows: $F\left(  \left[  x_{1},\ldots,x_{d}\right]
\mathbf{x}_{1},\ldots,\left[  x_{1},\ldots,x_{d}\right]  \mathbf{x}%
_{n}\right)  $ is the result of substituting $\sum_{i=1}^{d}B_{j,i}\left(
x_{i}\mid u_{r}\right)  $ for each indeterminate $\left(  x_{j}\mid
u_{r}\right)  $ in $F$.
\par
Hence, both forms $F\left(  \left[  x_{1},\ldots,x_{d}\right]  \mathbf{x}%
_{1},\ldots,\left[  x_{1},\ldots,x_{d}\right]  \mathbf{x}_{n}\right)  $ and
$\omega\left(  F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i},\ldots
,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  \right)  $ have been
characterized in the same way (namely, as the result of substituting
$\sum_{i=1}^{d}B_{j,i}\left(  x_{i}\mid u_{r}\right)  $ for each indeterminate
$\left(  x_{j}\mid u_{r}\right)  $ in $F$). Thus, these two forms are equal.
In other words, (\ref{pf.thm.3.2.2nd.omega-use}) is proven.}. Hence,
(\ref{pf.thm.3.2.2nd.bratF}) becomes%
\begin{align*}
\left[  x_{1},\ldots,x_{d}\right]  ^{t}\cdot F  &  =F\left(  \left[
x_{1},\ldots,x_{d}\right]  \mathbf{x}_{1},\ldots,\left[  x_{1},\ldots
,x_{d}\right]  \mathbf{x}_{n}\right) \\
&  =\omega\left(  F\left(  \sum_{i=1}^{d}\lambda_{1,i}\mathbf{x}_{i}%
,\ldots,\sum_{i=1}^{d}\lambda_{n,i}\mathbf{x}_{i}\right)  \right)  \in P^{B}.
\end{align*}
In other words, $\left[  x_{1},\ldots,x_{d}\right]  ^{t}\cdot F$ belongs to
the $K$-subalgebra of $P$ generated by all brackets (since $P^{B}$ is the
$K$-subalgebra of $P$ generated by all brackets). In other words, $\left[
x_{1},\ldots,x_{d}\right]  ^{t}\cdot F$ is a polynomial in the brackets.
Hence, Lemma \ref{lem.bracket2} (applied to $g=t$) shows that $F$ is a linear
combination of standard rectangular bideterminants. This completes the second
proof of Theorem \ref{thm.3.2}.
\end{proof}

\subsection{A note on homogeneous forms}

Theorem \ref{thm.3.2} quickly yields the following corollary:

\begin{corollary}
\label{cor.3.2.sameshape}Let $K$ be an infinite field. A homogeneous form in
$P$ is invariant iff it is a linear combination of standard rectangular
bideterminants, all of which have the same shape $\left(  d,\ldots,d\right)  $.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.3.2.sameshape}.]The \textquotedblleft
if\textquotedblright\ part of Corollary \ref{cor.3.2.sameshape} is clearly
true (since rectangular bideterminants are always invariant). Thus, it remains
to prove the \textquotedblleft only if\textquotedblright\ part.

Let $F$ be an invariant homogeneous form in $P$. We must show that $F$ is a
linear combination of standard rectangular bideterminants, all of which have
the same shape $\left(  d,\ldots,d\right)  $.

The form $F$ is homogeneous; let $t$ be its degree.

Theorem \ref{thm.3.2} shows that $F$ is a linear combination of standard
rectangular bideterminants. In other words, $F$ can be written in the form%
\begin{equation}
F=\sum_{i\in I}b_{i}\left(  U_{i}\mid U_{i}^{\prime}\right)  ,
\label{pf.cor.3.2.sameshape.1}%
\end{equation}
where the $\left[  U_{i},U_{i}^{\prime}\right]  $ are standard bitableaux such
that the tableaux $U_{i}^{\prime}$ are rectangular, and where the $b_{i}$ are
elements of $K$.

For each $i\in I$, we let $g_{i}$ be the number of rows of the tableau
$U_{i}^{\prime}$.

Now, let $i\in I$. Then, the tableau $U_{i}^{\prime}$ is rectangular and has
$g_{i}$ rows (by the definition of $g_{i}$). Consequently, this tableau must
have shape $\underbrace{\left(  d,\ldots,d\right)  }_{g_{i}\text{ entries}}$.
Hence, this tableau has $dg_{i}$ cells. Therefore,
\begin{equation}
\text{the bideterminant }\left(  U_{i}\mid U_{i}^{\prime}\right)  \in P\text{
is homogeneous of degree }dg_{i} \label{pf.cor.3.2.sameshape.hg}%
\end{equation}
(because the bideterminant $\left(  U\mid U^{\prime}\right)  $ of a bitableau
$\left[  U,U^{\prime}\right]  $ always is homogeneous of degree equal to the
number of cells of $U^{\prime}$).

Moreover, the tableau $U_{i}^{\prime}$ has shape $\underbrace{\left(
d,\ldots,d\right)  }_{g_{i}\text{ entries}}$. In other words,%
\begin{equation}
\text{the bideterminant }\left(  U_{i}\mid U_{i}^{\prime}\right)  \text{ has
shape }\underbrace{\left(  d,\ldots,d\right)  }_{g_{i}\text{ entries}}.
\label{pf.cor.3.2.sameshape.sh}%
\end{equation}


Now, forget that we fixed $i$. We thus have proven
(\ref{pf.cor.3.2.sameshape.hg}) and (\ref{pf.cor.3.2.sameshape.sh}) for each
$i\in I$.

Now, let $\pi_{t}:P\rightarrow P$ be the projection of the graded ring $P$
onto its $t$-th homogeneous component. Then, $\pi_{t}\left(  F\right)  =F$
(since $F$ is homogeneous of degree $t$). But applying the map $\pi_{t}$ to
both sides of the equality (\ref{pf.cor.3.2.sameshape.1}), we obtain%
\begin{align*}
\pi_{t}\left(  F\right)   &  =\pi_{t}\left(  \sum_{i\in I}b_{i}\left(
U_{i}\mid U_{i}^{\prime}\right)  \right)  =\sum_{i\in I}b_{i}\pi_{t}\left(
U_{i}\mid U_{i}^{\prime}\right) \\
&  =\sum_{\substack{i\in I;\\dg_{i}=t}}b_{i}\underbrace{\pi_{t}\left(
U_{i}\mid U_{i}^{\prime}\right)  }_{\substack{=\left(  U_{i}\mid U_{i}%
^{\prime}\right)  \\\text{(since (\ref{pf.cor.3.2.sameshape.hg}) shows that
}\left(  U_{i}\mid U_{i}^{\prime}\right)  \\\text{is homogeneous of degree
}dg_{i}=t\text{)}}}+\sum_{\substack{i\in I;\\dg_{i}\neq t}}b_{i}%
\underbrace{\pi_{t}\left(  U_{i}\mid U_{i}^{\prime}\right)  }%
_{\substack{=0\\\text{(since (\ref{pf.cor.3.2.sameshape.hg}) shows that
}\left(  U_{i}\mid U_{i}^{\prime}\right)  \\\text{is homogeneous of degree
}dg_{i}\neq t\text{)}}}\\
&  =\sum_{\substack{i\in I;\\dg_{i}=t}}b_{i}\left(  U_{i}\mid U_{i}^{\prime
}\right)  +\underbrace{\sum_{\substack{i\in I;\\dg_{i}\neq t}}b_{i}0}%
_{=0}=\sum_{\substack{i\in I;\\dg_{i}=t}}b_{i}\left(  U_{i}\mid U_{i}^{\prime
}\right)  =\sum_{\substack{i\in I;\\g_{i}=t/d}}b_{i}\left(  U_{i}\mid
U_{i}^{\prime}\right)  .
\end{align*}
Comparing this with $\pi_{t}\left(  F\right)  =F$, we obtain%
\begin{equation}
F=\sum_{\substack{i\in I;\\g_{i}=t/d}}b_{i}\left(  U_{i}\mid U_{i}^{\prime
}\right)  . \label{pf.cor.3.2.sameshape.fin}%
\end{equation}


But for each $i\in I$ satisfying $g_{i}=t/d$, the shape of the bideterminant
$\left(  U_{i}\mid U_{i}^{\prime}\right)  $ is $\underbrace{\left(
d,\ldots,d\right)  }_{g_{i}\text{ entries}}$ (by
(\ref{pf.cor.3.2.sameshape.sh})). In other words, for each $i\in I$ satisfying
$g_{i}=t/d$, the shape of the bideterminant $\left(  U_{i}\mid U_{i}^{\prime
}\right)  $ is $\underbrace{\left(  d,\ldots,d\right)  }_{t/d\text{ entries}}$
(since $g_{i}=t/d$). Thus, (\ref{pf.cor.3.2.sameshape.fin}) shows that $F$ is
a linear combination of standard rectangular bideterminants, all of which have
the same shape $\underbrace{\left(  d,\ldots,d\right)  }_{t/d\text{ entries}}%
$. This proves Corollary \ref{cor.3.2.sameshape}.
\end{proof}

\section{Appendix (Darij Grinberg)}

Let me add some remarks on the paper above.

This paper is one of the first studies in characteristic-free invariant theory
(i.e., invariant theory over a field $K$ of arbitrary characteristic). What is
called an \textit{invariant form} in this paper is more or less the same as
(what is nowadays known as) an invariant of the special linear group
$\operatorname*{SL}\left(  d\right)  $ acting on the coordinate ring
$\mathcal{O}\left(  \left(  K^{d}\right)  ^{n}\right)  $, where
$\operatorname*{SL}\left(  d\right)  $ acts on $K^{d}$ in the usual way. (This
situation is commonly known as \textquotedblleft invariants of
$\operatorname*{SL}\left(  d\right)  $ on $n$ vectors\textquotedblright.) An
equivalent version of Theorem \ref{thm.3.2} appears already in \cite[Theorem
4]{Igusa}, but the proof given there is far too brief to be considered
complete. Theorem \ref{thm.3.2} also appears in \cite[Section 9, Theorem 1]%
{3}, where it is proven more or less in the same way as in the paper above.
(Actually, the paper \cite{3} can be regarded as a preliminary version of the
above paper; it offers much motivation and neat applications to elementary
geometry, but is lacking in mathematical rigor.)

On the other hand, many more works have appeared since the publication of the
paper above. I shall mention only a few of them:

\begin{itemize}
\item The paper \cite{DCEiPr} by DeConcini, Eisenbud and Procesi studies the
algebra that we call $P$ in Theorem \ref{thm.3.2} in much more depth. In
particular, our Theorem \ref{thm.3.2} is (more or less) \cite[Corollary
3.5]{DCEiPr}\footnote{The proof of \cite[Corollary 3.5]{DCEiPr} is somewhat
similar to our first proof of Theorem \ref{thm.3.2}, but is slicker. As an
intermediate result, it also characterizes the forms $F\in P$ that are
\textquotedblleft invariant under lower-unitriangular
matrices\textquotedblright\ (i.e., that satisfy $LF=F$ for every
lower-unitriangular $d \times d$-matrix $L \in K^{d \times d}$); namely, these
forms are linear combinations of standard bideterminants $\left(  U\mid
U^{\prime}\right)  $ such that each row of the tableau $U^{\prime}$ has the
form $u_{i}u_{i+1}\cdots u_{d}$ for some $i\in\left\{  1,2,\ldots,d\right\}
$. (See \cite[Theorem 3.3 0)]{DCEiPr} for this result; but beware that what we
call $LF$ for an invertible $d \times d$-matrix $L \in K^{d \times d}$ and a
form $F\in P$ corresponds to what \cite{DCEiPr} would call $\left(
L^{T}\right)  ^{-1}F$, which is why our lower-triangular matrices correspond
to upper-triangular matrices in \cite{DCEiPr}.)}. Also, \cite[Theorem
2.1]{DCEiPr} combines the above paper's Theorem \ref{thm.1} and Corollary
\ref{cor.2.2.basis} (and generalizes them to the case where $K$ is an
arbitrary commutative ring).

\item The booklet \cite{GrRoSt} by Grosshans, Rota and Stein appears to extend
the above considerations to the \textquotedblleft superalgebra
case\textquotedblright\ (i.e., forms containing \textquotedblleft
positively\textquotedblright\ and \textquotedblleft
negatively\textquotedblright\ signed variables). (I have not read the booklet.)

\item Swan's preprint \cite{Swan} gives a different approach to the
straightening algorithm. Specifically, the above paper's Corollary
\ref{cor.2.2.basis} is covered by \cite[Corollary 5.1 and Theorem 5.3]{Swan}
(and it is not hard to derive Theorem \ref{thm.1} from \cite[Theorem
4.1]{Swan} as well).

\item Procesi's book \cite{Procesi} covers characteristic-free invariant
theory in its Chapter 13. In particular, \cite[Section 13.4, Theorem]{Procesi}
is exactly Corollary \ref{cor.2.2.basis} but using the base ring $\mathbb{Z}$
instead of the field $K$; furthermore, \cite[Section 13.6.3]{Procesi} comes
rather close to the Theorem \ref{thm.3.2} above.
\end{itemize}

Let me now briefly discuss the possibility of extending some of the results in
the above paper to the more general situation when $K$ is an arbitrary
commutative ring (rather than a field).

Assume from now on that $K$ (instead of being a field) is just a commutative
ring (with unity).

Then, all claims and proofs made in Section \ref{sect.straight} remain valid.
(Of course, we need to make obvious modifications, such as replacing the word
\textquotedblleft vector space\textquotedblright\ by \textquotedblleft
module\textquotedblright\ everywhere.)

Furthermore, all claims and proofs made in Section \ref{sect.basis} remain
valid, as long as the following modification is made: The claim of Theorem
\ref{thm.2.1} (i) must be replaced by \textquotedblleft The one-element family
$\left(  C\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)
\right)  $ is $K$-linearly independent (i.e., if an $a\in K$ satisfies
$aC\left(  T,T^{\prime}\right)  \left(  T\mid T^{\prime}\right)  =0$, then
$a=0$).\textquotedblright. (The proof of this new claim is exactly the proof
of Theorem \ref{thm.2.1} (i) we gave above.)

All claims and proofs made in Subsection \ref{subsect.invar.sft} remain valid
(except for the arguments in Remark \ref{rmk.3.1.old}, which are unimportant),
as long as the following modification is made: The equality
(\ref{pf.thm.3.1.fn1.obs4.eq}) must be replaced by \textquotedblleft The
one-element family $\left(  \phi\left(  T\mid T^{\prime}\right)  \right)  $ is
$K$-linearly independent (i.e., if an $a\in K$ satisfies $\phi\left(  T\mid
T^{\prime}\right)  =0$, then $a=0$).\textquotedblright. (The proof of this new
claim is exactly the proof of Lemma \ref{lem.3.1.o4} we gave above.)

Much more complicated is the situation with Subsections
\ref{subsect.invar.fft}, \ref{subsect.invar.fft1} and \ref{subsect.invar.fft2}%
. Theorem \ref{thm.3.2}, to begin with, doesn't even hold when $K$ is a finite
field\footnote{Indeed, let $K$ be the finite field $\mathbb{F}_{q}$ for some
prime power $q$. Set $n=1$. Then, the ring $P$ is the polynomial ring over $K$
in the indeterminates $\left(  x_{1}\mid u_{j}\right)  $ for all $1\leq j\leq
d$. Let us denote these indeterminates by $X_{j}$; thus, $P=K\left[
X_{1},X_{2},\ldots,X_{d}\right]  $. As before, the group of all invertible
$d\times d$-matrices $L\in K^{d\times d}$ acts on $P$ by algebra
endomorphisms. If the conclusion of Theorem \ref{thm.3.2} were to hold for our
field $K$, then, for any $d\geq2$, the only invariant forms in $P$ would be
constants (since the only standard rectangular bideterminant in this setting
is the empty one $\left(  \ \ \mid\ \ \right)  $, because $d\geq2>1=n$ ensures
that a rectangular bitableau cannot be standard). But this is not the case:
For each $d$-tuple $\left(  \alpha_{1},\alpha_{2},\ldots,\alpha_{d}\right)
\in\mathbb{N}^{d}$ of nonnegative integer, the determinant%
\[
A_{\alpha}=\det\left(
\begin{array}
[c]{cccc}%
X_{1}^{q^{\alpha_{1}}} & X_{2}^{q^{\alpha_{1}}} & \cdots & X_{d}%
^{q^{\alpha_{1}}}\\
X_{1}^{q^{\alpha_{2}}} & X_{2}^{q^{\alpha_{2}}} & \cdots & X_{d}%
^{q^{\alpha_{2}}}\\
\vdots & \vdots & \ddots & \vdots\\
X_{1}^{q^{\alpha_{d}}} & X_{2}^{q^{\alpha_{d}}} & \cdots & X_{d}%
^{q^{\alpha_{d}}}%
\end{array}
\right)
\]
is an invariant in the sense of the above paper (more precisely, it has the
property that every $d\times d$-matrix $L\in K^{d\times d}$ satisfies
$LA_{\alpha}=\left(  \det L\right)  A_{\alpha}$; see \cite[7th Variation]%
{Macdonald} for this). If the entries of the $d$-tuple $\alpha$ are distinct
(and if $d\geq1$), then $A_{\alpha}$ is not a constant. Thus, the requirement
that $K$ is infinite in Theorem \ref{thm.3.2} cannot be dropped!
\par
The phenomenon of these nonconstant invariants over finite fields is
interesting in its own right; it has been studied already by Dickson in 1911.
We can find nonconstant forms in $F$ that are not only invariant in the above
sense, but actually satisfy the stronger condition that each invertible
$d\times d$-matrix $L\in K^{d\times d}$ satisfies $LF=F$ (not just
$LF=a\left(  L\right)  F$ for some scalar $a\left(  L\right)  $). Indeed, such
invariants can be constructed from the above determinants $A_{\alpha}$;
namely, if we let $\delta$ be the $d$-tuple $\left(  d-1,d-2,\ldots,0\right)
$, then it can be shown that $A_{\alpha}/A_{\delta}$ is a polynomial for every
$d$-tuple $\alpha$, and this polynomial $A_{\alpha}/A_{\delta}$ satisfies the
stronger condition (i.e., if we set $F=A_{\alpha}/A_{\delta}$, then each
invertible $d\times d$-matrix $L\in K^{d\times d}$ satisfies $LF=F$).
Polynomials satisfying this condition have been characterized by Dickson (see,
e.g., \cite[Theorem 1.2]{Wilkerson} or \cite[Theorem A]{Steinberg}).}. But
merely requiring $K$ to be infinite is not sufficient when $K$ is not a
field.\footnote{Counterexamples can easily be obtained when $K$ is the
(infinite) ring $\mathbb{F}_{q}\times\mathbb{F}_{q}\times\mathbb{F}_{q}%
\times\cdots$ for a prime power $q$.} Instead, a meaningful way to salvage
Theorem \ref{thm.3.2} is to replace the concept of invariant by the concept of
an \textit{absolute invariant}:

A form $F\in P$ is said to be an \textit{absolute invariant}\ if it satisfies
$LF=a\left(  L\right)  F$ not just for invertible $d\times d$-matrices $L\in
K^{d\times d}$ defined over $K$, but also for invertible $d\times d$-matrices
$L\in A^{d\times d}$ defined over any commutative $K$-algebra $A$ (where
$a\left(  L\right)  $ is now required to be an element of $A$). In particular,
this allows for an invertible $d\times d$-matrix whose entries are distinct
indeterminates. This makes the $LF=a\left(  L\right)  F$ condition stronger,
and as one can see, this added strength suffices to make Theorem \ref{thm.3.2}
valid for every commutative ring $K$:

\begin{theorem}
[the first fundamental theorem of invariant theory]\label{thm.cring.3.2}Let
$K$ be a commutative ring. A form in $P$ is an absolute invariant iff it is a
linear combination of standard rectangular bideterminants.
\end{theorem}

The first proof of Theorem \ref{thm.3.2} given above actually can be extended
(with some additional work) to prove Theorem \ref{thm.cring.3.2}\footnote{The
exact place where we need the form to be an absolute invariant (instead of
just being invariant) is the definition of the matrix $L$ that acts on $P$ by
(\ref{pf.thm.3.2.L1.def1}) and (\ref{pf.thm.3.2.L1.def2}). In the original
proof of Theorem \ref{thm.3.2}, it was sufficient to let $c$ be a nonzero
scalar in $K$. However, if $K$ is not an infinite field, there may not be
enough nonzero scalars to make such a definition useful. (In particular, we
might not be able to derive the equality (\ref{pf.thm.3.2.L1.7}) from knowing
that every nonzero scalar $c\in K$ satisfies (\ref{pf.thm.3.2.L1.4}).) Thus,
instead of letting $c$ be a nonzero scalar in $K$, we have to set $c$ to be
the indeterminate $\mathbf{t}$ in the univariate polynomial ring $K\left[
\mathbf{t}\right]  $. The matrix $L$ will thus be a matrix in $\left(
K\left[  \mathbf{t}\right]  \right)  ^{d\times d}$, not in $K^{d\times d}$.
Assuming that $F$ is an absolute invariant, we shall then be able to deduce
(\ref{pf.thm.3.2.L1.7}) from (\ref{pf.thm.3.2.L1.4}) quickly: Every two
elements $s$ and $t$ satisfy $\alpha_{s}c^{b_{j}^{\left(  s\right)  }}%
=\alpha_{s}a\left(  L\right)  =\alpha_{s}c^{b_{j}^{\left(  t\right)  }}$; this
rewrites as $\alpha_{s}\mathbf{t}^{b_{j}^{\left(  s\right)  }}=\alpha
_{s}\mathbf{t}^{b_{j}^{\left(  t\right)  }}$; but since $\alpha_{s}$ is
nonzero and $\mathbf{t}$ is an indeterminate, this can only happen if
$b_{j}^{\left(  s\right)  }=b_{j}^{\left(  t\right)  }$.}. (The second proof
can probably not be extended, at least not easily.)

\begin{thebibliography}{99}                                                                                               %


\bibitem[1]{1}\textsc{C. De Concini and C. Procesi}, A characteristic free
approach to invariant theory, \textit{Advances in Math.} \textbf{21} (1976),
330--354.\newline\url{https://doi.org/10.1016/S0001-8708(76)80003-5}

\bibitem[2]{2}\textsc{P. Doubilet and G.-C. Rota}, Skew-symmetric invariant
theory, \textit{Advances in Math.} \textbf{21} (1976), 196--201.\newline\url{https://doi.org/10.1016/0001-8708(76)90075-X}

\bibitem[3]{3}\textsc{P. Doubilet, G.-C. Rota, and J. Stein}, On the
foundations of combinatorial theory: IX. Combinatorial methods in invariant
theory, \textit{Stud. Appl. Math.} \textbf{53} (1974), 185--216.\newline\url{https://doi.org/10.1002/sapm1974533185}

\bibitem[4]{4}\textsc{H. Weyl}, \textquotedblleft The Classical
Groups,\textquotedblright\ Princeton Univ. Press, Princeton, N.J., 1946. ISBN: 0691057567.

\bibitem[5]{5}\textsc{W. Whiteley}, Logic and invariant theory. I: Invariant
theory of projective properties, \textit{Trans. Amer. Math. Soc.}
\textbf{1977} (1973), 121--139.\newline\url{https://doi.org/10.1090/S0002-9947-1973-0446962-6}

\bibitem[6]{6}\textsc{J. D\'{e}sarm\'{e}nien and G.-C. Rota}, Th\'{e}orie
Combinatoire des Invariants Classiques, Series de Math\'{e}matique Pures et
Appliqu\'{e}es, IRMA, Strasbourg, 1977.

\bibitem[7]{7}\textsc{David G. Mead}, Determinantal ideals, identities and the
Wronskian, \textit{Pacific J. Math.} \textbf{42} (1972), 165--175.\newline\url{http://dx.doi.org/10.2140/pjm.1972.42.165}

\bibitem[8]{8}\textsc{W. Specht}, Die irreduciblen Darstellungen der
symmetrischen Gruppe, \textit{Math. Zeit.} \textbf{39} (1935),
696--711.\newline\url{http://eudml.org/doc/168580}
\end{thebibliography}

Papers mentioned in bibliographies of papers above are not repeated
here.\footnote{\textit{Correction:} The references below have been inserted by
me. Also, I have taken the liberty to fix two typos in the references \cite{7}
and \cite{8} above, as well as include URLs, DOIs and ISBNs whenever I could
find them.}

\renewcommand{\refname}{Additional references}

\begin{thebibliography}{9999}                                                                                             %


\bibitem[9]{DCEiPr}\textsc{C. DeConcini, David Eisenbud, and C. Procesi},
Young Diagrams and Determinantal Varieties, \textit{Inventiones mathematicae}
\textbf{56} (1980), 129--166.\newline\url{http://eudml.org/doc/142693}

\bibitem[10]{GrRoSt}\textsc{Frank D. Grosshans, Gian-Carlo Rota, and Joel A.
Stein}, \textquotedblleft Invariant theory and superalgebras\textquotedblright%
, CBMS Regional Conference Series in Mathematics, 69.

\bibitem[11]{Swan}\textsc{Richard G. Swan}, On the straightening law for
minors of a matrix, arXiv:1605.06696v1.\newline\url{https://arxiv.org/abs/1605.06696v1}

\bibitem[12]{Procesi}\textsc{Claudio Procesi}, \textquotedblleft Lie Groups:
An Approach through Invariants and Representations\textquotedblright,
Springer, 2007.

\bibitem[13]{Igusa}\textsc{Jun-ichi Igusa}, On the arithmetic normality of the
Grassmann variety, \textit{Proc. Nat. Acad. Sc. U.S.A.} \textbf{40} (1954),
309--313.\newline\url{https://www.ncbi.nlm.nih.gov/pmc/articles/PMC534128/}

\bibitem[14]{Steinberg}\textsc{Robert Steinberg}, On Dickson's theorem on
invariants, \textit{J. Fac. Sci. Univ. Tokyo Sect. IA Math.} \textbf{34}
(1987), no. 3, 699--707.\newline\url{http://jairo.nii.ac.jp/0021/00005736/en}

\bibitem[15]{Wilkerson}\textsc{Clarence Wilkerson}, A primer on the Dickson
invariants, corrected version of a paper published in \textit{Contemporary
Mathematics} \textbf{19} (1983), 421--434.\newline\url{http://www.math.purdue.edu/~wilker/papers/dickson.pdf}

\bibitem[16]{Macdonald}\textsc{I. G. Macdonald}, Schur functions : theme and
variations, \textit{Publ. I.R.M.A. Strasbourg, Actes 28 e S\'{e}minaire
Lotharingien}, \textbf{498} (1992), 5--39.\newline\url{https://eudml.org/doc/119412}

\bibitem[17]{KacGri}\textsc{Victor Kac}, \textquotedblleft Vertex algebras,
Lecture 3: Fundamentals of formal distributions\textquotedblright, scribe
notes by Darij Grinberg, version of 24 July 2017.\newline\url{http://www.cip.ifi.lmu.de/~grinberg/algebra/va3.pdf}
\end{thebibliography}


\end{document}