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\begin{document}
\shortauthor{A. Lascoux and P. Pragacz}
\shorttitle{Double Sylvester sums for subresultants}


\title{Double Sylvester sums for subresultants and multi-Schur functions}
\author[1]{Alain Lascoux}
\author[2]{Piotr Pragacz}
\address[1]{C.N.R.S., Institut Gaspard Monge,
Universit\'e de Marne-la-Vall\'ee,\\ 5 Bd Descartes,
Champs sur Marne, F 77454 Marne La Vall\'ee Cedex 2, France,
e-mail: Alain.Lascoux@univ-mlv.fr}
\address[2]{Institute of Mathematics, Polish Academy of Sciences,
\'Sniadeckich 8, P.O.Box~21, PL 00-956 Warszawa 10, Poland,
e-mail:  pragacz@impan.gov.pl}
\maketitle
%\subjclass{05E05, 13B25, 14M15}

\begin{abstract} J. J. Sylvester has announced formulas expressing
the subresultants (or
the successive polynomial remainders for the Euclidean division)
of two polynomials, in terms of some
double sums over the roots of the two polynomials. We prove
Sylvester formulas using the techniques of multivariate polynomials
involving multi-Schur functions and divided differences.
\end{abstract}


\setcounter{footnote}0
\section*{Introduction and statement of the main result}

The subresultants played a fundamental role in the theory of
polynomial equations in the 19-th century, cf. e.g. \citep{S1,S2,S3},
\citep{Bo}, and \citep{Sa}.
Recently they
also have found important applications in computer algebra, for example,
in devising efficient methods for computing greatest common
divisors of two polynomials \citep{C1,C2}, \citep{B}, and
\citep{BT},
for carrying out
quantifier elimination over complex of real numbers cf. e.g. \citep{C3},
and also for coding theory \citep{Sh}. They have been also extended
to some noncommutative polynomials \citep{Ch}, \citep{Li}, and \citep{H2}.
\smallskip

Suppose that two polynomials in one variable
$$
P(x)=x^m+\alpha_1 x^{m-1}+\cdots +\alpha_m
\qquad \hbox{and} \qquad Q(x)=x^n+\beta_1 x^{n-1}+\cdots +\beta_n
$$
with, say, complex coefficients are given. Then the
{\it subresultant of degree} $d$ associated with $P$ and $Q$,
denoted by $\Cal R^{(d)}$,
is defined by the following determinant:
$$
\Cal R^{(d)}=
\left|\
\CD
1 \quad & \quad \alpha_1 \quad & \quad \alpha_2  \quad & \quad \alpha_3 \ \
&  \quad \hdots  \quad & \quad \alpha_{m+n-2d-2} \quad & \quad x^{n-d-1}P \quad \\
0 & 1       & \alpha_1&\alpha_2 &  \hdots      & \alpha_{m+n-2d-3} &   x^{n-d-2}P \\
\vdots  & \vdots  &  \vdots &  \vdots & & \vdots & \vdots  \\
\cdot   & \cdot   &  \cdot  &  \cdot  & & \alpha_{m-d-1}& P \\
1 & \beta_1& \beta_2&\beta_3 &  \ \hdots  \ & \beta_{m+n-2d-2} \ & \ x^{m-d-1}Q \\
0 & 1       & \beta_1&\beta_2 &  \hdots      & \beta_{m+n-2d-3} &   x^{m-d-2}Q \\
\vdots  & \vdots  &  \vdots &  \vdots & & \vdots & \vdots  \\
\cdot   & \cdot   &  \cdot  &  \cdot  & & \beta_{n-d-1}& Q \\
\endCD\ \right|.
$$
Here we understand that $\alpha_i=0$ for $i>m$ and $\beta_j=0$ for $j>n$.
Of course, the subresultant $\Cal R^{(0)}$ of degree zero equals
the resultant. We refer to
\citep{BT}, \citep{C1,C2}, \citep{GLRR}, \citep{H1}, \citep{J1,J2},
\citep{L3}, \citep{LRS}, and \citep{Lo}
for a more detailed discussion of this notion and the properties of
subresultants.

We are interested in the problem of expressing  $\Cal R^{(d)}$
as polynomials in the roots of $P$ and $Q$,
generalizing the expression of the resultant $\Cal R^{(0)}$
as the product of the differences of the roots of $P$ and $Q$.

Recall the following well-known interpretation of the subresultants.
Suppose (for the rest of this paper) that
$$m=\deg(P) \ge n=\deg(Q)\,.$$
Let us look at the iterated division of $P$ and $Q$:
$$
P = * \ Q + c_1 \Cal R_1  \quad  , \quad \  Q= * \ \Cal R_1 + c_2\Cal R_2 \quad , \quad \
\Cal R_1= * \ \Cal R_2 + c_3 \Cal R_3 \quad , \quad \ldots \ .
$$
The successive coefficients ``\ ${*}$\ " are the unique polynomials
such that
$$
n>\deg \Cal R_1 > \deg \Cal R_2 > \deg \Cal R_3 > \cdots \ .
$$
Instead of the usual Euclidean division, where $c_1=c_2=c_3=\cdots=1$,
we choose the constants $c_i$ in such a way that the successive remainders
$\Cal R_i$ are equal to $\Cal R^{(n-i)}$, $i=0,1,\ldots,n-1$.
In particular,
instead of being rational functions in the roots of $P$ and $Q$
(as in the Euclidean division case), they are polynomials in the roots,
cf. Remark \ref{rem1.12}.

An interesting solution to the above problem was proposed
by Sylvester about 160 years ago
\citep{S1,S2,S3}, who found \
{``the successive residues,
divested of their allotrious factors \dots"} \
that is, who also normalized the remainders
in such a way as to obtain polynomials in the roots,
the last ``residue" being the resultant.
To state Sylvester's result, we need the following notation. For two finite sets $\A$ and $\B$
of elements in a commutative ring, we set
$$
R(\A,\B):= \prod _{a \in \A, b \in \B} (a-b)\,.
$$
Let now $\A=\{a_1,\ldots,a_m\}$, $\B=\{b_1,\ldots,b_n\}$ be two sequences
of (commuting) variables.
Let $0\le p\le m$, $0\le q\le n$ be two integers.
Define, after \citep{S2,S3}, the following double sum:
$$
\Sylv^{p,q}(\A,\B;x)
:=\sum_{\A'\subset \A}  \ \sum_{\B'\subset \B} R(x,\A')
R(x,\B') \frac{R(\A',\B')R(\A \smallsetminus \A',\B \smallsetminus \B')}
{R(\A',\A \smallsetminus \A')R(\B',\B \smallsetminus \B')}\,,
$$
the sum being over all subsets $\A'$ of cardinality $p$ and $\B'$ of cardinality
$q$. Sometimes, we will also use the notation $\Sylv^{p,q}(x)$ or $\Sylv^{p,q}$ for this sum.
The Sylvester sum $\Sylv^{p,q}(\A,\B;x)$ is a polynomial in $\A$, $\B$, and $x$;
moreover $\deg_x \Sylv^{p,q}=p+q$\,.
Observe that $\Sylv^{0,0}=R(\A,\B)$ and if $p=0$ or $q=0$ then the
Sylvester sum reduces to a single sum. More interesting Sylvester
sums are given in Example \ref{ex2.10} and \ref{ex2.12},
and Remark \ref{rem2.13}.
Summations of this type and their
generalizations (cf. \citep{LS2}) play nowadays an important
role e.g. in representation
theory, and in the description of Gysin maps
(called also ``integrations over a fibre")
for fibrations with homogeneous spaces as fibres
(cf. \citep{AC}, \citep{Br}, \citep{L1}, and \citep{P2,P3}).


\smallskip
The main result connecting the subresultants
and Sylvester sums is the following theorem.


\bth \label{th0.1} If $p+q<n$, then specializing $\A$ and $\B$
in the roots of the polynomials $P$ and $Q$ respectively, one has
$$
\Sylv^{p,q} = (-1)^{p(m-p-q)}\binom{p+q}p \Cal R^{(p+q)}\,.
$$
\eth

This result was stated without complete proof in
\citep{S2,S3} and we did not
find any in the literature.
All other expressions of subresultants that we have
found in the literature can be derived from either the above determinant
or the determinant \eqr(2.3), and this is why we wanted to comment
about Sylvester's formulas, from an algebraic point of view only.
We shall not get involved into elimination theory, but refer
for that to the papers
on subresultants quoted in the references.
The case of a single Sylvester sum $p=0$ or $q=0$ was already 
announced in an earlier paper
\citep{S1}, as a generalization of Lagrange interpolation,
and proved in \citep{Bo} (we discuss the
result of Borchardt in Remark \ref{rem2.13}.).
Let us stress that the case of a single  sum {\it does not} imply the
assertion of the theorem for the double Sylvester sums.

The objective of the present paper
is to give a proof of Theorem \ref{th0.1} with the help of techniques of
multivariate polynomials, involving multi-Schur functions
and divided differences. The assumption $p+q<n$ {\it is essential}. In fact, for the needs of the
proof, we also compute $\Sylv^{p,q}(\A,\B;x)$ when $p+q=n$. A ``closed form" formula
for Sylvester sum in the case $p+q>n$, remains an open problem.
The strategy of the present paper is to express all functions
as multi-Schur functions. This was already used in \cite{L3}.

\smallskip

The paper is organized as follows.

\smallskip

In the first section we recall some information on multi-Schur functions.

In the second section which is the core of the paper,
we provide a proof of Theorem \ref{th0.1}. This proof goes by
induction on $\deg(Q)$ and makes use of a formula which expresses
the first remainder $\Cal R_1$ with the help of
divided differences. An important ingredient of the proof
is an expression of a subresultant in the form of a multi-Schur
function given in Proposition 2.2. This last expression is proved
via specialization. We also discuss there the result of \citep{Bo}
on single Sylvester sums.

In the third section we show, using some symmetrizing operators,
another way to obtain the key identity \eqr(2.11) that relates
the double Sylvester sums with the multi-Schur functions.

These algebro-combinatorial computations can be interpreted in terms of\break
Gysin maps for the fibre product of two Grassmann bundles;
we discuss this in the fourth section.

\setcounter{footnote}0
\section{Review of symmetric polynomials and multi-Schur\\ functions}
Given a formal series in $z$, $\sum_{i=0}^{\infty} z^i S_i$\,,
with $S_0=1$ and with coefficients from an arbitrary commutative ring,
one associates
with the series an infinite matrix of their coefficients:
\begin{equation}
\bold S = [ S_{k-h} ]_{h,k\ge 1}
\label{eq1.1}
\end{equation}
(one has $S_i=0$ if $i<0$). For any natural number $l$ and any
$I=(i_1,\ldots,i_l)\in \Bb Z^l$ one associates with the series
the {\it Schur function} $S_I$ indexed by $I$: by definition
$S_I$ is the minor of $\bold S$ taken over the $l$ first rows
and over the columns $i_1+1$, $i_2+2$, \dots, $i_l+l$ \ ($S_I$ is zero
if one of these integers is negative).

More explicitly,
\begin{equation}
S_I = | S_{i_k+k-h} |_{1 \le h,k \le l}\,.
\label{eq1.2}
\end{equation}

We note that $S_I=\pm S_J$ or $0$, where $J=(0\le j_1\le \cdots \le j_l)$
is a {\it partition}.
For example,
$$
S_{5,7,2,0,3}
=\left|\,\CD
S_5 \ & \ S_8 \ & S_4 \ & \ S_3 \ &  \ S_7 \\
S_4 & S_7 & S_3 & S_2 & S_6 \\
S_3 & S_6 & S_2 & S_1 & S_5 \\
S_2 & S_5 & S_1 & 1 & S_4 \\
S_1 & S_4 & 1 & 0 & S_3 \\
\endCD\,\right|
=\left|\,\CD
S_3 \ & \ S_4 \ & S_5 \ & \ S_7 \ &  \ S_8 \\
S_2 & S_3 & S_4 & S_6 & S_7 \\
S_1 & S_2 & S_3 & S_5 & S_6 \\
1 & S_1 & S_2 & S_4 & S_5 \\
0 & 1 & S_1 & S_3 & S_4 \\
\endCD\,\right|=S_{3,3,3,4,4}\,.
$$

Given a (finite) {\it alphabet} $\A$, i.e. a multi-set of elements
in a commutative ring
(often indeterminates), we denote by $S_I(\A)$ the Schur functions
associated with the series
$$
\prod_{a\in \A}(1-za)^{-1}\,.
$$
This notion is extended to {\it virtual alphabets}%
\footnote{By a virtual alphabet, we understand a
formal finite $\Z$-linear combination of alphabets. The precise sense
of the operations ``$+$" and ``$-$" for alphabets, is explained in this
section. Of course, using these operations, any virtual alphabet has the form
$\A -\B$, for some alphabets $\A$ and $\B$. All these notions are
clarified by the use of $\lambda$-rings, which are the appropriate
structure for manipulations of symmetric functions.}
as follows.
Given two alphabets $\A, \B$,
we denote by $S_I(\A - \B)$ the Schur functions associated
with the series
$$
\prod_{b\in \B}(1-zb)/ \prod_{a\in \A}(1-za)\,.
$$
So $S_i(\A-\B)$ interpolates between $S_i(\A)$ -- the complete
homogeneous symmetric polynomial of degree $i$ in $\A$ 
and $S_i(-\B)$ -- the $i$-th
elementary polynomial in $\B$ times $(-1)^i$.

The notation $\A -\B$ is compatible with the multiplication
of series:
\begin{equation}
\sum z^i S_i(\A - \B) \cdot \sum z^j S_j(\A' - \B') =
\sum z^i S_i\bigl((\A+\A') - (\B+\B')\bigr)\,,
\label{eq1.3}
\end{equation}
the sum $\A + \A'$ denoting the union of two alphabets $\A$ and $\A'$.

\dfbn{Convention} We will often identify an alphabet $\A=\{a_1,\ldots,a_m\}$
with the sum $a_1+\cdots +a_m$ and perform usual algebraic operations
on such elements. We will give priority to the algebraic notation over
the set-theoretic one.
\enddfbn

We have $(\A+\C) - (\B+\C) = \A + \B$, and this corresponds to
simplification of the common factor for the rational series:
$$
\sum z^i S_i((\A + \C) - (\B + \C)) = \sum z^i S_i(\A-\B)\,.
$$

The cardinality of $\B$ being $n$, we can write the polynomial
$\prod_{b\in \B}(x-b)$ in the compact form $S_n(x-\B)$; after
the previous definition for $\A=\{x\}$ one has
\begin{equation}
S_n(x-\B)=x^n + x^{n-1}S_1(-\B)+ \cdots +x^0 S_n(-\B)\,.
\label{eq1.4}
\end{equation}
Moreover, one has for $i>0$,
\begin{equation}
S_{n+i}(-\B)=0 \qquad  \hbox{and} \qquad S_{n+i}(x-\B)=x^i S_n(x-\B)\,.
\label{eq1.5}
\end{equation}
The inverse of the rational series $\sum z^i S_i(\A-\B)$ is the series
$\sum z^i S_i(\B-\A)$; the inversion of series corresponds to the
inversion of the matrices {\bf S}, and for their minors which 
are Schur functions,
to the transposition of the partitions indexing them. 
In particular,
for a partition $I=(i_1,\ldots,i_l)$, one has $S_I(-\A)=
(-1)^{i_1+\cdots+i_l}S_J(\A)$ where $J$ is the transpose
of $I$.

Given $l$ virtual alphabets $\A^{(1)}, \ldots, \A^{(l)}$,
and a sequence $I \in \Z^l$, one associates with them the
{\it multi-Schur function}
\begin{equation}
S_I\bigl(\A^{(1)},\ldots,\A^{(l)}\bigr)
= | S_{i_k+k-h}(\A^{(k)}) |_{1\le h,k \le l}\,.
\label{eq1.6}
\end{equation}
Numerous applications of multi-Schur functions are discussed in \citep{LS1}.

\dfbn{Notation}
Suppose that $I=(i_1,i_2,\ldots)$ is a partition.
We will group the parts of $I$ which correspond
to the same argument. So, if $i_r=i_{r+1}=\cdots=i_t(=i)$ is the maximal
subsequence of equal parts of $I$ corresponding to the fixed argument
$\A^{(k)}$, then
we will write the multi-Schur function (1.6) as
$$
S_{\ldots ;i^{t-r}; \ldots}(\ldots ; \A^{(k)}; \ldots)\,.
$$
For example, the multi-Schur function $S_{1,1,1,1,2,2,4,4,4,4,5}$
associated with a sequence of virtual alphabets
$$
\A^{(1)},\A^{(2)},\A^{(2)},\A^{(3)},\A^{(3)},\A^{(3)},\A^{(4)},\A^{(4)},\A^{(4)},\A^{(5)},\A^{(5)}
$$
is written as
$$
S_{1;1^2;1;2^2;4^3;4;5}(\A^{(1)};\A^{(2)};\A^{(3)};\A^{(3)};\A^{(4)};\A^{(5)};\A^{(5)})\,.
$$
\enddfbn

By using the relation
$$
S_p(\A-x)=S_p(\A)-xS_{p-1}(\A)\,,
$$
where $\A$ is a virtual alphabet and $x$ is a single variable,
and performing elementary operations on the rows
(or columns) of a determinant of the form \eqr(1.6), we arrive at the following
transformation lemma:

\blm \label{lem1.7} Let $t\le l$ be a positive integer and $\X$ an alphabet of
cardinality $r\le t$.
\begin{itemize}
\item[\rm(i)] The replacement of $S_p(\A^{(k)})$ by $S_p(\A^{(k)}-\X)$
in the first $t-r$ rows of a sequence of $t$ successive rows of
the determinant \eqr(1.6), does not change the value of the determinant.
\item[\rm(ii)] Suppose that there exists $k$ such that $k+t\le l$,
$i_{k+1}=i_{k+2}=\cdots=i_{k+t}$, and $\A^{(k+1)}=\cdots =\A^{(k+t)}=\A$.
Then the replacement of $S_p(\A)$ by $S_p(\A-\X)$ in the
$(k+t)$-th, $(k+t-1)$-th, \ldots, $(k+r+2)$-th column
of the determinant \eqr(1.6),
does not change the value of the determinant.
\end{itemize}
\elm

\bex \label{ex1.8} We illustrate {\rm(i)}. For virtual alphabets $\A$, $\B$, \dots , and an alphabet $\X$
of cardinality $2$, the following equality holds:
\begin{multline*}
\left|\,\CD
\ S_6(\A)\ &\ S_9(\B)\ &\ S_5(\C)\ &\ S_4(\D)\ &\ S_8(\E)\ \\
S_5(\A) & S_8(\B) & S_4(\C) & S_3(\D) & S_7(\E) \\
S_4(\A) & S_7(\B) & S_3(\C) & S_2(\D) & S_6(\E) \\
S_3(\A) & S_6(\B) & S_2(\C) & S_1(\D) & S_5(\E) \\
S_2(\A) & S_5(\B) & S_1(\C) & 1 & S_4(\E) \\
\endCD\,\right|\\
\noalign{\vskip3pt}
=\left|\,\CD
S_6(\A) & S_9(\B) & S_5(\C) & S_4(\D) & S_8(\E) \\
\ S_5(\A-\X)\ &\ S_8(\B-\X)\ &\ S_4(\C-\X)\ &\ S_3(\D-\X)\ &\ S_7(\E-\X)\ \\
S_4(\A-\X) & S_7(\B-\X) & S_3(\C-\X) & S_2(\D-\X) & S_6(\E-\X) \\
S_3(\A) & S_6(\B) & S_2(\C) & S_1(\D) & S_5(\E) \\
S_2(\A) & S_5(\B) & S_1(\C) & 1 & S_4(\E) \\
\endCD\,\right|\,.
\end{multline*}
\eex

Let now $\A$ and $\B$ be two alphabets. The element
$R(\A,\B)=\prod_{a\in \A,b\in \B}(a-b)$ is
the resultant of the polynomials with roots $\A$ and $\B$.
Given an indeterminate $x$, the polynomial $S_m(x-\A)$ can be written
unambiguously as $R(x,\A)$. In general, we have the following expression:
\begin{equation}
R(\A,\B) = S_{n^m}(\A-\B)\,.
\label{eq1.9}
\end{equation}
See \citep{C1,C2}, \citep{J1,J2}, and \citep{L2} for more about resultants.

We will need the following two lemmas.

\blm \label{lem1.10} \ For positive integers $r$ and $t$, the following
equality holds:
$$
S_{t^r}(\A-\B)=\sum S_I(\A)S_{I^*}(-\B)\,,
$$
where the sum is over all partitions $I=(i_1,\ldots,i_r)$ with
$i_r \le t$, and $I^*$ stands for the partition $(t-i_r,\ldots,t-i_1)$.
\elm

\noindent
This a particular case of a well-known linearity formula,
cf. e.g. \citep{LS1}, \citep{L2}, and \citep{P2}.


\blm \label{lem1.11} For positive integers $d$, $r$, and $t$, and virtual alphabets
$\A$ and $\B$, the following equality holds:
$$
S_{1^d;t^r}(\A;\B) = \sum_{i=0}^d (-1)^i S_{1^{d-i}}(\A)S_{t^{r-1},t+i}(\B)\,.
$$
\elm

\begin{proof}
The Laplace expansion of the determinant $S_{1^d;t^r}(\A;\B)$ along the
first $d$ columns gives a sum of products of Schur functions of $\A$ and $\B$
respectively. One checks that their indices are as stated.
\end{proof}


For example, consider the determinant
of $S_{1,1;2,2,2}(\A;\B)$:
$$
\left|\,\CD
\ S_1(\A) \ & \ S_2(\A) \ & \ S_4(\B) \ & \ S_5(\B) \ & \ S_6(\B)\ \\
1& S_1(\A) & S_3(\B) & S_4(\B) & S_5(\B) \\
0& 1 & S_2(\B) & S_3(\B) & S_4(\B) \\
0 & 0 & S_1(\B) & S_2(\B) & S_3(\B) \\
0 & 0 & 1 & S_1(\B) & S_2(\B) \\
\endCD\,\right|\,.
$$
A term in the expansion is
$$
- \left|\,\CD \ S_1(\A) \ & \ S_2(\A)\ \\
    0 & 1        \\ \endCD\,\right|  \quad
\left|\,\CD
\ S_3(\B) \ & \ S_4(\B) \ & \ S_5(\B)\ \\
S_1(\B) & S_2(\B) & S_3(\B) \\
1 & S_1(\B) & S_2(\B) \\
\endCD\,\right|
$$
in which one recognizes  $-S_1(\A) \, S_{223}(\B)$, up to transpositions
along the anti-diagonals.

\brm \label{rem1.12} Let $P=S_m(x-\A)$ and $Q=S_n(x-\B)$ be two polynomials
in one variable (with $m\ge n$, as usual).
Algebraic operations on these two polynomials preserve the symmetry in the
roots of each of the two polynomials, and thus produce symmetric functions
in $\A$ and $\B$ separately.
The last nonzero remainder in the Euclidean division of $P$ by $Q$
is the GCD of $P$ and $Q$.
The GCD of $P$ and $Q$ is of degree $d$ iff the Schur functions
$S_{n^m}(\A-\B),\ldots, S_{(n-d+1)^{m-d+1}}(\A -\B)$
vanish and $S_{(n-d)^{m-d}}(\A -\B) \ne 0$, cf. \citep{T}, \citep{L2},
and \citep{P1,P2}.

We want the $n$-th normalized%
\footnote{By normalization, we mean here multiplication by a nonzero
scalar.}
remainder to be the resultant,
$\Cal R_n =\Cal R^{(0)}=R(\A,\B)=S_{n^m}(\A-\B)$.
We normalize the successive remainders by asking that
the $(n-d)$-th remainder $\Cal R_{n-d}$ has the leading term
$x^dS_{(n-d)^{m-d}}(\A-\B)$. Then it equals the
subresultant $\Cal R^{(d)}$. For more details see
\citep{C1,C2}, \citep{BT}, \citep{Lo}, \citep{L3}, and \citep{J2}.
In particular, if the GCD of $P$ and $Q$ is of degree $d$, then it
is given - up to a nonzero scalar - by the explicit determinants
\eqr(2.1) and \eqr(2.3) below.
\erm

\setcounter{footnote}0
\section{Proof of the main theorem}

Recall that the subresultant $\Cal R^{(d)}$ of degree $d$, associated
with a pair of polynomials $S_m(x-\A)$ and $S_n(x-\B)$,
is defined as the following
determinant:
\begin{equation}
\Cal R^{(d)} = \left|
\CD
S_0(-\A)  & \q \hdots \q & \q S_{m+n-2d-2}(-\A) \q
& \q S_{m+n-d-1}(x-\A)\q \\
\vdots  &   &  \vdots &  \vdots \\
\q S_{d-n+1}(-\A)\q &  \hdots & S_{m-d-1}(-\A)  & S_m(x-\A) \\
S_0(-\B)& \hdots & S_{m+n-2d-2}(-\B) & S_{m+n-d-1}(x-\B) \\
\vdots  &   &  \vdots &  \vdots \\
S_{d-m+1}(-\B) &  \hdots & S_{n-d-1}(-\B)  & S_n(x-\B) \\
\endCD\ \right|.
\label{eq2.1}
\end{equation}

In the following proposition we present the subresultant $\Cal R^{(d)}$
as a multi-Schur function.
We give a proof by specialization, using some simple transformations
of determinants. (See \citep{L3} for a connection
with Wro\'nski's work \citep{W}; cf. also \citep{L4}.)

\bpr\label{prop2.2} For $d=0,1,\ldots,n-1$, we have
$$
\Cal R^{(d)} = (-1)^{mn-d(m+n)} S_{1^d;(m-d)^{n-d}}(\B-x;\B-\A)\,.
$$
\epr

\begin{proof}
We recall that \ $S_{1^d;(m-d)^{n-d}}(\B-x;\B-\A)$ is the determinant:
{\footnotesize
\begin{equation}
\left|\CD
\q S_1(\B-x)\q &  \q \hdots \q &\q S_d(\B-x) \q & \q S_m(\B-\A)\q & \q  \hdots \q
&\q S_{m+n-d-1}(\B-\A)\q \\
\vdots &  &\vdots &\vdots &  &\vdots  \\
\q S_{2-n}(\B-x)\q & \hdots &\q S_{d-n+1}(\B-x)\q
 &\q S_{m-n+1}(\B-\A)\q & \hdots &\q S_{m-d}(\B-\A)\q
\endCD\right|\,.
\label{eq2.3}
\end{equation}}
It suffices to show that the $n$
 specializations $x =b\in \B$ \ of
$\Cal R^{(d)}$ and of the determinant \eqr(2.3)
coincide because these two polynomials have degree $<n$.

We specialize the determinant \eqr(2.3).
First, using the transformation Lemma \ref{lem1.7}(i), we subtract $\B-b$
in the top row.
We perform operations on alphabets only, and therefore will write only them.
So, starting from
$$
\left|\CD
\q \B-b \q & \q \hdots \q & \q \B-b \q & \q \B-\A \q & \q \hdots \q
& \q \B-\A\q \\
\vdots &  & \vdots & \vdots &  &\vdots \\
\B-b &  \hdots  & \B-b & \B-\A &  \hdots  & \B-\A \\
\endCD\right|\,,
$$
(where $\B-b$ appears in the first $d$ columns), we see
that the previous
determinant is equal to the following one:
$$
\left|\CD
 0  & \q \hdots \q &  0  &  b-\A  & \q \hdots \q &  b-\A \\
\q \B-b\q  & \q \hdots \q &\q \B-b \q & \q \B-\A\q & \q \hdots \q &\q \B-\A\q \\
\vdots &  & \vdots & \vdots &  &\vdots \\
\B-b &  \hdots  & \B-b & \B-\A &  \hdots & \B-\A \\
\endCD\right|
$$
(where the zeros appear in the first $d$ columns).
Now, using the transformation Lemma \ref{lem1.7}(ii), we subtract $b$
in the last $n-d-1$ columns; we see that
the previous determinant becomes
$$
\left|\CD
0 &  \hdots  & 0 & b-\A  &  -\A  &  \hdots &  -\A \\
\q \B-b\q &\q \hdots\q &\q \B-b\q &\q \B-\A\q &\q \B-b-\A\q &\q \hdots\q
&\q \B-b-\A\q \\
\vdots &  & \vdots & \vdots & \vdots &  &\vdots \\
\B-b & \hdots & \B-b & \B-\A & \B-b-\A & \hdots & \B-b-\A \\
\endCD\right|\,.
$$
This last determinant clearly factorizes, because in the first row, the elements
$$
S_{m+1}(-\A), \ldots, S_{m+n-d-1}(-\A)
$$
are null. It is equal to
$$
(-1)^{d} R(b,\A) S_{(m-d)^{n-d-1}}(\B-b-\A)\,.
$$
Similarly, we specialize the determinant \eqr(2.1)
$$
\Cal R^{(d)}=\left|\CD
\q -\A \q & \q \hdots \q & \q -\A \q & \q x-\A\q  \\
\vdots &  & \vdots & \vdots \\
-\A  &  \hdots  &  -\A  &  x-\A  \\
-\B  &  \hdots  &  -\B  &  x-\B  \\
\vdots & & \vdots & \vdots \\
-\B  &  \hdots  &  -\B  &  x-\B  \\
\endCD\right|\,,
$$
in $x=b\in \B$. We obtain
$$
\left|\CD
\q -\A \q & \q \hdots \q & \q -\A \q & \q b-\A \q \\
\vdots &  & \vdots & \vdots \\
-\A  &  \hdots  &  -\A  &  b-\A  \\
-\B  &  \hdots  &  -\B  &  0  \\
\vdots &  & \vdots & \vdots \\
-\B  &  \hdots  &  -\B  &  0  \\
\endCD\right|\,.
$$
The bottom part of the last column:
$$
\CD
S_{m+n-d-1}(x-\B)\\
\vdots \\
S_n(x-\B) \endCD
$$
is null,
and one can write $0$ instead of $b-\B$.

Now, use the transformation Lemma \ref{lem1.7}(i) and subtract $b$ in the first
$n-d-2$ rows; we see
that the previous determinant is equal to the following one:
$$
\left|\CD
\q -\A-b \q & \q \hdots \q & \q -\A-b \q & \q -\A \q \\
\vdots &  & \vdots & \vdots \\
-\A-b  &  \hdots  &  -\A-b  &  -\A  \\
-\A  &  \hdots  &  -\A  & \q b-\A \q \\
-\B  &  \hdots  &  -\B  &  0  \\
\vdots &  & \vdots & \vdots \\
-\B  &  \hdots  &  -\B  &  0  \\
\endCD\right|\,.
$$
Taking finally the Laplace expansion of the last determinant
along its last column, we see that this determinant is equal to
$$
(-1)^{(m-d)(n-d)} R(b,\A) S_{(m-d)^{n-d-1}}(\B-b-\A)\,.
$$
(Here we also use Lemma \ref{lem1.10} giving the expansion of
$S_{(m-d)^{n-d-1}}(\B-b-\A)$ as a combination of the products
of Schur functions in $-\B$ and $-\A-b$.)
\end{proof}

In particular, $\Cal R_1=\Cal R^{(n-1)}$ equals \
$(-1)^m S_{1^{n-1};m-n+1}(\B-x;\B-\A)$.

\smallskip

Let $\Cal R_1^E$ denote the first remainder in the Euclidean
division of $S_m(x-\A)$ by $S_n(x-\B)$ (i.e. we put $c_1=1$ in
the notation of the Introduction).

\blm\label{lem2.4} One has \
$
\Cal R_1^E=(-1)^{n-1} S_{1^{n-1};m-n+1}(\B-x;\B-\A)
$.
\elm
\begin{proof}
Write $x-\A = (x-\B) + (\B-\A)$.
Then, using (1.5), we have
$$
S_m(x-\A)=\sum_{i=0}^{n-1}S_i(x-\B)S_{m-i}(\B-\A) + S_n(x-\B)
\bigl(\sum_{i=n}^m S_{m-i}(\B-\A)x^{i-n}\bigr)
$$
$$
= \bigl(\sum_{i=n}^m S_{m-i}(\B-\A)x^{i-n}\bigr) S_n(x-\B) + (-1)^{n-1} S_{1^{n-1};m-n+1}(\B-x;\B-\A)\,,
$$
by the Laplace expansion of the determinant forming the last summand,
along its last column.
\end{proof}


Let now $\B=\{b_1,\dots,b_n\}$  be a set of indeterminates.
Divided differences, due to \citep{N}, are linear operators
$$
\d_i:\Bb Z[\B]\to \Bb Z[\B]\qquad (\hbox{of degree }-1)
$$
defined by
\begin{equation}
\d_i(f)=(f - \tau_if)/(b_i- b_{i+1}),\qquad i=1,\dots,n-1,
\label{eq2.5}
\end{equation}
where $\tau_i$ denotes the $i$-th simple transposition exchanging $i$ with
$i+1$. More generally, to every reduced decomposition
$\sigma=\tau_{i_1} \cdots \tau_{i_k}$\,,
one associates the product $\d_{i_1} \cdots \d_{i_k}$
which is independent of the choice of the reduced decomposition of $\sigma$
(because the $\d_i$'s satisfy the braid relations), and thus can be denoted unambiguously
$\d_{\sigma}$.

\dfbn{Notation} Given an alphabet $\B=\{b_1,b_2,\ldots \}$,
we write $\B_i:=\{b_1,\ldots,b_i\}$.
\enddfbn

\blm\label{lem2.6} For every $k\in \Z$, and a virtual alphabet $\A$ which is
independent%
\footnote{It suffices that $\tau_i$ preserves $\A$.} of $\B$, one has
$$
\d_i S_k(\B_i -\A) =S_{k-1}(\B_{i+1}-\A)\,.
$$
\elm

\begin{proof} Introducing an extra variable $z$, and writing
$\B_i-\A= -b_{i+1} + (\B_{i+1}-\A)$, one factorizes $\sum_{k=0}^{\infty}z^kS_k(\B_i -\A)$
into   $fg$, with
$$
f= 1-zb_{i+1} \quad \hbox{and} \quad g=\sum_{k=0}^{\infty} z^kS_k(\B_{i+1} -\A)\ .
$$
The second series is symmetrical in $b_i, b_{i+1}$, therefore commutes
with $\d_i$. The image of $1-zb_{i+1}$ under $\d_i$ being $z$,
the assertion follows.
\end{proof}

The first remainder $\Cal R_1^E$ can be conveniently presented through
divided differences (the same operator expresses the remainder 
in the Lagrange
interpolation of a function in one variable).

\blm\label{lem2.7} The first remainder $\Cal R_1^E$ in the Euclidean
division
of $P(x)=S_m(x-\A)$ by $S_n(x-\B)$ equals
$$
\Cal R_1^E = \d_{n-1} \cdots \d_1 \bigl( P(b_1)R(x,\B-b_1)\bigr)\,.
$$
\elm

\begin{proof} We have
\bml
R(x,\B-b_1)=(x-b_2) \cdots (x-b_n)= S_{n-1}(x-(\B-b_1)) \hfill\cr
\hfill{}=(-1)^{n-1} S_{1^{n-1}}((\B-x)-b_1)\,,
\end{multline*}
and by the transformation Lemma \ref{lem1.7}(i), the product $P(b_1)R(x,\B-b_1)$
equals
$$
(-1)^{n-1} S_{1^{n-1};m}(\B-x;b_1-\A)\,.
$$
Thus, using Lemma \ref{lem2.6}, we see that the image of $P(b_1)R(x,\B-b_1)$
under $\d_{n-1}\cdots \d_1$ is
$$
(-1)^{n-1} S_{1^{n-1};m-n+1}(\B-x;\B-\A)=\Cal R_1\,,
$$
which is $\Cal R_1^E$ by Lemma \ref{lem2.4}.
\end{proof}

See Remark \ref{rem2.13} for some comments about this lemma and
its generalizations.

\dfbn{Convention} In the following, by the ``top coefficient" of a polynomial in $x$,
we shall mean the coefficient of the highest power of $x$ in the polynomial.
\enddfbn

The following lemma will be used in the proof several times:

\blm\label{lem2.8} \
\begin{itemize}
\item[\rm(i)] The specialization $\Sylv^{p,q}(\A,\B;a)$,
$a\in \A$, is equal to
$$
(-1)^p R(a,\B) \ c \,,
$$
where $c$ is the top coefficient of $\Sylv^{p,q}(\A-a,\B;x)$.

\item[\rm(ii)] The specialization $\Sylv^{p,q}(\A,\B;b)$, $b\in \B$,
is equal to
$$
(-1)^{m-p+q} R(b,\A) \ c \,,
$$
where $c$ is the top coefficient of $\Sylv^{p,q}(\A,\B-b;x)$.
\end{itemize}
\elm

\begin{proof}
We consider, for example, (ii). We have
$$
\aligned
\Sylv^{p,q}(b)&=\sum_{\A'\subset \A, \B'\subset \B - b} R(b,\A')R(b,\B')\frac{R(\A',\B')R(\A-\A',\B-\B')}{R(\A',\A-\A')R(\B',\B-\B')} \cr
&=(-1)^{m-p+q} R(b,\A)
\sum_{\A'\subset \A, \B'\subset \B - b}
\frac{R(\A',\B')R(\A-\A',\B-b-\B')}{R(\A',\A-\A')R(\B',\B-b-\B')}\,,
\endaligned
$$
as claimed.
\end{proof}


We need also the following
proposition concerning the limit case when $p+q=n$.\footnote{In fact, in the proof of Theorem \ref{th0.1}, we will need only part~(i)
of the proposition. Part~(ii) is added for the need of induction in the proof of the
proposition.}

\bpr\label{pr2.9} \
\begin{itemize}
\item[\rm(i)] Suppose that $p+q=n<m$. Then we have
$$
\Sylv^{p,q}(\A,\B;x) = (-1)^{(m-n)p}\binom{n}p R(x,\B)\,.
$$
\item[\rm(ii)] Suppose that $p+q=n=m$. Then we have
$$
\Sylv^{p,q}(\A,\B;x) = \binom{n-1}q R(x,\A)
+ \binom{n-1}p R(x,\B)\,.
$$
\end{itemize}
\epr

\begin{proof} Both assertions depend on the four-tuple
$\bigl(m=\card(\A), n=\card(\B),p,q\bigr)$.
We prove (i) and (ii) simultaneously using induction on $m+n$.

(i)\hphantom{i} Let $m>n$. By Lemma \ref{lem2.8}, 
the specialization $\Sylv^{p,q}(\A,\B;a)$
is equal to $(-1)^p R(a,\B)$ times the top coefficient of $\Sylv^{p,q}(\A-a,\B;x)$.

If $m-1>n$, then by (i) for $(m-1,n,p,q)$, this specialization is equal to
$$
(-1)^{(m-1-n)p+p}\binom{n}p R(a,\B)=(-1)^{(m-n)p}\binom{n}p R(a,\B)
$$
and these $m$ specializations determine the polynomial.

If $m-1=n$, then according to (ii) for $(n,n,p,q)$, the top coefficient
of $\Sylv^{p,q}(\A-a,\B;x)$
is equal to $\binom{n-1}p+\binom{n-1}q = \binom np$, and therefore
$\Sylv^{p,q}(\A,\B;x)$ coincides with $(-1)^p \binom np R(x,\B)$.

(ii) Let $m=n=p+q$. By Lemma \ref{lem2.8}, the specialization
$\Sylv^{p,q}(\A,\B;a)$ is equal to $(-1)^p R(a,\B)$ times the top
coefficient of $\Sylv^{p,q}(\A-a,\B;x)$. Observe that this last coefficient
equals the top coefficient of $\Sylv^{p,q-1}(\B,\A-a;x)$.  Then by (i)
for $(n,n-1,p,q-1)$, the considered specialization is $\binom{n-1}p R(a,\B)$.

Similarly, the specialization $\Sylv^{p,q}(\A,\B;b)$ is equal
to $R(b,\A)$ times the top coefficient of $\Sylv^{p,q}(\A,\B-b;x)$.
Observe that this last coefficient is equal to $(-1)^q$ times the top
coefficient of $\Sylv^{q,p-1}(\A,\B-b;x)$. Then by (i) for $(n,n-1,q,p-1)$,
the considered specialization is $\binom{n-1}q R(b,\A)$.

The polynomial $\Sylv^{p,q}(\A,\B;x)$ is determined by these
specializations.
\end{proof}

\bex\label{ex2.10} \
\begin{itemize}
\item[\rm(i)] For $m=3$, $n=2$, and $p=q=1$,
$$
\aligned
 \frac{(a_1-b_1)(a_2-b_2)(a_3-b_2)}{(a_1-a_2)(a_1-a_3)(b_1-b_2)}
(x-a_1)(x-b_1) &\\
{}+\frac{(a_1-b_2)(a_2-b_1)(a_3-b_1)}{(a_1-a_2)(a_1-a_3)(b_2-b_1)}
(x-a_1)(x-b_2) &\\
{}+\frac{(a_2-b_1)(a_1-b_2)(a_3-b_2)}{(a_2-a_1)(a_2-a_3)(b_1-b_2)}
(x-a_2)(x-b_1) &\\
{}+\frac{(a_2-b_2)(a_1-b_1)(a_3-b_1)}{(a_2-a_1)(a_2-a_3)(b_2-b_1)}
(x-a_2)(x-b_2) &\\
{}+\frac{(a_3-b_1)(a_1-b_2)(a_2-b_2)}{(a_3-a_1)(a_3-a_2)(b_1-b_2)}
(x-a_3)(x-b_1) &\\
{}+\frac{(a_3-b_2)(a_1-b_1)(a_2-b_1)}{(a_3-a_1)(a_3-a_2)(b_2-b_1)}
(x-a_3)(x-b_2) &
\endaligned
$$
is equal to $-2(x-b_1)(x-b_2)$.

\item[\rm(ii)] For $m=n=3$, $p=2$, and $q=1$,
%\footnotesize
$$
\aligned
  \frac{(a_1 - b_1)(a_2 - b_1)(a_3 - b_2)(a_3  - b_3)}
 {(a_1 - a_3)(a_2 - a_3)(b_1 - b_2)(b_1 - b_3)}(x - a_1) (x - a_2)(x - b_1)&\\
{}+ \frac{(a_1 - b_1)(a_3 - b_1)(a_2 - b_2)(a_2  - b_3)}
 {(a_1 - a_2)(a_3 - a_2)(b_1 - b_2)(b_1 - b_3)}(x - a_1) (x - a_3)(x - b_1)&\\
{}+\frac{(a_2 - b_1)(a_3 - b_1)(a_1- b_2)(a_1 - b_3)}
  {(a_2 - a_1)(a_3 - a_1)(b_1 - b_2)(b_1 - b_3)}(x - a_2)(x - a_3)(x - b_1)&\\
{}+\frac{(a_1 - b_2)(a_2 - b_2)(a_3 - b_1)(a_3  - b_3)}
 {(a_1 - a_3)(a_2 - a_3)(b_2 - b_1)(b_2  - b_3)} (x - a_1)(x-a_2)(x-b_2)&\\
{}+\frac{(a_1 - b_2)(a_3 - b_2)(a_2 - b_1)(a_2  - b_3)}
 {(a_1 - a_2)(a_3 - a_2)(b_2 - b_1)(b_2  - b_3)}(x - a_1) (x - a_3)(x - b_2)&\\
{}+\frac{(a_2 - b_2)(a_3 - b_2)(a_1- b_1)(a_1 - b_3)}
 {(a_2 - a_1)(a_3 - a_1)(b_2 - b_1)(b_2 - b_3)}(x - a_2) (x - a_3)(x - b_2)&\\
{}+\frac{(a_1 - b_3)(a_2 - b_3)(a_3 - b_1)(a_3  - b_2)}
 {(a_1 - a_3)(a_2 - a_3)(b_3 - b_1)(b_3 - b_2)}(x - a_1) (x - a_2)(x - b_3)&\\
{}+\frac{(a_1 - b_3)(a_3 - b_3)(a_2 - b_1)(a_2  - b_2)}
 {(a_1 - a_2)(a_3 - a_2)(b_3 - b_1)(b_3 - b_2)}(x - a_1) (x - a_3)(x - b_3)&\\
{}+\frac{(a_2 - b_3)(a_3 - b_3)(a_1- b_1)(a_1 - b_2)}
{(a_2 - a_1)(a_3  - a_1)(b_3 - b_1)(b_3 - b_2)}(x - a_2) (x - a_3)(x - b_3)&
\endaligned
$$
%\normalsize
is equal to $2(x-a_1)(x-a_2)(x-a_3) + (x-b_1)(x-b_2)(x-b_3)$.
\end{itemize}
\eex

We now can pass to the

\begin{pf} Write $d:=p+q$.
The assertion
of the theorem is equivalent to the identity:
\begin{equation}
\Sylv^{p,q}(\A,\B;x) = (-1)^{(m-d)(n-q)+d} \binom{d}p
S_{1^{d};(m-d)^{n-d}}(\B-x;\B-\A)
\label{eq2.11}
\end{equation}
if $d<n$.

The proof goes by induction on $n$. For $n=1$, $p=q=0$, and any $m\ge 1$, both
sides of \eqr(2.11) are equal to $R(\A,b_1)$.


To perform the induction step, observe that $\Sylv^{p,q}$, as a polynomial in $x$,
is of degree $d<n$.
Hence it coincides with its first remainder $\Cal R_1^E$ 
modulo $R(x,\B)$, and one can
use Lemma \ref{lem2.7} to obtain it starting from $\Sylv^{p,q}(b_1)$:
$$
\Sylv^{p,q}(x)= \d_{n-1}\cdots \d_1 \bigl(\Sylv^{p,q}(b_1)
R(x,\B-b_1)\bigl)\,.
$$
By Lemma \ref{lem2.8}, the specialization of $\Sylv^{p,q}(\A,\B;x)$ in $x=b_1$ is equal to
$(-1)^{m-d} R(b_1,\A)$ times
the top coefficient in $\Sylv^{p,q}(\A,\B-b_1;x)$. If $d<n-1$ then we can
apply the induction assumption, and we see that this coefficient
is equal to
$$
(-1)^{(m-d)(n-1-q)} \binom{d}p S_{\square} (\B-b_1-\A)\,,
$$
with $\square = (m-d)^{n-d-1}$. If $d=n-1$ then the coefficient
is given by the same formula by Proposition 2.9.

We are left with the computation of the following expression:
$$
\d_{n-1}\cdots\d_1 \bigl(R(x,\B-b_1)R(b_1,\A) S_{\square}(\B-b_1-\A)
\bigl)\,.
$$
We rewrite
\bml
R(x,\B-b_1)=S_{n-1}(x-(\B-b_1))\\
{}=(-1)^{n-1}S_{1^{n-1}}((\B-x)-b_1)=(-1)^{n-1}
\Bigl(\sum_{i=0}^{n-1} (-1)^i S_{1^{n-i-1}}(\B-x)b_1^i\Bigr)\,.
\end{multline*}
We have by \eqr(1.5)
$$
b_1^iR(b_1,\A)=S_{m+i}(b_1-\A)\,,
$$
and by the transformation Lemma \ref{lem1.7}(i),
$$
S_{m+i}(b_1-\A) S_{\square}(\B-b_1-\A) = S_{\square ; m+i}(\B-\A;b_1-\A)\,.
$$
Finally, we get with $\epsilon=(-1)^{(m-d)(n-q)+n-1}$,
$$
\Sylv^{p,q}(b_1)R(x,\B-b_1)=\epsilon \binom{d}p \sum_{i=0}^{n-1}(-1)^i S_{1^{n-1-i}}(\B-x)
S_{\square;m+i}(\B-\A; b_1-\A)\,.
$$
By Lemma \ref{lem2.6}, its image under $\d_{n-1} \cdots \d_1$ is equal to
$$
\epsilon \binom{d}p \sum_{i=0}^{n-1}(-1)^i S_{1^{n-1-i}}(\B-x) S_{\square , m+i-(n-1)}(\B-\A)\,.
$$
In this last sum, the terms for $i=0, \ldots, n-d-2$ disappear because
they correspond to determinants having two identical columns. In the remaining
sum for $i=n-d-1,\ldots,n-1$:
$$
\sum_{j=0}^d (-1)^j S_{1^{d-j}}(\B-x) S_{\square , m-d+j}(\B-\A)\,,
$$
one recognizes the Laplace expansion of the determinant
$S_{1^d;(m-d)^{n-d}}(\B-x;\B-\A)$
along the first $d$ columns (cf. Lemma \ref{lem1.11}). One also checks that all the signs fit
to give the assertion.

This ends the proof of the theorem.
\end{pf}

We illustrate the theorem by the following example.

\bex\label{ex2.12} Let $m=5, n=4$. Then the Sylvester sum $\Sylv^{2,1}(\A,\B;x)$:
\footnotesize
$$
\aligned
&  \quad \   \frac{(a_3-b_2)(a_4-b_2)(a_5-b_2)(a_3-b_3)(a_4-b_3)(a_5-b_3)(a_3-b_4)(a_4-b_4)
(a_5-b_4)}{(a_1-a_3)(a_1-a_4)(a_1-a_5)(a_2-a_3)(a_2-a_4)(a_2-a_5)(b_1-b_2)(b_1-b_3)
(b_1-b_4)}\cr
&\quad \ \times (a_1-b_1)(a_2-b_1)(x-a_1)(x-a_2)(x-b_1)\cr
&+\frac{(a_3-b_1)(a_4-b_1)(a_5-b_1)(a_3-b_3)(a_4-b_3)(a_5-b_3)(a_3-b_4)(a_4-b_4)
(a_5-b_4)}{(a_1-a_3)(a_1-a_4)(a_1-a_5)(a_2-a_3)(a_2-a_4)(a_2-a_5)(b_2-b_1)(b_2-b_3)
(b_2-b_4)}\cr
&\quad \ \times (a_1-b_2)(a_2-b_2)(x-a_1)(x-a_2)(x-b_2)\cr
&+\frac{(a_3-b_1)(a_4-b_1)(a_5-b_1)(a_3-b_2)(a_4-b_2)(a_5-b_2)(a_3-b_4)(a_4-b_4)
(a_5-b_4)}{(a_1-a_3)(a_1-a_4)(a_1-a_5)(a_2-a_3)(a_2-a_4)(a_2-a_5)(b_3-b_1)(b_3-b_2)
(b_3-b_4)}\cr
&\quad \ \times (a_1-b_3)(a_2-b_3)(x-a_1)(x-a_2)(x-b_3)\cr
&+\frac{(a_3-b_1)(a_4-b_1)(a_5-b_1)(a_3-b_2)(a_4-b_2)(a_5-b_2)(a_3-b_4)(a_4-b_4)
(a_5-b_4)}{(a_1-a_3)(a_1-a_4)(a_1-a_5)(a_2-a_3)(a_2-a_4)(a_2-a_5)(b_4-b_1)(b_4-b_2)
(b_4-b_3)}\cr
&\quad \ \times (a_1-b_4)(a_2-b_4)(x-a_1)(x-a_2)(x-b_4)\cr
&+ 9\times 4 \ \hbox{similar terms  obtained  by  replacing }
(a_1,a_2) \hbox{ by } (a_i,a_j) \hbox{ for } i<j
\endaligned
$$
\normalsize
is equal to $3 \Cal R^{(3)}$.
\eex

\brm \label{rem2.13}
\stepcounter{equation}
If $p=0$ or $q=0$, then double Sylvester sums reduce to single ones;
these single sums appeared already in \citep{S1}
together with an announcement of the assertion of Theorem~\ref{th0.1}
in that case. For example,
\begin{equation}
\Sylv^{0,q}(\A,\B;x)= \sum_{\B'\subset \B} R(x,\B')\frac{R(\A,\B-\B')}{R(\B',\B-\B')}\,,
\label{eq2.14}
\end{equation}
the sum being over all subsets $\B'$ of cardinality $q$.
In this case, the assertion of Theorem~\ref{th0.1}
was proved in \citep{Bo}. Here is a short proof of it,
or equivalently of the identity \eqr(2.11), using a result of \citep{L1}.
We have, with $\B'':= \B-\B'$,
\bml
\Sylv^{0,q}(\A,\B;x)=
\sum_{\B'\subset \B} R(x,\B')\frac{R(\A,\B'')}{R(\B',\B'')}
=\d_{\B''\B'}\bigl(S_q(x-\B')S_{m^{n-q}}(\B''-\A)\bigr) \\
{}=\pm \d_{\B''\B'}\bigl(S_{1^q}(\B'-x)
S_{m^{n-q}}(\B''-\A)\bigr)
=(-1)^{(m-q)(n-q)+q} S_{1^q;(m-q)^{n-q}}(\B-x;\B-\A)\,.
\end{multline*}

Note that one can rewrite \eqr(2.14) as:
\begin{equation}
(-1)^{m(n-q)}\sum_{1\le j_1 < \cdots < j_q \le n}
 \ \prod_{i\ne j_1,\ldots,j_q} \ \prod_{k=1}^q {{P(b_{i})(x-b_{j_k})}
 \over {b_{j_k}-b_i}}\,,
\label{eq2.15}
\end{equation}
where $P(x)=S_m(x-\A)$. So the expression \eqr(2.15)
equals the subresultant
$\Cal R^{(q)}$ associated with $P(x)$ and $S_n(x-\B)$.
In particular, for $q=n-1$, one recovers Lagrange interpolation:
\begin{equation}
\Cal R_1^E=(-1)^{m-n+1}\Cal R_1=\sum_{i=1}^n P(b_i)\prod_{j \ne i}
{{x-b_j}\over{b_i-b_j}}\,.
\label{eq2.16}
\end{equation}

For example, for $n=3$,
$$
\Cal R_1^E=P(b_1)\frac{(x-b_2)(x-b_3)}{(b_1-b_2)(b_1-b_3)}+P(b_2)\frac{(x-b_1)(x-b_3)}
{(b_2-b_1)(b_2-b_3)}+P(b_3)\frac{(x-b_1)(x-b_2)}{(b_3-b_1)(b_3-b_2)}\,.
$$

In \citep{J2}, one can find another treatment of such interpolation
formulas of Lagrange, Sylvester, and Borchardt.
The expressions of interpolation for the subresultants given there
are different from the double Sylvester sums, and they are
based on the theory of the so-called ``Borchardtiens" developed
in loc.cit.

Jacobi, in his thesis in 1825, already considered the problem of 
generalizing the Lagrange interpolation. In particular, in the appendix 
to his thesis \citep{Jac} (Propositio generalis III, p.562), Jacobi 
expresses, for any symmetric polynomial $P$ in $p$ variables, 
the value of
$$
\sum_{\A'\subset \A} \frac{P(\A')}{R(\A',\A - \A')}\,,
$$
where $\card(\A')=p$, as a multi-residue.
\erm


\setcounter{footnote}0
\section{Another approach via symmetrizing operators}

We will now sketch another possible method for getting the identity
\eqr(2.11), using only
divided differences, and not specializations. It is based upon the following
two lemmas\footnote{We do not state these results in the greatest possible
generality, but in the form needed here.}:

\blm\label{lem3.1} Let $i\le n-1$  be a nonnegative integer and let
$\C$ be an alphabet such that $\card(\C) \le i$. Then for a nonnegative
integer $j$ and a virtual alphabet~$\D$,
$$
\d_{n-1}\cdots \d_1 \bigl(S_i(b_1-\C)S_j(b_1+\D)\bigr)=S_{i+j-n+1}(\B-\C+\D)\,,
$$
provided $\C$ and $\D$ are independent of $\B$.
\elm

\begin{proof}
We have
$$
S_i(b_1-\C)S_j(b_1+\D) = \sum_{k,l} b_1^{k+l}S_{i-k}(-\C)S_{j-l}(\D)\,.
$$
This element is sent via $\d_{n-1}\cdots \d_1$ to
$$
\sum_k S_{i-k}(-\C) S_{j+k-n+1}(\B+\D)
$$
because $k\le n-1$. Since $S_k(-\C)=0$ for $k\ge i$, this last sum is the
expression of $S_{i+j-n+1}(\B-\C+\D)$ and the assertion of the lemma follows.
\end{proof}

\dfbn{Notation} Given an alphabet $\B=\{b_1,b_2,\ldots \}$,
we write $\B^i:=\{b_i,b_{i+1},\ldots \}$\,.
\enddfbn

\blm\label{lem3.2} Let $i_0$ be a nonnegative integer and let
$\C^{(0)}$  be an alphabet such\break that $\card(\C^{(0)})\le i_0$.
Then for integers $i_1$, \dots, $i_{n-1}$,  and virtual alphabets\break
$\C^{(1)}$, \dots, $\C^{(n-1)}$,
\bml
\d_{n-1}\cdots \d_1 \bigl(S_{i_0}(b_1-\C^{(0)}) S_{i_1,\ldots,i_{n-1}}
(\B^2-\C^{(1)},\ldots,\B^2-\C^{(n-1)})\bigr) \\
{}=(-1)^{n-1} S_{i_0,i_1-1,\ldots,i_{n-1}-1}(\B-\C^{(0)},\B-\C^{(1)},
\ldots,\B-\C^{(n-1)})\,,
\end{multline*}
provided $\C^{(0)},\ldots,\C^{(n-1)}$ are independent of\/ $\B$.
\elm

\begin{proof}
Using the transformation Lemma \ref{lem1.7}(i), we have the factorization
\bml
S_{i_0}(b_1-\C^{(0)})S_{i_1,\ldots,i_{n-1}}(\B^2-\C^{(1)},\ldots,
\B^2-\C^{(n-1)})\\
{}=S_{i_1,\ldots,i_{n-1},i_0}(\B-\C^{(1)},\ldots,\B-\C^{(n-1)},b_1-\C^{(0)})\,.
\end{multline*}
The image of this element via $\d_{n-1}\cdots \d_1$ is
\bml
S_{i_1,\ldots,i_{n-1},i_0-n+1}(\B-\C^{(1)},\ldots,\B-\C^{(n-1)},
\B-\C^{(0)}) \\
{}=(-1)^{n-1} S_{i_0,i_1-1,\ldots,i_{n-1}-1}(\B-\C^{(0)},\B-\C^{(1)},
\ldots,\B-\C^{(n-1)})\,.
\end{multline*}
The lemma is proved.
\end{proof}

\dfbn{Notation}
We denote  \ $\d_{\B}:=\d_\omega$ where $\omega$ is the longest
permutation reordering $(b_1,b_2,\ldots ,b_n)$ to
$(b_n,b_{n-1},\dots,b_1)$.
For two complementary subsequences $\B'$ and $\B"$ of $\B$, we
write  \ $\d_{\B"\B'}:=\d_{\sigma}$ where the permutation $\sigma$
reorders the sequence $\B"\B'$ to $\B$. For example,
$\d_{n-1}\cdots \d_1$ is denoted by $\d_{\B^2b_1}$.
\enddfbn


In this computation, we fix $\A':=\A_p$, $\A'':=\A^{p+1}$, $\B':=\B_q$,
and $\B'':=\B^{q+1}$. (Here, $a_1,\ldots,a_m$ are variables.)

Recall (cf. e.g. \citep[$\S3$]{P3} and references there) 
that the divided
difference operator $\d_{\A''\A'}$ has the following interpretation as
a symmetrizing operator occurring in Sylvester sums.
For a polynomial $P(a_1,\ldots,a_m)$,
$$
\d_{\A''\A'}(P) = \frac1{p!(m-p)!}\sum_{\sigma} \Bigl(\frac P{R(\A_p,\A^{p+1})}\Bigr)^{\sigma}\,,
$$
the sum being over permutations $\sigma$ of $\{1,\ldots,m\}$.
Hence equation \eqr(2.11) is equivalent to
\begin{multline}
\d_{\A'' \A'}\d_{\B'' \B'}[R(x,\A'+\B')R(\A',\B')R(\A'',\B'')]\\
{}= (-1)^{(m-p-q)(n-q)+p+q} \binom{p+q}p 
S_{1^{p+q};(m-p-q)^{n-p-q}}(\B-x;\B-\A) \,.
\label{eq3.3}
\end{multline}

\dfbn{Notation} For an alphabet $\A=\{a_1,\ldots,a_m\}$ we write
$$\Delta(\A):=\prod_{i<j\le m}(a_i-a_j)\,,$$ and for an auxiliary alphabet $\B$,
we set $\Delta(\A+\B):=\Delta(\A) \Delta(\B) R(\A,\B)$.\footnote{Note that
$\Delta(\A+\B)=(-1)^{mn}\Delta(\B+\A)$, where $n=\card(\B)$.}
\enddfbn

Since
\bml
\d_{\A'}\d_{\B'} \bigl((1/p!)(1/q!) \Delta(x+\A')\Delta(x+\B')R(\A',\B')
R(\A'',\B'')\bigr)\\
{}=R(x,\A'+\B')R(\A',\B')R(\A'',\B'')\,,
\end{multline*}
equation \eqr(3.3) is equivalent to
\begin{multline}
\nabla \bigl(\Delta(x+\A'+\B')R(\A'',\B'')\bigr)\\
{}= \epsilon (p+q)!
S_{1^{p+q};(m-p-q)^{n-p-q}}(\B-x;\B-\A) \,,
\label{eq3.4}
\end{multline}
where
$$
\nabla:=\d_{\B^2b_1} \cdots \d_{\B^{q-p+1}b_{q-p}} \d_{\A^2a_1}
\d_{\B^{q-p+2}b_{q-p+1}} \cdots \d_{\A^pa_{p-1}} \d_{\B^qb_{q-1}}\d_{A''a_p}
\d_{B''b_q}
$$
(without loss of generality we assume that $p\le q$),
and $\epsilon=(-1)^{(m-p-q)(n-q)+p+q}$.
\smallskip

We will now show \eqr(3.4) for $m=\card(\A)=11$, $n=\card(\B)=9$,
$p=2$, and $q=3$.
We want to show
$$
\d_{\B^2b_1} \d_{\A^2a_1} \d_{\B^3b_2} \d_{\A^3a_2} \d_{\B^4b_3}
[\Delta(x+\A_2+\B_3)R(\A^3,\B^4)]
=-5! S_{1^5;6^4}(\B-x;\B-\A)\,.
$$
We shall disregard the signs, for simplicity.
We have
\bml
\!\!\Delta(\B_3+\A_2+x)R(\B^4,\A^3)
=\Delta(\B_2+\A_2+x)S_5(b_3-(\B_2+\A_2+x))S_{9^6}(\B^4-\A^3)\\
{}=\Delta(\B_2+\A_2+x)S_{9^6;5}(\B^3-\A^3;b_3-(\B_2+\A_2+x))
\end{multline*}
(the last equality follows from the transformation Lemma \ref{lem1.7}(i)).
By Lemma~\ref{lem3.2}, this element goes via $\d_{\B^4b_3}$ to
$$
\pm \Delta(\B_2+\A_2+x)S_{5;8^6}(\B^3-(\B_2+\A_2+x);\B^3-\A^3)\,.
$$
We explain the effect of applying $\d_{\A^3a_2}$ to this last element.
By the transformation Lemma \ref{lem1.7}(ii), we have
$$
S_{5;8^6}(\B^3-(\B_2+\A_2+x);\B^3-\A^3)
=S_{5;8;8^5}(\B^3-(\B_2+\A_2+x);\B^3-\A^3;\B^3-\A^2)\,,
$$
and by a well-known linearity formula this last element is
$$
S_{5;8;8^5}(\B^3-(\B_2+\A_1+x)-a_2;\B^3-\A^3;\B^3-\A^2)=S_{5;8;8^5} -a_2S_{4;8;8^5}\,,
$$
the Schur functions in $(\B^3-(\B_2+\A_1+x);\B^3-\A^3;\B^3-\A^2)$.
Factorizing the last $\Delta$ and using \eqr(1.5), to compute is the image
under $\d_{\A^3a_2}$ of
$$
\Delta(\B_2+\A_1+x)[S_4(a_2-(\B_2+\A_1+x))
S_{5;8;8^5}-S_5(a_2-(\B_2+\A_1+x))S_{4;8;8^5}]\,,
$$
the Schur functions in $(\B^3-(\B_2+\A_1+x);\B^3-\A^2+a_2;\B^3-\A^2)$ .
By using Lemma~\ref{lem3.1},
the first summand in the bracket $[ \quad ]$ is sent to
$$
S_{5,3;8^5}(\B^3-(\B_2+\A_1+x);\B^3-\A^2)
=-S_{4^2;8^5}(\B^3-(\B_2+\A_1+x);\B^3-\A^2)\,.
$$
The second summand in the bracket $[ \quad ]$ is sent to \
$-S_{4^2;8^5}(\B^3-(\B_2+\A_1+x);\B^3-\A^2)$.
In sum, the initial element goes via $\d_{\A^3a_2} \d_{\B^4b_3}$ to
$$
\pm 1 \cdot 2 \cdot \Delta(\B_2+\A_1+x) S_{4^2;8^5}
(\B^3-(\B_2+\A_1+x);\B^3-\A^2)\,.
$$
Similarly, we have
\bml
\d_{\B^3b_2}
[\Delta(\B_2+\A_1+x) S_{4^2;8^5}(\B^3-(\B_2+\A_1+x);\B^3-\A^2)]\\
\hfill{}=\pm 3\Delta(\B_1+\A_1+x) S_{3^3;7^5}
(\B^2-(\B^1+\A^1+x);\B^2 -\A^2)\,,
\quad\\
\quad \d_{\A^2a_1} [\Delta(\B_1+\A_1+x) S_{3^3;7^5}(\B^2-(\B^1+\A^1+x);
\B^2 -\A^2)]\hfill\\
=\pm 4\Delta(\B_1+x) S_{2^4;7^5}(\B^2-(\B^1+x);\B^2 -\A^1)\,,
\end{multline*}
and
$$
\d_{B^2b_1}[\Delta(\B_1+x) S_{2^4;7^5}(\B^2-(\B^1+x);\B^2 -\A^1)]
=\pm 5S_{1^4;5^4}(\B^1-x;\B^1 - \A^1)\,.
$$
In sum, we get
$$
\d_{\B_2b_1} \d_{\A_2a_1} \d_{\B_3b_2} \d_{\A_3a_2} \d_{\B_4b_3}
[\Delta(x+\A_2+\B_3) R(\A^3,\B^4)]=- 5! S_{1^5;6^4}(\B-x;\B-\A)\,,
$$
as wanted in \eqr(3.4).

\setcounter{footnote}0
\section{A glimpse of Gysin maps}

This section is just a by-product of the previous ones.
We translate here the above algebraic computations to a formula
for Gysin maps.

Let $X$ be a smooth variety, and
suppose that $E^{(1)},\ldots, E^{(l)}$ are $l$ virtual bundles on $X$.
Given a sequence $I=(i_1,\ldots,i_l) \in \Z^l$, we set
$$
S_I\bigl(E^{(1)},\ldots,E^{(l)}\bigr)
:=S_I\bigl(\A^{(1)},\ldots,\A^{(l)}\bigr)\,,
$$
where $\A^{(1)},\ldots, \A^{(l)}$ are the virtual alphabets of
Chern roots associated with $E^{(1)},\ldots, E^{(l)}$.

Let now $E$ and $F$ be vector bundles on $X$ of respective ranks
$m\ge n$.
For integers $0\le p \le m$ and $0\le q \le n$, let
$$
\pi_E: \Gr^p(E) \to X \quad \   \hbox{and} \quad \  \pi_F: \Gr^q(F) \to X
$$
be two Grassmann bundles on $X$ parametrizing $\rank p$ quotients of $E$
and $\rank q$ quotients of $F$. These Grassmannians are equipped with
the two tautological exact sequences:
$$
0 \to R_E \to E \to Q_E \to 0  \quad \hbox{and} \quad
0 \to R_F \to F \to Q_F \to 0 \,,
$$
where $\rank(Q_E)=p$ and $\rank(Q_F)=q$. (We use the same notation
for vector bundles and their pullbacks.)
We form a product over $X$:
$$
\pi := \pi_E \times_X \pi_F : \Gr^p(E)\times_X \Gr^q(F) \to X\,.
$$

Recall (cf. \citep{Br}, \citep[$\S3$]{P3}) that the operator
$\d_{\A''\A'}$
appearing in \eqr(3.3) induces the Gysin map $(\pi_E)_*$.
More precisely, for a polynomial $P(a_1,\ldots,a_m)$
symmetric in the first $p$ variables and in the last $m-p$ variables
separately, one has
\begin{equation}
(\pi_E)_ *(P(x_1,\ldots,x_m)=
\d_{\A''\A'}\bigl(P(a_1,\ldots,a_m)\bigr)|_{a_1=x_1,\ldots,a_m=x_m}\,,
\label{eq4.1}
\end{equation}
where $x_1,\ldots,x_p$ are the Chern roots of $Q_E$,
and $x_{p+1},\ldots,x_m$
are the ones of~$R_E$. By using \eqr(4.1),
equation \eqr(3.3)
is translated into the following formula about the top Chern classes
and the Gysin map \
$$
\pi_*: H( \Gr^p(E)\times_X \Gr^q(F)) \to H(X)\,.
$$

\bpr\label{prop4.2} Let $L$ be a line bundle on $X$. Then for $d=p+q<n$,
$$
\pi_*\Bigl(c_{\top}(Q_E^{\vee}\otimes L) \ c_{\top}(Q_F^{\vee}\otimes L)
\ c_{\top}(Q_F^{\vee}\otimes Q_E) \ c_{\top}(R_F^{\vee}\otimes R_E)\Bigr)
=\epsilon \binom{d}p S_{1^{d};(m-d)^{n-d}}\,,
$$
where $\epsilon=(-1)^{(m-d)(n-q)+d}$ and the Schur function is
in $(F-L;F-E)$.
\epr
Note that by comparing the coefficients of the maximal power
$c_1(L)^{d}$, we get
\begin{equation}
\pi_*\bigl(c_{\top}(Q_F^{\vee}\otimes Q_E) \ c_{\top}(R_F^{\vee}
\otimes R_E)\bigr)=\pm
\binom{d}p S_{(m-d)^{n-d}}(F-E)\,.
\label{eq4.3}
\end{equation}

We point out that neither the proposition nor equation \eqr(4.3)
follow {\it instantly} from a standard formula \citep{L1}, \citep{JLP}
for the Gysin map of a Grassmann bundle applied to $\Gr^p(E)$
and $\Gr^q(F)$.

\smallskip

In a similar way one can translate formulas given in Proposition 2.9.

\bigskip
\noindent{\bf Acknowledgements.}

Thanks are due to H. Hong, J.-P. Jouanolou, and A. Al Amrani
for sending us references concerning subresultants.

This article is a revised version of the Max-Planck-Institut
f\"ur Mathematik in Bonn Preprint
2001-30. The second author wishes to thank the MPIM for its hospitality
during the work on this paper, and to Professor F. Hirzebruch for his
encouragement.

The authors were partially
supported by grant No.~5031 of French-Polish cooperation C.N.R.S.-P.A.N.,
and while preparing the final version -- by Polish
KBN grant No.~2P03A 024 23.



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\end{document}
