%  OperatorCalc_Adv.tex                 Advances in Maths {\bf 140} (1998)  1--43.
\input amstex
\documentstyle {amsppt}
\magnification = \magstep 1

\font\footfont=cmr10 scaled \magstep 0

%\vsize 7.5in
%\hsize 5.7in


\def \hak  {^{\vee}}
\def \widQ {\widetilde Q}
\let \hoto =\hookrightarrow
\let \loto =\longrightarrow
\let \ol  =\overline
\let \setminus =\smallsetminus
\let \subne =\varsubsetneq
\let \tk  =\triangledown

\document
\topmatter
\title
  OPERATOR  CALCULUS  FOR  $\widetilde{\hbox{Q}}$-POLYNOMIALS  
  AND  SCHUBERT  POLYNOMIALS
\endtitle
\author 
Alain Lascoux \ and \ Piotr Pragacz
\endauthor

\endtopmatter

\centerline{( 31 May, 1997 )}
\vskip 60pt

\rightline{\it d\'edi\'e \`a Maria}

\vskip 40pt



\noindent
Contents:
{\parindent=25pt
\bigskip
\item{}  Introduction
\bigskip
\item{1.} Divided differences associated with the hyperoctahedral groups.
\bigskip
\item{2.} Reproducing kernels and a vanishing property.
\bigskip
\item{3.} Action of $\tk$ on the basis $\{\frak S_{\mu}\cdot\widQ_I\}$
- an inductive approach.
\bigskip
\item{4.} Action of $\tk$ on the basis $\{\frak S_{\mu}\cdot\widQ_I\}$
via the vanishing property.
\bigskip
\item{5.} Key formula and vertex operators.
\bigskip
Appendix A: (written in collaboration with J. Ratajski):
\medskip
\hskip 39pt Symplectic Schubert polynomials \`a la polonaise.
\bigskip
Appendix B: Three geometric applications.

\bigskip
References
\par}

\newpage

\head
{\bf Introduction.}
\endhead

\bigskip

The goal of the present paper is to develop convenient algebraic tools
to compute in the cohomology ring of Lagrangian flag manifolds
as well as with Lagrangian degeneracy loci. 
There is quite a long series of papers devoted to this goal overlapping
among others [B-G-G], [B-H], [B], [D1,2], [F-K], [F1,2], [H-B] and [P-R1,2].
The present paper will certainly not end this list. The cohomological
formulas related to a symplectic form are extremely useful for computing
in the cohomology (or Chow) rings of various moduli spaces including
the moduli spaces of curves and the moduli spaces of abelian varieties
in characteristic $p$. This is, perhaps, a raison d'\^etre for the
present paper.
\smallskip

Our basic tool here is a certain basis which is orthonormal
with respect to the scalar product defined by the top symplectic divided
difference. This basis has a natural geometric explanation. Consider
a vector bundle $V$ of rank $2n$, equipped with a nondegenerate symplectic
form. Then the Lagrangian flag bundle parametrizing all complete Lagrangian
flags in $V$ fibers over the Lagrangian Grassmannian bundle parametrizing
Lagrangian $n$-bundles with a fiber being the "usual" flag variety of
all complete flags in an $n$-dimensional vector space. The basis is a natural
"product-basis" associated with this fibration. It consists
of products of $\widQ$-polynomials from [P-R2] (which provide a basis
for the cohomology of the base of the fibration) times Schubert polynomials
(which provide a basis for the cohomology of the fiber).
We extensively investigate the action of different operators (notably
divided differences) on this basis. It appears that such tools as
a certain vanishing property (which is much in spirit of interpolation
theory) and vertex operators (which have their origin in the theory
of infinite-dimensional Lie algebras) are extremely useful to
achieve this goal. The main technical result of the present paper 
is the "key formula" (Theorem 5.1). 
Many formulas reduce to the computation of the image of products
of the type $x_1^k\widQ_J$ under divided differences.
The main algebraical upshot of this work is a study of the structure
of a polynomial ring as a free module over the ring of polynomials
symmetrical in the squares of the variables.

\smallskip

Even if the elements of the basis in question do not represent the
Schubert classes  in the Lagrangian flag manifold, they are very useful
in the study of such classes. Recall that the polynomial representatives for
those classes were defined already in [B-G-G] and [D1,2]. Recently,
some algebro-combinatorial constructions of such representatives
were studied in [B-H] and [F-K] and some other appeared implicitly in [F1,2].
Following [P-R2], we study in this paper still other polynomial
representatives of Lagrangian Schubert classes in Appendix A. 
They seem to be well suited
for the purposes of algebraic geometry. They are defined with the
help of divided differences starting from some very simple "top"
polynomial, and for maximal Grassmannian elements they are equal
to certain $\widQ$-polynomials - exactly those predicted for them
by the Giambelli-type formula from [P, Sect.6]. Also, as shown in
the present paper, these polynomials behave well w.r.t. different
operators which geometrically often correspond to Gysin maps.
Symplectic Schubert polynomials \`a la polonaise are not positive
in general (though involve much smaller quantities 
of negative monomials than the original polynomials of [B-G-G] and [D1,2]).
In general, these polynomials involve much fewer monomials
than the other known families of Lagrangian Schubert classes 
representatives; hence they are also better suited for numerical computations.
\smallskip

This paper contains also some simple proofs of previously known results:
a new operator proof of the Giambelli-type formula for maximal Lagrangian
Schubert classes from [P, Sect.6], and a purely algebraic
proof of the congruence modulo the corresponding ideal of two (competing)
expressions for diagonals from [F1,2] and [P-R2].

\smallskip

The article is organized as follows.
 
\smallskip

In Section 1 we collect needed information on the hyperoctahedral groups
and divided differences associated with them. Moreover we recall
two families of polynomials crucial for this paper: Schubert polynomials
and $\widQ$-polynomials. Several scalar products on polynomial
rings are also discussed together with their orthogonality properties.

\smallskip

In Section 2 we study two reproducing kernels for a certain scalar product
defined on the ring of symmetric polynomials with values in the ring
of symmetric polynomials in the squares of the variables. These kernels
stem from [F1,2] and [P-R2] and are motivated by geometry. More precisely
they are polynomial representatives of the classes of relative diagonals
in Lagrangian Grassmannian bundles. For a recent geometric development
of this idea, we refer the reader to [G]. We prove that these kernels
satisfy a certain vanishing property that characterizes them almost completely
(see [L-S] for the type $A_n$.)
 

\smallskip

In Section 3 we derive in an elementary way (invoking 
only simple properties of divided differences and the linearity formula
for $\widQ$-polynomials) how a certain key operator $\tk$ acts 
on our product-basis.

\smallskip

In Section 4 we achieve the samy goal using the vanishing property.
The proof is much shorter, but it relies on
the operator characterization of $\widQ$-polynomials.

\smallskip

Section 5 is the heart of the paper. The key formula is proven there
with the help of vertex operators associated with Schur's $P$-functions
and $Q'$-functions from [L-L-T2]. 
The starting point of our approach is a presentation of the "exceptional"
divided difference $\partial_0$ as an (infinite) differential
operator. This observation is, in fact, the main motivation of using
the vertex operators in various calculations in this section.

\smallskip

Appendix A contains the proofs of two crucial properties of Symplectic
Schubert polynomials \`a la polonaise i.e. stability and
the maximal Grassmannian property. 
Using the key formula, we establish in a purely algebraic way 
the above mentioned operator characterization of $\widQ$-polynomials.
The table of the polynomials
for $n=3$ is also included.

\smallskip

In Appendix B we apply the above calculations to give a new operator
proof of the Lagrangian Giambelli-type formula, to derive a new Gysin
formula as well as to provide a new formula for Lagrangian flagged
degeneracy loci.

\medskip

In the future, we plan to present the orthogonal variants 
of this theory, discuss the non-maximal Grassmannian case and
develop the corresponding Grothendieck polynomials. Hopefully,
the quantum version will be also included.

\bigskip
\noindent
{\bf Acknowledgements.} 
The authors started to work on this project during their stay in the
Max-Planck-Institut f\"ur Mathematik in Bonn in November 1995.
They gratefully acknowledge the hospitality and stimulating atmosphere
of the MPIfM.
\smallskip

The second named author thanks
the KBN grant No. 2PO3A 02711 for supporting this research. 
 
\bigskip\bigskip\bigskip

\head
{\bf 1. Divided differences associated with the hyperoctahedral groups.}
\endhead

\bigskip

Recall first the definition of the hyperoctahedral group 
i.e. the Weyl group of the symplectic group. Let $n$ be positive
integer. Let $S_n$ be the symmetric group (of all bijections of 
$\{1,\ldots,n\}$ ) generated by the simple transpositions 
$s_i=[1,\ldots,i-1,i+1,i,i+2,\ldots,n]$, $i=1,\ldots,n-1$. The 
hyperoctahedral group $W=W_n$ is an extension of $S_n$ by an element $s_0$
such that 
$$
s_0^2=1, \quad s_0\cdot s_1\cdot s_0\cdot s_1=s_1\cdot s_0\cdot s_1\cdot s_0,
\quad s_0\cdot s_i=s_i\cdot s_0 \quad \hbox{for} \quad i\ge 2.
\tag1
$$
More precisely, $W=S_n\ltimes \Bbb Z_2^n$ where $S_n$ acts on $\Bbb Z_2^n$
by permutations. Writing a typical element of $W$ as $w=(\sigma,\tau)$
with $\sigma\in S_n$ and $\tau \in \Bbb Z_2^n$, we have for $w'=(\sigma',
\tau')$,
$$
w\cdot w' = (\sigma \sigma',\delta),
$$
where $\delta_i=\tau_{\sigma'(i)}\cdot \tau'_i$. To represent elements of $W$
we use the standard "barred-permutation" notation, writing them as
permutations with bars in those places (numbered with "$i$")
where $\tau_i=-1$. We have $s_0=[\ol 1,2,\ldots,n]$, and the right 
multiplication by $s_i$ is given by
$$
[w_1,w_2,\ldots,w_n]\cdot s_0 = [\ol w_1,w_2,\ldots,w_n] 
$$
$$
\hbox{and} \quad [w_1,w_2,\ldots,w_n]\cdot s_i=
[w_1,\ldots,w_{i-1},w_{i+1},w_i,w_{i+2},\ldots,w_n].
$$
Let $\omega=\omega^{(n)}, w_0=w_0^{(n)}$ denote respectively the elements
of maximal length of $S_n$ and $W$. In addition, we set
$$
v_0=v_0^{(n)}=w_0\cdot \omega = \omega\cdot w_0.
$$
In the above barred permutation notation,
$$
\omega=[n,n-1,\ldots,1], \quad w_0=[\ol 1,\ldots,\ol n] \quad 
v_0=[\ol n,\ldots,\ol 1].
$$
Let $X=X_n=(x_1,\ldots,x_n)$ be a sequence of commuting independent variables.
The group algebra of $W$ acts on the ring $\Bbb Z[X]$ of polynomials in $X$:
$s_i$ exchanges $x_i$ and $x_{i+1}$, $1\le i\le n-1$, and $s_0$ replaces
$x_1$ by $-x_1$ (all remaining variables being invariant).
\bigskip

In the present paper we shall make an extensive use of the algebra of
divided differences, with generators $\partial_0,\partial_1,\ldots,
\partial_{n-1}$ acting on the ring $\Bbb Z[X]$ according to
$$ 
\partial_i(f) = (f-s_if)/(x_i-x_{i+1}), \quad i\ge 1
$$
$$
\hbox{and}\quad \partial_0 = (f-s_0f)/2x_1.
$$
These generators satisfy the braid relations (1). Consequently, for every
$w\in W$, there exists a well-defined divided difference $\partial_w=
\partial_{i_1}\circ \ldots \circ \partial_{i_l}$ which can be obtained from
any reduced decomposition $s_{i_1}\cdot\ldots\cdot s_{i_l}$ of $w$
(see [B-G-G] and [D1,2]).
In particular,
$$
\partial_{v_0}=\partial_0(\partial_1 \partial_0) \ldots (\partial_{n-1}\ldots \partial_1 \partial_0)
=(\partial_0\partial_1 \ldots \partial_{n-1})\ldots 
(\partial_0\partial_1)\partial_0
\tag 2
$$
and
$$
\partial_{w_0}=\partial_{v_0}\circ 
\partial_{\omega}=\partial_{\omega}\circ\partial_{v_0}.
\tag3
$$
Recall (see  e.g. [Mcd2]) that the operator $\partial_{\omega}$ 
coincides with the Jacobi symmetrizer
$$
f\mapsto \sum_{\mu \in S_n} \mu [f/\vartriangle(X)],
$$
where $\vartriangle(X)=\prod_{i<j\le n} (x_i-x_j)$.
\medskip
On the other hand, the operator $\partial_{w_0}$ (that appears in the 
Weyl character formula for type $C_n$) coincides with
$$
f\mapsto \sum_{w\in W} w[f/\square(X)],
$$
where 
$$
\square(X)=\vartriangle(X) \prod_{i\le j\le n}(x_i+x_j)=2^n x_1\cdot \ldots\cdot
x_n \prod_{i<j\le n}(x_i^2-x_j^2).
$$
Observe that these expressions of
$\partial_{\omega}$ and $\partial_{w_0}$ imply that for every symmetric
(in $X$) polynomial $f$, the operator $\partial_{v_0}$ acts as
$$
f\mapsto \sum_{w\in W/S_n} w[f/\prod_{i\le j\le n}(x_i+x_j)].
\tag4
$$
\noindent
(To see this, recall that the Jacobi symmetrizer is a surjection onto
the ring of symmetric polynomials in $X$).
\bigskip

Let now $T_i$, $i=1,\ldots,n,$ be the operator on $\Bbb Z[X]$ acting only
on $x_i$ by
$T_i(x_i)=-x_i$. Note that $T_i$ is conjugate to $s_0$:
$$
T_i=s_{i-1}\circ \ldots \circ s_1\circ s_0 \circ s_1 \circ 
\ldots \circ s_{i-1}. 
$$
With the help of the $T_i$'s, the operator $\partial_{v_0}$ has a third 
expression which respects the symmetry between the variables.
\proclaim{Lemma 1.1} One has the following equalities of operators
on $\Bbb Z[X]$:
$$\aligned
\sum_{w\in W} (-1)^{l(w)}w &= \sum_{\mu \in S_n} (-1)^{l(\mu)} \mu 
(1-T_n)\ldots
(1-T_1) \\
&=(1-T_n)\ldots (1-T_1) \sum_{\mu \in S_n} (-1)^{l(\mu)}\mu. \\
\endaligned
$$
\endproclaim
Indeed, these equalities follow immediately from $W=S_n\ltimes \Bbb Z_2^n$
invoking the following formula for the length $l(w)$ of $w=(\sigma,\tau)
\in W$:
$$
l(w)=\sum_{i=1}^n a_i + \sum_{j, \tau_j=-1}(2b_j+1),
\tag5
$$
where $a_i=\hbox{card} \{p \ | \ p>i \ \& \ \sigma_p < \sigma_i\}$ and 
$b_j=\hbox{card} \{ p \ | \ p<j \ \& \ \sigma_p<\sigma_j \}$. The equality
(5) can be proved by an easy induction on $l(w)$.
\bigskip
To simplify signs in many formulas of the present paper, we shall also
use $\partial_i'=-\partial_i$ ($1\le i \le n-1$), $\partial_{\mu}'$
($\mu\in S_n$) \ and instead of $\partial_{v_0}$,
$$
\tk:=\partial_0(\partial_1'\partial_0) \ldots (\partial_{n-1}'\ldots
\partial_1'\partial_0) 
=(\partial_0\partial_1' \ldots \partial_{n-1}')\ldots 
(\partial_0\partial_1')\partial_0=(-1)^{n(n-1)/2}\partial_{v_0}.
\tag 6
$$
With the help of the $T_i$'s the restriction of $\partial_{v_0}$ (or $\tk$)
to symmetric polynomials has a simple expression:
\proclaim{Lemma 1.2} For any symmetric polynomial in $X$, 
$$\aligned
\partial_{v_0}(f)&= {1\over {\square(X)}} 
(1-T_n)\ldots (1-T_1)(f\cdot \triangle (X)) \\
&={1\over {2x_n}} (1-T_n)\ldots {1\over {2x_1}}(1-T_1)
\bigl(f/\prod_{i<j} (x_i+x_j)\bigr). \\
\endaligned
$$
\endproclaim

Indeed, let us first record
$$
\bigl(\sum_{\mu \in S_n}(-1)^{l(\mu)}\mu\bigr) \triangle(X)=n!\triangle(X). 
\tag 7
$$

Since $f$ is symmetric, one has by (7)
$$
f={1\over n!} \partial_{\omega}(\triangle(X) \cdot f)
$$
and thus
$$\aligned
\partial_{v_0}(f)&={1\over n!} \partial_{w_0}(\triangle(X) \cdot f) \\
&={1\over {\square(X)}} 
(1-T_n)\ldots (1-T_1)(f\cdot \triangle (X)) \\
\endaligned
$$
by Lemma 1.1 and the equality (7) again.
\smallskip
 
The second equality is a restatement of the first one because $x_i^2$ are
scalars w.r.t. $T_j$.

\bigskip

Let $\Cal S \Cal P(X)$ denote the ring of symmetric polynomials in $X$.
It follows from Lemma 1.2 (e.g. by remarking that $(1-T_i)^2=0$)
that the operator $\tk$ sends $\Cal S \Cal P(X)$ to $\Cal S \Cal P(X^2)$
where $X^2:= (x_1^2,\ldots, x_n^2)$. More explicitly, the image of the
$\Bbb Z$-linear basis of Schur polynomials is given by the next
proposition (see [P-R, Theorem 5.13]).

We shall use the following notation. Given a sequence
$\alpha=(\alpha_1,\ldots ,\alpha_n) \in \Bbb N^n$, 
we will denote 
by $x^{\alpha}$
the monomial $x_1^{\alpha_1}\cdot \ldots \cdot x_n^{\alpha_n}$.
Given two sequences $\alpha,\beta\in \Bbb N^n$,  we shall
denote 
by $\alpha +\beta$
the sequence $(\alpha_1+\beta_1,\ldots,\alpha_n+\beta_n)$.
Moreover, set $\rho_k=(k,k-1,\ldots, 2,1)$.

\proclaim{Proposition 1.3}
The image of an $S$-polynomial under $\tk$, $\tk\bigl(s_I(X)\bigr)$\,,
is non-zero only if the partition $I$ is of the form $\rho_n+2J$ for some
partition $J$.
In this case,
$$
\tk\bigl(s_{\rho_n+2J}(X)\bigr)=\psi_2\bigl(s_J(X)\bigr)\,,
$$
where $\psi_2:\Bbb Z[X] \to \Bbb Z[X^2]$ is the ring 
homomorphism defined by 
$\psi_2 (x_i)=x_i^2$, $i=1,\ldots,n$ \,.
\endproclaim
\demo{Proof} Since
$$
s_I(X)=\partial_{\omega}(x^{I+\rho_{n-1}}),
$$
we have
$$
\tk(s_I(X))=\partial_{w_0}(x^{I+\rho_{n-1}}).
$$
Using the first equality in Lemma 1.1, we see that $\tk (s_I(X))\neq 0$
only if all the components of $I+\rho_{n-1}$ are even. In that case
$I=J+J+\rho_{n-1}=2J+\rho_{n-1}$ for some partition $J$. Since we have
$$
(1-T_n)\ldots (1-T_1) (x^{2J+\rho_n +\rho_{n-1}})=2^n x^{2J+\rho_n+\rho_{n-1}},
$$
we infer finally that
$$\aligned
\tk(s_{2J+\rho_n}(X))&=\partial_{w_0}(x^{2J+\rho_n+\rho_{n-1}}) \\
&=2^n \sum_{\mu \in S_n} \mu [x^{2J+\rho_n+\rho_{n-1}}/\square (X)] \\
&=\sum_{\mu \in S_n} [x^{2J+2\rho_{n-1}}/\vartriangle(X^2)] \\
&=\psi_2(s_J(X)) \\
\endaligned
$$
as asserted.
\qed
\enddemo
\medskip

It is, perhaps, in order to mention that D.E. Littlewood used
in [Li2] the notions of $2$-quotient, $2$-core and $2$-sign
(whose definitions can be found e.g. in loc.cit., [Mcd1] and [D-L-T]) 
to evaluate $\psi_2\bigl(s_J(X)\bigr)$. As a matter of fact,
he computed more generally $\psi_p\bigl(s_J(X)\bigr)$, where,
for any integer $p\ge 2$, $\psi_p(x_i)=x_i^p$ , $i=1,2,\ldots$ 
For another computation of $\psi_p\bigl(s_J(X)\bigr)$ as a
quotient of two $S$-polynomials see [P-R2, Proposition 5.11].

\bigskip

The expression of $\tk$ as a sum, with rational-function coefficients,
of elements of $W$ is not totally straightforward, but we shall only need the 
leading term described by the following lemma.

\proclaim{Lemma 1.4}
Write $\tk=\sum\limits_{w\in W} a_ww$\,, where $a_w$ are rational functions
in $X$.
Then there is a unique $w=v_0$ of maximal length such that $a_w\ne 0$. 
Moreover, $a_{v_0}=(-1)^{n(n+1)/2}\prod_{i\le j\le n}(x_i+x_j)^{-1}$. 
\endproclaim

Indeed, we have, using (2), the following presentation of $\tk$:
$$\aligned
\tk = {1\over {2x_1}}(1-s_0){1\over {x_2-x_1}}&(1-s_1)\ldots   
{1\over {x_n-x_{n-1}}}(1-s_{n-1}) \ldots \\ 
&\ldots {1\over {2x_1}}(1-s_0)
{1\over {x_1-x_2}}(1-s_1){1\over {2x_1}}(1-s_0) \\
\endaligned
$$
which implies, by an easy induction on $n$, that its term of maximal length
equals
$$\aligned
(-1)^{n(n+1)/2}
 {1\over {2x_1}}s_0{1\over {x_2-x_1}}&s_1\ldots   
{1\over {x_n-x_{n-1}}}s_{n-1} \ldots \\ 
&\ldots {1\over {2x_1}}s_0
{1\over {x_1-x_2}}s_1{1\over {2x_1}}s_0 \\
&=(-1)^{n(n+1)/2}{1\over 2^n x_1\cdot \ldots 
\cdot x_n \prod_{i<j\le n}(x_i+x_j)} \ v_0, \\
\endaligned
$$
as asserted.\qed

\proclaim{Corollary 1.5} With the notation of Lemma 1.4, 
the unique $w=(\sigma,\tau)$ for which $\hbox{card}\{j \ | \ \tau_j=-1\}=n$ 
and $a_w\neq 0$ is $w=v_0$.
\endproclaim
Indeed, this follows directly from the previous lemma once we observe that,
by (5), $v_0$ is the element of the smallest length among the
barred permutations with $n$ bars.
\medskip

The following "commutation property" will be frequently used in this article.
\proclaim{Lemma 1.6} 
For any $\mu\in S_n$\,,
$$
\tk\circ\partial_{\mu}=\partial_{\omega\mu\omega}\circ\tk\,.
$$
\endproclaim

Indeed, recalling that the symmetric group $S_n$ acts on the set of barred 
permutations from the \underbar{right} by permuting the places, and,
from the \underbar{left} by permuting the values, we get the equality
$$
[\ol n,\ol{n-1},\ldots,\ol 2,\ol 1]\cdot s_i=
s_{n-i}\cdot [\ol n,\ol{n-1},\ldots,\ol 2,\ol 1]\, ,
$$
whence
$$\tk \circ \partial_i = \partial_{n-i} \circ \tk
$$
which immediately implies the assertion.

\bigskip

All the operators introduced above will be used to describe the ring 
$\Bbb Z[X]$ as an $\Cal S\Cal P(X^2)$-module.
First we consider $\Bbb Z[X]$ as an $\Cal S \Cal P(X)$-module.

The ring $\Bbb Z[X]$ is a free $\Cal S\Cal P(X)$-module with 
a basis provided by Schubert 
polynomials $\frak S_{\mu},\ \mu \in S_n$, defined in the following way.
For a permutation $\mu\in S_n$\, we set 
$$
\frak S_{\mu}:=\frak S_{\mu}(X):=
\partial_{\mu^{-1}\omega}(x_1^{n-1}x_2^{n-2}\ldots x_{n-1})
$$
the {\it Schubert polynomial}
associated with $\mu$\,.
Observe that, equivalently, the polynomials $\frak S_{\mu}$ are defined
inductively by the equation
$$
\partial_i\frak S_{\mu}=\frak S_{\mu s_i}
$$
if $\mu(i)>\mu(i+1)$\,, the "top" polynomial being $\frak S_{\omega}=
x^{\rho_{n-1}}$\,.
Recall that the Schubert polynomials have the following stability property.
Let $S_n \hoto S_n$ be the embedding of the symmetric groups via the
first $n$ components ($n<m$). Then we have for $\mu \in S_n \subset S_m$,
$$
\frak S_{\mu}(X_m)|_{x_{n+1}=\ldots=x_m=0}=\frak S_{\mu}(X_n)\,.
$$
It follows from this equality that $\frak S_{\mu}$ is a well-defined
polynomial for each permutation $\mu \in S_{\infty} = \bigcup_{n} S_n$.
\bigskip

On $\Bbb Z[X]$ there is a scalar product
$$
(\ ,\ ):\Bbb Z[X]\times\Bbb Z[X]\to \Cal S\Cal P(X)
$$
defined for $f,g\in\Bbb Z[X]$ by
$$
(f,g)=\partial_{\omega}(f\cdot g)\,.
$$
Schubert polynomials are almost orthonormal basis w.r.t. ( , )\,.
Given $w\in W$, we set $X^w=(w(x_1),\dots, w(x_n))$
and $\ol X=X^{w_0}=(-x_1,\ldots,-x_n)$. 
In particular, $X^{\omega}=(x_n,\ldots,x_2,x_1)$ and $\ol X^{\omega}=
(-x_n,\ldots,-x_1)$.

\medskip

One has for any $\mu,\nu\in S_n$\,,
$$
\bigl(\frak S_{\mu}(X),\frak S_{\nu\omega}(\ol X^{\omega})\bigr)
=\delta_{\mu,\nu}
$$
(the Kronecker delta).
For more on Schubert polynomials see [Mcd2] and the references therein.
\bigskip

Now we will describe the ring $\Cal S\Cal P(X)$ 
as an $\Cal S\Cal P(X^2)$-module.
As is well-known, $\Cal S\Cal P(X)\otimes \Bbb Q$ is a free 
$\Cal S\Cal P(X^2)\otimes \Bbb Q$-module with a basis provided
by Schur's $Q$-polynomials $\{Q_I(X)\}$, 
where strict $I\subset\rho_n$.
In this article, we want a basis over $\Bbb Z$, and moreover geometry
imposes the need to take a basis better adapted to the scalar product 
$\langle\ ,\ \rangle$ defined below.
Therefore, we shall use the following basis of $\widetilde Q$-polynomials 
modelled on Schur's $Q$-polynomials.
The reader can find in [P-R2], where these polynomials were extensively
studied, another geometric motivation of their choice.
We set $\widQ_i:=\widQ_i(X):=e_i(X)$ - the $i$-th elementary
symmetric polynomial in $X$\,.
Given two nonnegative integers $i,j$, we define
$$
\widQ_{i,j}:=\widQ_{i,j}(X)=\widQ_i\cdot\widQ_j
+2\sum\limits^j_{p=1}(-1)^p\widQ_{i+p}\cdot\widQ_{j-p}\,.
$$
Finally, for any partition $I=(i_1\ge i_2\ge\ldots\ge i_{l(I)}>0)$\,,
the polynomial $\widQ_I=\widQ_I(X)$ is defined recurrently
on $l=l(I)$ by putting for odd $l$\,,
$$
\widQ_I:=\widQ_I(X)=\sum\limits^l_{j=1}(-1)^{j-1}
\widQ_{i_j}\cdot\widQ_{(i_1,\ldots,i_{j-1},i_{j+1},\ldots,i_l)}
$$
and for even $l$\,,
$$
\widQ_I:=\widQ_I(X)=\sum\limits^l_{j=2}(-1)^j
\widQ_{i_1,i_j}\cdot\widQ_{(i_2,\ldots,i_{j-1},i_{j+1},\ldots,
i_l)}
$$
The ring $\Cal S\Cal P(X)$ is a free $\Cal S\Cal P(X^2)$-module with a basis 
provided by $\widQ$-polynomials $\widQ_I(X)$\,, where strict 
$I\subset\rho_n$\,.
The polynomial $\widQ_I(X_n)$ can be defined in another recursive way
as follows.
For any strict partition $I$ one has
$$
\widetilde Q_I(X_n) = \sum_{j=0}^{l(I)} x_n^j \bigr( \sum_{|I|-|J|=j}
\widetilde Q_J(X_{n-1})\bigl),
\tag8
$$
where the sum is over all (i.e. not necessary strict) partitions $J\subset I$
such that $I/J$ has at most one box in every row
(see [P-R2, Proposition 4.1]).
Moreover, given a strict
partition $I'=(i_1,\ldots,j,j,\ldots,i_k)$ and denoting $I=(i_1,\ldots,i_k)$,
one has the factorization property
$$
\widetilde Q_{I'}=\widQ_{j,j}\cdot \widQ_I.
\tag9
$$
\bigskip

The following scalar product
$\langle\ ,\ \rangle:\Cal S\Cal P(X)\times \Cal S\Cal P(X)\to \Cal S
\Cal P(X^2)$ \ defined by $\langle f,g\rangle=\tk(f\cdot g)$,
will be of particular importance for the present paper.
We have, for strict $I,J\subset\rho_n$\,,
$$
\langle\widQ_I(X),\widQ_{\rho_n\setminus J}(X)\rangle=\delta_{I,J}\,,
\tag10
$$
where $\rho_n\setminus J$ denotes the strict partition whose parts complement
the parts of $J$ in $\{n,n-1,\ldots,2,1\}$ (see [P-R, Theorem 5.23]).
\medskip

Denote by 
$$\langle(\ ,\ )\rangle:\Bbb Z[X]\times\Bbb Z[X]
\to \Cal S\Cal P(X^2)
$$ 
the scalar product defined for $f,g\in\Bbb Z[X]$ by
$$
\langle(f,g)\rangle=(-1)^{n(n-1)/2} \partial_{w_0}(f\cdot g)\,.
$$

The $\Cal S\Cal P(X^2)$-basis $\{\frak S_{\mu}\cdot
\widQ_I\}$ of $\Bbb Z[X]$, where $\mu\in S_n$ and $I\subset\rho_n$
are strict,
is especially convenient with regard to the scalar product
$\langle(\ ,\ )\rangle$ because obviously
$$
\langle(\frak S_{\mu}(X)\cdot\widQ_I \ ,\ \frak S_{\nu\omega}(\ol X^{\omega})
\cdot \widQ_{\rho_n\setminus J})\rangle=\delta_{\mu,\nu}\delta_{I,J}\,.
$$

Note, however, that the elements of this basis 
are not representatives of Schubert
cycles in the manifold of complete Lagrangian flags, but form 
a natural "product-basis"
associated with a fibration of this variety over the Lagrangian Grassmannian 
of maximal isotropic subspaces, with the  "usual" complete flag variety
as its fiber (for more about that, see [P-R2, Sect.5]).
\medskip

The main theme of the present paper is to study the action of different
operators on this basis.
Among the operators in question the following divided-difference operators
will be of particular importance:
$$
\tk_k=(\partial_{n-k}'\ldots\partial_1'\partial_0)\ldots
(\partial_{n-2}'\ldots\partial_1'\partial_0)
(\partial_{n-1}'\ldots\partial_1'\partial_0)\,.
\tag11
$$
(thus $\tk=\tk_n$\,).
\bigskip


As a consequence of the main results of the present paper, we will obtain
a purely algebraic proof of the next proposition not invoking the
characteristic map and the result from [P] quoted below.
For details see Corollary A.8 in Appendix A.
Since, however, we will need this result in Section 4, we record it now
with a proof which uses proprieties available in already published literature.

\proclaim{Proposition 1.7}
For each strict partition $I\subset\rho_n$ there exists a divided-difference
operator $\partial^I:\Bbb Z[X]\to\Bbb Z[X]$ such that $\partial^I
(\widQ_I)=1$ and such that, for any strict $J\ne I$ with $|J|=|I|$,
$\partial^I(\widQ_J)=0$\,.
\endproclaim
\noindent
(Note that from the first assertion it follows that $\partial^I=\partial_w
\circ\partial_0$ for some $w$\,.)
\demo{Proof}
This proof uses the Borel characteristic map [B] and the Giambelli-type 
formula for Lagrangian Grassmannians from [P].
Recall (see [P]) that Schubert classes $\sigma(I)$ in the
Lagrangian Grassmannian $G$ of isotropic $n$-subspaces in 
$\Bbb C\,\strut^{2n}$ equipped with a nondegenerate symplectic form,
are indexed by strict partitions $I\subset\rho_n$\,.
There exists a ring homomorphism (the characteristic map)
$$
c:\Cal S\Cal P(X) \loto A(G)
$$
such that for a homogeneous polynomial $f$, 
$$
c(f)=\sum\limits_{I, |I|=deg f}\partial_{w_I}'(f)\sigma(I),
$$
where $w_I\in W$ is the barred permutation 
$$
w_I=[\ol{i_1}>\ldots>\ol{i_l},j_1<\ldots<j_{n-l}]
\tag12
$$
associated with $I=(i_1,\ldots,i_l)$ \ (for details see [B], [B-G-G]
and [D1,2]). Now [P, Theorem 6.17] says that
$$
c(\widQ_I)=\sigma(I)\ .
$$
This implies that $\partial^I:=\partial_{w_I}$ fulfills the assertion of the
proposition.
\qed
\enddemo



\bigskip\bigskip

\head
{\bf 2. Reproducing kernels and a vanishing property.}
\endhead
\bigskip

Let $X=(x_1,\ldots,x_n),\ Y=(y_1,\ldots,y_n)$ be two sequences of
independent variables (we fix $n$).

In the following, by $\equiv$ we will understand the congruence in the ring
$\Cal S\Cal P(X)\otimes \Cal S\Cal P(Y)$ modulo the ideal generated 
by all $f(X^2)-f(Y^2)$ where
$f$ symmetrical and $X^2$ stands for $(x_1^2,\ldots,x_n^2)$.
We define the following polynomials in 
$\Cal S\Cal P(X)\otimes \Cal S\Cal P(Y)$
(the definition of $F(X,Y)$ is inspired by [F1,2] and the one
of $\widQ(X,Y)$ - by [P-R2]).
$$\aligned
F(X,Y) & =\ |e_{n+1+j-2i}X+e_{n+1+j-2i}Y|_{1\le i,j\le n} \\
& \\
&=\ {\left|\ \CD
e_nX+e_nY & 0 &.\;.\;. & \\
e_{n-2}X+e_{n-2}Y\  &\ e_{n-1}X+e_{n-1}Y &\ .\;.\;.\ & \\
\vbox{\offinterlineskip\hbox{.}\vskip3pt\hbox{.}\vskip3pt\hbox{.}}
&\vbox{\offinterlineskip\hbox{.}\vskip3pt\hbox{.}\vskip3pt\hbox{.}}
&\vbox{\offinterlineskip\hbox{.}\vskip3pt\hbox{\ \ .}\vskip3pt
\hbox{\ \ \ \ .}} & \\
&& 2 & e_1X+e_1Y \\
\endCD\ \right|}\,, \\
\endaligned
$$
and
$$
\widQ(X,Y) 
=\sum\widQ_IX\cdot\widQ_{\rho_n
\setminus I}Y, 
\tag13
$$
where the summation is over all strict partitions $I\subset\rho_n$.
\bigskip

The following results will give us some reproducing kernels for the scalar
product
$$
\langle\ ,\ \rangle:\Bbb Z[X]\times\Bbb Z[X]\loto \Cal S\Cal P(X^2)
$$
defined in the previous section.

\proclaim{Proposition 2.1}
For any $f\in \Cal S\Cal P(X)$\,,
$$
\langle f(X),\widQ(X,Y)\rangle\equiv f(Y)\,.
$$
\endproclaim

\demo{Proof}
Since we are working modulo $\equiv$ which allow us to exchange 
$\Cal S\Cal P(X^2)$-coeffi\-cients into the corresponding $\Cal S\Cal P(Y^2)$-coefficients,
it suffices to check the equality ($I$ a strict partition):
$$
\langle\widQ_I(X),\widQ(X,Y)\rangle=\widQ_I(Y)\,,
$$
which follows immediately from the orthogonality of the $\widQ_I$'s
w.r.t. $\langle\ ,\,\rangle$ recalled in (10).
\qed
\enddemo
To prove a similar property for $F(X,Y)$ we need the following 
result.

\proclaim{Proposition 2.2}
(i) \ For $w\in S_n\subset W_n$\,,
$$
F(X,X^w)=F(X,X)=2^ns_{\rho_n}(X)\,.
$$
(ii) \ For $w\in W_n\setminus S_n$\,,
$$
F(X,X^w)=0\,.
$$
\endproclaim

\demo{Proof}
(i) This is obvious from a familiar expression for $s_{\rho_n}$
as a determinant in elementary symmetric functions.
(note that we use that the degree-zero-entries in the matrix defining $F$
are equal to 2).

\smallskip\noindent
(ii) Let $w\in W_n\setminus S_n$\,.
If $w$ has odd number of bars, then the (1,1)-entry $e_nX+e_nY$ becomes
zero for $Y=X^w$ and thus $F(X,X^w)=0$\,.
Hence we can assume that the number of bars is even, and (because of symmetry)
bars are placed in the last $2k$ places of $w$\,.
Let $X'=(x_1,\ldots,x_{n-2k})$ and $X''=(x_{n-2k+1},x_{n-2k+2},\ldots,x_n)$\,.
Then specializing $Y=X^w$\,, $e_iX+e_iY$ becomes
$$
2\sum\limits_{j\ge 0}e_{i-2j}(X')\cdot e_{2j}(X'')\,.
$$
We claim that if we replace the entries of the matrix of $F(X,Y)$ by putting
$\sum\limits_{j\ge 0}e_{i-2j}(X')e_{2j}(X'')$ instead of $e_iX+e_iY$, then
the determinant of the so-obtained matrix $M$ is zero.
This can be seen, e.g., as follows.
Take the (formal) $n\times n$-matrix $[e_{n+1+j-2i}]$ whose determinant gives 
$s_{\rho_n}$\,.
Of course if we substitute $e_i:=e_i(X')$\,, then the determinant vanishes.
But using some elementary operations on rows of this specialized matrix $N$
we arrive at $M$\,.
More precisely, this is done as follows:
\bigskip
{\parindent=25pt
\item{$\bullet$} add $e_{2l-2}(X'')$ times the $l$-th row to the top row 
(all $l$); 
\item{} this changes the top row of $N$ to the top row of $M$\,.
\item{$\bullet$} add $e_{2l-4}(X'')$ times the $l$-th row to the second row 
(all $l$);
\item{} this changes the second row of $N$ to the second row of $M$\,.
\par}
\bigskip
\noindent
Continuing down the rows, $N$ is changed into $M$\,.
The assertion follows.
\qed
\enddemo
\bigskip

We will refer to the assertions of the lemma as a "vanishing property".

\medskip

\proclaim{Proposition 2.3}
For any $f\in \Cal S\Cal P(X)$\,,
$$
\langle f(X),F(X,Y)\rangle\equiv f(Y)\,.
$$
\endproclaim

\demo{Proof}
Observe that the left-hand side belongs to $\Cal S\Cal P(X^2)\otimes \Cal S\Cal P(Y)$\,.
The assertion to be proved says: by replacing the coefficients from $\Cal S\Cal P(X^2)$
by the corresponding ones from $\Cal S\Cal P(Y^2)$\,, we get $f(Y)$\,.
Hence it suffices to check that under the specialization $Y=X$\,,
$$
\langle f(X),F(X,X)\rangle=f(X)\,.
$$
Recall now (see (4)) that for any $f\in \Cal S\Cal P(X)$\,,
$$
\tk(f)=\sum\limits_{x\in W_n/S_n} w\bigl[f/2^ns_{\rho_n}(X)\bigr]\,.
$$

Applying (4)
to $\tk\bigl(f(X)\cdot F(X,X)\bigr)$, we see, by 
Proposition 2.2, that only one summand on the right-hand side survives, giving
us $f(X)$\,, as claimed.
\qed
\enddemo
\bigskip

The content of the next proposition provides an algebraic link between
the formulas of [F1,2] and [P-R2].

\proclaim{Proposition 2.4}
$F(X,Y)\equiv \widQ(X,Y)$\,.
\endproclaim

\demo{Proof}
We can write
$$
F(X,Y)\equiv\sum f_I(Y)\widQ_I(X)\,,
$$
where the sum is over strict partitions $I$ and $f_I(Y)\in \Cal S\Cal P(Y)$\,.
By Proposition 2.2 and the orthogonality (10) of the $\widQ_I(X)$'s, we get for 
strict $I$:
$$\aligned
\widQ_I(Y) & \equiv \langle\widQ_I(X),F(X,Y)\rangle \\
& \equiv\langle\widQ_I(X),\sum f_J(Y)\widQ_J(X)\rangle \\
& =f_{\rho_n\setminus J}(Y) \\
\endaligned $$
This implies $f_I(Y)\equiv\widQ_{\rho_n\setminus I}(Y)$\.
Consequently
$$
F(X,Y)\equiv\sum\widQ_{\rho_n\setminus I}(Y)\widQ_I(X)=\widQ(X,Y)\,,
$$
as asserted.
\qed
\enddemo

\noindent
{\bf Remark 2.5.}
This result was known earlier from comparison of geometric formulas of [F1,2]
and [P-R2].
Here, we have given a purely algebraic proof of it.


\example{Example 2.6}
For $n=2,3\,\ F(X,Y)=\widQ(X,Y)$\,. For $n=4$\,,
$$
\widQ(X,Y)-F(X,Y)=(x_1^2+x_2^2+x_3^2+x_4^2-y_1^2-y_2^2-y_3^2-y_4^2)
(x_1^2x_2^2x_3^2x_4^2-y_1^2y_2^2y_3^2y_4^2)\,.
$$
\endexample

\bigskip

Finally, observe that Proposition 2.4 implies that 
$\widQ(X,Y)$ has also the vanishing property. The following variant of it
will be used in Section 4.

\proclaim {Lemma 2.7}
\noindent
(i) \ For any $w\in S_n$ (and $w_0=[\ol 1,\ol 2,\ldots,\ol n]$\,),
$$
\widQ(X^{w_0w},\ol X)=2^ns_{\rho_n}(\ol X)\,.
\noindent
$$
(ii) \ For $w\in W_n\setminus w_0S_n$\,,
$$
\widQ(X^w,\ol X)=0\,.
$$
\endproclaim



\bigskip\bigskip

\head
{\bf 3. Action of $\tk$ on the basis $\{\frak S_{\mu}\cdot\widQ_I\}$ 
- an inductive approach.}
\endhead
\bigskip

In this section we give an elementary inductive proof of the following

\proclaim {Theorem 3.1} (i) \ For $\mu\in S_n$, writing $X=X_n$ and
$\omega =\omega^{(n)}$, one has 
$$
\ \tk\bigl(\frak S_{\mu}(X)
\cdot\widQ_{\rho_n}(X)\bigr) = (-1)^{l(\mu)}\frak S_{\mu}(X^{\omega})
=\frak S_{\mu}(\ol X^{\omega}).
$$

\medskip

\noindent (ii) \ For strict $I\subne\rho_n,\ \mu\in S_n,\ 
\tk\bigl(\frak S_{\mu}(X)\cdot\widQ_I(X)\bigr) = 0$.
\endproclaim
\bigskip

We give first an outline of the proof.
For the need of induction we write:
$$
\tk^{(k)}=
(\partial_0\partial_1'\ldots\partial_{k-1}')(\partial_0\partial_1'\ldots
\partial_{k-2}')\ \ldots\ (\partial_0\partial_1')\partial_0
$$
We have:
$$
\tk=\tk^{(n)}=\partial_0\partial_1'\ldots\partial_{n-1}'\tk^{(n-1)}.
\tag14
$$
We first show that it suffices to prove assertions (i) and (ii) for 
$\mu=\omega$.
Then, we prove (i) and (ii) by a simultaneous induction on $n$ using (14).
In the course of the proof of (i), we use the linearity formula (8) for 
$\widQ_I$ and
$$
\partial_1\partial_2\ldots\partial_{n-1}(x_1 x_2^2\ldots x_{n-1}^{n-1}) =
x_3x_4^2\ldots x_n^{n-2}.
$$
In the course of the proof of (ii), we use the linearity formula for 
$\widQ_I$ and the vanishing property:
$$
\partial_0\partial_1\ldots\partial_{n-1}(x_n^j x_{n-1}^{n-2} x_{n-2}^{n-3}
\ \ldots\ x_2) = 0
$$
for $0\le j\le n-1$.
\bigskip

We now pass to a detailed proof.

\proclaim{Lemma 3.2} It suffices to verify assertions (i) and (ii) of the
theorem for $\mu=\omega$.
\endproclaim

\demo{Proof}
Denote $v=v_0$;
so that $\tk=\partial_v'$ and
$\frak S_{\mu}=\frak S_{\mu}(X_n)$, $\widQ_I=\widQ_I(X_n)$ for brevity.
Since $\widQ_I$ is symmetric in $X$, we have
\medskip
\halign{\hskip 20pt $#$ & $#$\hfil &\qquad #\hfil \cr
\tk(\frak S_{\mu}\widQ_I)&=(-1)^{n(n-1)/2} \partial_v(\frak S_{\mu}\widQ_I)& \cr
&&\cr
&=\pm \partial_{v\mu^{-1}\omega}(x^{\rho_{n-1}}\widQ_I) & \cr 
&&\cr
&=\pm \partial_{v\omega\nu^{-1}\omega\omega}(x^{\rho_{n-1}}\widQ_I) 
&$(\mu:=\omega\nu\omega)$ \cr
&&\cr
&=\pm \partial_{\nu^{-1}\omega v}(x^{\rho_{n-1}}\widQ_I) 
&($v\omega=\omega v$ is the element of highest length \cr
&&in $W$ and commutes with any element of it) \cr
&&\cr
&=\partial_{\nu^{-1}\omega}\tk(x^{\rho_{n-1}}\widQ_I) &\cr
}
$$\hskip 31pt
=\left\{{(-1)^{n(n-1)/2}\partial_{\nu^{-1}\omega}
\bigl(\omega (x^{\rho_{n-1}})\bigr) \hskip 40 pt
\hbox{for } I=\rho_n \hbox{ by the case } \mu=\omega
\atop
0 \hskip 108pt\hbox{for }I\ne\rho_n \hbox{ by the case }\mu=\omega}\right.
$$
\vskip 4pt
$$\hskip -102pt
=\left\{{(-1)^{l(\mu)} \frak S_{\mu}(X^{\omega})\hskip 40pt\hbox{for }I=\rho_n
\atop
0 \hskip 56pt\hbox{for } I\ne\rho_n.}\right.
$$
\qed
\enddemo

\bigskip

In the proof of the next proposition we will need the following Lemma.

\proclaim{Lemma 3.3}
$\partial_1\partial_2\ldots\partial_{n-1}(x_1x^2_2\ldots x^{n-1}_{n-1})=
x_3x^2_4\ldots x_n^{n-2}$.
\endproclaim
{\demo{Proof}
We show by descending induction on $i\ (i=n,n-1,\ldots,1)$ that after
applying $\partial_i$ and before applying $\partial_{i-1}$ the resulting sum 
has the properties:

\noindent
1) \ it contains a unique summand without $x_i$, namely
$$
x_1x^2_2\ldots x^{i-1}_{i-1}x^{i-1}_{i+1}x^i_{i+2}\ldots x_n^{n-2}
$$
2) \ all but one summands are monomials of the form
$$
x_1x^2_2\ldots x^{i-1}_{i-1}x^a_i x^{\bullet}_{i+1}\ \ldots\ ,
$$
where $0<a\le i-1$.

The beginning of the induction is trivial: before applying $\partial_{n-1}$
we have the unique summand $x_1x^2_2\ldots x^{n-1}_{n-1}$.
All monomials in the image of 2) under $\partial_{i-1}$ contain $x_{i-1}$, 
so the unique summand without $x_{i-1}$ in the image of $\partial_{i-1}$ 
stems by applying $\partial_{i-1}$ to 1) and equals
$$
x_1x^2_2\ldots x^{i-2}_{i-2}x_i^{i-2}x^{i-1}_{i+1}\ldots x_n^{n-2}.
$$
All other monomials in the image of 1) are of the form:
$$
x_1x^2_2\ldots x^{i-2}_{i-2}x^a_{i-1} x^{\bullet}_i \ \ldots\ ,
$$
where $0<a\le i-2$.
This is also the form of all monomials in the image of 2) under 
$\partial_{i-1}$, and the inductive step follows.
Consequently, after applying $\partial_2$ and before applying $\partial_1$,
all but one summands are of the form: $x_1x_2 x^{\bullet}_3 \ldots$, 
and they go to zero under $\partial_1$.
The unique summand not containing $x_2$ is:
$$
x_1x_3x^2_4\ \ldots\ x_n^{n-2}
$$
and it goes under $\partial_1$ to $x_3x^2_4\ldots x_n^{n-2}$.
This proves the lemma.
\qed
\enddemo

\medskip
\proclaim{Proposition 3.4}:
$\tk\bigl(x^{\rho_{n-1}}\cdot\widQ_{\rho_n}(X_n)\bigr)=(-1)^{n(n-1)/2}
\omega (x^{\rho_{n-1}}).$
\endproclaim

\demo{Proof}
We have
$$
\tk\bigl(x^{\rho_{n-1}}\widQ_{\rho_n}(X_n)\bigr)=
\tk^{(n)}\bigl(x^{\rho_{n-2}}\widQ_{\rho_{n-1}}(X_n)x_1^2\ldots 
x^2_{n-1} x_n\bigr)
$$
$$
=\partial_0\partial_1'\ldots\partial_{n-1}'\tk^{(n-1)}\bigl(x^{\rho_{n-2}}
\widQ_{\rho_{n-1}}(X_{n-1}+x_n) x_1^2\ldots x^2_{n-1}x_n\bigr)
$$
We apply the linearity formula (8) to $\widQ_{\rho_{n-1}}(X_{n-1}+x_n)$.
All summands different from $\widQ_{\rho_{n-1}}(X_{n-1})$ are of the 
form $\widQ_J(X_{n-1})x_n^{\bullet}$ where $J\subne\rho_{n-1}$.
After applying to $J$ the factorization property (9) (i.e. factoring out
$\widQ_{pp}(X_{n-1})$ corresponding to all pairs of equal rows), as well 
as factoring out $x_n^{\bullet}$, we apply $\tk^{(n-1)}$ to 
$x^{\rho_{n-2}}\widQ_{J'}(X_{n-1})$ where $J'\subne\rho_{n-1}$
is the resulting strict partition.
We get the vanishing by the induction assumption (ii) for $n-1$.
Thus the above expression is equal to:
$$\aligned
\partial_0\partial_1'&\ldots \partial_{n-1}'\tk^{(n-1)}\bigl(
x^{\rho_{n-2}}\widQ_{\rho_{n-1}}(X_{n-1})x_1^2\ldots 
x^2_{n-1}x_n\bigr)\\
&=\partial_0\partial_1'\ldots\partial_{n-1}'\Bigl(x_1^2\ldots x^2_{n-1}x_n
\tk^{(n-1)}\bigl(x^{\rho_{n-2}}\widQ_{\rho_{n-1}}(X_{n-1})\bigr)\Bigr)\\
&=\partial_0\partial_1'\ldots\partial_{n-1}'\bigl(x^2_1\ldots x^2_{n-1}x_n
(x_2x^2_3\ldots x^{n-2}_{n-1}\cdot x_1\ldots x_{n-1})\bigr) \\
&=(-1)^{(n-1)(n-2)/2} \partial_0\bigl(x_1\ldots x_n\ 
\partial_1'\ldots\partial_{n-1}'
(x_1x^2_2\ldots x^{n-1}_{n-1})\bigr)
\endaligned $$
by the induction assumption (i) for $n-1$.
\medskip

Now, using Lemma 3.3, the above expression is rewritten as
$(-1)^{n(n-1)/2}$ times 
$$
\partial_0(x_1x_2x_3^2\ldots x_n^{n-1})
= x_2x^2_3\ldots x_n^{n-1}=
\omega(x^{\rho_{n-1}}),
$$
and the proof of the proposition is complete.
\qed
\enddemo

\medskip

\proclaim{Proposition 3.5} :
$\tk\bigl(x^{\rho_{n-1}}\cdot\widQ_I(X_n)\bigr)=0$ \qquad for strict 
$I \subset\rho_n$.
\endproclaim

\demo{Proof}
We apply the linearity formula (8) to $\widQ_I(X_n)=\widQ_I(X_{n-1}+
x_n)$. 
Consider a typical summand $\widQ_J(X_{n-1})\cdot x_n^{\bullet}$.
Recall that $J$ is a partition (not necessary strict) gotten from $I$ by
subtracting boxes from $I$, but no more then two from one row.
Moreover, all parts of $J$ are less than or equal to $n-1$.
By dividing $x^{\rho_{n-1}}$ by $x_1\cdot\ldots\cdot x_{n-1}$ and multiplying
$\widQ_J(X_{n-1})\cdot x_n^{\bullet}$ by $x_1\cdot\ldots\cdot x_{n-1}$,
we get the summand:
$$
x^{\rho_{n-2}}\widQ_{(n-1)J}(X_{n-1})\cdot x_n^{\bullet}
$$
whose factor depending on $X_{n-1}$ is an element of the basis 
associated with  $X_{n-1}$.

\noindent
If $J=(n-1)J'$ then $J'\subne\rho_{n-1}$ ($J'=\rho_{n-1}$ implies 
$I=\rho_n$ which is impossible).
Then using the factorization property (9) to $\widQ_{(n-1)J}(X_{n-1})$
and factoring out all possible $\widQ_{p,p}(X_{n-1})$ for $p=n-1,\dots,1$ \ ,
we apply $\tk^{(n-1)}$ to $x^{\rho_{n-2}}\widQ_{J''}(X_{n-1})$
where $J''\subne\rho_{n-1}$ is strict, thus getting the vanishing by the 
induction assumption (ii) for $n-1$.

\noindent
Hence we can assume $j_1\le n-2$.
Observe that in $J$ no three consecutive rows are equal.
Therefore it is impossible to factor out from $J$ a pair of equal rows 
to get $\rho_{n-2}$.
Hence the only $J$ which after a possible subtraction of pairs of equal
parts and adding the part $(n-1)$, gives 
us $\rho_{n-1}$ is the partition $\rho_{n-2}$ itself.
The unique possibilities for the initial partition $I$ are then:
$$
I=(n-1,n-2,\ldots,j+2,j,j-1,\ldots,1) \qquad j=1,\ldots,n-1.
$$
(Pictorially:
$$\CD
\bullet\hfil &&&&&& &\\
\bullet\hfil & \bullet\hfil &&&&& & \\
\bullet\hfil & \bullet\hfil & \bullet\hfil &&&& & \\
\bullet\hfil & \bullet\hfil & \bullet\hfil & \bullet\hfil & *\hfil && & \\
\bullet\hfil & \bullet\hfil & \bullet\hfil & \bullet\hfil & \bullet\hfil & 
*\hfil & & \\
\quad\bullet\quad & \quad\bullet\quad & \quad\bullet\quad & \quad\bullet\quad 
& \quad\bullet\quad & \quad\bullet\quad & \quad* & \qquad \\
\endCD$$
\bigskip
\noindent
where the total number of rows is $n-2$ and the number of rows without
"*" equals $j$. In the above figure we have $n=8$ and $j=3$.) 
This brings us to the summand
$$
x^{\rho_{n-2}}\widQ_{\rho_{n-1}}(X_{n-1})x_n^j \qquad 1\le j\le n-1.
$$
We have by the induction assumption (i) for $n-1$:
$$\aligned
\tk\bigl(x^{\rho_{n-1}}\widQ_{\rho_{n-1}}(X_{n-1})x_n^j\bigr)
&=\partial_0\partial_1'\ldots\partial_{n-1}'\Bigl(x_n^j\tk^{(n-1)}
\bigl(x^{\rho_{n-2}}\widQ_{\rho_{n-1}}(X_{n-1})\bigr)\Bigr) \\
&=\pm \partial_0\partial_1\ldots\partial_{n-1}\bigl(x_n^jx_{n-1}^{n-2}x_{n-2}^{n-3}
\ldots x_2\bigr). \\
\endaligned $$
We show this expression is zero for $1\le j\le n-1$ by induction on $n$.
If $j=n$ or $j=n-1$, then
$$
\partial_0\partial_1\ldots\partial_n\bigl(x^j_{n+1}x_n^{n-1}x^{n-2}_{n-1}
\ldots x_2\bigr)=0
$$
by direct calculation.
Assume $1\le j\le n-1$.
Then $\partial_n$ sends $x^j_{n+1}x_n^{n-1}x^{n-2}_{n-1}\ldots$ to
$$
x^j_{n+1}x^j_n(x_{n}^{n-1-j-1}+x_{n}^{n-1-j-2}x_{n+1}+ \ldots)
x_{n-1}^{n-2}\ldots
$$
(for $i=n$, the third factor would be empty).
In every monomial, the variable $x_n$ appears with exponent
$\le j+n-1-j-1=n-2$.
Hence, by the induction assumption, $\partial_0\partial_1\ldots\partial_{n-1}$
annihilates this expression.
The proposition has been proved.
\qed
\enddemo

This completes the proof of Theorem 3.1.

\proclaim{Corollary 3.6}
Im $(\tk)$ is a free $\Cal S\Cal P(X^2)$-module with a basis 
$\{\frak S_{\mu}(X^{\omega})\}_{\mu\in S_n}$.
\endproclaim

\medskip

We end this section with the following, related formula.

\proclaim{Proposition 3.7}
For any $\mu\in S_n,\ \tk\bigl(\frak S_{\mu}\cdot s_I(X)\bigr)\ne 0$ iff $I$
is of the form $\rho_n+2J$.
In this case,
$$
\tk\bigl(\frak S_{\mu}(X)\cdot s_I(X)\bigr)=(-1)^{l(\mu)}
\frak S_{\mu}(X^{\omega})\cdot\psi_2
\bigl(s_J(X)\bigr),
$$
(where $\psi_2$ is the ring homomorphism sending $x_i$ to
$x_i^2$, $i=1,\dots,n$).
\endproclaim

\demo{Proof}
It follows from Theorem 3.1 (e.g. by developing $s_I(X)$ in the 
$\Cal S \Cal P(X^2)$-basis of
$\widQ$-polynomials) that
$$
\tk\bigl(\frak S_{\mu}(X) \cdot s_I(X)\bigr)=c\cdot \frak S_{\mu}(X^{\omega})
$$
for some $c\in \Cal S\Cal P(X^2)$.
The coefficient $c$ is computed in the following way using Lemma 1.6:
$$\aligned
c&=\partial'_{\omega\mu\omega}\tk\bigl(\frak S_{\mu}(X)\cdot s_I(X)\bigr)=
\tk \partial'_{\mu}\bigl(\frak S_{\mu}(X)\cdot s_I(X)\bigr) \\
&=(-1)^{l(\mu)} \tk\bigl(s_I(X)\bigr),
\endaligned$$
and the assertion follows from the formula recalled in Proposition 1.3.
\qed
\enddemo

\bigskip\bigskip

\head
{\bf 4. Action of $\tk$ on the basis $\{\frak S_{\mu}\cdot\widQ_I\}$
via the vanishing property.}
\endhead

\bigskip

The goal of this section is to derive the formula of Theorem 3.1 via the 
vanishing property of Section 2. 
The starting point of the approach of this section is the determination of
the image of $\tk:\Bbb Z[X]\to\Bbb Z[X]$.

\proclaim{Proposition 4.1}
Im $(\tk)$ is a free $\Cal S\Cal P(X^2)$-module with a basis 
$\{\frak S_{\mu}(X^{\omega})\}_{\mu\in S_n}$.
\endproclaim

\demo{Proof}
We write for brevity: 
$\widQ_I=\widQ_I(X),\ \Cal S\Cal P=\Cal S\Cal P(X)$ 
and $\Cal S\Cal P^2=\Cal S\Cal P(X^2)$.
Knowing that $\Bbb Z[X]$ is an $\Cal S\Cal P$-module with a basis
$\{\frak S_{\mu}(X^{\omega})\}_{\mu\in S_n}$ and $\Cal S\Cal P$ is 
an $\Cal S\Cal P^2$-module with 
a basis $\{\widQ_I\}_{strict\ I\subset\rho_n}$, we can write any 
element in $Im(\tk)$ as a unique expression
$$
\sum c_{\mu,I}\frak S_{\mu}(X^{\omega})\cdot\widQ_I\quad,
\tag15
$$
where $c_{\mu,I}\in \Cal S\Cal P^2$ and $\mu,I$ are as above.
Suppose that there exist $\nu$, $J$ with  $|J|\ge 1$ such that
$c_{\nu,J}\ne 0$. Let $\mu$ be a permutation of highest length 
among such $\nu$'s. The operator $\partial_{\omega\mu\omega}'$
sends $\frak S_{\mu}(X^{\omega})$ to $1$ and all 
$\frak S_{\nu}(X^{\omega})$, where
$\nu\ne\mu$ and $l(\nu)\le l(\mu)$, to $0$.
Applying this operator to (15),  we get
$$
\sum c_{\mu,J}\widQ_J+\bigl(\Cal S\Cal P^2 \hbox{-combination of the}
 \ \frak S_{\nu}(X^{\omega})\hbox{'s} \bigr).
\tag16
$$
Let $I$ be a partition of highest weight
among those $J$ such that $c_{\mu,J}\ne 0$.
Let $\partial^I$ be the divided-difference operator s.t.
$\partial^I(\widQ_I)=1$ (hence 
$\partial^I=\partial_w\circ\partial_0$ for some $w$) and
$\partial^I(\widQ_J)=0$ for all strict $J\ne I$, $|J|\le|I|$
(see Proposition 1.7).
Then the image of (16) under $\partial^I$ is $c_{\mu,I}\ne 0$.
(Observe that $\partial^I(\frak S_{\nu}(X^{\omega})=0$ 
because the $\frak S_{\nu}(X^{\omega})$'s
do not depend on $x_1$).

On the other hand, using $\tk\partial_{\omega\mu\omega}'
=\partial_{\mu}\tk$
(Lemma 1.6), we infer that (16) belongs to $Im(\tk)$.
Since $\partial_0\tk=0$, we have $\partial^I\tk=0$, 
and we get a contradiction with the above calculation.
\medskip

The proposition has been proved.
\qed
\enddemo

\medskip
 
We now give 
\medskip

{\bf Another proof of Theorem 3.1.}

\bigskip

By the preceding proposition we know that
$$
\tk\bigl(\frak S_{\mu}(X) \cdot\widQ_{\rho_n\setminus I}(X)\bigr)=
\sum c_{I,\nu} \frak S_{\nu}(X^{\omega})\ ,
$$
where $c_{I,\nu}\in \Cal S\Cal P(X^2)$. Invoking the definition (13)
of $\widQ(X,Y)$, we thus have, by linearity,
$$
\aligned
\tk\bigl(\frak S_{\mu}(X)\cdot\widQ(X,\ol Y)\bigr) &=
\tk\bigl(\frak S_{\mu}(X) \sum\limits_I(-1)^{|I|}\widQ_{\rho_n\setminus I}
(X)\cdot\widQ_I(Y)\bigr)\\
&=\sum\limits_{I,\nu}(-1)^{|I|} c_{I,\nu} \frak S_{\nu}(X^{\omega}) \cdot
\widQ_I(Y). \\
\endaligned$$
Specialize now $Y$ in $X$ (that is we put $y_1=x_1,\ldots,y_n=x_n$).
Since $\{\frak S_{\mu}(X^{\omega})\cdot\widQ_I(X)\}$ is an $\Cal S \Cal P(X^2)$-basis,
the value of the image of $ \tk\bigl(\frak S_{\mu}(X)\cdot\widQ(X,\ol Y)\bigr)$
under this specialization still determines the coefficients $c_{I,\nu}$.
\medskip

Write $\tk=\sum\limits_{w\in W}a_w w$, where $a_w$ are rational
functions in $X$. Then
$$
\tk\bigl(\frak S_{\mu}(X)\cdot\widQ(X,\ol Y)\bigr)=
\sum\limits_{w\in W} a_w\frak S_{\mu}(X^w)\cdot\widQ(X^w,\ol Y).
$$
where $W$ acts only on $X$-variables.
It follows, from Corollary 1.5 and Lemma 2.7, that 
among those $w$ for which $a_w\ne 0$
there is only one $w=v_0=[\ol n,\ol{n-1},\ldots,\ol 1]=
w_0\omega$ for which $\widQ(X^w,\ol X)\ne 0$; in fact, we have
$$
\widQ(X^{v_0},\ol X)=2^n s_{\rho_n}(\ol X)
$$
Since $a_{v_0}=\bigl(2^n s_{\rho_n}(\ol X)\bigr)^{-1}$ by Lemma 1.4, 
we get finally
$$\aligned
\tk\bigl(\frak S_{\mu}(X) \cdot\widQ(X,\ol Y)\bigr)\mid_{X=Y} 
&=\frak S_{\mu}(X^{v_0}) \\
&= (-1)^{l(\mu)}\frak S_{\mu}(X^{\omega}). \\
\endaligned $$
This completes the proof.
\qed


\bigskip\bigskip

\head
{\bf 5. Key formula and vertex operators}
\endhead

\bigskip

In this section we prove the following identity.

\proclaim{Theorem 5.1 {\rm (Key formula)}}
Suppose $n\ge k>0$. Let  $IkJ$ be a strict partition 
and $H$ a strict partition not 
containing $k$.
Then
$$
(-1)^{n-1}\partial_{n-1}\partial_{n-2}\ldots\partial_1\partial_0
\bigl(x_1^{n-k}\widQ_{IkJ}(X_n)\bigr)=
(-1)^{l(I)}\widQ_{IJ}(X_n)
$$
and
$$
\partial_{n-1}\partial_{n-2}\ldots\partial_1\partial_0
\bigl(x_1^{n-k}\widQ_H(X_n)\bigr)=0.
$$
\endproclaim

\noindent

Our proof of the theorem will use some differential operators (called
usually "vertex operators"). The vertex operators that we shall use
appeared already in the literature (see e.g. [D-K-M], [D-J-K-M1] 
and [D-J-K-M2]), 
but in the context different from
the one in the present paper. Our notation and conventions for 
vertex operators follow rather the exposition of [C-T]. 
\medskip

Besides $\widQ$-polynomials we will need Schur's $P$-functions and
$Q'$-functions from [L-L-T2]. (We use the term "function" in the sense
of [Mcd1] i.e., a function depends on an infinite
sequence of variables $(x_1,x_2,\ldots)$.) Given a symmetric 
function $f$, we denote
$$
f(X_n)= f_{|x_{n+1}=x_{n+2}=\ldots = 0},
$$  
the corresponding symmetric polynomial in the first $n$ variables.
\medskip

For the purposes of this paper it suffices to work with Schur $P$-functions 
indexed by sequences of positive integers (like in Schur's original paper [S]).
Recall that for $J\in\Bbb N^{*m}$ with odd $m$,
$$
P_J=P_{j_1}\cdot P_{j_2,\ldots,j_m}-P_{j_2}\cdot P_{j_1,j_3,\ldots,j_m}+\ldots
+P_{j_m}\cdot P_{j_1,\ldots,j_{m-1}}
$$
and with even $m$,
$$
P_J=P_{j_1,j_2}\cdot P_{j_3,\ldots,j_m}-P_{j_1,j_3}\cdot P_{j_2,j_4,\ldots,j_m}
+\ldots+P_{j_1,j_m}\cdot P_{j_2,\ldots,j_{m-1}}
$$
Here, $P_j=\sum s_L$, sum over all hook partitions $L$ of $j$, and
$$
P_{i,j}=
P_i\cdot P_j+2\sum\limits^{j-1}_{k=1}(-1)^kP_{i+k}P_{j-k}+(-1)^jP_{i+j}.
$$

For another expression of $P_J$ in the form of a quadratic polynomial
in the $S$-functions, see [L-L-T1].
\bigskip

If we restrict ourselves to a finite set of indeterminates $X_n$,
where $n>|J|$, then
$P_J(X_n)$
is given by the following symmetrization:
$$
P_J(X_n)=\pi\Bigl(x^J\prod\limits_{i<j\le n\atop i\le m}\bigl(1+{x_j \over x_i}
\bigr)\Bigr),
\tag17
$$
where $\pi$ is the symmetrizer defined for $f\in \Bbb Z[X_n]$ by
$$
\pi(f)=\sum\limits_{w\in S_n}w\bigl[f\,x^{\rho_{n-1}}/\triangle(X)]
=\partial_{\omega}(f\cdot x^{\rho_{n-1}}).
$$
Recall that $P_J$ is an antisymmetric function of the components of $J$;
in particular, $P_J=0$ if two components of $J$ are equal.
\medskip

 
Given a symmetric function $f$, let $D_f$ be the adjoint 
operator to the multiplication by $f$ w.r.t. the standard scalar product
on the ring of symmetric functions (see [Mcd1]).

\proclaim{Proposition 5.2} Let $k>0$.
Then
$$
P_{kJ}=\sum\limits_{j\ge 0}(-1)^jP_{k+j}\cdot D_{e_j}(P_J).
$$
\endproclaim

\demo{Proof}
Suppose that $n>>0$ (in fact, $n\ge k+|J|$ will do the job).
Using (17) and $\pi^2=\pi$, we get
$$
P_{kJ}(X_n)=\pi\Bigl(x_1^k(1+{x_2\over x_1})\ldots(1+{x_n\over x_1})
P_J(x_2,\ldots,x_n)\Bigr).
$$
Passing to the $\lambda$-ring setup, we have
$$\aligned
P_{kJ}(X_n) &= {1\over 2}\pi\Bigl(x_1^k(1+{x_1\over x_1})(1+{x_2\over x_1})\ldots
(1+{x_n\over x_1})P_J(X_n-x_1)\Bigr) \\
&={1\over 2}\pi\Bigl(\bigl(\sum\limits_{i=0}^n x_1^{k-i}e_i(X_n)\bigr)
\bigl(\sum\limits_{j\ge 0}(-x_1)^j\cdot (D_{e_j}P_J)(X_n)\bigr)\Bigr) \\
&={1\over 2} \sum\limits_i\sum\limits_j (-1)^j s_{k+j-i}(X_n)\cdot e_i(X_n)
\cdot 
(D_{e_j}P_J)(X_n) \\
&=\sum\limits_{j\ge 0}(-1)^jP_{k+j}(X_n)\cdot(D_{e_j}P_J)(X_n).
\endaligned 
$$
Above, we have used the equalities $\pi(x_1^i)=s_i(X_n)$ and
$$
\bigl(\sum e_i\bigr)\cdot\bigl(\sum s_j\bigr)=1+2\sum P_i. 
$$
\medskip
The proposition has been proved.
\qed 
\enddemo
\medskip

By introducing the following differential operator
on the ring of symmetric functions.
$$
V^e_k:=P_k-P_{k+1}D_{e_1}+P_{k+2}D_{e_2}-\ldots\ ,
$$
we can restate the formula in  Proposition 5.2  as the equation
$$
V^e_k(P_J)=P_{kJ}\ .
$$

We also need $Q'$-functions:
for a partition $I$, $Q_I'$ is the image of  
$\widQ_I$ under the involution 
which exchanges $e_i$ and $s_i$, $i\in \Bbb N$.
In the following remark, for the reader's convenience, we collect some
properties of these functions.
\medskip
\noindent
{\bf Remark 5.3.} (i) Let $Q_I'(q)$ be the Hall-Littlewood polynomial 
where the initial alphabet $X$ is replaced by $X/(1-q)$ (in the sense of 
$\lambda$-rings). Then $Q_I'$ equals $Q_I'(q)$ when 
$q$ is specialized to $-1$.

(ii) Equivalently, using raising operators $R_{ij}$ (see e.g. [Mcd1]), 
one has
$$
Q_I'=\prod\limits_{i<j}{1-R_{ij} \over 1+R_{ij}}h_I\ ,
$$
where $h_I$ is the product of complete symmetric functions associated
with the parts of $I$.

(iii) $Q'$-functions can be defined also using differential 
operators.
In fact, this approach extends their definition to all finite sequences 
of integers.
For $k\in\Bbb Z$ one has:
$$
Q_{kJ}'=(s_k-2s_{k+1}D_{P_1}+2s_{k+2}D_{P_2}-\ldots\,)(Q_J')\ .
$$

(iv) Littlewood [Li1] showed that the $P$ and $Q$-Hall-Littlewood bases
are adjoint w.r.t. a certain scalar product. Going back to the standard
scalar product on the ring of symmetric functions, his result says
that  $\{P_I(q)\}$ is adjoint to $\{Q'_I(q)\}$, where $I$ runs over
all partitions. Specializing $q$ to $-1$, one gets 
that $\{Q_J'\}$ and $\{P_J\}$, where $J$ runs over the set of strict
partitions, are two adjoint bases of some free
$\Bbb Z$-modules $A$ and $B$ of symmetric functions, w.r.t. the standard 
scalar product.
\medskip
\noindent
We refer the reader to [Li1], [Mcd1], [L-L-T2] and [D-L-T]
for details and precise references to the results described in this remark.
\bigskip

Introduce now two other differential operators on the ring of symmetric
functions:
$$
U^s_k:=D_{P_k}-s_1D_{P_{k+1}}+s_2D_{P_{k+2}}-\ldots
$$
and
$$
U^e_k:=D_{P_k}-e_1D_{P_{k+1}}+e_2D_{P_{k+2}}-\ldots \ .
$$
\medskip

In particular, by Remark 5.3(iii), we have for $k>0$,
$$
U^s_k(Q_J')=(-1)^k{1\over 2}Q_{-k,J}'\ .
$$
Our goal now is to investigate the action of the operator $U^s_k$ on
$\widQ$-functions.
We compute first the action of $U^e_k$ on $Q'$-functions.

\proclaim{Proposition 5.4}
Suppose $k>0$. Let $IkJ$ be a strict partition and $H$- 
a strict partition not containing $k$. Then
$$
U^e_k(Q_{IkJ}')=(-1)^{l(I)}Q_{IJ}'
$$
and
$$
U^e_k(Q_H')=0\ .
$$
\endproclaim

\demo{Proof}
It follows from Proposition 5.2 that $V^e_k(B)\subset B$ and hence 
$U^e_k(A)\subset A$.
Denote by [ , ] the standard scalar product on the 
ring of symmetric functions.
For strict partitions $K,L$ we must evaluate
$$
\bigl[ U^e_k(Q_K'),P_L \bigr]= \bigl[ Q_K',V^e_k(P_L)
\bigr]=
\bigl[ Q_K',P_{kL} \bigr]\,.
$$
This immediately yields both assertions.
\qed
\enddemo

Applying the involution on the ring of symmetric polynomials,
that exchanges $e_i$ with $s_i$ for $i\in \Bbb N$,
we get

\proclaim{Proposition 5.5}
Suppose $k>0$. Let $IkJ$ be a strict partition and $H$- a strict partition not
containing $k$\,. Then
$$
U^s_k(\widQ_{IkJ})=(-1)^{l(I)}\widQ_{IJ}
$$
and
$$
U^s_k(\widQ_H)=0\,.
$$
\endproclaim

We now want to link the above differential operators with
divided-difference operators.
To this end we give first the following presentation of the operator
induced by $s_0$ on the ring of symmetric functions.

\proclaim{Lemma 5.6}
For every symmetric function $f$,
$$
s_0(f)=f-2x_1D_{P_1}(f)+2x^2_1D_{P_2}(f)-2x^3_1D_{P_3}(f)+\ldots
$$
\endproclaim

\demo{Proof} We will use the $\lambda$-ring setup. For any symmetric
function $f$ and three alphabets $X,B,C$, one has
$$
f(X-B+C)=\sum s_I(-B+C) D_{s_{I}}(f)(X),
$$
where $C-B$ denotes the formal difference of the alphabets $C$ and $B$.
\smallskip

Take now $B=x_1$, $C=\ol x_1$; then $X-B+C$ is obtained from $X$ by
changing $x_1$ to $\ol x_1$. On another hand, the only non-zero
$s_I(-B+C)$ for those special $B$ and $C$, are those for which $I$
is a hook $I=(j+1,1^i)$.
\smallskip

In that case, for $I\ne 0$, one has the explicit value
$$
s_I(-B+C)=(\ol x_1-x_1)(\ol x_1)^j(-x_1)^i
$$
which specializes to $2(-x_1)^{|I|}$ under $\ol x_1 \mapsto x_1$.
Since the sum of all hook Schur $S$-functions of weight $k$ is equal to $P_k$,
collecting all hooks of the same weight, one finally gets
$$
f\bigl((-x_1),x_2,x_3,\ldots \bigr)=f-2x_1 D_{P_1}(f)+2x_1^2 D_{P_2}(f)-\ldots
,
$$
the right-hand side being evaluated in $x_1,x_2,\ldots $.
\qed
\enddemo

\proclaim{Corollary 5.7}
For every symmetric function $f$\,,
$$
\partial_0(f)=D_{P_1}(f)-x_1D_{P_2}(f)+x^2_1D_{P_3}(f)-\ldots
.$$
\endproclaim
\bigskip

Let us define the following operator $D_k^{(n)}$ on the ring of symmetric
functions by

$$
D_k^{(n)}(f):=\partial_{n-1}\ldots\partial_1\partial_0(x^k_1\cdot f)\,.
$$

\proclaim{Proposition 5.8} Fix $n\in \Bbb N$.
Let $0\le k<n$\,. If $f$ is a symmetric function, then one has the equality
$$
\Bigl((-1)^{n-1}D_k^{(n)}f\Bigr)(X_n)=(U^s_{n-k}f)(X_n)\,;
$$
in other words,
$$
\Bigl((-1)^{n-1}D_k^{(n)}f\Bigr)(X_n)=\Bigl(D_{P_{n-k}}f
-s_1\cdot D_{P_{n-k+1}}f+s_2\cdot D_{P_{n-k+2}}f-\ldots\Bigr)(X_n).
$$
\endproclaim

\demo{Proof}
We have
$$
\partial_0(x_1^{2j+1}\cdot f)= (-x_1)^{2j+1}\partial_0(f)+x_1^{2j}\cdot f
$$
and
$$
\partial_0(x_1^{2j}\cdot f)=x_1^{2j}\partial_0(f).
$$
If $2j+1<n$\,, then $\partial_n\ldots\partial_1(x_1^{2j}f)=0$\,.
The assertion follows from  Corollary 5.7  by using the 
well-known formula $\partial_{n-1}\ldots\partial_1(x_1^k)=s_{k-n+1}(X_n)$\,.
\qed
\enddemo

This completes the proof of Theorem 5.1. (Note that the case $k=n$ was proved 
earlier in [P-R2] by another method.)
\bigskip

We end this section by showing how the key formula explicates the 
action of the operators $\tk_k$ defined by (11)
on the functions specified in the following

\proclaim{Proposition 5.9}
Let $0<k\le n$ and $n-1\ge\alpha_1>\alpha_2>\ldots>\alpha_k\ge 0$\,.
Then the image under $\tk_k$ of
$$
\partial'_{\omega^{(k)}}\bigl(x_1^{\alpha_1}x_2^{\alpha_2}\ldots x_k^{\alpha_k}
\cdot \widQ_I(X_n)\bigr)=
\partial'_{\omega^{(k)}}\bigl(x_1^{\alpha_1}x_2^{\alpha_2}\ldots x_k^{\alpha_k}
\bigr) \cdot \widQ_I(X_n)
$$
is non-zero only if $n-\alpha_1,n-\alpha_2,\ldots,n-\alpha_k$ are 
parts of $I$\,.
In this case, the image is
$$
(-1)^{\sum p_j-k}\widQ_J(X_n)\,,
$$
where $J$ is the strict partition with parts $\{i_1,i_2,\ldots \} \setminus
\{n-\alpha_1,\ldots,n-\alpha_k\}$ 
and $p_j$ is such that $i_{p_j}=n-\alpha_j$ for $j=1,\ldots,k$.
\endproclaim

\medskip

We need some properties of the powers of $\tk_1$.

\proclaim{Lemma 5.10}
Let $\alpha_i\le n-i$\,.
Then for a symmetric function $f$\,,
$$
\tk_1^k(x_1^{\alpha_1}\cdot\ldots\cdot x_k^{\alpha_k}\cdot f)=
\tk_1\Bigl(x_1^{\alpha_k}\cdot\ldots\cdot\tk_1(x_1^{\alpha_2}\cdot\tk_1
(x_1^{\alpha_1}\cdot f)\bigr)\ldots\Bigr)\,.
$$
\endproclaim

In other words, by identifying a variable with the operator of multiplication
by this variable, the lemma asserts the following equality of operators 
on the ring of symmetric functions:
$$
\tk_1^kx_1^{\alpha_1}\ldots x_k^{\alpha_k}=
\tk_1x_1^{\alpha_1}\tk_1x_1^{\alpha_2}\ldots\tk_1x_1^{\alpha_k}\,.
$$

\demo{Proof}
The proof is by induction on $k$. For simplicity we only prove the case $k=2$
and show how the case $k=2$ implies the case 
$k=3$\,.
This is sufficient to see the general pattern of the inductive proof.
For $k=2$\,, we have
$$\aligned
\tk_1^2(x_1^{\alpha}&x_2^{\beta}\cdot f) = \\
&= \tk_1\partial_{n-1}\ldots\partial_1\partial_0(x_1^{\alpha}x_2^{\beta}
\cdot f) \\
&= \tk_1\partial_{n-1}\ldots\partial_1\bigl[\partial_0(x_1^{\alpha}\cdot f)
\cdot x_2^{\beta}\bigr] \\
&= \tk_1\partial_{n-1}\ldots\partial_2\bigl[\partial_1\partial_0(x_1^{\alpha}
\cdot f)\cdot x_1^{\beta}+\partial_0(x_1^{\alpha}\cdot f)\cdot\partial_1
(x_2^{\beta})\bigr].
\endaligned $$
Observe that
$$
\partial_{n-1}\ldots\partial_2\bigl(\partial_0(x_1^{\alpha}\cdot f)\cdot
\partial_1(x_2^{\beta})\bigr)=0
$$
because $\partial_0(x_1^{\alpha}\cdot f)$ is a scalar w.r.t.
$\partial_{n-1}\ldots\partial_3\partial_2$ and we have $\beta\le n-2$\,.
We conclude that
$$
\tk_1^2(x_1^{\alpha}x_2^{\beta}\cdot f)=
\tk_1\bigl(x_1^{\beta}\cdot\tk_1(x_1^{\alpha}\cdot f)\bigr)
$$
as asserted, because $x_1^{\beta}$ is a scalar w.r.t. 
$\partial_{n-1}\ldots\partial_2$.

For $k=3$\,, we have
$$ \aligned
\tk_1^3(x_1^{\alpha}&x_2^{\beta}x_3^{\gamma}\cdot f)= \\
&=\tk_1^2\partial_{n-1}\ldots\partial_1\bigl[\partial_0(x_1^{\alpha}x_2^{\beta}
\cdot f)\cdot x_3^{\gamma}\bigr] \\
&= \tk_1^2\partial_{n-1}\ldots\partial_2\bigl[\partial_1\partial_0
(x_1^{\alpha}x_2^{\beta}\cdot f)\cdot x_3^{\gamma}\bigr] \\
&= \tk_1^2\partial_{n-1}\ldots\partial_3\bigl[\partial_2\partial_1\partial_0
(x_1^{\alpha}x_2^{\beta}\cdot f)\cdot x_2^{\gamma}+\partial_1\partial_0
(x_1^{\alpha}x_2^{\beta}\cdot f)\cdot\partial_2(x_3^{\gamma})\bigr] \\
\endaligned $$
Observe that
$$
\partial_{n-1}\ldots\partial_3\bigl[\partial_1\partial_0(x_1^{\alpha}
x_2^{\beta}\cdot f)\cdot\partial_2(x_3^{\gamma})\bigr]=0
$$
because $\partial_1\partial_0(x_1^{\alpha}x_2^{\beta}\cdot f)$ is a scalar 
w.r.t. $\partial_{n-1}\ldots\partial_3$ and $\gamma\le n-3$\,.
Since $x_2^{\gamma}$ is a scalar w.r.t. $\partial_{n-1}\ldots\partial_3$\,,
the above expression can be rewritten as:
$$\aligned
\tk_1^2\bigl(x_2^{\gamma}\cdot&\tk_1(x_1^{\alpha}x_2^{\beta}\cdot f)\bigr) \\
&= \tk_1\partial_{n-1}\ldots\partial_1\bigl(x_2^{\gamma}\cdot\partial_0\tk_1
(x_1^{\alpha}x_2^{\beta}\cdot f)\bigr) \\
&= \tk_1\partial_{n-1}\ldots\partial_2\Bigl(\partial_1\partial_0\tk_1
(x_1^{\alpha}x_2^{\beta}\cdot f)x_1^{\gamma} + \partial_0\tk_1
(x_1^{\alpha}x_2^{\beta}\cdot f)\cdot\partial_1(x_2^{\gamma})\Bigr).
\endaligned $$
But $\partial_0\tk_1(x_1^{\alpha}x_2^{\beta}\cdot f)=\partial_0
\bigl(x_1^{\beta}\cdot\tk_1(x_1^{\alpha}\cdot f)\bigr)$ is a scalar w.r.t.
$\partial_{n-1}\ldots\partial_2$ and $\gamma\le n-3$\,; this implies
$$
\partial_{n-1}\ldots\partial_2\bigl[\partial_0\tk_1(x_1^{\alpha}x_2^{\beta}
\cdot f)\cdot\partial_1(x_2^{\gamma})\bigr]=0.
$$
Since $x_1^{\gamma}$ is a scalar w.r.t. $\partial_{n-1}\ldots\partial_2$,
the above expression equals
$$
\tk_1\bigl(x_1^{\gamma}\cdot\tk_1^2(x_1^{\alpha}x_2^{\beta}\cdot f)\bigr)=
\tk_1\Bigl(x_1^{\gamma}\cdot\tk_1\bigl(x_1^{\beta}\cdot\tk_1(x_1^{\alpha}\cdot
f)\bigr)\Bigr)\,,
$$
by the case $k=2$ proved above.
\qed
\enddemo

{\bf Proof of Proposition 5.9:}
The assertion follows directly from Lemma 5.10, the equality
$\tk_k\circ\partial'_{\omega^{(k)}}=\tk_1^k$ and the key formula
(Theorem 5.1).
\qed

\example{Example 5.11}
$n=6,\ k=3,\ \widQ_I=\widQ_I(X_6)$\,.
$$ \aligned
\tk_3\bigl(\partial_{\omega^{(3)}}'(x_1^4x_2^1x_3^0)
&\cdot\widQ_{654321}\bigr) \\
&= \pm\tk_1\Bigl(x_1^0\cdot\tk_1\bigl(x_1^1\cdot\tk_1(x_1^4\cdot
\widQ_{654321})\bigr)\Bigr) \\
&= \pm\tk_1\bigl(x_1^0\cdot\tk_1(x_1^1\cdot\widQ_{65431})\bigr) \\
&= \pm\tk_1(x_1^0\cdot\widQ_{6431}) \\
&= \widQ_{431}\,.\\
\endaligned $$
\endexample

We give now other useful consequences of the key formula. Given two barred
permutations $w$ and $v$, we say that $u$ transforms $w$ to $v$
(or we pass from $w$ to $v$ using $u$) if $w\cdot u^{-1}=v$.

\proclaim{Proposition 5.12}
Let $I=(i_1,\ldots,i_l)\subset \rho_n$ be a strict partition.
Then
$$
\partial_{i_1-1}'\ldots\partial_1'\partial_0\bigl(\widQ_I(X_n)\bigr)=
\widQ_{(i_2,\ldots,i_l)}(X_n)\,.
$$
\endproclaim

\demo{Proof}
Suppose that $\{i_1,\ldots,i_l,j_1<j_2<\ldots<j_k\}=\{1,\ldots,n\}$\,.
Let $w,\,v,\,u$ and $\mu$ be elements of $W_n$ which transform
the barred permutations as indicated in the following
square 
$$\CD
[\ol{j_1},\ldots,\ol{j_k},\ol{i_1},\ol{i_2},\ldots,
\ol{i_l}]  
@< \mu <<  
[\ol{j_1}<\ldots<\ol{j_{p-1}}<\ol{i_1}<\ol{j_p}<\ldots
<\ol{j_k},\ol{i_2},\ldots,\ol{i_l}] \\
@VV v V   @VV w V \\
[\ol{i_1},\ol{i_2},\ldots,\ol{i_l},j_1,\ldots,j_k]
@> u >>
[\ol{i_2},\ldots,\ol{i_l},j_1<\ldots<j_{p-1}<i_1<j_p<\ldots<j_k]
\endCD $$
for some $p$\,. Note that $\mu = s_k\cdot \ldots \cdot s_{p+1}\cdot s_p$.
\medskip

We have $w=u\circ v\circ\mu$\,, hence $\partial_w'=\partial_u'\circ
\partial_v'\circ\partial_{\mu}'$\,.
Observe that
$$
v=(s_l\ldots s_1s_0)\ldots(s_{n-2}\ldots s_1s_0)
(s_{n-1}\ldots s_1s_0)\circ\omega^{(k)}\,.
$$
Hence by Proposition 5.9, we have
$$\aligned
\partial_v'\bigl(x_1^{n-j_1}x_2^{n-j_2} & \ldots x_k^{n-j_k}
\widQ_{\rho_n}(X_n)\bigr)= \\
&= (-1)^{n-j_1+n-j_2+\ldots+n-j_k}\widQ_{(i_1,\ldots,i_l)}(X_n)\,. \\
\endaligned $$
In a similar way, we infer:
$$\aligned
\partial_w'\bigl(x_1^{n-j_1} & \ldots x_p^{n-i_1}\ldots x_{k+1}^{n-j_k}
\widQ_{\rho_n}(X_n)\bigr)= \\
&= (-1)^{n-j_1+\ldots+n-i_1+\ldots+n-j_k}
\widQ_{(i_2,\ldots,i_l)}(X_n)\,.
\endaligned $$
Since $\partial_{i_1-1}'\ldots\partial_1'\partial_0=\partial_u'$\,, to prove
the assertion it suffices to show that
$$
\partial_{\mu}'\bigl(x_1^{n-j_1}\ldots x_p^{n-i_1}\ldots x_{k+1}^{n-j_k}\bigr)
=(-1)^{n-i_1}x_1^{n-j_1}\ldots x_k^{n-j_k}\,.
$$
But this is an obvious equality for {\it dominant} Schubert polynomials:
$$\aligned
\partial_k\ldots\partial_{p+1}\partial_p
& \frak S_{(n+1-j_1,\ldots,n+1-i_1,\ldots,n+1-j_k,n+1-i_2,\ldots,n+1-i_l)}= \\
& = \frak S_{(n+1-j_1,\ldots,n+1-j_k,n+1-i_1,\ldots,n+1-i_l)}\,,
\endaligned $$
where $n+1-i_1$ is placed in the $p$-th place, in the first expression.
\medskip

Thus the proposition has been proved.
\qed
\enddemo

\example{Example 5.13}
$n=9,\ I=(6,4,3,2)$\,.
The elements $w,\,v,\,u$ and $\mu$ of $W_9$ are :
$$\CD
[\ol 1,\ol 5,\ol 7,\ol 8,\ol 9,\ol 6,\ol 4,\ol 3,\ol 2]
@<\mu<<
[\ol 1,\ol 5,\ol 6,\ol 7,\ol 8,\ol 9,\ol 4,\ol 3,\ol 2] \\
@VV v V   @VV w V \\
[\ol 6,\ol 4,\ol 3,\ol 2,1,5,7,8,9]
@>u>>
[\ol 4,\ol 3,\ol 2,1,5,6,7,8,9]
\endCD $$
\medskip

We have, with $\widQ_I=\widQ_I(X_9)$,

$$\aligned
\partial_v'(x_1^8x_2^4x_3^2x_4^1x_5^0\cdot\widQ_{\rho_9})
&=\pm\widQ_{6432}\,, \\
\partial_w'(x_1^8x_2^4x_3^3x_4^2x_5^1x_6^0\cdot\widQ_{\rho_9})
&=\pm\widQ_{432} \hbox{ and} \\
\partial_5\partial_4\partial_3
(x_1^8x_2^4x_3^3x_4^2x_5^1x_6^0)
&=x_1^8x_2^4x_3^2x_4^1x_5^0\,.
\endaligned $$
\endexample
\bigskip

\proclaim{Corollary 5.14}
Let $I=(i_1,\ldots,i_k,\ldots,i_l)$ be a strict partition. Then
$$
\partial_{i_k-1}'\ldots\partial_1'\partial_0\ldots\partial_{i_1-1}'\ldots
\partial_1'\partial_0\bigl(\widQ_I(X_n)\bigr)
=\widQ_{(i_{k+1},i_{k+2},\ldots,i_l)}(X_n)\,.
$$
In particular ( invoking the notation of (12) ), we get
$$
\partial_{w_I}'\bigl(\widQ_I(X_n)\bigr)=1\,.
$$

\endproclaim

\proclaim{Corollary 5.15}
Let $I=(i_1,\ldots,i_l)$ be a strict partition.
Then, for $j>i_1$\,,
$$
\partial_{j-1}\ldots\partial_1\partial_0\bigl(\widQ_I(X_n)\bigr)=0\,.
$$
\endproclaim

Indeed, we have
$$
\partial_{j-1}\ldots\partial_{i_1}\partial_{i_1-1}\ldots\partial_1\partial_0
\bigl(\widQ_I(X_n)\bigr)=\partial_{j-1}\ldots\partial_{i_1}
\widQ_{(i_2,\ldots)}(X_n)=0\,.
$$
Observe that this implies that for any strict partition $J\ne I$ which
precedes $I$ in the reverse lexicographic order (see [Mcd1]),
$$
\partial_{w_J}\bigl(\widQ_I(X_n)\bigr)=0.
$$

\bigskip

\noindent
{\bf Remark 5.16.}
Some of the above results for "trapezoidal" partitions $I=(n,n-1,\ldots,k+1)$
were originally proved in [P-R2] by different methods.
\bigskip

%{\bf Remark 5.17.}
%Associated with the situation studied in this section is the following
%remarkable Clifford-type algebra.
%Let $W^s_k$ be the following differential operator acting on the ring
%of symmetric functions:
%$$
%W_k^s:=e_k-2e_{k+1}D_{P_1}+2e_{k+2}D_{P_2}-\ldots\ .
%$$
%Then we have $W_k^s(\widQ_I)=\widQ_{kI}$ if $kI$ is strict, and for other
%$k\in \Bbb Z$ we can use this equality as the definition of $\widQ_{kI}$\,.
%The functions symmetrical in the squares of variables are scalars w.r.t.
%$U_k^s$ and $W_k^s$. (??? check)
%These operators satisfy the following identities:
%$$
%U_k^sU_l^s+U_l^sU_k^s=0\,,\qquad W_k^sW_l^s+W_l^sW_k^s=0
%$$
%$$
%\hbox{and}\qquad\qquad U_k^sW_l^s+W_k^sU_l^s=2\delta_{kl}\,.
%$$
%Moreover, $U_k^s$ and $W_k^s$ are adjoint w.r.t. to the Hermitian 
%for which $\{\widQ_I\}$ is an orthonormal basis.
%Hence $U_k^s$ and $W_k^s$ span a Clifford algebra.
%What is the role of this algebra in the story described in this section?


\bigskip\bigskip\medskip

\head
\hfil
{\bf Appendix A} {\rm (written in collaboration with J. Ratajski):}\hfil\break
{\bf Symplectic Schubert polynomials \`a la polonaise.}
\endhead

\bigskip

Billey and Haiman [B-H] and Fomin and Kirillov [F-K]
have defined various Schubert polynomials for type $C_n$. Following
[P-R2, Appendix B], we propose
here other candidates to represent the Schubert classes in the 
variety of complete Lagrangian flags.

\bigskip

Let $(x_1,x_2\ldots)$ be a sequence of independent variables.
Let $w_0=w_0^{(n)}$ be the longest element in the Weyl 
group $W_n$ of type $C_n$\,.
Define
$$
\Cal C_{w_0}=\Cal C_{w_0}(X_n):=(-1)^{n(n-1)/2}x^{\rho_{n-1}}\widQ_{\rho_n}
(X_n)\,,
$$
and for an arbitrary $w\in W_n$\,,
$$
\Cal C_w:=\Cal C_w(X_n)=\partial_{w^{-1}w_0}'(\Cal C_{w_0})\,,
$$
the {\it symplectic Schubert polynomial} associated with $w$.

\medskip

\example{Example A.1}
Symplectic Schubert polynomials for $n=2$ are:
\medskip

$$\aligned
\Cal C_{[\overline{1},\overline{2}]} = & -x_1^3x_2-x_1^2x_2^2, \\
\Cal C_{[1,\overline{2}]}=-x_1^2x_2\qquad
& ,\qquad \Cal C_{[\overline{2},\overline{1}]}=x_1^2x_2+x_1x_2^2, \\
\Cal C_{[\overline{2},1]}=x_1x_2\qquad
& ,\qquad \Cal C_{[2,\overline{1}]}=x_2^2, \\
\Cal C_{[2,1]}=x_2 \qquad & ,\qquad \Cal C_{[\overline{1},2]}=x_1+x_2, \\
\Cal C_{[1,2]} & =1\,. 
\endaligned$$
\medskip

Their expression in terms of the $\frak S \widQ$-basis is:
\medskip

$$\aligned
\Cal C_{[\overline{1},\overline{2}]} = & -\frak S_{[2,1]}\widQ_{21} \ , \\
\Cal C_{[1,\overline{2}]}=-\frak S_{[2,1]} \widQ_2\qquad
& ,\qquad \Cal C_{[\overline{2},\overline{1}]}=\widQ_{21} \ , \\
\Cal C_{[\overline{2},1]}=\widQ_2\qquad
& ,\qquad \Cal C_{[2,\overline{1}]}=-\frak S_{[2,1]}\widQ_1 +(x_1^2+x_2^2)
+\widQ_2 \ , \\
\Cal C_{[2,1]}=-\frak S_{[2,1]} + \widQ_1 \qquad 
& ,\qquad \Cal C_{[\overline{1},2]}= \widQ_1 \ , \\
\Cal C_{[1,2]} & =1\,. 
\endaligned$$

\endexample
\medskip

\proclaim{Theorem A.2}{\rm (Stability)} Suppose that $m>n$\,.
Let $W_n\hoto W_m$ be the embedding via the first $n$ components.
Then, for any $w\in W_n$\,, the following equality holds:
$$
\Cal C_w(X_m)|_{x_{n+1}=\ldots=x_m=0}=\Cal C_w(X_n)\,.
$$
\endproclaim

\demo{Proof}
To show the assertion we must prove that the above definition is consistent
with the divided-difference recurences.
Taking into account the equality
$$
w_0^{(n)}=w_0^{(n-1)}s_{n-1}s_{n-2}\ldots s_1s_0s_1\ldots s_{n-2}s_{n-1}\,,
$$
and our sign conventions, the wanted assertion is an immediate consequence
of the equality:
$$
\partial_{n-1}'\ldots\partial_2'\partial_1'\partial_0\partial_1\partial_1
\ldots\partial_{n-1}(f_n)=f_{n-1}\,,
$$
where $f_n=x^{\rho_{n-1}}\widQ_{\rho_n}(x_n)$\,.

\noindent
We need a simple:
\enddemo

\proclaim{Lemma A.3}
For any polynomial $f=f(X_n)$\,,
$$
\partial_{n-1}\ldots\partial_1\partial_0(x_1^2x_2\ldots x_n\cdot 
f)|_{x_n=0}=0
$$
\endproclaim

Indeed, after applying $\partial_0$ we get a (possibly zero) sum of monomials
divisible by $x_1\cdot\ldots\cdot x_n$\,.
Since $x_1\cdot\ldots\cdot x_n$ is a scalar w.r.t. $\partial_{n-1}\ldots
\partial_1$\,, the final substitution $x_n=0$ gives the vanishing.

Suppose now that 
$$
\partial_1\partial_2\ldots\partial_{n-1}(x^{\rho_{n-1}})=
\sum\limits_J a_Jx^J\,,
\tag18
$$
where $a_J\in\Bbb Z$\,, so that
$$
\partial_1\partial_2\ldots\partial_{n-1}(f_n)=\Bigl(\sum a_jx^J\Bigr)
\widQ_{\rho_n}(X_n)\,.
$$
We claim that if for $J$ with $a_j\ne 0$\,, we have $j_1\ge 1$\,, then
$$
\partial_{n-1}\ldots\partial_1\partial_0\bigl(x^JQ_{\rho_n}(X_n)\bigr)\,
|\,_{x_n=0}=0\,.
$$
This follows from the lemma since $Q_{\rho_n}(X_n)=x_1\cdot\ldots\cdot x_n
Q_{\rho_{n-1}}(X_n)$\,.
We next observe:

\proclaim{Lemma A.4}
With the notation of (18) if $a_J\ne 0$ and $j_1=0$\,, then 
$J=(0,n-2,n-3,\ldots,2,1,0)$\,.
\endproclaim

Indeed, let $M$ be a monomial appearing in the decomposition of
$$
\partial_2\ldots\partial_{n-1}(x_2^{n-2}x_3^{n-3}\ldots x_{n-1})
$$
as a sum of monomials.
To get in $\partial_1(x_1^{n-1}M)$ a monomial without $x_1$\,, the monomial
$M$ must be independent of $x_2$ ($x_1x_2$ is a scalar w.r.t. $\partial_1$).
By induction we can assume $M=x_3^{n-3}\ldots x_{n-1}$\,.
But then, the unique monomial in $\partial_1(x_1^{n-1}x_3^{n-3}\ldots 
x_{n-1})$\, that does not depend on $x_1$,
is $x_2^{n-2}x_3^{n-3}\ldots x_{n-1}$, as claimed.
\medskip

Hence we must compute:
$$\aligned
\partial_{n-1}' & \ldots\partial_1'\partial_0\bigl(x_2^{n-2}\cdot x_3^{n-3}
\cdot\ldots\cdot x_{n-1}\cdot\widQ_{\rho_n}(X_n)\bigr)\,|\,_{x_n=0} \\
& = \partial_{n-1}'\ldots\partial_1'\partial_0\bigl(x_2^{n-2}\cdot\ldots\cdot
x_{n-1}\cdot x_1\cdot\ldots\cdot x_n\cdot\widQ_{\rho_{n-1}}(X_n)\bigr)\,
|\,_{x_n=0} \\
& = \partial_{n-1}'\ldots\partial_1'\partial_0\bigl(x_2^{n-2}\cdot\ldots\cdot
x_{n-1}\cdot x_1\cdot\ldots\cdot x_n\cdot\widQ_{\rho_{n-1}}(x_2,\ldots,x_n)
\bigr)\,|\,_{x_n=0} \\
\endaligned $$
by Lemma A.3. The last expression can be rewritten as
$$
\partial_{n-1}'\ldots\partial_1'\bigl(x_2^{n-1}\ldots x_{n-1}^2x_n\cdot
\widQ_{\rho_{n-1}}(x_2,\ldots,x_n)\bigr)\,|\,_{x_n=0}.
\tag 19
$$

Now we need

\proclaim{Lemma A.5}
Let $x,y$ be indeterminates and $f=f(x,y)$ a polynomial function depending on
$x,y$ and perhaps on some other variables.
Then denoting by $\partial_{xy}$ the divided difference
$f\mapsto {f(x,y)-f(y,x) \over x-y}$, we have
$$
\partial_{xy}\bigl(y\cdot f(x,y)\bigr)\,|\,_{y=0}=\partial_{xy}\bigl(y\cdot
f(0,y)\bigr)\,|\,_{y=0}\,.
$$
\endproclaim

Indeed, $y\cdot f(x,y)-y\cdot f(0,y)=xy\cdot g$ for some polynomial $g$\,.
Then 
$$
\partial_{xy}\bigl(y\cdot f(x,y)-yf(0,y)\bigr)=\partial_{xy}
(xy\cdot g)=xy\partial_{xy}(g)
$$
vanishes after the substitution $y=0$\,.
\medskip

Coming back to our expression (19), we see that before applying
$\partial_{n-1}'$ we can substitute $x_{n-1}=0$\,, so before applying 
$\partial_{n-2}'$ we can substitute $x_{n-2}=0$ etc.
In other words, in the course of computing (19) we can after applying
$\partial_1'$ substitute $x_2=0$\,, after applying $\partial_2'$
substitute $x_3=0,\ldots$\,, after applying $\partial_{n-2}'$ substitute
$x_{n-1}=0$\,, and at the end substitute $x_n=0$\,.
\medskip

Using the Leibniz rule, we get 
%(in the following $s_1=(2,1,3,\ldots,n)$ )
$$\aligned
&\partial_1'\bigl((x_2^{n-1}\ldots 
x_{n-1}^2x_n\widQ_{\rho_{n-1}}(x_2,\ldots,x_n)\bigr)\,|\,_{x_2=0} \\
& = \Bigl[\partial_1'(x_2^{n-1}\ldots x_n)\cdot s_1\widQ_{\rho_{n-1}}
(x_2,\ldots,x_n)\Bigr]\,|\,_{x_2=0} \\
& \hskip 80pt + \Bigl[x_2^{n-1}\ldots x_n\partial_1'
\bigl(\widQ_{\rho_{n-1}}(x_2,\ldots,x_n)\bigr)\Bigr]\,|\,_{x_2=0} \\
& = x_1^{n-2}x_3^{n-2}\ldots x_n\cdot\widQ_{\rho_{n-1}}(x_1,x_3,\ldots,x_n)+0.
\endaligned $$

Similarly, applying $\partial_2'$ to the obtained polynomial and substituting
$x_3=0$\,, we get
$$
x_1^{n-2}x_2^{n-3}x_4^{n-3}x_5^{n-4}\ldots x_n\widQ_{\rho_{n-1}}
(x_1,x_2,x_4,\ldots,x_n)\,.
$$

Continuing in this way, we get, after applying $\partial_{n-1}'$\,:
$$
x_1^{n-2}x_2^{n-3}\ldots x_{n-2}\widQ_{\rho_{n-1}}(x_1,\ldots,x_{n-1})
=f_{n-1}\,.
$$
This completes the proof of the stability property.
\qed
\bigskip

\proclaim{Theorem A.6}
{\rm(the maximal Grassmannian property)} Let $I=(i_1>\ldots>i_l>0)$ 
be a strict partition contained in $\rho_n$\,. Then, invoking the notation
of (12), one has the equality
$$
\Cal C_{w_I}(X_n)=\widQ_I(X_n)\,.
$$
\endproclaim

\demo{Proof}
The proof is very similar to the one of Proposition 5.12, but we give all
details for the reader's convenience.
We first explain how, in the course of the proof, we pass from $w_0$ to
$w_I$\,.
Write $k=n-l$\,.
First, by using an appropriate permutation $\sigma$\,, we pass from
$w_0=[\ol 1,\ol 2,\ldots,\ol n]$ to
$$
[\ol{j_1},\ol{j_2},\ldots,\ol{j_k},\ol{i_1},\ldots,\ol{i_l}]\,.
$$
Secondly, using $\omega^{(k)}$ we pass to
$$
[\ol{j_k},\ol{j_{k-1}},\ldots,\ol{j_1},\ol{i_1},\ldots,\ol{i_l}]\,.
$$
Thirdly, using
$$
(s_{n-k}\ldots s_1s_0)\ldots(s_{n-2}\ldots s_1s_0)(s_{n-1}\ldots s_1s_0)\,,
$$
we pass to
$$
[\ol{i_1},\ldots,\ol{i_l},j_1,j_2,\ldots,j_k]=w_I\,.
$$
Observe that after applying to $\Cal C_{w_0}(X_n)$ the divided difference
corresponding to the first permutation, we get
$$\aligned
\Cal C & _{[\ol{j_1},\ol{j_2},\ldots,\ol{j_k},\ol{i_1},\ldots,\ol{i_l}]} \\
& =(-1)^{n(n-1)/2+l(\sigma)}\frak S_{(n+1-j_1,\ldots,n+1-j_k,n+1-i_1,\ldots,
n+1-i_l)}\cdot\widQ_{\rho_n}(X_n)\,,
\endaligned $$
where $\frak S_{(n+1-j_1,\ldots,n+1-j_k,n+1-i_1,\ldots,n+1-i_l)}$ is the
monomial $x_1^{n-j_1}x_2^{n-j_2}\ldots x_k^{n-j_k}$ by [Mcd2].
The operator corresponding to the composition of the second permutation
and the third barred permutation is $\tk_1^k=\tk_k\circ
\partial'_{\omega^{(k)}}$\,, and using Proposition 5.9 we get
$$
\tk_1^k \bigl(x_1^{n-j_1}x_2^{n-j_2}\ldots x_k^{n-j_k}\widQ_{\rho_n}(X_n)
\bigr) 
=(-1)^{n-j_1+n-j_2+\ldots+n-j_k}\widQ_{(i_1,\ldots,i_l)}(X_n)\,.
$$
Since $n-j_1+\ldots+n-j_k$ is the length of the permutation passing from
$[n,n-1,\ldots,2,1]$ to 
$[j_1,j_2,\ldots,j_k,i_1,\ldots,i_l]$ and $\sigma$
passes from $[1,2,\ldots,n]$ to 
\hfill\break
$[j_1,j_2,\ldots,j_k,i_1,\ldots,i_l]$\,,
we see that the above signs fit to give us:
$$
\tk_k'\partial_{\omega^{(k)}}'\partial_{\sigma}'
\bigl((-1)^{n(n-1)/2}x^{\rho_{n-1}}
\widQ_{\rho_n}(X_n)\bigr)=\widQ_I(X_n)\,,
$$
as asserted.
\qed
\enddemo

\example{Example A.7} $n=6\,,\ I=(4,3,1)$\,. We compute, with $\widQ_I=
\widQ_I(X_6)$,
$$\aligned
\Cal C_{[\ol 1,\ol 2,\ol 3,\ol 4,\ol 5,\ol 6)]} @>\ \ \partial_{\sigma}\ \ >>\ 
& \Cal C_{[\ol 2,\ol 5,\ol 6,\ol 4,\ol 3,\ol 1]} \\
& =\pm\frak S_{(5,2,1,3,4,6)}\cdot\widQ_{\rho_6}
=\pm x_1^4x_2^1x_3^0\cdot\widQ_{\rho_6}\,.
\endaligned $$
Then to compute the image under $\tk_3\circ\partial'_{\omega^{(3)}}$ of the
last element, we follow Example 5.11 to get finally $\widQ_{(4,3,1)}$\,.
\endexample

\medskip

>From the maximal Grassmannian property it follows the following 
operator characterization of $\widetilde Q$-polynomials.

\proclaim{Corollary A.8} For any strict partition $I$, 
$w=w_I$ is the unique element
of $W$ such that $l(w)=|I|$ and $\partial_w(\widetilde Q_I(X)\ne 0$;
in fact, $\partial_{w_I}(\widetilde Q_I(X)=1$. 
\endproclaim

\newpage


We end this Appendix with the table of symplectic Schubert polynomials for
$n=3$ computed with the help of $ACE$ (see [V1,2]).

\medskip


[1, 2, 3]   : $
1
$

$[\ol 1, 2, 3]   : 
x_{1}+x_{2}+x_{3}
$

$[2, 1, 3]   : 
x_{2}+x_{3}
$

$[\ol 2, 1, 3]   : 
x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3}
$

$[2, \ol 1, 3]   : 
x_{2}^{2}+x_{2}x_{3}+x_{3}^{2}
$

$[\ol 2, \ol 1, 3]   : 
x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+x_{1}x_{2}x_{3}
+x_{1}x_{3}^{2}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2}
$

$[1, \ol 2, 3]   :
-(x_{2}+x_{3})x_{1}^{2}
$

$[\ol 1, \ol 2, 3]   : 
-(x_{1}x_{2}+x_{1}x_{3}+x_{2}^{2}+x_{3}^{2})x_{1}^{2}
$

$[1, 3, 2]   : 
x_{3}
$

$[\ol 1, 3, 2]   : 
(x_{1}+x_{2}+x_{3})x_{3}
$

$[3, 1, 2]   : 
x_{2}x_{3}
$

$[\ol 3, 1, 2]   : 
x_{1}x_{2}x_{3}
$

$[3, \ol 1, 2]   : 
(x_{2}+x_{3})x_{2}x_{3}
$

$[\ol 3, \ol 1, 2]   : 
(x_{1}+x_{2}+x_{3})x_{1}x_{2}x_{3}
$

$[1, \ol 3, 2]   : 
-x_{1}^{2}x_{2}x_{3}
$

$[\ol 1, \ol 3, 2]   : 
-(x_{1}+x_{2}+x_{3})x_{1}^{2}x_{2}x_{3}
$


$[2, 3, 1]   : 
x_{3}^{2}
$

$[\ol 2, 3, 1]   : 
(x_{1}+x_{2})x_{3}^{2}
$

$[3, 2, 1]   : 
x_{2}x_{3}^{2}
$

$[\ol 3, 2, 1]   : 
x_{1}x_{2}x_{3}^{2}
$

$[3, \ol 2, 1]   : 
x_{2}^{2}x_{3}^{2}
$
 
$[\ol 3, \ol 2, 1]   : 
(x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})x_{1}x_{2}x_{3}
$

$[2, \ol 3, 1]   :
-(x_{2}+x_{3})x_{1}^{2}x_{2}x_{3}
$

$[\ol 2, \ol 3, 1]   : 
-(x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})x_{1}^{2}x_{2}x_{3}
$

$[2, 3, \ol 1]   : 
x_{3}^{3}
$

$[\ol 2, 3, \ol 1]   : 
(x_{1}^{2}+x_{1}x_{3}+x_{2}^{2}+x_{2}x_{3})x_{3}^{2}
$

$[3, 2, \ol 1]   : 
x_{2}x_{3}^{3}
$

$[\ol 3, 2, \ol 1]   : 
-(x_{1}x_{2}-x_{3}^{2})x_{1}x_{2}x_{3}
$

$[3, \ol 2, \ol 1]   : 
(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2}
)x_{2}x_{3}
$

$[\ol 3, \ol 2, \ol 1]   : 
(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+x_{1}x_{2}x_{3
}+x_{1}x_{3}^{2}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2})x_{1}x_{2}x_{
3}
$

$[2, \ol 3, \ol 1]   : 
-(x_{2}^{2}+x_{2}x_{3}+x_{3}^{2})x_{1}^{2}x_{2}x_{3}
$

$[\ol 2, \ol 3, \ol 1]   : 
-(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+x_{1}x_{2}x_{
3}+x_{1}x_{3}^{2}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2})x_{1}^{2}x_{
2}x_{3}
$

$[1, 3, \ol 2]   : 
-(x_{1}^{2}+x_{2}^{2})x_{3}^{2}
$

$[\ol 1, 3, \ol 2]   : 
-(x_{1}^{3}x_{3}+2x_{1}^{2}x_{2}^{2}+x_{1}^{2}x_{2}x_{3}+x_{
1}^{2}x_{3}^{2}+x_{1}x_{2}^{2}x_{3}+x_{2}^{3}x_{3}+x_{2}^{2}
x_{3}^{2})x_{3}
$

$[3, 1, \ol 2]   : 
-(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{2}^{2}x_{3})x_{2}x_{3}
$

$[\ol 3, 1, \ol 2]   : 
-(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+x_{2}^{2}x_{3
})x_{1}x_{2}x_{3}
$

$[3, \ol 1, \ol 2]   : 
-(2x_{1}^{2}x_{2}^{2}+2x_{1}^{2}x_{2}x_{3}+x_{1}^{2}x_{3}^{2
}+x_{2}^{3}x_{3}+x_{2}^{2}x_{3}^{2})x_{2}x_{3}
$

$[\ol 3, \ol 1, \ol 2]   : 
-(x_{1}+x_{2})(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+
x_{1}x_{2}x_{3}+x_{1}x_{3}^{2}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2}
)x_{1}x_{2}x_{3}
$

$[1, \ol 3, \ol 2]   : 
(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2}
)x_{1}^{2}x_{2}x_{3}
$

$[\ol 1, \ol 3, \ol 2]   : 
(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+x_{1}x_{2}x_{3
}+x_{1}x_{3}^{2}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2})x_{1}^{3}x_{2
}x_{3}
$

$[1, 2, \ol 3]   : 
x_{1}^{2}x_{2}^{2}x_{3}
$

$[\ol 1, 2, \ol 3]   : 
(x_{1}+x_{2}+x_{3})x_{1}^{2}x_{2}^{2}x_{3}
$

$[2, 1, \ol 3]   : 
x_{1}^{2}x_{2}^{3}x_{3}
$

$[\ol 2, 1, \ol 3]   : 
(x_{1}x_{2}-x_{3}^{2})x_{1}^{2}x_{2}^{2}x_{3}
$

$[2, \ol 1, \ol 3]   : 
(x_{2}^{2}+x_{2}x_{3}+x_{3}^{2})x_{1}^{2}x_{2}^{2}x_{3}
$

$[\ol 2, \ol 1, \ol 3]   : 
(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{1}x_{2}^{2}+x_{1}x_{2}x_{3
}+x_{1}x_{3}^{2}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2})x_{1}^{2}x_{2
}^{2}x_{3}
$

$[1, \ol 2, \ol 3]   : 
-(x_{1}^{2}x_{2}+x_{1}^{2}x_{3}+x_{2}^{2}x_{3}+x_{2}x_{3}^{2
})x_{1}^{2}x_{2}^{2}x_{3}
$

$[\ol 1, \ol 2, \ol 3]   : 
-(x_{1}^{2}x_{2}^{2}x_{3}^{2}+(x_{1}+x_{2})x_{1}^{2}x_{2}^{2
}x_{3}+(x_{1}^{2}+x_{2}^{2})x_{1}x_{2}x_{3}^{2}+(x_{1}^{2}x_{
2}+x_{1}x_{2}^{2})x_{3}^{3})x_{1}^{2}x_{2}
$


\bigskip
\noindent
The tables expressing the $\Cal C_w$'s in the basis
$\{\frak S_{\mu}\cdot \widQ_I \}$ can be obtained from the authors upon
the request.
\bigskip

It is natural, in spirit of the present paper, to define the 
{\it double} symplectic Schubert polynomial associated with $w\in W_n$
as the following expression
$$
\partial'_{w^{-1}w_0} \Bigl((-1)^{n(n-1)/2} \widQ(X,Y)
\prod_{i+j\le n}(x_i+y_j)\Bigr),
$$
where $Y=(y_1,\ldots,y_n)$ is a second sequence of independent variables.
The properties od such double polynomials will be discussed
in the next version of this paper.

\bigskip\bigskip\medskip


\head
{\bf Appendix B: Three geometric applications}
\endhead

\bigskip

\noindent {\bf 1.} \ In Corollary A.8 we have obtained a purely
algebraic proof (based on the key formula) of the following result.
For any strict partition $J$ with $|J|=|I|$\,, we have $\partial_{w_J}
(\widQ_I)\ne 0$ only if $I=J$\,, and, in this case, we get 
$\partial_{w_I}'(\widQ_I)=1$\,.
Applying this to the characteristic map
$$
c:\Cal S\Cal P(X) \loto A^*(G)
$$
given for a homogeneous polynomial $f$ by 
$$
c(f)=\sum\limits_{I,|I|=deg f} \partial_{w_I}'(f)\sigma(I)
$$ 
(the notation as in Section 1), this implies
$$
c(\widQ_I)=\sigma(I)
\tag20
$$
which is the Giambelli-type formula for the Lagrangian Grassmannian, originally
obtained in [P, Sect.6] by a different method.
Note that this purely algebraic operator proof of (20) gives us,
after invoking a Pieri-type formula for Hall-Littlewood functions from [Mo],
a Pieri-type formula for Lagrangian Schubert classes obtained
by other methods in [H-B] and [P-R1].

\bigskip
\noindent {\bf 2.} \ The key formula admits the following interpretation in
terms of Gysin maps.
Let $X$ be a scheme and $V$ a rank $2n$ vector bundle on $X$ equipped with 
a nondegenerate symplectic form.
Let $V_n$ be a rank $n$ isotropic subbundle of $V$\,.
By $p:\Cal F\to X$ we denote the flag bundle parametrizing pairs $A\subset B$
of subbundles of $V$ such that rank $A=n-1$\,, rank $B=n$\,, $A\subset V_n$
and $B$ is Lagrangian.
Let $C\subset D$ be the tautological bundles of respective ranks $n-1$ and
$n$\,, on $\Cal F$\,.
Set $S=D/C$\,.

Suppose  $n\ge k>0$. Let $IkJ$ be a strict partition and 
$H$ a strict partition not containing $k$\,.
Then the key formula says:
$$
p_*\bigl(c_1(S\hak)^{n-k}\cdot\widQ_{IkJ}(D\hak)\bigr)
=(-1)^{l(I)}\widQ_{IJ}(V_n\hak) 
\tag21
$$
and
$$
p_*\bigl(c_1(S\hak)^{n-k}\cdot\widQ_H(D\hak)\bigr)
=0\,. 
$$
For $k=n$\,, this result agrees with Proposition 3.1 in [P-R2], which was
proved using geometric arguments.
Does it exist a geometric proof for the key formula in general?
One more comment may be here in order.
Note that the arguments in both sides of (21) are different
whereas in the key formula, they are the same.
As a matter of fact, in the above translation of the key formula, we have used
a result from [P-R2, Sect.8] allowing us to make an appropriate exchange of
the sequences of the Chern roots of the bundles involved under the action
of $p_*$.


\bigskip
\noindent {\bf 3.} \  The results of [P-R2] and the 
present paper combined with some results of [F1,2] allow us to
generalize the formulas for the Schubert classes in Lagrangian
Grassmannian bundles from [P-R2] to the case of arbitrary Lagrangian
flag bundles and flagged symplectic degeneracy loci. Here, for simplicity,
we describe just the difference between our formulas and the ones obtained
by Fulton in [F1,2]. 
Let $\Cal X_w$ \ ($w\in W_n$) \ be the degeneracy locus 
defined as in [F1, p.243] with the help of a nondegenerate symplectic form
on a rank $2n$ vector bundle $V$ which is 
equipped with two complete filtrations:
$$
E_1 \subset E_2 \subset \ldots \subset E_n \subset V,
$$
$$ 
F_1 \subset F_2 \subset \ldots \subset F_n \subset V
$$
by isotropic subbundles. Then, it follows from  Proposition 2.4 of the 
present paper (or its geometric counterpart in [P-R2]), 
and the results of [F1,2], that the fundamental 
class of $\Cal X_w$
is evaluated by the expression given in [F1, Theorem ($C_n$) on p.243], where
instead of $F(X,Y)$ (denoted by $\triangle$ in loc.cit.) 
one uses $\widQ(X,Y)$. More precisely, if the
ambient variety is Cohen-Macaulay and $\Cal X_w$ is of pure codimension
$n^2-l(w)$, then
$$
[\Cal X_w]=
\partial_{w^{-1}} \bigl(\prod_{i+j\le n}(x_i-y_j) \widQ(X,Y)\bigr)\cap [X],
$$
where we specialize $x_i:=-c_1(E_i/E_{i-1})$ and $y_i:=
-c_1(F_{n+1-i}/F_{n-i})$
for $i=1,\ldots,n$.
\medskip 

We point out this difference between the approach of [F1,2] and ours
because it appears that for numerical computations 
(e.g. with the help of $ACE$ of [V1,2]) the reproducing
kernel $\widQ(X,Y)$ is more economical than $F(X,Y)$. 
   
\newpage



\widestnumber\key{D-J-K-M1}
\Refs


\bigskip

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\smallskip

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\smallskip

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\smallskip

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\by      Ch. Carr\'e, J.-Y. Thibon
\paper   Plethysm and vertex operators 
\jour    Adv. Appl. Math. \vol 13 \yr 1992 \pages 390--403
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\smallskip

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\by      E. Date, M. Kashiwara, T. Miwa 
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\jour    Proc. Japan Acad., Ser. \vol A 57 \yr 1981 \pages 387--392
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\smallskip

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\by      E. Date, M. Jimbo, M. Kashiwara, T. Miwa 
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\jour    Physica  \vol D 4 \yr 1982 \pages 343--365
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\smallskip

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\jour    (M. Jimbo, T. Miwa eds.), World Scientific \yr 1983 \pages 41-119
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\smallskip

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\smallskip

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\smallskip

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\by     J. D\'esarm\'enien, B. Leclerc, J.-Y. Thibon
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in representation theory
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\smallskip


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\by     S. Fomin, A.N. Kirillov
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\smallskip

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\by      W. Fulton
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\jour    Proceedings of the Hirzebruch 65 Conference on Algebraic
Geometry, IMCP \vol 9 \yr 1996 \pages 241--262
\endref

\smallskip

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\by      W. Fulton
\paper   Determinantal Formulas for Orthogonal and Symplectic 
Degeneracy Loci
\jour    J. Diff. Geometry \vol 43 \yr 1996 \pages 276-290 
\endref

\smallskip

\ref\key G
\by     W. Graham
\paper  The class of the diagonal of flag bundles
\jour   preprint, IAS Princeton \yr 1996
\endref

\smallskip



\ref\key H-B
\by     H. Hiller, B. Boe
\paper  Pieri formula for $SO_{2n+1}/U_n$ and $Sp_n/U_n$
\jour   Adv. in Math. \vol 62 \yr 1986 \pages 49--67
\endref

\smallskip

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\by     A. Lascoux, B. Leclerc, J.-Y. Thibon
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\smallskip

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\by     A. Lascoux, B. Leclerc, J.-Y. Thibon
\paper  Fonctions de Hall-Littlewood et polyn\^omes de Kostka-Foulkes
aux racines de l'unit\'e
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\smallskip

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\by    A. Lascoux, M.P. Sch\"utzenberger
\paper Alg\`ebre des diff\'erences divis\'ees
\jour  Discrete Math. \vol 99 \yr 1992 \pages 165--179
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\smallskip

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\by    D.E. Littlewood
\paper On certain symmetric functions
\jour  Proc. London Math. Soc. \vol (3) \yr 1961 \pages 485-498
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\smallskip


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\by    D.E. Littlewood
\paper Modular representations of symmetric groups
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\smallskip

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\publ   Oxford Univ. Press \yr 1995
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\smallskip

\ref\key Mcd2
\by      I.G. Macdonald
\book    Notes on Schubert polynomials
\publ    Publ. LACIM-UQUAM, Montr\'eal  \yr 1991 
\endref

\smallskip

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\by     A.O. Morris
\paper  A note on the multiplication of Hall-Littlewood polynomials
\jour   Journal London Math. Soc. \vol 39 \yr 1964 \pages 481--488
\endref
    
\smallskip

\ref\key P
\by     P. Pragacz
\paper  Algebro-geometric applications of Schur $S$- and $Q$-polynomials
\jour   S\'eminare d'Alg\`ebre Dubreil--Malliavin 1989--1990
        (M.-P. Malliavin - ed.),
	Springer Lecture Notes in Math. \vol {1478}  
	\yr 1991 \pages 130--191
\endref

\smallskip

\ref\key P-R1
\by      P. Pragacz, J. Ratajski
\paper   Pieri type formula for isotropic Grassmannians; the operator 
approach
\jour   Manuscripta Math. \vol 79 \yr 1993 \pages 127--151 
\endref

\smallskip

\ref\key P-R2
\by      P. Pragacz, J. Ratajski
\paper   Formulas for Lagrangian and orthogonal degeneracy loci;
$\widetilde Q$-polynomial approach 
\jour    alg-geom /9602019, to appear in Compositio Math., June 1997
\endref

\smallskip

\ref\key S
\by      I. Schur
\paper   \"Uber die Darstellung der symmetrischen und der alternierenden
Gruppe durch gebrochene lineare Substitutionen
\jour    J. reine angew. Math. \vol 139 \yr 1911 \pages 155-250 
\endref

\smallskip

\ref\key V1
\by     S. Veigneau 
\paper  ACE, an Algebraic Combinatorics Environment for the computer
algebra system MAPLE: User's reference manual
\jour   Version 2.0 , http://phalanstere.univ-mlv.fr/$\sim$sv/ACE,
IGM \vol 97-05 \yr 1996
\endref

\smallskip

\ref\key V2
\by     S. Veigneau 
\paper  Calcul Symbolique et Calcul Distribu\'e en Combinatoire 
Alg\'ebrique
\jour   Universit\'e de Marne-la-Vall\'ee , Thesis, IGM \vol 96-39 \yr 1996
\endref




\endRefs

{\eightrm
\bigskip\bigskip\bigskip
\noindent
A.L. : Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee,
2 rue de la Butte-Verte, 93166 Noisy-le-Grand cedex, France
\bigskip\medskip
\noindent
P.P. : Mathematical Institute of Polish Academy of Sciences, Chopina 12,
87-100 Toru\'n, Poland
\bigskip\medskip
\noindent
J.R. : Mathematical Institute of Polish Academy of Sciences, \'Sniadeckich 8,
00-950 Warszawa, Poland}



\enddocument
\bye



