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\headline={\normal   \normal  $m$-Quasi-Invariants \hfill $\ess\ess\ess\ess\ess$  $\ess\ess\ess$\today 
$\ess\ess\ess$ \folio } \footline={\hfil}


\def \sij {s_{ij}}
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\def \QSS {{\cal QI}}
\def \SYM {{\cal SYM}}
\def \orb {{[a\, ] }}
\def   \CJ {{\cal J}}
\def \ida {\CJ_{\orb}\hskip -.05in(m)} 
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\sa 
\centerline{\bol Some old methods}

\centerline{\bol in the }
\centerline{\bol   Theory of $\bf m$-Quasi-Invariants}

 \centerline{  by} 

 \centerline{\bol  A.M. Garsia and N. Wallach }
\sap 

{\narrower\smallskip 
 \noindent{ \binerm ABSTRACT} 

We  show here that some representation theoretical methods
developed in the early nineties for the study of the Garsia-Haiman modules
may successfully be used in the theory of $m$-Quasi-Invariants.
Using the non-degeneracy of the fundamental bilinear  $m$-form we show
here that the Cohen Macauliness of the ring ${\cal QI}_m$ of $m$-Quasi-Invariants
may be derived with the greatest of ease. In fact, in addition,
this approach directly yields that the quotient of ${\cal QI}_m$ by the ideal generated by
the elementary symmetric functions affords a graded version of the
left regular representation. These results had been previously
obtained by combining work of Etingof-Ginzburg [], Feigin-Veselov []
and  Felder-Veselov []. Our presentation here focuses on the $S_n$ case and
it is completely self-contained except in what concerns the 
 bilinear  $m$-form.  These methods are also applicable in the case of 
general reflection groups.      
  
\smallskip}

\def \SR {{\cal SR}}


\noindent{\bol Introduction}
\sa

\noindent{\bol 1. Preliminaries}

In this section we recall some definitions and state some basic results about $m$-differential operators
and the corresponding bilinear $m$-form. This will be the only portion of this writing that is not self contained.
This material constitutes the  background  necessary to work with $m$-quasi-invariants. Up to quite recently 
proofs had to be found scattered throughout two decades of litterature. For this reason we found it
necessary to combine all this background in a single set of lecture notes [] and make it available to researchers
in the field. The reader will find in [] the proofs of every result stated here.
\sas

To begin let us call $\SR_n(x)$ the ring of special rational functions generated by the variables $\xon \,$  and all the
fractions ${1\over x_i-x_j}$. In symbols
$$
\SR_n(x)\ses \Q\Big[\xon ;\textstyle{1\over x_i-x_j  }\, ;\, 1\le i<j\le n\, \Big]
$$
This ring has a natural grading which decomposes it into  homogeneous subspaces yielding the direct sum decomposition
$$
\SR_n(x )\ses \bigoplus_{r=-\infty}^{+\infty}  \CH_{r }\Big(\SR_n(x )\Big)
$$
Where $\CH_{r }\big(\SR_n(x )\big)$ is spanned by rational function of the form 
$$
f(x )\ses 
{x_1^{p_1}  x_2^{p_2}\cdots   x_n^{p_n}   \over \prod_{1\le i<j\le n}(x_i-x_j)^{q_{ij}}} 
$$
with
$$
\sum_{i=1}^n p_i\, - \hskip -.14in\sum_{1\le i<j\le n}q_{ij}= r
 \, .
$$
We call the elements of $\CH_{r }\big(\SR_n(x )\big)$   homogeneous of  degree $r$. 
\sas

To be brief, $S_n$-$m$-quasi-invariants naturally arise in the search of differential operators which commute 
with the deformed laplacian  
$$
L_m\ses \DD_2\sms 2m\sum_{1\le i<j \le n}{1\over x_i-x_j}(\del_{x_i}-\del_{x_j})\, .
\eqno 1.1
$$
It was shown by Chalykh and Veselov  [], [] that for each $q\in \QI$, homogeneous of degree $d$, there is a unique operator $\ggg_q$ which commutes with
$L_m$ of the form
$$
\ggg_q\ses \ggg_q(x,\del_x)\ses q(\del_x)\sps \sum_{|p|<d} { a_p(x)} \,\, \del_x^p
\eqno 1.2
$$
where 
$$
q(\del_x)=q(\del_{x_1},\del_{x_2},\ldots ,\del_{x_n}) \scs\ess\ess\ess \del_x^p=\del_{x_1}^{p_1}\del_{x2}^{p_2}\cdots \del_{x_n}^{p_n}\, ,
$$
$$
|p|=p_1+p_2+\cdots +p_n
$$
and
$$
a_p(x)\in \CH_{|p|-d}\big( \SR_n(x)\big)\, .
\eqno 1.3
$$
Now it can be shown that the  commutativity with $L_m$ and the special form in 1.2 yields that $\ggg_q$ 
has the explicit expression due to Berest []:

\def \uq {{\underline q}}
\def \up {{\underline p}}
$$
\ggg_q\ses {1\over2^dd!}\sum_{r=0}^d{\textstyle {d\choose r}}(-1)^rL_m^{d-r}\uq(x)L_m^r
\eqno 1.4
$$
where $\uq(x)$ denotes the operator ``{\ita multiplication by $q(x)$}''. 
\sas

Note that if $\CA$ is an algebra of operators and for $U,V\in\CA$ we write
$$
D_UV\ses [U,V]\ses UV-VU
$$
then 1.4 may simply be written in the form
$$
\ggg_q\ses {1\over2^dd!}D_{L_m}^d\, \uq(x)\, .
\eqno 1.5
$$
It follows from all this that the linear extension of the map $q\,  \RA\,  \gg_q$ is an algebra
isomorphism of $\QI$ onto the algebra of operators that commute with $L_m$.
We should  point out that the multiplicativity of this map is a simpe consequence of the
commutattivity with $L_m$ and
the Leibnitz formula
$$
D_W^d U\, V\ses \sum_{r=0}^d {\textstyle {d\choose r}}D_W^rU\,\, D_W^{d-r}V\, .
\eqno 1.6
$$
To  see this, note that if $p,q\in\QI$ are respectively homogeneous of degrees $a$ and $b$ with $a+b=d$
then 1.6 gives
$$
\eqalign{
\ggg_{pq}
 \ses
 {1\over2^dd!}D^d_{L_m}\up(x)\uq(x)
&\ses \sum_{r=0}^d {\textstyle {d\choose r}}D_{L_m}^r\up(x)\,\, D_{L_m}^{d-r}\uq(x)
\cr
&\ses {{d\choose a}\over2^dd!}\ssp  D_{L_m}^a\up(x)\ssp  D_{L_m}^b\uq(x)\ses \ggg_p\ggg_q\, .
\cr
}\eqno 1.7
$$
where the third equal sign is due to the fact that since
$$
D_{L_m}^a\up(x)\ses 2^aa! \ggg_p \ess\ess \hbox{and}\ess\ess D_{L_m}^b\uq(x)\ses 2^aa! \ggg_q
$$
the commutativity of $\ggg_p$ and $\ggg_q$ with $L_m$ gives
$$
D_{L_m}^{a+1}\up(x)\ses 0\ess\ess \hbox{and}\ess\ess D_{L_m}^{b+1}\uq(x)\ses 0\, .
$$
The operator $L_m$ is truly remarkable. In fact, not only it acts on $\QI$ 
but it has the additional property that for two polynomials $P,Q$ we have
$$
L_mP=Q \ess\ess \hbox{with }\ess Q\in \QI \ess\ess  \Longleftrightarrow\ess\ess P\in \QI
\eqno 1.8
$$
Since $\QI$ is an algebra it immediately follows from 1.4 that all the operators $\ggg_q$ act on $\QI$. 
We thus see that if $p,q\in\QI$ then $\ggg_p\, q$ is in particular a polynomial and it makes sense to
set
$$
\LL p\scs q\RR_m\ses  \ggg_p\, q\, \big|_{x=0}\, .
\eqno 1.9
$$
The linear extension of this formula to $\QI\otimes  \QI$ yields the bilinear $m$-form we 
mentioned in the introduction. All of this has a completely analogous conterpart for every
reflection group and the non degeneracy of this form in this generality is a highly non-trivial result
due to Opdam. Nevertheless we should mention that in the case of $S_n$ we show in []
that the non degeneracy can be deduced from the classical evaluation of the Metha integral.
\sas

The following basic properties of the operators $\ggg_p$ and bilinear form will play a crucial role in the sequel.
\sas

\noindent{\bol Proposition 1.1}

{\ita For any $\sig\in S_n$ and any $p\in \QI$ we have
$$
\sig \,\ggg_p\sig^{-1}\ses \ggg_{\sig p}
\eqno 1.10
$$
and for any $p,q\in \QI$ it follows that  
$$
\LL \sig p\scs q\RR_m\ses \LL  p\scs \sig^{-1}q\RR_m
\eqno 1.11
$$}    
\noindent{\bol Proof}
 
It is easily seen from the definition in 1.1 that if 
$$
\sig L_m \sig^{-1}\ses L_m
\eqno 1.12
$$
Keeping this in mind, note that we need only verify 1.10 and 1.11 when $p$ and $q$ are homogeneous. This given,
when $p$ is homogewneous of degree $d$, the identity in 1.4, with $q$ replaced by $\sig p$, gives
$$
\eqalign{
\ggg_{\sig p}
&\ses {1\over2^dd!}\sum_{r=0}^d{\textstyle {d\choose r}}(-1)^rL_m^{d-r}{\underline \sig p}(x)L_m^r 
\cr
\hbox{(by 1.12)}
&\ses {1\over2^dd!}\sum_{r=0}^d{\textstyle {d\choose r}}(-1)^r\sig L_m^{d-r}\sig^{-1}{\underline \sig p}(x)\sig L_m^r \sig^{-1}
\cr
&\ses \sig {1\over2^dd!}\sum_{r=0}^d{\textstyle {d\choose r}}(-1)^r  L_m^{d-r} {\underline   p}(x)  L_m^r \sig^{-1}\ses \sig \ggg_p \sig^{-1}
\cr}
$$
This proves 1.10. Now using 1.10 the definition in 1.19 gives
$$
\LL \ggg_{\sig p}\scs q\RR_m\ses \sig \ggg_p\sig^{-1}q\, \big|_{x=0}\ses \ggg_p\sig^{-1}q\, \big|_{x=0}\ses \LL \ggg_{  p}\scs \sig^{-1}q\RR_m\, .
$$ 
This completes our proof.
\sa

\noindent{\bol 2. The orbit Ring}

We begin by selecting a point  
$$
a=(a_1,a_2,\ldots ,a_n )
$$ 
with distinct components and denote by $[a]_{S_n}$, (or briefly by $[a]$) the $S_n$ orbit of $a$. In symbols
$$
\orb\ses \{ a\sig\, :\, \sig\in S_n\}
$$
Next we construct a polynomial $\phi_a(x)$ such that
$$
\phi_a(x)\ses
\cases{
1 & if $x=a$\cr\cr
0 & if $x=b$ with $b\in \orb$ and $b\ne a$
\cr}
\eqno 2.1
$$
To get the best results it is important that we are as economical as possible
with the degree of $\phi_a(x)$. To this end note that the polynomial
$$
(x_1-a_2)(x_1-a_3)\cdots (x_1-a_n)
$$
vanishes at all points $a\sig=(a_{\sig_1}a_{\sig_2},\ldots ,a_{\sig_n})$ where $\sig_1\ne 1$.
It is clear that once we place this factor in our $\phi_a(x)$, the only points of the orbit
we need to worry about are those where $\sig_1=1$. But on those we can use the same trick 
to force $\sig_2=2$ by placing in $\phi_a$ the also the factor
$$
(x_2-a_3)(x_2-a_4)\cdots (x_2-a_n)
$$
Proceeding in this manner we can easily see that the polynomial
$$
\phi_a(x)=\prod_{i=1}^{n-1} \prod_{j=i+1}^n{(x_i-a_j)\over(a_i-a_j) }
$$
will satisfy the requirement in (1). We should note that we have
$$
degree(\phi_a(x))\ses \textstyle{n\choose 2}
$$
It can be shown that this is the least degree needed to satisfy 2.2.


\sa

Now for each $b\in \orb$ with $b=a\sig$ we set
$$
\Phi_b(x)\ses \phi_a(x\sig^{-1})\scs\ess\ess\ess  \eee(b)\ses sign(\sig)\, .
\eqno  2.2
$$
Note that from (1) it immediately follows that we have
$$
\phi_b (x)=
\cases{
1 & if $x =a\sig$\cr\cr
0 & if $x =b\sig$ with $b\in \orb$ and $b\sig\ne a\sig$
\cr}
$$
this may be rewritten as
$$
\phi_b (x)=\cases{
1 & if $x=b$\cr\cr
0 & if $x=b'$ with $b'\in \orb$ and $b'\ne b$
\cr}
\eqno  2.3
$$
Next we let $\ida  $ denote the ideal of $m$-quasi-invariants that vanish on $\orb$.
In symbols
$$
\ida \ses\big\{ P(x)\in \QI \, :\, P(b)=0\ess\forall \ess b\in \orb\big\}
\eqno  2.4
$$
This given,  it follows that 

\sas

\noindent{\bol Proposition 2.1}

{\ita The quotient ring
$$
\BR_{\orb}\ses \QI/\ida
\eqno  2.5
$$
has dimension $n!$ and affords the left regular representation of $S_n$}

\noindent{\bol Proof}

Let $\Pi(x)$ denote the Vandermonde determinant in $\xon$, and set
$$
\psi_b(x)\ses \phi_b(x) \textstyle {\Pi(x)^{2m}\over \Pi(a)^{2m} }\, .
$$
Note that since $\Pi(x)^{2m}$ is $S_n$-invariant we derive from 2.3 that
$$
\psi_b (x)=\cases{
1 & if $x=b$\cr\cr
0 & if $x=b'$ with $b'\in \orb$ and $b'\ne b$
\cr}
\eqno  2.6
$$
moreover we also have
$$
\psi_b (x)\in \QI\ess\ess\ess \forall\ess\ess b\in \orb\, .
\eqno  2.7
$$
Now note that for any $P\in\QI$ 2.6  gives
$$
P(x)\, - \sum_{b\in \orb} P(b)\, \psi_b(x)\in \ida \, .
$$
It will be convenient to express this relation by writing
$$
P(x)\, \cong_{[a]}\,  \sum_{b\in \orb}  P(b)\, \psi_b(x)\,  .
\eqno  2.8
$$
Thus the collection
$$
\big\{\psi_b(x)\big\}_{b\in \orb}
\eqno  2.9
$$
is a basis for the quotient $\BR_{\orb}$. 

\def \congo {\ssp \cong_{\orb} \ssp }

Since the ideal $\CJ_{\orb }$ is $S_n$-invariant it immediately follows that
$S_n$ acts on $\BR_{\orb}$. Thus to complete our proof we only need to compute the  
character of this action. To this end note that for each $\sig\,  \&\,  \bb \in S_n$ 
from 2.8 we derive that
$$
\eqalign{
\sig\psi_{a\bb }(x)
 \ses 
\psi_a(x\sig\bb^{-1})
&\congo    \sum_{\aa\in S_n} \psi_a(a\aa\sig\bb^{-1} )\psi_{a\aa}(x)
\cr
&
\congo    \sum_{\aa\in S_n} \chi\big (\aa\sig\bb^{-1} =id\big )\psi_{a\aa}(x)
\cr
&
\congo    \sum_{\aa\in S_n} \chi\big (\aa\sig  =\bb \big )\psi_{a\aa}(x)
}
$$ 
Thus the action of $S_n$ on the basis $\big\{\psi_b(x)\big\}_{b\in \orb}$ is given by the matrix $A(\sig)=\|\chi\big (\aa\sig  =\bb \big )\|_{\aa,\bb\in S_n}$. 
It follows that the character of the $S_n$ action on  $\BR_{\orb}$ is given by
$$
trace\, A(\sig)\ses \sum_{\aa\in S_n} \chi\big (\aa\sig  =\aa \big )\ses \cases {n! & if $\sig=id$\cr\cr
0 & if $\sig\neq id$}
$$
and this is the character of the left regular representation of $S_n$ precisely as asserted.
\sa

\noindent{\bol Remark 2.1}

The ring $\BR_{\orb}$ is not graded but it has a filtration given by the subspaces 
$$
\CH_{\le k}(\BR_{\orb})\ses \CL_\orb\big [ P \, : \,  P\in \CH_{\le k}\big(\QI\big) \big]
\eqno  2.10
$$
where ``$\CL_\orb$'' denotes ``{\ita Linear Span}'' modulo $\CJ_\orb$ and $\CH_{\le k}\big(\QI\big)$ is the subspace of
$\QI$ spanned by its elements of degree $\le k$. Now note that since by construction
we have
$$
degree \, \psi_b(x)\ses (2m+1) {\textstyle { n\choose 2}}
\eqno  2.11
$$
it immediately follows from the expansion in 2.8 that
$$
\BR_{\orb}\ses \CH_{\le d_{m,n}}\big (\BR_{\orb}\big )
\eqno  2.12
$$
where for convenience we have set
$$
 d_{m,n}\ses (2m+1) {\textstyle { n\choose 2}}
\eqno  2.13
$$

\def \grJ {{\bf gr\, \CJ_{ \orb}}}
\def \Ra {{ \BR_\orb}}
\def \grRa  {{\bf gr \BR_\orb}} 
\vbox{

\noindent{\bol 3. The graded version of $\BR_\orb$ }
\sas

For each polynomial $P$ we shall here and after denote by $h(P)$ the homogeneous component of
highest degree in $P$. This given, we let $\grJ$ be the ideal in $\QI$ generated by the highest degree
components of elements of $\CJ_\orb$ in symbols
$$
\grJ\ses \big(h(P)\, :\, P\in \CJ_\orb\, \big)_{\QI}\, .
\eqno 3.1
$$
}

\noindent
This brings us to define the ``{\ita graded version }'' of $\Ra$ as the quotient
$$
\grRa \ses \QI/\grJ \,  .
\eqno 3.2
$$
Since $\grJ$ is generated by homogeneous polynomials, $\grRa$ is necessarily a graded ring.  Note further that
if $P$ is any homogeneous polynomial of degree $>d_{m,n}$ then the fact that
$$
P(x)\sms \sum_{b\in \orb}P(b)\, \psi_b(x) \in J_\orb
$$
immediately implies that $P\in \grJ$. Thus it follows that $\grRa$ has the direct sum decomposition
$$
\grRa\ses \bigoplus_{k=0}^{ d_{m,n}}\CH_{=k}(\grRa)
\eqno 3.3
$$
where $\CH_{=k}(\grRa)$ denotes the subspace of $ \grRa $ spanned by its homogeneous elements of \hbox{ degree $k$}. 
It will be convenient to choose once and for all a  basis $\CA^{=k} $ for each subspace
$\CH_{=k}(\grRa)$.   
 This given, we have the following useful fact
\sas

\noindent{\bol Proposition 3.1}

{\ita The collection 
$$
\CB^{\le k}\ses \CA^{=0}\sps \CA^{=1}\sps \cdots \sps \CA^{=k}
$$
is also a basis for the subspace  $\CH_{\le k}(\Ra)$. In particular $\CB_{\le d_{m,n}} $ is
a basis of $\Ra$} 

\noindent{\bol Proof}

Clearly the result holds true for $k=0$ since $\CA_{=0}$ reduces to the single constant $\bf 1$.
So we may proceed by induction on $k$. Let us then assume that each $P\in \CH_{\le k-1}(\Ra)$ 
has an expansion in terms of $\CB^{\le k-1}$. This given, note that any $P\in \CH_{\le k}(\QI)$,
viewed as a representative of an element of $\grRa$,  may be expanded in terms of the basis $\CB^{\le k}$.
In other words we will have constants $c_\phi$ such that
$$
P\ses \sum_{\phi\in \CB^{\le k}}c_\phi \, \phi\sps R
\eqno 3.4
$$
for a suitable $R\in\grJ$. Now the definition of $\grJ $ yields that there must be elements $f_i\in J_\orb$ 
and $A_i\in \QI$ giving
$$
R\ses \sum_{i }A_i\,  h(f_i)
\eqno 3.5
$$
Since $R$ is necessarily of degree $\le k$ we see that there is no loss in assuming that we have
$$
degree \, A_i\, \le\,  k- degree \, h(f_i)\,
\eqno 3.6
$$
Using 3.5 we may rewrite 3.4 in the form
$$
P\ses \sum_{\phi\in \CB^{\le k}}c_\phi \, \phi\sps  \sum_{i }A_i\,  f_i\sms  \sum_{i }A_i\,  (f_i-h(f_i))\, .
\eqno 3.7
$$
and this implies that 
$$
P\sms \sum_{\phi\in \CB^{\le k}}c_\phi \,\phi\ssp \cong_{\orb}\sms  \sum_{i }A_i\,  (f_i-h(f_i))\, .
\eqno 3.8
$$
But now 3.6 and the definition of $h(f_i)$ yield that
$$
degree \, \sum_{i } A_i\,  (f_i-h(f_i))\ssp \le\ssp k-1 
\eqno 3.9
$$
Thus our inductive hypothesis yields that we have constants $d_\phi$ giving
$$
 \sum_{i }A_i\,  (f_i-h(f_i))\ssp \cong_{\orb}  \sum_{\phi\in \CB^{\le k-1}}d_\phi \,\phi
$$
and this combined with 3.8 completes the induction.  This proves the first assertion.
The second assertion follows from the identity in 2.13.
\sas

Proposition 3.1 has the following remarkable corollary.
\sa

\noindent{\bol Theorem 3.1}

{\ita The ring $\grRa$ yields a graded version of the regular representation. In fact for every 
$0\le k\le d_{m,n}$  we have the character relation 
$$
char\ssp \CH_{=k}(\grRa)\ses char\ssp \CH_{\le k}(\Ra) \sms char\ssp \CH_{\le k-1}(\Ra)
\eqno 3.10
$$
In particular the characters of the subspaces $\CH_{\le k}(\Ra)$ are related to the graded character of  $\grRa$ by the following identity}
$$
\sum_{k= 0}^{d_{m,n}}q^k char\ssp \CH_{=k}(\grRa)\ses q^{d_{m,n}} char\ssp \Ra\sps  (1-q)\sum_{k=0}^{d_{m,n}-1} q^k\, char\ssp \CH_{\le k}(\Ra) 
\eqno 3.11
$$



\noindent{\bol Proof}

Set  $\CB^{\le k}=\{  \phi_{i} \}_{i=1}^{m_k} $ so that $\CA^{= k}=\{  \phi_{i} \}_{m_{k-1}<i\le {m_k}}  $
  and  let $A (\sig)\ses \| a_{i,j} (\sig)\|_{i,j=1}^{n!}$
be the matrix expressing the action of $S_n$ on the basis  $\CB^{\le d_{m,n}}$ as elements of $\Ra$ . 
Since $\CH_{\le k}(\Ra)$ is $S_n$ invariant it follows  that for any  $j\le m_k$ we have the 
expansion 
$$
\sig \phi_j \ses\sum_{i=1}^{m_k}\phi_i \, a_{i,j} (\sig)\sps R\bigsp \hbox{ (with $R\in \CJ_\orb$ of degree $\le k$)}
\eqno 3.12
$$
This means that the action of $S_n$ on the subspace $\CH_{\le k}(\Ra)$ induces a induces a representation 
given by the matrix
$$
 A^{\le k}(\sig)\ses\big\|  a_{i,j} (\sig) \big\|_{1<i,j\le m_k}  
\eqno 3.13
$$
Note further that when $m_{k-1}<j\le m_k$, then $\phi_j$ is homogeneous of degree $k$ and 
equating homogeneous terms of degree $k$ in 3.12 we get
$$
\sig \phi_j \ses\sum_{m_{k-1}<i\le m_k} \phi_i \, a_{i,j} (\sig)\sps 
\cases{h(R) & if $degree \ssp R=k$
\cr\cr
0 & if $degree \ssp R<k$
}\bigsp\hbox{(with $R\in \CJ_\orb$  ).}
$$
In either case we derive that 
$$
\sig \phi_j \ssp \cong\ssp \sum_{m_{k-1}<i\le m_k} \phi_i \, a_{i,j} (\sig) \bigsp\hbox{(modulo $\grJ$)}
$$
This shows that the action of $S_n$ on the subspace $\CH_{=k}\big(\grRa\big)$ induces a representation 
given by the matrix
$$
 A^{=k}(\sig)\ses\big\|  a_{i,j} (\sig) \big\|_{m_{k-1}<i,j\le m_k}  
$$
Thus 
$$
char\ssp \CH_{=k}\big(\grRa\big)\ses trace\ssp A^{=k}\ses trace\ssp   A^{\le k}\sms trace \ssp  A^{\le k-1} \, .
$$ 
This proves 3.10.  This given, multiplying 3.10 by $q^k$ and summing for $0\le k\le d_{m,n}$,
the identity in 2.12 yields 3.12.                 
\sa

\noindent{\bol 4. The orbit $\bf m$-Harmonics}

For a given orbit $[a]$ we set
$$
\BH_{[a]}(m)\ses \big\{ P\in \QI\, : \LL Q\scs P\RR_m\ses 0\ess\ess \forall Q\in \grJ\big\}\, .
\eqno 4.1
$$ 
We call the elements of $\BH_{[a]}(m) $ ``{\ita Orbit Harmonics}''.  We have the following 
characterization of this remarkable space.
\sa
\noindent{\bol Theorem 4.1}

$$
\BH_{[a]}(m)\ses \big\{P\in \QI\, :\, \ggg_QP\ses 0\ess \forall\ess\ess Q\in \grJ\big\}
\eqno 4.2
$$
\noindent{\bol Proof}

Clearly, the condition $ \ggg_QP\ses 0$ is stronger than $\LL Q\scs P\RR_m= 0$. Thus we need only stablish the containment
$$
\BH_{[a]}(m)\con \big\{P\in \QI\, :\, \ggg_QP\ses 0\ess \forall\ess\ess Q\in \grJ\big\}
\eqno 4.3
$$
To this end note that since $\grJ$ is an ideal of $\QI$, the defining condition in 4.1 may be also written in the form
$$
\LL RQ\scs  P\RR_m\ses 0\ess\ess\forall\ess\ess Q\in \grJ \ess\ess\hbox{and}\ess\ess R\in \QI
$$
On the other hand from 1.7 and the definition in 1.9 of the bilinear form $\LL \scs  \RR_m $ we derive that
$$
\LL RQ\scs  P\RR_m\ses \LL R\scs \ggg_QP\RR_m
$$
Thus
$$
P\in \BH_{[a]}(m)\ess\ess   \Longrightarrow \ess\ess \LL R\scs \ggg_QP\RR_m=0\ess\ess \ess\ess\forall\ess\ess Q\in \grJ \ess\ess\hbox{and}\ess\ess R\in \QI
$$
But then the non-degeneracy of $\LL \scs  \RR_m $ yields that
$$
P\in \BH_{[a]}(m) \ess\ess   \Longrightarrow \ess\ess   \ggg_QP=0   \ess\ess  \forall\ess\ess  Q\in \grJ\, .  
$$
This proves 4.3 and completes the proof of the theorem.
\sa

It immediately follows from 4.2 that $\BH_\orb(m)$ is a graded vector space. However we can establish
a considerably stronger result.
\sas

\noindent{\bol Theorem 4.2}

{\ita $\BH_\orb(m)$ is a graded $S_n$ module equivalent to  $\grRa$. In particular it also
has $n!$ dimension and carries the left regular representation of $S_n$.} 

\noindent{\bol Proof}

For a given $1\le k\le d_{m,n}$ let $\{\psi_1,\psi_2,\ldots ,\psi_s\}$ be $m$-quasi-invariants yielding
a basis for $\CH_{=k}\big(\grJ)$, and let $\{\phi_1,\phi_2,\ldots ,\phi_r\}\con \QI$ be constructed
so that, together with  $\{\psi_1,\psi_2,\ldots ,\psi_s\}$ they yield a basis for $\CH_{=k}\big(\QI  \big)$.
In particular, $\{\phi_1,\phi_2,\ldots ,\phi_r\}\con \QI$ are independent modulo $\grJ$ and therefore
the elements of $\grRa$ they represent are a basis for $\CH_{=k}\big( \grRa \big)$.

Now note that the non-degeneracy of the form $\LL\scs \RR_m$ on the subspace $\CH_{=k}\big(\QI  \big)$
assures the  non-singularity of the block matrix
$$
M\ses \left[
\matrix{
\|\LL \phi_i,\phi_j\RR_m \|_{1\le i,j\le r} &\|\LL \phi_i,\psi_j\RR_m\|_{\multi{1\le i\le r\cr 1\le j\le s}} 
\cr
\|\LL \psi_j,\phi_i\RR_m\|_{\multi{1\le i\le r\cr 1\le j\le s}} &\|\LL \psi_i,\psi_j\RR_m\|_{1\le i,j\le s}
\cr}\right].
$$
It then follows the matrix product
$$
\LL \phi_1,\phi_2,\ldots ,\phi_r,\psi_1,\psi_2,\ldots ,\psi_s\RR M^{-1}
$$
yields a basis for  $\CH_{=k}\big(\QI  \big)$ that is ``{\ita dual }'' to $\LL \phi_1,\phi_2,\ldots ,\phi_r,\psi_1,\psi_2,\ldots ,\psi_s\RR $
 with respect to the bilinear form $<\scs >_m$.     That means that if we denote this new  basis
$$
\LL \eta_1,\eta_2,\ldots ,\eta_r,\ggg_1,\ggg_2,\ldots ,\ggg_s\RR 
$$
we shall have
$$
a)\ess\ess \LL \phi_i,\eta_j\RR_m\ses \cases {1 & if $i=j$ \cr 0 & if $i\ne j$\cr}\scs \ess\ess b)\ess\ess  
\LL \psi_i,\ggg_j\RR_m\ses \cases {1 & if $i=j$ \cr 0 & if $i\ne j$\cr}
\eqno 4.4
$$            
and
$$
a)\ess\ess  \LL \phi_i,\ggg_j\RR_m =0\ess  \forall\ess i,j\scs \ess\ess  
b)\ess\ess  \LL \psi_i,\eta_j\RR_m =0\ess  \forall\ess i,j
\eqno 4.5
$$ 
Note that since   $\psi_1,\psi_2,\ldots ,\psi_s$ span $\CH_{=k} \big(\grJ\big)$ the relations in 4.5 b) imply that
$\eta_1,\eta_2,\ldots ,\eta_r$ lie in $\CH_{=k} \big(\CH_\orb(m)\big)$. We claim that $\LL \eta_1,\eta_2,\ldots ,\eta_r\RR$
are in fact a basis of $\CH_{=k} \big( \BH_\orb(m)\big)$. To show this note that the duality relations in 4.4 and 4.5 yield
that for any $m$-quasi-invariant polynomial $P\in\CH_{=k}\big(\QI \big)$ we have the expansions
$$
P\ses \sum_{i=1}^r \LL P,\eta_i\RR_m \phi_i \sps \sum_{j=1}^s \LL P,\ggg_j\RR_m \psi_i
\eqno 4.6
$$
$$
P\ses \sum_{i=1}^r \LL P,\phi_i\RR_m \eta_i \sps \sum_{j=1}^s \LL P,\psi_j\RR_m \ggg_i
\eqno 4.7
$$
In particular  if $P\in \CH_{=k} \big( \BH_\orb(m)\big) $ the latter  expansion reduces to
$$
P\ses \sum_{i=1}^r \LL P,\phi_i\RR_m \eta_i  \, .
\eqno 4.8
$$
This shows that  $\LL \eta_1,\eta_2,\ldots ,\eta_r\RR$ span $\CH_{=k} \big( \BH_\orb(m)\big)$. Since they are 
independent, by construction, it follows that they are a basis as asserted. This given, let $A(\sig)=\|a_{i,j }(\sig)\|_{i,j=1}^r $
be the matrix that expresses the action of $S_n$ on $\LL \eta_1,\eta_2,\ldots ,\eta_r\RR$. That is for all $\sig\in S_n$ we have
$$
\sig\,  \eta_i\ses \sum_{u=1}^r \, \eta_u\,\, a_{u,i}(\sig)\bigsp\hbox{for $1\le i\le r$} .
\eqno 4.9
$$
Using the expansion in 4.6 for $P=\sig \phi_v$ we get
$$
\eqalign{
\sig \phi_v
&\ses  \sum_{i=1}^r \LL \sig \phi_v ,\eta_i\RR_m \phi_i \sps \sum_{j=1}^s \LL \sig \phi_v,\ggg_j\RR_m \psi_i
\cr
&\ses  \sum_{i=1}^r \LL  \phi_v ,\sig^{-1}\eta_i\RR_m \phi_i \sps \sum_{j=1}^s \LL \sig \phi_v,\ggg_j\RR_m \psi_i
\cr
&\cong \ssp  \sum_{i=1}^r \LL  \phi_v ,\sig^{-1}\eta_i\RR_m \phi_i \bigsp \hbox{(modulo $\grJ$)}
}
\eqno 4.10
$$
But using 4.9 we derive (from 4.4 a)):
$$
\LL  \phi_v ,\sig^{-1}\eta_i\RR_m\ses \sum_{u=1}^r \LL  \phi_v , \eta_u\RR_m a_{u,i}(\sig^{-1})\ses a_{v,i}(\sig^{-1})
$$
and thus 4.10 reduces  to
$$
\sig \phi_v\ssp \cong\ssp   \sum_{i=1}^r    a_{v,i} (\sig^{-1})\phi_i \bigsp \hbox{(modulo $\grJ$)}.
$$
In other words the action of $S_n$ on the basis $\LL \phi_1,\phi_2,\ldots ,\phi_r\RR$ of $\CH_{=k}\big( \grRa \big)$ is given by the matrix
$$
A^\top  \big(\sig^{-1}\big )\, .
$$
Since
$$
trace\ssp A(\sig)\ses trace \ssp A^\top  \big(\sig^{-1}\big )
$$
we can thereby conclude that
$$
char\ssp \CH_{=k}\big(\BH_\orb(m) \big)\ses char\ssp \CH_{=k}\big( \grRa \big)  \big)\, .
$$
This proves that  $ \CH_{=k}\big(\BH_\orb(m) \big)$ and $\CH_{=k}\big( \grRa \big)  \big)$ are equivalent as
graded $S_n$ modules and completes the proof of the Theorem.
\sa

\noindent{\bol 5. The $\bf m$-Harmonics of $\bf S_n$}
\sas

In the classical case the space $\BH_{\bf S_n}$ of ``{\ita Harmonics }'' of $S_n$ is   defined as the orthogonal complement of the 
ideal generated by the elementary symmetric functions in thge polynomial ring $\Q[X_n]$. For $m$-harmonics the definition is entirely analogous.
we simply let $\CJ_{S_n}$ denote the ideal generated by $e_1,e_2,\ldots ,e_n$ in $\QI$. In symbols 
$$
\CJ_{S_n}\ses \big(e_1,e_2,\ldots ,e_n\big)_{\QI}
$$
and set
$$
\BH_{\bf S_n}(m)\ses \big\{P\in \QI \ssp : \ssp \LL Q\scs P\RR_m =0 \ess\ess\forall Q\in  \CJ_{S_n}\big\}
\eqno 5.1
$$ 
AS in the case of orbit harmonics we have the following characterization of $m$-harmonics.
\sas

\noindent{\bol Theorem 5.1}

$$
\BH (m)\ses \big\{P\in \QI\, :\, \ggg_QP\ses 0\ess \forall\ess\ess Q\in \CJ_{S_n}\big\}
\eqno 5.2
$$
\noindent{\bol Proof}

Clearly, the condition $ \ggg_QP\ses 0$ is stronger than $\LL Q\scs P\RR_m= 0$. Thus we need only stablish the containment
$$
\BH (m)\con \big\{P\in \QI\, :\, \ggg_QP\ses 0\ess \forall\ess\ess Q\in  \CJ_{S_n}\big\}
\eqno 5.3
$$
To this end note that since $ \CJ_{S_n}$ is an ideal of $\QI$, the defining condition in 5.1 may be also written in the form
$$
\LL RQ\scs  P\RR_m\ses 0\ess\ess\forall\ess\ess Q\in \CJ_{S_n} \ess\ess\hbox{and}\ess\ess R\in \QI
$$
On the other hand from   1.7 and the definition in 1.9 of the bilinear form $\LL \scs  \RR_m $ we derive that
$$
\LL RQ\scs  P\RR_m\ses \LL R\scs \ggg_QP\RR_m
$$
Thus
$$
P\in \BH (m)\ess\ess   \Longrightarrow \ess\ess \LL R\scs \ggg_QP\RR_m=0\ess\ess \ess\ess\forall\ess\ess Q\in\CJ_{S_n}\ess\ess\hbox{and}\ess\ess R\in \QI
$$
But then the non-degeneracy of $\LL \scs  \RR_m $ yields that
$$
P\in \BH (m) \ess\ess   \Longrightarrow \ess\ess   \ggg_QP=0   \ess\ess  \forall\ess\ess  Q\in \CJ_{S_n}\, .  
$$
This proves 5.3 and completes the proof of the theorem.
\sa

It immediately follows from 5.2 that $\BH (m)$ is a graded vector space. This given, to be consistent with previous notation,  let us denote by
$\CH_{=k}\big(\BH(m)\big)$ the subspace of $\BH(m)$ spanned by its homogeneous elements of degree $k$. At this stage we are not yet in a position
to  bound the degrees of elements of $\BH(m)$. However, there is a very natural bound for the dimension of $\BH(m)$ that is provided by the
following  truly remarkable result of  Feigin-Veselov []:
\sas

\noindent{\bol Theorem 5.2}
$$
\dim\, \BH(m)\ssp\le \ssp n!
\eqno 5.4
$$
\noindent{\bol Proof}

Notice to the reader. The following very simple argument was provided to us by E-mail by Feigin and Veselov. We feel compelled
to reproduce it here since it is not available in the present litterature. Although this result is stated in many places (see [],[] [])
proofs (when not omitted) give no indication that it could be established in such a simple and elementary manner.
\sa

The idea is to pick once and for all a point $a=(a_1,a_2,\ldots ,a_n)$ with distinct components and show that each element $P\in \BH(m)$
is completely determined by  $n!$ of its derivatives at the point $a$.  Note further that since the every homogeneous symmetric
polynomial is a polynomial in $e_1,e_2,\ldots,e_n$ it follows from 5.2 that if $A(x)$ is  homogeneous and symmetric then
$$
P\in \BH(m)\ess\Longrightarrow\ess\ggg_A P= 0
\eqno 5.5
$$
Now it is well known [] that the collection of monomials
$$
Art(n)\ses \big\{x^\eee= x_2^{\eee_1}x_3^{\eee_3}\cdots x_n^{\eee_n}\, :\, 0\le \eee_i\le i-1\, \big \}
$$
is a basis for the quotient of the polynomial ring $\Q[X_n]$ by the ideal $(e_1,e_2,\ldots,e_n)_{\Q[X_n]}$.
This implies that each monomial $x^q$ has an expansion of the form
$$
x^q\ses \sum_{x^\eee\in Art(n)} x^\eee\ssp A_{\eee,q}(x) 
\eqno 5.6
$$
with $A_{\eee,q}(x) $ a symmetric polynomial. In fact, the coefficients $A_{\eee,q}(x)$ are uniquely determined from 5.6,
but the latter fact is not needed here. For our purposes it is good to separate terms in 5.6 where
the $A_{\eee,q}(x) $ reduce to scalars. Thus if $x^q$ is of degree $d$ can write 5.6 in the form
$$
x^q\ses \sum_{x^\eee\in Art(n)}\chi(|\eee|=d) \ssp a_{\eee,q} \ssp x^\eee \sps  \sum_{x^\eee\in Art(n)}\hskip-.06in \chi(|\eee|<d) \ssp x^\eee  A_{\eee,q}(x)  \ssp
\eqno 5.7
$$
and here $A_{\eee,q}(x)$ is symmetric, homogenous and 
$$
 degree A_{\eee,q}(x)\ses d-|\eee| .
\eqno 5.8
$$
Thus for any polynomial $P(x)$ we may write
$$
\del_x^q\, P(x)\ses \sum_{x^\eee\in Art(n)}\hskip-.06in\chi(|\eee|=d)\ssp a_{\eee,q}\ssp   \del_x^\eee  P(x)   \sps  
\sum_{x^\eee\in Art(n)}\hskip-.06in \chi(|\eee|<d)\ssp  \del_x^\eee\ssp A_{\eee,q}(\del_x) P(x) 
\eqno 5.9
$$
Now if $P\in \BH(m)$ then from 5.5 we derive that
$$
\ggg_{A_{\eee,q}}P(x)\ses 0.
\eqno 5.10
$$
On the other hand the expansion in 1.2 gives that
$$
\ggg_{A_{\eee,q}}\ses A_{\eee,q}(\del_x)\sps  \sum_{|p|<d-|\eee|} a_{p,\eee,q}(x)\del_x^p
$$
and so  5.10  gives
$$
 A_{\eee,q}(\del_x)P(x)\ses - \sum_{|p|<d-|\eee|} a_{p,\eee,q}(x)\del_x^pP(x)
$$
Using this in 5.9 we finally obtain that
$$
\eqalign{
\del_x^q\, P(x)\ses \sum_{x^\eee\in Art(n)}\hskip-.06in&\chi(|\eee|=d)\ssp  a_{\eee,q}\ssp   \del_x^\eee  P(x)    
\cr 
&\sps \sum_{x^\eee\in Art(n)}\hskip-.06in \chi(|\eee|<d) \hskip-.06in  \ssp\sum_{|p|<d-|\eee|}\hskip-.06in \del_x^\eee\,  \big(a_{p,\eee,q}(x)\, \del_x^pP(x)\big)
\cr
}
\eqno 5.11
$$
Now it follows from 1.3 that   $a_{p,\eee,q}(x)$ a rational function whose denominator contains only
powers of the linear forms $x_i-x_j$. Thus it is perfectly safe to evaluate both sides of 5.11 at our chosen point $a$ and obtain
$$
\eqalign{
\del_x^q\, P(x)\big|_{x=a}\ses \sum_{x^\eee\in Art(n)}\hskip-.06in&\chi(|\eee|=d)\ssp  a_{\eee,q}\ssp   \del_x^\eee  P(x)\big|_{x=a}    
\cr 
&\sps \sum_{x^\eee\in Art(n)}\hskip-.06in \chi(|\eee|<d) \hskip-.06in  \ssp\sum_{|p|<d-|\eee|}\hskip-.06in \del_x^\eee\,  \big(a_{p,\eee,q}(x)\, \del_x^pP(x)\big)\big|_{x=a}
\cr
}
\eqno 5.12
$$
Now we can clearly see that the second sum here will only produce derivatives of $P$ of order stritly less than $d$.
This given, we may view 5.12 as the crucial step in a recursive algorithm for expressing all the derivatives of $P(x)$ in terms
of the collection of derivatives
$$
\big\{  \del_x^\eee  P(x)\big|_{x=a}\ssp :\ssp x^\eee\in Art(n)\big\}
$$
Since this collection has $n!$ the dimension bound in 5.4 necessarily follows. This completes our proof.
\sa

We are now in a position to establish the crucial result of this paper.
\sas

\noindent{\bol Theorem 5.3}

{\ita The two subspaces $\BH_\orb(m)$ and $\BH(m)$ are identical in particular  $\BH(m)$ carries a graded
version of the left regular representation of $S_n$.}

\noindent{\bol Proof}

Note that if $Q$ is a symmetric  polynomial then the polynomial $P(x)-P(a)$ vanishe at all points of the 
orbit $\orb$. Thus if $Q$ is also homogeneous it necessarily follows  that 
$$
Q(x)\in \grJ
$$
and that, together with 4.2. implies that we will have
$$
\ggg_Q P\ses 0\ess\ess \forall \ess\ess P\in \BH_\orb(m)\, .
\eqno 5.13
$$
since by definition $\CJ_{S_n}$ is generated by $e_1,e_2,\ldots ,e_n$ it follows that 5.13 will hold true for all 
$Q\in \CJ_{S_n}$ as well. Thus if  $P$ is homogeneous of degree $k$ we necessarily have that $P\in \CH_{=k}\big(\BH(m)\big)$.
we have thus established that
$$
\CH_{=k}\big(\BH_\orb(m)\big)\con \CH_{=k}\big(\BH(m)\big)\ess\ess\ess \forall\ess\ess 0\le k\le d_{m,n}
\eqno 5.14
$$
and in particular
$$
\dim \CH_{=k}\big(\BH_\orb(m)\big)\ssp \le\ssp\dim  \CH_{=k}\big(\BH(m)\big)\ess\ess\ess \forall\ess\ess 0\le k\le d_{m,n}
\eqno 5.15
$$
Thus
$$
\sum_{k=0}^{d_{m,n}}\dim \CH_{=k}\big(\BH_\orb(m)\big)\ssp \le\ess  \dim\, \BH(m)
$$ 
since from Theorems 4.2  and 5.2 both sides of this inequality evaluate to $n!$ it follows that this is an equality and in fact both
4.14 and 5.15 must be equalities as well. This proves that the two spaces $\BH (m)$ and $\BH_\orb(m)$ are identical and completes the
proof of the theorem.

\end


end

$$
\ggg_{RQ}P\ses 0\ess \forall\ess\ess Q\in \grJ\big\ess\ess\hbox{and}\ess\ess R\in \QI
$$
