A basis for the Diagonal Harmonic Alternants

by

A. M. Garsia††$(\dag)$ $Garsia\enskip\enskip is\enskip supported\enskip
by\enskip an\enskip NSF\enskip Grant$ and M. Zabrocki

Abstract

It will be shown here that there are differential operators $E,F$ and
$H=[E,F]$ for each $n\geq 1$, acting on Diagonal Harmonics, yielding that
$DH_{n}$ is a representation of $sl[2]$ (see [3] Chapter 3). Our main effort
here is to use $sl[2]$ theory to predict a basis for the Diagonal Harmonic
Alternants, $DHA_{n}$. It can be shown that the irreducible representations
$sl[2]$ are all of the form $P,EP,E^{2}P,\cdots,E^{k}P$, with $FP=0$ and
$E^{k+1}P=0$. The polynomial $P$ is known to be called a “String Starter”.
From $sl[2]$ theory it follows that $DHA_{n}$ is a direct sum of strings. Our
main result so far is a formula for the number of string starters. A recent
paper by Carlsson and Oblomkov (see [2]) constructs a basis for the space of
Diagonal Coinvariants by Algebraic Geometrical tools. It would be interesting
to see if any our results can be derived from theirs.

Introduction

We set $X_{n}=x_{1},x_{2},\ldots,x_{n}$ and $Y_{n}=y_{1},y_{2},\ldots,y_{n}$,
we will be working here with polynomials $P(X_{n};Y_{n})$ with rational
coefficients, that is $P(X_{n};Y_{n})\in{\bf
Q}[x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{n}]$.

The diagonal action of $S_{n}$ is defined by setting for any $\sigma\in S_{n}$

$\sigma\,P(X_{n};Y_{n})\,=\,P(x_{\sigma_{1}},x_{\sigma_{2}},\ldots,x_{\sigma_{n}};y_{\sigma_{1}},y_{\sigma_{2}},\ldots,y_{\sigma_{n}})\,.$
$None$

Another important tool in studying $S_{n}$ modules that are invariant under
the diagonal action is the scalar product

$\big{\langle}P\hskip 2.5pt,\hskip 2.5ptQ\big{\rangle}\,=\,L_{o}P(\partial
X_{n};\partial Y_{n})Q(X_{n};Y_{n}).$ $None$

where the differential operator $P(\partial X_{n};\partial Y_{n})$ is obtained
by the replacements $x_{i}{\rightarrow}\partial_{x_{i}}$ and
$y_{i}{\rightarrow}\partial_{y_{i}}$. It is easy to see that we have

$\big{\langle}\sigma P\hskip 2.5pt,\hskip 2.5pt\sigma
Q\big{\rangle}\,=\,\big{\langle}P\hskip 2.5pt,\hskip
2.5ptQ\big{\rangle}\enskip\enskip\enskip\enskip\enskip\enskip\enskip\hbox{(for
all $\sigma\in S_{n}$ )}.$ $None$

In this paper we will study the $S_{n}$ module $DH_{n}$ of Diagonal Harmonic
polynomials. This module was originally defined as the orthogonal complement,
with respect to the scalar product in I.2, of the ideal of polynomials that
are invariant under the diagonal action. By a result of Hermann Weyl (see
[16]) it follows that $P(X_{n};Y_{n})\in DH_{n}$ if and only if

$\sum_{i=1}^{n}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}\,P(X_{n};Y_{n})\,=\,0\hskip
36.135pt\hbox{(for all $p+q\geq 1$)}\,.$ $None$

This simpler definition makes it obvious that $DH_{n}$ is invariant under the
diagonal action.

It also immediately follows from I.4 that if $P\in DH_{n}$ then all the bi-
homgeneous components of $P$ are in $DH_{n}$. This implies that we have the
direct sum decomposition

$DH_{n}\,=\,\bigoplus_{0\leq r+s\leq{n\choose 2}}{\cal
H}_{r,s}\big{(}DH_{n}\big{)},$ $None$

where ${\cal H}_{r,s}(DH_{n})$ is the subspace of diagonal Harmonics
polynomials which are bi-homogeneous of degree $r$ in the $x^{\prime}s$ and
degree $s$ in the $y^{\prime}s$. It then follows that the character resulting
from the diagonal action of $S_{n}$ on $DH_{n}$ can be written in the form

$\chi^{DH_{n}}\,=\,\sum_{0\leq r+s\leq{n\choose
2}}t^{r}q^{s}\chi^{r,s},\enskip\enskip\enskip\enskip\enskip\enskip\hbox{\big{(}where
$\chi^{r,s}$ is the character of ${\cal H}_{r,s}(DH_{n})$
{\hskip-7.22743pt}\big{)}.}$ $None$

The upper bound ${n\choose 2}$ in I.5 follows from the operator conjecture
(proved by Mark Haiman in [13]). $DH_{n}$ can be obtained by applying
differential operators to the Vandermonde determinant $\prod_{1\leq i<j\leq
n}(x_{i}-x_{j})$.

The Frobenius map “${\cal F}$ ”(see [3]) considerably simplifies the operation
of computing the character of an $S_{n}$ representation. Frobenius uses the
dimension equality between the class functions of $S_{n}$ and the space
$\Lambda^{=n}$ of homogeneous symmetric functions of degree $n$. This given,
${\cal F}$ maps Class Functions onto the power basis by the formula

${\cal F}C_{\mu}\,=\,p_{\mu}/z_{\mu}.$ $None$

where $C_{\mu}$ is the sum of all the permutations of cycle structure $\mu$,
$p_{\mu}=p_{1}p_{2}\cdots p_{\l(\mu)}$, where $\l(\mu)$ denotes the length of
$\mu$ and

$z_{\mu}\,=\,1^{a_{1}}2^{a_{2}}\cdots n^{a_{n}}a_{1}!a_{2}!\cdots
a_{n}!\enskip\enskip\enskip\enskip\enskip\enskip\enskip\enskip\hbox{(when
$\mu=1^{a_{1}}2^{a_{2}}\cdots n^{a_{n}}\vdash n$)}$ $None$

It follows from I.7 that Young’s irreducible character $\chi^{\lambda}$ is
given by the identity

$\chi^{\lambda}\,=\,{\cal F}^{-1}s_{\lambda}$ $None$

If ${\cal M}_{n}$ is an $S_{n}$ module with character $\chi^{{\cal M}_{n}}$
then the Frobenius Characteristic of ${\cal M}_{n}$ is the symmetric
polynomial ${\cal F}\,\chi^{{\cal M}_{n}}$. It was conjectured (in [6]) and
shown by Mark Haiman using tools of Algebraic Geometry (in [12]) that the
Frobenius characteristic of $DH_{n}$ is the symmetric rational function

${\cal F}\,\chi^{DH_{n}}\,=\,\sum_{\mu\vdash
n}{T_{\mu}{\widetilde{H}}_{\mu}(X;q,t)MB_{\mu}(q,t)\Pi_{\mu}(q,t)\over
w_{\mu}(q,t)}.$ $None$

To define the ingredients that appear in this formula it is convenient to use
Macdonald’s notation [14] yet identify partitions with their French Ferrers
diagram. Given a partition $\mu$ and a cell $c\in\mu$, as indicated in the
above display, we will introduce four parameters $l_{\mu}(c)$,
$l^{\prime}_{\mu}(c)$, $a_{\mu}(c)$ and $a^{\prime}_{\mu}(c)$ called leg,
coleg, arm  and coarm  which give the number of lattice cells of $\mu$
strictly North, South, East and West of $c$. Denoting by $\mu^{\prime}$ the
conjugate of $\mu$, we set

$\eqalign{n(\mu)\,=\,\sum_{c\in\mu}{l^{\prime}_{\mu}(c)},\enskip\enskip\enskip\enskip\enskip
T_{\mu}=t^{n(\mu)}q^{n(\mu^{\prime})},&\enskip\enskip\enskip\enskip
M=(1-t)(1-q),\enskip\enskip\enskip\enskip
B_{\mu}(q,t)=\sum_{c\in\mu}t^{l^{\prime}_{\mu}(c)}q^{a^{\prime}_{\mu}(c)},\cr\Pi_{\mu}(q,t)=\hskip-14.45377pt\prod_{c\in\mu;c\neq(0,0)}\hskip-14.45377pt(1-{l^{\prime}_{\mu}(c)}q^{a^{\prime}_{\mu}(c)}),&w_{\mu}(q,t)=\prod_{c\in\mu}(q^{a_{\mu}(c)}-t^{l_{\mu}(c)+1})(t^{l_{\mu}(c)}-q^{a_{\mu}(c)+1}).\cr}$
$None$

This accounts for every thing that occurs in I.10 except for the modified
Macdonald Basis element ${\widetilde{H}}_{\mu}(X;q,t)$. This symmetric
polynomial was conjectured in [5] and proved by Mark Haiman in [12] to be the
Frobenius Characteristic of the linear span of derivatives of the alternant
that corresponds to the partition $\mu$.

Our first goal is to construct operators that preserve $DH_{n}$. We will prove
this property for all the differential operators

$a)\enskip\enskip
F_{r,s}\,=\,\sum_{i=1}^{n}\underline{x}_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{s}\,,\hskip
36.135pt\enskip\enskip b)\enskip\enskip
E_{r,s}\,=\,\sum_{i=1}^{n}y_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{s}\,,\enskip\enskip\enskip\enskip\enskip\hbox{(for
$r+s\geq 1$ ) }.$ $None$

We will also prove the relations

$\eqalign{a)\enskip\enskip\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptF_{r,s}\big{]}&\,=\,(p-r)F_{p+r-1,q+s}\,,\cr
b)\enskip\enskip\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}&\,=\,qF_{p+r,q+s-1}\,-\,rE_{p+r-1,q+s}\,,\cr
c)\enskip\enskip\big{[}E_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}&\,=\,(q-s)E_{p+r,q+s-1}\,.\cr}$ $None$

Furthermore by setting

$a)\enskip\enskip F\,=\,F_{0,1}\,,\hskip 36.135ptb)\enskip\enskip
E\,=\,E_{1,0}\,,\hskip 36.135ptc)\enskip\enskip
H\,=\,[E,F]\,=\,\sum_{i=1}^{n}\big{(}\underline{y}_{i}\partial_{y_{i}}-\underline{x}_{i}\partial_{x_{i}}\big{)}$
$None$

we derive that $DH_{n}$ is a direct sum of irreducible representations of
$sl[2]$.

Finally, using the results of the $q,t$-Catalan paper (see [8],[9]) we derive
that the Hilbert polynomial

$c_{n}(q,t)\,=\,{\cal F}\,\chi^{DHA_{n}}\Big{|}_{s[1^{n}]}$ $None$

of the Diagonal Alternants can be obtained by a purely combinatorial
construction. To describe this construction we need further definitions. By a
Dyck path in the $n\times n$ lattice square ${\cal L}_{n}$ we mean a path
which proceeds by $n$ unit North steps and $n$ unit East steps and goes from
$(0,0)$ to $(n,n)$ always remaining weakly

above the main diagonal of ${\cal L}_{n}$. This is the straight line that
joins $(0,0)$ to $(n,n)$. In the illustration on the right we colored light
green all the cells bisected by the main diagonal of ${\cal L}_{8}$. A Dyck
path is depicted there in light blue. It is clear that we only need to give
the abscissas of the North steps of the Dyck path $D$. Those are the integers
that are on the left of the rows of ${\cal L}_{8}$. Thus
$D=[0,0,1,1,2,3,3,5]$. Each $Dyck$ path has two statistics which we call
$area(D)$ and $bounce(D)$. The area statistic is quite simple, its formula is
$area(D)=[0,1,\ldots,n-1]-[0,d_{1},\ldots,d_{n-1}]$.

This is the number of cells between the Dyck path and the lattice diagonal
(the green cells). In our case $area(D)=13$. The bounce statistic is the sum
of the places where the bounce path hits the main diagonal of ${\cal L}_{n}$.
In our display we depicted the bounce path in red using a thinner line. In the
general case it starts straight North until it touches the West end of an East
step. Then it goes straight East until it touches the diagonal. Then goes
straight North until it touches the West end of an East step… alternating
straight North and straight East until it reaches $(n,n)$. We place in the
possible diagonal touching points the labels $1,2,\ldots,n-1$ as indicated in
our display. In our example, $bounce(D)=3+6$. We must emphasize that the
bounce path does not change direction by touching the East end of an East
step. In our display that happens in the $5^{th}$ and $8^{th}$ rows.

In particular we obtain

$c_{n}(q,t)\,=\,\sum_{D\in{\cal
D}_{n}}t^{bounce(D)}q^{area(D)}\,=\,\sum_{D\in{\cal
D}_{n}}t^{area(D)}q^{dinv(D)}$ $None$

The first identity was conjectured by Jim Haglund the second was conjectured
by Mark Haiman. The dinv statistic has also a purely combinatorial definition.
Let $dinv_{a}(D)=\sum_{1\leq i<j\leq n}\chi(u_{i}=u_{j})$ where $u_{i}$ is the
contribution to the area statistic by the $i^{th}$ north step. Similarly we
let $dinv_{b}(D)=\sum_{1\leq i<j\leq n}\chi(u_{i}=u_{j}+1)$, then set
$dinv(D)=dinv_{a}(D)+dinv_{b}(D)$. The problem to construct a further
statistic that combined with area gives I.16 was stated in [6], these two
solutions were discovered quite a few years later.

For $n=3$ we get

${\cal F}DH_{3}\Big{|}_{s[1^{3}]}\,=\,\sum_{D\in{\cal
D}_{3}}t^{bounce(D)}q^{area(D)}\,=\,t^{3}\,+\,\
t^{2}q\,+\,tq\,+\,tq^{2}\,+\,q^{3}.$ $None$

This identity reveals that the alternating character $\chi^{[1,1,1]}$ occurs
in bi-degrees $(3,0),(2,1),(1,2),(0,3)$ and $(1,1)$. In this particular case
these are Frobenius images of the $sl[2]$ strings generated by the following
two alternants

$a)\enskip\enskip\Delta_{1,1,1}\,=\,\det\pmatrix{1&1&1\cr x_{1}&x_{2}&x_{3}\cr
x_{1}^{2}&x_{2}^{2}&x_{3}^{2}\cr},\hskip 36.135pt\enskip\enskip
b)\enskip\enskip\Delta_{2,1}\,=\,\det\pmatrix{1&1&1\cr x_{1}&x_{2}&x_{3}\cr
y_{1}&y_{2}&y_{3}\cr}.$ $None$

It stands to reason that we should be able to construct a basis for the
Diagonal Harmonics Alternants using the operators in I.12, $sl[2]$ theory and
the combinatorics of the $q,t$-Catalan.

Mark Haiman’s proof in [13] of the Operator Conjecture implies the following
result

Theorem I.1

For any $n\geq 1$ let $m$ be the dimension of $DHA_{n}$ in bi-degree $(a,b)$,
then for this bi-degree, it is always possible to construct $m$ sequences
$1\leq r_{1}\leq r_{2}\leq\cdots\leq r_{b}\leq n$ such that the Diagonal
Harmonic Alternants $E_{r_{1},0}E_{r_{2},0}\cdots E_{r_{b},0}\Delta_{1^{n}}$
are linearly independent. This requires that ${n\choose
2}-(r_{1}+r_{2}+\cdots+r_{b})=a$

The only significant results we will prove in this paper, besides introducing
$sl[2]$ theory and proving the invariance under the diagonal action of the
differential operators in I.12 and proving the properties in I.13, are a
formula for the number of starters and an algorithm that gives that number for
every $n$.

1\. The differential operators.

We start with an auxiliary fact concerning the interaction between
multiplication operators and differential operators.

Propositiom 1.1

For any variable $y$ and integer exponent $q\geq 1$ we have

$\partial_{y}^{q}\,\underline{y}\,=\,q\,\partial_{y}^{q-1}+\,\underline{y}\,\partial_{y}^{q},\enskip\enskip\enskip(\hbox{where
``$\underline{y}$'' is the multiplication by $y$ operator})$ $None$

Proof

Suppose that $P(y)$ is a polynomial in $y$. Then for $q=1$ we get

$\partial_{y}\,\underline{y}\,P(y)\,=\,P(y)\,+\,\underline{y}\,\partial_{y}P(y).$
$None$

Thus 1.1 is true for $q=1$. Proceeding by induction on $q$, suppose that 1.1
is true up to $q-1$. Then we have

$\eqalign{\partial_{y}^{q}\,y\,P(y)&\,=\,\partial_{y}\,\partial_{y}^{q-1}y\,P(y)\,=\,\partial_{y}\,(q-1)\partial_{y}^{q-2}P(y)\,+\,\partial_{y}\,y\,\partial_{y}^{q-1}P(y)\,=\cr&\,=\,(q-1)\partial_{y}^{q-1}P(y)\,+\,\partial_{y}^{q-1}P(y)\,+\,y\,\partial_{y}^{q}P(y)\,=\,q\,\partial_{y}^{q-1}P(y)\,+\,y\,\partial_{y}^{q}P(y).\cr}$
$None$

this proves 1.1.

As an example we will show that

Theorem 1.1

The $sl[2]$ operators

$a)\enskip\enskip
F\,=\,\sum_{i=1}^{n}\underline{x}_{i}\partial_{y_{i}}\,,\hskip
36.135pt\enskip\enskip b)\enskip\enskip
E\,=\,\sum_{i=1}^{n}\underline{y}_{i}\partial_{x_{i}}\,,$ $None$

preserve $DH_{n}[X_{n};Y_{n}]$.

Proof

To this end we will first compute the bracket

$\big{[}\Pi_{p,q}^{n}\hskip 2.5pt,\hskip
2.5ptE\big{]}\,=\,\sum_{i=1}^{n}\sum_{j=1}^{n}\big{[}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}\hskip
2.5pt,\hskip
2.5pt\underline{y}_{j}\partial_{x_{j}}\big{]}\,,\enskip\enskip\enskip\enskip\enskip\enskip\hbox{\big{(}where
$\displaystyle\Pi_{p,q}^{n}=\sum_{i=1}^{n}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}$
\big{)}.}$ $None$

Since for $j\neq i$ the differential and multiplication operators commute, we
only need to work with

$\big{[}\Pi_{p,q}^{n}\hskip 2.5pt,\hskip
2.5ptE\big{]}\,=\,\sum_{i=1}^{n}\big{[}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}\hskip
2.5pt,\hskip 2.5pt\underline{y}_{i}\partial_{x_{i}}\big{]}\,.$ $None$

Using 1.1 for $q\geq 1$ we obtain

$\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}\underline{y}_{i}\partial_{x_{i}}\,=\,\partial_{x_{i}}^{p}\big{(}q\,\partial_{y_{i}}^{q-1}+\,\underline{y}_{i}\,\partial_{y_{i}}^{q}\big{)}\partial_{x_{i}}\,=\,q\,\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q-1}\,+\,\underline{y}_{i}\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q}.$
$None$

We also have

$\underline{y}_{i}\partial_{x_{i}}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}\,=\,\underline{y}_{i}\,\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q}$
$None$

so 1.6 becomes

$\big{[}\Pi_{p,q}^{n}\hskip 2.5pt,\hskip
2.5ptE\big{]}\,=\,q\sum_{i=1}^{n}\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q-1}\,+\,\sum_{i=1}^{n}\underline{y}_{i}\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q}\,-\,\sum_{i=1}^{n}\underline{y}_{i}\,\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q}\,=\,q\sum_{i=1}^{n}\partial_{x_{i}}^{p+1}\partial_{y_{i}}^{q-1}\,,$
$None$

or equivalently

$\Pi_{p,q}^{n}\,E\,=\,E\,\Pi_{p,q}^{n}\,+\,q\,\Pi_{p+1,q-1}^{n}.$ $None$

Thus applying $\Pi_{p,q}^{n}\,E$ to a diagonal harmonic polynomial
$P(X_{n};Y_{n})$, gives

$\Pi_{p,q}^{n}\,EP(X_{n};Y_{n})\,=\,E\,\Pi_{p,q}^{n}P(X_{n};Y_{n})\,+\,q\,\Pi_{p+1,q-1}^{n}P(X_{n};Y_{n})\,=\,0.$
$None$

The case $q=0$ is trivial. Proving that the operator $E$ preserves diagonal
harmonics. Working with $F$ we will reach the same result using analogous
steps.

We can use the same idea on the operators

$a)\enskip\enskip
F_{r,s}\,=\,\sum_{i=1}^{n}\underline{x}_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{s}\,,\hskip
36.135pt\enskip\enskip b)\enskip\enskip
E_{r,s}\,=\,\sum_{i=1}^{n}y_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{s}\,.$
$None$

As before we can reduce the calculation to $j=i$ and work with

$\Pi_{p,q}^{n}F_{r,s}\,=\,\sum_{i=1}^{n}\partial_{x_{i}}^{p}\underline{x}_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{q+s}\,=\,\sum_{i=1}^{n}\big{(}p\,\partial_{x_{i}}^{p+r-1}\textstyle\partial_{y_{i}}^{q+s}\,+\,\underline{x}_{i}\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{q+s}\big{)}.$
$None$

Likewise we have

$F_{r,s}\Pi_{p,q}^{n}\,=\,\sum_{i=1}^{n}\underline{x}_{i}\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{q+s}\,.$
$None$

Thus

$\Pi_{p,q}^{n}\,F_{r,s}\,=\,F_{r,s}\Pi_{p,q}^{n}\,+\,p\,\Pi_{p+r-1,q+s}^{n}$
$None$

Now if $P(X_{n};Y_{n})$ is Diagonal Harmonic then

$\Pi_{p,q}^{n}\,F_{r,s}P(X_{n};Y_{n})\,=\,F_{r,s}\Pi_{p,q}^{n}P(X_{n};Y_{n})\,+\,p\,\Pi_{p+r-1,q+s}^{n}P(X_{n};Y_{n})\,=\,0\,.$
$None$

Proving that $F_{r,s}$ preserves Diagonal Harmonics. An analogous argument
yields the same result for $E_{r,s}$.

Theorem 1.2

The following identities hold true for all $p+q\geq 1$ and $r+s\geq 1$.

$\eqalign{a)\enskip\enskip\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptF_{r,s}\big{]}&\,=\,(p-r)F_{p+r-1,q+s}\,,\cr
b)\enskip\enskip\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}&\,=\,qF_{p+r,q+s-1}\,-\,rE_{p+r-1,q+s}\,,\cr
c)\enskip\enskip\big{[}E_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}&\,=\,(q-s)E_{p+r,q+s-1}\,.\cr}$ $None$

Proof

Reducing again to the case $j=i$ we can write

$\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptF_{r,s}\big{]}\,=\,\sum_{i=1}^{n}\big{[}\underline{x}_{i}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q}\hskip
2.5pt,\hskip
2.5pt\underline{x}_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{s}\big{]}\,=\,\sum_{i=1}^{n}\underline{x}_{i}(\partial_{x_{i}}^{p}\underline{x}_{i})\partial_{x_{i}}^{r}\partial_{y_{i}}^{q+s}\,-\,\sum_{i=1}^{n}\underline{x}_{i}(\partial_{x_{i}}^{r}\underline{x}_{i})\partial_{x_{i}}^{p}\partial_{y_{i}}^{q+s}\,.$
$None$

Since we have

$a)\enskip\enskip\partial_{x_{i}}^{p}\underline{x}_{i}=p\,\partial_{x_{i}}^{p-1}+\underline{x}_{i}\partial_{x_{i}}^{p}\hskip
36.135pt\hskip
36.135ptb)\enskip\enskip\partial_{x_{i}}^{r}\underline{x}_{i}=r\,\partial_{x_{i}}^{r-1}+\underline{x}_{i}\partial_{x_{i}}^{r}$
$None$

The identity in 1.18 becomes, using 1.19

$\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptF_{r,s}\big{]}\,=\,\sum_{i=1}^{n}\underline{x}_{i}(p\,\partial_{x_{i}}^{p-1}+\underline{x}_{i}\partial_{x_{i}}^{p})\partial_{x_{i}}^{r}\partial_{y_{i}}^{q+s}\,-\,\sum_{i=1}^{n}\underline{x}_{i}(r\,\partial_{x_{i}}^{r-1}+\underline{x}_{i}\partial_{x_{i}}^{r})\partial_{x_{i}}^{p}\partial_{y_{i}}^{q+s}\,.$
$None$

Now this can be rearranged to

$\eqalign{\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptF_{r,s}\big{]}&\,=\,\sum_{i=1}^{n}\underline{x}_{i}(p\,\partial_{x_{i}}^{p-1})\partial_{x_{i}}^{r}\partial_{y_{i}}^{q+s}\,-\,\sum_{i=1}^{n}\underline{x}_{i}(r\,\partial_{x_{i}}^{r-1})\partial_{x_{i}}^{p}\partial_{y_{i}}^{q+s}\,+\,\sum_{i=1}^{n}(\underline{x}_{i}^{2}-\underline{x}_{i}^{2})\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{q+s}\,.\cr}$

From which we derive that

$\eqalign{\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptF_{r,s}\big{]}&\,=\,(p-r)\sum_{i=1}^{n}\underline{x}_{i}\partial_{x_{i}}^{p+r-1}\partial_{y_{i}}^{q+s}\,=\,(p-r)F_{p+r-1,q+s}\,.\cr}$
$None$

This proves a) of 1.17.

Next we work on b) of 1.17. Reducing to $j=i$ we can write

$\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}\,=\,\sum_{i=1}^{n}\big{[}\underline{x}_{i}\partial_{x_{i}}^{p}\partial_{y_{i}}^{q},\underline{y}_{i}\partial_{x_{i}}^{r}\partial_{y_{i}}^{s}\big{]}\,=\,\sum_{i=1}^{n}\underline{x}_{i}(\partial_{y_{i}}^{q}\underline{y}_{i})\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{s}\,-\,\sum_{i=1}^{n}\underline{y}_{i}(\partial_{x_{i}}^{r}\underline{x}_{i})\partial_{x_{i}}^{p}\partial_{y_{i}}^{q+s}$
$None$

Using 1.19 we get

$\eqalign{\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}&\,=\,\sum_{i=1}^{n}\underline{x}_{i}(q\,\partial_{y_{i}}^{q-1}+\underline{y}_{i}\partial_{y_{i}}^{q})\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{s}\,-\,\sum_{i=1}^{n}\underline{y}_{i}(r\,\partial_{x_{i}}^{r-1}+\underline{x}_{i}\partial_{x_{i}}^{r})\partial_{x_{i}}^{p}\partial_{y_{i}}^{q+s}\cr&\,=\,q\sum_{i=1}^{n}\underline{x}_{i}\,\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{q+s-1}\,-\,r\sum_{i=1}^{n}\underline{y}_{i}\partial_{x_{i}}^{p+r-1}\partial_{y_{i}}^{q+s}\,+\,\sum_{i=1}^{n}(\underline{x}_{i}\underline{y}_{i}-\underline{y}_{i}\underline{x}_{i})\partial_{x_{i}}^{p+r}\partial_{y_{i}}^{q+s}\cr&\,=\,q\,F_{p+r,q+s-1}\,-\,r\,E_{p+r-1,q+s}\,.}$
$None$

This proves 1.17 b). The identity in 1.17 c) is proved the same way we proved
a).

Remark 1.1

Using b) of 1.17 we can prove the result in c) of I.14. In fact, since
$E=E_{1,0}$ and $F=F_{0,1}$, setting $p=0,q=1,r=1,s=0$ in

$\big{[}F_{p,q}\hskip 2.5pt,\hskip
2.5ptE_{r,s}\big{]}\,=\,qF_{p+r,q+s-1}\,-\,rE_{p+r-1,q+s}\,,$

we obtain

$\big{[}F_{0,1}\hskip 2.5pt,\hskip
2.5ptE_{1,0}\big{]}\,=\,F_{1,0}\,-\,E_{0,1}\,,$

Using 1.12 this gives that the $sl[2]$ operator in I.14

$H\,=\,\big{[}E\hskip 2.5pt,\hskip
2.5ptF\big{]}\,=\,\sum_{i=1}^{n}y_{i}\partial_{y_{i}}\,-\,\sum_{i=1}^{n}\underline{x}_{i}\partial_{x_{i}}\,,$
$None$

on Diagonal Harmonics is none other than the Euler operator in the
$y^{\prime}s$ minus the Euler operator in the $x^{\prime}s$.

Remark 1.2

We will make multiple uses of the operators $E_{r,0}$, we must point out that
that these operators commute regardless of the values of $r$. This is one of
the consequences of the identities in 1.17. In fact, if $q=s$ in c) that will
immediately force the commutativity of $E_{p,q}$ and $E_{r,s}$. The analogous
result can also be obtained for the differential operators pairs in 1.17 a).

Proof of Theorem I.1

For $A=(a_{1},a_{2},\ldots,a_{n})$ with $a_{i}\in Z_{\geq
0}=\\{0,1,2,\ldots\\}$ set

$E^{A}\,=\,E_{1,0}^{a_{1}}E_{2,0}^{a_{2}}\cdots E_{n,0}^{a_{n}}.$ $None$

The proof in [13] of the operator conjecture implies that if we set ${\cal
H}[x_{1},x_{2},\ldots,x_{n}]={\cal L}_{\partial}[\Delta_{1^{n}}]$ (the
ordinary Harmonics of $S_{n}$), then

$\sum_{A}E^{A}{\cal H}[x_{1},x_{2},\ldots,x_{n}]\,=\,{\cal
H}[x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{n}],$ $None$

the space of Diagonal Harmonics. This implies that if
$\\{h_{1},h_{2},\ldots,h_{n!}\\}$ is any basis for ${\cal
H}[x_{1},x_{2},\ldots,x_{n}]$ then the elements
$\\{E^{A_{1}}{h_{1}},E^{A_{2}}h_{2},\ldots,E^{A_{n!}}h_{n!}\\}$ span the
Diagonal Harmonics, in particular they span the Diagonal Harmonic Alternants.
If the basis is isotypical that is there are as many independent elements
generating the irreducible representation character $\chi^{\lambda}$ as the
dimension of this representation. Since the alternating element occurs with
multiplicity one and is the Vandermonde in $x_{1},x_{2},\ldots,x_{n}$ the
basis elements must be all of the form

$E^{A}\Delta_{1^{n}}\,=\,E_{r_{1},0}E_{r_{2},0}\cdots
E_{r_{b},0}\Delta_{1^{n}},$ $None$

with $0\leq r_{1}\leq r_{2}\leq\cdots\leq r_{b}$. This proves the theorem.

2\. Analysing the computer data for $\bf n=5,6$.

Although our differential operators are constructing Diagonal Harmonic
alternants we will be guided

by the Frobenius characteristic of the alternants in $DH_{n}$

$c_{n}(q,t)\,=\,\sum_{D\in{\cal D}_{n}}t^{bounce(D)}q^{area(D)}.$ $None$

We will do that for $n=5$ and $n=6$.

In the display on the right we have depicted a visual image of this polynomial
for $n=5$. For instance the term of $c_{5}(q,t)$ that is in the fifth row and
fifth column corresponds to the Dyck path $D\in{\cal D}_{5}$ for which
$t^{bounce(D)}q^{area(D)}=2\,t^{4}q^{4}$. We can easily locate the $sl[2]$
strings in this display. In $DH_{n}$ they are of the form
$P(X_{n};Y_{n}){\rightarrow}EP(X_{n};Y_{n}){\rightarrow}\cdots{\rightarrow}E^{k}P(X_{n};Y_{n}),$
where $P(X_{n};Y_{n})$ is killed by $F$ and is a homogeneous polynomial of bi-
degree $(a,b)$. The operator $E$ diminishes degree in $x^{\prime}s$ by one and
increases degree in $y^{\prime}s$ by one, thus the polynomial
$E^{i}P(X_{n};Y_{n})$ is homogeneous of bi-degree $(a-i,b+i)$. The final
element is of

bi-degree $(b,a)$ and is killed by $E$. The circled entries correspond to
alternants indexed by partitions. The one corresponding to $t^{10}$ is the
Vandermonde $\Delta_{1^{5}}$. The one corresponding to $t^{6}q^{1}$ is
$\Delta_{21^{3}}$. $\Delta_{221},\Delta_{311}$ and $\Delta_{32}$ form an
$sl[2]$ string. Our list ends with $\Delta_{41}$ and the Vandermonde
$\Delta_{5}$. This display shows that the alternants in $DH_{5}$ factor into a
direct sum of $7$ of $sl[2]$ strings. The one of highest total degree is
started by $\Delta_{1^{5}}$. Let us call it String $1$. String $2$ starts with
bi-degree $(8,1)$ and ends in bi-degree $(1,8)$. String $3$ starts with bi-
degree $(7,1)$ and ends in bi-degree $(1,7)$. String $4$ starts with bi-degree
$(6,2)$ and ends in bi-degree $(2,6)$. That accounts for the $2^{\prime}s$ in
the image of its path. String $5$ starts with bi-degree $(6,1)$ and ends in
bi-degree $(1,6)$. These two ends are $\Delta_{21^{3}}$ and $\Delta_{41}$.
String $6$ starts with bi-degree $(6,2)$ and ends in bi-degree $(2,6)$. That
accounts for all the $2^{\prime}s$ in the image of its path. String $7$ starts
with bi-degree $(4,2)$ and ends in bi-degree $(2,4)$.

Our computations using MAPLE yielded the following seven starters:

$\Delta_{1^{5}}$, $E_{2,0}\Delta_{1^{5}}$, $E_{3,0}\Delta_{1^{5}}$,
$E_{4,0}\Delta_{1^{5}}$,
$3E_{2,0}E_{2,0}\Delta_{1^{5}}+2EE_{3,0}\Delta_{1^{5}}$,
$2E_{2,0}E_{2,0}E_{2,0}\Delta_{1^{5}}-3EEE_{4,0}\Delta_{1^{5}}$,
$E_{4,0}E_{2,0}\Delta_{1^{5}}$.

Notice that for $n<7$ there is at most one starter at any bi-degree. If we
find one that starts at bi-degree $(a,b)$ and ends at $(b,a)$ we can safely
complete the string in a construction of a basis. However, the result is not a
basis that is consistent with Theorem I.1. This is what happens with all the
elements of the strings started by the $5^{th}$ and $6^{th}$ starters.

It will be good to make a few observations before we work on $n=6$. Firstly,
Remark 1.2 tells us that the operators $E_{r,0}$ commute whatever is the value
of $r$. From Theorem I.1 we know that to obtain an alternant of bi-degree
$(a,b)$ in $DH_{n}$ using the polynomial $E_{r_{1},0}E_{r_{2},0}\cdots
E_{r_{b},0}\Delta_{1^{n}}$ we must require the equality

$r_{1}+r_{2}+\cdots+r_{b}\,=\,{\textstyle{n\choose 2}}-a$ $None$

Due to the commutativity fact, there is no loss in assuming that our sequences
are weakly decreasing ($r_{1}\geq r_{2}\geq\cdots\geq r_{b}$) and satisfy 2.2.
This is the number of $b$ parts partitions of ${\textstyle{n\choose 2}}-a$.

Notice that our display at the row indexed by $t^{5}$ and column indexed by
$q^{2}$ reveals that there is only one alternant in $DH_{5}$ with bi-degree
$(5,2)$. Yet the $2$ parts partitions of $5$ are $32$ and $41$. Thus when we
reduce the two polynomials $E_{3,0}E_{2,0}\Delta_{1^{5}}$ and
$E_{4,0}E_{1,0}\Delta_{1^{5}}$ to have leading monomial with coefficient $1$
the resulting alternants must be identical!

This fact shows that we can only use partitions as an upper bound to the
number of strings that start at a given bi-degree. However, in our computer
experimentation with $DHA_{6}$ we discovered that pairs that were discarded in
$DHA_{5}$ had also to be discarded as factors in the construction of string
starters in $DHA_{6}$. This suggests that the construction of a basis for
$DHA_{n}$ might demand a recursion on $n$.

These findings suggest that it might be possible to obtain a recursive
construction of a basis. The following result gives us a tool for not using
factors that have been discarded for $n-1$ in the construction of basis
elements for $n$. For simplicity we will state it in the simplest useful case.

Theorem 2.1

Suppose that $C_{n-1}(q,t)\big{|}_{t^{c}q^{b-1}}=k-1$ and
$C_{n}(q,t)\big{|}_{t^{a}q^{b}}=k$, then the polynomial

$P(X_{n};Y_{n})\,=\,E_{r_{1},0}E_{r_{2},0}\cdots E_{r_{b},0}\Delta_{1^{n}}$
$None$

cannot be used as a starter in bi-degree $(a,b)$ if

$a\,=\,n-1+c-r_{b}$ $None$

for any discarded pairs of solutions of

$r_{1}+r_{2}+\cdots+r_{b-1}\,=\,{\textstyle{n-1\choose 2}}\,-\,c\,.$ $None$

Proof

Since our conjecture requires that

$r_{1}+r_{2}+\cdots+r_{b}\,=\,1+2+\cdots+n-1\,-\,a\,.$ $None$

this can be rewritten in an inductive way by relating for $n$ what we already
obtained for $n-1$

$0\,=\,r_{1}+r_{2}+\cdots+r_{b-1}\,-\,({\textstyle{n-1\choose
2}}\,-\,c)\,=\,n-1\,-\,a\,+\,c\,-\,r_{b}\,.$ $None$

This proves 2.4. Our task is now to explore $n=6$

On the right we have a visual display of a portion of the Frobenius
characteristic of the alternants in $DHA_{6}$. This is the polynomial
$c_{6}(q,t)$, The number of strings start at bi-degree $(a,b)$ is again

$c_{6}(q,t)\big{|}_{t^{a+1}q^{b-1}}-\,c_{6}(q,t)\big{|}_{t^{a}q^{b}}\leq 1$
$None$

Without the 0’s the string starters are $({\bf
1}{\rightarrow})\Delta_{1^{6}}$, $({\bf 2}{\rightarrow})E_{2}\Delta_{1^{6}}$,

$({\bf 3}{\rightarrow})E_{3}\Delta_{1^{6}}$, $({\bf
4}{\rightarrow})11E_{2}E_{2}\Delta_{1^{6}}+4E_{3}E_{1}\Delta_{1^{6}}$

$({\bf 5}{\rightarrow})E_{4}\Delta_{1^{6}}$, $({\bf
6}{\rightarrow})2E_{3}E_{2}\Delta_{1^{6}}+E_{4}E_{1}\Delta_{1^{6}}$, $({\bf
7}{\rightarrow})12E_{2}E_{2}E_{2}\Delta_{1^{6}}+18E_{3}E_{2}E_{1}\Delta_{1^{6}}+5E_{4}E_{1}E_{1}\Delta_{1^{6}}$

$({\bf 8}{\rightarrow})E_{5}\Delta_{1^{6}}$ $({\bf
9}{\rightarrow})3E_{3}E_{3}\Delta_{1^{6}}+2E_{5}E_{1}\Delta_{1^{6}}$, $({\bf
10}{\rightarrow})28E_{3}E_{2}E_{2}\Delta_{1^{6}}-19E_{5}E_{1}E_{1}\Delta_{1^{6}}$,

$({\bf
11}{\rightarrow})5E_{2}E_{2}E_{2}E_{2}\Delta_{1^{6}}+24E_{3}E_{2}E_{2}E_{1}\Delta_{1^{6}}+4E_{4}E_{2}E_{1}E_{1}\Delta_{1^{6}}-8E_{5}E_{1}E_{1}E_{1}\Delta_{1^{6}}$

$({\bf 12}{\rightarrow})E_{4}E_{3}\Delta_{1^{6}}$, $({\bf
13}{\rightarrow})3E_{3}E_{3}E_{2}\Delta_{1^{6}}+2E_{4}E_{3}E_{1}\Delta_{1^{6}}$,

$({\bf
14}{\rightarrow})20E_{3}E_{2}E_{2}E_{2}\Delta_{1^{6}}+42E_{3}E_{2}E_{2}E_{1}\Delta_{1^{6}}+39E_{4}E_{2}E_{2}E_{1}\Delta_{1^{6}}$

$({\bf 15}{\rightarrow})E_{4}E_{4}\Delta_{1^{6}}$, $({\bf
16}{\rightarrow})5E_{3}E_{3}E_{3}\Delta_{1^{6}}-9E_{4}E_{4}E_{1}\Delta_{1^{6}}$

$({\bf 17}{\rightarrow})E_{5}E_{2}E_{2}\noindent E_{2}\Delta_{1^{6}}$.

We can derive an algorithm for obtaining the number of starters for any $n$,
that MAPLE permits, from the following calculation. Since every string is of
the form

$P(X_{n};Y_{n}){\rightarrow}EP(X_{n};Y_{n}){\rightarrow}\cdots{\rightarrow}E^{k}P(X_{n};Y_{n}).$
$None$

If $P(X_{n};Y_{n})$ is an alternant homogeneous of bi-degree $(u,v)$ then the
alternant $E^{i}P(X_{n};Y_{n})$ is homogeneous of bi-degree $(u-i,v+i)$, with
$E^{k}P(X_{n};Y_{n})$ homegeous of bi-degree $(v,u)$. The contribution of this
string to the Hilbert series of $DHA_{n}$ is the polynomial
$\sum_{i=0}^{u-v}t^{u-i}q^{v+i}$. Note that

$t^{u}q^{v}\,=\,(qt)^{u+v\over 2}({\textstyle{q\over t}})^{v-u\over 2},$
$None$

so we can write

$\eqalign{t^{u}q^{v}\Big{(}1+&({\textstyle{q\over
t}})+\cdots+({\textstyle{q\over t}})^{u-v}\Big{)}\,=\,(qt)^{u+v\over
2}({\textstyle{t\over q}})^{u-v\over 2}\Big{(}1+({\textstyle{q\over
t}})+\cdots+({\textstyle{q\over t}})^{u-v}\Big{)}\cr&\,=\,(qt)^{u+v\over
2}({\textstyle{t\over q}})^{u-v\over 2}{1\,-\,({\textstyle{q\over
t}})^{{u-v}+1}\over 1\,-\,({\textstyle{q\over t}})}\,=\,(qt)^{u+v\over
2}{({\textstyle{t\over q}})^{u-v\over 2}\,-\,({\textstyle{q\over
t}})^{{u-v\over 2}+1}\over 1\,-\,({\textstyle{q\over
t}})}\cr&\,=\,(qt)^{u+v\over 2}{({\textstyle{t\over q}})^{u-v\over
2}({\textstyle{t\over q}})^{1\over 2}\,-\,({\textstyle{q\over t}})^{{u-v\over
2}+1}({\textstyle{t\over q}})^{1\over 2}\over({\textstyle{t\over q}})^{1\over
2}\,-\,({\textstyle{q\over t}})({\textstyle{t\over q}})^{1\over
2}}\,=\,(qt)^{u+v\over 2}{({\textstyle{t\over q}})^{u-v+1\over
2}\,-\,({\textstyle{q\over t}})^{{u-v+1\over 2}}\over({\textstyle{t\over
q}})^{1\over 2}\,-\,({\textstyle{q\over t}})^{1\over 2}}\cr}$ $None$

Thus the Hilbert series of $DHA_{n}$ is the polynomial

$h(q,t)\,=\,\sum_{u=0}^{n\choose 2}\sum_{v=0}^{n\choose
2}b_{u,v}\,\,(qt)^{u+v\over 2}{({\textstyle{t\over q}})^{u-v+1\over
2}\,-\,({\textstyle{q\over t}})^{{u-v+1\over 2}}\over({\textstyle{t\over
q}})^{1\over 2}\,-\,({\textstyle{q\over t}})^{1\over 2}}$ $None$

where $b_{u,v}$ is the number of strings that start in bi-degree $(u,v)$.
Making the specialization $t{\rightarrow}q^{-1}$ gives

$h(q,q^{-1})\,=\,\sum_{u=0}^{n\choose 2}\sum_{v=0}^{n\choose
2}b_{u,v}\,\,{q^{u-v+1}\,-\,q^{{-(u-v+1)}}\over q\,-\,q^{-1}}$

or better

$\enskip\enskip\enskip(q\,-\,q^{-1})h(q,q^{-1})\,=\,\sum_{u=0}^{n\choose
2}\sum_{v=0}^{n\choose 2}b_{u,v}\,\,\big{(}q^{u-v+1}\,-\,q^{-(u-v+1)}\big{)}$
$None$

We can rewrite the identity in 2.11 in the form

$\enskip\enskip\enskip(q\,-\,q^{-1})h(q,q^{-1})\,=\,\sum_{u=0}^{n\choose
2}\sum_{v=0}^{n\choose
2}\chi\big{(}u-v+1=r\big{)}\big{(}q^{r}-q^{-r}\big{)}\sum_{u-v+1=r}b_{u,v}$
$None$

If we let $c_{r}=\sum_{u-v+1=r}b_{u,v}$ then 2.12 becomes

$\eqalign{{\hskip-7.22743pt}{\hskip-7.22743pt}{\hskip-7.22743pt}{\hskip-7.22743pt}(q\,-\,q^{-1})h(q,q^{-1})&\,=\,\sum_{u=0}^{n\choose
2}\sum_{v=0}^{n\choose
2}\chi\big{(}u-v+1=r\big{)}\,\big{(}q^{r}-q^{-r}\big{)}\,c_{r}\cr&\,=\,\sum_{r=1}^{n\choose
2}\big{(}q^{r}-q^{-r}\big{)}\,c_{r}\sum_{u=0}^{n\choose 2}\sum_{v=0}^{n\choose
2}\chi\big{(}r=1-u+v\big{)}\,=\,\sum_{r=1}^{n\choose
2}\big{(}q^{r}\,-\,q^{-r}\big{)}\,c_{r}\,.\cr}$ $None$

Haiman has proved in [12] that $h(q,t)$ is exactly the Garsia-Haiman
$q,t$-Catalan. In [6] the specialization $t{\rightarrow}q^{-1}$ is derived
from Macdonald identities to be related to the none other than the q-analogue
of the number of Dyck paths in the $n\times n$ lattice rectangle, more
precisely we have:

$q^{n\choose 2}h(q,q^{-1})\,=\,{1\over[n+1]_{q}}\Big{[}{2\,n\atop
n}\Big{]}_{q}\,.$ $None$

Thus the left hand side of 2.13 is

$(q\,-\,q^{-1})q^{-{n\choose 2}}{q-1\over
q^{n+1}-1}{(q^{n+1}-1)\cdots(q^{2\,n}-1)\over(q-1)\cdots(q^{n}-1)}\,=\,q^{-{n\choose
2}-1}{(q^{n+2}-1)\cdots(q^{2\,n}-1)\over(q^{3}-1)\cdots(q^{n}-1)}$ $None$

and 2.13 becomes

$q^{-{n\choose
2}-1}{(q^{n+2}-1)\cdots(q^{2\,n}-1)\over(q^{3}-1)\cdots(q^{n}-1)}\,=\,\sum_{r=1}^{n\choose
2}\,\,c_{r}\,\,q^{r}\,-\,\sum_{r=1}^{n\choose 2}c_{r}\,\,q^{-r}\,$ $None$

Since the number of starters is $\sum_{r=1}^{n\choose 2}c_{r}$ all we need is
to sum the positive coefficients of the powers of $q$. Of course this is
really only an algorithm, but we can also obtain a formula from 2.16.

To do this we first apply an odd power of the Euler operator $q\partial_{q}$
to both side of 2.16 :

$\big{(}q\partial_{q}\big{)}^{2k+1}\,\left(q^{-{n\choose
2}-1}{(q^{n+2}-1)\cdots(q^{2\,n}-1)\over(q^{3}-1)\cdots(q^{n}-1)}\right)\,=\,\sum_{r=1}^{n\choose
2}\,\,r^{2k+1}\,c_{r}\,(q^{r}+q^{-r})$ $None$

The determinant of the matrix
$\big{\|}z_{r}^{2k}\big{\|}_{r,k=1}^{{n(n-1)/2}+1}$ factorized into a product
of $(z_{r}^{2}-z_{s}^{2})$ and each is different from zero as long as
$z_{r}\neq z_{s}$. this proves that the matrix
$\big{\|}r^{2k+1}\big{\|}_{r,k=1}^{{n(n-1)/2}+1}$ has non zero determinant.
Denoting by $\big{\|}d_{s,r}\big{\|}_{s,r=1}^{{n(n-1)/2}+1}$ the inverse, it
follows that

$\sum_{r=1}^{n\choose
2}d_{s,r}\big{(}q\partial_{q}\big{)}^{2k+1}\,\left(q^{-{n\choose
2}-1}{(q^{n+2}-1)\cdots(q^{2\,n}-1)\over(q^{3}-1)\cdots(q^{n}-1)}\right)\,=\,\sum_{r=1}^{n\choose
2}\,\,c_{r}\,(q^{r}+q^{-r})$ $None$

Thus

${\textstyle{1\over 2}}\sum_{r=1}^{n\choose
2}d_{s,r}\big{(}q\partial_{q}\big{)}^{2k+1}\,\left(q^{-{n\choose
2}-1}{(q^{n+2}-1)\cdots(q^{2\,n}-1)\over(q^{3}-1)\cdots(q^{n}-1)}\right)\Big{|}_{q=1}\,=\,\sum_{r=1}^{n\choose
2}\,\,c_{r}\,$ $None$

as desired.

Using the algorithm in 2.16, we obtain the following sequence giving the
number of string starters for each $n\geq 2$.

2.20

The Encyclopedia of Integer Sequences was not aware of the existence of this
sequence.

Conjecture in [1] and Theorem 2.1, give us an algorithm for constructing
starters, this done a basis for $DHA_{n}$ is easily constructed. From the
display in 2.20, we see that the number of starters for $n=7$ is $44$ and for
$n=8$ is already $120$. With some work we may be able to construct the
starters for $n=7$, with even more work we may be able to do it for $n=8$.
However, a paper [1] published in 1998 by Ed Allen yields us a basis for every
$n$. To see how we can use [1] we need only to depict an example.

On the right we have the lattice square ${\cal L}_{6}$. The green cells are
the lattice diagonal. We have depicted in yellow the cells of the english
Ferrers diagram of the partition $[4,2,1]$. Starting from $(0,0)$ we have $3$
black North steps, followed by $3$ red North steps, delimiting a Dyck path $D$
in ${\cal L}_{6}$. The $7$ yellow cells give the co-area of $D$. We will call
the yellow partition the “co-partition” of $D$. In [6] we conjecture that the
dimension of $DHA_{n}$ is given by the number Dyck paths in ${\cal L}_{n}$.
Accordingly in [1] it is conjectured that the following succession of steps
constructs

a basis of $DHA_{n}$. First step, compute the power function expansion of the
Schur function indexed by the co-partition for each Dyck path in ${\cal
L}_{n}$. Second step, replace each $p_{r}$ by the operator $E_{r}$. Of course
that means $p_{r}^{k}$ gets replaced by $E_{r}$ repeated $k$ times. This done,
apply the resulting differential operator to the Vandermonde determinant in
$x_{1},x_{2},\ldots,x_{n}$. This sequence of steps should yield a basis for
$DHA_{n}$.

Since the power function expansion of a Schur function involves many terms the
resulting alternant may not be bi-homogeneous. For instance for the co-
partition $[4,2,1]$ the power basis expansion of the corresponding Schur
function is the sum of the following terms

$\big{[}{\textstyle{1\over 144}}p_{1}^{7},{\textstyle{-1\over
72}}p_{1}^{4}p_{3},{\textstyle{-1\over
48}}p_{1}^{3}p_{2}^{2},{\textstyle{-1\over
24}}p_{1}^{3}p_{4},{\textstyle{-1\over
12}}p_{3}p_{1}^{2}p_{2},{\textstyle{-1\over
18}}p_{1}p_{3}^{2},{\textstyle{-1\over 24}}p_{3}p_{2}^{2},{\textstyle{-1\over
24}}p_{3}p_{2}^{2},{\textstyle{-1\over 24}}p_{3}p_{2}^{2},{\textstyle{-1\over
24}}p_{3}p_{2}^{2},{\textstyle{-1\over 24}}p_{3}p_{2}^{2},{\textstyle{-1\over
24}}p_{3}p_{2}^{2}\big{]}$ $None$

The next result reinforces Conjecture in [1].

Theorem 2.2

On the validity of the conjecture in [1] it is possible to construct a bi-
homogeneous basis for $DHA_{n}$.

Proof

It suffices to construct such a basis by an algorithm that works for all $n$.
A fundamental fact is that $DHA_{n}$ is a bi-graded vector space. Moreover it
is shown in [12] that $DHA_{n}$ has dimention equal to the number of Dyck
paths in ${\cal L}_{n}$. The conjecture in [1] starts with the power basis
expansion of Schur functions indexed by co-partitions in ${\cal L}_{n}$. In
the next step we replace each $p_{r}$ in the resulting expansion by the
operator $Er$ repeated as many times as the exponent of $p_{r}$. This done we
apply the resulting operator to the Vandermonde in $x_{1},x_{2},\ldots,x_{n}$.
The resulting polynomial will be a bi-homogeneous element of $DHA_{n}$, for
every term in the power function expansion.

To construct a basis we proceed with $0\leq i\leq{n\choose 2}$, starting by
the diagonal of the string generated by the Vandermonde in
$x_{1},x_{2},\ldots,x_{n}$ in bi-degree ${n\choose 2},0$ and ending with the
Vandermonde in $y_{1},y_{2},\ldots,y_{n}$ in bi-degree $0,{n\choose 2}$. Then
process all the bi-degrees in the shorter diagonals as $i$ becomes
progressively smaller.

Say that in bi-degree $a,b$ the Frobenius characteristic of $DHA_{n}$ tells us
that there are $m$ independent bi-homogeneous polynomials. To construct such a
set we process from smallest to biggest the co-partition expansions breaking
ties in the lex order of the partition sequence. We must eventually find $m$
independent polynomials of bi-degree $a,b$ if the basis conjectured in [1] is
valid. This completes our proof.

The display above and on the right we exhibited the Frobenius characteristic
of $DHA_{7}$. The integers immediately to the left of the display give the
degrees in $x_{i}^{\prime}s$ of the polynomials exhibited the rows. The
integers at the bottom of the display give the degrees in $y_{i}^{\prime}s$ of
the polynomials exhibited in the columns. The smaller integers under the
diagonals are the number of independent polynomials that occupy that
particular bi-degree. the circles indicate the polynomials indexed by the
partitions of $n=7$. For instance the circle in bi-degree $21,0$ is the
Vandermonde in the $x_{i}^{\prime}s$ and the circle in bi-degree $0,21$ is the
Vandermonde in the $y_{i}^{\prime}s$. Since our guide is the Frobenius
characteristic of $DHA_{n}$, we can see that each of the four polynomials in
bi-degrees $(7,4),(6,5),(5,6),(4,7)$ will be picked up, in the construction of
a basis, by our algorithm on the validity of the conjecture in [1] for $n=7$.
We mention this fact since N. Wallach (see [15]) that predicts, for any number
$k$ of distinct variables, the polynomials that will occur in any basis. What
we are witnessing here, for $k=2$ and $n=7$, the validity of a general result
of this general result.

Theorem 2.3

In $DHA_{n}$ there is a basis of $ker\,F$ of the form

$E_{2}^{a_{2}}E_{3}^{a_{3}}\cdots
E_{n-1}^{a_{n-1}}\Delta_{1^{n}}\,+\,E_{1}{\cal L}_{\bf
Q}[E_{1},E_{2},\cdots,E_{n-1}]\Delta_{1^{n}}$ $None$

where $a_{i}\in Z_{\geq 0}$ and ${\cal L}_{\bf Q}$ denotes the linear span
with coefficients in $\bf Q$.

Proof

It is well known from $sl[2]$ theory that If $W$ is a subspace of ${\bf
C}[x_{1},x_{2},\ldots,x_{n};y_{1},y_{2},\ldots,y_{n}]$ that is invariant under
$E$ and $F$ then

$W\,=\,E_{1}W\,\oplus\,\ker\,F|_{W}.$ $None$

Moreover any irreduble representation of dimension $k+1$ starts with a
polynomial $v_{0}$ such that

$a)\enskip\enskip F\,v_{0}=0,\hskip 36.135pt\hskip 36.135ptb)\enskip\enskip
H\,v_{0}\,=\,-k\,v_{0}.$ $None$

Then a basis of the representation is

$v_{0},Ev_{0},E^{2}v_{0},\cdots,E^{k}v_{0}$ $None$

Thus if $f\in DHA_{n}$ then

$f\,=\,{\cal L}_{\bf
Q}[E_{1},E_{2},\cdots,E_{n-1}]\Delta_{1^{n}}\,+\,E_{1}{\cal L}_{\bf
Q}[E_{1},E_{2},\cdots,E_{n-1}]\Delta_{1^{n}}.$ $None$

From the operator theorem it follows that there is a basis of the form

${\cal L}_{\bf Q}[E_{2},E_{3},\cdots,E_{n-1}]\Delta_{1^{n}}.$ $None$

If $F\,f=0$ then we can assure that

$f\,=\,{\cal L}_{\bf
Q}[E_{2},E_{3},\cdots,E_{n-1}]\Delta_{1^{n}}\,+\,E_{1}{\cal L}_{\bf
Q}[E_{1},E_{2},\cdots,E_{n-1}]\Delta_{1^{n}}.$ $None$

This proves the Theorem.

Acknowledgement: The authors would like to thank Nolan Wallach for his
contributions to the mathematics in this paper. The authors would also like to
thank Marino Romero for numerous conversations about the contents of the
paper.

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