# The defining ideals of conjugacy classes of nilpotent matrices and a
conjecture of Weyman

Riccardo Biagioli Institut Camille Jordan, UMR 5208 du CNRS, Université de
Lyon, Université Lyon 1<EMAIL_ADDRESS>Sara Faridi Department of
Mathematics, Dalhousie University, Halifax, Canada<EMAIL_ADDRESS>(research supported by NSERC) Mercedes Rosas Departamento de Álgebra,
Universidad de Sevilla<EMAIL_ADDRESS>(research supported by a Ramón y Cajal
grant, MEC)

###### Abstract

Tanisaki introduced generating sets for the defining ideals of the schematic
intersections of the closure of conjugacy classes of nilpotent matrices with
the set of diagonal matrices. These ideals are naturally labeled by integer
partitions. Given such a partition $\lambda$, we define several methods to
produce a reduced generating set for the associated ideal
${\mathcal{I}}_{\lambda}$. For particular shapes we find nice generating sets.
By comparing our sets with some generating sets of ${\mathcal{I}}_{\lambda}$
arising from a work of Weyman, we find a counterexample to a related
conjecture of Weyman.

##  1 Introduction

Let $X$ be the set of $n\times n$ matrices over a field $k$ of characteristic
$0$. In his paper Kostant [K] showed that the ideal of polynomial functions
vanishing on the set of nilpotent matrices in $X$, is given by the invariants
of the action by conjugation of $GL(n)$ on $X$. Let $C_{\lambda}$ be the
conjugacy class of nilpotent matrices in $X$ having Jordan block sizes
$\lambda^{\prime}_{1},\ldots,\lambda^{\prime}_{h}$, with $\lambda$ a partition
of $n$ and $\lambda^{\prime}$ its transpose. Let $\overline{C}_{\lambda}$ be
the nilpotent orbit variety defined as the Zariski closure of $C_{\lambda}$.
De Concini and Procesi [DP] asked for a description of the ideal
${\mathcal{J}}_{\lambda}$ of polynomial functions vanishing on $C_{\lambda}$,
for a general partition $\lambda$. They were interested in a refinement of
Kostant’s result, which corresponds to the case $\lambda=(1^{n})$. De Concini
and Procesi described a set of elements of ${\mathcal{J}}_{\lambda}$ that they
conjectured to be a generating set. Later, Tanisaki [T] conjectured a simpler
generating set, and Eisenbud and Saltman [ES] generalized Tanisaki’s
conjecture to rank varieties. Finally, in 1989 Weyman [W1] used geometric
methods to show that the three conjectures hold, and conjectured a minimal
generating set $\mathcal{W}_{\lambda}$ for these ideals.

In the present paper we focus on a related family of ideals that we denote by
${\mathcal{I}}_{\lambda}$ and call _De Concini-Procesi ideals_. These are the
ideals of the scheme-theoretic intersection of nilpotent orbit varieties
$\overline{C}_{\lambda}$ with the set of diagonal matrices. De Concini and
Procesi [DP] produced a set of generators for these ideals that was later
simplified by Tanisaki [T]. In both cases, the sets of generators are highly
nonminimal. In the case $\lambda=(1^{n})$, Kostant’s theorem implies that the
elementary symmetric functions of the eigenvalues of the matrices give a
minimal set of generators for ${\mathcal{I}}_{(1^{n})}$.

Our work in this paper is motivated by the search for a minimal generating set
for De Concini–Procesi ideals. To this end, we simplify the generating set
described by Tanisaki using elementary facts of the theory of symmetric
functions. We provide several reduction methods. The obtained sets are minimal
in special cases, and are generally much smaller. The main tool we use is a
special filling of the Young diagram of the partition $\lambda$ which we call
the _regular filling_.

Clearly, by adding the defining ideal of the diagonal matrices to any
generating set for the ideal ${\mathcal{J}}_{\lambda}$, we obtain a generating
set for ${\mathcal{I}}_{\lambda}$. The following question is natural: Is it
true that, after adding these generators to Weyman’s conjectured minimal
generating set for ${\mathcal{J}}_{\lambda}$, a minimal generating set for
${\mathcal{I}}_{\lambda}$ is obtained ? We give a negative answer to this
question and provide some infinite families of counterexamples. With the help
of Macaulay 2 we verify that one of these counterexamples is also a
counterexample to the original conjecture of Weyman on a minimal generating
set of ${\mathcal{J}}_{\lambda}$. This has been a well studied problem that
has been open for the past seventeen years. We hope that our methods together
with those of Weyman will eventually lead to a complete solution of the
problem of finding a minimal generating set for both ideals
${\mathcal{I}}_{\lambda}$ and ${\mathcal{J}}_{\lambda}$.

Our paper is organized as follows. In Section 2 we introduce some basic tools
from the theory of symmetric functions. In Section 3, we introduced Tanisaki’s
generating set for the De Concini-Procesi ideal, and derive a simple
combinatorial description for it. This leads to a simple rule to read a set of
generators of the ideal directly from a special filling of the Young diagram
of the partition that call the _regular filling_. In Section 4 we show that
only generators read from the top entries of the regular filling are necessary
in order to construct a generating set for ${\mathcal{I}}_{\lambda}$. The
resulting generating set is in a one-to-one correspondence with a generating
set that arises from the work of Weyman [W1]. In the case where the partition
$\lambda$ is a hook, our result coincides with the minimal generating set we
introduced in [BFR]. For a general shape though, this generating set could be
far from minimal. In Section 5 we reduce the number of generators coming from
each column of the Young diagram. Finally in Section 6, we provide many
examples and counterexamples to the modified version of Weyman’s conjecture,
and discuss classes where our reductions work best. Inside those families we
are able to find a counterexample to the original conjecture of Weyman on a
minimal generating set for the ideal ${\mathcal{J}}_{\lambda}$. Throughout the
paper, we raise new questions whose answers could help illuminate the problem
of finding minimal generating sets for ${\mathcal{I}}_{\lambda}$ and
${\mathcal{J}}_{\lambda}$.

##  2 Basic Tools

We will be working in the polynomial ring $R=k[x_{1},\ldots,x_{n}]$, where $k$
may be an arbitrary field of characteristic $0$.

We define a partition of $n\in\mathbb{N}$ to be a finite sequence
$\lambda=(\lambda_{1},\ldots,\lambda_{k})\in\mathbb{N}^{k}$, such that
$\sum_{i=1}^{k}\lambda_{i}=n$ and $\lambda_{1}\geq\ldots\geq\lambda_{k}$. If
$\lambda$ is a partition of $n$ we write $\lambda\vdash n$. The nonzero terms
$\lambda_{i}$ are called parts of $\lambda$. The number of parts of $\lambda$
is called the length of $\lambda$, denoted by $\ell(\lambda)$, so
$\lambda_{i}=0$ if $i>\ell(\lambda)$.

Let $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ be a partition of $n$. The
Young diagram of a partition $\lambda$ is the left-justified array with
$\lambda_{i}$ squares in the $i$-th row, from bottom to top. We use the symbol
$\lambda$ for both a partition and its associated Young diagram. For example,
the diagram of $\lambda=(4,4,2,1)$ is illustrated in Figure 1 on the left.

For a partition $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ we define its
conjugate partition as
$\lambda^{\prime}=(\lambda_{1}^{\prime},\ldots,\lambda_{h}^{\prime})$, where
for each $i\geq 1$, $\lambda_{i}^{\prime}$ is the number of parts of $\lambda$
that are bigger than or equal to $i$. The diagram of $\lambda^{\prime}$ is
obtained by flipping the diagram of $\lambda$ across the diagonal.

${{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip 3.0pt plus 1.0pt minus
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height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&}\kern-0.3pt\cr}}\end{matrix}$ Figure 1: The partition
$\lambda=(4,4,2,1)$ and its conjugate $\lambda^{\prime}=(4,3,2,2)$.

We shall need some basic definitions from the theory of symmetric functions.
First, we introduce the generating series for the elementary and the complete
symmetric polynomials (denoted respectively by $E(S,z)$ and $H(S,z)$). These
series are defined as:

$\displaystyle E(S,z)=\sum_{i\geq 0}z^{i}e_{i}(S)=\prod_{a\in S}(1+za),$ and
$\displaystyle H(S,z)=\sum_{i\geq 0}z^{i}h_{i}(S)=\prod_{a\in
S}\frac{1}{1-za},$ (1)

where $S$ is a set of variables, and $z$ is a formal variable. Therefore, the
elementary symmetric polynomial $e_{r}(S)$ is the sum of all square free
monomials of degree $r$ in the variables of $S$, and the complete symmetric
polynomial $h_{r}(S)$ is the sum of all monomials of degree $r$ in the
variables of $S$.

In order to introduce the monomial symmetric polynomials $m_{\lambda}(S)$, we
say that a monomial ${\bf x}^{s}=x_{1}^{s_{1}}x_{2}^{s_{2}}\cdots
x_{n}^{s_{n}}$ has type $\lambda$, if the partition $\lambda$ is obtained by
rearranging the sequence $(s_{1},s_{2},\ldots,s_{n})$ in weakly descending
order. Given a partition $\lambda$, the monomial symmetric polynomial
$m_{\lambda}=m_{\lambda}(S)$ is defined as

$m_{\lambda}(S)=\sum\mathbf{x}^{s}$

where the sum is taken over all different monomials $\mathbf{x}^{s}$ of type
$\lambda$ and with all variables in $S$.

If $f\in k[x_{1},\ldots,x_{n}]$ is a symmetric polynomial, and
$S\subseteq\\{x_{1},\ldots,x_{n}\\}$, we define $f(S)$ as the evaluation of
$f$ at the set $S$, by setting all variables
$x\in\\{x_{1},\ldots,x_{n}\\}\setminus S$ to be equal to $0$ in $f$. For
instance, $e_{2}(x_{1},x_{3})=x_{1}x_{3}$. The polynomial $f(S)$ is called a
partially symmetric polynomial. In general, it is no longer invariant under
the action of the symmetric group on $n$ letters.

For simplicity, given a symmetric polynomial $f\in k[x_{1},\ldots,x_{n}]$, for
all $1\leq k\leq n$, we will denote by $f(k)$ the following set of partially
symmetric polynomials,

$f(k)=\\{f(S)\mid S\subseteq\\{x_{1},\ldots,x_{n}\\},\ |S|=k\\}.$

For example, let $n=4$, then $e_{2}(3)=\\{x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3},\
x_{1}x_{2}+x_{1}x_{4}+x_{2}x_{4},\ x_{1}x_{3}+x_{1}x_{4}+x_{3}x_{4},\
x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4}\\}.$ Note that if $r>k$ we have
$e_{r}(k)=\emptyset$.

######  Notation.

Let $S\subseteq\\{x_{1},\ldots,x_{n}\\}$. For $x\in S$, and
$I=\\{x_{i_{1}},\ldots,x_{i_{k}}\\}\subseteq S$, we let

$\displaystyle S_{x}$ $\displaystyle=S\setminus\\{x\\}\ \ {\rm and}\ \
S_{i_{1},\ldots,i_{k}}=S\setminus I.$

We shall be using the following elementary lemma later in the paper.

######  Lemma 2.1 (Basic Lemma).

Let $S\subseteq\\{x_{1},\ldots,x_{n}\\}$, $|S|=s$, and let $j\leq s$. Then

  1. 1.

$e_{j}(S)=e_{j}(S_{x})+xe_{j-1}(S_{x})$ for all $x\in S$;

  2. 2.

$\displaystyle\sum_{x\in S}e_{j}(S_{x})=(s-j)e_{j}(S)$;

  3. 3.

$\displaystyle\sum_{x\in S}xe_{j-1}(S_{x})=je_{j}(S)$.

###### Proof.

  1. 1.

Clear.

  2. 2.

Fix a square-free monomial $M$ of degree $j$ appearing in $e_{j}(S)$. Without
loss of generality, assume $M=x_{1}\cdots x_{j}$ and
$S=\\{x_{1},\ldots,x_{s}\\}$. Then each $e_{j}(S_{x_{t}})$ contains exactly
one copy of $M$, for $t=j+1,\ldots,s$. There are exactly $s-j$ such indices
$t$, so $M$ appears $s-j$ times in the left-hand sum.

  3. 3.

We use the equation in Part 1, and sum over all elements of $S$ : $\sum_{x\in
S}e_{j}(S)=\sum_{x\in S}e_{j}(S_{x})+\sum_{x\in S}xe_{j-1}(S_{x})$ so by Part
2 we have $se_{j}(S)=(s-j)e_{j}(S)+\sum_{x\in S}xe_{j-1}(S_{x})$ and hence
$je_{j}(S)=\sum_{x\in S}xe_{j-1}(S_{x}).$

∎

######  Proposition 2.2 (Another presentation of the partially symmetric
polynomials).

Let $S=\\{x_{1},\ldots,x_{n}\\}$, $i\leq n$, and define the ideal
${\mathcal{E}}_{i}(S)=(e_{1}(S),\ldots,e_{i}(S))$ in the polynomial ring
$k[x_{1},\ldots,x_{n}]$. Let $U\subseteq S$ be a subset of cardinality $u$.
Then for $i\leq n-u$ we have

$e_{i}(S\setminus U)=(-1)^{i}h_{i}(U)\mbox{ mod }{\mathcal{E}}_{i}(S).$ (2)

###### Proof.

This result follows from a formal manipulation of the generating functions in
(1). We have

$E(S\setminus U,z)=\prod_{\begin{subarray}{c}a\in S\\\ a\not\in
U\end{subarray}}(1+za)=\frac{\prod_{a\in S}(1+za)}{\prod_{a\in
U}(1+za)}=E(S,z)H(U,-z).$

Therefore, extracting the coefficient of $z^{i}$ from both sides of the
resulting equation $E(S\setminus U,z)=E(S,z)H(U,-z)$ we obtain

$e_{i}(S\setminus U)=\sum_{j=0}^{i}e_{j}(S)(-1)^{i-j}h_{i-j}(U).$

By hypothesis $e_{j}(S)$ is in the ideal for $j=1,\ldots,i$. Since
$e_{0}(S)=1$, the result follows. ∎

##  3 A new combinatorial description of Tanisaki’s generating set for
${\mathcal{I}}_{\lambda}$

In this section, we define a family of ideals ${\mathcal{I}}_{\lambda}$ in the
polynomial ring $R=k[x_{1},\ldots,x_{n}]$ indexed by partitions $\lambda$ of
$n$. The ideal ${\mathcal{I}}_{\lambda}$ was first introduced by De Concini
and Procesi [DP] in order to describe the coordinate ring of the schematic
intersection of the Zariski closure of the conjugacy class of nilpotent
matrices of shape $\lambda$, with the set of diagonal matrices.

In order to manipulate De Concini-Procesi ideals, we use a generating set
defined by Tanisaki [T]. A nice feature of Tanisaki’s generating set is that
its elements are elementary partially symmetric polynomials. Furthermore,
Tanisaki’s proof of the correctness of his generating set is both elegant and
elementary, and it is based on standard linear algebra facts. Finally,
Tanisaki’s generating set has proven to be very fruitful in algebraic
combinatorics, see for example [AB, BG, GP].

Let $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ be a partition of $n$. For the
purpose of the next formula, we add enough zeroes to the end of $\lambda$ so
that it has $n$ terms: $\lambda=(\lambda_{1},\ldots,\lambda_{n})$. For any
$1\leq k\leq n$, we define

$\delta_{k}(\lambda)=\lambda^{\prime}_{n}+\lambda^{\prime}_{n-1}+\ldots+\lambda^{\prime}_{n-k+1}.$
(3)

It is clear that
$\delta_{n}(\lambda)\geq\delta_{n-1}(\lambda)\geq\ldots\geq\delta_{1}(\lambda)$,
and that $\delta_{n}(\lambda)=n$.

######  Theorem 3.1 (Tanisaki’s generating set [T]).

The ideal $\mathcal{I}_{\lambda}$ is generated by the following collection of
elementary partially symmetric polynomials

$\mathcal{I}_{\lambda}=\big{(}e_{r}(k)\mid k=1,\ldots,n,\;{\rm and}\ \ k\geq
r>k-\delta_{k}(\lambda)\big{)}.$ (4)

######  Definition 3.2 (De Concini-Procesi ideal).

We call the ideal $\mathcal{I}_{\lambda}$ defined in Theorem 3.1 the De
Concini-Procesi ideal of the partition $\lambda$.

Since for any partition $\lambda$ of $n$, $\delta_{n}(\lambda)=n$, when we set
$k=n$ in (4) we conclude that $\mathcal{I}_{\lambda}$ contains all the
elementary symmetric polynomials in all the variables $x_{1},\ldots,x_{n}$.

######  Example 3.3.

Let $\lambda=(4,4,2,1,0,0,0,0,0,0,0)\vdash 11$ be the partition appearing in
Figure 1. Then
$(\delta_{1}(\lambda),\ldots,\delta_{11}(\lambda))=(0,0,0,0,0,0,0,2,4,7,11)$.
Hence

$(1-\delta_{1}(\lambda),\ldots,11-\delta_{11}(\lambda))=(1,2,3,4,5,6,7,6,5,3,0).$

Here $n=11$. For $k=1,\ldots,7$ there is no admissible $e_{r}(k)$ in the
generating set described in (4). So the generating set of
$\mathcal{I}_{(4421)}$ consists of the following elements

We now give a simple combinatorial description of the set of generators for
${\mathcal{I}}_{\lambda}$ described in Theorem 3.1, and then demonstrate how
to shorten it so that one can read a reduced generating set for
${\mathcal{I}}_{\lambda}$ directly from the diagram of the partition
$\lambda$. In order to do so we introduce the notion of regular filling.

######  Definition 3.4 (The regular filling of a partition).

Let $\lambda$ be a partition of $n$. Draw its Young diagram and then fill its
cells with the numbers $1,2,\ldots,n$ from top to bottom and from left to
right, skipping the cells in the bottom row, which should be filled at the end
from right to left. This is called the regular filling of $\lambda$, denoted
rf.

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width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to10.0pt{\hrule height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 10$\hss}
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to10.0pt{\hss$\scriptstyle 8$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr}}\end{matrix}$
Figure 2: The regular filling of $(4,4,2,1)$.

######  Definition 3.5 (The reading process).

We associate to any filling $f$ of the Young diagram of $\lambda$ a set of
partial symmetric polynomials, denoted by $\mathcal{G}_{f}(\lambda)$. We read
the elements of this set from the filling as follows. For a given column of
$\lambda$ we add to $\mathcal{G}_{f}(\lambda)$ all the elements of the sets
$e_{r}(k)$, where $k$ is the entry in the bottom cell of the column, and the
degrees $r$’s are given by all the entries in that column.

######  Notation.

From now on, we enumerate columns and rows of a Young diagram from left to
right by starting from zero. So the “first” column will be the $0$-th column;
similarly for rows.

######  Example 3.6.

For the partition $\lambda=(4,4,2,1)$, the regular filling rf is illustrated
in Figure 2. The reading process of this filling gives the set
$\mathcal{G}_{\it rf}(\lambda)$ consisting of: the elementary symmetric
polynomials $e_{1}(x_{1},\ldots,x_{11})$, $e_{2}(x_{1},\ldots,x_{11})$,
$e_{3}(x_{1},\ldots,x_{11})$, $e_{11}(x_{1},\ldots,x_{11})$, coming from the
$0$-th column; the partially symmetric polynomials of the sets
$e_{4}(10),e_{5}(10),e_{10}(10)$ read from the first column,
$e_{6}(9),e_{9}(9)$ from the second column, and $e_{7}(8),e_{8}(8)$ from the
last column.

By using this reading process, we are going to read Tanisaki’s generators from
a special filling.

######  Definition 3.7 (The antidiagonal filling).

Let $\lambda$ be a partition of $n$. Compute the partition $\delta(\lambda)$

$\delta(\lambda)=\delta_{n}(\lambda)\geq\delta_{n-1}(\lambda)\geq\ldots\geq\delta_{1}(\lambda),$

where $\delta_{k}(\lambda)$ is defined as in (3), and draw the Young diagram
of its conjugate $\delta^{\prime}(\lambda)$. Now fill the $0$-th column of
$\delta^{\prime}(\lambda)$ by $1,2,\ldots,n$ from top to bottom, and then fill
the remainder of the diagram so that the filling is constant following each
antidiagonal. We call this the antidiagonal filling of
$\delta^{\prime}(\lambda)$ and denote it by af.

For our running example $\lambda=(4,4,2,1,0^{7})$, we have
$\delta(\lambda)=(11,7,4,2,0^{7})$; the antidiagonal filling of
$\delta^{\prime}(\lambda)$ is given in Figure 3. Note that the bottom entry of
the $k$-th column of $\delta^{\prime}(\lambda)$ is $n-k$.

${{{{{{{{{{{{{{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip 3.0pt plus
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\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
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height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to10.0pt{\hrule height=0.3pt\vss\hbox
to10.0pt{\hss$\scriptstyle 8$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to10.0pt{\hrule height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 7$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to10.0pt{\hrule
height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 11$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to10.0pt{\hrule height=0.3pt\vss\hbox
to10.0pt{\hss$\scriptstyle 10$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to10.0pt{\hrule height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 9$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to10.0pt{\hrule
height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 8$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr}}\end{matrix}$
Figure 3: The antidiagonal filling of $\delta^{\prime}(\lambda)$.

Let $\lambda$ be a partition of $n$. Compute the set $\mathcal{G}_{\it
af}(\delta^{\prime}(\lambda))$ by applying the reading process to the
antidiagonal filling af of $\delta^{\prime}(\lambda)$. We have the following
lemma.

######  Lemma 3.8.

Let $\lambda$ be a partition of $n$. Then Tanisaki’s set of generators is
$\mathcal{G}_{\it af}(\delta^{\prime}(\lambda))$. In particular,

${\mathcal{I}}_{\lambda}=(\mathcal{G}_{\it af}(\delta^{\prime}(\lambda))).$

###### Proof.

Let $\lambda=(\lambda_{1},\ldots,\lambda_{n})$. Compute
$\delta^{\prime}(\lambda)$ and fill its diagram with the antidiagonal filling.
According to Theorem 3.1, to compute Tanisaki’s generating set, we need to
find for which $k$ the interval $[k-\delta_{k}(\lambda)+1,\ldots,k-1,k]$ is
nonempty; clearly this happens when $\delta_{k}(\lambda)>0$.

From the definition of $\delta_{k}(\lambda)$, the only times
$\delta_{k}(\lambda)>0$ is when $k=n-\lambda_{1}+1,\ldots,n$. So we are
considering values $e_{r}(S)$ for sets $S$ such that
$n-\lambda_{1}+1\leq|S|\leq n$. This is an interval of length $\lambda_{1}$,
and the numbers $k=|S|$ we are considering are exactly the entries in the
first row of $\delta^{\prime}(\lambda)$.

Now, fix a column $t$ that has entry $n-t$ in its bottom cell. The generating
set described in Theorem 3.1 has $e_{r}(S)$, where $|S|=n-t$ and
$r=n-t-\delta_{n-t}(\lambda)+1,\ldots,n-t$. Note that there exactly
$\delta_{n-t}(\lambda)$ values that $r$ takes, and that is exactly the size of
the $t$-th column of $\delta^{\prime}(\lambda)$. The mentioned values of $r$
are exactly the entries of the $t$-th column of the antidiagonal filling of
$\delta^{\prime}(\lambda)$. ∎

One can easily check that this procedure applied to the antidiagonal filling
in Figure 3 produces the generators given in the table of Example 3.3.

We are now able to show the main result of this section, namely, that
${\mathcal{I}}_{\lambda}$ is the sum of three simpler ideals. In order to do
so we will use the regular filling.

######  Theorem 3.9.

Let $\lambda$ be a partition of $n$. Fill the diagram of $\lambda$ with the
regular filling, and compute the set $\mathcal{G}_{\it rf}(\lambda)$ by using
the reading process described in Definition 3.5. Then

$\mathcal{I}_{\lambda}=(\mathcal{G}_{\it rf}(\lambda)).$

###### Proof.

Compute the partition $\delta^{\prime}(\lambda)$, fill its diagram with the
antidiagonal filling and read off all of Tanisaki’s generators. By Part 2 of
Lemma 2.1, if $e_{r}(x_{1},\ldots,x_{j})\neq 0$ belongs to the ideal, so does
$e_{r}(x_{1},\ldots,x_{J})$ for any $J>j$. Therefore, for each entry
$r=1,\ldots,n$, we only need to keep the generators coming from the rightmost
occurrence of that $r$ in the antidiagonal filling of
$\delta^{\prime}(\lambda)$. So we delete all other occurrences of $r$ in that
filling, and the corresponding cell. We obtain a filling that contains exactly
one occurrence of each of the numbers from $1$ to $n$. Now observe that the
differences of heights between adjacent columns of $\delta^{\prime}(\lambda)$
are given by the sequence
$\lambda^{\prime}_{1},\ldots,\lambda^{\prime}_{\lambda_{1}}$. So after the
deletion process, explained above, the remaining diagram will have columns of
height $\lambda^{\prime}_{1},\ldots,\lambda^{\prime}_{\lambda_{1}}$. Hence it
is the diagram of our partition $\lambda$. Moreover the resulting is the
regular filling, and we are done. The case of the partition
$\lambda=(4,4,2,1)$ is displayed in Figure 4. ∎

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to10.0pt{\hss$\scriptstyle 5$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
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height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to10.0pt{\hrule height=0.3pt\vss\hbox
to10.0pt{\hss$\scriptstyle 11$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to10.0pt{\hrule height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 10$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to10.0pt{\hrule
height=0.3pt\vss\hbox to10.0pt{\hss$\scriptstyle 9$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to10.0pt{\hrule height=0.3pt\vss\hbox
to10.0pt{\hss$\scriptstyle 8$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr}}\end{matrix}\end{array}$ Figure 4:
From the antidiagonal to the regular filling.

######  Remark 3.10.

Observe that $e_{j}(S)$ for $S$ of cardinality $j$ is a square free monomial
of degree $j$. So once we have all square-free monomials of degree
$n-\lambda_{1}+1$ in our ideal, then we have the ones of higher degree. These
monomials are obtained when we read the generators coming from the rightmost
entry of the bottom row.

The following statement follows easily from the previous remark and Theorem
3.9.

######  Corollary 3.11 (First reduction of Tanisaki’s generating set for
${\mathcal{I}}_{\lambda}$).

Let $\lambda$ be a partition of $n$. Then ${\mathcal{I}}_{\lambda}$ can be
described as the sum of the following three ideals:

${\mathcal{I}}_{\lambda}={\mathcal{M}}_{\lambda}+{\mathcal{E}}_{\lambda}+{\mathcal{K}}_{\lambda},$

where

  * •

${\mathcal{M}}_{\lambda}$ is generated by all square-free monomials of degree
$n-\lambda_{1}+1$;

  * •

${\mathcal{E}}_{\lambda}$ is generated by the elementary symmetric polynomials
$e_{1}(x_{1},\ldots,x_{n}),\ldots,e_{\ell(\lambda)-1}(x_{1},\ldots,x_{n})$;

  * •

${\mathcal{K}}_{\lambda}$ is generated by the partially symmetric polynomials
in $e_{r}(k)$, where $n-1\geq k\geq n-\lambda_{1}+1$, and $r$ in an entry of
the regular filling of $\lambda$, in the same column as $k$, and strictly
above it.

In the particular case where the indexing partition $\lambda$ is a hook, we
recover the minimal generating set for $\mathcal{I}_{\lambda}$ described in
[BFR, Proposition 3.4].

##  4 Second reduction of the generating set for ${\mathcal{I}}_{\lambda}$

Our goal in the rest of the paper is to shave off as many redundant generators
as possible from the generating set given in Corollary 3.11 . It turns out
that only partially symmetric polynomials coming from the top value of each
column are required in the generating set. This finding already gives a large
reduction in the number of generator needed in the generating set of Tanisaki.
Several other reductions will be obtained in the following sections.

Suppose we have a partition $\lambda$ of an integer $n$, and fill the diagram
of $\lambda$ with the regular filling defined in Definition 3.4. For $k\geq 1$
we label the value in the top cell of the $k$-th column with $b_{k}$, as long
as the height of the $k$-th column is $\geq 2$. If the right-most column of
$\lambda$ has height 1, then we label its entry $b_{s}$. This is reflected in
the diagram in Figure 5. Note that with this notation we have

$b_{1}=\lambda^{\prime}_{1},\
b_{2}=\lambda^{\prime}_{1}+\lambda^{\prime}_{2}-1,\ \ldots,\
b_{k}=\lambda^{\prime}_{1}+\ldots+\lambda^{\prime}_{k}-k+1\mbox{ for }k\leq
t,\ b_{s}=n-s,$

where we set

$t=\lambda_{2}-1,\ \mbox{ and }\ s=\lambda_{1}-1.$ (6)

Clearly if $\lambda_{1}=\lambda_{2}$, then $t=s$ and $b_{s}$ does not exist.

${{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{\tiny{\begin{matrix}\vbox{\vskip
3.0pt plus 1.0pt minus
1.0pt\offinterlineskip\halign{&\vbox{#}\kern-\Thickness\cr\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
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width=0.3pt} \kern-0.3pt}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to19.0pt{\hrule height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle 2$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
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height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle b_{1}$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to19.0pt{\hrule height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle b_{2}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle b_{t}$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to19.0pt{\hrule height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle b_{1}-1$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle b_{2}-1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to19.0pt{\hrule height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle b_{3}-1$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
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width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle n-s-1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to19.0pt{\hrule height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle n$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle n-1$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle n-2$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to19.0pt{\hrule height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox to19.0pt{\hss$\scriptstyle n-t$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to19.0pt{\hrule height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to19.0pt{\hrule
height=0.3pt\vss\hbox
to19.0pt{\hss$\scriptstyle\stackrel{{\scriptstyle\scriptscriptstyle{b_{s}=}}}{{\scriptscriptstyle{n-s}}}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr}}\end{matrix}}$ Figure 5: Diagram of a partition $\lambda$ of
$n$ with the regular filling.

By Corollary 3.11) the reduced form of Tanisaki’s generating set for
${\mathcal{I}}_{\lambda}$ is the union of the following sets:

(13)

Our goal here is to show that it is enough to pick only one set of generators
in each column, other than the $0$-th column; namely, the ones coming from the
top values in each column.

######  Theorem 4.1 (Principal reduction of the generating set for
${\mathcal{I}}_{\lambda}$).

Let $\lambda$ be a partition of $n$, and suppose that the diagram of $\lambda$
has been filled as in Figure 5. Then a generating set for
${\mathcal{I}}_{\lambda}$ is

(20)

If $\lambda=(1^{n})$ is the one-column partition, then we also need to add the
element $e_{n}(n)=x_{1}\cdots x_{n}$ to this generating set. If $\lambda=(n)$
is the one-row partition, we only need generators from the last column, in
other words ${\mathcal{I}}_{(n)}=(x_{1},\ldots,x_{n})$.

###### Proof.

We need to show that having in the ideal all generators read from the top
index of each column implies that the other partially symmetric functions
coming from the larger indices in that column also belong to the ideal. We go
column by column, and build a new ideal $I_{\lambda}$ by adding generators
described in (20) for each column of $\lambda$. We show, each time, that
$I_{\lambda}$ contains all the other generators described in (13) (coming from
the same column), and therefore $I_{\lambda}={\mathcal{I}}_{\lambda}$.

  * _Col. 0._

There is nothing to prove here, as we are keeping all the generators
$e_{1}(n),\ldots,e_{b_{1}-1}(n)$.

  * _Col. 1._

Assume that we have $e_{b_{1}}(S)\in I_{\lambda}$ for all $S$ with $|S|=n-1$.
By Part 2 of Lemma 2.1, setting $j=b_{1}$, we see that we have
$e_{b_{1}}(n)\in I_{\lambda}$.

For each $i>b_{1}$, we can assume by induction on $i$ that

$e_{1}(n),\ldots,e_{i-1}(n)\in I_{\lambda}\mbox{ and }\hfill
e_{b_{1}}(n-1),\ldots,e_{i-1}(n-1)\in I_{\lambda}.$

Apply Part 3 of Lemma 2.1 with $j=i$, to see that $e_{i}(n)\in I_{\lambda}$.

Fix a set $S$ with $|S|=n-1$ and $x\notin S$. Let $S^{x}=S\cup\\{x\\}$. Part 1
of Lemma 2.1 implies that

$e_{i}(S)=e_{i}(S^{x})-xe_{i-1}(S)$

which demonstrates that $e_{i}(S)\in I_{\lambda}$. Hence $e_{i}(n-1)\in
I_{\lambda}$.

The fact that the generators $e_{b_{1}}(n-1)$ can be replaced by the powers
$x_{1}^{b_{1}},\ldots,x_{n}^{b_{1}}$ follows directly from Proposition 2.2.
Note that, in particular, we have $e_{i}(n-1)\in I_{\lambda}$, for all $i\geq
b_{1}$.

  * _Col. j._

Suppose $I_{\lambda}$ contains all generators from the previous columns
$0,\ldots,j-1$ as described in (20). Let $|S|=n-j$, and suppose $x\notin S$,
so that $|S^{x}|=n-j+1$, ($S^{x}=S\cup\\{x\\}$). We know by induction that
$I_{\lambda}$ contains $e_{h}(S^{x})$ for all $h\geq b_{j-1}$. Therefore,
since $b_{j}>b_{j-1}$, for $i\geq b_{j}$ we have by Part 1 of Lemma 2.1

$\begin{array}[]{lll}e_{i}(S)&=e_{i}(S^{x})-xe_{i-1}(S)&=-xe_{i-1}(S)\\\
&=-x(e_{i-1}(S^{x})-xe_{i-2}(S))&=x^{2}e_{i-2}(S)\\\
&=x^{2}(e_{i-2}(S^{x})-xe_{i-3}(S))&=-x^{3}e_{i-3}(S)\\\ &\hskip
14.45377pt\vdots&\\\ &=(-1)^{i-b_{j}}x^{i-b_{j}}e_{b_{j}}(S)&(mod\ Col.\
j-1)\end{array}$

This means that once we include $e_{b_{j}}(S)$ in $I_{\lambda}$, we will have
all $e_{i}(S)\in I_{\lambda}$ for $i\geq b_{j}$.

∎

In the case where $\lambda$ is a hook, the generating set described in Theorem
4.1 coincides with the minimal generating set for ${\mathcal{I}}_{\lambda}$
introduced in our earlier work [BFR].

######  Example 4.2.

Let $\lambda=(5,4,4,3)$. Then, the regular filling of $\lambda$ is

${{{{{{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip 3.0pt plus 1.0pt
minus 1.0pt\offinterlineskip\halign{&\vbox{#}\kern-\Thickness\cr\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 4$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
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height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 2$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 5$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 8$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 10$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 3$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 6$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 9$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 11$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 16$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 15$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 14$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 13$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 12$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr}}\end{matrix}$

So the generators of ${\mathcal{I}}_{\lambda}$ are

Later in Example 6.4 we shall further reduce the generating set of this
particular partition.

###  4.1 Remarks on a related work and conjecture of Weyman

We end this section by showing some relations between the generating set of
Theorem 4.1 and two generating sets for ${\mathcal{I}}_{\lambda}$ arising in
the work of Weyman [W1].

In [W1] Weyman uses the representation theory of the general linear group to
construct and study generating sets for the ideal ${\mathcal{J}}_{\lambda}$ of
polynomial functions vanishing on the conjugacy class $\mathcal{C}_{\lambda}$.
The generators in the first family, denoted by $V_{\lambda}$, are expressed as
sums of minors, and come from reducible representations of $GL(n)$. The second
set of generators $U_{\lambda}$, on the other hand, arises from the
irreducible representations of $GL(n)$. The set $U_{\lambda}$ is smaller than
$V_{\lambda}$, but how to compute its elements is not explicit in the paper.

The set $V_{\lambda}$ (respectively $U_{\lambda}$) is given by the disjoint
union of sets $V_{i,p}$ (respectively $U_{i,p}$), where the family of indices
$(i,p)$ can be read off from a special diagram introduced by Weyman; see [W1,
Example (4.5)]. We call this diagram the Weyman diagram of $\lambda$. It is
possible to construct the Weyman diagram of a partition starting from the
antidiagonal filling (see Definition 3.7) as follows. First, consider the
antidiagonal filling of $\delta^{\prime}(\lambda)$, and justify its columns in
such a way that equal entries are now in same rows. Then, replace any entry of
this diagram by an $X$. The resulting picture is the Weyman diagram. In Figure
6 we illustrate the Weyman diagram corresponding to the partition
$\lambda=(4,4,2,1)$. Compare this diagram to the one in Figure 3. Note that if
the top $X$ in the $i$-th column of Weyman diagram of $\lambda$ has
coordinates $(i,p)$, then the top cell of the $i$-th column of the regular
filling of $\lambda$ is filled by $p$.

$\displaystyle\begin{matrix}p=1&\underline{X}&&&&\\\ p=2&\underline{X}&&&&\\\
p=3&\underline{X}&&&&\\\ p=4&\underline{X}&\underline{X}&&&\\\
p=5&\underline{X}&X&&&\\\ p=6&\underline{X}&X&\underline{X}&&\\\
p=7&\underline{X}&X&X&\underline{X}&\\\ p=8&\underline{X}&X&X&X&\\\
p=9&\underline{X}&X&X&&\\\ p=10&\underline{X}&X&&&\\\
p=11&\underline{X}&&&&\\\ i=&0&1&2&3\\\ \end{matrix}$ Figure 6: Weyman diagram
for $\lambda=(4,4,2,1)$.

We would like to remark that Weyman follows a convention opposite to ours when
labelling the ideals ${\mathcal{I}}_{\lambda}$ and ${\mathcal{J}}_{\lambda}$:
he labels $\mathcal{J}_{\lambda}$ the ideal of polynomial functions vanishing
on all nilpotent matrices with Jordan blocks $\lambda_{1},\ldots,\lambda_{n}$,
while we use the transpose. On the other hand, he associates to a partition
$\lambda$ what in our setting would be the Weyman diagram of
$\lambda^{\prime}$. These two facts cancel out, and we do not need to take any
transpose when reading statements involving his diagrams.

######  Definition 4.3 (Weyman’s generating set for $\mathcal{J}_{\lambda}$).

In [W1, Theorem (4.6)] Weyman shows that the ideal ${\mathcal{J}}_{\lambda}$
is generated by the $U_{i,p}$, where the $(i,p)$’s are the coordinates of the
top cells of the columns ($i\geq 1$) of the Weyman diagram of $\lambda$,
together with the invariants $U_{0,p}$ with $1\leq p\leq n$. This result
implies that the ideal ${\mathcal{J}}_{\lambda}$ is also generated by the
$V_{i,p}$ coming from the same set of indices $(i,p)$.

######  Example 4.4.

For the partition $\lambda=(4,4,2,1)$, whose Weyman diagram is in Figure 6,
Weyman’s set $U_{\lambda}$ consists of $U_{0,p}$, with $1\leq p\leq 11$,
$U_{1,4}$, $U_{2,6}$, and $U_{3,7}$ (and similarly for the set $V_{\lambda}$).
The cells $X$ whose coordinates label this generating set are underlined.

After adding the generators for the ideal defining the diagonal matrices to
the two sets $V_{\lambda}$ and $U_{\lambda}$, one gets two generating sets for
${\mathcal{I}}_{\lambda}$; we denote these two generating sets by
$\tilde{V_{\lambda}}$ and $\tilde{U_{\lambda}}$.

Instead of going into the definitions of $V_{\lambda}$ and $U_{\lambda}$ that
can be found in [W1, Section 4], we explicitly state the cardinalities of
their components in order to compare them with our generating set. We
emphasize the fact that Tanisaki’s generators (the ones we use) are easier to
handle than Weyman’s generators. We have that

$|V_{i,p}|={n\choose i}^{2}\ \text{and}\ \ |\tilde{V}_{i,p}|={n\choose i},$

and

$|U_{i,p}|={n\choose i}^{2}-{n\choose i-1}^{2}\ \text{and}\ \
|\tilde{U}_{i,p}|={n\choose i}-{n\choose i-1}.$

It turns out that the cardinalities of the generating set for
${\mathcal{I}}_{\lambda}$ given by the $\tilde{V}_{i,p}$’s and the generating
set given in Theorem 4.1 are the same. Moreover, it is not difficult to
describe a one-to-one correspondence between the two generating sets. Under
this correspondence Weyman’s $V_{i,p}$ generators correspond to our generators
read from the top cell of the $i$-th column of the regular filling, as
described in Theorem 4.1.

Weyman conjectured that a special subset of $U_{\lambda}$ gives a minimal
generating set of ${\mathcal{J}}_{\lambda}$; see Conjecture 5.1 and Remark 5.3
of [W1].

######  Conjecture 4.5 (Weyman’s original conjecture).

Let $\lambda$ be a partition. The set consisting of $U_{0,p}$ for $1\leq
p\leq\ell(\lambda)$, and $U_{i,p}$, where $(i,p)$ labels a top cell of the
$i$-th row (in the Weyman diagram of $\lambda$), such that there are no $X$’s
to the right of or on the line segment joining $(i,p)$ with $(0,1)$, is a
minimal set of generators $\mathcal{W}_{\lambda}$ of
${\mathcal{J}}_{\lambda}$.

A very interesting question is the following.

######  Question 4.6 (Diagonal version of Weyman’s conjecture).

Is the generating set $\tilde{\mathcal{W}}_{\lambda}$ for
${\mathcal{I}}_{\lambda}$ arising from Weyman’s conjecture minimal ?

In the following sections we show that the the answer to this question is
negative. Indeed, we provide some infinite families of counterexamples. These
observations, together with the help of Macaulay 2 led us to the discovery
that even the original conjecture of Weyman (Conjecture 4.5) fails already for
one of the smallest elements in these families.

##  5 Reducing generators of ${\mathcal{I}}_{\lambda}$ of a fixed degree

The aim of this section is to consider the generating set of
${\mathcal{I}}_{\lambda}$ described in Theorem 4.1, and eliminate as many
redundant generators as possible from each column.

######  Proposition 5.1 (Columns of height $>1$).

Let $\lambda$ be a partition whose diagram is represented in Figure 5. For
$k\geq 2$, if the height of the $(k-1)$-st column is $>1$, then we can
eliminate ${n-1\choose k-1}+1$ generators of ${\mathcal{I}}_{\lambda}$ (as
described in (20)) that come from the $k$-th column. Indeed, if $S$ denotes
the set of variables $x_{1},\ldots,x_{n}$, we can eliminate the elements in
the set $\\{e_{b_{k}}(S_{1,i_{2},\ldots,i_{k}})\mid 1<i_{2}<\ldots<i_{k}\leq
n\\}$ and $e_{b_{k}}(S_{2,3,\ldots,k+1})$.

###### Proof.

Let $k>1$, by using Part 2 of Lemma 2.1 we write

$\displaystyle\sum_{j\notin\\{i_{1},\ldots,i_{k-1}\\}}e_{b_{k}}(S_{i_{1},\ldots,i_{k-1},j})=(n-b_{k}-k+1)e_{b_{k}}(S_{i_{1},\ldots,i_{k-1}})\equiv
0\ ({\rm mod}\ \mathcal{I}_{k-1})$ (22)

where $\mathcal{I}_{k-1}$ is the ideal of generators coming from columns $0$
to $k-1$.

So we have a system of $n\choose k-1$ linear homogeneous equations, in
$n\choose k$ variables. In fact we have one equation for each choice of a
$(k-1)$-subset $\\{i_{1},\ldots,i_{k-1}\\}$, and one variable
$e_{b_{k}}(S_{i_{1},\ldots,i_{k-1},j})$ for each $k$-subset
$\\{i_{1},\ldots,i_{k-1},j\\}$.

The matrix associated to this system has columns $J$ indexed by the
$k$-subsets of $\\{1,2,\ldots,n\\}$, and rows $I$ indexed by $k-1$-subsets of
$\\{1,2,\ldots,n\\}$. Equation (22), says that at position $(I,J)$ the entry
will be $1$ if $I\subseteq J$ and $0$ if $I\not\subseteq J$.

We claim that we can drop from the generating set of Theorem 4.1
$e_{b_{k}}(S_{J})$, for all $J$ of cardinality $k$ containing $1$, and
$e_{b_{k}}(S_{2,\ldots,k+1}).$ To prove this it suffices to show that the
submatrix corresponding to these columns has full rank ${n-1\choose k-1}+1$.

We order the columns of this submatrix in this way: we put first the the
columns indexed by a $J$ containing $1$ in alphabetical order, and then column
indexed by $\\{2,\ldots,k+1\\}$. Similarly, we order the rows starting with
those indexed by subsets $I$ that do not contain $1$, in alphabetical order,
and then the row indexed by $\\{1,\ldots,k-1\\}$, and then the other rows in
any order. In Figure 7 two examples are displayed.

The square submatrix given by the first ${n-1\choose k-1}+1$ rows consists of
two blocks. An identity ${n-1\choose k-1}$-matrix together with an additional
row: $(1,\ldots,1,0,\ldots,0)$, with $n-k+1$ ones. In fact, this last row is
indexed by $\\{1,\ldots,k-1\\}$, and the entries are $1$ at columns indexed by
$\\{1,2,\ldots,k-1,j\\}$ for $j>k$, and zero otherwise. By Gauss elimination,
it is easy to see that this submatrix has full rank. ∎

$\begin{array}[]{c|cccccc}&12&13&14&23&24&34\\\ \hline\cr 2&{\bf 1}&{\bf
0}&{\bf 0}&{\bf 1}&1&0\\\ 3&{\bf 0}&{\bf 1}&{\bf 0}&{\bf 1}&0&1\\\ 4&{\bf
0}&{\bf 0}&{\bf 1}&{\bf 0}&1&1\\\ 1&{\bf 1}&{\bf 1}&{\bf 1}&{\bf 0}&0&0\\\
\end{array}\hskip 56.9055pt\begin{array}[]{c|cccccc}&123&124&134&234\\\
\hline\cr 23&{\bf 1}&{\bf 0}&{\bf 0}&{\bf 1}\\\ 24&{\bf 0}&{\bf 1}&{\bf
0}&{\bf 1}\\\ 34&{\bf 0}&{\bf 0}&{\bf 1}&{\bf 1}\\\ 12&{\bf 1}&{\bf 1}&{\bf
0}&{\bf 0}\\\ \hline\cr 13&{\bf 1}&{\bf 0}&{\bf 1}&{\bf 0}\\\ 14&{\bf 0}&{\bf
1}&{\bf 1}&{\bf 0}\\\ \end{array}$ Figure 7: The non-singular submatrices for
$n=4$, $k=2$, and $n=4$, $k=3$.

######  Remark 5.2.

The system (22) has ${n\choose k-1}$ linear equations and ${n\choose k}$
variables. If all the equations are independent, then ${n\choose k-1}$
variables are redundant. Hence only ${n\choose k}-{n\choose k-1}$ of them are
necessary. Then using Gauss elimination we would obtain an explicit generating
set of the same size as Weyman’s $\tilde{U}_{k,p}$. We note that there is no
explicit construction for the generators in $U_{\lambda}$ in Weyman’s paper
[W1].

######  Remark 5.3.

Let $\lambda$ be a partition of $n$ different than $(n)$. As a consequence of
Proposition 5.1, the number of generators coming from the top cell of column
$k$ in our generating set for ${\mathcal{I}}_{\lambda}$ is ${n\choose
k}-{n-1\choose k-1}-1$. On the other hand, and as discussed in Section 4.1 the
corresponding $\tilde{U}_{k,p}$ in Weyman’s generating set consists of
${n\choose k}-{n\choose k-1}$ elements. Since for all partitions other than
$(n)$, we have that $n>k$, we conclude that the difference between the two
sets is ${n-1\choose k-2}-1$, for each $k>2$. For columns $0$, $1$, and $2$
their cardinalities coincide.

We now focus on eliminating generators from a column of height 1.

######  Proposition 5.4 (Columns of height 1).

Let $\lambda$ be a diagram represented in Figure 5. If $s>t\geq 1$, then we
can eliminate ${n-s+t\choose t}$ square-free monomial generators of
${\mathcal{I}}_{\lambda}$ coming from the last column.

###### Proof.

Note that as $n-s>b_{t}$ (see Figure 5), from the proof of Theorem 4.1 we know
that $e_{n-s}(n-t)\in{\mathcal{I}}_{\lambda}$. We now claim that we can drop
monomial generators of the form

$\begin{array}[]{ll}e_{n-s}(S_{1,2,\ldots,s-t,i_{1},\ldots,i_{t}}),&\
s-t<i_{1}<i_{2}<\ldots<i_{t}\leq n\end{array}$

from the generating set for ${\mathcal{I}}_{\lambda}$. Since there are
${n-s+t\choose t}$ such choices for sets $\\{i_{1},\ldots,i_{t}\\}$, this will
settle the statement of the proposition. But this follows from the trivial
identity

$e_{k}(A)=\sum_{\begin{subarray}{c}J\subseteq A\\\
|J|=k\end{subarray}}e_{k}(J),$

which implies

$e_{n-s}(S_{1,2,\ldots,s-t,i_{1},\ldots,i_{t}})=e_{n-s}(S_{i_{1},\ldots,i_{t}})-\sum_{\stackrel{{\scriptstyle\scriptstyle{\mbox{$\scriptstyle\\{j_{1},\ldots,j_{s-t}\\}\neq\\{1,\ldots,s-t\\}$}}}}{{\scriptstyle{\mbox{$\scriptstyle\\{j_{1},\ldots,j_{s-t}\\}\cap\\{i_{1},\ldots,i_{t}\\}=\emptyset$}}}}}e_{n-s}(S_{j_{1},\ldots,j_{s-t},i_{1},\ldots,i_{t}})\in{\mathcal{I}}_{\lambda}.$

∎

Therefore using Propositions 5.1 and 5.4, we have reduced our generating set
to that in the table in Figure 8, using the Vandermonde identity ${n\choose
k}={n-1\choose k-1}+{n-1\choose k}$.

$\begin{array}[]{l|l|l}{\bf Column}&{\bf Generators}&{\bf Number}\\\
\hline\cr&&\\\
0&e_{1}(n),\ldots,e_{b_{1}-1}(n)&b_{1}-1=\lambda^{\prime}_{1}-1\\\
1&x_{1}^{b_{1}},\ldots,x_{n}^{b_{1}}&{n\choose 1}\hfill={n-1\choose 1}+1\\\
2&e_{b_{2}}(n-2)&{n\choose 2}-{n-1\choose 1}-1\hfill={n-1\choose 2}-1\\\
\vdots&\hskip 18.06749pt\vdots&\hskip 25.29494pt\vdots\\\
t&e_{b_{t}}(n-t)&{n\choose t}-{n-1\choose t-1}-1\hfill={n-1\choose t}-1\\\ s\
(\mbox{if }s>t)&e_{n-s}(n-s)&{n\choose s}-{n-s+t\choose t}\end{array}$ Figure
8: Number of generators in each degree in the reduced generating set for
${\mathcal{I}}_{\lambda}$

######  Example 5.5.

Consider the partition $\lambda=(4,4,2,1)$ in Figure 2. Our formula gives 177
generators, but in fact, Macaulay2 verifies that 168 generators are enough.
The extra generators are in degree 7 (see table in Figure 8):

Degrees | Number of generators from Table 8 | Actual number of generators required
---|---|---
1, 2, 3 | 1 in each degree | 1 in each degree
4 | 11 | 11
6 | 44 | 44
7 | 119 | 110

While in many examples such as the previous one, the predictions of the
diagonal version of Weyman’s conjecture are correct, this is not always the
case.

######  Example 5.6.

Consider the partition $\lambda=(5,4,1)$.

Figure 9: The partition $\lambda=(5,4,1)$

We denote by
$\mathcal{I}_{01}=(e_{1}(10),e_{2}(10),x_{1}^{3},\ldots,x_{10}^{3})$ the ideal
generated by the elements of the $0$-th and $1$-st column. Now consider
$e_{4}(8)$ coming from the second column. Let $A\subseteq\\{1,\ldots,n\\}$ be
a subset of of cardinality $8$, and let $B$ be its complement ($|B|=2$). By
Proposition 2.2, we have mod ${\mathcal{E}}_{3}(10)$

$\displaystyle e_{4}(A)\equiv h_{4}(B)=m_{(4)}(B)+m_{(3,1)}(B)+m_{(2,2)}(B).$
(24)

Among the monomial symmetric polynomials appearing in (24), $m_{(4)}$, and
$m_{(3,1)}$ are already in the ${\mathcal{I}}_{01}$, since it contains
$x_{1}^{3},\ldots,x_{n}^{3}.$ So from the second column we only need to add
the set $m_{(2,2)}(2)$ to the generators of $\mathcal{I}_{01}$ to obtain a
bigger ideal denoted $\mathcal{I}_{012}$ included in
${\mathcal{I}}_{\lambda}$. That is, we need to add all generators of the form
$(x_{i}x_{j})^{2}$ for $i<j$.

Now let us consider $e_{5}(A)$, where $|A|=7$ and $B$ is its complement. From
the third column

$-e_{5}(A)\equiv
h_{5}(B)=m_{(5)}(B)+m_{(3,2)}(B)+m_{(4,1)}(B)+m_{(3,1,1)}(B)+m_{(2,2,1)}(B).$
(25)

It is clear that each one of these monomial symmetric polynomials is already
in the ideal ${\mathcal{I}}_{012}$. In fact, every monomial in the first four
summands in (25) contains a power $x_{i}^{3}$, and each element in
$m_{(2,2,1)}(B)$ can be obtained as a combination of elements in
$m_{(2,2)}(2)$. Hence the third column will not contribute any new generator.
The same happens for the last column. Let $|A|=6$ and $B$ be its complement,
$|B|=4$. Then

$\displaystyle e_{6}(A)=h_{6}(B)$ $\displaystyle=$ $\displaystyle
m_{(6)}(B)+m_{(5,1)}(B)+m_{(4,2)}(B)+m_{(3,3)}(B)$ $\displaystyle+$
$\displaystyle m_{(4,1,1)}(B)+m_{(3,2,1)}(B)+m_{(2,2,2)}(B)$ $\displaystyle+$
$\displaystyle m_{(3,1,1,1)}(B)+m_{(2,2,1,1)}(B),$

and all monomials in this sum are already in the ideal, since they contain
either a power $x_{i}^{3}$, or a monomial $(x_{i}x_{j})^{2}$. So we have
${\mathcal{I}}_{\lambda}={\mathcal{I}}_{012}$.

######  Counterexample 5.7 (Counterexample to the diagonal version of Weyman’s
conjecture).

Example 5.6 proves that the generating set $\tilde{\mathcal{W}}_{\lambda}$ for
${\mathcal{I}}_{\lambda}$ coming from the minimal generating set for
${\mathcal{J}}_{\lambda}$ conjectured by Weyman is not in general minimal (see
Question 4.6). More precisely, according to his diagram in Figure 10, some
generators of degree $5$ and $6$ should be needed, while they are not, as we
just showed. In Figure 10 the coordinates of the underlined $X$’s label the
generators of ${\mathcal{I}}_{\lambda}$ arising from the diagonal version of
Weyman’s conjecture. The generators coming from the shaded $X$’s are not
needed. This is the convention that we shall use later as well.

$\displaystyle\begin{matrix}p=1&\underline{X}&&&&\\\ p=2&\underline{X}&&&&\\\
p=3&\underline{X}&\underline{X}&&&\\\ p=4&X&X&\underline{X}&&\\\
p=5&X&X&X&\hbox{\pagecolor{shade}$\underline{X}$}&\\\
p=6&X&X&X&X&\hbox{\pagecolor{shade}$\underline{X}$}\\\ p=7&X&X&X&X&\\\
p=8&X&X&X&&\\\ p=9&X&X&&&\\\ p=10&X&&&&\\\ i=&0&1&2&3&4\\\ \end{matrix}$
Figure 10: Weyman diagram for $\lambda=(5,4,1)$.

It might be possible to generalize the reasoning used in Example 5.6 with an
algorithm, as explained below.

######  Algorithm 5.8.

Consider the Young diagram of $\lambda$ filled with the regular filling. Let
$b_{1},\ldots,b_{s}$ be the top-cell entries of $\lambda$ as in Figure 5. Set
$\mathcal{G}_{0}=\\{e_{1}(n),\ldots,e_{b_{1}-1}(n)\\}$, and create a list of
partitions $L_{0}=\emptyset.$ For all $k\geq 1$, define

$U_{k}=\\{\mu\vdash b_{k}\,|\,\ell(\mu)\leq k\ \text{ and}\
\nu\not\subseteq\mu,\ \mbox{for any}\ \nu\in L_{k-1}\\},$

where $\nu\subseteq\mu$ means that the Young diagram of $\nu$ is contained in
that of $\mu$.

  * 1)

If $|U_{k}|=1$, say $U_{k}=\\{\theta\\}$, then $L_{k}=L_{k-1}\cup\\{\theta\\}$
and $\mathcal{G}_{k}=\mathcal{G}_{k-1}\cup m_{\theta}(k)$.

  * 2)

If $|U_{k}|=0$, then $\mathcal{G}_{k}=\mathcal{G}_{k-1}$ and $L_{k}=L_{k-1}$.

  * 3)

If $|U_{k}|>1$, then
$\mathcal{G}_{k}=\mathcal{G}_{k-1}\bigcup\big{(}\bigcup_{l\geq
k}h_{b_{l}}(l)\big{)}$, and stop.

Denote by $\mathcal{G}$ the set produced by the algorithm at the last step.

######  Question 5.9.

Is the set $\mathcal{G}$ a generating set for ${\mathcal{I}}_{\lambda}$?

Clearly this algorithm produces a subset of the generating set given by the
Theorem 4.1. All generators coming from cells labeled $b_{k}$ satisfying
condition $2)$ in the above algorithm would become redundant.

We used this algorithm to produce generating sets for all families of examples
and counterexamples considered in the next section. Then, we proceeded to
prove their correctness on a one by one basis. A proof of the correctness of
the algorithm would be greatly welcomed.

##  6 Families of examples and a counterexample to Weyman’s conjecture

We conclude the paper by producing simple generating sets for some particular
families of shapes. In particular, this allows us to construct two infinite
families of counterexamples to the diagonal version of Weyman’s conjecture
(Question 4.6), as well as a counterexample to the original conjecture of
Weyman for a minimal generating set of the ideal $\mathcal{J}_{\lambda}$ (see
Conjecture 4.5).

######  Example 6.1 (The case of two-column partitions).

As mentioned above a partition of $n$ of the form $\lambda=(2^{a},1^{c})$,
where $a+c=\ell=\ell(\lambda)$ the length of the partition,
${\mathcal{I}}_{\lambda}$ is generated by $e_{1}(n),\ldots,e_{\ell-1}(n),$
$x_{1}^{\ell},\ldots,x_{n}^{\ell}$.

######  Theorem 6.2 (The case of partially-rectangular partitions).

Let $\lambda$ be a partition of $n$, and let $k>2$ be any integer. If columns
$0,1,\ldots,k-1$ of the Young diagram have the same height, then in the
generating set for the ideal ${\mathcal{I}}_{\lambda}$ described in Theorem
4.1 generators coming from columns $2,\ldots,k$ are redundant.

###### Proof.

The regular filling of the partition $\lambda$ has the following form.

${{{{{{{{{{{{{{{{{{{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip 3.0pt
plus 1.0pt minus
1.0pt\offinterlineskip\halign{&\vbox{#}\kern-\Thickness\cr\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle 1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle g+1$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle 2g+1$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&}\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle 2$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle g+2$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle 2g+2$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle kg+1$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle g$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle 2g$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle 3g$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle n$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to20.0pt{\hrule
height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to20.0pt{\hrule height=0.3pt\vss\hbox
to20.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to20.0pt{\hrule height=0.3pt\vss\hbox to20.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr}\kern-0.3pt\cr}}\end{matrix}$

By Theorem 4.1 and Proposition 2.2, modulo the previous columns, the
generators coming from Column $k$ are of the form

$h_{kg+1}=\sum_{a_{1}+\ldots+a_{k}=kg+1}x_{j_{1}}^{a_{1}}\ldots
x_{j_{k}}^{a_{k}}$

where $1\leq j_{1}\leq\ldots\leq j_{k}\leq n$.

Consider a term $x_{j_{1}}^{a_{1}}\ldots x_{j_{k}}^{a_{k}}$ in the sum above.
We claim that for at least one power $a_{i}$, $a_{i}\geq g+1$, making this
monomial redundant in the presence of the second column generators, which are
the $(g+1)$-st powers of the variables.

To see this, suppose $a_{1}\leq g,\ldots,a_{k}\leq g$. Then we should have
that

$kg+1=a_{1}+\ldots+a_{k}\leq kg$

which is a contradiction. ∎

######  Remark 6.3.

Drawing the Weyman diagram associated to partially rectangular partitions
considered in Theorem 6.2, one can see that the points $(0,1),$ $(1,g+1),$
$(2,2g+1),\ldots,(k,kg+1)$ are collinear because they can successively
obtained by adding the vector $(1,g)$. Therefore, the diagonal version of
Weyman’s conjecture predicts that the generators coming from cells
$(2,2g+1),\ldots,(k,kg+1)$ are redundant. This is true: in fact these are
precisely the redundant cells according to Theorem 6.2.

######  Example 6.4.

Let $\lambda=(5,4,4,3)$ be the partition in Example 4.2. Theorem 6.2 implies
that the generating set for ${\mathcal{I}}_{\lambda}$ consists of the elements
in the second column in the table below (compare with Example 4.2), and the
reduced number from the table in Figure 8 is in the third column. No $7$ and
$10$-degree generators are needed in the generating set. In this case the
prediction of the diagonal version of Weyman’s conjecture was correct: cells
$(2,7)$ and $(3,10)$ are redundant; see Figure 11.

Column | Generators | Numbers from Figure 8
---|---|---
0 | $e_{1}(16),e_{2}(16),e_{3}(16)$ | 3
1 | $x_{1}^{4},\ldots,x_{16}^{4}$ | 16
2 | redundant | –
3 | redundant | –
4 | $e_{12}(12)$ | 1365
| Total | 1384

$\displaystyle\begin{matrix}1&\underline{X}&&&&\\\ 2&\underline{X}&&&&\\\
3&\underline{X}&&&&\\\ 4&\underline{X}&\underline{X}&&&\\\ 5&X&X&&&\\\
6&X&X&&&\\\ 7&X&X&\hbox{\pagecolor{shade}$\underline{X}$}&&\\\ 8&X&X&X&&\\\
9&X&X&X&&\\\ 10&X&X&X&\hbox{\pagecolor{shade}$\underline{X}$}&\\\
11&X&X&X&X&\\\ 12&X&X&X&X&\underline{X}\\\ 13&X&X&X&X&\\\ 14&X&X&X&&\\\
15&X&X&&&\\\ 16&X&&&&\\\ i=&0&1&2&3&4\end{matrix}$ Figure 11: An example of a
partially–rectangular partition $\lambda=(5,4,4,3)$.

######  Corollary 6.5 (The case of rectangular partitions).

For a rectangular partition of $n$ of the form $\lambda=(u^{\ell})$, the
generating set of ${\mathcal{I}}_{\lambda}$ will simply be
$e_{1}(n),\ldots,e_{\ell-1}(n),x_{1}^{\ell},\ldots,x_{n}^{\ell}$, where
$n=u\,\ell.$

######  Corollary 6.6 (The case of two-row partitions).

For a two-row partition of $n$ of the form $\lambda=(u,v)$, a generating set
is given by $e_{1}(n)$, $x_{1}^{2},\ldots,x_{n}^{2}$, and $e_{u}(u)$.

######  Theorem 6.7.

Let $\lambda$ be a partition of $n$.

  1. 1.

If $\lambda=(u^{a},(u-1)^{c})$ with $g=a+c$, then a generating set of
${\mathcal{I}}_{\lambda}$ is given by

$e_{1}(n),\ldots,e_{g-1}(n),x_{1}^{g},\ldots,x_{n}^{g}.$

  2. 2.

If $\lambda=(u^{a},(u-1)^{c},1)$ with $u\geq 3$ and $g=a+c>1$, then
${\mathcal{I}}_{\lambda}$ is generated by

$e_{1}(n),\ldots,e_{g}(n),x_{1}^{g+1},\ldots,x_{n}^{g+1},(x_{1}x_{2})^{g},(x_{1}x_{3})^{g},\ldots,(x_{n-1}x_{n})^{g}.$

  3. 3.

If $\lambda=(u^{a},(u-1)^{c},1,1)$ with $u\geq 4$ and $g=a+c+1>2$, then
${\mathcal{I}}_{\lambda}$ is generated by

$\displaystyle
e_{1}(n),\ldots,e_{g}(n),x_{1}^{g+1},\ldots,x_{n}^{g+1},(x_{i}+x_{j})(x_{i}x_{j})^{g-1}\text{
for all $i\neq j$},\text{and }(x_{i}x_{j}x_{k})^{g-1}\text{ for all $i<j<k$}.$

###### Proof.

  1. 1.

This is an easy consequence of Theorem 6.2.

  2. 2.

The regular filling of $(u^{a},(u-1)^{c},1)$ will be of the form:

${{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip
3.0pt plus 1.0pt minus
1.0pt\offinterlineskip\halign{&\vbox{#}\kern-\Thickness\cr\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle 1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle 2$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle g+1$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle 2g$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle 3g-1$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\stackrel{{\scriptstyle\scriptscriptstyle{lg-l}}}{{\scriptscriptstyle{+2}}}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&}\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\stackrel{{\scriptstyle\scriptscriptstyle{(u-1)g}}}{{\scriptscriptstyle{-u+3}}}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle g$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle n$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle n-1$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle n-2$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle n-3$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle n-l$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\stackrel{{\scriptstyle\scriptscriptstyle{n-u}}}{{\scriptscriptstyle{+1}}}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr}\kern-0.3pt\cr}}\end{matrix}$

Columns $0$ and $1$ clearly provide the generators
$e_{1}(n),\ldots,e_{g}(n),x_{1}^{g+1},\ldots x_{n}^{g+1}$. By Proposition 2.2,
Column $2$ provides generators of the form

$h_{2g}=\sum_{a+b=2g}x_{i}^{a}x_{j}^{b}$

for $1\leq i<j\leq n$. Since we already have $x_{i}^{g+1}$ and $x_{j}^{g+1}$
in the ideal, this sum reduces to the monomial $x_{i}^{g}x_{j}^{g}$. Hence the
third column provides the remaining generators
$(x_{1}x_{2})^{g},(x_{1}x_{3})^{g},\ldots,(x_{n-1}x_{n})^{g}$.

It remains to show that the generators coming from Columns $3,\ldots,u-1$ are
redundant. Let $l$ be any integer such that $3\leq l\leq u-1$. The generators
from Column $l$, by Proposition 2.2 and the fact that we have all $(g+1)$-st
powers of the variables in the ideal, are of the form

$h_{lg-l+2}=\sum_{\stackrel{{\scriptstyle\scriptstyle{a_{1}+\cdots+a_{l}=lg-l+2}}}{{\scriptstyle{a_{1},\ldots,a_{l}\leq
g}}}}x_{i_{1}}^{a_{1}}\ldots x_{i_{l}}^{a_{l}}$

where $1\leq i_{1}<i_{2}<\ldots<i_{l}\leq n$, and in each monomial
$x_{i_{1}}^{a_{1}}\ldots x_{i_{l}}^{a_{l}}$ at most one of the powers $a_{u}$
is equal to $g$. For such a monomial in the sum, we therefore have

$a_{1}+\cdots+a_{l}\leq(l-1)(g-1)+g=lg-l+1\Longrightarrow lg-l+2\leq lg-l+1$

which is a contradiction. So there is no generator from Column $l$ if $l\geq
3$.

  3. 3.

The regular filling of $(u^{a},(u-1)^{c},1,1)$ will be of the following form.

${{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip
3.0pt plus 1.0pt minus
1.0pt\offinterlineskip\halign{&\vbox{#}\kern-\Thickness\cr\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle 1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle 2$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle 3$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle g+1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle 2g-1$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle 3g-3$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\stackrel{{\scriptstyle\scriptscriptstyle{lg-2l}}}{{\scriptscriptstyle{\scriptscriptstyle{+3}}}}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt}\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle g$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\stackrel{{\scriptstyle\scriptscriptstyle{(u-1)g}}}{{\scriptscriptstyle{-2u+5}}}$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle n$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to21.0pt{\hrule height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to21.0pt{\hrule
height=0.3pt\vss\hbox to21.0pt{\hss$\scriptstyle\cdots$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to21.0pt{\hrule height=0.3pt\vss\hbox
to21.0pt{\hss$\scriptstyle$\hss} \vss\hrule height=0.3pt} \vrule width=0.3pt}
\kern-0.3pt }\kern-0.3pt\cr}\kern-0.3pt\cr}}\end{matrix}$

Again Columns $0$ and $1$ provide the generators
$e_{1}(n),\ldots,e_{g}(n),x_{1}^{g+1},\ldots x_{n}^{g+1}$.

By Proposition 2.2, Column $2$ provides generators of the form

$h_{2g-1}=\sum_{a+b=2g-1}x_{i}^{a}x_{j}^{b}$

for $1\leq i<j\leq n$. Since we already have $x_{i}^{g+1}$ and $x_{j}^{g+1}$
in the ideal, we can additionally assume that $a,b\leq g$ for each monomial
$x_{i}^{a}x_{j}^{b}$ in the sum, and so at least one of $a$ or $b$ would have
to be $g-1$ and the other $g$. This produces a generator of the form
$x_{i}^{g}x_{j}^{g-1}+x_{i}^{g-1}x_{j}^{g}=(x_{i}+x_{j})(x_{i}x_{j})^{g-1}$.

Similarly, Column $3$ will produce generators of the form

$h_{3g-3}=\sum_{a+b+c=3g-3}x_{i}^{a}x_{j}^{b}x_{k}^{c}$

for $1\leq i<j<k\leq n$. Once more, we can assume that $a,b,c\leq g$, which
reduces the sum above to

$\begin{array}[]{l}x_{i}^{g-1}x_{j}^{g-1}x_{k}^{g-1}+x_{i}^{g-2}(x_{j}^{g}x_{k}^{g-1}+x_{j}^{g-1}x_{k}^{g})+x_{j}^{g-2}(x_{i}^{g}x_{k}^{g-1}+x_{i}^{g-1}x_{k}^{g})+x_{k}^{g-2}(x_{i}^{g}x_{j}^{g-1}+x_{i}^{g-1}x_{j}^{g})\\\
\\\
=x_{i}^{g-1}x_{j}^{g-1}x_{k}^{g-1}+x_{i}^{g-2}x_{j}^{g-1}x_{k}^{g-1}(x_{j}+x_{k})+x_{j}^{g-2}x_{i}^{g-1}x_{k}^{g-1}(x_{i}+x_{k})+x_{k}^{g-2}x_{i}^{g-1}x_{j}^{g-1}(x_{i}+x_{j}).\end{array}$

The last three summands are in the ideal already (coming from Column $2$), so
the generators from Column $3$ can all be written as
$x_{i}^{g-1}x_{j}^{g-1}x_{k}^{g-1}$ for $1\leq i<j<k\leq n$.

We now need to show that generators coming from Column $l$, where $4\leq l\leq
u-1$ are redundant. The generators from Column $l$, by Proposition 2.2 and the
fact that we have all $(g+1)$-st powers of the variables in the ideal, are of
the form

$h_{lg-2l+3}=\sum_{\stackrel{{\scriptstyle\scriptstyle{a_{1}+\cdots+a_{l}=lg-2l+3}}}{{\scriptstyle{a_{1},\ldots,a_{l}\leq
g}}}}x_{i_{1}}^{a_{1}}\ldots x_{i_{l}}^{a_{l}}$

where $1\leq i_{1}<i_{2}<\ldots<i_{l}\leq n$.

Suppose that $M=x_{i_{1}}^{a_{1}}\ldots x_{i_{l}}^{a_{l}}$ is a monomial in
this sum.

If one of the powers, say $a_{1}$, is equal to $g$, then we must have another
power among $a_{2},\ldots,a_{l}$ that is $g$ or $g-1$. If not, all of
$a_{2},\ldots,a_{l}$ are $\leq g-2$, and we have

$lg-2l+3=a_{1}+\cdots+a_{l}\leq g+(l-1)(g-2)=lg-2l+2$

which is a contradiction. So there is at least another power, say $a_{2}$,
such that $a_{2}\geq g-1$.

     * •

$a_{1}=a_{2}=g$. In this case, we can write

$\begin{array}[]{ll}x_{i_{1}}^{g}x_{i_{2}}^{g}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}&=(x_{i_{1}}+x_{i_{2}})(x_{i_{1}}x_{i_{2}})^{g-1}[1/2x_{i_{2}}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}+1/2x_{i_{1}}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}]\\\
&-1/2x_{i_{1}}^{g+1}x_{i_{2}}^{g-1}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}-1/2x_{i_{1}}^{g-1}x_{i_{2}}^{g+1}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}\end{array}$

All the terms on the right-hand side are already in the ideal, and hence so is
$x_{i_{1}}^{g}x_{i_{2}}^{g}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}$.

     * •

$a_{1}=g$ and $a_{2}=g-1$. In this case, there is another monomial
$M^{\prime}=x_{i_{1}}^{g-1}x_{i_{2}}^{g}x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}$
in the sum as well, and there is exactly one copy of $M$ and one copy of
$M^{\prime}$ in the sum. Now we have

$M+M^{\prime}=(x_{i_{1}}+x_{i_{2}})x_{i_{1}}^{g-1}x_{i_{2}}^{g-1}(x_{i_{3}}^{a_{3}}x_{i_{4}}^{a_{4}}...x_{i_{l}}^{a_{l}}).$

So each such monomial $M$ is paired with a unique monomial $M^{\prime}$ in the
sum, and their sum is already in the ideal.

Now assume that all the powers $a_{1},\ldots,a_{l}$ are $\leq g-1$. If $l-2$
of the powers $a_{1},\ldots,a_{l}$ are $\leq g-2$, then we have

$lg-2l+3=a_{1}+\cdots+a_{l}\leq(l-2)(g-2)+2(g-1)=lg-2l+2$

which is a contradiction. So there are at least 3 powers among
$a_{1},\ldots,a_{l}$ that are equal to $g-1$. But then the monomial
$x_{i_{1}}^{a_{1}}\ldots x_{i_{l}}^{a_{l}}$ is already in
${\mathcal{I}}_{\lambda}$, because it is a multiple of a generator coming from
Column $3$.

∎

######  Corollary 6.8.

Suppose that the first $l+1$ columns of a partition $\lambda$ belong to one of
the three families of shapes described in Theorem 6.7. Then

  * a)

In cases 1 and 2, the generators coming from Columns $3,\ldots,l$ are
redundant. For Columns $0,1,2$ we can use the generators described in Theorem
6.7.

  * b)

In Case 3, the generators coming from columns $4,\ldots,l$ are redundant. For
Columns $0,1,2,3$ we can use the generators described in Theorem 6.7.

######  Counterexample 6.9 (Counterexamples to the diagonal version of
Weyman’s conjecture).

The two infinite families of partitions described in parts 2 and 3 of Theorem
6.7 are counterexamples to the diagonal version of Weyman’s conjecture.
Indeed, according to it, all generators coming from each of the top cells of
their diagrams should be necessary because for $k>0$, the top cells are
collinear (for the first family we can move from one top cell to the next one
by adding the vector $(1,g-1)$, and for the second family, by adding the
vector $(1,g-2)$). But the line containing those points does not pass through
$(0,1)$. Instead it passes through $(0,2)$ for the first family, and through
$(0,3)$ for the second family.

Let $\lambda$ be a partition such that its first $l$ columns belong to one of
the two families of shapes described above, with $l>2$ for the first family
and $l>3$ for the second one. The preceding corollary shows that the
generators coming from Column $k$, with $3<k\leq l$ are redundant. We conclude
that each such $\lambda$ is a counterexample to the diagonal version of
Weyman’s conjecture. A first counterexample was shown in Counterexample 5.7.

######  Example 6.10.

Consider the partition $(5,5,1,1)$ that fits inside one of the families in
Theorem 6.7. As proved in that theorem, the cell containing 7 is redundant.
Translated into the Weyman diagram, this means that the ${\underline{X}}$ in
position $(4,7)$ is redundant (see Figure 12).

${{{{{{{{{{{{{\displaystyle\begin{matrix}\vbox{\vskip 3.0pt plus 1.0pt minus
1.0pt\offinterlineskip\halign{&\vbox{#}\kern-\Thickness\cr\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 1$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 2$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 3$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 4$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 5$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 6$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 7$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt\cr\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 12$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 11$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule
width=0.3pt\vbox to12.0pt{\hrule height=0.3pt\vss\hbox
to12.0pt{\hss$\scriptstyle 10$\hss} \vss\hrule height=0.3pt} \vrule
width=0.3pt} \kern-0.3pt }\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox
to12.0pt{\hrule height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 9$\hss}
\vss\hrule height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt&\hbox{\vrule width=0.3pt\vbox to12.0pt{\hrule
height=0.3pt\vss\hbox to12.0pt{\hss$\scriptstyle 8$\hss} \vss\hrule
height=0.3pt} \vrule width=0.3pt} \kern-0.3pt
}\kern-0.3pt\cr}\kern-0.3pt\cr}}\end{matrix}\hskip
28.45274pt\begin{matrix}1&\underline{X}&&&&&\\\ 2&\underline{X}&&&&&\\\
3&\framebox{$\underline{X}$}&&&&&\\\
4&\underline{X}&\framebox{$\underline{X}$}&&&&\\\
5&X&X&\framebox{$\underline{X}$}&&&\\\ 6&X&X&X&\framebox{$\underline{X}$}&&\\\
7&X&X&X&X&{\framebox{\hbox{\pagecolor{shade}$\underline{X}$}}}&\\\
8&X&X&X&X&X&\\\ 9&X&X&X&X&&\\\ 10&X&X&X&&&\\\ 11&X&X&&&&\\\ 12&X&&&&&\\\
i=&0&1&2&3&4\end{matrix}$ Figure 12: The regular filling and the Weyman
diagram of $\lambda=(5,5,1,1)$.

The following table, computed with Macaulay2, confirms our prediction that the
$275$ degree $7$ generators that should be in the generating set according to
the diagonal version of the conjecture, are not needed.

Degrees | Minimal number of generators
---|---
1, 2, 3 | 1 in each degree
4 | 12
5 | 54
6 | 154
7 | redundant

Theorems 6.2 and 6.7 can be reformulated in a suggestive geometrical way as
special instances of the following statement.

######  Question 6.11.

Let $\lambda$ be a partition and draw the Weyman diagram of $\lambda$. If the
$X^{\prime}s$ at the top of columns $1,2,\ldots,r$ are collinear, and the line
containing them passes through the point $(0,k)$, then are the generators
coming from columns $k+1,\ldots,r$ redundant ?

We have evidence that suggests that this statement is true: it was proven to
be true when $k=1$ in Theorem 6.2, for $k=2$ in Theorem 6.7 Part 1, and for
$k=3$ in Theorem 6.7 Part 2, (see Figure 12: the collinear $X$’s have been
surrounded). For $k=4$, we used Macaulay2 to verify whether the statement is
still true for the smallest possible member of this family, the partition
(6,5,1,1,1) (see Figure 13). As predicted, all degree $9$ generators are
redundant.

$\displaystyle\begin{matrix}1&\underline{X}&&&&&\\\ 2&\underline{X}&&&&&\\\
3&\underline{X}&&&&&\\\ 4&\framebox{$\underline{X}$}&&&&&\\\
5&\underline{X}&\framebox{$\underline{X}$}&&&&\\\
6&X&X&\framebox{$\underline{X}$}&&&\\\ 7&X&X&X&\framebox{$\underline{X}$}&&\\\
8&X&X&X&X&\framebox{$\underline{X}$}&\\\
9&X&X&X&X&X&{\framebox{\hbox{\pagecolor{shade}$\underline{X}$}}}\\\
10&X&X&X&X&X&\\\ 11&X&X&X&X&&\\\ 12&X&X&X&&&\\\ 13&X&X&&&&\\\ 14&X&&&&&\\\
i=&0&1&2&3&4&5\end{matrix}$ Figure 13: An evidence regarding the statement in
Question 6.11 for $\lambda=(6,5,1,1,1)$

Degrees | Minimal number of generators
---|---
1, 2, 3, 4 | 1 in each degree
5 | 14
6 | 77
7 | 273
8 | 637
9 | redundant

###  6.1 Weyman’s original conjecture

To finish our work, we focus our attention at the original conjecture of
Weyman. It seems plausible that those partitions that give counterexamples to
the diagonal version of Weyman’s conjecture are also counterexamples to
Weyman’s original conjecture. We used Macaulay2 to verify if this was the case
for the smallest shape in the families described in Counterexample 6.9.

######  Counterexample 6.12 (Counterexample to Weyman’s original conjecture).

Consider the partition $(4,3,1)$ whose Weyman diagram is represented in Figure
14. The points $(1,3),(2,4)$ and $(3,5)$ are collinear, but the line that
contains them does not pass through $(0,1)$. So according to Weyman’s
conjecture, all these cells contribute generators to a minimal generating set
of $\mathcal{J}_{(4,3,1)}$. However, Theorem 6.7 suggests that the generators
coming from cell $(3,5)$ may be redundant.

$\displaystyle\begin{matrix}1&\underline{X}&&&&&\\\
2&\framebox{$\underline{X}$}&&&&&\\\
3&\underline{X}&\framebox{$\underline{X}$}&&&&\\\
4&X&X&\framebox{$\underline{X}$}&&&\\\
5&X&X&X&{\framebox{\hbox{\pagecolor{shade}$\underline{X}$}}}&&\\\
6&X&X&X&&&\\\ 7&X&X&&&&\\\ 8&X&&&&&\\\ i=&0&1&2&3\end{matrix}$ Figure 14:  A
counterexample to Weyman’s original conjecture: $(4,3,1)$.

Using Macaulay 2, we computed the minimal generating set for
$\mathcal{J}_{(4,3,1)}$ and verified that this is indeed the case. We conclude
that $(4,3,1)$ is a counterexample to Weyman’s original conjecture.

Degrees | Weyman’s conjecture | Minimal number of generators
---|---|---
1 | 1 | 1
2 | 1 | 1
3 | 64 | 64
4 | 720 | 720
5 | 2352 | redundant
Total | 3138 | 786

To summarize, in this particular case, Weyman’s conjecture predicts that we
need 3138 generators, but only $786$ of them are really necessary.

Unfortunately, even large servers were not able to handle slightly larger
examples, so at this point we do not know if other partitions in the families
described earlier are counterexamples to Weyman’s original conjecture.

We end the paper with a natural question.

######  Question 6.13.

Does the statement of Question 6.11 hold for $\mathcal{J}_{\lambda}$ ?

## Acknowledgments

We wish to thank Jerzy Weyman for many interesting conversations and
suggestions, as well as his interest in our project. At the same time that we
were working on this project, he showed independently that $(4,3,1)$ is indeed
a counterexample to his Conjecture 5.1 using geometric reasoning [W3].
Moreover, he has shown that Conjecture 5.1 holds for all the other partitions
of $n\leq 9$ except for the three partitions of $9$ belonging to the shapes
described in Counterexample 6.9. We also wish to Mark Shimozono pointing out
some references and Emmanuel Briand for his help during this project.

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