# The Canonical Component of the nilfibre for Parabolic adjoint action in type
$A$

Yasmine Fittouhi and Anthony Joseph

Department of Mathematics

The University of Haifa

Haifa, 3498838, Israel

<EMAIL_ADDRESS>
Donald Frey Professional Chair

Department of Mathematics

The Weizmann Institute of Science

Rehovot, 7610001, Israel

<EMAIL_ADDRESS>
Key Words: Invariants, Parabolic adjoint action, the nilfibre.

AMS Classification: 17B35

Abstract.

This work is a continuation of [Y. Fittouhi and A. Joseph, Parabolic adjoint
action, Weierstrass Sections and components of the nilfibre in type $A$].

Let $P$ be a parabolic subgroup of an irreducible simple algebraic group $G$.
Let $P^{\prime}$ be the derived group of $P$ and let $\mathfrak{m}$ be the Lie
algebra of the nilradical of $P$.

A theorem of Richardson implies that the subalgebra
$\mathbb{C}[\mathfrak{m}]^{P^{\prime}}$, spanned by the $P$ semi-invariants in
$\mathbb{C}[\mathfrak{m}]$, is polynomial.

A linear subvariety $e+V$ of $\mathfrak{m}$ is is called a Weierstrass section
for the action of $P^{\prime}$ on $\mathfrak{m}$, if the restriction map
induces an isomorphism of $\mathbb{C}[\mathfrak{m}]^{P^{\prime}}$ onto
$\mathbb{C}[e+V]$. Thus a Weierstrass section can exist only if the latter is
polynomial, but even when this holds its existence is far from assured.

Let $\mathscr{N}$ be zero locus of the augmentation
$\mathbb{C}[\mathfrak{m}]^{P^{\prime}}_{+}$. It is called the nilfibre
relative to this action.

Suppose $G=SL(n,\mathbb{C})$, and $P$ a parabolic subgroup.

In [Y. Fittouhi and A. Joseph, loc. cit.], the existence of a Weierstrass
section $e+V$ in $\mathfrak{m}$ was established by a general combinatorial
construction. Notably $e\in\mathscr{N}$ and is a sum of root vectors with
linearly independent roots. The Weierstraass section $e+V$ looks very
different for different choices of parabolics but nevertheless has a uniform
construction and exists in all cases. It is called the “canonical Weierstrass
section”.

Through [Y. Fittouhi and A. Joseph, loc. cit. Prop. 6.9.2, Cor. 6.9.8], there
is always a “canonical” component $\mathscr{N}^{e}$ of $\mathscr{N}$
containing $e$.

It was announced in [Y. Fittouhi and A. Joseph, loc. cit., Prop. 6.10.4] that
one may augment $e$ to an element $e_{VS}$ by adjoining root vectors. Then the
linear span $E_{VS}$ of these root vectors lies in $\mathscr{N}^{e}$ and its
closure is just $\mathscr{N}^{e}$.

Yet this same result shows that $\mathscr{N}^{e}$ need not admit a dense $P$
orbit [Y. Fittouhi and A. Joseph, loc. cit., Lemma 6.10.7].

For the above [Y. Fittouhi and A. Joseph, loc. cit., Theorem 6.10.3] was
needed. However thhis theorem was only verified in the special case needed to
obtain the example showing that $\mathscr{N}^{e}$ may fail to admit a dense
$P$ orbit. Here a general proof is given (Theorem 4.4.5).

Finally a map from compositions to the set of distinct non-negative integers
is defined. Its image is shown to determine the canonical Weierstrass section.

One may anticipate that the remaining components of $\mathscr{N}$ can be
similarly described. However this is a long story and will be postponed for a
subsequent paper.

These results should form a template for general type.

##  1\. Introduction

The base field is the field of complex numbers $\mathbb{C}$. For every
positive integer $n$, we set $[1,n]=\\{1,2,,\ldots\,n\\}$.

### 1.1.

Let $G$ be a simple, connected algebraic group over $\mathbb{C}$ and
$\mathfrak{g}$ its Lie algebra. Let $\mathfrak{h}$ be a Cartan subalgebra of
$\mathfrak{g}$. Let $H$ be the unique connected closed subgroup of $G$ with
Lie algebra $\mathfrak{h}$.

Let $\Delta\subset\mathfrak{h}^{*}$ be the set of roots for the pair
$(\mathfrak{g},\mathfrak{h})$. Let $\Delta^{+}$ be a choice of positive roots
and $\pi$ the resulting set of simple roots. Let $\mathfrak{n}$ (resp.
$\mathfrak{n}^{-}$) be the subalgebra of $\mathfrak{g}$ spanned by the
positive (resp. negative) root vectors. Then
$\mathfrak{b}:=\mathfrak{h}+\mathfrak{n}$ is a Borel subalgebra of
$\mathfrak{g}$. It will remain fixed throughout.

The Weyl group of $G$ is the subgroup $W$ of $\operatorname{Aut}\mathfrak{h}$
generated by the reflections $s_{\alpha}:\alpha\in\Delta$, where
$s_{\alpha}h=h-\alpha(h)\alpha^{\vee}$, for all $h\in\mathfrak{h}$, with
$\alpha^{\vee}$ the coroot to $\alpha$.

### 1.2.

A (standard) parabolic subalgebra of $\mathfrak{g}$ is one containing
$\mathfrak{b}$.

For any subset $\pi^{\prime}\subset\pi$, let $\mathfrak{r}_{\pi^{\prime}}$ be
the reductive Lie subalgebra of $\mathfrak{g}$ generated over $\mathfrak{h}$
by the root vectors $x_{\alpha},x_{-\alpha};\alpha\in\pi^{\prime}$. It is the
Levi factor of a unique standard parabolic subalgebra
$\mathfrak{p}_{\pi^{\prime}}$. It is complemented by the nilradical
$\mathfrak{m}_{\pi^{\prime}}$ of $\mathfrak{p}_{\pi^{\prime}}$ and indeed
$\mathfrak{m}_{\pi^{\prime}}$ is the span of the root subspaces
$x_{\beta}:\beta\in\Delta^{+}\setminus\mathbb{N}\pi^{\prime}\cap\Delta^{+}$.
All standard parabolics so obtain.

In what follows the $\pi^{\prime}$ subscript will often be omitted.

Let $\mathfrak{p}^{\prime}$ denote the derived algebra of $\mathfrak{p}$. Let
$P$ (resp. $P^{\prime}$) denote the unique closed subgroup of $G$ with Lie
algebra $\mathfrak{p}$ (resp. $\mathfrak{p}^{\prime}$).

### 1.3.

As a consequence of a theorem of Richardson,
$\mathbb{C}[\mathfrak{m}]^{P^{\prime}}$ is a polynomial algebra - [4, 2.2.3].

The nilfibre $\mathscr{N}$ for the action of $P$ on $\mathfrak{m}$ is defined
to be the zero locus of the augmentation
$\mathbb{C}[\mathfrak{m}]^{P^{\prime}}_{+}$.

Given $e\in\mathfrak{m}$ and $V$ a linear subspace of $\mathfrak{m}$, one
calls $e+V$ a linear subvariety of $\mathfrak{m}$.

A Weierstrass section for the action of $P$ on $\mathfrak{m}$ is a linear
subvariety $e+V\subset\mathfrak{m}$ such that restriction $\varphi$ of
functions induces an isomorphism of $\mathbb{C}[\mathfrak{m}]^{P^{\prime}}$
onto $\mathbb{C}[e+V]$.

A Weierstrass section realizes rather explicitly the polynomiality of
$\mathbb{C}[\mathfrak{m}]^{P^{\prime}}$. It also has the significant geometric
interpretation that every $P^{\prime}$ orbit meeting $e+V$, meets the latter
at just one point, so providing a canonical representative in that orbit.
Polynomiality is not enough to ensure the existence of a Weierstrass section
[4, 1.2.7].

The main result in [5, 5.4.12] was the existence of a Weierstrass section
$e+V$ for the action of a parabolic on its nilradical in type $A$. The proof
was rather combinatorial but nevertheless essentially canonical. We call $e+V$
the canonical Weierstrass section.

A main point is that $e$ belongs to a canonical component $\mathscr{N}^{e}$ of
$\mathscr{N}$ which can be determined. We described $\mathscr{N}^{e}$ in some
detail in [5, Cor. 6.9.8], which we reproduce here in 2.7. Here we start by
extending this description proving in particular [5, Prop. 6.10.4], given here
as Prop. 4.5.3, which we previously only verified in some needed special
cases.

### 1.4.

In order to place our results in a form that may more easily generalize to all
$G$ simple, let us describe how the canonical Weierstrass section $e+V$, noted
in 1.3, determines the canonical component $\mathscr{N}^{e}$. Then one can ask
whether from a component of $\mathscr{N}$ one may construct a Weierstrass
section.

Call $e\in\mathscr{N}$ regular if it generates a dense $P$ orbit in a
component of $\mathscr{N}$ thereby determining this component.

Let $\mathscr{N}_{reg}$ denote the subset of all regular elements in
$\mathscr{N}$.

Of course a Weierstrass section is by no means unique. However when
$e\in\mathscr{N}_{reg}$, then any other Weierstrass section with this property
is determined up to a choice of component and up to conjugation.

Thus an “abstract” description of all Weierstrass sections should start by
describing the $P$ orbits in $\mathscr{N}_{reg}$. Then given
$e\in\mathscr{N}_{reg}$ one should choose $V$ as a vector space complement to
$\mathfrak{p}.e$ in $\mathfrak{m}$. This is how the “Kostant Section” is
obtained when $\mathfrak{p}=\mathfrak{g}$. In that case $\mathscr{N}_{reg}$ is
irreducible.

A first difficulty is that in the construction of a Weierstrass section $e+V$
of [5, 5.4.12] the resulting $e$ may be “too small” to be regular.

This is partly overcome by augmenting $e$ to an element $e_{VS}$ by adjoining
[5, 6.10.4] canonically chosen co-ordinate vectors to $e$, which we call $VS$
elements following the work of Victoria Sevostyanova [9].

A second (serious) difficulty is that $\mathscr{N}_{reg}$ may be empty. In [5,
Lemma 6.10.7] it was shown that a component of $\mathscr{N}$ may contain no
regular orbit.

Yet in our construction, $e_{VS}$ is a sum of weight vectors. Call $e_{VS}$
quasi-regular if the corresponding direct sum $E_{VS}$ of weight subspaces has
the property that $P.E_{VS}$ is dense in some necessarily unique component
$\mathscr{N}^{e}$ of $\mathscr{N}$. Then $\mathscr{N}^{e}$ is determined by
$e_{VS}$ and so by $e$. It is called the canonical component.

In Section 3, we give a proof of the quasi-regularity $e_{VS}$. It is by no
means easy.

The main point in the distinction between regular and quasi-regular elements
is that $H.e_{VS}$ need not be dense in $E_{VS}$ \- [5, 6.10.7].

Our aim is to show that every component of $\mathscr{N}$ admits a quasi-
regular element. However it is by no means obvious that the existence of a
quasi-regular element should lead to a Weierstrass section, nor that such a
construction would extend to all types. For the moment it remains a signpost
along the road to the solution of an immensely difficult problem.

### 1.5.

An orbital variety closure is the closure of a set of the form
$B.(\mathfrak{n}\cap w(\mathfrak{n})):w\in W$. It is called a hypersurface
orbital variety if it is also a hypersurface in some nilradical of a parabolic
subalgebra. As noted in [4, Lemma 2.3.4] the generators of
$\mathbb{C}[\mathfrak{m}]^{P^{\prime}}$ are put into bijection with the
hypersurface orbital varieties in $\mathfrak{m}$, by taking zero loci. This
holds for all $\mathfrak{g}$ simple, but only in type $A$ can one easily
determine the hypersurface orbital varieties [6, Prop. 2.17]. They have been
described in classical types by Perelman [8] but the details are far from
transparent and so difficult to translate to a general case-free form. Here as
briefly discussed in [4, 3.4], the real difficulty lies in the subtle nature
of the Springer construction of simple Weyl group modules associated to non-
trivial representations of the component group of a given nilpotent orbit and
as a consequence we do not know how to determine such modules arising from
hypersurface orbital varieties (in type $E$) - see [4, Remark 3.4.9].

### 1.6.

More generally let $\mathfrak{u}$ be a subalgebra of $\mathfrak{n}$. The $B$
saturation set defined by $\mathfrak{u}$ is just the closure
$\overline{B.\mathfrak{u}}$ of $B.\mathfrak{u}$. Remarkably the canonical
component $\mathscr{C}$ takes this form [5, Cor. 6.9.8] and moreover [5,
6.9.7] has dimension $d:=\dim\mathfrak{m}-g$. The proof is far from easy.

Our eventual aim is to show that every irreducible component of the nilfibre
is a $B$ saturation set defined by a subalgebra of $\mathfrak{n}$ and to
determine this subalgebra (which need not be unique). In general this $B$
saturation set is not an orbital variety closure. As noted in [5, 6.9.9], this
holds if and only if $\dim G.\mathfrak{u}=2\dim B.\mathfrak{u}$. Examples
$3,4$ of [5, 6.10.9] show that this can hold and this can fail.

##  2\. Type $A$ \- The Zero Locus of the Benlolo-Sanderson Invariant.

### 2.1.

From now on we assume that $G$ is simple of type $A$, that is isomorphic to
$SL(n,\mathbb{C})$ for some integer $n>1$. In this case the Weyl group $W$ is
just the symmetric group $S_{n}$ on $n$ letters. Again the component group pf
any orbit is trivial and so Springer theory recovers the classical fact that
there is a bijection between nilpotent orbits and simple Weyl group modules
(in type $A$).

Let $\textbf{M}_{n}$, or simply M, denote the set of $n\times n$ matrices over
$\mathbb{C}$ and in this we write we write $x_{i,j}:i,j\in[1,n]$, for the
standard matrix units. It is sometimes called the $(i,j)^{th}$ co-ordinate
function, or co-ordinate vector, which assigns to M its value at its
$(i,j)^{th}$ entry.

Then the Lie algebra $\mathfrak{sl}(n,\mathbb{C})$ of $SL(n,\mathbb{C})$ is
just the subspace $\textbf{M}_{n}$ of traceless matrices over $\mathbb{C}$
given a Lie bracket through the commutator.

Let $\alpha_{i}:i\in[1,n-1]$ be the simple roots in the Bourbaki notation [2]
and set $\alpha_{i,j}:=\alpha_{i}+\cdots+\alpha_{j-1}$. If $i<j$, then
$x_{i,j}$ is the root vector $x_{\alpha_{i,j}}\in\mathfrak{n}$.

### 2.2.

Let $\pi^{\prime}$ be a subset of $\pi$. Let $W_{\pi^{\prime}}$ be the
subgroup of $W$ generated by the simple reflections
$s_{\alpha}:\alpha\in\pi^{\prime}$ and $w_{\pi^{\prime}}$ it unique longest
element.

Recall that the Levi factor $\mathfrak{r}_{\pi^{\prime}}$ of
$\mathfrak{p}_{\pi^{\prime}}$ is given by a set of $c_{i}\times c_{i}$ blocks
$\\{B_{i}\\}_{i=1}^{k}$ on the diagonal of $\textbf{M}_{n}$, where the
$c_{i}-1$ are the cardinalities of the connected components of $\pi^{\prime}$.
One has $\sum_{i=1}^{k}c_{i}=n$. It is complemented in
$\mathfrak{p}_{\pi^{\prime}}$ by the nilradical $\mathfrak{m}_{\pi^{\prime}}$
of $\mathfrak{p}_{\pi^{\prime}}$, lying in $\textbf{M}_{n}$ above this set of
blocks.

Often $\pi^{\prime}$ will be viewed as being fixed and the $\pi^{\prime}$
subscript omitted.

Let $\textbf{C}_{i}$ denote the rectangular block in $\mathfrak{m}$ lying
strictly above $B_{i}$. We call it the $i^{th}$ column block. Its width is
$c_{i}$ and its height $\sum_{j<i}c_{j}$. In this presentation,
$\mathfrak{m}=\oplus_{i=2}^{k}\textbf{C}_{i}$.

### 2.3.

We regard $\mathfrak{p}$ as being defined by the composition
$(c_{1},c_{2},\ldots,c_{k})$, hence by a diagram $\mathscr{D}_{\mathfrak{m}}$
formed by a set of columns $\\{C_{i}\\}_{i=1}^{k}$ labelled from left to right
with ht $C_{i}=c_{i}$.

Let $\\{R_{j}\\}_{j=1}^{\ell}$ be the rows of $\mathscr{D}_{\mathfrak{m}}$
labelled with increasing positive integers from top to bottom. For all
$i\in\mathbb{N}^{+}$, let $R^{i}$ denote the union of the rows
$\\{R_{j}\\}_{j=1}^{i}$.

All columns are deemed to start from the top so meeting row $R_{1}$.

Let $\mathscr{T}_{\mathfrak{m}}$, be the tableau obtained from
$\mathscr{D}_{\mathfrak{m}}$ in which the integers $1,2,\ldots,n$ are inserted
sequentially, first down the columns and then on going from left to right. In
this we often drop the $\mathfrak{m}$ subscript, since $\mathfrak{m}$ will
usually be fixed and to avoid cumbersome notation.

This presentation has the following property. Let $b,b^{\prime}$ be boxes of
$\mathscr{T}$ with $b^{\prime}$ strictly to the right of $b$. Let $i$ (resp.
$j$) be the entry of $b$ (resp. $b^{\prime}$). Then $x_{i,j}\in\mathfrak{m}$
and these elements form a basis of $\mathfrak{m}$. Let $\ell_{b,b^{\prime}}$
be the line joining $b,b^{\prime}$. In the above labelling it is sometimes
written as $\ell_{i,j}$ and said to join $i,j$. It defines the co-ordinate
vector $x_{i,j}$.

$\mathscr{T}$ need not be a standard tableau because column heights need not
be increasing, in other words there may be gaps in the rows of $\mathscr{T}$.
Yet $\mathscr{T}$ becomes a standard tableau when we shift boxes in a given
row from right to left to close up gaps. Indeed the content of each row does
not change and in the new tableau entries still increase down the columns.
Then through the Robinson-Schensted correspondence and a result of
Spaltenstein, $\mathscr{T}$ corresponds to an orbital variety for
$\mathfrak{sl}(n)$ and every orbital variety is so obtained exactly once - see
[6, 2.1], for example.

One may recover $\mathscr{D}$ from $\mathscr{T}$ by forgetting the entries.

### 2.4.

As in [4, 4.1.2] we say that two columns of height $s$ of
$\mathscr{T}_{\mathscr{m}}$ are neighbouring if there are no columns of height
$s$ strictly between them. To any pair of neighbouring columns
$C_{v},C_{v^{\prime}}$, there is a Benlolo-Sanderson invariant
$M(v,v^{\prime})$ [4, 4.1.3] which is a polynomial in the co-ordinate
functions $x_{i,j}$, where $\\{i,j|i<j\\}$ are the entries of the columns
between $C_{v},C_{v^{\prime}}$. We will recall the description of this
invariant in 2.6.

The zero set of $M(v,v^{\prime})$ is a hypersurface orbital variety
$\mathscr{V}(v,v^{\prime})$ in $\mathfrak{m}_{\pi^{\prime}}$. It is described
in [5, 2.6] as a $B$ saturation set, specifically as
$\overline{B.\mathfrak{u}_{v,v^{\prime}}}$, with
$\mathfrak{u}_{v,v^{\prime}}=\mathfrak{n}\cap w_{v.v^{\prime}}(\mathfrak{n})$
for a suitable choice of $w_{v,v^{\prime}}\in W$.

### 2.5.

In the above $w_{v,v^{\prime}}$ is obtained by modifying [5, 2.4] the tableau
$\mathscr{T}_{\mathfrak{m}}$. Here the details will be briefly recalled below.
In this the columns outside $C_{v},C_{v^{\prime}}$, play no role.

Being neighbouring columns of height $s$, means that there are no columns of
height $s$ strictly between $C_{v},C_{v^{\prime}}$. Let $m$ be the entry in
the box $C_{v^{\prime}}\cap R_{s}$.

Then as in [5, 2.6], ignoring columns of height $<s$ shift $m$ into the
rightmost column of height $\geq s$ and shift simultaneously the parts of the
columns strictly below $R_{s}$, one position to the left.

This defines the modified tableau $\mathscr{T}_{\mathfrak{m}}(v,v^{\prime})$.
For example, if every column strictly between $C_{v},C_{v^{\prime}}$ has
height strictly less than $s$, then $\mathscr{T}_{\mathfrak{m}}(v,v^{\prime})$
is obtained by removing $m$ from $C_{v^{\prime}}$ and placing it in the box
$R_{s+1}\cap C_{v}$.

Now define $w_{v,v^{\prime}}$ in word form by reading the entries of
$\mathscr{T}_{\mathfrak{m}}(n)$ from bottom to top and then from left to
right.

One always has $\mathfrak{u}_{v,v^{\prime}}\subset\mathfrak{m}$ via [5, Lemma
2.5].

If there are no columns of height $>s$ between $C_{v},C_{v^{\prime}}$, then in
the matrix presentation of $\mathfrak{m}$ we obtain
$\mathfrak{u}_{v,v^{\prime}}$ by removing the last column from $\mathfrak{m}$,
when $C_{v^{\prime}}$ is the last column of $\mathscr{D}$ and if not in a
suitable truncation of $\mathfrak{m}$.

The general description of $\mathfrak{u}_{v,v^{\prime}}$ is more complicated.
It is determined through the root vectors of $\mathfrak{n}$ that do not appear
in $\mathfrak{u}$ \- called the excluded root vectors. They are given in [5,
2.7].

### 2.6.

Let $C_{v},C_{v^{\prime}}$ are neighbouring columns of height $s$. If they are
not the first and last columns of $\mathscr{D}$, then we may truncate M to
make this so. (See also the reduction in [4, 3.2].)

Let $M_{s}(v,v^{\prime})$ be the $n-s\times n-s$ minor in the bottom left hand
corner of the truncated M.

Let $P^{-}$ denote the opposed parabolic to $P$. Then $M_{s}(v,v^{\prime})$ is
a $P^{-}$ semi-invariant for co-adjoint action.

We consider $M_{s}(v,v^{\prime})$ as a polynomial function on
$\mathfrak{m}+\operatorname{Id}$, through the Killing form. In particular it
is not homogeneous and it is not necessarily a $P^{\prime}$ invariant for
adjoint action. Remarkably its leading term
$\operatorname{gr}M_{s}(v,v^{\prime})$ is a $P^{\prime}$ invariant [1] for
adjoint action, which we call the Benlolo-Sanderson invariant111A suggestion
as to how this should generalize for all types is given in [4, 3.2, 3.5,
3.6].. As noted in [1], its degree $d(\operatorname{gr}M_{s}(v,v^{\prime}))$
is given by

$d(\operatorname{gr}M_{s}(v,v^{\prime}))=\sum_{i=v}^{v^{\prime}-1}\min(c_{i},s).$

The following is proved in [5, 2.6] by an argument similar to that of [6,
Prop. 3.8].

######  Proposition.

The closure of $B.(\mathfrak{n}\cap w_{v,v^{\prime}}(\mathfrak{n}))$ is the
zero variety of $\operatorname{gr}M_{s}(v,v^{\prime})$ in $\mathfrak{m}$.

### 2.7.

Fix a parabolic $\mathfrak{p}$ and as in 2.3, let $\mathscr{D}$ denote the
diagram it defines. Let $\mathscr{P}$ be the set of all pairs of neighbouring
columns $C_{v},C_{v^{\prime}}$ of $\mathscr{D}$. For ease of noatation we
write $C_{v},C_{v^{\prime}}$ as $v,v^{\prime}$.

By Proposition 2.6 and [4, 4.1.2], the nilfibre $\mathscr{N}$ for the action
of $\mathfrak{p}$ on its nilradical is given by

$\mathscr{N}=\cap_{{v,v^{\prime}}\in\mathscr{P}}\overline{B.\mathfrak{u}_{v,v^{\prime}}}.$

Obviously this contains

$\mathscr{C}:=\overline{B.\cap_{{v,v^{\prime}}\in\mathscr{P}}\mathfrak{u}_{v,v^{\prime}}}.$

We showed in [5, 6.9.8] that rather remarkably, the latter is an irreducible
component of $\mathscr{N}$ and is just the canonical component
$\mathscr{N}^{e}$ referred to in 1.4.

##  3\. The Description of the Canonical Weierstrass Section

### 3.1.

Start from the tableau $\mathscr{T}$ defined by the nilradical $\mathfrak{m}$
of a standard parabolic as in 2.3.

We may describe a (rather particular) linear subvariety $e+V$ of
$\mathfrak{m}$ by a family of lines $\ell_{i,j}$ joining boxes in
$\mathscr{T}$ with labels $i<j$.

By 2.3, the line $\ell_{i,j}$ defines a co-ordinate vector
$x_{i,j}\in\mathfrak{m}$.

Then $e+V$ is defined by two families of lines. Those of the first family
carry the label $1$ and specify the sum of root vectors giving $e$ and those
of the second family are labelled by $\ast$ and specify the sum of root
subspaces giving $V$.

N.B. We say interchangeably that a line $\ell$ is labelled by a $1$ (or a
$\ast$) or carries a $1$ (or a $\ast$).

In this we call the support of $e$, denoted $\operatorname{supp}{e}$, (resp.
support of $V$, denoted $\operatorname{supp}{V}$) those co-ordinate vectors
(resp. subspaces) defined by the lines carrying a $1$ (resp. a $\ast$).

A line carrying a $\ast$ may be “gated” (see [5, 4.1].

The actual choice of the labelled lines defining $e+V$ leading to it being a
Weierstrass section is via a complicated combinatorial construction which is
nevertheless essentially canonical. The details are given in [5, Sect. 5].
They will not be repeated here, but in 3.4 we describe the resulting lines by
an algorithm which is surprisingly simple.

In 3.6 we show that these lines with their labels can be recovered very simply
from the image of a map from compositions to the set of distinct non-negative
integers together with the set of column heights. This “composition” map may
well be of independent interest.

Recall the notation of 2.1. We may view $e+V$ as a linear subvariety of
$\mathfrak{m}$, by putting $1$ (resp. $\ast$) at the $(i,j)^{th}$ place of M,
whenever in our previous convention the line $\ell_{i,j}$ carries a $1$ (resp.
$\ast$).

We call $e+V$ the canonical Weierstrass section, since it has a uniform
construction for all parabolics.

### 3.2.

#### 3.2.1.

Following 2.3, let $\mathscr{C}=(C_{1},C_{2},\ldots,C_{k})$ be the set of
columns in $\mathscr{D}$ defining the Levi factor of a parabolic subalgebra
$\mathfrak{p}$ (in $\mathfrak{sl}(n)$), whose block sizes are the
$\\{c_{i}\\}_{i=1}^{k}$.

Let $\mathscr{T}$ be the tableau obtained from $\mathscr{D}$ by the
prescription given in 2.3.

We denote the set of labelled lines in $\mathscr{T}$ between columns in
$\mathscr{C}$ given through [5, 5.4] by $\ell(\mathscr{C})$.

#### 3.2.2.

By [5, 5.4.9] $\ell(\mathscr{C})$ behaves well under removing (or adjoining)
columns on the right.

Indeed let $S$ denote the set of (ordered) columns obtained from $\mathscr{C}$
by deleting the last column $C_{k}$. Then

######  Lemma.

$\ell(S)$ is obtained from $\ell(\mathscr{C})$ by deleting $C_{k}$ and all the
lines from $S$ to $C_{k}$.

#### 3.2.3.

This fails for removal of columns from the left but we have the next best
thing described below.

To make the comparison with 3.2.2 less confusing we adjoin a column $C_{0}$ on
the left of $\mathscr{C}$ and denote the resulting set of (ordered) columns by
$T$.

######  Lemma.

Suppose $C_{0}$ has height $i$. Concerning only the lines lying entirely in
$R^{i-1}$, one obtains $\ell(\mathscr{C})$ from $\ell(T)$ by deleting $C_{0}$
and all the lines from $\mathscr{C}$ to $C_{0}$.

This is proved in a slightly stronger form in [5, 5.4.10].

### 3.3.

To complete our description of $e$ we need to describe how our construction
adjoins lines in the last column to the earlier ones. This is achieved in the
following subsections.

Let $b_{i,j}$ denote the box in $\mathscr{D}$ lying in $R_{i}\cap C_{j}$. We
may also denote a box by its entry $l$, that is as $b(l)$.

#### 3.3.1.

Recall the notion of 3.2. In particular $C_{k}$ denotes the last column of
$\mathscr{D}$. Recall the notation of 2.1, 2.2.

Since we are considering the height $s$ of $C_{k}$ to be a variable (taking
non-negative integer values), it is convenient to write $C_{k}$ (resp.
$\textbf{C}_{k}$) more precisely as $C_{k}(s)$ (resp. $\textbf{C}_{k}(s)$)
when this height is $s\in\mathbb{N}$. Similarly we write M as $\textbf{M}(s)$,
when its last column block is $\textbf{C}_{k}(s)$.

We may note that $n$ is the largest entry of $C_{k}$ and occurs in its lowest
row.

Recall that $S$ denotes the set of columns (resp. column blocks) excluding the
last, that is $S=\\{C_{1},C_{2},\ldots,C_{k-1}\\}$. Let S denote the
corresponding set of column blocks, that is
$\\{\textbf{C}_{1},\textbf{C}_{2},\ldots,\textbf{C}_{k-1}\\}$.

Recall 3.1 that M has entries in $\\{1,*\\}$ which determine the Weierstrass
section $e+V$. These entries all occur in the sub-matrix defining
$\mathfrak{m}$.

Let $\textbf{M}^{k}$ be the $n-s\times n-s$ matrix formed by omitting the last
$s$ rows and columns of M. The blocks $\\{B_{i}\\}_{i=1}^{k-1}$ (resp. the
column blocks $\\{\textbf{C}_{i}\\}_{i=2}^{k-1}$) define the Levi factor
$\mathfrak{l}^{\kappa}$ and nilradical $\mathfrak{m}^{\kappa}$ of a standard
parabolic subalgebra $\mathfrak{p}^{\kappa}$ of $\mathfrak{sl(n-s)}$.

By Lemma 3.2.2, the entries in S are unchanged if we omit the last column
$C_{k}(s)$ or increase its height $s$. Thus if we define $e^{\kappa}$ (resp.
$V^{\kappa}$) by dropping all terms which obtain from lines joining $C_{k}$ to
$S$, then $e^{\kappa}+V^{\kappa}$ is a Weierstrass section for the action of
the derived algebra of $\mathfrak{p}^{\kappa}$ on $\mathfrak{m}^{\kappa}$.

N.B.. The notation $C_{k}$ was already used in [5, 2.1] for the last column of
$\mathscr{D}$. However in Sections 4.4 and 4.5. it also convenient to use $k$
as a dummy index for the co-ordinates. This could cause some confusion (for
example in Lemma 4.4.3). Thus we have used $\kappa$ instead of $k$ as the
superscript above.

#### 3.3.2.

Let $\ell_{b,b^{\prime}}$ be a line given by the construction of a Weierstrass
section, that is by [5, 5.4]. It may carry a $1$ or a $\ast$. Suppose $b$
occurs in $R_{i}$ and labelled in $\mathscr{D}$ by $r$. Then a $1$ or a $\ast$
appears in the $r^{th}$ row of M. Yet $i,r$ are only vaguely related, so it is
convenient to denote the row in which a $1$ or a $\ast$ appears in M using
boldface to avoid confusion, that is by r, which will always refer to a row of
M.

Let $C,C^{\prime}$ be a pair of neighbouring columns. Then $\ast$ always
appears in the rightmost column of the column block $\textbf{C}^{\prime}$, [5,
6.6.2]. The row of M in which it appears is also specified in [5, 6.6.2],
however this is more complicated. It is through the canonical row associated
to the pair which will now be denoted in boldface, that is by
$\textbf{r}_{C,C^{\prime}}$. It is the row of M in which $\ast$ ought to
appear and in fact does appear if the pair $C,C^{\prime}$ does not “surround”
any other pair of neighbouring columns in the sense of [5, 6.6.2], as recalled
below.

We say that a pair $C_{1},C_{1}^{\prime}$ of neighbouring columns surround the
pair $C,C^{\prime}$, if ht$C_{1}^{\prime}=\text{ht}C+1$, and $C_{1}^{\prime}$
lies strictly to the right of $C^{\prime}$ with no columns of height $\geq s$
between. We do not require $C_{1}$ to lie to the left of $C$ \- see examples
below. In this case and if there is a pair $C_{1},C_{1}^{\prime}$ which do
surround the pair $C,C^{\prime}$, then the $\ast$ corresponding to the latter
pair again occurs in $\textbf{r}_{C,C^{\prime}}$.

This process gives rise to a string of $\ast$’s in
$\textbf{r}_{C,C^{\prime}}$. This string may or may not end in a $1$ \- see
[5, 6.6.2, Examples].

One may ask how $\textbf{r}_{C,C^{\prime}}$ and
$\textbf{r}_{C_{1},C_{1}^{\prime}}$ are related. The latter is strictly less
than the former (that is to say the second $\ast$ is pushed downwards in M) if
$C_{1}$ lies to the left of $C$, for example in the composition $(2,1,1,2)$.
Yet the latter is strictly greater than the former (that is to say the second
$\ast$ is pushed upwards in M) if $C_{1}$ lies to right of $C$, for example in
the composition $(1,2,1,2)$.

###  3.4. Increasing the height $s$ of $C_{k}$

Given a column $C$ of height $s$ and $t\leq s$, let $\textbf{C}^{t}$ denote
the $t^{th}$ column of the corresponding column block C. Again the labelling
of the columns in $\mathscr{D}$ and in M is different but related.

####  3.4.1. Two Preliminaries

Recall the notation of 3.3.1. By 3.3.2, if $C_{k}(s)$ has a left neighbour in
$S$, then $\textbf{C}_{k}(s)$ has just one $*$ entry and this occurs in
$\textbf{C}_{k}(s)^{s}$, which is the rightmost column of $\textbf{C}_{k}(s)$,
and has no $\ast$ entries otherwise. Again by [5, Lemma 5.4.8(vi)] every
column and every row of $\textbf{C}_{k}(s)$ has at most a single $1$ entry.

Recall [5, 5.4.1] that a right extremal box of $S$ is one with no adjacent
right neighbour in $\mathscr{C}$. Such a box exists exactly when $S$ admits a
rightmost column $C$ of height $>s$. In (b) below, we let $b$ denote the box
$R_{s+1}\cap C$, which is clearly right extremal and denote its entry by $m$.

Suppose $C_{k}$ has height one. This means that $\textbf{C}_{k}$ is a single
column. Let $t$ be the height of $C_{k-1}$. Then either

(a). If $S$ has no column of height one, then the only entry of
$\textbf{C}_{k}$ is $1$ on its $(n-t)^{th}$ row.

(b). If $S$ has a column of height one, then $\textbf{C}_{k}$ has an $*$ on
its $(n-t)^{th}$ row. Moreover with $b,m$ given as above, then
$\textbf{C}_{k}$ has a $1$ on row m (in its unique column).

For (a) and the first part of (b), observe that step two of [5, 5.3] gives a
horizontal line $\ell$ carrying a $1$ (resp $\ast$) in the first row $R_{1}$
joining the highest box $b^{\prime}$ in $C_{k-1}$ to a unique box in $C_{k}$.
Moreover by [5, 5.4] in step $3$, the horizontal lines in $R_{1}$ and their
labels are left unchanged.

The last part of (b) follows from [5, 5.4.7(i)], which gives a line
$\ell_{b,b_{s,k}}\in\ell(\mathscr{D})$ carrying a $1$, if there is a right
extremal box $b$ in $S\cap R_{s+1}$.

#### 3.4.2.

As we increase the height $s$ of $C_{k}$, then by Lemma 3.2.2, the only
changes which occur are in the entries of $\textbf{C}_{k}$ (and not in the
entries of S). Moreover these behave according to a surprisingly simple set of
rules which we now describe.

The proofs will be given in 3.5.

(i). Suppose $t\in[1,s-1]$. The entries of $\textbf{C}_{k}(s+1)^{t}$ are those
of $\textbf{C}_{k}(s)^{t}$.

We now describe the entries of the last two columns of $\textbf{C}_{k}(s+1)$,
that is when $t$ above is in $\\{s,s+1\\}$.

(ii). Suppose that $S$ has a column of height $s$. Then there is a $\ast$ in
$\textbf{C}_{k}(s)^{s}$. It is replaced by a $1$ in $\textbf{C}_{k}(s+1)^{s}$,
in the same row.

In other words the position of $\ast$ does not change but it becomes a $1$.

(iii). Suppose that $1$ appears in $\textbf{C}_{k}(s)^{s}$ in some row r of M,
then either (A) or (B) below holds.

(A). Suppose there is a $\ast$ in $\textbf{C}_{k}(s)^{s}$ and in row
$\textbf{r}^{\prime}$ (different of course to r).

(This occurs exactly when $S$ has a column of height $s$ and a column of
height $>s$. In this the line $\ell$ carrying $\ast$ is gated and the line
$\ell^{\prime}$ carrying $1$ is constructed by [5, 5.4.7(i) or(ii)] at stage
$s+1$ and both lines have $b_{s,k}$ as their right hand end point.)

Then a $\ast$ (resp. $1$) appears in $\textbf{C}_{k}(s+1)^{s+1}$ and in row r
if $S$ has (resp. does not have) a column of height $s+1$.

In other words $1$ is shifted to the right becoming a $\ast$ or a $1$.

(B). Suppose there is no $\ast$ in $\textbf{C}_{k}(s)$. This occurs exactly
when $S$ has no column of height $s$.

Then a $1$ appears in $\textbf{C}_{k}(s+1)^{s}$ in row r.

In other words this $1$ entry stays fixed as $s$ increases by one.

If $S$ admits a column of height $s+1$, a $\ast$ appears
$\textbf{C}_{k}(s+1)^{s+1}$, at the end of the lowest string (cf 3.3.2) of
$\ast$’s (not ending in a $1$). It is replaced by a $1$ at that place if $S$
meets $R_{s+1}$, but does not admit a column of height $s+1$.

In other words a string of $\ast$’s on a given row of M may be augmented in
its last column by a $\ast$ or a $1$, as prescribed.

Examples. Consider the composition $(3,2,1,1,2,2)$. Then the line
$\ell:=\ell_{9,11}$ carries a $\ast$ and the line $\ell^{\prime}:=\ell_{6,11}$
(obtained by the joining of lose ends [5, 5.4.7(ii)]) carries a $1$. Thus $A$
holds.

Increasing the height of the last column by one, gives a line (namely
$\ell_{6,12})$ carrying a $\ast$. Had the first column been of height four,
this line would have carried a $1$.

By contrast for the composition $(1,2,1)$, the line $\ell:=\ell_{2,4}$ carries
a $\ast$ and the line $\ell^{\prime}:=\ell_{3,4}$ (obtained by [5, 5.4.7(i)])
carries a $1$. Increasing the height of the last column by one, gives a line
(namely $\ell_{3,5})$ carrying a $\ast$. Had the first column been of height
three, this line would have carried a $1$.

Consider the composition $(3,1,1,2)$. There is no $\ast$ in the last column;
but there is a $1$ given by the line $\ell_{4,7}$. Thus $B$ holds.

When the last column is increased in height by one, a $\ast$ appears in
$\textbf{C}_{k}(3)^{3}$. It is replaced by a $1$ if the first column is
increased in height by one.

Suppose that $s$ has been increased by one and (i)-(iii) have be carried out.

N.B. Recall [5, 5.4.8(viii)] that a $1$ in a given row of M cannot be followed
by a $\ast$ in a subsequent column.

Suppose $S$ does have a column of height $s+1$. Then (for $s\geq 1$) the rules
(i)-(iii) may not give a $1$ in the last column of $\textbf{M}(s+1)$.

In this case the construction of [5, 5.4] can add a $1$ to the last column of
$\textbf{C}_{k}(s+1)$ and this follows the rule (iv) given below.

Let $\mathscr{R}$ denote the set of rows of $\textbf{M}(s+1)$ either not
having a $1$ except perhaps in the last column or not having a $\ast$ in its
last column.

Now recall that the elements of Weyl group of the Levi factor of $P$
interchanges the rows of $\textbf{M}(s+1)$ and so induce equivalence classes
in $\mathscr{R}$. If $\mathscr{R}$ is not empty, let $\textbf{r}^{\prime}$ be
the unique lowest row in $\mathscr{R}$ (i.e. that with the largest subscript)
and let r be the unique highest row in the equivalence class of
$\textbf{r}^{\prime}$.

(iv). Suppose $\mathscr{R}$ is not empty. Then $1$ is placed in
$\textbf{C}_{k}(s+1)^{s+1}$ in row r, or already occurs there through
$(i)-(iii)$.

Thus up to the action of the Levi, $1$ in the last column of $\text{M}(s+1)$
occurs in the row below the empty rows.

Remark 1. The complication of having to consider equivalence classes comes
about because our construction though canonical is by induction on rows in
which the highest free box is linked. It is perfectly possible that we would
obtain a Weierstrass section $e+V$ with $e$ regular by putting $1$ in the
intersection of the last column of $\textbf{C}_{k}(s+1)$ with the lowest row
of $\mathscr{R}$, that is with $\textbf{r}^{\prime}$.

Remark 2. In the subsequent application of (iv), the above N.B. is of prime
importance.

Remark 3. Our presentation is perhaps a convoluted way of saying that if $1$
already appears in the last column by (i)-(iii) above, then it must also
follow the rule, set out by (iv).

(v). If the last column of $\textbf{C}_{k}(s)$ is empty, so are the last two
columns of $\textbf{C}_{k}(s+1)$.

### 3.5.

In order to prove the assertions in 3.4, we must recall the construction in
[5, Sect. 5.4]. Unfortunately the reader may have to study loc. cit. first but
we shall give explicit references to every required result.

Recall the notation of 3.3.

Recall that $s$ is the height of $C_{k}(s)$.

On increasing $s$ by one, a left going horizontal line from $b_{s+1,k}$ may be
introduced at step $2$.

Then we apply the $(s+1)^{th}$ stage of [5, 5.4]. Briefly this does just three
things.

$(\textbf{X})$. (Some) horizontal lines in $R_{s+1}$, required by step $2$,
are deleted [5, 5.4.6].

$(\textbf{Y})$. (Some) lines in $R^{s}\setminus R^{s-1}$ carrying a $1$ given
by applying [5, 5.4.7(ii)] at stage $s$, are deleted. (These are the lines
given by “joining loose ends”.) No lines in $R^{s}$ carrying a $\ast$ are
deleted, but may be relabelled by a $1$.

$(\textbf{Z})$. New lines in $R^{s+1}\setminus R^{s}$ are introduced according
to the rule in [5, 5.4.7]. They may carry either a $1$ or a $\ast$.

Recall that since only the last column $C_{k}$ is being changed, only lines
meeting $C_{k}$ are changed (Lemma 3.2.2).

Example. Let $\mathscr{D}$ be given by the composition $(2,1,1,1)$, so then
$k=4,s=1$.

The only lines joining $C_{4}$ to $S$ are $\ell_{4,5}$ which carries a $\ast$
and $\ell_{3,5}$ which carries a $1$ (the latter joins loose ends).

When $s$ is increased by one, a horizontal line $\ell_{2,6}$ is introduced at
step $2$.

At stage $2$, $\ell_{3,5}$ is deleted and the $\ast$ on $\ell_{4,5}$ is
replaced by a $1$, illustrating $(\textbf{Y})$ and $\ell_{2,6}$ is deleted,
illustrating $(\textbf{X})$. The latter is replaced by $\ell_{3,6}$, which
carries a $\ast$, illustrating $(\textbf{Z})$. (Strictly speaking $\ell_{2,4}$
also replaces $\ell_{2,6}$ but is already present when $s=2$ as predicted by
Lemma 3.2.2.)

It remains to consider the assertions of 3.4.2.

Consider $(i)$.

By [5, 5.4.8 (iii),(ix)] a right going line can only go down by one row and
then it must carry a $1$ and $b^{\prime}$ must also be the right end point of
a line carrying a $\ast$. Then by [5, 6.6.2, second paragraph] $b^{\prime}$
must be the box $C_{k}(s)\cap R_{s}$. Thus

$(*)$. Every line with right end point in $R^{s-1}\cap C_{k}(s)$ lies entirely
in $R^{s-1}$.

On the other hand when the height of $C_{k}$ is increased from $s$ to $s+1$
and stage $s+1$ is implemented, the lines lying entirely in $R^{s-1}$ are left
unchanged by $(\textbf{X})-(\textbf{Z})$. Then by $(*)$, the lines joined to a
$b_{t,k}:t<s$ are unchanged (and so are their labels).

This gives the assertion of (i).

Consider (ii).

By hypothesis $S$ has a column of height $s$.

Then the left going line $\ell_{b,b_{s,k}}$ from $b_{s,k}$ to that box $b\in
S$ determined in stage $s$ by [5, 5.4.7(i)], carries a $\ast$ by [5, 5.4.8].
Moreover it lies entirely in $R^{s}$, so by $(\textbf{Y})$, it is not deleted
but now carries a $1$ because $C_{k}(s+1)$ has height $s+1$. This establishes
(ii).

Consider (iii)A.

The hypothesis means that there is a line $\ell$ from $b\in S$ to $b_{s,k}\in
C_{k}(s)$ carrying a $\ast$ and a second line $\ell^{\prime}$ from
$b^{\prime}\in S$ to $b_{s,k}\in C_{k}(s)$ carrying a $1$. This means that
$\ell$ must be gated and the construction in [5, 5.4.7(i)or(ii)] gives a line
$\ell^{\prime}$ carrying a $1$ whose left end point lies in $S$.

As $s$ is increased by one, $\ell$ remains with the same end points but now
carries a $1$ (as noted in (ii)) so can no longer be gated. This forces
$\ell^{\prime}$ to be replaced by a right going line $\ell^{\prime\prime}$ to
$b_{s+1,k}$, which of course is in the same column of $\mathscr{D}$ as
$b_{s,k}$. Then by the minimal distance criterion of [5, 5.4.7], the line
$\ell^{\prime\prime}$ must have the same left-hand point $b^{\prime}$ as had
$\ell^{\prime}$. It carries a $\ast$ (resp. $1$) if $S$ has (resp. does not
have) a column of height $s+1$. The new entry to which $\ell^{\prime\prime}$
gives rise, lies in the same row of M, yet now in the last column of
$C_{k}(s+1)$, because its right hand end point is $b_{s+1,k}$.

This establishes the last part of $(iii)A$.

Examples. In the case of the parabolic defined by the composition $(1,2,2,1)$,
there is a gated line $\ell=\ell_{4,6}$. Then $\ell^{\prime}=\ell_{5,6}$ being
given by (i) of [5, 5.4.7]. When the last column is increased by one in
height, $\ell$ carries a $1$ and $\ell^{\prime}$ is replaced by a line
$\ell^{\prime\prime}=\ell_{5,7}$ and in this case carries a $\ast$.

On the other hand for the composition $(2,3,1,3,2,1,2)$ the gated line $\ell$
is that joining $(10,14)$. There is a further gated line joining $(8,11)$.
Then $\ell^{\prime}$ is given by [5, 5.4.7(ii)] and joins $(8,14)$. When the
last column is increased in height by one, $\ell$ carries a $1$ and
$\ell^{\prime}$ is replaced by a line $\ell^{\prime\prime}$ with the same
beginning point as $\ell^{\prime}$ but ending in $b_{s+1,k}$, that is joining
$(8,15)$ and in this case carries a $\ast$.

Consider (iii)B.

Under the hypothesis there is no line going from $S$ to $b_{s,k}$ carrying a
$\ast$. Consequently the line $\ell$ from $b\in S$ to $b_{s,k}\in C_{k}(s)$
carrying a $1$ is obtained at stage $s$ and moreover by using (i) and not by
(ii) of [5, 5.4.7]. It lies entirely in $R^{s}$, though it need not be
horizontal (viz. $\ell_{4,7}$ of the example below). Thus it remains at stage
$s+1$ by $(\textbf{Y})$.

This proves the first part of (iii)B.

Example. Consider the line joining $(4,7)$ in the composition $(3,1,1,2)$
obtained by applying [5, 5.4.7(i)] at stage $2$. It is carries a $1$ since
there is no column in $S$ of height $2$. When $s$ is increased by $1$, it
remains.

Consider the second part of (iii)(B).

Under the hypothesis, at step $2$ there is a left going horizontal line $\ell$
from $b_{s+1,k}$ if $S$ meets $R_{s+1}$ and no such line otherwise. In the
former case it carries a $\ast$ (resp. a $1$) if $S$ admits (resp. does not
admit) a column of height $s+1$. In step $3$, the right end point of $\ell$ is
unchanged by the way boxes are rejoined [5, 5.4.7(i)] and the labelling is
left unchanged [5, 5.4.8]. Thus a $\ast$ (resp. $1$) appears in
$\textbf{C}_{k}(s+1)^{s+1}$, as asserted.

Example. Consider again the composition $(3,1,1,2)$. When $s$ is increased by
$1$, the new line $\ell^{\prime}$ joins $(3,8)$ and carries a $\ast$. Had the
first column had height $>3$, then $\ell^{\prime}$ would carry a $1$. Either
appear in $R_{3}$ at the end of the lowest string of $\ast$’s not ending in
$1$ (which is actually an empty string in this case).

Consider (iv). (This is the most difficult part.)

There can be at most a single $1$ in $\textbf{C}_{k}(s+1)^{s+1}$. It
corresponds to a line $\ell_{b,b_{k}(s+1)}$ in $\ell(S\sqcup C_{k}(s+1))$,
carrying a $1$. Let $r$ be the entry of $b$. Then $1$ appears in row r of M.
Set $\mathscr{R}^{-}=\mathscr{R}\setminus\textbf{r}$.

Condition (iv) holds if every element of
$\mathscr{R}^{-}\cap\textbf{C}_{k}(s+1)$ lies above r, or below but reached by
the action of the Levi factor. This holds for example if $\mathscr{R}^{-}$ is
empty.

Let $C_{1},C_{2},\ldots,C_{m-1}$ be the columns of height $s+1$ in $S$ and set
$C_{m}:=C_{k}(s+1)$.

We now establish (iv) under

$(*)$. Suppose $S$ has no column of height $>s+1$.

Claim. Under $(*)$, $\mathscr{R}^{-}$ is empty.

Adjoin a column $C_{0}$ to the extreme left of $\mathscr{D}$ of height $s+1$.
Then through the construction of [5, 5.4.7(i)] and the labelling in [5,
5.4.8], there exists for all $i\in[1,m]$ a box $b^{\prime}_{i}\in C_{i}\cap
R_{s+1}$, a box $b_{i}\in R^{s+1}$, which by adjacency [5, 5.4.3], must lie to
the right of $C_{i-1}$ and a line $\ell_{b_{i},b_{i}^{\prime}}$ carrying a
$\ast$.

On the other hand by [5, 5.4.8(v)] every box in $R^{s+1}$, with the exception
(of course) of those in $C_{m}$ and the possible exception of the
$b_{i}:i\in[1,m]$ has a right going line carrying a $1$.

When $C_{0}$ is suppressed, then $b_{1}\in C_{0}$ is also suppressed. If
$b_{1}\notin C_{0}$, one has $\ell_{b_{1},b_{1}^{\prime}}\in\ell(\mathscr{D})$
carrying a $1$, by the joining of loose ends [5, 5.4.7(ii)], and likewise line
$\ell_{b_{i},b_{i+1}^{\prime}}\in\ell(\mathscr{D}):i\in[2,m-1]$ carries a $1$
with the exception of $b_{m}:m\geq 2$, which carries a $\ast$ appearing in the
last column of $\textbf{M}(s+1)$. Up to this exception every box $R^{s+1}$
hence in $\mathscr{D}$, not of course in $C_{m}$, has a right going line
carrying a $1$. This proves the claim.

Remark. $\mathscr{R}$ itself is empty if $m=1$. For the composition $(1,2,2)$
it is empty, for the composition $(1,1,2,2)$ it is not empty.

We now establish (iv) under

$(**)$. There is a column of height $>s+1$ in $S$.

Choose a column $C$ of height $>s+1$ in $S$ nearest to the right. Retain our
previous notation for the columns of height $s+1$.

Choose $t\in[1,m]$ minimal such that $C_{t}$ lies to the right of $C$. Set
$b^{\prime}=C\cap R_{s+2}$ which is an extremal box. By our chosen labelling
$b_{t}^{\prime}$ lies strictly to the right of $C$. If $t\geq 2$, then
$\ell_{b_{t},b_{t}^{\prime}}$ is gated and since [5, 5.4.7(i)] is carried out
before [5, 5.4.7(ii)], there is a line $\ell_{b^{\prime},b_{t}^{\prime}}$
carrying a $1$ and this precludes the line $\ell_{b_{t-1},b_{t}^{\prime}}$.

$(a)$. Suppose $t\leq m-2$.

Then by the joining of loose ends [5, 5.4.7(ii)] there is a line
$\ell_{b_{m-1},b_{m}^{\prime}}$ carrying a $1$. Let $r$ be the entry of
$b_{m-1}$. This puts a $1$ on the last column of $\textbf{M}(s+1)$ in row r.

Yet any box with entry $u>r$ lies between the columns $C_{m-2},C_{m}$ and so
(as in $(*)$) has a right going line carrying a $1$, with the exception of
$\ell_{b_{m-1},b^{\prime}_{m}}$ which carries a $\ast$ appearing in
$\textbf{C}_{k}(s+1)^{s+1}$.

This establishes (iv) if $(a)$ holds.

$(b)$. Suppose $t=m-1$.

The gated line $\ell_{b_{m-1},b^{\prime}_{m-1}}$ cannot cross $C$ by adjacency
[5, 5.4.8(vii)]. Thus $b_{m-1}$ lies in a column to the right of $C$.

Now by the joining of loose ends [5, 5.4.7(ii)] there is a line
$\ell_{b_{m-1},b_{m}^{\prime}}$ carrying a $1$. If $b_{m-1}$ lies in a column
strictly to the right of $C$, then (iv) holds exactly as in $(a)$.

Suppose $b_{m-1}\in C$. Then there may be empty rows meeting the last column
of $\textbf{M}(s+1)$ below r indexed by the entries $r^{\prime}>r$ of the
boxes in $C$ strictly below $b^{\prime}$. However these rows are equivalent to
r via the action of the Levi factor.

This establishes (iv) if $(b)$ holds.

$(c)$. Suppose $t=m:m>1$.

In this case since $\ell_{b_{m},b^{\prime}_{m}}$ is gated, so there is a line
$\ell_{b^{\prime},b^{\prime}_{m}}$ carrying a $1$ and we conclude as in $(b)$.

This establishes (iv) if $(c)$ holds.

$(d)$. Suppose $t=m=1$.

Adjoin a column $C_{0}$ of height $s+1$ to the left of $\mathscr{D}$. Then by
the joining of loose ends [5, 5.4.7(ii)], there is a line
$\ell_{b_{1},b_{1}^{\prime}}$ which carries a $\ast$ and therefore cannot
cross $C$ by adjacency [5, 5.4.8(vii)]. Then by [5, 5.4.10] it remains with
the same end points when $C_{0}$ is removed but $\ast$ is replaced by a $1$.
Then we may conclude as in $(b)$.

This completes the proof of (iv).

The following exemplifies $(*)$.

Example 1. Take the column heights in $S\sqcup C_{k}(s)$ to be given by the
composition $(1,2,2,2,1,2)$, with $s=1$.

In the example, $i=6,j=5$ and [5, 5.4.7(ii)] gives the line $\ell_{5,10}$
carrying a $1$. This line would not have been given by $(1)-(3)$ of 3.4.

The following exemplifies $(**)$ in case $(a)$.

Example 2. Consider the composition $(2,3,2,2,2)$, with $s=1$. Then $1$ occurs
in row 8 and a $\ast$ in row 9 both in the last column of $\textbf{M}(s+1)$.
Through the joining of loose ends $1$ occurs in row 7. This would not have
been given by $(1)-(3)$ of 3.4.

The following exemplifies $(**)$ in case $(b)$, when $b_{m-1}$ lies in a
column strictly to the right of $C$.

Example 3. Consider the composition $(3,1,1,2)$, with $s=1$. The line with
right end point $b_{2,4}$ obtained from step $2$ is just $\ell_{2,7}$. Yet
since $\ell_{4,5}$ is gated, this line is modified in step $3$ to
$\ell_{4,7}$, via [5, 5.4.7(i)]. The resulting $1$ in M lies in row 4, so
below the empty row 3.

The following exemplifies $(**)$ in case $(c)$.

Consider the composition $(4,2,1,3)$. The line with right end point $b_{3,4}$
obtained from step $2$ is just $\ell_{3,10}$. It is not modified in step $3$.
The resulting $1$ in M lies in row 3 so above the empty row 4. Yet these rows
are interchanged by that part of the Levi factor coming from the first column.

Consider (v). Let $|S|$ denote the greatest height of a column in $S$.

Observe that (v) implies that $b_{s,k}$ is a left extremal box and so $s>|S|$.

Again (v) also means that there is no gated line $\ell_{b,b_{s-1,k}}$ in
$R^{s-1}$ in stage $s$. Otherwise there would be a line $\ell_{b,b_{s,k}}$
carrying a $1$ by the very first part of [5, 5.4.7]. Moreover if $s>|S|$ then
by [5, 5.4.8] no lines in $R^{s}$ carrying a $\ast$ are adjoined in stage
$s+1$.

By the first paragraph $s+1>|S|$ and by the second paragraph there are no
gated lines in $R^{s}$ at stage $s+1$. Thus the conclusion of (v) results by a
second use of the very first part of [5, 5.4.7].

###  3.6. Stabilisation

####  3.6.1. A Column Existence Lemma

The following is proved as part of “Claim” in the proof of [5, Cor. 6.5.4].
For the convenience of the reader we shall repeat the argument.

######  Lemma.

Let $C_{1}<C_{2}$ be columns of $\mathscr{D}$ and a choice of boxes $b_{i}\in
C_{i}:i=1,2$. Let $R_{t}$ be the row in which $b_{2}$ occurs. Assume that
$\ell_{b_{1},b_{2}}\in\ell(\mathscr{D})$. Then there exists a column
$C^{\prime}:C_{1}\leq C^{\prime}<C_{2}$ of height at least t -1.

###### Proof.

This is immediate unless $b_{1}\in R^{t-2}$.

If not, $\ell:=\ell_{b_{1},b_{2}}$ must be left and downward going by at least
two rows. This can only happen if there is a line $\ell^{\prime}$ carrying a
$\ast$, ungated at the $(t-1)^{th}$ stage [5, 5.4.3.5.4.4], with the same left
hand end-point $b_{1}$ as $\ell$, so allowing $\ell$ to be drawn [5, 5.4.7].
By up-going linkage [5, 5.4.3] the right hand end point $b^{\prime}$ of
$\ell^{\prime}$ lies in $R_{t-1}$ and since $\ell^{\prime}$ carries a $\ast$,
the box $b^{\prime}$ lies at the bottom of the column $C^{\prime}$ in which it
is contained, so $C^{\prime}$ has height $t-1$. Yet $C_{2}$ has height $\geq
t$ so by adjacency [5, 5.4.3], $C^{\prime}$ lies strictly to the left of
$C_{2}$ and of course to the right of $C_{1}$.

∎

#### 3.6.2.

Recall that a $\ast$ can only appear at most once in $\textbf{C}_{k}(s)$ and
then necessarily in $\textbf{C}_{k}(s)^{s}$. Recall also that, for $t\leq s$,
at most one $1$ can appear in $\textbf{C}_{k}(s)^{t}$, by 3.3.2. We shall use
$\textbf{C}_{k}(s)^{t}=\phi$ to mean that neither a $\ast$ nor a $1$ appear in
$\textbf{C}_{k}(s)^{t}$ and we say that this column is empty.

######  Lemma.

For all $t\leq s$ one has $\textbf{C}_{k}(s)^{t}=\phi$ if and only if
$\textbf{C}_{k}(s+1)^{t}=\phi$.

###### Proof.

For $t<s$, this is immediate from 3.4.2 (i). For $t=s$, “only if” follows from
3.4.2 (v). For “if” suppose $\textbf{C}_{k}(s+1)^{s}=\phi$. Then by 3.4.2(ii),
there is no $\ast$ in $\textbf{C}_{k}(s+1)^{s}=\phi$. Then by 3.4.2(iii),
there is no $1$ in $\textbf{C}_{k}(s+1)^{s}=\phi$. ∎

#### 3.6.3.

######  Lemma.

(i). Take $t>|S|$. Then $\textbf{C}_{k}(s)^{t+1}=\phi$, for all
$s\in\mathbb{N}^{+}$.

(ii). Take $t\leq|S|$. Then $\textbf{C}_{k}(s)^{t}\neq\phi$, for all
$s\in\mathbb{N}^{+}:s\geq t$.

###### Proof.

Consider (i).

Take $m$ to be an integer $\geq 0$. Recall (3.3) that $b_{(|S|+m+2),k}$ is the
lowest box in the last column $C_{k}(|S|+m+2)$ deemed to have height
$|S|+m+2$.

If the last column of $\textbf{C}_{k}(|S|+m+2):m\geq 0$ is not empty, then
there is a line $\ell_{b,b_{(|S|+m+2),k}}$ for some $b\in S$. Then by Lemma
3.6.1, there is a column of height $|S|+m+1$ in $S$. This contradiction proves
that $\textbf{C}_{k}(|S|+m+2)^{|S|+m+2}=\phi$ for all $m\in\mathbb{N}$.

Applying Lemma 3.6.2, we conclude that $\textbf{C}_{k}(s)^{|S|+m+2}=\phi$, for
all $s\geq|S|+m+2$, whilst for $s<|S|+m+2$, the assertion is trivial. Finally
the condition $t>|S|$, means that we can write $t=|S|+m+1$, for some
$m\in\mathbb{N}$, whilst we have just shown that
$\textbf{C}_{k}(s)^{|S|+m+2}=\phi$, for all $s\in\mathbb{N}$. Hence (i).

Consider (ii).

By definition $S$ has a column $C$ of height $|S|$. Choose that nearest to the
right. If $s=|S|$, set $C^{\prime}=C_{k}(s)$. Otherwise adjoin a column
$C^{\prime}$ of height $|S|$ to $\mathscr{D}$, on its right. By Lemma 3.2.2
this does not change the left going lines from $C_{k}(s)$, nor their labels.

Take $t\leq|S|$. Since $C_{k}(s)$ lies between $C,C^{\prime}$ every box in
$R^{|S|}\cap C_{k}(s)$ has a left going line by [5, 5.4.8(v)]. One may remark
that they all carry a $1$ except if this line is left going from
$C_{k}(s)^{t}:s=t=|S|$.

Consequently $\textbf{C}_{k}(s)^{t}\neq\phi$ if $s\geq t$ and $t\leq|S|$, as
required.

∎

Remark. Take $s\in\mathbb{N}^{+}$. Then $\textbf{C}_{k}(s)^{|S|+1}$ may or may
not be empty.

#### 3.6.4.

######  Lemma.

For all $s>|S|$, the column block $\textbf{C}_{k}(s+1)$ obtains from
$\textbf{C}_{k}(s)$ by adjoining an empty column on the right.

###### Proof.

By 3.4.2(i) one has $\textbf{C}_{k}(s+1)^{u}=\textbf{C}_{k}(s)^{u}$ for $u<s$.
For $u=s$, this equality again holds by 3.4.2(v) if the right hand side is
empty. Otherwise for $s>|S|$ it holds by 3.4.2(iii)B.

Finally by Lemma 3.6.3(i), one has $\textbf{C}_{k}(s+1)^{s+1}=\phi$, for all
$s>|S|$.

∎

#### 3.6.5.

######  Corollary.

$(i)$. There exists an “asymptotic” last column block $\textbf{C}_{k}(\infty)$
which obtains from $\textbf{C}_{k}(|S|+1)$ by adjoining an infinite block of
empty columns on the right.

(ii) $\textbf{C}_{k}(\infty)^{t}\neq\phi$, for all $t\leq|S|$ and has a $1$ in
some unique row, $\textbf{r}_{t}$, which are moreover pairwise distinct.

###### Proof.

∎

(i). For all $l\geq|S|$, the last column block $\textbf{C}_{k}(l+1)$ obtains
from $\textbf{C}_{k}(|S|+1)$ by adjoining $l-|S|$ empty columns on the right.
Hence the assertion.

(ii). The first part follows from Lemma 3.6.3(ii). Since $S$ has no column of
height $>|S|$, this non-trivial entry can only be a $1$. Yet by [5, 5.4.8(vi)]
a $1$ can occur at most once on any given column (or row—) of C, for any
column $C$ of $\mathscr{D}$. The first part gives the uniqueness of
$\textbf{r}_{t}$, the second part that they are pairwise distinct.

Remark. Following the remark of 3.6.3, we note that
$\textbf{C}_{k}(\infty)^{|S|+1}$ need not be empty and then has a $1$ in some
row $\textbf{r}_{|S|+1}$. Otherwise we set $\textbf{r}_{|S|+1}=0$.

#### 3.6.6.

The above result defines a map $S\mapsto\\{\textbf{r}_{t}\\}_{t=1}^{|S|+1}$
from compositions to the set of all pairwise distinct $|S|+1$ non-negative
integers. We call it the composition map.

#### 3.6.7.

One might hope that the image of composition map determines the last column
block $\textbf{C}_{k}(s)$ for all $s\in\mathbb{N}$. By definition it
determines $\textbf{C}_{k}(s)$ for all $s|s>|S|$.

To do this one should show that $\textbf{C}_{k}(s+1)$ determines
$\textbf{C}_{k}(s)$. Yet this fails as the following example shows.

Example.

Consider the compositions $(2,1,1,3);(1,1,1,1,3)$ labelling the last column by
$C_{k}(s+1):s=2$.

In both cases the last column block $\textbf{C}_{k}(s+1)$ has a $1$ at co-
ordinates $x_{4,5},x_{3,6}$ and its last column is empty.

Now replace $s+1$ by $s$. In the first case the $x_{3,6}$ acquires a $\ast$,
whilst in the second case there is no change.

#### 3.6.8.

Yet we can realise the goal of 3.6.7 by including information on column
heights in $S$. This is described in the lemma below using the following
notation.

Let $\mathscr{S}$ denote the set of heights of $S$ counted without
multiplicities.

Let $\textbf{r}_{s}$ be defined through the composition map (3.6.6).

Recall 3.6.5. In particular $\textbf{r}_{|S|+1}$ may or may not be empty.

######  Lemma.

Take $s\leq|S|+1$.

(i). Suppose $s\in\mathscr{S}$. Then $\ast$ occurs in $\textbf{C}_{k}(s)^{s}$
in row $\textbf{r}_{s}$. Moreover $1$ occurs in $\textbf{C}_{k}(s)^{s}$ in row
$\textbf{r}_{s+1}$, if and only if the latter is non-empty.

(ii). Suppose $s\notin\mathscr{S}$. Then $1$ occurs in $\textbf{C}_{k}(s)^{s}$
in row $\textbf{r}_{s}$.

In particular the image of the composition map and $\mathscr{S}$ determine the
last column block $\textbf{C}_{k}(s)$, for all $s\in\mathbb{N}^{+}$.

###### Proof.

By 3.4.2(i), it is enough to show that the conclusions give the correct
entries for $\textbf{C}_{k}(s+1)^{t}:t=s,s+1$ as described in 3.4.2.

Consider (i).

By 3.4.2(ii) a $\ast$ in $\textbf{C}_{k}(s)^{s}$ gives a $1$ in
$\textbf{C}_{k}(s+1)^{s}$ in the same row. Then by 3.4.2(i), this entry
remains in $\textbf{C}_{k}(\infty)^{s}$ again in the same row but becoming or
staying a $1$. It therefore must belong to $\textbf{r}_{s}$, by definition of
the latter. This gives the first part of (i).

If a $1$ also occurs in $\textbf{C}_{k}(s)^{s}$, then 3.4.2(iii)A gives a
$\ast$ (resp. a $1$) in $\textbf{C}_{k}(s+1)^{s+1}$ if $s+1\in\mathscr{S}$
(resp. $s+1\notin\mathscr{S}$), in the same row. In the first case by
3.4.2(i,ii), this entry remains in $\textbf{C}_{k}(\infty)^{s+1}$ again in the
same row but becoming a $1$. It therefore must belong to $\textbf{r}_{s+1}$,
by definition of the latter. Notice by 3.6.3(ii), $s+1\in\mathscr{S}$ implies
$\textbf{r}_{s+1}\neq\phi$.

In the second case a $1$ appears in $\textbf{C}_{k}(s+1)^{s+1}$. Then by
3.4.2(i) and 3.4.2(iii)B, this $1$ remains in $\textbf{C}_{k}(\infty)^{s+1}$
again in the same row, thus in $\textbf{r}_{s+1}$ by definition of the latter.
However in this second case it can happen that $\textbf{r}_{s+1}=\phi$. This
gives a contradiction which implies that $1$ does not occur in
$\textbf{C}_{k}(s)^{s}$.

Similarly (ii) follows from 3.4.2(iii)B.

∎

Remarks. Noting that in the example of 3.6.7 that $S$ has no column of height
$>s=2$, one may check that the conclusion of the lemma is in accordance with
this example. Again consider the composition $(2,1,1,s)$. In this case
$\textbf{r}_{1}=\textbf{4},\textbf{r}_{2}=\textbf{3},\textbf{r}_{1}=\phi$.
Correspondingly $1,\ast$ occur in $\textbf{C}_{4}(1)^{1}$, whilst only $\ast$
occurs in $\textbf{C}_{4}(2)^{2}$.

Considering the complexity of the rules in 3.4.2, the assertion of Lemma 3.6.8
is extraordinarily simple.

Clearly the composition map is of independent interest.

##  4\. Regularity

We continue to take $e+V$ to be the canonical Weierstrass section constructed
in [5, 5.4] and described rather precisely in 3.4 above.

We would have liked a solution which starting from the ordered set of heights
of the columns directly gives the lines determining $e,V$.

This solution described in 3.5 is perhaps not ideal, but it does behave well
under adjoining of columns on the right Lemma 3.2.2 and increasing the height
of the last column. Then by Lemma 3.6.8, the solution is completely determined
by the images of the composition maps and the set of column heights.

By [5, Lemma 6.1], $e$ belongs to the nilfibre $\mathscr{N}$.

We now would like to show that $e\in\mathscr{N}_{reg}$. Unfortunately this can
fail because $e$ may give rise in VS pairs. This is explained in 4.3 and 4.4.

###  4.1. Pseudo-regularity

Recall that e is a sum of co-ordinate vectors $\\{e_{i}\\}_{1=1}^{r}$, none of
which lie on the same row or same column of M. In particular the roots of the
$\\{e_{i}\\}_{1=1}^{r}$ are linearly independent. Let $E$ denote the
corresponding sum of root subspaces. As an immediate consequence of linear
independence

$None$ $\mathfrak{h}.e=E.$

Recall that $e\in\mathscr{N}$. Then via [5, Lemma 3.1(ii)] the property that
$e$ is regular, is equivalent to

$None$ $\mathfrak{p}.e\oplus V=\mathfrak{m}.$

Consider $\sum_{i=1}^{r}\mathfrak{p}.e_{i}$. Obviously

$None$ $\mathfrak{p}.E:=\sum_{i=1}^{r}\mathfrak{p}.e_{i}\supset\mathfrak{p}.e$

As we shall see the inclusion may be strict.

Definition $e$ is said to pseudo-regular if

$None$ $\mathfrak{p}.E+V=\mathfrak{m}.$

It should be carefully noted that this is different to $e$ being quasi-
regular. Indeed from the definition of pseudo-regularity it follows that
$\dim\mathfrak{p}.E\geq\dim\mathfrak{m}-\dim V$ and this inequality can be
strict. On the other hand quasi-regularity means that
$\overline{P.E}=\mathscr{N}^{c}$ and as we shall see (Lemma 4.5.3(ii)) it
implies that $\dim P.E=\dim\mathfrak{m}-\dim V$. Quasi-regularity generally
fails and to recover it we must augment $E$ to a larger subspace $E_{VS}$ to
be constructed in 4.4.4 below.

Pseudo-regularity may be visualized as follows.

Consider a chess board with squares being the co-ordinate places of
$\mathfrak{m}$ with those of $V$ removed. Partition this board by vertical
(resp. horizontal) lines extending the sides of the blocks.

Consider $e$ as a set of rooks or castles222Apparently this strange dual
terminology came about from a confusion of Rukh (Chariot in Persian) with
Rocca (Italian for fortress). at each co-ordinate place
$e_{i}:i=1,2,\ldots,s$. Allow each rook to move leftwards not crossing a
vertical line, arbitrarily rightwards and similarly downwards not crossing a
horizontal line and arbitrarily upwards.

Then pseudo-regularity means that the set of rooks cover every square on this
chess-board. In this we may note that, unlike chess, the rooks cover the
places at which they stand through the action of $\mathfrak{h}$. This can fail
if we adjoin rooks and the corresponding root vectors are no longer linearly
independent. This is the source of the failure of $\mathscr{C}$ to admit a
dense orbit [5, Lemma 6.10.7].

Below an example is given when the rooks also cover some elements of $V$. This
contradicts [5, 3.1] and arises because $(4)$ is in general a strict
inclusion. As we shall see this is caused by VS pairs (4.3 and 4.4).

Example. Consider the nilradical $\mathfrak{m}$ given by the composition
$(1,2,2,1)$. Then our procedure [5, 5.4] gives $e=x_{1,2}+x_{2,4}+x_{5,6}$,
whilst $V=\mathbb{C}x_{3,5}+\mathbb{C}x_{4,6}$. Thus the rooks cover all the
squares in $\mathfrak{m}$ excepting the co-ordinate $(3,5)$ which contradicts
[5, 3.1(ii)]. Yet we cannot simultaneously obtain $x_{2,5}$ and $x_{4,6}$
through the action of $x_{4,5}$ since it would have to be used twice. Moreover
we recover pseudo-regularity since $x_{4,5}\in V$.

###  4.2. Pseudo-regularity

#### 4.2.1.

######  Proposition.

$e$ is pseudo-regular.

###### Proof.

The proof is by induction on the number of columns. Define
$e^{\kappa}\in\textbf{S}$ as in 3.3.1. Then we may write $e=e^{\kappa}+e_{k}$,
with $e_{k}\in\textbf{C}_{k}$. Recalling Lemma 3.2.2 we may assume by the
induction hypothesis that $e^{\kappa}$ is pseudo-regular in S.

Thus we only need to show that the rooks coming from $e^{\kappa}$ and those
coming from $e_{k}$, cover every square in $\textbf{C}_{k}$.

The proof of the latter is by induction on the height $s$ of
$\textbf{C}_{k}(s)$.

Suppose $s=1$ and let $C_{k-1}$ have height $t$. If (a) of 3.4.1 holds, then
the last column is covered by the rook on the $(n-t)^{th}$ row, noting that
the downward motion need not cross a horizontal line (though the upward motion
may).

For (b), we apply (iv) of 3.4.2 as in the general case below.

By (i) of 3.4 the first $s-1$ columns of $\textbf{C}_{k}(s+1)$ are those of
$\textbf{C}_{k}(s)$ and so by the induction hypothesis the rooks cover all the
first $s-1$ columns of $\textbf{C}_{k}(s+1)$.

Let us show that this also hold for $\textbf{C}_{k}(s+1)^{s}$.

If this column has no $\ast$ then by $(iii)B$ of 3.4, this column coincides
with $\textbf{C}_{k}(s)^{s}$ and so again by the induction hypothesis on last
column height, the rooks cover all the entries of this column in
$\textbf{C}_{k}(s+1)$.

If this column does have a $\ast$ then by (ii) of 3.4, it becomes a $1$ in the
same place and by (iii) of 3.4 all other changes concern only
$\textbf{C}_{k}(s+1)^{s+1}$. Thus we are again reduced to the previous case.

It thus remains to show that all the entries of $\textbf{C}_{k}(s+1)^{s+1}$
are covered. This follows easily from (iv) of 3.4.

∎

####  4.2.2. A Comment

The following result is not used; but it indicates what holds for entries in
the last column block.

Let $s$ be the height of $C_{k}$ and set $m=\min{(|S|,s-1)}$.

######  Lemma.

Each of the first $m$ columns of $\textbf{C}_{k}$ admits a $1$ occurring (of
course) on the different rows.

###### Proof.

This follows from Lemma 3.6.3(ii) noting that just a $\ast$ may appear in
$\textbf{C}_{k}(s)^{s}$. A more direct proof may also be given using 3.4.2. ∎

Remark. These $1$’s may be inserted sequentially in the columns going from
left to right; but they still may not appear in the lowest possible column.
For example consider the composition $(3,2,1,4)$. Then the $1$ in column $9$
occupies row 3 instead of row 4.

###  4.3. Abstract VS Pairs

Here it is convenient to use $k$ as a dummy index for co-ordinates. Yet $k$
was also used as the label of the last column of $\mathscr{D}$ in 3.2 and
already in our previous paper [2.2] [5]. Although this should not cause any
confusion, we introduced in N.B. of 3.3.1 the device of replacing the $k$
superscript by $\kappa$ in reference to the last column of $\mathscr{D}$.

Following the work of Victoria Sevostyanova [9] a pair of co-ordinates vectors
$x_{i,j},x_{k,l}$ in $\mathfrak{m}$ is called a VS pair if $x_{j,k}$, lies in
$\mathfrak{n}$ or is a root vector of the Levi factor. Notice here that we
must already have $i<j,k<l$.

Given a VS pair, commutation of $x_{i,j}$ with $x_{j,k}$ gives $x_{i,k}$,
which lies in the same row as $x_{i,j}$ whilst commutation of $x_{j,k}$ with
$x_{k,l}$ gives $x_{j,l}$, which lies in the same column as $x_{k,l}$.

Consider the action of $\mathfrak{p}$. We cannot simultaneously obtain the
entire row to the right of $x_{i,j}$ and the entire column above $x_{k,\ell}$
because the same element of $\mathfrak{p}$, namely $x_{j,k}$, is being used
twice.

In fact this is exactly the source of the strict inclusion in $(4)$. Otherwise
the desired regularity of $e$ would result from Proposition 4.2.1.

In the example of 4.1, both the co-ordinates $(2,5)$ and $(4,6)$ (where an
element of $V$ lies) cannot be simultaneously covered. Of course the action of
$\mathfrak{p}$ on $e$ gives a linear combination of these two vectors. Thus a
cat zapped by rooks at $(2,4)$ and $(5,6)$ would end up like that of
Schroedinger’s cat, a linear combination of a live cat and a dead cat. This
can get far more complicated and just imagine trying to play chess taking
account of VS pairs.

A main result of our present paper is Theorem 4.4.5, which severely limits the
number of VS pairs which need to be considered, those which we shall call
“bad”.

In this we shall not attempt a complete classification of bad pairs for the
reasons given in Remark 4.4.3.

###  4.4. VS Pairs Associated to $e+V$

####  4.4.1. VS pairs relative to the canonical Weierstrass section.

Recall 3.1 and the notation used there.

Not all abstract VS pairs are needed.

Definition. A pair $x_{i,j},x_{k,l}\in\operatorname{supp}{e}$, with $j\neq k$,
is called a VS pair associated to the Weierstrass section $e+V$ if neither
$x_{i,k}$, nor $x_{j,l}$ belong to $\operatorname{supp}{V}$ and in addition
$x_{j,k}$ is a root vector of $\mathfrak{p}$, called the connecting element of
the VS pair.

Notice the definition incorporates the fact that by [5, 3.1(ii)], we cannot
expect nor do we need to recover $V$.

In particular in the example of 4.1 that we are not considering
$x_{2,4},x_{5,6}$ to be a VS pair, since $x_{4,6}\in\operatorname{supp}V$.

Recall that $E=\sum_{x_{(i,j)}\in\operatorname{supp}e}\mathbb{C}x_{i,j}$ and
Equation $(1)$.

####  4.4.2. VS Quadruplets

If $C,C^{\prime}$ are columns of $\mathscr{T}_{\mathfrak{m}}$ , we write
$C<C^{\prime}$ is $C$ lies strictly to the left of $C^{\prime}$.

Definition. A VS pair $x_{i,j},x_{k,l}$ gives a quadruplet $(i,j,k,l)$, which
we call a VS quadruplet if $x_{j,k}\in\operatorname{supp}{V}$.

Given $t\in\\{1,2,\ldots,n\\}$, let $b(t)$ be the box containing $t$ and
$C_{(t)}$ the column containing $b(t)$.

Suppose $(i,j,k,l)$ is a quadruplet, then $C_{(i)}<C_{(j)}\leq
C_{(k)}<C_{(l)}$, by definition of $e$.

Let $c_{(t)}$ denote the height of $C_{(t)}$ for any $t\in\\{i,j,k,l\\}$.

Note that the condition $C_{(j)}=C_{(k)}$ means that $j,k$ are in the same
column and so the connecting element is a root vector lying in the Levi factor
$\mathfrak{l}$ of $\mathfrak{p}$.

As noted in 4.3 we cannot obtain both $x_{i,k}$ and $x_{j,l}$ as elements of
$\mathfrak{p}.e$ (only a linear combination). Then for example, the co-
ordinate subspace $E_{j,l}$ need not lie in $\mathfrak{p}.e+V$.

####  4.4.3. Bad VS Pairs

We shall define the notion of a bad VS pair relative to $e+V$ inductively on
the number of columns. In this we recall the notation and observation of
3.3.1.

As 3.3.1, let $S$ denote the (ordered) set of columns in $\mathscr{D}$
excluding its rightmost column $C$. We shall sometimes denote $C$ by $C(s)$ to
specify its height $s$. We shall sometimes denote $\mathscr{D}$ by $S\sqcup
C$.

Let $x_{i,j},x_{k,l}$ be a VS pair defined by $e+V$.

$(*)$. It can happen that there is either a co-ordinate vector
$x_{j,z}\in\operatorname{supp}e$ or a co-ordinate vector
$x_{y,k}\in\operatorname{supp}e$, so that the action of
$x_{z,l}\in\mathfrak{p}$ (resp. $x_{i,y}\in\mathfrak{p}$) gives $x_{j,l}$
(resp. $x_{i,k}$). Consequently either $x_{j,l}$ or $x_{i,k}$ is recovered in
a second fashion.

[N.B.. In the above we also must check that the given element of
$\mathfrak{p}$ is not used in a second commutator $[x_{z,l},x_{l,u}]$ (resp.
$[x_{h,i},x_{i,y}]$).

In the first case if $l\in C$, that is labels an element in the last column,
then $x_{l,u}$ cannot lie in $\mathfrak{m}$, so this holds trivially. In the
second case we show that the labels of the labels of a quadruplet $(h,i,y,k)$
lie entirely in $S$ and use the induction hypothesis. Specifically this check
is carried out in the last part of 4.4.6 for the Levi factor case and
following the claim in 4.4.7 for the general case.]

When $(*)$ does not happen, we say that $x_{i,j},x_{k,l}$ is a bad VS pair
defined by $e+V$.

Remark. Checking that a pair is not bad can be relatively easy. Checking it is
bad is more difficult and this is why we do not attempt to classify all bad
pairs.

######  Lemma.

Let $x_{i,j},x_{k,l}$ be a bad VS pair defined by $e+V$. If $l$ does not
belong to $C$, it is a bad VS pair relative to $e^{\kappa}+V^{\kappa}$.

###### Proof.

This is immediate from Lemma 3.2.2.

∎

####  4.4.4. The Extended Elements

Following the above lemma we inductively define $e_{VS}$, (resp. $E_{VS}$) by
adjoining the co-ordinate vector $x_{j,l}$ (resp. co-ordinate subspace
$\mathbb{C}e_{j,l}$) to $e$ (resp. $E$), for each bad VS pair
$x_{i,j},x_{k,l}$ as $l$ runs through the columns of $\mathscr{D}$ going from
left to right.

Extending the notation of 3.3.1 we let $e_{VS}^{\kappa}$ (resp.
$E_{VS}^{\kappa}$) denote the element (resp. subspace $E_{VS}$) obtained as
above but in which the contributions coming from the last column $C_{k}$ are
omitted.

By definition $E_{VS}$ compensates for the possible strict inclusion in $(4)$,
so Proposition 4.2.1, we obtain

######  Lemma.

$\mathfrak{p}.e+E_{VS}+V=\mathfrak{m}$.

Remark. We would really like that $\mathfrak{p}.e_{VS}+V=\mathfrak{m}$. As we
shall see this fails. The basic difficulty is that the roots occurring in
$e_{VS}$, as opposed to those occurring in $e$ may not be linearly independent
as elements of $\mathfrak{h}^{*}$. Thus as noted in 4.1 the rooks defined by
$e_{VS}$ need not cover themselves.

####  4.4.5. Towards Bad VS Pairs

A key result in excluding a VS pair from being bad is the following.

######  Theorem.

Let $x_{i,j},x_{k,l}$ be a bad VS pair. Then the connecting element $x_{j,k}$
comes from a line in $\mathscr{T}_{\mathfrak{m}}$ carrying a $*$, equivalently
$x_{j,k}\in\operatorname{supp}V$. In particular $(i,j,k,l)$ is a VS
quadruplet.

The proof of the theorem is given below. We retain its hypothesis and
notation.

In view of lemma 4.4.3 we can adopt the notation of 4.4.3 that
$\mathscr{D}=S\sqcup C$ with $C$ the last column of $\mathscr{D}$ and $l$ an
element of $C$. This notation will be retained until the end of Section 4.5.

####  4.4.6. The Levi Factor Case

Suppose that in the quadruplet $(i,j,k,l)$, the boxes $b(j),b(k)$ lie in the
same column, that is $C_{(j)}=C_{(k)}$. (This means that $x_{j,k}$ belongs to
the Levi factor of $\mathfrak{p}$.)

######  Lemma.

If $C_{(j)}=C_{(k)}$, then $x_{i,j},x_{k,l}$ is not a bad VS pair. In other
words the assertion of the theorem holds in this case.

###### Proof.

Let $t$ be the height of $C_{(j)}=C_{(k)}$.

Suppose first that $j<k$. By hypothesis $b(k)$ has a right going line $\ell$
to $b(l)$ labelled by a $1$. Then by [5, 5.4.8(v)], the height of $C_{(l)}$ is
at most one less than the height of $C_{(k)}$. Yet $b(j)$ lies above $b(k)$ in
$C_{(j)}=C_{(k)}$, so $b(j)$ is not right extremal. Thus there exists a right
going line $\ell^{\prime}$ from $b(j)$.

Let $b(z)$ be the right end-point of $\ell^{\prime}$. Suppose $\ell^{\prime}$
is labelled by a $\ast$, then by [5, 6.5.4], one has $b(z)=b(\ell)$. Thus the
line $j,l$ is labelled by a $\ast$ and so the pair $(i,j),(k,l)$ is not a VS
pair (Definition 4.4.2).

Thus we can suppose that $\ell^{\prime}$ is labelled by a $1$, in other words
$(j,z)\in\operatorname{supp}{e}$. Moreover either $z\in C$ and $x_{z,l}$
belongs to the Levi factor or $z<l$. In both cases $x_{z,l}$ is a root vector
of $\mathfrak{p}$ and we obtain $x_{j,l}$ in a second fashion as
$[x_{j,z},x_{z,l}]$. Again $x_{z,l}$ cannot be used in a commutator
$[x_{z,l},x_{l,u}]$ because $l$ labels a column in the last column block C and
so $x_{l,u}$ cannot lie in $\mathfrak{m}$.

Suppose now that $j>k$. By hypothesis $b(j)$ has a left going line $\ell$ to
$b(i)$, labelled by a $1$ and $b(k)$ lies above $b(j)$ in $C_{(j)}=C_{(k)}$.
Then by [5, 6.5.4], there exists a left going line $\ell^{\prime}$ from $b(k)$
labelled by a $1$. Let $b(y)$ be the left-hand end-point of $\ell^{\prime}$.
Since $y$ occurs in a column strictly to the left of $C_{(j)}$, necessarily
$y<k$.

Let $R_{t}$ be the row in which $b(j)$ occurs. Take $b_{1}=b(i),b_{2}=b(j)$ in
Lemma 3.6.1. By its conclusion there is a column $C^{\prime}$ of height $\geq
t-1$ to the right of $C(i)$ and strictly to the left of $C(j)$. Since the only
line carrying a $\ast$ having a box in $C_{(j)}$ as its right hand end point
must lie strictly below $b(j)$, every left-going line from a box in $C_{(j)}$
above $b(j)$ must be downward going by [5, 5.4.8(ix)] and hence meet
$C^{\prime}$ or a column between $C^{\prime},C_{(j)}$. In both cases this
column lies to the right of $C_{(i)}$, though not necessarily strictly.

Thus either $i<y$ or both $b(y),b(i)$ lie in $C_{(i)}$ and in this case
$x_{i,y}$ lies in the Levi factor. In both cases we recover $x_{i,k}$ as the
commutator $[x_{i,y},x_{y,k}]$.

Finally suppose $x_{i,y}$ is used in a second commutator
$[x_{h,i},x_{i,y}]=x_{h,y}$, coming from a second VS pair $x_{h,i},x_{y,u}$.
Then the line $\ell_{y,u}$ carries a $1$. However the line $\ell_{y,k}$ is
also labelled by a $1$ and by [5, 5.4.8(vi)] this forces $u=k$. Then the
resulting quadruplet $(h,i,y,k)$ lies entirely in $S$.

Thus if this second VS pair is bad we can assume by the inductive construction
4.4.4 that the subspace $\mathbb{C}x_{i,k}$ has already been adjoined to
$E^{\kappa}_{VS}$ and thus we recover $x_{h,y}$, as required.

∎

####  4.4.7. The General Case.

The proof of general case is similar, but one has to be more careful in view
of the possible appearance of $\ast$ which allows the VS pair to be a VS
quadruplet.

Let $(i,j),(k,l)$ be a VS pair, with $l\in C$, which does not form a VS
quadruplet and that $b(j),b(k)$ are in distinct columns, so then $i<j<k<l$. By
definition of a VS pair, $b(j)$ (resp. $b(k)$) admits a left (resp. right)
going line to $b(i)$ (resp. $b(l)$). Under these conditions one has the

Claim. Either $b(j)$ admits a right going line labelled by a $1$, to
$b(u):j<u<l$, or $b(k)$ admits a left going line labelled by a $1$, to
$b(v):i<v<k$.

In the above, the inequality $j<u$ (resp $v<k$) follows because the box joined
to $b(j)$ (resp. $b(k)$) lies strictly to the right (resp. left) of $C_{(j)}$
(resp. $C_{(k)}$). The second pair of inequalities is more delicate.

Assume the claim holds.

In the first case we obtain $x_{j,l}$ as $[x_{j,u},x_{u,l}]$ in the second
case we obtain $x_{i,k}$ as $[x_{i,v},x_{v,k}]$.

As in the case when $b(j),b(k)$ are in the same column, we cannot use
$x_{u,l}$ again with respect to a further VS pair $x_{j,u},x_{l,t}$ because
$l\in C$ and so $x_{l,t}\notin\mathfrak{m}$.

As in the case when $b(j),b(k)$ are in the same column, one may use $x_{i,v}$
for a second VS pair $x_{f,i},x_{v,k}$; but this will lie entirely in $S$. As
in the proof of Lemma 4.4.6, if this pair is bad, we can assume that the
subspace $\mathbb{C}x_{i,k}$ has already been adjoined to $E^{\kappa}_{VS}$. .

Thus the theorem results from the claim, which we now proceed to prove.

####  4.4.8. Proof of claim.

Suppose first that $t:=c_{(j)}=c_{(k)}$. Then by [5, 5.4.8(v)] every box in
$R^{t}$ lying in $C_{(j)}$ (resp. $C_{(k)}$) has a right (resp. left) going
line to a box between these two columns. In particular this box may be $b(j)$
(resp. $b(k)$) and writing the other end point of this line as $b(u)$ (resp.
$b(v)$) we obtain the remaining inequality $u<l$ (resp. $i<v$). One of these
two lines carry a $1$ with the following possible exception. Namely, the right
going line from $b(j)$ is the same as the left going line from $b(k)$ and this
line carries a $\ast$. However this means that $(i,j,k,l)$ is a VS quadruplet.

Thus we can assume $c_{(j)}\neq c_{(k)}$.

Suppose that $c_{(j)}>c_{(k)}$ and Set $t=c_{(j)}$. Then $C_{(j)}$ admits a
nearest column $C^{\prime}$ of height $t$ strictly to the right of $C_{(k)}$
or we can adjoin one to the extreme right of $\mathscr{D}$. By Lemma 3.2.2,
this adjunction will not change the left going lines from boxes in $C_{(k)}$.

Let $C^{\prime\prime}$ be a left neighbour to $C^{\prime}$. It lies to the
right of $C_{(j)}$ and strictly to the left of $C_{(k)}$.

By [5, 5.4.8(v)] every box in $C_{(k)}$ (so particularly $b(k)$) has a left
going line $\ell$ to a box lying to the right of $C^{\prime\prime}$ labelled
by a $1$, unless $\ell$ goes leftwards from the box $C^{\prime}\cap R_{t}$.
The latter is impossible because $C^{\prime}$ lies strictly to the right of
$C_{(k)}$. Finally the box $b(v)$ joined to $b(k)$ by $\ell$ lies to the right
of $C^{\prime\prime}$, so to the right of $C_{(j)}$, so strictly to the right
of $C_{(i)}$. This gives the required inequality $i<v$.

It remains to consider the case $c_{(j)}<c_{(k)}$, which is similar but a
little more complicated.

Set $t=c_{(k)}$. Then $C_{(k)}$ admits a column $C^{\prime\prime}$ of height
$t$ strictly to the left but nearest to $C_{(j)}$ or we can adjoin it as a
left neighbour to $C_{(k)}$ to the extreme left of $\mathscr{D}$. By Lemma
3.2.3, and recalling that $c_{(j)}<c_{(k)}$, this adjunction will not change
the right going lines from boxes in $C_{(j)}$, though it may change their
labels.

By [5, 5.4.8(v)] the box $b(j)$ in $C_{(j)}$ has a right going line $\ell$ to
a box at the left of $C_{(k)}$, hence strictly to the left of $C_{(l)}$. If it
is labelled by a $1$, the claim holds in this case as in the previous case.

Otherwise by [5, 5.4.8(v)], there is exactly a right going line
$\ell^{\prime}$ from $b(j)$ labelled by a $\ast$ whose right end point is the
box $b^{\prime}$ in $C_{(k)}\cap R_{t}$. Notice that this also means that
$C_{(k)}$ must already have a left neighbour $C^{\prime\prime}$ in
$\mathscr{D}$, so adjoining a left neighbour is unnecessary. Moreover by
adjacency [5, 5.4.3,5.4,11], $\ell^{\prime}$ cannot cross a column of height
$\geq t$, so there is no column of height $t$ strictly between
$C_{(j)},C_{(k)}$. Thus $C^{\prime\prime}$ is a left neighbour to $C_{(k)}$ in
$\mathscr{D}$.

If $b^{\prime}=b(k)$, then $(i,j,k,l)$ is a VS quadruplet and we are done.

However by definition $b(k)\in C_{(k)}$ and so if $b(k)\neq b^{\prime}$, then
$b(k)$ must appear strictly above $b^{\prime}$, that is as $C_{(k)}\cap
R_{t^{\prime}}$ for some $t^{\prime}<t$.

Yet by Lemma 3.6.1 there is a column $C^{\prime}$ of height $\geq t-1$ with
$C_{(j)}\leq C^{\prime}<C_{(k)}$. Then exactly as in the proof of the last
part of Lemma 4.4.7 we conclude by [5, 5.4.8(v)] that $b(k)$ admits a left
going line $\ell^{\prime\prime}$ labelled with a $1$ to some box $b(v)$ to the
right of $C^{\prime}$. Yet $C^{\prime}$ lies to the right of $C_{(j)}$ and
thus strictly to the right of $C_{(i)}$. Consequently $i<v$, as required.

This completes the proof of the claim and taking account of Lemma 4.4.7 the
theorem.

Remark. This result cuts down drastically the number of bad VS pairs. Indeed
by [5, 5.4.8(vi)] there is at most left (resp) right going line carrying a $1$
from a box in $\mathscr{T}_{\mathfrak{m}}$. We conclude that a line, say
$\ell_{j,k}$, carrying a $\ast$ uniquely determines a VS quadruplet
$(i,j,k,l)$ (if one exists). Thus the number of bad VS pairs is at most $g$.

Examples.

$(i)$. Consider the array $(1,2,1,1,2,2,3)$. Take $i=3,j=4,k=9,l=11$. Then
$(3,4),(9,11)$ is a VS pair. Now the lines $\ell_{4,7}$ and $\ell_{7,9}$ are
the obvious choices for the lines of the claim but both carry a $\ast$, so at
first sight one seems to get a contradiction. However there is also a line
$\ell_{4,9}$ carrying a $1$ obtained by the joining of loose ends [5,
5.4.7(ii)]. Such joinings are essential to validate [5, 5.4.8(v)] used in the
above proof.

$(ii)$. Consider the array $(1,2,1,1,2,3)$. Then the VS pair $(3,4),(7,9)$
forms a VS quadruplet. Actually since the line $(4,10)$ carries a $1$ and $10$
lies in the same column as $9$, $x_{4,9}$ can be recovered in a second fashion
as $[x_{4,10},x_{10,9}]$. Thus this VS pair is not bad.

$(iii)$. Consider the array $(3,2,1,1,2,3)$. There are just two VS quadruplets
$(4,6,7.8)$ and $(4,6,9,11)$. One may easily check that the first is not bad.
Thus the second being not bad would imply that $e$ is regular. One can check
that the second is indeed bad, but a better method is to show that $e$ cannot
be regular by computing its nilpotency class as in [5, 6.10.9, Example 1].

Again as in loc. cit. one may adjoin the co-ordinate vector $(6,11)$ to $e$.
The resulting element $e_{VS}$ is regular because the graph defined by the
roots of the root vectors in $e_{VS}$ has no cycles and so these roots are
linearly independent. More details of this example are in section 5 figure 6.

###  4.5. Weierstrass Sections

####  4.5.1. A Preliminary lemma

Recall the subalgebra $\mathfrak{u}_{\pi^{\prime}}$ of $\mathfrak{m}$ defined
in [5, 2.5]. It is spanned by root subspaces and those which do not appear in
$\mathfrak{m}$ are called the excluded root vectors [5, 2.7]. The set of all
such excluded root vectors in $\mathfrak{m}$is denoted by $X$ in [5, 6.9.5].
In the matrix presentation, that is in M, they are encircled by $O$ [5,
6.9.5]. (Here we shall not use this diagrammatic presentation but a reader
working through examples is urged to do so.)

We construct $E_{VS}$ by adjoining $\mathbb{C}x_{j,l}$ for each bad VS pair
$x_{i,j},x_{k,l}$. In this Theorem 4.4.5 allows us to assume that $(i,j,k,l)$
is a VS quadruplet.

We show that $E_{VS}\subset\mathfrak{u}_{\pi^{\prime}}$. To do this we shall
not need to require that $(i,j,k,l)$ is bad, as a VS quadruplet.

Further to the definition of $X$ above, we recall [5, 6.9.3] that every $\ast$
is encircled. Following [5, 6.9.5] we let that $Y$ be the set of co-ordinate
vectors with a $O$ encircling a $\ast$ and $Z$ the set of co-ordinate vectors
which are just encircled by an $O$.

Finally set $Z=X\setminus Y$.

The hypothesis of the following lemma holds if for example $(i,j,k,l)$ is a VS
quadruplet.

######  Lemma.

Suppose $x_{j,k}\in Y$. If $x_{k,l}$ is a co-ordinate vector of $e$, for some
$l>k$, then $x_{j,l}\notin X$. In other words $x_{j,k}$ is encircled but its
“neighbour” $x_{j,l}$ is not.

###### Proof.

The set $X$ is the union (not necessarily disjoint) of the sets
$X_{C^{\prime},C}$ of elements excluded by a pair of neighbouring columns
$C^{\prime},C$ of height, say $s$. The elements of a given $X_{C^{\prime},C}$
have co-ordinates from the entries of the boxes in the columns between
$C^{\prime},C$ and moreover are determined by the distribution of these
entries through a simple combinatorial procedure [5, 2.7].

Again by Lemma 3.2.2, the hypothesis of the lemma does not depend on the
columns strictly to the right of the column containing $l$. Thus we may assume
without loss of generality that $C$ is the rightmost column of $\mathscr{D}$.

By [5, 2.7], the subset $X_{C^{\prime},C}$ of elements of $X$ obtained from
the neighbouring pair $C^{\prime},C$ is described in the following fashion.

Let $C_{1},C_{2},\ldots,C_{u}$ be the columns of height strictly greater than
$s$ between $C^{\prime},C$. For all $i=1,2,\ldots,u$, let $c_{i}$ denote the
height of $C_{i}$.

Recall the notation of 2.2. After [5, 2.7] we have the following result.

The excluded root vectors in M from set $X_{C^{\prime},C}$ lie in the
rectangles $R_{j^{\prime}}:j^{\prime}\in[1,u+1]$ formed from the last
$c_{j^{\prime}}-s:j^{\prime}\leq u$ columns (or last column if
$j^{\prime}=u+1$) of $\textbf{C}_{j^{\prime}}$ starting from the highest row
of the block $B_{{j^{\prime}}-1}$ to the row just above the highest row of
$B_{j^{\prime}}$.

Moreover each such rectangle $R_{j^{\prime}}:j^{\prime}\in[1,u+1]$ has a gap
between the $(s+1)^{th}$ and $c_{j^{\prime}-1}^{th}$ row relative to
$B_{j^{\prime}-1}$, where exactly no excluded roots lie.

For $(3)$ below, one should keep in mind that the origin of this gap is in the
columns of height $<s$ lying strictly between $C_{j^{\prime}-1}$ and
$C_{j^{\prime}}$.

Furthermore

Remark 1. Recall that by [5, 6.6.2] an element of $Y$ in a column block always
lies in its last column.

Remark 2. No co-ordinate of $e$ is excluded, equivalently no $1$ in M is
encircled [5, Cor. 6.9.3].

There are three cases to consider.

In the first and second of these, the gap noted above will cause no problems,
in the last case it might.

$Case(1)$. The element $x_{j,k}\in Y$, lies in a column block of a column of
height $>s$, that is some $\textbf{C}_{j^{\prime}}:j^{\prime}\in[1,u]$ in the
above notation.

By Remark $1$, $x_{j,k}$ lies in the last column of $\textbf{C}_{j^{\prime}}$.
Thus an entry $x_{j,l}:l>k$ must lie in one of the column blocks
$\textbf{C}_{j^{\prime\prime}}:j^{\prime\prime}>j^{\prime}$ and in particular
strictly above the rows of the $B_{j^{\prime}}$. Thus it cannot lie in any of
the rectangles $R_{j^{\prime\prime}}:j^{\prime\prime}\in[1,u]$, for
$j^{\prime\prime}>j^{\prime}$ and nor of course for $j^{\prime\prime}\leq
j^{\prime}$. Thus $x_{j,l}$ is not an excluded co-ordinate and so not
encircled.

$Case(2)$. The element of $x_{j,k}\in Y$ lies in a column block of a column of
height $s$, that is to say from $\textbf{C}_{k}$.

In this case the co-ordinate vector $x_{j,l}$ strictly to right of $x_{j,k}$
does not even exist.

$Case(3)$. The element of $x_{j,k}\in Y$ lies in a column block of a column of
height $s^{\prime}<s$.

This means that $x_{j,k}$ belongs to a column $C^{\prime\prime}$ of height
$<s$, so lying strictly between $C_{j^{\prime}-1}$ and $C_{j^{\prime}}$, for
some $j^{\prime}\in[2,u]$.

Recall that the element $x_{j,k}\in Y$ given by [5, Sect. 5.4.8] obtains by a
two-fold process. First in step $2$ of [5, 5.3], one starts from a horizontal
line carrying a $\ast$ going rightwards to the box in $R_{s^{\prime}}\cap
C^{\prime\prime}$. Then at Step $3$, this line may have its left hand end
point moved down to the top of a column $C^{\prime\prime\prime}$ of even lower
height than $C^{\prime\prime}$ [5, 5.4.7,5.4.8], but without crossing a column
of height by $\geq s$ via adjacency [5, Lemma 5.4.8 (vii)].

We conclude that $C^{\prime\prime\prime}$ also lies strictly between
$C_{j^{\prime}-1}$ and $C_{j^{\prime}}$.

As an immediate consequence

$(\alpha)$. The row containing $x_{j,k}$ meets $R_{j^{\prime}}$ strictly below
its gap.

Yet by the hypothesis the line $\ell_{k,l}$ carries a $1$ and so by Remark 2,
the root vector $x_{k,l}$ is not excluded, equivalently $x_{k,l}$ is not in
$X$. Then

either no co-ordinate vector $x_{k^{\prime},l}:k^{\prime}\leq k$ is excluded
and in particular $x_{j,l}\notin X$,

or $x_{j,l}\in X$, and $x_{k,l}$ lies in a gap of some
$R_{j^{\prime\prime}}:j^{\prime\prime}\in[1,u]$.

$(\beta)$. In the second case, $x_{j,l}$ lies in that part of
$R_{j^{\prime\prime}}$ strictly above its gap.

Since the rows of distinct rectangles are disjoint, $(\alpha),(\beta)$ force
$j^{\prime}=j^{\prime\prime}$.

Then $(\alpha),(\beta)$ become contradictory excluding the second option.

Hence the conclusion of the lemma.

∎

####  4.5.2. VS Quadruplets

Let $(i,j,k,l)$ be a VS quadruplet. Recall the definition of $E_{VS}$ given in
4.4.4.

######  Corollary.

There is no element of $Z$ at the co-ordinate $j,l$. In particular
$E_{VS}\subset\mathfrak{u}_{\pi^{\prime}}$.

###### Proof.

Since by definition of a VS quadruplet, an element of $Y$ occurs at the co-
ordinate $(j,k)$, whilst $1$ occurs at $(k,l)$. Then the hypothesis of Lemma
4.5.1 and by its conclusion there is no element of $Z$ at the co-ordinate
$(j,l)$, as required.

The second part is immediate.

∎

Remark. Return to the parabolic defined by the composition $(3,2,1,1,2,3)$,
that is Example (iii) of 4.4.8. In this the VS quadruplet $(4,6,9,11)$ is bad.
According to the corollary, $x_{6,11}\notin X$, as may be checked. Thus this
root subspace may be included in $E_{VS}$ to retain the property that
$E_{VS}\subset\mathfrak{u}_{\pi^{\prime}}$. Yet $x_{4,9}\in X$, so this
property would be destroyed had we instead required this second root subspace
to lie in $E_{VS}$.

This is why we chose to adjoin the “right-hand” element $x_{j,l}$ to $e_{VS}$
and not the “left-hand” element $x_{i,k}$. It is remarkable that this choice
always works.

####  4.5.3. VS Quadruplets

Recall the definition of $\mathscr{N}$ given in 1.3. The following key result
was announced in [5, Prop. 6.10.]. Here it is proved/

######  Proposition.

(i).
$P.E_{VS}\subset\overline{B.\mathfrak{u}_{\pi^{\prime}}}:=\mathscr{N}^{e}$.

(ii). $\dim P.E_{VS}=\dim\mathfrak{m}-g$.

(iii). $\overline{P.E_{VS}}=\overline{B.\mathfrak{u}_{\pi^{\prime}}}$.

###### Proof.

By [5, Cor. 6.9.8], $\mathscr{C}$ is a component of $\mathscr{N}$, hence $P$
stable. Then (i) follows from Corollary 4.5.2.

Let $F\subset E_{VS}$ be a vector space complement to $\mathfrak{p}.e$ in
$\mathfrak{p}.e+E_{VS}:=U$. So then

$None$ $U=\mathfrak{p}.e\oplus F.$

Now $e,F$ both belong to the vector space $E_{VS}$, so $P.(e+F)\subset
P.E_{VS}\subset\mathscr{C}$, by (i).

Following the (standard) argument given on p. 262 of [3, Lemma 8.1.1], let
$(y_{1},\ldots,y_{r})$ be a basis for $F$. Then we have a morphism
$\psi:P\times F\mapsto\mathscr{C}$ defined by
$\psi(p,\xi_{1},\ldots,\xi_{r})=p.(e+\sum_{i=1}^{r}\xi_{i}y_{i})$.

Through $(6)$ it follows as in [3, Lemma 8.1.1] that image of the tangent map
at $0\in\mathbb{C}^{r}$ is $U$. Thus $\dim P.E_{VS}\geq\dim P.(e+F)=\dim U$.

On the other hand $U+V=\mathfrak{m}$, by Lemma 4.4.4, and so $\dim
U\geq\dim\mathfrak{m}-\dim V$. Thus

$None$ $\dim P.E_{VS}\geq\mathfrak{m}-\dim V.$

Yet $\dim\overline{B.\mathfrak{u}_{\pi^{\prime}}}=\dim\mathfrak{m}-\dim V$, by
[5, Thm. 6.9.7].

Combined with (i), this gives the opposite inequality to $(7)$. Hence (ii).

Finally (iii) follows from (i),(ii) and the irreducibility of the right hand
side.

∎

Remarks. This proves [5, Prop. 6.10.4]. On the other hand by [5, 6.10.7] it
cannot be that $P.e_{VS}$ is always dense in $\mathscr{C}$. Again as we saw in
4.1, even the dimension of $\mathfrak{p}.E$ can strictly exceed
$\dim\mathfrak{m}-g$.

####  4.5.4. Weierstrass Sections

Fix a standard parabolic subalgebra $\mathfrak{p}$ of $\mathfrak{g}$, with
$\mathfrak{m}$ its nilradical and $\mathscr{T}_{\mathfrak{m}}$ the tableau it
defines.

Let $C,C^{\prime}$ be a pair of neighbouring columns of
$\mathscr{T}_{\mathfrak{m}}$ having height $s$ and $f_{C,C^{\prime}}$ the
corresponding Benlolo-Sanderson invariant and let
$\varphi:(\mathbb{C}[\mathfrak{m}])^{P^{\prime}}\rightarrow\mathbb{C}[e+V]$
(resp.
$\varphi_{VS}:(\mathbb{C}[\mathfrak{m}])^{P^{\prime}}\rightarrow\mathbb{C}[e_{VS}+V]$)
be defined by restriction of functions.

######  Proposition.

$(i)$. For every pair of neighbouring columns $C,C^{\prime}$ of
$\mathscr{T}_{\mathfrak{m}}$, one has
$\varphi(f_{C,C^{\prime}})=\varphi_{VS}(f_{C,C^{\prime}})$.

$(ii)$. $e_{VS}+V$ is a Weierstrass section for the action of $P^{\prime}$ on
$\mathfrak{m}$.

###### Proof.

Recall [4, 4.2.5] that every monomial in $f_{C,C^{\prime}}$ corresponds to a
disjoint union of composite lines passing though every element in $R^{s}$
between $C,C^{\prime}$. We shall refer to this last property by saying the
disjoint union is complete.

Recall that our construction in [5, 5.4.11] gives one particular complete
disjoint union of composite lines, formed from the lines defining $e+V$.

Moreover in loc. cit. we also showed that there are no other possible complete
disjoint union formed from the lines defining $e+V$.

Observation. To prove (i) it is obviously enough to show that the additional
lines given by adjoining the lines $\ell_{j,l}$ for every VS quadruplet
$(i,j,k,l)$ does not introduce a new complete disjoint union of composite
lines.

Recall (Remark 4.4.8) that a line, say $\ell_{j,k}$, carrying a $\ast$
uniquely determines a VS quadruplet $(i,j,k,l)$ (if one exists). Moreover the
line $\ell_{j,l}$ introduced by this quadruplet is uniquely the composition of
the lines $\ell_{j,k},\ell_{k,l}$ and so uniquely determined by $\ell_{j,k}$.

$(*)$. Thus there can be only new composite line $\mathscr{L}^{\prime}$ which
simply bypasses the box with entry $k$. Moreover the value it gives to the
monomial, changes from $x_{j,k}$ to $1$.

Yet as we shall see below even this extra scalar factor cannot appear!

Recall that in [5, 5.4.11] that we proved that there is only one complete
disjoint union of composite lines joining $C,C^{\prime}$, formed from the
lines defining $e+V$. As noted in loc cit, this was practically obvious if we
forbid passage across gated lines (carrying a $\ast$), Now in this the only
line, call it $\ell_{j,k}$, carrying a $\ast$ which is not gated, is exactly
the unique line whose right hand end point is the box $C^{\prime}\cap R_{s}$.

Moreover going leftwards from this box there is a unique composite line
$\mathscr{L}$ reaching a box in $C$. The boxes through which $\mathscr{L}$
passes cannot, by definition, appear in the passage of a further composite
line $\mathscr{L}^{\prime}$ disjoint to $\mathscr{L}$. As note in [5, 5.4.11]
this forbids passage across all the lines carrying a $\ast$ gated at the $s-1$
stage. Then by downward induction (cf [5, 5.4.11], one concludes that there is
only one complete disjoint union of composite lines joining $C,C^{\prime}$.
Moreover amongst the set of individual lines in these composite lines, there
is only one carrying a $\ast$ namely $\ell_{j,k}$.

We conclude that the (unique) complete disjoint union of composite lines
determined by $e+V$ is unchanged, when we pass to $e_{VS}$ because no line
carrying a $\ast$ is used in the first union except $\ell_{j,k}$ and this
cannot be replaced by $\ell_{j,l}$ because $k$ is an entry of $C^{\prime}$ and
therefore $l$ is in a box strictly to the right of $C^{\prime}$. Thus the
potentially new composite line $\mathscr{L}^{\prime}$ mentioned in $(*)$ above
does not even exist. Hence (i).

(ii) follows from (i) and [5, 5.4.12].

∎

##  5\. Figures

The following figures illustrate the assertions of 3.4, for different
assignments of column heights of $S$. In each case three pictures are drawn.
On the left (resp. right) of the top picture, we give the last (resp. last
two) columns of $\textbf{M}(s)$ (resp. $\textbf{M}(s+1)$). In the middle
picture, a composition realizing these assignments of height is given. In this
$s$ is its last entry. The bottom picture describes the resulting matrix
presentation of $\textbf{M}(s)\rightarrow\textbf{M}(s+1)$.

###  5.1. Case $1$

$\left.\begin{array}[]{|c|}\hline\cr\\\ \hline\cr\vdots\\\ \hline\cr\\\
\hline\cr{1}\\\ \hline\cr\\\ \hline\cr\vdots\\\ \hline\cr\\\ \hline\cr\\\
\hline\cr\vdots\\\ \hline\cr\end{array}\right.$
$\left.\begin{array}[]{|c|c|}\hline\cr&\\\ \hline\cr\vdots&\vdots\\\
\hline\cr&\\\ \hline\cr{1}&\\\ \hline\cr&\\\ \hline\cr\vdots&\vdots\\\
\hline\cr&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}\\\
\hline\cr&\\\ \hline\cr\vdots&\vdots\\\ \hline\cr\end{array}\right.$
Figure 1. $S$ has a column of height $>s+2$ but none of height $s,s+1$. This
illustrates the last part of $(iii)B$

Example: Consider the parabolic defined by the composition $(3,1)$:

${1}$${4}$${2}$${3}$$\scriptstyle{1}$${1}$${4}$${2}$${5}$${3}$$\scriptstyle{1}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$

${1}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${1}$$\left(\vbox{\hrule
height=29.75389pt,depth=29.75389pt,width=0.0pt}\right.$$\left.\vbox{\hrule
height=29.75389pt,depth=29.75389pt,width=0.0pt}\right)$${1}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${1}$$\left(\vbox{\hrule
height=36.30911pt,depth=36.30911pt,width=0.0pt}\right.$$\left.\vbox{\hrule
height=36.30911pt,depth=36.30911pt,width=0.0pt}\right)$

###  5.2. Case $2$

$\left.\begin{array}[]{|c|}\hline\cr\\\ \hline\cr\vdots\\\ \hline\cr\\\
\hline\cr{1}\\\ \hline\cr\\\ \hline\cr\vdots\\\ \hline\cr\\\
\hline\cr\vdots\\\ \hline\cr\\\ \hline\cr\vdots\\\
\hline\cr\end{array}\right.$ $\left.\begin{array}[]{|c|c|}\hline\cr&\\\
\hline\cr\vdots&\vdots\\\ \hline\cr&\\\ \hline\cr{1}&\\\ \hline\cr&\\\
\hline\cr\vdots&\vdots\\\
\hline\cr&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\ast}\\\
\hline\cr\vdots&\vdots\\\
\hline\cr&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}\\\
\hline\cr\vdots&\vdots\\\ \hline\cr\end{array}\right.$
Figure 2.  $S$ has no column of height $s$, a column of height $s+1$ and a
column of height $>s+1$. This illustrates the first part of $(iii)B$ and
$(iv)$.

Example: Consider the parabolic defines by the composition $(2,3,1)$ to
$(2,3,2)$:

${1}$${3}$${6}$${2}$${4}$${5}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$

${1}$${3}$${6}$${2}$${4}$${7}$${5\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{}}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\ast}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$

${1}$${0}$${1}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$$\left(\vbox{\hrule
height=42.86433pt,depth=42.86433pt,width=0.0pt}\right.$$\left.\vbox{\hrule
height=42.86433pt,depth=42.86433pt,width=0.0pt}\right)$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\ast}}$${0}$${0}$${0}$${0}$${1}$${0}$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$$\left(\vbox{\hrule
height=49.41956pt,depth=49.41956pt,width=0.0pt}\right.$$\left.\vbox{\hrule
height=49.41956pt,depth=49.41956pt,width=0.0pt}\right)$

###  5.3. Case $3$

$\left.\begin{array}[]{|c|}\hline\cr\\\ \hline\cr\vdots\\\ \hline\cr\\\
\hline\cr\vdots\\\ \hline\cr 1\\\ \hline\cr\vdots\\\ \hline\cr\ast\\\
\hline\cr\vdots\\\ \hline\cr\\\ \hline\cr\vdots\\\
\hline\cr\end{array}\right.$ $\left.\begin{array}[]{|c|c|}\hline\cr&\\\
\hline\cr\vdots&\vdots\\\
\hline\cr&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}\\\
\hline\cr\vdots&\vdots\\\
\hline\cr&\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\ast}\\\
\hline\cr\vdots&\vdots\\\
\hline\cr\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}&\\\
\hline\cr\vdots&\vdots\\\ \hline\cr&\\\ \hline\cr\vdots&\vdots\\\
\hline\cr\end{array}\right.$
Figure 3. $S$ has a column of height $s$ and two columns of height $s+1$. This
illustrates $(ii),(iii)A$ and $(iv)$.

Example: Consider the parabolic defines by the composition $(2,3,2,1,1)$ to
$(2,3,2,1,2)$:

${1}$${3}$${6}$${8}$${9}$${2}$${4}$${7}$${5\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{}}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{\ast}$$\scriptstyle{1}$$\scriptstyle{\ast}$$\scriptstyle{1}$$\scriptstyle{1}$

${1}$${3}$${6}$${8}$${9}$${2}$${4}$${7}$${10}$${5\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{}}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{\definecolor[named]{.}{rgb}{0,0,1}\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}1}$$\scriptstyle{1}$$\scriptstyle{\ast}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\ast}$$\scriptstyle{1}$

${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${\ast}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${\ast}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$$\left(\vbox{\hrule
height=62.53pt,depth=62.53pt,width=0.0pt}\right.$$\left.\vbox{\hrule
height=62.53pt,depth=62.53pt,width=0.0pt}\right)$

${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${\ast}$${0}$${0}$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{1}}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}{\ast}}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}{1}}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${0}$${1}$$\left(\vbox{\hrule
height=69.08522pt,depth=69.08522pt,width=0.0pt}\right.$$\left.\vbox{\hrule
height=69.08522pt,depth=69.08522pt,width=0.0pt}\right)$

The three following figures illustrate the VS quadruplet of the examples of
4.4.8.
Examples: $(i)$. Consider the parabolic defines by the composition
$(1,2,1,1,2,2,3)$:

${1}$${2}$${4}$${5}$${6}$${8}$${10}$${3}$${7}$${9}$${11}$${12}$$\scriptstyle{1}$$\scriptstyle{*}$$\scriptstyle{1}$$\scriptstyle{*}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$$\scriptstyle{\definecolor[named]{.}{rgb}{0,0,1}\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}1}$$\scriptstyle{*}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{*}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$
Figure 4. The VS pair $(3,4,9,11)$ in red is not bad thanks to the blue line
$\ell_{4,9}$.

$(ii)$. Let the parabolic defines by the composition $(1,2,1,1,2,3)$:

${1}$${2}$${4}$${5}$${6}$${8}$${3}$${7}$${9}$${10}$$\scriptstyle{1}$$\scriptstyle{*}$$\scriptstyle{1}$$\scriptstyle{*}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$$\scriptstyle{\definecolor[named]{.}{rgb}{0,0,1}\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}1}$$\scriptstyle{*}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$
Figure 5. the VS pair $(3,4,7,9)$ in red is not bad. the line $\ell_{4,9}$
representing $x_{4,9}\in\operatorname{supp}e$ lies in the same column as
$x_{4,9}$.

$(iii)$. Let the parabolic defines by the composition $(3,2,1,1,2,3)$:

${1}$${4}$${6}$${7}$${8}$${10}$${2}$${5}$${9}$${11}$${3}$${12}$$\scriptstyle{1}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$$\scriptstyle{*}$$\scriptstyle{*}$$\scriptstyle{\definecolor[named]{.}{rgb}{0,1,0}\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}1}$$\scriptstyle{*}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{1}$$\scriptstyle{\definecolor[named]{.}{rgb}{1,0,0}\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}1}$$\scriptstyle{1}$
Figure 6. The VS pair $(3,4,7,9)$ in red is bad, we augment the $e$ with the
green line $\ell_{4,10}$.

Recall [5, 4.2] that $e$ is given by the co-ordinate vectors defined by the
lines with label $1$. In figure 6, these form three disjoint lines $(1,4,6)$,
$(2,5,7,8,10)$, $(3,9,11)$, so $e$ has nilpotency class is $(5,3,3,1)$ giving
rise to column heights $(4,3,3,1,1)$. Thus $\frac{1}{2}\dim|G\cdot
e=\dim\mathfrak{n}-\frac{1}{2}(4\times 3+3\times 2+3\times
2)=\dim\mathfrak{n}-12$, whilst $\dim\mathfrak{m}=\dim\mathfrak{n}-8$.
Consequently $\frac{1}{2}\dim G\cdot e=\dim P\cdot e=\dim\mathfrak{m}-4$. Yet
$e\in\mathscr{N}$ is regular if and only if $\dim P\cdot
e=\dim\mathfrak{m}-\dim V=\dim\mathfrak{m}-3$.

The lines in red represent the bad VS pair $(3,4,7,9)$ in figure 6. By 4.4.4
we must adjoin the green line, that is $\ell_{6,11}$, giving
$e_{VS}=e+x_{6,11}$, which by Prop. 4.5.3 lies in $\mathscr{N}^{e}$. The
nilpotency class of $e_{VS}$ is $(5,4,2,1)$ giving rise to column heights
$(4,3,2,2,1)$. Thus $\frac{1}{2}\dim G\cdot e_{VS}=\dim\mathfrak{n}-11$ and
$\frac{1}{2}\dim G\cdot e_{VS}=\dim P\cdot
e_{VS}=\dim\mathfrak{m}-3=\dim\mathscr{N}^{e}$, the last equality by Prop.
4.5.3. Thus $Pe_{VS}$ is dense orbit in $\mathscr{N}^{e}$ and so $e_{VS}$ is a
regular element in $\mathscr{N}$.

##  6\. Index of Notation

Symbols used frequently are given below in the order in which they appear.

1. $\mathbb{C},[1,n]$.

1.1.
$G,\mathfrak{g},\mathfrak{h},H,\Delta,\Delta^{+},\pi,\mathfrak{n},\mathfrak{n}^{-},\mathfrak{b},W,s_{\alpha},\alpha^{\vee}$.

1.2.
$\mathfrak{r}_{\pi^{\prime}},\mathfrak{p}_{\pi^{\prime}},\mathfrak{m}_{\pi^{\prime}},P,P^{\prime},\mathfrak{p}^{\prime}$.

1.3. $\mathscr{N},e+V,\mathscr{N}^{e}$.

1.4. $\mathscr{N}_{reg},e_{VS},E_{VS}$.

1.6. $B.\mathfrak{u},d$.

2.1. $\textbf{M}_{n},x_{i,j},\alpha_{i,j}$.

2.2. $W_{\pi^{\prime}},w_{\pi^{\prime}},c_{i},B_{i},\textbf{C}_{i}$.

2.3.
$\mathscr{D}_{\mathfrak{m}},R_{i},R^{i},C_{i},\mathscr{T}_{\mathfrak{m}},\ell_{b,b^{\prime}},\ell_{i,j}$.

2.4 $M_{v,v^{\prime}},\mathfrak{u}_{v,v^{\prime}},w_{u,u^{\prime}}$.

2.6. $M_{s}(v,v^{\prime}),\mathscr{T}^{t},M_{s},M_{s}(\mathfrak{m})$.

2.7. $\mathscr{P}$.

3.2.1. $\mathscr{C},\ell(\mathscr{C})$.

3.3. $b_{i,j}$.

3.3.1.
$\textbf{C}(s),\textbf{M}(s),s,\textbf{S},\textbf{M}^{k},e^{\kappa},V^{\kappa}$.

3.3.2. $\textbf{r}_{C,C^{\prime}}$.

3.4. $\textbf{C}^{t}$.

3.4.2. $\mathscr{R},|S|$.

3.6.8. $\mathscr{S}$.

4.4.2. $C<C^{\prime},b(t),C_{(t)},c_{(t)}$.

4.4.4. $e^{\kappa}_{VS},E^{\kappa}_{VS}$.

Acknowledgements.

This work was the subject of a joint talk on Zoom hosted by the Weizmann
Institute held during 25 April to 6 May 2021, on the occasion of the $60^{th}$
birthday of Vera Serganova. We would like to thank the organizers (Inna
Entova-Aizenbud, Maria Gorelik and Dimitar Grantcharov) for the invitation to
speak. Our talk may be accessed on

https://innaentova.wixsite.com/springfest2021/recording-links, using the
password $362880$, which is $9!$.

The first author was supported at the Weizmann Institute, arranged by Ronen
Basri and Gal Binyamini on ISRAEL SCIENCE FOUNDATION (grant No. 1167/17) and
the MINERVA Stiftung with the funds from the BMBF of the Federal Republic of
Germany.

During the academic year $2021-2022$ the first author was supported through
ISF grant 1017/19 as a postdoctoral student of Anna Melnikov at the University
of Haifa.

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