#  Jordan structures of nilpotent matrices in the centralizer of a nilpotent
matrix with two Jordan blocks of the same size

Duško Bogdanić  Faculty of Natural Sciences and Mathematics, University of
Banja Luka, Mladena Stojanovića 2, 78000 Banja Luka, Bosnia and Herzegovina
<EMAIL_ADDRESS>,  Alen Đurić  Université Paris Cité, CNRS, Inria,
IRIF, F-75013, Paris, France<EMAIL_ADDRESS>,  Sara Koljančić
Faculty of Natural Sciences and Mathematics, University of Banja Luka, Mladena
Stojanovića 2, 78000 Banja Luka, Bosnia and Herzegovina
<EMAIL_ADDRESS>,  Polona Oblak  Faculty of Computer and
Information Science, University of Ljubljana, Večna pot 113, SI-1000
Ljubljana, Slovenia; Faculty of Mathematics and Physics, University of
Ljubljana and Institute of Mathematics, Physics, and Mechanics, Jadranska
ulica 19, 1000 Ljubljana, Slovenia<EMAIL_ADDRESS>and  Klemen
Šivic  Faculty of Mathematics and Physics, University of Ljubljana, and
Institute of Mathematics, Physics, and Mechanics, Jadranska ulica 19, 1000
Ljubljana, Slovenia<EMAIL_ADDRESS>
###### Abstract.

In this paper we characterize all nilpotent orbits under the action by
conjugation that intersect the nilpotent centralizer of a nilpotent matrix $B$
consisting of two Jordan blocks of the same size. We list all the possible
Jordan canonical forms of the nilpotent matrices that commute with $B$ by
characterizing the corresponding partitions.

###### Key words and phrases:

nilpotent matrix, centralizer, nilpotent orbit

###### 2020 Mathematics Subject Classification:

15A27, 14L30, 15A21

The authors acknowledge the financial support from the bilateral project no.
BI-BA/19-20-044 funded by Slovenian Research Agency and Ministry of Civil
Affairs, Bosnia and Herzegovina. Polona Oblak and Klemen Šivic have received
funding from Slovenian Research Agency (research core funding no. P1-0222 and
projects no. J1-3004 (PO&KŠ) and N1-0103 (KŠ)) and Alen Đurić has received
funding from the European Union’s Horizon 2020 research and innovation
programme under the Marie Skłodowska-Curie grant agreement No. 754362.

##  1\. Introduction and background

Let $F$ be an arbitrary field and $N$ a positive integer. By $M_{N}(F)$ we
denote the set of all $N\times N$ matrices over $F$, and by $\mathcal{N}_{N}$
we denote the set of all nilpotent elements in $M_{N}(F)$. Note that the group
$GL_{N}(F)$ acts on $\mathcal{N}_{N}$ by conjugation, i.e., $g\cdot
X=gXg^{-1}$ for $g\in GL_{N}(F)$ and $X\in\mathcal{N}_{N}$. Recall that each
nilpotent $N\times N$ matrix over $F$ is similar to a nilpotent matrix $J$ in
the Jordan canonical form. If we assume that Jordan blocks of $J$ are ordered
by non-increasing size, then such $J$ is uniquely defined. Hence the set of
all the partitions ${\bf{\lambda}}=(\lambda_{1},\ldots,\lambda_{t})$,
$\lambda_{1}\geq\ldots\geq\lambda_{t}$, $\sum_{i=1}^{t}\lambda_{i}=N$, of the
integer $N$ is in a bijective correspondence with the set of all orbits of
$\mathcal{N}_{N}$ under the action of the group $GL_{N}(F)$. The orbit
corresponding to the partition ${\bf{\lambda}}$ is denoted by
$\mathcal{O}_{{\bf{\lambda}}}$. It was first proved by Gerstenhaber [13] that
the dominance ordering on the set of partitions of number $N$ coincides with
the inclusion ordering of orbits of $\mathcal{N}_{N}$. The algebraic and
combinatorial properties of nilpotent orbits over various fields and in other
reductive Lie algebras have been exhaustively studied, see e.g., [9, 12, 27].

In the last decade, there has been an extensive investigation of Jordan
structures of multiplication maps on finite-dimensional commutative algebras.
If either the algebra is local and $x$ is an element of the maximal ideal or
the algebra is graded and $x$ has a positive degree (for example, $x$ is a
linear form), then the multiplication map by $x$ is a nilpotent linear map on
the algebra, whose Jordan structure is called the Jordan type of the element
$x$. The “generic” Jordan type is an important invariant of the algebra, which
is tightly connected to the Lefschetz properties, see e.g., [1, 10, 16, 20].

Recently, the pairs of commuting nilpotent operators have been extensively
explored, see e.g., [2, 3, 4, 15, 22]. Let $B\in\mathcal{N}_{N}$ be an
arbitrary nilpotent matrix. Let $\mathcal{C}(B)$ denote the _centralizer of
$B$_, i.e., $\mathcal{C}(B)=\\{A\in M_{N}(F);AB=BA\\}$, and let

$\mathcal{N}(B)=\mathcal{C}(B)\cap\mathcal{N}_{N}$

denote the _nilpotent centralizer of $B$_, i.e., the set of all nilpotent
elements of $\mathcal{C}(B)$. It is known that $\mathcal{N}(B)$ is an
irreducible variety, [4].

One of the classical questions in the matrix theory is to determine the
intersection of the nilpotent centralizer with the nilpotent orbits.

######  Question 1.

For a nilpotent matrix $B\in M_{N}(F)$, determine which of the
$GL_{N}(F)$-orbits of $\mathcal{N}_{N}$ intersect the nilpotent centralizer of
$B$ or, equivalently, determine all possible partitions ${\bf{\lambda}}$
corresponding to Jordan canonical forms of the nilpotent matrices that commute
with $B$.

In general, this seems to be a difficult question, which has been answered so
far only for some specific cases of matrices $B$.

For example, the nilpotent centralizer of a single Jordan block
$J_{(N)}\in{\mathcal{O}_{(N)}}$ intersects only $N$ nilpotent orbits. This is
the consequence of the fact that the only matrices commuting with $J_{(N)}$
are polynomials in $J_{(N)}$, see e.g., [17, Theorem 3.2.4.2]. Thus the
partitions ${\bf{\lambda}}$ that correspond to the matrices commuting with
$J_{(N)}$ are precisely the ones whose parts differ by at most 1, i.e.,
$\lambda_{1}-\lambda_{t}\leq 1$. Such partitions are called _almost
rectangular_ , [21]. Following [19], we denote the unique almost rectangular
partition of $N$ into $k$ parts by

$[N]^{k}=\left((q+1)^{r},q^{k-r}\right),$

where $q=\left\lfloor\frac{N}{k}\right\rfloor$, $r=N\mod k$, and $m^{t}$
abbreviates $t$ copies of $m$ in the partition. On the other hand, the
matrices over a finite field or over an algebraically closed field of
characteristic 0 with largest intersections with nilpotent orbits were
characterized in [8, 26]; if $N\geq 4$, then $B^{2}=0$ if and only if
$\mathcal{N}(B)\cap\mathcal{O}_{{\bf{\lambda}}}\neq\emptyset$ for all
partitions ${\bf{\lambda}}$ of $N$.

For a general matrix $B$, only partial answers to Question 1 were given. The
lower bound for the number of parts of ${\bf{\lambda}}$ was determined in [4]
and the upper bound for $\lambda_{1}$ was given in [25]. Moreover, it was
proved in [2] that if ${\bf{\mu}}^{\prime}$ is the conjugate partition of the
Jordan canonical form of $B$, then
$\mathcal{N}(B)\cap\mathcal{O}_{{\bf{\mu}}^{\prime}}\neq\emptyset$.

Since $\mathcal{N}(B)$ is an irreducible variety, there exists a partition
${\bf{\lambda}}$ of $N$ such that
$\mathcal{O}_{{\bf{\lambda}}}\cap\mathcal{N}(B)$ is a dense open subset of
$\mathcal{N}(B)$. If ${\bf{\mu}}$ is the partition of $N$ that corresponds to
the Jordan canonical form of $B$, then $\mathcal{Q}({\bf{\mu}})$ is defined to
be the unique partition ${\bf{\lambda}}$ such that
$\mathcal{O}_{{\bf{\lambda}}}\cap\mathcal{N}(B)$ is a dense open subset of
$\mathcal{N}(B)$, which is the maximal partition in the dominance ordering,
for which its nilpotent orbit intersects $\mathcal{N}(B)$. Recently, the
conjectured recursive algorithm to compute $\mathcal{Q}({\bf{\mu}})$ (see
results and discussions in [6, 7, 18, 21]) has been proved to work for general
${\bf{\mu}}$, [5]. Furthermore, the preimage $\mathcal{Q}^{-1}(\lambda)$ was
completely determined for partitions $\lambda$ with two parts which differ by
at least $2$ [19]. For generalization of $\mathcal{Q}(\mu)$ to other simple
Lie algebras, see [29].

Motivated by these partial results and open problems, this paper considers and
completely resolves the following special case of Question 1.

######  Question 2.

Let $B\in\mathcal{N}_{2n}$ be a nilpotent matrix with the Jordan canonical
form corresponding to the partition $(n,n)$. Characterize all possible
partitions ${\bf{\lambda}}$ corresponding to Jordan canonical forms of the
nilpotent matrices that commute with $B$.

There are some known partitions ${\bf{\lambda}}$, for which the answer to
Question 2 is positive. Namely, since the centralizers of square-zero matrices
intersect all nilpotent orbits, any partition of the form
${\bf{\lambda}}=(2^{t},1^{s})$, $2t+s=2n$, gives rise to a nilpotent orbit
that intersects ${\mathcal{N}}(B)$, [8, 26]. Moreover, we can assume that $B$
is already in its Jordan canonical form, thus $B=J_{(n)}\oplus J_{(n)}$. Since
every matrix commutes with its polynomial, it follows that $B$ commutes with
$p(J_{(n)})\oplus q(J_{(n)})$, for any polynomials $p,q\in F[X]$. By the above
results, it follows that ${\bf{\lambda}}=([n]^{t},[n]^{s})$, where $1\leq
t,s\leq n$, gives a positive answer to Question 2 as well. Also, in [26,
Theorem 3.6(c)] it was proved that any nilpotent orbit corresponding to an
almost rectangular partition of $(2n)$ intersects ${\mathcal{N}}(B)$.
Furthermore, it was proved in the same paper that all nilpotent orbits that
give a positive answer to Question 2 with the corresponding partition
${\bf{\lambda}}\neq(2n)$ have $\lambda_{1}\leq n+1$, and
${\mathcal{O}}_{(n+1,n-1)}$ has a non-empty intersection with
$\mathcal{N}(B)$. It was noted in [26, Example 3.7] that these partitions do
not constitute the complete list of nilpotent orbits intersecting
${\mathcal{N}}_{B}$. Some of the partitions ${\bf{\lambda}}$ obtained by block
antidiagonal and block upper triangular nilpotent matrices commuting with $B$
were listed in [24, Theorems 6.7, 6.19].

The main objective of this paper is to completely resolve Question 2, and to
give a complete list of nilpotent orbits $\mathcal{O}_{{\bf{\lambda}}}$ having
a non-empty intersection with ${\mathcal{N}}(B)$, where $B$ has the Jordan
canonical form $(n,n)$. We will prove the following theorem.

######  Theorem 3.

Let $B\in\mathcal{N}_{2n}$ be a nilpotent matrix with the Jordan canonical
form corresponding to the partition $(n,n)$. If $\mathcal{N}(B)$ intersects
the nilpotent orbit $\mathcal{O}_{{\bf{\lambda}}}$, then $\lambda$ is one of
the partitions given in Table 1. The partitions from the cases (P1)-(P7)
always belong to some nilpotent orbits intersecting $\mathcal{N}(B)$, while
occurrence of the partitions from (P8)-(P9) depends on the field $F$. In
particular, if $F=\mathbb{C}$, then ${\mathcal{N}}(B)$ intersects exactly the
nilpotent orbits ${\mathcal{O}}_{{\bf{\lambda}}}$, where ${\bf{\lambda}}$ is
given in Table 1.

|  ${\bf{\lambda}}$ |  constraints
---|---|---
(P1) |  $[2n]^{s}$ |  $1\leq s\leq 2n$.
(P2) |  $([n+m]^{m+z},[n-m]^{l-m+z})$ |  $0\leq z\leq n-1$, $0<2m\leq l<n-z$, $\lceil\frac{n+m}{m+z}\rceil\geq\lceil\frac{n-m}{l-m+z}\rceil+1$
|  |  or
|  |  $0\leq z\leq n-1$, $0<m<n-z$, $l=n-z$
(P3) |  $([n+\alpha^{\prime}]^{m+z},[n-\alpha^{\prime}]^{l-m+z})$ |  $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$, $n+l-m>\alpha(l+2z)$, $\alpha^{\prime}=l-m+(2m-l)\alpha$
(P4) |  $([n+\alpha^{\prime\prime}]^{l-m+z}),[n-\alpha^{\prime\prime}]^{m+z})$ |  $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$, $n+m+2z\leq\alpha(l+2z)<n+l+z$,
|  |  $\alpha^{\prime\prime}=m-(2m-l)\alpha$
(P5) |  $([n]^{z+1},[n]^{m+z})$ |  $0\leq z\leq n-1$, $2\leq m\leq n-z$
(P6) |  $((2\alpha)^{l-\beta},(2\alpha-1)^{2z},(2\alpha-2)^{\beta})$ |  $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$, $\max\\{l-m-z,0\\}\leq\beta<\min\\{m+z,l\\}$
|  |  or
|  |  $0\leq z\leq n-1$, $0\leq\beta<l<m\leq n-z$
(P7) |  $((2\alpha-1)^{2l-m+z-\beta},(2\alpha-2)^{2m-l+2z},(2\alpha-3)^{\beta-m-z})$ |  $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$, $\max\\{m+z,l\\}\leq\beta<\min\\{2l-m+z,l+2z\\}$
(P8) |  $([n+\gamma^{\prime}]^{m+z-t},[n-\gamma^{\prime}]^{m+z+t})$ |  $0\leq z\leq n-1$, $0<m<\frac{n-z}{2}$, $\lceil\frac{n+m+2z}{m+z}\rceil=\lceil\frac{n+m}{m+z}\rceil\geq 5$, $1\leq t\leq\min\\{m,m-n+\lceil\frac{n-m}{m+z}\rceil(m+z)\\}$, $\gamma^{\prime}=m-\lceil\frac{n-z}{m+z}\rceil t$
(P9) |  $([n+\gamma^{\prime\prime}]^{m+z+t},[n-\gamma^{\prime\prime}]^{m+z-t})$ |  $0\leq z\leq n-1$, $0<m<\frac{n-z}{2}$, $\lceil\frac{n-m}{m+z}\rceil\leq\lceil\frac{n-z}{m+z}\rceil$, $\lceil\frac{n+m}{m+z}\rceil\geq 4$, $t\geq 1$, $n+z+t\leq\lceil\frac{n+m}{m+z}\rceil(m+z)\leq n+2m+z-t$,
|  |  $\gamma^{\prime\prime}=m+\lceil\frac{n-m-2z}{m+z}\rceil t$
Table 1. A complete list of the nilpotent orbits
$\mathcal{O}_{{\bf{\lambda}}}$ having a non-empty intersection with
${\mathcal{N}}(B)$, $B\in\mathcal{O}_{(n,n)}$. Here, we use the notation
$\alpha=\left\lceil\frac{n+z}{l+2z}\right\rceil$ and
$\beta=\alpha(l+2z)-(n+z)$.

Note that the cases (P1)-(P9) are not complementary and that some of them
coincide. Since there are nine different cases, it is expected that the proof
will be technical. This is why we first give some auxiliary results in Section
2, which will be used in Section 3, completely dedicated to proving Theorem 3.
We complete the paper with some examples in Section 4. In particular, we show
that the partitions in the cases (P8)-(P9) may not occur when $F=\mathbb{Q}$.

##  2\. Preliminaries

Let $0_{n}$ and $I_{n}$ be the zero and the identity matrix in $M_{n}(F)$,
respectively, where we omit the subscripts when the order of the matrix is
clear from the context.

Throughout the paper, let $B\in\mathcal{N}_{2n}$ be a nilpotent matrix with
the Jordan canonical form $(n,n)$. There is no loss of generality if we assume
that $B$ is in the Jordan canonical form, i.e., $B=J_{n}\oplus J_{n}$ where
$J_{n}$ denotes the $n\times n$ Jordan block. A matrix $A$ then commutes with
$B$ if and only if it is of the form

$A=\left[\begin{array}[]{cccccccc}a_{0}&a_{1}&\cdots&a_{n-1}&b_{0}&b_{1}&\cdots&b_{n-1}\\\
&\ddots&\ddots&\vdots&&\ddots&\ddots&\vdots\\\
&&\ddots&a_{1}&&&\ddots&b_{1}\\\ &&&a_{0}&&&&b_{0}\\\
c_{0}&c_{1}&\cdots&c_{n-1}&d_{0}&d_{1}&\cdots&d_{n-1}\\\
&\ddots&\ddots&\vdots&&\ddots&\ddots&\vdots\\\
&&\ddots&c_{1}&&&\ddots&d_{1}\\\ &&&c_{0}&&&&d_{0}\end{array}\right]$

for some $a_{i},b_{i},c_{i},d_{i}\in F$ (see e.g., [14]), and with the omitted
entries all equal to zero. Apart from this, there are two other useful
descriptions of the matrices $A\in{\mathcal{N}}(B)$: the Weyr form [11, 28]
and the polynomial notation [23]. In general, the notation is complicated, but
in the case when two Jordan blocks of $B$ are of the same size, the notation
can be simplified to the following:

  * •

(Weyr form) Using a suitable permutation of the basis of $F^{2n}$ we can see
that the matrix $J_{n}\oplus J_{n}$ is similar to the matrix

(1) $\left[\begin{array}[]{ccccc}0_{2}&I_{2}&0_{2}&\cdots&0_{2}\\\
&\ddots&\ddots&\ddots&\vdots\\\ &&\ddots&\ddots&0_{2}\\\ &&&\ddots&I_{2}\\\
&&&&0_{2}\end{array}\right],$

where all blocks are of size $2\times 2$. Such a matrix is said to be in the
Weyr canonical form and it is very useful when the commutativity questions are
considered (see [28]). By abuse of the notation we will denote matrix (1)
again by $B$, since it represents the same linear transformation. In this
basis, the matrix of the linear transformation $A$ that commutes with $B$ has
the form

(2) $A=\left[\begin{array}[]{cccc}A_{0}&A_{1}&\cdots&A_{n-1}\\\
&\ddots&\ddots&\vdots\\\ &&\ddots&A_{1}\\\ &&&A_{0}\end{array}\right],$

where $A_{i}=\left[\begin{array}[]{cc}a_{i}&b_{i}\\\
c_{i}&d_{i}\end{array}\right]$ for each $i=0,1,\ldots,n-1$. Clearly, $A$ is
nilpotent if and only if $A_{0}$ is nilpotent, which was already observed in
[4, Lemma 2.3].

  * •

(Polynomial notation) Denote the polynomial ring $F[X]$ by $R$ and define the
following polynomials in $R$:

$a(X)=a_{n-1}X^{n-1}+\cdots+a_{1}X+a_{0},\quad
b(X)=b_{n-1}X^{n-1}+\cdots+b_{1}X+b_{0},$
$c(X)=c_{n-1}X^{n-1}+\cdots+c_{1}X+c_{0},\quad
d(X)=d_{n-1}X^{n-1}+\cdots+d_{1}X+d_{0}.$

Then

(3) $A=\left[\begin{array}[]{cc}a(J_{n})&b(J_{n})\\\
c(J_{n})&d(J_{n})\end{array}\right].$

The map

$\left[\begin{array}[]{cc}a(X)&b(X)\\\
c(X)&d(X)\end{array}\right]\mapsto\left[\begin{array}[]{cc}a(J_{n})&b(J_{n})\\\
c(J_{n})&d(J_{n})\end{array}\right]$

is a surjective homomorphism of algebras $M_{2}(R)\to\mathcal{C}(B)$ with the
kernel consisting of all $2\times 2$ matrices with entries in the ideal
$(X^{n})$, therefore, the algebra $\mathcal{C}(B)$ is isomorphic to
$M_{2}(F[X]/(X^{n}))$ (see [23]). For simplicity, we denote the quotient ring
$F[X]/(X^{n})$ by $S$. Again, we will use the same notation for the matrices
from $M_{2}(S)$ and from $\mathcal{C}(B)$. We will also use the same notation
for polynomials from $R$ and for the corresponding elements of the quotient
ring $S$. This should not cause any confusion, since we explicitly state which
ring is considered at any definition of a new polynomial. We also note that in
the polynomial notation the matrix $B$ is equal to
$\left[\begin{array}[]{cc}X&0\\\ 0&X\end{array}\right]$.

First, we state a lemma that will help us understand the powers of structured
polynomial matrices.

######  Lemma 4.

Let $C\in M_{2}(R)$ be an arbitrary matrix of the form
$C=\left[\begin{array}[]{cc}0&1\\\ c(X)&d(X)\end{array}\right]$ for some
polynomials $c(X),d(X)\in R$. Define the sequence of polynomials
$s_{0}(X),s_{1}(X),\ldots\in R$ by

$s_{0}(X):=0,s_{1}(X):=1,\text{ and }s_{k+1}(X):=d(X)s_{k}(X)+c(X)s_{k-1}(X)$

for $k\geq 1$. Then for each positive integer $k$, the following equalities
hold:

  1. (A)

$C^{k}=\left[\begin{array}[]{cc}c(X)s_{k-1}(X)&s_{k}(X)\\\
c(X)s_{k}(X)&s_{k+1}(X)\end{array}\right],$

  2. (B)

$s_{k-1}(X)s_{k+1}(X)-s_{k}(X)^{2}=(-1)^{k}c(X)^{k-1},$

  3. (C)

$s_{2k-1}(X)=\sum\limits_{i=1}^{k}{2k-i-1\choose i-1}c(X)^{i-1}d(X)^{2k-2i},$

  4. (D)

$s_{2k}(X)=\sum\limits_{i=0}^{k-1}{2k-i-1\choose i}c(X)^{i}d(X)^{2k-2i-1}$.

(Here, the convention $0^{0}=1$ is used if the corresponding polynomial is
zero.)

###### Proof.

The equality for $C^{k}$ in (A) can be proved by an easy induction using the
recursive formula for the polynomials $s_{k}(X)$. If $c(X)$ is non-zero,
equality (B) follows immediately from the identity

$(-1)^{k}c(X)^{k}=(\det
C)^{k}=\det(C^{k})=c(X)s_{k-1}(X)s_{k+1}(X)-c(X)s_{k}(X)^{2}.$

On the other hand, if $c(X)=0$, then $s_{k}(X)=d(X)^{k-1}$ for each $k\geq 1$,
which completes the proof of part (B). Moreover, (C) and (D) can be shown by
simultaneous induction as well. ∎

######  Lemma 5.

If $B\in\mathcal{N}_{2n}$ is a nilpotent matrix corresponding to the Jordan
canonical form $(n,n)$ and $A\in{\mathcal{N}}(B)$ a non-zero matrix, then
there exist a maximal non-negative integer $z$ and a matrix
$C\in{\mathcal{N}}(B)$ such that $A=B^{z}C$. If $B$ is in the Weyr form and
$A$ is of the form $(2)$, then

$C=\left[\begin{array}[]{cccccc}C_{0}&\cdots&C_{n-z-1}&0&\cdots&0\\\
&\ddots&&\ddots&\ddots&\vdots\\\ &&\ddots&&\ddots&0\\\ &&&\ddots&&C_{n-z-1}\\\
&&&&\ddots&\vdots\\\ &&&&&C_{0}\end{array}\right]\in{\mathcal{N}}(B),$

with $C_{i}$ of the form $C_{i}=\left[\begin{array}[]{cc}a_{i}&b_{i}\\\
c_{i}&d_{i}\end{array}\right]$ for $i\in\\{0,1,\ldots,n-z-1\\}$.

###### Proof.

Let $A\in{\mathcal{N}}(B)$ be as in (2) and let $z^{\prime}$ be the smallest
non-negative integer satisfying $A_{z^{\prime}}\neq 0$. If $A_{z^{\prime}}$ is
nilpotent, we take $z=z^{\prime}$ and

$C=\left[\begin{array}[]{ccccccc}A_{z^{\prime}}&A_{z^{\prime}+1}&\cdots&A_{n-1}&0&\cdots&0\\\
&\ddots&\ddots&&\ddots&\ddots&\vdots\\\ &&\ddots&\ddots&&\ddots&0\\\
&&&\ddots&\ddots&&A_{n-1}\\\ &&&&\ddots&\ddots&\vdots\\\
&&&&&\ddots&A_{z^{\prime}+1}\\\ &&&&&&A_{z^{\prime}}\end{array}\right].$

If $A_{z^{\prime}}$ is not nilpotent, then $z^{\prime}\geq 1$, since $A$ is
nilpotent. So we take $z=z^{\prime}-1$ and

$C=\left[\begin{array}[]{ccccccc}0&A_{z^{\prime}}&\cdots&A_{n-1}&0&\cdots&0\\\
&\ddots&\ddots&&\ddots&\ddots&\vdots\\\ &&\ddots&\ddots&&\ddots&0\\\
&&&\ddots&\ddots&&A_{n-1}\\\ &&&&\ddots&\ddots&\vdots\\\
&&&&&\ddots&A_{z^{\prime}}\\\ &&&&&&0\end{array}\right].\hfill{\qed}$

##  3\. Proof of Theorem 3

Let $B\in\mathcal{N}_{2n}$ be a nilpotent matrix with the Jordan canonical
form $(n,n)$. By Lemma 5, any $A\in{\mathcal{N}}(B)$ is of the form $A=B^{z}C$
for some non-negative integer $z$ and $C\in{\mathcal{N}}(B)$. If $A=0$, then
the corresponding partition is $[2n]^{2n}$, which is listed in Table 1(P1). In
the rest of the proof we therefore assume that $A$ is non-zero, and hence we
assume that $z$ is maximal possible such that $C$ is nilpotent. We will use
the Weyr form and write $B$ as in (1) and $C,C_{i}$ as in Lemma 5, or in the
polynomial notation and write

$B=\left[\begin{array}[]{cc}X&0\\\
0&X\end{array}\right],C=\left[\begin{array}[]{cc}a(X)&b(X)\\\
c(X)&d(X)\end{array}\right],\text{ and
}A=\left[\begin{array}[]{cc}X^{z}a(X)&X^{z}b(X)\\\
X^{z}c(X)&X^{z}d(X)\end{array}\right],$

where $a(X)=a_{0}+\cdots+a_{n-z-1}X^{n-z-1}$,
$b(X)=b_{0}+\cdots+b_{n-z-1}X^{n-z-1}$,
$c(X)=c_{0}+\cdots+c_{n-z-1}X^{n-z-1},$ and
$d(X)=d_{0}+\cdots+d_{n-z-1}X^{n-z-1}$ by Lemma 5.

In the proof we consider two possibilities for $C_{0}$, whether $C_{0}$ is a
non-zero or a zero matrix.

###  3.1. Case 1: $C_{0}$ is non-zero.

Conjugating matrices $A$, $B$ and $C$ in the Weyr form by
$G=\left[\begin{array}[]{ccc}g\\\ &\ddots\\\ &&g\end{array}\right]\in
GL_{2n}(F)$ for any $g\in GL_{2}(F)$ changes neither the matrix $B$ nor the
partitions corresponding to nilpotent orbits in which the matrices lie. On the
other hand, each diagonal block of $G^{-1}CG$ is equal to $g^{-1}C_{0}g$.
Since $C$ is nilpotent, $C_{0}$ is nilpotent as well [4, Lemma 2.3]. Therefore
$C_{0}$ is similar to the nilpotent $2\times 2$ Jordan block over any field
$F$, and hence we can assume that $C_{0}=\left[\begin{array}[]{cc}0&1\\\
0&0\end{array}\right]$. In polynomial notation this means that $a(X),c(X),$
and $d(X)$ are divisible by $X$ and the constant term of $b(X)$ is 1. Choose
now $G=\left[\begin{array}[]{cc}b(X)&0\\\ -a(X)&1\end{array}\right]$, which is
invertible in $M_{2}(S)$, since the constant term of $b(X)$ is non-zero. Then

$G^{-1}AG=\left[\begin{array}[]{cc}0&X^{z}\\\
X^{z}\left(b(X)c(X)-a(X)d(X)\right)&X^{z}\left(a(X)+d(X)\right)\end{array}\right],$

so without any loss of generality we can assume that $a(X)=0$ and $b(X)=1$.
Note that the assumption that $c(X)$ and $d(X)$ are both divisible by $X$ does
not change after the above conjugation, and that after conjugation we still
assume that the degrees of the polynomials $c(X)$ and $d(X)$ are bounded by
$n-z-1$.

Let $l,m\leq n-z$ be the largest positive integers such that $X^{l}$ divides
$c(X)$ and $X^{m}$ divides $d(X)$ and let $p(X),q(X)\in R$ be unique
polynomials satisfying $c(X)=X^{l}p(X)$ and $d(X)=X^{m}q(X)$. Polynomials
$p(X)$ and $q(X)$ have non-zero constant terms if $l,m<n-z$, while for $l=n-z$
(respectively, $m=n-z$) the corresponding polynomial $p(X)$ (respectively,
$q(X)$) is zero.

We first consider the special case when $p(X)$ is zero. If $p(X)=q(X)=0$, then
$A^{k}=0$ for each $k\geq 2$ and $\mathrm{rank}\,A=n-z$, so $A$ corresponds to
the partition $(2^{n-z},1^{2z})=[2n]^{n+z}$. This shows that Table 1(P1) is
possible whenever we decompose $2n$ into an almost rectangular partition with
at least $n$ parts. As noted in the introduction and proved in [8, 26], the
corresponding nilpotent orbit intersects every nilpotent centralizer. If
$p(X)=0$ and $q(X)$ is non-zero, then an easy induction shows that for each
$k\geq 1$ we have

$A^{k}=\left[\begin{array}[]{cc}0&X^{kz+(k-1)m}q(X)^{k-1}\\\
0&X^{kz+km}q(X)^{k}\end{array}\right],$

which has rank $n-kz-(k-1)m$ whenever $k\leq\frac{n+m}{z+m}$, and 0 otherwise.
Let $\alpha=\lceil\frac{n+m}{z+m}\rceil$ and $\beta=\alpha(z+m)-(n+m)$ (so
that $0\leq\beta<z+m$). Then

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}n+z&k=0,\\\
m+z&1\leq k\leq\alpha-2,\\\ m+z-\beta&k=\alpha-1,\\\
0&k\geq\alpha,\end{cases}$

which implies that $A$ corresponds to the partition
$(\alpha^{m+z-\beta},(\alpha-1)^{\beta},1^{n-m})=([n+m]^{m+z},1^{n-m})$. These
partitions belong to Table 1(P2, $l=n-z$), and it follows from the above proof
that for each of these partitions ${\bf{\lambda}}$ there exists
$A\in\mathcal{N}(B)\cap{\mathcal{O}}_{{\bf{\lambda}}}$.

In the rest of Case 1 we assume that $p(X)$ is non-zero, or equivalently that
$l<n-z$. We determine the partition corresponding to $A$ using the same
argument as above. First we compute the powers of $A$. By Lemma 4, we have

$A^{k}=\left[\begin{array}[]{cc}X^{kz}c(X)s_{k-1}(X)&X^{kz}s_{k}(X)\\\
X^{kz}c(X)s_{k}(X)&X^{kz}s_{k+1}(X)\end{array}\right]$

for each positive integer $k$, where $s_{0}(X)=0$, $s_{1}(X)=1$, and

(4) $s_{k+1}(X)=d(X)s_{k}(X)+c(X)s_{k-1}(X)\quad\mathrm{for}\,k\geq 1.$

Moreover,

(5) $\displaystyle s_{2k-1}(X)$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{k}{2k-i-1\choose
i-1}X^{2km+i(l-2m)-l}p(X)^{i-1}q(X)^{2k-2i}\quad\mathrm{and}$ (6)
$\displaystyle s_{2k}(X)$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{k-1}{2k-i-1\choose
i}X^{2km+i(l-2m)-m}p(X)^{i}q(X)^{2k-2i-1}\quad\mathrm{for}\,k\geq 1.$

Now we consider two subcases.

####  3.1.1. Case 1.1: $l<2m$

In this case, (5) and (6) imply that the highest powers of $X$ dividing
$s_{2k-1}(X)$ and $s_{2k}(X)$ are $X^{kl-l}$ and $X^{kl-l+m}$, respectively,
unless $q(X)=0$ when $s_{2k}(X)=0$ for each $k\geq 1$. For each $k\geq 1$, we
can write $s_{2k-1}(X)=X^{kl-l}r_{2k-1}(X)$ and
$s_{2k}(X)=X^{kl-l+m}r_{2k}(X)$ for some polynomials $r_{2k-1}(X),r_{2k}(X)\in
R$, where $r_{2k-1}(X)$ is uniquely determined and has non-zero constant term,
while $r_{2k}(X)$ is such if $q(X)\neq 0$. Lemma 4 then implies that the
equalities

(7) $\displaystyle r_{2k-1}(X)r_{2k+1}(X)-X^{2m-l}r_{2k}(X)^{2}$
$\displaystyle=$ $\displaystyle p(X)^{2k-1},$ (8) $\displaystyle
X^{2m-l}r_{2k}(X)r_{2k+2}(X)-r_{2k+1}(X)^{2}$ $\displaystyle=$
$\displaystyle-p(X)^{2k}$

hold in $R$ for each $k\geq 1$. Furthermore, the last equality holds also for
$k=0$ if we define $r_{0}(X)=0$.

Now we first consider the odd powers of $A$:

$A^{2k^{\prime}-1}=\left[\begin{array}[]{cc}X^{(2k^{\prime}-1)z+k^{\prime}l-l+m}p(X)r_{2k^{\prime}-2}(X)&X^{(2k^{\prime}-1)z+k^{\prime}l-l}r_{2k^{\prime}-1}(X)\\\
X^{(2k^{\prime}-1)z+k^{\prime}l}p(X)r_{2k^{\prime}-1}(X)&X^{(2k^{\prime}-1)z+k^{\prime}l-l+m}r_{2k^{\prime}}(X)\end{array}\right].$

By definition, the polynomial $r_{2k^{\prime}-1}(X)$ has non-zero constant
term, so it is invertible in the quotient ring $S=R/(X^{n})$. Let
$r_{2k^{\prime}-1}(X)^{-1}$ denote its inverse in $S$. Applying Gaussian
elimination to $A^{2k^{\prime}-1}$ and using (8) we now get

$\displaystyle\left[\begin{array}[]{cc}r_{2k^{\prime}-1}(X)^{-1}&0\\\
-X^{m}r_{2k^{\prime}-1}(X)^{-1}r_{2k^{\prime}}(X)&1\end{array}\right]\cdot
A^{2k^{\prime}-1}\cdot\left[\begin{array}[]{cc}r_{2k^{\prime}-1}(X)&0\\\
-X^{m}p(X)r_{2k^{\prime}-2}(X)&1\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}0&X^{(2k^{\prime}-1)z+k^{\prime}l-l}\\\
X^{(2k^{\prime}-1)z+k^{\prime}l}p(X)\Big{(}r_{2k^{\prime}-1}(X)^{2}-X^{2m-l}r_{2k^{\prime}-2}(X)r_{2k^{\prime}}(X)\Big{)}&0\end{array}\right]=$
$\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}0&X^{(2k^{\prime}-1)z+k^{\prime}l-l}\\\
X^{(2k^{\prime}-1)z+k^{\prime}l}p(X)^{2k^{\prime}-1}&0\end{array}\right].$

Matrices $\left[\begin{array}[]{cc}r_{2k^{\prime}-1}(X)^{-1}&0\\\
-X^{m}r_{2k^{\prime}-1}(X)^{-1}r_{2k^{\prime}}(X)&1\end{array}\right]$ and
$\left[\begin{array}[]{cc}r_{2k^{\prime}-1}(X)&0\\\
-X^{m}p(X)r_{2k^{\prime}-2}(X)&1\end{array}\right]$ are invertible in
$M_{2}(S)\cong\mathcal{C}(B)$, therefore,

(14) $\displaystyle\mathrm{rank}\,A^{2k^{\prime}-1}$ $\displaystyle=$
$\displaystyle\mathrm{rank}\left[\begin{array}[]{cc}0&X^{(2k^{\prime}-1)z+k^{\prime}l-l}\\\
X^{(2k^{\prime}-1)z+k^{\prime}l}p(X)^{2k^{\prime}-1}&0\end{array}\right]$ (15)
$\displaystyle=$ $\displaystyle\begin{cases}2n-(2k^{\prime}-1)(2z+l)&1\leq
k^{\prime}\leq\frac{n+z}{l+2z},\\\
n-(2k^{\prime}-1)z-k^{\prime}l+l&\frac{n+z}{l+2z}\leq
k^{\prime}\leq\frac{n+l+z}{l+2z},\\\ 0&\frac{n+l+z}{l+2z}\leq
k^{\prime}.\end{cases}$

For the further use we denote

$\alpha:=\left\lceil\frac{n+z}{l+2z}\right\rceil\text{ and
}\beta:=\alpha(l+2z)-(n+z),$

and in this case $0\leq\beta<l+2z$ and $\alpha>1$. Moreover,
$\frac{n+z}{l+2z}\leq\frac{n+l+z}{l+2z}\leq\frac{n+z}{l+2z}+1$, so

$\left\lceil\frac{n+l+z}{l+2z}\right\rceil\in\\{\alpha,\alpha+1\\}.$

Now we consider the even powers of $A$:

$A^{2k^{\prime}}=\left[\begin{array}[]{cc}X^{2k^{\prime}z+k^{\prime}l}p(X)r_{2k^{\prime}-1}(X)&X^{2k^{\prime}z+k^{\prime}l-l+m}r_{2k^{\prime}}(X)\\\
X^{2k^{\prime}z+k^{\prime}l+m}p(X)r_{2k^{\prime}}(X)&X^{2k^{\prime}z+k^{\prime}l}r_{2k^{\prime}+1}(X)\end{array}\right]$

for $k^{\prime}\geq 1$. To compute the rank of such a matrix we have to
distinguish between two cases.

Case 1.1(a): $m\leq l$. In this case $q(X)$ is non-zero as well. With the same
argument as above we see that $r_{2k^{\prime}}(X)$ is invertible in $S$ for
$k^{\prime}\geq 1$, and let $r_{2k^{\prime}}(X)^{-1}$ be its inverse. Using
(7), for each $k^{\prime}\geq 1$ we compute

$\displaystyle\left[\begin{array}[]{cc}r_{2k^{\prime}}(X)^{-1}&0\\\
-X^{l-m}r_{2k^{\prime}}(X)^{-1}r_{2k^{\prime}+1}(X)&1\end{array}\right]\cdot
A^{2k^{\prime}}\cdot\left[\begin{array}[]{cc}r_{2k^{\prime}}(X)&0\\\
-X^{l-m}p(X)r_{2k^{\prime}-1}(X)&1\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}0&X^{2k^{\prime}z+k^{\prime}l-l+m}\\\
X^{2k^{\prime}z+k^{\prime}l+l-m}p(X)\Big{(}X^{2m-l}r_{2k^{\prime}}(X)^{2}-r_{2k^{\prime}-1}(X)r_{2k^{\prime}+1}(X)\Big{)}&0\end{array}\right]=$
$\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}0&X^{2k^{\prime}z+k^{\prime}l-l+m}\\\
-X^{2k^{\prime}z+k^{\prime}l+l-m}p(X)^{2k^{\prime}}&0\end{array}\right],$

therefore

(21) $\displaystyle\mathrm{rank}\,A^{2k^{\prime}}$ $\displaystyle=$
$\displaystyle\mathrm{rank}\,\left[\begin{array}[]{cc}0&X^{2k^{\prime}z+k^{\prime}l-l+m}\\\
-X^{2k^{\prime}z+k^{\prime}l+l-m}p(X)^{2k^{\prime}}&0\end{array}\right]$ (22)
$\displaystyle=$
$\displaystyle\begin{cases}2n-2k^{\prime}(2z+l)&k^{\prime}\leq\frac{n+m-l}{l+2z},\\\
n-k^{\prime}(2z+l)+l-m&\frac{n+m-l}{l+2z}\leq
k^{\prime}\leq\frac{n+l-m}{l+2z},\\\ 0&\frac{n+l-m}{l+2z}\leq
k^{\prime}.\end{cases}$

Moreover, it is obvious that equality (22) holds for $k^{\prime}=0$, too.

We now compare fractions in (22) with $\alpha$. Since
$\frac{n+z}{l+2z}-1\leq\frac{n+m-l}{l+2z}\leq\frac{n+z}{l+2z}$, we get

$\left\lceil\frac{n+m-l}{l+2z}\right\rceil\in\\{\alpha-1,\alpha\\}.$

Similarly, $\frac{n+m-l}{l+2z}\leq\frac{n+l-m}{l+2z}\leq\frac{n+l+z}{l+2z}$,
so

$\left\lceil\frac{n+l-m}{l+2z}\right\rceil\in\\{\alpha-1,\alpha,\alpha+1\\}.$

We consider various cases.

  1. (i)

Assume first that $\lceil\frac{n+l-m}{l+2z}\rceil=\alpha+1$, or equivalently
$\beta<l-m-z$, and hence in particular $l>m+z$. Then it is clear that
$\lceil\frac{n+l+z}{l+2z}\rceil=\alpha+1$ as well. Furthermore, since $l<2m$,
we have $\frac{n+l-m}{l+2z}-1\leq\frac{n+m-l}{l+2z}$, so
$\lceil\frac{n+m-l}{l+2z}\rceil=\alpha$. It follows from (15) and (22) that

$\mathrm{rank}\,A^{2k^{\prime}}=\begin{cases}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-1,\\\
l-m-z-\beta&k^{\prime}=\alpha,\\\ 0&k^{\prime}\geq\alpha+1\end{cases}$

and

$\mathrm{rank}\,A^{2k^{\prime}-1}=\begin{cases}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
l-\beta&k^{\prime}=\alpha,\\\ 0,&k^{\prime}\geq\alpha+1.\end{cases}$

In particular, $A^{2\alpha+1}=0$ and $\mathrm{rank}\,A^{2\alpha}=l-m-z-\beta$
which is strictly positive by assumption, so $2\alpha+1$ is the nilpotency
index of $A$. Moreover,

$\mathrm{rank}A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq
2\alpha-3,\\\ l+2z-\beta&k=2\alpha-2,\\\ m+z&k=2\alpha-1,\\\
l-m-z-\beta&k=2\alpha,\\\ 0&k\geq 2\alpha+1,\end{cases}$

which implies that $A$ corresponds to the partition Table 1(P3)

$\displaystyle((2\alpha+1)^{l-m-z-\beta},(2\alpha)^{2m-l+2z+\beta},(2\alpha-1)^{l-m+z-\beta},(2\alpha-2)^{\beta})$
$\displaystyle=$
$\displaystyle([n+l-m+\alpha(2m-l)]^{m+z},[n+m-l-\alpha(2m-l)]^{l-m+z}),$

where $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$, $\alpha>1$ and
$0\leq\beta<l-m-z$, where $0<m\leq l$ is redundant. We note that
$n+m-l-\alpha(2m-l)$ is indeed positive for $\alpha\geq 1$, as it is equal to
$\alpha(2l-2m+2z)+m-l-z-\beta\geq l-m+z-\beta>0$.

  2. (ii)

Assume that $\lceil\frac{n+l-m}{l+2z}\rceil=\alpha$, or equivalently
$l-m-z\leq\beta<2l-m+z$. We know that
$\lceil\frac{n+m-l}{l+2z}\rceil\in\\{\alpha-1,\alpha\\}$ and
$\lceil\frac{n+l+z}{l+2z}\rceil\in\\{\alpha,\alpha+1\\}$. We consider all
options.

     * •

Assume $\lceil\frac{n+m-l}{l+2z}\rceil=\alpha$ and
$\lceil\frac{n+l+z}{l+2z}\rceil=\alpha+1$, which is equivalent to $m+z>\beta$
and $l>\beta$. In this case

$\mathrm{rank}\,A^{2k^{\prime}}=\begin{cases}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha\end{cases}$

and

$\mathrm{rank}\,A^{2k^{\prime}-1}=\begin{cases}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
l-\beta&k^{\prime}=\alpha,\\\ 0&k^{\prime}\geq\alpha+1.\end{cases}$

We therefore get

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq
2\alpha-3,\\\ l+2z-\beta&k=2\alpha-2,\\\ l-\beta&k=2\alpha-1,\\\ 0&k\geq
2\alpha,\end{cases}$

so $A\in{\mathcal{O}}_{{\bf{\lambda}}}$, where

${\bf{\lambda}}=((2\alpha)^{l-\beta},(2\alpha-1)^{2z},(2\alpha-2)^{\beta}),$

corresponds to the partition Table 1(P6), and $0\leq z\leq n-1$, $0<m\leq
l<n-z$, $l<2m$ and $\max\\{l-m-z,0\\}\leq\beta<\min\\{m+z,l\\}$. (Again, the
condition $\beta<2l-m+z$ is redundant.)

     * •

Assume $\lceil\frac{n+m-l}{l+2z}\rceil=\lceil\frac{n+l+z}{l+2z}\rceil=\alpha$,
which is equivalent to $l\leq\beta<m+z$. Then

$\mathrm{rank}\,A^{2k^{\prime}}=\begin{cases}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{cases}$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\begin{cases}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{cases}$

and

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq
2\alpha-3,\\\ 2l+2z-2\beta&k=2\alpha-2,\\\ 0&k\geq 2\alpha-1.\end{cases}$

Consequently, $A$ corresponds to the partition Table 1(P1)

$\left((2\alpha-1)^{2l+2z-2\beta},(2\alpha-2)^{2\beta-l}\right)=[2n]^{l+2z},$

where $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$ and $l\leq\beta<m+z$ (while
the conditions $l-m-z\leq\beta<2l-m+z$ and $\beta<l+2z$ are redundant).

     * •

Assume $\lceil\frac{n+m-l}{l+2z}\rceil=\alpha-1$ and
$\lceil\frac{n+l+z}{l+2z}\rceil=\alpha+1$, which is equivalent to
$m+z\leq\beta<l$. It follows that

$\mathrm{rank}\,A^{2k^{\prime}}=\begin{cases}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-2,\\\
2l-m+z-\beta&k^{\prime}=\alpha-1,\\\ 0&k^{\prime}\geq\alpha,\end{cases}$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\begin{cases}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
l-\beta&k^{\prime}=\alpha,\\\ 0&k^{\prime}\geq\alpha+1\end{cases}$

and

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq
2\alpha-4,\\\ l+m+3z-\beta&k=2\alpha-3,\\\ l-m+z&k=2\alpha-2,\\\
l-\beta&k=2\alpha-1,\\\ 0&k\geq 2\alpha.\end{cases}$

Consequently, $A$ corresponds to the partition Table 1(P4)

$\displaystyle((2\alpha)^{l-\beta},(2\alpha-1)^{z-m+\beta},(2\alpha-2)^{2m+2z-\beta},(2\alpha-3)^{\beta-
m-z})$ $\displaystyle=$
$\displaystyle([n+m-\alpha(2m-l)]^{l-m+z},[n-m+\alpha(2m-l)]^{m+z}),$

where $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$ and $m+z\leq\beta<l$ (while
the conditions $l-m-z\leq\beta<2l-m+z$ and also $m\leq l$ are redundant).

     * •

It remains to consider the case when $\lceil\frac{n+m-l}{l+2z}\rceil=\alpha-1$
and $\lceil\frac{n+l+z}{l+2z}\rceil=\alpha$, which is equivalent to
$\beta\geq\max\\{m+z,l\\}$. Now we have

$\mathrm{rank}\,A^{2k^{\prime}}=\begin{cases}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-2,\\\
2l-m+z-\beta&k^{\prime}=\alpha-1,\\\ 0&k^{\prime}\geq\alpha,\end{cases}$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\begin{cases}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{cases}$

and

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq
2\alpha-4,\\\ l+m+3z-\beta&k=2\alpha-3,\\\ 2l-m+z-\beta&k=2\alpha-2,\\\
0&k\geq 2\alpha-1.\end{cases}$

Hence, $A$ corresponds to the partition Table 1(P7)

$\left((2\alpha-1)^{2l-m+z-\beta},(2\alpha-2)^{2m-l+2z},(2\alpha-3)^{\beta-
m-z}\right),$

where $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$ and
$\max\\{m+z,l\\}\leq\beta<\min\\{2l-m+z,l+2z\\}$ (and the condition $\beta\geq
l-m-z$ is redundant).

  3. (iii)

Assume now that $\lceil\frac{n+l-m}{l+2z}\rceil=\alpha-1$, or equivalently
$\beta\geq 2l-m+z$, so in particular we have $z>l-m$, since $\beta<l+2z$.
Since $l\geq m$, we get also $\lceil\frac{n+m-l}{l+2z}\rceil=\alpha-1$.
Moreover, the conditions $m\leq l$ and $z\geq 0$ imply $2l-m+z\geq l$, so
$\beta\geq l$, which is equivalent to $\lceil\frac{n+l+z}{l+2z}\rceil=\alpha$.
We thus get

$\mathrm{rank}\,A^{2k^{\prime}}=\begin{cases}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-2,\\\
0&k^{\prime}\geq\alpha-1,\end{cases}$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\begin{cases}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{cases}$

and therefore

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq
2\alpha-4,\\\ 2n-(2\alpha-3)(l+2z)&k=2\alpha-3,\\\ 0&k\geq
2\alpha-2.\end{cases}$

It follows that $A\in{\mathcal{O}}_{{\bf{\lambda}}}$, where

${\bf{\lambda}}=((2\alpha-2)^{2n-(2\alpha-3)(l+2z)},(2\alpha-3)^{(2\alpha-2)(l+2z)-2n})=[2n]^{l+2z}$

as in Table 1(P1), for $0\leq z\leq n-1$, $0<m\leq l<n-z$, $l<2m$ and
$2l-m+z\leq\beta<l+2z$.

At the end of Case 1.1(a) we observe that it immediately follows from the
above proof that for each of the partitions ${\bf{\lambda}}$ obtained in
(i)-(iii) there exists $A\in\mathcal{N}(B)\cap{\mathcal{O}}_{{\bf{\lambda}}}$.

Note that in the special case when $l=1$, the condition (a) forces $m=1$.
Moreover, the option (i) above and the last two cases in (ii) are not
possible, and in all other cases we get the partition $[2n]^{2z+1}$. Hence, in
particular we get all odd cases in the first set of partitions listed in the
theorem, except $[2n]^{2n-1}$, since $1<n-z$. However, the existence of this
partition was shown at the beginning of the proof in the case when
$p(X)=q(X)=0$. The existence of all even cases will be proved in Case 2 below,
but we note here that we get almost all of them also from the partitions above
in the case $l=2$. Indeed, in this case the condition $m\leq l<2m$ implies
$m=2$, again the option (i) and the last two cases in (ii) are not possible,
and the only case when we get a partition different from $[2n]^{2z+2}$ is when
$\beta=1$. Note also that for $n\geq 2$ the partition $[2n]^{2n-2}$ is not
covered by the above cases if $l=2$, but it has been established when we
considered the case $p(X)=q(X)=0$.

Case 1.1(b): $n-z\geq m>l$. We proceed as in Case 1.1(a). As above, the
polynomial $r_{2k^{\prime}+1}(X)$ has non-zero constant term, therefore it is
invertible in $S=R/(X^{n})$ for each $k^{\prime}\geq 0$. Let
$r_{2k^{\prime}+1}(X)^{-1}$ be its inverse in $S$. For $k^{\prime}\geq 1$ we
compute, using (7), that

$\displaystyle\left[\begin{array}[]{cc}1&-X^{m-l}r_{2k^{\prime}+1}(X)^{-1}r_{2k^{\prime}}(X)\\\
0&r_{2k^{\prime}+1}(X)^{-1}\end{array}\right]\cdot
A^{2k^{\prime}}\cdot\left[\begin{array}[]{cc}r_{2k^{\prime}+1}(X)&0\\\
X^{m}p(X)r_{2k^{\prime}}(X)&1\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}X^{2k^{\prime}z+k^{\prime}l}p(X)\Big{(}r_{2k^{\prime}-1}(X)r_{2k^{\prime}+1}(X)-X^{2m-l}r_{2k^{\prime}}(X)^{2}\Big{)}&0\\\
0&X^{2k^{\prime}z+k^{\prime}l}\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}X^{2k^{\prime}z+k^{\prime}l}p(X)^{2k^{\prime}}&0\\\
0&X^{2k^{\prime}z+k^{\prime}l}\end{array}\right],$

so

(28) $\displaystyle\mathrm{rank}\,A^{2k^{\prime}}$ $\displaystyle=$
$\displaystyle\mathrm{rank}\,\left[\begin{array}[]{cc}X^{2k^{\prime}z+k^{\prime}l}p(X)^{2k^{\prime}}&0\\\
0&X^{2k^{\prime}z+k^{\prime}l}\end{array}\right]=$ (31) $\displaystyle=$
$\displaystyle\left\\{\begin{array}[]{ll}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\frac{n}{l+2z},\\\
0&k^{\prime}\geq\frac{n}{l+2z},\end{array}\right.$

and clearly $\mathrm{rank}\,A^{0}=2n$, so (31) holds for $k^{\prime}=0$. Since
$\frac{n+z}{l+2z}-1\leq\frac{n}{l+2z}\leq\frac{n+z}{l+2z}$, we get

$\left\lceil\frac{n}{l+2z}\right\rceil\in\\{\alpha-1,\alpha\\}.$

On the other hand, we know that
$\lceil\frac{n+l+z}{l+2z}\rceil\in\\{\alpha,\alpha+1\\}$. Moreover,
$\frac{n+l+z}{l+2z}-\frac{n}{l+2z}\leq 1$, therefore
$\lceil\frac{n+l+z}{l+2z}\rceil-\lceil\frac{n}{l+2z}\rceil\in\\{0,1\\}$. We
consider all options.

  1. (i)

Assume first that $\lceil\frac{n+l+z}{l+2z}\rceil=\alpha+1$, which is
equivalent to $\beta<l$. Then it is clear that
$\lceil\frac{n}{l+2z}\rceil=\alpha$. Furthermore, we get from (15) and (31)
that

$\mathrm{rank}\,A^{2k^{\prime}}=\left\\{\begin{array}[]{ll}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{array}\right.$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\left\\{\begin{array}[]{ll}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
l-\beta&k^{\prime}=\alpha,\\\ 0&k^{\prime}\geq\alpha+1,\end{array}\right.$

and

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}l+2z&k\leq
2\alpha-3,\\\ l+2z-\beta&k=2\alpha-2,\\\ l-\beta&k=2\alpha-1,\\\ 0&k\geq
2\alpha,\end{array}\right.$

and the corresponding partition is Table 1(P6)

$\left((2\alpha)^{l-\beta},(2\alpha-1)^{2z},(2\alpha-2)^{\beta}\right),$

for $0\leq z\leq n-1$, $0<l<m\leq n-z$ and $0\leq\beta<l$.

  2. (ii)

Assume now that
$\lceil\frac{n+l+z}{l+2z}\rceil=\lceil\frac{n}{l+2z}\rceil=\alpha$, which is
equivalent to $l\leq\beta<l+z$. It follows from (15) and (31) that

$\mathrm{rank}\,A^{2k^{\prime}}=\left\\{\begin{array}[]{ll}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{array}\right.$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\left\\{\begin{array}[]{ll}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{array}\right.$

and

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}l+2z&k\leq
2\alpha-3,\\\ 2n-2(\alpha-1)(l+2z)&k=2\alpha-2,\\\ 0&k\geq
2\alpha-1,\end{array}\right.$

so the corresponding partition is Table 1(P1)

$\left((2\alpha-1)^{2n-2(\alpha-1)(l+2z)},(2\alpha-2)^{-2n+(2\alpha-1)(l+2z)}\right)=[2n]^{l+2z},$

where $0\leq z\leq n-1$, $0<l<m\leq n-z$ and $l\leq\beta<l+z$.

  3. (iii)

It remains to consider the case when $\lceil\frac{n}{l+2z}\rceil=\alpha-1$
(and hence $\lceil\frac{n+l+z}{l+2z}\rceil=\alpha$), which is equivalent to
$\beta\geq l+z$. In this case we have

$\mathrm{rank}\,A^{2k^{\prime}}=\left\\{\begin{array}[]{ll}2n-2k^{\prime}(l+2z)&k^{\prime}\leq\alpha-2,\\\
0&k^{\prime}\geq\alpha-1,\end{array}\right.$
$\mathrm{rank}\,A^{2k^{\prime}-1}=\left\\{\begin{array}[]{ll}2n-(2k^{\prime}-1)(l+2z)&k^{\prime}\leq\alpha-1,\\\
0&k^{\prime}\geq\alpha,\end{array}\right.$

and

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}l+2z&k\leq
2\alpha-4,\\\ 2n-(2\alpha-3)(l+2z)&k=2\alpha-3,\\\ 0&k\geq
2\alpha-2,\end{array}\right.$

and the corresponding partition is Table 1(P1)

$\left((2\alpha-2)^{2n-(2\alpha-3)(l+2z)},(2\alpha-3)^{-2n+(2\alpha-2)(l+2z)}\right)=[2n]^{l+2z},$

where $0\leq z\leq n-1$, $0<l<m\leq n-z$ and $\beta\geq l+z$.

As before, it follows from the above proof that for each of the partitions
obtained in (i)-(iii) there indeed exists $A\in\mathcal{N}(B)$ that
corresponds to the partition.

####  3.1.2. Case 1.2: $l\geq 2m$

Then $q(X)$ has non-zero constant term. Moreover, it follows from (5) and (6)
that, for each $k\geq 1$, the polynomial $s_{k}(X)$ is divisible by
$X^{km-m}$. For each $k\geq 1$, we write $s_{k}(X)=X^{km-m}r_{k}(X)$ for some
uniquely determined polynomial $r_{k}(X)\in R$, and let $r_{0}(X)=0$. Then

(32) $r_{k+1}(X)=q(X)r_{k}(X)+X^{l-2m}p(X)r_{k-1}(X)$

for $k\geq 1$ by (4). Furthermore, Lemma 4 implies that

$A^{k}=\left[\begin{array}[]{cc}X^{kz+km-2m+l}p(X)r_{k-1}(X)&X^{kz+km-m}r_{k}(X)\\\
X^{kz+km-m+l}p(X)r_{k}(X)&X^{kz+km}r_{k+1}(X)\end{array}\right]$

for $k\geq 1$, and

(33) $r_{k-1}(X)r_{k+1}(X)-r_{k}(X)^{2}=(-1)^{k}X^{(k-1)(l-2m)}p(X)^{k-1}$

holds in $R$ for each $k\geq 2$. Clearly, (33) holds also for $k=1$. If
$l>2m$, then it is also clear from (5) and (6) that $r_{k}(X)$ has non-zero
constant term for each $k\geq 1$.

As we will see below, it is convenient to define the numbers

$\alpha=\left\lceil\frac{n+m}{z+m}\right\rceil,\beta=\alpha(z+m)-(n+m),\gamma=\left\lceil\frac{n-m}{z+l-m}\right\rceil,\mathrm{and}\,\delta=\gamma(z+l-m)-(n-m).$

Then $0\leq\beta<z+m$ and $0\leq\delta<z+l-m$. Moreover, since $l\geq 2m$, we
get $z+l-m\geq z+m$, hence $\frac{n+m}{z+m}\geq\frac{n-m}{z+l-m}$, and
consequently $\alpha\geq\gamma.$ Furthermore, $l<n-z$ implies $\gamma>1$.

Suppose first that $k\geq 1$ is such that $r_{k}(X)$ is not divisible by $X$.
(As noted above, this happens always if $l>2m$.) Then it is invertible in
$S=R/(X^{n})$, and let $r_{k}(X)^{-1}$ be its inverse in $S$. Using (33) we
now compute

$\displaystyle\left[\begin{array}[]{cc}r_{k}(X)^{-1}&0\\\
-X^{m}r_{k}(X)^{-1}r_{k+1}(X)&1\end{array}\right]\cdot
A^{k}\cdot\left[\begin{array}[]{cc}r_{k}(X)&0\\\
-X^{l-m}p(X)r_{k-1}(X)&1\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}0&X^{kz+km-m}\\\
X^{kz+km+l-m}p(X)\Big{(}r_{k}(X)^{2}-r_{k-1}(X)r_{k+1}(X)\Big{)}&0\end{array}\right]=$
$\displaystyle=$ $\displaystyle\left[\begin{array}[]{cc}0&X^{kz+km-m}\\\
(-1)^{k+1}X^{kz+kl-km+m}p(X)^{k}&0\end{array}\right].$

We can therefore compute

$\displaystyle\mathrm{rank}\,A^{k}$ $\displaystyle=$
$\displaystyle\mathrm{rank}\,\left[\begin{array}[]{cc}0&X^{kz+km-m}\\\
(-1)^{k+1}X^{kz+kl-km+m}p(X)^{k}&0\end{array}\right]$ $\displaystyle=$
$\displaystyle\begin{cases}2n-k(l+2z)&k\leq\frac{n-m}{z+l-m},\\\
n+m-k(m+z)&\frac{n-m}{z+l-m}\leq k\leq\frac{n+m}{z+m},\\\
0&k\geq\frac{n+m}{z+m}.\end{cases}$

If $k\geq 1$ and $r_{k}(X)$ is not divisible by $X$, we therefore get

(38)
$\mathrm{rank}\,A^{k}=\left\\{\begin{array}[]{ll}2n-k(l+2z)&k\leq\gamma-1,\\\
n+m-k(m+z)&\gamma\leq k\leq\alpha-1,\\\ 0&k\geq\alpha,\end{array}\right.$

and it is clear that the above holds also for $k=0$.

Suppose now that $X$ divides $r_{k}(X)$ for some $k\geq 1$. We already know
that this forces $l=2m$, hence $\gamma=\lceil\frac{n-m}{z+m}\rceil$, and in
particular $\alpha\leq\gamma+2$. Moreover, if $X$ divides $r_{k-1}(X)$, too,
then (32) implies that $r_{j}(X)$ is divisible by $X$ for any $j$. (Recall
that $p(X)$ is not divisible by $X$.) However, this contradicts $r_{1}(X)=1$.
It follows that no two consecutive polynomials $r_{j}(X)$ and $r_{j-1}(X)$ are
divisible by $X$.

Let $t$ be the largest positive integer smaller than or equal to $m+1$ such
that $X^{t}$ divides $r_{k}(X)$, and let $\widetilde{r_{k}}(X)\in R$ be the
unique polynomial satisfying $r_{k}(X)=X^{t}\widetilde{r_{k}}(X)$. If $t\leq
m$, then the polynomial $\widetilde{r_{k}}(X)$ has non-zero constant term, so
it is invertible in $S=R/(X^{n})$. Let $\widetilde{r_{k}}(X)^{-1}$ be its
inverse in $S$. For $t\leq m$, we get by (33) that

$\displaystyle\left[\begin{array}[]{cc}\widetilde{r_{k}}(X)^{-1}&0\\\
-X^{m-t}\widetilde{r_{k}}(X)^{-1}r_{k+1}(X)&1\end{array}\right]\cdot
A^{k}\cdot\left[\begin{array}[]{cc}\widetilde{r_{k}}(X)&0\\\
-X^{m-t}p(X)r_{k-1}(X)&1\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}0&X^{kz+km-m+t}\\\
X^{kz+km+m-t}p(X)\Big{(}r_{k}(X)^{2}-r_{k-1}(X)r_{k+1}(X)\Big{)}&0\end{array}\right]=$
$\displaystyle=$ $\displaystyle\left[\begin{array}[]{cc}0&X^{kz+km-m+t}\\\
(-1)^{k+1}X^{kz+km+m-t}p(X)^{k}&0\end{array}\right],$

so

(42) $\mathrm{rank}\,A^{k}=\begin{cases}2n-2k(m+z)&k\leq\frac{n-m+t}{m+z},\\\
n+m-t-k(m+z)&\frac{n-m+t}{m+z}\leq k\leq\frac{n+m-t}{m+z},\\\
0&k\geq\frac{n+m-t}{m+z}.\end{cases}$

On the other hand, for $t=m+1$ we use the fact that $r_{k+1}(X)$ has non-zero
constant term, which was shown above. Hence $r_{k+1}(X)$ is invertible in $S$
and we obtain

$\displaystyle\left[\begin{array}[]{cc}1&-X\widetilde{r_{k}}(X)r_{k+1}(X)^{-1}\\\
0&r_{k+1}(X)^{-1}\end{array}\right]\cdot
A^{k}\cdot\left[\begin{array}[]{cc}r_{k+1}(X)&0\\\
-X^{m}p(X)r_{k}(X)&1\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}X^{kz+km}p(X)\Big{(}r_{k-1}(X)r_{k+1}(X)-r_{k}(X)^{2}\Big{)}&0\\\
0&X^{kz+km}\end{array}\right]=$ $\displaystyle=$
$\displaystyle\left[\begin{array}[]{cc}X^{kz+km}p(X)^{k}&0\\\
0&X^{kz+km}\end{array},\right]$

hence

$\mathrm{rank}\,A^{k}=\begin{cases}2n-2k(m+z)&k\leq\frac{n}{m+z},\\\
0&k\geq\frac{n}{m+z},\end{cases}$

which is a special case of (42) when $t=m$. If $t=m+1$, we therefore get no
additional partitions corresponding to orbits intersecting $\mathcal{N}(B)$,
so we do not consider this case below.

Suppose that $k\leq\gamma-1$. If $r_{k}(X)$ is not divisible by $X$, then (38)
shows that $\mathrm{rank}\,A^{k}=2n-k(l+2z)$. On the other hand, if $X^{t}$ is
the highest power of $X$ that divides $r_{k}(X)$ for some $1\leq t\leq m$,
then $l=2m$ and it is clear that $k\leq\frac{k-m+t}{m+z}$, so (42) implies
that again we have $\mathrm{rank}\,A^{k}=2n-2k(m+z)=2n-k(l+2z)$. Similarly,
the rank of $A^{k}$ in (42) is zero if $k\geq\alpha$. It follows that the
ranks of $A^{k}$ in (38) and in (42) are different only if $\gamma\leq
k\leq\alpha-1$. Now we consider two cases.

Case 1.2(a): $r_{k}(X)$ is not divisible by $X$ for any
$k\in\\{\gamma,\ldots,\alpha-1\\}$. As noted above, this case in particular
includes the case when $l>2m$. We consider various cases with respect to
$\alpha$ and $\gamma$:

  1. (i)

If $\alpha\geq\gamma+2$, then (38) implies

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}l+2z&k\leq\gamma-2,\\\
l+2z-\delta&k=\gamma-1,\\\ m+z&\gamma\leq k\leq\alpha-2,\\\
m+z-\beta&k=\alpha-1,\\\ 0&k\geq\alpha,\end{array}\right.$

hence the matrix $A$ belongs to ${\mathcal{O}}_{\bf{\lambda}}$, where

${\bf{\lambda}}=(\alpha^{m+z-\beta},(\alpha-1)^{\beta},\gamma^{l-m+z-\delta},(\gamma-1)^{\delta})=([n+m]^{m+z},[n-m]^{l-m+z}),$

as in Table 1(P2), for $0\leq z\leq n-1$, $0<2m\leq l<n-z$ and
$\alpha\geq\gamma+2$.

  2. (ii)

If $\alpha=\gamma+1$, then we again use (38) to get

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\begin{cases}l+2z&k\leq\gamma-2,\\\
l+2z-\delta&k=\gamma-1,\\\ m+z-\beta&k=\gamma,\\\ 0&k\geq\gamma+1.\end{cases}$

It follows that $A\in{\mathcal{O}}_{\bf{\lambda}}$, where

${\bf{\lambda}}=((\gamma+1)^{m+z-\beta},\gamma^{l-m+z+\beta-\delta},(\gamma-1)^{\delta})=([n+m]^{m+z},[n-m]^{l-m+z}),$

as in Table 1(P2), for $0\leq z\leq n-1$, $0<2m\leq l<n-z$ and
$\alpha=\gamma+1$.

  3. (iii)

If $\alpha=\gamma$, then

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}l+2z&k\leq\gamma-2,\\\
2n-(\gamma-1)(l+2z)&k=\gamma-1,\\\ 0&k\geq\gamma,\end{array}\right.$

so $A$ corresponds to the partition Table 1(P1)

$(\gamma^{2n-(\gamma-1)(l+2z)},(\gamma-1)^{\gamma(l+2z)-2n})=[2n]^{l+2z},$

where $0\leq z\leq n-1$, $0<2m\leq l<n-z$ and $\alpha=\gamma$.

Conversely, it follows from the above proof that for each partition obtained
in (i)-(iii) there indeed exists $A\in\mathcal{N}(B)$ that corresponds to this
partition. The only case of existence that might not be obvious is when $l=2m$
and $r_{k}(X)$ is not divisible by $X$ for any $k\geq 1$. However, in this
case we can define $p(X)=q(X)=1$, and then (32) implies that $r_{k}(X)$ is a
strictly positive constant for each $k\geq 1$, and hence non-zero.

Case 1.2(b): $l=2m$ and there exists $j\in\\{\gamma,\ldots,\alpha-1\\}$ such
that $r_{j}(X)$ is divisible by $X$. Since $r_{2}(X)=q(X)$ has non-zero
constant term, we immediately get $j\geq 3$, and consequently $\alpha\geq 4$.
As shown above, we may assume that $r_{j}(X)=X^{t}\widetilde{r_{j}}(X)$ for
some $t\leq m$ and some polynomial $\widetilde{r_{j}}(X)\in R$ with non-zero
constant term. We have shown above also that $r_{j-1}(X)$ and $r_{j+1}(X)$ are
not divisible by $X$. Moreover, since $l=2m$, we have
$\frac{n-m}{l-m+z}+2=\frac{n+m+2z}{m+z}\geq\frac{n+m}{m+z}$, so
$\gamma+2\geq\alpha$. It follows that $j\in\\{\alpha-2,\alpha-1\\}$. We
consider each of the two options.

  1. (i)

Assume first that $j=\alpha-2$. Then $\gamma=\alpha-2$ and $\delta=\beta-2z$,
so $\beta\geq 2z$ and $\delta<m-z$. (Recall that $\beta<m+z$.) Furthermore,
from (42) we get

$\mathrm{rank}\,A^{\alpha-2}=\left\\{\begin{array}[]{ll}2n-2(\alpha-2)(m+z)&m\geq
t>\delta,\\\ n+m-t-(\alpha-2)(m+z)&t\leq\delta.\end{array}\right.$

Note that the third option in (42) is not possible, since it would imply
$t\geq 2m-\delta>m+z\geq m$, which contradicts our assumption on $t$. If
$t\leq\delta$, then (38) and (42) imply

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-4,\\\
2m+2z+t-\delta&k=\alpha-3,\\\ m+z-t&k=\alpha-2,\\\ m+z-\beta&k=\alpha-1,\\\
0&k\geq\alpha,\end{array}\right.$

so $A$ corresponds to the partition Table 1(P8),

$\displaystyle{\bf{\lambda}}$ $\displaystyle=$
$\displaystyle\left(\alpha^{m-z-\delta},(\alpha-1)^{\delta+2z-t},(\alpha-2)^{m+z+2t-\delta},(\alpha-3)^{\delta-t}\right)=$
$\displaystyle=$
$\displaystyle\left([n+m-(\alpha-1)t]^{m+z-t},[n-m+(\alpha-1)t]^{m+z+t}\right),$

where $0\leq z\leq n-1$, $0<2m<n-z$, $\alpha\geq 5$, $\gamma=\alpha-2$ and
$t\leq\delta$. On the other hand, if $m\geq t>\delta$, then

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-3,\\\
m+z-\delta&k=\alpha-2,\\\ m+z-\beta&k=\alpha-1,\\\
0&k\geq\alpha,\end{array}\right.$

so $A$ corresponds to the partition Table 1(P8),
$(\alpha^{m-z-\delta},(\alpha-1)^{2z},(\alpha-2)^{m+z+\delta})$, which is a
special case (for $t=\delta$) of the partition considered in the case when
$t\leq\delta$.

  2. (ii)

Assume now that $j=\alpha-1$, which is (because of the assumption $\gamma\leq
j\leq\alpha-1$) equal either to $\gamma$ or to $\gamma+1$. Using (42), we get

$\mathrm{rank}\,A^{\alpha-1}=\left\\{\begin{array}[]{ll}2n-2(\alpha-1)(m+z)&t>m-z+\beta,\\\
m+z-t-\beta&t\leq\min\\{m-z+\beta,m+z-\beta\\},\\\ 0&m\geq
t>m+z-\beta.\end{array}\right.$

We consider two cases:

     * •

Assume first that $j=\gamma+1$. Then, as in (i), we get $\gamma=\alpha-2$ and
$\delta=\beta-2z\geq 0$. In particular, we get $t\leq m\leq m-z+\beta$.

If $t\leq m+z-\beta$, then we get from (38) and (42) that

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-4,\\\
2m+2z-\delta&k=\alpha-3,\\\ m+z+t&k=\alpha-2,\\\ m+z-t-\beta&k=\alpha-1,\\\
0&k\geq\alpha,\end{array}\right.$

so $A\in{\mathcal{O}}_{\bf{\lambda}}$, where

$\displaystyle{\bf{\lambda}}$ $\displaystyle=$
$\displaystyle\left(\alpha^{m+z-t-\beta},(\alpha-1)^{\beta+2t},(\alpha-2)^{m+z-t-\delta},(\alpha-3)^{\delta}\right)=$
$\displaystyle=$ $\displaystyle\left([n+m+\gamma t]^{m+z+t},[n-m-\gamma
t]^{m+z-t}\right)$

as in Table 1(P9) for $0\leq z\leq n-1$, $0<2m<n-z$, $\gamma=\alpha-2\geq 2$,
and $t\leq m+z-\beta$.

On the other hand, if $m\geq t>m+z-\beta$, then

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-4,\\\
2m+2z-\delta&k=\alpha-3,\\\ 2m-\delta&k=\alpha-2,\\\
0&k\geq\alpha-1,\end{array}\right.$

so $A$ corresponds to the partition
$\left((\alpha-1)^{2m-\delta},(\alpha-2)^{2z},(\alpha-3)^{\delta}\right)$,
which is a special case (for $t=m+z-\beta$) of the partition considered above.

     * •

Assume now that $j=\gamma$, so $\gamma=\alpha-1$, or, equivalently,
$z-m\leq\beta<2z$. More precisely, we have $\beta=z-m+\delta$.

If $t\leq\min\\{m-z+\beta,m+z-\beta\\}$, then we get from (38) and (42) that

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-3,\\\
m+3z+t-\beta&k=\alpha-2,\\\ m+z-t-\beta&k=\alpha-1,\\\
0&k\geq\alpha,\end{array}\right.$

so $A\in{\mathcal{O}}_{\bf{\lambda}}$, where

$\displaystyle{\bf{\lambda}}$ $\displaystyle=$
$\displaystyle\left(\alpha^{m+z-t-\beta},(\alpha-1)^{2z+2t},(\alpha-2)^{m-z-t+\beta}\right)=$
$\displaystyle=$
$\displaystyle\left(\alpha^{m+z-t-\beta},(\alpha-1)^{2t+\beta},(\alpha-1)^{2z-\beta},(\alpha-2)^{m-z-t+\beta}\right)=$
$\displaystyle=$
$\displaystyle\left([n+m+(\alpha-2)t]^{m+z+t},[n-m-(\alpha-2)t]^{m+z-t}\right)$

corresponds to the partition Table 1(P9) for $0\leq z\leq n-1$, $0<2m<n-z$,
$\gamma=\alpha-1\geq 3$ and $t\leq\min\\{m+z-\beta,m-z+\beta\\}$.

If $m\geq t>m+z-\beta$, then

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-3,\\\
2n-2(\alpha-2)(m+z)&k=\alpha-2,\\\ 0&k\geq\alpha-1,\end{array}\right.$

so $A$ corresponds to the partition

$\left((\alpha-1)^{2n-2(\alpha-2)(m+z)},(\alpha-2)^{2(\alpha-1)(m+z)-2n}\right)=[2n]^{2(m+z)},$

which is of the form Table 1(P1) and whose existence will be shown in Case 2.

Similarly, if $m\geq t>m-z+\beta$, then

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(m+z)&k\leq\alpha-2,\\\
2n-2(\alpha-1)(m+z)&k=\alpha-1,\\\ 0&k\geq\alpha,\end{array}\right.$

so $A$ corresponds to the partition
$(\alpha^{2n-2(\alpha-1)(m+z)},(\alpha-1)^{2\alpha(m+z)-2n})$, which is again
of the form $[2n]^{2(m+z)}$.

To finish the proof of Case 1 it remains to show that when $F=\mathbb{C}$ the
partitions obtained in the case 1.2(b) are indeed all possible. First we set
the notation. Given polynomials $p(X),q(X)\in R$, we define the sequence of
polynomials $r_{k}(X)\in R$ by

$r_{0}(X)=0,\quad r_{1}(X)=1,\quad\mathrm{and}\quad
r_{k+1}(X)=q(X)r_{k}(X)+p(X)r_{k-1}(X)\quad\mathrm{for}\quad k\geq 1.$

Note that Lemma 4 then holds for the polynomials $p(X)$, $q(X)$, and
$r_{0}(X),r_{1}(X),\ldots$.

Now let $z\leq n-1$ be a non-negative integer, let $m<\frac{n-z}{2}$ be a
positive integer, and let $\alpha=\left\lceil\frac{n+m}{m+z}\right\rceil$ and
$\beta=\alpha(m+z)-(n+m)$. To show that all partitions obtained in the case
1.2(b) are indeed possible it suffices to show that for $j=\alpha-2\geq 3$ and
$1\leq t\leq\min\\{2z+\beta,m\\}$, and for $j=\alpha-1\geq 3$ and $1\leq
t\leq\min\\{m+z-\beta,m-z+\beta\\}$ there exist polynomials $p(X),q(X)\in R$
with non-zero constant terms such that $X$ does not divide $r_{k}(X)$ for any
$k<\alpha$ satisfying $k\neq j$ and that $X^{t}$ is the highest power of $X$
that divides $r_{j}(X)$. We will show that this happens if we define $q(X)=1$
and $p(X)=X^{t}+u$ for a suitable non-zero $u\in\mathbb{C}$. In this case
Lemma 4 implies that for $k\geq 1$ we have

$r_{2k-1}(X)=\sum_{i=1}^{k}{2k-i-1\choose
i-1}u^{i-1}+\sum_{i=1}^{k}(i-1){2k-i-1\choose i-1}u^{i-2}X^{t}+\sum_{i\geq
1}v_{i}X^{t+i}$

and

$r_{2k}(X)=\sum_{i=0}^{k-1}{2k-i-1\choose
i}u^{i}+\sum_{i=0}^{k-1}i{2k-i-1\choose i}u^{i-1}X^{t}+\sum_{i\geq
1}v^{\prime}_{i}X^{t+i}$

for some $v_{i},v_{i}^{\prime}\in\mathbb{C}$, and let $r_{0}(X)=0$. We define
$h_{0}(u):=0$ and for $k\geq 1$ let

$h_{2k-1}(u):=\sum_{i=1}^{k}{2k-i-1\choose i-1}u^{i-1}\text{ and
}h_{2k}(u):=\sum_{i=0}^{k-1}{2k-i-1\choose i}u^{i}.$

It follows that

(46)
$r_{k}(X)=h_{k}(u)+h_{k}^{\prime}(u)X^{t}+\,\mathrm{higher}\,\mathrm{terms}$

for all $k\geq 0$ (where $h_{k}^{\prime}$ is the derivative of $h_{k}$). Since
$r_{k}(X)=r_{k-1}(X)+(X^{t}+u)r_{k-2}(X)$ for $k\geq 2$, we therefore get

$h_{0}(u)=0,\quad h_{1}(u)=1,\quad\mathrm{and}\quad
h_{k}(u)=h_{k-1}(u)+uh_{k-2}(u)\,\,\mathrm{for}\,k\geq 2.$

Note that we have assumed that $u\neq 0$. Assume now that $u\neq-\frac{1}{4}$
as well. Then we can show, either by solving the above difference equation or
by induction, that

$h_{k}(u)=\frac{1}{\sqrt{1+4u}}\left(\left(\frac{1+\sqrt{1+4u}}{2}\right)^{k}-\left(\frac{1-\sqrt{1+4u}}{2}\right)^{k}\right)$

for each $k\geq 0$. Here we define the square root as the one that has
argument within $[0,\pi)$. An easy computation shows that

(47)
$h_{k}(u)=0\Leftrightarrow\sqrt{1+4u}=i\,\mathrm{tg}\,\frac{k^{\prime}\pi}{k}\quad\text{for
some }\,k^{\prime}\in\\{1,\ldots,k-1\\}\setminus\left\\{\frac{k}{2}\right\\}.$

(Note that $k^{\prime}=0$ is not possible, since we have assumed that
$u\neq-\frac{1}{4}$.) Fix now $j\geq 3$ and assume that either $j=\alpha-2$
and $t\leq\min\\{2z+\beta,m\\}$ or $j=\alpha-1$ and
$t\leq\min\\{m+z-\beta,m-z+\beta\\}$ (and hence $\alpha\geq 4$ in both cases).
Let $u_{j}=-\frac{1}{4\cos^{2}\frac{\pi}{j}}$. Then
$\sqrt{1+4u_{j}}=i\,\mathrm{tg}\,\frac{\pi}{j}$ and $h_{j}(u_{j})=0$ by (47).

We now consider $k\neq j$. We will show that $h_{k}(u_{j})\neq 0$. Assume
first that $k<j$. If $k^{\prime}>\lfloor\frac{k}{2}\rfloor$, then
$\mathrm{tg}\,\frac{k^{\prime}\pi}{k}<0$ and therefore
$i\,\mathrm{tg}\,\frac{k^{\prime}\pi}{k}$ cannot be a square root and hence
equal to $\sqrt{1+4u_{j}}$. On the other hand, if
$k^{\prime}\leq\lfloor\frac{k}{2}\rfloor$, then
$\frac{k^{\prime}}{k}\geq\frac{1}{k}>\frac{1}{j}$ and hence
$i\,\mathrm{tg}\,\frac{k^{\prime}\pi}{k}\neq
i\,\mathrm{tg}\,\frac{\pi}{j}=\sqrt{1+4u_{j}}$. On the other hand, if $k>j$,
then the only possibility is $j=\alpha-2$ and $k=\alpha-1$, as $k<\alpha$. If
$k-1\geq k^{\prime}>\lfloor\frac{\alpha-1}{2}\rfloor$, then
$\mathrm{tg}\,\frac{k^{\prime}\pi}{k}<0$ and hence
$i\,\mathrm{tg}\,\frac{k^{\prime}\pi}{k}\neq\sqrt{1+4u_{j}}$. On the other
hand, it is clear that
$i\,\mathrm{tg}\,\frac{\pi}{\alpha-1}\neq\sqrt{1+4u_{j}}$ and that for $2\leq
k^{\prime}\leq\lfloor\frac{\alpha-1}{2}\rfloor$ we have
$\frac{k^{\prime}}{\alpha-1}>\frac{1}{\alpha-2}$, which again implies
$i\,\mathrm{tg}\,\frac{k^{\prime}\pi}{k}\neq\sqrt{1+4u_{j}}$. Using (47) we
therefore see that $h_{k}(u_{j})=0$ if and only if $k=j$, and (46) then
implies that $r_{k}(X)$ is not divisible by $X$ for $k\neq j$ and that
$r_{j}(X)$ is divisible by $X^{t}$ if we define $p(X)=X^{t}+u_{j}$.

It remains to show that $r_{j}(X)$ is not divisible by $X^{t+1}$, which is by
(46) equivalent to $h_{j}^{\prime}(u_{j})\neq 0$. In the neighborhood of
$1-\frac{1}{\cos^{2}\frac{\pi}{j}}$ the square root is a holomorphic function,
so we can compute

$\displaystyle h_{j}^{\prime}(u)=$
$\displaystyle-\frac{2}{(1+4u)^{\frac{3}{2}}}\left(\left(\frac{1+\sqrt{1+4u}}{2}\right)^{j}-\left(\frac{1-\sqrt{1+4u}}{2}\right)^{j}\right)+$
$\displaystyle+\frac{j}{1+4u}\left(\left(\frac{1+\sqrt{1+4u}}{2}\right)^{j-1}+\left(\frac{1-\sqrt{1+4u}}{2}\right)^{j-1}\right)=$
$\displaystyle=$
$\displaystyle-\frac{2}{1+4u}h_{j}(u)+\frac{j}{1+4u}\left(\left(\frac{1+\sqrt{1+4u}}{2}\right)^{j-1}+\left(\frac{1-\sqrt{1+4u}}{2}\right)^{j-1}\right).$

In particular,

$h_{j}^{\prime}(u_{j})=\frac{j}{1+4u_{j}}\left(\left(\frac{1+i\,\mathrm{tg}\,\frac{\pi}{j}}{2}\right)^{j-1}+\left(\frac{1-i\,\mathrm{tg}\,\frac{\pi}{j}}{2}\right)^{j-1}\right)=\frac{j\cos\frac{(j-1)\pi}{j}}{2^{j-2}(1+4u_{j})\left(\cos\frac{\pi}{j}\right)^{j-1}}\neq
0$

for $j\geq 3$, which had to be proved. It follows that all the partitions
obtained in case 1.2(b) indeed correspond to nilpotent orbits in
$\mathcal{N}(B)$.

###  3.2. Case 2: $C_{0}$ is a zero matrix

Since $A$ is non-zero and $z$ is maximal possible, we get that $C_{1}$ is not
nilpotent. We consider two subcases.

####  3.2.1. Case 2.1: $C_{1}$ is invertible

In this case it is clear that $\mathrm{rank}\,A^{k}=2n-2k(z+1)$ for
$k<\frac{n}{z+1}$ and $A^{k}=0$ for $k\geq\frac{n}{z+1}$. Let
$\alpha=\left\lceil\frac{n}{z+1}\right\rceil$. Since we have assumed at the
beginning of the proof that $A\neq 0$, we have $\alpha\geq 2$, and hence

$\mathrm{rank}\,A^{k}-\mathrm{rank}\,A^{k+1}=\left\\{\begin{array}[]{ll}2(z+1)&k\leq\alpha-2,\\\
2n-2(\alpha-1)(z+1)&k=\alpha-1,\\\ 0&k\geq\alpha,\end{array}\right.$

so $A$ corresponds to the partition Table 1(P1)

$\left(\alpha^{2n-2(\alpha-1)(z+1)},(\alpha-1)^{2\alpha(z+1)-2n}\right)=[2n]^{2(z+1)}.$

Conversely, it is clear from the above proof that for each partition
$[2n]^{2(z+1)}$ where $0\leq z\leq n-1$ there exists $A\in\mathcal{N}(B)$
corresponding to this partition. This concludes the proof for the even cases
in the first set of partitions stated in the theorem.

####  3.2.2. Case 2.2: $C_{1}$ is not invertible

Since $C_{1}$ is not nilpotent and the multiplication with a non-zero constant
has no influence on the Jordan canonical form of a nilpotent matrix, we may
assume that $C_{1}$ is idempotent, so $C_{1}^{2}=C_{1}$. Moreover, we may
conjugate $C$ by an invertible matrix in $\mathcal{C}(B)$ to assume that
$C_{1}=\left[\begin{array}[]{cc}1&0\\\ 0&0\end{array}\right]$ or,
equivalently, $a_{1}=1$ and $b_{1}=c_{1}=d_{1}=0$. We write matrix $C$ as
$C=\left[\begin{array}[]{cc}Xa^{\prime}(X)&X^{2}b^{\prime}(X)\\\
X^{2}c^{\prime}(X)&X^{2}d^{\prime}(X)\end{array}\right]$ for some polynomials
$a^{\prime}(X),b^{\prime}(X),c^{\prime}(X),d^{\prime}(X)$. As the constant
term of $a^{\prime}(X)$ is 1, it is invertible in $S$. Therefore, by defining
first the constant term, then the linear term, and so on, we may find
$f(X),r(X)\in S$ such that the equalities

$\displaystyle a^{\prime}(X)\left(f(X)-d^{\prime}(X)\right)$ $\displaystyle=$
$\displaystyle-Xc^{\prime}(X)b^{\prime}(X)-Xf(X)d^{\prime}(X)+Xf(X)^{2},$
$\displaystyle a^{\prime}(X)r(X)$ $\displaystyle=$ $\displaystyle
X\left(c^{\prime}(X)+d^{\prime}(X)r(X)-b^{\prime}(X)r(X)^{2}\right)$

are satisfied in $S$. We furthermore define

$e(X)=a^{\prime}(X)+Xb^{\prime}(X)r(X)\text{ and
}q(X)=-Xb^{\prime}(X)\left(a^{\prime}(X)-Xf(X)\right)^{-1}$

in $S$, and the matrices $P=\left[\begin{array}[]{cc}1&q(X)\\\
r(X)&1\end{array}\right]$ and $D=\left[\begin{array}[]{cc}Xe(X)&0\\\
0&X^{2}f(X)\end{array}\right]$. Note that $e(X)$ is invertible in $S$, while
$q(X)$ and $r(X)$ are divisible by $X$, which implies that matrix $P$ is
invertible in $M_{2}(S)$. Furthermore, a short calculation shows that $CP=PD$,
so we may replace $C$ by a diagonal matrix
$D=P^{-1}CP=\left[\begin{array}[]{cc}Xe(X)&0\\\ 0&X^{m}g(X)\end{array}\right]$
for some $m\geq 2$ where $e(X),g(X)\in S$ are invertible or $g(X)=0$ and
$m=n$. Consequently, $P^{-1}AP=\left[\begin{array}[]{cc}X^{z+1}e(X)&0\\\
0&X^{z+m}g(X)\end{array}\right]$ and it is clear that such a matrix
corresponds to the partition Table 1(P5)

$([n]^{z+1},[n]^{z+m})$

and that all such partitions are possible. Note that here we have $m\geq 2$;
the case $m=1$ was considered in the case 2.1. ∎

##  4\. Examples and open questions

We conclude our paper with a few examples that illustrate the methods and
results obtained by our work.

######  Example 6.

Assume that $F=\mathbb{C}$. The list of orbits
${\mathcal{O}}_{{\bf{\lambda}}}$ that have non-empty intersection with the
nilpotent centralizer of a nilpotent matrix $B\in{\mathcal{O}}_{(6,6)}$ were
not completely characterized yet, as noted in [26, Example 3.7]. Among the
undecided partitions ${\bf{\lambda}}$ (listed in [24, Example 6.25]) there is
only one partition ${\bf{\lambda}}=(5,4,3)$ (of type (P3)), which has a non-
empty intersection with ${\mathcal{N}}_{B}$. The complete list is presented in
Table 2.

|  ${\bf{\lambda}}$, such that $\mathcal{O}_{{\bf{\lambda}}}$ has a non-empty
intersection with ${\mathcal{N}}(B)$, $B\in\mathcal{O}_{(6,6)}$
---|---
(P1) |  $(12)$, $(6^{2})$, $(4^{3})$, $(3^{4})$, $(3^{2},2^{3})$, $(2^{6})$, $(2^{5},1^{2})$, $(2^{4},1^{4})$, $(2^{3},1^{6})$, $(2^{2},1^{8})$, $(2,1^{10})$, $(1^{12})$
(P2) |  $(7,5)$, $(7,3,2)$, $(7,2^{2},1)$, $(7,2,1^{3})$, $(7,1^{5})$, $(4^{2},2^{2})$, $(4^{2},2,1^{2})$, $(4^{2},1^{4})$, $(4,3^{2},2)$, $(4,3,2^{2},1)$, $(4,3,2,1^{3})$, $(4,3,1^{5})$, $(3^{3},1^{3})$, $(3^{2},2^{2},1^{2})$, $(3^{2},2,1^{4})$, $(3,2^{4},1)$, $(3,2^{3},1^{3})$, $(3,2^{2},1^{5})$
(P3) |  $(5,4,3)$
(P4) |  $(4,2^{4})$
(P5) |  $(6,3^{2})$, $(6,2^{3})$, $(6,2^{2},1^{2})$, $(6,2,1^{4})$, $(6,1^{6})$, $(3^{2},1^{6})$
Table 2. A complete list of the nilpotent orbits intersecting the nilpotent
centralizer of $B\in\mathcal{O}_{(6,6)}$. Partitions listed for each type are
only the ones that do not appear in lines above it. The underlined partition
had not been previously resolved.

Moreover, we also provide the complete list of
${\mathcal{O}}_{{\bf{\lambda}}}$ that have non-empty intersection with the
nilpotent centralizer of a nilpotent matrix $B\in{\mathcal{O}}_{(7,7)}$ in
Table 3. It shows that also partitions of type (P6), (P8), and (P9) appear for
$n=7$.

|  ${\bf{\lambda}}$, such that $\mathcal{O}_{{\bf{\lambda}}}$ has a non-empty
intersection with ${\mathcal{N}}(B)$, $B\in\mathcal{O}_{(7,7)}$
---|---
(P1) |  $(14)$, $(7^{2})$, $(5^{2},4)$, $(4^{2},3^{2})$, $(3^{4},2)$, $(3^{2},2^{4})$, $(2^{7})$, $(2^{6},1^{2})$, $(2^{5},1^{4})$, $(2^{4},1^{6})$, $(2^{3},1^{8})$, $(2^{2},1^{10})$, $(2,1^{12})$, $(1^{14})$.
(P2) |  $(8,6)$, $(8,3^{2})$, $(8,2^{3})$, $(8,2^{2},1^{2})$, $(8,2,1^{4})$, $(8,1^{6})$, $(5,4,3,2)$, $(5,4,2^{2},1)$, $(5,4,2,1^{3})$, $(5,4,1^{5})$, $(4^{2},2^{3})$, $(4^{2},2^{2},1^{2})$, $(4^{2},2,1^{4})$, $(4^{2},1^{6})$, $(4,3^{2},2,1^{2})$, $(4,3^{2},1^{4})$, $(3^{3},2^{2},1)$, $(3^{3},2,1^{3})$, $(3^{3},1^{5})$, $(3^{2},2^{3},1^{2})$, $(3^{2},2^{2},1^{4})$, $(3^{2},2,1^{6})$, $(3,2^{5},1)$, $(3,2^{4},1^{3})$, $(3,2^{3},1^{5})$,
(P3) |
(P4) |  $(6,4^{2})$, $(4,3,2^{3},1)$, $(4,2^{5})$
(P5) |  $(7,4,3)$, $(7,3,2^{2})$, $(7,2^{3},1)$, $(7,2^{2},1^{3})$, $(7,2,1^{5})$, $(7,1^{7})$, $(4,3^{2},2^{2})$, $(4,3,2^{2},1^{3})$, $(4,3,2,1^{5})$, $(4,3,1^{7})$, $(3,2^{2},1^{7})$
(P6) |  $(4^{3},2)$
(P7) |
(P8) |  $(5,3^{3})$
(P9) |  $(3^{4},1^{2})$
Table 3. A complete list of the nilpotent orbits intersecting the nilpotent
centralizer of $B\in\mathcal{O}_{(7,7)}$. Partitions listed for each type are
only the ones that do not appear in lines above it.

Note that examples $n=6$ and $n=7$ show different structures of partitions
$\lambda$ for which the nilpotent orbit ${\mathcal{O}}_{\lambda}$ intersects
${\mathcal{N}}(B)$. Although partitions of type (P7) do not appear in these
two cases, note that for example for $n=22$ ($z=1$, $m=3$, $l=5$, $\alpha=4$,
$\beta=5$) the partition $(7^{3},6^{3},5)$ is of the form (P7) but not
(P1)-(P6). This shows that, although types (P1)-(P9) of partitions in Table 1
are not complementary, they are all significant to describe
${\mathcal{N}}(B)$.

######  Example 7.

Let $n=11$ and consider the nilpotent orbit $\mathcal{O}_{(7,5^{3})}$. It can
be checked that this orbit is of type (P8) with the corresponding parameters
$z=0$, $m=2$, $t=1$, and that it does not belong to any of the types (P1)-(P7)
or (P9). We will show that it does not intersect with the nilpotent
centralizer of a nilpotent matrix $B\in\mathcal{O}_{(11,11)}$ if
$F=\mathbb{Q}$.

As shown in the proof of Theorem 3, we may assume that the corresponding
matrix in $\mathcal{N}(B)$ is of the form $A=\left[\begin{array}[]{cc}0&1\\\
X^{4}p(X)&X^{2}q(X)\end{array}\right]$, and let $p_{0}$ and $q_{0}$ be the
(non-zero) constant terms of $p(X)$ and $q(X)$, respectively. A short
calculation shows that

$A^{5}=\left[\begin{array}[]{cc}X^{10}\Big{(}2p(X)^{2}q(X)+p(X)q(X)^{3}\Big{)}&X^{8}\Big{(}p(X)^{2}+3p(X)q(X)^{2}+q(X)^{4}\Big{)}\\\
0&X^{10}\Big{(}3p(X)^{2}q(X)+4p(X)q(X)^{3}+q(x)^{5}\Big{)}\end{array}\right].$

By the assumption the matrix $A$ corresponds to the partition $(7,5^{3})$,
which implies that $\mathrm{rank}\,A^{5}=2$, and consequently the constant
term of $p(X)^{2}+3p(X)q(X)^{2}+q(X)^{4}$ has to be zero. However, the
equation $p_{0}^{2}+3p_{0}q_{0}^{2}+q_{0}^{4}=0$ has no non-zero solutions in
$\mathbb{Q}$, which shows that the nilpotent orbit $\mathcal{O}_{(7,5^{3})}$
does not intersect the nilpotent centralizer of $B$ if $F=\mathbb{Q}$.

######  Example 8.

Let $n=9$ and consider the nilpotent orbit $\mathcal{O}_{(5^{3},3)}$, which is
of type (P9) with parameters $z=0$, $m=2$, $t=1$, and it does not belong to
any of the types (P1)-(P8). Let $A=\left[\begin{array}[]{cc}0&1\\\
X^{4}p(X)&X^{2}q(X)\end{array}\right]$ be the corresponding matrix in
$\mathcal{N}(B)$. Then

$A^{5}=\left[\begin{array}[]{cc}0&X^{8}\Big{(}p(X)^{2}+3p(X)q^{2}(X)+q(X)^{4}\Big{)}\\\
0&0\end{array}\right],$

which has to be the zero matrix, as $A$ corresponds to the partition
$(5^{3},3)$. As in the previous example we get the equation
$p_{0}^{2}+3p_{0}q_{0}^{2}+q_{0}^{4}=0$, which has no non-zero solutions in
$\mathbb{Q}$. It follows that the nilpotent orbit $\mathcal{O}_{(5^{3},3)}$ of
type (P9) does not intersect the nilpotent centralizer of
$B\in\mathcal{O}_{(9,9)}$ if $F=\mathbb{Q}$.

Although we were able to characterize nilpotent orbits
${\mathcal{O}}_{{\bf{\lambda}}}$ that have non-empty intersection with the
nilpotent centralizer of a nilpotent matrix $B\in{\mathcal{O}}_{(n,n)}$, the
partitions ${\bf{\lambda}}$ in Table 1 are presented in a rather scattered
way.

######  Question 9.

Is there a combinatorial description of partitions obtained in Table 1?

Moreover, there are plenty of possible generalizations to Theorem 3 that would
be interesting to resolve.

######  Question 10.

Do any of the above methods apply to the case when:

  1. (1)

$B\in{\mathcal{O}}_{(\mu_{1},\mu_{2})}$, $\mu_{1}\neq\mu_{2}$, i.e., in the
case when the matrix $B$ still has just two Jordan blocks, but different in
size,

  2. (2)

$B\in{\mathcal{O}}_{(n^{\ell})}$, $\ell\geq 3$, i.e., in the case when the
matrix $B$ has at least three Jordan blocks, all of the same size,

  3. (3)

we consider some special nilpotent orbits over other simple Lie algebras?

Or, alternatively, could we use the results from Theorem 3 to further extend
the partial answers to open questions mentioned in this paper?

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