# Centres of centralizers of nilpotent elements in Lie superalgebras
$\mathfrak{sl}(m|n)$ or $\mathfrak{osp}(m|2n)$

Leyu Han

###### Abstract.

Let $\bar{G}$ be the simple algebraic supergroup $\mathrm{SL}(m|n)$ or
$\mathrm{OSp}(m|2n)$ over $\mathbb{C}$. Let
$\mathfrak{g}=\mathrm{Lie}(\bar{G})=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$
and let $G=\bar{G}(\mathbb{C})$ where $\mathbb{C}$ is considered as a
superalgebra concentrated in even degree. Suppose $e\in\mathfrak{g}_{\bar{0}}$
is nilpotent. We describe the centralizer $\mathfrak{g}^{e}$ of $e$ in
$\mathfrak{g}$ and its centre $\mathfrak{z}(\mathfrak{g}^{e})$. In particular,
we give bases for $\mathfrak{g}^{e}$, $\mathfrak{z}(\mathfrak{g}^{e})$ and
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$. We also determine the
labelled Dynkin diagram $\varDelta$ with respect to $e$ and subsequently
describe the relation between
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ and $\varDelta$.

##  1\. Introduction

Let $\bar{G}$ be the simple algebraic supergroup $\mathrm{SL}(m|n)$ or
$\mathrm{OSp}(m|2n)$ over $\mathbb{C}$. Let
$\mathfrak{g}=\mathrm{Lie}(\bar{G})=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$
and let $G=\bar{G}(\mathbb{C})$ where $\mathbb{C}$ is considered as a
superalgebra concentrated in even degree. Let $e\in\mathfrak{g}_{\bar{0}}$ be
nilpotent and write $\mathfrak{g}^{e}=\\{x\in\mathfrak{g}:[e,x]=0\\}$ (resp.
$G^{e}=\\{g\in G:geg^{-1}=e\\}$) for the centralizer of $e$ in $\mathfrak{g}$
(resp. $G$). The main result of this paper is an explicit description of the
centre of centralizer
$\mathfrak{z}(\mathfrak{g}^{e})=\\{x\in\mathfrak{g}^{e}:[x,y]=0\text{ for all
}y\in\mathfrak{g}^{e}\\}$ of $e$ in $\mathfrak{g}$ and its structure
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ under the adjoint action
of $G^{e}$ where
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\\{x\in\mathfrak{z}(\mathfrak{g}^{e}):gxg^{-1}=x\text{
for all }g\in G^{e}\\}$. The present paper is one of two papers in which we
calculate bases for $\mathfrak{g}^{e}$, $\mathfrak{z}(\mathfrak{g}^{e})$ and
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ for $\mathfrak{g}$ a
basic classical Lie superalgebra except $\mathfrak{psl}(n|n)$. In this paper,
we deal with Lie superalgebras $\mathfrak{sl}(m|n)$ and
$\mathfrak{osp}(m|2n)$, while the other deals with exceptional Lie
superalgebras $D(2,1;\alpha)$, $G(3)$ and $F(4)$.

Research on the centralizer and the centre of centralizer of nilpotent
elements in simple Lie algebras has been developed over the years since
Springer [20] considered the centralizer $G^{u}$ of a unipotent element $u$ in
a simple algebraic group $G$. For $G$ is of exceptional type, further study of
$G^{u}$ was carried out in [2], [21], [18], [19], [12] and[13]. A description
of the structure of the centralizer $\mathfrak{g}^{e}$ of nilpotent elements
in classical Lie algebras can be found in Jantzen’s monograph, [8]. In [23],
Yakimova identified the centre of $\mathfrak{g}^{e}$ for classical Lie algebra
$\mathfrak{g}$ over a field of characteristic zero. Note that the structure of
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ described in this paper
is a new result not only in the case of Lie superalgebras, but also in the
case of Lie algebras. In [16] and [17], Seitz further considered $Z(G^{u})$.
Lawther–Testerman studied the centralizer $G^{u}$ and its centre $Z(G^{u})$
over a field of characteristic $0$ or a good prime in [10]. They subsequently
determined the dimension of the Lie algebra of $Z(G^{u})$. In particular,
Lawther–Testerman [10] used work of Yakimova in [23] to deal with classical
cases. In recent work, Liebeck and Seitz developed a new approach to classify
unipotent and nilpotent orbits in [11].

To our best knowledge, there is a lot less study in this direction in the case
of Lie superalgebras. Finite-dimensional simple Lie superalgebras over an
algebraically closed field of characteristic zero were classified by V. G. Kac
in [9]. After this classification, a wide range of relevant problems have
drawn the attention of mathematicians. Determining the centralizers
$\mathfrak{g}^{e}$ of nilpotent even elements in
$\mathfrak{g}=\mathfrak{gl}(m|n)$ and $\mathfrak{osp}(m|2n)$ for a field of
zero or prime characteristic formed part of the results of recent work of Wang
and Zhao [22] and Hoyt [7]. However, more than forty years after Kac’s
classification, the dimension of the centre of $\mathfrak{g}^{e}$ where
$\mathfrak{g}$ is a simple Lie superalgebra still remains a mystery. We
attempt to discover this mystery in the present paper so that our knowledge
about $\mathfrak{z}(\mathfrak{g}^{e})$ can be broaden to a similar level to
that in the area of Lie algebras.

In the remaining part of this introduction, we give a more detailed survey of
our results.

In this paper, note that $\mathfrak{g}=\mathfrak{sl}(m|n)$ and
$G=\\{(A,B):A\in\mathrm{GL}_{m}(\mathbb{C}),B\in\mathrm{GL}_{n}(\mathbb{C})\text{
and }\mathrm{det}(A)=\mathrm{det}(B)\\}$, or
$\mathfrak{g}=\mathfrak{osp}(m|2n)$ and
$G=\mathrm{O}_{m}(\mathbb{C})\times\mathrm{Sp}_{2n}(\mathbb{C})$. More
generally, if $\mathfrak{g}$ is a direct sum of Lie superalgebras of the form
$\mathfrak{sl}(m_{i}|n_{i})$ or $\mathfrak{osp}(m_{i}|n_{i})$, we define $G$
similarly.

In Subsection 2.3, we give a full definition of the labelled Dynkin diagram
with respect to $e$. We classify the labelled Dynkin diagrams using Dynkin
pyramids. Pyramids are defined in [5] for non-super case and generalized in
[7] for super case. Each pyramid determines a nilpotent element $e$ and a
semisimple element $h$ such that $\\{e,h\\}$ can be extended to an
$\mathfrak{sl}(2)$-triple in $\mathfrak{g}_{\bar{0}}$. We use $h$ to determine
the labelled Dynkin diagram with respect to $e$. In contrast to the case of
non-super Lie algebras, in general $e$ determines more than one labelled
Dynkin diagram.

Our results can be viewed as Lie superalgebra versions of those obtained by
Lawther and Testerman in [10]. They obtain four theorems as direct results of
their findings. In this paper, we obtain analogues of Theorems 2–4 in [10] for
Lie superalgebras $\mathfrak{sl}(m|n)$ and $\mathfrak{osp}(m|2n)$. We take
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ as the correct
replacement for $Z(G^{e})$ since
$\mathrm{Lie}(Z(G^{e}))=\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$
for a field of characteristic $0$. Note that for
$\mathfrak{g}=\mathfrak{sl}(m|n)$, we have
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\mathfrak{z}(\mathfrak{g}^{e})$.

Fix $\varDelta$ to be a labelled Dynkin diagram with respect to $e$. Denote by
$n_{i}(\varDelta)$ the number of nodes which have labels equal to $i$ in
$\varDelta$. As a consequence of our calculations, we observe that the choice
of $\varDelta$ does not affect the following theorems and labels in
$\varDelta$ can only be $0,1$ or $2$.

Our first theorem concerns the case where labels in $\varDelta$ are even.

######  Theorem 1.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{sl}(m|n)$
or $\mathfrak{osp}(m|2n)$ and $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent.
Assume $\varDelta$ has no label equal to $1$, then

(1)
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=n_{2}(\varDelta)=\dim\mathfrak{z}(\mathfrak{g}^{h})$
for $\mathfrak{g}=\mathfrak{sl}(m|n),m\neq n$ or $\mathfrak{osp}(m|2n)$;

(2)
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}-1=n_{2}(\varDelta)=\dim\mathfrak{z}(\mathfrak{g}^{h})$
for $\mathfrak{g}=\mathfrak{sl}(n|n)$.

Our next result gives a more general result relating
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ and $\varDelta$.

######  Theorem 2.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{sl}(m|n)$
or $\mathfrak{osp}(m|2n)$ and $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent. Let
$a_{1},\dots,a_{l}$ be the labels in $\varDelta$. Then

$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\left\lceil\frac{1}{2}\sum_{i=1}^{l}a_{i}\right\rceil+\varepsilon$

where the value of $\varepsilon$ is equal to $0$ except when
$\mathfrak{g}=\mathfrak{sl}(n|n)$, we have $\varepsilon=1$.

Theorem 1 is subsumed by a more general result as stated in Theorem 3 and the
proof of which involves more techniques. In order to state Theorem 3, we
require some notations first. Define the sub-labelled Dynkin diagram
$\varDelta_{0}$ to be the $2$-free core of $\varDelta$ such that
$\varDelta_{0}$ is obtained by removing all nodes with labels equal to $2$
from $\varDelta$. For $\mathfrak{g}=\mathfrak{sl}(m|n)$ (resp.
$\mathfrak{g}=\mathfrak{osp}(m|2n)$), let $\lambda$ be a partition of $(m|n)$
(resp. $(m|2n)$) and let $P$ be the Dynkin pyramid (resp. ortho-symplectic
Dynkin pyramid) of shape $\lambda$ which will be defined in Subsection 3.1
(resp. Subsection 4.2). Let $r_{i}$ (resp. $s_{i}$) be the number of boxes on
the $i$th column with parity $\bar{0}$ (resp. $\bar{1}$) in $P$ and $k\geq 0$
be minimal such that the $k$th column in $P$ contains no boxes. Then we define
$\tau$ to be:

(1) the number of $i$ such that $r_{i}=s_{i}\neq 0$ for all $i>k$ or $i<-k$
when $\mathfrak{g}=\mathfrak{sl}(m|n)$;

(2) the number of $i$ such that $r_{i}=s_{i}\neq 0$ for all $i>k$ when
$\mathfrak{g}=\mathfrak{osp}(m|2n)$.

We also define

$\nu_{0}=\begin{cases}0&\text{if
}\mathfrak{g}=\mathfrak{sl}(m|n),\sum_{-k<i<k}r_{i}\neq\sum_{-k<i<k}s_{i}\text{
or }\mathfrak{g}=\mathfrak{osp}(m|2n);\\\ 1&\text{if
}\mathfrak{g}=\mathfrak{sl}(m|n),\sum_{-k<i<k}r_{i}=\sum_{-k<i<k}s_{i}.\end{cases}$

######  Theorem 3.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{sl}(m|n)$
or $\mathfrak{osp}(m|2n)$ and $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent. Let
$\varDelta_{0}$ be the 2-free core of $\varDelta$. Let $\mathfrak{g}_{0}$ be
the subalgebra of $\mathfrak{g}$ generated by the root vectors corresponding
to the simple roots in $\varDelta_{0}$, then $\mathfrak{g}_{0}$ is a direct
sum of simple Lie superalgebras. Let $G_{0}$ be the subgroup of $G$ defined as
above. There exists a nilpotent $G_{0}$-orbit in
$(\mathfrak{g}_{0})_{\bar{0}}$ having labelled Dynkin diagram $\varDelta_{0}$.
Suppose $e_{0}\in(\mathfrak{g}_{0})_{\bar{0}}$ is a representative of this
orbit, then

(1). $\dim\mathfrak{g}^{e}-\dim\mathfrak{g}_{0}^{e_{0}}=n_{2}(\varDelta)$ for
all $\mathfrak{g}$;

(2). When $\mathfrak{g}=\mathfrak{sl}(m|n),m\neq n$ or $\mathfrak{osp}(m|2n)$,
then
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}-\dim\left(\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})\right)^{G_{0}^{e_{0}}}=n_{2}(\varDelta)-\tau-\nu_{0}$;

(3). When $\mathfrak{g}=\mathfrak{sl}(n|n)$, then
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}-\dim\left(\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})\right)^{G_{0}^{e_{0}}}=n_{2}(\varDelta)+1-\tau-\nu_{0}$.

This paper is organized as follows. In Section 2, we recall some fundamental
concepts of Lie superalgebras such as basic classical Lie superalgebras, root
system and labelled Dynkin diagrams. In Sections 3–4, we recall the concept of
Dynkin pyramid and ortho-symplectic Dynkin pyramid. We use this to determine a
nilpotent element $e\in\mathfrak{g}_{\bar{0}}$ and the labelled Dynkin diagram
with respect to $e$ explicitly, and to calculate bases of the centralizers
$\mathfrak{g}^{e}$ and centres $\mathfrak{z}(\mathfrak{g}^{e})$. In Subsection
4.9, we find a basis for $\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$
where $\mathfrak{g}=\mathfrak{osp}(m|2n)$ and
$G=\mathrm{O}_{m}(\mathbb{C})\times\mathrm{Sp}_{2n}(\mathbb{C})$.

##  2\. Preliminaries

###  2.1. Basic classical Lie superalgebras

Let $\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$ be a
finite-dimensional simple Lie superalgebra over $\mathbb{C}$. Recall that
$\mathfrak{g}$ is called a basic classical Lie superalgebra if
$\mathfrak{g}_{\bar{0}}$ is a reductive Lie algebra and $\mathfrak{g}$ has a
non-degenerate even invariant supersymmetric bilinear form $(\cdotp,\cdotp)$.
Finite-dimensional complex simple Lie superalgebras were classified by V. G.
Kac in [9, Theorem 5]. The simple basic classical Lie superalgebras that are
not Lie algebras consist of classical types which are infinite families and
three exceptional types. The infinite series are $\mathfrak{sl}(m|n)$,
$\mathfrak{psl}(n|n)$ and $\mathfrak{osp}(m|2n)$. In this paper, we consider
Lie superalgebras $\mathfrak{sl}(m|n)$ and $\mathfrak{osp}(m|2n)$.

The following definitions can be found in [3, Chapter 1]. Let $\mathfrak{h}$
be a Cartan subalgebra of $\mathfrak{g}$. There exists a root space
decomposition
$\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\mathfrak{h}^{*}}\mathfrak{g_{\alpha}}$
where $\mathfrak{g_{\alpha}}:=\\{x\in\mathfrak{g}:[h,x]=\alpha(h)x\ \text{for\
all}\ h\in\mathfrak{\mathfrak{h}}\\}$ is the root space corresponding to
$\alpha$ and we have $\mathfrak{h}=\mathfrak{g}_{0}$. The set
$\Phi=\\{\alpha\in\mathfrak{h^{*}:\alpha\neq}0,\mathfrak{g}_{\alpha}\neq 0\\}$
is called the root system of $\mathfrak{g}$. The set of even (resp. odd) roots
is defined to be
$\Phi_{\bar{0}}=:\\{\alpha\in\Phi:\mathfrak{g}_{\alpha}\subseteq\mathfrak{g}_{\bar{0}}\\}$
(resp.
$\Phi_{\bar{1}}=:\\{\alpha\in\Phi:\mathfrak{g}_{\alpha}\subseteq\mathfrak{g}_{\bar{1}}\\}$).
A set of positive roots for a root system is a set $\Phi^{+}\subseteq\Phi$
such that all root $\alpha\in\Phi$ there is exactly one of $\alpha$,$-\alpha$
contained in $\Phi^{+}$; and for any two distinct roots
$\alpha,\beta\in\Phi^{+}$, $\alpha+\beta\in\Phi$ implies that
$\alpha+\beta\in\Phi^{+}$. Note that
$\Phi^{+}=\Phi_{\bar{0}}^{+}\cup\Phi_{\bar{1}}^{+}$. The set of simple roots
$\varPi=\\{\alpha_{1},...,\alpha_{l}\\}\subseteq\Phi^{+}$ consists of the
positive roots that cannot be written as sums of positive roots. Note that $l$
does not depend on choice of $\varPi$ and we call it the rank of
$\mathfrak{g}$.

A Borel subalgebra $\mathfrak{b}$ of $\mathfrak{g}$ is defined to be a
solvable subalgebra such that
$\mathfrak{b}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi^{+}}\mathfrak{g}_{\alpha}$.
Note that there are in general many inequivalent conjugacy classes of Borel
subalgebras and every Borel subalgebra containing $\mathfrak{h}$ determines a
corresponding system of positive roots $\Phi^{+}$. Consequently $\mathfrak{b}$
determines a system of simple roots $\varPi$. Recall that the Weyl group $W$
of $\mathfrak{g}$ is generated by the Weyl reflections
$w_{\alpha}(\beta)=\beta-2\frac{(\alpha,\beta)}{(\alpha,\alpha)}\alpha$ with
$\alpha\in\Phi_{\bar{0}},\beta\in\Phi$. For each conjugacy class of Borel
subalgebras of $\mathfrak{g}$, a simple root system can be transformed into an
equivalent one under the transformation of the Weyl group $W$ of
$\mathfrak{g}$, see [6, Subsection 2.3].

###  2.2. Dynkin diagrams

In this subsection, we continue working with $\mathfrak{g}$ as in Subsection
2.1. We next recall the concept of the Dynkin diagram as defined for example
in [6, Section 2.2].

We know that there exists a non-degenerate even invariant supersymmetric
bilinear form $(\cdotp,\cdotp)$ on $\mathfrak{g}$. One can check that
$(\cdotp,\cdotp)$ restricts to a non-degenerate symmetric bilinear form on
$\mathfrak{h}$. Therefore, there exists an isomorphism from $\mathfrak{h}$ to
$\mathfrak{h}^{*}$ which provides a symmetric bilinear form on
$\mathfrak{h}^{*}$. Then the Dynkin diagram of a Lie superalgebra
$\mathfrak{g}$ with a simple root system $\varPi$ is a graph where the
vertices are labelled by $\varPi$ and there are $\mu_{\alpha\beta}$ lines
between the vertices labelled by simple roots $\alpha_{i}$ and $\alpha_{j}$
such that:

(2.1) $\mu_{\alpha\beta}=\begin{cases}|(\alpha_{i},\alpha_{j})|&\text{if
}(\alpha_{i},\alpha_{i})=(\alpha_{j},\alpha_{j})=0,\\\
\frac{2|(\alpha_{i},\alpha_{j})|}{min\\{|(\alpha_{i},\alpha_{i})|,|(\alpha_{j},\alpha_{j})|\\}}&\text{if
}(\alpha_{i},\alpha_{i})(\alpha_{j},\alpha_{j})\neq 0,\\\
\frac{2|(\alpha_{i},\alpha_{j})|}{min_{(\alpha_{k},\alpha_{k})\neq
0}|(\alpha_{k},\alpha_{k})|}&\text{if }(\alpha_{i},\alpha_{i})\neq 0,\
(\alpha_{j},\alpha_{j})=0\ \text{\text{and}\ }\alpha_{k}\in\Phi.\end{cases}$

We say a root $\alpha\in\Phi$ is isotropic if $(\alpha,\alpha)=0$ and is non-
isotropic if $(\alpha,\alpha)\neq 0$. We associate a white node $\ocircle$ to
each even root, a grey node $\varotimes$ to each odd isotropic root and a
black node $\newmoon$ to each odd non-isotropic root. Moreover, when
$\mu_{\alpha\beta}>1$, we put an arrow pointing from the vertex labelled by
$\alpha_{i}$ to the vertex labelled by $\alpha_{j}$ if
$(\alpha_{i},\alpha_{i})(\alpha_{j},\alpha_{j})\neq 0$ and
$(\alpha_{i},\alpha_{i})>(\alpha_{j},\alpha_{j})$ or if
$(\alpha_{i},\alpha_{i})=0,(\alpha_{j},\alpha_{j})\neq 0$ and
$|(\alpha_{j},\alpha_{j})|<2$, or pointing from the vertex labelled by
$\alpha_{j}$ to the vertex labelled by $\alpha_{i}$ if
$(\alpha_{i},\alpha_{i})=0,(\alpha_{j},\alpha_{j})\neq 0$ and
$|(\alpha_{j},\alpha_{j})|>2$. If the value of $\mu_{\alpha\beta}$ is not a
natural number, then we label the edge between vertices corresponding to roots
$\alpha$ and $\beta$ with $\mu_{\alpha\beta}$ instead of drawing multiple
lines between them. Note that the Dynkin diagram depends on $\varPi$ up to
conjugacy by $W$, thus Dynkin diagrams of $\mathfrak{g}$ for different choices
of simple roots can be different.

######  Remark 4.

In this paper, the Dynkin diagram for $\mathfrak{sl}(n|n)$ is by convention
the one for $\mathfrak{gl}(n|n)$.

###  2.3. Labelled Dynkin diagrams

Let $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent. There exists an
$\mathfrak{sl}(2)$-triple $\\{e,h,f\\}\subseteq\mathfrak{g}_{\bar{0}}$ by the
Jacobson–Morozov Theorem, see for example [4, Theorem 3.3.1]. An
$\mathfrak{sl}(2)$-triple determines a grading on $\mathfrak{g}$ according to
the eigenvalues of ad$h,$ thus we can decompose $\mathfrak{g}$ into its
ad$h$-eigenspaces $\mathfrak{g}=\bigoplus_{j\in\mathbb{Z}}\mathfrak{g}(j)$
where $\mathfrak{g}(j)=\\{x\in\mathfrak{g}:[h,x]=jx\\}$. In order to construct
the labelled Dynkin diagram of $e$, we first fix
$\mathfrak{h}\subseteq\mathfrak{g}(0)$ to be a Cartan subalgebra of
$\mathfrak{g}$ containing $h$. Then we choose a system $\Phi^{+}(0)$ of
positive roots for $\mathfrak{g}(0)$ to get a Borel subalgebra
$\mathfrak{b}(0)$ of $\mathfrak{g}(0)$. Let $\mathfrak{b}$ be the Borel
subalgebra such that $\mathfrak{b}=\mathfrak{b}(0)\oplus\mathfrak{g}(j>0)$,
then we obtain the corresponding system of positive roots $\Phi^{+}$ and a
system of simple roots $\varPi=\\{\alpha_{1},...,\alpha_{l}\\}$ which will
give a Dynkin diagram of $\mathfrak{g}$. Furthermore, for each $i=1,...,l$,
note that $\mathfrak{g}_{\alpha_{i}}$ is the root space corresponding to
$\alpha_{i}$ and $\mathfrak{g}_{\alpha_{i}}\subseteq\mathfrak{g}(j_{i})$ for
some $j_{i}\geq 0$. Hence, we have $\alpha_{i}(h)\geq 0$ for $i=1,...,l$.

######  Definition 5.

The labelled Dynkin diagram $\varDelta$ of $e$ determined by $\varPi$ is given
by taking the Dynkin diagram of $\mathfrak{g}$ and labelling each node
$\alpha$ with $\alpha(h)$.

##  3\. Lie superalgebras $\mathfrak{sl}(m|n)$

###  3.1. Dynkin pyramids and labelled Dynkin diagram for $\mathfrak{sl}(m|n)$

Let $V=V_{\bar{0}}\oplus V_{\bar{1}}$ be a finite-dimensional vector
superspace such that $\dim V_{\bar{0}}=m$ and $\dim V_{1}=n$. Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{sl}(m|n)=\mathfrak{sl}(V)$.
Note that the nilpotent orbits in $\mathfrak{g}_{\bar{0}}$ are parametrized by
the partitions of $(m|n)$. Let $\lambda$ be a partition of $(m|n)$ such that

(3.1) $\lambda=(p|q)=(p_{1},\dots,p_{r}|q_{1},\dots,q_{s})$

where $p$ (resp. $q$) is a partition of $m$ (resp. $n$) and
$p_{1}\geq\dots\geq p_{r},q_{1}\geq\dots\geq q_{s}$. By rearranging the order
of numbers in $(p|q)$, we write

(3.2) $\lambda=(\lambda_{1},\dots,\lambda_{r+s})$

such that $\lambda_{1}\geq\dots\geq\lambda_{r+s}$. For $i=1,\dots,r+s$, we
define $\left|i\right|$ such that for $c\in\mathbb{Z}$, we have that
$\left|\\{i:\lambda_{i}=c,\left|i\right|=\bar{0}\\}\right|=\left|\\{j:p_{j}=c\\}\right|$,
$\left|\\{i:\lambda_{i}=c,\left|i\right|=\bar{1}\\}\right|=\left|\\{j:q_{j}=c\\}\right|$
and if $\lambda_{i}=\lambda_{j}$,
$\left|i\right|=\bar{0},\left|j\right|=\bar{1}$, then $i<j$. i.e.
$\sum_{\left|i\right|=\bar{0}}\lambda_{i}=m$ and
$\sum_{\left|i\right|=\bar{1}}\lambda_{i}=n$.

For the purpose of describing the dimension of $\mathfrak{g}^{e}$ and the
labelled Dynkin diagram for each nilpotent orbit $e\in\mathfrak{g}_{\bar{0}}$,
we recall that a Dynkin pyramid $P$ of shape $\lambda$ is defined to be a
finite collection of boxes of size $2\times 2$ in the $xy$-plane which are
centred at integer coordinates, see [5, Section 4] and [7, Section 7]. Write
$\lambda=(\lambda_{1},\dots,\lambda_{r+s})$ as defined in (3.2), we number
rows of the Dynkin pyramid $P$ from $1,\dots,r+s$ from bottom to top, then the
$j$th row of $P$ has length $\lambda_{j}$ and we mark boxes in the row $j$
with parity $\bar{0}$ or $\bar{1}$ according to $|j|$. We associate a
numbering $1,2,...,m+n$ for the boxes of the Dynkin pyramid from top to bottom
and from left to right. The row number of the $i$th box is denoted by
$\mathrm{row}(i)$ and the parity of $\mathrm{row}(i)$ is denoted by
$\left|\mathrm{row}(i)\right|$. We say that the column number
$\mathrm{col}(i)$ of $i$ is the $x$-coordinate of the centre of the $i$th box.

Define a basis $\\{v_{1},\dots,v_{m+n}\\}$ of $V=V_{\bar{0}}\oplus
V_{\bar{1}}$ where $v_{i}\in V_{\left|\mathrm{row}(i)\right|}$. According to
[7, Section 7], $P$ determines a nilpotent element
$e\in\mathfrak{g}_{\bar{0}}$ such that $e$ sends $v_{i}$ to $v_{j}$ if the box
labelled by $j$ is the left neighbour of $i$ and sends $v_{i}$ to zero if the
box labelled by $v_{i}$ has no left neighbour. Write $e_{ij}$ for the
$ij$-matrix unit, then

$e=\sum_{\mathrm{row}(i)=\mathrm{row}(j),\mathrm{col}(j)=\mathrm{col}(i)-2}e_{ij}$

for all $1\leq i,j\leq m+n$. We know that $P$ also determines a semisimple
element $h\in\mathfrak{g}_{\bar{0}}$ such that
$h=\sum_{i=1}^{m+n}-\text{col}(i)e_{ii}$. Note that $\\{e,h\\}$ can be
extended to an $\mathfrak{sl}(2)$-triple $\\{e,h,f\\}$ in
$\mathfrak{g}_{\bar{0}}$ according to [7, Section 7].

Next we recall a construction of the root system for $\mathfrak{g}$ according
to [14, Section 2.2]. Let $\mathfrak{h}$ be the Cartan subalgebra of
$\mathfrak{g}$ consisting of all diagonal matrices in $\mathfrak{g}$ and let
$a=\mathrm{diag}(a_{1},\dots,a_{m+n})\in\mathfrak{h}$. For the basis
$\\{v_{1},\dots,v_{m+n}\\}$ of $V$, we define
$\\{\varepsilon_{1},\dots,\varepsilon_{m+n}\\}\subseteq\mathfrak{h}^{*}$ such
that $\varepsilon_{i}(a)=a_{i}$ for $i=1,\dots,m+n$ and the parity of
$\varepsilon_{i}$ is equal to $\left|\mathrm{row}(i)\right|$. Then the root
system of $\mathfrak{g}$ with respect to $\mathfrak{h}$ is
$\Phi=\Phi_{\bar{0}}\cup\Phi_{\bar{1}}$ where
$\Phi_{\bar{0}}=\\{\varepsilon_{i}-\varepsilon_{j}:i\neq j,|i|=|j|\\}$ and
$\Phi_{\bar{1}}=\\{\varepsilon_{i}-\varepsilon_{j}:i\neq j,|i|\neq|j|\\}$ and
$(\varepsilon_{i},\varepsilon_{j})=(-1)^{|i|}\delta_{ij}$. By computing
$(\varepsilon_{i}-\varepsilon_{j},\varepsilon_{i}-\varepsilon_{j})$ for all
$i\neq j,|i|\neq|j|$, we have that all odd roots are isotropic.

The labelled Dynkin diagram for a nilpotent orbit $e\in\mathfrak{g}_{\bar{0}}$
is constructed as follows. First, draw the Dynkin pyramid of shape $\lambda$.
For $i=1,...,m+n-1,$ we associate a white node $\ocircle$ (resp. a grey node
$\varotimes$) for root $\alpha_{i}=\varepsilon_{i}-\varepsilon_{i+1}$ if
$\left|\mathrm{row}(i+1)\right|=\left|\mathrm{row}(i)\right|$ (resp.
$\left|\mathrm{row}(i+1)\right|\neq\left|\mathrm{row}(i)\right|$). We label
the $i$th node with the value $\mathrm{col}(i+1)-\mathrm{col}(i)$.

######  Example 6.

For a partition $\lambda=(5,1|3)$, the Dynkin pyramid of shape $\lambda$ is

We calculate that $h=\text{diag}(4,2,2,0,0,0,-2,-2,-4)$. Then the
corresponding labelled Dynkin diagram is

######  Remark 7.

Different numberings within columns and different choices of parities of rows
are possible and would lead to different labelled Dynkin diagrams. In this way
one can get all possible labelled Dynkin diagrams. In this paper, the way we
define the partition $\lambda$ in (3.2) leads to a unique labelled Dynkin
diagram.

The following example shows that for different choices of parities of rows of
a given pyramid, we can get different Dynkin diagrams.

######  Example 8.

For a partition $\lambda=(3,2|2,1)$, the corresponding Dynkin pyramid and
labelled Dynkin diagram are shown below:

However, if we allow different parities for rows, then there exist another
Dynkin pyramid and therefore a different labelled Dynkin diagram:

###  3.2. Centralizer of nilpotent elements $e\in\mathfrak{g}_{\bar{0}}$

Let $e\in\mathfrak{g}_{\bar{0}}$ be a nilpotent element with Jordan type
$\lambda=(\lambda_{1},\dots,\lambda_{r+s})$ which is defined as in (3.2). In
order to calculate the dimension of $\mathfrak{\mathfrak{g}}^{e}$, we first
recall a basis for $\mathfrak{gl}(m|n)^{e}$ based on [23, Section 1] and [7,
Section 3.2].

Let
$\mathfrak{\bar{g}}=\bar{\mathfrak{g}}_{\bar{0}}\oplus\bar{\mathfrak{g}}_{\bar{1}}=\mathfrak{gl}(m|n)$.
Let $u_{1},\dots,u_{r+s}\in V$ such that $u_{i}=v_{k}$ for
$\mathrm{\mathrm{row}}(k)=i$ and $\mathrm{col}(k)=\lambda_{k}$, then the
vectors $e^{j}u_{i}$ with $0\leq j\leq\lambda_{i}-1,\ \left|i\right|=\bar{0}$
form a basis for $V_{\bar{0}}$ and the vectors $e^{j}u_{i}$ with $0\leq
j\leq\lambda_{i}-1,\ \left|i\right|=\bar{1}$ form a basis for $V_{\bar{1}}$.
Note that $e^{\lambda_{i}}u_{i}=0$ for $1\leq i\leq r+s$.

With the above notation, Hoyt worked out a basis of $\mathfrak{\bar{g}}^{e}$
in [7, Section 3.2.1] which we recall below. For
$\xi\in\mathfrak{\bar{g}}^{e}$, we have $\xi(e^{j}u_{i})=e^{j}\xi(u_{i})$.
Hence, each $\xi$ is determined by $\xi(u_{i})$ for $i=1,\dots,r+s$ and we can
write

(3.3)
$\xi(u_{i})=\sum_{j=1}^{r+s}\sum_{k=\max\\{\lambda_{j}-\lambda_{i},0\\}}^{\lambda_{j}-1}c_{i}^{j,k}e^{k}u_{j}$

where $c_{i}^{j,k}\in\mathbb{C}$ are coefficients. Then
$\mathfrak{\bar{g}}^{e}$ has a basis

(3.4) $\\{\xi_{i}^{j,k}:1\leq i,j\leq r+s\text{ and
}\max\\{\lambda_{j}-\lambda_{i},0\\}\leq k\leq\lambda_{j}-1\\}$

such that $\xi_{i}^{j,k}(u_{t})=\delta_{it}e^{k}u_{j}$.

Instead of using the formula for $\dim\mathfrak{\bar{g}}^{e}$ in [7, Section
3.2.1], we obtain an alternative formula for $\dim\mathfrak{\bar{g}}^{e}$
below. Note that formulas in [7, Section 3.2.1] and Proposition 9 are
equivalent, but the formula in Proposition 9 is more convenient to use in
Subsection 3.4.

######  Proposition 9.

Let $\lambda$ be a partition of $(m|n)$ denoted as in (3.2). Denote by $P$ the
Dynkin pyramid of $\lambda$ and $e\in\mathfrak{g}_{\bar{0}}$ be a nilpotent
element determined by $P$. Let $c_{i}$ be the number of boxes in the $i$th
column of $P$. Then
$\dim\mathfrak{\bar{g}}^{e}=\sum_{i\in\mathbb{Z}}c_{i}^{2}+\sum_{i\in\mathbb{Z}}c_{i}c_{i+1}$.

###### Proof.

Let $\mathfrak{\bar{g}}=\bigoplus_{j\in\mathbb{Z}}\mathfrak{\bar{g}}(j)$ as
defined in Subsection 2.3. According to [7, Definition 4.1 and Theorem 7.2],
we have that the map
$\text{ad}e:\mathfrak{\bar{g}}(j\geq-1)\rightarrow\mathfrak{\bar{g}}(j\geq 1)$
is surjective and
$\ker(\text{ad}e)=\mathfrak{\bar{g}}^{e}\subseteq\mathfrak{\bar{g}}(j\geq-1)$,
$\dim\ker(\text{ad}e)+\dim\text{im}(\text{ad}e)=\dim\mathfrak{\bar{g}}(j\geq-1)$,
so we have that
$\dim\mathfrak{\bar{g}}^{e}=\dim\mathfrak{\bar{g}}(j\geq-1)-\dim\mathfrak{\bar{g}}(j\geq
1)=\dim\mathfrak{\bar{g}}(0)+\dim\mathfrak{\bar{g}}(-1)$. We also calculate
that $[h,e_{kl}]=\left(\mathrm{col}(l)-\mathrm{col}(k)\right)e_{kl}$, which
means $e_{kl}\in\mathfrak{\bar{g}}(j)$ if $j=\mathrm{col}(l)-\mathrm{col}(k)$.
This implies that $\mathfrak{\bar{g}}(0)=\langle
e_{kl}:\mathrm{col}(l)=\mathrm{col}(k)\rangle\cong\bigoplus_{i\in\mathbb{Z}}\mathrm{End}(\mathbb{C}^{c_{i}})$
and $\mathfrak{\bar{g}}(-1)=\langle
e_{kl}:\mathrm{col}(l)-\mathrm{col}(k)=-1\rangle=\bigoplus\mathrm{Hom}(\mathbb{C}^{c_{i}},\mathbb{C}^{c_{i+1}})$.
Therefore, we obtain that
$\dim\mathfrak{\bar{g}}^{e}=\sum_{i\in\mathbb{Z}}c_{i}^{2}+\sum_{i\in\mathbb{Z}}c_{i}c_{i+1}$.∎

We also use an alternative notation for $\lambda$ such that

(3.5) $\lambda=(c^{m_{c}+n_{c}},\dots,1^{m_{1}+n_{1}})$

where $m_{i}=\left|\\{j:\lambda_{j}=i,\left|j\right|=\bar{0}\\}\right|$ and
$n_{i}=\left|\\{j:\lambda_{j}=i,\left|j\right|=\bar{1}\\}\right|$. We define
$M^{j}=\langle u_{i}:\lambda_{i}=j\rangle$ for $1\leq j\leq c$. By [7, Theorem
3.4], we have that
$\mathfrak{\bar{g}}^{e}(0)\cong\mathfrak{gl}(M^{1})\oplus\mathfrak{gl}(M^{2})\oplus\dots\oplus\mathfrak{gl}(M^{c})$.
In addition, we have $M^{j}=M_{\bar{0}}^{j}\oplus M_{\bar{1}}^{j}$ where $\dim
M_{\bar{0}}^{j}=m_{j}$, $\dim M_{\bar{1}}^{j}=n_{j}$ and thus
$\mathfrak{gl}(M^{j})\cong\mathfrak{gl}(m_{j}|n_{j})$ for each $j$. Therefore,
we obtain that
$\mathfrak{\bar{g}}^{e}(0)\cong\bigoplus_{j=1}^{c}\mathfrak{gl}(m_{j}|n_{j})$.

Next we move on to calculate $\dim\mathfrak{g}^{e}$.

######  Theorem 10.

Let $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent, we have that
$\dim\mathfrak{g}^{e}=\dim\mathfrak{\bar{g}}^{e}-1$.

###### Proof.

We first look at a Lie superalgebra homomorphism
$\text{str}:\mathfrak{\bar{g}}\rightarrow\mathbb{C}.$ Note that
$\ker(\text{str})=\mathfrak{g}$ and $\dim\text{im}(\text{str})=1$. By
restricting we get
$\text{str}^{e}:\mathfrak{\bar{g}}^{e}\rightarrow\mathbb{C}$ and
$\ker(\text{str}^{e})=\mathfrak{g}^{e}$. Then by the rank-nullity theorem, we
get
$\dim\ker(\text{str}^{e})+\dim\text{im}(\text{str}^{e})=\dim\mathfrak{\bar{g}}^{e}$.

Let

$e=\begin{pmatrix}e_{1}&0\\\
0&e_{2}\end{pmatrix}\in\mathfrak{g}_{\bar{0}}\subseteq\mathfrak{\bar{g}}_{\bar{0}}$

where $e_{1}$ and $e_{2}$ are $m\times m$ and $n\times n$ matrices
respectively.

Let $I_{m}\in\mathfrak{gl}(m)$ be the $m\times m$ identity matrix and
$0_{n\times n}\in\mathfrak{gl}(n)$ be the $n\times n$ zero matrix. Let
$I(m|0)$ be the $(m+n)\times(m+n)$ matrix

(3.6) $I(m|0)=\begin{pmatrix}I_{m}&0_{m\times n}\\\ 0_{n\times
m}&0_{n}\end{pmatrix}.$

Then we calculate $[I(m|0),e]=0$. Hence, we have
$I(m|0)\in\mathfrak{\bar{g}}^{e}$.

Therefore, we have that $\mathrm{str}^{e}(I(m|0))=\mathrm{trace}(I_{m})=m$,
thus $\mathrm{str}^{e}$ is non-zero and $\dim\text{im}(\text{str}^{e})=1$.
Therefore, we have $\dim\mathfrak{g}^{e}=\dim\mathfrak{\bar{g}}^{e}-1$.∎

###  3.3. Centre of centralizer of nilpotent elements
$e\in\mathfrak{g}_{\bar{0}}$

In order to give a basis for $\mathfrak{z}(\mathfrak{g}^{e})$, we determine a
basis for $\mathfrak{z}(\mathfrak{\bar{g}}^{e})$ first.

######  Proposition 11.

Let $e\in\mathfrak{\bar{g}}_{\bar{0}}$ be nilpotent, we have that
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})\subseteq\mathfrak{\bar{g}}_{\bar{0}}$.

###### Proof.

Suppose $x\in\mathfrak{z}(\mathfrak{\bar{g}}^{e})$, then
$x\in\mathfrak{\bar{g}}^{e}$ and $[x,y]=0$ for all
$y\in\mathfrak{\bar{g}}^{e}$. Since $I(m|0)\in\mathfrak{\bar{g}}^{e}$ where
$I(m|0)$ is defined in (3.6), we have that $[x,I(m|0)]=0$ . Therefore, we have
that
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})\subseteq(\mathfrak{\bar{g}}^{e})^{I(m|0)}\subseteq\mathfrak{\bar{g}}^{I(m|0)}$.

Let $x=\begin{pmatrix}A&B\\\ C&D\end{pmatrix}\in\mathfrak{\bar{g}}$. Notice
that $[I(m|0),x]=0$ if and only if $B=C=0$. This implies that
$x\in\mathfrak{\bar{g}}_{\bar{0}}$. Hence, we have
$\mathfrak{\bar{g}}^{I(m|0)}=\mathfrak{\bar{g}}_{\bar{0}}$. Therefore, we
deduce that
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})\subseteq\mathfrak{\bar{g}}_{\bar{0}}$.∎

Yakimova shows that $\mathfrak{z}(\mathfrak{gl}(m+n)^{e})=\langle
I,e,...,e^{l}\rangle$ where $l=\lambda_{1}-1$ in [23, Theorem 2]. Now we use
the above result to work out a basis of
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})$.

######  Theorem 12.

$\mathfrak{z}(\mathfrak{\bar{g}}^{e})$ has a basis
$\\{I,e,...,e^{l}:l=\lambda_{1}-1\\}$.

###### Proof.

Firstly, we denote the Lie algebra $\mathfrak{gl}(m+n)$ by
$\mathfrak{g}^{\prime}$. Note that $\mathfrak{gl}(m+n)$ is isomorphic to
$\mathfrak{\bar{g}}$ as a vector space. We also have that
$\mathfrak{\bar{g}}^{e}=(\mathfrak{g}^{\prime})^{e}$ as a vector space because
$[e,x]=[e,x^{\prime}]$ for $x\in\mathfrak{\bar{g}}$ and
$x^{\prime}\in\mathfrak{g}^{\prime}$. Denote by

$\mathfrak{g}^{\prime}_{0}=\left\\{\begin{pmatrix}A^{\prime}&0\\\
0&D^{\prime}\end{pmatrix}\in\mathfrak{g}^{\prime}:A^{\prime}\in\mathfrak{gl}(m)\text{
and }D^{\prime}\in\mathfrak{gl}(n)\right\\}$

and

$\mathfrak{g}^{\prime}_{1}=\left\\{\begin{pmatrix}0&B^{\prime}\\\
C^{\prime}&0\end{pmatrix}\in\mathfrak{g}^{\prime}:B^{\prime}\text{is a
}m\times n\text{ matrix and }C^{\prime}\text{ is a }n\times m\text{
matrix}\right\\}.$

Then we let
$\mathfrak{z}((\mathfrak{g}^{\prime})^{e})_{0}=\mathfrak{z}((\mathfrak{g}^{\prime})^{e})\cap\mathfrak{g}_{0}^{\prime}$
and
$\mathfrak{z}((\mathfrak{g}^{\prime})^{e})_{1}=\mathfrak{z}((\mathfrak{g}^{\prime})^{e})\cap\mathfrak{g}^{\prime}_{1}$.
We further observe that
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})_{\bar{0}}=\mathfrak{z}((\mathfrak{g}^{\prime})^{e})_{0}$
by definition. Since we already found that
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})\subseteq\mathfrak{\bar{g}}_{\bar{0}}$,
thus $\mathfrak{z}(\mathfrak{\bar{g}}^{e})_{\bar{1}}=0$. Using the same
argument as in Proposition 11, we have that
$\mathfrak{z}((\mathfrak{g}^{\prime})^{e})_{1}=0$. Hence,
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})=\mathfrak{z}(\mathfrak{\bar{g}}^{e})_{\bar{0}}=\mathfrak{z}((\mathfrak{g}^{\prime})^{e})_{0}=\mathfrak{z}((\mathfrak{g}^{\prime})^{e})$.
Therefore, $\mathfrak{z}(\mathfrak{\bar{g}}^{e})=\langle
I,e,...,e^{l}:l=\lambda_{1}-1\rangle$.∎

Now we give a basis for $\mathfrak{z}(\mathfrak{g}^{e})$. We divide our
analysis into two cases: $m\neq n$ and $m=n>1$.

######  Theorem 13.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{sl}(m|n)$
and $\lambda$ be a partition of $(m|n)$ denoted as (3.2). Let
$e\in\mathfrak{g}_{\bar{0}}$ be a nilpotent element determined by the Dynkin
pyramid of $\lambda$, then $\mathfrak{z}(\mathfrak{g}^{e})=\langle
e,...,e^{l}:l=\lambda_{1}-1\rangle$ except for $m=n,n>1$, in which case
$\mathfrak{z}(\mathfrak{g}^{e})=\langle I,e,...,e^{l}:l=\lambda_{1}-1\rangle$.

###### Proof.

When $m\neq n$, we know that
$\mathfrak{\bar{g}}^{e}=\mathfrak{g}^{e}\oplus\mathbb{C}I$ and thus
$\mathfrak{z}(\mathfrak{\bar{g}}^{e})\subseteq\mathfrak{z}(\mathfrak{g}^{e})\oplus\mathbb{C}I$.
Let $x\in\mathfrak{z}(\mathfrak{g}^{e})$, then $[x,y]=0$ for all
$y\in\mathfrak{g}^{e}$. We also know that $x\in\mathfrak{\bar{g}}^{e}$ since
$\mathfrak{g}^{e}\subseteq\mathfrak{\bar{g}}^{e}$. Moreover, we have that
$[x,I]=0$ and $[x,y]=0$ for all $y\in\mathfrak{g}^{e}\oplus\mathbb{C}I$. Thus
we have that
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\mathfrak{z}(\mathfrak{\bar{g}}^{e})$.
Hence, a basis of $\mathfrak{z}(\mathfrak{g}^{e})$ consists of all basis
vectors of $\mathfrak{z}(\mathfrak{\bar{g}}^{e})$ except the identity matrix
$I$. Therefore, we have that $\mathfrak{z}(\mathfrak{g}^{e})=\langle
e,...,e^{l}:l=\lambda_{1}-1\rangle$ and
$\dim\mathfrak{z}(\mathfrak{g}^{e})=\lambda_{1}-1$.

When $m=n,n>1$, for a partition $\lambda=(\lambda_{1},\dots,\lambda_{r+s})$,
we have elements
$\lambda_{j}\xi_{i}^{i,0}-(-1)^{\bar{i}}\lambda_{i}\xi_{j}^{j,0}$ with $i\neq
j$ lie in the Cartan subalgebra $\mathfrak{h}\subseteq\mathfrak{g}$.

Let $S=\langle I,e,...,e^{l}:l=\lambda_{1}-1\rangle$. Clearly
$S\subseteq\mathfrak{z}(\mathfrak{g}^{e})$. We know that
$e^{k}=\sum_{i=1}^{r+s}\xi_{i}^{i,k}$ and $e^{k}\in\mathfrak{g}$ for all
$0\leq k\leq\lambda_{1}-1$. Suppose $x\in\mathfrak{z}(\mathfrak{g}^{e})$ is of
the form

$x=\sum_{1\leq t,h\leq r+s,\max\\{\lambda_{t}-\lambda_{h}\\}\leq
k\leq\lambda_{t}-1}c_{h}^{t,k}\xi_{h}^{t,k}$

where $c_{h}^{t,k}\in\mathbb{C}$ are coefficients. If $r+s\geq 3$, then $x$
commutes with
$\lambda_{j}\xi_{i}^{i,0}-(-1)^{\bar{i}}\lambda_{i}\xi_{j}^{j,0}$ for all
$i,j$. By computing
$[\lambda_{j}\xi_{i}^{i,0}-\left(-1\right)^{\bar{i}}\lambda_{i}\xi_{j}^{j,0},x]$
for $i\neq j$ we have

$[\lambda_{j}\xi_{i}^{i,0}-(-1)^{\bar{i}}\lambda_{i}\xi_{j}^{j,0},x]=\lambda_{j}(\sum_{h,k}c_{h}^{i,k}\xi_{h}^{i,k}-\sum_{t,k}c_{i}^{t,k}\xi_{i}^{t,k})+(-1)^{\bar{i}}\lambda_{i}(\sum_{t,k}c_{j}^{t,k}\xi_{j}^{t,k}-\sum_{h,k}c_{h}^{j,k}\xi_{h}^{j,k}).$

This is equal to $0$, which forces
$\sum_{h,k}c_{h}^{i,k}\xi_{h}^{i,k}=\sum_{t,k}c_{i}^{t,k}\xi_{i}^{t,k}$ and
$\sum_{h,k}c_{h}^{j,k}\xi_{h}^{j,k}=\sum_{t,k}c_{j}^{t,k}\xi_{j}^{t,k}$. This
implies that $c_{h}^{t,k}=0$ for all $h\neq t$ and thus
$x\in\langle\xi_{t}^{t,k}\rangle$.

If $r+s=2$, we have $\lambda_{1}=\lambda_{2}=n>1$. In this case we only deal
with $\xi_{h}^{t,k}$ for $h,t\in\\{1,2\\}$. Note that a basis of
$\mathfrak{g}^{e}$ is
$\\{\xi_{1}^{1,0}+\xi_{2}^{2,0},\xi_{1}^{1,j},\xi_{2}^{2,j},\xi_{1}^{2,k},\xi_{2}^{1,k}:j=1,\dots,n-1,k=0,1,\dots,n-1\\}$.
Hence, an element $y\in\mathfrak{g}^{e}$ is of the form $y=\sum_{1\leq t,h\leq
2,0\leq k\leq n-1}c_{h}^{t,k}\xi_{h}^{t,k}+c(\xi_{1}^{1,0}+\xi_{2}^{2,0})$.
Now suppose $y\in\mathfrak{z}(\mathfrak{g}^{e})$, by computing
$[\xi_{1}^{1,1},y]=\sum_{k=0}^{n-1}c_{2}^{1,k}\xi_{2}^{1,k+1}\pm\sum_{k=0}^{n-1}c_{1}^{2,k}\xi_{1}^{2,k+1}$,
we obtain that $c_{1}^{2,k}=0$ and $c_{2}^{1,k}=0$ for all $k=0,\dots,n-2$.
Then we calculate
$[\xi_{2}^{1,0},y]=c_{1}^{2,n-1}(\xi_{1}^{1,n-1}\pm\xi_{2}^{2,n-1})$, which
implies that $c_{1}^{2,n-1}=0$. Similarly we have that $c_{2}^{1,n-1}=0$.
Therefore, we obtain that
$x\in\langle\xi_{1}^{1,0}+\xi_{2}^{2,0},\xi_{t}^{t,k}:k>0\rangle$.

From above we have that $x=\sum_{t,k}c_{t}^{t,k}\xi_{t}^{t,k}$. Adding an
element of $S$ we may assume that $c_{1}^{1,k}(x)=0$ for all $k$. Suppose
$x\notin S$, then there exist some $c_{i}^{i,k}\neq 0$. Next considering
$\xi_{1}^{i,0}\in\mathfrak{g}^{e}$, we have
$[x,\xi_{1}^{i,0}]=\sum_{t,k}c_{t}^{t,k}(x)[\xi_{t}^{t,k},\xi_{1}^{i,0}]=\sum_{k}c_{i}^{i,k}(x)\xi_{1}^{i,k}\neq
0$, thus $x\notin\mathfrak{z}(\mathfrak{g}^{e})$. Hence
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq S$. Therefore, we deduce that
$\mathfrak{z}(\mathfrak{g}^{e})=S$ as required. This implies that
$\dim\mathfrak{z}(\mathfrak{g}^{e})=\lambda_{1}$. ∎

###  3.4. Proof of theorems

Let $r_{i}$ (resp. $s_{i}$) be the number of boxes with parity $\bar{0}$
(resp. $\bar{1}$) in the $i$th column of the Dynkin pyramid $P$ and denote
$c_{i}=r_{i}+s_{i}$.

In order to prove Theorem 1 for $\mathfrak{g}=\mathfrak{sl}(m|n)$, we first
look at the case that the corresponding $\varDelta$ only has even labels. Note
that labels in labelled Dynkin diagram $\varDelta$ are the horizontal
difference between consecutive boxes in the pyramid. Hence, there is no label
equal to $1$ in $\varDelta$ if and only if $\lambda_{i}-\lambda_{i+1}$ are
even for all $i=1,\dots,r+s$, i.e. $\lambda_{i}$ are all even or all odd.
Based on the way that the labelled Dynkin diagram is constructed, we have that
$n_{2}(\varDelta)=\lambda_{1}-1$. Therefore, we have that
$\dim\text{$\mathfrak{z}$}(\mathfrak{g}^{e})=n_{2}(\varDelta)$ for $m\neq n$
and $\dim\text{$\mathfrak{z}$}(\mathfrak{g}^{e})=n_{2}(\varDelta)+1$ for
$m=n>1$.

Next we turn to look for $\text{$\mathfrak{z}$}(\mathfrak{g}^{h})$. Note that
an element in $\mathfrak{g}^{h}$ is of the form

$\begin{pmatrix}x_{-\lambda_{1}+1}&\cdots&0\\\ \vdots&\ddots&\vdots\\\
0&\cdots&x_{\lambda_{1}-1}\end{pmatrix}$

where $x_{i}\in\mathfrak{gl}(r_{i}|s_{i})$ are block matrices for
$i=-\lambda_{1}+1,-\lambda_{1}+3\dots,\lambda_{1}-3,\lambda_{1}-1$ such that
$\sum_{i}\text{str}(x_{i})=0$. Thus we have that an element in
$\text{$\mathfrak{z}$}(\mathfrak{g}^{h})$ is of the form

$\begin{pmatrix}d_{-\lambda_{1}+1}I(r_{-\lambda_{1}+1}|s_{-\lambda_{1}+1})&\cdots&0\\\
\vdots&\ddots&\vdots\\\
0&\cdots&d_{\lambda_{1}-1}I(r_{\lambda_{1}-1}|s_{\lambda_{1}-1})\end{pmatrix}$

for some $d_{i}\in\mathbb{Z}$ such that $\sum d_{i}(r_{i}-s_{i})=0$. Hence, we
deduce that
$\dim\text{$\mathfrak{z}$}(\mathfrak{g}^{h})=\lambda_{1}-1=n_{2}(\varDelta)$
for $m\neq n$ and
$\dim\text{$\mathfrak{z}$}(\mathfrak{g}^{h})=n_{2}(\varDelta)+1$ for $m=n>1$.

Next we prove Theorem 2 for Lie superalgebras $\mathfrak{sl}(m|n)$. Based on
the way that the labelled Dynkin diagram is constructed, any labelled Dynkin
diagram for $e\in\mathfrak{g}_{\bar{0}}$ is the same as the labelled Dynkin
diagram for $e\in\mathfrak{gl}(m+n)$ except some of the vertices are
$\otimes$, i.e. given a nilpotent element $e\in\mathfrak{g}_{\bar{0}}$, all
the labels $a_{i}$ in the labelled Dynkin diagram with respect to
$\mathfrak{sl}(m|n)$ are the same as that in the labelled Dynkin diagram with
respect to $\mathfrak{gl}(m+n)$ so that $\sum a_{i}$ is also the same. We also
have $a_{i}=\text{col}(i+1)-\text{col}(i)$, thus $\sum
a_{i}=\sum_{i=1}^{m+n-1}\left(\text{col}(i+1)-\text{col}(i)\right)=\text{col}(m+n)-\text{col}(1)=2\lambda_{1}-2=2\dim\text{$\mathfrak{z}$}(\mathfrak{g}^{e})-2$.
Therefore, we have that
$\dim\text{$\mathfrak{z}$}(\mathfrak{g}^{e})=\left\lceil\frac{1}{2}\sum
a_{i}\right\rceil+\varepsilon$ where $\varepsilon=0$ for $m\neq n$ and
$\varepsilon=1$ for $m=n>1$.

To prove Theorem 3 for $\mathfrak{sl}(m|n)$, we define $\mathfrak{g}_{0}$ to
be the subalgebra generated by the root spaces $\mathfrak{g}_{-\alpha}$ and
$\mathfrak{g}_{\alpha}$ for $\alpha$ a simple root with label $0$ or $1$ in
$\varDelta$. We consider two general cases: the labelled Dynkin diagram
$\varDelta$ has no label equal to $1$ and $\varDelta$ has some labels equal to
$1$.

When $\varDelta$ has no label equal to $1$. Note that $e_{0}=0$ since
$\varDelta_{0}$ has all labels equal to $0$. We also have
$\mathfrak{g}_{0}^{e_{0}}=\mathfrak{g}_{0}=\bigoplus_{i\in\mathbb{Z}}\mathfrak{sl}(r_{i}|s_{i})$.
Then
$\dim\mathfrak{g}_{0}^{e_{0}}=\dim\mathfrak{g}_{0}=\sum_{i\in\mathbb{Z}}\dim\mathfrak{sl}(r_{i}|s_{i})=\sum_{i\in\mathbb{Z}:c_{i}>0}(c_{i}^{2}-1)$.
Hence,
$\dim\mathfrak{g}^{e}-\dim\mathfrak{g}_{0}^{e_{0}}=(\sum_{i\in\mathbb{Z}:c_{i}>0}c_{i}^{2}-1)-\sum_{i\in\mathbb{Z}:c_{i}>0}(c_{i}^{2}-1)=-1+\sum_{i\in\mathbb{Z}:c_{i}>0}1=\lambda_{1}-1=n_{2}(\varDelta)$
since there are in total $\lambda_{1}$ columns in $P$ with non-zero boxes.
Moreover, we have that $\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=\tau$ where
$\tau$ is the number of $i$ such that $r_{i}=s_{i}$. Therefore, we obtain that
$\dim\mathfrak{z}(\mathfrak{g}^{e})-\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=n_{2}(\varDelta)-\tau$
for $m\neq n$ and
$\dim\mathfrak{z}(\mathfrak{g}^{e})-\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=n_{2}(\varDelta)+1-\tau$
for $m=n>1$.

When $\varDelta$ has some labels equal to $1$. There are in total
$2\lambda_{1}+1$ columns with labels from $-\lambda_{1}$ to $\lambda_{1}$ in
the Dynkin pyramid $P$. Let $k>0$ be minimal such that $c_{k}=0$ and thus we
know that $n_{2}(\varDelta)=\lambda_{1}-k$. Then we have that

$\displaystyle\mathfrak{g}_{0}\cong\mathfrak{sl}(r_{-\lambda_{1}+1}$
$\displaystyle|s_{-\lambda_{1}+1})\oplus\dots\oplus\mathfrak{sl}(r_{-k-1}|s_{-k-1})\oplus\mathfrak{sl}\left(\sum_{i=-k+1}^{k-1}r_{i}|\sum_{i=-k+1}^{k-1}s_{i}\right)$
$\displaystyle\oplus\mathfrak{sl}(r_{k+1}|s_{k+1})\dots\oplus\mathfrak{sl}(r_{\lambda_{1}-1}|s_{\lambda_{1}-1}).$

Note that the projection of $e_{0}$ in each $\mathfrak{sl}(r_{i}|s_{i})$ is
equal to $0$ for $i>k$ and $i<-k$, thus
$e_{0}\in\mathfrak{sl}\left(\sum_{i=-k+1}^{k-1}r_{i}|\sum_{i=-k+1}^{k-1}s_{i}\right)$.
We know that

$\dim\mathfrak{sl}\left(\sum_{i=-k+1}^{k-1}r_{i}|\sum_{i=-k+1}^{k-1}s_{i}\right)^{{}^{e_{0}}}=\sum_{i=-k+1}^{k-1}c_{i}^{2}+\sum_{i=-k+1}^{k-1}c_{i}c_{i+1}-1$

and $\dim\mathfrak{sl}(r_{i}|s_{i})=c_{i}^{2}-1$. We also know that $P$ is
symmetric, thus

$\dim\mathfrak{g}_{0}^{e_{0}}=\sum_{i=-k+1}^{k-1}c_{i}^{2}+\sum_{i=-k+1}^{k-1}c_{i}c_{i+1}-1+2(c_{k+1}^{2}-1)+\dots+2(c_{\lambda_{1}-1}^{2}-1).$

Therefore, we have
$\dim\mathfrak{g}^{e}-\dim\mathfrak{g}_{0}^{e_{0}}=\lambda_{1}-k=n_{2}(\varDelta).$

Observe that when $r_{i}\neq s_{i}$ for all $i>k$ and $i<-k$ and
$\sum_{i=-k+1}^{k-1}r_{i}\neq\sum_{i=-k+1}^{k-1}s_{i}$, then
$\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=k-1$ as
$\mathfrak{z}(\mathfrak{sl}(r_{i}|s_{i}))=0$. However, when there exist some
$i$ for $i>k$ or $i<-k$ such that $r_{i}=s_{i}$, then
$\dim\mathfrak{z}(\mathfrak{sl}(r_{i}|r_{i}))=1$. Moreover,
$\dim\mathfrak{z}\left(\mathfrak{sl}\left(\sum_{i=-k+1}^{k-1}r_{i}\mid\sum_{i=-k+1}^{k-1}s_{i}\right)^{e_{0}}\right)=k$
if $\sum_{i=-k+1}^{k-1}r_{i}=\sum_{i=-k+1}^{k-1}s_{i}$ and
$\dim\mathfrak{z}\left(\mathfrak{sl}\left(\sum_{i=-k+1}^{k-1}r_{i}\mid\sum_{i=-k+1}^{k-1}s_{i}\right)^{e_{0}}\right)=k-1$
if $\sum_{i=-k+1}^{k-1}r_{i}\neq\sum_{i=-k+1}^{k-1}s_{i}$ by Subsection 3.3.
Let

$\nu_{0}=\begin{cases}0&\text{if
}\sum_{i=-k+1}^{k-1}r_{i}\neq\sum_{i=-k+1}^{k-1}s_{i};\\\ 1&\text{if
}\sum_{i=-k+1}^{k-1}r_{i}=\sum_{i=-k+1}^{k-1}s_{i}.\end{cases}$

Then we have that
$\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=k-1+\tau+\nu_{0}$ where $\tau$ is
the number of $i$ such that $i>k$ or $i<-k$ and $r_{i}=s_{i}$. Therefore, we
have
$\dim\mathfrak{z}(\mathfrak{g}^{e})-\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=n_{2}(\varDelta)-\tau-\nu_{0}$
for $m\neq n$ and
$\dim\mathfrak{z}(\mathfrak{g}^{e})-\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=n_{2}(\varDelta)+1-\tau-\nu_{0}$
for $m=n>1$.

##  4\. The ortho-symplectic Lie superalgebras

###  4.1. Matrix expression of ortho-sympletic Lie superalgebras

Suppose $V=V_{\bar{0}}\oplus V_{\bar{1}}$ is a $\mathbb{Z}_{2}$-graded vector
space over $\mathbb{C}$. Let $B:V\times V\rightarrow V$ be a non-degenerate
even supersymmetric bilinear form on $V$, i.e. $B(V_{\bar{i}},V_{\bar{j}})=0$
unless $\bar{i}+\bar{j}=\bar{0}$, the restriction of $B$ to $V_{\bar{0}}$ is
symmetric and the restriction of $B$ to $V_{\bar{1}}$ is skew-symmetric.
Recall that the ortho-symplectic Lie superalgebra $\mathfrak{osp}(V)$ is
defined to be
$\mathfrak{osp}(V)=\mathfrak{osp}(V)_{\bar{0}}\oplus\mathfrak{osp}(V)_{\bar{1}}$
where

(4.1)
$\mathfrak{osp}(V)_{\bar{i}}:=\\{x\in\mathfrak{gl}(V)_{\bar{i}}:B(x(v),w)=-(-1)^{\bar{i}\bar{v}}B(v,x(w))\text{
for homogeneous }v,w\in V\\}$

for $\bar{i}\in\\{\bar{0},\bar{1}\\}$ and $\bar{v}$ is the parity of $v$. We
write $\mathfrak{osp}(m|2n)$ for $\mathfrak{osp}(V)$ when $\dim V_{\bar{0}}=m$
and $\dim V_{\bar{1}}=2n$. Note that the even part
$\mathfrak{g}_{\bar{0}}=\mathfrak{o}(m)\oplus\mathfrak{sp}(2n)$.

We next explain how to represent $\mathfrak{osp}(m|2n)$ using matrices with
respect to certain choices of basis of $V$. Let
$l=\left\lfloor\frac{m}{2}\right\rfloor$. Given any sequence
$\eta=(\eta_{1},\dots,\eta_{l+n})\in\\{\bar{0},\bar{1}\\}$ such that
$\left|\\{i:\eta_{i}=\bar{0}\\}\right|=l$ and
$\left|\\{i:\eta_{i}=\bar{1}\\}\right|=n$. Then when $m$ is odd, we define the
standard basis $\mathfrak{B}$ of $V$ with respect to $\eta$ to be
$\mathfrak{B}=\\{v_{1}^{\eta_{1}},\dots,v_{l+n}^{\eta_{l+n}},v_{0}^{\bar{0}},v_{-(l+n)}^{\eta_{l+n}},\dots,v_{-1}^{\eta_{1}}\\}$,
where $v_{i}^{\eta_{i}},v_{-i}^{\eta_{i}}\in V_{\bar{0}}$ if
$\eta_{i}=\bar{0}$ and $v_{i}^{\eta_{i}},v_{-i}^{\eta_{i}}\in V_{\bar{1}}$ if
$\eta_{i}=\bar{1}$ for each $i$. When $m$ is even,
$\mathfrak{B}=\\{v_{1}^{\eta_{1}},\dots,v_{l+n}^{\eta_{l+n}},v_{-(l+n)}^{\eta_{l+n}},\dots,v_{-1}^{\eta_{1}}\\}$
is the standard basis of $V$. With the above basis, the non-degenerate even
supersymmetric bilinear form $B$ on $V$ is given by

(4.2) $B(v_{i}^{\eta_{i}},v_{j}^{\eta_{j}})=\begin{cases}0&\text{\text{if}\
}i\neq-j;\\\ 1&\text{\text{if}\ }i=-j,\ \eta_{i}=\bar{0}\text{ or
}\eta_{i}=\bar{1}\text{ and }i>0\\\ -1&\text{\text{if}\ }i=-j,\
\eta_{i}=\bar{1},\ i<0.\end{cases}$

Thus the matrix of $B$ with respect to basis $\mathfrak{B}$ is

$\begin{pmatrix}&&&&&1\\\ &&&&\diagup\\\ &&&1\\\
&&\left(-1\right)^{\eta_{1}}\\\ &\diagup\\\
\left(-1\right)^{\eta_{k+n}}\end{pmatrix}$

Next we explain a basis of $\mathfrak{g}$ corresponding to the above basis
$\mathfrak{B}$ of $V$. Define $E_{j,k}$ to be the linear transformation
sending $v_{k}^{\eta_{k}}$ to $v_{j}^{\eta_{j}}$. Note that
$\mathfrak{osp}(V)_{\bar{0}}$ is spanned by elements of the form $E_{j,-j}$
for $\eta_{j}=\bar{1}$ and $E_{j,k}+\gamma_{-k,-j}E_{-k,-j}$ for
$\eta_{j}=\eta_{k}$ and $j\neq-k$ and $\mathfrak{osp}(V)_{\bar{1}}$ is spanned
by elements of the form $E_{j,k}+\gamma_{-k,-j}E_{-k,-j}$ for
$\eta_{j}\neq\eta_{k}$ where $\gamma_{-k,-j}=\pm 1$ as specified below. We use
equation (4.1) to determine $\gamma_{-k,-j}$. For
$i\in\\{\bar{0},\bar{1}\\},j,k\neq 0$, we have

(4.3)
$B((E_{j,k}+\gamma_{-k,-j}E_{-k,-j})v_{k}^{\eta_{k}},v_{-j}^{\eta_{j}})=-(-1)^{i\eta_{k}}B(v_{k}^{\eta_{k}},(E_{j,k}+\gamma_{-k,-j}E_{-k,-j})v_{-j}^{\eta_{j}}).$

according to (4.1). Then by (4.2) we have that

$\text{LHS of
\eqref{eq:B()}}=B(v_{j}^{\eta_{j}},v_{-j}^{\eta_{j}})=\begin{cases}1&\text{\text{if}\
}\eta_{j}=\bar{0}\text{ or }\eta_{j}=\bar{1},\ j>0;\\\ -1&\text{\text{if}\
}\eta_{j}=\bar{1},\ j<0.\end{cases}$

and

RHS of (4.3)
$\displaystyle=-(-1)^{i\eta_{k}}\gamma_{-k,-j}B(v_{k}^{\eta_{k}},v_{-k}^{\eta_{k}})$
$\displaystyle=\begin{cases}-(-1)^{i\eta_{k}}\gamma_{-k,-j}&\text{\text{if}\
}\eta_{k}=\bar{0}\text{ or }\eta_{k}=\bar{1},\ k>0;\\\
(-1)^{i\eta_{k}}\gamma_{-k,-j}&\text{\text{if}\ }\eta_{k}=\bar{1},\
k<0.\end{cases}$

Hence, for $E_{j,k}+\gamma_{-k,-j}E_{-k,-j}\in\mathfrak{osp}(V)_{\bar{0}}$, we
have that

$\gamma_{-k,-j}=\begin{cases}1&\text{if }\eta_{j}=\eta_{k}=\bar{1}\text{ and
}jk<0,\\\ -1&\text{if }\eta_{j}=\eta_{k}=\bar{0}\text{ or
}\eta_{j}=\eta_{k}=\bar{1}\text{ and }jk>0.\end{cases}$

For $E_{j,k}+\gamma_{-k,-j}E_{-k,-j}\in\mathfrak{osp}(V)_{\bar{1}}$, we have
that $\eta_{j}\neq\eta_{k}$. Let $\text{sign}(j)=1$ for $j>0$ and
$\text{sign}(j)=-1$ for $j<0$. We deduce that

(4.4) $\gamma_{-k,-j}=-\eta_{j}\text{sign}(j)+\eta_{k}\text{sign}(k)\text{ for
all }\eta_{j}\neq\eta_{k}.$

Note that $\eta_{j},\eta_{k}$ in equation (4.4) are viewed as
$0,1\in\mathbb{Z}$.

We further calculate signs in basis elements $e_{0,k}+\gamma_{-k,0}e_{-k,0}$
for $k>0$ and $e_{k,0}+\gamma_{0,-k}e_{0,-k}$ for $k>0$ if they exist.
Applying a similar argument we get that for $k>0$,

(4.5) $\gamma_{-k,0}=\begin{cases}-1&\text{\text{if}\ }\eta_{k}=\bar{0},\\\
1&\text{\text{if}\ }\eta_{k}=\bar{1}.\end{cases}\text{ and }\gamma_{0,-k}=-1.$

Therefore, for the matrix expression of $\mathfrak{osp}(m|2n)$, we fix all
entries above the skew diagonal to have positive sign and then use the above
rules to determine signs of entries below the skew diagonal. More precisely, a
basis of $\mathfrak{osp}(m|2n)$ is

(4.6) $\displaystyle\\{e_{i,-i},e_{j,k}+\gamma_{-k,-j}e_{-k,-j}$
$\displaystyle:\eta_{i}=\bar{1},0<j,k\leq l+n,\text{ or }j=0,k>0,$
$\displaystyle\text{ or }j>0,k=0\text{ or }jk<0,j+k<0\\}$

where $\gamma_{-k,-j}$ is determined as above.

######  Example 14.

For $\mathfrak{osp}(3|2)$, take the basis to be
$\\{v_{1}^{\bar{0}},v_{2}^{\bar{1}},v_{0}^{\bar{0}},v_{-2}^{\bar{1}},v_{-1}^{\bar{0}}\\}$,
then we have

$\mathfrak{osp}(3|2)=\left\\{\begin{pmatrix}a&b&c&d&0\\\ e&f&g&h&-d\\\
k&l&0&-g&-c\\\ r&s&l&-f&b\\\
0&r&-k&-e&-a\end{pmatrix}:a,b,c,d,e,f,g,h,k,l,r,s\in\mathbb{C}\right\\}.$

###  4.2. The ortho-symplectic Dynkin pyramid

Note that the nilpotent $G$-orbits in $\mathfrak{g}_{\bar{0}}$ are
parametrized by the partitions of $(m|2n)$. Let $\lambda$ be a partition of
$(m|2n)$ such that

(4.7) $\lambda=(p|q)=(p_{1},\dots,p_{r}|q_{1},\dots,q_{s})$

where $p$ (resp. $q$) is a partition of $m$ (resp. $2n$), $p_{1}\geq\dots\geq
p_{r},q_{1}\geq\dots\geq q_{s}$ and all even parts of $p$ and all odd parts of
$q$ have even multiplicity. Write
$\lambda=(c^{m_{c}+n_{c}},\dots,1^{m_{1}+n_{1}})$ such that
$m_{i}=|\\{j:p_{j}=i\\}|$ and $n_{i}=|\\{j:q_{j}=i\\}|$. In this subsection,
we recall the ortho-symplectic Dynkin pyramid $P$ for $\lambda$ which is given
in [7, Section 8]. We use the ortho-symplectic Dynkin pyramid to give a nice
representative of the nilpotent orbit and determine the labelled Dynkin
diagram with respect to this orbit. Note that we use a numbering which is
different from that in [7, Section 8].

Recall that $P$ consists of $(m+2n)$ boxes with size $2\times 2$ in the
$xy$-plane and is centrally symmetric about $(0,0)$. Define the row number
(resp. column number) of a box to be the $y$-coordinate (resp. $x$-coordinate)
of the centre of the box. Below we describe the rule to place boxes in the
upper half plane and the rest of the boxes are added to the lower half plane
in a centrally symmetric way.

Firstly, we set the zeroth row to be empty if $m$ is even. If $m$ is odd,
there exist some odd parts appearing with odd multiplicity and we let $a_{1}$
be the largest such part in $\lambda$ for all $p_{i}$. Then we put $a_{1}$
boxes into the zeroth row in the columns $1-a_{1},3-a_{1},\dots,a_{1}-1$. Next
we remove one part of $a_{1}$ from $\lambda$.

Then $p$ becomes a partition that contains an even number of odd parts with
odd multiplicity. Denote these representatives by
$c_{1}>b_{1}>\dots>c_{N}>b_{N}$. In the upper half plane, the rest of the
boxes are added inductively to the next row following the rules below.

Suppose $a_{2}$ is the largest part remaining in $\lambda$. When $m_{a_{2}}$
is odd, then $a_{2}=c_{k}$ for some $k\in\\{1,\dots,N\\}$. We add an even skew
row of length $\frac{c_{k}+b_{k}}{2}$ and put boxes in this row in the columns
$1-b_{k},3-b_{k},\dots,c_{k}-1$ with even parity $\bar{0}$. After that we
remove $c_{k}$ and $b_{k}$ from the partition. Then add
$\left\lfloor\frac{m_{a_{2}}}{2}\right\rfloor$ rows of length $a_{2}$ and
boxes in these rows are placed in the columns $1-a_{2},3-a_{2},\dots,a_{2}-1$
with parity $\bar{0}$. When $m_{a_{2}}$ is even, we only add
$\left\lfloor\frac{m_{a_{2}}}{2}\right\rfloor$ rows in columns
$1-a_{2},3-a_{2},\dots,a_{2}-1$ with parity $\bar{0}$. When $n_{a_{2}}$ is
odd, then an odd skew row of length $\frac{a_{2}}{2}$ is added in the columns
$1,\dots,a_{2}-1$ with boxes labelled by parity $\bar{1}$. We draw a box with
a cross through it to represent each missing box in skew rows. Then we add
$\left\lfloor\frac{n_{a_{2}}}{2}\right\rfloor$ rows of length $a_{2}$ and put
boxes in the columns $1-a_{2},3-a_{2},\dots,a_{2}-1$ with parity $\bar{1}$.
When $n_{a_{2}}$ is even, we only add
$\left\lfloor\frac{n_{a_{2}}}{2}\right\rfloor$ rows in the columns
$1-a_{2},3-a_{2},\dots,a_{2}-1$ with parity $\bar{1}$. Then remove
$a_{2}^{m_{a_{2}}+n_{a_{2}}}$ from $\lambda$.

Let $l=\left\lfloor\frac{m}{2}\right\rfloor$. In this paper, we label boxes in
$P$ down columns from left to right with numbers $1,\dots,l+n,-(l+n),\dots,-1$
such that boxes labelled by $i$ and $-i$ are central symmetrically. For the
case where $m$ is odd we have an additional central box which is labelled by
$0$.

######  Example 15.

The ortho-symplectic Dynkin pyramids for the partitions $(5,3,1|3,3)$ and
$(3,3|4)$ are:

Let $\mathrm{row}(i)$ (resp. $\mathrm{col}(i)$) be the row number (resp.
column number) of the $i$th box. Let
$\left|\mathrm{row}(i)\right|\in\\{\bar{0},\bar{1}\\}$ be the parity of
$\mathrm{row}(i)$ for $i=\pm 1,\dots,\pm(l+n)$ (resp. $i=\pm
1,\dots,\pm(l+n),0$) if $m$ is even (resp. $m$ is odd). Note that the above
numbering of the Dynkin pyramid determines a basis $\mathfrak{B}$ of $V$ which
is given in Subsection 4.1 and $\eta_{i}=\left|\mathrm{row}(i)\right|$.

According to [7, Section 8], the ortho-symplectic pyramid $P$ determines a
nilpotent element $e\in\mathfrak{g}_{\bar{0}}$ and $e=\sum\gamma_{i,j}E_{i,j}$
where the sum is over all $i$ and $j$ such that

1\. $\mathrm{row}(i)=\mathrm{row}(j)$ and $\mathrm{col}(j)=\mathrm{col}(i)-2;$

2\. $\mathrm{row}(i)=-\mathrm{row}(j),\left|\mathrm{row}(i)\right|=\bar{0}$
and the box labelled by $i$ is in an skew-row in the upper half plane,
$\mathrm{col}(i)=2$ and $\mathrm{col}(j)=0$ or $\mathrm{col}(i)=0$ and
$\mathrm{col}(j)=-2$;

3\. $\mathrm{row}(i)=-\mathrm{row}(j),\left|\mathrm{row}(i)\right|=\bar{1}$
and the box labelled by $i$ is in an skew-row in the upper half plane,
$\mathrm{col}(i)=1$ and $\mathrm{col}(j)=-1$.

The pyramid $P$ also defines a semisimple element $h\in\mathfrak{g}_{\bar{0}}$
such that $h$ is the $(m+2n)$-diagonal matrix where the $i$th entry is
$-\mathrm{col}(i)$. Note that $\\{e,h\\}$ can be extended to an
$\mathfrak{sl}(2)$-triple $\\{e,h,f\\}$ in $\mathfrak{g}_{\bar{0}}$ according
to [7, Section 8].

###  4.3. Root system and and labelled Dynkin diagram for
$\mathfrak{osp}(m|2n)$

Let
$\mathfrak{g}=\mathfrak{osp}(m|2n)=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$
and $\mathfrak{h}$ be the set consisting of all diagonal matrices in
$\mathfrak{g}$. A basis of $\mathfrak{h}^{*}$ is given by
{$\varepsilon_{1}^{\eta_{1}}$, $\dots$, $\varepsilon_{l+n}^{\eta_{l+n}}$}
where $\eta_{i}$ are the parities as defined in Subsection 4.1 and
$(\varepsilon_{i}^{\eta_{i}},\varepsilon_{j}^{\eta_{j}})=(-1)^{\eta_{i}}\delta_{ij}$.

According to [14, Section 2.3], for odd $m\geq 1$, $n\geq 1$, the root system
for $\mathfrak{g}$ is given by $\Phi=\Phi_{\bar{0}}\cup\Phi_{\bar{1}}$ such
that

$\Phi_{\bar{0}}=\\{\pm\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}}:\eta_{i}=\eta_{j}\\}\cup\\{\pm\varepsilon_{i}^{\bar{0}}\\}\cup\\{\pm
2\varepsilon_{i}^{\bar{1}}\\},$
$\Phi_{\bar{1}}=\\{\pm\varepsilon_{i}^{\bar{1}}\\}\cup\\{\pm\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}}:\eta_{i}\neq\eta_{j}\\};$

and a choice of positive roots is
$\Phi^{+}=\\{\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}},\varepsilon_{k}^{\eta_{k}},2\varepsilon_{t}^{\bar{1}}\\}$.

For $m=2$, $n\geq 1$, the root system for $\mathfrak{g}$ is given by
$\Phi=\Phi_{\bar{0}}\cup\Phi_{\bar{1}}$ such that

$\Phi_{\bar{0}}=\\{\pm\varepsilon_{i}^{\bar{1}}\pm\varepsilon_{j}^{\bar{1}}\\}\cup\\{\pm
2\varepsilon_{i}^{\bar{1}}\\},\Phi_{\bar{1}}=\\{\pm\varepsilon_{1}^{\eta_{1}}\pm\varepsilon_{j}^{\eta_{j}}:\eta_{1}\neq\eta_{j}\\};$

and a choice of positive roots is
$\Phi^{+}=\\{\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}},2\varepsilon_{t}^{\bar{1}}\\}$.

For even $m>2$, $n\geq 1$, the root system for $\mathfrak{g}$ is given by
$\Phi=\Phi_{\bar{0}}\cup\Phi_{\bar{1}}$ such that

$\Phi_{\bar{0}}=\\{\pm\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}}:\eta_{i}=\eta_{j}\\}\cup\\{\pm
2\varepsilon_{i}^{\bar{1}}\\},\Phi_{\bar{1}}=\\{\pm\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}}:\eta_{i}\neq\eta_{j}\\};$

and a choice of positive roots is
$\Phi^{+}=\\{\varepsilon_{i}^{\eta_{i}}\pm\varepsilon_{j}^{\eta_{j}},2\varepsilon_{t}^{\bar{1}}\\}$.

We have that odd roots $\pm\varepsilon_{i}^{\bar{1}}$ are non-isotropic and
all other odd roots are isotropic.

The labelled Dynkin diagram with respect to $e\in\mathfrak{g}_{\bar{0}}$ is
constructed as follows: firstly, draw the ortho-symplectic Dynkin pyramid $P$
of $\lambda$ following Subsection 4.2. For boxes labelled by
$i=1,\dots,l+n-1$, we associate a white node $\fullmoon$ (resp. a grey node
$\otimes$) to the root $\alpha_{i}$ if
$\left|\mathrm{row}(i+1)\right|=\left|\mathrm{row}(i)\right|$ (resp.
$\left|\mathrm{row}(i+1)\right|\neq\left|\mathrm{row}(i)\right|$) and connect
the $(i-1)$th and $i$th node with a single line. We label the $i$th node with
$a_{i}=\mathrm{col}(i+1)-\mathrm{col}(i)$. For $i=l+n$, we need to consider
different cases.

When $m$ is odd, we associate a white node $\fullmoon$ (resp. a black node
$\newmoon$) to root $\alpha_{l+n}$ if
$\left|\mathrm{row}(l+n)\right|=\left|\mathrm{row}(0)\right|$ (resp.
$\left|\mathrm{row}(l+n)\right|\neq\left|\mathrm{row}(0)\right|$). We connect
the $(l+n-1)$th and $(l+n)$th node with $2$ lines and put an arrow pointing
from the $(l+n-1)$th node to the $(l+n)$th node. The $(l+n)$th node is
labelled by $a_{l+n}=\mathrm{col}(0)-\mathrm{col}(l+n)$.

When $m=2$.

  * •

If $|\mathrm{row}(n+1)|=\bar{1}$, we associate a grey node $\otimes$ to root
$\alpha_{n+1}$. We connect the $n$th and $(n+1)$th node with 2 lines and
connect the $(n-1)$th and $(n+1)$th node with a single line. The $(n+1)$th
node is labelled by $a_{n+1}=-\mathrm{col}(n+1)-\mathrm{col}(n)$.

  * •

If $|\mathrm{row}(n+1)|=\bar{0}$, we associate a white node $\fullmoon$ to
root $\alpha_{n+1}$. We connect the $n$th and $(n+1)$th node with $2$ lines
and put an arrow pointing from the $(n+1)$th node to the $n$th node. The
$(n+1)$th node is labelled by $a_{n+1}=-2\mathrm{col}(n+1)$.

When $m>2$ is even.

  * •

If $|\mathrm{row}(l+n)|=\bar{0}$ and $|\mathrm{row}(l+n-1)|=\bar{1}$, we
associate a grey node $\otimes$ to root $\alpha_{l+n}$. We connect the
$(l+n)$th and the $(l+n-1)$th node with $2$ lines and connect the $(l+n-2)$th
and $(l+n)$th node with a single line. The $(l+n)$th node is labelled by
$-\mathrm{col}(l+n)-\mathrm{col}(l+n-1)$.

  * •

If $|\mathrm{row}(l+n)|=|\mathrm{row}(l+n-1)|=\bar{0}$, we associate a white
node $\fullmoon$ to root $\alpha_{l+n}$ and put a single line between the
$(l+n)$th and the $(l+n-2)$th node. The $(l+n)$th node is labelled by
$a_{l+n}=-2\mathrm{col}(l+n)$.

  * •

If $|\mathrm{row}(l+n)|=\bar{1}$, we associate a white node $\fullmoon$ to
root $\alpha_{l+n}$ and connect the $(l+n)$th and the $(l+n-1)$th node with
$2$ lines. An arrow is pointing from $(l+n)$th node to the $(l+n-1)$th node.
The $(l+n)$th node is labelled by $-\mathrm{col}(l+n)-\mathrm{col}(l+n-1)$.

We describe the corresponding simple roots and draw the labelled Dynkin
diagram for each case in Table LABEL:tab:LDD-osp. Note that the symbol  in
Table LABEL:tab:LDD-osp represents either a white or grey node can appear.

Table 1. Labelled Dynkin diagrams for $\mathfrak{osp}(m|2n)$ |  |
---|---|---
|  sets of simple roots |  labelled Dynkin diagram
$m\geq 1$ is odd |  $\\{\varepsilon_{1}^{\eta_{1}}-\varepsilon_{2}^{\eta_{2}},\dots,\varepsilon_{l+n-1}^{\eta_{l+n-1}}-\varepsilon_{l+n}^{\eta_{l+n}},\varepsilon_{l+n}^{\eta_{l+n}}\\}$ |
$m=2$ |  $\\{\varepsilon_{1}^{\eta_{1}}-\varepsilon_{2}^{\eta_{2}},\dots,\varepsilon_{n}^{\eta_{n}}-\varepsilon_{n+1}^{\eta_{n+1}},2\varepsilon_{n+1}^{\bar{1}}\\}$ |
$m=2$ |  $\\{\varepsilon_{1}^{\eta_{1}}-\varepsilon_{2}^{\eta_{2}},\dots,\varepsilon_{n}^{\bar{1}}-\varepsilon_{n+1}^{\bar{0}},\varepsilon_{n}^{\bar{1}}+\varepsilon_{n+1}^{\bar{0}}\\}$ |
$m>2$ is even |  $\\{\varepsilon_{1}^{\eta_{1}}-\varepsilon_{2}^{\eta_{2}},\dots,\varepsilon_{l+n-1}^{\eta_{l+n-1}}-\varepsilon_{l+n}^{\bar{1}},2\varepsilon_{l+n}^{\bar{1}}\\}$ |
|  $\\{\varepsilon_{1}^{\eta_{1}}-\varepsilon_{2}^{\eta_{2}},\dots,\varepsilon_{l+n-1}^{\bar{1}}-\varepsilon_{l+n}^{\bar{0}},\varepsilon_{l+n-1}^{\bar{1}}+\varepsilon_{l+n}^{\bar{0}}\\}$ |
|  $\\{\varepsilon_{1}^{\eta_{1}}-\varepsilon_{2}^{\eta_{2}},\dots,\varepsilon_{l+n-1}^{\bar{0}}-\varepsilon_{l+n}^{\bar{0}},\varepsilon_{l+n-1}^{\bar{0}}+\varepsilon_{l+n}^{\bar{0}}\\}$ |

######  Example 16.

For the partitions $(5,3,1|3,3)$ and $(3,3|4)$, the corresponding Dynkin
pyramids are shown in Example 15. Then the corresponding labelled Dynkin
diagrams are:

and

respectively.

######  Remark 17.

Similar to Dynkin pyramid, different numberings within columns for a ortho-
symplectic Dynkin pyramid are possible and would lead to different labelled
Dynkin diagrams. In this way one can get all labelled Dynkin diagram. The
following example shows we get different labelled Dynkin diagram if we allow
different numberings within columns.

######  Example 18.

For the partition $(5,3,1|3,3)$, if we follow the principle of numbering
within columns given in this subsection, the corresponding labelled Dynkin
diagram is given in Example 15. However, if we choose a different numbering as
shown below, the corresponding labelled Dynkin diagram shown below is
different from that is in Example 15.

###  4.4. Alternative Dynkin pyramid for $\lambda$

In this subsection, we use an alternative notation for $\lambda$ and rewrite

(4.8)
$\lambda=(\lambda_{1},\dots,\lambda_{a},\lambda_{a+1},\lambda_{-(a+1)},\dots,\lambda_{b},\lambda_{-b})$

where $\lambda_{1},\dots,\lambda_{a}$ are the parts with odd multiplicity,
$\lambda_{1}>\lambda_{2}>\dots>\lambda_{a}$ and
$\lambda_{a+1}=\lambda_{-(a+1)}\geq\dots\geq\lambda_{b}=\lambda_{-b}$. We
define $\left|i\right|\in\\{\bar{0},\bar{1}\\}$ such that for
$c\in\mathbb{Z}$, we have
$\left|\\{i:\lambda_{i}=c,\left|i\right|=\bar{0}\\}\right|=\left|\\{j:p_{j}=c\\}\right|$
and
$\left|\\{i:\lambda_{i}=c,\left|i\right|=\bar{1}\\}\right|=\left|\\{j:q_{j}=c\\}\right|$
for some $j$. Next we establish a Dynkin pyramid $\tilde{P}$ which is
different from that we used in the previous section. We use $\tilde{P}$ to
determine a basis for $\mathfrak{g}^{e}$ in Subsection 4.5.

The new version of Dynkin pyramid $\tilde{P}$ consists of $(m+2n)$ boxes with
size $2\times 2$ in the $xy$-plane and is centred on $(0,0)$. We label rows
from $1$ to $b$ in the upper half plane which is different from the way we
used in Subsection 4.2. For each $\lambda_{i}$ with $i>0$, we put
$\lambda_{i}$ boxes both into the $i$th row and $-i$th row in the columns
$1-\lambda_{i},3-\lambda_{i},\dots,\lambda_{i}-1$. We start with the parts
$\lambda_{1},\dots,\lambda_{a}$. For $1\leq i\leq a$, we cross out
$\lfloor\frac{\lambda_{i}}{2}\rfloor$ boxes in the $i$th row from left to
right and cross out $\lceil\frac{\lambda_{i}}{2}\rceil$ boxes in the $-i$th
row from right to left. If $\lambda_{i}$ is odd (resp. even), we label boxes
without cross in the $i$th row from left to right with
$i_{0},i_{2},\dots,i_{\lambda_{i}-1}$ (resp.
$i_{1},i_{3},\dots,i_{\lambda_{i}-1}$) and boxes without cross in the $-i$th
row from left to right with $i_{-(\lambda_{i}-1)},\dots,i_{-2}$ (resp.
$i_{-(\lambda_{i}-1)},\dots,i_{-1}$). Then we deal with the parts
$\lambda_{a+1},\lambda_{-(a+1)},\dots,\lambda_{b},\lambda_{-b}$. For $a+1\leq
i\leq b$, we label boxes in the $i$th row with
$i_{1-\lambda_{i}},i_{3-\lambda_{i}},\dots,i_{\lambda_{i}-1}$ and boxes in the
$-i$th row are labelled by
$-i_{1-\lambda_{i}},-i_{3-\lambda_{i}},\dots,-i_{\lambda_{i}-1}$. Let
$\left|\mathrm{row}(i)\right|$ be the parity of the $i$th row such that
$\left|\mathrm{row}(i)\right|=\left|i\right|$ and $\left|i\right|$ is defined
in (4.8).

Note that from the above Dynkin pyramid $\tilde{P}$ we get a basis

(4.9) $\\{v_{i_{j}}:i_{j}\text{ is a box in }\tilde{P}\\}$

of $V$. More precisely, basis elements $\\{v_{i_{j}}:1\leq i\leq a,0\leq
j\leq\lambda_{i}-1\text{ for odd }\lambda_{i}\text{ and }1\leq
j\leq\lambda_{i}\text{ for even }\lambda_{i}\\}\cup\\{v_{i_{j}}:a+1\leq i\leq
b\\}$ (resp. $\\{v_{i_{j}}:1\leq i\leq a,1-\lambda_{i}\leq j<0\text{ for odd
}\lambda_{i}\text{ and }-\lambda_{i}\leq j\leq-1\text{ for even
}\lambda_{i}\\}\cup\\{v_{-i_{j}}:a+1\leq i\leq b\\}$) correspond to the boxes
in the upper (lower) half of $\tilde{P}$. The bilinear form $B(\ldotp,\ldotp)$
on $V$ is given by

$B(v_{i_{j}},v_{k_{l}})=\begin{cases}\pm 1&\text{if }i=\pm k,j=-l\\\
0&\text{otherwise,}\end{cases}$

and signs will be given explicitly later in (4.10) and (4.11). Note that
$\tilde{P}$ also gives a nilpotent element $e\in\mathfrak{g}_{\bar{0}}$ such
that $e$ sends $v_{i_{j}}$ to $v_{i_{j-2}}$ if there exists a box labelled by
$v_{i_{j-2}}$ and sends $v_{i_{j}}$ to zero if there no such box.

######  Example 19.

The alternative Dynkin pyramid $\tilde{P}$ for the partition
$\lambda=(5,3|2^{2})$ is

Figure 4.1. Dynkin pyramid for the partition $\lambda=(5,3|2^{2})$

Define $u_{i}=v_{i_{\lambda_{i}-1}}\in V$ and thus
$e^{j}u_{i}=v_{i_{\lambda_{i}-2j-1}}$. Then the vectors $e^{j}u_{i}$ with
$\left|i\right|=\bar{0},\ 0\leq j\leq\lambda_{i}-1$ form a basis for
$V_{\bar{0}}$ and $e^{j}u_{i}$ with $\left|i\right|=\bar{1},\ 0\leq
j\leq\lambda_{i}-1$ form a basis for $V_{\bar{1}}$. Moreover, we have
$e^{\lambda_{i}}u_{i}=0$ for all $i$ and they satisfy the following
conditions:

(1) For $i=1,\dots,a$, we have

(4.10) $B(e^{k}u_{i},e^{h}u_{j})=\begin{cases}(-1)^{k}&\text{if }i=j\text{ and
}k+h=\lambda_{i}-1,\\\ 0,&\text{otherwise},\end{cases}$

(2) For $i=\pm(a+1),\dots,\pm b$, then there exists $\theta_{i}\in\\{-1,1\\}$
such that

(4.11) $B(e^{k}u_{i},e^{h}u_{j})=\begin{cases}(-1)^{k}\theta_{i}&\text{if
}-i=j\text{ and }k+h=\lambda_{i}-1,\\\ 0,&\text{otherwise}.\end{cases}$

###  4.5. Centralizer of nilpotent element $e\in\mathfrak{g}_{\bar{0}}$

In this subsection, we give a basis for $\mathfrak{g}^{e}$ based on [23,
Chapter 1] and then state an alternative formula to the formula given in [7,
Subsection 3.2.2] for $\dim\mathfrak{g}^{e}$. In order to describe a basis for
$\mathfrak{g}^{e}$, we define $i^{*}=i$ for $i=1,\dots,a$ and $i^{*}=-i$ for
$i=\pm(a+1),\dots,\pm b$.

Recall that a basis for $\mathfrak{gl}(m|n)^{e}$ is known in terms of
$\xi_{i}^{j,k}$ such that $\xi_{i}^{j,k}$ sends $u_{i}$ to $e^{k}u_{j}$ and
all other $u_{t}$ to $0$. We know that
$\mathfrak{g}^{e}=\mathfrak{g}\cap\mathfrak{gl}(m|2n)^{e}$. Therefore, the
elements in a basis of $\mathfrak{g}_{\bar{0}}^{e}$ are of the form:

$\xi_{i}^{i^{*},\lambda_{i}-1-k}\text{ for }0\leq k\leq\lambda_{i}-1,\ k\text{
is odd if }\left|i\right|=\bar{0}\text{ and }k\text{ is even if
}\left|i\right|=\bar{1};$

$\xi_{i}^{j,\lambda_{j}-1-k}+\varepsilon_{i}^{j,\lambda_{j}-1-k}\xi_{j^{*}}^{i^{*},\lambda_{i}-1-k}\text{
for all }0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}-1,\left|i\right|=\left|j\right|$

where $\varepsilon_{i}^{j,\lambda_{j}-1-k}\in\\{\pm 1\\}$ can be determined by
using [8, Section 3.2]. More precisely, we have that

$\varepsilon_{i}^{j,\lambda_{j}-1-k}=(-1)^{\lambda_{j}-k}\theta_{j}\theta_{i}\text{
for all }0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}-1,\left|i\right|=\left|j\right|.$

Elements in a basis of $\mathfrak{g}_{\bar{1}}^{e}$ are of the form

$\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{j^{*}}^{i^{*},\lambda_{i}-1-k}$

with appropriate choices of signs for $0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}-1$, $\left|i\right|\neq\left|j\right|$.

Write $e=e_{\mathfrak{o}}+e_{\mathfrak{sp}}$ where
$e_{\mathfrak{o}}\in\mathfrak{o}(m)$ and
$e_{\mathfrak{sp}}\in\mathfrak{sp}(2n)$, we know that
$\dim\mathfrak{g}_{\bar{0}}^{e}=\dim\mathfrak{o}(m)^{e_{\mathfrak{o}}}+\dim\mathfrak{sp}(2n)^{e_{\mathfrak{sp}}}$.
By [8, Section 3.2], we have that
$\dim\mathfrak{o}(m)^{e_{\mathfrak{o}}}=\frac{1}{2}\dim\mathfrak{gl}(m)^{e_{\mathfrak{o}}}-\frac{1}{2}\left|\\{i:\lambda_{i}\text{
is odd},\left|i\right|=\bar{0}\\}\right|$ and
$\dim\mathfrak{sp}(2n)^{e_{\mathfrak{sp}}}=\frac{1}{2}\dim\mathfrak{gl}(2n)^{e_{\mathfrak{sp}}}+\frac{1}{2}\left|\\{i:\lambda_{i}\text{
is odd},\left|i\right|=\bar{1}\\}\right|$. Hence, we obtain that

$\dim\mathfrak{g}_{\bar{0}}^{e}=\frac{1}{2}\dim\mathfrak{gl}(m|2n)_{\bar{0}}^{e}-\frac{1}{2}\left|\\{i:\lambda_{i}\text{
is
odd},\left|i\right|=\bar{0}\\}\right|+\frac{1}{2}\left|\\{i:\lambda_{i}\text{
is odd},\left|i\right|=\bar{1}\\}\right|.$

In addition, Hoyt argues that
$\dim\mathfrak{g}_{\bar{1}}^{e}=\frac{1}{2}\dim\mathfrak{gl}(m|2n)_{\bar{1}}^{e}$
in [7, Section 3.2.2]. Therefore, we have that

(4.12)
$\dim\mathfrak{g}^{e}=\frac{1}{2}\dim\mathfrak{gl}(m|2n)^{e}-\frac{1}{2}\left|\\{i:\lambda_{i}\text{
is
odd},\left|i\right|=\bar{0}\\}\right|+\frac{1}{2}\left|\\{i:\lambda_{i}\text{
is odd},\left|i\right|=\bar{1}\\}\right|.$

We obtain an alternative formula for $\dim\mathfrak{g}^{e}$ below.

######  Proposition 20.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{osp}(m|2n)$
and $\lambda$ be a partition of $(m|2n)$. Denote by $P$ the ortho-symplectic
Dynkin pyramid of $\lambda$. Then $P$ determines an $\mathfrak{sl}(2)$-triple
$\\{e,h,f\\}$ in $\mathfrak{g}_{\bar{0}}$. Let $c_{i}$ be the number of boxes
in the $i$th column of $P$ and $r_{i}$ (resp. $s_{i}$) be the number of boxes
with parity $\bar{0}$ (resp. $\bar{1}$) in the $i$th column of $P$. We have
that $\dim\mathfrak{g}^{e}=\frac{1}{2}(\sum c_{i}^{2}+\sum
c_{i}c_{i+1})-\frac{r_{0}}{2}+\frac{s_{0}}{2}$.

###### Proof.

Let $\mathfrak{g}=\bigoplus_{j\in\mathbb{Z}}\mathfrak{g}(j)$. Using the same
argument as in the proof of Proposition 9, we get
$\dim\mathfrak{g}^{e}=\dim\mathfrak{g}(0)+\dim\mathfrak{g}(-1)$. For an
element $E_{k,l}+\gamma_{l,k}E_{-l,-k}\in\mathfrak{g}$, we calculate that
$[h,E_{k,l}+\gamma_{l,k}E_{-l,-k}]=\left(\mathrm{col}(l)-\mathrm{col}(k)\right)(E_{k,l}+\gamma_{l,k}E_{-l,-k})$,
this implies that $E_{k,l}+\gamma_{l,k}E_{-l,-k}\in\mathfrak{g}(j)$ if
$j=\mathrm{col}(l)-\mathrm{col}(k)$. Hence, we have that

$\mathfrak{g}(0)=\langle
E_{k,l}+\gamma_{l,k}E_{-l,-k}:\mathrm{col}(l)=\mathrm{col}(k)\rangle\cong(\bigoplus_{i<0}\mathfrak{gl}(r_{i}|s_{i}))\oplus\mathfrak{osp}(r_{0}|s_{0}).$

Moreover, we know that

$\dim\mathfrak{osp}(r_{0}|s_{0})=\frac{(r_{0}+s_{0})^{2}-(r_{0}+s_{0})}{2}+s_{0}=\frac{c_{0}^{2}-r_{0}+s_{0}}{2}.$

Thus we have that
$\dim\mathfrak{g}(0)=\sum_{i<0}c_{i}^{2}+\frac{c_{0}^{2}}{2}-\frac{r_{0}}{2}+\frac{s_{0}}{2}=\frac{1}{2}\sum
c_{i}^{2}-\frac{r_{0}}{2}+\frac{s_{0}}{2}$. Observe that

$\mathfrak{g}(-1)=\langle
E_{k,l}+\gamma_{l,k}E_{-l,-k}:\mathrm{col}(l)-\mathrm{col}(k)=-1\rangle\cong\bigoplus_{i<0}\mathrm{Hom}(\mathbb{C}^{c_{i}},\mathbb{C}^{c_{i+1}}).$

Hence, we have that $\dim\mathfrak{g}(-1)=\sum_{i<0}c_{i}c_{i+1}$.

For each row which corresponds to an odd $\lambda_{i}$, there must exist a box
in the $0$th column. Thus we have that $r_{0}=\left|\\{i:\lambda_{i}\text{ is
odd},\left|i\right|=\bar{0}\\}\right|$ and $s_{0}=\left|\\{i:\lambda_{i}\text{
is odd},\left|i\right|=\bar{1}\\}\right|$. Therefore, we deduce that
$\dim\mathfrak{g}^{e}=\frac{1}{2}(\sum c_{i}^{2}+\sum
c_{i}c_{i+1})-\frac{r_{0}}{2}+\frac{s_{0}}{2}$. ∎

###  4.6. Centre of centralizer of nilpotent element
$e\in\mathfrak{g}{}_{\bar{0}}$ with Jordan type $\lambda$ such that all parts
of $\lambda$ have even multiplicity

We know that there is a root space decomposition
$\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in\Phi}\mathfrak{g}_{\alpha}$
and thus
$\mathfrak{z}(\mathfrak{g})\subseteq\mathfrak{g}^{\mathfrak{h}}=\mathfrak{h}$.
Construct a Dynkin pyramid $\tilde{P}$ following the rules described in
Subsection 4.4. We know that $\tilde{P}$ determines a nilpotent element
$e\in\mathfrak{g}_{\bar{0}}$ and we can embed $e$ into an
$\mathfrak{sl}(2)$-triple $\\{e,h,f\\}\subseteq\mathfrak{g}_{\bar{0}}$ by the
Jacobson–Morozov theorem. Then we have that the centralizer $\mathfrak{h}^{e}$
of $e$ in $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}^{e}$
according to [1, Section 3]. Moreover, [1, Lemma 13] shows that the set of
weights of $\mathfrak{h}^{e}$ on $\mathfrak{g}^{e}$ is equivalent to the set
of weights of $\mathfrak{h}^{e}$ on $\mathfrak{g}$. Then we can obtain the
following decomposition for $\mathfrak{g}^{e}$:

$\mathfrak{g}^{e}=(\mathfrak{g}^{e})^{\mathfrak{h}^{e}}\oplus\bigoplus_{\alpha\in\Phi^{e}}\mathfrak{g}_{\alpha}^{e}$

where $\Phi^{e}\subseteq(\mathfrak{h}^{e})^{*}$ is defined as the set of non-
zero weights of $\mathfrak{h}^{e}$ on $\mathfrak{g}^{e}$ and
$\mathfrak{g}_{\alpha}^{e}=\\{x\in\mathfrak{g}^{e}:[h,x]=\alpha(h)x\ \text{for
all }h\in\mathfrak{h}^{e}\\}$. Hence, we have
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq(\mathfrak{g}^{e})^{\mathfrak{h}^{e}}.$

We first consider the case where all parts of the Jordan type
$\lambda=(\lambda_{1},\lambda_{-1},\dots,\lambda_{b},\lambda_{-b})$ with
respect to $e$ have even multiplicity. Then we know that there are $2b$ rows
in the ortho-symplectic Dynkin pyramid $P$ and we label rows in the upper half
plane from bottom to top by $1,2,\dots,b$ and rows in the lower half plane in
a symmetric way. Let $S$ be the set spanned by the odd powers of $e$, i.e.
$S=\langle
e,e^{3},\dots,e^{t}:t=2\lfloor\frac{\lambda_{1}}{2}\rfloor-1\rangle$.

######  Theorem 21.

Let
$\mathfrak{g}=\mathfrak{osp}(m|2n)=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}$
and the Jordan type with respect to $e\in\mathfrak{g}_{\bar{0}}$ is
$\lambda=(\lambda_{1},\lambda_{-1},\dots,\lambda_{b},\lambda_{-b})$ such that
$\lambda_{1}\geq\dots\geq\lambda_{b}$ and $\lambda_{i}=\lambda_{-i}$. Then
$\mathfrak{z}(\mathfrak{g}^{e})=S$ except when $\lambda_{1}$ is odd and
$\lambda_{1}>\lambda_{i}$ for $i\neq\pm 1$ and $|1|=\bar{0}$. In which case,
we have that
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}\rangle$.

###### Proof.

This proof will proceed in steps. It is clear that
$S\subseteq\mathfrak{z}(\mathfrak{g}^{e})$. We know that
$e^{l}=\sum_{j=1}^{b}(\xi_{j}^{j,l}\pm\xi_{-j}^{-j,l})$ and
$e^{l}\in\mathfrak{g}_{\bar{0}}$ for all odd $l$ with $0\leq l<\lambda_{1}$.

Step 1: Deduce that
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\langle\xi_{j}^{j,\lambda_{j}-1-k}+\varepsilon_{j}^{j,\lambda_{j}-1-k}\xi_{-j}^{-j,\lambda_{j}-1-k}:1\leq
j\leq b,0\leq k\leq\lambda_{j}-1\rangle$.

Define $h_{i}=\xi_{i}^{i,0}-\xi_{-i}^{-i,0}$ for all $0\leq i\leq b$. Then we
have $\mathfrak{h}^{e}=\langle h_{i}:0\leq i\leq b\rangle$. We define
$\beta_{i}\in(\mathfrak{h}^{e})^{*}$ by

$\beta_{i}(h_{j})=\begin{cases}1&\text{if }i=j;\\\ 0&\text{if }i\neq
j,\end{cases}$

and $\beta_{-i}=-\beta_{i}$ for all $i$. As in the proof of [23, Theorem 2],
we can calculate that
$[\xi_{i}^{i,0},\xi_{j}^{t,s}]=\delta_{it}\xi_{j}^{i,s}-\delta_{ij}\xi_{i}^{t,s}.$
Thus for $h\in\mathfrak{h}^{e}$, computing the commutator between $h$ and the
basis element
$\xi_{l}^{j,\lambda_{j}-1-k}+\varepsilon_{l}^{j,\lambda_{j}-1-k}\xi_{-j}^{-l,\lambda_{l}-1-k}$
( $l,j\geq 0,$ $l\neq j$ ) of $\mathfrak{g}^{e}$ we obtain:

(4.13)
$[h,\xi_{l}^{j,\lambda_{j}-1-k}+\varepsilon_{l}^{j,\lambda_{j}-1-k}\xi_{-j}^{-l,\lambda_{l}-1-k}]=(\beta_{j}-\beta_{l})(h)(\xi_{l}^{j,\lambda_{j}-1-k}+\varepsilon_{l}^{j,\lambda_{j}-1-k}\xi_{-j}^{-l,\lambda_{l}-1-k}).$

We also compute the commutator between $h$ and the basis element
$\xi_{l}^{-j,\lambda_{j}-1-k}+\varepsilon_{l}^{-j,\lambda_{j}-1-k}\xi_{j}^{-l,\lambda_{l}-1-k}$
( $l,j\geq 0,$ $l\neq j$ ) of $\mathfrak{g}^{e}$:

(4.14)
$[h,\xi_{l}^{-j,\lambda_{j}-1-k}+\varepsilon_{l}^{-j,\lambda_{j}-1-k}\xi_{j}^{-l,\lambda_{l}-1-k}]=(-\beta_{j}-\beta_{l})(h)(\xi_{l}^{-j,\lambda_{j}-1-k}+\varepsilon_{l}^{-j,\lambda_{j}-1-k}\xi_{j}^{-l,\lambda_{l}-1-k})$

and the commutator between $h$ and the basis element
$\xi_{-l}^{j,\lambda_{j}-1-k}+\varepsilon_{-l}^{j,\lambda_{j}-1-k}\xi_{-j}^{l,\lambda_{l}-1-k}$
( $l,j\geq 0,$ $l\neq j$ ) of $\mathfrak{g}^{e}$:

$[h,\xi_{-l}^{j,\lambda_{j}-1-k}+\varepsilon_{-l}^{j,\lambda_{j}-1-k}\xi_{-j}^{l,\lambda_{l}-1-k}]=(\beta_{j}+\beta_{l})(h)(\xi_{-l}^{j,\lambda_{j}-1-k}+\varepsilon_{-l}^{j,\lambda_{j}-1-k}\xi_{-j}^{l,\lambda_{l}-1-k}).$

Hence, the coefficient of
$\xi_{l}^{j,\lambda_{j}-1-k}+\varepsilon_{l}^{j,\lambda_{j}-1-k}\xi_{-j}^{-l,\lambda_{l}-1-k}$
in an element of $(\mathfrak{g}^{e})^{\mathfrak{h}^{e}}$ can be nonzero if and
only if $(\beta_{j}-\beta_{l})(h)=0$ for all $h\in\mathfrak{h}^{e}$, the
coefficient of
$\xi_{l}^{-j,\lambda_{j}-1-k}+\varepsilon_{l}^{-j,\lambda_{j}-1-k}\xi_{j}^{-l,\lambda_{l}-1-k}$
in an element of $(\mathfrak{g}^{e})^{\mathfrak{h}^{e}}$ can be nonzero if and
only if $(-\beta_{j}-\beta_{l})(h)=0$ for all $h\in\mathfrak{h}^{e}$ and the
coefficient of
$\xi_{-l}^{j,\lambda_{j}-1-k}+\varepsilon_{-l}^{j,\lambda_{j}-1-k}\xi_{-j}^{l,\lambda_{l}-1-k}$
in an element of $(\mathfrak{g}^{e})^{\mathfrak{h}^{e}}$ can be nonzero if and
only if $(\beta_{j}+\beta_{l})(h)=0$. Take $h=h_{j}$, we obtain that
$(\beta_{j}-\beta_{l})(h)=1$ and $(-\beta_{j}-\beta_{l})(h)=-1$ for $l\neq j$.
Therefore, we deduce that

$(\mathfrak{g}^{e})^{\mathfrak{h}^{e}}=\langle\xi_{j}^{j,\lambda_{j}-1-k}+\varepsilon_{j}^{j,\lambda_{j}-1-k}\xi_{-j}^{-j,\lambda_{j}-1-k}\rangle$

and thus
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\langle\xi_{j}^{j,\lambda_{j}-1-k}+\varepsilon_{j}^{j,\lambda_{j}-1-k}\xi_{-j}^{-j,\lambda_{j}-1-k}\rangle$.

We now have that an element $x$ in $\mathfrak{z}(\mathfrak{g}^{e})$ is of the
form
$\sum_{j,k}c_{j}^{j,k}(\xi_{j}^{j,\lambda_{j}-1-k}+\varepsilon_{j}^{j,\lambda_{j}-1-k}\xi_{-j}^{-j,\lambda_{j}-1-k})$
from Step 1. Fix $j$ and $k$ and let $l=\lambda_{j}-1-k$.

Step 2: Show that $c_{j}^{j,l}=0$ whenever $l$ is even except in one special
case.

According to Equation (1) in [8, Section 3.2], we have that

$\varepsilon_{j}^{j,l}=\begin{cases}1&\text{if }l\text{ is odd;}\\\
-1&\text{if }l\text{ is even.}\end{cases}$

Next we consider an element $\xi_{i}^{-i,0}$. Note that
$\xi_{i}^{-i,0}\in\mathfrak{g}^{e}$ if $\lambda_{i}$ is even and
$\left|i\right|=\bar{0}$, or $\lambda_{i}$ is odd and
$\left|i\right|=\bar{1}$. Hence, when $\xi_{i}^{-i,0}\in\mathfrak{g}^{e}$, the
commutator between $x$ and $\xi_{i}^{-i,0}$ is:

$[\xi_{i}^{-i,0},\sum_{j,l}c_{j}^{j,l}(\xi_{j}^{j,l}+\varepsilon_{j}^{j,l}\xi_{-j}^{-j,l})]=\sum
c_{j}^{j,l}(\xi_{j}^{-j,l}-\varepsilon_{j}^{j,l}\xi_{j}^{-j,l}).$

Hence, we deduce that $c_{j}^{j,l}=0$ whenever $l$ is even. When
$\xi_{i}^{-i,0}\notin\mathfrak{g}^{e}$, i.e. $\lambda_{i}$ is odd and
$\left|i\right|=\bar{0}$, or $\lambda_{i}$ is even and
$\left|i\right|=\bar{1}$. We take commutator between $x$ and $\xi_{i}^{-i,1}$:

$[\xi_{i}^{-i,1},\sum_{j,l}c_{j}^{j,l}(\xi_{j}^{j,l}+\varepsilon_{j}^{j,l}\xi_{-j}^{-j,l})]=\sum
c_{j}^{j,l}(\xi_{j}^{-j,l+1}-\varepsilon_{j}^{j,l}\xi_{j}^{-j,l+1}).$

Hence, we deduce that $c_{j}^{j,l}=0$ for all $l$ is even, except when
$l=\lambda_{j}-1$ and $\left|i\right|=\bar{0}$. For $l$ even and
$l=\lambda_{j}-1$, suppose that there exist some $\lambda_{i}$ with $i>0$ such
that $\lambda_{i}\geq\lambda_{j}$, then we can compute

$[\xi_{i}^{j,0}+\varepsilon_{i}^{j,0}\xi_{-j}^{-i,0},\sum_{j}c_{j}^{j,\lambda_{j}-1}(\xi_{j}^{j,\lambda_{j}-1}-\xi_{-j}^{-j,\lambda_{j}-1})]=-\sum_{j}c_{j}^{j,\lambda_{j}-1}(\varepsilon_{i}^{j,0}\xi_{-j}^{-i,\lambda_{j}-1}+\xi_{i}^{j,\lambda_{j}-1}),$

which implies that $c_{j}^{j,\lambda_{j}-1}=0$. Thus
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\langle\xi_{j}^{j,l}+\xi_{-j}^{-j,l}:l\text{
is odd}\rangle$ except for $\lambda_{1}$ is odd with $\left|1\right|=\bar{0}$
such that $\lambda_{1}>\lambda_{i}$ for $i\neq\pm 1$, in which case we cannot
show that $c_{1}^{1,\lambda_{1}-1}=0$ and
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\langle\xi_{j}^{j,l}+\xi_{-j}^{-j,l}:l\text{
is
odd}\rangle\oplus\langle\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}\rangle$.

Step 3: Show that $c_{i}^{i,l}=c_{t}^{t,l}$ for all $i,t$ whenever $l$ is odd.

Now consider
$\xi_{i}^{t,0}+\varepsilon_{i}^{t,0}\xi_{-t}^{-i,\lambda_{i}-\lambda_{t}}\in\mathfrak{g}^{e}$
with $i<t$ and $i,t>0$. Then we compute that

$\displaystyle\left[\xi_{i}^{t,0}+\varepsilon_{i}^{t,0}\xi_{-t}^{-i,\lambda_{i}-\lambda_{t}},x\right]$
$\displaystyle=\left[\xi_{i}^{t,0}+\varepsilon_{i}^{t,0}\xi_{-t}^{-i,\lambda_{i}-\lambda_{t}},\sum_{j,l}c_{j}^{j,l}(\xi_{j}^{j,l}+\xi_{-j}^{-j,l})\right]$
$\displaystyle=\sum_{l}(c_{i}^{i,l}-c_{t}^{t,l})\xi_{i}^{t,l}+\sum_{l}(c_{t}^{t,l}-c_{i}^{i,l})\varepsilon_{i}^{t,0}\xi_{-t}^{-i,\lambda_{i}-\lambda_{t}+l}.$

This equals to zero if and only if $c_{i}^{i,l}=c_{t}^{t,l}$ for all $i$ and
$t$. Hence, $\mathfrak{z}(\mathfrak{g}^{e})\subseteq S$ and therefore
$\mathfrak{z}(\mathfrak{g}^{e})=S$ except for $\lambda_{1}>\lambda_{i}$ for
$i\neq\pm 1$ and $\left|1\right|=\bar{0}$.

Step 4: Show that
$\xi_{1}^{1,\lambda_{1}-1}-\xi_{1^{*}}^{1^{*},\lambda_{1}-1}\in\mathfrak{z}(\mathfrak{g}^{e})$
when $\lambda_{1}$ is odd with $\left|1\right|=\bar{0}$ such that
$\lambda_{1}>\lambda_{i}$ for $i\neq\pm 1$.

Suppose $\lambda_{1}$ is odd with $\left|1\right|=\bar{0}$ and
$l=\lambda_{1}-1$. Suppose $\lambda_{1}>\lambda_{i}$ for $i\neq\pm 1$. Clearly
$\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}$ commutes with all
basis elements in $\mathfrak{g}^{e}$ of the form
$\xi_{i}^{-i,\lambda_{i}-1-k}$ for $i\neq\pm 1$ and
$\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{-i,\lambda_{i}-1-k}$ for $i,j\neq\pm
1$.

It remains to check whether
$\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}$ commutes with
$\xi_{1}^{-1,\lambda_{1}-1-k}$, $\xi_{-1}^{1,\lambda_{1}-1-k}$ for $0\leq
k\leq\lambda_{1}-1$, $k$ is odd,
$\xi_{1}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{-1,\lambda_{1}-1-k}$,
$\xi_{-1}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{1,\lambda_{1}-1-k}$ for $0\leq
k\leq\lambda_{j}-1$ and
$\xi_{i}^{1,\lambda_{1}-1-k}\pm\xi_{-1}^{-i,\lambda_{i}-1-k}$,
$\xi_{i}^{-1,\lambda_{1}-1-k}\pm\xi_{1}^{-i,\lambda_{i}-1-k}$ for $0\leq
k\leq\lambda_{i}-1$. Note that

$[\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1},\xi_{1}^{-1,\lambda_{1}-1-k}]=-2\xi_{1}^{-1,2\lambda_{1}-2-k}$

and

$[\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1},\xi_{1}^{-1,\lambda_{1}-1-k}]=2\xi-_{1}^{1,2\lambda_{1}-2-k}.$

We know that $\xi_{1}^{-1,2\lambda_{1}-2-k}=0$ and
$\xi-_{1}^{1,2\lambda_{1}-2-k}=0$ because $2\lambda_{1}-2-k\geq\lambda_{1}$.
Similarly we can compute that
$[\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1},\xi_{1}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{-1,\lambda_{1}-1-k}]=0$,
$[\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1},\xi_{-1}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{1,\lambda_{1}-1-k}]=0$,
$[\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1},\xi_{i}^{1,\lambda_{1}-1-k}\pm\xi_{-1}^{-i,\lambda_{i}-1-k}]=0$
and
$[\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1},\xi_{i}^{-1,\lambda_{1}-1-k}\pm\xi_{1}^{-i,\lambda_{i}-1-k}]=0$.
Hence, we have that $\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}$
commutes with all basis elements in $\mathfrak{g}^{e}$. Therefore, we have
that
$\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}\in\mathfrak{z}(\mathfrak{g}^{e})$
in this case and
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{1,\lambda_{1}-1}-\xi_{-1}^{-1,\lambda_{1}-1}\rangle$.
∎

###  4.7. Centre of centralizer of nilpotent element
$e\in\mathfrak{g}{}_{\bar{0}}$ with Jordan type $\lambda$ such that all parts
of $\lambda$ have multiplicity one

Next we consider the case where all parts of the Jordan type
$\lambda=(\lambda_{1},\dots,\lambda_{a})$ of $e$ have multiplicity one, which
implies $\lambda_{i}$ is odd for $\left|i\right|=\bar{0}$ and $\lambda_{i}$ is
even for $\left|i\right|=\bar{1}$. Note that when $m=0$ or $n=0$, then
$\mathfrak{g}$ is either an orthogonal or symplectic Lie algebra, a basis of
$\mathfrak{z}(\mathfrak{g}^{e})$ has been given in [23, Theorem 4]. Recall
that $S=\langle
e,e^{3},\dots,e^{t}:t=2\lfloor\frac{\lambda_{1}}{2}\rfloor-1\rangle$. Below we
give a general result for $\mathfrak{z}(\mathfrak{g}^{e})$.

######  Theorem 22.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{osp}(m|2n)$
and let $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent with a partition
$\lambda=(\lambda_{1},\dots,\lambda_{a})$ which is defined in (4.8) and
$\lambda_{1}>\dots>\lambda_{a}$.

(1) If $m=0$ or $n=0$, we have $\mathfrak{z}(\mathfrak{g}^{e})=S$ except when
$n=0$, $\lambda_{2}>\lambda_{3}$ and both $\lambda_{1}$ and $\lambda_{2}$ are
odd, in which case we have
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\rangle$;

(2) If $m,n\neq 0$, we have $\mathfrak{z}(\mathfrak{g}^{e})=S$ except when
$a\geq 3$, $\left|1\right|=\left|2\right|=\bar{0}$, or $a=2$ and
$\left|1\right|\neq\left|2\right|$, in which cases we have
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\rangle$.

###### Proof.

For $m=0$ or $n=0$, then $\mathfrak{g}=\mathfrak{sp}(2n)$ or
$\mathfrak{so}(m)$, the detailed proof can be found in [23, Theorem 4].

For $m,n\neq 0$, it is clear that $S\subseteq\mathfrak{z}(\mathfrak{g}^{e})$.
We know that $e^{l}=\sum_{i=1}^{a}\xi_{i}^{i,l}$ and $e^{l}\in\mathfrak{g}$
for all odd $l$ and $0\leq l\leq\lambda_{1}-1$. Note that a basis of
$\mathfrak{g}^{e}$ contains elements of the form:

$\xi_{i}^{i,k}\text{ for all }1\leq i\leq a\text{ and odd }k\text{ with
}0<k\leq\lambda_{i}-1,$
$\xi_{i}^{j,\lambda_{j}-1-k}+\varepsilon_{i}^{j,\lambda_{j}-1-k}\xi_{j}^{i,\lambda_{i}-1-k}\text{
for all }1\leq i<j\leq a,\ 0\leq k\leq\lambda_{j}-1$

where $\varepsilon_{i}^{j,\lambda_{j}-1-k}\in\\{\pm 1\\}$ can be determined.
Thus an element $x\in\mathfrak{z}(\mathfrak{g}^{e})$ is of the form

$x=\sum_{i,k}c_{i}^{i,k}\xi_{i}^{i,k}+\sum_{i,j,k}c_{i}^{j,k}(\xi_{i}^{j,\lambda_{j}-1-k}+\varepsilon_{i}^{j,\lambda_{j}-1-k}\xi_{j}^{i,\lambda_{i}-1-k})$

where $c_{i}^{j,k}\in\mathbb{C}$ are coefficients.

Assume $a\geq 3$. For $1\leq t\leq a$, we have that $\xi_{t}^{t,1}$ commutes
with $\sum_{i,k}c_{i}^{i,k}\xi_{i}^{i,k}$. By taking the commutator between
$\xi_{t}^{t,1}$ and $x$ we obtain that

$\displaystyle[\xi_{t}^{t,1},x]$
$\displaystyle=\sum_{i<t}\sum_{k=0}^{\lambda_{t}-1}c_{i}^{t,k}(\xi_{i}^{t,\lambda_{t}-k}-\varepsilon_{i}^{t,\lambda_{t}-1-k}\xi_{t}^{i,\lambda_{i}-k})$
$\displaystyle+\sum_{t<i}\sum_{k=0}^{\lambda_{i}-1}c_{t}^{i,k}(\varepsilon_{t}^{i,\lambda_{i}-1-k}\xi_{i}^{t,\lambda_{t}-k}-\xi_{t}^{i,\lambda_{i}-k}).$

This is equal to $0$ for all $t$ if and only if $c_{i}^{j,k}=0$ for all
$0<k\leq\lambda_{j}-1$. Now we have that
$x=\sum_{i,k}c_{i}^{i,k}\xi_{i}^{i,k}+\sum_{i,j}c_{i}^{j,0}(\xi_{i}^{j,\lambda_{j}-1}+\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{j}^{i,\lambda_{i}-1})$.

For $1\leq l<h\leq a$, by taking the commutator between
$\xi_{l}^{h,0}+\varepsilon_{l}^{h,0}\xi_{h}^{l,\lambda_{l}-\lambda_{h}}$ and
$x$ we obtain that

$\displaystyle[\xi_{l}^{h,0}+\varepsilon_{l}^{h,0}\xi_{h}^{l,\lambda_{l}-\lambda_{h}},x]$
$\displaystyle=\sum_{k}(c_{l}^{l,k}-c_{h}^{h,k})(\xi_{l}^{h,k}-\varepsilon_{l}^{h,0}\xi_{h}^{l,\lambda_{l}-\lambda_{h}+k})$
$\displaystyle+\sum_{i}c_{i}^{l,0}(\xi_{i}^{h,\lambda_{l}-1}\pm\varepsilon_{l}^{h,0}\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{h}^{i,\lambda_{l}-\lambda_{h}+\lambda_{i}-1})$
$\displaystyle+\sum_{j}c_{l}^{j,0}(\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{j}^{h,\lambda_{l}-1}\pm\varepsilon_{l}^{h,0}\xi_{h}^{j,\lambda_{l}-\lambda_{h}+\lambda_{j}-1})$
$\displaystyle+\sum_{i}c_{i}^{h,0}(\varepsilon_{l}^{h,0}\xi_{i}^{l,\lambda_{l}-1}\pm\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{l}^{i,\lambda_{i}-1})$
$\displaystyle+\sum_{j}c_{h}^{j,0}(\varepsilon_{l}^{h,0}\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{j}^{l,\lambda_{l}-1}\pm\xi_{l}^{j,\lambda_{j}-1}).$

Note that
$\xi_{i}^{h,\lambda_{l}-1}=\xi_{j}^{h,\lambda_{l}-1}=\xi_{h}^{i,\lambda_{l}-\lambda_{h}+\lambda_{i}-1}=\xi_{h}^{j,\lambda_{l}-\lambda_{h}+\lambda_{j}-1}=0$
since $\lambda_{l}>\lambda_{h}$. Hence, we have that

$\displaystyle[\xi_{l}^{h,0}+\varepsilon_{l}^{h,0}\xi_{h}^{l,\lambda_{l}-\lambda_{h}},x]$
$\displaystyle=\sum_{k}(c_{l}^{l,k}-c_{h}^{h,k})(\xi_{l}^{h,k}-\varepsilon_{l}^{h,0}\xi_{h}^{l,\lambda_{l}-\lambda_{h}+k})$
$\displaystyle+\sum_{i}c_{i}^{h,0}(\varepsilon_{l}^{h,0}\xi_{i}^{l,\lambda_{l}-1}\pm\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{l}^{i,\lambda_{i}-1})$
(4.15)
$\displaystyle+\sum_{j}c_{h}^{j,0}(\varepsilon_{l}^{h,0}\varepsilon_{i}^{j,\lambda_{j}-1}\xi_{j}^{l,\lambda_{l}-1}\pm\xi_{l}^{j,\lambda_{j}-1}).$

This is equal to $0$ if and only if $c_{l}^{l,k}=c_{h}^{h,k}$ for all $1\leq
l<h\leq a$ and $c_{i}^{j,0}=0$ for all $1\leq i<j\leq a$ except when
$(i,j)=(1,2)$ and $(\left|1\right|,\left|2\right|)=(\bar{0},\bar{0})$. When
$\left|1\right|=\left|2\right|=\bar{0}$, the commutator between
$\xi_{l}^{h,0}+\varepsilon_{l}^{h,0}\xi_{h}^{l,\lambda_{l}-\lambda_{h}}$ and
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}$ gives terms
$\varepsilon_{l}^{2,0}\xi_{1}^{l,\lambda_{l}-1}+\xi_{l}^{1,\lambda_{1}-1}-\varepsilon_{l}^{1,0}\xi_{2}^{l,\lambda_{l}-1}-\xi_{l}^{2,\lambda_{2}-1}$
and we check this is equal to zero for $l<h$. Hence, when $a\geq 3$, we have
that $\mathfrak{z}(\mathfrak{g}^{e})\subseteq S$ and thus
$\mathfrak{z}(\mathfrak{g}^{e})=S$ except when
$\left|1\right|=\left|2\right|=\bar{0}$, in which case we have
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq
S\oplus\left\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\right\rangle$.

Next suppose $a\geq 3$ and $\left|1\right|=\left|2\right|=\bar{0}$, we know
that $\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}$ commutes with all
basis elements in $\mathfrak{g}_{\bar{0}}^{e}$ by [23, Theorem 4]. Hence, it
remains to check that $\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}$
commutes with basis elements
$\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{j}^{i,\lambda_{i}-1-k}$ for
$\left|i\right|=\bar{0},\left|j\right|=\bar{1}$ and $0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}-1$. Computing

(4.16)
$\displaystyle[\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1},\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{j}^{i,\lambda_{i}-1-k}]=\pm\xi_{j}^{2,\lambda_{2}-1+\lambda_{1}-1-k}$
$\displaystyle\pm\xi_{j}^{1,\lambda_{1}-1+\lambda_{2}-1-k}-\xi_{1}^{j,\lambda_{2}-1+\lambda_{j}-1-k}+\xi_{2}^{j,\lambda_{1}-1+\lambda_{j}-1-k}.$

We have that all terms in (4.16) are equal to $0$ for $0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}-1$ as $\xi_{h}^{l,r}=0$ for all $h,l$
and $r>\lambda_{l}-1$. Hence, we have that
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\in\mathfrak{z}(\mathfrak{g}^{e})$
and
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\left\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\right\rangle$
in this case.

When $a=2$ and $\left|1\right|\neq\left|2\right|$, i.e. the Jordan type of $e$
is $(m,2n)$ such that $m$ is odd. Assume that $m\geq 2n$, in this case a basis
of $\mathfrak{g}^{e}$ only contains elements of the form:

$\xi_{1}^{1,k}\text{ for }k\text{ is odd},1\leq k\leq m-1;\
\xi_{2}^{2,k}\text{ for }k\text{ is odd},\ 1\leq k\leq 2n-1;$ $\text{and
}\xi_{1}^{2,2n-1-k}+\varepsilon_{1}^{2,2n-1-k}\xi_{2}^{1,m-1-k}\text{ for
}0\leq k\leq 2n-1,\varepsilon_{1}^{2,2n-1-k}\in\\{-1,1\\}.$

By applying the similar argument to that used in the case $r+s\geq 3$, we get
that $c_{1}^{1,k}=c_{2}^{2,k}$ and $c_{1}^{2,k}=0$ for all $1\leq k\leq 2n-1$.
The only remaining element to check is $\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1}$.
Note that $\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1}$ commutes with $\xi_{1}^{1,k}$ and
$\xi_{2}^{2,k}$ for all $k$ is odd. The element
$\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1}$ also commutes with
$\xi_{1}^{2,2n-1-k}+\varepsilon_{1}^{2,2n-1-k}\xi_{2}^{1,m-1-k}$ for all
$k=0,1,\dots,2n-2$. Thus we only need to check that
$[\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1},\xi_{1}^{2,0}+\varepsilon_{1}^{2,0}\xi_{2}^{1,m-2n}]=0$.
We calculate

(4.17)
$\displaystyle[\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1},\xi_{1}^{2,0}+\varepsilon_{1}^{2,0}\xi_{2}^{1,m-2n}]$
$\displaystyle=\varepsilon_{1}^{2,0}\xi_{2}^{2,m-1}-\xi_{1}^{1,m-1}-\xi_{2}^{2,m-1}+\varepsilon_{1}^{2,0}\xi_{1}^{1,m-1}$
$\displaystyle=(\varepsilon_{1}^{2,0}-1)\xi_{1}^{1,m-1}$

since $\xi_{2}^{2,m-1}=0$ as $\xi_{2}^{2,k}=0$ for $k>2n-1$.

Next we want to know the value of $\varepsilon_{1}^{2,0}$. Let
$\mathfrak{osp}(m|2n)=\mathfrak{osp}(V)$ and $V=V_{\bar{0}}\oplus
V_{\bar{1}}$. Then we know that there exist $u_{1},u_{2}\in V$ such that
$u_{1},eu_{1},\dots,e^{m-1}u_{1}$ (resp. $u_{2},eu_{2},\dots,e^{2n-1}u_{2}$)
is a basis for $V_{\bar{0}}$ (resp. $V_{\bar{1}}$) according to Subsection
4.5. Moreover, we know that $B(u_{1},e^{m-1}u_{1})=1$ and
$B(u_{2},e^{2n-1}u_{2})=1$ by equations (4.10) and (4.11). By using equation
(4.1), we have that

$B((\xi_{1}^{2,0}+\varepsilon_{1}^{2,0}\xi_{2}^{1,m-2n})u_{1},e^{2n-1}u_{2})=(u_{2},e^{2n-1}u_{2})=1$

and

$\displaystyle
B((\xi_{1}^{2,0}+\varepsilon_{1}^{2,0}\xi_{2}^{1,m-2n})u_{1},e^{2n-1}u_{2})$
$\displaystyle=B(u_{1},(\xi_{1}^{2,0}+\varepsilon_{1}^{2,0}\xi_{2}^{1,m-2n})e^{2n-1}u_{2})$
$\displaystyle=B(u_{1},\varepsilon_{1}^{2,0}e^{m-1}u_{1})=\varepsilon_{1}^{2,0}.$

Therefore, we obtain that $\varepsilon_{1}^{2,0}=1$. Therefore, we have that
$[\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1},\xi_{1}^{2,0}+\varepsilon_{1}^{2,0}\xi_{2}^{1,m-2n}]=0$
and thus $\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1}\in\mathfrak{z}(\mathfrak{g}^{e})$.

Therefore, we deduce that
$\xi_{1}^{2,2n-1}-\xi_{2}^{1,m-1}\in\mathfrak{z}(\mathfrak{g}^{e})$ and
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\rangle$
as required.

When $m<2n$, we obtain same result by applying a similar argument. ∎

###  4.8. Centre of centralizer of general nilpotent element
$e\in\mathfrak{g}_{\bar{0}}$

For Lie superalgebras $\mathfrak{g}=\mathfrak{osp}(m|2n)$, we already know the
construction of $\mathfrak{z}(\mathfrak{g}^{e})$ if: (1) the Jordan type of
$e$ has all parts with even multiplicity; (2) the Jordan type of $e$ has all
parts with multiplicity $1$. Now we want to use Theorems 21 and 22 to deduce a
basis of $\mathfrak{z}(\mathfrak{g}^{e})$ for a general nilpotent element $e$.

Let $e\in\mathfrak{g}_{\bar{0}}$ be a nilpotent element with Jordan type
$\lambda$ as denoted in (4.8). Take a Dynkin pyramid $\tilde{P}$ following
Subsection 4.4 and $\\{v_{i_{j}}\\}$ in (4.9) form a basis for $V$ with
respect to $\tilde{P}$. Now we write $V=V_{1}\oplus V_{2}$ where
$\\{v_{i_{j}}:1\leq i\leq a\\}$ form a basis for $V_{1}$ and
$\\{v_{i_{j}},v_{-i_{j}}:a+1\leq i\leq b\\}$ form a basis for $V_{2}$. Define
$\mathfrak{g^{\prime}=g}_{1}\oplus\mathfrak{g}_{2}$ where
$\mathfrak{g}_{1}=\mathfrak{osp}(V_{1})$ and
$\mathfrak{g}_{2}=\mathfrak{osp}(V_{2})$. Then the nilpotent element
$e\in\mathfrak{g}_{\bar{0}}$ can also be written as $e=e_{1}+e_{2}$ and
$e_{i}\in\mathfrak{osp}(V_{i})$ such that the Jordan type of $e_{1}$ has all
parts with multiplicity $1$ and the Jordan type of $e_{2}$ has all parts with
even multiplicity. That is to say, $(\lambda_{1},\dots,\lambda_{a})$ is the
Jordan type of $e_{1}$ in descending order and
$(\lambda_{a+1},\lambda_{-\left(a+1\right)},\dots,\lambda_{b},\lambda_{-b})$
is the Jordan type of $e_{2}$.

Now consider a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$, we know
that a basis for $\mathfrak{h}^{e}$ is
$\\{h_{i}=\xi_{i}^{i,0}-\xi_{-i}^{-i,0}:i=a+1,\dots,b\\}$. Define
$U_{i}=\mathrm{Span}\\{v_{i_{j}}:i_{j}\text{ is a box in }P\\}$ for $1\leq
i\leq a$ and $U_{i}=U_{i}^{+}\oplus U_{i}^{-}$ for $a+1\leq i\leq b$ where
$U_{i}^{+}=\mathrm{Span}\\{v_{i_{j}}:i_{j}\text{ is a box in }P\\}$ and
$U_{i}^{-}=\mathrm{Span}\\{v_{-i_{j}}:-i_{j}\text{ is a box in }P\\}$. We also
define $U_{i}^{\bot}=\left\\{v\in V:(v,u)=0\text{ for any }u\in
U_{i}\right\\}$. Then we have that
$\mathfrak{g}^{h_{i}}\cong\mathfrak{gl}\left(U_{i}^{+}\right)\oplus\mathfrak{osp}\left(U_{i}^{\bot}\right)$
where elements of $\mathfrak{gl}(U_{i}^{+})$ can be viewed as elements of
$\mathfrak{osp}(U_{i}^{+}\oplus U_{i}^{-})$. Let $H=\sum_{i=a+1}^{b}h_{i}$,
with the above basis of $V$ we have that $H$ is of the form

$H=\begin{pmatrix}1\\\ &\ddots\\\ &&1\\\ &&&0\\\ &&&&\ddots\\\ &&&&&0\\\
&&&&&&-1\\\ &&&&&&&\ddots\\\ &&&&&&&&-1\end{pmatrix},$

Hence, we have that

$\mathfrak{g}^{H}\cong\mathfrak{gl}(\bigoplus_{i=a+1}^{b}U_{i}^{+})\oplus\mathfrak{osp}((\bigoplus_{i=1}^{a}U_{i})\subseteq\mathfrak{g}^{\prime}.$

Therefore, we have that
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\mathfrak{g}^{H}\subseteq\mathfrak{g}^{\prime}$.
Let $x\in\mathfrak{z}(\mathfrak{g}^{e})$, then
$x\in(\mathfrak{g}^{\prime})^{e}$ and $\left[x,y\right]=0$ for all
$y\in\mathfrak{g}^{e}$. Since
$(\mathfrak{g}^{\prime})^{e}\subseteq\mathfrak{g}^{e}$, we have that $[x,y]=0$
for all $y\in(\mathfrak{g}^{\prime})^{e}$. Therefore, we deduce that
$\mathfrak{z}(\mathfrak{g}^{e})\subseteq\mathfrak{z}((\mathfrak{g}^{\prime})^{e})=\mathfrak{z}(\mathfrak{g}_{1}^{e_{1}})\oplus\mathfrak{z}(\mathfrak{g}_{2}^{e_{2}})$.

Now we are ready to combine Theorems 21 and 22 to obtain a basis for
$\mathfrak{z}(\mathfrak{g}^{e})$.

######  Corollary 23.

Suppose that
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{osp}(m|2n)$
and $e=e_{1}+e_{2}$ is given by a partition
$\lambda=(\lambda_{1},\dots,\lambda_{a},\lambda_{a+1},\lambda_{-(a+1)},\dots,\lambda_{b},\lambda_{-b})$.
Let $S=\langle e^{k}:k\text{ is odd and }1\leq
k\leq\max\\{\lambda_{1},\lambda_{a+1}\\}-1\rangle$. Then
$\mathfrak{z}(\mathfrak{g}^{e})=S$ except for two special cases:

Case 1: If $a\geq 3$, $\lambda_{2}>\lambda_{a+1}$,
$\left|1\right|=\left|2\right|=\bar{0}$ or
$a=2,\left|1\right|\neq\left|2\right|$, then we have that
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\rangle$;

Case 2: If $\lambda_{1}<\lambda_{a+1}$, $\lambda_{a+1}>\lambda_{a+2}$,
$\left|a+1\right|=\bar{0}$ and $\lambda_{a+1}$ is odd, we have that
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{a+1}^{a+1,\mu_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\mu_{a+1}-1}\rangle$.

###### Proof.

It is clear that $S\subseteq\mathfrak{z}(\mathfrak{g}^{e})$. We first consider
the case that
$\mathfrak{z}(\mathfrak{g}_{1}^{e_{1}})=\mathrm{Span}\\{e_{1}^{k}:k\text{ is
odd and }1\leq k\leq\lambda_{1}-1\\}$ and
$\mathfrak{z}(\mathfrak{g}_{2}^{e_{2}})=\mathrm{Span}\\{e_{2}^{k}:k\text{ is
odd and }1\leq k\leq\lambda_{a+1}-1\\}$. According to Theorem 21 and 22, we
have that elements in a basis of $\mathfrak{z}((\mathfrak{g}^{\prime})^{e})$
can be written as:

$\sum_{t=1}^{a}\xi_{t}^{t,k}\text{ for }k\text{ is odd and }1\leq
k\leq\lambda_{1}-1;$ $\sum_{t=a+1}^{b}(\xi_{t}^{t,k}+\xi_{-t}^{-t,k})\text{
for }k\text{ is odd and }1\leq k\leq\lambda_{a+1}-1.$

Thus an element $x\in\mathfrak{z}(\mathfrak{g}^{e})$ is of the form

(4.18) $x=\sum_{k\text{ is
odd;}k=1}^{\lambda_{1}-1}a_{k}\left(\sum_{t=1}^{a}\xi_{t}^{t,k}\right)+\sum_{k\text{
is
odd;}k=1}^{\lambda_{a+1}-1}b_{k}\left(\sum_{t=a+1}^{b}(\xi_{t}^{t,k}+\xi_{-t}^{-t,k})\right)$

for coefficient $a_{k},b_{k}\in\mathbb{C}$. We assume that
$\lambda_{1}\geq\lambda_{a+1}$, then take the commutator with
$\xi_{1}^{a+1,0}+\varepsilon_{1}^{a+1,0}\xi_{-(a+1)}^{1,\lambda_{1}-\lambda_{a+1}}$:

$[\xi_{1}^{a+1,0}+\varepsilon_{1}^{a+1,0}\xi_{-(a+1)}^{1,\lambda_{1}-\lambda_{a+1}},x]=\sum_{k\text{
is
odd;}k=1}^{\lambda_{a+1}-1}\left(a_{k}-b_{k}\right)\left(\xi_{1}^{a+1,k}\pm\xi_{-(a+1)}^{1,\lambda_{1}-\lambda_{a+1}+k}\right).$

This is equal to $0$ if and only if $a_{k}=b_{k}$ for all $k$ is odd and
$1\leq k\leq\lambda_{a+1}-1$. If $\lambda_{1}<\lambda_{a+1}$, then by taking
the commutator between
$\xi_{1}^{a+1,\lambda_{a+1}-\lambda_{1}}+\varepsilon_{1}^{a+1,\lambda_{a+1}-\lambda_{1}}\xi_{-(a+1)}^{1,0}$
and $x$, we obtain that $a_{k}=b_{k}$ for all $k$ is odd and $1\leq
k\leq\mu_{1}-1$. Hence, we have that $\mathfrak{z}(\mathfrak{g}^{e})\subseteq
S$ for this case and thus $\mathfrak{z}(\mathfrak{g}^{e})=S$. Therefore, we
have that
$\dim\mathfrak{z}(\mathfrak{g}^{e})=\lceil\frac{\max\left\\{\lambda_{1},\lambda_{a+1}\right\\}-1}{2}\rceil$
for this case.

Now we look at the special cases:

For the special case when
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\in\mathfrak{z}(\mathfrak{g}_{1}^{e_{1}})$
and
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}\notin\mathfrak{z}(\mathfrak{g}_{2}^{e_{2}})$,
then an element $y\in\mathfrak{z}(\mathfrak{g}^{e})$ is of the form
$y=x+c_{1}^{2,\lambda_{2}-1}(\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1})$
where $x$ is defined in (4.18) and $c_{1}^{2,\lambda_{2}-1}\in\mathbb{C}$ is
the coefficient. By calculating
$[y,\xi_{1}^{a+1,0}+\varepsilon_{1}^{a+1,0}\xi_{-(a+1)}^{1,\lambda_{1}-\lambda_{a+1}}]=0$
for $\lambda_{1}>\lambda_{a+1}$ and
$[y,\xi_{1}^{a+1,\lambda_{a+1}-\lambda_{1}}+\varepsilon_{1}^{a+1,\lambda_{a+1}-\lambda_{1}}\xi_{-(a+1)}^{1,0}]=0$
for $\lambda_{1}\leq\lambda_{a+1}$, we obtain that $a_{k}=b_{k}$ for all $k$
is odd and $c_{1}^{2,\lambda_{2}-1}=0$ when $\lambda_{1}\leq\lambda_{a+1}$. If
$\lambda_{2}\leq\lambda_{a+1}<\lambda_{1}$, then computing

$[x+c_{1}^{2,\lambda_{2}-1}(\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}),\xi_{2}^{a+1,\lambda_{a+1}-\lambda_{2}}\pm\xi_{-(a+1)}^{2,0}]=0$

implies that $c_{1}^{2,\lambda_{2}-1}=0$. However, if
$\lambda_{2}>\lambda_{a+1}$, we calculate the commutator between
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}$ and basis elements of
the form $\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{i,\lambda_{i}-1-k}$ for
$1\leq i\leq a$, $a+1\leq j\leq b$,$0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}$:

(4.19)
$\displaystyle[\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1},\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{i,\lambda_{i}-1-k}]$
$\displaystyle=\pm\xi_{-j}^{2,\lambda_{2}-1+\lambda_{1}-1-k}\pm\xi_{-j}^{1,\lambda_{1}-1+\lambda_{2}-1-k}$
$\displaystyle\ \
\pm\xi_{1}^{j,\lambda_{j}-1-k+\lambda_{2}-1}\pm\xi_{2}^{j,\lambda_{j}-1-k+\lambda_{1}-1}.$

We have that all terms in (4.19) are equal to $0$ for $0\leq
k\leq\min\\{\lambda_{i},\lambda_{j}\\}$. This implies that
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}$ commutes with all other
basis elements in $\mathfrak{g}^{e}$. Therefore, we deduce that
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\rangle$
in this case and
$\dim\mathfrak{z}(\mathfrak{g}^{e})=\lceil\frac{\max\left\\{\lambda_{1},\lambda_{a+1}\right\\}-1}{2}\rceil+1$.

For the special case when
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\notin\mathfrak{z}(\mathfrak{g}_{1}^{e_{1}})$
and
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}\in\mathfrak{z}(\mathfrak{g}_{2}^{e_{2}})$,
then an element $z\in\mathfrak{z}(\mathfrak{g}^{e})$ is of the form
$z=x+c_{a+1}^{a+1,\lambda_{a+1}-1}(\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1})$
where $x$ is defined in (4.18) and
$c_{a+1}^{a+1,\lambda_{a+1}-1}\in\mathbb{C}$ is the coefficient. If
$\lambda_{1}\geq\lambda_{a+1}$, then computing
$[x,\xi_{1}^{a+1,0}+\varepsilon_{1}^{a+1,0}\xi_{-(a+1)}^{1,\lambda_{1}-\lambda_{a+1}}]=0$
gives that $a_{k}=b_{k}$ for all $k$ is odd and $1\leq k\leq\lambda_{a+1}-1$
and $c_{a+1}^{a+1,\lambda_{a+1}-1}=0$. However, if
$\lambda_{1}<\lambda_{a+1}$, computing
$[x,\xi_{1}^{a+1,\lambda_{a+1}-\lambda_{1}}+\varepsilon_{1}^{a+1,\lambda_{a+1}-\lambda_{1}}\xi_{-(a+1)}^{1,0}]=0$
gives that $a_{k}=b_{k}$ for all $k$ is odd and $1\leq k\leq\lambda_{1}-1$. It
remains to check that
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}\in\mathfrak{z}(\mathfrak{g}^{e})$.
It is obvious that
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}$
commutes with all elements of the form
$\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{i,\lambda_{i}-1-k}$ for $1\leq
i,j\leq a$ or $a+1\leq i,j\leq b$. We now calculate commutators between
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}$ and
$\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{i,\lambda_{i}-1-k}$ for $1\leq i\leq
a$, $a+1\leq j\leq b$, $1\leq k\leq\min${$\lambda_{i},\lambda_{j}$}:

(4.20)
$\displaystyle[\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1},\xi_{i}^{j,\lambda_{j}-1-k}\pm\xi_{-j}^{i,\lambda_{i}-1-k}]=\xi_{i}^{a+1,2\lambda_{a+1}-2-k}$
$\displaystyle\pm\xi_{-(a+1)}^{i,\lambda_{a+1}-1+\lambda_{i}-1-k}.$

We have that all terms in (4.20) are equal to $0$. This implies that
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}$
commutes with all other basis elements in $\mathfrak{g}^{e}$. Therefore, we
deduce that
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}\in\mathfrak{z}(\mathfrak{g}^{e})$
in this case and
$\dim\mathfrak{z}(\mathfrak{g}^{e})=\lceil\frac{\max\left\\{\lambda_{1},\lambda_{a+1}\right\\}-1}{2}\rceil+1$.

Moreover, when
$\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\in\mathfrak{z}(\mathfrak{g}_{1}^{e_{1}})$
and
$\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}\in\mathfrak{z}(\mathfrak{g}_{2}^{e_{2}})$,
applying a similar argument to above we also have that
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{1}^{2,\lambda_{2}-1}-\xi_{2}^{1,\lambda_{1}-1}\rangle$
for $\lambda_{2}>\lambda_{a+1}$ and
$\mathfrak{z}(\mathfrak{g}^{e})=S\oplus\langle\xi_{a+1}^{a+1,\lambda_{a+1}-1}-\xi_{-(a+1)}^{-(a+1),\lambda_{a+1}-1}\rangle$
for $\lambda_{1}<\lambda_{a+1}$. ∎

###  4.9. Adjoint action on $\mathfrak{osp}(m|2n)$

Recall that $G=\mathrm{O}_{m}(\mathbb{C})\times\mathrm{Sp}_{2n}(\mathbb{C})$
and $e\in\mathfrak{g}_{\bar{0}}$ be nilpotent. Recall that the adjoint action
of $G$ on $\mathfrak{g}$ is given by $g\cdot x=gxg^{-1}$ for all $g\in G$,
$x\in\mathfrak{g}$. Then the centralizer $G^{e}$ of $e$ in $G$ acts on
$\mathfrak{g}^{e}$ by the adjoint action. Consider
$(\mathfrak{g}^{e})^{G^{e}}$ which is defined to be
$(\mathfrak{g}^{e})^{G^{e}}=\\{x\in\mathfrak{g}^{e}:gxg^{-1}=x\text{ for all
}g\in G^{e}\\}$. The restriction of the adjoint action of $G^{e}$ to
$(\mathfrak{g}^{e})^{G^{e}}$ is the trivial action, so
$\mathrm{Ad}_{\mid(\mathfrak{g}^{e})^{G^{e}}}:G^{e}\rightarrow\mathrm{GL}((\mathfrak{g}^{e})^{G^{e}})$
is the trivial map and thus
$\mathrm{ad}_{\mid(\mathfrak{g}^{e})^{G^{e}}}:\mathfrak{g}^{e}\rightarrow\mathfrak{gl}((\mathfrak{g}^{e})^{G^{e}})$
is the trivial map. Hence, the adjoint action of $\mathfrak{g}^{e}$ on
$(\mathfrak{g}^{e})^{G^{e}}$ is trivial. Therefore, we have that
$(\mathfrak{g}^{e})^{G^{e}}\subseteq\mathfrak{z}(\mathfrak{g}^{e})$. Since
$\mathfrak{z}(\mathfrak{g}^{e})$ is stable under any automorphism of
$\mathfrak{g}^{e}$, it is stable under the adjoint action of $G^{e}$. Take any
$x\in\mathfrak{z}(\mathfrak{g}^{e})$, $g\in G^{e}$ and $y\in\mathfrak{g}^{e}$,
then $[g\cdotp x,y]=[g\cdotp x,g\cdot(g^{-1}\cdot y)]=g\cdotp[x,g^{-1}\cdotp
y]$. Since $g^{-1}\cdotp y\in\mathfrak{g}^{e}$, we have that $[g\cdotp x,y]=0$
as required. Therefore, we deduce that
$(\mathfrak{g}^{e})^{G^{e}}=\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$.

According to [8, Section 3.12], we have that $G^{e}$ is the semidirect product
of the subgroup $C^{e}$ and the connected normal subgroup $R^{e}$, i.e.
$G^{e}\cong C^{e}\ltimes R^{e}$. Denote the connected component of $G^{e}$
(resp. $C^{e}$) containing the identity by $(G^{e})^{\circ}$ (resp.
$(C^{e})^{\circ}$), we have that $G^{e}/(G^{e})^{\circ}\cong
C^{e}/(C^{e})^{\circ}$ based on [8, Section 3.13].

######  Theorem 24.

Let
$\mathfrak{g}=\mathfrak{g}_{\bar{0}}\oplus\mathfrak{g}_{\bar{1}}=\mathfrak{osp}(m|2n)$
and $G=\mathrm{O}_{m}(\mathbb{C})\times\mathrm{Sp}_{2n}(\mathbb{C})$. Let
$e=e_{1}+e_{2}\in\mathfrak{g}_{\bar{0}}$ be nilpotent with Jordan type
$\lambda$ denoted as in (4.8). Let $S=\langle e^{l}:l\text{ is odd and
}l\leq\max\\{\lambda_{1},\lambda_{a+1}\\}-1\rangle$, then
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=S$.

###### Proof.

Recall that $\\{v_{i_{j}}\\}$ is a basis of $V$ according to (4.9) and the
bilinear form $B(\ldotp,\ldotp)$ on $V$ is given in Subsection 4.4. We define
an involution $\rho$ on the labels of the basis of $V$ by

$\rho(i_{k})=\begin{cases}i_{-k}&\text{if }i=1,\dots,a;\\\ -i_{-k}&\text{if
}i=\pm(a+1),\dots,\pm b.\end{cases}$

Let $e_{i_{t},d_{l}}$ be the $(i_{t},d_{l})$-matrix unit. Then
$\\{e_{i_{t},d_{l}}\pm e_{\rho(d_{l}),\rho(i_{t})}\\}$ is a basis for
$\mathfrak{g}$ with appropriate signs and conditions on $i_{t},d_{l}$.

Let $l=\lfloor\frac{m}{2}\rfloor$. By reordering boxes on the right hand half
of $\tilde{P}$ from $1,\ldots,l+n$ and boxes on the left hand half of
$\tilde{P}$ from $-(l+n),\ldots,-1$ (note that there exists a box labelled by
$0$ if $m$ is odd), we get a basis $\\{e_{j,k}+\gamma_{-k,-j}e_{-k,-j}\\}$ for
$\mathfrak{g}$ as defined in (4.6). There exists an isomorphism between basis
element $e_{j,k}+\gamma_{-k,-j}e_{-k,-j}$ and $e_{i_{t},d_{l}}\pm
e_{\rho(d_{l}),\rho(i_{t})}$.

It is clear that
$S\subseteq\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$. When
$\mathfrak{z}(\mathfrak{g}^{e})=S$, then
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}\subseteq\mathfrak{z}(\mathfrak{g}^{e})=S$
and thus we obtain $\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=S$.
The following part of this proof deals with two special cases in Corollary 23
where $\mathfrak{z}(\mathfrak{g}^{e})\neq S$.

Case 1: When $a\geq 3$, $\lambda_{2}>\lambda_{a+1}$,
$\left|1\right|=\left|2\right|=\bar{0}$ or $a=2$,
$\left|1\right|\neq\left|2\right|$.

In this case, the extra basis element $x$ can be written as
$e_{1_{-\lambda_{1}+1},2_{\lambda_{2}-1}}-e_{2_{-\lambda_{2}+1},1_{\lambda_{1}-1}}$.
We consider a matrix $Q$ which sends each $v_{1_{k}}$ to $-v_{1_{k}}$ and all
other $v_{i_{{}_{k}}}$ for $i\neq 1$ to itself. Obviously
$S\subseteq\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}\subseteq\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}$.
Then we have that

$B(Qv_{i_{k}},Qv_{j_{l}})=\begin{cases}(-1)^{k}\delta_{i_{k},\rho(j_{l})}&\text{if
}1\leq i,j\leq a;\\\ \delta_{i_{k},\rho(j_{l})}&\text{if }\pm(a+1)\leq
i,j\leq\pm b;\\\ 0&\text{otherwise}.\end{cases}$

Thus $Q$ preserves the form $B$ on $V$. Hence, we have that $Q\in G$. For any
$i_{k},j_{l}$, observe that

$Qe_{i_{k},j_{l}}Q^{-1}=\begin{cases}-e_{i_{k},j_{l}}&\text{if }i\neq j\text{
and }i=1\text{ or }j=1;\\\ e_{i_{k},j_{l}}&\text{otherwise.}\end{cases}$

we have that

$\displaystyle Q\cdotp e_{1}$
$\displaystyle=Q\cdotp(\sum_{t;i=1}^{a}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}})=\sum_{t;i=1}^{a}\varepsilon_{i_{t},i_{t-2}}Qe_{i_{t},i_{t-2}}Q^{-1}$
$\displaystyle=\sum_{t}\varepsilon_{1_{t},1_{t-2}}e_{1_{t},1_{t-2}}+\sum_{t;i\neq
1}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}}=\sum_{t;i=1}^{a}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}}=e_{1}.$

Furthermore, based on the way we defined $Q,$ it fixes $e_{2}$. This implies
that $Q\in G^{e}$ and thus we know that $Q\cdotp e^{l}=e^{l}$ for all $l$ is
odd. Next we have that

$Q\cdotp
x=QxQ^{-1}=-e_{1_{-\lambda_{1}+1},2_{\lambda_{2}-1}}+e_{2_{-\lambda_{2}+1},1_{\lambda_{1}-1}}=-x.$

Thus we deduce that $x\notin\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}$
and $\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}\subseteq S$. Therefore,
we have that
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}=S$
and
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\lceil\frac{\lambda_{1}-1}{2}\rceil$.

Case 2: When $\lambda_{1}<\lambda_{a+1}$, $\lambda_{a+1}>\lambda_{a+2}$,
$\left|a+1\right|=\bar{0}$ and $\lambda_{a+1}$ is odd.

In this case, the extra basis element $x$ can be written as

$x=e_{(a+1)_{-\lambda_{a+1}+1},(a+1)_{\lambda_{a+1}-1}}-e_{-(a+1)_{-\lambda_{a+1}+1},-(a+1)_{\lambda_{a+1}-1}}.$

We consider a matrix $Q$ which sends $v_{(a+1)_{k}}$ to $v_{-(a+1)_{k}}$ and
all other $v_{i_{{}_{k}}}$ for $i\neq\pm(a+1)$ to itself. Then we have that

$B(Qv_{i_{k}},Qv_{j_{l}})=\begin{cases}B(v_{i_{k}},v_{j_{l}})=(-1)^{k}\delta_{i_{k},\rho(j_{l})}&\text{if
}1\leq i,j\leq a;\\\
B(v_{-i_{k}},v_{-j_{l}})=\delta_{-i_{k},\rho(-j_{l})}=\delta_{i_{k},\rho(j_{l})}&\text{if
}i,j=\pm(a+1);\\\ B(v_{i_{k}},v_{j_{l}})=\delta_{i_{k},\rho(j_{l})}&\text{if
}\pm(a+2)\leq i,j\leq\pm b;\\\ 0&\text{otherwise.}\end{cases}$

Thus $Q$ preserves the form on $V$ and $Q\in G$. For any $i_{k},j_{l}$,
observe that

(4.21) $Qe_{i_{k},j_{l}}Q^{-1}=\begin{cases}e_{i_{k},j_{l}}&\text{if
}i,j\neq\pm(a+1);\\\ e_{-i_{k},j_{l}}&\text{if }i=\pm(a+1),j\neq\pm(a+1);\\\
e_{i_{k},-j_{l}}&\text{if }i\neq\pm(a+1),j=\pm(a+1);\\\
e_{-i_{k},-j_{l}}&\text{if }i,j=\pm(a+1).\end{cases}$

Note that we can write $e_{2}$ to be
$\sum_{\begin{subarray}{c}t;i=\pm(a+1)\end{subarray}}^{b}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}}$.
Then we have that

$\displaystyle Q\cdotp e_{2}$
$\displaystyle=Q\cdotp(\sum_{\begin{subarray}{c}t;i=\pm(a+1)\end{subarray}}^{b}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}})=\sum_{\begin{subarray}{c}t;i=\pm(a+1)\end{subarray}}^{b}\varepsilon_{i_{t},i_{t-2}}Qe_{i_{t},i_{t-2}}Q^{-1}$
$\displaystyle=\sum_{t}\varepsilon_{(a+1)_{t},(a+1)_{t-2}}e_{-(a+1)_{t},-(a+1)_{t-2}}+\sum_{t}\varepsilon_{-(a+1)_{t},-(a+1)_{t-2}}e_{(a+1)_{t},(a+1)_{t-2}}$
$\displaystyle+\sum_{t;i\neq\pm(a+1)}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}}$
$\displaystyle=\sum_{\begin{subarray}{c}t;i=\pm(a+1)\end{subarray}}^{b}\varepsilon_{i_{t},i_{t-2}}e_{i_{t},i_{t-2}}=e_{2}.$

Furthermore, based on the way we defined $Q,$ it fixes $e_{1}$. This implies
that $Q\in G^{e}$ and thus we know that $Q\cdotp e^{l}=e^{l}$ for all $l$ is
odd. Next we have that

$Q\cdotp
x=QxQ^{-1}=e_{-(a+1)_{-\lambda_{a+1}+1},-(a+1)_{\lambda_{a+1}-1}}-e_{(a+1)_{-\lambda_{a+1}+1},(a+1)_{\lambda_{a+1}-1}}=-x.$

Thus we know that $x\notin\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}$ and
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}\subseteq S$. Therefore, we
have that
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{Q}=S$
and
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\lceil\frac{\lambda_{a+1}-1}{2}\rceil$.∎

######  Remark 25.

If we choose
$G^{\prime}=\mathrm{SO}_{m}(\mathbb{C})\times\mathrm{Sp}_{2n}(\mathbb{C})$,
applying a similar argument we obtain that
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{\prime
e}}=\mathfrak{z}(\mathfrak{g}^{e})$ if $\lambda_{2}>\lambda_{a+1}$ and
$\lambda_{i}$ is even for $3\leq i\leq b$, or $\lambda_{1}<\lambda_{a+1}$,
$\lambda_{a+1}\neq\lambda_{a+2}$ and $\lambda_{\pm(a+1)}$ are the only odd
parts for $1\leq i\leq b$. For other cases, we have
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{\prime e}}=S$.

###  4.10. Proof of the Theorems

Let $\lambda=(\lambda_{1},\dots,\lambda_{r+s})$ be the Jordan type of $e$ as
defined in (3.2). Based on Subsections 4.6–4.9, we have that
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\lceil\frac{\lambda_{1}-1}{2}\rceil$.

In order to prove Theorem 1 for $\mathfrak{g}=\mathfrak{osp}(m|2n)$, we
calculate the number of labels in the labelled Dynkin diagram $\varDelta$
which are equal to $2$. Given a ortho-symplectic Dynkin pyramid $P$ and a
partition $\lambda$ as defined in (3.5), let $r_{i}$ (resp. $s_{i}$) be the
number of boxes with parity $\bar{0}$ (resp. $\bar{1}$) on the $i$th column.
We observe that $\varDelta$ has no label equal to $1$ if and only if all parts
of $\lambda$ are odd or all parts of $\lambda$ are even. Now assume that all
labels in $\varDelta$ equal to $0$ or $2$. Then based on the way that labelled
Dynkin diagram is constructed, we observe that
$n_{2}(\varDelta)=\lfloor\frac{\lambda_{1}}{2}\rfloor$. Note that when
$\lambda_{1}$ is even, we have that
$\lceil\frac{\lambda_{1}-1}{2}\rceil=\frac{\lambda_{1}}{2}$ and
$n_{2}(\varDelta)=\lfloor\frac{\lambda_{1}}{2}\rfloor=\frac{\lambda_{1}}{2}$,
when $\lambda_{1}$ is odd, we have that
$\lceil\frac{\lambda_{1}-1}{2}\rceil=\frac{\lambda_{1}-1}{2}$ and
$n_{2}(\varDelta)=\frac{\lambda_{1}-1}{2}$. Therefore, we deduce that
$\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=n_{2}(\varDelta)$.

To calculate $\mathfrak{z}(\mathfrak{g}^{h})$, we first consider the following
example:

######  Example 26.

For a nilpotent element $e\in\mathfrak{g}_{\bar{0}}$ with Jordan type
$(5,3,1|3,3)$, the corresponding Dynkin pyramid is shown in Example 15. Hence,
we have $n_{2}(\varDelta)=\lfloor\frac{5}{2}\rfloor=2$. We can calculate that
$\mathfrak{g}^{h}=\mathfrak{gl}(1|0)\oplus\mathfrak{gl}(2|2)\oplus\mathfrak{osp}(3|2)$.
Note that $\mathfrak{z}(\mathfrak{gl}(1|0))=I_{1}$,
$\mathfrak{z}(\mathfrak{gl}(2|2))=I_{4}$ and
$\mathfrak{z}(\mathfrak{osp}(3|2))=0$. Therefore, we have that
$\dim\mathfrak{z}(\mathfrak{g}^{h})=2$.

Assume that there is no label equal to $1$ in $\varDelta$. When all parts of
$\lambda$ are odd, we observe that

$\mathfrak{g}^{h}=\bigoplus_{i>0}\mathfrak{gl}(r_{i}|s_{i})\oplus\mathfrak{osp}(r_{0}|s_{0}).$

When all parts of $\lambda$ are even , then $\mathfrak{g}^{h}$ is just the
direct sum of $\mathfrak{gl}(r_{i}|s_{i})$ for $i>0$. Note that
$\mathfrak{z}(\mathfrak{gl}(r_{i}|s_{i}))=I_{r_{i}+s_{i}}$ and
$\mathfrak{z}(\mathfrak{osp}(r_{0}|s_{0}))=0$. Hence, we have that
$\mathfrak{z}(\mathfrak{g}^{h})$ is the direct sum of $I_{r_{i}+s_{i}}$ for
$i>0$. Since there are in total $\lambda_{1}$ columns in $P$ with non-zero
boxes, we deduce that
$\dim\mathfrak{z}(\mathfrak{g}^{h})=\lfloor\frac{\lambda_{1}}{2}\rfloor=n_{2}(\varDelta)$.

Next we describe the relation between
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}$ and the sum of
labels $\sum a_{i}$ in $\varDelta$. Note that
$a_{i}=\mathrm{col}(i+1)-\mathrm{col}(i)$ for $i=1,\dots,l+n-1$. When $m$ is
even, we have

$a_{l+n}=\begin{cases}-2\mathrm{col}(l+n)&\text{if
}\left|l+n\right|=\bar{1};\\\ -\mathrm{col}(l+n)-\mathrm{col}(l+n-1)&\text{if
}\left|l+n\right|=\bar{0}.\end{cases}$

Then

$\displaystyle\sum a_{i}$
$\displaystyle=\sum_{i=1}^{l+n-1}\left(\mathrm{col}(i+1)-\mathrm{col}(i)\right)+\begin{cases}-2\mathrm{col}(l+n)&\text{if
}\left|l+n\right|=\bar{1};\\\
(\mathrm{col}(-l-n)-\mathrm{col}(l+n-1))&\text{if
}\left|l+n\right|=\bar{0}\end{cases}$
$\displaystyle=\begin{cases}\mathrm{col}(-l-n)-\mathrm{col}(1)&\text{if
}\left|l+n\right|=\bar{1};\\\ -\mathrm{col}(l+n-1)-\mathrm{col}(1)&\text{if
}\left|l+n\right|=\bar{0}.\end{cases}$

Hence, we have that $\sum a_{i}=\lambda_{1}-1$ or $\lambda_{1}$.

When $m$ is odd, then there exists a box that is labelled by $0$, we have
$a_{l+n}=\mathrm{col}(0)-\mathrm{col}(l+n)$. Then

$\displaystyle\sum a_{i}$
$\displaystyle=\sum_{i=1}^{l+n-1}\left(\mathrm{col}(i+1)-\mathrm{col}(i)\right)+(\mathrm{col}(0)-\mathrm{col}(l+n))$
$\displaystyle=\mathrm{col}(0)-\mathrm{col}(1)=\lambda_{1}-1.$

Hence, we deduce that

$\lceil\frac{1}{2}\sum
a_{i}\rceil=\begin{cases}\lceil\frac{\lambda_{1}}{2}\rceil&\text{\text{if} all
parts of the Jordan type of }e\text{ are even;}\\\
\lceil\frac{\lambda_{1}-1}{2}\rceil&\text{otherwise.}\end{cases}$

Therefore, we deduce that
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}=\lceil\frac{1}{2}\sum
a_{i}\rceil$.

In order to prove Theorem 3 for $\mathfrak{g}$, we consider two general cases
below.

Case 1: When $\varDelta$ has no label equal to $1$, i.e. all labels are equal
to $0$ or $2$. In this case, we know that either all parts of $\lambda$ are
odd or all parts of $\lambda$ are even. Note that $e_{0}=0$ since
$\varDelta_{0}$ has all labels equal to $0$. Thus we have that

$\mathfrak{g}_{0}^{e_{0}}=\mathfrak{g}_{0}=\left(\bigoplus_{i>0}\mathfrak{sl}(r_{i}|s_{i})\right)\oplus\mathfrak{osp}(r_{0}|s_{0})$

We denote $c_{i}=r_{i}+s_{i}$. Then

$\dim\mathfrak{g}_{0}^{e_{0}}=\dim\mathfrak{g}_{0}=\sum_{i>0}\dim\mathfrak{sl}(r_{i}|s_{i})+\dim\mathfrak{osp}(r_{0}|s_{0}).$

Now if all parts of $\lambda$ are even, then we have
$\dim\mathfrak{osp}(r_{0}|s_{0})=0$. This implies that
$\dim\mathfrak{g}_{0}^{e_{0}}=\sum_{i>0}(c_{i}^{2}-1).$ We also have that
$\arrowvert\\{i:\lambda_{i}\text{ is
odd},\left|i\right|=\bar{0}\\}\mid=\arrowvert\\{i:\lambda_{i}\text{ is
odd},\left|i\right|=\bar{1}\\}\mid=0$. Thus
$\dim\mathfrak{g}^{e}=\frac{1}{2}\dim\mathfrak{gl}(m|2n)^{e}=\frac{1}{2}(\sum_{i\in\mathbb{Z}}c_{i}^{2})$
by Subsection 4.5. Hence, we have that

$\dim\mathfrak{g}^{e}-\dim\mathfrak{g}_{0}^{e_{0}}=\frac{1}{2}(\sum_{i\in\mathbb{Z}}c_{i}^{2})-\sum_{i>0}(c_{i}^{2}-1)=\sum_{i>0:c_{i}>0}1=n_{2}(\varDelta).$

Next we consider when all parts of $\lambda$ are odd, we have that

$\dim\mathfrak{osp}(r_{0}|s_{0})=\frac{(r_{0}+s_{0})^{2}-(r_{0}+s_{0})}{2}+s_{0}=\frac{c_{0}^{2}-r_{0}+s_{0}}{2}.$

Note that in this case
$\dim\mathfrak{g}^{e}=\frac{1}{2}\dim\mathfrak{gl}(m|n)^{e}-\frac{r}{2}+\frac{s}{2}$
by Subsection 4.5 where $r$ (resp. $s$) is the total number of
$\lambda_{i},\left|i\right|=\bar{0}$ (resp.
$\lambda_{i},\left|i\right|=\bar{1}$). Since all parts of $\lambda$ are odd,
then each row which corresponds to a certain part of $\lambda$ has a box in
the $0$th column. Hence, the number of $\lambda_{i}$ with
$\left|i\right|=\bar{0}$ (resp. $\left|i\right|=\bar{1}$) is equal to the
number of even (resp. odd) boxes in the $0$th column, i.e. $r=r_{0}$ and
$s=s_{0}$. Thus we can calculate

$\dim\mathfrak{g}^{e}=\frac{1}{2}(\sum_{i\in\mathbb{Z}}c_{i}^{2}+c_{0}^{2})-\frac{r_{0}}{2}+\frac{s_{0}}{2}.$

Therefore, we have that

$\displaystyle\dim\mathfrak{g}^{e}-\dim$
$\displaystyle\mathfrak{g}_{0}^{e_{0}}=\left(\frac{1}{2}(\sum_{i\in\mathbb{Z}}c_{i}^{2}+c_{0}^{2})-\frac{r_{0}}{2}+\frac{s_{0}}{2}\right)-\left(\sum_{i>0}(c_{i}^{2}-1)+\frac{c_{0}^{2}-r_{0}+s_{0}}{2}\right)$
$\displaystyle=\sum_{i>0:c_{i}>0}1=n_{2}(\varDelta).$

When $r_{i}\neq s_{i}$ for all $i>0$, we have that
$\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=\dim\mathfrak{z}(\mathfrak{g}_{0})=0$
because $\mathfrak{z}(\mathfrak{sl}(r_{i}|s_{i}))=0$ for all $i$ and
$\mathfrak{z}(\mathfrak{osp}(r_{0}|s_{0}))=0$. However, if there exists
$r_{i}=s_{i}$ for some $i$, we have that $I_{r_{i}\mid
r_{i}}\in\mathfrak{z}(\mathfrak{sl}(r_{i}|s_{i}))$ and thus
$\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=\tau$ where $\tau$ is the number
of $i>0$ for which $r_{i}=s_{i}$. Hence, we have that
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}-\dim\left(\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})\right)^{G_{0}^{e_{0}}}=n_{2}(\varDelta)-\tau$.

Case 2: When there exist some labels equal to $1$ in $\varDelta$. Note that
there are in total $2\lambda_{1}+1$ columns in $P$ and let
$n_{2}(\varDelta)=t$. Observe that $\varDelta$ has some labels equal to $2$ if
there exist some $c_{k_{j}}=0$ for $j=1,\dots,t$ and $t=n_{2}(\varDelta)$. Let
$k>0$ be the minimal column number such that $c_{k}=0$ and $k+2t=\lambda_{1}$.
Note that once a label equal to $1$ occurs, say $a_{k}$, then there is no
label equal to $2$ for all $a_{h}$ with $h>k$. Then we have that

$\mathfrak{g}_{0}\cong\left(\bigoplus_{i=1}^{t}\mathfrak{sl}(r_{k+2i-1}|s_{k+2i-1})\right)\oplus\mathfrak{osp}(\sum_{i=1-k}^{k-1}r_{i}\mid\sum_{i=1-k}^{k-1}s_{i}).$

Hence,

$\dim\mathfrak{g}_{0}^{e_{0}}=\sum_{i=k+1}^{\lambda_{1}}\left(c_{i}^{2}+c_{i}c_{i+1}-1\right)+\dim\mathfrak{osp}(\sum_{i=1-k}^{k-1}r_{i}\mid\sum_{i=1-k}^{k-1}s_{i}).$

Denote the Jordan type of $e_{0}$ to be $\lambda^{0}$ which is defined the
similar way to (3.2). We also observe that $\left|\\{i:\lambda_{i}\text{ is
odd},\left|i\right|=\bar{0}\\}\right|=\left|\\{i:\lambda_{i}^{0}\text{ is
odd},\left|i\right|=\bar{0}\\}\right|$ and $\left|\\{i:\lambda_{i}\text{ is
odd},\left|i\right|=\bar{1}\\}\right|=\left|\\{i:\lambda_{i}^{0}\text{ is
odd},\left|i\right|=\bar{1}\\}\right|$. Note that

$\displaystyle\dim\mathfrak{osp}(\sum_{i=1-k}^{k-1}r_{i}|\sum_{i=1-k}^{k-1}s_{i})$
$\displaystyle=\frac{1}{2}(\sum_{i=1-k}^{k-1}(c_{i}^{2}+c_{i}c_{i+1}))$
$\displaystyle-\frac{1}{2}\left|\\{i:\lambda_{i}^{0}\text{ is
odd},\left|i\right|=\bar{0}\\}\right|$
$\displaystyle+\frac{1}{2}\left|\\{i:\lambda_{i}^{0}\text{ is
odd},\left|i\right|=\bar{1}\\}\right|.$

Therefore, we have that
$\dim\mathfrak{g}^{e}-\dim\mathfrak{g}_{0}^{e_{0}}=\sum_{i=1}^{t}1=t=n_{2}(\varDelta)$.
Moreover, let
$\mathfrak{\hat{g}}_{0}=\mathfrak{osp}(\sum_{i=1-k}^{k-1}r_{i}\mid\sum_{i=1-k}^{k-1}s_{i})$.
We observe that the projection of $e_{0}$ in each
$\mathfrak{sl}(r_{k+2i-1}|s_{k+2i-1})$ is $0$ and so
$e_{0}\in\mathfrak{\hat{g}}_{0}$. Thus

$\dim\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})=\sum_{i=1}^{t}\dim\mathfrak{z}(\mathfrak{sl}(r_{k+2i-1}|s_{k+2i-1}))+\text{dim}\mathfrak{z}(\mathfrak{\hat{g}}_{0}^{e_{0}}).$

Similar to Case 1, $\mathfrak{z}(\mathfrak{sl}(r_{k+2i-1}|s_{k+2i-1}))=0$ when
$r_{k+2i-1}\neq s_{k+2i-1}$ for all $i$. If there exist some $i>0$ such that
$r_{k+2i-1}=s_{k+2i-1}$, then
$I_{r_{k+2i-1}|r_{k+2i-1}}\in\mathfrak{z}(\mathfrak{sl}(r_{k+2i-1}|s_{k+2i-1}))$.
We know that
$\dim\left(\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})\right)^{G_{0}^{e_{0}}}=\lceil\frac{k_{1}-1}{2}\rceil+\tau$
where $\tau$ is the number of positive column number $i$ such that
$r_{i}=s_{i}$. Hence, we deduce that
$\dim\left(\mathfrak{z}(\mathfrak{g}^{e})\right)^{G^{e}}-\dim\left(\mathfrak{z}(\mathfrak{g}_{0}^{e_{0}})\right)^{G_{0}^{e_{0}}}=n_{2}(\varDelta)-\tau$
in this case.

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