# On split Leibniz superalgebras

Antonio J. Calderón Martín
Departamento de Matemáticas.
Universidad de Cádiz. 11510 Puerto Real, Cádiz, Spain.
e-mail<EMAIL_ADDRESS>

José M. Sánchez-Delgado
Departamento Álgebra, Geometría y Topología.
Universidad de Málaga. 29071 Málaga, Spain.
e-mail<EMAIL_ADDRESS>Supported by the PCI of the UCA ‘Teoría de Lie y
Teoría de Espacios de Banach’, by the PAI with project numbers FQM298, FQM7156
and by the project of the Spanish Ministerio de Educación y Ciencia
MTM2010-15223.

###### Abstract

We study the structure of arbitrary split Leibniz superalgebras. We show that
any of such superalgebras ${\mathfrak{L}}$ is of the form
${\mathfrak{L}}={\mathcal{U}}+\sum_{j}I_{j}$ with ${\mathcal{U}}$ a subspace
of an abelian (graded) subalgebra $H$ and any $I_{j}$ a well described
(graded) ideal of ${\mathfrak{L}}$ satisfying $[I_{j},I_{k}]=0$ if $j\neq k$.
In the case of ${\mathfrak{L}}$ being of maximal length, the simplicity of
${\mathfrak{L}}$ is also characterized in terms of connections of roots.

Keywords: Leibniz superalgebras, roots, root spaces, structure theory.

2010 MSC: 17A32, 17A70, 17B05.

##  1 Introduction and previous definitions

Throughout this paper, Leibniz superalgebras ${\mathfrak{L}}$ are considered
of arbitrary dimension and over an arbitrary base field 𝕂. It is worth to
mention that, unless otherwise stated, there is not any restriction on
$\dim{\mathfrak{L}}_{\alpha}$, the products
$[{\mathfrak{L}}_{\alpha},{\mathfrak{L}}_{-\alpha}]$, or $\\{k\in\hbox{\hueca
K}:k\alpha\in\Lambda\\}$, where ${\mathfrak{L}}_{\alpha}$ denotes the root
space associated to the root $\alpha$, and $\Lambda$ the set of nonzero roots
of ${\mathfrak{L}}$. The class of Leibniz superalgebras appears as an
extension of the one of Leibniz algebras (see [5, 7, 15, 20, 21, 22, 23, 24]),
in a similar way that Lie superalgebras extends to Lie algebras, motivated in
part for its applications in physics. In fact, the class of Leibniz
superalgebras also extends the one of Lie superalgebras by removing the skew-
supersymmetry, which is of interest in the formalism of mechanics of Nambu
[17].

In $\S 2$ we develop techniques of connections of roots in the framework of
split Leibniz superalgebras so as to show that ${\mathfrak{L}}$ is of the form
${\mathfrak{L}}={\mathcal{U}}+\sum_{j}I_{j}$ with ${\mathcal{U}}$ a subspace
of an abelian (graded) subalgebra $H$ and any $I_{j}$ a well described
(graded) ideal of ${\mathfrak{L}}$ satisfying $[I_{j},I_{k}]=0$ if $j\neq k$.
In $\S 3$ we focuss on those of maximal length and characterize the simplicity
of this class of superalgebras in terms of connections in the set of their
nonzero roots.

######  Definition 1.1.

A Leibniz superalgebra ${\mathfrak{L}}$ is a ${\hbox{\hueca Z}}_{2}$-graded
algebra
${\mathfrak{L}}={\mathfrak{L}}_{\bar{0}}\oplus{\mathfrak{L}}_{\bar{1}}$ over
an arbitrary base field ${\mathbb{K}}$, endowed with a bilinear product
$[\cdot,\cdot]$, whose homogenous elements
$x\in{\mathfrak{L}}_{\bar{i}},y\in{\mathfrak{L}}_{\bar{j}},{\bar{i}},{\bar{j}}\in{\hbox{\hueca
Z}}_{2}$, satisfy

$[x,y]\subset{\mathfrak{L}}_{{\bar{i}}+{\bar{j}}}$
$[x,[y,z]]=[[x,y],z]-(-1)^{{\bar{j}}{\bar{k}}}[[x,z],y]\hbox{ (Super Leibniz
identity)}$

for any homogenous element
$z\in{\mathfrak{L}}_{\bar{k}},{\bar{k}}\in{\hbox{\hueca Z}}_{2}$.

######  Remark 1.1.

Note that Super Leibniz identity is considered by the right side in the sense
that the multiplication operators on the right by elements in
${\mathfrak{L}}_{\bar{0}}$ are derivations on the homogeneous elements. In
this way is fixed in the references [5, 7, 15, 20, 24]. However, we could have
considered a Super Leibniz identity in which the multiplication operators on
the left by elements in ${\mathfrak{L}}_{\bar{0}}$ would act as derivations on
the homogeneous elements, as it is the case in the references [21, 22, 23]. Of
course, the development of the present job would have been similar in this
case.

Clearly ${\mathfrak{L}}_{\bar{0}}$ is a Leibniz algebra. Moreover, if the
identity $[x,y]=-(-1)^{{\bar{i}}{\bar{j}}}[y,x]$ holds we have that Super
Leibniz identity becomes Super Jacobi identity and so Leibniz superalgebras
generalize both Leibniz algebras and Lie superalgebras.

We also recall that the center of a Leibniz superalgebra ${\mathfrak{L}}$ is
the set

${\mathcal{Z}}({\mathfrak{L}}):=\\{x\in{\mathfrak{L}}:[x,{\mathfrak{L}}]+[{\mathfrak{L}},x]=0\\}.$

The usual regularity concepts will be understood in the graded sense. That is,
a supersubalgebra $A$ of ${\mathfrak{L}}$ is a graded subspace
$A=A_{\bar{0}}\oplus A_{\bar{1}}$ satisfying $[A,A]\subset A$; and an ideal
$I$ of ${\mathfrak{L}}$ is a graded subspace $I=I_{\bar{0}}\oplus I_{\bar{1}}$
of ${\mathfrak{L}}$ such that

$[I,{\mathfrak{L}}]+[{\mathfrak{L}},I]\subset I.$

The (graded) ideal ${\mathfrak{I}}$ generated by

$\\{[x,y]+(-1)^{\bar{i}\bar{j}}[y,x]:x\in{\mathfrak{L}}_{\bar{i}},y\in{\mathfrak{L}}_{\bar{j}},\bar{i},\bar{j}\in{\hbox{\hueca
Z}}_{2}\\}$

plays an important role in the theory since it determines the (possible) non-
super Lie character of ${\mathfrak{L}}$. From Super Leibniz identity, it is
straightforward to check that this ideal satisfies

$[{\mathfrak{L}},{\mathfrak{I}}]=0.$ (1)

Here we note that the usual definition of simple superalgebra lack of interest
in the case of Leibniz superalgebras because would imply the ideal
${\mathfrak{I}}={\mathfrak{L}}$ or ${\mathfrak{I}}={0}$, being so
${\mathfrak{L}}$ an abelian (product zero) or a Lie superalgebra respectively
(see Equation (1)). Abdykassymova and Dzhumadil’daev introduced in [1, 2] an
adequate definition in the case of Leibniz algebras $(L,[\cdot,\cdot])$ by
calling simple to the ones satisfying that its only ideals are $\\{0\\}$, $L$
and the one generated by the set $\\{[x,x]:x\in L\\}$. Following this vain, we
consider the next definition.

######  Definition 1.2.

A Leibniz superalgebra ${\mathfrak{L}}$ will be called simple if
$[{\mathfrak{L}},{\mathfrak{L}}]\neq 0$ and its only (graded) ideals are
$\\{0\\},{\mathfrak{I}}$ and ${\mathfrak{L}}$.

Let us introduce the class of split algebras in the framework of Leibniz
superalgebras in a similar way to the cases of Lie algebras, Leibniz algebras
and Lie superalgebras, (see [8], [13] and [11] respectively), among other
classes of algebras.

We begin by considering a maximal abelian graded subalgebra
$H=H_{\bar{0}}\oplus H_{\bar{1}}$ among the abelian graded subalgebras of
${\mathfrak{L}}$.

######  Definition 1.3.

Denote by $H=H_{\bar{0}}\oplus H_{\bar{1}}$ a maximal abelian (graded)
subalgebra of a Leibniz superalgebra ${\mathfrak{L}}$. For a linear functional

$\alpha:H_{\bar{0}}\longrightarrow\hbox{\hueca K},$

we define the root space of ${\mathfrak{L}}$ associated to $\alpha$ as the
subspace

${\mathfrak{L}}_{\alpha}:=\\{v_{\alpha}\in{\mathfrak{L}}:[v_{\alpha},h_{\bar{0}}]=\alpha(h_{\bar{0}})v_{\alpha}\hskip
5.69046pt{\it for}\hskip 5.69046pt{\it any}\hskip 5.69046pth_{\bar{0}}\in
H_{\bar{0}}\\}.$

The elements $\alpha\in(H_{\bar{0}})^{*}$ satisfying
${\mathfrak{L}}_{\alpha}\neq 0$ are called roots of ${\mathfrak{L}}$ respect
to $H$ and we denote
$\Lambda:=\\{\alpha\in(H_{\bar{0}})^{*}\setminus\\{0\\}:{\mathfrak{L}}_{\alpha}\neq
0\\}$. We say that ${\mathfrak{L}}$ is a split Leibniz superalgebra if

${\mathfrak{L}}=H\oplus(\bigoplus\limits_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha}).$

We also say that $\Lambda$ is the root system of ${\mathfrak{L}}$.

Split Lie algebras, split Leibniz algebras and split Lie superalgebras are
examples of split Leibniz superalgebras. Hence, the present paper extends the
results in [8], [11] and [13]. We also have to note that the results in the
above references are given for split Lie algebras, split Leibniz algebras and
split Lie superalgebras with a symmetric root system, that is, satisfying that
if $\alpha$ is a nonzero root then $-\alpha$ it is also. However, our results
are developed for split Leibniz superalgebras with a non-necessarily symmetric
root system. Hence, the results in the present paper not only extend the
previous ones for the classes of split Lie algebras, split Leibniz algebras
and split Lie superalgebras to a more general class of algebras, (the one of
split Leibniz superalgebras), but also extend the results on these classes of
algebras given in [8, 11, 13] to split Lie algebras, split Leibniz algebras
and split Lie superalgebras with a non-necessarily symmetric root system.

We would finally like to mention the works of Albeverio, Ayupov, Gago, Ladra,
Liu, Omirov and Turdibaev in the finite-dimensional setup, where Cartan
subalgebras, roots and root spaces of Leibniz and $n$-Leibniz algebras are
studied, [4, 25, 26, 27]; and also to the papers [16, 28] in the framework of
Leibniz algebras.

######  Example 1.1.

Consider the 5-dimensional ${\mathbb{Z}}_{2}$-graded vector space
${\mathfrak{L}}={\mathfrak{L}}_{\bar{0}}\oplus{\mathfrak{L}}_{\bar{1}}$, over
a base field ${\mathbb{K}}$ of characteristic different from 2, with basis
$\\{u_{1},u_{2},u_{3}\\}$ of ${\mathfrak{L}}_{\bar{0}}$ and
$\\{e_{1},e_{2}\\}$ of ${\mathfrak{L}}_{\bar{1}};$ where the products on these
elements are given by:

$[u_{2},u_{1}]=-u_{3},\hskip 8.5359pt[u_{1},u_{2}]=u_{3},\hskip
8.5359pt[u_{1},u_{3}]=-2u_{1},$ $[u_{3},u_{1}]=2u_{1},\hskip
8.5359pt[u_{3},u_{2}]=-2u_{2},\hskip 8.5359pt[u_{2},u_{3}]=2u_{2},$
$[e_{1},u_{2}]=e_{2},\hskip 8.5359pt[e_{1},u_{3}]=-e_{1},\hskip
8.5359pt[e_{2},u_{1}]=e_{1},\hskip 8.5359pt[e_{2},u_{3}]=e_{2},$

and where the omitted products are equal to zero. Then
${\mathfrak{L}}={\mathfrak{L}}_{\bar{0}}\oplus{\mathfrak{L}}_{\bar{1}}$
becomes a (non-Lie) split Leibniz superalgebra

${\mathfrak{L}}=H\oplus{\mathfrak{L}}_{\alpha}\oplus{\mathfrak{L}}_{-\alpha}\oplus{\mathfrak{L}}_{\beta}\oplus{\mathfrak{L}}_{-\beta}$

where $H=\langle u_{3}\rangle,$ ${\mathfrak{L}}_{\alpha}=\langle
u_{2}\rangle,$ ${\mathfrak{L}}_{-\alpha}=\langle u_{1}\rangle,$
${\mathfrak{L}}_{\beta}=\langle e_{1}\rangle$ and
${\mathfrak{L}}_{-\beta}=\langle e_{2}\rangle,$ being
$\alpha,\beta:H\to{\mathbb{K}}$ defined by $\alpha(\lambda u_{3})=2\lambda$
and $\beta(\lambda u_{3})=-\lambda$, $\lambda\in{\mathbb{K}}.$

######  Example 1.2.

Consider the $(n+4)$-dimensional complex ${\mathbb{Z}}_{2}$-graded vector
space ${\mathfrak{L}}={\mathfrak{L}}_{\bar{0}}\oplus{\mathfrak{L}}_{\bar{1}}$
with basis $\\{h,u,v\\}$ of ${\mathfrak{L}}_{\bar{0}}$ and
$\\{e_{0},e_{1},...,e_{n}\\}$ of ${\mathfrak{L}}_{\bar{1}};$ with the
following table of multiplication:

$[u,h]=2u,\hskip 8.5359pt[h,u]=-2u,\hskip 8.5359pt[v,h]=-2v,$ $[h,v]=2v,\hskip
8.5359pt[u,v]=h,\hskip 8.5359pt[v,u]=-h,$ $[e_{k},h]=(n-2k)e_{k},$ for $0\leq
k\leq n;$ $[e_{k},v]=e_{k+1},$ for $0\leq k\leq n-1;$
$[e_{k},u]=k(k-n-1)e_{k-1},$ for $1\leq k\leq n;$

and where the omitted products are equal to zero. Then
${\mathfrak{L}}={\mathfrak{L}}_{\bar{0}}\oplus{\mathfrak{L}}_{\bar{1}}$ is a
(non-Lie) split Leibniz superalgebra

${\mathfrak{L}}=H\oplus{\mathfrak{L}}_{\alpha}\oplus{\mathfrak{L}}_{-\alpha}\oplus(\bigoplus\limits_{k=0}^{n}{\mathfrak{L}}_{\beta_{k}})$

where $H=\langle h\rangle,$ ${\mathfrak{L}}_{\alpha}=\langle u\rangle,$
${\mathfrak{L}}_{-\alpha}=\langle v\rangle$ and
${\mathfrak{L}}_{\beta_{k}}=\langle e_{k}\rangle,$ $k=0,...,n$; being
$\alpha,\beta_{k}:H\to{\mathbb{K}}$ defined by $\alpha(\lambda h)=2\lambda$
and $\beta_{k}(\lambda h)=(n-2k)\lambda$ for $k=0,...,n$ and
$\lambda\in{\mathbb{C}}.$

By returning to a general split Leibniz superalgebra ${\mathfrak{L}}$,
(Definition 1.3), it is clear that the root space associated to the zero root
satisfies $H\subset{\mathfrak{L}}_{0}.$ Conversely, given any
$v_{0}\in{\mathfrak{L}}_{0}$ we can write
$v_{0}=h+\sum\limits_{i=1}^{n}v_{\alpha_{i}}$ with $h\in H$ and
$v_{\alpha_{i}}\in{\mathfrak{L}}_{\alpha_{i}}$ for $i\in\\{1,...,n\\}$, being
$\alpha_{i}\in\Lambda$ with $\alpha_{i}\neq\alpha_{j}$ if $i\neq j$. Hence
$0=[h+\sum\limits_{i=1}^{n}v_{\alpha_{i}},h_{\bar{0}}]=\sum\limits_{i=1}^{n}\alpha_{i}(h_{\bar{0}})v_{\alpha_{i}}$
for any $h_{\bar{0}}\in H_{\bar{0}}$. So, taking into account the direct
character of the sum and that $\alpha_{i}\neq 0$, we have that any
$v_{\alpha_{i}}=0$ and then $v_{0}\in H$. Consequently

$H={\mathfrak{L}}_{0}.$

By the grading of ${\mathfrak{L}}$ we have that given any
$v_{\alpha}\in{\mathfrak{L}}_{\alpha},\alpha\in\Lambda\cup\\{0\\}$, expressed
in the form $v_{\alpha}=v_{\alpha,{\bar{0}}}+v_{{\alpha},{\bar{1}}}$ with
$v_{\alpha,{\bar{i}}}\in{\mathfrak{L}}_{\bar{i}},{\bar{i}}\in{\hbox{\hueca
Z}}_{2}$, then

$[v_{\alpha,{\bar{i}}},h_{\bar{0}}]=\alpha(h_{\bar{0}})v_{\alpha,{\bar{i}}}$

for any $h_{\bar{0}}\in H_{\bar{0}}$. From here,
${\mathfrak{L}}_{\alpha}=({\mathfrak{L}}_{\alpha}\cap{\mathfrak{L}}_{\bar{0}})\oplus({\mathfrak{L}}_{\alpha}\cap{\mathfrak{L}}_{\bar{1}})$.
Hence, by denoting

${\mathfrak{L}}_{\alpha,{\bar{i}}}:={\mathfrak{L}}_{\alpha}\cap{\mathfrak{L}}_{\bar{i}}$

we can write

${\mathfrak{L}}_{\alpha}={\mathfrak{L}}_{\alpha,{\bar{0}}}\oplus{\mathfrak{L}}_{\alpha,{\bar{1}}}$
(2)

for any $\alpha\in\Lambda\cup\\{0\\}$.

From the above,

$H_{\bar{0}}={\mathfrak{L}}_{0,{\bar{0}}}$ and
$H_{\bar{1}}={\mathfrak{L}}_{0,{\bar{1}}}$,

and also

${\mathfrak{L}}_{\bar{0}}=H_{\bar{0}}\oplus(\bigoplus\limits_{\alpha\in\Lambda_{\bar{0}}}{\mathfrak{L}}_{\alpha,{\bar{0}}}),$
${\mathfrak{L}}_{\bar{1}}=H_{\bar{1}}\oplus(\bigoplus\limits_{\alpha\in\Lambda_{\bar{1}}}{\mathfrak{L}}_{\alpha,{\bar{1}}})$,

where

$\Lambda_{\bar{i}}=\\{\alpha\in\Lambda:{\mathfrak{L}}_{\alpha,{\bar{i}}}\neq
0\\}$

for any ${\bar{i}}\in{\mathbb{Z}}_{2}$.

Taking into account the above expression of ${\mathfrak{L}}_{\bar{0}}$, the
direct character of the sum and the fact that $\alpha\neq 0$ for any
$\alpha\in\Lambda$, we have that $H_{\bar{0}}$ is a maximal abelian subalgebra
of the Leibniz algebra ${\mathfrak{L}}_{\bar{0}}$. Hence
${\mathfrak{L}}_{\bar{0}}$ is a split Leibniz algebra respect to $H_{\bar{0}}$
(see [13]).

The below lemma in an immediate consequence of Super Leibniz identity.

######  Lemma 1.1.

If $[{\mathfrak{L}}_{\alpha,\bar{i}},{\mathfrak{L}}_{\beta,\bar{j}}]\neq 0$
with $\alpha,\beta\in\Lambda\cup\\{0\\}$ and
$\bar{i},\bar{j}\in{\mathbb{Z}}_{2}$, then
$\alpha+\beta\in\Lambda_{\bar{i}+\bar{j}}\cup\\{0\\}$ and
$[{\mathfrak{L}}_{\alpha,\bar{i}},{\mathfrak{L}}_{\beta,\bar{j}}]\subset{\mathfrak{L}}_{\alpha+\beta,\bar{i}+\bar{j}}$.

From Lemma 1.1 and Equation (2) we also can assert that

$[{\mathfrak{L}}_{\alpha},{\mathfrak{L}}_{\beta}]\subset{\mathfrak{L}}_{\alpha+\beta}$
(3)

for any $\alpha,\beta\in\Lambda\cup\\{0\\}$.

##  2 Connections of Roots. Decompositions

In the following, ${\mathfrak{L}}$ denotes a split Leibniz superalgebra and
${\mathfrak{L}}=H\oplus(\bigoplus\limits_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha})$
the corresponding root spaces decomposition. We begin by developing
connections of roots techniques in this framework.

Given a linear functional $\alpha:H_{\bar{0}}\to{\hbox{\hueca K}}$, we denote
by $-\alpha:H_{\bar{0}}\to{\hbox{\hueca K}}$ the element in $H_{\bar{0}}^{*}$
defined by $(-\alpha)(h):=-\alpha(h)$ for all $h\in H_{\bar{0}}$. We will also
denote by $-\Lambda=\\{-\alpha:\alpha\in\Lambda\\}$ and by
$\pm\Lambda=\Lambda\cup-\Lambda$.

######  Definition 2.1.

Let $\alpha\in\Lambda$ and $\beta\in\Lambda$ be two nonzero roots. We say that
$\alpha$ is connected to $\beta$ if there exists a family
$\alpha_{1},\alpha_{2},...,\alpha_{n}\in\pm\Lambda$ satisfying the following
conditions:

  1. 1.

$\alpha_{1}=\alpha.$

  2. 2.

$\\{\alpha_{1}+\alpha_{2},\alpha_{1}+\alpha_{2}+\alpha_{3},...,\alpha_{1}+\cdots+\alpha_{n-1}\\}\subset\pm\Lambda.$

  3. 3.

$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\in\\{\beta,-\beta\\}$.

We also say that $\\{\alpha_{1},...,\alpha_{n}\\}$ is a connection from
$\alpha$ to $\beta$.

Trivially, for any $\alpha\in\Lambda$ we have that the set $\\{\alpha\\}$ is a
connection from $\alpha$ to itself, and that in case $-\alpha$ also belongs to
$\Lambda$ the same set is a connection from $\alpha$ to $-\alpha$.

Next, given $\alpha$ and $\beta$ two elements of $\Lambda$, we write
$\alpha\sim\beta$ when $\alpha$ is connected to $\beta$. Note that the
observation in the above paragraph gives us $\alpha\sim\alpha$ for any
$\alpha\in\Lambda$. That is, the relation $\sim$ is reflexive. If
$\alpha\sim\beta$ there exists a connection
$\\{\alpha_{1},\alpha_{2},\alpha_{3},...,\alpha_{n-1},\alpha_{n}\\}\subset\pm\Lambda$
from $\alpha$ to $\beta$ satisfying, in particular,
$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=\epsilon\beta$ for some
$\epsilon\in\\{1,-1\\}$. If $n=1$ then $\alpha_{1}=\alpha$ and
$\beta\in\\{\alpha,-\alpha\\}$. So $\\{\beta\\}$ is a connection from $\beta$
to $\alpha$. If $n>1$, then we can verify in a straightforward way that

$\\{\beta,-\epsilon\alpha_{n},-\epsilon\alpha_{n-1},...,-\epsilon\alpha_{3},-\epsilon\alpha_{2}\\}$

is a connection from $\beta$ to $\alpha$. That is, $\sim$ is symmetric.
Finally, suppose $\alpha\sim\beta$ and $\beta\sim\gamma$, and write
$\\{\alpha_{1},\alpha_{2},...,\alpha_{n}\\}$ for a connection from $\alpha$ to
$\beta$, which satisfies
$\alpha_{1}+\alpha_{2}\cdots+\alpha_{n}=\epsilon\beta$ for some
$\epsilon\in\\{1,-1\\}$; and $\\{\beta_{1},\beta_{2},...,\beta_{m}\\}$ for a
connection from $\beta$ to $\gamma$. If $m=1$, then
$\gamma\in\pm\\{\beta,-\beta\\}$ and so
$\\{\alpha_{1},\alpha_{2},...,\alpha_{n}\\}$ is a connection from $\alpha$ to
$\gamma$. If $m>1$, then it is easy to show that
$\\{\alpha_{1},\alpha_{2},...,\alpha_{n},\epsilon\beta_{2},...,\epsilon\beta_{m}\\}$
is a connection from $\alpha$ to $\gamma$. That is, $\sim$ is transitive.
Summarizing, (see also [13, Proposition 2.1]), we can assert:

######  Proposition 2.1.

The relation $\sim$ in $\Lambda$, defined by $\alpha\sim\beta$ if and only if
$\alpha$ is connected to $\beta$ is an equivalence relation.

By Proposition 2.1 the connection relation is an equivalence relation in
$\Lambda$ and so we can consider the quotient set

$\Lambda/\sim:=\\{[\alpha]:\alpha\in\Lambda\\},$

becoming $[\alpha]$ the set of nonzero roots of ${\mathfrak{L}}$ which are
connected to $\alpha$. By the definition of $\sim$, it is clear that if
$\beta\in[\alpha]$ and $-\beta\in\Lambda$ then $-\beta\in[\alpha]$.

Our next goal is to associate an ideal $I_{[\alpha]}$ of ${\mathfrak{L}}$ to
any $[\alpha]$. Fix $\alpha\in\Lambda$, we start by defining the set
$H_{[\alpha]}\subset H$ as follows

$H_{[\alpha]}:=span_{\hbox{\hueca
K}}\\{[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]:\beta\in[\alpha]\\}=$
$=(\sum\limits_{\beta\in[\alpha]}([{\mathfrak{L}}_{\beta,{\bar{0}}},{\mathfrak{L}}_{-\beta,{\bar{0}}}]+[{\mathfrak{L}}_{\beta,{\bar{1}}},{\mathfrak{L}}_{-\beta,{\bar{1}}}]))\oplus(\sum\limits_{\beta\in[\alpha]}([{\mathfrak{L}}_{\beta,{\bar{0}}},{\mathfrak{L}}_{-\beta,{\bar{1}}}]+[{\mathfrak{L}}_{\beta,{\bar{1}}},{\mathfrak{L}}_{-\beta,{\bar{0}}}]))\subset$
$\subset H_{\bar{0}}\oplus H_{\bar{1}},$

last equality being consequence of Equation (3). Next, we define

$V_{[\alpha]}:=\bigoplus\limits_{\beta\in[\alpha]}{\mathfrak{L}}_{\beta}=\bigoplus\limits_{\beta\in[\alpha]}({\mathfrak{L}}_{\beta,{\bar{0}}}\oplus{\mathfrak{L}}_{\beta,{\bar{1}}}).$

Finally, we denote by ${\mathfrak{L}}_{[\alpha]}$ the (graded) subspace given
by the direct sum of the two subspaces above, that is,

${\mathfrak{L}}_{[\alpha]}:=H_{[\alpha]}\oplus V_{[\alpha]}.$

######  Proposition 2.2.

For any $\alpha\in\Lambda$, the linear subspace ${\mathfrak{L}}_{[\alpha]}$ is
a Leibniz supersubalgebra of ${\mathfrak{L}}$.

###### Proof.

We have to check that ${\mathfrak{L}}_{[\alpha]}$ satisfies

$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}_{[\alpha]}]\subset{\mathfrak{L}}_{[\alpha]}.$

Taking into account $H={\mathfrak{L}}_{0}$ and Equation (3), we have

$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}_{[\alpha]}]=[H_{[\alpha]}\oplus
V_{[\alpha]},H_{[\alpha]}\oplus V_{[\alpha]}]\subset
V_{[\alpha]}+\sum\limits_{\beta,\delta\in[\alpha]}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{\delta}].$
(4)

If $\delta=-\beta$ then

$[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{\delta}]\subset H_{[\alpha]}.$ (5)

If $\delta\neq-\beta$, by Equation (3) we have that in case
$[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{\delta}]\neq 0$ then
$\beta+\delta\in\Lambda$. From here, $\\{\beta,\delta\\}$ is a connection from
$\beta$ to $\beta+\delta$. Hence $\beta+\delta\in[\alpha]$ and so

$[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{\delta}]\subset V_{[\alpha]}.$ (6)

From Equations (4), (5) and (6) we conclude
$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}_{[\alpha]}]\subset{\mathfrak{L}}_{[\alpha]}$.
∎

######  Proposition 2.3.

For any $[\alpha]\neq[\beta]$ we have
$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}_{[\beta]}]=0.$

###### Proof.

We have

$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}_{[\beta]}]=[H_{[\alpha]}\oplus
V_{[\alpha]},H_{[\beta]}\oplus V_{[\beta]}]\subset$
$\subset[H_{[\alpha]},V_{[\beta]}]+[V_{[\alpha]},H_{[\beta]}]+[V_{[\alpha]},V_{[\beta]}].$
(7)

Consider the above third summand $[V_{[\alpha]},V_{[\beta]}]$ and suppose
there exist $\eta\in[\alpha]$ and $\gamma\in[\beta]$ such that
$[{\mathfrak{L}}_{\eta},{\mathfrak{L}}_{\gamma}]\neq 0$. As necessarily
$\eta\neq-\gamma$, then $\eta+\gamma\in\Lambda$. So $\\{\eta,\gamma,-\eta\\}$
is a connection from $\eta$ to $\gamma$. By the transitivity of the connection
relation we have $\alpha\sim\beta$, a contradiction. Hence
$[{\mathfrak{L}}_{\eta},{\mathfrak{L}}_{\gamma}]=0$ and so

$[V_{[\alpha]},V_{[\beta]}]=0.$ (8)

Consider now the second summand $[V_{[\alpha]},H_{[\beta]}]$ from (7) and
suppose exist $\eta\in[\alpha],\gamma\in[\beta]$ such that
$[{\mathfrak{L}}_{\eta},[{\mathfrak{L}}_{\gamma},{\mathfrak{L}}_{-\gamma}]]\neq
0$. Hence, there exist $\bar{i},\bar{j},\bar{k}\in{\mathbb{Z}}_{2}$ such that

$[{\mathfrak{L}}_{\eta,\bar{i}},[{\mathfrak{L}}_{\gamma,\bar{j}},{\mathfrak{L}}_{-\gamma,\bar{k}}]]\neq
0.$

By Super Leibniz identity we get either
$[{\mathfrak{L}}_{\eta,\bar{i}},{\mathfrak{L}}_{\gamma,\bar{j}}]\neq 0$ or
$[{\mathfrak{L}}_{\eta,\bar{i}},{\mathfrak{L}}_{-\gamma,\bar{k}}]\neq 0$. From
here $[V_{[\alpha]},V_{[\beta]}]\neq 0$ in any case, what contradicts Equation
(8).

Finally, consider the first summand $[H_{[\alpha]},V_{[\beta]}]$ from (7) and
suppose there exist $\eta\in[\alpha],\gamma\in[\beta]$ such that
$[[{\mathfrak{L}}_{\eta},{\mathfrak{L}}_{-\eta}],{\mathfrak{L}}_{\gamma}]\neq
0$. We have as above that we can take
$\bar{i},\bar{j},\bar{k}\in{\mathbb{Z}}_{2}$ satisfying

$[[{\mathfrak{L}}_{\eta,{\bar{i}}},{\mathfrak{L}}_{-\eta,{\bar{j}}}],{\mathfrak{L}}_{\gamma,{\bar{k}}}]\neq
0,$

and that by Super Leibniz identity either
$[{\mathfrak{L}}_{\eta,\bar{i}},{\mathfrak{L}}_{\gamma,\bar{k}}]\neq 0$ or
$[{\mathfrak{L}}_{-\eta,\bar{j}},{\mathfrak{L}}_{\gamma,\bar{k}}]\neq 0$. From
here $[V_{[\alpha]},V_{[\beta]}]\neq 0$ in any case, what contradicts Equation
(8). Hence

$[V_{[\alpha]},V_{[\beta]}]=0.$

By Equation (7) we conclude
$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}_{[\beta]}]=0$. ∎

Proposition 2.2 allows us to assert that for any
$\alpha\in\Lambda,{\mathfrak{L}}_{[\alpha]}$ is a Leibniz supersubalgebra of
${\mathfrak{L}}$ that we call the Leibniz supersubalgebra of ${\mathfrak{L}}$
associated to $[\alpha]$.

######  Theorem 2.1.

The following assertions hold.

  1. 1.

For any $\alpha\in\Lambda$, the Leibniz supersubalgebra

${\mathfrak{L}}_{[\alpha]}=H_{[\alpha]}\oplus V_{[\alpha]}$

of ${\mathfrak{L}}$ associated to $[\alpha]$ is an ideal of ${\mathfrak{L}}$.

  2. 2.

If ${\mathfrak{L}}$ is simple, then there exists a connection between any two
nonzero roots of ${\mathfrak{L}}$ and
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$.

###### Proof.

1\. We have to check that

$[{\mathfrak{L}}_{[\alpha]},{\mathfrak{L}}]+[{\mathfrak{L}},{\mathfrak{L}}_{[\alpha]}]\subset{\mathfrak{L}}_{[\alpha]}.$

Since ${\mathfrak{L}}_{[\alpha]}=H_{[\alpha]}\oplus V_{[\alpha]}$ and
${\mathfrak{L}}=H\oplus\big{(}\bigoplus\limits_{\beta\in\Lambda}{\mathfrak{L}}_{\beta}\big{)}$,
we are going to consider several steps. First note that

$[H_{[{\alpha}]},H]+[H,{H}_{[{\alpha}]}]\subset[H,H]=0.$

Next, given $\beta\in\Lambda$ an application of Propositions 2.2 and 2.3 gives

$[{\mathfrak{L}}_{[{\alpha}]},{\mathfrak{L}}_{\beta}]+[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{[{\alpha}]}]\subset{\mathfrak{L}}_{[{\alpha}]}.$

Therefore,

$[{\mathfrak{L}}_{[{\alpha}]},{\mathfrak{L}}]+[{\mathfrak{L}},{\mathfrak{L}}_{[{\alpha}]}]\subset{\mathfrak{L}}_{[{\alpha}]}.$

2\. The simplicity of ${\mathfrak{L}}$ applies to get that
${\mathfrak{L}}_{[\alpha]}\in\\{{\mathfrak{I}},{\mathfrak{L}}\\}$ for any
$\alpha\in\Lambda$. If ${\mathfrak{L}}_{[{\alpha_{0}}]}={\mathfrak{L}}$ for
some $\alpha_{0}\in\Lambda$ then ${[{\alpha_{0}}]}=\Lambda$; which implies
that any nonzero root of ${\mathfrak{L}}$ is connected to $\alpha_{0}$, and
therefore, any nonzero roots of ${\mathfrak{L}}$ are connected. Assume now
that ${\mathfrak{L}}_{[\alpha]}={\mathfrak{I}}$ for every $\alpha\in\Lambda$.
Then $[\alpha]=[\beta]$ for any $\alpha,\beta\in{\mathfrak{L}}$. Since
$\Lambda=\bigcup\limits_{[\gamma]\in\Lambda\diagup\sim}[\gamma]$ we get
$[\alpha]=\Lambda$. That is, all nonzero roots of ${\mathfrak{L}}$ are
connected. To finish, observe that the fact
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$
follows in any case. ∎

######  Notation 2.1.

Let us denote

$H_{\Lambda}:=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$

In what follows, we will use the terminology
$I_{[\alpha]}:={\mathfrak{L}}_{[\alpha]}$ where ${\mathfrak{L}}_{[\alpha]}$ is
one of the ideals of ${\mathfrak{L}}$ described in Theorem 2.1-1.

######  Theorem 2.2.

For a vector space complement $\mathcal{U}$ of $H_{\Lambda}$ in $H$, it
follows

${\mathfrak{L}}=\mathcal{U}+\sum_{[\alpha]\in\Lambda/\sim}I_{[\alpha]}.$

Moreover,

$[I_{[\alpha]},I_{[\beta]}]=0,$

whenever $[\alpha]\neq[\beta]$.

###### Proof.

From

${\mathfrak{L}}=H\oplus\big{(}\bigoplus\limits_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha}\big{)}=(\mathcal{U}\oplus{H_{\Lambda}})\oplus(\bigoplus_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha}),$

it follows

$\bigoplus\limits_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha}=\bigoplus\limits_{[\alpha]\in\Lambda^{/\sim}}V_{[\alpha]},\quad\quad
H_{\Lambda}=\sum_{[\alpha]\in\Lambda^{/\sim}}H_{[\alpha]},$

which implies

${\mathfrak{L}}=(\mathcal{U}\oplus
H_{\Lambda})\oplus(\bigoplus_{\alpha\in\Lambda}{{\mathfrak{L}}_{\alpha}})=\mathcal{U}+\sum\limits_{[\alpha]\in\Lambda^{/\sim}}I_{[\alpha]},$

where each $I_{[\alpha]}$ is an ideal of ${\mathfrak{L}}$ by Theorem 2.1. Now,
given $[\alpha]\neq[\beta]$, the assertion

$[I_{[\alpha]},I_{[\beta]}]=0,$

follows of Proposition 2.3. ∎

######  Corollary 2.1.

If ${\mathcal{Z}}({\mathfrak{L}})=0$ and
${\mathfrak{L}}=[{\mathfrak{L}},{\mathfrak{L}}]$, then

${\mathfrak{L}}=\bigoplus\limits_{[\alpha]\in\Lambda/\sim}I_{[\alpha]}.$

###### Proof.

Since $[{\mathfrak{L}},{\mathfrak{L}}]={\mathfrak{L}}$, Theorem 2.2 applies to
get

$\left[\mathcal{U}+\sum_{[\alpha]\in\Lambda/\sim}I_{[\alpha]},\,\,\,\mathcal{U}+\sum_{[\alpha]\in\Lambda/\sim}I_{[\alpha]}\right]=\mathcal{U}+\sum_{[\alpha]\in\Lambda/\sim}I_{[\alpha]}.$

Keeping in mind that

$\mathcal{U}\subset H\quad\mbox{ and
}\quad[I_{[\alpha]},I_{[\beta]}]=0,\,\,\mbox{ if }\,\,[\alpha]\neq[\beta],$

we get $\mathcal{U}=0$, that is,

${\mathfrak{L}}=\sum_{[\alpha]\in\Lambda/\sim}I_{[\alpha]}.$

To finish, we show the direct character of the sum. Given $x\in
I_{[\alpha]}\cap\sum\limits_{\tiny{\begin{array}[]{c}[\beta]\in\Lambda/\sim\\\
\beta\nsim\alpha\\\ \end{array}}}I_{[\beta]}$ then using again the equation
$\,\,[I_{[\alpha]},I_{[\beta]}]=0$, for $[\alpha]\neq[\beta]$, we obtain

$\displaystyle\big{[}x,I_{[\alpha]}\big{]}+\left[x,\sum_{\tiny{\begin{array}[]{c}[\beta]\in\Lambda/\sim\\\
\beta\nsim\alpha\\\ \end{array}}}I_{[\beta]}\right]$ $\displaystyle=0,$
$\displaystyle\big{[}I_{[\alpha]},x\big{]}+\left[\sum_{\tiny{\begin{array}[]{c}[\beta]\in\Lambda/\sim\\\
\beta\nsim\alpha\\\ \end{array}}}I_{[\beta]},x\right]$ $\displaystyle=0.$

It implies $[x,{\mathfrak{L}}]+[{\mathfrak{L}},x]=0$, that is,
$x\in{\mathcal{Z}}({\mathfrak{L}})=0$. Thus $x=0$, as desired. ∎

##  3 Split Leibniz superalgebras of maximal length. The simple components

In this section we focus on the simplicity of split Leibniz superalgebras by
centering our attention in those of maximal length. This terminology is taking
borrowed from the theory of gradations of Lie and Leibniz algebras, (see [3,
6, 18, 19]). See also [6, 8, 9, 11, 13, 14, 29] for examples.

######  Definition 3.1.

We say that a split Leibniz superalgebra ${\mathfrak{L}}$ is of maximal length
if $\dim({{\mathfrak{L}}}_{\alpha,\bar{i}})\in\\{0,1\\}$ for any
$\alpha\in\Lambda,$ $\bar{i}\in{\hbox{\hueca Z}}_{2}.$

Our target is to characterize the simplicity of ${\mathfrak{L}}$ in terms of
connectivity properties in $\Lambda$. Therefore we would like to attract
attention to the definition of simple Leibniz superalgebra given in Definition
1.2, and the previous discussion.

The following lemma is consequence of the fact that the set of multiplications
by elements in $H_{\bar{0}}$ is a commuting set of diagonalizable
endomorphisms and $I$ is invariant under this set.

######  Lemma 3.1.

Let
${\mathfrak{L}}=H\oplus(\bigoplus\limits_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha})$
be a split Leibniz superalgebra. If $I$ is an ideal of ${\mathfrak{L}}$ then
$I=(I\cap
H)\oplus(\bigoplus\limits_{\alpha\in\Lambda}(I\cap{\mathfrak{L}}_{\alpha})).$

From now on
${\mathfrak{L}}=H\oplus(\bigoplus\limits_{\alpha\in\Lambda}{\mathfrak{L}}_{\alpha})$
denotes a split Leibniz superalgebra of maximal length without further
mention. We begin by observing that in this case Lemma 3.1 allows us to assert
that given any nonzero (graded) ideal $I=I_{\bar{0}}\oplus I_{{\bar{1}}}$ of
${\mathfrak{L}}$ then

$I=(I\cap
H)\oplus(\bigoplus\limits_{\alpha\in\Lambda}I\cap{\mathfrak{L}}_{\alpha})=$
$(I\cap
H)\oplus(\bigoplus\limits_{\alpha\in\Lambda}((I_{\bar{0}}\cap{\mathfrak{L}}_{\alpha,{\bar{0}}})\oplus(I_{{\bar{1}}}\cap{\mathfrak{L}}_{\alpha,{\bar{1}}})))=$

$(I\cap
H)\oplus(\bigoplus\limits_{\alpha\in\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\alpha,\bar{0}})\oplus(\bigoplus\limits_{\beta\in\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\beta,\bar{1}})$
(9)

where
$\Lambda_{\bar{i}}^{I}:=\\{\gamma\in\Lambda:I_{\bar{i}}\cap{\mathfrak{L}}_{\gamma,\bar{i}}\neq
0\\}$ for $\bar{i}\in{\hbox{\hueca Z}}_{2}$. In the particular, (an important)
case $I={\mathfrak{I}}$, we get

${\mathfrak{I}}=({\mathfrak{I}}\cap
H)\oplus(\bigoplus\limits_{\alpha\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}}{\mathfrak{L}}_{\alpha,\bar{0}})\oplus(\bigoplus\limits_{\beta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}}{\mathfrak{L}}_{\beta,\bar{1}})$
(10)

with

$\Lambda_{\bar{i}}^{\mathfrak{I}}:=\\{\gamma\in\Lambda:{\mathfrak{I}}_{\bar{i}}\cap{\mathfrak{L}}_{\gamma,\bar{i}}\neq
0\\}=\\{\gamma\in\Lambda:0\neq{\mathfrak{L}}_{\gamma,\bar{i}}\subset{\mathfrak{I}}_{\bar{i}}\\},$

$\bar{i}\in{\hbox{\hueca Z}}_{2}$.

From here, we can write

$\Lambda=\underbrace{(\Lambda_{\bar{0}}^{{\mathfrak{I}}}\dot{\cup}\Lambda_{\bar{0}}^{\neg{\mathfrak{I}}})}_{\Lambda_{\bar{0}}}\cup\underbrace{(\Lambda_{\bar{1}}^{{\mathfrak{I}}}\dot{\cup}\Lambda_{\bar{1}}^{\neg{\mathfrak{I}}})}_{\Lambda_{\bar{1}}}$
(11)

where

$\Lambda_{\bar{i}}^{\neg{\mathfrak{I}}}:=\\{\gamma\in\Lambda:{\mathfrak{L}}_{\gamma,\bar{i}}\neq
0$ and ${\mathfrak{I}}_{\bar{i}}\cap{\mathfrak{L}}_{\gamma,\bar{i}}=0\\}$

for $\bar{i}\in{\hbox{\hueca Z}}_{2}$. We will also denote

$\Lambda^{\Upsilon}:=\Lambda_{\bar{0}}^{{\Upsilon}}\cup\Lambda_{\bar{1}}^{{\Upsilon}}$,
for $\Upsilon\in\\{{\mathfrak{I}},\neg{\mathfrak{I}}\\}$.

Hence, we can write

${\mathfrak{L}}=(H_{\bar{0}}\oplus
H_{\bar{1}})\oplus(\bigoplus\limits_{\alpha\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}}{\mathfrak{L}}_{\alpha,\bar{0}})\oplus(\bigoplus\limits_{\beta\in\Lambda_{\bar{0}}^{\neg{\mathfrak{I}}}}{\mathfrak{L}}_{\beta,\bar{0}})\oplus(\bigoplus\limits_{\gamma\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}}{\mathfrak{L}}_{\gamma,\bar{1}})\oplus(\bigoplus\limits_{\delta\in\Lambda_{\bar{1}}^{\neg{\mathfrak{I}}}}{\mathfrak{L}}_{\delta,\bar{1}}).$
(12)

######  Example 3.1.

Consider the 5-dimensional split Leibniz superalgebra

${\mathfrak{L}}=H\oplus{\mathfrak{L}}_{\alpha}\oplus{\mathfrak{L}}_{-\alpha}\oplus{\mathfrak{L}}_{\beta}\oplus{\mathfrak{L}}_{-\beta}$

given in Example 1.1, where
${\mathfrak{L}}_{\bar{0}}=H\oplus{\mathfrak{L}}_{\alpha}\oplus{\mathfrak{L}}_{-\alpha}$
and
${\mathfrak{L}}_{\bar{1}}={\mathfrak{L}}_{\beta}\oplus{\mathfrak{L}}_{-\beta}$.
This is a split Leibniz superalgebra of maximal length such that
${\mathfrak{I}}=\langle e_{1},e_{2}\rangle$. Hence
$\Lambda^{\neg\mathfrak{I}}_{\bar{0}}=\\{\alpha,-\alpha\\},\Lambda^{\mathfrak{I}}_{\bar{1}}=\\{\beta,-\beta\\}$
and
$\Lambda^{\mathfrak{I}}_{\bar{0}}=\Lambda^{\neg\mathfrak{I}}_{\bar{1}}=\emptyset$.

######  Example 3.2.

Consider the $(n+4)$-dimensional split Leibniz superalgebra

${\mathfrak{L}}=H\oplus{\mathfrak{L}}_{\alpha}\oplus{\mathfrak{L}}_{-\alpha}\oplus(\bigoplus\limits_{k=0}^{n}{\mathfrak{L}}_{\beta_{k}})$

given in Example 1.2, where
${\mathfrak{L}}_{\bar{0}}=H\oplus{\mathfrak{L}}_{\alpha}\oplus{\mathfrak{L}}_{-\alpha}$
and
${\mathfrak{L}}_{\bar{1}}=\bigoplus\limits_{k=0}^{n}{\mathfrak{L}}_{\beta_{k}}$.
This is a split Leibniz superalgebra of maximal length satisfying
${\mathfrak{I}}=\langle e_{0},...,e_{n}\rangle$. From here,
$\Lambda^{\neg\mathfrak{I}}_{\bar{0}}=\\{\alpha,-\alpha\\}$ and
$\Lambda^{\mathfrak{I}}_{\bar{1}}=\\{e_{0},...,e_{n}\\}$.

######  Remark 3.1.

Since our aim is this section to characterize the simplicity of
${\mathfrak{L}}$, in terms of connectivity of roots, Theorem 2.1-2 gives us
that we have to center our attention in those split Leibniz superalgebras
satisfying
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$.
This is for instance the case whence
${\mathfrak{L}}=[{\mathfrak{L}},{\mathfrak{L}}]$. We would like to note that
if
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$,
then the decomposition given by Equation (12) and Equation (1) show

$\small{H}=$
$\small{(\underbrace{\sum\limits_{\alpha\in\Lambda_{\bar{0}}^{\neg{\mathfrak{I}}}}[{\mathfrak{L}}_{-\alpha,\bar{0}},{\mathfrak{L}}_{\alpha,\bar{0}}]+\sum\limits_{\alpha\in\Lambda_{\bar{1}}^{\neg{\mathfrak{I}}}}[{\mathfrak{L}}_{-\alpha,\bar{1}},{\mathfrak{L}}_{\alpha,\bar{1}}]}_{H_{\bar{0}}})\oplus(\underbrace{\sum\limits_{\alpha\in\Lambda_{\bar{0}}^{\neg{\mathfrak{I}}}}[{\mathfrak{L}}_{-\alpha,\bar{1}},{\mathfrak{L}}_{\alpha,\bar{0}}]+\sum\limits_{\alpha\in\Lambda_{\bar{1}}^{\neg{\mathfrak{I}}}}[{\mathfrak{L}}_{-\alpha,\bar{0}},{\mathfrak{L}}_{\alpha,\bar{1}}]}_{H_{\bar{1}}})}.$
(13)

Now, observe that the concept of connectivity of nonzero roots given in
Definition 2.1 is not strong enough to detect if a given $\alpha\in\Lambda$
belongs to $\Lambda_{\bar{i}}^{{{\mathfrak{I}}}}$ or to
$\Lambda_{\bar{i}}^{\neg{{\mathfrak{I}}}}$ for some
${\bar{i}}\in{\mathbb{Z}}_{2}$. Consequently we lose the information respect
to whether a given root space ${{\mathfrak{L}}}_{\alpha}$ intersects to
${{\mathfrak{I}}}$ in a non-trivial way or not, which is fundamental to study
the simplicity of ${{\mathfrak{L}}}$. So, we are going to refine the previous
concept of connections of non-zero roots.

######  Definition 3.2.

Let $\alpha\in\Lambda_{\bar{i}}^{\Upsilon}$ and
$\beta\in\Lambda_{\bar{j}}^{\Upsilon}$ with
$\Upsilon\in\\{{\mathfrak{I}},\neg{\mathfrak{I}}\\}$ and
${\bar{i}},{\bar{j}}\in{\mathbb{Z}}_{2}$. We say that $\alpha$ is
${\neg{\mathfrak{I}}}$-connected to $\beta,$ denoted by
$\alpha\sim_{\neg{\mathfrak{I}}}\beta,$ if either
$\beta\in\\{\alpha,-\alpha\\}$ or there exists a family of nonzero roots
$\alpha_{1},\alpha_{2},...,\alpha_{n}$ such that

$\alpha_{k}\in\Lambda_{{\bar{i}}_{k}}^{\neg{\mathfrak{I}}}$

for some ${\bar{i}}_{k}\in{\mathbb{Z}}_{2}$ and for any $k=2,...,n$; and such
that

  * 1.

$\alpha_{1}=\alpha$.

  * 2.

$\alpha_{1}+\alpha_{2}\in\Lambda_{{\bar{i}}+{\bar{i}_{2}}}^{\Upsilon},$
$\alpha_{1}+\alpha_{2}+\alpha_{3}\in\Lambda_{{\bar{i}}+{\bar{i}_{2}}+{\bar{i}_{3}}}^{\Upsilon},$
$\vdots$
$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}\in\Lambda_{{\bar{i}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}}^{\Upsilon},$
$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}+\alpha_{n}\in\Lambda_{{\bar{i}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}+{\bar{i}_{n}}}^{\Upsilon}$

and

  * 3.

$\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}+\alpha_{n}=\beta$ and
${\bar{i}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}+{\bar{i}_{n}}={\bar{j}}$.

The set $\\{\alpha_{1},\alpha_{2},...,\alpha_{n}\\}$ is called a
${\neg{\mathfrak{I}}}$-connection from $\alpha$ to $\beta$.

Let us introduce the notion of root-multiplicativity in the framework of split
Leibniz superalgebras of maximal length, in a similar way to the ones for
split Lie algebras, split Lie superalgebras and split Leibniz algebras among
other split algebraic structures, (see [8, 10, 11, 12, 13] for these notions
and examples).

######  Definition 3.3.

We say that a split Leibniz superalgebra of maximal length ${\mathfrak{L}}$ is
root-multiplicative if the below conditions hold.

  1. 1.

Given
$\alpha\in{\Lambda}_{\bar{i}}^{{\neg{\mathfrak{I}}}},\beta\in{\Lambda}_{\bar{j}}^{{\neg{\mathfrak{I}}}}$
such that $\alpha+\beta\in\Lambda$ then
$[{\mathfrak{L}}_{\alpha,\bar{i}},{\mathfrak{L}}_{\beta,\bar{j}}]\neq 0.$

  2. 2.

Given $\alpha\in{\Lambda}_{\bar{i}}^{\neg{\mathfrak{I}}}$ and
$\gamma\in{\Lambda}_{\bar{j}}^{{\mathfrak{I}}}$ such that
$\alpha+\gamma\in{\Lambda}^{{\mathfrak{I}}}$ then
$[{\mathfrak{L}}_{\gamma,\bar{j}},{\mathfrak{L}}_{\alpha,\bar{i}}]\neq 0.$

######  Notation 3.1.

We will say that $\Lambda^{\Upsilon}$,
$\Upsilon\in\\{{\mathfrak{I}},\neg{\mathfrak{I}}\\}$, has all of its roots
${\neg{\mathfrak{I}}}$-connected if for any
${\bar{i}},{\bar{j}}\in{\mathbb{Z}}_{2}$ we have that
$\Lambda^{\Upsilon}_{\bar{i}}$ has all of its roots connected to any root in
$\Lambda^{\Upsilon}_{\bar{j}}$.

######  Proposition 3.1.

Suppose
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$
and ${\mathfrak{L}}$ is root-multiplicative. If $\Lambda^{\neg{\mathfrak{I}}}$
has all of its roots ${\neg{\mathfrak{I}}}$-connected and
$|\Lambda^{\neg{\mathfrak{I}}}|>2$, then any nonzero ideal $I$ of
${\mathfrak{L}}$ such that $I\nsubseteq H+{\mathfrak{I}}$ satisfies that
$I={\mathfrak{L}}$

###### Proof.

Since $I\nsubseteq H+{\mathfrak{I}}$, there exists $\alpha_{0}\in\Lambda^{\neg
J}_{{\bar{i}}_{0}}$ such that

$0\neq{\mathfrak{L}}_{\alpha_{0},{\bar{i}}_{0}}\subset I.$ (14)

for some ${\bar{i}}_{0}\in{\hbox{\hueca Z}}_{2}$, (see Equation (12)). Given
now any
$\alpha\in\Lambda^{\neg{\mathfrak{I}}}_{\bar{j}}\setminus\\{\alpha_{0},-\alpha_{0}\\}$,
being then $0\neq{\mathfrak{L}}_{\alpha,{\bar{j}}}$, the fact that
$\alpha_{0}$ and $\alpha$ are ${\neg{\mathfrak{I}}}$-connected gives us a
${\neg{\mathfrak{I}}}$-connection $\\{\delta_{1},\delta_{2},...,\delta_{n}\\}$
from $\alpha_{0}$ to $\alpha$ such that

$\delta_{1}=\alpha_{0}\in\Lambda_{{\bar{i}}_{0}}^{\neg{\mathfrak{I}}},\hskip
2.84544pt\delta_{k}\in\Lambda_{{\bar{i}}_{k}}^{\neg{\mathfrak{I}}}\hskip
1.42271pt{\rm for}\hskip 1.42271ptk=2,...,n,$
$\delta_{1}+\delta_{2}\in\Lambda_{{\bar{i}_{0}}+{\bar{i}_{2}}}^{\neg{\mathfrak{I}}},...,\delta_{1}+\delta_{2}+\cdots+\delta_{n-1}\in\Lambda_{{\bar{i}_{0}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}}^{\neg{\mathfrak{I}}},\hskip
2.84544pt{\rm and}$
$\delta_{1}+\delta_{2}+\cdots+\delta_{n-1}+\delta_{n}\in\Lambda_{{\bar{i}_{0}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}+{\bar{i}_{n}}}^{\neg{\mathfrak{I}}}$

with $\delta_{1}+\delta_{2}+\cdots+\delta_{n-1}+\delta_{n}=\alpha$ and
${\bar{i}_{0}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}+{\bar{i}_{n}}={\bar{j}}$.

Consider $\delta_{1},\delta_{2}$ and $\delta_{1}+\delta_{2}$. Since
$\delta_{1}=\alpha_{0}\in\Lambda_{{\bar{i}}_{0}}^{\neg{\mathfrak{I}}}$,
$\delta_{2}\in\Lambda_{{\bar{i}}_{2}}^{\neg{\mathfrak{I}}}$ and
$\delta_{1}+\delta_{2}\in\Lambda$, the root-multiplicativity and maximal
length of ${\mathfrak{L}}$ show

$0\neq[{\mathfrak{L}}_{\delta_{1},{{\bar{i}}_{0}}},{\mathfrak{L}}_{\delta_{2},{{\bar{i}}_{2}}}]={\mathfrak{L}}_{\delta_{1}+\delta_{2},{\bar{i}}_{0}}+{{\bar{i}}_{2}},$

and by Equation (14)

$0\neq{\mathfrak{L}}_{\delta_{1}+\delta_{2},{\bar{i}}_{0}+{\bar{i}}_{2}}\subset
I.$

We can argue in a similar way from $\delta_{1}+\delta_{2},\delta_{3}$ and
$\delta_{1}+\delta_{2}+\delta_{3}$. That is,
$\delta_{1}+\delta_{2}\in\Lambda_{{\bar{i}}_{0}+{\bar{i}}_{2}}^{\neg{\mathfrak{I}}},\delta_{3}\in\Lambda_{{\bar{i}}_{3}}^{\neg{\mathfrak{I}}}$
and $\delta_{1}+\delta_{2}+\delta_{3}\in\Lambda.$ Hence

$0\neq[{\mathfrak{L}}_{\delta_{1}+\delta_{2},{\bar{i}}_{0}+{\bar{i}}_{2}},{\mathfrak{L}}_{\delta_{3},{\bar{i}}_{3}}]={\mathfrak{L}}_{\delta_{1}+\delta_{2}+\delta_{3},{\bar{i}}_{0}+{\bar{i}}_{2}+{\bar{i}}_{3}},$

and by the above

$0\neq{\mathfrak{L}}_{\delta_{1}+\delta_{2}+\delta_{3},{\bar{i}}_{0}+{\bar{i}}_{2}+{\bar{i}}_{3}}\subset
I.$

Following this process with the ${\neg{\mathfrak{I}}}$-connection
$\\{\delta_{1},...,\delta_{n}\\}$ we obtain that

$0\neq{\mathfrak{L}}_{\delta_{1}+\delta_{2}+\delta_{3}+\cdots+\delta_{n},{\bar{i}_{0}}+{\bar{i}_{2}}+\cdots+{\bar{i}_{n-1}}+{\bar{i}_{n}}}\subset
I$

and so ${\mathfrak{L}}_{\alpha,{\bar{j}}}\subset I$. That is, for any
$\alpha\in\Lambda^{\neg{\mathfrak{I}}}_{\bar{j}}\setminus\\{\alpha_{0},-\alpha_{0}\\}$
and for each ${\bar{j}}\in{\mathbb{Z}}_{2}$, we have

${\mathfrak{L}}_{\alpha,{\bar{j}}}\subset I.$ (15)

Let us now verify that in case
$0\neq{\mathfrak{L}}_{\epsilon\alpha_{0},\bar{k}}$ for some
$\epsilon\in\\{1,-1\\}$ and ${\bar{k}}\in{\mathbb{Z}}_{2}$, we have
$0\neq{\mathfrak{L}}_{\epsilon\alpha_{0},\bar{k}}\subset I.$ Indeed, since
$|\Lambda^{\neg{\mathfrak{I}}}|>2$, we can take
$\beta\in\Lambda^{\neg\mathfrak{I}}_{\bar{i}}$, for some
$\bar{i}\in{\mathbb{Z}}_{2}$, such that
$\beta\notin\\{\alpha_{0},-\alpha_{0}\\}$ and, by Equation (15), satisfying
$0\neq{\mathfrak{L}}_{\alpha,{\bar{j}}}\subset I$. Hence we can argue as above
with the root-multiplicativity and maximal length of ${\mathfrak{L}}$ from
$\beta$ instead of $\alpha_{0}$, to get that in case some
$\epsilon\alpha_{0}\in\Lambda^{\neg\mathfrak{I}}_{\bar{k}}$ for some
$\epsilon\in\\{1,-1\\}$ and ${\bar{k}}\in{\mathbb{Z}}_{2}$, then

$0\neq{\mathfrak{L}}_{\epsilon\alpha_{0},\bar{k}}\subset I.$ (16)

Since
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$,
Remark 3.1 and equations (15), (16) give us

$H\subset I.$ (17)

Now, for any $\Upsilon\in\\{{\mathfrak{I}},\neg{\mathfrak{I}}\\}$ and
${\bar{k}}\in{\mathbb{Z}}_{2}$, given any
$\gamma\in\Lambda^{\Upsilon}_{\bar{k}}$ since $\gamma\neq 0$ we have
${\mathfrak{L}}_{\gamma,{\bar{k}}}=[{\mathfrak{L}}_{\gamma,{\bar{k}}},H_{\bar{0}}]$
and then Equation (17) shows ${\mathfrak{L}}_{\gamma,{\bar{k}}}\subset I$. The
decomposition of ${\mathfrak{L}}$ in Equation (12) finally gives us
$I={\mathfrak{L}}$. ∎

Let us introduce an interesting notion related to a split Leibniz superalgebra
of maximal length ${\mathfrak{L}}$. We wish to distinguish those elements of
${\mathfrak{L}}$ which annihilate the “Lie type roots” of ${\mathfrak{I}}$, so
we have the following definition.

######  Definition 3.4.

The Lie-annihilator of a split Leibniz superalgebra of maximal length
${\mathfrak{L}}$ is the set

$\hbox{${\mathcal{Z}}_{Lie}({\mathfrak{L}}):=\\{v\in{\mathfrak{L}}:[v,{\mathfrak{L}}_{\alpha}]+[{\mathfrak{L}}_{\alpha},v]=0$
for any $\alpha\notin\Lambda^{\mathfrak{I}}\\}$}.$

Observe that
${\mathcal{Z}}({\mathfrak{L}})\subset{\mathcal{Z}}_{Lie}({\mathfrak{L}}).$

######  Proposition 3.2.

Suppose
$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}]$,
${\mathcal{Z}}_{Lie}({\mathfrak{L}})=0$ and ${\mathfrak{L}}$ is root-
multiplicative. If ${\Lambda}^{\mathfrak{I}}$ has all of its elements
${\neg{\mathfrak{I}}}$-connected and $|\Lambda^{{\mathfrak{I}}}|>2$, then any
nonzero ideal $I$ of ${\mathfrak{L}}$ such that $I\subset{\mathfrak{I}}$
satisfies $I={\mathfrak{I}}$.

###### Proof.

By Equation (9) we can write

$I=(I\cap
H)\oplus(\bigoplus\limits_{\beta\in{\Lambda}^{{I}}_{\bar{0}}}{{\mathfrak{L}}_{\beta,\bar{0}}})\oplus(\bigoplus\limits_{\gamma\in{\Lambda}^{I}_{\bar{1}}}{{\mathfrak{L}}_{\gamma,\bar{1}}})$

where

$\Lambda_{\bar{i}}^{I}:=\\{\delta\in\Lambda:{I}_{\bar{i}}\cap{\mathfrak{L}}_{\delta,\bar{i}}\neq
0\\}=\\{\delta\in\Lambda:0\neq{\mathfrak{L}}_{\delta,\bar{i}}\subset{I}_{\bar{i}}\\}$

and with
${\Lambda}^{{I}}_{\bar{i}}\subset{\Lambda}^{{\mathfrak{I}}}_{\bar{i}}$ for any
${\bar{i}}\in{\mathbb{Z}}_{2}$. We begin by showing that

${\mathfrak{I}}\cap H\subset{\mathcal{Z}}_{Lie}({\mathfrak{L}}).$ (18)

Indeed, fixed some ${\bar{i}}\in{\mathbb{Z}}_{2}$ and for any
$\alpha\notin\Lambda^{\mathfrak{I}}$ and $\bar{j}\in{\mathbb{Z}}_{2}$ we have

$[{\mathfrak{I}}\cap
H_{\bar{i}},{\mathfrak{L}}_{\alpha,\bar{j}}]+[{\mathfrak{L}}_{\alpha,\bar{j}},{\mathfrak{I}}\cap
H_{\bar{i}}]\subset{\mathfrak{L}}_{\alpha,\bar{i}+\bar{j}}\subset{\mathfrak{I}}.$

Hence, in case $[{\mathfrak{I}}\cap
H_{\bar{i}},{\mathfrak{L}}_{\alpha,\bar{j}}]+[{\mathfrak{L}}_{\alpha,\bar{j}},{\mathfrak{I}}\cap
H_{\bar{i}}]\neq 0$ we have $\alpha\in\Lambda^{\mathfrak{I}}$, a
contradiction. Hence $[{\mathfrak{I}}\cap
H_{\bar{i}},{\mathfrak{L}}_{\alpha,\bar{j}}]+[{\mathfrak{L}}_{\alpha,\bar{j}},{\mathfrak{I}}\cap
H_{\bar{i}}]=0$ for each ${\bar{i}},{\bar{j}}\in{\mathbb{Z}}_{2}$ and so
${\mathfrak{I}}\cap H\subset{\mathcal{Z}}_{Lie}({\mathfrak{L}}).$

From the above ${\mathfrak{I}}\cap
H\subset{\mathcal{Z}}_{Lie}({\mathfrak{L}})=0$ and we can write

${\mathfrak{I}}=(\bigoplus\limits_{\alpha\in{\Lambda}^{{{\mathfrak{I}}}}_{\bar{0}}}{{\mathfrak{L}}_{\alpha,\bar{0}}})\oplus(\bigoplus\limits_{\delta\in{\Lambda}^{{\mathfrak{I}}}_{\bar{1}}}{{\mathfrak{L}}_{\delta,\bar{1}}}).$
(19)

Taking into account ${I}\cap H\subset{\mathfrak{I}}\cap H=0$, we also can
write

$I=(\bigoplus\limits_{\beta\in{\Lambda}^{{I}}_{\bar{0}}}{{\mathfrak{L}}_{\beta,\bar{0}}})\oplus(\bigoplus\limits_{\gamma\in{\Lambda}^{I}_{\bar{1}}}{{\mathfrak{L}}_{\gamma,\bar{1}}})$

with any ${\Lambda}^{I}_{\bar{i}}\subset{\Lambda}^{{\mathfrak{I}}}_{\bar{i}}$.
Hence, we can take some $\beta_{0}\in{\Lambda}^{I}_{\bar{i}}$ such that

$0\neq{\mathfrak{L}}_{\beta_{0},{\bar{i}}}\subset I.$ (20)

Now, we can argue with the root-multiplicativity and the maximal length of
${\mathfrak{L}}$ as in Proposition 3.1 to conclude that given any
$\beta\in{\Lambda}^{\mathfrak{I}}_{\bar{j}}\setminus\\{\beta_{0},-\beta_{0}\\}$,
$\bar{j}\in{\mathbb{Z}}_{2}$, there exists a ${\neg{\mathfrak{I}}}$-connection

$\\{\delta_{1},\delta_{2},...,\delta_{r-1},\delta_{r}\\}$ (21)

from $\beta_{0}$ to $\beta$ such that

$0\neq[[...[{\mathfrak{L}}_{\beta_{0},\bar{i}},{\mathfrak{L}}_{\delta_{2},\bar{i}_{2}}],...],{\mathfrak{L}}_{\delta_{r},\bar{i}_{r}}]={\mathfrak{L}}_{\beta,\bar{j}}\subset
I.$ (22)

Now we have to study whether in case
$\epsilon\beta_{0}\in\Lambda^{\mathfrak{I}}_{\bar{k}}$ for some
$\epsilon\in\\{1,-1\\}$ and $\bar{k}\in{\mathbb{Z}}_{2}$ then
${\mathfrak{L}}_{\epsilon\beta_{0},{\bar{k}}}\subset I$. To do that, observe
that the fact $|\Lambda^{{\mathfrak{I}}}|>2$ allows us to take an element
$\beta\in\Lambda^{\mathfrak{I}}_{\bar{i}}$, for some
$\bar{i}\in{\mathbb{Z}}_{2}$, such that
$\beta\notin\\{\beta_{0},-\beta_{0}\\}$ and, by Equation (22), satisfying
$0\neq{\mathfrak{L}}_{\beta,{\bar{j}}}\subset I$. Hence we can argue as above
with the root-multiplicativity and maximal length of ${\mathfrak{L}}$ from
$\beta$ instead of $\beta_{0}$, to get that in case
$\epsilon\beta_{0}\in\Lambda^{\mathfrak{I}}_{\bar{k}}$ for some
$\epsilon\in\\{1,-1\\}$ and ${\bar{k}}\in{\mathbb{Z}}_{2}$, then
$0\neq{\mathfrak{L}}_{\epsilon\beta_{0},\bar{k}}\subset I.$

The decomposition of ${\mathfrak{I}}$ in Equation (19) finally gives us
$I={\mathfrak{I}}$. ∎

######  Theorem 3.1.

Suppose

$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}],\hskip
2.84544pt{\mathcal{Z}_{Lie}}(\mathfrak{L})=0,$

${\mathfrak{L}}$ is root-multiplicative and
$|\Lambda^{\neg{\mathfrak{I}}}|>2,|\Lambda^{{\mathfrak{I}}}|>2$. Then
${\mathfrak{L}}$ is simple if and only if $\Lambda^{\mathfrak{I}}$ and
$\Lambda^{\neg{\mathfrak{I}}}$ have (respectively) all of their elements
${\neg{\mathfrak{I}}}$-connected.

###### Proof.

Fix some $\alpha\in\Lambda^{\neg{\mathfrak{I}}}_{\bar{i}}$ with
${\bar{i}}\in{\mathbb{Z}}_{2}$, and denote by
$I({\mathfrak{L}}_{\alpha,{\bar{i}}})$ the (graded) ideal of ${\mathfrak{L}}$
generated by ${\mathfrak{L}}_{\alpha,{\bar{i}}}$. By simplicity
$I({\mathfrak{L}}_{\alpha,{\bar{i}}})={\mathfrak{L}}.$ Observe that Remark 3.1
together with Super Leibniz identity allow us to assert that
$I({\mathfrak{L}}_{\alpha,{\bar{i}}})$ is contained in the linear span of the
set

$\\{[[\cdots[v_{\alpha},v_{\beta_{1}}],...],v_{\beta_{n}}];\hskip
2.84544pt[v_{\beta_{n}},[...[v_{\beta_{1}},v_{\alpha}],...]];$
$[[...[v_{\beta_{1}},v_{\alpha}],...],v_{\beta_{n}}];\hskip
2.84544pt[v_{\beta_{n}},[...[v_{\alpha},v_{\beta_{1}}],...]]$ with $0\neq
v_{\alpha}\in{\mathfrak{L}}_{\alpha,{\bar{i}}}$, $0\neq
v_{\beta_{i}}\in{\mathfrak{L}}_{\beta_{i},{\bar{j}_{i}}},\beta_{i}\in\Lambda$,
${\bar{j}_{i}}\in{\mathbb{Z}}_{2}$ and $n\in{\hbox{\hueca N}}$ }

being the partial sums
$\alpha+\beta_{1},\alpha+\beta_{1}+\beta_{2},...,\alpha+\beta_{1}+\beta_{2}+\cdots+\beta_{n}$
nonzero. From here, given any
$\alpha^{\prime}\in\Lambda^{\neg{\mathfrak{I}}}_{\bar{j}}$, with
${\bar{j}}\in{\mathbb{Z}}_{2}$, the above observation gives us that we can
write $\alpha^{\prime}=\alpha+\beta_{1}+\cdots+\beta_{n}$ with any
$\beta_{i}\in\Lambda^{\neg\mathfrak{I}}_{\bar{j}_{i}}$ and being the partial
sums
$\alpha+\beta_{1}+\cdots+\beta_{k}\in\Lambda^{\neg\mathfrak{I}}_{\bar{i}+\bar{j}_{1}+\cdots+\bar{j}_{k}},$
(observe that in case some $\beta_{i}\in\Lambda^{\mathfrak{I}}_{\bar{j}_{i}}$
or some partial sum a root belonging to $\Lambda^{\mathfrak{I}}_{\bar{n}}$,
then either the product involving $v_{\beta_{i}}$ or the partial sum is zero,
or implies $\alpha^{\prime}\in\Lambda^{\mathfrak{I}}_{\bar{k}}$. Hence any
$\beta_{i}\in\Lambda^{\neg{\mathfrak{I}}}_{\bar{j}_{i}}$ and the partial sums
are in $\Lambda^{\neg{\mathfrak{I}}}_{\bar{n}}$). From here, we have that
$\\{\alpha,\beta_{1},...,\beta_{n}\\}$ is a ${\neg{\mathfrak{I}}}$-connection
from $\alpha$ to $\alpha^{\prime}$ and we can assert that
$\Lambda^{\neg{\mathfrak{I}}}$ has all of its elements
${\neg{\mathfrak{I}}}$-connected.

If $\Lambda^{\mathfrak{I}}\neq\emptyset$ and we take some
$\beta\in\Lambda^{\mathfrak{I}}_{\bar{i}}$ with
${\bar{i}}\in{\mathbb{Z}}_{2}$, a similar above argument gives us
${\Lambda}^{\mathfrak{I}}$ has all of its elements
${\neg{\mathfrak{I}}}$-connected.

Let us see the converse. Consider $I$ a nonzero ideal of ${\mathfrak{L}}$ and
let us show that necessarily either $I={\mathfrak{I}}$ or $I={\mathfrak{L}}$.
Let us distinguish two possibilities:

  * •

If $I\nsubseteq H+{\mathfrak{I}}$, Proposition 3.1 gives us
$I={\mathfrak{L}}$.

  * •

If $I\subset H+{\mathfrak{I}}$, observe that $I\cap
H\subset{\mathcal{Z}_{Lie}}(\mathfrak{L})$. Indeed, for any
$\alpha\notin\Lambda^{\mathfrak{I}}$ and $\bar{i},\bar{j}\in{\mathbb{Z}}_{2}$
we have $[I\cap
H_{\bar{i}},{\mathfrak{L}}_{\alpha,\bar{j}}]+[{\mathfrak{L}}_{\alpha,\bar{j}},I\cap
H_{\bar{i}}]\subset
I\cap{\mathfrak{L}}_{\alpha,\bar{i}+\bar{j}}\subset{\mathfrak{I}}$ and so
$[I\cap
H_{\bar{i}},{\mathfrak{L}}_{\alpha,\bar{j}}]+[{\mathfrak{L}}_{\alpha,\bar{j}},I\cap
H_{\bar{i}}]=0$. From here

$I\cap H\subset{\mathcal{Z}_{Lie}}(\mathfrak{L})=0.$

Hence Lemma 3.1 gives us $I\subset{\mathfrak{I}}$. Finally Proposition 3.2
implies $I={\mathfrak{I}}$ and the proof is complete.

∎

It remains to study the cases in which either
$|\Lambda^{\neg{\mathfrak{I}}}|\leq 2$ or $|\Lambda^{{\mathfrak{I}}}|\leq 2.$

######  Proposition 3.3.

Suppose

$H=\sum\limits_{\beta\in\Lambda}[{\mathfrak{L}}_{\beta},{\mathfrak{L}}_{-\beta}],\hskip
2.84544pt{\mathcal{Z}_{Lie}}(\mathfrak{L})=0,$

${\mathfrak{L}}$ is root-multiplicative and $\Lambda^{\neg{\mathfrak{I}}}$,
$\Lambda^{\mathfrak{I}}$ have (respectively) all of their elements
${\neg{\mathfrak{I}}}$-connected. If either
$|\Lambda^{\neg{\mathfrak{I}}}|\leq 2$ or $|\Lambda^{{\mathfrak{I}}}|\leq 2$,
then one of the following assertions holds.

  1. (1)

${\mathfrak{L}}$ is a simple split Leibniz superalgebra.

  2. (2)

${\rm char}({\mathbb{K}})\neq 2$ and
${\mathfrak{L}}=H\oplus(\bigoplus\limits_{\alpha\in\Lambda_{\bar{0}}^{\neg{\mathfrak{I}}}}{\mathfrak{L}}_{\alpha,\bar{0}})\oplus(\bigoplus\limits_{\beta\in\Lambda_{\bar{1}}^{\neg{\mathfrak{I}}}}{\mathfrak{L}}_{\beta,\bar{1}})\oplus
I\oplus K$ with ${\mathfrak{I}}=I\oplus K$ and where either

     * (i)

$I={\mathfrak{L}}_{\gamma,\bar{i}}\oplus{\mathfrak{L}}_{-\gamma,\bar{j}}$ is a
two-dimensional ideal of ${\mathfrak{L}}$ and
$K={\mathfrak{L}}_{\gamma,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\gamma,\bar{j}+\bar{1}}$
is a one or two-dimensional (abelian) subalgebra of ${\mathfrak{L}}$
(depending on either ${\mathfrak{L}}_{\gamma,\bar{i}+\bar{1}}=0$ or
${\mathfrak{L}}_{-\gamma,\bar{j}+\bar{1}}=0$), for
${\bar{i}},{\bar{j}}\in{\mathbb{Z}}_{2}$; or

     * (ii)

$I={\mathfrak{L}}_{\gamma,\bar{0}}\oplus{\mathfrak{L}}_{\gamma,\bar{1}}\oplus{\mathfrak{L}}_{-\gamma,\bar{i}}$
is a three-dimensional ideal of ${\mathfrak{L}}$ and
$K={\mathfrak{L}}_{-\gamma,\bar{i}+\bar{1}}$ a one-dimensional (abelian)
subalgebra of ${\mathfrak{L}}$, for ${\bar{i}}\in{\mathbb{Z}}_{2}$.

  3. (3)

${\rm char}({\mathbb{K}})=2$ and ${\mathfrak{L}}=I\oplus
K\oplus(\bigoplus\limits_{\gamma\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\gamma,\bar{0}})\oplus(\bigoplus\limits_{\delta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\delta,\bar{1}})$
where
$I=([{\mathfrak{L}}_{\alpha,\bar{i}},{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}]+[{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}},{\mathfrak{L}}_{\alpha,\bar{i}}])\oplus{\mathfrak{L}}_{\alpha,\bar{i}}\oplus(\bigoplus\limits_{\tau\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\tau,\bar{0}})\oplus(\bigoplus\limits_{\eta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\eta,\bar{1}})$
is a nonzero ideal of ${\mathfrak{L}}$, $I\neq{\mathfrak{I}}$; and
$K=[{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}},{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}]\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}$
is a two-dimensional subalgebra of ${\mathfrak{L}}$, for
${\bar{i}}\in{\mathbb{Z}}_{2}$.

  4. (4)

${\rm char}({\mathbb{K}})\neq 2$ and ${\mathfrak{L}}=I\oplus K$ where either

     * (i)

$I=H_{\bar{1}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}}\oplus(\bigoplus\limits_{\tau\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\tau,\bar{0}})\oplus(\bigoplus\limits_{\eta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\eta,\bar{1}})$
is an ideal of ${\mathfrak{L}}$ and
$K=H_{\bar{0}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}+\bar{1}}\oplus(\bigoplus\limits_{\gamma\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\gamma,\bar{0}})\oplus(\bigoplus\limits_{\delta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\delta,\bar{1}})$
with ${\rm
dim}({\mathfrak{L}}_{\alpha,\bar{i}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}+\bar{1}})=3$,
for ${\bar{i}}\in{\mathbb{Z}}_{2}$, or

     * (ii)

$I=(I\cap H_{\bar{0}})\oplus
H_{\bar{1}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}}\oplus(\bigoplus\limits_{\tau\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\tau,\bar{0}})\oplus(\bigoplus\limits_{\eta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\eta,\bar{1}})$
is an ideal of ${\mathfrak{L}}$ and

$K=K_{\bar{0}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}+\bar{1}}\oplus(\bigoplus\limits_{\gamma\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\gamma,\bar{0}})\oplus(\bigoplus\limits_{\delta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\delta,\bar{1}}),$

where

$K_{\bar{0}}=\sum\limits_{\tiny{[{\mathfrak{L}}_{\epsilon\alpha,\bar{i}+\bar{1}},{\mathfrak{L}}_{-\epsilon\alpha,\bar{i}+\bar{1}}]\cap
I=0;\epsilon\in\\{1,-1\\}}}[{\mathfrak{L}}_{\epsilon\alpha,\bar{1}+\bar{1}},{\mathfrak{L}}_{-\epsilon\alpha,\bar{1}+\bar{1}}]$

and ${\rm
dim}({\mathfrak{L}}_{\alpha,\bar{i}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}+\bar{1}})=4$,
for ${\bar{i}}\in{\mathbb{Z}}_{2}$, or

     * (iii)

$I=(I\cap H_{\bar{0}})\oplus(I\cap
H_{\bar{1}})\oplus{\mathfrak{L}}_{\alpha,\bar{i}}\oplus(\bigoplus\limits_{\tau\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\tau,\bar{0}})\oplus(\bigoplus\limits_{\eta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\cap\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\eta,\bar{1}})$
is an ideal of ${\mathfrak{L}}$ and

$K=K_{\bar{0}}\oplus
K_{\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}+\bar{1}}\oplus(\bigoplus\limits_{\gamma\in\Lambda_{\bar{0}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{0}}^{I}}{\mathfrak{L}}_{\gamma,\bar{0}})\oplus(\bigoplus\limits_{\delta\in\Lambda_{\bar{1}}^{{\mathfrak{I}}}\setminus\Lambda_{\bar{1}}^{I}}{\mathfrak{L}}_{\delta,\bar{1}})$

where any
$K_{\bar{j}}=\sum\limits_{\tiny{[{\mathfrak{L}}_{\epsilon\alpha,\bar{k}},{\mathfrak{L}}_{-\epsilon\alpha,\bar{k}+\bar{j}}]\cap
I=0;\epsilon\in\\{1,-1\\}};{\bar{k}}\in{\mathbb{Z}}_{2}}[{\mathfrak{L}}_{\epsilon\alpha,\bar{k}},{\mathfrak{L}}_{-\epsilon\alpha,\bar{k}+\bar{j}}]$
and ${\rm
dim}({\mathfrak{L}}_{\alpha,\bar{i}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}}\oplus{\mathfrak{L}}_{\alpha,\bar{i}+\bar{1}}\oplus{\mathfrak{L}}_{-\alpha,\bar{i}+\bar{1}})=4$,
for ${\bar{i}}\in{\mathbb{Z}}_{2}$.

###### Proof.

Now, suppose ${\mathfrak{L}}$ is not simple, then there exists a nonzero ideal
$I$ of ${\mathfrak{L}}$ such that
$I\notin\\{{\mathfrak{I}},{\mathfrak{L}}\\}$. Let us distinguish three cases:

1\. $|\Lambda^{{\mathfrak{I}}}|\leq 2$ and $|\Lambda^{\neg{\mathfrak{I}}}|>2.$
If $I\subsetneq H+{\mathfrak{I}}$, we have as in Proposition 3.1 that
$I={\mathfrak{L}}$. From here $I\subset H+{\mathfrak{I}}.$ Since we get $I\cap
H=0$ as in Theorem 3.1 we have

$I\subset{\mathfrak{I}}.$

Let us show that necessarily

$|\Lambda^{{\mathfrak{I}}}|\neq 1.$

Indeed, if $|\Lambda^{{\mathfrak{I}}}|=1$, then
$|\Lambda^{{\mathfrak{I}}}|=\\{\gamma\\}$ and so either
$\gamma\in\Lambda^{\mathfrak{I}}_{\bar{0}}\cup\Lambda^{\mathfrak{I}}_{\bar{1}}$
or $\gamma\in\Lambda^{\mathfrak{I}}_{\bar{i}}$ with
$\gamma\notin\Lambda^{\mathfrak{I}}_{\bar{i}+1}.$ We have as in Proposition
3.2 that some $0\neq{\mathfrak{L}}_{\gamma,\bar{j}}\subset I$, (in the second
case $\bar{j}=\bar{i}$). Since ${\mathcal{Z}}_{Lie}({\mathfrak{L}})=0$, there
exists $\alpha\in\Lambda_{\bar{k}}^{\neg{\mathfrak{I}}}$ such that
$[{\mathfrak{L}}_{\gamma,\bar{j}},{\mathfrak{L}}_{\alpha,\bar{k}}]+[{\mathfrak{L}}_{\alpha,\bar{k}},{\mathfrak{L}}_{\gamma,\bar{j}}]\neq
0$ and so $\gamma+\alpha\in\Lambda_{\bar{j}+\bar{k}}^{{\mathfrak{I}}}.$ From
here $\gamma+\alpha=\gamma$ and so $\alpha=0$, a contradiction. Hence the
possibility $|\Lambda^{{\mathfrak{I}}}|=1$ never happens. From here, we just
have to consider the case in which $|\Lambda^{{\mathfrak{I}}}|=2.$ Then either
$\Lambda^{{\mathfrak{I}}}=\\{\gamma,\delta\\}$ with $\delta\neq-\gamma$ or
$\Lambda^{{\mathfrak{I}}}=\\{\gamma,-\gamma\\}$ and ${\rm
char}({\mathbb{K}})\neq 2$. In the first possibility we can argue as in
Proposition 3.2 to get $I={\mathfrak{I}}$ and so we just have to center on the
second one, that is, $\Lambda^{{\mathfrak{I}}}=\\{\gamma,-\gamma\\}$ and ${\rm
char}({\mathbb{K}})\neq 2$. As in the case in which
$|\Lambda^{{\mathfrak{I}}}|=1$, we have that some
$0\neq{\mathfrak{L}}_{\epsilon\gamma,\bar{j}}\subset I$ with
$\epsilon\in\\{1,-1\\}$ and that there exists
$\alpha\in\Lambda_{\bar{k}}^{\neg{\mathfrak{I}}}$ such that
$[{\mathfrak{L}}_{\epsilon\gamma,\bar{j}},{\mathfrak{L}}_{\alpha,\bar{k}}]+[{\mathfrak{L}}_{\alpha,\bar{k}},{\mathfrak{L}}_{\epsilon\gamma,\bar{j}}]\neq
0$ being then
$\epsilon\gamma+\alpha\in\Lambda_{\bar{j}+\bar{k}}^{{\mathfrak{I}}}.$ From
here, either $\epsilon\gamma+\alpha=\epsilon\gamma$ or
$\epsilon\gamma+\alpha=-\epsilon\gamma.$ Since the first possibility never
happens we get $\alpha=-2\epsilon\gamma$ and so
$0\neq[{\mathfrak{L}}_{\epsilon\gamma,\bar{j}},{\mathfrak{L}}_{\alpha,\bar{k}}]+[{\mathfrak{L}}_{\alpha,\bar{k}},{\mathfrak{L}}_{\epsilon\gamma,\bar{j}}]={\mathfrak{L}}_{-\epsilon\gamma,\bar{j}+\bar{k}}\subset
I$. Hence, if $\epsilon\gamma\notin\Lambda^{\mathfrak{I}}_{\bar{j}+\bar{1}}$
and $-\epsilon\gamma\notin\Lambda^{\mathfrak{I}}_{\bar{j}+\bar{k}+\bar{1}}$ we
get $I={\mathfrak{I}}$. From here, we just have to study the opposite case in
which it is straightforward to verify that we necessarily have the option (2)
of the Proposition.

2\. $|\Lambda^{\neg{\mathfrak{I}}}|\leq 2$ and $|\Lambda^{{\mathfrak{I}}}|>2.$
If $I\subset H+{\mathfrak{I}}$, we have as in item 1. that
$I\subset{\mathfrak{I}}$ and then $I={\mathfrak{I}}$ by Proposition 3.2. From
here,

$I\subsetneq H+{\mathfrak{I}}.$

Since Equation (13) gives us $|\Lambda^{\neg{\mathfrak{I}}}|\neq 0$, (in the
opposite case $H=0$), we get $|\Lambda^{\neg{\mathfrak{I}}}|\in\\{1,2\\}$ and
so we can distinguish two possibilities:

  * •

If $|\Lambda^{\neg{\mathfrak{I}}}|=1$, Equation (13) implies that necessarily
${\rm char}({\mathbb{K}})=2$. By taking also into account Lemma 3.1, the fact
$\alpha(H_{\bar{0}})\neq 0$ for any nonzero root $\alpha$; and that
${\mathcal{Z}_{Lie}}(\mathfrak{L})=0$ we get in a straightforward way the case
(3) of the Proposition.

  * •

If $|\Lambda^{\neg{\mathfrak{I}}}|=2$, then either
$\Lambda^{\neg{\mathfrak{I}}}=\\{\alpha,\beta\\}$ with $\beta\neq-\alpha$ or
$\Lambda^{\neg{\mathfrak{I}}}=\\{\alpha,-\alpha\\}$ and ${\rm
char}({\mathbb{K}})\neq 2$. In the first possibility we have as in Proposition
3.1 that $I={\mathfrak{L}}$ and so we just have to consider the second one,
that is, $\Lambda^{\neg{\mathfrak{I}}}=\\{\alpha,-\alpha\\}$ and ${\rm
char}({\mathbb{K}})\neq 2$. By a routinary case by case study on ${\rm
dim}(\bigoplus\limits_{\gamma\in\Lambda^{\neg{\mathfrak{I}}}}{\mathfrak{L}}_{\gamma})\in\\{2,3,4\\}$,
taking into account Equation (13), Lemma 3.1, the fact
$\alpha(H_{\bar{0}})\neq 0$ for any nonzero root $\alpha$ and that
${\mathcal{Z}_{Lie}}(\mathfrak{L})=0$, we obtain the possibility (4) of the
Proposition.

3\. $|\Lambda^{{\mathfrak{I}}}|\leq 2$ and $|\Lambda^{\neg{\mathfrak{I}}}|\leq
2.$ If $I\subset H+{\mathfrak{I}}$ we have as in item 1. that
$I\subset{\mathfrak{I}}$ and $|\Lambda^{\mathfrak{I}}|\neq 1$, being then
$|\Lambda^{\mathfrak{I}}|=2$. By arguing as in item 1. we get that either
$I={\mathfrak{I}}$ or we are in the case (2) of the Proposition.

If $I\nsubseteq H+{\mathfrak{I}}$, by arguing as in item 2. we obtain that
necessarily either possibility (3) or possiblity (4) of the Proposition holds.
∎

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