# On naturally graded Lie and (non-Lie) Leibniz superalgebras

L.M. Camacho  Luisa María Camacho
Dpto. Matemática Aplicada I, Universidad de Sevilla, Sevilla (Spain)
<EMAIL_ADDRESS>,  R.M. Navarro  Rosa María Navarro
Dpto. de Matemáticas, Universidad de Extremadura, Cáceres (Spain)
<EMAIL_ADDRESS>and  J.M. Sánchez  José M. Sánchez
Dpto. de Matemáticas, Universidad de Cádiz, Campus de Puerto Real, Cádiz
(Spain<EMAIL_ADDRESS>
###### Abstract.

In general, the study of gradations has always represented a cornerstone in
algebra theory. In particular, naturally graded seems to be the first and the
most relevant gradation when it comes to nilpotent algebras, a large class of
solvable ones. In fact, many families of relevant solvable algebras have been
obtained by extensions of naturally graded nilpotent algebras, i.e. solvable
algebras with a well-structured nilradical. Thus, the aim of this work is
introducing the concept of naturally graded for superalgebra structures such
as Lie and (non-Lie) Leibniz. After having defined naturally graded Lie and
Leibniz superalgebras, we characterize natural gradations on a very important
class of each of them, that is, those with maximal super-nilindex.

This work has been supported by Agencia Estatal de Investigación (Spain),
grant MTM2016-79661-P (European FEDER support included, UE), by the PCI of the
UCA ‘Teoría de Lie y Teoría de Espacios de Banach’ and by the PAI with project
number FQM298.

2010 MSC: 17A32; 17B30; 17B70; 17A70

Key-Words: Lie (super)algebras , cohomology, deformation, Leibniz
(super)algebras , nilpotent, filiform, naturally graded.

##  1\. Introduction

In general, the study of graded algebras have always played a fundamental role
into Lie theory. Moreover, in nilpotent Lie/Leibniz algebras among all the
possible gradations to be considered, natural gradation seems to be the most
important due to the fact that it is defined through the filtration which
derives from the descending central sequence (see [10, 15]). Recently, the
importance of naturally graded Lie/Leibniz algebras has been increased by
means of the use of them as nilradical of relevant families of solvable ones
(see [1, 5, 13, 28]). Therefore, it remains as an inmediate and future work
the use of the results of the present paper to obtain important families of
solvable Lie/Leibniz superalgebras by extensions of non-nilpotent outer
derivations, for instance. Recall that nowadays the classification of solvable
algebras is an open and unsolved problem.

Filiform Lie algebras, on one hand, were firstly introduced in [34] by Vergne
and they have important properties such as every filiform Lie algebra can be
obtained by a deformation of the model filiform algebra $L_{n}$. Furthermore
this type of nilpotent Lie algebras is considered to be the least nilpotent,
that is, with the longest possible descending central sequence and maximal
nilindex. The generalization of filiform Lie algebras for Lie superalgebras
has already been obtained (see [8]) and, in the same way as occurs for Lie
algebras, all filiform Lie superalgebras can be obtained by infinitesimal
deformations of the model Lie superalgebra $L^{n,m}$ and have maximal super-
nilindex. Thus, we begin the present work introducing the concept of naturally
graded for Lie/Leibniz superalgebras and secondly we tackle the problem of
determining which filiform Lie superalgebras are naturally graded. For
approaching this second problem we will consider some special infinitesimal
deformations of $L^{n,m}$ which are defined by even 2-cocycles in
$Z^{2}_{0}(L^{n,m},L^{n,m})$. Throughout this work we give a complete
classification (up to isomorphism) of naturally graded filiform Lie
superalgebras of dimension less or equal to $7$ as well as some other
classifications with either $n$ or $m$ arbitrary.

On the other hand, the notion of Leibniz superalgebras as a generalization of
Leibniz algebras was firstly introduced in [2], general graded Leibniz
algebras was considered before in work [29], though. Since Leibniz algebras
are a generalization of Lie algebras [30], consequently many of the features
of Leibniz superalgebras are generalization of Lie superalgebras [3, 11, 12,
19]. Likewise, the study of nilpotent Leibniz algebras [2, 6, 7] can be very
useful to study nilpotent Leibniz superalgebras. The latter constitute the
context of the last part of our work, nilpotent Leibniz superalgebras. Thus,
as naturally graded Leibniz algebras have been already obtained in [6], our
main purpouse throughout the last part of this paper is obtaining the analog
for the case of superalgebras. More precisely, we characterize naturally
graded Leibniz superalgebras with maximal super-nilindex. Note that the
techniques used for Leibniz superalgebras are totally different from the ones
used for Lie superalgebras and involve a huge amount of computation. Remark
that Mathematica software has been a truly useful tool to successfully
accomplish this task.

Throughout the present paper we will consider vector spaces and algebras over
the field of complex numbers $\mathbb{C}$.

##  2\. Preliminary results

###  2.1. Preliminary results for Lie superalgebras

A vector space $V$ is said to be $\mathbb{Z}_{2}$-graded if it admits a
decomposition in direct sum, $V=V_{\bar{0}}\oplus V_{\bar{1}}$, where
${\bar{0}},{\bar{1}}\in\mathbb{Z}_{2}$, and such that
$V_{\bar{i}}V_{\bar{j}}\subset V_{\overline{i+j}}$ for any
${\bar{i}},{\bar{j}}\in\mathbb{Z}_{2}$. An element $X\in V$ is called
homogeneous of degree $\bar{i}$ if it is an element of
$V_{\bar{i}},{\bar{j}}\in\mathbb{Z}_{2}$. In particular, the elements of
$V_{\bar{0}}$ (resp. $V_{\bar{1}}$) are also called even (resp. odd). A Lie
superalgebra (see [14, 32]) is a $\mathbb{Z}_{2}$-graded vector space
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$, with
an even bilinear commutation operation (or “supercommutation”)
$[\cdot,\cdot]$, which satisfies the conditions

  1. 1.

$[X,Y]=-(-1)^{\bar{i}\bar{j}}[Y,X],$

  2. 2.

$(-1)^{\bar{k}\bar{i}}[X,[Y,Z]]+(-1)^{\bar{i}\bar{j}}[Y,[Z,X]]+(-1)^{\bar{j}\bar{k}}[Z,[X,Y]]=0$
(super Jacobi identity)

for all
$X\in{\mathfrak{g}}_{\bar{i}},Y\in{\mathfrak{g}}_{\bar{j}},Z\in{\mathfrak{g}}_{\bar{k}}$
with $\bar{i},\bar{j},\bar{k}\in\mathbb{Z}_{2}$.

Thus, ${\mathfrak{g}}_{\bar{0}}$ is an ordinary Lie algebra, and
${\mathfrak{g}}_{\bar{1}}$ is a module over ${\mathfrak{g}}_{\bar{0}}$; the
Lie superalgebra structure also contains the symmetric pairing
$S^{2}{\mathfrak{g}}_{\bar{1}}\longrightarrow{\mathfrak{g}}_{\bar{0}}$.
Hereafter, any time we write $[X_{i},X_{j}]$ or $[X_{i},Y_{j}]$ with $X_{i}$
even and $Y_{j}$ odd elements, the products $[X_{j},X_{i}]$, $[Y_{j},X_{i}]$
are asumed to be obtained by the skew-symmetric property as it is usual in Lie
theory. Also, for simplicity the symmetric products
$[Y_{i},Y_{j}]=[Y_{j},Y_{i}]$ are denoted by $(Y_{i},Y_{j})$. Note that when
we deal with Leibniz superalgebras we do not have skew-symmetric nor symmetric
property and each single bracket product must be specified.

In general, the descending central sequence of a Lie superalgebra
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ is
defined in the same way as for Lie algebras:
${\mathcal{C}}^{0}({\mathfrak{g}}):={\mathfrak{g}}$,
${\mathcal{C}}^{k+1}({\mathfrak{g}}):=[{\mathcal{C}}^{k}({\mathfrak{g}}),{\mathfrak{g}}]$
for all $k\geq 0$. Consequently, if
${\mathcal{C}}^{k}({\mathfrak{g}})=\\{0\\}$ for some $k$, then the Lie
superalgebra is called nilpotent. Nevertheless, there are also defined two
others descending sequences called
${\mathcal{C}}^{k}({\mathfrak{g}}_{\bar{0}})$ and
${\mathcal{C}}^{k}({\mathfrak{g}}_{\bar{1}})$ which will be important in our
study. They are defined by
${\mathcal{C}}^{0}({\mathfrak{g}}_{\bar{i}}):={\mathfrak{g}}_{\bar{i}},{\mathcal{C}}^{k+1}({\mathfrak{g}}_{\bar{i}}):=[{\mathfrak{g}}_{\bar{0}},{\mathcal{C}}^{k}({\mathfrak{g}}_{\bar{i}})],\quad
k\geq 0,{\bar{i}}\in\mathbb{Z}_{2}.$

If ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ is
a nilpotent Lie superalgebra, then ${\mathfrak{g}}$ has super-nilindex or
s-nilindex $(p,q)$ if satisfies

$({\mathcal{C}}^{p-1}({\mathfrak{g}}_{\bar{0}}))\neq
0,\qquad({\mathcal{C}}^{q-1}({\mathfrak{g}}_{\bar{1}}))\neq
0,\qquad{\mathcal{C}}^{p}({\mathfrak{g}}_{\bar{0}})={\mathcal{C}}^{q}({\mathfrak{g}}_{\bar{1}})=0.$

It can be noted also that a module $A=A_{\bar{0}}\oplus A_{\bar{1}}$ of the
Lie superalgebra $\mathfrak{g}$ is an even bilinear map $\mathfrak{g}\times
A\to A$ satisfying $X(Ya)-(-1)^{\bar{i}\bar{j}}Y(Xa)=[X,Y]a$ for
$X\in\mathfrak{g}_{\bar{i}}$, $Y\in\mathfrak{g}_{\bar{j}}$,
$\bar{i},\bar{j}\in\mathbb{Z}_{2}$, and $a\in A$.

Lie superalgebra cohomology is defined in the well-known way ( see e.g. [14,
33] ). Thus, we have the cohomology groups
$H^{q}_{p}({\mathfrak{g}};A):=Z^{q}_{p}({\mathfrak{g}};A)\left/B^{q}_{p}({\mathfrak{g}};A)\right.$
where, in particular, the elements of $Z^{q}_{0}({\mathfrak{g}};A)$ and
$Z^{q}_{1}({\mathfrak{g}};A)$ are called even q-cocycles and odd q-cocycles,
respectively.

We denote the variety of nilpotent Lie superalgebras
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ with
$\dim({\mathfrak{g}}_{\bar{0}})=n+1$ and $\dim({\mathfrak{g}}_{\bar{1}})=m$
with the notation ${\mathcal{N}}^{n+1,m}$, in complete analogy to Lie algebras
(see [23, 24]). Then, any nilpotent Lie superalgebra
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}\in{\mathcal{N}}^{n+1,m}$
with s-nilindex $(n,m)$ is called filiform [16]. We denote by
${\mathcal{F}}^{n+1,m}$ the subset of ${\mathcal{N}}^{n+1,m}$ consisting of
all the filiform Lie superalgebras. Likewise, it can be seen that any filiform
Lie superalgebra can be always expressed in a suitable basis or so-called
adapted basis. Therefore, if
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}\in{\mathcal{F}}^{n+1,m}$,
then there exists an adapted basis of ${\mathfrak{g}}$, namely
$\\{X_{0},X_{1},\dots,X_{n},Y_{1},\dots,Y_{m}\\}$, with
$\\{X_{0},X_{1},\dots,X_{n}\\}$ a basis of ${\mathfrak{g}}_{\bar{0}}$ and
$\\{Y_{1},\dots,Y_{m}\\}$ a basis of ${\mathfrak{g}}_{\bar{1}}$, such that

$\begin{array}[]{lll}[X_{0},X_{i}]=X_{i+1},&1\leq i\leq
n-1,&[X_{0},X_{n}]=0\\\\[2.84526pt] [X_{0},Y_{j}]=Y_{j+1},&1\leq j\leq
m-1,&[X_{0},Y_{m}]=0\end{array}$

From this result it can be observed that the simplest filiform Lie
superalgebra or so-called the model filiform Lie superalgebra, denoted by
$L^{n,m}$, will be defined by the only non-zero bracket products

$L^{n,m}:\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},&1\leq i\leq
n-1\\\\[2.84526pt] [X_{0},Y_{j}]=Y_{j+1},&1\leq j\leq m-1\end{array}\right.$

$L^{n,m}$ will be the most important filiform Lie superalgebra, in complete
analogy to Lie algebras, since all the other filiform Lie superalgebras can be
obtained from it by deformations. These infinitesimal deformations will be
given by the even 2-cocycles, $Z^{2}_{0}(L^{n,m},L^{n,m})$, which can be
decomposed in the way

$\begin{array}[]{rl}Z^{2}_{0}(L^{n,m},L^{n,m})=&\underbrace{Z^{2}(L^{n,m},L^{n,m})\cap{\rm
Hom}({\mathfrak{g}}_{\bar{0}}\wedge{\mathfrak{g}}_{\bar{0}},{\mathfrak{g}}_{\bar{0}})}_{A}\oplus\underbrace{Z^{2}(L^{n,m},L^{n,m})\cap{\rm
Hom}({\mathfrak{g}}_{\bar{0}}\wedge{\mathfrak{g}}_{\bar{1}},{\mathfrak{g}}_{\bar{1}})}_{B}\oplus\\\
&\underbrace{Z^{2}(L^{n,m},L^{n,m})\cap{\rm
Hom}(S^{2}{\mathfrak{g}}_{\bar{1}},{\mathfrak{g}}_{\bar{0}})}_{C}\ =\ A\oplus
B\oplus C\\\ \end{array}$

where ${\mathfrak{g}}_{\bar{0}}=L^{n,m}_{\bar{0}}$ and
${\mathfrak{g}}_{\bar{1}}=L^{n,m}_{\bar{1}}$. The component $C$ has been
determined in [8, 17], and components $A,B$ have been determined in [26] and
[25], respectively. Moreover, using these deformations it has been obtained a
complete classification (up to isomorphism) of complex filiform Lie
superalgebras of dimension less or equal to $7$ (for more details see [31]).
The main result used for this classification is the fact that every filiform
Lie superalgebra can be obtained by $L^{n,m}+\Psi$ with $\Psi$ an
infinitesimal deformation verifying
$\Psi\circ\Psi(x,y,z)=\Psi(\Psi(x,y),z)+\Psi(\Psi(z,x),y)+\Psi(\Psi(y,z),x)=0$.

###  2.2. Preliminary results for Leibniz superalgebras

Many results and definitions of the above section can be extended for Leibniz
superalgebras.

######  Definition 2.1.

[2]. A $\mathbb{Z}_{2}$-graded vector space $L=L_{\bar{0}}\oplus L_{\bar{1}}$
is called a Leibniz superalgebra if it is equipped with a product
$[\cdot,\cdot]$ which satisfies the condition

$[x,[y,z]]=[[x,y],z]-(-1)^{\bar{i}\bar{j}}[[x,z],y]\hskip
11.38092pt\hbox{{\it(super Leibniz identity)}}$

for all $x\in L$, $y\in L_{\bar{i}}$, $z\in L_{\bar{j}}$,
$\bar{i},\bar{j}\in\mathbb{Z}_{2}$.

Note that if a Leibniz superalgebra $L$ satisfies the identity
$[x,y]=-(-1)^{\bar{i}\bar{j}}[y,x]$ for $x\in L_{\bar{i}}$, $y\in
L_{\bar{j}}$, then the super Leibniz identity becomes the super Jacobi
identity. Therefore Leibniz superalgebras are a generalization of Lie
superalgebras. In the same way as for Lie superalgebras, isomorphisms are
assumed to be consistent with the $\mathbb{Z}_{2}$-graduation.

If we denote by $R_{x}$ the right multiplication operator, i.e.
$R_{x}:L\rightarrow L$ given as $R_{x}(y):=[y,x]$ for $y\in L$, then the super
Leibniz identity can be expressed as
$R_{[x,y]}=R_{y}R_{x}-(-1)^{\bar{i}\bar{j}}R_{x}R_{y}$ where $x\in
L_{\bar{i}},y\in L_{\bar{j}}$.

We denote by $R(L)$ the set of all right multiplication operators. It is not
difficult to prove that $R(L)$ with the multiplication

(2.1) $<R_{a},R_{b}>:=R_{a}R_{b}-(-1)^{\bar{i}\bar{j}}R_{b}R_{a}$

for $R_{a}\in{R(L)_{\bar{i}}}$, $R_{b}\in{R(L)_{\bar{j}}}$, becomes a Lie
superalgebra.

Note that the concepts of descending central sequence, the variety of Leibniz
superalgebras and Engel’s theorem are natural extensions from Lie theory.

Let $V=V_{\bar{0}}\oplus V_{\bar{1}}$ be the underlying vector space of $L$,
$L=L_{\bar{0}}\oplus L_{\bar{1}}\in Leib^{n,m}$, being $Leib^{n,m}$ the
variety of Leibniz superalgebras, and let $G(V)$ be the group of the
invertible linear mappings of the form $f=f_{\bar{0}}+f_{\bar{1}}$, such that
$f_{\bar{0}}\in GL(n,\mathbb{C})$ and $f_{\bar{1}}\in GL(m,\mathbb{C})$ (then
$G(V)=GL(n,\mathbb{C})\oplus GL(m,\mathbb{C}))$. The action of $G(V)$ on
$Leib^{n,m}$ induces an action on the Leibniz superalgebras variety: two laws
$\lambda_{1},\lambda_{2}$ are isomorphic if there exists a linear mapping
$f=f_{\bar{0}}+f_{\bar{1}}\in G(V)$, such that

$\lambda_{2}(x,y)=f_{\bar{i}+\bar{j}}^{-1}(\lambda_{1}(f_{\bar{i}}(x),f_{\bar{j}}(y))),\hskip
5.69046pt\hbox{for any}x\in V_{\bar{i}},y\in V_{\bar{j}}.$

The description of the variety of any class of algebras or superalgebras is a
difficult problem. Different papers (for example, [4, 9, 20, 21]) are
concerning the applications of algebraic groups theory to the description of
the variety of Lie/Leibniz algebras.

######  Definition 2.2.

For a Leibniz superalgebra $L=L_{\bar{0}}\oplus L_{\bar{1}}$ we define the
right annihilator of $L$ as the set $Ann(L):=\\{x\in L:[L,x]=0\\}$.

It is easy to see that $Ann(L)$ is a two-sided ideal of $L$ and $[x,x]\in
Ann(L)$ for any $x\in L_{\bar{0}}$. This notion is good and compatible with
the right annihilator in Leibniz algebras. If we consider the ideal generated
as $I:=ideal<[x,y]+(-1)^{\bar{i}\bar{j}}[y,x]:x\in L_{\bar{i}},y\in
L_{\bar{j}}>$, then $I\subset Ann(L).$

Let $L=L_{\bar{0}}\oplus L_{\bar{1}}$ be a nilpotent Leibniz superalgebra with
$\dim L_{\bar{0}}=n$ and $\dim L_{\bar{1}}=m.$ From Equation (2.1) we have
that $R(L)$ is a Lie superalgebra, and in particular $R(L_{\bar{0}})$ is a Lie
algebra. As $L_{\bar{1}}$ has $L_{\bar{0}}$-module structure we can consider
$R(L_{\bar{0}})$ as a subset of $GL(V_{\bar{1}})$ , where $V_{\bar{1}}$ is the
underlying vector space of $L_{\bar{1}}$. So, we have a Lie algebra formed by
nilpotent endomorphisms of $V_{\bar{1}}$. Applying the Engel’s theorem [22] we
have the existence of a sequence of subspaces of $V_{\bar{1}}$, $V_{0}\subset
V_{1}\subset V_{2}\subset\dots\subset V_{m}=V_{\bar{1}},$ with
$R(L_{\bar{0}})(V_{\overline{i+1}})\subset V_{\bar{i}}.$ Then, it can defined
the descending sequences $C^{k}(L_{\bar{0}})$ and $C^{k}(L_{\bar{1}})$ and the
super-nilindex in the same way as for Lie superalgebras.

Similar to the case of null-filiform Leibniz algebras, it is easy to check
that a Leibniz superalgebra is null-filiform if and only if it is single-
generated. Moreover, a null-filiform superalgebra has the maximal super-
nilindex (see [27]).

##  3\. Naturally graded Lie/Leibniz superalgebras

In terms of algebras, recall that for nilpotent Lie/Leibniz algebras
${\mathfrak{g}}$, the descending central sequence defines a filtration over
the algebra and therefore there is associated a graded Lie/Leibniz algebra
structure that we denote as

$gr{\mathfrak{g}}:=\color[rgb]{0,0,0}\sum{\mathcal{C}}^{i-1}({\mathfrak{g}})/{\mathcal{C}}^{i}({\mathfrak{g}})$

If ${\mathfrak{g}},gr{\mathfrak{g}}$ are isomorphic, then ${\mathfrak{g}}$ is
said to be naturally graded. In particular, there are only two non-isomorphic
naturally graded filiform Lie algebras called $L_{n}$ and $Q_{n}$, whose laws
can be expressed in an adapted basis $\\{X_{0},X_{1},\dots,X_{n}\\}$ by the
following non-zero bracket products (for more details it can be consulted [20,
34])

$\begin{array}[]{ll}L_{n}:\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},&\
1\leq i\leq n-1\\\\[2.84526pt]
\end{array}\right.&Q_{n}:\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},&\
1\leq i\leq n-1\\\\[2.84526pt] [X_{i},X_{n-i}]=(-1)^{i}X_{n},&\ 1\leq i\leq
n-1\\\\[2.84526pt] \end{array}\right.\end{array}$

Note that both algebras are $(n+1)$-dimensional, although $Q_{n}$ only appears
in the cases $n$ odd and $n\geq 5$.

Analogously, for superalgebra structures, i.e. nilpotent Lie/Leibniz
superalgebras
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$, can
be considered the descending sequences
${\mathcal{C}}^{k}({\mathfrak{g}}_{\bar{i}}):={\mathfrak{g}}_{\bar{i}}$,
${\mathcal{C}}^{k+1}({\mathfrak{g}}_{\bar{i}}):=[{\mathfrak{g}}_{\bar{0}},{\mathcal{C}}^{k}({\mathfrak{g}}_{\bar{i}})]$,
with $k\geq 0$, $\bar{i}\in\mathbb{Z}_{2}$.

It can be seen that these sequences define a filtration over
${\mathfrak{g}}_{\bar{0}}$ and ${\mathfrak{g}}_{\bar{1}}$, respectively. Thus,
we have on one hand a structure of graded Lie/Leibniz algebra

$gr{\mathfrak{g}}_{\bar{0}}=\sum{\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{0}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{0}})$

and on the other hand a structure of graded ${\mathfrak{g}}_{\bar{0}}$-module

$gr{\mathfrak{g}}_{\bar{1}}=\sum{\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{1}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{1}})$

If we denote
${\mathfrak{g}}^{i}_{\bar{0}}:={\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{0}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{0}})$
and
${\mathfrak{g}}^{i}_{\bar{1}}:={\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{1}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{1}})$,
then it is verified that

$[{\mathfrak{g}}^{i}_{\bar{0}},{\mathfrak{g}}^{j}_{\bar{0}}]\subset{\mathfrak{g}}^{i+j}_{\bar{0}}\qquad\mbox{
and
}\qquad[{\mathfrak{g}}^{i}_{\bar{0}},{\mathfrak{g}}^{j}_{\bar{1}}]\subset{\mathfrak{g}}^{i+j}_{\bar{1}}$

######  Definition 3.1.

Given a nilpotent Lie (resp. Leibniz) superalgebra
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$,
consider
${\mathfrak{g}}^{i}={\mathfrak{g}}^{i}_{\bar{0}}\oplus{\mathfrak{g}}^{i}_{\bar{1}}$,
with
${\mathfrak{g}}^{i}_{\bar{0}}={\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{0}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{0}})\mbox{
and
}{\mathfrak{g}}^{i}_{\bar{1}}={\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{1}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{1}})$.
Thus, ${\mathfrak{g}}$ is said to be naturally graded if the following
conditions hold:

  * $1.$

$gr({\mathfrak{g}})=\sum_{i\in\mathbb{N}}{\mathfrak{g}}^{i}$ is a graded Lie
(resp. Leibniz) superalgebra
($[{\mathfrak{g}}^{i},{\mathfrak{g}}^{j}]\subset{\mathfrak{g}}^{i+j}$),

  * $2.$

${\mathfrak{g}},gr({\mathfrak{g}})$ are isomorphic.

######  Remark 3.1.

The condition $1$ is in fact required in the above definition. Note that in
contrast with Lie/Leibniz algebras, $gr({\mathfrak{g}})$ could not be graded.
Indeed, if we consider the nilpotent Leibniz superalgebra (which is also a Lie
superalgebra) expressed in the adapted basis
$\\{X_{1},X_{2},X_{3},X_{4},Y_{1}\\}$ as

$\left\\{\begin{array}[]{lll}[X_{2},X_{1}]=X_{3},&[X_{3},X_{1}]=X_{4},&[X_{1},X_{2}]=-X_{3}\\\\[2.84526pt]
[X_{1},X_{2}]=-X_{3},&[X_{1},X_{3}]=-X_{4},&[Y_{1},Y_{1}]=X_{4}\\\\[2.84526pt]
\end{array}\right.$

we have
${\mathfrak{g}}^{1}={\mathfrak{g}}^{1}_{\bar{0}}\oplus{\mathfrak{g}}^{1}_{\bar{1}}=({\mathcal{C}}^{0}({\mathfrak{g}}_{\bar{0}})/{\mathcal{C}}^{1}({\mathfrak{g}}_{\bar{0}}))\oplus({\mathcal{C}}^{0}({\mathfrak{g}}_{\bar{1}})/{\mathcal{C}}^{1}({\mathfrak{g}}_{\bar{1}}))=<X_{1},X_{2}>\oplus<Y_{1}>$
Analogously, $L^{2}=<X_{3}>$ and ${\mathfrak{g}}^{3}=<X_{4}>$. Nevertheless,
as $(Y_{1},Y_{1})=X_{4}$ then it is not verified that
$[{\mathfrak{g}}^{1},{\mathfrak{g}}^{1}]\subset{\mathfrak{g}}^{2}$ and
consequently it is not graded.

##  4\. Naturally graded filiform Lie superalgebras

Among all the nilpotent Lie superalgebras, the filiform ones (those with
maximal super-nilindex) constitutes one of the most important types due to its
properties and applications. Thanks to both the theorem of adapted basis and
the definition itself of filiform Lie superalgebras it can be seen that if
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ is a
filiform Lie superalgebra, then its graded superalgebra associated
$gr({\mathfrak{g}})=\sum_{i\in\mathbb{N}}{\mathfrak{g}}^{i}$ with
$[{\mathfrak{g}}^{i},{\mathfrak{g}}^{j}]\subset{\mathfrak{g}}^{i+j}$, is
exactly

$\begin{array}[]{ccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\dots\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&\end{array}$

where for simplicity
${\mathfrak{g}}^{1}={\mathfrak{g}}^{1}_{\bar{0}}\oplus{\mathfrak{g}}^{1}_{\bar{1}}=<X_{0},X_{1}>\oplus<Y_{1}>$
has been replaced by $<X_{0},X_{1},Y_{1}>$ as there can not be any possible
confusion between even elements, called $X_{i}$, and odd ones, called $Y_{j}$.
Similarly, the same replacement has been applied to every
${\mathfrak{g}}^{i}$. The last terms of $gr({\mathfrak{g}})$ depend on three
possibilities:

1\. If $n<m$, then

$\begin{array}[]{ccccccccc}\dots&\oplus&\underbrace{<X_{n},Y_{n}>}&\oplus&\underbrace{<Y_{n+1}>}&\oplus&\dots&\oplus&\underbrace{<Y_{m}>}\\\
&&{\mathfrak{g}}^{n}&&{\mathfrak{g}}^{n+1}&&&&{\mathfrak{g}}^{m}\\\
\end{array}$

2\. If $n=m$, then

$\begin{array}[]{ccccc}\dots&\oplus&\underbrace{<X_{n-1},Y_{n-1}>}&\oplus&\underbrace{<X_{n},Y_{n}>}\\\
&&{\mathfrak{g}}^{n-1}&&{\mathfrak{g}}^{n}\\\ \end{array}$

3\. If $n>m$, then

$\begin{array}[]{ccccccccc}\dots&\oplus&\underbrace{<X_{m},Y_{m}>}&\oplus&\underbrace{<X_{m+1}>}&\oplus&\dots&\oplus&\underbrace{<X_{n}>}\\\
&&{\mathfrak{g}}^{m}&&{\mathfrak{g}}^{m+1}&&&&{\mathfrak{g}}^{n}\\\
\end{array}$

######  Remark 4.1.

It can be seen that the even elements of $gr({\mathfrak{g}})$, that is
$\sum{\mathfrak{g}}^{i}_{\bar{0}}$, constitutes a naturally graded filiform
Lie algebra. Furthermore, the definition of naturally graded Lie superalgebras
is a generalization of the one of naturally graded Lie algebras. Thus, if
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ is a
naturally graded Lie superalgebra, then ${\mathfrak{g}}_{\bar{0}}$ is a
naturally graded Lie algebra. Indeed, ${\mathfrak{g}}^{i}_{\bar{0}}$ is
equivalent to
${\mathcal{C}}^{i-1}({\mathfrak{g}}_{\bar{0}})/{\mathcal{C}}^{i}({\mathfrak{g}}_{\bar{0}})$.
Recall that the latter is used in the definition of naturally graded Lie
algebras.

Thanks to this remark we can divide our study into two cases: either
${\mathfrak{g}}_{\bar{0}}=L_{n}$ or ${\mathfrak{g}}_{\bar{0}}=Q_{n}$.
Nevertheless, before doing that we are going to obtain all the naturally
graded filiform Lie superalgebras for low dimensions. We can accomplish such
classification due to the results of [31].

######  Theorem 4.1.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=n+1$ and $\dim({\mathfrak{g}}_{\bar{1}})=m$.
If $\dim({\mathfrak{g}})\leq 7$ and ${\mathfrak{g}}$ is naturally graded then
the law of ${\mathfrak{g}}$ will be isomorphic to one law of the following
list.

List of laws

Pair of dimensions: $n=2$, $m=1$. There is only one naturally graded filiform
Lie superalgebra whose law can be expressed in an adapted basis by

$L^{2,1}+\varphi_{1,2}:\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2},&(Y_{1},Y_{1})=X_{2}\end{array}\right.$

Pair of dimensions: $n=2$, $m=2$. There is only one naturally graded filiform
Lie superalgebra whose law can be expressed in an adapted basis as

$L^{2,2}+\varphi_{1,2}:\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2},&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{1}]=Y_{2},&\end{array}\right.$

Pair of dimensions: $n=2$, $m=3$. There is only one naturally graded filiform
Lie superalgebra whose law can be expressed in an adapted basis by

$L^{2,3}+\varphi_{1,2}:\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2},&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{i}]=Y_{i+1},&1\leq i\leq 2\end{array}\right.$

Pair of dimensions: $n=2$, $m=4$. There is only one naturally graded filiform
Lie superalgebra whose law can be expressed in an adapted basis by

$L^{2,4}+\varphi_{1,2}:\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2},&[X_{0},Y_{i}]=Y_{i+1},\
1\leq i\leq 3\\\\[2.84526pt] (Y_{1},Y_{1})=X_{2}&\\\\[2.84526pt]
\end{array}\right.$

Pair of dimensions: $n=3$, $m=2$. There are two non-isomorphic naturally
graded filiform Lie superalgebras whose laws can be expressed in an adapted
basis by

$\begin{array}[]{ll}L^{3,2}+\varphi_{1,2}:&L^{3,2}+\Psi_{1,1}^{2}+\varphi_{1,2}:\\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},\ 1\leq i\leq
2&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{1}]=Y_{2}&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\end{array}\right.&\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 2&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{1}]=Y_{2}&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt]
[X_{1},Y_{1}]=Y_{2}&\end{array}\right.\end{array}$

Pair of dimensions: $n=4$, $m=2$. There is only one naturally graded filiform
Lie superalgebra whose law can be expressed in an adapted basis by

$L^{4,2}+\overline{\varphi}_{2,4}:\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 3&(Y_{1},Y_{2})=X_{3}\\\\[2.84526pt]
[X_{0},Y_{1}]=Y_{2}&(Y_{2},Y_{2})=X_{4}\\\\[2.84526pt]
(Y_{1},Y_{1})=2X_{2}&\end{array}\right.$

Pair of dimensions: $m=3$, $n=3$. There are two non-isomorphic naturally
graded filiform Lie superalgebras whose laws can be expressed in an adapted
basis as

$\begin{array}[]{ll}L^{3,3}+\varphi_{1,2}:&L^{3,3}+\Psi_{1,1}^{2}+\varphi_{1,2}:\\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},\ 1\leq i\leq
2&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt] [X_{0},Y_{j}]=Y_{j+1},\ 1\leq j\leq
2&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\end{array}\right.&\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 2&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt] [X_{0},Y_{j}]=Y_{j+1},\ 1\leq
j\leq 2&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt]
[X_{1},Y_{1}]=Y_{2}&\\\\[2.84526pt]
[X_{1},Y_{2}]=Y_{3}&\end{array}\right.\end{array}$

where the usual bracket products $[\ ,\ ]$ are skew-symmetrical and the
products deriving from the symmetric pairing
$S^{2}{\mathfrak{g}}_{\bar{1}}\longrightarrow{\mathfrak{g}}_{\bar{0}}$ and
denoted by $(\ ,\ )$ are symmetrical.

######  Remark 4.2.

There is no any naturally graded filiform Lie superalgebra for the cases
$n=3,m=1$; $n=4,m=1$ and $n=5,m=1$. Note also that all the naturally graded
filiform Lie superalgebras of the aforementioned theorem derive from the
naturally graded filiform Lie algebra $L_{n}$. There is no-one deriving from
$Q_{n}$.

Next, we start with $\dim({\mathfrak{g}}_{\bar{0}})=3=n+1$.

###  4.1. $n=2$ and $m$ arbitrary

Throughout this subsection we are going to study naturally graded filiform Lie
superalgebras
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ with
$\dim({\mathfrak{g}}_{\bar{0}})=n+1=3$ and $m\geq 5$ since the cases $1\leq
m\leq 4$ have already been studied.

######  Theorem 4.2.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=3$ and $\dim({\mathfrak{g}}_{\bar{1}})=m$ with
$m\geq 5$. If ${\mathfrak{g}}$ is naturally graded then the law of
${\mathfrak{g}}$ is isomorphic to

$L^{2,m}+\varphi_{1,2}:\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2}&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{i}]=Y_{i+1},&1\leq i\leq m-1\\\\[2.84526pt] \end{array}\right.$

###### Proof.

It can be seen that ${\mathfrak{g}}_{\bar{0}}=L_{2}$ since
$\dim({\mathfrak{g}}_{\bar{0}})=3$ and ${\mathfrak{g}}_{\bar{0}}$ is a
naturally graded filiform Lie algebra. Therefore $gr({\mathfrak{g}})$ is
exactly

$\begin{array}[]{ccccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<Y_{3}>}&\oplus&\dots&\oplus&\underbrace{<Y_{m}>}\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&&&{\mathfrak{g}}^{m}\end{array}$

Thus, the bracket products can be considered to be

$\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2},&[X_{1},Y_{i}]=b_{1i}Y_{i+1},\
1\leq i\leq m-1\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
m-1,&[X_{2},Y_{i}]=b_{2i}Y_{i+2},\ 1\leq i\leq m-2\\\\[2.84526pt]
(Y_{1},Y_{1})=c_{11}X_{2}&\end{array}\right.$

with $c_{11}\neq 0$ in order not to have a degenerated Lie superalgebra, that
is, a Lie algebra. By means of a simple change of scale it can be seen that
there is no loss of generality in considering $c_{11}=1$. It can be seen that
$(Y_{1},Y_{1})=X_{2}$ corresponds with the cocycle $\varphi_{1,2}$.
Additionally, the remaining structure constants $b_{1i},b_{2i}$ indeed will be
determined by the cocycles $\Psi_{1,1}^{2},\Psi_{2,1}^{3},$ respectively (more
details it can be consulted [25]) given as

$\begin{array}[]{lll}\Psi_{1,1}^{2}:\ [X_{1},Y_{j}]=Y_{j+1},\ 1\leq j\leq
m-1&&\Psi_{2,1}^{3}:\left\\{\begin{array}[]{l}[X_{1},Y_{j}]=-(j-1)Y_{j+1},\
2\leq j\leq m-1\\\\[2.84526pt] [X_{2},Y_{j}]=Y_{j+2},\ 1\leq j\leq
m-2\\\\[2.84526pt] \end{array}\right.\end{array}$

These cocycles, or so-called infinitesimal deformations, are the only one
“naturally graded”, that is, of weight $0$. Recall that the weight of
$\Psi_{k,1}^{s}$ is $s-k-1$.

Next, we apply the method of classification used in [31]: every filiform Lie
superalgebra can be expressed by $L^{n,m}+\Psi$, with $L^{n,m}$ the law of the
model filiform Lie superalgebra and $\Psi$ a linear (infinitesimal)
deformation that verifies $\Psi\circ\Psi=0$, being
$\Psi\circ\Psi(x,y,z)=\Psi(\Psi(x,y),z)+\Psi(\Psi(z,x),y)+\Psi(\Psi(y,z),x)$.
Thus, we can consider

$L^{2,m}+\Psi:\left\\{\begin{array}[]{ll}[X_{0},X_{1}]=X_{2},&[X_{1},Y_{j}]=[a-b(j-1)]Y_{j+1},\
2\leq i\leq m-1\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
m-1,&[X_{2},Y_{j}]=bY_{j+2},\ 1\leq i\leq m-2\\\\[2.84526pt]
[X_{1},Y_{1}]=aY_{2},&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt] \end{array}\right.$

where $\Psi=\varphi_{1,2}+a\Psi_{1,1}^{2}+b\Psi_{2,1}^{3}$. From the condition
$\Psi\circ\Psi(X_{1},X_{2},Y_{1})=0$ it is obtained that $b=0$. Then applying
the isomorphism, or so-called change of basis, defined by
$X^{\prime}_{0}=X_{0}$, $X^{\prime}_{1}=X_{1}-aX_{0}$, $X^{\prime}_{2}=X_{2}$
and $Y^{\prime}_{j}=Y_{j}$ for any $j\in\mathbb{N}$, it can seen that there is
no loss of generality in supposing $a=0$. Consequently, it remains only
$L^{2,m}+\varphi_{1,2}$. ∎

###  4.2. $n=3$ and $m=4$

In this case we have the following results:

######  Theorem 4.3.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=\dim({\mathfrak{g}}_{\bar{1}})=4$. If
${\mathfrak{g}}$ is naturally graded then the law of ${\mathfrak{g}}$ is
isomorphic to one of the following list of pairwise non-isomorphic laws

$\begin{array}[]{lll}\begin{array}[]{l}L^{3,4}+\varphi_{1,2}:\\\ \\\
\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1}\\\\[2.84526pt] (Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{1}{2}X_{3}\end{array}\right.\end{array}&\begin{array}[]{ll}L^{3,4}+\varphi_{1,2}+\Psi_{3,1}^{4}:\\\
\\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1}&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1}&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt]
[X_{1},Y_{3}]=Y_{4}&\\\\[2.84526pt] [X_{2},Y_{2}]=-Y_{4}&\\\\[2.84526pt]
[X_{3},Y_{1}]=Y_{4}&\end{array}\right.\end{array}&\begin{array}[]{l}L^{3,4}+\varphi_{1,2}+\Psi_{1,1}^{2}:\\\
\\\ \left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1}\\\\[2.84526pt] [X_{1},Y_{j}]=Y_{j+1}\\\\[2.84526pt]
(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{1}{2}X_{3}\end{array}\right.\end{array}\end{array}$

with $1\leq i\leq 2$ and $1\leq j\leq 3.$

###### Proof.

We consider naturally graded filiform Lie superalgebras
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ with
$\dim({\mathfrak{g}}_{\bar{0}})=4$ and $\dim({\mathfrak{g}}_{\bar{1}})=4.$
Therefore ${\mathfrak{g}}_{\bar{0}}=L_{3}$ and $gr({\mathfrak{g}})$ is

$\begin{array}[]{ccccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\underbrace{<Y_{4}>}\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&{\mathfrak{g}}^{4}&&\end{array}$

Thus, the bracket products can be considered to be

$\begin{array}[]{ll}\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1},\
i=1,2\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq 3\\\\[2.84526pt]
[X_{1},Y_{i}]=b_{1i}Y_{i+1},\ 1\leq i\leq 3\\\\[2.84526pt]
[X_{2},Y_{i}]=b_{2i}Y_{i+2},\ 1\leq i\leq 2\\\\[2.84526pt]
\end{array}\right.&\begin{array}[]{ll}[X_{3},Y_{1}]=b_{31}Y_{4},\\\\[2.84526pt]
(Y_{1},Y_{1})=c_{11}X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=c_{12}X_{3}\\\\[2.84526pt] \end{array}\end{array}$

In a similar way to the previous case the structure constants can be
considered to be determined by the cocycles $\Psi_{1,1}^{2},\Psi_{2,1}^{3}$,
$\Psi_{3,1}^{4}$ and $\varphi_{1,2}$ (see [8, 25]). In general,
$\Psi_{1,1}^{2}$ and $\Psi_{2,1}^{3}$ have already been described in the above
case, and $\Psi_{3,1}^{4}$ and $\varphi_{1,2}$ are as follows

$\begin{array}[]{ll}\begin{array}[]{l}\Psi_{3,1}^{4}:\left\\{\begin{array}[]{l}[X_{1},Y_{3}]=Y_{4}\\\\[2.84526pt]
[X_{2},Y_{2}]=-Y_{4}\\\\[2.84526pt] [X_{3},Y_{1}]=Y_{4}\\\\[2.84526pt]
\end{array}\right.\end{array}&\begin{array}[]{l}\varphi_{1,2}:\left\\{\begin{array}[]{l}(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt]
\end{array}\right.\end{array}\end{array}$

Thus, in order not to have a degenerated case, that is, a Lie algebra, the
coefficient of $\varphi_{1,2}$ must be different from zero and then, it can be
easily seen that there is no loss of generality in considering this
coefficient equals $1$. Therefore we can consider

$\begin{array}[]{ll}&\\\
L^{3,4}+\Psi:\left\\{\begin{array}[]{l}[X_{0},X_{1}]=X_{2},\\\\[2.84526pt]
[X_{0},X_{2}]=X_{3},\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
3,\\\\[2.84526pt] [X_{1},Y_{1}]=a_{1}Y_{2},\\\\[2.84526pt]
[X_{1},Y_{2}]=(a_{1}-a_{2})Y_{3},\\\\[2.84526pt]
[X_{1},Y_{3}]=(a_{1}-2a_{2}+a_{3})Y_{4},\\\\[2.84526pt]
\end{array}\right.&\begin{array}[]{ll}[X_{2},Y_{1}]=a_{2}Y_{3}\\\\[2.84526pt]
[X_{2},Y_{2}]=(a_{2}-a_{3})Y_{4}\\\\[2.84526pt]
[X_{3},Y_{1}]=a_{3}Y_{4}\\\\[2.84526pt] (Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt] \end{array}\end{array}$

being
$\Psi=\varphi_{1,2}+a_{1}\Psi_{1,1}^{2}+a_{2}\Psi_{2,1}^{3}+a_{3}\Psi_{3,1}^{4}$.
From the condition $\Psi\circ\Psi(Y_{1},Y_{1},Y_{1})=0$ we obtain $a_{2}=0,$
and on account of $\Psi\circ\Psi(X_{1},X_{2},Y_{1})=0$ we conclude
$a_{1}a_{3}=0$.

  * •

If $a_{1}=0$ we distinguish two cases: $a_{3}=0$ or $a_{3}\neq 0.$

    * –

In the case of $a_{3}=0$ we obtain $L^{3,4}+\varphi_{1,2}$.

    * –

If $a_{3}\neq 0$ without loss of generality we can consider $a_{3}=1$ (by
means of a simple change of scale) obtaining
$L^{3,4}+\varphi_{1,2}+\Psi_{3,1}^{4}$.

  * •

If $a_{1}\neq 0$ then $a_{3}=0.$ Therefore, applying the isomorphism (change
of scale) defined by $X^{\prime}_{0}=a_{1}X_{0}$, $X^{\prime}_{1}=X_{1}$,
$X^{\prime}_{2}=a_{1}X_{2}$, $X^{\prime}_{3}=a_{1}^{2}X_{3}$,
$Y^{\prime}_{1}=\sqrt{a_{1}}Y_{1},$ $Y^{\prime}_{2}=a_{1}\sqrt{a_{1}}Y_{2}$
and $Y^{\prime}_{3}=a_{1}^{2}\sqrt{a_{1}}Y_{3}$, it can seen that there is no
loss of generality in supposing $a_{1}=1,$ and we obtain
$L^{3,4}+\varphi_{1,2}+\Psi_{1,1}^{2}$.

∎

###  4.3. $n=3$ and $m\geq 5$

Throughout this subsection we are going to study naturally graded filiform Lie
superalgebras
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ with
$\dim({\mathfrak{g}}_{\bar{0}})=n+1=4$ and $m\geq 5$ since the cases $1\leq
m\leq 4$ have already been studied.

######  Theorem 4.4.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=4$ and $\dim({\mathfrak{g}}_{\bar{1}})=m$ with
$m\geq 5$. If ${\mathfrak{g}}$ is naturally graded then the law of
${\mathfrak{g}}$ is isomorphic to one of the two non-isomorphic laws:

$\begin{array}[]{ll}\begin{array}[]{l}L^{3,m}+\varphi_{1,2}:\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1}\\\\[2.84526pt] (Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{1}{2}X_{3}\end{array}\right.\end{array}&\begin{array}[]{ll}L^{3,m}+\varphi_{1,2}+\Psi_{1,1}^{2}:\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1}&(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1}&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt]
[X_{1},Y_{j}]=Y_{j+1}&\end{array}\right.\end{array}\end{array}$

with $1\leq i\leq 2,$ $1\leq j\leq m-1.$

###### Proof.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a naturally graded filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=4$ and $\dim({\mathfrak{g}}_{\bar{1}})=m\geq
5.$ Therefore, ${\mathfrak{g}}_{\bar{0}}=L_{4}$ and $gr({\mathfrak{g}})$ is

$\begin{array}[]{ccccccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\underbrace{<Y_{4}>}&\oplus&\dots&\oplus&\underbrace{<Y_{m}>}\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&{\mathfrak{g}}^{4}&&&&{\mathfrak{g}}^{m}\end{array}$

So the bracket products are

$\begin{array}[]{lll}[X_{0},X_{1}]=X_{2},&&[X_{2},Y_{i}]=b_{2i}Y_{i+2},\ 1\leq
i\leq m-2\\\\[2.84526pt] [X_{0},X_{2}]=X_{3},&&[X_{3},Y_{i}]=b_{3i}Y_{i+3},\
1\leq i\leq m-3\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
m-1,&&(Y_{1},Y_{1})=c_{11}X_{2}\\\\[2.84526pt] [X_{1},Y_{i}]=b_{1i}Y_{i+1},\
1\leq i\leq m-1,&&(Y_{1},Y_{2})=c_{12}X_{3}\end{array}$

As in the above case and in order not to have any degenerated case the
coefficient of $\varphi_{1,2}$ must be different from zero and then, it can be
easily seen that there is no loss of generality in considering this
coefficient equals $1$. Therefore we can consider $L^{3,m}+\Psi,$ with
$\Psi=\varphi_{1,2}+a_{1}\Psi_{1,1}^{2}+a_{2}\Psi_{2,1}^{3}+a_{3}\Psi_{3,1}^{4},$
and we have the following products:

$L^{3,m}+\Psi:\left\\{\begin{array}[]{l}[X_{0},X_{1}]=X_{2}\\\\[2.84526pt]
[X_{0},X_{2}]=X_{3}\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
m-1\\\\[2.84526pt] [X_{1},Y_{1}]=a_{1}Y_{2},\\\\[2.84526pt]
[X_{1},Y_{i}]=\bigl{(}a_{1}-(i-1)a_{2}+\frac{(i-1)(i-2)}{2}a_{3}\bigr{)}Y_{i+1},\
2\leq i\leq m-1\\\\[2.84526pt]
[X_{2},Y_{i}]=\bigl{(}a_{2}-(i-1)a_{3}\bigr{)}Y_{i+2},\ 1\leq i\leq
m-2\\\\[2.84526pt] [X_{3},Y_{i}]=a_{3}Y_{i+3},\ 1\leq i\leq m-3\\\\[2.84526pt]
(Y_{1},Y_{1})=X_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt] \end{array}\right.$

The condition $\Psi\circ\Psi(Y_{1},Y_{1},Y_{1})=0$ leads to $a_{2}=0$ and
then, from the condition $\Psi\circ\Psi(X_{1},X_{3},Y_{1})=0$ it is obtained
that $a_{3}=0$. So if $a_{1}=0$ we have $L^{3,m}+\varphi_{1,2}$. Otherwise we
have $L^{3,4}+\varphi_{1,2}+\Psi_{1,1}^{2}$ after applying a simple change of
scale. ∎

###  4.4. $n$ and $m$ arbitrary with $n>2m$

######  Theorem 4.5.

There is not any non-degenerated naturally graded filiform Lie superalgebra
${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ with
the condition $n>2m$, being $\dim({\mathfrak{g}}_{\bar{0}})=n+1$ and
$\dim({\mathfrak{g}}_{\bar{1}})=m$.

###### Proof.

This result derives from the fact that there is not any cocycle
$\varphi_{k,s}$ naturally graded, that is, of weight equals $0$ since we have
always $\max\\{\frac{n-2m-1}{2},n-2m\\}\leq$ weight$(\varphi_{k,s})$ (see [8],
Proposition 6.4). ∎

######  Remark 4.3.

We have as particular cases of the above theorem $m=1,n\geq 3$; $m=2,n\geq 5$
and $m=3,n\geq 7$. Consequently, and in order to study all the posibilities
for $m\leq 3,$ only remains to consider the cases $4\leq n\leq 6$ and $m=3$.

###  4.5. $n=4$ and $m=3$

######  Theorem 4.6.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=5$ and $\dim({\mathfrak{g}}_{\bar{1}})=3$. If
${\mathfrak{g}}$ is naturally graded then the law of ${\mathfrak{g}}$ is
isomorphic to one of the following non-isomorphic laws:

$\begin{array}[]{ll}\begin{array}[]{l}L^{4,3}+\varphi_{2,4}:\\\ \\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1}&(Y_{1},Y_{3})=-X_{4}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1}&(Y_{2},Y_{2})=X_{4}\end{array}\right.\end{array}&\begin{array}[]{l}L^{4,3}+\varphi_{2,4}+\Psi_{1,1}^{2}+2\Psi_{2,1}^{3}:\\\
\\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},&[X_{2},Y_{1}]=Y_{3}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1},&(Y_{1},Y_{3})=-X_{4}\\\\[2.84526pt]
[X_{1},Y_{1}]=Y_{2}&(Y_{2},Y_{2})=X_{4}\\\\[2.84526pt]
[X_{1},Y_{2}]=-Y_{3}&\\\\[2.84526pt] \end{array}\right.\end{array}\end{array}$
$\begin{array}[]{ll}\begin{array}[]{l}L^{4,3}+\varphi_{1,2}+t\varphi_{2,4}:\\\
\\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},&(Y_{1},Y_{2})=\frac{1}{2}X_{3}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1},&(Y_{1},Y_{3})=(\frac{1}{2}-t)X_{4}\\\\[2.84526pt]
(Y_{1},Y_{1})=X_{2},&(Y_{2},Y_{2})=tX_{4}\end{array}\right.\end{array}&\begin{array}[]{l}L^{4,3}+4\varphi_{1,2}+\varphi_{2,4}+\Psi_{1,1}^{2}:\\\
\\\
\left\\{\begin{array}[]{ll}[X_{0},X_{i}]=X_{i+1},&(Y_{1},Y_{1})=4X_{2}\\\\[2.84526pt]
[X_{0},Y_{j}]=Y_{j+1},&(Y_{1},Y_{2})=2X_{3}\\\\[2.84526pt]
[X_{1},Y_{1}]=Y_{2},&(Y_{1},Y_{3})=X_{4}\\\\[2.84526pt]
[X_{1},Y_{2}]=Y_{3},&(Y_{2},Y_{2})=X_{4}\end{array}\right.\end{array}\end{array}$

with $1\leq i\leq 3$, $1\leq j\leq 2$ and $t\in\mathbb{C}$ a parameter.

###### Proof.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a naturally graded filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=5$ and $\dim({\mathfrak{g}}_{\bar{1}})=3.$
Therefore, ${\mathfrak{g}}_{\bar{0}}=L_{4}$ and $gr({\mathfrak{g}})$ is

$\begin{array}[]{ccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\underbrace{<X_{4}>}\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&{\mathfrak{g}}^{4}\end{array}$

In an analogous way to previous cases, we can consider $L^{4,3}+\Psi,$ with
$\Psi=c\varphi_{1,2}+d\varphi_{2,4}+a_{1}\Psi_{1,1}^{2}+a_{2}\Psi_{2,1}^{3},$
and we have the following products:

$\begin{array}[]{ll}L^{3,4}+\Psi:\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 3,\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
2,\\\\[2.84526pt] [X_{1},Y_{1}]=a_{1}Y_{2},\\\\[2.84526pt]
[X_{1},Y_{2}]=(a_{1}-a_{2})Y_{3},\\\\[2.84526pt]
[X_{2},Y_{1}]=a_{2}Y_{3}\end{array}\right.&\begin{array}[]{ll}(Y_{1},Y_{1})=cX_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{c}{2}X_{3}\\\\[2.84526pt]
(Y_{1},Y_{3})=(\frac{c}{2}-d)X_{4}\\\\[2.84526pt]
(Y_{2},Y_{2})=dX_{4}\end{array}\end{array}$

where at least $c$ or $d$ is non-zero. The condition
$\Psi\circ\Psi(Y_{1},Y_{1},Y_{1})=0$ give us $a_{2}c=0$ and, since
$\Psi\circ\Psi(X_{1},Y_{1},Y_{2})=0,$ we have
$a_{1}d+(a_{2}-a_{1})(\frac{c}{2}-d)=0$.

  * •

If $c=0$ then $d\neq 0$, and from $a_{1}d+(a_{2}-a_{1})(\frac{c}{2}-d)=0$ we
have $a_{2}=2a_{1}$.

    * –

For $a_{1}=0$ is $a_{2}=0$ and $d\neq 0,$ so using a change of scale $d=1$ and
we obtain $L^{4,3}+\varphi_{2,4}$.

    * –

For $a_{1}\neq 0$ again $d\neq 0$, and with a similar change of scale
$a_{1}=d=1$ and $a_{2}=2,$ obtaining
$L^{4,3}+\varphi_{2,4}+\Psi_{1,1}^{2}+2\Psi_{2,1}^{3}$.

  * •

In the case of $c\neq 0$ then necessarily $a_{2}=0$ and from
$a_{1}d+(a_{2}-a_{1})(\frac{c}{2}-d)=0$ is $4a_{1}d=a_{1}c$.

    * –

For $a_{1}=0$ by means of the change of scale $c=1,d=t,$ being $t$ a
parameter, we obtain $L^{4,3}+\varphi_{1,2}+t\varphi{2,4}$.

    * –

For $a_{1}\neq 0$ is $d\neq 0$ and with the change of scale
$a_{1}=d=1,a_{2}=0$ and $c=4$ we obtain
$L^{4,3}+4\varphi_{1,2}+\varphi_{2,4}+\Psi_{1,1}^{2}$.

∎

###  4.6. $n=5$ and $m=3$

######  Theorem 4.7.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=6$ and $\dim({\mathfrak{g}}_{\bar{1}})=3$. If
${\mathfrak{g}}$ is naturally graded then the law of ${\mathfrak{g}}$ will be
isomorphic to the following law

$\begin{array}[]{ll}L^{5,3}+\overline{\varphi}_{2,4}:\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 4,\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
2,\\\\[2.84526pt] (Y_{1},Y_{1})=3X_{2},\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{3}{2}X_{3},\end{array}\right.&\begin{array}[]{ll}(Y_{1},Y_{3})=\frac{1}{2}X_{4}\\\\[2.84526pt]
(Y_{2},Y_{2})=X_{4}\\\\[2.84526pt]
(Y_{2},Y_{3})=\frac{1}{2}X_{5}\\\\[2.84526pt] \end{array}\end{array}$

###### Proof.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a naturally graded filiform Lie superalgebra with $dim({\mathfrak{g}}_{0})=6$
and $dim({\mathfrak{g}}_{1})=3.$ Then $gr({\mathfrak{g}})$ is

$\begin{array}[]{ccccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\underbrace{<X_{4}>}&\oplus&\underbrace{<X_{5}>}\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&{\mathfrak{g}}^{4}&&{\mathfrak{g}}^{5}\end{array}$

In this case we have to distinguish two separate sub-cases:
${\mathfrak{g}}_{\bar{0}}=L_{5}$ or ${\mathfrak{g}}_{\bar{0}}=Q_{5}$, the two
possible naturally graded filiform Lie algebras for this dimension.

  * •

If ${\mathfrak{g}}_{\bar{0}}=L_{5}$, and as in previous cases we get
$L^{5,3}+\Psi,$ with
$\Psi=c\overline{\varphi}_{2,4}+a_{1}\Psi_{1,1}^{2}+a_{2}\Psi_{2,1}^{3},$
being $\overline{\varphi}_{2,4}=3\varphi_{1,2}+\varphi_{2,4}$ and $c\neq 0$.
So we have

$\begin{array}[]{ll}L^{5,3}+\Psi:\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 4,\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
2,\\\\[2.84526pt] [X_{1},Y_{1}]=a_{1}Y_{2},\\\\[2.84526pt]
[X_{1},Y_{2}]=(a_{1}-a_{2})Y_{3},\\\\[2.84526pt]
[X_{2},Y_{1}]=a_{2}Y_{3},\end{array}\right.&\begin{array}[]{ll}(Y_{1},Y_{1})=3cX_{2}\\\\[2.84526pt]
(Y_{1},Y_{2})=\frac{3}{2}cX_{3}\\\\[2.84526pt]
(Y_{1},Y_{3})=\frac{c}{2}X_{4}\\\\[2.84526pt]
(Y_{2},Y_{2})=cX_{4}\\\\[2.84526pt]
(Y_{2},Y_{3})=\frac{c}{2}X_{5}\\\\[2.84526pt] \end{array}\end{array}$

The condition $\Psi\circ\Psi(X_{1},Y_{1},Y_{2})=0$ give us $a_{1}=0$. Since
$\Psi\circ\Psi(Y_{1},Y_{1},Y_{1})=0$ we have $a_{2}=0.$ With a change of scale
we can consider $c=1$ and we obtain $L^{5,3}+\overline{\varphi}_{2,4}$.

  * •

If ${\mathfrak{g}}_{\bar{0}}=Q_{5}$ we have to add the brackets
$[X_{1},X_{4}]=-X_{5}$ and $[X_{2},X_{3}]=X_{5}$. Indeed, these brackets stem
from cocycles of [26]. All the above reasoning is also valid but from the
condition $\Psi\circ\Psi(X_{1},Y_{2},Y_{2})=0$ we obtain that
$[X_{1},X_{4}]=0$, which is clearly a contradiction. Thus, there is not any
non-degenerated filiform Lie superalgebra deriving from the Lie algebra
$Q_{5}$.

∎

###  4.7. $n=6$ and $m=3$

######  Theorem 4.8.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a non-degenerated filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=7$ and $\dim({\mathfrak{g}}_{\bar{1}})=3$. If
${\mathfrak{g}}$ is naturally graded then the law of ${\mathfrak{g}}$ is
isomorphic to the following law

$\begin{array}[]{ll}L^{6,3}+\overline{\varphi}_{3,6}:\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 5,\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
2,\\\\[2.84526pt] (Y_{1},Y_{1})=6X_{2},\\\\[2.84526pt]
(Y_{1},Y_{2})=3X_{3},\end{array}\right.&\begin{array}[]{ll}(Y_{1},Y_{3})=X_{4}\\\\[2.84526pt]
(Y_{2},Y_{2})=2X_{4}\\\\[2.84526pt] (Y_{2},Y_{3})=X_{5}\\\\[2.84526pt]
(Y_{3},Y_{3})=X_{6}\end{array}\end{array}$

###### Proof.

Let ${\mathfrak{g}}={\mathfrak{g}}_{\bar{0}}\oplus{\mathfrak{g}}_{\bar{1}}$ be
a naturally graded filiform Lie superalgebra with
$\dim({\mathfrak{g}}_{\bar{0}})=5$ and $\dim({\mathfrak{g}}_{\bar{1}})=3.$
Therefore ${\mathfrak{g}}_{\bar{0}}=L_{6}$ and $gr({\mathfrak{g}})$ is

$\begin{array}[]{ccccccccccc}\underbrace{<X_{0},X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\underbrace{<X_{4}>}&\oplus&\underbrace{<X_{5}>}&\oplus&\underbrace{<X_{6}>}\\\
{\mathfrak{g}}^{1}&&{\mathfrak{g}}^{2}&&{\mathfrak{g}}^{3}&&{\mathfrak{g}}^{4}&&{\mathfrak{g}}^{5}&&{\mathfrak{g}}^{6}\end{array}$

As in previous cases, since the products we get $L^{6,3}+\Psi,$ with
$\Psi=c\overline{\varphi}_{3,6}+a_{1}\Psi_{1,1}^{2}+a_{2}\Psi_{2,1}^{3},$
where $\overline{\varphi}_{3,6}=6\varphi_{1,2}+2\varphi_{2,4}+\varphi_{3,6}$,
and we have these products:

$\begin{array}[]{ll}L^{6,3}+\Psi:\left\\{\begin{array}[]{l}[X_{0},X_{i}]=X_{i+1},\
1\leq i\leq 5,\\\\[2.84526pt] [X_{0},Y_{i}]=Y_{i+1},\ 1\leq i\leq
2,\\\\[2.84526pt] [X_{1},Y_{1}]=a_{1}Y_{2},\\\\[2.84526pt]
[X_{1},Y_{2}]=(a_{1}-a_{2})Y_{3},\\\\[2.84526pt]
[X_{2},Y_{1}]=a_{2}Y_{3},\\\\[2.84526pt]
(Y_{1},Y_{1})=6cX_{2},\end{array}\right.&\begin{array}[]{ll}(Y_{1},Y_{2})=3cX_{3}\\\\[2.84526pt]
(Y_{1},Y_{3})=cX_{4}\\\\[2.84526pt] (Y_{2},Y_{2})=2cX_{4}\\\\[2.84526pt]
(Y_{2},Y_{3})=cX_{5}\\\\[2.84526pt]
(Y_{3},Y_{3})=cX_{6}\end{array}\end{array}$

with $c$ non-zero to avoid the trivial case. The condition
$\Psi\circ\Psi(Y_{1},Y_{1},Y_{1})=0$ give us $a_{2}=0$, and since
$\Psi\circ\Psi(X_{1},Y_{1},Y_{2})=0$ we have $a_{1}=0.$ By using a change of
scale $c=1$ we obtain $L^{6,3}+\overline{\varphi}_{3,6}$. ∎

##  5\. Naturally Graded (non-Lie) Leibniz Superalgebras with maximal super-
nilindex

Throughout this section we are going to study the Leibniz superalgebras with
super-nilindex $(n,m)$ because of the fact that in this case the even part
$L_{\bar{0}}$ is a null-filiform Leibniz algebra and the odd part
$L_{\bar{1}}$ has structure of filiform $L_{\bar{0}}-$module. Consequently, it
seems to be the first case to consider the “naturally graded” structure. Note
that all these Leibniz superalgebras with super-nilindex $(n,m)$ are non-Lie
ones and contain in particular the only type of Leibniz superalgebra single
generated or so-called null-filiform Leibniz superalgebras (for more details
see [27]). It is not difficult to see that these last superalgebras, in the
case of non-degenerated, are not naturally graded.

Next, we are going to recall the general expression of the Leibniz
superalgebras with super-nilindex $(n,m)$ (see [18]).

######  Theorem 5.1.

If $L=L_{\bar{0}}\oplus L_{\bar{1}}$ is a Leibniz superalgebra of super-
nilindex $(n,m)$, then there exists an adapted basis of $L$, namely
$\\{X_{1},X_{2},\dots,X_{n},Y_{1},Y_{2},\dots,Y_{m}\\}$, with
$\\{X_{1},X_{2},\dots,X_{n}\\}$ a basis of $L_{\bar{0}}$ and
$\\{Y_{1},Y_{2},\dots,Y_{m}\\}$ a basis of $L_{\bar{1}}$, such that

$\begin{array}[]{llc}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[X_{n},X_{1}]=0\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq
m-1,&[Y_{m},X_{1}]=0\\\\[2.84526pt] \end{array}$

Moreover, $[Y_{j},X_{k}]=0$ for $1\leq j\leq m$ and $2\leq k\leq n$, and the
omitted products of $L_{\bar{0}}=<X_{1},X_{2},\dots,X_{n}>$ vanish.

Thus, it can be seen that if $L=L_{\bar{0}}\oplus L_{\bar{1}}$ is a Leibniz
superalgebra of s-nilindex $(n,m)$, then its graded superalgebra associated
$gr(L)=\sum_{i\in\mathbb{N}}L^{i}$, satisfying $[L^{i},L^{j}]\subset L^{i+j}$,
is exactly:

$\begin{array}[]{ccccccc}\underbrace{<X_{1},Y_{1}>}&\oplus&\underbrace{<X_{2},Y_{2}>}&\oplus&\underbrace{<X_{3},Y_{3}>}&\oplus&\dots\\\
L^{1}&&L^{2}&&L^{3}&&\end{array}$

where for simplicity $L^{1}=L^{1}_{\bar{0}}\oplus
L^{1}_{\bar{1}}=<X_{1}>\oplus<Y_{1}>$ has been replaced by $<X_{1},Y_{1}>$ as
there can not be any possible confusion between even elements, called $X_{i}$,
and odd ones, called $Y_{j}$. Analogously, the same replacement has been
applied to every $L^{i}$. The last terms of $gr(L)$ depend on three
possibilities:

1\. If $n<m$, then

$\begin{array}[]{ccccccccc}\dots&\oplus&\underbrace{<X_{n},Y_{n}>}&\oplus&\underbrace{<Y_{n+1}>}&\oplus&\dots&\oplus&\underbrace{<Y_{m}>}\\\
&&L^{n}&&L^{n+1}&&&&L^{m}\\\ \end{array}$

2\. If $n=m$, then

$\begin{array}[]{ccccc}\dots&\oplus&\underbrace{<X_{n-1},Y_{n-1}>}&\oplus&\underbrace{<X_{n},Y_{n}>}\\\
&&L^{n-1}&&L^{n}\\\ \end{array}$

3\. If $n>m$, then

$\begin{array}[]{ccccccccc}\dots&\oplus&\underbrace{<X_{m-1},Y_{m}>}&\oplus&\underbrace{<X_{m}>}&\oplus&\dots&\oplus&\underbrace{<X_{n}>}\\\
&&L^{m}&&L^{m+1}&&&&L^{n}\\\ \end{array}$

######  Remark 5.1.

It can be seen that the even elements of $gr(L)$, that is
$\sum_{i\in\mathbb{N}}L^{i}_{\bar{0}}$, constitutes a naturally graded null-
filiform Leibniz algebra. That is,

$NF^{n}:[X_{i},X_{1}]=X_{i+1},\quad 1\leq i\leq n-1,$

Furthermore, the definition of naturally graded Leibniz superalgebras is a
generalization of naturally graded Leibniz algebras. Thus, if
$L=L_{\bar{0}}\oplus L_{\bar{1}}$ is a naturally graded Leibniz superalgebra,
then $L_{\bar{0}}$ is a naturally graded Leibniz algebra. Indeed,
$L^{i}_{\bar{0}}$ is equivalent to
${\mathcal{C}}^{i-1}(L_{\bar{0}})/{\mathcal{C}}^{i}(L_{\bar{0}})$. Recall that
the latter is used in the definition of naturally graded Leibniz algebras.

###  5.1. Low dimensions

Next we are going to obtain all the naturally graded Leibniz superalgebras
with super-nilindex $(n,m)$ for low dimensions of either the even part or the
odd part. We can accomplish such classification due to the results of [18].

######  Theorem 5.2.

Let $L=L_{\bar{0}}\oplus L_{\bar{1}}$ be a non-degenerated naturally graded
Leibniz superalgebras with s-nilindex $(n,m)$, then the law of $L$ is
isomorphic to a law of the following list where the omitted products are equal
to zero.

List of laws.

Pair of dimensions: $n=2$, $m=2$. There is a one-parametric family of
naturally graded Leibniz superalgebras of s-nilindex $(2,2)$ whose law can be
expressed in an adapted basis $\\{X_{1},X_{2},Y_{1},Y_{2}\\}$ as

$\mu_{1}^{{\alpha}}=\left\\{\begin{array}[]{lll}[X_{1},X_{1}]=X_{2},&[X_{1},Y_{1}]={\alpha}Y_{2},&{\alpha}\in\mathbb{C}\\\\[2.84526pt]
[Y_{1},X_{1}]=Y_{2},&[Y_{1},Y_{1}]=X_{2}&\end{array}\right.$

Pair of dimensions: $n$ arbitray with $n\geq 3$, $m=2$. The only sub-case in
which there are naturally graded superalgebras is exactly $n=3$. In this case
there exist two non-isomorphic naturally graded Leibniz superalgebras of
s-nilindex $(3,2)$ whose laws can be expressed in an adapted basis
$\\{X_{1},X_{2},X_{3},Y_{1},Y_{2}\\}$ by:

$\begin{array}[]{ll}\mu_{2}=\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq
i\leq 2\\\\[2.84526pt] [Y_{1},X_{1}]=Y_{2}&\\\\[2.84526pt]
[Y_{1},Y_{1}]=X_{2}&\\\\[2.84526pt]
[Y_{2},Y_{1}]=X_{3}&\end{array}\right.&\mu_{3}=\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},1\leq
i\leq 2,&[Y_{1},X_{1}]=Y_{2}\\\\[2.84526pt]
[X_{1},Y_{1}]=-Y_{2},&[Y_{1},Y_{2}]=X_{3}\\\\[2.84526pt]
[Y_{1},Y_{1}]=X_{2},\end{array}\right.\end{array}$

Pair of dimensions: $n=2$, $m=3$. There are two non-isomorphic naturally
graded Leibniz superalgebras of s-nilindex $(2,3)$ whose laws can be expressed
in an adapted basis $\\{X_{1},X_{2},Y_{1},Y_{2},Y_{3}\\}$ by:

$\begin{array}[]{ll}\mu_{1}=\left\\{\begin{array}[]{ll}[X_{1},X_{1}]=X_{2},&\\\\[2.84526pt]
[Y_{1},X_{1}]=Y_{2}&\\\\[2.84526pt] [Y_{2},X_{1}]=Y_{3}&\\\\[2.84526pt]
[Y_{1},Y_{1}]=X_{2}&\end{array}\right.&\mu_{3}=\left\\{\begin{array}[]{ll}[X_{1},X_{1}]=X_{2},&[Y_{1},X_{1}]=Y_{2}\\\\[2.84526pt]
[X_{1},Y_{1}]=-Y_{2},&[Y_{2},X_{1}]=Y_{3}\\\\[2.84526pt]
[X_{1},Y_{2}]=-Y_{3},&[Y_{1},Y_{1}]=X_{2}\end{array}\right.\end{array}$

Pair of dimensions: $n=3$, $m=3$. There are two non-isomorphic naturally
graded Leibniz superalgebras of s-nilindex $(3,3)$ whose laws can be expressed
in an adapted basis $\\{X_{1},X_{2},X_{3},Y_{1},Y_{2},Y_{3}\\}$ by:

$\begin{array}[]{ll}\mu_{1}=\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq
i\leq 2\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq 2\\\\[2.84526pt]
[Y_{1},Y_{2}]=X_{3}&\\\\[2.84526pt] [Y_{2},Y_{1}]=-X_{3}&\\\\[2.84526pt]
\end{array}\right.&\mu_{8}=\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq
i\leq 2\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq 2\\\\[2.84526pt]
[X_{1},Y_{2}]=-Y_{3}&\\\\[2.84526pt] [Y_{1},Y_{1}]=X_{2}&\\\\[2.84526pt]
[Y_{1},Y_{2}]=X_{3}&\\\\[2.84526pt] \end{array}\right.\end{array}$

Pair of dimensions: $n=4$, $m=3$. There are two non-isomorphic naturally
graded Leibniz superalgebras of s-nilindex $(4,3)$ whose laws can be expressed
in an adapted basis $\\{X_{1},X_{2},X_{3},X_{4},Y_{1},Y_{2},Y_{3}\\}$ by:

$\begin{array}[]{ll}\mu_{9}=\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},1\leq
i\leq 3,&[Y_{1},Y_{3}]=X_{4}\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},1\leq j\leq
2,&[Y_{2},Y_{2}]=-X_{4}\\\\[2.84526pt]
[X_{1},Y_{1}]=-Y_{2},&[Y_{3},Y_{1}]=X_{4}\\\\[2.84526pt]
[X_{1},Y_{2}]=-Y_{3}&\end{array}\right.&\mu_{12}=\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq
i\leq 3\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq 2\\\\[2.84526pt]
[Y_{1},Y_{1}]=X_{2}&\\\\[2.84526pt] [Y_{2},Y_{1}]=X_{3}&\\\\[2.84526pt]
[Y_{3},Y_{1}]=X_{4}&\end{array}\right.\end{array}$

Pair of dimensions: $n$ arbitrary with $n\geq 5$, $m=3$. There is not any
naturally graded Leibniz superalgebra of s-nilindex $(n,3)$.

Pair of dimensions: $n=2$, $m$ arbitrary with $m\geq 4$. There are two non-
isomorphic naturally graded Leibniz superalgebras of s-nilindex $(2,m)$ whose
laws can be expressed in an adapted basis
$\\{X_{1},X_{2},Y_{1},Y_{2},Y_{3},Y_{4},\dots,Y_{m}\\}$ by

$\begin{array}[]{ll}\mu_{m-1}=\left\\{\begin{array}[]{ll}[X_{1},X_{1}]=X_{2}&\\\\[2.84526pt]
[Y_{i},X_{1}]=Y_{i+1},&1\leq i\leq m-1\\\\[2.84526pt]
[Y_{1},Y_{1}]=X_{2}&\\\\[2.84526pt]
\end{array}\right.&\mu_{m+1}=\left\\{\begin{array}[]{ll}[X_{1},X_{1}]=X_{2}&\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&2\leq j\leq m-1\\\\[2.84526pt]
[Y_{i},X_{1}]=Y_{i+1},&1\leq i\leq m-1\\\\[2.84526pt]
[Y_{1},Y_{1}]=X_{2}&\\\\[2.84526pt] \end{array}\right.\end{array}$

###  5.2. Cases $n\geq 3$ and $m\geq 4$

Throughout this sub-section we deal with the problem of determining the
structure of the aforementioned naturally graded Leibniz superalgebras for the
remaining dimensions. Notice that the naturally graded Leibniz superalgebra
with maximal s-nilindex named $NG^{n,m}$ appears in every pair of arbitrary
dimensions $n$ and $m$.

$NG^{n,m}:\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1\\\\[2.84526pt]
[Y_{i},Y_{1}]=X_{i+1},&1\leq i\leq\min\\{n-1,m\\}\end{array}\right.$

with the vectors $\\{X_{1},X_{2},\dots,X_{n},Y_{1},Y_{2},\dots,Y_{m}\\}$ as
adapted basis. Also, we have the following result.

######  Theorem 5.3.

Let $L=L_{\bar{0}}\oplus L_{\bar{1}}$ be a non-degenerated naturally graded
Leibniz superalgebras with s-nilindex $(n,m)$ and adapted basis
$\\{X_{1},X_{2},\dots,X_{n},Y_{1},Y_{2},\dots,Y_{m}\\}$. Then the law of $L$
will be isomorphic either to the law named $NG^{n,m}$ or to a law belonging to
the following families of laws. The precise expression of these last families
of laws depends on two possibilities, that is:

  * 1.

If $n\leq m$,

$\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[Y_{1},Y_{1}]=\gamma_{1}X_{2},&\\\\[2.84526pt]
[Y_{i},Y_{1}]=\gamma_{i}X_{i+1},&3\leq i\leq n-1,\\\\[2.84526pt]
[Y_{i},Y_{j}]=\displaystyle\sum_{s=0}^{j-1}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{i+s}X_{i+j},&1\leq i\leq n-2,\ 2\leq j\leq
n-i-1.\end{array}\right.$

with $\gamma_{2}=0$ and
$\gamma_{2j}=(-1)^{j}\displaystyle\sum_{s=0}^{j-2}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{j+s+1},\ \mbox{ for }2\leq
j\leq\lfloor\frac{n-1}{2}\rfloor.$

  * 2.

If $n>m$,

$\left\\{\begin{array}[]{ll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,\\\\[2.84526pt] [Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[Y_{1},Y_{1}]=\gamma_{1}X_{2},&\\\\[2.84526pt]
[Y_{i},Y_{1}]=\gamma_{i}X_{i+1},&3\leq i\leq m,\\\\[2.84526pt]
[Y_{i},Y_{j}]=\displaystyle\sum_{s=0}^{min\\{j-1,m-i\\}}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{i+s}X_{i+j},&2\leq j\leq m,\ 1\leq i\leq
min\\{m,n-j\\}.\end{array}\right.$

with $\gamma_{2}=0$ and

$\gamma_{2j}=(-1)^{j}\sum_{s=0}^{j-2}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{j+s+1},\ 2\leq
j\leq\lfloor\frac{m}{2}\rfloor\qquad\qquad\gamma_{i}=\sum_{s=1}^{m-i}(-1)^{s+1}\left(\begin{array}[]{c}m\\\
s\end{array}\right)\gamma_{i+s}.$

$\lfloor\ \rfloor$ denotes the whole part of the number.

###### Proof.

An adequate use of Mathematica software has been truly helpful to accomplish
the proof of this theorem. It has been a very useful tool to both complete all
the calculations involved in many particular pair of dimensions and deduce the
process to follow in arbitrary dimensions. Next, we will distinguish two
cases. We only prove case 1 in details and omit case 2 completely in order not
to repeat the same ideas.

Case 1:  $n\leq m$. We consider the adapted basis
$\\{X_{1},X_{2},\dots,X_{n},Y_{1},Y_{2},\dots,Y_{m}\\}$ and the gradation

$<X_{1},Y_{1}>\oplus<X_{2},Y_{2}>\oplus\dots<X_{n},Y_{n}>\oplus<Y_{n+1}>\oplus<Y_{n+2}>\oplus\dots\oplus<Y_{m}>$

Using their properties

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[X_{1},Y_{j}]=\alpha_{j}Y_{j+1},&1\leq j\leq
m-1,&[X_{i},Y_{j}]=\beta_{i\,j}Y_{i+j},&2\leq i\leq n,\quad 1\leq j\leq
m-i\\\\[2.84526pt] [Y_{i},Y_{j}]=\gamma_{i\,j}X_{i+j},&1\leq i\leq m,\quad
1\leq j\leq n-i.&\end{array}$

We have that $[X_{1},Y_{1}]+[Y_{1},X_{1}]\in Ann(L),$ then
$(\alpha_{1}+1)Y_{2}\in Ann(L).$

Case 1.1:  $1+\alpha_{1}\neq 0$, then $Y_{2}\in Ann(L).$ Taking into account
that $[Y_{2},X_{1}]=Y_{3}$, it is easy to prove that $Y_{3}\in Ann(L).$
Analogously, $Y_{4},Y_{5},\dots,Y_{m}\in Ann(L).$ Thus, we have

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[X_{1},Y_{1}]=\beta_{1}Y_{2},&&[X_{i},Y_{1}]=\beta_{i}Y_{i+1},&2\leq i\leq
n,\\\\[2.84526pt] [Y_{i},Y_{1}]=\gamma_{i}X_{i+1},&1\leq i\leq
n-1.&&\end{array}$

We make the super Leibniz identity on the triples $\\{Y_{i},Y_{1},X_{1}\\}$
and we obtain $\gamma_{i}=\gamma_{1}=\gamma$ with $1\leq i\leq n-1.$ If
$\gamma=0$ then, the superalgebra is degenerated, thus $\gamma\neq 0$ and
making $Y^{\prime}_{1}=\frac{1}{\sqrt{\gamma}}Y_{1}$ allows us to consider
$\gamma=1.$

Now, from the super Leibniz identity on the triples $\\{X_{i},Y_{1},Y_{1}\\}$
for $1\leq i\leq n-2$ we get $\beta_{i}=0.$ From $\\{X_{n-2},Y_{1},X_{1}\\}$
and $\\{X_{n-1},Y_{1},X_{1}\\}$ we obtain $\beta_{n-1}=0$ and $\beta_{n}=0,$
respectively.

Thus, we obtain the naturally graded Leibniz superalgebra $NG^{n,m}$.

Case 1.2:  $1+\alpha_{1}=0.$ We can distinguish two cases:

(a) $1+\alpha_{i}=0$ for $2\leq i\leq m-1.$ We have:

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq
m-1,&[X_{i},Y_{j}]=\beta_{i\,j}Y_{i+j},&2\leq i\leq n,\quad 1\leq j\leq
m-i\\\\[2.84526pt] [Y_{i},Y_{j}]=\gamma_{i\,j}X_{i+j},&1\leq i\leq m,\quad
1\leq j\leq n-i.&&\end{array}$

The super Leibniz identity on the following triples imposes further
constraints on the above family.

Super Leibniz identity |  | Constraint
---|---|---
$\\{X_{1},X_{1},Y_{j},\\},\ 1\leq j\leq m-2$ |  $\Rightarrow$ | $\beta_{2\,j}=0$
$\\{X_{1},X_{i},Y_{j}\\},\ 3\leq i\leq n,\ 1\leq j\leq m-i-1$ |  $\Rightarrow$ |  $\beta_{i\,j}=0$ and
|  | $[X_{i},Y_{m-i}]=\beta_{i\,m-i}Y_{m}=\beta_{i}Y_{m},\ \ 3\leq i\leq n$
$\\{X_{i},X_{1},Y_{m-i-1},\\},\ 2\leq i\leq n-1$ |  $\Rightarrow$ |  $\beta_{3}=0$ and $\beta_{i+1}=(-1)^{i+1}\beta_{3}=0$
$\\{Y_{i},X_{1},Y_{j-1}\\},\ 1\leq i\leq n-2,\ 2\leq j\leq n-i$ |  $\Rightarrow$ | $\gamma_{i\,j}=\gamma_{i\,j-1}-\gamma_{i+1\,j-1}$

Thus, we prove by the induction method that:

$\gamma_{i\,1}=\gamma_{i},\ 1\leq i\leq
n-1,\qquad\gamma_{i\,j}=\displaystyle\sum_{k=0}^{j-1}(-1)^{k}\left(\begin{array}[]{c}j-1\\\
k\end{array}\right)\gamma_{i+k},\ \ 1\leq i\leq n-2,\ 2\leq j\leq n-i$

Applying the above relations, we can write

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq
m-1,&[Y_{i},Y_{1}]=\gamma_{i}X_{i+1},&1\leq i\leq n-1\\\\[2.84526pt]
[Y_{i},Y_{j}]=\displaystyle\sum_{k=0}^{j-1}(-1)^{k}\left(\begin{array}[]{c}j-1\\\
k\end{array}\right)\gamma_{i+k}X_{i+j},&1\leq i\leq n-2,&2\leq j\leq
n-i\end{array}$

For instance, the super Leibniz identity on the triples
$[X_{1},[Y_{1},Y_{1}]]$ and $[X_{1},[Y_{j},Y_{j}]]$ gives respectively
$\gamma_{2}=0$ and

$\gamma_{2j}=(-1)^{j}\displaystyle\sum_{s=0}^{j-2}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{j+s+1},\ 2\leq j\leq\lfloor\frac{n-1}{2}\rfloor.$

We obtain the family of naturally graded Leibniz superalgebras with maximal
s-nilindex of the theorem statement.

(b) There exists $k$ such that $1+\alpha_{i}=0$ for $2\leq i\leq k-2$, and
$1+\alpha_{k-1}\neq 0.$ We have

$[X_{1},Y_{k-1}]+[Y_{k-1},X_{1}]=(1+\alpha_{k-1})Y_{k}\in Ann(L).$

Then $Y_{k}\in Ann(L)\mbox{ and }Y_{k+1},Y_{k+2},\dots,Y_{m}\in Ann(L).$
Hence, we can consider

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1,\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq k-2,&[X_{1},Y_{k-1}]=\alpha
Y_{k},&1+\alpha\neq 0\\\\[2.84526pt] [X_{i},Y_{j}]=\beta_{i\,j}Y_{i+j},&1\leq
j\leq k-1,\quad 2\leq i\leq min\\{n,m-j\\}&&\\\\[2.84526pt]
[Y_{i},Y_{j}]=\gamma_{i\,j}X_{i+j},&1\leq j\leq k-1,\quad 1\leq i\leq
n-k+1.&&\end{array}$

Consider the super Leibniz identity on the triple $\\{Y_{i},X_{1},Y_{j}\\}$
with $2\leq j\leq k-1$ and $1\leq i\leq n-j$. We get

$\gamma_{i\,j}=\gamma_{i\,j-1}-\gamma_{i+1\,j-1}.$

For $j=k,$ we have that $\gamma_{i\,k-1}=\gamma_{1\,k-1}.$ If we rename
$\gamma_{i\,1}:=\gamma_{i}$ with $1\leq i\leq n-1$, it is easy to prove by
induction method on $j,$ that

(5.1)
$\gamma_{i\,j}=\displaystyle\sum_{s=0}^{j-1}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{i+s},\quad\quad 2\leq j\leq k-1,\ \ 1\leq i\leq
n-j$

Thus, we have that

$\begin{array}[]{ll}[Y_{i},Y_{1}]=\gamma_{i}X_{i+1},&1\leq i\leq
n-1\\\\[2.84526pt]
[Y_{i},Y_{j}]=\displaystyle\sum_{s=0}^{j-1}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{i+s}X_{i+j},&2\leq j\leq k-1,\ \ 1\leq i\leq
n-j\end{array}$

From the super Leibniz identity on the triple $\\{X_{1},Y_{1},Y_{1}\\}$ we get
$\gamma_{2}=0.$

Using the induction method together with super Leibniz identity on the triple
$\\{X_{i},X_{1},Y_{j}\\}$, we prove that

$\beta_{i\,j}=\left\\{\begin{array}[]{ll}0,&3\leq i+j\leq k-1\\\
(-1)^{j-k+1}\left(\left(\begin{array}[]{c}i-2\\\
j-k+i\end{array}\right)+\left(\begin{array}[]{c}i-1\\\
j-k+i\end{array}\right)\alpha\right),&k\leq i+j\leq m\end{array}\right.$

with $2\leq i\leq\min\\{n,m-j\\}.$

$(1)$ $\bf m\geq n+3.$ If we make the super Leibniz identity on the triples
$\\{X_{n},X_{1},Y_{1}\\}$ and $\\{X_{n},X_{1},Y_{2}\\}$ we get
$\alpha=-\displaystyle\frac{k-2}{n}$ and $\alpha=-\displaystyle\frac{k-3}{n},$
respectively. But that is a contradiction, then there is not any superalgebra.

$(2)$ $\bf m=n+2.$

Super Leibniz identity |  | Constraint
---|---|---
$\\{X_{n},X_{1},Y_{1},\\}$ | $\Rightarrow$ | $\alpha=-\displaystyle\frac{k-2}{n}.$
$\\{Y_{i},Y_{1},Y_{n+1-i}\\}$ | $\Rightarrow$ |  $\left\\{\begin{array}[]{l}\gamma_{i}\beta_{i+1\,n+1-i}=0,\\\ \beta_{i+1\,n+1-i}\neq 0,\\\ i\geq n+2-k\end{array}\right.$ $\Rightarrow$ $\gamma_{i}=0,\ n+2-k\leq i\leq n-1.$
$\\{Y_{i},Y_{1},Y_{1}\\}$ | $\Rightarrow$ |  $\left\\{\begin{array}[]{l}\gamma_{i}\beta_{i+1\,1}=0,\\\ \beta_{i+1\,1}\neq 0,\\\ k-2\leq i\leq n\end{array}\right.$ $\Rightarrow$ $\gamma_{i}=0,\ k-2\leq i\leq n-1.$
$\\{X_{1},Y_{k-1},Y_{j}\\}$ | $\Rightarrow$ |  $\alpha\gamma_{k\,j}-\gamma_{j+1\,k-1}=0,$ $1\leq j\leq n-k.$

By other hand, recall that $\gamma_{j}=0$, for $k-2\leq j\leq n-1$, and by
Equation (5.1) we have that $\gamma_{k\,j}=0.$ Accordingly
$\gamma_{j+1\,k-1}=0$ for $1\leq j\leq n-k,$ that is,

$\sum_{s=0}^{k-2}(-1)^{s}\left(\begin{array}[]{c}k-2\\\
s\end{array}\right)\gamma_{j+1+s}=0,\quad 1\leq j\leq n-k$

Then we have a system of $(n-k)$ linear equations and $(n-k)$ variables
admitting one unique solution $\gamma_{i}=0$, $2\leq i\leq n-k+1.$

Only rest to prove that $\gamma_{1}=0.$ For this, it is sufficient to consider
the super Leibniz identity on the triple $\\{Y_{1},Y_{1},Y_{k-1}\\}.$ We
conclude that the superalgebra is degenerated.

$(3)$ $\bf m=n+1.$ From the super Leibniz identity on
$\\{Y_{i},Y_{1},Y_{1}\\}$ we get $\gamma_{i}\beta_{i+1\,1}=0$. If $k-2\leq
i\leq n$ then $\beta_{i+1\,1}\neq 0.$ Thus $\gamma_{i}=0$ for $k-2\leq i\leq
n-1.$

Analogously, from $\\{Y_{k-3},Y_{1},Y_{3}\\},\\{X_{1},Y_{1},Y_{j}\\}$ we
conclude that $\gamma_{k-3}=0$, and $\gamma_{2\,j}+\gamma_{j+1}=0,$
respectively. By other hand, remenber that $\gamma_{j}=0$ with $k-2\leq j\leq
n-1$, which implies that

$\sum_{s=1}^{j-2}(-1)^{s}\left(\begin{array}[]{c}j-1\\\
s\end{array}\right)\gamma_{2+s}+((-1)^{j-1}+1)\gamma_{j+1}=0,\quad 4\leq j\leq
k-3$

Then we have a system of $(k-6)$ linear equations and $(k-6)$ variables
admiting one unique solution $\gamma_{i}=0$ with $3\leq i\leq k-4.$

Only rest to prove that $\gamma_{1}=0.$ For this, it is sufficient to consider
the super Leibniz identity on the triples $\\{Y_{1},Y_{1},Y_{k-1}\\}$ if $k$
is odd and $\\{Y_{1},Y_{1},Y_{k-2}\\}$ if $k$ is even. We conclude then, that
the superalgebra is degenerated.

Case 2:  $n>m$. We consider the adapted basis
$\\{X_{1},X_{2},\dots,X_{n},Y_{1},Y_{2},\dots,Y_{m}\\}$ and the gradation

$<X_{1},Y_{1}>\oplus<X_{2},Y_{2}>\oplus\dots<X_{m},Y_{m}>\oplus<X_{m+1}>\oplus<X_{m+2}>\oplus\dots\oplus<X_{n}>$

By using the adapted basis and the properties of natural gradation

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1\\\\[2.84526pt]
[X_{1},Y_{j}]=\alpha_{j}Y_{j+1},&1\leq j\leq
m-1,&[X_{i},Y_{j}]=\beta_{i\,j}Y_{i+j},&1\leq j\leq m-2,\quad 2\leq i\leq
m-j\\\\[2.84526pt] [Y_{i},Y_{j}]=\gamma_{i\,j}X_{i+j},&1\leq i\leq m,&1\leq
j\leq\min\\{n-i,\ m\\}\end{array}$

Case 2.1 If $1+\alpha_{1}\neq 0,$ then $Y_{2},Y_{3},\dots,Y_{m}\in Ann(L).$ We
have

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1\\\\[2.84526pt]
[X_{1},Y_{1}]=\alpha Y_{2},&&[X_{i},Y_{1}]=\beta_{i}Y_{i+1},&2\leq i\leq
m-1\\\\[2.84526pt] [Y_{i},Y_{1}]=\gamma_{i}X_{i+1},&1\leq i\leq m\end{array}$

Similar to the previous cases, using the super Leibniz identity, we get the
superalgebra $NG^{n,m}.$

Case 2.2:  $1+\alpha_{1}=0.$ We distinguish the following two cases:

(a) $1+\alpha_{i}=0$ for all $i,\ 2\leq i\leq m-1.$ We have:

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq
m-1,&[X_{i},Y_{j}]=\beta_{i\,j}Y_{i+j},&1\leq j\leq m-2,\quad 2\leq i\leq
m\\\\[2.84526pt] [Y_{i},Y_{j}]=\gamma_{i\,j}X_{i+j},&1\leq i\leq m,&1\leq
j\leq\min\\{m,n-i\\}\end{array}$

We obtain the family of naturally graded Leibniz superalgebras with maximal
s-nilindex of the theorem statement.

(b) There exists $k$, with $2\leq k\leq m-1,$ such that $1+\alpha_{i}=0$ for
$2\leq i\leq k-2$ and $1+\alpha_{k-1}\neq 0.$ We have

$[X_{1},Y_{k-1}]+[Y_{k-1},X_{1}]=(1+\alpha_{k-1})Y_{k}\in Ann(L).$

Then $Y_{k}\in Ann(L)\mbox{ and }Y_{k+1},Y_{k+2},\dots,Y_{m}\in Ann(L).$ Thus,

$\begin{array}[]{llll}[X_{i},X_{1}]=X_{i+1},&1\leq i\leq
n-1,&[Y_{j},X_{1}]=Y_{j+1},&1\leq j\leq m-1\\\\[2.84526pt]
[X_{1},Y_{j}]=-Y_{j+1},&1\leq j\leq k-2,&[X_{1},Y_{k-1}]=\alpha
Y_{k},&1+\alpha\neq 0\\\\[2.84526pt] [X_{i},Y_{j}]=\beta_{i\,j}Y_{i+j},&1\leq
j\leq k-1,\quad 2\leq i\leq m-j&\\\\[2.84526pt]
[Y_{i},Y_{j}]=\gamma_{i\,j}X_{i+j},&1\leq j\leq k-1,\quad 1\leq i\leq
min\\{m,n-j\\}&&\end{array}$

By using the same techinques as in the case $1$, we conclude that the obtained
superalgebra is degenerated.

∎

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# A point-plane incidence theorem in matrix rings

Nguyen Van The University of Science, Vietnam National University - Hanoi,
Email<EMAIL_ADDRESS>Le Anh Vinh Vietnam National University -
Hanoi, Email<EMAIL_ADDRESS>Vietnam Institute of Educational Sciences.
Email<EMAIL_ADDRESS>
###### Abstract

In this paper, we study a point-hyper plane incidence theorem in matrix rings,
which generalizes all previous works in literature of this direction.

##  1 Introduction

Let $\mathbb{F}_{q}$ be a finite field of order $q$ where $q$ is an odd prime
power. Let $M_{n}(\mathbb{F}_{q})$ be the set of $n\times n$ matrices with
entries in $\mathbb{F}_{q}$ and $GL_{n}(\mathbb{F}_{q})$ be the set of
invertible matrices in $M_{n}(\mathbb{F}_{q})$. For
$A_{1},A_{2},\dots,A_{d},B\in M_{n}(\mathbb{F}_{q}),$ we define a hyper-plane
in $M_{n}(\mathbb{F}_{q})^{d+1}=M_{n}(\mathbb{F}_{q})\times\dots\times
M_{n}(\mathbb{F}_{q})$ as the set of points $(X_{1},\dots,X_{d},Y)\in
M_{n}(\mathbb{F}_{q})^{d+1}$ satisfying

$A_{1}X_{1}+\dots A_{d}X_{d}+B=Y.$ (1)

Note that, this definition is a formal definition, however, it is associated
to an $d$-dimensional affine plane over the module
$M_{n}(\mathbb{F}_{q})^{d+1}$. When $d=1$, we say the set of points satisfying
(1) is a line in $M_{n}(\mathbb{F}_{q})^{2}$ formed by $A_{1}$ and $B$. In the
first version of this note, we prove the following point-line incidence with
$d=1$ and $n=2$.

######  Theorem 1.1.

Let $\mathcal{P}$ be a set of points and $\mathcal{L}$ be a set of lines in
$M_{2}(\mathbb{F}_{q})\times M_{2}(\mathbb{F}_{q})$, then we have

$I(\mathcal{P},\mathcal{L})\leq\dfrac{|\mathcal{P}||\mathcal{L}|}{q^{4}}+\sqrt{2}q^{7/2}\sqrt{|\mathcal{P}||\mathcal{L}|}.$

During revision process, Xie and Ge [5] extended this results by considering
the equation $AX+BY=C+D$ where $A,B,C,D,X,Y\in M_{n}(\mathbb{F}_{q})$, which
also generalizes the work of Mohammadi, Pham and Wang [2] for $n=2$. Some
similar results could be found in [1, 3], in which they consider the sum-
product equation.

In this note, we prove a generalization of all above mentioned results in [1,
2, 3, 5]. Particularly, we consider the equation (1) for all $d\geq 1$ and
$n\geq 2$.

######  Theorem 1.2.

For $d\geq 1,n\geq 2,$ let
$\mathcal{A}_{1},\dots,\mathcal{A}_{d},\mathcal{B}_{1},\dots,\mathcal{B}_{d},\mathcal{E},\mathcal{F}\subset
M_{n}(\mathbb{F}_{q})$ and $N$ be the number of solutions to the sum-product
equation

$A_{1}B_{1}+A_{2}B_{2}+\dots+A_{d}B_{d}=E+F,\quad
A_{i}\in\mathcal{A}_{i},B_{i}\in\mathcal{B}_{i},E\in\mathcal{E},F\in\mathcal{F}.$

Then we have

$\left|N-\dfrac{|\mathcal{E}||\mathcal{F}||\mathcal{A}_{1}||\mathcal{B}_{1}|\dots|\mathcal{A}_{n}||\mathcal{B}_{n}|}{q^{n^{2}}}\right|\ll
q^{dn^{2}-(d-1)n/2-1/2}\sqrt{|\mathcal{E}||\mathcal{F}|\prod_{i=1}^{n}|\mathcal{A}_{i}||\mathcal{B}_{i}|}.$

Theorem 1.2 will be proved via spectral graph theory, in particular, the
expander mixing lemma of regular directed graph given by Vu [4]. Firstly, we
recall some definitions from graph theory as the following subsection, which
is extracted from Section 4 in [3].

##  2 Tools from spectral graph theory

Let $G$ be a directed graph (digraph) on $n$ vertices where the in-degree and
out-degree of each vertex are both $d.$
Let $A_{G}$ be the adjacency matrix of $G$, i.e., $a_{ij}=1$ if there is a
directed edge from $i$ to $j$ and zero otherwise. Suppose that
$\lambda_{1}=d,\lambda_{2},...,\lambda_{n}$ are the eigenvalues of $A_{G}.$
These eigenvalues can be complex, so we cannot order them, but it is known
that $|\lambda_{i}|\leq d$ for all $1\leq i\leq n.$ Define
$\lambda(G):=\max_{|\lambda_{i}|\neq d}|\lambda_{i}|.$ This value is called
the second largest eigenvalue of $A_{G}.$ We say that the $n\times n$ matrix
$A$ is normal if $A^{t}A=AA^{t}$ where $A^{t}$ is the transpose of $A.$ The
graph $G$ is normal if $A_{G}$ is normal. There is a simple way to check
whenever $G$ is normal or not. Indeed, for any two vertices $x$ and $y,$ let
$N^{+}(x,y)$ be the set of vertices $z$ such that
$\overrightarrow{xz},\overrightarrow{yz}$ are edges, and $N^{-}(x,y)$ be the
set of vertices $z$ such that $\overrightarrow{zx},\overrightarrow{zy}$ are
edges. By a direct computation, we have $A_{G}$ is normal if and only if
$|N^{+}(x,y)|=|N^{-}(x,y)|$ for any two vertices $x$ and $y.$

A digraph $G$ is called an $(n,d,\lambda)-digraph$ if $G$ has $n$ vertices,
the in-degree and out-degree of each vertex are both $d,$ and
$\lambda(G)\leq\lambda.$ Let $G$ be an $(n,d,\lambda)-digraph.$ We have the
following expander mixing lemma given by Vu [4].

######  Lemma 2.1 (Expander Mixing Lemma, [4]).

Let $G=(V,E)$ be an $(n,d,\lambda)-digraph.$ For any two sets $B,C\subset V,$
we have

$\left|e(B,C)-\frac{d}{n}|B||C|\right|\leq\lambda\sqrt{|B||C|}$

where $e(B,C)$ be the number of ordered pairs $(u,w)$ such that $u\in B,w\in
C,$ and $\overrightarrow{uw}\in e(G).$

As $d|B||C|/n$ is the expected number of edges from $B$ to $C$, the above
lemma gives us a bound for the gap between this number and the number of edges
between $B$ and $C$.

##  3 Proof of Theorem 1.2

To prove Theorem 1.2, we define the sum-product digraph $G=(V,E)$ with the
vertex set

$V=M_{n}(\mathbb{F}_{q})^{d+1}=M_{n}(\mathbb{F}_{q})\times
M_{n}(\mathbb{F}_{q})\times\dots\times M_{n}(\mathbb{F}_{q})$

. There exists a directed edge from $(A_{1},\dots,A_{d},E)$ to
$(B_{1},\dots,B_{d},F)$ if

$A_{1}B_{1}+A_{2}B_{2}+\dots+A_{d}B_{d}=E+F.$

Theorem 1.2 will be directly followed from Lemma 2.1 and the following
proposition.

######  Proposition 3.1.

The sum-product digraph $G=(V,E)$ is an
$\left(q^{(d+1)n^{2}},q^{dn^{2}},Cq^{dn^{2}-(d-1)n/2-1/2}\right)-digraph$ for
some positive constant $C$.

To prove Proposition 3.1, we need some basic facts from linear algebra via the
following lemmas.

######  Lemma 3.2.

Denote $M_{n\times t}(\mathbb{F}_{q})$ be the set of $n\times t$ matrices with
entries restricted in $\mathbb{F}_{q}$ with $n<t$. Then the number of matrices
of rank $k\leq n$ is less than

$\binom{t}{k}\left(q^{n}-1\right)\left(q^{n}-q\right)\dots\left(q^{n}-q^{k-1}\right)q^{(t-k)k}\leq
C_{t,k}q^{nk+k(t-k)}$

for some positive constant $C_{t,k}$.

###### Proof.

It is observed that the number of $k$ independent column vectors
$\mathbf{c}_{1},\dots,\mathbf{c}_{k}$ in $\mathbb{F}_{q}^{n}$ is

$\left(q^{n}-1\right)\left(q^{n}-q\right)\dots\left(q^{n}-q^{k-1}\right).$

For each matrix $A\in M_{n\times t}(\mathbb{F}_{q})$ of rank $k$, there exist
$k$ independent column vectors $\mathbf{c}_{1},\dots,\mathbf{c}_{k}$ in $A$.
Furthermore, the $t-k$ other columns can be written as a linear combination of
$\left\\{\mathbf{c}_{i}\right\\}_{i=1}^{k},$ each has $q^{k}$ possibilities.
We can choose $k$ of $t$ column vectors, which are linearly independent. Since
two different ways can be just one matrix, the number of matrices of rank
$k\leq n$ is bounded by

$\displaystyle\binom{t}{k}\left(q^{n}-1\right)\left(q^{n}-q^{2}\right)\dots\left(q^{n}-q^{k-1}\right)q^{k(t-k)}\leq
Cq^{nk+k(t-k)}$

for some positive constant $C$. ∎

######  Lemma 3.3.

For $n\geq m\geq k\geq 0$, let $\mathcal{T}_{m,k}$ be the number of pair
$(M,C)$ with $M\in M_{n\times t}\left(\mathbb{F}_{q}\right),C\in
M_{n}(\mathbb{F}_{q})$ satisfying $\mathrm{rank}(M)=m,\mathrm{rank}(C)=k$, and
the system $MZ=C$ has solution. We have

$\mathcal{T}_{m,k}\ll q^{nm+m(t-m)+mk+k(n-k)}$

###### Proof.

For each matrix $M\in M_{n\times t}$ of rank $m$, we will bound the number of
matrices $C\in M_{n}(\mathbb{F}_{q})$ of rank $k$ that satisfy the equation
$MZ=C$ has solution. It follows from Kronecker-Capelli theorem that the
equation $MZ=C$ has solution if and only if
$\mathrm{rank}(M)=\mathrm{rank}(\overline{M})$ where
$\overline{M}=\left(M\,C\right)$ is the matrix obtained by adding $C$ to $M$
on the right.

Let $\mathbf{r}_{M_{1}},\dots,\mathbf{r}_{M_{n}}$ be row vectors of $M$ and
$\mathbf{r}_{C_{1}},\dots,\mathbf{r}_{C_{n}}$ be row vectors of $C$,
respectively. Without loss of generality, we assume that
$\mathbf{r}_{M_{1}},\dots,\mathbf{r}_{M_{k}}$ are linearly independent and

$\mathbf{r}_{M_{i}}=\alpha_{1_{i}}\mathbf{r}_{M_{1}}+\alpha_{2_{i}}\mathbf{r}_{M_{2}}+\dots+\alpha_{m_{i}}\mathbf{r}_{M_{m}},\,i=m+1,m+2,\dots,n$

for some $\alpha_{j_{i}}\in\mathbb{F}_{q}$ for all $j=1,\dots,m$. Since
$\mathrm{rank}(M)=\mathrm{rank}\left(\overline{M}\right)$, we have

$\mathbf{r}_{C_{i}}=\alpha_{1_{i}}\mathbf{r}_{C_{1}}+\alpha_{2_{i}}\mathbf{r}_{C_{2}}+\dots+\alpha_{m_{i}}\mathbf{r}_{C_{m}},\,i=m+1,m+2,\dots,n.$

This means the $n-m$ later row vectors of $C$ are uniquely determined by first
$m$ row vectors $\mathbf{r}_{C_{1}},\dots,\mathbf{r}_{C_{m}}$ for a given
matrix $M$. Therefore, we only need to count number of $m$ row vectors
$\mathbf{r}_{c_{1}},\dots,\mathbf{r}_{c_{m}}$ satisfying
$\mathrm{rank}\left(\mathbf{r}_{c_{1}},\dots,\mathbf{r}_{c_{m}}\right)=k$, or
equivalently, the number of $m\times n$ matrices of rank $k$. Hence, it
follows from Lemma 3.2 that the number of matrices $C$ satisfying $MZ=C$ has
solution is bounded by $dq^{mk+k(n-k)}$ for some positive constant $d$.

Again, applying Lemma 3.2, there are at most $dq^{nm+m(t-m)}$ matrices $M$ of
rank $m$ in $M_{n\times t}$. Thus, we obtain

$\mathcal{T}_{m,k}\ll q^{nm+m(t-m)+mk+k(n-k)},$

which completes the proof of Lemma 3.3. ∎

We are now ready to prove Proposition 3.1.

###### Proof of Proposition 3.1.

It is obvious that the order of $G$ is $q^{(d+1)n^{2}},$ because
$\left|M_{n}(\mathbb{F}_{q})\right|=q^{n^{2}}$ and so
$\left|M_{n}(\mathbb{F}_{q})^{d+1}\right|=q^{(d+1)n^{2}}.$ Next, we observe
that $G$ is a regular digraph of in-degree and out-degree $q^{dn^{2}}.$
Indeed, for any vertex $(A_{1},\dots,A_{d},E)\in V,$ if we choose each
$d$-tuple $(B_{1},\dots,B_{d})\in M_{n}(\mathbb{F}_{q})^{d},$ there exists a
unique $F=A_{1}B_{1}+\dots+A_{d}B_{d}-E$ such that

$A_{1}B_{1}+\dots+A_{d}B_{d}=E+F.$

Hence, the out-degree of any vertex in $G$ is
$\left|M_{n}(\mathbb{F}_{q})^{d}\right|,$ which is $q^{dn^{2}}.$ The same
holds for the in-degree of each vertex. Therefore, to prove Proposition 3.1,
we only need to bound the second largest eigenvalue of $G$.

To this end, we first need to show that $G$ is a normal digraph. Let $A_{G}$
be the adjacency matrix of $G$. It is known that if $A_{G}$ is a normal matrix
and $\beta$ is an eigenvalue of $A_{G},$ then the complex conjugate
$\overline{\beta}$ is an eigenvalue of $A_{G}^{t}.$ Hence, $|\beta|^{2}$ is an
eigenvalue of $A_{G}A_{G}^{t}$ and $A_{G}^{t}A_{G}.$ In other words, in order
to bound $\beta,$ it is enough to bound the second largest eigenvalue of
$A_{G}A_{G}^{t}.$

Firstly, we will show that $G$ is a normal graph. Let
$(A_{1},A_{2},\dots,A_{d},E)$ and
$(A^{\prime}_{1},A^{\prime}_{2},\dots,A^{\prime}_{d},E^{\prime})$ be two
different vertices, we now count the of the neighbors $(B_{1},\dots,B_{d},F)$
such that there are directed edges from $(A_{1},\dots,A_{d},E)$ and
$(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})$ to $(B_{1},\dots,B_{d},F).$
This number is
$N^{+}((A_{1},\dots,A_{d},E),(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})).$
We have

$\displaystyle A_{1}B_{1}+\dots+A_{d}B_{d}=E+F\text{ and
}A^{\prime}_{1}B_{1}+\dots+A^{\prime}_{d}B_{d}=E^{\prime}+F,$ (2)

which implies

$\displaystyle(A_{1}-A^{\prime}_{1})B_{1}+\dots+(A_{d}-A^{\prime}_{d})B_{d}=A-A^{\prime}.$
(3)

Note that if we fix a solution $(B_{1},\dots,B_{d})$ to the equation (3), then
$F$ in (2) is uniquely determined. Let
$M=\left(A_{1}-A^{\prime}_{1}\,\,\,A_{1}-A^{\prime}_{1}\,\,\dots\,\,A_{d}-A^{\prime}_{d}\right)$,
$X=\left(B_{1}\,\,\,B_{2}\,\,\dots\,\,B_{d}\right)$ and $Y=E-E^{\prime}$, then
the equation can be rewritten as the following matrix equation.

$MX=Y$ (4)

with $M\in M_{n\times dn}(\mathbb{F}_{q}),Y\in M_{n}(\mathbb{F}_{q})$ and
$X\in M_{dn\times n}(\mathbb{F}_{q})$. We now fall into the following cases.

  * •

Case 1. If $\mathrm{rank}(M)=n$, then there exists unique $X$ such that
$MX=Y$. Thus the system (2) has only one solution in this case.

  * •

Case 2. If $\mathrm{rank}(M)=m$ and $\mathrm{rank}(Y)=k$ with $m<k\leq n$,
then the equation $MX=Y$ has no solution.

  * •

Case 3. If $\mathrm{rank}(M)=m$ and $\mathrm{rank}(Y)=k$ with $n>m\geq k$, we
need to further consider different situations as follows.

    * -

Case 3.1. If $\mathrm{rank}(M)=\mathrm{rank}(Y)=0$, then
$E=E^{\prime},A_{i}=A^{\prime}_{i}$ for all $i=1,\dots,d$, which contracts
with our assumption that $(A_{1},A_{2},\dots,A_{d},E)$ and
$(A^{\prime}_{1},A^{\prime}_{2},\dots,A^{\prime}_{d},E^{\prime})$ are two
different vertices. Thus, we can rule out this case.

    * -

Case 3.2. If $\mathrm{rank}\left(\overline{M}\right)>m$ where
$\overline{M}=\left(M\,\,Y\right)$, it follows from Kronecker-Capelli theorem
that the equation $MX=E$ has no solution.

    * -

Case 3.3. Suppose that $\mathrm{rank}\left(\overline{M}\right)=m$ where
$\overline{M}=\left(M\,\,Y\right)$, we have known that the equation $MX=Y$ has
solution by the Kronecker Capelli theorem. Moreover, the number of its
solutions is equal to the number of solutions of $MX=0.$ Now, we are ready to
count the number of solutions of $MX=0$.

Put $X=[x_{1}\,\,x_{2}\,\dots\,x_{n}]$ for some column vectors
$x_{1},x_{2},\dots,x_{n}\in\mathbb{F}_{q}^{dn},$ we just need to estimate the
number of solutions of $Mx_{1}=0$ because the number of solutions of $MX=0$ is
equal to $n^{th}$ power of the number of solutions of $Mx_{1}=0.$ It is known
that the set $L$ of all solutions of $Ax_{1}=0$ is a vector subspace of
$\mathbb{F}_{q}^{dn}$ and has the dimension $\dim L=dn-\mathrm{rank}(M)=dn-m.$
Hence, we have

$|L|=q^{dn-m}.$

Therefore, the equation $MX=0$ has $q^{n(dn-m)}$ solutions if this equation
has solutions.

Since the same argument works for the case of
$N^{-}((A_{1},\dots,A_{d},E),(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})),$
we obtain the same value for
$N^{-}((A_{1},\dots,A_{d},e),(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})).$
In short, $A_{G}$ is normal.

As we discussed above, in order to bound the second largest eigenvalue of $M,$
it is enough to bound the second largest value of $MM^{t}.$ Note that each
entry of $A_{G}A_{G}^{t}$ can be interpreted as counting the number of common
outgoing neighbors between two vertices. Based on previous calculations, we
have

$\displaystyle A_{G}A_{G}^{t}=\left(q^{dn^{2}}-1\right)I+J-\sum_{0\leq m<k\leq
n}F_{m,k}-\sum_{\begin{subarray}{c}0\leq k\leq m<n:\\\
(m,k)\neq(0,0)\end{subarray}}H_{m,k}+\left(q^{n(n-m)}-1\right)\sum_{\begin{subarray}{c}0\leq
k\leq m<n:\\\ (m,k)\neq(0,0)\end{subarray}}E_{m,k}$

where $I$ is the identity matrix, $J$ denotes the all-one matrix and the
others defined as follows.

$F_{m,k}$ is the adjacency matrix of the graph $\mathcal{G}_{m,k},0\leq
m<k\leq n$ with the vertex set
$V(F_{m,k})=M_{n}\left(\mathbb{F}_{q}\right)^{d+1}$ and there is an edge
between $(A_{1},\dots,A_{d},E)$ and
$(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})$ if

$\mathrm{rank}\left(A_{1}-A^{\prime}_{1}\,\,\dots\,\,A_{d}-A^{\prime}_{d}\right)=m\text{
and }\mathrm{rank}\left(E-E^{\prime}\right)=k.$

$H_{m,k}$ is the adjacency matrix of the graph $\mathcal{G}_{m,k},0\leq k\leq
m<n,(m,k)\neq(0,0)$ with the vertex set
$V(H_{m,k})=M_{n}\left(\mathbb{F}_{q}\right)^{d+1}$ and there is an edge
between $(A_{1},\dots,A_{d},E)$ and
$(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})$ if
$\mathrm{rank}\left(A_{1}-A^{\prime}_{1}\,\,\dots\,\,A_{d}-A^{\prime}_{d}\right)=m,\,\mathrm{rank}\left(E-E^{\prime}\right)=k$
and

$\mathrm{rank}\left(A_{1}-A^{\prime}_{1}\,\,\dots\,\,A_{d}-A^{\prime}_{d}\,\,\,E-E^{\prime}\right)>m.$

$E_{m,k}$ is the adjacency matrix of the graph $\mathcal{GP}_{m,k},0\leq k\leq
m<n,(m,k)\neq(0,0)$ with the vertex set
$V(E_{m,k})=M_{n}\left(\mathbb{F}_{q}\right)^{d+1}$ and there is an edge
between $(A_{1},\dots,A_{d},E)$ and
$(A^{\prime}_{1},\dots,A^{\prime}_{d},E^{\prime})$ if
$\mathrm{rank}\left(A_{1}-A^{\prime}_{1}\,\,\dots\,\,A_{d}-A^{\prime}_{d}\right)=m,\,\mathrm{rank}\left(E-E^{\prime}\right)=k$
and

$\mathrm{rank}\left(A_{1}-A^{\prime}_{1}\,\,\dots\,\,A_{d}-A^{\prime}_{d}\,\,\,E-E^{\prime}\right)=m.$

Using the Lemma 3.2, one can easily to check that for any $0\leq k\leq n,0\leq
m<n$ and $(m,k)\neq(0,0),$ the graph $\mathcal{G}_{mk}$ is
$d_{mk}-\text{regular}$ for some $d_{mk}$ where

$d_{mk}\ll q^{nm+m(dn-m)}q^{nk+k(n-k)}=q^{(d+1)nm+2nk-m^{2}-k^{2}}\leq
q^{(d+1)n^{2}-(d-1)n-1}.$

For the graph $\mathcal{GP}_{m,k},0\leq k\leq m<n,(m,k)\neq(0,0),$ it follows
directly from Lemma 3.3 that $\mathcal{G}_{mk}$ is $y_{mk}-\text{regular}$ for
some $y_{mk}$ where

$y_{m,k}\ll q^{nm+m(dn-m)+mk+k(n-k)}.$

Suppose $\lambda_{2}$ is the second largest eigenvalue of $A_{G}$ and
$\overrightarrow{v_{2}}$ is the corresponding eigenvector. Since $G$ is a
regular graph, we have $J\cdot\overrightarrow{v_{2}}=0.$ (Indeed, since $G$ is
regular, it always has $(1,1,\dots,1)$ as an eigenvector with eigenvalue being
its regular-degree. Moreover, since the graph $G$ is connected, this
eigenvalue has multiplicity one. Thus any other eigenvectors will be
orthogonal to $(1,1,\dots,1)$ which in turns gives us
$J\cdot\overrightarrow{v_{2}}=0).$ Since
$A_{G}A_{G}^{t}\overrightarrow{v_{2}}=|\lambda_{2}|^{2}\overrightarrow{v_{2}},$
we get

$|\lambda_{2}|^{2}\overrightarrow{v_{2}}=\left[\left(q^{dn^{2}}-1\right)-\sum_{0\leq
m<k\leq n}F_{m,k}-\sum_{\begin{subarray}{c}0\leq k\leq m<n:\\\
(m,k)\neq(0,0)\end{subarray}}H_{m,k}+\left(q^{n(n-m)}-1\right)\sum_{\begin{subarray}{c}0\leq
k\leq m<n:\\\
(m,k)\neq(0,0)\end{subarray}}E_{m,k}\right]\overrightarrow{v_{2}}.$ (5)

One easily to check that $y_{m,k}q^{n(dn-m)}\ll q^{2dn^{2}-(d-1)n-1}$ for all
$0\leq k\leq m<n,(m,k)\neq 0$. Therefore, all previous calculations and the
equation (5) give us

$|\lambda_{2}|^{2}\ll q^{2dn^{2}-(d-1)n-1},$

since eigenvalues of a sum of matrices are bounded by the sum of largest
eigenvalue of the summands. We complete the proof of Proposition 3.1. ∎

## References

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