∎  ∎

11institutetext:  Abdennour Kitouni 22institutetext: Division of Mathematics
and Physics, School of Education, Culture and Communication, Mälardalen
University, Box 883, 72123 Västerås, Sweden.
22email<EMAIL_ADDRESS>33institutetext: Sergei Silvestrov
(corresponding author) 44institutetext: Division of Mathematics and Physics,
School of Education, Culture and Communication, Mälardalen University, Box
883, 72123 Västerås, Sweden.
44email<EMAIL_ADDRESS>
# On properties and classification of a class of $4$-dimensional $3$-Hom-Lie
algebras with a nilpotent twisting map

Abdennour Kitouni  Sergei Silvestrov

###### Abstract

The aim of this work is to investigate the properties and classification of an
interesting class of $4$-dimensional $3$-Hom-Lie algebras with a nilpotent
twisting map $\alpha$ and eight structure constants as parameters. Derived
series and central descending series are studied for all algebras in this
class and are used to divide it into five non-isomorphic subclasses. The
levels of solvability and nilpotency of the $3$-Hom-Lie algebras in these five
classes are obtained. Building up on that, all algebras of this class are
classified up to Hom-algebra isomorphism. Necessary and sufficient conditions
for multiplicativity of general $(n+1)$-dimensional $n$-Hom-Lie algebras as
well as for algebras in the considered class are obtained in terms of the
structure constants and the twisting map. Furthermore, for some algebras in
this class, it has been determined whether the terms of the derived and
central descending series are weak subalgebras, Hom-subalgebras, weak ideals
or Hom-ideals.
2020 Mathematics Subject Classification: 17B61,17A40,17A42,17B30

###### Keywords:

Hom-algebra, $n$-Hom-Lie algebra, classification

##  1 Introduction

Hom-Lie algebras and more general quasi-Hom-Lie algebras where introduced
first by Hartwig, Larsson and Silvestrov in HLS , where the general quasi-
deformations and discretizations of Lie algebras of vector fields using more
general $\sigma$-derivations (twisted derivations) and a general method for
construction of deformations of Witt and Virasoro type algebras based on
twisted derivations have been developed, initially motivated by the
$q$-deformed Jacobi identities observed for the $q$-deformed algebras in
physics, $q$-deformed versions of homological algebra and discrete
modifications of differential calculi AizawaSaito ; ChaiElinPop ;
ChaiIsLukPopPresn ; ChaiKuLuk ; ChaiPopPres ; CurtrZachos1 ; DamKu ;
DaskaloyannisGendefVir ; Hu ; Kassel92 ; LiuKQuantumCentExt ;
LiuKQCharQuantWittAlg ; LiuKQPhDthesis . The general abstract quasi-Lie
algebras and the subclasses of quasi-Hom-Lie algebras and Hom-Lie algebras as
well as their general colored (graded) counterparts have been introduced in
HLS ; LarssonSilvJA2005:QuasiHomLieCentExt2cocyid ; LSGradedquasiLiealg ;
Czech:witt ; LS2 . Subsequently, various classes of Hom-Lie admissible
algebras have been considered in ms:homstructure . In particular, in
ms:homstructure , the Hom-associative algebras have been introduced and shown
to be Hom-Lie admissible, that is leading to Hom-Lie algebras using commutator
map as new product, and in this sense constituting a natural generalization of
associative algebras, as Lie admissible algebras leading to Lie algebras via
commutator map as new product. In ms:homstructure , moreover, several other
interesting classes of Hom-Lie admissible algebras generalizing some classes
of non-associative algebras, as well as examples of finite-dimensional Hom-Lie
algebras have been described. Hom-algebras structures are very useful since
Hom-algebra structures of a given type include their classical counterparts
and open more possibilities for deformations, extensions of cohomological
structures and representations. Since these pioneering works HLS ;
LarssonSilvJA2005:QuasiHomLieCentExt2cocyid ; LS2 ; LSGradedquasiLiealg ; LS3
; ms:homstructure , Hom-algebra structures have developed in a popular broad
area with increasing number of publications in various directions (see for
example AmmarEjbehiMakhlouf:homdeformation ; BenMakh:Hombiliform ;
ElchingerLundMakhSilv:BracktausigmaderivWittVir ;
LarssonSilvJA2005:QuasiHomLieCentExt2cocyid ; LarssonSigSilvJGLTA2008 ;
LarssonSilvestrovGLTMPBSpr2009:GenNComplTwistDer ; MakhSil:HomHopf ;
MakhSilv:HomAlgHomCoalg ; MakhSilv:HomDeform ;
MakSil:HomLieAdmissibleHomCoalgHomHopf ; RichardSilvestrovJA2008 ; shenghomrep
; SigSilvGLTbdSpringer2009 ; YauHomEnv ; YauHomHom  and references therein).

Ternary Lie algebras appeared first in generalization of Hamiltonian mechanics
by Nambu Nambu:GenHD . Besides Nambu mechanics, $n$-Lie algebras revealed to
have many applications in physics. The mathematical algebraic foundations of
Nambu mechanics have been developed by Takhtajan in
Takhtajan:foundgenNambuMech . Filippov, in Filippov:nLie  independently
introduced and studied structure of $n$-Lie algebras and Kasymov Kasymov:nLie
investigated their properties. Properties of $n$-ary algebras, including
solvability and nilpotency, were studied in Carlsson:n-aryalgebras ;
Kasymov:nLie ; VainermanKerner:n-aryalgebras . Kasymov Kasymov:nLie  pointed
out that $n$-ary multiplication allows for several different definitions of
solvability and nilpotency in $n$-Lie algebras, and studied their properties.
Further properties, classification, and connections of $n$-ary algebras to
other structures such as bialgebras, Yang-Baxter equation and Manin triples
for $3$-Lie algebras were studied in BaiRBaiCWang:realizations3Liealg ;
BaiGuoSheng:BialgYangBaxtereqManintri3Liealg ; BaiWuLiZhou:Constrnplus1nLiealg
; BaiRSongZhang:nLieclas ; BaiRWangXiaoAn:nLieclaschar2 ;
BaiRAnLi:CentStrnLiealg ; BaiRChenmeng:FrattinisubalgnLiealg ;
BaiRMengD:CentrextnLiealg ; BaiRMengD:CentrnLiealg ;
BaiRZhangLiShi:Innerderivalgnplus1dimnLiealg ; Kasymov:nLie . The structure of
$3$-Lie algebras induced by Lie algebras, classification of $3$-Lie algebras
and application to constructions of B.R.S. algebras have been considered in
Abramov2018:WeilAlg3LiealgBRSalg ; AbramovLatt2016:classifLowdim3Liesuperalg ;
Abramov2017:Super3LiealgebrasinducedsuperLiealg . Interesting constructions of
ternary Lie superalgebras in connection to superspace extension of Nambu-
Hamilton equation is considered in AbramovLatt2020TernaryLieSuper . In
CasasLodayPirashvili:Leibniznalg , Leibniz $n$-algebras have been studied. The
general cohomology theory for $n$-Lie algebras and Leibniz $n$-algebras was
established in DalTakh ; Rotkiewicz:cohomringnLiealg ; Takhtajan:cohomology .
The structure and classification of finite-dimensional $n$-Lie algebras were
considered in BaiRSongZhang:nLieclas ; Filippov:nLie ;
LingW:StrnLiealgPhThes1993  and many other authors. For more details of the
theory and applications of $n$-Lie algebras, see
DeAzcarragaIzquierdo:nAryalgrevappl  and references therein.

Classifications of $n$-ary or Hom generalizations of Lie algebras have been
considered, either in very special cases or in low dimensions. The
classification of $n$-Lie algebras of dimension up to $n+1$ over a field of
characteristic $p\neq 2$ has been completed by Filippov Filippov:nLie  using
the specific properties of $(n+1)$-dimensional $n$-Lie algebras that make it
possible to represent their bracket by a square matrix in a similar way as
bilinear forms, the number of cases obtained depends on the properties of the
base field, the list is ordered by ascending dimension of the derived ideal,
and among them, one nilpotent algebra, and a class of simple algebras which
are all isomorphic in the case of an algebraically closed field, the remaining
algebras are $k$-solvable for some $2\leq k\leq n$ depending on the algebra.
These simple algebras are proven to be the only simple finite-dimensional
$n$-Lie algebras in LingW:StrnLiealgPhThes1993 . The classification of
$(n+1)$-dimensional $n$-Lie algebras over a field of characteristic $2$ has
been done by Bai, Wang, Xiao, and An BaiRWangXiaoAn:nLieclaschar2  by finding
and using a similar result in characteristic $2$. Bai, Song and Zhang
BaiRSongZhang:nLieclas  classify the $(n+2)$-dimensional $n$-Lie algebras over
an algebraically closed field of characteristic 0 using the fact that an
$(n+2)$-dimensional $n$-Lie algebra has a subalgebra of codimension $1$ if the
dimension of its derived ideal is not $3$, thus constructing most of the cases
as extensions of the $(n+1)$-dimensional $n$-Lie algebras listed by Filippov.
In SimpleLinearlyCompact , Cantarini and Kac classified all simple linearly
compact $n$-Lie superalgebras, which turned out to be $n$-Lie algebras, by
finding a bijective correspondence between said algebras and a special class
of transitive $\mathbb{Z}$-graded Lie superalgebras, the list they obtained
consists of four representatives, one of them is the $(n+1)$-dimensional
vector product $n$-Lie algebra, and the remaining three are infinite-
dimensional $n$-Lie algebras.

Classifications of $n$-Lie algebras in higher dimensions have only been
studied in particular cases. Metric $n$-Lie algebras, that is $n$-Lie algebras
equipped with a non-degenerate compatible bilinear form, have been considered
and classified, first in dimension $n+2$ by Ren, Chen and Liang metric_n+2
and dimension $n+3$ by Geng, Ren and Chen metric_n+3 , and then in dimensions
$n+k$ for $2\leq k\leq n+1$ by Bai, Wu and Chen metric_n+k . The
classification is based on the study of the Levi decomposition, the center and
the isotropic ideals and properties around them. Another case that has been
studied is the case of nilpotent $n$-Lie algebras, more specifically nilpotent
$n$-Lie algebras of class $2$. Eshrati, Saeedi and Darabi lowdimnil-nLie
classify $(n+3)$-dimensional nilpotent $n$-Lie algebras and
$(n+4)$-dimensional nilpotent $n$-Lie algebras of class $2$ using properties
introduced in char_nil_nLie ; multiplier_nLie . Similarly Hoseini, Saeedi and
Darabi 2s-nil-dim:n+5  classify $(n+5)$-dimensional nilpotent $n$-Lie algebras
of class $2$. In 2s-nil-dim:n+6 , Jamshidi, Saeedi and Darabi classify
$(n+6)$-dimensional nilpotent $n$-Lie algebras of class $2$ using the fact
that such algebras factored by the span of a central element give
$(n+5)$-dimensional nilpotent $n$-Lie algebras of class $2$, which were
classified before. There has been a study of the classification of
$3$-dimensional $3$-Hom-Lie algebras with diagonal twisting maps by Ataguema,
Makhlouf and Silvestrov in AtMaSi:GenNambuAlg .

Hom-type generalization of $n$-ary algebras, such as $n$-Hom-Lie algebras and
other $n$-ary Hom algebras of Lie type and associative type, were introduced
in AtMaSi:GenNambuAlg , by twisting the defining identities by a set of linear
maps. The particular case, where all these maps are equal and are algebra
morphisms has been considered and a way to generate examples of $n$-ary Hom-
algebras from $n$-ary algebras of the same type have been described. Further
properties, construction methods, examples, representations, cohomology and
central extensions of $n$-ary Hom-algebras have been considered in
AmmarMabMakh:naryhomrep ; ams:ternary ; ams:n ; kms:nhominduced ; YauGenCom ;
YauHomNambuLie . These generalizations include $n$-ary Hom-algebra structures
generalizing the $n$-ary algebras of Lie type including $n$-ary Nambu
algebras, $n$-ary Nambu-Lie algebras and $n$-ary Lie algebras, and $n$-ary
algebras of associative type including $n$-ary totally associative and $n$-ary
partially associative algebras. In kms:narygenBiHomLieBiHomassalgebras2020 ,
constructions of $n$-ary generalizations of BiHom-Lie algebras and BiHom-
associative algebras have been considered. Generalized derivations of
$n$-BiHom-Lie algebras have been studied in
BenAbdeljElhamdKaygorMakhl201920GenDernBiHomLiealg . Generalized derivations
of multiplicative $n$-ary Hom-$\Omega$ color algebras have been studied in
BeitesKaygorodovPopov . Cohomology of Hom-Leibniz and $n$-ary Hom-Nambu-Lie
superalgebras has been considered in AbdaouiMabroukMakhlouf  Generalized
derivations and Rota-Baxter operators of $n$-ary Hom-Nambu superalgebras have
been considered in
MabroukNcibSilvestrov2020:GenDerRotaBaxterOpsnaryHomNambuSuperalgs . A
construction of $3$-Hom-Lie algebras based on $\sigma$-derivation and
involution has been studied in AbramovSilvestrov:3homLiealgsigmaderivINvol .
Multiplicative $n$-Hom-Lie color algebras have been considered in
BakyokoSilvestrov:MultiplicnHomLiecoloralg .

In almy:quantnambu , Awata, Li, Minic and Yoneya introduced a construction of
$(n+1)$-Lie algebras induced by $n$-Lie algebras using combination of bracket
multiplication with a trace in their work on quantization of the Nambu
brackets. Further properties of this construction, including solvability and
nilpotency, were studied in BaiRBaiCWang:realizations3Liealg ; akms:ternary ;
km:nary . In ams:ternary ; ams:n , this construction was generalized using the
brackets of general Hom-Lie algebra or $n$-Hom-Lie and trace-like linear forms
satisfying conditions depending on the twisting linear maps defining the Hom-
Lie or $n$-Hom-Lie algebras. In AbdelkhaderSamiOthmen , a method was
demonstrated of how to construct $n$-ary multiplications from the binary
multiplication of a Hom-Lie algebra and a $(n-2)$-linear function satisfying
certain compatibility conditions. Solvability and nilpotency for $n$-Hom-Lie
algebras and $(n+1)$-Hom-Lie algebras induced by $n$-Hom-Lie algebras have
been considered in kms:solvnilpnhomlie2020 . In cln1dimnHomLieniltw ,
properties and classification of $n$-Hom-Lie algebras in dimension $n+1$ were
considered, and $4$-dimensional $3$-Hom-Lie algebras for various special cases
of the twisting map have been computed in terms of structure constants as
parameters and listed in classes in the way emphasizing the number of free
parameters in each class.

The $n$-Hom-Lie algebras are fundamentally different from the $n$-Lie algebras
especially when the twisting maps are not invertible or not diagonalizable.
When the twisting maps are not invertible, the Hom-Nambu-Filippov identity
becomes less restrictive since when elements of the kernel of the twisting
maps are used, several terms or even the whole identity might vanish.
Isomorphisms of Hom-algebras are also different from isomorphisms of algebras
since they need to intertwine not only the multiplications but also the
twisting maps. All of this make the classification problem different,
interesting, rich and not simply following from the case of $n$-Lie algebras.
In this work, we consider $n$-Hom-Lie algebras with a nilpotent twisting map
$\alpha$, which means in particular that $\alpha$ is not invertible.

To our knowledge, the classification of $4$-dimensional $3$-Hom-Lie algebras
up to Hom-algebras isomorphism has not been achieved previously in the
literature. The aim of this work is to investigate the properties and
classification of an interesting class of $4$-dimensional $3$-Hom-Lie algebras
with a nilpotent twisting map $\alpha$ and eight structure constants as
parameters, namely $4_{3,N(2),6}$ given in cln1dimnHomLieniltw . All
$3$-dimensional $3$-Hom-Lie algebras with diagonal twisting maps have been
listed unclassified in AtMaSi:GenNambuAlg . The algebras considered in our
article are $4$-dimensional, and the twisting maps are of a different type,
namely nilpotent. Nilpotent linear maps are neither invertible nor
diagonalizable, which makes the object of our study fundamentally different
from the case of $n$-Hom-Lie algebras with diagonal twisting maps in the sense
that when the twisting maps are not invertible, the Hom-Nambu-Filippov
identity becomes less restrictive since when elements of the kernel of the
twisting maps are used in the identity, several terms or even the whole
identity might vanish, and when the twisting maps are not diagonalizable, the
change induced by introducing them in the identity is more significant. In
this work, we achieved a complete classification up to isomorphism of Hom-
algebras of the considered class of $4$-dimensional $3$-Hom-Lie algebras with
a nilpotent twisting map, computed derived series and central descending
series for all of the $3$-Hom-Lie algebras of this class, studied solvability
and nilpotency, characterized the multiplicative $3$-Hom-Lie algebras among
them and studied the ideal properties of the terms of derived series and
central descending series of some chosen examples of the Hom-algebras from the
classification. These results improve understanding of the rich structure of
$n$-ary Hom-algebras and in particular the important class of $n$-Hom-Lie
algebras. It is also a step towards the complete classification of
$4$-dimensional $3$-Hom-Lie algebras and in general $(n+1)$-dimensional
$n$-Hom-Lie algebras. Moreover, our results contribute to in-depth study of
the structure and important properties and sub-classes of $n$-Hom-Lie
algebras.

In Section 2, definitions and properties of $n$-Hom-Lie algebras that are used
in the study are recalled, and new results characterizing nilpotency as well
as necessary and sufficient conditions for multiplicativity of general
$(n+1)$-dimensional $n$-Hom-Lie algebras and for algebras in the considered
class are obtained in terms of the structure constants and the twisting map.
In Section 4, Derived series and central descending series are studied for all
algebras in this class and are used to divide it into five non-isomorphic
subclasses. The levels of solvability and nilpotency of the $3$-Hom-Lie
algebras in these five classes are obtained. In Section 5, building up on the
previous sections, all algebras of this class are classified up to Hom-algebra
isomorphism. In Section 6, for some algebras in this class, it has been
determined whether the terms of the derived and central descending series are
weak subalgebras, Hom-subalgebras, weak ideals or Hom-ideals.

##  2 Definitions and properties of n-Hom-Lie algebras

In this section, we present the basic definitions and properties of $n$-Hom-
Lie algebras needed for our study. Throughout this article, it is assumed that
all linear spaces are over a field $\mathbb{K}$ of characteristic $0$, and for
any subset $S$ of a linear space, $\langle S\rangle$ denotes the linear span
of $S$. The arity of all the considered algebras is assumed to be greater than
or equal to $2$. Hom-Lie algebras are a generalization of Lie algebras
introduced in HLS  while studying $\sigma$-derivations. The $n$-ary case was
introduced in AtMaSi:GenNambuAlg .

######  Definition 1 (HLS ; ms:homstructure )

A Hom-Lie algebra $(A,[\cdot,\cdot],\alpha)$ is a linear space $A$ together
with a bilinear map $[\cdot,\cdot]:A\times A\to A$ and a linear map
$\alpha:A\to A$ satisfying, for all $x,y,z\in A$:

$\displaystyle[x,y]$ $\displaystyle=-[y,x],$ Skew-symmetry
$\displaystyle\sum_{\circlearrowleft{(x,y,z)}}[\alpha(x),[y,z]]$
$\displaystyle=[\alpha(x),[y,z]]+[\alpha(y),[z,x]]+[\alpha(z),[x,y]]=0.$
$\displaystyle\begin{array}[c]{c}\text{Hom-Jacobi}\\\ \text{identity }\\\
\text{(cyclic form)}\end{array}$

In Hom-Lie algebras, by skew-symmetry, the Hom-Jacobi identity is equivalent
to

$[\alpha(x),[y,z]]=[[x,y],\alpha(z)]+[\alpha(y),[x,z]]\quad\begin{array}[t]{c}\text{Hom-
Jacobi identity}\\\ \text{(Hom-derivation form)}\end{array}$ (1)

Hom-algebras satisfying just the Hom-algebra identity (1), without requiring
the skew-symmetry identity, are called Hom-Leibniz algebras LS2 ;
ms:homstructure . Thus, Hom-Lie algebras are skew-symmetric Hom-Leibniz
algebras. There are many Hom-Leibniz algebras which are not skew-symmetric and
thus not Hom-Lie algebras. When the twisting map is the identity map
$\alpha=\operatorname{Id}_{A}$ on $A$, Hom-Leibniz algebras become (left)
Leibniz algebras, and Hom-Lie algebras become Lie algebras. A Hom-Leibniz
algebra is also a Leibniz algebra, or a Hom-Lie algebra is also a Lie algebra,
if and only if the map $\operatorname{Id}_{A}$ belongs to the set of all
linear maps $\alpha$ for which the identity (1) holds. Whether the map
$\operatorname{Id}_{A}$ belongs to the set of all linear maps $\alpha$ for
which the identity (1) holds or not depends on the underlying algebra. The
Hom-algebra identity (1) is linear with respect to $\alpha$ in the linear
space of all linear maps on the algebra, and hence, the set of all such
$\alpha$, for which the identity (1) holds, is a linear subspace of the linear
space of all linear maps on the algebra. There are many Hom-Leibniz algebra
which are not Leibniz algebras, or Hom-Lie algebras which are not Lie
algebras.

######  Definition 2 (HLS ; LarssonSilvJA2005:QuasiHomLieCentExt2cocyid )

Hom-Lie algebra morphisms from Hom-Lie algebra
$\mathcal{A}=(A,[\cdot,\cdot]_{\mathcal{A}},\alpha)$ to Hom-Lie algebra
$\mathcal{B}=(B,[\cdot,\cdot]_{\mathcal{B}},\beta)$ are linear maps $f:A\to B$
satisfying, for all $x,y\in A$,

$\displaystyle f([x,y]_{\mathcal{A}})$ $\displaystyle=$
$\displaystyle[f(x),f(y)]_{\mathcal{B}},$ (2) $\displaystyle f\circ\alpha$
$\displaystyle=$ $\displaystyle\beta\circ f.$ (3)

Linear maps $f:A\to B$ satisfying only condition (2) are called weak morphisms
of Hom-Lie algebras.

######  Definition 3 (BenMakh:Hombiliform ; ms:homstructure )

A Hom-Lie algebra $(A,[\cdot,\cdot],\alpha)$ is said to be multiplicative if
$\alpha$ is an algebra morphism, and it is said to be regular if $\alpha$ is
an isomorphism.

###### Definition 4

An $n$-ary Hom-algebra $(A,[\cdot,\dots,\cdot],\\{\alpha_{i}\\}_{1\leq i\leq
n-1})$ is a linear space $A$ together with an $n$-ary operation, that is an
$n$-linear map $[\cdot,\dots,\cdot]:A^{n}\to A$ and $(n-1)$ linear maps
$\alpha_{i}:A\to A,1\leq i\leq n-1$. An $n$-ary Hom-algebra is said to be
skew-symmetric if its $n$-ary operation is skew-symmetric, that is satisfying
for all $x_{1},\dots,x_{n-1},y_{1},\dots,y_{n}\in A$,

$\displaystyle[x_{\sigma(1)},\dots,x_{\sigma(n)}]=sgn(\sigma)[x_{1},\dots,x_{n}].$
Skew-symmetry (4)

The $n$-Hom-Lie algebras are an $n$-ary generalization of Hom-Lie algebras to
$n$-ary algebras satisfying a generalisation of the Hom-algebra identity (1)
involving $n$-ary product and $n-1$ linear maps.

######  Definition 5 (AtMaSi:GenNambuAlg )

An $n$-Hom-Lie algebra $(A,[\cdot,\dots,\cdot],\\{\alpha_{i}\\}_{1\leq i\leq
n-1})$ is a skew-symmetric $n$-ary Hom-algebra satisfying, for all
$x_{1},\dots,x_{n-1},y_{1},\dots,y_{n}\in A$

$\begin{array}[]{l}\text{Hom-Nambu-Filippov identity}\\\\[5.69054pt]
\left[\alpha_{1}(x_{1}),\dots,\alpha_{n-1}(x_{n-1}),[y_{1},\dots,y_{n}]\right]=\\\\[2.84544pt]
\quad\displaystyle{\sum_{i=1}^{n}}[\alpha_{1}(y_{1}),\dots,\alpha_{i-1}(y_{i-1}),[x_{1},\dots,x_{n-1},y_{i}],\alpha_{i}(y_{i+1}),\dots,\alpha_{n-1}(y_{n})].\end{array}$
(5)

###### Remark 1

If $\alpha_{i}=Id_{A}$ for all $1\leq i\leq n-1$, then one gets an $n$-Lie
algebra (Filippov:nLie ). Therefore, the class of $n$-Lie algebras is included
in the class of $n$-Hom-Lie algebras. For any linear space $A$, if
$\left[x_{1},\dots,x_{n}\right]_{0}=0$ for all $x_{1},\dots,x_{n}\in A$ and
any linear maps $\alpha_{1},\dots,\alpha_{n-1}$, then
$(A,\left[\cdot,\dots,\cdot\right]_{0},\alpha_{1},\dots,\alpha_{n-1})$ is an
$n$-Hom-Lie algebra.

######  Definition 6 (AtMaSi:GenNambuAlg ; YauHomNambuLie )

$n$-Hom-Lie algebra morphisms of $n$-Hom-Lie algebras
$\mathcal{A}=(A,[\cdot,\dots,\cdot]_{\mathcal{A}},\\{\alpha_{i}\\}_{1\leq
i\leq n-1})$ and
$\mathcal{B}=(B,[\cdot,\dots,\cdot]_{\mathcal{B}},\\{\beta_{i}\\}_{1\leq i\leq
n-1})$ are linear maps $f:A\to B$ satisfying, for all $x_{1},\dots,x_{n}\in
A$,

$\displaystyle f([x_{1},\dots,x_{n}]_{\mathcal{A}})$ $\displaystyle=$
$\displaystyle[f(x_{1}),\dots,f(x_{n})]_{\mathcal{B}},$ (6) $\displaystyle
f\circ\alpha_{i}$ $\displaystyle=$ $\displaystyle\beta_{i}\circ
f,\quad\text{for all}\quad 1\leq i\leq n-1.$ (7)

Linear maps satisfying only condition (6) are called weak morphisms of
$n$-Hom-Lie algebras.

The $n$-Hom-Lie algebras $(A,[\cdot,\dots,\cdot],\\{\alpha_{i}\\}_{1\leq i\leq
n-1})$ with $\alpha_{1}=\dots=\alpha_{n-1}=\alpha$ will be denoted by
$(A,\left[\cdot,\dots,\cdot\right],\alpha)$.

######  Definition 7 (YauHomNambuLie )

An $n$-Hom-Lie algebra $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ is called
multiplicative if $\alpha$ is an algebra morphism, and regular if $\alpha$ is
an algebra isomorphism.

The following proposition, providing a way to construct an $n$-Hom-Lie algebra
from an $n$-Lie algebra and an algebra morphism, was first introduced in the
case of Lie algebras and then generalized to the $n$-ary case in
AtMaSi:GenNambuAlg . A more general version of this theorem, given in
YauHomNambuLie , states that the category of $n$-Hom-Lie algebras is closed
under twisting by weak morphisms.

######  Proposition 1 (AtMaSi:GenNambuAlg ; YauHomNambuLie )

Let $\beta:\mathcal{A}\to\mathcal{A}$ be a weak morphism of $n$-Hom-Lie
algebra
$\mathcal{A}=\left(A,\left[\cdot,\dots,\cdot\right],\\{\alpha_{i}\\}_{1\leq
i\leq n-1}\right)$, and multiplication
$\left[\cdot,\dots,\cdot\right]_{\beta}$ is defined by
$\left[x_{1},\dots,x_{n}\right]_{\beta}=\beta\left(\left[x_{1},\dots,x_{n}\right]\right).$
Then,
$\left(A,\left[\cdot,\dots,\cdot\right]_{\beta},\\{\beta\circ\alpha_{i}\\}_{1\leq
i\leq n-1}\right)$ is an $n$-Hom-Lie algebra. Moreover, if
$\left(A,\left[\cdot,\dots,\cdot\right],\alpha\right)$ is multiplicative and
$\beta\circ\alpha=\alpha\circ\beta$, then
$\left(A,\left[\cdot,\dots,\cdot\right]_{\beta},\beta\circ\alpha\right)$ is
multiplicative.

The following particular case of Proposition 1 is obtained if $\alpha=Id_{A}$.

###### Corollary 1

Let $\left(A,\left[\cdot,\dots,\cdot\right]\right)$ be an $n$-Lie algebra,
$\beta:A\to A$ an algebra morphism, and
$\left[\cdot,\dots,\cdot\right]_{\beta}$ is defined by
$\left[x_{1},\dots,x_{n}\right]_{\beta}=\beta\left(\left[x_{1},\dots,x_{n}\right]\right).$
Then, $\left(A,\left[\cdot,\dots,\cdot\right]_{\beta},\beta\right)$ is a
multiplicative $n$-Hom-Lie algebra.

The following definition is a specialization of the standard definition of a
subalgebra in general algebraic structures to the case of $n$-Hom-Lie algebras
and $n$-ary skew-symmetric Hom-algebras considered in this paper.

###### Definition 8

A Hom-subalgebra of an $n$-Hom-Lie algebra or more generally an $n$-ary skew-
symmetric Hom-algebra
$\mathcal{A}=(A,[\cdot,\dots,\cdot]_{\mathcal{A}},\alpha_{1},\dots,\alpha_{n-1})$
is an $n$-ary Hom-algebra
$\mathcal{B}=(B,[\cdot,\dots,\cdot]_{\mathcal{B}},\beta_{1},\dots,\beta_{n-1})$
consisting of a subspace $B$ of $A$ satisfying, for all $x_{1},\dots,x_{n}\in
B$,

  1. 1)

$\alpha_{i}(B)\subseteq B$ for all $1\leq i\leq n-1$,

  2. 2)

$[x_{1},\dots,x_{n}]_{\mathcal{A}}\in B$,

with the restricted from $A$ multiplication
$[\cdot,\dots,\cdot]_{\mathcal{B}}=[\cdot,\dots,\cdot]_{\mathcal{A}}$ and
linear maps $\beta_{i}=\alpha_{i},1\leq i\leq n-1$ on $B$.

The following definition is a direct extension of the corresponding definition
in BenMakh:Hombiliform ; ms:homstructure ; YauHomNambuLie  to arbitrary
$n$-ary skew-symmetric Hom-algebras.

###### Definition 9

An ideal of an $n$-Hom-Lie algebra or more generally of an $n$-ary skew-
symmetric Hom-algebra $(A,[\cdot,\dots,\cdot],\alpha_{1},\dots,\alpha_{n-1})$
is a subspace $I$ of $A$ satisfying, for all $x_{1},\dots,x_{n-1}\in A$, $y\in
I$:

  1. 1)

$\alpha_{i}(I)\subseteq I$ for all $1\leq i\leq n-1$.

  2. 2)

$[x_{1},\dots,x_{n-1},y]\in I$ (or equivalently $[y,x_{1},\dots,x_{n-1}]\in
I$).

The following definitions are a direct extension of the corresponding
definitions in kms:solvnilpnhomlie2020  to arbitrary $n$-ary skew-symmetric
Hom-algebras.

###### Definition 10

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha_{1},\dots,\alpha_{n-1})$ be an
$n$-Hom-Lie algebra or more generally an $n$-ary skew-symmetric Hom-algebra,
and let $I$ be an ideal of $A$. For $2\leq k\leq n$ and $p\in\mathbb{N}$, we
define the $k$-derived series of the ideal $I$ by

$D_{k}^{0}(I)=I\text{ and
}D_{k}^{p+1}=\langle\big{[}\underset{k}{\underbrace{D_{k}^{p}(I),\dots,D_{k}^{p}(I)}},\underset{n-k}{\underbrace{A,\dots,A}}\big{]}\rangle.$

We define the $k$-central descending series of $I$ by

$C_{k}^{0}(I)=I\text{ and
}C_{k}^{p+1}(I)=\langle\big{[}C_{k}^{p}(I),\underset{k-1}{\underbrace{I,\dots,I}},\underset{n-k}{\underbrace{A,\dots,A}}\big{]}\rangle.$

###### Definition 11

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha_{1},\dots,\alpha_{n-1})$ be an
$n$-Hom-Lie algebra or more generally an $n$-ary skew-symmetric Hom-algebra,
and let $I$ be an ideal of $A$. For $2\leq k\leq n$, the ideal $I$ is said to
be $k$-solvable (resp. $k$-nilpotent) if there exists $r\in\mathbb{N}$ such
that $D_{k}^{r}(I)=\\{0\\}$ (resp. $C_{k}^{r}(I)=\\{0\\}$), and the smallest
$r\in\mathbb{N}$ satisfying this condition is called the class of
$k$-solvability (resp. the class of nilpotency) of $I$.

The following direct extension of the corresponding result in
kms:solvnilpnhomlie2020  to arbitrary $n$-ary skew-symmetric Hom-algebras is
proved in the same way as in kms:solvnilpnhomlie2020  since the proof does not
involve the Hom-Nambu-Filippov identity.

###### Lemma 1

Let $\mathcal{A}=(A,\left[\cdot,\dots,\cdot\right]_{A},(\alpha_{i})_{1\leq
i\leq n})$ and
$\mathcal{B}=(B,\left[\cdot,\dots,\cdot\right]_{B},(\beta_{i})_{1\leq i\leq
n})$ be two $n$-ary skew-symmetric Hom-algebras, $f:\mathcal{A}\to\mathcal{B}$
be a surjective $n$-Hom-Lie algebras morphism and $I$ an ideal of
$\mathcal{A}$. Then for all $r\in\mathbb{N}$ and $2\leq k\leq n$:

$f\left(D_{k}^{r}(I)\right)=D_{k}^{r}\left(f(I)\right)\text{ and
}f\left(C_{k}^{r}(I)\right)=C_{k}^{r}\left(f(I)\right).$

This lemma also implies that if two $n$-Hom-Lie algebras are isomorphic, they
would also have isomorphic terms of the derived series and central descending
series, which also means that if two algebras have a significant difference in
the derived series or the central descending series, for example different
dimensions of given corresponding terms, then these algebras cannot be
isomorphic.

######  Lemma 2 (cln1dimnHomLieniltw )

Let $A$ be a linear space, let $\left[\cdot,\dots,\cdot\right]$ be an
$n$-linear skew-symmetric map ($n\geq 2$) and let
$\alpha_{1},\dots,\alpha_{n-1}$ be linear maps on $A$. If the $(n-1)$-linear
map

$(x_{1},\dots,x_{n-1})\mapsto\left[\alpha_{1}(x_{1}),\dots,\alpha_{n-1}(x_{n-1}),d\right]$

is skew-symmetric for all $d\in\left[A,\dots,A\right]$, then the
$(2n-1)$-linear map $H$ defined by

$\displaystyle H$
$\displaystyle(x_{1},\dots,x_{n-1},y_{1},\dots,y_{n})=[\alpha_{1}(x_{1}),\dots,\alpha_{n-1}(x_{n-1}),[y_{1},\dots,y_{n}]]$
$\displaystyle\quad-\sum_{k=1}^{n}[\alpha_{1}(y_{1}),\dots,\alpha_{k-1}(y_{k-1}),[x_{1},\dots,x_{n-1},y_{k}],\alpha_{k}(y_{k+1}),\dots,\alpha_{n-1}(y_{n})],$

for all $x_{1},\dots,x_{n-1},y_{1},\dots,y_{n}\in A$, is skew-symmetric in its
first $n-1$ arguments and in its last $n$ arguments.

######  Proposition 2 (cln1dimnHomLieniltw )

Let $A$ be an $n$-dimensional linear space ($n\geq 2$), and $(e_{i})_{1\leq
i\leq n}$ a basis of $A$. Any skew-symmetric $n$-linear map
$\left[\cdot,\dots,\cdot\right]$ on $A$ is fully defined by giving
$\left[e_{1},\dots,e_{n}\right]=d\in A$. Let $\alpha_{1},\dots,\alpha_{n-1}$
be linear maps on $A$. If the $(n-1)$-linear map

$(x_{1},\dots,x_{n-1})\mapsto\left[\alpha_{1}(x_{1}),\dots,\alpha_{n-1}(x_{n-1}),d\right]$

is skew-symmetric, then
$(A,\left[\cdot,\dots,\cdot\right],\alpha_{1},\dots,\alpha_{n-1})$ is an
$n$-Hom-Lie algebra.

######  Corollary 2 (cln1dimnHomLieniltw )

Let $A$ be an $n$-dimensional linear space ($n\geq 2$), and $(e_{i})_{1\leq
i\leq n}$ a basis of $A$. Any skew-symmetric $n$-linear map
$\left[\cdot,\dots,\cdot\right]$ on $A$ is fully defined by giving
$\left[e_{1},\dots,e_{n}\right]=d\in A$. For any linear map $\alpha$ on $A$,
$(A,\left[\cdot,\dots,\cdot\right],\alpha)$ is an $n$-Hom-Lie algebra.

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an $n$-ary skew-symmetric
algebra of dimension $n+1$ with a linear map $\alpha$. Given a linear basis
$(e_{i})_{1\leq i\leq n+1}$ of $A$, the linear map $\alpha$ is fully
determined by its matrix determined by action of $\alpha$ on the basis, and a
skew-symmetric $n$-ary multi-linear bracket is fully determined by
$\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]$ for all $1\leq i\leq
n+1$ represented by a matrix $B=(b(i,j))_{1\leq i,j\leq n+1}$ as follows:

$\displaystyle\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]$
$\displaystyle=$ $\displaystyle(-1)^{n+1+i}w_{i},$ (8) $\displaystyle
w_{i}=\sum_{p=1}^{n+1}b(p,i)e_{p},\ (w_{1},\dots,w_{n+1})$ $\displaystyle=$
$\displaystyle(e_{1},\dots,e_{n+1})B.$

######  Proposition 3 (cln1dimnHomLieniltw )

Let $\mathcal{A}_{1}=(A,\left[\cdot,\dots,\cdot\right]_{1},\alpha_{1})$ and
$\mathcal{A}_{2}=(A,\left[\cdot,\dots,\cdot\right]_{2},\alpha_{2})$ be two
$(n+1)$-dimensional $n$-ary skew-symmetric Hom-algebras represented by
matrices $\left[\alpha_{1}\right]$, $B_{1}$ and $\left[\alpha_{2}\right]$,
$B_{2}$ respectively. The Hom-algebras $\mathcal{A}_{1}$ and $\mathcal{A}_{2}$
are isomorphic if and only if there exists an invertible matrix $T$ satisfying
the following conditions:

$\displaystyle
B_{2}=\det(T)^{-1}TB_{1}T^{T},\quad\quad[\alpha_{2}]=T[\alpha_{1}]T^{-1}.$

######  Proposition 4 (cln1dimnHomLieniltw )

Let $(e_{i})_{1\leq i\leq n+1}$ be a basis of a linear space $A$, let $\sigma$
be a permutation of the set $\\{1,\dots,n+1\\}$ of $n+1$ elements, and let
$B=(b_{i,j})_{1\leq i,j\leq n+1}$ be a matrix representing a skew-symmetric
$n$-ary bracket in this basis, then the matrix representing the same bracket
in the basis $(e_{\sigma(i)})_{1\leq i\leq n+1}$ is given by the matrix
$sgn(\sigma)(b_{\sigma^{-1}(i),\sigma^{-1}(j)})_{1\leq i,j\leq n+1}$.

###### Remark 2

(cln1dimnHomLieniltw ). Let $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an
$(n+1)$-dimensional $n$-Hom-Lie algebra and let $B$ be the matrix representing
its bracket. $D^{1}_{n}(A)=\left[A,\dots,A\right]$ is generated by
$\\{w_{1},\dots,w_{n+1}\\}$. Which means that $Rank(B)=\dim D^{1}_{n}(A)$.

If $Rank(B)\leq n$ or equivalently $\det(B)=0$ then $D^{1}_{n}(A)$ has
dimension at most $n$, which means that $D^{2}_{n}(A)$ has dimension at most
$1$ and then $D^{3}_{n}(A)=0$.

###### Remark 3

(cln1dimnHomLieniltw ). For the whole algebra $A$, all the $k$-central
descending series, for all $2\leq k\leq n$, are equal. Therefore all the
notions of $k$-nilpotency, for all $2\leq k\leq n$, are equivalent, and we
denote $C^{p}_{k}(A)$ for any $2\leq k\leq n$ by $C^{p}(A)$.

###### Definition 12

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha_{1},\dots,\alpha_{n-1})$ be an
$n$-Hom-Lie algebra or more generally an $n$-ary skew-symmetric Hom-algebra.
Define $Z(A)$, the center of $A$, by

$\displaystyle Z(A)=\\{z\in A:\left[x_{1},\dots,x_{n-1},z\right]=0,\forall
x_{1},\dots,x_{n-1}\in A\\}.$

######  Lemma 3 (cln1dimnHomLieniltw )

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an $n$-Hom-Lie algebra with
$A\neq\\{0\\}$. If $A$ is $k$-nilpotent, for any $2\leq k\leq n$, then the
center $Z(A)$ of $A$ is not trivial ($Z(A)\neq\\{0\\}$).

###### Lemma 4

Let $\mathcal{A}=(A,\left[\cdot,\dots,\cdot\right],(\alpha_{i})_{1\leq i\leq
n-1})$ be an $n$-ary skew-symmetric Hom-algebra with $A\neq\\{0\\}$.

  1. (i)

If $\mathcal{A}$, is nilpotent then $Z(\mathcal{A})$ is not trivial
($Z(A)\neq\\{0\\}$).

  2. (ii)

If $\dim A=n+1$, then $\dim Z(\mathcal{A})=0$ or $\dim Z(\mathcal{A})=1$ or
$Z(\mathcal{A})=A$.

###### Proof

(i) The first statement is a generalization of Lemma 3 to the case of $n$-ary
skew-symmetric Hom-algebras, and is proved in the same way, since the original
proof does not use the Hom-Nambu-Filippov identity.
(ii) Suppose that $\dim A=n+1$ and that $\dim Z(\mathcal{A})>1$. Let $(e_{i})$
be a basis of $A$ such that $e_{1},e_{2}\in Z(\mathcal{A}$, then
$\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]=0$ for all $1\leq
i\leq n+1$, which means that $\left[x_{1},\dots,x_{n}\right]=0$ for all
$x_{1},\dots,x_{n}\in A$.∎

The following direct extension of the corresponding result in
cln1dimnHomLieniltw  to arbitrary $n$-ary skew-symmetric Hom-algebras is
proved in the same way as in cln1dimnHomLieniltw  since the proof does not
involve the Hom-Nambu-Filippov identity.

###### Proposition 5

Let $\mathcal{A}=(A,[\cdot,\dots,\cdot],\\{\alpha_{i}\\}_{1\leq i\leq n-1})$
be an $(n+1)$-dimensional $n$-ary skew-symmetric algebra. The algebra
$\mathcal{A}$ is nilpotent and non abelian if and only if $\dim
Z(\mathcal{A})=1$ and $\left[A,\dots,A\right]=Z(\mathcal{A})$.

###### Proposition 6

Let $\mathcal{A}=(A,[\cdot,\dots,\cdot],\\{\alpha_{i}\\}_{1\leq i\leq n-1})$
be an $n$-Hom-Lie algebra or more generally an $n$-ary skew-symmetric Hom-
algebra with $A\neq\\{0\\}$. $\mathcal{A}$ is nilpotent of class $p$ if and
only if $\\{0\\}\subsetneq C^{p-1}(A)\subseteq Z(A)$.

###### Proof

The statement holds, since $\mathcal{A}$ is nilpotent of class $p$ if and only
if $C^{p}(A)=\\{0\\}$ and $C^{p-1}(A)\neq\\{0\\}$, and

$\displaystyle C^{p}(A)=\\{0\\}$
$\displaystyle\iff\left[C^{p-1}(A),A,\dots,A\right]=\\{0\\}$
$\displaystyle\iff\forall\ c\in C^{p-1}(A),\forall\ x_{1},\dots,x_{n-1}\in
A,\left[c,x_{1},\dots,x_{n-1}\right]=0$ $\displaystyle\iff\forall\ c\in
C^{p-1}(A),c\in Z(\mathcal{A})$ $\displaystyle\iff C^{p-1}(A)\subseteq
Z(\mathcal{A}).$ ∎

###### Proposition 7

Let $\mathcal{A}=(A,\left[\cdot,\dots,\cdot\right]_{\mathcal{A}},\alpha)$ and
$\mathcal{B}=(B,\left[\cdot,\dots,\cdot\right]_{\mathcal{B}},\beta)$ be
$n$-ary Hom-algebras. Let $f:\mathcal{A}\to\mathcal{B}$ be an $n$-ary Hom-
algebras homomorphism, then if $\mathcal{A}$ is multiplicative then
$\mathcal{B}$ is multiplicative. Moreover, if $f$ is an isomorphism, then
$\mathcal{A}$ is multiplicative if and only if $\mathcal{B}$ is
multiplicative.

###### Proof

Let $f:\mathcal{A}\to\mathcal{B}$ be a surjective homomorphism, then for all
$y_{1},\dots,y_{n}\in B$ there exists $x_{1},\dots,x_{n}\in A$ such that
$f(x_{i})=y_{i}$ for $1\leq i\leq n$, and $\beta\circ f=f\circ\alpha$. Suppose
that $\mathcal{A}$ is multiplicative, then we have,

$\displaystyle\beta\left(\left[y_{1},\dots,y_{n}\right]_{\mathcal{B}}\right)$
$\displaystyle=\beta\left(\left[f(x_{1}),\dots,f(x_{n})\right]_{\mathcal{B}}\right)$
$\displaystyle=\beta\circ
f\left(\left[x_{1},\dots,x_{n}\right]_{\mathcal{A}}\right)=f\circ\alpha\left(\left[x_{1},\dots,x_{n}\right]_{\mathcal{A}}\right)$
$\displaystyle=\left[f\circ\alpha(x_{1}),\dots,f\circ\alpha(x_{n})\right]_{\mathcal{B}}=\left[\beta\circ
f(x_{1}),\dots,\beta\circ f(x_{n})\right]_{\mathcal{B}}$
$\displaystyle=\left[\beta(y_{1}),\dots,\beta(y_{n})\right]_{\mathcal{B}}.$

If $f$ is an isomorphism, then the converse can be proved by applying the same
argument using $f^{-1}$ instead of $f$.

######  Proposition 8 (cln1dimnHomLieniltw )

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an $n$-ary Hom-algebra with
$\dim A=n+1$, $\left[\cdot,\dots,\cdot\right]$ skew-symmetric, $\alpha$
nilpotent, $\dim\ker\alpha=2$ and the bracket is represented by the matrix
$B=(b_{i,j})$ as in (8), in a basis where $\alpha$ is in Jordan normal form.
The bracket $\left[\cdot,\dots,\cdot\right]$ satisfies the Hom-Nambu-Filippov
identity if and only if

$b_{i_{0}-1,j}b_{p,n+1}-b_{n+1,j}b_{p,i_{0}-1}=0,\forall\ 1\leq j,p\leq
n+1,j\neq 1,j\neq i_{0},$

where $i_{0}$ is such that $\ker\alpha=\langle e_{1},e_{i_{0}}\rangle$.

###### Remark 4

Let us compare the polynomial equations obtained from the Nambu-Filippov
identity and the Hom-Nambu-Filippov identity in dimension $n+1$ with various
types of twisting maps:
Diagonalizable and invertible with eigenvalues $\lambda_{i},1\leq i\leq n+1$:

$\displaystyle(\lambda_{i}b_{j,i}-\lambda_{j}b_{i,j})b_{p,k}+(\lambda_{k}b_{i,k}-\lambda_{i}b_{k,i})b_{p,j}+(\lambda_{j}b_{k,j}-\lambda_{k}b_{j,k})b_{p,i}=0,$
(9) $\displaystyle\forall\ 1\leq i,j,k,p\leq n+1;i<j<k;$

Diagonalizable with $\dim\ker\alpha=1$ with eigenvalues $\lambda_{i},1\leq
i\leq n+1$:

$\lambda_{k}b_{1,k}w_{j}-\lambda_{k}b_{j,k}w_{1}-\lambda_{j}b_{1,j}w_{k}+\lambda_{j}b_{k,j}w_{1}=0,\quad\forall\
1<j<k\leq n+1;$ (10)

Diagonalizable with $\dim\ker\alpha=2$ with eigenvalues $\lambda_{i},1\leq
i\leq n+1$:

$b_{1,k}w_{2}-b_{2,k}w_{1}=0,\quad\forall\ 3\leq k\leq n+1;$ (11)

Nilpotent with $\dim\ker\alpha=1$:

$\displaystyle(b_{k-1,i}-b_{i-1,k})b_{p,n+1}-b_{n+1,i}b_{p,k-1}+b_{n+1,k}b_{p,i-1}=0,$
(12) $\displaystyle\forall\ 1\leq i,k,p\leq n+1,i<k;$

Nilpotent with $\dim\ker\alpha=2$:

$b_{i_{0}-1,j}b_{p,n+1}-b_{n+1,j}b_{p,i_{0}-1}=0,\quad\forall\ 1\leq j,p\leq
n+1,j\neq 1,j\neq i_{0}.$ (13)

These different cases are separate from each other, and the case of $n$-Lie
algebras is the special case of (9) where all the $\lambda_{i}$ are equal.
Notice that the higher the dimension of $\ker\alpha$ the less equation we have
and the less terms we have in each equation, that is, in these cases, the Hom-
Nambu-Filippov identity is considerably less restrictive. Another difference
from the case of $n$-Lie algebras is that the isomorphisms in Hom-algebras
intertwine the multiplications and the twisting maps, which leads to
different, more restrictive isomorphism conditions and, in general, more
isomorphism classes.

###### Lemma 5

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an $n$-ary Hom-algebra with
$\dim A=n+1$, $\left[\cdot,\dots,\cdot\right]$ skew-symmetric and $\alpha$
nilpotent. Let $(e_{i})_{1\leq i\leq n+1}$ be a basis of $A$ where $\alpha$ is
in its Jordan form, and consider $\left[\cdot,\dots,\cdot\right]$ to be
defined as in (8).

If $\dim\ker\alpha\geq 2$, then $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ is
multiplicative if and only if $\left[A,\dots,A\right]\subseteq\ker\alpha$.

If $\dim\ker\alpha=1$, then $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ is
multiplicative if and only if $\alpha(w_{1})=(-1)^{n}w_{n+1}$ and
$w_{i}\in\ker\alpha,\forall\ 2\leq i\leq n+1$, where $(w_{i})$ are defined in
(8).

###### Proof

Suppose that $\dim\ker\alpha\geq 2$, then for all $1\leq i\leq n+1$,

$\displaystyle\alpha(w_{i})$
$\displaystyle=(-1)^{n+1+i}\alpha\left(\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]\right)$
$\displaystyle=(-1)^{n+1+i}\left[\alpha\left(e_{1}\right),\dots,\widehat{\alpha\left(e_{i}\right)},\dots,\alpha\left(e_{n+1}\right)\right]=0,$

since $e_{i}\in\ker\alpha$ for at least two different indices $i$, that is at
least one of the
$\alpha\left(e_{1}\right),\dots,\widehat{\alpha\left(e_{i}\right)},\dots,\alpha\left(e_{n+1}\right)$
is zero. Thus, $\left[A,\dots,A\right]=\langle
w_{1},\dots,w_{n+1}\rangle\subseteq\ker\alpha$.

Suppose now that $\dim\ker\alpha=1$, then we have $\alpha(e_{1})=0$ and
$\alpha(e_{i})=e_{i-1}$ for $2\leq i\leq n+1$. We get

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

For $i\neq 1$ we have,

$\displaystyle\alpha(w_{i})$
$\displaystyle=(-1)^{n+1+i}\alpha\left(\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]\right)$
$\displaystyle=(-1)^{n+1+i}\left[\alpha\left(e_{1}\right),\dots,\widehat{\alpha\left(e_{i}\right)},\dots,\alpha\left(e_{n+1}\right)\right]$
$\displaystyle=(-1)^{n+1+i}\left[0,e_{1},\dots,\widehat{e_{i-1}},\dots,e_{n}\right]=0,$

that is $\alpha(w_{i})=0$ for $i\neq 1$.

###### Proposition 9

Let $\mathcal{A}=(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an
$(n+1)$-dimensional $n$-Hom-Lie algebra. If $\dim\ker\alpha\geq 2$ then
$\mathcal{A}$ is multiplicative if and only if $[\alpha]B=0$, where $[\alpha]$
and $B$ are the matrices representing the twisting map $\alpha$ and the
bracket in any given basis.

###### Proof

Let $(e_{i})_{1\leq i\leq n+1}$ be a basis of $A$ containing a basis of
$\ker\alpha$. Then $\mathcal{A}$ is multiplicative if and only if

$\alpha\left(\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]\right)=\left[\alpha(e_{1}),\dots,\widehat{\alpha(e_{i})},\dots,\alpha(e_{n+1})\right]\
\mbox{for all}\ 1\leq i\leq n+1.$

On the other hand,
$\left[\alpha(e_{1}),\dots,\widehat{\alpha(e_{i})},\dots,\alpha(e_{n+1})\right]=0$
since at least one of the elements $e_{1},\dots,e_{i-1},e_{i+1},\dots,e_{n+1}$
is in $\ker\alpha$. Moreover $[\alpha]B$ is the matrix whose columns are the
coordinates of
$(-1)^{n+i+1}\alpha\left(\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]\right)$.
Thus $\alpha$ is an algebra morphism if and only if $[\alpha]B=0$.

Let now $[\alpha]_{2}$ and $B_{2}$ be the matrices representing $\alpha$ and
$\left[\cdot,\dots,\cdot\right]$ in another basis $(e^{\prime}_{i})$, then
there exists an invertible matrix $P$ such that $[\alpha]_{2}=P[\alpha]P^{-1}$
and $B_{2}=(\det P)^{-1}PBP^{T}$, and we get

$\displaystyle[\alpha]_{2}B_{2}$ $\displaystyle=(P[\alpha]P^{-1})((\det
P)^{-1}PBP^{T})$ $\displaystyle=(\det P)^{-1}(P[\alpha]P^{-1}PBP^{T})=(\det
P)^{-1}(P[\alpha]BP^{T}).$

Therefore $[\alpha]_{2}B_{2}=0$ if and only if $[\alpha]B=0$, since $P$ is
invertible.

###### Corollary 3

Let $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be an $n$-ary Hom-algebra with
$\dim A=n+1$, $\left[\cdot,\dots,\cdot\right]$ skew-symmetric and $\alpha$
nilpotent. Let $(e_{i})_{1\leq i\leq n+1}$ be a basis of $A$ where $\alpha$ is
in its Jordan form, and consider $\left[\cdot,\dots,\cdot\right]$ to be
defined by its structure constants in this basis, that is,
$\left[e_{i_{1}},\dots,e_{i_{n}}\right]=\sum\limits_{k=1}^{\dim
A}c_{i_{1},\dots,i_{n}}^{k}e_{k}$.

If $\dim\ker\alpha\geq 2$, then $(A,\left[\cdot,\dots,\cdot\right],\alpha)$ is
multiplicative if and only if $c_{i_{1},\dots,i_{n}}^{k}=0$, for all $1\leq
i_{1},\dots,i_{n}\leq\dim A$ and $k$ such that $e_{k}\notin\ker\alpha$.

###### Remark 5

Note that when $\dim A=n+1$, it is sufficient to define the bracket by its
structure constants as
$\left[e_{1},\dots,\widehat{e_{i}},\dots,e_{n+1}\right]=\sum\limits_{k=1}^{\dim
A}c_{1,\dots,i-1,i+1,\dots,n+1}^{k}e_{k}$. The parameters $b(p,i)$ in (8) are
$b(p,i)=(-1)^{n+1+i}c_{1,...,i-1,i+1,...,n+1}^{p}$.

##  3 Class $4_{3,N(2),6}$ of $4$-dimensional $3$-Hom-Lie algebras

An interesting class of $4$-dimensional $3$-Hom-Lie algebras
$4_{3,N(2),6}=(A,\left[\cdot,\dots,\cdot\right],\alpha)$ is defined according
to (8) on the basis $(e_{i})_{1\leq i\leq 4}$ by

$\left[\alpha\right]=\begin{pmatrix}0&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\
0&0&0&0\\\ \end{pmatrix},\quad
B=\left(\begin{array}[]{cccc}0&c(1,3,4,1)&-c(1,2,4,1)&0\\\
0&c(1,3,4,2)&-c(1,2,4,2)&0\\\ 0&c(1,3,4,3)&-c(1,2,4,3)&0\\\
0&c(1,3,4,4)&-c(1,2,4,4)&0\\\ \end{array}\right),$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$

where $c(i_{1},\dots,i_{n},k)=c_{i_{1},\dots,i_{n}}^{k}$ are the structure
constants according to

$\left[e_{i_{1}},\dots,e_{i_{n}}\right]=\sum\limits_{k=1}^{\dim
A}c_{i_{1},\dots,i_{n}}^{k}e_{k}=\sum\limits_{k=1}^{\dim
A}c(i_{1},\dots,i_{n},k)e_{k}.$

Applying Lemma 5 to the class of $3$-Hom-Lie algebras $4_{3,N(2),6}$, we get
the following result describing all multiplicative $3$-Hom-Lie algebras in the
class $4_{3,N(2),6}$.

###### Corollary 4

The $3$-Hom-Lie algebra from $4_{3,N(2),6}$ is multiplicative if and only if

$c(1,2,4,3)=0,\ c(1,2,4,4)=0,\ c(1,3,4,3)=0,\ c(1,3,4,4)=0.$

###### Proof

By Lemma 5, the $3$-Hom-Lie algebra $4_{3,N(2),6}$ is multiplicative if and
only if
$\left[e_{1},e_{2},e_{4}\right],\left[e_{1},e_{3},e_{4}\right]\in\ker\alpha$
which is $\langle\\{e_{1},e_{2}\\}\rangle$, and this is the case if and only
if $c(1,2,4,3)=0$, $c(1,2,4,4)=0$, $c(1,3,4,3)=0$, $c(1,3,4,4)=0.$ ∎

So, the $3$-Hom-Lie algebra from $4_{3,N(2),6}$ is in the subclass
$4_{3,N(2),6,M}$ of multiplicative $3$-Hom-Lie algebras, if and only if the
multiplication (bracket) is defined by

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0,$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2},$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2},$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0.$

##  4 Derived series and central descending series for $4_{3,N(2),6}$

A consequence of Lemma 1 is that the derived series and the central descending
series of an $n$-Hom-Lie algebra are algebraic invariants. Here, we divide the
considered class of $3$-Hom-Lie algebras into five subclasses following their
derived series and central descending series. Two $3$-Hom-Lie algebras in two
different subclasses will necessarily be non-isomorphic, and we use this as an
intermediate step towards the full classification up to isomorphism of the
algebras in this class.

In the case of $n$-Hom-Lie algebras, the terms of the derived series and the
central descending series are in general not ideals as in the case of $n$-Lie
algebras. In the most general case, they are weak subalgebras, and they can be
subalgebras or ideals if the twisting maps are algebra morphisms or surjective
algebra morphisms respectively, as it has been shown in
kms:solvnilpnhomlie2020 . For the case of $4_{3,N(2),6,M}$, we have the
following result.

###### Theorem 4.1

Consider $\mathcal{A}=(A,\left[\cdot,\cdot,\cdot\right],\alpha)=4_{3,N(2),6}$.
Suppose that $B\neq 0$ and define
$d(p,q)=c(1,2,4,p)c(1,3,4,q)-c(1,2,4,q)c(1,3,4,p)$ with $1\leq p,q\leq 4$,
that is, $d(p,q)$ are all the potentially non-zero $2\times 2$ subdeterminants
of the matrix $B$ defining the bracket of $\mathcal{A}$. Then $\mathcal{A}$ is
$3$-solvable of class $2$.

$\mathcal{A}$ is $2$-solvable if and only if $d(1,4)=0$, this implies moreover
that there exists
$(\lambda,\lambda^{\prime})\in\mathbb{K}^{2}\setminus\\{(0,0)\\}$ such that
$\lambda d(2,4)+\lambda^{\prime}d(1,2)=0$ and $\lambda
d(3,4)+\lambda^{\prime}d(1,3)=0$, or equivalently that
$Rank\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}<2$ which is
equivalent to $\begin{vmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{vmatrix}=0$.

If $RankB=2$ or equivalently, there exists $1\leq p<q\leq 4$ such that
$d(p,q)\neq 0$, then

  1. 1)

$Z(\mathcal{A})=\\{0\\}$. This also means that $4_{3,N(2),6}$ is not
nilpotent.

  2. 2)

If $\mathcal{A}$ is $2$-solvable, then

    1. 2.a)

If $\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}\neq 0$, then
$\mathcal{A}$ is $2$-solvable of class $3$.

    2. 2.b)

If $\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}=0$ then,
$\mathcal{A}$ is $2$-solvable of class $2$.

If $RankB=1$ or equivalently $d(p,q)=0$, for all $1\leq p<q\leq 4$, then
$4_{3,N(2),6}$ is $2$-solvable of class $2$, and also $\dim Z(\mathcal{A})=1$,
and

$Z(\mathcal{A})=\langle\\{c(1,3,4,p)e_{2}-c(1,2,4,p)e_{3}\\}\rangle,$

where $c(1,2,4,p)\neq 0$ or $c(1,3,4,p)\neq 0$. Moreover, the algebra is
nilpotent if and only if $Z(\mathcal{A})=\left[A,A,A\right]$, or equivalently
if and only if $c(1,2,4,1)=c(1,2,4,4)=c(1,3,4,1)=c(1,3,4,4)=0$ and
$c(1,3,4,p)c(1,2,4,3)+c(1,2,4,p)c(1,2,4,2)=0$ and
$c(1,3,4,p)c(1,3,4,3)+c(1,2,4,p)c(1,3,4,2)=0$.

###### Proof

By Remark 2, we know that $4_{3,N(2),6}$ is $3$-solvable. The derived series
of $\mathcal{A}$ are given by

$\displaystyle D^{1}_{3}(\mathcal{A})$
$\displaystyle=\langle\\{c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4},$
$\displaystyle\quad
c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4}\\}\rangle,$

and
$D^{2}_{3}(\mathcal{A})=\left[D^{1}_{3}(\mathcal{A}),D^{1}_{3}(\mathcal{A}),D^{1}_{3}(\mathcal{A})\right]=\\{0\\}$
by skew-symmetry, since $\dim D^{1}_{3}(A)$ is less than $3$ (the arity). We
compute now the $2$-derived series,

$\displaystyle D^{1}_{2}(\mathcal{A})$
$\displaystyle=\langle\\{c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4},$
$\displaystyle\quad
c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}\\}\rangle$

We have $0\leq\dim D^{1}_{2}(A)\leq 2$. If $\dim D^{1}_{2}(A)=2$, then

$\displaystyle D^{2}_{2}(\mathcal{A})$
$\displaystyle=\langle\\{\left[e_{1},w_{2},w_{3}\right],\left[e_{2},w_{2},w_{3}\right],\left[e_{3},w_{2},w_{3}\right],\left[e_{4},w_{2},w_{3}\right]\\}\rangle$
(14)
$\displaystyle=\langle\\{(c(1,3,4,2)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,2))w_{3}$
$\displaystyle\hskip
85.35826pt-(c(1,3,4,3)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,3))w_{2},$
$\displaystyle\quad-(c(1,3,4,1)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,1))w_{3},$
$\displaystyle\quad-(c(1,3,4,1)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,1))w_{2},$
$\displaystyle\quad(c(1,3,4,1)c(1,2,4,2)-c(1,3,4,2)c(1,2,4,1))w_{3}$
$\displaystyle\hskip
76.82234pt-(c(1,3,4,1)c(1,2,4,3)-c(1,3,4,3)c(1,2,4,1))w_{2}\\}\rangle.$

If $\dim D^{2}_{2}(\mathcal{A})=2$, then
$D^{2}_{2}(\mathcal{A})=D^{1}_{2}(\mathcal{A})$ since
$D^{2}_{2}(\mathcal{A})\subseteq D^{1}_{2}(\mathcal{A})$ and has the same
dimension. We conclude in this case that $A$ is not $2$-solvable.

If $\dim D^{2}_{2}(\mathcal{A})=1$, then
$D^{2}_{2}(\mathcal{A})=\langle\\{v\\}\rangle$ with $v\in A,v\neq 0$. In this
case, $D^{3}_{2}(\mathcal{A})=\langle\\{[e_{i},v,v],1\leq i\leq 4\\}\rangle$,
that is $D^{3}_{2}(\mathcal{A})=\\{0\\}$ and $A$ is $2$-solvable of class $3$.
This occurs if and only if the rank of the family of generators of
$D_{2}^{2}(\mathcal{A})$ listed in (14) is $1$, that is if and only if, for
some $\lambda,\lambda^{\prime}\in\mathbb{K}$,

$\displaystyle(c(1,3,4,1)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,1))=0,$
$\displaystyle\lambda$
$\displaystyle(c(1,3,4,2)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,2))$
$\displaystyle+\lambda^{\prime}(c(1,3,4,1)c(1,2,4,2)-c(1,3,4,2)c(1,2,4,1))=0,$
$\displaystyle\lambda$
$\displaystyle(c(1,3,4,3)c(1,2,4,4)-c(1,3,4,4)c(1,2,4,3))$
$\displaystyle+\lambda^{\prime}(c(1,3,4,1)c(1,2,4,3)-c(1,3,4,3)c(1,2,4,1))=0.$

On the other hand, we have that

$\displaystyle\det\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}$
$\displaystyle=(c(1,2,4,3)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,3))\times$
$\displaystyle\qquad\times(c(1,2,4,4)c(1,3,4,1)-c(1,2,4,1)c(1,3,4,4)$
$\displaystyle=d(2,3)d(1,4),$

which means that $\det\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}=0$ if and only if $d(2,3)=0$ or $d(1,4)=0$. This
means also that the condition $\det\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}=0$ and $d(1,4)=0$ is equivalent to only saying that
$d(1,4)=0$.

The coefficients appearing in the generators of $D^{2}_{2}(A)$ in (14) are the
entries of the matrix $\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}$, that is $D^{2}_{2}(A)=\\{0\\}\ \text{if and only
if}\ \begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}=0.$

If $\dim D^{1}_{2}(A)=1$, then all the coefficients appearing in the
generators of $D^{2}_{2}(A)$ are zero, since they are $2\times 2$
subdeterminants of the matrix $B$ which is of rank $1$. This means that
$D^{2}_{2}(A)=\\{0\\}$ and $A$ is $2$-solvable of class $2$.

We know that an $(n+1)$-dimensional $n$-Hom-Lie algebra is nilpotent and non-
abelian, if and only if $[A,\dots,A]=Z(A)$ and $\dim Z(A)=1$ (See
cln1dimnHomLieniltw  Proposition 9). Therefore, if $\dim[A,\dots,A]=2$, $A$
cannot be nilpotent. In this case
$C_{k}^{r}(A)=\langle\\{w_{2},w_{3}\\}\rangle$, for all $r\geq 1$. Consider
now the center of $\mathcal{A}$,

$\displaystyle Z(A)$ $\displaystyle=\\{z=\sum_{k=1}^{4}z_{k}e_{k}\mid\forall\
x,y\in A,\left[x,y,z\right]=0\\}$
$\displaystyle=\\{z=\sum_{k=1}^{4}z_{k}e_{k}\mid\forall\ 1\leq i<j\leq
4,\left[e_{i},e_{j},z\right]=0\\}$

and we get the following system of equations

$\displaystyle c(1,2,4,1)z_{1}=0;\ c(1,2,4,2)z_{1}=0;\ c(1,2,4,3)z_{1}=0;\
c(1,2,4,4)z_{1}=0;\ $ $\displaystyle c(1,3,4,1)z_{1}=0;\ c(1,3,4,2)z_{1}=0;\
c(1,3,4,3)z_{1}=0;\ c(1,3,4,4)z_{1}=0;\ $ $\displaystyle
c(1,2,4,1)z_{2}+c(1,3,4,1)z_{3}=0;\ c(1,2,4,2)z_{2}+c(1,3,4,2)z_{3}=0;\ $
$\displaystyle c(1,2,4,3)z_{2}+c(1,3,4,3)z_{3}=0;\
c(1,2,4,4)z_{2}+c(1,3,4,4)z_{3}=0;\ $ $\displaystyle c(1,2,4,1)z_{4}=0;\
c(1,2,4,2)z_{4}=0;\ c(1,2,4,3)z_{4}=0;\ c(1,2,4,4)z_{4}=0;\ $ $\displaystyle
c(1,3,4,1)z_{4}=0;\ c(1,3,4,2)z_{4}=0;\ c(1,3,4,3)z_{4}=0;\
c(1,3,4,4)z_{4}=0.$

Then we get, $z_{1}\neq 0$ or $z_{4}\neq 0$ if and only if the algebra is
abelian, that is $c(1,2,4,i)=c(1,3,4,i)=0$, for all $1\leq i\leq 4$. Excluding
this case, we get the following system

$\displaystyle c(1,2,4,1)z_{2}+c(1,3,4,1)z_{3}=0;\
c(1,2,4,2)z_{2}+c(1,3,4,2)z_{3}=0,$ $\displaystyle
c(1,2,4,3)z_{2}+c(1,3,4,3)z_{3}=0;\ c(1,2,4,4)z_{2}+c(1,3,4,4)z_{3}=0.$

which is equivalent to $z_{2}w_{3}+z_{3}w_{2}=0$. Therefore $\dim
Z(\mathcal{A})=1$ if and only if
$RankB=\dim\langle\\{w_{2},w_{3}\\}\rangle=1$. In this case,

$\displaystyle Z(A)$ $\displaystyle=\\{z=\sum_{k=1}^{4}z_{k}e_{k}\in
A:z_{1}=z_{4}=0\text{ and }c(1,2,4,p)z_{2}+c(1,3,4,p)z_{3}=0\\}$
$\displaystyle=\\{z_{2}e_{2}-\frac{z_{2}c(1,2,4,p)}{c(1,3,4,p)}e_{3}:z_{2}\in\mathbb{K}\\}$
$\displaystyle=\\{z_{2}(c(1,3,4,p)e_{2}-c(1,2,4,p)e_{3}):z_{2}\in\mathbb{K}\\}$

if there exists $1\leq p\leq 4$ such that $c(1,3,4,p)\neq 0$, and

$\displaystyle Z(A)$ $\displaystyle=\\{z=\sum_{k=1}^{4}z_{k}e_{k}\in
A:z_{1}=z_{4}=0\text{ and }c(1,2,4,p)z_{2}+c(1,3,4,p)z_{3}=0\\}$
$\displaystyle=\\{-z_{3}\frac{c(1,3,4,p)}{c(1,2,4,p)}e_{2}+z_{3}e_{3}:z_{3}\in\mathbb{K}\\}$
$\displaystyle=\\{z_{3}(c(1,3,4,p)e_{2}-c(1,2,4,p)e_{3}):z_{3}\in\mathbb{K}\\}$
$\displaystyle=\\{z_{3}e_{3}:z_{3}\in\mathbb{K}\\}$

otherwise. By Proposition 5, $\mathcal{A}$ is nilpotent if and only if
$Z(\mathcal{A})=\left[A,A,A\right]$, since $\dim Z(\mathcal{A})=1$. Now, we
prove that this is equivalent to

$\displaystyle c(1,2,4,1)=c(1,2,4,4)=c(1,3,4,1)=c(1,3,4,4)$ $\displaystyle=0,$
$\displaystyle c(1,3,4,p)c(1,2,4,3)+c(1,2,4,p)c(1,2,4,2)$ $\displaystyle=0,$
(15) $\displaystyle c(1,3,4,p)c(1,3,4,3)+c(1,2,4,p)c(1,3,4,2)$
$\displaystyle=0.$

$Z(\mathcal{A})=[A,A,A]$ if and only if
$\dim\langle\\{w_{2},w_{3},c(1,3,4,p)e_{2}-c(1,2,4,p)e_{3}\\}\rangle=1$, which
is equivalent to $Rank\begin{pmatrix}c(1,3,4,1)&-c(1,2,4,1)&0\\\
c(1,3,4,2)&-c(1,2,4,2)&c(1,3,4,p)\\\ c(1,3,4,3)&-c(1,2,4,3)&-c(1,2,4,p)\\\
c(1,3,4,4)&-c(1,2,4,4)&0\\\ \end{pmatrix}=1$, that is all the $2\times 2$
minors of this matrix are zero, which gives the system (15). ∎

###### Corollary 5

The class of $3$-Hom-Lie algebras $4_{3,N(2),6}$ with $B\neq 0$ can be split
into five non-isomorphic subclasses:

  1. 1)

$3$-solvable of class $2$, non-$2$-solvable, non-nilpotent, with trivial
center:

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

with $d(1,4)\neq 0$, in that case we have $Rank\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}=2$.

  2. 2)

$3$-solvable of class $2$, $2$-solvable of class $3$, non-nilpotent, with
trivial center:

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=\lambda
c(1,2,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+\lambda c(1,2,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

with $(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$ or

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

such that $Rank\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}=1$.

  3. 3)

$3$-solvable of class $2$, $2$-solvable of class $2$, non-nilpotent, with
trivial center:

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}\\\
\left[e_{1},e_{3},e_{4}\right]&=c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad\text{with}\ d(2,3)\neq
0.$

  4. 4)

$3$-solvable of class $2$, $2$-solvable of class $2$, non-nilpotent, with
$1$-dimensional center:

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=\lambda
c(1,2,4,1)e_{1}+\lambda c(1,2,4,2)e_{2}+\lambda c(1,2,4,3)e_{3}$
$\displaystyle\hskip 113.81102pt+\lambda c(1,2,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

with $\left[e_{1},e_{2},e_{4}\right]\neq 0$ (that is not all $c(1,2,4,1)$,
$c(1,2,4,2)$, $c(1,2,4,3)$, $c(1,2,4,4)$ are zero), or

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

  5. 5)

$3$-solvable of class $2$, $2$-solvable of class $2$, nilpotent of class $2$,
with $1$-dimensional center:

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}\\\
\left[e_{1},e_{3},e_{4}\right]&=\frac{-c(1,2,4,2)^{2}}{c(1,2,4,3)}e_{2}-c(1,2,4,2)e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\\\ \end{array}\quad,\quad c(1,2,4,3)\neq 0$
or $\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c(1,2,4,2)e_{2}+\frac{-c(1,2,4,2)^{2}}{c(1,3,4,2)}e_{3}\\\
\left[e_{1},e_{3},e_{4}\right]&=c(1,3,4,2)e_{2}-c(1,2,4,2)e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\\\ \end{array}\quad,\quad c(1,3,4,2)\neq 0$

###### Remark 6

In the last case above, either $c(1,3,4,2)\neq 0$ or $c(1,2,4,3)\neq 0$, if
both are zero, then the bracket is zero.

###### Corollary 6

In the subclasses presented in Corollary 5, cases 1 and 3 cannot be
multiplicative. All the multiplicative $3$-Hom-Lie algebras in the considered
class are contained in the remaining subclasses:

  1. 2m)

$3$-solvable of class $2$, $2$-solvable of class $3$, non-nilpotent, with
trivial center

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

with $d(1,2)=c(1,2,4,1)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,1)\neq 0$.

  2. 4m)

$3$-solvable of class $2$, $2$-solvable of class $2$, non-nilpotent, with
$1$-dimensional center

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=\lambda
c(1,2,4,1)e_{1}+\lambda c(1,2,4,2)e_{2}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0$

  3. 5m)

$3$-solvable of class $2$, $2$-solvable of class $2$, nilpotent of class $2$,
with $1$-dimensional center

$\begin{array}[]{lll}\left[e_{1},e_{2},e_{3}\right]&=&0\\\
\left[e_{1},e_{2},e_{4}\right]&=&0\\\
\left[e_{1},e_{3},e_{4}\right]&=&c(1,3,4,2)e_{2}\\\
\left[e_{2},e_{3},e_{4}\right]&=&0\end{array}\quad,\quad c(1,3,4,2)\neq 0.$

##  5 Isomorphism classes for $4_{3,N(2),6}$

The following theorem gives the classification up to isomorphism of the class
of $3$-Hom-Lie algebras $4_{3,N(2),6}$. Note that isomorphisms are considered
in the sense of Hom-algebras, that is they are required to intertwine not only
the multiplications, but also the twisting maps.

###### Theorem 5.1

Any $3$-Hom-Lie algebra $\mathcal{A}$ in the class of $3$-Hom-Lie algebras
$4_{3,N(2),6}$ with $B\neq 0$ is isomorphic to one of the following:

  1. 1)

$\dim D^{1}_{3}(\mathcal{A})=2$, non-$2$-solvable, non-nilpotent, with trivial
center:

    1. 1.a)

$c(1,2,4,4)\neq 0$.

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c^{\prime}(1,3,4,1)e_{1}+c^{\prime}(1,3,4,3)e_{3}+c^{\prime}(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$
$\displaystyle c^{\prime}(1,3,4,1)$
$\displaystyle=\frac{-d(1,4)}{c(1,2,4,4)}\neq 0,\quad
c^{\prime}(1,3,4,3)=\frac{-d(3,4)}{c(1,2,4,4)^{2}},$ $\displaystyle
c^{\prime}(1,3,4,4)$ $\displaystyle=\frac{c(1,2,4,3)+c(1,3,4,4)}{c(1,2,4,4)}.$

Two such algebras, given by the structure constants $(c^{\prime}(i,j,k,p))$
and $(c^{\prime\prime}(i,j,k,p))$ respectively, are isomorphic if and only if
$c^{\prime}(1,3,4,3)=c^{\prime\prime}(1,3,4,3)$,
$c^{\prime}(1,3,4,4)=c^{\prime\prime}(1,3,4,4)$ and
$\frac{c^{\prime}(1,3,4,1)}{c^{\prime\prime}(1,3,4,1)}$ is a square in
$\mathbb{K}$. Thus, this family of algebras up to isomorphism is parametrized
by
$\frac{\mathbb{K}^{*}}{(\mathbb{K}^{*})^{2}}\times\mathbb{K}\times\mathbb{K}$,
where$\frac{\mathbb{K}^{*}}{(\mathbb{K}^{*})^{2}}$ is the factor group of
$\mathbb{K}^{*}$ by $(\mathbb{K}^{*})^{2}=\\{x^{2}|x\in\mathbb{K}^{*}\\}$.

    2. 1.b)

$c(1,2,4,4)=0$, $c(1,2,4,3)\neq 0$ and $c(1,2,4,3)\neq c(1,3,4,4)$. In this
case $c(1,2,4,1)$ and $c(1,3,4,4)$ are non-zero since $d(1,4)\neq 0$.

$\displaystyle\begin{array}[]{lll}\left[e_{1},e_{2},e_{3}\right]&=&0\\\
\left[e_{1},e_{2},e_{4}\right]&=&c^{\prime}(1,2,4,1)e_{1}+e_{3}\\\
\left[e_{1},e_{3},e_{4}\right]&=&c^{\prime}(1,3,4,4)e_{4}\\\
\left[e_{2},e_{3},e_{4}\right]&=&0,\end{array}\quad,\quad\begin{array}[c]{ll}c^{\prime}(1,2,4,1)&=c(1,2,4,1)\neq
0,\\\ c^{\prime}(1,3,4,4)&=\frac{c(1,3,4,4)}{c(1,2,4,3)}\neq 0.\end{array}$

Two such algebras, given by the structure constants $(c^{\prime}(i,j,k,p))$
and $(c^{\prime\prime}(i,j,k,p))$ respectively, are isomorphic if and only if
$\frac{c^{\prime}(1,2,4,1)}{c^{\prime\prime}(1,2,4,1)}$ is a square in
$\mathbb{K}$.

    3. 1.c)

$c(1,2,4,4)=0$, $c(1,2,4,3)\neq 0$ and $c(1,2,4,3)=c(1,3,4,4)$. In this case
also $c(1,2,4,1)$ and $c(1,3,4,4)$ are non-zero since $d(1,4)\neq 0$.

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c^{\prime}(1,3,4,3)e_{3}+e_{4}\\\
\left[e_{1},e_{3},e_{4}\right]&=c^{\prime}(1,2,4,1)e_{1}+e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0,\end{array}\quad,\quad\begin{array}[c]{ll}c^{\prime}(1,3,4,3)&=\frac{c(1,3,4,4)}{c(1,2,4,3)}\neq
0,\\\ c^{\prime}(1,2,4,1)&=c(1,2,4,1)\neq 0.\end{array}$

Two such algebras, given by the structure constants $(c^{\prime}(i,j,k,p))$
and $(c^{\prime\prime}(i,j,k,p))$ respectively are isomorphic if and only if
$\frac{c^{\prime}(1,2,4,1)}{c^{\prime\prime}(1,2,4,1)}$ is a square in
$\mathbb{K}$.

    4. 1.d)

$c(1,2,4,4)=0$ and $c(1,2,4,3)=0$. Similarly, in this case $c(1,2,4,1)$ and
$c(1,3,4,4)$ are non-zero since $d(1,4)\neq 0$.

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c^{\prime}(1,2,4,1)e_{1}\\\
\left[e_{1},e_{3},e_{4}\right]&=e_{4}\\\
\left[e_{2},e_{3},e_{4}\right]&=0,\end{array}\quad,\quad
c^{\prime}(1,2,4,1)=c(1,2,4,1).$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ are isomorphic if and only if
$\frac{c^{\prime}(1,2,4,1)}{c^{\prime\prime}(1,2,4,1)}$ is a square in
$\mathbb{K}$. In particular, If $c(1,2,4,1)$ is a square in $\mathbb{K}$, we
get the following bracket

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{1}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0.$

  2. 2)

$\dim D^{1}_{3}(\mathcal{A})=2$, $2$-solvable of class $3$, non-nilpotent,
with trivial center, that is $d(1,4)=0$, $\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}\neq 0$, equivalent to
$(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$ and
$(c(1,3,4,1),c(1,3,4,4))=\lambda(c(1,2,4,1),c(1,2,4,4))$ for some
$\lambda\in\mathbb{K}$ or $(c(1,2,4,1),c(1,2,4,4))=(0,0)$ and
$(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$:

    1. 2.a)

$c(1,2,4,4)\neq 0$, hence $(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c^{\prime}(1,3,4,2)e_{2}+c^{\prime}(1,3,4,3)e_{3}+c^{\prime}(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$
$\displaystyle c^{\prime}(1,3,4,2)$ $\displaystyle=\frac{\lambda
c(1,2,4,3)^{2}-\lambda c(1,2,4,2)c(1,2,4,4)}{c(1,2,4,4)^{2}}$
$\displaystyle\quad+\frac{-c(1,3,4,3)c(1,2,4,3)+c(1,2,4,4)c(1,3,4,2)}{c(1,2,4,4)^{2}},$
$\displaystyle c^{\prime}(1,3,4,3)$ $\displaystyle=\frac{c(1,3,4,3)-\lambda
c(1,2,4,3)}{c(1,2,4,4)},$ $\displaystyle c^{\prime}(1,3,4,4)$
$\displaystyle=\frac{\lambda c(1,2,4,4)+c(1,2,4,3)}{c(1,2,4,4)}.$

Any two different brackets of this form give non-isomorphic $3$-Hom-Lie
algebras.

    2. 2.b)

$(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$ and $c(1,2,4,4)=0$, which means that
$c(1,2,4,1)\neq 0$ (else the algebra would be $2$-solvable of class $2$). For
$c(1,2,4,3)\neq 0$,

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c^{\prime}(1,2,4,1)e_{1}+e_{3}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=\lambda^{\prime}c^{\prime}(1,2,4,1)e_{1}+c^{\prime}(1,3,4,2)e_{2}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$
$\displaystyle c^{\prime}(1,2,4,1)$ $\displaystyle=c(1,2,4,1)\neq
0,\quad\lambda^{\prime}=\frac{\lambda c(1,2,4,3)-c(1,3,4,3)}{c(1,2,4,3)},$
$\displaystyle c^{\prime}(1,3,4,2)$ $\displaystyle=\frac{-\lambda
c(1,2,4,3)c(1,3,4,3)-\lambda c(1,2,4,2)c(1,2,4,3)}{c(1,2,4,3)^{2}}$
$\displaystyle\hskip
56.9055pt+\frac{c(1,3,4,3)^{2}+c(1,2,4,3)c(1,3,4,2)}{c(1,2,4,3)^{2}}.$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$\frac{c^{\prime}(1,2,4,1)}{c^{\prime\prime}(1,2,4,1)}$ is a square in
$\mathbb{K}$.

    3. 2.c)

$(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$ and $c(1,2,4,4)=0$, which means that
$c(1,2,4,1)\neq 0$ (else the algebra would be $2$-solvable of class $2$). For
$c(1,2,4,3)=0$ and $c(1,3,4,3)\neq 0$,

$\displaystyle\begin{array}[]{l}\left[e_{1},e_{2},e_{3}\right]=0\\\
\left[e_{1},e_{2},e_{4}\right]=c^{\prime}(1,2,4,1)e_{1}\\\
\left[e_{1},e_{3},e_{4}\right]=\lambda^{\prime}c^{\prime}(1,2,4,1)e_{1}+e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]=0,\end{array}$ $\displaystyle\lambda^{\prime}$
$\displaystyle=$ $\displaystyle\frac{-\lambda c(1,2,4,2)+\lambda
c(1,3,4,3)+c(1,3,4,2)}{c(1,3,4,3)},$ $\displaystyle
c^{\prime}(1,2,4,1)=c(1,2,4,1).$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ are isomorphic if and only if
$\frac{c^{\prime}(1,2,4,1)}{c^{\prime\prime}(1,2,4,1)}$ is a square in
$\mathbb{K}$.

    4. 2.d)

$(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$ and $c(1,2,4,4)=0$, which means that
$c(1,2,4,1)\neq 0$ (else the algebra would be $2$-solvable of class $2$). We
consider $c(1,2,4,3)=0$ and $c(1,3,4,3)=0$. In this case, the algebra is
multiplicative.

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c^{\prime}(1,2,4,1)e_{1}\\\
\left[e_{1},e_{3},e_{4}\right]&=e_{2}\\\
\left[e_{2},e_{3},e_{4}\right]&=0,\end{array}\quad,\quad c^{\prime}(1,2,4,1)$
$\displaystyle=c(1,2,4,1)\neq 0.$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$\frac{c^{\prime}(1,2,4,1)}{c^{\prime\prime}(1,2,4,1)}$ is a square in
$\mathbb{K}$.

    5. 2.e)

$(c(1,2,4,1),c(1,2,4,4))=(0,0)$ and $(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$,
$c(1,2,4,3)\neq 0$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{3}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c^{\prime}(1,3,4,3)e_{3}+c^{\prime}(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$
$\displaystyle c^{\prime}(1,3,4,3)$
$\displaystyle=\frac{c(1,2,4,2)c(1,2,4,3)+c(1,3,4,3)c(1,2,4,3)}{c(1,2,4,3)^{2}}$
$\displaystyle\quad-\frac{c(1,2,4,2)c(1,3,4,4)}{c(1,2,4,3)^{2}},$
$\displaystyle c^{\prime}(1,3,4,4)$
$\displaystyle=\frac{c(1,3,4,4)}{c(1,2,4,3)}.$

Any two different brackets of this form give non-isomorphic $3$-Hom-Lie
algebras.

    6. 2.f)

$(c(1,2,4,1),c(1,2,4,4))=(0,0)$ and $(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$,
$c(1,2,4,3)=0$, $c(1,3,4,4)\neq 0$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c^{\prime}(1,2,4,2)e_{2}\\\
\left[e_{1},e_{3},e_{4}\right]&=e_{4}\\\
\left[e_{2},e_{3},e_{4}\right]&=0,\end{array}\quad,\quad c^{\prime}(1,2,4,2)$
$\displaystyle=\frac{c(1,2,4,2)}{c(1,3,4,4)}.$

    7. 2.g)

$(c(1,2,4,1),c(1,2,4,4))=(0,0)$ and $(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$,
$c(1,2,4,3)=0$, $c(1,3,4,4)=0$ and $c(1,3,4,3)\neq 0$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c^{\prime}(1,2,4,2)e_{2}\\\
\left[e_{1},e_{3},e_{4}\right]&=c^{\prime}(1,3,4,1)e_{1}+e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0,\end{array}\quad,\quad\begin{array}[]{ll}c^{\prime}(1,2,4,2)&=\frac{c(1,2,4,2)}{c(1,3,4,3)},\\\
c^{\prime}(1,3,4,1)&=c(1,3,4,1).\end{array}$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$c^{\prime}(1,2,4,2)=c^{\prime\prime}(1,2,4,2)$ and
$\frac{c^{\prime}(1,3,4,1)}{c^{\prime\prime}(1,3,4,1)}$ is a square in
$\mathbb{K}$.

    8. 2.h)

$(c(1,2,4,1),c(1,2,4,4))=(0,0)$ and $(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$,
$c(1,2,4,3)=0$, $c(1,3,4,4)=0$ and $c(1,3,4,3)=0$. This algebra is
multiplicative,

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=e_{2}\\\
\left[e_{1},e_{3},e_{4}\right]&=c^{\prime}(1,3,4,1)e_{1}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad
c^{\prime}(1,3,4,1)=c(1,3,4,1)\neq 0.$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$\frac{c^{\prime}(1,3,4,1)}{c^{\prime\prime}(1,3,4,1)}$ is a square in
$\mathbb{K}$.

  3. 3)

$\dim D^{1}_{3}(\mathcal{A})=2$, $2$-solvable of class $2$, non-nilpotent,
with trivial center:

    1. 3.a)

$c(1,2,4,3)\neq 0$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c^{\prime}(1,2,4,2)e_{2}+e_{3}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=c^{\prime}(1,3,4,2)e_{2}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$
$\displaystyle c^{\prime}(1,3,4,2)$
$\displaystyle=\frac{c(1,2,4,3)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,3)}{c(1,2,4,3)^{2}},$
$\displaystyle c^{\prime}(1,2,4,2)$
$\displaystyle=\frac{-c(1,2,4,2)-c(1,3,4,3)}{c(1,2,4,3)}.$

    2. 3.b)

$c(1,2,4,3)=0$, $c(1,3,4,3)\neq 0$ and $c(1,3,4,3)\neq c(1,2,4,2)$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=c^{\prime}(1,2,4,2)e_{2}\\\
\left[e_{1},e_{3},e_{4}\right]&=e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad c^{\prime}(1,2,4,2)$
$\displaystyle=\frac{c(1,2,4,2)}{c(1,3,4,3)}.$

    3. 3.c)

$c(1,2,4,3)=0$, $c(1,3,4,3)\neq 0$ and $c(1,3,4,3)=c(1,2,4,2)$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=e_{2}\\\
\left[e_{1},e_{3},e_{4}\right]&=c^{\prime}(1,3,4,2)e_{2}+e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad c^{\prime}(1,3,4,2)$
$\displaystyle=\frac{c(1,3,4,2)}{c(1,3,4,3)}.$

  4. 4)

$\dim D^{1}_{3}(\mathcal{A})=1$, $2$-solvable of class $2$, non-nilpotent,
with $1$-dimensional center:

    1. 4.a)

$w_{3}\neq 0$, $w_{2}=\lambda w_{3}$ with $\lambda\in\mathbb{K}$ and
$c(1,2,4,4)\neq 0$

$\begin{array}[]{lll}\left[e_{1},e_{2},e_{3}\right]&=&0\\\
\left[e_{1},e_{2},e_{4}\right]&=&e_{4}\\\
\left[e_{1},e_{3},e_{4}\right]&=&\lambda^{\prime}e_{4}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad\lambda^{\prime}=\frac{c(1,2,4,3)}{c(1,2,4,4)}+\lambda.$

Two such brackets with parameters $\lambda^{\prime}$ and
$\lambda^{\prime\prime}$ define isomorphic $3$-Hom-Lie algebras if and only if
$\lambda^{\prime}=\lambda^{\prime\prime}.$

    2. 4.b)

$w_{3}\neq 0$, $w_{2}=\lambda w_{3}$ with $\lambda\in\mathbb{K}$ and
$c(1,2,4,4)=0$, $c(1,2,4,3)\neq 0$, $c(1,2,4,1)\neq 0$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=e_{1}+c^{\prime}(1,2,4,3)e_{3}\\\
\left[e_{1},e_{3},e_{4}\right]&=0\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad c^{\prime}(1,2,4,3)$
$\displaystyle=c(1,2,4,3)\neq 0.$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$\frac{c^{\prime}(1,2,4,3)}{c^{\prime\prime}(1,2,4,3)}$ is a square in
$\mathbb{K}$.

    3. 4.c)

$w_{3}\neq 0$, $w_{2}=\lambda w_{3}$ with $\lambda\in\mathbb{K}$ and
$c(1,2,4,4)=0$, $c(1,2,4,3)\neq 0$, $c(1,2,4,1)=0$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{3}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=\lambda^{\prime}e_{3}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$

    4. 4.d)

$w_{3}\neq 0$, $w_{2}=\lambda w_{3}$ with $\lambda\in\mathbb{K}$ and
$c(1,2,4,4)=0$, $c(1,2,4,3)=0$, $c(1,2,4,1)\neq 0$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=c^{\prime}(1,2,4,1)e_{1}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$

    5. 4.e)

$w_{3}\neq 0$, $w_{2}=\lambda w_{3}$ with $\lambda\in\mathbb{K}$ and
$c(1,2,4,4)=0$, $c(1,2,4,3)=0$, $c(1,2,4,1)=0$, $c(1,2,4,2)\neq 0$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{2}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$

    6. 4.f)

$w_{3}=0$, $c(1,3,4,4)\neq 0$

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$ $\displaystyle=e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$

    7. 4.g)

$w_{3}=0$, $c(1,3,4,4)=0$, $c(1,3,4,1)\neq 0$, $c(1,3,4,3)\neq 0$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=0\\\
\left[e_{1},e_{3},e_{4}\right]&=c^{\prime}(1,3,4,1)e_{1}+e_{3}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad c^{\prime}(1,3,4,1)$
$\displaystyle=c(1,3,4,1).$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$\frac{c^{\prime}(1,3,4,1)}{c^{\prime\prime}(1,3,4,1)}$ is a square in
$\mathbb{K}$.

    8. 4.h)

$w_{3}=0$, $c(1,3,4,4)=0$, $c(1,3,4,1)\neq 0$, $c(1,3,4,3)=0$

$\displaystyle\begin{array}[]{ll}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=\\\
\left[e_{1},e_{3},e_{4}\right]&=c^{\prime}(1,3,4,1)e_{1}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad,\quad c^{\prime}(1,3,4,1)$
$\displaystyle=c(1,3,4,1)$

Two such brackets given by the structure constants $(c^{\prime}(i,j,k,p))$ and
$(c^{\prime\prime}(i,j,k,p))$ define isomorphic algebras if and only if
$\frac{c^{\prime}(1,3,4,1)}{c^{\prime\prime}(1,3,4,1)}$ is a square in
$\mathbb{K}$. This bracket defines a multiplicative algebra.

  5. 5)

$\dim D^{1}_{3}(\mathcal{A})=1$, $2$-solvable of class $2$, nilpotent of class
$2$, with $1$-dimensional center:

5.a)  $\begin{array}[t]{rl}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=0\\\ \left[e_{1},e_{3},e_{4}\right]&=e_{2}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}$ 5.b)
$\begin{array}[t]{rl}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=e_{3}\\\ \left[e_{1},e_{3},e_{4}\right]&=0\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}$

###### Proof

Let $\mathcal{A}=(A,\left[\cdot,\dots,\cdot\right],\alpha)$ be a $3$-Hom-Lie
algebra in one of the classes presented in Corollary 5 and consider the matrix
$B$ defining its bracket in a basis $(e_{i})$ where $\alpha$ is in its Jordan
normal form. Any $3$-Hom-Lie algebra isomorphic to $\mathcal{A}$ has its
bracket given by a matrix $B^{\prime}=\frac{1}{\det(P)}PBP^{T}$ where $P$ is
an invertible matrix that commutes with $[\alpha]$, the matrix representing
$\alpha$ in the basis $(e_{i})$. A matrix $P=(p(i,j))_{1\leq i,j\leq 4}$
commutes with $[\alpha]=\begin{pmatrix}0&0&0&0\\\ 0&0&1&0\\\ 0&0&0&1\\\
0&0&0&0\end{pmatrix}$ if and only if it is of the form
$P=\begin{pmatrix}p(1,1)&0&0&p(1,4)\\\ p(2,1)&p(3,3)&p(2,3)&p(2,4)\\\
0&0&p(3,3)&p(2,3)\\\ 0&0&0&p(3,3)\\\ \end{pmatrix},$ with $\det(P)\neq 0$ that
is $p(1,1)p(3,3)^{3}\neq 0$ which is equivalent to $p(1,1)\neq 0$ and
$p(3,3)\neq 0$. We denote by $c^{\prime}(i,j,k,p)$ the structure constants of
the bracket after the transformation by $P$.

In the following, in the matrix $B^{\prime}$, there appear structure constants
of the form $c^{\prime}(i,j,k,l)=\frac{c(i,j,k,l)}{p(1,1)p(3,3)}$ or
$\frac{c(i,j,k,l)}{p(3,3)^{2}}$. Note that, since $p(1,1)\neq 0$ and
$p(3,3)\neq 0$,

$\frac{c(i,j,k,l)}{p(1,1)p(3,3)}=0\text{ or
}\frac{c(i,j,k,l)}{p(3,3)^{2}}=0\iff c(i,j,k,l)=0,$ (17)

and thus in such a case, the algebras given by the bracket with $c(i,j,k,l)=0$
and the bracket with $c(i,j,k,l)\neq 0$ cannot be isomorphic.

  1. 1)

$\dim D^{1}_{3}(\mathcal{A})=2$, non-$2$-solvable, non-nilpotent, with trivial
center, that is

$B=\begin{pmatrix}0&c(1,3,4,1)&-c(1,2,4,1)&0\\\ 0&c(1,3,4,2)&-c(1,2,4,2)&0\\\
0&c(1,3,4,3)&-c(1,2,4,3)&0\\\ 0&c(1,3,4,4)&-c(1,2,4,4)&0\end{pmatrix},$

with $d(1,4)=c(1,2,4,1)c(1,3,4,4)-c(1,2,4,4)c(1,3,4,1)\neq 0$.

$\displaystyle B^{\prime}=\frac{1}{\det(P)}PBP^{T}=$
$\displaystyle\left(\begin{array}[]{cccc}0&b^{\prime}(1,2)&\frac{-c(1,2,4,1)p(1,1)-c(1,2,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0\\\
0&b^{\prime}(2,2)&b^{\prime}(2,3)&0\\\
0&b^{\prime}(3,2)&\frac{-c(1,2,4,4)p(2,3)-c(1,2,4,3)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,3,4,4)p(3,3)^{2}-c(1,2,4,4)p(2,3)p(3,3)}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,4)}{p(1,1)p(3,3)}&0\end{array}\right),$
$\displaystyle b^{\prime}(1,2)$
$\displaystyle=c^{\prime}(1,3,4,1)=\frac{p(2,3)(-c(1,2,4,1)p(1,1)-c(1,2,4,4)p(1,4))}{p(1,1)p(3,3)^{3}}$
$\displaystyle\hskip
56.9055pt+\frac{p(3,3)(c(1,3,4,1)p(1,1)+c(1,3,4,4)p(1,4))}{p(1,1)p(3,3)^{3}},$
$\displaystyle b^{\prime}(2,2)$ $\displaystyle=c^{\prime}(1,3,4,2)$
$\displaystyle=\frac{p(2,3)(-c(1,2,4,1)p(2,1)-c(1,2,4,3)p(2,3)-c(1,2,4,4)p(2,4)-c(1,2,4,2)p(3,3))}{p(1,1)p(3,3)^{3}}$
$\displaystyle\qquad+\frac{p(3,3)(c(1,3,4,1)p(2,1)+c(1,3,4,3)p(2,3)+c(1,3,4,4)p(2,4)+c(1,3,4,2)p(3,3))}{p(1,1)p(3,3)^{3}},$
$\displaystyle b^{\prime}(2,3)$
$\displaystyle=-c^{\prime}(1,2,4,2)=\frac{-c(1,2,4,1)p(2,1)-c(1,2,4,3)p(2,3)}{p(1,1)p(3,3)^{2}}$
$\displaystyle\hskip
59.75095pt+\frac{-c(1,2,4,4)p(2,4)-c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}},$
$\displaystyle b^{\prime}(3,2)$
$\displaystyle=c^{\prime}(1,3,4,3)=\frac{p(2,3)(-c(1,2,4,4)p(2,3)-c(1,2,4,3)p(3,3))}{p(1,1)p(3,3)^{3}}$
$\displaystyle\hskip
56.9055pt+\frac{p(3,3)(c(1,3,4,4)p(2,3)+c(1,3,4,3)p(3,3))}{p(1,1)p(3,3)^{3}}.$

and notice that $\frac{c(1,2,4,4)}{p(1,1)p(3,3)}=0$ if and only if
$c(1,2,4,4)=0$, therefore a bracket with $c(1,2,4,4)=0$ and a bracket with
$c(1,2,4,4)\neq 0$ cannot define isomorphic $3$-Hom-Lie algebras. If
$c(1,2,4,4)\neq 0$, then choosing

$\displaystyle P=P_{1,1}=$
$\displaystyle\begin{pmatrix}\frac{c(1,2,4,4)}{p(3,3)}&0&0&-\frac{c(1,2,4,1)}{p(3,3)}\\\
p(2,1)&p(3,3)&-\frac{c(1,2,4,3)p(3,3)}{c(1,2,4,4)}&p(2,4)\\\
0&0&p(3,3)&-\frac{c(1,2,4,3)p(3,3)}{c(1,2,4,4)}\\\ 0&0&0&p(3,3)\end{pmatrix}$
$\displaystyle p(2,1)$
$\displaystyle=-\frac{(-c(1,2,4,4)c(1,3,4,3)c(1,2,4,3)+c(1,3,4,4)c(1,2,4,3)^{2})p(3,3)}{-c(1,2,4,4)d(1,4)}$
$\displaystyle\qquad+\frac{(c(1,2,4,4)^{2}c(1,3,4,2)-c(1,2,4,2)c(1,2,4,4)c(1,3,4,4))p(3,3)}{-c(1,2,4,4)d(1,4)}$
$\displaystyle p(2,4)$
$\displaystyle=\frac{\big{(}c(1,2,4,1)c(1,2,4,4)c(1,3,4,2)-c(1,2,4,1)c(1,3,4,3)c(1,2,4,3)\big{)}p(3,3)}{-c(1,2,4,4)d(1,4)}$
$\displaystyle\qquad+\frac{\big{(}-c(1,2,4,2)c(1,2,4,4)c(1,3,4,1)+c(1,3,4,1)c(1,2,4,3)^{2}\big{)}p(3,3)}{-c(1,2,4,4)d(1,4)}$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,2,4,4)c(1,3,4,1)-c(1,2,4,1)c(1,3,4,4)}{c(1,2,4,4)p(3,3)^{2}}&0&0\\\
0&0&0&0\\\
0&\frac{c(1,2,4,4)c(1,3,4,3)-c(1,2,4,3)c(1,3,4,4)}{c(1,2,4,4)^{2}}&0&0\\\
0&\frac{c(1,2,4,3)+c(1,3,4,4)}{c(1,2,4,4)}&-1&0\end{array}\right).$ If
$c(1,2,4,4)=0$, then

$B^{\prime}=\begin{pmatrix}0&b^{\prime}(1,2)&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&b^{\prime}(2,2)&b^{\prime}(2,3)&0\\\
0&b^{\prime}(3,2)&-\frac{c(1,2,4,3)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,3,4,4)}{p(1,1)p(3,3)}&0&0\end{pmatrix}$ $\displaystyle
b^{\prime}(1,2)=c^{\prime}(1,3,4,1)=\frac{p(3,3)(c(1,3,4,1)p(1,1)+c(1,3,4,4)p(1,4))-c(1,2,4,1)p(1,1)p(2,3)}{p(1,1)p(3,3)^{3}}$
$\displaystyle
b^{\prime}(2,2)=c^{\prime}(1,3,4,2)=\frac{p(2,3)(-c(1,2,4,1)p(2,1)-c(1,2,4,3)p(2,3)-c(1,2,4,2)p(3,3))}{p(1,1)p(3,3)^{3}}$
$\displaystyle\hskip
56.9055pt+\frac{p(3,3)(c(1,3,4,1)p(2,1)+c(1,3,4,3)p(2,3)+c(1,3,4,4)p(2,4)+c(1,3,4,2)p(3,3))}{p(1,1)p(3,3)^{3}}$
$\displaystyle
b^{\prime}(2,3)=-c^{\prime}(1,2,4,2)=\frac{-c(1,2,4,1)p(2,1)-c(1,2,4,3)p(2,3)-c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}$
$\displaystyle
b^{\prime}(3,2)=c^{\prime}(1,3,4,3)=\frac{p(3,3)(c(1,3,4,4)p(2,3)+c(1,3,4,3)p(3,3))-c(1,2,4,3)p(2,3)p(3,3)}{p(1,1)p(3,3)^{3}}.$

Using the same argument, consider the cases where each of the structure
constants $c(1,3,4,4)$, $c(1,2,4,3)$ and $c(1,2,4,1)$ are zero or non-zero.

If $c(1,3,4,4)=0$, then $d(1,4)=0$ and

${\scriptsize\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}}=\\\
{\scriptsize\left(\begin{array}[]{cc}0,&0\\\
c(1,2,4,1)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,1),&c(1,2,4,1)c(1,3,4,3)-c(1,2,4,3)c(1,3,4,1)\end{array}\right)}$

has rank less than or equal to $1$, which means that the algebra is
$2$-solvable.

If $c(1,2,4,1)=0$, then $d(1,4)=0$ and

${\scriptsize\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}=\begin{pmatrix}c(1,2,4,2)c(1,3,4,4),&c(1,2,4,3)c(1,3,4,4)\\\
-c(1,2,4,2)c(1,3,4,1),&-c(1,2,4,3)c(1,3,4,1)\end{pmatrix},}$

has also rank less than or equal to $1$, which means that the algebra is
$2$-solvable.

If $c(1,2,4,3)=0$, $c(1,3,4,4)\neq 0$ and $c(1,2,4,1)\neq 0$ then $d(1,4)\neq
0$ and

${\scriptsize\begin{pmatrix}d(2,4)&d(3,4)\\\
d(1,2)&d(1,3)\end{pmatrix}=\begin{pmatrix}c(1,2,4,2)c(1,3,4,4),&0\\\
c(1,2,4,1)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,1),&c(1,2,4,1)c(1,3,4,3)\end{pmatrix}}$

has rank $2$ if and only if determinant
$c(1,2,4,1)c(1,2,4,2)c(1,3,4,3)c(1,3,4,4)\neq 0$, that is, if and only if,
$c(1,2,4,2)\neq 0$ and $c(1,3,4,3)\neq 0$.

If $c(1,2,4,4)=0$, $c(1,2,4,1)\neq 0$, $c(1,3,4,4)\neq 0$, $c(1,2,4,3)\neq 0$,
$c(1,2,4,3)\neq c(1,3,4,4)$, then choosing

$\displaystyle P=P_{1,2}=$
$\displaystyle\begin{pmatrix}\frac{c(1,2,4,3)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&\frac{c(1,3,4,3)p(3,3)}{c(1,2,4,3)-c(1,3,4,4)}&p(2,4)\\\
0&0&p(3,3)&\frac{c(1,3,4,3)p(3,3)}{c(1,2,4,3)-c(1,3,4,4)}\\\
0&0&0&p(3,3)\end{pmatrix}$ $\displaystyle
p(1,4)=-\frac{c(1,2,4,3)\big{(}c(1,2,4,3)c(1,3,4,1)-c(1,3,4,4)c(1,3,4,1)-c(1,2,4,1)c(1,3,4,3)\big{)}}{(c(1,2,4,3)-c(1,3,4,4))c(1,3,4,4)p(3,3)}$
$\displaystyle
p(2,1)=-\frac{(c(1,2,4,2)c(1,2,4,3)+c(1,3,4,3)c(1,2,4,3)-c(1,2,4,2)c(1,3,4,4))p(3,3)}{c(1,2,4,1)(c(1,2,4,3)-c(1,3,4,4))}$
$\displaystyle
p(2,4)=\frac{\big{(}-c(1,2,4,1)c(1,3,4,3)^{2}+c(1,2,4,3)c(1,3,4,1)c(1,3,4,3)+c(1,2,4,2)c(1,2,4,3)c(1,3,4,1)\big{)}p(3,3)}{c(1,2,4,1)(c(1,2,4,3)-c(1,3,4,4))c(1,3,4,4)}$
$\displaystyle+\frac{\big{(}-c(1,2,4,1)c(1,2,4,3)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,1)c(1,3,4,4)+c(1,2,4,1)c(1,3,4,2)c(1,3,4,4)\big{)}p(3,3)}{c(1,2,4,1)(c(1,2,4,3)-c(1,3,4,4))c(1,3,4,4)}$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&0&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&0&-1&0\\\ 0&\frac{c(1,3,4,4)}{c(1,2,4,3)}&0&0\end{array}\right).$

If $c(1,2,4,4)=0$, $c(1,2,4,1)\neq 0$, $c(1,3,4,4)\neq 0$, $c(1,2,4,3)\neq 0$,
$c(1,2,4,3)=c(1,3,4,4)$, then choosing

$\displaystyle P=P_{1,3}=$
$\displaystyle\left(\begin{array}[]{cccc}\frac{c(1,3,4,4)}{p(3,3)}&0&0&\frac{c(1,2,4,1)p(2,3)-c(1,3,4,1)p(3,3)}{p(3,3)^{2}}\\\
\frac{-c(1,3,4,4)p(2,3)-c(1,2,4,2)p(3,3)}{c(1,2,4,1)}&p(3,3)&p(2,3)&p(2,4)\\\
0&0&p(3,3)&p(2,3)\\\ 0&0&0&p(3,3)\end{array}\right)$ $\displaystyle p(2,4)=$
$\displaystyle\frac{-c(1,2,4,1)c(1,3,4,3)p(2,3)+c(1,3,4,1)c(1,3,4,4)p(2,3)}{c(1,2,4,1)c(1,3,4,4)}$
$\displaystyle+\frac{c(1,2,4,2)c(1,3,4,1)p(3,3)-c(1,2,4,1)c(1,3,4,2)p(3,3)}{c(1,2,4,1)c(1,3,4,4)}$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&0&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&\frac{c(1,3,4,3)}{c(1,3,4,4)}&-1&0\\\ 0&1&0&0\end{array}\right).$

If $c(1,2,4,4)=0$, $c(1,2,4,1)\neq 0$, $c(1,3,4,4)\neq 0$ and $c(1,2,4,3)=0$
then choosing

$\displaystyle P=P_{1,4}=$
$\displaystyle\left(\begin{array}[]{cccc}\frac{c(1,3,4,4)}{p(3,3)}&0&0&\frac{-c(1,2,4,1)c(1,3,4,3)-c(1,3,4,1)c(1,3,4,4)}{c(1,3,4,4)p(3,3)}\\\
-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,1)}&p(3,3)&-\frac{c(1,3,4,3)p(3,3)}{c(1,3,4,4)}&p(2,4)\\\
0&0&p(3,3)&-\frac{c(1,3,4,3)p(3,3)}{c(1,3,4,4)}\\\ 0&0&0&p(3,3)\\\
\end{array}\right)$ $\displaystyle
p(2,4)=-\frac{\big{(}-c(1,2,4,1)c(1,3,4,3)^{2}-c(1,2,4,2)c(1,3,4,1)c(1,3,4,4)+c(1,2,4,1)c(1,3,4,2)c(1,3,4,4)\big{)}p(3,3)}{c(1,2,4,1)c(1,3,4,4)^{2}}$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&0&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&0&0&0\\\ 0&1&0&0\end{array}\right)$

  2. 2)

$\dim D^{1}_{3}(\mathcal{A})=2$, $2$-solvable of class $3$, non-nilpotent,
with trivial center, which is equivalent to $d(1,4)=0$ and
$\begin{pmatrix}d(2,4)&d(3,4)\\\ d(1,2)&d(1,3)\end{pmatrix}\neq 0$, thus $B$
takes the form
$B=\begin{pmatrix}0&\lambda c(1,2,4,1)&-c(1,2,4,1)&0\\\
0&c(1,3,4,2)&-c(1,2,4,2)&0\\\ 0&c(1,3,4,3)&-c(1,2,4,3)&0\\\ 0&\lambda
c(1,2,4,4)&-c(1,2,4,4)&0\end{pmatrix}$ if $(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$,
or
$B=\begin{pmatrix}0&c(1,3,4,1)&0&0\\\ 0&c(1,3,4,2)&-c(1,2,4,2)&0\\\
0&c(1,3,4,3)&-c(1,2,4,3)&0\\\ 0&c(1,3,4,4)&0&0\end{pmatrix}$ if
$(c(1,2,4,1),c(1,2,4,4))=(0,0)$.

Consider first the case where $(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&b^{\prime}(1,2)&-\frac{c(1,2,4,1)p(1,1)+c(1,2,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0\\\
0&b^{\prime}(2,2)&b^{\prime}(2,3)&0\\\
0&b^{\prime}(3,2)&-\frac{c(1,2,4,4)p(2,3)+c(1,2,4,3)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,2,4,4)(\lambda
p(3,3)-p(2,3))}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,4)}{p(1,1)p(3,3)}&0\end{array}\right)$
$\displaystyle b^{\prime}(1,2)$
$\displaystyle=c^{\prime}(1,3,4,1)=\frac{(c(1,2,4,1)p(1,1)+c(1,2,4,4)p(1,4))(\lambda
p(3,3)-p(2,3))}{p(1,1)p(3,3)^{3}},$ $\displaystyle b^{\prime}(2,2)$
$\displaystyle=c^{\prime}(1,3,4,2)$ $\displaystyle=\frac{p(3,3)\big{(}\lambda
c(1,2,4,1)p(2,1)+\lambda
c(1,2,4,4)p(2,4)+c(1,3,4,3)p(2,3)+c(1,3,4,2)p(3,3)\big{)}}{p(1,1)p(3,3)^{3}}$
$\displaystyle\quad+\frac{-p(2,3)\big{(}c(1,2,4,1)p(2,1)+c(1,2,4,3)p(2,3)+c(1,2,4,4)p(2,4)+c(1,2,4,2)p(3,3)\big{)}}{p(1,1)p(3,3)^{3}},$
$\displaystyle b^{\prime}(3,2)$ $\displaystyle=c^{\prime}(1,3,4,3)$
$\displaystyle=\frac{c(1,2,4,4)p(2,3)(\lambda
p(3,3)-p(2,3))+p(3,3)(c(1,3,4,3)p(3,3)-c(1,2,4,3)p(2,3))}{p(1,1)p(3,3)^{3}},$
$\displaystyle b^{\prime}(2,3)$ $\displaystyle=-c^{\prime}(1,2,4,2)$
$\displaystyle=-\frac{c(1,2,4,1)p(2,1)+c(1,2,4,3)p(2,3)+c(1,2,4,4)p(2,4)+c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}.$

If $c(1,2,4,4)\neq 0$ then choosing

$P=P_{2,1}=\begin{pmatrix}\frac{c(1,2,4,4)}{p(3,3)}&0&0&-\frac{c(1,2,4,1)}{p(3,3)}\\\
p(2,1)&p(3,3)&-\frac{c(1,2,4,3)p(3,3)}{c(1,2,4,4)}&\frac{\scriptsize\begin{array}[]{c}c(1,2,4,3)^{2}p(3,3)\\\
-c(1,2,4,1)c(1,2,4,4)p(2,1)\\\
-c(1,2,4,2)c(1,2,4,4)p(3,3)\end{array}}{c(1,2,4,4)^{2}}\\\
0&0&p(3,3)&-\frac{c(1,2,4,3)p(3,3)}{c(1,2,4,4)}\\\ 0&0&0&p(3,3)\end{pmatrix}$

we get $B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}\lambda c(1,2,4,3)^{2}-\lambda
c(1,2,4,2)c(1,2,4,4)\\\
-c(1,3,4,3)c(1,2,4,3)+c(1,2,4,4)c(1,3,4,2)\end{array}}{c(1,2,4,4)^{2}}&0&0\\\
0&\frac{c(1,3,4,3)-\lambda c(1,2,4,3)}{c(1,2,4,4)}&0&0\\\ 0&\frac{\lambda
c(1,2,4,4)+c(1,2,4,3)}{c(1,2,4,4)}&-1&0\end{array}\right)$

If $c(1,2,4,4)=0$ and $c(1,2,4,3)\neq 0$ then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)p(1,1)+c(1,3,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}c(1,3,4,1)p(2,1)-c(1,2,4,2)p(2,3)+c(1,3,4,3)p(2,3)\\\
+c(1,3,4,4)p(2,4)+c(1,3,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,3,4,4)p(2,3)+c(1,3,4,3)p(3,3)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,4)}{p(1,1)p(3,3)}&0&0\end{array}\right)$

By choosing

$P=P_{2,2}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,3)}{p(3,3)}&0&0&p(1,4)\\\
-\frac{(c(1,2,4,2)+c(1,3,4,3))p(3,3)}{c(1,2,4,1)}&p(3,3)&\frac{c(1,3,4,3)p(3,3)}{c(1,2,4,3)}&p(2,4)\\\
0&0&p(3,3)&\frac{c(1,3,4,3)p(3,3)}{c(1,2,4,3)}\\\
0&0&0&p(3,3)\end{array}\right),$

we get

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,2,4,1)(\lambda
c(1,2,4,3)-c(1,3,4,3))}{c(1,2,4,3)p(3,3)^{2}}&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}-\lambda c(1,2,4,3)c(1,3,4,3)-\lambda
c(1,2,4,2)c(1,2,4,3)\\\
+c(1,3,4,3)^{2}+c(1,2,4,3)c(1,3,4,2)\end{array}}{c(1,2,4,3)^{2}}&0&0\\\
0&0&-1&0\\\ 0&0&0&0\end{array}\right).$

If $c(1,2,4,4)=0$, in which case $c(1,2,4,1)\neq 0$ (else the algebra would be
$2$-solvable of class $2$ by Theorem 4.1). We consider $c(1,2,4,3)=0$ and
$c(1,3,4,3)\neq 0$. We have

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,2,4,1)(\lambda
p(3,3)-p(2,3))}{p(3,3)^{3}}&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}p(3,3)\big{(}\lambda
c(1,2,4,1)p(2,1)+c(1,3,4,3)p(2,3)\\\ +c(1,3,4,2)p(3,3)\big{)}-p(2,3)\times\\\
\times(c(1,2,4,1)p(2,1)+c(1,2,4,2)p(3,3))\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,1)p(2,1)+c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,3,4,3)}{p(1,1)p(3,3)}&0&0\\\ 0&0&0&0\end{array}\right),$

choosing

$\displaystyle P=P_{2,3}$
$\displaystyle=\left(\begin{array}[]{cccc}\frac{c(1,3,4,3)}{p(3,3)}&0&0&p(1,4)\\\
-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,1)}&p(3,3)&\frac{p(3,3)(\lambda
c(1,2,4,2)-c(1,3,4,2))}{c(1,3,4,3)}&p(2,4)\\\ 0&0&p(3,3)&\frac{p(3,3)(\lambda
c(1,2,4,2)-c(1,3,4,2))}{c(1,3,4,3)}\\\ 0&0&0&p(3,3)\end{array}\right)$

we get $B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,2,4,1)(-\lambda
c(1,2,4,2)+\lambda
c(1,3,4,3)+c(1,3,4,2))}{c(1,3,4,3)p(3,3)^{2}}&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&1&0&0\\\ 0&0&0&0\end{array}\right).$
Consider now $(c(1,2,4,1),c(1,2,4,4))\neq(0,0)$ and $c(1,2,4,4)=0$, meaning
that $c(1,2,4,1)\neq 0$. Suppose also that $c(1,2,4,3)=0$ and $c(1,3,4,3)=0$.
Then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)p(1,1)p(3,3)-c(1,2,4,1)p(1,1)p(2,3)}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}p(3,3)(c(1,3,4,1)p(2,1)+c(1,3,4,2)p(3,3))\\\
+p(2,3)(-c(1,2,4,1)p(2,1)-c(1,2,4,2)p(3,3))\end{array}}{p(1,1)p(3,3)^{3}}&\frac{-c(1,2,4,1)p(2,1)-c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&0&0&0\end{array}\right)$

$P=P_{2,4}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,1)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,1)}{c(1,2,4,1)p(3,3)}&0&0&p(1,4)\\\
-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,1)}&p(3,3)&\frac{c(1,3,4,1)p(3,3)}{c(1,2,4,1)}&p(2,4)\\\
0&0&p(3,3)&\frac{c(1,3,4,1)p(3,3)}{c(1,2,4,1)}\\\
0&0&0&p(3,3)\end{array}\right)$

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&1&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{array}\right).$

In this case, the algebra is multiplicative by Corollary 4.

If $(c(1,2,4,1),c(1,2,4,4))=(0,0)$ and $(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$,
then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)p(1,1)+c(1,3,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}p(3,3)(c(1,3,4,1)p(2,1)-c(1,2,4,2)p(2,3)\\\
+c(1,3,4,3)p(2,3)+c(1,3,4,4)p(2,4)\\\
+c(1,3,4,2)p(3,3))-c(1,2,4,3)p(2,3)^{2}\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,3)p(2,3)+c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}-c(1,2,4,3)p(2,3)+c(1,3,4,4)p(2,3)\\\
+c(1,3,4,3)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,3)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,3,4,4)}{p(1,1)p(3,3)}&0&0\end{array}\right).$

If $c(1,2,4,3)\neq 0$, choosing

$P=P_{2,5}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,3)}{p(3,3)}&0&0&-\frac{c(1,2,4,3)c(1,3,4,1)}{c(1,3,4,4)p(3,3)}\\\
p(2,1)&p(3,3)&-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,3)}&\frac{\scriptsize\begin{array}[]{c}-c(1,2,4,3)c(1,3,4,1)p(2,1)\\\
-c(1,2,4,3)c(1,3,4,2)p(3,3)\\\
+c(1,2,4,2)c(1,3,4,3)p(3,3)\end{array}}{c(1,2,4,3)c(1,3,4,4)}\\\
0&0&p(3,3)&-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,3)}\\\
0&0&0&p(3,3)\end{array}\right),$

we get $B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\
0&\frac{c(1,2,4,3)c(1,3,4,3)+c(1,2,4,2)(c(1,2,4,3)-c(1,3,4,4))}{c(1,2,4,3)^{2}}&-1&0\\\
0&\frac{c(1,3,4,4)}{c(1,2,4,3)}&0&0\end{array}\right).$

If $c(1,2,4,3)=0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)p(1,1)+c(1,3,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}c(1,3,4,1)p(2,1)-c(1,2,4,2)p(2,3)\\\
+c(1,3,4,3)p(2,3)+c(1,3,4,4)p(2,4)+c(1,3,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,3,4,4)p(2,3)+c(1,3,4,3)p(3,3)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,4)}{p(1,1)p(3,3)}&0&0\end{array}\right).$

If $c(1,3,4,4)\neq 0$, choosing

$P=P_{2,6}=\left(\begin{array}[]{cccc}\frac{c(1,3,4,4)}{p(3,3)}&0&0&-\frac{c(1,3,4,1)}{p(3,3)}\\\
p(2,1)&p(3,3)&-\frac{c(1,3,4,3)p(3,3)}{c(1,3,4,4)}&\frac{\scriptsize\begin{array}[]{c}c(1,3,4,3)^{2}p(3,3)\\\
-c(1,2,4,2)c(1,3,4,3)p(3,3)\\\ -c(1,3,4,1)c(1,3,4,4)p(2,1)\\\
-c(1,3,4,2)c(1,3,4,4)p(3,3)\end{array}}{c(1,3,4,4)^{2}}\\\
0&0&p(3,3)&-\frac{c(1,3,4,3)p(3,3)}{c(1,3,4,4)}\\\
0&0&0&p(3,3)\end{array}\right),$

we get $B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&0&-\frac{c(1,2,4,2)}{c(1,3,4,4)}&0\\\ 0&0&0&0\\\ 0&1&0&0\end{array}\right).$

If $c(1,3,4,4)=0$, since $(c(1,3,4,1),c(1,3,4,4))\neq(0,0)$, then
$c(1,3,4,1)\neq 0$,

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}c(1,3,4,1)p(2,1)-c(1,2,4,2)p(2,3)\\\
+c(1,3,4,3)p(2,3)+c(1,3,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,3,4,3)}{p(1,1)p(3,3)}&0&0\\\ 0&0&0&0\end{array}\right).$

If $c(1,3,4,3)\neq 0$, then taking

$P=P_{2,7}=\left(\begin{array}[]{cccc}\frac{c(1,3,4,3)}{p(3,3)}&0&0&p(1,4)\\\
\frac{\scriptsize\begin{array}[]{c}c(1,2,4,2)p(2,3)-c(1,3,4,3)p(2,3)\\\
-c(1,3,4,2)p(3,3)\end{array}}{c(1,3,4,1)}&p(3,3)&p(2,3)&p(2,4)\\\
0&0&p(3,3)&p(2,3)\\\ 0&0&0&p(3,3)\end{array}\right),$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&0&-\frac{c(1,2,4,2)}{c(1,3,4,3)}&0\\\ 0&1&0&0\\\ 0&0&0&0\end{array}\right).$

If $c(1,3,4,3)=0$, then the algebra is now multiplicative, and we have

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&\frac{p(3,3)(c(1,3,4,1)p(2,1)+c(1,3,4,2)p(3,3))-c(1,2,4,2)p(2,3)p(3,3)}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&0&0&0\\\ 0&0&0&0\end{array}\right).$

As in the previous case, $c(1,3,4,1\neq 0$, moreover $c(1,2,4,2)\neq 0$
because otherwise we would have $\dim D_{3}^{1}(\mathcal{A})=1$. Choosing

$\displaystyle P=P_{2,8}$
$\displaystyle=\left(\begin{array}[]{cccc}\frac{c(1,2,4,2)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&\frac{c(1,3,4,1)p(2,1)+c(1,3,4,2)p(3,3)}{c(1,2,4,2)}&p(2,4)\\\
0&0&p(3,3)&\frac{c(1,3,4,1)p(2,1)+c(1,3,4,2)p(3,3)}{c(1,2,4,2)}\\\
0&0&0&p(3,3)\end{array}\right),$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&0&-1&0\\\ 0&0&0&0\\\ 0&0&0&0\end{array}\right).$

  3. 3)

$\dim D^{1}_{3}(\mathcal{A})=2$, $2$-solvable of class $2$, non-nilpotent,
with trivial center. In this case, the matrix defining the bracket is given by

$B=\begin{pmatrix}0&0&0&0\\\ 0&c(1,3,4,2)&-c(1,2,4,2)&0\\\
0&c(1,3,4,3)&-c(1,2,4,3)&0\\\ 0&0&0&0\end{pmatrix},$
$B^{\prime}=\begin{pmatrix}0&0&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}p(2,3)(-c(1,2,4,3)p(2,3)-c(1,2,4,2)p(3,3))\\\
+p(3,3)(c(1,3,4,3)p(2,3)+c(1,3,4,2)p(3,3))\end{array}}{p(1,1)p(3,3)^{3}}&\frac{-c(1,2,4,3)p(2,3)-c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,3,4,3)p(3,3)^{2}-c(1,2,4,3)p(2,3)p(3,3)}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,3)}{p(1,1)p(3,3)}&0\\\
0&0&0&0\end{pmatrix}.$

Note that $c^{\prime}(1,2,4,3)=0$ if and only if $c(1,2,4,3)=0$. Thus the
cases where $c(1,2,4,3)=0$ and $c(1,2,4,3)\neq 0$ cannot be isomorphic.

    1. 3.a)

If $c(1,2,4,3)\neq 0$ then taking

$P=P_{3,1}=\begin{pmatrix}\frac{c(1,2,4,3)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&\frac{c(1,3,4,3)p(3,3)}{c(1,2,4,3)}&p(2,4)\\\
0&0&p(3,3)&\frac{c(1,3,4,3)p(3,3)}{c(1,2,4,3)}\\\ 0&0&0&p(3,3)\end{pmatrix},$

for arbitrary $p(2,1),p(1,4),p(2,4)$ and $p(3,3)\neq 0$, gives the following
matrix defining the bracket

$\displaystyle B^{\prime}=\begin{pmatrix}0&0&0&0\\\
0&c^{\prime}(1,3,4,2)&-c^{\prime}(1,2,4,2)&0\\\ 0&0&-1&0\\\
0&0&0&0\end{pmatrix},$ $\displaystyle
c^{\prime}(1,3,4,2)=\frac{c(1,2,4,3)c(1,3,4,2)-c(1,2,4,2)c(1,3,4,3)}{c(1,2,4,3)^{2}},$
$\displaystyle c^{\prime}(1,2,4,2)=\frac{-c(1,2,4,2)-c(1,3,4,3)}{c(1,2,4,3)}.$

    2. 3.b)

If $c(1,2,4,3)=0$ then consider $c(1,3,4,3)\neq 0$ and $c(1,2,4,2)\neq 0$,
since otherwise the center of the algebra would become non-zero (Theorem 4.1)

$B^{\prime}=\begin{pmatrix}0&0&0&0\\\
0&\frac{(c(1,3,4,3)-c(1,2,4,2))p(2,3)+c(1,3,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,3,4,3)}{p(1,1)p(3,3)}&0&0\\\ 0&0&0&0\end{pmatrix}.$

Taking

$\displaystyle P=P_{3,2}=$
$\displaystyle\begin{pmatrix}\frac{c(1,3,4,3)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&-\frac{(c(1,3,4,3)-c(1,3,4,2))p(3,3)}{c(1,2,4,2)-c(1,3,4,3)}&p(2,4)\\\
0&0&p(3,3)&-\frac{(c(1,3,4,3)-c(1,3,4,2))p(3,3)}{c(1,2,4,2)-c(1,3,4,3)}\\\
0&0&0&p(3,3)\end{pmatrix}$

for arbitrary $p(2,1),p(1,4),p(2,4)$ and $p(3,3)\neq 0$ and for
$c(1,3,4,3)\neq c(1,2,4,2)$, gives the following matrix defining the bracket
$B^{\prime}=\begin{pmatrix}0&0&0&0\\\ 0&1&-c^{\prime}(1,2,4,2)&0\\\ 0&1&0&0\\\
0&0&0&0\end{pmatrix},$ with
$c^{\prime}(1,2,4,2)=\frac{c(1,2,4,2)}{c(1,3,4,3)}$.

Consider now two such algebras with different parameters
$c^{\prime}(1,2,4,2)=a$ and $c^{\prime\prime}(1,2,4,2)=b$, and denote the
matrices defining the brackets by $B^{\prime}_{1}$ and $B^{\prime}_{2}$
respectively. Those algebras are isomorphic if and only if

$\frac{1}{\det(P)}PB^{\prime}_{1}P^{T}-B^{\prime}_{2}=\begin{pmatrix}0&0&0&0\\\
0&\frac{(a+1)p(2,3)}{p(1,1)p(3,3)^{2}}&\frac{a}{p(1,1)p(3,3)}-b&0\\\
0&\frac{1}{p(1,1)p(3,3)}-1&0&0\\\ 0&0&0&0\end{pmatrix}=0.$

    3. 3.c)

If $(c(1,3,4,3)=c(1,2,4,2)$, then $B^{\prime}=\begin{pmatrix}0&0&0&0\\\
0&\frac{c(1,3,4,2)}{p(1,1)p(3,3)}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0&0\\\ 0&0&0&0\end{pmatrix}.$
Taking $P=P_{3,3}=\left(\begin{array}[]{cccc}p(1,1)&0&0&p(1,4)\\\
p(2,1)&\frac{c(1,3,4,3)}{p(1,1)}&p(2,3)&p(2,4)\\\
0&0&\frac{c(1,3,4,3)}{p(1,1)}&p(2,3)\\\
0&0&0&\frac{c(1,3,4,3)}{p(1,1)}\end{array}\right)$ gives

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&\frac{c(1,3,4,2)}{c(1,3,4,3)}&-1&0\\\ 0&1&0&0\\\ 0&0&0&0\end{array}\right).$

  4. 4)

$\dim D^{1}_{3}(\mathcal{A})=1$, $\mathcal{A}$ is $2$-solvable of class $2$,
non-nilpotent, with $1$-dimensional center.

In this case $w_{2}$ and $w_{3}$ are linearly dependent.

If $w_{3}\neq 0$, $w_{2}=\lambda w_{3}$, $\lambda\in\mathbb{K}$, then
$B=\begin{pmatrix}0&\lambda c(1,2,4,1)&-c(1,2,4,1)&0\\\ 0&\lambda
c(1,2,4,2)&-c(1,2,4,2)&0\\\ 0&\lambda c(1,2,4,3)&-c(1,2,4,3)&0\\\ 0&\lambda
c(1,2,4,4)&-c(1,2,4,4)&0\end{pmatrix}.$

If $w_{3}=0$ and $w_{2}\neq 0$, then $B=\begin{pmatrix}0&c(1,3,4,1)&0&0\\\
0&c(1,3,4,2)&0&0\\\ 0&c(1,3,4,3)&0&0\\\ 0&c(1,3,4,4)&0&0\end{pmatrix}.$

    1. 4.a)

We consider first the case when $w_{3}\neq 0$ and $w_{2}=\lambda w_{3}$ where
$\lambda\in\mathbb{K}$, then
$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{\scriptsize\begin{array}[]{c}(\lambda
p(3,3)-p(2,3))(c(1,2,4,1)p(1,1)\\\
+c(1,2,4,4)p(1,4))\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,1)p(1,1)+c(1,2,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}(\lambda
p(3,3)-p(2,3))\big{(}c(1,2,4,1)p(2,1)\\\ +c(1,2,4,3)p(2,3)+c(1,2,4,4)p(2,4)\\\
+c(1,2,4,2)p(3,3)\big{)}\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{\scriptsize\begin{array}[]{c}c(1,2,4,1)p(2,1)+c(1,2,4,3)p(2,3)\\\
+c(1,2,4,4)p(2,4)+c(1,2,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}(\lambda
p(3,3)-p(2,3))(c(1,2,4,4)p(2,3)\\\
+c(1,2,4,3)p(3,3))\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,4)p(2,3)+c(1,2,4,3)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,2,4,4)(\lambda
p(3,3)-p(2,3))}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,4)}{p(1,1)p(3,3)}&0\end{array}\right).$
If $c(1,2,4,4)\neq 0$ then taking

$P=P_{4,1}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,4)}{p(3,3)}&0&0&-\frac{c(1,2,4,1)}{p(3,3)}\\\
p(2,1)&p(3,3)&-\frac{c(1,2,4,3)p(3,3)}{c(1,2,4,4)}&\frac{\scriptsize\begin{array}[]{c}c(1,2,4,3)^{2}p(3,3)-\\\
c(1,2,4,1)c(1,2,4,4)p(2,1)-\\\
c(1,2,4,2)c(1,2,4,4)p(3,3)\end{array}}{c(1,2,4,4)^{2}}\\\
0&0&p(3,3)&-\frac{c(1,2,4,3)p(3,3)}{c(1,2,4,4)}\\\
0&0&0&p(3,3)\end{array}\right)$

we get $B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
0&\frac{c(1,2,4,3)}{c(1,2,4,4)}+\lambda&-1&0\end{array}\right)$

    2. 4.b)

If $c(1,2,4,4)=0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,2,4,1)(\lambda
p(3,3)-p(2,3))}{p(3,3)^{3}}&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}(\lambda
p(3,3)-p(2,3))\big{(}c(1,2,4,1)p(2,1)\\\
+c(1,2,4,3)p(2,3)+c(1,2,4,2)p(3,3)\big{)}\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{\scriptsize\begin{array}[]{c}c(1,2,4,1)p(2,1)+c(1,2,4,3)p(2,3)\\\
+c(1,2,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,2,4,3)(\lambda
p(3,3)-p(2,3))}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,3)}{p(1,1)p(3,3)}&0\\\
0&0&0&0\end{array}\right).$

If $c(1,2,4,3)\neq 0$ and $c(1,2,4,1)\neq 0$, taking

$P=P_{4,2}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,3)}{p(3,3)}&0&0&p(1,4)\\\
\frac{-\lambda c(1,2,4,3)p(3,3)-c(1,2,4,2)p(3,3)}{c(1,2,4,1)}&p(3,3)&\lambda
p(3,3)&p(2,4)\\\ 0&0&p(3,3)&\lambda p(3,3)\\\ 0&0&0&p(3,3)\end{array}\right),$

we get
$B^{\prime}=\left(\begin{array}[]{cccc}0&0&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&0&-1&0\\\ 0&0&0&0\end{array}\right).$

    3. 4.c)

If $c(1,2,4,3)=0$ and $c(1,2,4,1)\neq 0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,2,4,1)(\lambda
p(3,3)-p(2,3))}{p(3,3)^{3}}&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&\frac{\scriptsize\begin{array}[]{c}(\lambda p(3,3)-p(2,3))\times\\\
\times(c(1,2,4,1)p(2,1)+c(1,2,4,2)p(3,3))\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{c(1,2,4,1)p(2,1)+c(1,2,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&0&0&0\end{array}\right),$

$P=P_{4,3}=\left(\begin{array}[]{cccc}p(1,1)&0&0&p(1,4)\\\
-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,1)}&p(3,3)&\lambda p(3,3)&p(2,4)\\\
0&0&p(3,3)&\lambda p(3,3)\\\ 0&0&0&p(3,3)\end{array}\right),$

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&-\frac{c(1,2,4,1)}{p(3,3)^{2}}&0\\\
0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{array}\right).$

    4. 4.d)

If $c(1,2,4,3)\neq 0$ and $c(1,2,4,1)=0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}(\lambda p(3,3)-p(2,3))\times\\\
\times(c(1,2,4,3)p(2,3)+c(1,2,4,2)p(3,3))\end{array}}{p(1,1)p(3,3)^{3}}&-\frac{\scriptsize\begin{array}[]{c}c(1,2,4,3)p(2,3)\\\
+c(1,2,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&0\\\
0&\frac{c(1,2,4,3)(\lambda
p(3,3)-p(2,3))}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,3)}{p(1,1)p(3,3)}&0\\\
0&0&0&0\end{array}\right),$

$P=P_{4,4}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,3)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,3)}&p(2,4)\\\
0&0&p(3,3)&-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,3)}\\\
0&0&0&p(3,3)\end{array}\right),$

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\
0&\frac{c(1,2,4,2)}{c(1,2,4,3)}+\lambda&-1&0\\\ 0&0&0&0\\\
\end{array}\right).$

    5. 4.e)

If $c(1,2,4,3)=0$ and $c(1,2,4,1)=0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&\frac{c(1,2,4,2)(\lambda
p(3,3)-p(2,3))}{p(1,1)p(3,3)^{2}}&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&0&0&0\\\ 0&0&0&0\end{array}\right),$

$P_{4,5}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,2)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&\lambda p(3,3)&p(2,4)\\\ 0&0&p(3,3)&\lambda p(3,3)\\\
0&0&0&p(3,3)\end{array}\right),$

$B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&-1&0\\\ 0&0&0&0\\\
0&0&0&0\end{array}\right).$

    6. 4.f)

Now we consider the case where $w_{3}=0$ and $w_{2}\neq 0$, we have

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)p(1,1)+c(1,3,4,4)p(1,4)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{\scriptsize\begin{array}[]{c}c(1,3,4,1)p(2,1)+c(1,3,4,3)p(2,3)\\\
+c(1,3,4,4)p(2,4)+c(1,3,4,2)p(3,3)\end{array}}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,4)p(2,3)+c(1,3,4,3)p(3,3)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,4)}{p(1,1)p(3,3)}&0&0\end{array}\right).$

If $c(1,3,4,4)\neq 0$, then choosing

$P=P_{4,6}=\left(\begin{array}[]{cccc}\frac{c(1,3,4,4)}{p(3,3)}&0&0&-\frac{c(1,3,4,1)}{p(3,3)}\\\
p(2,1)&p(3,3)&-\frac{c(1,3,4,3)p(3,3)}{c(1,3,4,4)}&\frac{\scriptsize\begin{array}[]{c}c(1,3,4,3)^{2}p(3,3)\\\
-c(1,3,4,1)c(1,3,4,4)p(2,1)\\\
-c(1,3,4,2)c(1,3,4,4)p(3,3)\end{array}}{c(1,3,4,4)^{2}}\\\
0&0&p(3,3)&-\frac{c(1,3,4,3)p(3,3)}{c(1,3,4,4)}\\\
0&0&0&p(3,3)\end{array}\right)$

we get $B^{\prime}={\small\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\
0&0&0&0\\\ 0&1&0&0\end{array}\right)}.$

    7. 4.g)

If $c(1,3,4,4)=0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,1)p(2,1)+c(1,3,4,3)p(2,3)+c(1,3,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,3)}{p(1,1)p(3,3)}&0&0\\\ 0&0&0&0\end{array}\right).$

If $c(1,3,4,3)\neq 0$ then choosing

$P_{4,7}={\small\left(\begin{array}[]{cccc}\frac{c(1,3,4,3)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&\frac{-c(1,3,4,1)p(2,1)-c(1,3,4,2)p(3,3)}{c(1,3,4,3)}&p(2,4)\\\
0&0&p(3,3)&\frac{-c(1,3,4,1)p(2,1)-c(1,3,4,2)p(3,3)}{c(1,3,4,3)}\\\
0&0&0&p(3,3)\end{array}\right),}$

we get

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&0&0&0\\\ 0&1&0&0\\\ 0&0&0&0\end{array}\right).$

    8. 4.h)

If $c(1,3,4,3)=0$, then

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&\frac{c(1,3,4,1)p(2,1)+c(1,3,4,2)p(3,3)}{p(1,1)p(3,3)^{2}}&0&0\\\ 0&0&0&0\\\
0&0&0&0\end{array}\right).$

If $c(1,3,4,1)\neq 0$, then choosing

$P_{4,9}=\left(\begin{array}[]{cccc}p(1,1)&0&0&p(1,4)\\\
-\frac{c(1,3,4,2)p(3,3)}{c(1,3,4,1)}&p(3,3)&p(2,3)&p(2,4)\\\
0&0&p(3,3)&p(2,3)\\\ 0&0&0&p(3,3)\\\ \end{array}\right)$

gives

$B^{\prime}=\left(\begin{array}[]{cccc}0&\frac{c(1,3,4,1)}{p(3,3)^{2}}&0&0\\\
0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\ \end{array}\right).$

If $c(1,3,4,1)=0$ then the algebra becomes nilpotent.

  5. 5)

$\dim D^{1}_{3}(\mathcal{A})=1$, $2$-solvable of class $2$, nilpotent of class
$2$, with $1$-dimensional center. In this case, the matrix defining the
bracket of $\mathcal{A}$ takes the following form
$B_{5,1}=\begin{pmatrix}0&0&0&0\\\
0&\frac{-c(1,2,4,2)^{2}}{c(1,2,4,3)}&-c(1,2,4,2)&0\\\
0&-c(1,2,4,2)&-c(1,2,4,3)&0\\\ 0&0&0&0\\\ \end{pmatrix},$ where
$c(1,2,4,3)\neq 0$, or
$B_{5,2}=\begin{pmatrix}0&0&0&0\\\ 0&c(1,3,4,2)&-c(1,2,4,2)&0\\\
0&-c(1,2,4,2)&-\frac{-c(1,2,4,2)^{2}}{c(1,3,4,2)}&0\\\ 0&0&0&0\\\
\end{pmatrix},$ where $c(1,3,4,2)\neq 0$.

Consider the first form, then

$\displaystyle B^{\prime}=\frac{1}{\det(P)}PB_{5,1}P^{T}$
$\displaystyle={\small\left(\begin{array}[]{cccc}0,&0,&0,&0\\\
0,&-\frac{c(1,2,4,2)^{2}}{c(1,2,4,3)p(1,1)p(3,3)}-\frac{2c(1,2,4,2)p(2,3)}{p(1,1)p(3,3)^{2}}-\frac{c(1,2,4,3)p(2,3)^{2}}{p(1,1)p(3,3)^{3}},&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}-\frac{c(1,2,4,3)p(2,3)}{p(1,1)p(3,3)^{2}},&0\\\
0,&-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}-\frac{c(1,2,4,3)p(2,3)}{p(1,1)p(3,3)^{2}},&-\frac{c(1,2,4,3)}{p(1,1)p(3,3)},&0\\\
0,&0,&0,&0\\\ \end{array}\right)}$

where $c(1,2,4,3)\neq 0$. Taking

$P=P_{5,1}=\left(\begin{array}[]{cccc}\frac{c(1,2,4,3)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,3)}&p(2,4)\\\
0&0&p(3,3)&-\frac{c(1,2,4,2)p(3,3)}{c(1,2,4,3)}\\\ 0&0&0&p(3,3)\\\
\end{array}\right)$

we get $B^{\prime}={\small\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\
0&0&-1&0\\\ 0&0&0&0\\\ \end{array}\right)}.$

For the second form, we have

$\displaystyle B^{\prime}=\frac{1}{\det(P)}PB_{5,2}P^{T}=$
$\displaystyle{\small\left(\begin{array}[]{cccc}0&0&0&0\\\
0&\frac{c(1,2,4,2)^{2}p(2,3)^{2}}{c(1,3,4,2)p(1,1)p(3,3)^{3}}-\frac{2c(1,2,4,2)p(2,3)}{p(1,1)p(3,3)^{2}}+\frac{c(1,3,4,2)}{p(1,1)p(3,3)}&\frac{c(1,2,4,2)^{2}p(2,3)}{c(1,3,4,2)p(1,1)p(3,3)^{2}}-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&0\\\
0&\frac{c(1,2,4,2)^{2}p(2,3)}{c(1,3,4,2)p(1,1)p(3,3)^{2}}-\frac{c(1,2,4,2)}{p(1,1)p(3,3)}&\frac{c(1,2,4,2)^{2}}{c(1,3,4,2)p(1,1)p(3,3)}&0\\\
0&0&0&0\\\ \end{array}\right),}$

where $c(1,3,4,2)\neq 0$. If $c(1,2,4,2)\neq 0$, then by taking

$P=P_{5,2}=\left(\begin{array}[]{cccc}-\frac{c(1,2,4,2)^{2}}{c(1,3,4,2)p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&\frac{c(1,3,4,2)p(3,3)}{c(1,2,4,2)}&p(2,4)\\\
0&0&p(3,3)&\frac{c(1,3,4,2)p(3,3)}{c(1,2,4,2)}\\\ 0&0&0&p(3,3)\\\
\end{array}\right)$

we get $B^{\prime}={\small\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\
0&0&-1&0\\\ 0&0&0&0\\\ \end{array}\right)}.$

If $c(1,2,4,2)=0$, then $B^{\prime}=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&\frac{c(1,3,4,2)}{p(1,1)p(3,3)}&0&0\\\ 0&0&0&0\\\ 0&0&0&0\\\
\end{array}\right),$ and by choosing

$P=P_{5,3}=\left(\begin{array}[]{cccc}\frac{c(1,3,4,2)}{p(3,3)}&0&0&p(1,4)\\\
p(2,1)&p(3,3)&p(2,3)&p(2,4)\\\ 0&0&p(3,3)&p(2,3)\\\ 0&0&0&p(3,3)\\\
\end{array}\right)$

we get $B^{\prime}={\small\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&1&0&0\\\
0&0&0&0\\\ 0&0&0&0\\\ \end{array}\right)}.$ ∎

##  6 Examples and remarks

In this section, we consider some examples that show specific properties not
following from results proved above, and that may lead to further
investigations of the properties of $n$-Hom-Lie algebras. The following result
is a consequence of (kms:nhominduced, , Lemma 6.2).

###### Proposition 10

Let $\mathcal{A}=(A,\left[\cdot,\dots,\cdot\right],(\alpha_{i})_{1\leq i\leq
n-1})$ be an $n$-Hom-Lie algebra and let $I$ be an ideal of $\mathcal{A}$.
Then, for all $p\in\mathbb{N}$, $2\leq k\leq n$, $D_{k}^{p+1}(I)$ is a weak
ideal of $D_{k}^{p}(I)$ and $C_{k}^{p+1}(I)$ is a weak ideal of
$C_{k}^{p}(I)$. In particular, $D_{k}^{1}(A)$ and $C_{k}^{1}(A)$ are weak
ideals of $\mathcal{A}$. Moreover if all the $\alpha_{i},1\leq i\leq n-1$ are
Hom-algebra morphisms, then $D_{k}^{p+1}(I)$ is an ideal of $D_{k}^{p}(I)$ and
$C_{k}^{p+1}(I)$ is an ideal of $C_{k}^{p}(I)$.

A consequence of this is that all the multiplicative algebras in the above
classification are not simple since they have at least one non-trivial ideal
($D_{3}^{1}(A)$).

The element of the derived series and central descending series of $A$ for the
above algebras are given by

$\displaystyle D^{1}_{3}(\mathcal{A})$
$\displaystyle=\langle\\{c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4},$
$\displaystyle\quad\quad
c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}\\}\rangle,$
$\displaystyle D^{1}_{2}(\mathcal{A})$
$\displaystyle=\langle\\{c(1,2,4,1)e_{1}+c(1,2,4,2)e_{2}+c(1,2,4,3)e_{3}+c(1,2,4,4)e_{4},$
$\displaystyle\quad\quad
c(1,3,4,1)e_{1}+c(1,3,4,2)e_{2}+c(1,3,4,3)e_{3}+c(1,3,4,4)e_{4}\\}\rangle.$

For the cases 1.a) and 2.a), $D_{3}^{1}(A)$ is not invariant under $\alpha$,
that is, it is not an ideal.

Case 2.a) In this case,

$\displaystyle D^{2}_{2}(\mathcal{A})=\langle\\{$
$\displaystyle(c^{\prime}(1,3,4,2)c^{\prime}(1,2,4,4)-c^{\prime}(1,3,4,4)c^{\prime}(1,2,4,2))w_{3}-$
$\displaystyle\quad-(c^{\prime}(1,3,4,3)c^{\prime}(1,2,4,4)-c^{\prime}(1,3,4,4)c^{\prime}(1,2,4,3))w_{2},$
$\displaystyle\quad-(c^{\prime}(1,3,4,1)c^{\prime}(1,2,4,4)-c^{\prime}(1,3,4,4)c^{\prime}(1,2,4,1))w_{3},$
$\displaystyle\quad-(c^{\prime}(1,3,4,1)c^{\prime}(1,2,4,4)-c^{\prime}(1,3,4,4)c^{\prime}(1,2,4,1))w_{2},$
$\displaystyle(c^{\prime}(1,3,4,1)c^{\prime}(1,2,4,2)-c^{\prime}(1,3,4,2)c^{\prime}(1,2,4,1))w_{3}-$
$\displaystyle\quad-(c^{\prime}(1,3,4,1)c^{\prime}(1,2,4,3)-c^{\prime}(1,3,4,3)c^{\prime}(1,2,4,1))w_{2}\\}\rangle$
$\displaystyle=\langle\\{c^{\prime}(1,3,4,2)w_{3}-c^{\prime}(1,3,4,3)w_{2}\\}\rangle\neq\\{0\\}$

since in case 2) $\dim D_{3}^{1}(\mathcal{A})=2$. Denote by $v$ the generator
of $D^{2}_{2}(\mathcal{A})$:

$\displaystyle v=c^{\prime}(1,3,4,2)w_{3}-c^{\prime}(1,3,4,3)w_{2}$
$\displaystyle=-c^{\prime}(1,3,4,2)e_{4}-c^{\prime}(1,3,4,3)(c^{\prime}(1,3,4,2)e_{2}+c^{\prime}(1,3,4,3)e_{3}+c^{\prime}(1,3,4,4)e_{4})$
$\displaystyle=-c^{\prime}(1,3,4,2)e_{4}-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}$
$\displaystyle\quad-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)e_{4}$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)$
$\displaystyle\quad+c^{\prime}(1,3,4,2))e_{4}.$

In general, $D_{2}^{2}(\mathcal{A})$ is a weak subalgebra of $\mathcal{A}$. We
study whether $D_{2}^{2}(\mathcal{A})$ can be a Hom-subalgebra in this class.
To this end, we calculate the image by $\alpha$ of $D_{2}^{2}(\mathcal{A})$:

$\displaystyle\alpha(v)$
$\displaystyle=\alpha(-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}$
$\displaystyle\quad\quad\quad-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4})$
$\displaystyle=-\alpha(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2})-\alpha(c^{\prime}(1,3,4,3)^{2}e_{3})$
$\displaystyle\quad\quad\quad-\alpha((c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4})$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)\alpha(e_{2})-c^{\prime}(1,3,4,3)^{2}\alpha(e_{3})$
$\displaystyle\quad\quad\quad-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))\alpha(e_{4})$
$\displaystyle=-c^{\prime}(1,3,4,3)^{2}e_{2}-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{3}.$

In the case when
$(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))\neq 0$, the two
elements $\alpha(c^{\prime}(1,3,4,2)w_{3}-c^{\prime}(1,3,4,3)w_{2})$ and
$c^{\prime}(1,3,4,2)w_{3}-c^{\prime}(1,3,4,3)w_{2}$ are linearly independent,
which means that $D_{2}^{2}(\mathcal{A})$ is not invariant under $\alpha$ and
thus $D_{2}^{2}(\mathcal{A})$ is a weak subalgebra but not a Hom-subalgebra of
$\mathcal{A}$.

If $(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))=0$, then

$\displaystyle c^{\prime}(1,3,4,2)$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4),$
$\displaystyle\alpha(v)$ $\displaystyle=-c^{\prime}(1,3,4,3)^{2}e_{2},$
$\displaystyle v$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}$
$\displaystyle\quad-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4}$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}$
$\displaystyle=+c^{\prime}(1,3,4,3)^{2}c^{\prime}(1,3,4,4)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}$
$\displaystyle=c^{\prime}(1,3,4,3)^{2}(c^{\prime}(1,3,4,4)e_{2}-e_{3})$

If $c^{\prime}(1,3,4,4)\neq 0$, then in this case $c^{\prime}(1,3,4,3)\neq 0$
because otherwise $c^{\prime}(1,3,4,2)=0$ too which contradicts to the
assumption $\dim D^{1}_{3}(\mathcal{A})=2$. If $c^{\prime}(1,3,4,4)=0$ then
$c^{\prime}(1,3,4,2)=0$, and thus $c^{\prime}(1,3,4,3)\neq 0$ because
otherwise $\dim D_{1}^{3}(\mathcal{A})\neq 2$. Thus these elements are
linearly independent since $e_{2}$ and $e_{3}$ are linearly independent. Thus
in the case 2.a), $D_{2}^{2}(\mathcal{A})$ cannot be invariant under $\alpha$
and hence $D_{2}^{2}(\mathcal{A})$ is a weak subalgebra but not a Hom-
subalgebra of $\mathcal{A}$.

Since $D_{2}^{2}(\mathcal{A})$ is not a Hom-subalgebra of $\mathcal{A}$, it is
not a Hom-ideal either. Let us study now whether $D^{2}_{2}(\mathcal{A})$ is a
weak ideal of $\mathcal{A}$. We have

$\displaystyle\left[e_{1},e_{2},v\right]=$
$\displaystyle\left[e_{1},e_{2},-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)e_{2}-c^{\prime}(1,3,4,3)^{2}e_{3}\right.$
$\displaystyle\hskip
113.81102pt\left.-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4}\right]$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)\left[e_{1},e_{2},e_{2}\right]-c^{\prime}(1,3,4,3)^{2}\left[e_{1},e_{2},e_{3}\right]$
$\displaystyle\hskip
113.81102pt-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))\left[e_{1},e_{2},e_{4}\right]$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,2)0-c^{\prime}(1,3,4,3)^{2}0$
$\displaystyle\hskip
113.81102pt-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4}$
$\displaystyle=-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4}.$

If $(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))\neq 0$, when
$c^{\prime}(1,3,4,3)\neq 0$, $\left[e_{1},e_{2},v\right]$ and $v$ are linearly
independent. Thus $D_{2}^{2}(\mathcal{A})$ is a weak subalgebra, but not a
weak ideal of $\mathcal{A}$. If $c^{\prime}(1,3,4,3)=0$, then
$c^{\prime}(1,3,4,2)\neq 0$, since $\dim D^{1}_{3}(\mathcal{A})\neq 2$
otherwise, which contradicts the assumptions of the case 2.a). We get

$\displaystyle\left[e_{1},e_{3},v\right]=\left[e_{1},e_{3},-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))e_{4}\right]$
$\displaystyle=-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=-(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))(c^{\prime}(1,3,4,2)e_{2}+c^{\prime}(1,3,4,4)e_{4}).$

This element is linearly independent from $v$, and hence it is not in
$D_{2}^{2}(\mathcal{A})$. Thus $D_{2}^{2}(\mathcal{A})$ is not a weak ideal of
$\mathcal{A}$. If
$(c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)+c^{\prime}(1,3,4,2))=0$, then
$v=c^{\prime}(1,3,4,3)^{2}(c^{\prime}(1,3,4,4)e_{2}-e_{3})$. In this case,
$\left[e_{j},e_{k},v\right]\neq 0$ if and only if $(j,k)=(1,4)$ or
$(j,k)=(4,1)$. Therefore, we compute only $\left[e_{1},e_{4},v\right]$,

$\displaystyle\left[e_{1},e_{4},v\right]$
$\displaystyle=\left[e_{1},e_{4},c^{\prime}(1,3,4,3)^{2}(c^{\prime}(1,3,4,4)e_{2}-e_{3})\right]$
$\displaystyle=c^{\prime}(1,3,4,3)^{2}(\left[e_{1},e_{4},c^{\prime}(1,3,4,4)e_{2}\right]-\left[e_{1},e_{4},e_{3}\right])$
$\displaystyle=c^{\prime}(1,3,4,3)^{2}(-c^{\prime}(1,3,4,4)\left[e_{1},e_{2},e_{4}\right]+\left[e_{1},e_{3},e_{4}\right])$
$\displaystyle=c^{\prime}(1,3,4,3)^{2}\Big{(}-c^{\prime}(1,3,4,4)e_{4}+c^{\prime}(1,3,4,2)e_{2}$
$\displaystyle\hskip
113.81102pt+c^{\prime}(1,3,4,3)e_{3}+c^{\prime}(1,3,4,4)e_{4}\Big{)}$
$\displaystyle=-c^{\prime}(1,3,4,3)^{2}(-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)e_{2}+c^{\prime}(1,3,4,3)e_{3})$
$\displaystyle=c^{\prime}(1,3,4,3)^{3}(c^{\prime}(1,3,4,4)e_{2}-e_{3})$
$\displaystyle=c^{\prime}(1,3,4,3)v.$

Therefore $D_{2}^{2}(\mathcal{A})$ is a weak ideal of $\mathcal{A}$. In this
case the bracket of $\mathcal{A}$ is given by

$\displaystyle\left[e_{1},e_{2},e_{3}\right]$ $\displaystyle=0$
$\displaystyle\left[e_{1},e_{2},e_{4}\right]$ $\displaystyle=e_{4}$
$\displaystyle\left[e_{1},e_{3},e_{4}\right]$
$\displaystyle=-c^{\prime}(1,3,4,3)c^{\prime}(1,3,4,4)e_{2}+c^{\prime}(1,3,4,3)e_{3}+c^{\prime}(1,3,4,4)e_{4}$
$\displaystyle\left[e_{2},e_{3},e_{4}\right]$ $\displaystyle=0,$

where $c^{\prime}(1,3,4,3)\neq 0$

###### Example 1

If we take $\mathbb{K}=\mathbb{C}$, $c(1,3,4,4)=\pm i$ and $c(1,3,4,3)=-2$
then we get the following two examples where $D_{2}^{2}(\mathcal{A})$ is a
weak ideal of $\mathcal{A}$:

$\begin{array}[]{rl}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=e_{4}\\\
\left[e_{1},e_{3},e_{4}\right]&=2ie_{2}-2e_{3}+ie_{4}\\\
\left[e_{2},e_{3},e_{4}\right]&=0\end{array}\quad\quad\text{or}\quad\quad\begin{array}[]{rl}\left[e_{1},e_{2},e_{3}\right]&=0\\\
\left[e_{1},e_{2},e_{4}\right]&=e_{4}\\\
\left[e_{1},e_{3},e_{4}\right]&=-2ie_{2}-2e_{3}-ie_{4}\\\
\left[e_{2},e_{3},e_{4}\right]&=0.\end{array}$

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