The algebraic and geometric classification of
nilpotent binary and mono Leibniz algebras111 The first part of this work is
supported by FCT UIDB/MAT/00212/2020 and UIDP/MAT/00212/2020; grant
FZ-202009269, Ministry of higher education, science and innovations of the
Republic of Uzbekistan. The second part of this work is supported by the
Russian Science Foundation under grant 22-71-10001.  222Corresponding author:
Ivan Kaygorodov<EMAIL_ADDRESS>

Kobiljon Abdurasulov333CMA-UBI, Universidade da Beira Interior, Covilhã,
Portugal; Institute of Mathematics Academy of Sciences of Uzbekistan,
Tashkent, Uzbekistan<EMAIL_ADDRESS>Ivan Kaygorodov444CMA-UBI,
Universidade da Beira Interior, Covilhã, Portugal; Moscow Center for
Fundamental and Applied Mathematics, Moscow, Russia; Saint Petersburg
University, Russia<EMAIL_ADDRESS>& Abror
Khudoyberdiyev555Institute of Mathematics Academy of Sciences of Uzbekistan,
Tashkent, Uzbekistan; National University of Uzbekistan, Tashkent, Uzbekistan,
<EMAIL_ADDRESS>

Abstract: This paper is devoted to the complete algebraic and geometric
classification of complex $5$-dimensional nilpotent binary Leibniz and
$4$-dimensional nilpotent mono Leibniz algebras. As a corollary, we have the
complete algebraic and geometric classification of complex $4$-dimensional
nilpotent algebras of nil-index $3$.

Keywords: Leibniz algebra, binary Leibniz algebra, mono Leibniz algebras,
nilpotent algebra, algebraic classification, geometric classification.

MSC2020: 17A30, 17A32, 14D06, 14L30.

## Introduction

The algebraic classification (up to isomorphism) of algebras of dimension $n$
from a certain variety defined by a certain family of polynomial identities is
a classic problem in the theory of non-associative algebras. There are many
results related to the algebraic classification of small-dimensional algebras
in the varieties of Jordan, Lie, Leibniz, Zinbiel, and many other algebras
[30, 2, 5, 39, 14, 6] and references in [31]. Geometric properties of a
variety of algebras defined by a family of polynomial identities have been an
object of study since 1970’s (see, [48, 49, 32, 6, 17, 12, 22, 23, 5, 29, 35]
and references in [31]). Gabriel described the irreducible components of the
variety of $4$-dimensional unital associative algebras [17]. Grunewald and
O’Halloran calculated the degenerations for the variety of $5$-dimensional
nilpotent Lie algebras [22]. Degenerations have also been used to study a
level of complexity of an algebra [49, 48]. The study of degenerations of
algebras is very rich and closely related to deformation theory, in the sense
of Gerstenhaber [20].

Let $\Omega$ be a variety of algebras defined by a family of polynomial
identities, then we say that an algebra ${\rm A}\in\Omega_{i}$ if and only if
each $i$-generated subalgebra of ${\rm A}$ gives an algebra from $\Omega.$ In
particular, if ${\rm A}\in\Omega_{1},$ then ${\rm A}$ is a mono-$\Omega$
algebra, if ${\rm A}\in\Omega_{2},$ then ${\rm A}$ is a binary-$\Omega$
algebra. For example, let ${\rm Ass}$ be the class of associative algebras,
then by Artin’s theorem, the class ${\rm Ass}_{2}$ coincides with the class of
alternative algebras. Albert’s theorem follows that the class ${\rm Ass}_{1}$
coincides with the class of power-associative algebras [8]. It is easy to see
that ${\rm Lie}_{1}$ coincides with anticommutative algebras, i.e., they
satisfy the identity $x^{2}=0$ and the identities of ${\rm Lie}_{2}$ are
described by Gainov [18]. The main non-trivial example of binary Lie algebras
is the class of Malcev algebras, defined by Malcev in [38]. The algebraic
theory of binary Lie algebras was developed in some papers by Kuzmin,
Filippov, and Grishkov (see, for example, [37, 21, 24] and references
therein). So, Kuzmin proved Engel’s theorem for binary Lie algebras [37].
Grishkov described complex semisimple finite-dimensional binary-Lie algebras
[21] Filippov characterized prime binary-Lie algebras [24]. Umirbaev proved
that the variety of complex metabelian binary-Lie algebras is Spechtian (i.e.,
every subvariety of it has a finite basis of identities) in [47]. Chupina
proved that two complex finite-dimensional binary Lie algebras are isomorphic
if their lattices of subalgebras [15]. The question of specialty of binary-Lie
algebras is considered in [10]. Arenas and Arenas-Carmona studied the
universal Poisson envelope for binary-Lie algebras in [9]. On the other hand,
the theory of binary $(-1,1)$-algebras is also under intensive consideration
(the variety of $(-1,1)$-algebras is also known as Lie-admissible right
alternative algebras). So, the identities of binary $(-1,1)$-algebras are
described by Kleinfeld, Smith and Pchelintsev in [41, 36]. Later they
developed the theory of binary $(-1,1)$-algebras. For example, Pchelintsev
proved that if a complex binary $(-1,1)$-algebra is a nil algebra of bounded
index, then it is locally nilpotent [42] and described irreducible binary
$(-1,1)$-bimodules over simple finite-dimensional algebras [43]. Hentzel and
Smith proved that each complex simple binary $(-1,1)$ nil algebra is
associative [26]. Recently, defining identities for mono and binary Zinbiel
algebras are described in [28] and defining identities for mono and binary
Leibniz algebras are described in [27, 19]. On the other hand, the defining
identities for mono symmetric Zinbiel and mono symmetric Leibniz algebras
coincide with nil-algebras of nil-index 3 [11].

An algebra ${\bf A}$ is called a Leibniz algebra if it satisfies the identity

$(xy)z=(xz)y+x(yz).$

Leibniz algebras present a ”non antisymmetric” generalization of Lie algebras.
It appeared in some papers of Bloh [in the 1960s] and Loday [in 1990s].
Recently, they appeared in many geometric and physics applications (see, for
example, [16, 40, 44, 46] and references therein). A systematic study of
algebraic properties of Leibniz algebras is started from the Loday paper. So,
several classical theorems from Lie algebras theory have been extended to the
Leibniz algebras case; many classification results regarding nilpotent,
solvable, simple, and semisimple Leibniz algebras are obtained (see, for
example, [13, 50, 7, 4, 16, 35, 40, 33, 44] and references therein). Leibniz
algebras is a particular case of terminal algebras and, on the other hand,
symmetric Leibniz algebras are Poisson admissible algebras. In the present
paper, based on a known classification of $5$-dimensional nilpotent Leibniz
algebras [3], we give the algebraic and geometric classification of complex
$5$-dimensional nilpotent binary Leibniz and complex $4$-dimensional mono
Leibniz algebras.

Our method for classifying nilpotent Leibniz algebras is based on the
calculation of central extensions of nilpotent algebras of smaller dimensions
from the same variety. The algebraic study of central extensions of algebras
has been an important topic for years [25, 45]. First, Skjelbred and Sund used
central extensions of Lie algebras to obtain a classification of nilpotent Lie
algebras [45]. Note that the Skjelbred-Sund method of central extensions is an
important tool in the classification of nilpotent algebras. Using the same
method, $4$-dimensional nilpotent (bicommutative, commutative, terminal, and
so on) algebras, $5$-dimensional nilpotent (Zinbiel, symmetric Leibniz and so
on) algebras, $6$-dimensional nilpotent (anticommutative, binary Lie [1], Lie
[14], Tortkara and so on) algebras, $8$-dimensional dual Mock-Lie algebras
[12], and some others have been described. Our main results related to the
algebraic classification of cited varieties are summarized below.

######  Theorem A.

Up to isomorphism, there are infinitely many isomorphism classes of complex
$5$-dimensional nilpotent (non-Leibniz) binary Leibniz algebras, described
explicitly in section 1.3.2 in terms of $1$ one-parameter family and $13$
additional isomorphism classes.

######  Theorem B.

Up to isomorphism, there are infinitely many complex $4$-dimensional nilpotent
(non-binary Leibniz) mono Leibniz algebras, described explicitly in section
1.4.6 in terms of $10$ one-parameter families and $12$ additional isomorphism
classes.

######  Theorem C.

Up to isomorphism, there are infinitely many complex $4$-dimensional nilpotent
(non-$2$-step nilpotent) algebras of nil-index $3$, described explicitly in
section 1.5 in terms of $4$ one-parameter families and $11$ additional
isomorphism classes.

The degenerations between the (finite-dimensional) algebras from a certain
variety $\mathfrak{V}$ defined by a set of identities have been actively
studied in the past decade. The description of all degenerations allows one to
find the so-called rigid algebras and families of algebras, i.e. those whose
orbit closures under the action of the general linear group form irreducible
components of $\mathfrak{V}$ (with respect to the Zariski topology). We list
here some works in which the rigid algebras of the varieties of all
$4$-dimensional nilpotent commutative algebras, all $6$-dimensional nilpotent
anticommutative algebras, all $8$-dimensional dual Mock Lie algebras [12] have
been found. A full description of degenerations has been obtained for
$2$-dimensional algebras, for $3$-dimensional anticommutative algebras, for
$3$-dimensional Leibniz algebras, for $4$-dimensional binary Lie algebras in
[34], for $4$-dimensional Zinbiel and $4$-dimensional nilpotent Leibniz
algebras in [35], for $5$-dimensional nilpotent Malcev algebras in [34], for
$6$-dimensional nilpotent Lie algebras in [22], for $8$-dimensional $2$-step
nilpotent anticommutative algebras and so on. Our main results related to the
geometric classification of cited varieties are summarized below.

######  Theorem D.

The variety of complex $5$-dimensional nilpotent binary Leibniz algebras has
dimension 24 and it has 10 irreducible components (in particular, there is
only one rigid algebra in this variety).

######  Theorem E.

The variety of complex $4$-dimensional nilpotent algebras of nil-index $3$ has
dimension 15 and it has 2 irreducible components (in particular, there are no
rigid algebras in this variety).

######  Theorem F.

The variety of complex $4$-dimensional nilpotent mono Leibniz algebras has
dimension 15 and it has 3 irreducible components (in particular, there is only
one rigid algebra in this variety).

##  1\. The algebraic classification of nilpotent binary and mono Leibniz
algebras

###  1.1. Preliminaries and basic definitions

Further we use the notation

${\mathcal{L}}(x,y,z)$ | $=$ | $x(yz)-(xy)z+(xz)y,$
---|---|---
${\mathcal{L}_{\theta}}(x,y,z)$ | $=$ | $\theta(x,yz)-\theta(xy,z)+\theta(xz,y),$

where $\theta\colon{\bf A}\times{\bf A}\longrightarrow{\mathbb{V}},$
${\mathbb{V}}$ is a vector space over ${\mathbb{C}}$, and $\bf A$ is an
algebra over $\mathbb{C}$.

######  Definition.

A complex vector space is called a Leibniz algebra if it satisfies
${\mathcal{L}}(x,y,z)=0.$ A complex vector space is called a binary Leibniz
algebra if every two-generated subalgebra is a Leibniz algebra. A complex
vector space is called a mono Leibniz algebra if every one-generated
subalgebra is a Leibniz algebra.

In [27] it is proved that the algebra ${\bf A}$ is binary Leibniz if and only
if it satisfies the identities

${\mathcal{L}}(x,y,z)+{\mathcal{L}}(y,x,z)$ | $=$ | $0,$ | ${\mathcal{L}}(x,y,z)+{\mathcal{L}}(z,y,x)$ | $=$ | $0,$
---|---|---|---|---|---
${\mathcal{L}}(x,y,zt)+{\mathcal{L}}(x,t,zy)+{\mathcal{L}}(z,y,xt)+{\mathcal{L}}(z,t,xy)$ | $=$ | $0.$

In [27] it is proved that the algebra ${\bf A}$ is mono Leibniz if and only if
it satisfies the identities

${\mathcal{L}}(a,a,a)$ | $=$ |  $0,$ ${\mathcal{L}}(aa,a,a)$ | $=$ | $0.$
---|---|---|---|---

By linearization these identities, we have

${\mathcal{L}}(x,y,z)+{\mathcal{L}}(y,x,z)+{\mathcal{L}}(y,z,x)+{\mathcal{L}}(x,z,y)+{\mathcal{L}}(z,x,y)+{\mathcal{L}}(z,y,x)$ | $=$ | $0,$
---|---|---
${\mathcal{L}}(xy,z,t)+{\mathcal{L}}(xy,t,z)+{\mathcal{L}}(xz,y,t)+{\mathcal{L}}(xt,y,z)+{\mathcal{L}}(xz,t,y)+{\mathcal{L}}(xt,z,y)$ | $+$ |
${\mathcal{L}}(yx,z,t)+{\mathcal{L}}(yx,t,z)+{\mathcal{L}}(zx,y,t)+{\mathcal{L}}(tx,y,z)+{\mathcal{L}}(zx,t,y)+{\mathcal{L}}(tx,z,y)$ | $+$ |
${\mathcal{L}}(yz,x,t)+{\mathcal{L}}(yt,x,z)+{\mathcal{L}}(zy,x,t)+{\mathcal{L}}(ty,x,z)+{\mathcal{L}}(zt,x,y)+{\mathcal{L}}(tz,x,y)$ | $+$ |
${\mathcal{L}}(yz,t,x)+{\mathcal{L}}(yt,z,x)+{\mathcal{L}}(zy,t,x)+{\mathcal{L}}(ty,z,x)+{\mathcal{L}}(zt,y,x)+{\mathcal{L}}(tz,y,x)$ | $=$ | $0.$

From the definition of binary and mono Leibniz algebras we can conclude the
following:

  1. (1)

_For binary Leibniz algebras_ :

     * •

There are no nontrivial $1$-dimensional nilpotent binary Leibniz algebras.

     * •

Two-dimensional and three-dimensional nilpotent binary Leibniz algebras are
Leibniz algebras.

     * •

Two-generated binary Leibniz algebras are Leibniz algebra.

     * •

A binary Leibniz algebra $\mathfrak{L},$ such that for $\mathfrak{L}^{3}=0,$
is a Leibniz algebra.

Thus, non-Leibniz binary Leibniz algebras should be at least three generated.
Consequently, we have that any nilpotent binary Leibniz algebra with a
dimension less than five is a Leibniz algebra.

  2. (2)

_For mono Leibniz algebras_

     * •

There are no nontrivial $1$-dimensional nilpotent mono Leibniz algebras.

     * •

One-generated mono Leibniz algebras are Leibniz algebras.

     * •

Two and three-dimensional nilpotent mono Leibniz algebras are Leibniz
algebras.

     * •

A mono Leibniz algebra $\mathfrak{L},$ such that for $\mathfrak{L}^{3}=0,$ is
a Leibniz algebra.

Thus, we conclude that any nilpotent non-Leibniz mono Leibniz algebra has at
least two generators and $\mathfrak{L}^{3}\neq 0$.

In this work, we classify five-dimensional nilpotent non-Leibniz binary
Leibniz algebras and four-dimensional nilpotent non-Leibniz mono Leibniz
algebras.

###  1.2. Method of classification of nilpotent algebras

Throughout this paper, we use the notations and methods well written in [25],
which we have adapted for the binary Leibniz and mono Leibniz algebras case
with some modifications. Further in this section, we give some important
definitions.

As we know, the central extension is formed by the second cohomology space of
a given algebras. We now define the second cohomology space for binary and
unary Leibniz algebras.

  * •

Let $({\bf A},\cdot)$ be a binary Leibniz algebra over $\mathbb{C}$ and
$\mathbb{V}$ be a vector space over ${\mathbb{C}}$. The $\mathbb{C}$-linear
space ${\rm Z}_{\rm BL}^{2}\left(\bf A,\mathbb{V}\right)$ is defined as the
set of all bilinear maps $\theta\colon{\bf A}\times{\bf
A}\longrightarrow{\mathbb{V}}$ such that

${\mathcal{L}_{\theta}}(x,y,z)+{\mathcal{L}_{\theta}}(y,x,z)$ | $=$ | $0,$ | ${\mathcal{L}_{\theta}}(x,y,z)+{\mathcal{L}_{\theta}}(z,y,x)$ | $=$ | $0,$
---|---|---|---|---|---
${\mathcal{L}_{\theta}}(x,y,zt)+{\mathcal{L}_{\theta}}(x,t,zy)+{\mathcal{L}_{\theta}}(z,y,xt)+{\mathcal{L}_{\theta}}(z,t,xy)$ | $=$ | $0.$

These elements will be called binary Leibniz cocycles. For a linear map $f$
from $\bf A$ to $\mathbb{V}$, if we define $\delta f\colon{\bf A}\times{\bf
A}\longrightarrow{\mathbb{V}}$ by $\delta f(x,y)=f(xy)$, then $\delta f\in{\rm
Z}_{\rm BL}^{2}\left({\bf A},{\mathbb{V}}\right)$. We define ${\rm
B}^{2}\left({\bf A},{\mathbb{V}}\right)=\left\\{\theta=\delta f\ :f\in{\rm
Hom}\left({\bf A},{\mathbb{V}}\right)\right\\}$. We define the second
cohomology space ${\rm H}_{\rm BL}^{2}\left({\bf A},{\mathbb{V}}\right)$ as
the quotient space ${\rm Z}_{\rm BL}^{2}\left({\bf
A},{\mathbb{V}}\right)\big{/}{\rm B}^{2}\left({\bf A},{\mathbb{V}}\right)$.

  * •

Let $({\bf A},\cdot)$ be a mono Leibniz algebra over $\mathbb{C}$ and
$\mathbb{V}$ be a vector space over ${\mathbb{C}}$. The $\mathbb{C}$-linear
space ${\rm Z}_{\rm ML}^{2}\left(\bf A,\mathbb{V}\right)$ is defined as the
set of all bilinear maps $\theta\colon{\bf A}\times{\bf
A}\longrightarrow{\mathbb{V}}$ such that

${\mathcal{L}_{\theta}}(x,y,z)+{\mathcal{L}_{\theta}}(y,x,z)+{\mathcal{L}_{\theta}}(y,z,x)+{\mathcal{L}_{\theta}}(x,z,y)+{\mathcal{L}_{\theta}}(z,x,y)+{\mathcal{L}_{\theta}}(z,y,x)$ | $=$ | $0,$
---|---|---
${\mathcal{L}_{\theta}}(xy,z,t)+{\mathcal{L}_{\theta}}(xy,t,z)+{\mathcal{L}_{\theta}}(xz,y,t)+{\mathcal{L}_{\theta}}(xt,y,z)+{\mathcal{L}_{\theta}}(xz,t,y)+{\mathcal{L}_{\theta}}(xt,z,y)$ | $+$ |
${\mathcal{L}_{\theta}}(yx,z,t)+{\mathcal{L}_{\theta}}(yx,t,z)+{\mathcal{L}_{\theta}}(zx,y,t)+{\mathcal{L}_{\theta}}(tx,y,z)+{\mathcal{L}_{\theta}}(zx,t,y)+{\mathcal{L}_{\theta}}(tx,z,y)$ | $+$ |
${\mathcal{L}_{\theta}}(yz,x,t)+{\mathcal{L}_{\theta}}(yt,x,z)+{\mathcal{L}_{\theta}}(zy,x,t)+{\mathcal{L}_{\theta}}(ty,x,z)+{\mathcal{L}_{\theta}}(zt,x,y)+{\mathcal{L}_{\theta}}(tz,x,y)$ | $+$ |
${\mathcal{L}_{\theta}}(yz,t,x)+{\mathcal{L}_{\theta}}(yt,z,x)+{\mathcal{L}_{\theta}}(zy,t,x)+{\mathcal{L}_{\theta}}(ty,z,x)+{\mathcal{L}_{\theta}}(zt,y,x)+{\mathcal{L}_{\theta}}(tz,y,x)$ | $=$ | $0.$

These elements will be called mono Leibniz cocycles. We define the second
cohomology space ${\rm H}_{\rm ML}^{2}\left({\bf A},{\mathbb{V}}\right)$ as
the quotient space ${\rm Z}_{\rm ML}^{2}\left({\bf
A},{\mathbb{V}}\right)\big{/}{\rm B}^{2}\left({\bf A},{\mathbb{V}}\right)$.

Now when we say ${\bf A}$ algebra we mean binary or mono Leibniz algebra. Let
$\operatorname{Aut}({\bf A})$ be the automorphism group of ${\bf A}$ and let
$\phi\in\operatorname{Aut}({\bf A})$. For $\theta$ cocycles of the algebra
${\bf A}$, define the action of the group $\operatorname{Aut}({\bf A})$ on
space of cocycles of the algebra ${\bf A}$ by
$\phi\theta(x,y)=\theta\left(\phi\left(x\right),\phi\left(y\right)\right)$. It
is easy to verify that ${\rm B}^{2}\left({\bf A},{\mathbb{V}}\right)$ is
invariant under the action of $\operatorname{Aut}({\bf A}).$ So, we have an
induced action of $\operatorname{Aut}({\bf A})$ on ${\rm H}_{\rm
BL}^{2}\left({\bf A},{\mathbb{V}}\right)$ (${\rm H}_{\rm ML}^{2}\left({\bf
A},{\mathbb{V}}\right)$).

Let $\bf A$ be a algebra of dimension $m$ over $\mathbb{C}$ and ${\mathbb{V}}$
be a $\mathbb{C}$-vector space of dimension $k$. For the bilinear map
$\theta$, define on the linear space ${\bf A}_{\theta}={\bf
A}\oplus{\mathbb{V}}$ the bilinear product “ $\left[-,-\right]_{{\bf
A}_{\theta}}$” by $\left[x+x^{\prime},y+y^{\prime}\right]_{{\bf
A}_{\theta}}=xy+\theta(x,y)$ for all $x,y\in{\bf
A},x^{\prime},y^{\prime}\in{\mathbb{V}}$. The algebra ${\bf A}_{\theta}$ is
called a $k$-dimensional central extension of ${\bf A}$ by ${\mathbb{V}}$. One
can easily check that ${\bf A_{\theta}}$ is a binary (mono) Leibniz algebra if
and only if $\theta$ binary (mono) Leibniz cocycle of the algebra ${\bf A}$.

Call the set $\operatorname{Ann}(\theta)=\left\\{x\in{\bf
A}:\theta\left(x,{\bf A}\right)+\theta\left({\bf A},x\right)=0\right\\}$ the
annihilator of $\theta$. We recall that the annihilator of an algebra ${\bf
A}$ is defined as the ideal $\operatorname{Ann}({\bf A})=\left\\{x\in{\bf
A}:x{\bf A}+{\bf A}x=0\right\\}$. Observe that $\operatorname{Ann}\left({\bf
A}_{\theta}\right)=(\operatorname{Ann}(\theta)\cap\operatorname{Ann}({\bf
A}))\oplus{\mathbb{V}}$.

A well known result talks that every algebra with a non-zero annihilator is a
central extension of a smaller-dimensional algebra.

######  Definition 1.

Let ${\bf A}$ be an algebra and $I$ be a subspace of $\operatorname{Ann}({\bf
A})$. If ${\bf A}={\bf A}_{0}\oplus I$ then $I$ is called an annihilator
component of ${\bf A}$. A central extension of an algebra $\bf A$ without
annihilator component is called a non-split central extension.

Our task is to find all central extensions of an algebra $\bf A$ by a space
${\mathbb{V}}$. In order to solve the isomorphism problem we need to study the
action of $\operatorname{Aut}({\bf A})$ on ${\rm H^{2}}\left({\bf
A},{\mathbb{V}}\right)$. To do that, let us fix a basis $e_{1},\ldots,e_{s}$
of ${\mathbb{V}}$, and $\theta\in{\rm Z^{2}}\left({\bf
A},{\mathbb{V}}\right)$. Then $\theta$ can be uniquely written as
$\theta\left(x,y\right)=\displaystyle\sum_{i=1}^{s}\theta_{i}\left(x,y\right)e_{i}$,
where $\theta_{i}\in{\rm Z^{2}}\left({\bf A},\mathbb{C}\right)$. Moreover,
$\operatorname{Ann}(\theta)=\operatorname{Ann}(\theta_{1})\cap\operatorname{Ann}(\theta_{2})\cap\ldots\cap\operatorname{Ann}(\theta_{s})$.
Furthermore, $\theta\in{\rm B^{2}}\left({\bf A},{\mathbb{V}}\right)$ if and
only if all $\theta_{i}\in{\rm B^{2}}\left({\bf A},\mathbb{C}\right)$. It is
not difficult to prove (see [25, Lemma 13]) that given a binary (resp. mono)
Leibniz algebra ${\bf A}_{\theta}$, if we write as above
$\theta\left(x,y\right)=\displaystyle\sum_{i=1}^{s}\theta_{i}\left(x,y\right)e_{i}\in{\rm
Z^{2}}\left({\bf A},{\mathbb{V}}\right)$ and
$\operatorname{Ann}(\theta)\cap\operatorname{Ann}\left({\bf A}\right)=0$, then
${\bf A}_{\theta}$ has an annihilator component if and only if
$\left[\theta_{1}\right],\left[\theta_{2}\right],\ldots,\left[\theta_{s}\right]$
are linearly dependent in ${\rm H^{2}}\left({\bf A},\mathbb{C}\right)$.

Let ${\mathbb{V}}$ be a finite-dimensional vector space over $\mathbb{C}$. The
Grassmannian $G_{k}\left({\mathbb{V}}\right)$ is the set of all
$k$-dimensional linear subspaces of ${\mathbb{V}}$. Let $G_{s}\left({\rm
H^{2}}\left({\bf A},\mathbb{C}\right)\right)$ be the Grassmannian of subspaces
of dimension $s$ in ${\rm H^{2}}\left({\bf A},\mathbb{C}\right)$. There is a
natural action of $\operatorname{Aut}({\bf A})$ on $G_{s}\left({\rm
H^{2}}\left({\bf A},\mathbb{C}\right)\right)$. Let
$\phi\in\operatorname{Aut}({\bf A})$. For
$W=\left\langle\left[\theta_{1}\right],\left[\theta_{2}\right],\dots,\left[\theta_{s}\right]\right\rangle\in
G_{s}\left({\rm H^{2}}\left({\bf A},\mathbb{C}\right)\right)$ define $\phi
W=\left\langle\left[\phi\theta_{1}\right],\left[\phi\theta_{2}\right],\dots,\left[\phi\theta_{s}\right]\right\rangle$.
We denote the orbit of $W\in G_{s}\left({\rm H^{2}}\left({\bf
A},\mathbb{C}\right)\right)$ under the action of $\operatorname{Aut}({\bf A})$
by $\operatorname{Orb}(W)$. Given

$W_{1}=\left\langle\left[\theta_{1}\right],\left[\theta_{2}\right],\dots,\left[\theta_{s}\right]\right\rangle,W_{2}=\left\langle\left[\vartheta_{1}\right],\left[\vartheta_{2}\right],\dots,\left[\vartheta_{s}\right]\right\rangle\in
G_{s}\left({\rm H^{2}}\left({\bf A},\mathbb{C}\right)\right),$

we easily have that if $W_{1}=W_{2}$, then
$\bigcap\limits_{i=1}^{s}\operatorname{Ann}(\theta_{i})\cap\operatorname{Ann}\left({\bf
A}\right)=\bigcap\limits_{i=1}^{s}\operatorname{Ann}(\vartheta_{i})\cap\operatorname{Ann}({\bf
A})$, and therefore we can introduce the set

${\bf T}_{s}({\bf
A})=\left\\{W=\left\langle\left[\theta_{1}\right],\left[\theta_{2}\right],\dots,\left[\theta_{s}\right]\right\rangle\in
G_{s}\left({\rm H^{2}}\left({\bf
A},\mathbb{C}\right)\right):\bigcap\limits_{i=1}^{s}\operatorname{Ann}(\theta_{i})\cap\operatorname{Ann}({\bf
A})=0\right\\},$

which is stable under the action of $\operatorname{Aut}({\bf A})$.

Now, let ${\mathbb{V}}$ be an $s$-dimensional linear space and let us denote
by ${\bf E}\left({\bf A},{\mathbb{V}}\right)$ the set of all non-split
$s$-dimensional central extensions of ${\bf A}$ by ${\mathbb{V}}$. By above,
we can write

${\bf E}\left({\bf A},{\mathbb{V}}\right)=\left\\{{\bf
A}_{\theta}:\theta\left(x,y\right)=\sum_{i=1}^{s}\theta_{i}\left(x,y\right)e_{i}\
\ \text{and}\ \
\left\langle\left[\theta_{1}\right],\left[\theta_{2}\right],\dots,\left[\theta_{s}\right]\right\rangle\in{\bf
T}_{s}({\bf A})\right\\}.$

We also have the following result, which can be proved as in [25, Lemma 17].

######  Lemma 2.

Let ${\bf A}_{\theta},{\bf A}_{\vartheta}\in{\bf E}\left({\bf
A},{\mathbb{V}}\right)$. Suppose that
$\theta\left(x,y\right)=\displaystyle\sum_{i=1}^{s}\theta_{i}\left(x,y\right)e_{i}$
and
$\vartheta\left(x,y\right)=\displaystyle\sum_{i=1}^{s}\vartheta_{i}\left(x,y\right)e_{i}$.
Then the binary (resp. mono) Leibniz algebras ${\bf A}_{\theta}$ and ${\bf
A}_{\vartheta}$ are isomorphic if and only if

$\operatorname{Orb}\left\langle\left[\theta_{1}\right],\left[\theta_{2}\right],\dots,\left[\theta_{s}\right]\right\rangle=\operatorname{Orb}\left\langle\left[\vartheta_{1}\right],\left[\vartheta_{2}\right],\dots,\left[\vartheta_{s}\right]\right\rangle.$

This shows that there exists a one-to-one correspondence between the set of
$\operatorname{Aut}({\bf A})$-orbits on ${\bf T}_{s}\left({\bf A}\right)$ and
the set of isomorphism classes of ${\bf E}\left({\bf A},{\mathbb{V}}\right)$.
Consequently, we have a procedure that allows us, given a binary (resp. mono)
Leibniz algebra ${\bf A}^{\prime}$ of dimension $n-s$, to construct all non-
split central extensions of ${\bf A}^{\prime}$. This procedure is:

  1. (1)

For a given binary (resp. mono) Leibniz algebra ${\bf A}^{\prime}$ of
dimension $n-s$, determine ${\rm H^{2}}({\bf A}^{\prime},\mathbb{C})$,
$\operatorname{Ann}({\bf A}^{\prime})$ and $\operatorname{Aut}({\bf
A}^{\prime})$.

  2. (2)

Determine the set of $\operatorname{Aut}({\bf A}^{\prime})$-orbits on ${\bf
T}_{s}({\bf A}^{\prime})$.

  3. (3)

For each orbit, construct the binary (resp. mono) Leibniz algebra associated
with a representative of it.

The above-described method gives all (Leibniz and non-Leibniz) binary (resp.
mono) Leibniz algebras. But we are interested in developing this method in
such a way that it only gives non-Leibniz binary (resp. mono) Leibniz algebras
because the classification of all five-dimensional nilpotent Leibniz algebras
is given in [3]. Clearly, any central extension of a non-Leibniz binary (resp.
mono) Leibniz is non-Leibniz. But a Leibniz algebra may have extensions that
are not binary (resp. mono) Leibniz algebras. More precisely, let
${\mathcal{L}}$ be a Leibniz algebra and $\theta\in{\rm Z_{\rm
BL}^{2}}({\mathcal{L}},{\mathbb{C}})$ (resp. $\theta\in{\rm Z_{\rm
ML}^{2}}({\mathcal{L}},{\mathbb{C}})$). Then ${\mathcal{L}}_{\theta}$ is a
Leibniz algebra if and only if

$\theta(xy,z)=\theta(xz,y)+\theta(x,yz)$

for all $x,y,z\in{\mathcal{L}}.$ Define the subspace ${\rm
Z_{\mathcal{L}}^{2}}(\mathcal{L},{\mathbb{C}})$ of ${\rm
Z_{\rm{BL}}^{2}}({\mathcal{L}},{\mathbb{C}})$ (resp. of ${\rm
Z_{\rm{ML}}^{2}}({\mathcal{L}},{\mathbb{C}})$) by

${\rm Z_{\mathcal{L}}^{2}}({\mathcal{L}},{\mathbb{C}})=\\{\theta\in{\rm Z_{\rm
BL}^{2}}({\mathcal{L}},{\mathbb{C}})$ (resp. ${\rm Z_{\rm
ML}^{2}}({\mathcal{L}},{\mathbb{C}})$)$:\theta(xy,z)=\theta(xz,y)+\theta(x,yz),\text{
for all }x,y,z\in{\mathcal{L}}\\}.$

Observe that ${\rm B^{2}}({\mathcal{L}},{\mathbb{C}})\subseteq{\rm
Z_{\rm{BL}}^{2}}({\mathcal{L}},{\mathbb{C}})$ (resp. ${\rm
Z_{\rm{ML}}^{2}}({\mathcal{L}},{\mathbb{C}})$). Let ${\rm
H_{\mathcal{L}}^{2}}({\mathcal{L}},{\mathbb{C}})={\rm
Z_{\mathcal{L}}^{2}}({\mathcal{L}},{\mathbb{C}})\big{/}{\rm
B^{2}}({\mathcal{L}},{\mathbb{C}}).$ Then ${\rm
H_{\mathcal{L}}^{2}}({\mathcal{L}},{\mathbb{C}})$ is a subspace of ${\rm
H_{\rm{BL}}^{2}}({\mathcal{L}},{\mathbb{C}}).$ Define

$\displaystyle{\bf R}_{s}({\mathcal{L}})$ $\displaystyle=$
$\displaystyle\left\\{{\mathcal{W}}\in{\bf
T}_{s}({\mathcal{L}}):{\mathcal{W}}\in G_{s}({\rm
H_{\mathcal{L}}^{2}}({\mathcal{L}},{\mathbb{C}}))\right\\},$
$\displaystyle{\bf U}_{s}({\mathcal{L}})$ $\displaystyle=$
$\displaystyle\left\\{{\mathcal{W}}\in{\bf
T}_{s}({\mathcal{L}}):{\mathcal{W}}\notin G_{s}({\rm
H_{\mathcal{L}}^{2}}({\mathcal{L}},{\mathbb{C}}))\right\\}.$

Then ${\bf T}_{s}({\mathcal{L}})={\bf
R}_{s}({\mathcal{L}})\mathbin{\mathaccent 0{\cdot}\cup}{\bf
U}_{s}({\mathcal{L}}).$ The sets ${\bf R}_{s}({\mathcal{L}})$ and ${\bf
U}_{s}({\mathcal{L}})$ are stable under the action of
$\operatorname{Aut}({\mathcal{L}}).$ Thus, the binary (resp. mono) Leibniz
algebras corresponding to the representatives of
$\operatorname{Aut}({\mathcal{L}})$-orbits on ${\bf R}_{s}({\mathcal{L}})$ are
Leibniz algebras, while those corresponding to the representatives of
$\operatorname{Aut}({\mathcal{L}})$-orbits on ${\bf U}_{s}({\mathcal{L}})$ are
non-Leibniz binary (resp. mono) Leibniz algebras. Hence, we may construct all
non-split non-Leibniz binary (resp. mono) Leibniz algebras $\bf{A}$ of
dimension $n$ with $s$-dimensional annihilator from a given binary (resp.
mono) Leibniz algebra $\bf{A}^{\prime}$ of dimension $n-s$ in the following
way:

  1. (1)

If $\bf{A}^{\prime}$ is non-Leibniz, then apply the procedure.

  2. (2)

Otherwise, do the following:

    1. (a)

Determine ${\bf U}_{s}(\bf{A}^{\prime})$ and
$\operatorname{Aut}(\bf{A}^{\prime}).$

    2. (b)

Determine the set of $\operatorname{Aut}(\bf{A}^{\prime})$-orbits on ${\bf
U}_{s}(\bf{A}^{\prime}).$

    3. (c)

For each orbit, construct the binary (resp. mono) Leibniz algebra
corresponding to one of its representatives.

####  1.2.1. Notations

Let us introduce the following notations. Let ${\bf A}$ be a nilpotent algebra
with a basis $e_{1},e_{2},\ldots,e_{n}.$ Then by $\Delta_{ij}$ we will denote
the bilinear form $\Delta_{ij}:{\bf A}\times{\bf A}\longrightarrow\mathbb{C}$
with $\Delta_{ij}(e_{l},e_{m})=\delta_{il}\delta_{jm}.$ The set
$\left\\{\Delta_{ij}:1\leq i,j\leq n\right\\}$ is a basis for the linear space
of bilinear forms on ${\bf A},$ so every $\theta\in{\rm Z}_{\rm BL}^{2}({\bf
A},\bf\mathbb{V})$ (resp. $\theta\in{\rm Z}_{\rm ML}^{2}({\bf
A},\bf\mathbb{V})$ ) can be uniquely written as
$\theta=\displaystyle\sum_{1\leq i,j\leq n}c_{ij}\Delta_{{i}{j}}$, where
$c_{ij}\in\mathbb{C}$. Let us fix the following notations for our nilpotent
algebras:

$\begin{array}[]{lll}{\mathcal{N}}_{j}&\mbox{---}&j\mbox{th
}3\mbox{-dimensional $2$-step nilpotent algebra.}\\\
{\mathfrak{N}}_{j}&\mbox{---}&j\mbox{th }4\mbox{-dimensional $2$-step
nilpotent algebra.}\\\ \mathbb{M}_{j}&\mbox{---}&j\mbox{th
}4\mbox{-dimensional nilpotent (non-Leibniz) mono Leibniz algebra.}\\\ {\bf
B}_{j}&\mbox{---}&j\mbox{th }5\mbox{-dimensional nilpotent (non-Leibniz)
binary Leibniz algebra.}\\\ \end{array}$

###  1.3. Classification of 5-dimensional nilpotent binary Leibniz algebras

When we construct algebras by the central extension, the number of generators
does not change. So the generators of our algebra that we are looking at
through the central extension must be equal to 3. So we construct non-Leibniz
binary Leibniz algebras through one-dimensional central extensions of 4
generated four dimensional nilpotent Leibniz algebras. Such algebras are as
follows.

${\mathfrak{N}}_{01}$ | $:$ | $e_{1}e_{1}=e_{2}$ |  |  |  |
---|---|---|---|---|---|---
${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{01})$ | $=$ | $\Big{\langle}[\Delta_{13}],[\Delta_{14}],[\Delta_{21}],[\Delta_{31}],[\Delta_{33}],[\Delta_{34}],[\Delta_{41}],[\Delta_{43}],[\Delta_{44}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{01})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{01})$
${\mathfrak{N}}_{03}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ |  |  |
${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{03})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{14}],[\Delta_{22}],[\Delta_{24}],[\Delta_{13}-\Delta_{31}],[\Delta_{23}-\Delta_{32}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{03})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{03})\oplus\langle[\Delta_{34}-\Delta_{43}]\rangle$
${\mathfrak{N}}_{04}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ |  |
${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{04}^{\alpha\neq 0})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{14}],[\Delta_{21}],[\Delta_{22}],[\Delta_{24}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{04}^{0})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{14}],[\Delta_{21}],[\Delta_{22}],[\Delta_{24}],[\Delta_{31}+\Delta_{32}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{04}^{\alpha})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{04}^{\alpha})$
${\mathfrak{N}}_{05}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{3}$ |  |
${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{05})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{14}],[\Delta_{21}],[\Delta_{22}],[\Delta_{24}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{05})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{05})$
${\mathfrak{N}}_{06}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$ |  |  |
${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{06})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{06})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{06})$
${\mathfrak{N}}_{09}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=\alpha e_{4}$ | $e_{2}e_{1}=-\alpha e_{4}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{3}=e_{4}$
${\rm H}_{L}^{2}({\mathfrak{N}}_{09}^{\alpha})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{09}^{\alpha})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{09}^{\alpha})$
${\mathfrak{N}}_{10}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$
${\rm H}_{L}^{2}({\mathfrak{N}}_{10})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{10})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{10})$
${\mathfrak{N}}_{11}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |
${\rm H}_{L}^{2}({\mathfrak{N}}_{11})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{11})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{11})$
${\mathfrak{N}}_{15}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |
${\rm H}_{L}^{2}({\mathfrak{N}}_{15})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}_{\rm BL}^{2}({\mathfrak{N}}_{15})$ | $=$ | ${\rm H}_{\mathcal{L}}^{2}({\mathfrak{N}}_{15})$

From the previous table, we obtain that only algebra $\mathfrak{N}_{03}$ has a
non-Leibniz binary Leibniz central extension.

####  1.3.1. Central extensions of ${\mathfrak{N}}_{03}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{11}],$ | $\nabla_{2}=[\Delta_{12}],$ | $\nabla_{3}=[\Delta_{13}-\Delta_{31}],$ | $\nabla_{4}=[\Delta_{14}],$
---|---|---|---
$\nabla_{5}=[\Delta_{22}],$ | $\nabla_{6}=[\Delta_{23}-\Delta_{32}],$ | $\nabla_{7}=[\Delta_{24}],$ | $\nabla_{8}=[\Delta_{41}],$
$\nabla_{9}=[\Delta_{42}],$ | $\nabla_{10}=[\Delta_{44}],$ | $\nabla_{11}=[\Delta_{34}-\Delta_{43}].$ |

Take $\theta=\sum\limits_{i=1}^{11}\alpha_{i}\nabla_{i}\in{\rm H}_{\rm
BL}^{2}(\mathfrak{N}_{03}).$ The automorphism group of $\mathfrak{N}_{03}$
consists of invertible matrices of the form

$\phi=\left(\begin{array}[]{cccc}x&z&0&0\\\ y&t&0&0\\\ u&q&xt-yz&w\\\
v&p&0&r\end{array}\right).$

Since

$\phi^{T}\begin{pmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}&\alpha_{4}\\\
0&\alpha_{5}&\alpha_{6}&\alpha_{7}\\\ -\alpha_{3}&-\alpha_{6}&0&\alpha_{11}\\\
\alpha_{8}&\alpha_{9}&-\alpha_{11}&\alpha_{10}\\\
\end{pmatrix}\phi=\begin{pmatrix}\alpha_{1}^{*}&-\alpha^{*}+\alpha_{2}^{*}&\alpha_{3}^{*}&\alpha_{4}^{*}\\\
\alpha^{*}&\alpha_{5}^{*}&\alpha_{6}^{*}&\alpha_{7}^{*}\\\
-\alpha_{3}^{*}&-\alpha_{6}^{*}&0&\alpha_{11}^{*}\\\
\alpha_{8}^{*}&\alpha_{9}^{*}&-\alpha_{11}^{*}&\alpha_{10}^{*}\\\
\end{pmatrix},$

we have that the action of ${\rm Aut}(\mathfrak{N}_{03})$ on the subspace
$\langle\sum\limits_{i=1}^{11}\alpha_{i}\nabla_{i}\rangle$ is given by
$\langle\sum\limits_{i=1}^{11}\alpha_{i}^{*}\nabla_{i}\rangle,$ where

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1}+xy\alpha_{2}+vx\alpha_{4}+y^{2}\alpha_{5}+vy\alpha_{7}+vx\alpha_{8}+vy\alpha_{9}+v^{2}\alpha_{10},$
---|---|---
$\alpha^{*}_{2}$ | $=$ | $2xz\alpha_{1}+(tx+yz)\alpha_{2}+(px+vz)\alpha_{4}+2ty\alpha_{5}+$
|  | $(tv+py)\alpha_{7}+(px+vz)\alpha_{8}+(tv+py)\alpha_{9}+2pv\alpha_{10}$
$\alpha^{*}_{3}$ | $=$ | $(xt-yz)(x\alpha_{3}+y\alpha_{6}-v\alpha_{11}),$
$\alpha_{4}^{*}$ | $=$ | $r(x\alpha_{4}+y\alpha_{7}+v\alpha_{10}+u\alpha_{11})+w(x\alpha_{3}+y\alpha_{6}-v\alpha_{11}),$
$\alpha_{5}^{*}$ | $=$ | $z^{2}\alpha_{1}+tz\alpha_{2}+pz\alpha_{4}+t^{2}\alpha_{5}+pt\alpha_{7}+pz\alpha_{8}+pt\alpha_{9}+p^{2}\alpha_{10},$
$\alpha_{6}^{*}$ | $=$ | $(tx-yz)(z\alpha_{3}+t\alpha_{6}-p\alpha_{11}),$
$\alpha_{7}^{*}$ | $=$ | $w(z\alpha_{3}+t\alpha_{6}-p\alpha_{11})+r(z\alpha_{4}+t\alpha_{7}+p\alpha_{10}+q\alpha_{11}),$
$\alpha_{8}^{*}$ | $=$ | $-wx\alpha_{3}-wy\alpha_{6}+rx\alpha_{8}+ry\alpha_{9}+rv\alpha_{10}-ru\alpha_{11}+vw\alpha_{11},$
$\alpha_{9}^{*}$ | $=$ | $-wz\alpha_{3}-tw\alpha_{6}+rz\alpha_{8}+rt\alpha_{9}+pr\alpha_{10}-qr\alpha_{11}+pw\alpha_{11},$
$\alpha_{10}^{*}$ | $=$ | $r^{2}\alpha_{10},$
$\alpha_{11}^{*}$ | $=$ | $r(tx-yz)\alpha_{11}.$

Since ${\rm H}_{\rm BL}^{2}(\mathfrak{N}_{03})={\rm
H}_{\mathcal{L}}^{2}(\mathfrak{N}_{03})\oplus\langle[\Delta_{34}-\Delta_{43}]\rangle$
and we are interested only in new algebras, we have $\alpha_{11}\neq 0.$

Then putting

$\begin{array}[]{lcllcl}v&=&(x\alpha_{3}+y\alpha_{6})\alpha_{11}^{-1},&u&=&-((x\alpha_{3}+y\alpha_{6})\alpha_{10}+(x\alpha_{4}+y\alpha_{7})\alpha_{11})\alpha_{11}^{-2},\\\
p&=&(z\alpha_{3}+t\alpha_{6})\alpha_{11}^{-1},&q&=&-((z\alpha_{3}+t\alpha_{6})\alpha_{10}+\left(z\alpha_{4}+t\alpha_{7}\right)\alpha_{11})\alpha_{11}^{-2},\end{array}$

we have

$\alpha^{*}_{3}=\alpha^{*}_{4}=\alpha_{6}^{*}\alpha_{7}^{*}=0.$

This, without loss of generality, we can suppose
$\alpha_{3}=\alpha_{4}=\alpha_{6}=\alpha_{7}=0$. Then $p=v=u=q=0$ and we have
the following relations:

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1}+xy\alpha_{2}+y^{2}\alpha_{5},$
---|---|---
$\alpha_{2}^{*}$ | $=$ | $2xz\alpha_{1}+(tx+yz)\alpha_{2}+2ty\alpha_{5},$
$\alpha_{5}^{*}$ | $=$ | $z^{2}\alpha_{1}+tz\alpha_{2}+t^{2}\alpha_{5},$
$\alpha_{8}^{*}$ | $=$ | $r\left(x\alpha_{8}+y\alpha_{9}\right),$
$\alpha_{9}^{*}$ | $=$ | $r\left(z\alpha_{8}+t\alpha_{9}\right),$
$\alpha_{10}^{*}$ | $=$ | $r^{2}\alpha_{10},$
$\alpha_{11}^{*}$ | $=$ | $r(tx-yz)\alpha_{11}.$

  1. (1)

Let $\alpha_{1}=\alpha_{2}=\alpha_{5}=\alpha_{8}=\alpha_{9}=0$. Then we have
the following subcases:

    1. (a)

if $\alpha_{10}=0,$ then choosing $r=\frac{1}{(tx-yz)\alpha_{11}}$, we have
the representative $\langle\nabla_{11}\rangle$;

    2. (b)

if $\alpha_{10}\neq 0,$ then choosing $r=\frac{(tx-
yz)\alpha_{11}}{\alpha_{10}},$ we have the representative
$\langle\nabla_{10}+\nabla_{11}\rangle$.

  2. (2)

Let $\alpha_{1}=\alpha_{2}=\alpha_{5}=0,(\alpha_{8},\alpha_{9})\neq(0,0)$.
Then without loss of generality, we can assume $\alpha_{8}\neq 0$ and consider
the following subcases:

    1. (a)

if $\alpha_{10}=0,$ then choosing $z=-\frac{\alpha_{9}}{\alpha_{11}}$ and
$t=\frac{\alpha_{8}}{\alpha_{11}}$, we have the representative
$\langle\nabla_{8}+\nabla_{11}\rangle$;

    2. (b)

if $\alpha_{10}\neq 0,$ then choosing $x=0,$ $y=1,$
$r=\frac{\alpha_{9}}{\alpha_{10}},$ $z=-\frac{\alpha_{9}}{\alpha_{11}}$ and
$t=\frac{\alpha_{8}}{\alpha_{11}},$ we have the representative
$\langle\nabla_{8}+\nabla_{10}+\nabla_{11}\rangle$.

  3. (3)

Let $(\alpha_{1},\alpha_{5})=(0,0).$ Then $\alpha_{2}\neq 0$ and taking $y=0,\
z=0$ and we get that:

$\alpha^{*}_{1}$ | $=$ | $0,$ | $\alpha_{2}^{*}$ | $=$ | $tx\alpha_{2},$ | $\alpha_{5}^{*}$ | $=$ | $0,$ | $\alpha_{8}^{*}$ | $=$ | $rx\alpha_{8},$
---|---|---|---|---|---|---|---|---|---|---|---
$\alpha_{9}^{*}$ | $=$ | $rt\alpha_{9},$ | $\alpha_{10}^{*}$ | $=$ | $r^{2}\alpha_{10},$ | $\alpha_{11}^{*}$ | $=$ | $rtx\alpha_{11}.$ |  |  |

Then we have the following cases:

    1. (a)

$\alpha_{8}=\alpha_{9}=\alpha_{10}=0,$ then choosing
$r=\frac{\alpha_{2}}{\alpha_{11}},$ we have the representative
$\langle\nabla_{2}+\nabla_{11}\rangle$;

    2. (b)

$\alpha_{8}=\alpha_{9}=0,\alpha_{10}\neq 0,$ then choosing $t=1,$
$r=\frac{\alpha_{2}}{\alpha_{11}}$ and
$x=\frac{\alpha_{2}\alpha_{10}}{\alpha_{11}^{2}},$ we have the representative
$\langle\nabla_{2}+\nabla_{10}+\nabla_{11}\rangle$;

    3. (c)

$\alpha_{8}\neq 0,\alpha_{9}=0,\alpha_{10}=0,$ then choosing
$r=\frac{\alpha_{2}}{\alpha_{11}}$ and $t=\frac{\alpha_{8}}{\alpha_{11}},$ we
have the representative $\langle\nabla_{2}+\nabla_{8}+\nabla_{11}\rangle$;

    4. (d)

$\alpha_{8}=0,\alpha_{9}\neq 0,\alpha_{10}=0,$ in this case choosing the
suitable automorphism we can obtain $\alpha_{8}\neq
0,\alpha_{9}=0,\alpha_{10}=0,$ which is the case considered above;

    5. (e)

$\alpha_{8}\neq 0,\alpha_{9}=0,\alpha_{10}\neq 0,$ then choosing
$t=\frac{\alpha_{8}}{\alpha_{11}},$ $r=\frac{\alpha_{2}}{\alpha_{11}}$ and
$x=\frac{\alpha_{2}\alpha_{10}}{\alpha_{8}\alpha_{11}},$ we have the
representative $\langle\nabla_{2}+\nabla_{8}+\nabla_{10}+\nabla_{11}\rangle$;

    6. (f)

$\alpha_{8}=0,\alpha_{9}\neq 0,\alpha_{10}\neq 0,$ in this case choosing the
suitable automorphism we can obtain $\alpha_{8}\neq
0,\alpha_{9}=0,\alpha_{10}\neq 0,$ which is the case considered above;

    7. (g)

$\alpha_{8}\neq 0,\alpha_{9}\neq 0,$ then choosing
$r=\frac{\alpha_{2}}{\alpha_{11}},$ $t=\frac{\alpha_{8}}{\alpha_{11}}$ and
$x=\frac{\alpha_{9}}{\alpha_{11}},$ we have the family of representatives
$\langle\nabla_{2}+\nabla_{8}+\nabla_{9}+\alpha\nabla_{10}+\nabla_{11}\rangle$.

  4. (4)

Let $(\alpha_{1},\alpha_{5})\neq(0,0)$. Then without loss of generality, we
can assume $\alpha_{1}\neq 0.$ If $\alpha_{2}^{2}\neq 4\alpha_{1}\alpha_{5}$,
then choosing

$x=-\alpha_{2}+\sqrt{\alpha_{2}^{2}-4\alpha_{1}\alpha_{5}},\quad
z=-\alpha_{2}-\sqrt{\alpha_{2}^{2}-4\alpha_{1}\alpha_{5}},\quad
y=2\alpha_{1},\quad t=2\alpha_{1},$

we have $\alpha_{1}^{*}=0$ and $\alpha_{5}^{*}=0$ and it is the case
considered above.

Therefore, we can suppose $(\alpha_{1},\alpha_{5})\neq(0,0)$ and
$\alpha_{2}^{2}=4\alpha_{1}\alpha_{5}$. By taking
$z=-\frac{t\alpha_{2}}{2\alpha_{1}},$ we have the following

$\alpha^{*}_{1}$ | $=$ | $\frac{\left(2x\alpha_{1}+y\alpha_{2}\right){}^{2}}{4\alpha_{1}},$ | $\alpha_{2}^{*}$ | $=$ | $0,$ | $\alpha_{5}^{*}$ | $=$ | $0,$
---|---|---|---|---|---|---|---|---
$\alpha_{8}^{*}$ | $=$ | $r\left(x\alpha_{8}+y\alpha_{9}\right),$ | $\alpha_{9}^{*}$ | $=$ | $\frac{rt\left(2\alpha_{1}\alpha_{9}-\alpha_{2}\alpha_{8}\right)}{2\alpha_{1}},$ |  |  |
$\alpha_{10}^{*}$ | $=$ | $r^{2}\alpha_{10},$ | $\alpha_{11}^{*}$ | $=$ | $r\left(tx+\frac{ty\alpha_{2}}{2\alpha_{1}}\right)\alpha_{11}.$ |  |  |

Without loss of generality, one can assume $\alpha_{2}=0$ obtain

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1},$ | $\alpha_{8}^{*}$ | $=$ | $r\left(x\alpha_{8}+y\alpha_{9}\right),$ | $\alpha_{9}^{*}$ | $=$ | $rt\alpha_{9},$
---|---|---|---|---|---|---|---|---
$\alpha_{10}^{*}$ | $=$ | $r^{2}\alpha_{10},$ | $\alpha_{11}^{*}$ | $=$ | $rtx\alpha_{11}.$ |  |  |

Now consider following cases:

    1. (a)

$\alpha_{9}=0,\alpha_{8}=0,\alpha_{10}=0,$ then choosing $z=0,$ $x=1,$ $r=1$
and $t=\frac{\alpha_{1}}{\alpha_{11}},$ we have the representative
$\langle\nabla_{1}+\nabla_{11}\rangle$;

    2. (b)

$\alpha_{9}=0,\alpha_{8}=0,\alpha_{10}\neq 0,$ then choosing $z=0,$
$x=\sqrt{\alpha_{10}},$ $t=\frac{\sqrt{\alpha_{1}\alpha_{10}}}{\alpha_{11}}$
and $r=\sqrt{\alpha_{1}},$ we have the representative
$\langle\nabla_{1}+\nabla_{10}+\nabla_{11}\rangle$;

    3. (c)

$\alpha_{9}=0,\alpha_{8}\neq 0,$ then choosing $z=0,$ $x=1,$
$t=\frac{\alpha_{8}}{\alpha_{11}}$ and $r=\frac{\alpha_{1}}{\alpha_{8}},$ we
have the family of representatives
$\langle\nabla_{1}+\nabla_{8}+\alpha\nabla_{10}+\nabla_{11}\rangle$;

    4. (d)

$\alpha_{9}\neq 0,\alpha_{10}=0,$ then choosing $z=0,$
$x=\frac{\alpha_{9}}{\alpha_{11}},$ $r=1,$ $y=-\frac{\alpha_{8}}{\alpha_{11}}$
and $t=\frac{\alpha_{1}\alpha_{9}}{\alpha_{11}^{2}},$ we have the
representative $\langle\nabla_{1}+\nabla_{9}+\nabla_{11}\rangle$;

    5. (e)

$\alpha_{9}\neq 0,\alpha_{10}\neq 0,$ then choosing $z=0,$
$x=\frac{\alpha_{9}}{\alpha_{11}},$ $y=-\frac{\alpha_{8}}{\alpha_{11}},$
$t=\frac{\sqrt{\alpha_{1}\alpha_{10}}}{\alpha_{11}},$ and
$r=\frac{\alpha_{9}\sqrt{\alpha_{1}}}{\alpha_{11}\sqrt{\alpha}_{10}},$ we have
$\langle\nabla_{1}+\nabla_{9}+\nabla_{10}+\nabla_{11}\rangle$.

Summarizing, we have the following distinct orbits

$\langle\nabla_{11}\rangle,$ $\langle\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{8}+\nabla_{11}\rangle,$
$\langle\nabla_{8}+\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{2}+\nabla_{11}\rangle,$
$\langle\nabla_{2}+\nabla_{8}+\nabla_{11}\rangle,$
$\langle\nabla_{2}+\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{2}+\nabla_{8}+\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{2}+\nabla_{8}+\nabla_{9}+\alpha\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{1}+\nabla_{11}\rangle,$
$\langle\nabla_{1}+\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{1}+\nabla_{8}+\alpha\nabla_{10}+\nabla_{11}\rangle,$
$\langle\nabla_{1}+\nabla_{9}+\nabla_{11}\rangle,$
$\langle\nabla_{1}+\nabla_{9}+\nabla_{10}+\nabla_{11}\rangle.$

####  1.3.2. The classification theorem

Now we are ready to summarize all results related to the algebraic
classification of complex $5$-dimensional nilpotent binary Leibniz algebras.

######  Theorem A.

Let ${\bf B}$ be a complex $5$-dimensional nilpotent binary Leibniz algebra.
Then ${\bf B}$ is a Leibniz algebra or isomorphic to one algebra from the
following list:

${\bf B}_{01}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
---|---|---|---|---|---|---|---
${\bf B}_{02}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=e_{5}$ |
${\bf B}_{03}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{1}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |
${\bf B}_{04}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{1}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=e_{5}$
${\bf B}_{05}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
${\bf B}_{06}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{1}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |
${\bf B}_{07}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=e_{5}$ |
${\bf B}_{08}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{1}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=e_{5}$
${\bf B}_{09}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{1}=e_{5}$ |  |
|  | $e_{4}e_{2}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=\alpha e_{5}$ |  |  |
${\bf B}_{10}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |
${\bf B}_{11}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=e_{5}$
${\bf B}_{12}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ |  |
|  | $e_{4}e_{1}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=\alpha e_{5}$ |  |  |
${\bf B}_{13}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{2}=e_{5}$ | $e_{4}e_{3}=-e_{5}$
${\bf B}_{14}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{4}=e_{5}$ |  |
|  | $e_{4}e_{2}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ | $e_{4}e_{4}=e_{5}$ |  |  |

###  1.4. Classification of $4$-dimensional nilpotent mono Leibniz algebras

For algebras constructed by the method of central extension, the number of
generators do not change. So the generators of our algebra that we are looking
for must be equal to $2$. So we construct nilpotent non-Leibniz mono Leibniz
algebras through one-dimensional central extensions of $3$-dimensional
$2$-generated nilpotent Leibniz algebras. Such algebras are as follows.

The list of 2-step nilpotent 3-dimensional mono Leibniz algebras
---
${\mathcal{N}}_{01}$ | $:$ | $e_{1}e_{1}=e_{2}$ |  |  |  |
${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{01})$ | $=$ | $\Big{\langle}[\Delta_{13}],[\Delta_{21}],[\Delta_{31}],[\Delta_{33}]\Big{\rangle}$
${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{01})$ | $=$ | ${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{01})\oplus\langle[\Delta_{23}]\rangle$
${\mathcal{N}}_{02}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ |  |  |
${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{02})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{13}-\Delta_{31}],[\Delta_{22}],[\Delta_{23}-\Delta_{32}]\Big{\rangle}$
${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{02})$ | $=$ | ${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{02})\oplus\langle[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\rangle$
${\mathcal{N}}_{03}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ |  |
${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{03}^{\alpha\neq 0})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{21}],[\Delta_{22}]\Big{\rangle}$
${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{03}^{\alpha\neq 0})$ | $=$ | ${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{03}^{\alpha\neq 0})\oplus\langle[\Delta_{31}+\Delta_{32}],[\Delta_{32}]\rangle$
${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{03}^{0})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{21}],[\Delta_{22}],[\Delta_{31}+\Delta_{32}]\Big{\rangle}$
${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{03}^{0})$ | $=$ | ${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{03}^{0})\oplus\langle[\Delta_{32}]\rangle$
${\mathcal{N}}_{04}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$ |  |  |
${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{04})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{21}],[\Delta_{22}]\Big{\rangle}$
${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{04})$ | $=$ | ${\rm H}_{\rm BL}^{2}({\mathcal{N}}_{04})\oplus\langle[\Delta_{31}],[\Delta_{32}]\rangle$

####  1.4.1. Central extensions of ${\mathcal{N}}_{01}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{13}],$ | $\nabla_{2}=[\Delta_{21}],$ | $\nabla_{3}=[\Delta_{31}],$ | $\nabla_{4}=[\Delta_{23}],$ | $\nabla_{5}=[\Delta_{33}].$
---|---|---|---|---

Take $\theta=\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\in{\rm H_{\rm
ML}^{2}}({\mathcal{N}}_{01}).$ The automorphism group of ${\mathcal{N}}_{01}$
consists of invertible matrices of the form

$\phi=\begin{pmatrix}x&0&0\\\ y&x^{2}&z\\\ u&0&t\\\ \end{pmatrix}.$

Since

$\phi^{T}\begin{pmatrix}0&0&\alpha_{1}\\\ \alpha_{2}&0&\alpha_{4}\\\
\alpha_{3}&0&\alpha_{5}\\\
\end{pmatrix}\phi=\begin{pmatrix}\alpha^{*}&0&\alpha_{1}^{*}\\\
\alpha_{2}^{*}&0&\alpha_{4}^{*}\\\ \alpha_{3}^{*}&0&\alpha_{5}^{*}\\\
\end{pmatrix},$

we have that the action of ${\rm Aut}({\mathcal{N}}_{01})$ on the subspace
$\langle\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\rangle$ is given by
$\langle\sum\limits_{i=1}^{5}\alpha_{i}^{*}\nabla_{i}\rangle,$ where

$\alpha^{*}_{1}$ | $=$ | $t(x\alpha_{1}+y\alpha_{4}+u\alpha_{5}),$
---|---|---
$\alpha^{*}_{2}$ | $=$ | $x^{2}(x\alpha_{2}+u\alpha_{4}),$
$\alpha^{*}_{3}$ | $=$ | $xz\alpha_{2}+tx\alpha_{3}+uz\alpha_{4}+tu\alpha_{5},$
$\alpha_{4}^{*}$ | $=$ | $tx^{2}\alpha_{4},$
$\alpha_{5}^{*}$ | $=$ | $t(z\alpha_{4}+t\alpha_{5}).$

Since ${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{01})={\rm H}_{\rm
BL}^{2}({\mathcal{N}}_{01})\oplus\langle[\Delta_{23}]\rangle$ and we are
interested only in new algebras, we have $\alpha_{4}\neq 0.$ It is easy to see
that, choosing
$y={x(\alpha_{2}\alpha_{5}-\alpha_{1}\alpha_{4})}\alpha_{4}^{-2}$,
$u=-{x\alpha_{2}}{\alpha_{4}^{-1}}$ and $z=-{t\alpha_{5}}{\alpha_{4}^{-1}},$
we have $\alpha^{*}_{1}=\alpha^{*}_{2}=\alpha^{*}_{5}=0.$ Hence, we can
suppose $\alpha_{1}=\alpha_{2}=\alpha_{5}=0$ and we have the following

$\alpha^{*}_{3}=tx\alpha_{3},\quad\alpha_{4}^{*}=tx^{2}\alpha_{4}.$

  1. (1)

If $\alpha_{3}=0,$ then we have the representative $\langle\nabla_{4}\rangle.$

  2. (2)

If $\alpha_{3}\neq 0,$ then by choosing $x=\frac{\alpha_{3}}{\alpha_{4}},$ we
have the representative $\langle\nabla_{3}+\nabla_{4}\rangle.$

Thus, we have the following distinct orbits

$\langle\nabla_{4}\rangle,\quad\langle\nabla_{3}+\nabla_{4}\rangle,$

which gives the following new algebras:

$\mathbb{M}_{01}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{2}e_{3}=e_{4}$ |
---|---|---|---|---
$\mathbb{M}_{02}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$

####  1.4.2. Central extensions of ${\mathcal{N}}_{02}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{11}],$ | $\nabla_{2}=[\Delta_{12}],$ | $\nabla_{3}=[\Delta_{13}-\Delta_{31}],$ | $\nabla_{4}=[\Delta_{22}],$
---|---|---|---
$\nabla_{5}=[\Delta_{23}-\Delta_{32}],$ |  $\nabla_{6}=[\Delta_{31}]$, |  $\nabla_{7}=[\Delta_{32}]$, | $\nabla_{8}=[\Delta_{33}].$

Take $\theta=\sum\limits_{i=1}^{8}\alpha_{i}\nabla_{i}\in{\rm H_{\rm
ML}^{2}}({\mathcal{N}}_{02}).$ The automorphism group of ${\mathcal{N}}_{02}$
consists of invertible matrices of the form

$\phi=\left(\begin{array}[]{ccc}x&z&0\\\ y&t&0\\\ u&q&xt-
yz\end{array}\right).$

Since

$\phi^{T}\begin{pmatrix}\alpha_{1}&\alpha_{2}&\alpha_{3}\\\
0&\alpha_{4}&\alpha_{5}\\\
\alpha_{6}-\alpha_{3}&\alpha_{7}-\alpha_{5}&\alpha_{8}\\\
\end{pmatrix}\phi=\begin{pmatrix}\alpha_{1}^{*}&\alpha_{2}^{*}-\alpha^{*}&\alpha_{3}^{*}\\\
\alpha^{*}&\alpha_{4}^{*}&\alpha_{5}^{*}\\\
\alpha_{6}^{*}-\alpha_{3}^{*}&\alpha_{7}^{*}-\alpha_{5}^{*}&\alpha_{8}^{*}\\\
\end{pmatrix},$

we have that the action of ${\rm Aut}({\mathcal{N}}_{02})$ on the subspace
$\langle\sum\limits_{i=1}^{8}\alpha_{i}\nabla_{i}\rangle$ is given by
$\langle\sum\limits_{i=1}^{8}\alpha_{i}^{*}\nabla_{i}\rangle,$ where

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1}+xy\alpha_{2}+y^{2}\alpha_{4}+ux\alpha_{6}+uy\alpha_{7}+u^{2}\alpha_{8},$ |  |
---|---|---|---|---
$\alpha^{*}_{2}$ | $=$ | $2xz\alpha_{1}+(tx+yz)\alpha_{2}+2ty\alpha_{4}+qx\alpha_{6}+uz\alpha_{6}+tu\alpha_{7}+qy\alpha_{7}+2qu\alpha_{8}$ |  |
$\alpha^{*}_{3}$ | $=$ |  $(xt-yz)\left(x\alpha_{3}+y\alpha_{5}+u\alpha_{8}\right)$, |  |
$\alpha_{4}^{*}$ | $=$ |  $z^{2}\alpha_{1}+tz\alpha_{2}+t^{2}\alpha_{4}+qz\alpha_{6}+qt\alpha_{7}+q^{2}\alpha_{8}$, |  |
$\alpha_{5}^{*}$ | $=$ |  $(xt-yz)\left(z\alpha_{3}+t\alpha_{5}+q\alpha_{8}\right)$, |  |
$\alpha_{6}^{*}$ | $=$ |  $(xt-yz)\left(x\alpha_{6}+y\alpha_{7}+2u\alpha_{8}\right)$, |  |
$\alpha_{7}^{*}$ | $=$ |  $(xt-yz)\left(z\alpha_{6}+t\alpha_{7}+2q\alpha_{8}\right)$, |  |
$\alpha_{8}^{*}$ | $=$ |  $(xt-yz)^{2}\alpha_{8}$. |  |

Since ${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{02})={\rm H}_{\rm
BL}^{2}({\mathcal{N}}_{02})\oplus\langle[\Delta_{31}],[\Delta_{32}],[\Delta_{33}]\rangle$
and we are interested only in new algebras, we have
$(\alpha_{6},\alpha_{7},\alpha_{8})\neq(0,0,0).$

  1. (1)

Let $\alpha_{8}=0,$ then $(\alpha_{6},\alpha_{7})\neq 0$ and without loss of
generality (maybe with an action of a suitable $\phi$), we can suppose
$\alpha_{6}\neq 0$. Then we consider following subcases:

    1. (a)

if $\alpha_{5}=0,\alpha_{4}=0,\alpha_{3}=0,$ then choosing

$x=1,$ $y=0,$ $u=-{\alpha_{1}}{\alpha^{-1}_{6}}$ and
$q=t(2\alpha_{1}\alpha_{7}-\alpha_{2}\alpha_{6})\alpha_{6}^{-2},$

we have the representatives $\langle\nabla_{4}+\nabla_{6}\rangle$ and
$\langle\nabla_{6}\rangle,$ depending on
$\alpha_{7}(\alpha_{1}\alpha_{7}-\alpha_{2}\alpha_{6})\neq 0$ or not;

    2. (b)

if $\alpha_{5}=0,\alpha_{4}=0,\alpha_{3}\neq 0,\alpha_{7}=0,$ then choosing

$x=\alpha_{6},$ $y=0,$ $z=0,$ $t=\alpha_{6},$ $u=-\alpha_{1}$ and
$q=-\alpha_{2},$

we have the family of representatives
$\langle\alpha\nabla_{3}+\nabla_{6}\rangle_{\alpha\neq 0};$

    3. (c)

if $\alpha_{5}=0,\alpha_{4}=0,\alpha_{3}\neq 0,\alpha_{7}\neq 0,$ then
choosing

$x=0,$ $z=y\alpha_{3}^{-1}\alpha_{7},$ $t=-y\alpha_{3}^{-1}\alpha_{6},$ $u=0$
and $q=-y\alpha_{3}^{-1}\alpha_{2},$

we have the representatives $\langle\nabla_{4}+\nabla_{5}+\nabla_{6}\rangle$
and $\langle\nabla_{5}+\nabla_{6}\rangle,$ depending on
$\alpha_{1}\alpha_{7}\neq\alpha_{2}\alpha_{6}$ or not:

    4. (d)

if $\alpha_{5}=0,\alpha_{4}\neq 0,\alpha_{3}=0,$ then choosing $z=0,$
$t=-\alpha_{7}$ and $q=\alpha_{4},$ we have
$\alpha_{4}^{*}=\alpha_{5}^{*}=\alpha_{8}^{*}=0$ and it gives us previous
considered case;

    5. (e)

if $\alpha_{5}=0,\alpha_{4}\neq 0,\alpha_{3}\neq 0,$ then choosing $x=1,$
$y=0,$ $z=1,$ and $t=1,$ we have $\alpha_{5}^{*}\neq 0$ and this case will be
considered below;

    6. (f)

if $\alpha_{5}\neq 0,\alpha_{3}=0,$ then choosing $y=0,$
$z=-x\alpha_{7}\alpha_{5}^{-1},$ $t=x\alpha_{6}\alpha_{5}^{-1},$
$u=-x\alpha_{1}\alpha_{6}^{-1}$ and
$q=-x(2\alpha_{1}\alpha_{7}-\alpha_{2}\alpha_{6})\alpha_{5}^{-1}\alpha_{6}^{-1},$
we have the representatives $\langle\nabla_{4}+\nabla_{5}+\nabla_{6}\rangle$
and $\langle\nabla_{5}+\nabla_{6}\rangle,$ depending on
$\alpha_{4}\alpha_{6}^{2}\neq\alpha_{7}(\alpha_{2}\alpha_{6}-\alpha_{1}\alpha_{7})$
or not;

    7. (g)

if $\alpha_{5}\neq 0,\alpha_{3}\neq
0,\alpha_{5}\alpha_{6}=\alpha_{3}\alpha_{7},$ then choosing $x=\alpha_{6},$
$y=0,$ $z=-t\alpha_{5}\alpha_{3}^{-1},$ $u=-\alpha_{1}$ and
$q=-t(2\alpha_{1}\alpha_{5}-\alpha_{2}\alpha_{3})\alpha_{3}^{-1}\alpha_{6}^{-1},$
we have the families of representatives
$\langle\alpha\nabla_{3}+\nabla_{4}+\nabla_{6}\rangle_{\alpha\neq 0}$ and
$\langle\alpha\nabla_{3}+\nabla_{6}\rangle_{\alpha\neq 0},$ depending on
$\alpha_{3}^{2}\alpha_{4}+\alpha_{1}\alpha_{5}^{2}\neq\alpha_{2}\alpha_{3}\alpha_{5}$
or not;

    8. (h)

if $\alpha_{5}\neq 0,\alpha_{3}\neq
0,\alpha_{5}\alpha_{6}\neq\alpha_{3}\alpha_{7},$ then by choosing
$x=-y\alpha_{5}\alpha_{3}^{-1},$ we have $\alpha_{3}^{*}=0,$ which gives the
previously considered case.

  2. (2)

Let $\alpha_{8}\neq 0$. Then putting
$u=-\frac{x\alpha_{6}+y\alpha_{7}}{2\alpha_{8}},$
$q=-\frac{z\alpha_{6}+t\alpha_{7}}{2\alpha_{8}},$ we have
$\alpha^{*}_{6}=\alpha^{*}_{7}=0.$ Thus, without loss of generality, we can
suppose $\alpha_{6}=\alpha_{7}=0$ and obtain

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1}+xy\alpha_{2}+y^{2}\alpha_{4},$ | $\alpha^{*}_{2}$ | $=$ | $2xz\alpha_{1}+(tx+yz)\alpha_{2}+2ty\alpha_{4},$
---|---|---|---|---|---
$\alpha^{*}_{3}$ | $=$ | $(tx-yz)\left(x\alpha_{3}+y\alpha_{5}\right),$ | $\alpha_{4}^{*}$ | $=$ | $z^{2}\alpha_{1}+zt\alpha_{2}+t^{2}\alpha_{4},$
$\alpha_{5}^{*}$ | $=$ | $(tx-yz)\left(z\alpha_{3}+t\alpha_{5}\right),$ | $\alpha_{8}^{*}$ | $=$ | $(xt-yz)^{2}\alpha_{8}.$

    1. (a)

Let $(\alpha_{3},\alpha_{5})=(0,0)$. Then we have the following subcases:

      1. (i)

$(\alpha_{1},\alpha_{2},\alpha_{4})=(0,0,0),$ then we have the representative
$\langle\nabla_{8}\rangle$.

      2. (ii)

$(\alpha_{1},\alpha_{2},\alpha_{4})\neq(0,0,0),$ then without loss of
generality (maybe with an action of a suitable $\phi$), we can suppose
$\alpha_{1}\neq 0$.

        1. (A)

$\alpha_{2}^{2}\neq 4\alpha_{1}\alpha_{4},$ then choosing
$x=-\frac{\alpha_{2}+\sqrt{\alpha_{2}^{2}-4\alpha_{1}\alpha_{4}}}{2\alpha_{1}},$
$y=1,$
$z=\frac{-\alpha_{2}+\sqrt{\alpha_{2}^{2}-4\alpha_{1}\alpha_{4}}}{2\alpha_{8}}$,
$t=\frac{\alpha_{1}}{\alpha_{8}},$ we have the representative
$\langle\nabla_{2}+\nabla_{8}\rangle.$

        2. (B)

$\alpha_{2}^{2}=4\alpha_{1}\alpha_{4},$ then choosing
$z=-\frac{\alpha_{2}}{2\alpha_{1}}\sqrt{\frac{\alpha_{1}}{{\alpha_{8}}}}$,
$t=\sqrt{\frac{\alpha_{1}}{{\alpha_{8}}}},$ we have the representative
$\langle\nabla_{1}+\nabla_{8}\rangle.$

    2. (b)

Let $(\alpha_{3},\alpha_{5})\neq(0,0)$, then without loss of generality (maybe
with an action of a suitable $\phi$) one can suppose $\alpha_{3}\neq 0$ and
choosing $z=-\frac{t\alpha_{5}}{\alpha_{3}},$ we have $\alpha^{*}_{5}=0$.

      1. (i)

$\alpha_{4}=0,\alpha_{2}=0,$ then choosing $t=\alpha_{3}\alpha_{8}^{-1},$ we
have the family of representatives
$\langle\alpha\nabla_{1}+\nabla_{3}+\nabla_{8}\rangle$;

      2. (ii)

$\alpha_{4}=0,\alpha_{2}\neq 0,$ then choosing $x=\alpha_{2}\alpha_{3}^{-1},$
$y=-\alpha_{1}\alpha_{2}^{-1},$ $t=\alpha_{3}\alpha_{8}^{-1},$ we have the
representative $\langle\nabla_{2}+\nabla_{3}+\nabla_{8}\rangle$.

      3. (iii)

$\alpha_{4}\neq 0,$ then choosing $x=\sqrt{\frac{\alpha_{4}}{\alpha_{8}}},$
$y=-\frac{\alpha_{2}}{2\sqrt{\alpha_{4}\alpha_{8}}}$ and
$t=\frac{\alpha_{3}}{\alpha_{8}},$ we have the family of representatives
$\langle\alpha\nabla_{1}+\nabla_{3}+\nabla_{4}+\nabla_{8}\rangle$.

Summarizing, all considered case we have the following distinct orbits

$\langle\alpha\nabla_{3}+\nabla_{6}\rangle,$
$\langle\alpha\nabla_{3}+\nabla_{4}+\nabla_{6}\rangle,$
$\langle\nabla_{5}+\nabla_{6}\rangle,$
$\langle\nabla_{4}+\nabla_{5}+\nabla_{6}\rangle,$ $\langle\nabla_{8}\rangle,$
$\langle\nabla_{2}+\nabla_{8}\rangle,$ $\langle\nabla_{1}+\nabla_{8}\rangle,$
$\langle\alpha\nabla_{1}+\nabla_{3}+\nabla_{8}\rangle,$
$\langle\nabla_{2}+\nabla_{3}+\nabla_{8}\rangle,$
$\langle\alpha\nabla_{1}+\nabla_{3}+\nabla_{4}+\nabla_{8}\rangle,$

which gives the following new algebras:

$\mathbb{M}_{03}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=\alpha e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=(1-\alpha)e_{4}$ |
---|---|---|---|---|---|---
$\mathbb{M}_{04}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=\alpha e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=(1-\alpha)e_{4}$
$\mathbb{M}_{05}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=-e_{4}$ |
$\mathbb{M}_{06}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=-e_{4}$
$\mathbb{M}_{07}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |  |
$\mathbb{M}_{08}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |  |
$\mathbb{M}_{09}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathbb{M}_{10}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$
$\mathbb{M}_{11}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |
$\mathbb{M}_{12}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ |  |
|  | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |  |

####  1.4.3. Central extensions of ${\mathcal{N}}_{03}^{\alpha\neq 0}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{12}],$ | $\nabla_{2}=[\Delta_{21}],$ | $\nabla_{3}=[\Delta_{22}],$ | $\nabla_{4}=[\Delta_{31}+\Delta_{32}],$ | $\nabla_{5}=[\Delta_{32}].$
---|---|---|---|---

Take $\theta=\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\in{\rm H_{\rm
ML}^{2}}({\mathcal{N}}_{03}^{\alpha\neq 0}).$ The automorphism group of
${\mathcal{N}}_{03}^{\alpha\neq 0}$ consists of invertible matrices of the
form

$\phi=\begin{pmatrix}x&-\alpha y&0\\\ y&x+y&0\\\ z&t&x^{2}+xy+\alpha y^{2}\\\
\end{pmatrix}.$

Since

$\phi^{T}\begin{pmatrix}0&\alpha_{1}&0\\\ \alpha_{2}&\alpha_{3}&0\\\
\alpha_{4}&\alpha_{4}+\alpha_{5}&0\\\
\end{pmatrix}\phi=\begin{pmatrix}\alpha^{*}&\alpha_{1}^{*}+\alpha^{*}&0\\\
\alpha_{2}^{*}&\alpha_{3}^{*}+\alpha\alpha^{*}&0\\\
\alpha_{4}^{*}&\alpha_{4}^{*}+\alpha_{5}^{*}&0\\\ \end{pmatrix},$

we have that the action of ${\rm Aut}({\mathcal{N}}_{03}^{\alpha\neq 0})$ on
the subspace $\langle\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\rangle$ is
given by $\langle\sum\limits_{i=1}^{5}\alpha_{i}^{*}\nabla_{i}\rangle,$ where

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1}-y(x+y\alpha)\alpha_{2}+xy\alpha_{3}-yz\alpha\alpha_{4}+xz\alpha_{5},$
---|---|---
$\alpha^{*}_{2}$ | $=$ | $-y^{2}\alpha\alpha_{1}+(x+y)(x\alpha_{2}+y\alpha_{3})+t(x+y)\alpha_{4}+ty\alpha_{5},$
$\alpha^{*}_{3}$ | $=$ | $-y(2x+y))\alpha(\alpha_{1}+\alpha_{2})+((x+y)^{2}-y^{2}\alpha)\alpha_{3}+$
|  | $(t(x+y)-(ty+xz+yz)\alpha)\alpha_{4}+(t(x+y)-yz\alpha)\alpha_{5},$
$\alpha_{4}^{*}$ | $=$ | $(x^{2}+xy+y^{2}\alpha)((x+y)\alpha_{4}+y\alpha_{5}),$
$\alpha_{5}^{*}$ | $=$ | $(x^{2}+xy+y^{2}\alpha)(x\alpha_{5}-y\alpha\alpha_{4}).$

Since ${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{03}^{\alpha\neq 0})={\rm H}_{\rm
BL}^{2}({\mathcal{N}}_{03}^{\alpha\neq
0})\oplus\langle[\Delta_{31}],[\Delta_{32}]\rangle$ and we are interested only
in new algebras, we have $(\alpha_{4},\alpha_{5})\neq(0,0).$ Without loss of
generality, one can assume $\alpha_{4}\neq 0.$

  1. (1)

$\alpha\alpha_{4}^{2}+\alpha_{4}\alpha_{5}+\alpha_{5}^{2}\neq 0,$ then
choosing

$x=-\frac{y(\alpha_{4}+\alpha_{5})}{\alpha_{4}},\
t=\frac{y(\alpha\alpha_{1}\alpha_{4}(\alpha_{4}+2\alpha_{5})+\alpha\alpha_{2}\alpha_{4}(\alpha_{4}+2\alpha_{5})+\alpha_{3}(\alpha_{5}^{2}-\alpha\alpha_{4}^{2}))}{\alpha_{4}(\alpha\alpha_{4}^{2}+\alpha_{4}\alpha_{5}+\alpha_{5}^{2})},$
$z=\frac{y(\alpha_{1}(\alpha_{4}+\alpha_{5})^{2}-\alpha_{4}(\alpha_{2}((\alpha-1)\alpha_{4}-\alpha_{5})+\alpha_{3}(\alpha_{4}+\alpha_{5})))}{\alpha_{4}(\alpha\alpha_{4}^{2}+\alpha_{4}\alpha_{5}+\alpha_{5}^{2})},$

we have

$\alpha^{*}_{1}=\alpha^{*}_{3}=\alpha^{*}_{4}=0,$
$\alpha^{*}_{2}=\frac{y^{2}(\alpha_{5}(\alpha_{2}(\alpha_{4}+\alpha_{5})-\alpha_{3}\alpha_{4})-\alpha\alpha_{1}\alpha_{4}^{2})}{\alpha_{4}^{2}},$
$\alpha^{*}_{5}=-\frac{y^{3}(\alpha\alpha_{4}^{2}+\alpha_{4}\alpha_{5}+\alpha_{5}^{2}){}^{2}}{\alpha_{4}^{3}}.$

Hence, we have two representatives $\langle\nabla_{5}\rangle$ and
$\langle\nabla_{2}+\nabla_{5}\rangle,$ depending on
$\alpha\alpha_{1}\alpha_{4}^{2}=\alpha_{5}(\alpha_{2}(\alpha_{4}+\alpha_{5})-\alpha_{3}\alpha_{4})$
or not.

  2. (2)

$\alpha\alpha_{4}^{2}+\alpha_{4}\alpha_{5}+\alpha_{5}^{2}=0,$ i.e.,
$\alpha_{5}=-\frac{1}{2}(1\pm\sqrt{1-4\alpha})\alpha_{4},$ then choosing

$y=0,$ $t=-\frac{x\alpha_{2}}{\alpha_{4}},$
$z=\frac{2x\alpha_{1}}{\alpha_{4}(1\pm\sqrt{1-4\alpha})},$

we have

$\alpha^{*}_{1}=\alpha^{*}_{2}=0,$
$\alpha^{*}_{3}=-\frac{x^{2}(2\alpha\alpha_{1}+2\alpha\alpha_{2}-\alpha_{3}(1\pm\sqrt{1-4\alpha}))}{1\pm\sqrt{1-4\alpha}},$
$\alpha^{*}_{4}=x^{3}\alpha_{4},$
$\alpha^{*}_{5}=-\frac{x^{3}\alpha_{4}(1\pm\sqrt{1-4\alpha})}{2}.$

Hence, we have four families of representatives

$\langle\nabla_{3}+\nabla_{4}-\frac{1}{2}(1\pm\sqrt{1-4\alpha})\nabla_{5}\rangle$
and $\langle\nabla_{4}-\frac{1}{2}(1\pm\sqrt{1-4\alpha})\nabla_{5}\rangle,$

depending on
$2\alpha\alpha_{4}(\alpha_{1}+\alpha_{2})\neq\alpha_{3}(\alpha_{4}\pm\sqrt{(1-4\alpha)\alpha_{4}^{2}})$
or not.

####  1.4.4. Central extensions of ${\mathcal{N}}_{03}^{0}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{12}],$ | $\nabla_{2}=[\Delta_{21}],$ | $\nabla_{3}=[\Delta_{22}],$ | $\nabla_{4}=[\Delta_{31}+\Delta_{32}],$ | $\nabla_{5}=[\Delta_{32}]$
---|---|---|---|---

Take $\theta=\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\in{\rm H_{\rm
ML}^{2}}({\mathcal{N}}_{03}^{0}).$ The automorphism group of
${\mathcal{N}}_{03}^{0}$ consists of invertible matrices of the form

$\phi=\begin{pmatrix}x&0&0\\\ y-x&y&0\\\ z&t&xy\\\ \end{pmatrix}.$

Since

$\phi^{T}\begin{pmatrix}0&\alpha_{1}&0\\\ \alpha_{2}&\alpha_{3}&0\\\
\alpha_{4}&\alpha_{4}+\alpha_{5}&0\\\
\end{pmatrix}\phi=\begin{pmatrix}\alpha^{*}&\alpha_{1}^{*}+\alpha^{*}&0\\\
\alpha_{2}^{*}&\alpha_{3}^{*}&0\\\
\alpha_{4}^{*}&\alpha_{4}^{*}+\alpha_{5}^{*}&0\\\ \end{pmatrix},$

we have that the action of ${\rm Aut}({\mathcal{N}}_{03}^{0})$ on the subspace
$\langle\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\rangle$ is given by
$\langle\sum\limits_{i=1}^{5}\alpha_{i}^{*}\nabla_{i}\rangle,$ where

$\alpha^{*}_{1}$ | $=$ | $x(x\alpha_{1}+(x-y)\alpha_{2}-x\alpha_{3}+y\alpha_{3}+z\alpha_{5}),$
---|---|---
$\alpha^{*}_{2}$ | $=$ | $xy\alpha_{2}+y(y-x)\alpha_{3}+t(y\alpha_{4}+(y-x)\alpha_{5}),$
$\alpha^{*}_{3}$ | $=$ | $y(y\alpha_{3}+t(\alpha_{4}+\alpha_{5})),$
$\alpha_{4}^{*}$ | $=$ | $xy(y\alpha_{4}+(y-x)\alpha_{5}),$
$\alpha_{5}^{*}$ | $=$ | $x^{2}y\alpha_{5}.$

Since ${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{03}^{0})={\rm H}_{\rm
BL}^{2}({\mathcal{N}}_{03}^{0})\oplus\langle[\Delta_{32}]\rangle$ and we are
interested only in new algebras, we have $\alpha_{5}\neq 0.$ Then choosing
$z=(y(\alpha_{2}-\alpha_{3})-x(\alpha_{1}+\alpha_{2}-\alpha_{3}))\alpha_{5}^{-1},$
we have $\alpha^{*}_{1}=0.$

  1. (1)

If $\alpha_{5}+\alpha_{4}=0,$ then choosing
$t=y\left(x\alpha_{2}+(y-x)\alpha_{3}\right)x^{-1}\alpha^{-1}_{5},$ we have

$\alpha^{*}_{1}=\alpha^{*}_{2}=0,$ $\alpha^{*}_{3}=y^{2}\alpha_{3},$
$\alpha^{*}_{4}=-x^{2}y\alpha_{5},$ $\alpha^{*}_{5}=x^{2}y\alpha_{5}.$

Hence, we have two representatives $\langle\nabla_{4}-\nabla_{5}\rangle$ and
$\langle\nabla_{3}+\nabla_{4}-\nabla_{5}\rangle,$ depending on $\alpha_{3}=0$
or not.

  2. (2)

If $\alpha_{5}+\alpha_{4}\neq 0,$ then choosing
$y=\frac{x\alpha_{5}}{\alpha_{4}+\alpha_{5}},\
t=-\frac{x\alpha_{3}\alpha_{5}}{(\alpha_{4}+\alpha_{5})^{2}},$ we have

$\alpha^{*}_{1}=0,$
$\alpha^{*}_{2}=\frac{x^{2}\alpha_{2}\alpha_{5}}{\alpha_{4}+\alpha_{5}},$
$\alpha^{*}_{3}=\alpha_{4}^{*}=0,$
$\alpha^{*}_{5}=\frac{x^{3}\alpha_{5}^{2}}{\alpha_{4}+\alpha_{5}}.$

Hence, we have two representatives $\langle\nabla_{5}\rangle$ and
$\langle\nabla_{2}+\nabla_{5}\rangle,$ depending on $\alpha_{2}=0$ or not.

Summarizing all cases of the central extension of the algebra
${\mathcal{N}}_{03}^{\alpha}$, we have the following distinct orbits

$\langle\nabla_{5}\rangle,$
$\langle\nabla_{2}+\nabla_{5}\rangle,\quad\langle\nabla_{4}-\frac{1}{2}(1+\sqrt{1-4\alpha})\nabla_{5}\rangle,$
$\langle\nabla_{3}+\nabla_{4}-\frac{1}{2}(1+\sqrt{1-4\alpha})\nabla_{5}\rangle,$
$\langle\nabla_{4}-\frac{1}{2}(1-\sqrt{1-4\alpha})\nabla_{5}\rangle_{\alpha\neq
0},$
$\langle\nabla_{3}+\nabla_{4}-\frac{1}{2}(1-\sqrt{1-4\alpha})\nabla_{5}\rangle_{\alpha\neq
0},$

which gives the following new algebras:

$\mathbb{M}_{13}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{2}=e_{4}$ |
---|---|---|---|---|---|---
$\mathbb{M}_{14}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{2}=e_{4}$
$\mathbb{M}_{15}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1-\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{16}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}+e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1-\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{17}^{\alpha\neq 0}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1+\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{18}^{\alpha\neq 0}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}+e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1+\sqrt{1-4\alpha})e_{4}$

Note that algebras $\mathbb{M}_{17}^{4}(0)$ and $\mathbb{M}_{18}^{4}(0)$ are
Leibniz algebras with one dimensional annihilator.

####  1.4.5. Central extensions of ${\mathcal{N}}_{04}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{12}],$ | $\nabla_{2}=[\Delta_{21}],$ | $\nabla_{3}=[\Delta_{22}],$ | $\nabla_{4}=[\Delta_{31}],$ | $\nabla_{5}=[\Delta_{32}].$
---|---|---|---|---

Take $\theta=\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\in{\rm H_{\rm
ML}^{2}}({\mathcal{N}}_{04}).$ The automorphism group of ${\mathcal{N}}_{04}$
consists of invertible matrices of the form

$\phi_{1}=\begin{pmatrix}x&y&0\\\ -y&x&0\\\ z&t&x^{2}+y^{2}\\\
\end{pmatrix},\quad\phi_{2}=\begin{pmatrix}x&y&0\\\ y&-x&0\\\
z&t&x^{2}+y^{2}\\\ \end{pmatrix}.$

Since

$\phi_{1}^{T}\begin{pmatrix}0&\alpha_{1}&0\\\ \alpha_{2}&\alpha_{3}&0\\\
\alpha_{4}&\alpha_{5}&0\\\
\end{pmatrix}\phi_{1}=\begin{pmatrix}\alpha^{*}&\alpha_{1}^{*}&0\\\
\alpha_{2}^{*}&\alpha_{3}^{*}+\alpha^{*}&0\\\
\alpha_{4}^{*}&\alpha_{5}^{*}&0\\\ \end{pmatrix},$

we have that the action of ${\rm Aut}({\mathcal{N}}_{04})$ on the subspace
$\langle\sum\limits_{i=1}^{5}\alpha_{i}\nabla_{i}\rangle$ is given by
$\langle\sum\limits_{i=1}^{5}\alpha_{i}^{*}\nabla_{i}\rangle,$ where

$\alpha^{*}_{1}$ | $=$ | $x^{2}\alpha_{1}-y(y\alpha_{2}-z\alpha_{4})-x(y\alpha_{3}-z\alpha_{5}),$
---|---|---
$\alpha^{*}_{2}$ | $=$ | $-y^{2}\alpha_{1}+x(x\alpha_{2}-y\alpha_{3}+t\alpha_{4})-ty\alpha_{5},$
$\alpha^{*}_{3}$ | $=$ | $2xy(\alpha_{1}+\alpha_{2})+(x^{2}-y^{2})\alpha_{3}+(ty-xz)\alpha_{4}+(tx+yz)\alpha_{5},$
$\alpha_{4}^{*}$ | $=$ | $(x^{2}+y^{2})(x\alpha_{4}-y\alpha_{5}),$
$\alpha_{5}^{*}$ | $=$ | $(x^{2}+y^{2})(y\alpha_{4}+x\alpha_{5}).$

Since ${\rm H}_{\rm ML}^{2}({\mathcal{N}}_{04})={\rm H}_{\rm
BL}^{2}({\mathcal{N}}_{04})\oplus\langle[\Delta_{31}],[\Delta_{32}]\rangle$
and we are interested only in new algebras, we have
$(\alpha_{4},\alpha_{5})\neq(0,0).$ Moreover, without loss of generality, one
can assume $\alpha_{4}\neq 0$. Then we have the following cases.

  1. (1)

If $\alpha_{4}^{2}+\alpha_{5}^{2}\neq 0,$ then choosing

$y=-\frac{x\alpha_{5}}{\alpha_{4}},$
$t=-\frac{x\alpha_{2}\alpha_{4}^{2}+x\alpha_{3}\alpha_{4}\alpha_{5}-x\alpha_{1}\alpha_{5}^{2}}{\alpha_{4}(\alpha_{4}^{2}+\alpha_{5}^{2})}$,
$z=\frac{x(\alpha_{3}(\alpha_{4}^{2}-\alpha_{5}^{2})-2(\alpha_{1}+\alpha_{2})\alpha_{4}\alpha_{5})}{\alpha_{4}(\alpha_{4}^{2}+\alpha_{5}^{2})},$

we have

$\alpha_{1}^{*}=\frac{x^{2}(\alpha_{1}\alpha_{4}^{2}+\alpha_{5}(\alpha_{3}\alpha_{4}-\alpha_{2}\alpha_{5}))}{\alpha_{4}^{2}},$
$\alpha^{*}_{2}=\alpha_{3}^{*}=0,$
$\alpha^{*}_{4}=\frac{x^{3}(\alpha_{4}^{2}+\alpha_{5}^{2})^{2}}{\alpha_{4}^{3}},$
$\alpha^{*}_{5}=0.$

Hence, we have two representatives $\langle\nabla_{4}\rangle$ and
$\langle\nabla_{1}+\nabla_{4}\rangle,$ depending on
$\alpha_{1}\alpha_{4}^{2}+\alpha_{5}\alpha_{3}\alpha_{4}=\alpha_{2}\alpha_{5}^{2}$
or not.

  2. (2)

If $\alpha_{4}^{2}+\alpha_{5}^{2}=0,$ then choosing

$t=\frac{y^{2}\alpha_{1}-x^{2}\alpha_{2}+xy\alpha_{3}}{\alpha_{4}(x\pm iy)},\
z=\frac{y\alpha_{1}\alpha_{4}((2x^{2}+y^{2})\pm
ixy)+\alpha_{3}\alpha_{4}(x^{3}\mp iy^{3})+x\alpha_{2}\alpha_{4}(xy\pm
i(x^{2}+2y^{2}))}{\alpha_{4}^{2}(x\pm iy)^{2}},$

we have

$\alpha_{1}^{*}=\frac{(x^{2}+y^{2})^{2}(\alpha_{1}+\alpha_{2}\mp\alpha_{3})}{(x\pm
iy)^{2}},$ $\alpha^{*}_{2}=\alpha_{3}^{*}=0,$
$\alpha^{*}_{4}=(x^{2}+y^{2})(x\pm iy)\alpha_{4},\ \alpha_{5}^{*}=\mp
i(x^{2}+y^{2})(x\mp iy)\alpha_{4}.$

Hence, we have two representatives $\langle\nabla_{4}\pm i\nabla_{5}\rangle$
and $\langle\nabla_{1}+\nabla_{4}\pm i\nabla_{5}\rangle,$ depending on
$\alpha_{1}+\alpha_{2}\mp\alpha_{3}=0$ or not.

Since the automorphism $\phi_{2}=diag(1,-1,1)$ acts as

$\phi_{2}(\nabla_{4}+i\nabla_{5})=\nabla_{4}-i\nabla_{5}$ and
$\phi_{2}(\nabla_{1}+\nabla_{4}+i\nabla_{5})=\nabla_{1}+\nabla_{4}-i\nabla_{5},$

we get the following representatives of distinct orbits
$\langle\nabla_{4}+i\nabla_{5}\rangle$ and
$\langle\nabla_{1}+\nabla_{4}+i\nabla_{5}\rangle.$

Summarizing, all considered cases we have the following distinct orbits

$\langle\nabla_{4}\rangle,$ | $\langle\nabla_{1}+\nabla_{4}\rangle,$ | $\langle\nabla_{4}+i\nabla_{5}\rangle,$ | $\langle\nabla_{1}+\nabla_{4}+i\nabla_{5}\rangle,$
---|---|---|---

which gives the following new algebras:

$\mathbb{M}_{19}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |  |
---|---|---|---|---|---|---
$\mathbb{M}_{20}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |
$\mathbb{M}_{21}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=ie_{4}$ |
$\mathbb{M}_{22}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=ie_{4}$

Now we are ready to summarize all results related to the algebraic
classification of complex $4$-dimensional nilpotent mono Leibniz algebras.

####  1.4.6. The classification theorem

######  Theorem B.

Let ${\mathbb{L}}$ be a complex $4$-dimensional nilpotent mono Leibniz
algebra. Then ${\mathbb{L}}$ is a binary Leibniz algebra or isomorphic to one
algebra from the following list:

$\mathbb{M}_{01}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{2}e_{3}=e_{4}$ |  |  |
---|---|---|---|---|---|---
$\mathbb{M}_{02}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ |  |
$\mathbb{M}_{03}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=\alpha e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=(1-\alpha)e_{4}$
$\mathbb{M}_{04}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=\alpha e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=(1-\alpha)e_{4}$
$\mathbb{M}_{05}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=-e_{4}$
$\mathbb{M}_{06}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=e_{4}$ |  |
|  | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=-e_{4}$ |  |
$\mathbb{M}_{07}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathbb{M}_{08}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathbb{M}_{09}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |
$\mathbb{M}_{10}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ |  |
|  | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathbb{M}_{11}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$
$\mathbb{M}_{12}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ |
|  | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathbb{M}_{13}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{2}=e_{4}$ |
$\mathbb{M}_{14}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{2}=e_{4}$
$\mathbb{M}_{15}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1-\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{16}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}+e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1-\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{17}^{\alpha\neq 0}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1+\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{18}^{\alpha\neq 0}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\alpha e_{3}+e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=\frac{1}{2}(1+\sqrt{1-4\alpha})e_{4}$
$\mathbb{M}_{19}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |  |
$\mathbb{M}_{20}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |
$\mathbb{M}_{21}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=ie_{4}$ |
$\mathbb{M}_{22}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=ie_{4}$

Note that algebras $\mathbb{M}_{17}^{4}(0)$ and $\mathbb{M}_{18}^{4}(0)$ are
Leibniz algebras.

###  1.5. Classification of $4$-dimensional nilpotent algebras with nil-index
$3$

Thanks to [11], the intersection of left mono Leibniz and right mono Leibniz
algebras gives the variety of nil-algebras of nil-index $3.$ Hence, each
$4$-dimensional nilpotent algebra with nil-index $3$ is in the classification
given in Theorem B. Obviously, each $2$-step nilpotent algebra has nil-index
$3.$ Using the classification of $4$-dimensional Leibniz algebras [35] and
Theorem B, we can choose all Leibniz algebras with nil-index $3.$ Let us note
that the linearization of the identity $x^{3}=0$ (i.e. $x^{2}x=0=xx^{2}$)
gives two identities

$\sum\limits_{\sigma\in\mathbb{S}_{3}}(x_{\sigma(1)}x_{\sigma(2)})x_{\sigma(3)}=0$
and
$\sum\limits_{\sigma\in\mathbb{S}_{3}}x_{\sigma(1)}(x_{\sigma(2)}x_{\sigma(3)})=0$.

Hence, we will choose only algebras satisfying the last two identities.
Summarizing, we have the following Theorem.

######  Theorem C.

Let ${\mathfrak{n}}$ be a complex $4$-dimensional nilpotent algebra of nil-
index $3$. Then ${\mathfrak{n}}$ is a $2$-step nilpotent algebra or isomorphic
to one algebra from the following list:

$\mathfrak{L}_{01}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ |  |
---|---|---|---|---|---|---|---
$\mathfrak{L}_{09}$ | $:$ | $e_{1}e_{2}=-e_{3}+e_{4}$ | $e_{1}e_{3}=-e_{4}$ | $e_{2}e_{1}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |  |
$\mathfrak{L}_{10}$ | $:$ | $e_{1}e_{2}=-e_{3}$ | $e_{1}e_{3}=-e_{4}$ | $e_{2}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$ |
$\mathfrak{L}_{11}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=-e_{3}$ | $e_{1}e_{3}=-e_{4}$ | $e_{2}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$
$\mathfrak{L}_{12}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=-e_{3}$ | $e_{1}e_{3}=-e_{4}$ | $e_{2}e_{1}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |
$\mathbb{M}_{03}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=\alpha e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=(1-\alpha)e_{4}$ |
$\mathbb{M}_{04}^{\alpha}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=\alpha e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=(1-\alpha)e_{4}$
$\mathbb{M}_{05}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=-e_{4}$ |
$\mathbb{M}_{06}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=e_{4}$ | $e_{3}e_{2}=-e_{4}$
$\mathbb{M}_{07}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |  |
$\mathbb{M}_{08}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |  |
$\mathbb{M}_{09}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathbb{M}_{10}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$
$\mathbb{M}_{11}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |
$\mathbb{M}_{12}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$ |  |
|  | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |  |

##  2\. The geometric classification of nilpotent binary and mono Leibniz
algebras

###  2.1. Definitions and notation

Given an $n$-dimensional vector space $\mathbb{V}$, the set ${\rm
Hom}(\mathbb{V}\otimes\mathbb{V},\mathbb{V})\cong\mathbb{V}^{*}\otimes\mathbb{V}^{*}\otimes\mathbb{V}$
is a vector space of dimension $n^{3}$. This space has the structure of the
affine variety $\mathbb{C}^{n^{3}}.$ Indeed, let us fix a basis
$e_{1},\dots,e_{n}$ of $\mathbb{V}$. Then any $\mu\in{\rm
Hom}(\mathbb{V}\otimes\mathbb{V},\mathbb{V})$ is determined by $n^{3}$
structure constants $c_{ij}^{k}\in\mathbb{C}$ such that $\mu(e_{i}\otimes
e_{j})=\sum\limits_{k=1}^{n}c_{ij}^{k}e_{k}$. A subset of ${\rm
Hom}(\mathbb{V}\otimes\mathbb{V},\mathbb{V})$ is Zariski-closed if it can be
defined by a set of polynomial equations in the variables $c_{ij}^{k}$ ($1\leq
i,j,k\leq n$).

Let $T$ be a set of polynomial identities. The set of algebra structures on
$\mathbb{V}$ satisfying polynomial identities from $T$ forms a Zariski-closed
subset of the variety ${\rm Hom}(\mathbb{V}\otimes\mathbb{V},\mathbb{V})$. We
denote this subset by $\mathbb{L}(T)$. The general linear group ${\rm
GL}(\mathbb{V})$ acts on $\mathbb{L}(T)$ by conjugations:

$(g*\mu)(x\otimes y)=g\mu(g^{-1}x\otimes g^{-1}y)$

for $x,y\in\mathbb{V}$, $\mu\in\mathbb{L}(T)\subset{\rm
Hom}(\mathbb{V}\otimes\mathbb{V},\mathbb{V})$ and $g\in{\rm GL}(\mathbb{V})$.
Thus, $\mathbb{L}(T)$ is decomposed into ${\rm GL}(\mathbb{V})$-orbits that
correspond to the isomorphism classes of algebras. Let $O(\mu)$ denote the
orbit of $\mu\in\mathbb{L}(T)$ under the action of ${\rm GL}(\mathbb{V})$ and
$\overline{O(\mu)}$ denote the Zariski closure of $O(\mu)$.

Let $\bf A$ and $\bf B$ be two $n$-dimensional algebras satisfying the
identities from $T$, and let $\mu,\lambda\in\mathbb{L}(T)$ represent $\bf A$
and $\bf B$, respectively. We say that $\bf A$ degenerates to $\bf B$ and
write $\bf A\to\bf B$ if $\lambda\in\overline{O(\mu)}$. Note that in this case
we have $\overline{O(\lambda)}\subset\overline{O(\mu)}$. Hence, the definition
of degeneration does not depend on the choice of $\mu$ and $\lambda$. If $\bf
A\not\cong\bf B$, then the assertion $\bf A\to\bf B$ is called a proper
degeneration. We write $\bf A\not\to\bf B$ if
$\lambda\not\in\overline{O(\mu)}$.

Let $\bf A$ be represented by $\mu\in\mathbb{L}(T)$. Then $\bf A$ is rigid in
$\mathbb{L}(T)$ if $O(\mu)$ is an open subset of $\mathbb{L}(T)$. Recall that
a subset of a variety is called irreducible if it cannot be represented as a
union of two non-trivial closed subsets. A maximal irreducible closed subset
of a variety is called an irreducible component. It is well known that any
affine variety can be represented as a finite union of its irreducible
components in a unique way. The algebra $\bf A$ is rigid in $\mathbb{L}(T)$ if
and only if $\overline{O(\mu)}$ is an irreducible component of
$\mathbb{L}(T)$.

###  2.2. Method of the description of degenerations of algebras

In the present work we use the methods applied to Lie algebras in [22, 23].
First of all, if $\bf A\to\bf B$ and $\bf A\not\cong\bf B$, then
$\mathfrak{Der}(\bf A)<\mathfrak{Der}(\bf B)$, where $\mathfrak{Der}(\bf A)$
is the Lie algebra of derivations of $\bf A$. We compute the dimensions of
algebras of derivations and check the assertion $\bf A\to\bf B$ only for such
$\bf A$ and $\bf B$ that $\mathfrak{Der}(\bf A)<\mathfrak{Der}(\bf B)$.

To prove degenerations, we construct families of matrices parametrized by $t$.
Namely, let $\bf A$ and $\bf B$ be two algebras represented by the structures
$\mu$ and $\lambda$ from $\mathbb{L}(T)$ respectively. Let $e_{1},\dots,e_{n}$
be a basis of $\mathbb{V}$ and $c_{ij}^{k}$ ($1\leq i,j,k\leq n$) be the
structure constants of $\lambda$ in this basis. If there exist
$a_{i}^{j}(t)\in\mathbb{C}$ ($1\leq i,j\leq n$, $t\in\mathbb{C}^{*}$) such
that $E_{i}^{t}=\sum\limits_{j=1}^{n}a_{i}^{j}(t)e_{j}$ ($1\leq i\leq n$) form
a basis of $\mathbb{V}$ for any $t\in\mathbb{C}^{*}$, and the structure
constants of $\mu$ in the basis $E_{1}^{t},\dots,E_{n}^{t}$ are such rational
functions $c_{ij}^{k}(t)\in\mathbb{C}[t]$ that $c_{ij}^{k}(0)=c_{ij}^{k}$,
then $\bf A\to\bf B$. In this case $E_{1}^{t},\dots,E_{n}^{t}$ is called a
parametrized basis for $\bf A\to\bf B$. To simplify our equations, we will use
the notation $A_{i}=\langle e_{i},\dots,e_{n}\rangle,\ i=1,\ldots,n$ and write
simply $A_{p}A_{q}\subset A_{r}$ instead of $c_{ij}^{k}=0$ ($i\geq p$, $j\geq
q$, $k<r$).

Since the varieties of $4$-dimensional nilpotent mono Leibniz and
$5$-dimensional nilpotent binary Leibniz algebras contain infinitely many non-
isomorphic algebras, we have to do some additional work. Let ${\bf
A}(*):=\\{{\bf A}(\alpha)\\}_{\alpha\in I}$ be a series of algebras, and let
$\bf B$ be another algebra. Suppose that for $\alpha\in I$, $\bf A(\alpha)$ is
represented by the structure $\mu(\alpha)\in\mathbb{L}(T)$ and
$B\in\mathbb{L}(T)$ is represented by the structure $\lambda$. Then we say
that $\bf A(*)\to\bf B$ if
$\lambda\in\overline{\\{O(\mu(\alpha))\\}_{\alpha\in I}}$, and $\bf
A(*)\not\to\bf B$ if $\lambda\not\in\overline{\\{O(\mu(\alpha))\\}_{\alpha\in
I}}$.

Let $\bf A(*)$, $\bf B$, $\mu(\alpha)$ ($\alpha\in I$) and $\lambda$ be as
above. To prove $\bf A(*)\to\bf B$ it is enough to construct a family of pairs
$(f(t),g(t))$ parametrized by $t\in\mathbb{C}^{*}$, where $f(t)\in I$ and
$g(t)\in{\rm GL}(\mathbb{V})$. Namely, let $e_{1},\dots,e_{n}$ be a basis of
$\mathbb{V}$ and $c_{ij}^{k}$ ($1\leq i,j,k\leq n$) be the structure constants
of $\lambda$ in this basis. If we construct
$a_{i}^{j}:\mathbb{C}^{*}\to\mathbb{C}$ ($1\leq i,j\leq n$) and
$f:\mathbb{C}^{*}\to I$ such that
$E_{i}^{t}=\sum\limits_{j=1}^{n}a_{i}^{j}(t)e_{j}$ ($1\leq i\leq n$) form a
basis of $\mathbb{V}$ for any $t\in\mathbb{C}^{*}$, and the structure
constants of $\mu_{f(t)}$ in the basis $E_{1}^{t},\dots,E_{n}^{t}$ are such
rational functions $c_{ij}^{k}(t)\in\mathbb{C}[t]$ that
$c_{ij}^{k}(0)=c_{ij}^{k}$, then $\bf A(*)\to\bf B$. In this case
$E_{1}^{t},\dots,E_{n}^{t}$ and $f(t)$ are called a parametrized basis and a
parametrized index for $\bf A(*)\to\bf B$, respectively.

We now explain how to prove $\bf A(*)\not\to\mathcal{B}$. Note that if
$\mathfrak{Der}\ \bf A(\alpha)>\mathfrak{Der}\ \bf B$ for all $\alpha\in I$
then $\bf A(*)\not\to\bf B$. One can also use the following Lemma, whose proof
is the same as the proof of Lemma 1.5 from [22].

######  Lemma 3.

Let $\mathfrak{B}$ be a Borel subgroup of ${\rm GL}(\mathbb{V})$ and
$\mathcal{R}\subset\mathbb{L}(T)$ be a $\mathfrak{B}$-stable closed subset. If
$\bf A(*)\to\bf B$ and for any $\alpha\in I$ the algebra $\bf A(\alpha)$ can
be represented by a structure $\mu(\alpha)\in\mathcal{R}$, then there is
$\lambda\in\mathcal{R}$ representing $\bf B$.

###  2.3. The geometric classification of $5$-dimensional nilpotent binary
Leibniz algebras

The main result of the present section is the following theorem.

######  Theorem D.

The variety of $5$-dimensional nilpotent binary Leibniz algebras has dimension
$24$ and it has $10$ irreducible components defined by

$\mathcal{C}_{1}=\overline{\\{\mathcal{O}({\mathfrak{V}}_{4+1})\\}},$
$\mathcal{C}_{2}=\overline{\\{\mathcal{O}({\mathfrak{V}}_{3+2})\\}},$
$\mathcal{C}_{3}=\overline{\\{\mathcal{O}(\mathbb{S}_{21}^{\alpha,\beta})\\}},$
$\mathcal{C}_{4}=\overline{\\{\mathcal{O}(\mathbb{S}_{41}^{\alpha})\\}},$
$\mathcal{C}_{5}=\overline{\\{\mathcal{O}(\mathbb{L}_{28}^{\alpha})\\}},$
$\mathcal{C}_{6}=\overline{\\{\mathcal{O}(\mathbb{L}_{47}^{\alpha,\beta})\\}},$
$\mathcal{C}_{7}=\overline{\\{\mathcal{O}(\mathbb{L}_{52}^{\alpha,\beta})\\}},$
$\mathcal{C}_{8}=\overline{\\{\mathcal{O}(\mathbb{L}_{79}^{\alpha})\\}},$
$\mathcal{C}_{9}=\overline{\\{\mathcal{O}(\mathbb{L}_{82}))\\}},$
$\mathcal{C}_{10}=\overline{\\{\mathcal{O}({\bf B}_{09}^{\alpha})\\}},$

In particular, there is only one rigid algebra in this variety.

###### Proof.

Thanks to [3] the variety of $5$-dimensional nilpotent Leibniz algebras has
only $10$ irreducible components defined by

${\mathfrak{V}}_{4+1}$ | $:$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=\lambda e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=\mu e_{5}$ |
---|---|---|---|---|---|---
${\mathfrak{V}}_{3+2}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=\mu_{1}e_{5}$ | $e_{1}e_{3}=\mu_{2}e_{5}$ | $e_{2}e_{1}=\mu_{3}e_{5}$ | $e_{2}e_{2}=\mu_{4}e_{5}$
|  | $e_{2}e_{3}=\mu_{5}e_{5}$ | $e_{3}e_{1}=\mu_{6}e_{5}$ | $e_{3}e_{2}=\lambda e_{4}+\mu_{7}e_{5}$ | $e_{3}e_{3}=e_{5}$
$\mathbb{S}_{21}^{\alpha,\beta}$ | $:$ | $e_{1}e_{1}=\alpha e_{5}$ | $e_{1}e_{2}=e_{3}+e_{4}+\beta e_{5}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{3}$ |
|  | $e_{2}e_{2}=e_{5}$ | $e_{2}e_{3}=e_{4}$ | $e_{3}e_{1}=-e_{5}$ | $e_{3}e_{2}=-e_{4}$ |
$\mathbb{S}_{22}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{3}$ |
|  | $e_{2}e_{2}=\alpha e_{5}$ | $e_{2}e_{4}=e_{5}$ | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{4}=e_{5}$ |
$\mathbb{S}_{41}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=\alpha e_{5}$
|  | $e_{2}e_{3}=e_{4}$ | $e_{2}e_{4}=e_{5}$ | $e_{3}e_{1}=-e_{5}$ | $e_{3}e_{2}=-e_{4}$ | $e_{4}e_{2}=-e_{5}$
$\mathbb{L}_{28}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=\alpha e_{5}$ | $e_{2}e_{2}=e_{5}$ |
|  | $e_{3}e_{1}=e_{5}$ | $e_{3}e_{2}=e_{5}$ | $e_{4}e_{4}=e_{5}$ |  |
$\mathbb{L}_{47}^{\alpha,\beta}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-\alpha e_{3}$ |  |
|  | $e_{2}e_{2}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ | $e_{4}e_{2}=\beta e_{5}$ |  |
$\mathbb{L}_{52}^{\alpha,\beta}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=-e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{3}=\beta e_{5}$
|  | $e_{2}e_{4}=-\beta e_{5}$ | $e_{3}e_{1}=e_{5}$ | $e_{3}e_{2}=e_{5}$ | $e_{4}e_{1}=\alpha e_{5}$ | $e_{4}e_{2}=\beta e_{5}$
$\mathbb{L}_{79}^{\alpha}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=e_{3}$ |  |
|  | $e_{2}e_{2}=e_{4}+e_{5}$ | $e_{3}e_{1}=e_{4}+\alpha e_{5}$ | $e_{3}e_{2}=e_{5}$ | $e_{4}e_{1}=e_{5}$ |
$\mathbb{L}_{82}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{2}e_{1}=e_{3}$ | $e_{3}e_{1}=e_{4}$ | $e_{4}e_{1}=e_{5}$ |

After carefully checking the dimensions of orbit closures of the more
important for us algebras, we have

$\dim\mathcal{O}({\mathfrak{V}}_{3+2})$ | $=$ | $24,$
---|---|---
$\dim\mathcal{O}(\mathbb{L}_{28}^{\alpha})=\dim\mathcal{O}(\mathbb{L}_{47}^{\alpha,\beta})=\dim\mathcal{O}(\mathbb{L}_{52}^{\alpha,\beta})=\dim\mathcal{O}(\mathbb{L}_{79}^{\alpha})=\dim\mathcal{O}({\bf B}_{09}^{\alpha})$ | $=$ | $22,$
$\dim\mathcal{O}(\mathbb{S}_{21}^{\alpha,\beta})$ | $=$ | $21,$
$\dim\mathcal{O}(\mathbb{S}_{41}^{\alpha})=\dim\mathcal{O}({\mathfrak{V}}_{4+1})=\dim\mathcal{O}(\mathbb{L}_{82})$ | $=$ | $20.$

Hence,
$\mathbb{L}_{28}^{\alpha},\mathbb{L}_{47}^{\alpha,\beta},\mathbb{L}_{52}^{\alpha},\mathbb{L}_{79}^{\alpha}$,
${\mathfrak{V}}_{3+2}$ and ${\bf B}_{09}^{\alpha}$ give $6$ irreducible
components. Moreover, the algebra $\mathbb{L}_{82}$ is a only one one-
generated $5$-dimensional nilpotent Leibniz algebra. It is known that if
$A(*)\rightarrow B$ and $B$ is one-generated, then $A$ is one-generated, then
$\mathbb{L}_{82}$ is rigid and its closure gives us an irreducible component.

Below we have listed all the important reasons for necessary non-
degenerations.

Non-degenerations reasons
---
${\bf B}_{09}^{\alpha}$ | $\not\rightarrow$ | ${\mathfrak{V}}_{4+1},\mathbb{S}_{41}^{\alpha},\mathbb{S}_{21}^{\alpha,\beta}$ | $\mathcal{R}=\left\\{\begin{array}[]{ll}\mbox{new basis}\ f_{1}=e_{1},\ f_{2}=e_{2},\ f_{3}=e_{4},\ f_{4}=e_{3},\ f_{5}=e_{5}\\\ A_{1}^{2}\subseteq A_{4},\ A_{1}A_{5}+A_{5}A_{1}=0,\\\ c_{41}^{5}=-c_{14}^{5},\quad c_{42}^{5}=-c_{24}^{5},\quad c_{43}^{5}=-c_{34}^{5},\quad c_{44}^{5}=0\\\ \end{array}\right\\}$

The rest of the degenerations is given in the following two tables and it
completes the proof of the Theorem.

${\bf B}_{09}^{\frac{4\alpha(1+t^{2})-1}{4\alpha+t^{2}-1}}$ | $\to$ | $\mathbb{S}_{22}^{\alpha}$ | $E_{1}^{t}=\frac{t(1+t^{2})}{t^{2}+\sqrt{1-4\alpha(1+t^{2})}}e_{1}+\frac{t^{3}+t\sqrt{1-4\alpha(1+t^{2})}}{1-4\alpha-t^{2}}e_{2}+\frac{t\sqrt{1-4\alpha(1+t^{2})}}{1-4\alpha-t^{2}}e_{3}-\frac{t}{\sqrt{1-4\alpha(1+t^{2})}}e_{4}$
---|---|---|---
$E_{2}^{t}=\frac{t^{2}(1-\sqrt{1-4\alpha(1+t^{2})})}{2t^{2}+2\sqrt{1-4\alpha(1+t^{2})}}e_{1}+\frac{t^{2}(1-4\alpha+\sqrt{1-4\alpha(1+t^{2})})}{2(1-4\alpha-t^{2})}e_{2}+\frac{t^{2}\sqrt{1-4\alpha(1+t^{2})}}{1-4\alpha-t^{2}}e_{3}$
$E_{3}^{t}=\frac{t^{3}\sqrt{1-4\alpha(1+t^{2})}}{1-4\alpha-t^{2}}e_{3}+\frac{t^{3}(1+t^{2})(1-4\alpha+\sqrt{1-4\alpha(1+t^{2}))}}{2(1-4\alpha-t^{2})(t^{2}+\sqrt{1-4\alpha(1+t^{2})})}e_{5}$ | $E_{4}^{t}=\frac{t^{2}}{\sqrt{1-4\alpha(1+t^{2})}}e_{4}$ | $E_{5}^{t}=\frac{t^{4}}{1-4\alpha-t^{2}}e_{5}$

For the rest of degenerations, in case of $E_{1}^{t},\dots,E_{5}^{t}$ is a
parametric basis for ${\bf A}\to{\bf B},$ it will be denoted by ${\bf
A}\xrightarrow{(E_{1}^{t},\dots,E_{5}^{t})}{\bf B}$.

${\bf B}_{02}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-1}e_{4},t^{-3}e_{5})}$ | ${\bf B}_{01}$ | ${\bf B}_{04}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-2}e_{4},t^{-4}e_{5})}$ | ${\bf B}_{02}$
---|---|---|---|---|---
${\bf B}_{04}$ | $\xrightarrow{(t^{-2}e_{1},e_{2},t^{-2}e_{3},t^{-1}e_{4},t^{-3}e_{5})}$ | ${\bf B}_{03}$ | ${\bf B}_{08}$ | $\xrightarrow{(t^{-1}e_{1},e_{2},t^{-1}e_{3},t^{-1}e_{4},t^{-2}e_{5})}$ | ${\bf B}_{04}$
${\bf B}_{08}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},e_{4},t^{-2}e_{5})}$ | ${\bf B}_{05}$ | ${\bf B}_{08}$ | $\xrightarrow{(t^{-2}e_{1},e_{2},t^{-2}e_{3},e_{4},t^{-2}e_{5})}$ | ${\bf B}_{06}$
${\bf B}_{09}^{\frac{1}{t^{2}}}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},e_{4},t^{-2}e_{5})}$ | ${\bf B}_{07}$ | ${\bf B}_{09}^{\frac{1}{t}}$ | $\xrightarrow{(t^{-1}e_{1},e_{2},t^{-1}e_{3},e_{4},t^{-1}e_{5})}$ | ${\bf B}_{08}$
${\bf B}_{11}$ | $\xrightarrow{(te_{1},t^{-1}e_{2},e_{3},t^{2}e_{4},t^{2}e_{5})}$ | ${\bf B}_{10}$ | ${\bf B}_{14}$ | $\xrightarrow{(t^{-1}e_{1},e_{2},t^{-1}e_{3},t^{-1}e_{4},t^{-2}e_{5})}$ | ${\bf B}_{11}$
${\bf
B}_{09}^{\frac{\alpha}{t}}\quad\xrightarrow{(t^{-1}e_{1}+t^{-2}e_{2}+t^{-1}e_{3},t^{-1}e_{2},t^{-2}e_{3},t^{-1}e_{4},t^{-3}e_{5})}\quad{\bf
B}_{12}^{\alpha}$
${\bf B}_{09}^{0}$ | $\xrightarrow{(e_{1}-e_{2},-te_{2},-te_{3},t^{-1}e_{4},-e_{5})}$ | ${\bf B}_{13}$ | ${\bf B}_{09}^{-t^{2}}$ | $\xrightarrow{(e_{1}-e_{2},-te_{2},-te_{3},t^{-1}e_{4},-e_{5})}$ | ${\bf B}_{14}$

∎

It is easy to see, that each one-generated binary Leibniz algebra is a Leibniz
algebra. On the other side, there is only one one-generated $n$-dimensional
Leibniz algebra. Let us also give a trivial observation.

######  Lemma 4.

The variety of $n$-dimensional nilpotent binary Leibniz algebras has at least
one rigid algebra.

The present lemma provides a question.

######  Open question 1.

Are there non-one-generated rigid algebras in the variety of $n$-dimensional
nilpotent binary Leibniz algebras?

###  2.4. The geometric classification of $4$-dimensional nilpotent algebras
of nil-index 3

The main result of the present section is the following theorem.

######  Theorem E.

The variety of complex $4$-dimensional nilpotent algebras of nil-index $3$ has
dimension 15 and it has two irreducible components

$\mathcal{C}_{1}=\overline{\\{\mathcal{O}(\mathfrak{N}_{2}(\gamma))\\}},$
$\mathcal{C}_{2}=\overline{\\{\mathcal{O}(\mathbb{M}_{12}^{\alpha})\\}}.$

In particular, there are no rigid algebras in this variety.

###### Proof.

Thanks to [35], the variety of $2$-step nilpotent algebras has two irreducible
components defined by the following families of algebras

$\mathfrak{N}_{2}(\gamma)$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-\gamma e_{3}$ | $e_{2}e_{2}=-e_{4}$ |
---|---|---|---|---|---
$\mathfrak{N}_{3}(\alpha)$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=\alpha e_{4}$ | $e_{2}e_{1}=-\alpha e_{4}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{3}=e_{4}$

Thanks to [35], we have
$\mathfrak{L}_{11}\to\mathfrak{L}_{01},\mathfrak{L}_{09},\mathfrak{L}_{10},\mathfrak{L}_{12}.$
In the following tables, we have listed all necessary degenerations.

$\mathbb{M}_{06}$ | $\to$ | $\mathfrak{L}_{11}$ | $E_{1}^{t}=te_{1}-e_{2}+\frac{t^{2}-1}{t}e_{3}$ | $E_{2}^{t}=-te_{2}-2e_{3}$
---|---|---|---|---
|  |  | $E_{3}^{t}=t^{2}e_{3}-(t^{2}+t+1)e_{4}$ | $E_{4}^{t}=t^{2}e_{4}$
$\mathbb{M}_{06}$ | $\to$ | $\mathbb{M}_{03}^{\alpha}$ | $E_{1}^{t}=e_{1}+\alpha e_{2}-\alpha^{2}e_{3}$ | $E_{2}^{t}=te_{2}-2\alpha te_{3}$
|  |  | $E_{3}^{t}=te_{3}-(\alpha^{2}-\alpha)te_{4}$ | $E_{4}^{t}=te_{4}$
$\mathbb{M}_{06}$ | $\to$ | $\mathbb{M}_{04}^{\alpha}$ | $E_{1}^{t}=te_{1}+\alpha te_{2}-\alpha^{2}te_{3}$ | $E_{2}^{t}=t^{2}e_{2}-2\alpha t^{2}e_{3}$
|  |  | $E_{3}^{t}=t^{3}e_{3}-(\alpha^{2}-\alpha)t^{3}e_{4}$ | $E_{4}^{t}=t^{4}e_{4}$
$\mathbb{M}_{12}^{\frac{t}{1-t}}$ | $\to$ | $\mathbb{M}_{06}$ | $E_{1}^{t}=\frac{it}{\sqrt{1-t}}e_{1}+\frac{t}{t-1}e_{2}-\frac{it}{\sqrt{1-t}}e_{3}$ | $E_{2}^{t}=-\frac{2it}{\sqrt{1-t}}e_{1}-\frac{2t^{2}}{t-1}e_{2}$
|  |  | $E_{3}^{t}=-\frac{2it^{2}}{\sqrt{1-t}}e_{3}-\frac{2t^{2}}{t-1}e_{4}$ | $E_{4}^{t}=\frac{4t^{3}}{t-1}e_{4}$

For the rest of degenerations, in case of $E_{1}^{t},\dots,E_{4}^{t}$ is a
parametric basis for ${\bf A}\to{\bf B},$ it will be denoted by ${\bf
A}\xrightarrow{(E_{1}^{t},\dots,E_{4}^{t})}{\bf B}$.

$\mathbb{M}_{12}^{\frac{1}{t^{2}}}$ | $\xrightarrow{(te_{1},e_{2},e_{3}-\alpha t^{-1}e_{4},e_{4})}$ | $\mathfrak{N}_{3}(\alpha)$ | $\mathbb{M}_{06}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-3}e_{4})}$ | $\mathbb{M}_{05}$
---|---|---|---|---|---
$\mathbb{M}_{11}$ | $\xrightarrow{(e_{1}-e_{2}+e_{3},t^{-1}e_{2},t^{-1}e_{3},t^{-2}e_{4})}$ | $\mathbb{M}_{07}$ | $\mathbb{M}_{11}$ | $\xrightarrow{(te_{1}-te_{2}+te_{3},t^{-1}e_{2},e_{3},e_{4})}$ | $\mathbb{M}_{08}$
$\mathbb{M}_{11}$ | $\xrightarrow{(e_{1}+(t^{-2}-1)e_{2}+e_{3},t^{-1}e_{2},t^{-1}e_{3},t^{-2}e_{4})}$ | $\mathbb{M}_{09}$ | $\mathbb{M}_{11}$ | $\xrightarrow{(t^{-1}e_{1}+\alpha t^{-1}e_{2},e_{2},t^{-1}e_{3},t^{-2}e_{4})}$ | $\mathbb{M}_{10}^{\alpha}$
$\mathbb{M}_{12}^{-\frac{1}{4t^{2}}}$
$\xrightarrow{(t^{-1}e_{1}+\frac{1}{2}t^{-1}e_{2},e_{2},t^{-1}e_{3}-\frac{1}{2}t^{-2}e_{4},t^{-2}e_{4})}$
$\mathbb{M}_{11}$

After carefully checking the dimensions of orbit closures of the rest of the
algebras, we have

$\dim\mathcal{O}(\mathfrak{N}_{2}(\gamma))=12,$
$\dim\mathcal{O}(\mathbb{M}_{12}^{\alpha})=15.$

Non-degenerations reasons are given below.

${\mathbb{M}}_{12}^{\alpha}$ | $\not\rightarrow$ | $\mathfrak{N}_{2}(\gamma)$ | $\mathcal{R}=\left\\{\begin{array}[]{ll}A_{1}^{2}\subset A_{3},\ c_{11}^{3}=c_{22}^{3}=0,\ c_{12}^{3}=-c_{21}^{3}\\\ \end{array}\right\\}$
---|---|---|---

Hence, $\mathfrak{N}_{2}(\gamma)$ and $\mathbb{M}_{12}^{\alpha}$ give
irreducible components. ∎

Let us remember that the variety of $n$-dimensional nilpotent algebras of nil-
index 2 is irreducible (see, [32]). The last theorem gives an example of a
variety of nilpotent algebras of nil-index $3,$ that has two irreducible
components. The last observation is motivating the following question.

######  Open question 2.

Find the number of irreducible components of the variety of $n$-dimensional
nilpotent algebras of nil-index $3.$

###  2.5. The geometric classification of $4$-dimensional nilpotent mono
Leibniz algebras

The main result of the present section is the following theorem.

######  Theorem F.

The variety of complex $4$-dimensional nilpotent mono Leibniz algebras has
dimension 15 and it has 3 irreducible components:

$\mathcal{C}_{1}=\overline{\\{\mathcal{O}(\mathfrak{L}_{2})\\}},$
$\mathcal{C}_{2}=\overline{\\{\mathcal{O}(\mathbb{M}_{12}^{\alpha})\\}},$
$\mathcal{C}_{3}=\overline{\\{\mathcal{O}(\mathbb{M}_{14}^{\alpha})\\}}.$

In particular, there is only one rigid algebra in this variety.

###### Proof.

Thanks to [26] the variety of $4$-dimensional nilpotent Leibniz algebras has
only $4$ irreducible components defined by

$\mathfrak{N}_{3}(\alpha)$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=\alpha e_{4}$ | $e_{2}e_{1}=-\alpha e_{4}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{3}=e_{4}$ |
---|---|---|---|---|---|---|---
$\mathfrak{L}_{2}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{2}e_{1}=e_{3}$ | $e_{3}e_{1}=e_{4}$ |  |  |
$\mathfrak{L}_{5}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$ |  |
$\mathfrak{L}_{11}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=-e_{3}$ | $e_{1}e_{3}=-e_{4}$ | $e_{2}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$

In the Theorem Theorem E variety of complex $4$-dimensional nilpotent algebras
of nil-index $3$ has one irreducible component

$\mathbb{M}_{12}^{\alpha}$ | $:$ | $e_{1}e_{1}=\alpha e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{3}$
---|---|---|---|---|---
|  | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |

From Theorem Theorem E, we obtain that the algebra $\mathbb{M}_{12}^{\alpha}$
is degeneration to the algebra $\mathfrak{L}_{11}$.

After carefully checking the dimensions of orbit closures of the more
important for us algebras, we have

$\dim\mathcal{O}(\mathfrak{L}_{2})=12,$
$\dim\mathcal{O}(\mathbb{M}_{12}^{\alpha})=\dim\mathcal{O}(\mathbb{M}_{14}^{\alpha})=15.$

Hence, $\mathbb{M}_{12}^{\alpha}$ and $\mathbb{M}_{14}^{\alpha}$ give $2$
irreducible components. Moreover, since the algebra $\mathfrak{L}_{2}$ is only
one one-generated $4$-dimensional nilpotent Leibniz algebra, then
$\mathfrak{L}_{2}$ is rigid and its closure gives us an irreducible component.

The rest of the degenerations is given below in the following tables and it
completes the proof of the Theorem:

$\mathbb{M}_{14}^{t+1}$ | $\to$ | $\mathfrak{L}_{5}$ | $E_{1}^{t}=-2t^{2}e_{1}+(t^{2}-t)e_{2}+t^{2}e_{3}$ | $E_{2}^{t}=(t^{2}-t^{3})e_{2}-(2t^{3}+t^{2})e_{3}$
---|---|---|---|---
|  |  | $E_{3}^{t}=2t^{4}e_{3}-(t^{4}-t^{3})e_{4}$ | $E_{4}^{t}=2(t^{6}-t^{5})e_{4}$
$\mathbb{M}_{14}^{\alpha}$ | $\to$ | $\mathbb{M}_{16}^{\alpha}$ |  [let us fix $x=t(\sqrt{1-4\alpha}-t)$]
$E_{1}^{t}=xe_{1}-\frac{2x}{1-\sqrt{1-4(\alpha+x)}}e_{2}-\frac{x(2\alpha(\alpha+x)+x(1+\sqrt{1-4(\alpha+x)}))}{2(\alpha+x)^{2}}e_{3}$
$E_{2}^{t}=-\frac{2x\sqrt{1-4(\alpha+x)}}{1-\sqrt{1-4(\alpha+x)}}e_{2}-\frac{x}{2}(1+\sqrt{1-4(\alpha+x)})e_{3}$
$E_{3}^{t}=\frac{x^{3}(2(\alpha+x)-1-\sqrt{1-4(\alpha+x)})}{2(\alpha+x)^{2}}e_{3}-\frac{x^{3}(2(\alpha+x)-1-\sqrt{1-4(\alpha+x)})(1+\sqrt{1-4(\alpha+x)})}{4(\alpha+x)^{3}}e_{4}$
$E_{4}^{t}=\frac{x^{4}(1+\sqrt{1-4(\alpha+x)-2(\alpha+x)})(1+\sqrt{1-4(\alpha+x)})}{4(\alpha+x)^{3}}e_{4}$
$\mathbb{M}_{14}^{\alpha\neq 0}$ | $\to$ | $\mathbb{M}_{18}^{\alpha\neq 0}$ |  [let us fix $x=-t(\sqrt{1-4\alpha}+t)$]
$E_{1}^{t}=xe_{1}-\frac{2x}{1-\sqrt{1-4(\alpha+x)}}e_{2}-\frac{x(2\alpha(\alpha+x)+x(1+\sqrt{1-4(\alpha+x)}))}{2(\alpha+x)^{2}}e_{3}$
$E_{2}^{t}=-\frac{2x\sqrt{1-4(\alpha+x)}}{1-\sqrt{1-4(\alpha+x)}}e_{2}-\frac{x}{2}(1+\sqrt{1-4(\alpha+x)})e_{3}$
$E_{3}^{t}=\frac{x^{3}(2(\alpha+x)-1-\sqrt{1-4(\alpha+x)})}{2(\alpha+x)^{2}}e_{3}-\frac{x^{3}(2(\alpha+x)-1-\sqrt{1-4(\alpha+x)})(1+\sqrt{1-4(\alpha+x)})}{4(\alpha+x)^{3}}e_{4}$
$E_{4}^{t}=\frac{x^{4}(1+\sqrt{1-4(\alpha+x)-2(\alpha+x)})(1+\sqrt{1-4(\alpha+x)})}{4(\alpha+x)^{3}}e_{4}$
$\mathbb{M}_{20}$ | $\to$ | $\mathbb{M}_{22}$ | $E_{1}^{t}=t\sqrt{1-t}e_{1}+i(t^{2}-t)e_{2}-i(t^{3}-t)e_{3}$ | $E_{2}^{t}=it\sqrt{1-t}e_{1}+te_{2}+(t^{2}-t)e_{3}$
|  |  | $E_{3}^{t}=t^{3}e_{3}+it^{3}\sqrt{1-t}e_{4}$ | $E_{4}^{t}=t^{4}\sqrt{1-t}e_{4}$

and

$\mathbb{M}_{02}$ | $\xrightarrow{({t}^{-1}e_{1},t^{-2}e_{2},e_{3},t^{-2}e_{4})}$ | $\mathbb{M}_{01}$ | $\mathbb{M}_{14}^{0}$ | $\xrightarrow{(e_{1},e_{3},te_{2},te_{4})}$ | $\mathbb{M}_{02}$
---|---|---|---|---|---
$\mathbb{M}_{14}^{\alpha}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-3}e_{4})}$ | $\mathbb{M}_{13}^{\alpha}$ | $\mathbb{M}_{16}^{\alpha}$ | $\xrightarrow{(t^{-}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-3}e_{4})}$ | $\mathbb{M}_{15}^{\alpha}$
$\mathbb{M}_{18}^{\alpha\neq 0}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-3}e_{4})}$ | $\mathbb{M}_{17}^{\alpha\neq 0}$ | $\mathbb{M}_{20}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-3}e_{4})}$ | $\mathbb{M}_{19}$
$\mathbb{M}_{14}^{\frac{1}{t^{2}}}$ | $\xrightarrow{(te_{2},e_{1},e_{3},te_{4})}$ | $\mathbb{M}_{20}$ | $\mathbb{M}_{22}$ | $\xrightarrow{(t^{-1}e_{1},t^{-1}e_{2},t^{-2}e_{3},t^{-3}e_{4})}$ | $\mathbb{M}_{21}$

∎

It is easy to see, that each one-generated mono Leibniz algebra is a Leibniz
algebra. On the other side, there is only one one-generated $n$-dimensional
Leibniz algebra. Let us also give a trivial observation.

######  Lemma 5.

The variety of $n$-dimensional nilpotent mono Leibniz algebras has at least
one rigid algebra.

The present lemma provides a question.

######  Open question 3.

Are there non-one-generated rigid algebras in the variety of $n$-dimensional
nilpotent mono Leibniz algebras?

Compliance with ethical standard

Author contributions All authors contributed to the study, conception and
design. All authors read and approved the final manuscript.

Conflict of interest There is no potential conflict of ethical approval,
conflict of interest, and ethical standards.

Data Availibility Data sharing is not applicable to this article as no
datasets were generated or analyzed during the current study.

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