11institutetext: Pattern Recognition Lab, Friedrich-Alexander-University
Erlangen-Nuremberg, Erlangen, Germany
22institutetext: Learning Approaches for Vascular Analysis Group, Friedrich-
Alexander-University Erlangen-Nuremberg, Erlangen, Germany

# Segmentation of the Carotid Lumen and Vessel Wall using Deep Learning and
Location Priors

Florian Thamm 1122 Felix Denzinger 1122 Leonhard Rist 1122 Celia Martin
Vicario 1122 Florian Kordon 11 Andreas Maier 11

###### Abstract

In this report we want to present our method and results for the Carotid
Artery Vessel Wall Segmentation Challenge. We propose an image-based pipeline
utilizing the U-Net architecture and location priors to solve the segmentation
problem at hand.

###### Keywords:

Vessel Wall Segmentation Magnetic Resonance Imaging Deep Learning

##  1 Introduction

One of the leading causes of death worldwide is atherosclerosis, a disease
which is defined as the build-up of plaque deposits and formations narrowing
the lumen, affecting medium and large-sized arteries [2, 4]. This is
especially critical for the carotids, which supply the brain and face together
with the vertebral arteries. Occlusion in those vessels can thus cause severe
irreversible damage if not treated immediately. Therefore, the diagnosis of
atherosclerotic vessels is of crucial importance. The vessel wall imaging
(VWI) done with magnetic resonance imaging (MRI) is a modality designed to
identify and localize stenosis of the atherosclerotic kind. In particular, the
fast 3D carotid black blood MRI sequence (3D Motion Sensitized Driven
Equilibrium prepared Rapid Gradient Echo, short 3D-MERGE) is designed to
visualize blood vessels with sub-millimeter isotropic resolution and at the
same time atherosclerotic malformations, their severity, size, and the overall
stenosis [1]. Despite the good visibility using the 3D-MERGE sequence, expert
knowledge is still required to read such an image. As this is a tedious and
time consuming task, the Vessel-Wall-Segmentation Challenge hosted by the
Vascular Imaging Laboratory from the University of Washington has been
organized to foster and encourage automated methods addressing the
segmentation of the outer and the inner (lumen) wall of the left and right
external (ECAL, ECAR) and internal carotids (ICAL, ICAR). In this report, we
present our approach for the automated vessel wall segmentation and discuss
the achieved results.

##  2 Methods

The following section describes the methods we used to solve the segmentation
task at hand. First, we propose a processing pipeline utilizing the U-Net [5]
architecture which receives the MRI slices and returns the contours based on
the segmentation. Furthermore, we describe all hyperparameters of the network
and its training. Secondly, we describe certain characteristics in the
application of said pipeline and introduce two concepts our pipeline follows
to increase the segmentation accuracy.

###  2.1 Pipeline

The illustration in Fig. 1 shows both, training and testing phase of the
proposed pipeline. During training, a 2D U-Net model receives a region of
interest (RoI) of the input slices and predicts a binary segmentation mask of
the carotids, the lumen, and wall in separate channels. The RoI extraction is
described in Sec. 2.2. The internal and external carotid arteries were treated
separately with two different models but the same architecture. All ground
truth (GT) masks were given as contour formats in the CASCADE file format [3],
thus it is necessary to convert the contours into a binary mask format,
yielding the ground truth mask. This mask is used to supervise the
architecture with the binary cross-entropy loss $L$. During the inference and
thus testing, the predicted binary mask is transferred back into a contour-
based format in a post-processing step. This format is used for the final
evaluation. As described later in Sec. 2.2, the model predicts both sides
(left and right) separately.

Figure 1: Proposed Pipeline. Training includes all paths leading to the Loss
$L$. Inference (testing) only includes the bottom path from the input to the
predicted contour.

The U-Net architecture, consisting of 4 downsampling and hence 4 upsampling
steps, beginning with 64 channels and doubling channels up to the deepest
stage of 1024 channels, was trained using Adam for 1500 epochs using a
learning rate of $10^{-4}$ with a batch size of 32. Max-pooling was used for
the downsampling path and transposed Convolutions for the upsampling path.

###  2.2 Concepts

#### Contour-Mask Consistency

All ground truth masks were given as contours in the CASCADE file format [3].
As described in 2.1 the pipeline we propose is image-based, which means, the
contours are predicted based on a binary segmentation of the internal and
external carotid inner and outer walls. As the ground truth masks need to be
converted to masks, and the predictions vice-versa into contours, it is hence
of importance that both transforming routines are cycle consistent, explained
exemplary in the following. Suppose some contour $A$ is given and the task is
now to transform that contour into its mask representation using a function
Contour2Mask. If this mask is now used to be transformed back into the contour
representation using a function Mask2Contour returning contour $B$, the
Contour-Mask Consistency states that $B=A$. To enforce this consistency
contour points are forced to lie on the image grid, before converting them
into binary masks.

#### Reduction to Region of Interest (RoI)

To counteract class imbalance, as it is inherent in the problem at hand, we
propose to reduce the input image to the region where carotids are expected.
This can be done for the left and right carotids respectively. To evaluate
this region, the smallest bounding box was determined containing all ground
truth segmentation of one side. This bounding box was additionally enlarged to
ensure outliers in the test data are covered as well, to a final size of
$160\times 160$ visualized as RoI in 1. The U-Net segments both bounding
boxes, left and right, separately from each other. This has the advantage that
either one side can be flipped always to the opposite side making left and
right boxes visually look the same, since a symmetrical anatomy can be
assumed, or, to augment and flip left and right randomly. Small experiments
have shown the latter results in better validation performances.

##  3 Evaluation

The final evaluation was done based on 20 datasets with 720 voxels in height
(x), width (y), and depth (z). Five datasets had 640 voxels in all dimensions
instead. In total 25 datasets were therefore available. Due to technical
issues, the initial submission was not robust against resolutions different to
7203 voxels, thus failed for five datasets in total 107 slices. This has been
fixed and the evaluation restarted without retraining, which leads to the
performance shown in Tab. 3.

Metric |  | Initial
---
Submission
| Resolution
---
Robust
Dice Score | 0.672±0.248 | 0.707±0.238
Lumen area difference | 0.142±0.239 | 0.110±0.208
Wall area difference | 0.162±0.224 | 0.139±0.205
Normalized wall index difference | 0.141±0.212 | 0.135±0.233
| Hausdorff distance on
---
lumen normalized by radius
1.095±4.813 | 1.058±4.705
| Hausdorff distance on
---
wall normalized by radius
0.810±2.953 | 0.845±3.007
Quantitative score | 0.633±0.335 | 0.691±0.309
Number of unmatched slices | 371 | 264

##  4 Summary

For the Carotid Artery Vessel Wall Segmentation Challenge, we have proposed a
U-Net-based pipeline that automatically performs a segmentation of the
internal and external carotids on both sides based on 3D-MERGE MRI sequences.
The network is supervised in the image domain, while the transfer from- and to
contours is cycle consistent. To counteract class imbalances in the
segmentation, the pipeline automatically, and dynamically extracts bounding
boxes, at the regions where the carotids are expected. In the final test
evaluation, our approach achieved a dice score of 0.707 and a quantitative
measure of 0.691.

## Author Contribution Statement

The authors confirm contribution to the work as follows. Conception and
management: F.T., F.D., Manuscript draft: F.T., Manuscript review: F.D., L.R.,
C.V., A.M., Preparation and recording of presentations: F.T., Training of
models: F.D., Pipeline implementation: F.D., L.R., F.T., F.K., C.V.,
Deployment: F.T., L.R., Evaluations: F.T., F.D., Various experiments: F.D.,
F.K., F.T., L.R., C.V.

## References

  * [1] Balu, N., Yarnykh, V.L., Chu, B., Wang, J., Hatsukami, T., Yuan, C.: Carotid plaque assessment using fast 3d isotropic resolution black-blood mri. Magnetic resonance in medicine 65(3), 627–637 (2011)
  * [2] Herrington, W., Lacey, B., Sherliker, P., Armitage, J., Lewington, S.: Epidemiology of atherosclerosis and the potential to reduce the global burden of atherothrombotic disease. Circulation research 118(4), 535–546 (2016)
  * [3] Kerwin, W., Xu, D., Liu, F., Saam, T., Underhill, H., Takaya, N., Chu, B., Hatsukami, T., Yuan, C.: Magnetic resonance imaging of carotid atherosclerosis: plaque analysis. Topics in Magnetic Resonance Imaging 18(5), 371–378 (2007)
  * [4] Mitrovska, S., Jovanova, S., Matthiesen, I., Libermans, C.: Atherosclerosis: Understanding pathogenesis and challenge for treatment (2009)
  * [5] Ronneberger, O., Fischer, P., Brox, T.: U-net: Convolutional networks for biomedical image segmentation. In: International Conference on Medical image computing and computer-assisted intervention. pp. 234–241. Springer (2015)


The algebraic and geometric classification of antiassociative algebras

Renato Fehlberg Júnior,a Ivan Kaygorodovb $\&$ Crislaine Kusterc

a Departamento de Matemática, Universidade Federal do Espírito Santo, Vitória
- ES, Brazil

b Centro de Matemática e Aplicações, Universidade da Beira Interior, Covilhã,
Portugal

c Instituto de Matemática Pura e Aplicada, Rio de Janeiro - RJ, Brazil

E-mail addresses:

Renato Fehlberg Júnior<EMAIL_ADDRESS>
Ivan Kaygorodov<EMAIL_ADDRESS>
Crislaine Kuster<EMAIL_ADDRESS>

Abstract: This paper is devoted to the complete algebraic and geometric
classification of complex $4$ and $5$-dimensional antiassociative algebras. In
particular, we proved that the variety of complex $4$-dimensional
antiassociative algebras has dimension $12$ and it is defined by three
irreducible components (in particular, there is only $1$ rigid algebra in this
variety); the variety of complex $5$-dimensional antiassociative algebras has
dimension $24$ and it is defined by $8$ irreducible components (in particular,
there are only $4$ rigid algebras in this variety).

Keywords: antiassociative algebra, algebraic classification, central
extension, geometric classification, degeneration.

MSC2020: 17A30, 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 [23,
20, 13, 24]. Geometric properties of a variety of algebras defined by a family
of polynomial identities have been an object of study since 1970’s (see, [33,
11, 6, 2, 15, 9, 12, 31, 18, 19, 1, 26]). Gabriel described the irreducible
components of the variety of $4$-dimensional unital associative algebras [15].
Cibils considered rigid associative algebras with $2$-step nilpotent radical
[12]. Burde and Steinhoff constructed the graphs of degenerations for the
varieties of $3$-dimensional and $4$-dimensional Lie algebras [6]. Grunewald
and O’Halloran calculated the degenerations for the variety of $5$-dimensional
nilpotent Lie algebras [18]. Chouhy proved that in the case of finite-
dimensional associative algebras, the $N$-Koszul property is preserved under
the degeneration relation [11]. Degenerations have also been used to study a
level of complexity of an algebra [33]. The study of degenerations of algebras
is very rich and closely related to deformation theory, in the sense of
Gerstenhaber [17].

In the present paper, we give the algebraic and geometric classification of
complex $4$-dimensional and $5$-dimensional antiassociative algebras. An
antiassociative algebra is defined by the following identity $(xy)z+x(yz)=0.$
Thanks to [29], each antiassociative algebra is a $3$-nilpotent algebra and we
can use the standard method of classification of nilpotent algebras first
appeared in [32]. On the other hand, the variety of antiassociative algebras
is a particular case of $(\alpha,\beta,\gamma)$-algebras considered by Bergen
and Grzeszczuk in [5, 4], and by Calderón in [7]. It is also a particular case
of Lie type algebras considered by Makarenko in [27], which is a particular
case of $(\alpha,\beta,\gamma)$-algebras. It looks that antiassociative
algebras were initially introduced in [34]. Later, they appeared in several
articles, both with a study of theirs properties, see [29, 21], as well as
with theirs relations with others algebras, like Jordan–Lie superalgebras or
CB-algebras [29, 30]. In particular, an anticommutative algebra $A$ is a CB-
algebra if, and only if, $A$ is an antiassociative algebra [30]. García-
Martínez and Van der Linden proved that an algebraically coherent variety of
anti-commutative algebras is either a variety of Lie algebras or a variety of
antiassociative algebras [16]. They are also related to anticommutative
$g$-alternative algebras [10]. Moreover, given an antiassociative algebra $A$,
the new algebra $A^{(+)}$ is a Mock-Lie algebra, which is a Jordan algebra of
nilindex 3. A Dual Mock-Lie algebra is an anticommutative antiassociative
algebra [35]. Algebraic and geometric classifications of small dimensional
dual Mock-Lie algebras are given in [9]. Another entertaining fact about these
algebras is that Markl and Remm showed the non-Koszulness of the corresponding
operad for antiassociative algebras [28], hence the deformation cohomology
differs from the “standard” one, more precisely, the deformation cohomology
coincides with the triple cohomology. Futhermore, antiassocitive algebras
playing an important role in gauge theory, in particular, nonlinear gauge
theory for two or more spin-2 fields [3].

Since any product involving four or more elements of an antiassociative
algebra is identically zero, (see, [29, Lemma 1.1]), we can describe this
variety in two subsets: algebras that satisfies the identity $xyz=0$, called
2-step nilpotent algebras, and algebras that do not satisfies the identity
$xyz=0$, called non-2-step nilpotent algebras. Note that 2-step nilpotent
algebras are contained in the intersection of varieties of algebras defined by
families of polynomial identities of degree $3$, for example, Leibniz,
associative and Zinbiel algebras. Then, the classification of antiassociative
2-step nilpotent algebras of dimension up to $4$ can be extracted from the
list of algebras presented in [26]. The algebraic classification of all
$5$-dimensional $2$-step nilpotent algebras is a wild problem and it is not
concluded until now, but there is a geometric classification of these algebras
[22], which will be used in the present work for a geometric classification of
$5$-dimensional antiassociative algebras. The method for classifying
antiassociative 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
[8, 25, 20, 32]. First, Skjelbred and Sund used central extensions of Lie
algebras to obtain a classification of nilpotent Lie algebras [32]. Note that
the Skjelbred-Sund method of central extensions is an important tool in the
classification of nilpotent algebras. Using the same method, small dimensional
nilpotent (terminal [24], Lie [13], anticommutative [23], dual Mock-Lie [9])
algebras 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 is only $1$ complex $3$-dimensional (non-2-step
nilpotent) antiassociative algebras, described explicitly in section 1.2. Up
to isomorphism, there are only $3$ complex $4$-dimensional (non-2-step
nilpotent) antiassociative algebras, described explicitly in section 1.2. Up
to isomorphism, there are infinitely many isomorphism classes of complex
$5$-dimensional (non-2-step nilpotent) antiassociative algebras, described
explicitly in section 1.3 in terms of $4$ one-parameter families and $26$
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 varieties in which the rigid algebras of the varieties of all
$4$-dimensional Leibniz algebras, all $4$-dimensional nilpotent terminal
algebras [24], all $4$-dimensional nilpotent commutative algebras, all
$6$-dimensional nilpotent binary Lie algebras, all $6$-dimensional nilpotent
anticommutative algebras [23], all $8$-dimensional dual Mock Lie algebras [9]
have been found. A full description of degenerations has been obtained for
$2$-dimensional algebras, for $4$-dimensional Lie algebras in [6], for
$4$-dimensional Zinbiel and $4$-dimensional nilpotent Leibniz algebras in
[26], for $5$-dimensional nilpotent Lie algebras in [18], for $8$-dimensional
$2$-step nilpotent anticommutative algebras [1], for $(n+1)$-dimensional
$n$-Lie algebras, and some others. Our main results related to the geometric
classification of cited varieties are summarized below.

######  Theorem B.

The variety of complex $4$-dimensional antiassociative algebras has dimension
12 and it has 3 irreducible components (in particular, there is only one rigid
algebra in this variety). The variety of complex $5$-dimensional
antiassociative algebras has dimension 24 and it has $8$ irreducible
components (in particular, there are only $4$ rigid algebras in this variety).

##  1\. The algebraic classification of antiassociative algebras

###  1.1. Method of classification of nilpotent algebras

Throughout this paper, we use the notations and methods well written in [20],
which we have adapted for the antiassociative case with some modifications.
Further in this section we give some important definitions.

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

$\theta(xy,z)+\theta(x,yz)=0.$

These elements will be called 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 $\delta f\in{\rm Z^{2}}\left({\bf A},{\mathbb{V}}\right)$.
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 ${\rm H^{2}}\left({\bf A},{\mathbb{V}}\right)$ as the
quotient space ${\rm Z^{2}}\left({\bf A},{\mathbb{V}}\right)\big{/}{\rm
B^{2}}\left({\bf A},{\mathbb{V}}\right)$.

Let $\operatorname{Aut}({\bf A})$ be the automorphism group of ${\bf A}$ and
let $\phi\in\operatorname{Aut}({\bf A})$. For $\theta\in{\rm Z^{2}}\left({\bf
A},{\mathbb{V}}\right)$ define the action of the group
$\operatorname{Aut}({\bf A})$ on ${\rm Z^{2}}\left({\bf
A},{\mathbb{V}}\right)$ 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 ${\rm H^{2}}\left({\bf
A},{\mathbb{V}}\right)$.

Let $\bf A$ be a antiassociative 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 antiassociative algebra if and
only if $\theta\in{\rm Z^{2}}({\bf A},{\mathbb{V}})$.

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 $\operatorname{Ann}\left({\bf
A}_{\theta}\right)=(\operatorname{Ann}(\theta)\cap\operatorname{Ann}({\bf
A}))\oplus{\mathbb{V}}$.

The following result shows that every algebra with a non-zero annihilator is a
central extension of a smaller-dimensional algebra.

######  Lemma 1.

Let ${\bf A}$ be an $n$-dimensional antiassociative algebra such that
$\dim(\operatorname{Ann}({\bf A}))=m\neq 0$. Then there exists, up to
isomorphism, a unique $(n-m)$-dimensional antiassociative algebra ${\bf
A}^{\prime}$ and a bilinear map $\theta\in{\rm Z^{2}}({\bf
A^{\prime}},{\mathbb{V}})$ with $\operatorname{Ann}({\bf
A^{\prime}})\cap\operatorname{Ann}(\theta)=0$, where $\mathbb{V}$ is a vector
space of dimension m, such that ${\bf A}\cong{{\bf A}^{\prime}}_{\theta}$ and
${\bf A}/\operatorname{Ann}({\bf A})\cong{\bf A}^{\prime}$.

###### Proof.

Let ${\bf A}^{\prime}$ be a linear complement of $\operatorname{Ann}({\bf A})$
in ${\bf A}$. Define a linear map $P\colon{\bf A}\longrightarrow{\bf
A}^{\prime}$ by $P(x+v)=x$ for $x\in{\bf A}^{\prime}$ and
$v\in\operatorname{Ann}({\bf A})$, and define a multiplication on ${\bf
A}^{\prime}$ by $[x,y]_{{\bf A}^{\prime}}=P(xy)$ for $x,y\in{\bf A}^{\prime}$.
For $x,y\in{\bf A}$, we have

$P(xy)=P((x-P(x)+P(x))(y-P(y)+P(y)))=P(P(x)P(y))=[P(x),P(y)]_{{\bf
A}^{\prime}}.$

Since $P$ is a homomorphism $P({\bf A})={\bf A}^{\prime}$ is an
antiassociative algebra and ${\bf A}/\operatorname{Ann}({\bf A})\cong{\bf
A}^{\prime}$, which gives us the uniqueness. Now, define the map
$\theta\colon{\bf A}^{\prime}\times{\bf
A}^{\prime}\longrightarrow\operatorname{Ann}({\bf A})$ by
$\theta(x,y)=xy-[x,y]_{{\bf A}^{\prime}}$. Thus, ${\bf A}^{\prime}_{\theta}$
is ${\bf A}$ and therefore $\theta\in{\rm Z^{2}}({\bf
A^{\prime}},{\mathbb{V}})$ and $\operatorname{Ann}({\bf
A^{\prime}})\cap\operatorname{Ann}(\theta)=0$. ∎

######  Definition 2.

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 [20, Lemma 13]) that given a antiassociative
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 $G_{s}\left({\rm
H^{2}}\left({\bf A},\mathbb{C}\right)\right)$, given by $\phi
W=\left\langle\left[\phi\theta_{1}\right],\left[\phi\theta_{2}\right],\dots,\left[\phi\theta_{s}\right]\right\rangle$,
for each $\phi\in\operatorname{Aut}({\bf A})$ and each
$W=\left\langle\left[\theta_{1}\right],\left[\theta_{2}\right],\dots,\left[\theta_{s}\right]\right\rangle\in
G_{s}\left({\rm H^{2}}\left({\bf A},\mathbb{C}\right)\right)$. 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 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 [20, Lemma 17].

######  Lemma 3.

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 antiassociative 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 antiassociative
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 antiassociative 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 antiassociative algebra associated with a
representative of it.

####  1.1.1. 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 $\theta\in{\rm Z^{2}}({\bf
A},\bf\mathbb{V})$ can be uniquely written as $\theta=\displaystyle\sum_{1\leq
i,j\leq n}c_{ij}\Delta_{{i}{j}}$, where $c_{ij}\in\mathbb{C}$. We will use the
following notations for our algebras:

$\begin{array}[]{lll}{\mathcal{A}}_{i,j}&\mbox{---}&j\mbox{th
}i\mbox{-dimensional $2$-step nilpotent algebra;}\\\
{{\mathbb{A}}}_{i,j}&\mbox{---}&j\mbox{th }i\mbox{-dimensional non-2-step
nilpotent antiassociative algebra.}\end{array}$

###  1.2. The algebraic classification of $4$-dimensional non-2-step nilpotent
antiassociative algebras

Thanks to [14], we have the classification of all 3-dimensional nilpotent
algebras. From this list we are choosing only $3$-dimensional antiassociative
algebras:

${\mathcal{A}}_{3,1}$ | $:$ | $e_{1}e_{1}=e_{2}$ |  |
---|---|---|---|---
${\mathcal{A}}_{3,2}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ |
${\mathcal{A}}_{3,3}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$ |
${\mathcal{A}}^{\lambda}_{3,4}$ | $:$ | $e_{1}e_{1}=\lambda e_{3}$ | $e_{2}e_{1}=e_{3}$ | $e_{2}e_{2}=e_{3}$
${{\mathbb{A}}}_{3,1}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$

Note that the central extension with annihilator component of the non-$2$-step
nilpotent algebra ${{\mathbb{A}}}_{3,1}$, give us a four dimensional algebra
denoted by ${{\mathbb{A}}}_{4,1}$.

In the following table we give the description of the second cohomology space
of $3$-dimensional $2$-step nilpotent algebras.

${\rm H}^{2}({\mathcal{A}}_{3,1})$ | $=$ | $\Big{\langle}[\Delta_{12}]-[\Delta_{21}],[\Delta_{13}],[\Delta_{31}],[\Delta_{33}]\Big{\rangle}$
---|---|---
${\rm H}^{2}({\mathcal{A}}_{3,2})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{22}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{3,3})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{21}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}^{\lambda}_{3,4})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{21}]\par\Big{\rangle}$

Since $\operatorname{Ann}({\mathcal{A}}_{3,i})=\langle e_{3}\rangle$ we have
${\bf T}_{s}({\mathcal{A}}_{3,i})=\emptyset$, $i=2,3,4$, and, therefore, there
is no central extension without annihilator component of these algebras.

####  1.2.1. $1$-dimensional central extensions of ${\mathcal{A}}_{3,1}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{12}]-[\Delta_{21}],$ | $\nabla_{2}=[\Delta_{13}],$ | $\nabla_{3}=[\Delta_{31}],$ | $\nabla_{4}=[\Delta_{33}].$
---|---|---|---

Take $\theta=\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\in{\rm
H^{2}}({\mathcal{A}}_{3,1})$. Note that we can not have
$\alpha_{2}=\alpha_{3}=\alpha_{4}=0$ or $\alpha_{1}=0$. The automorphism group
of ${\mathcal{A}}_{3,1}$ consists of invertible matrices of the form

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

Since

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

we have that the action of ${\rm Aut}({\mathcal{A}}_{3,1})$ on the subspace
$\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\Big{\rangle}$ is given
by $\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}^{*}\nabla_{i}\Big{\rangle},$
where

$\alpha^{*}_{1}=\alpha_{1}x^{3},$ | $\alpha^{*}_{2}=\alpha_{1}xr+\alpha_{2}xy+\alpha_{4}yt,$ | $\alpha^{*}_{3}=\alpha_{3}xy-\alpha_{1}xr+\alpha_{4}yt,$ | $\alpha_{4}^{*}=\alpha_{4}y^{2}.$
---|---|---|---

Since $y\neq 0$, $\alpha_{4}\neq 0$ if, and only if, $\alpha_{4}^{\ast}\neq
0$, we need to consider only the following cases:

  1. (1)

Consider $\alpha_{4}=0$.

    1. (a)

If $\alpha_{3}=-\alpha_{2}$, by choosing $x=y=1$ and
$r=-\frac{\alpha_{2}}{\alpha_{1}}$, we get the representative
$\langle\nabla_{1}\rangle$, that we do not consider since the obtained algebra
would have $2$-dimensional annihilator.

    2. (b)

If $\alpha_{3}\neq-\alpha_{2}$, by choosing
$x=\frac{1}{\sqrt[3]{\alpha_{1}}}$,
$y=\frac{\sqrt[3]{\alpha_{1}}}{\alpha_{2}+\alpha_{3}}$ and
$r=\frac{\alpha_{3}}{(\alpha_{2}+\alpha_{3})\sqrt[3]{\alpha_{1}^{2}}}$, we
obtain the representative $\langle\nabla_{1}+\nabla_{2}\rangle$.

  2. (2)

If $\alpha_{4}\neq 0$, by choosing $x=\alpha_{1}\alpha_{4}$,
$y=\alpha_{1}^{2}\alpha_{4}$,
$r=\frac{\alpha_{1}\alpha_{4}(\alpha_{3}-\alpha_{2})}{2}$ and
$t=-\frac{\alpha_{1}(\alpha_{2}+\alpha_{3})}{2},$ we obtain the representative
$\langle\nabla_{1}+\nabla_{4}\rangle$.

Summarizing, we have the following distinct orbits:

$\langle\nabla_{1}+\nabla_{2}\rangle$ and
$\langle\nabla_{1}+\nabla_{4}\rangle$.

These orbits give us the following algebras:

${{\mathbb{A}}}_{4,2}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$
---|---|---|---|---|---
${{\mathbb{A}}}_{4,3}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$

###  1.3. The algebraic classification of $5$-dimensional non-2-step nilpotent
antiassociative algebras

It is easy to see that there are no non-2-step nilpotent $3$-dimensional
central extensions of antiassociative algebras of dimension $2$. The split
central extension of $4$-dimensional algebras obtained in Section 1.2 will be
denoted, respectively, by ${{\mathbb{A}}}_{5,1}$, ${{\mathbb{A}}}_{5,2}$ and
${{\mathbb{A}}}_{5,3}$. Hence, to classify $5$-dimensional non-split
antiassociative algebras, we need to find only $2$-dimensional central
extensions of ${\mathcal{A}}_{3,1}$ and $1$-dimensional central extensions of
$4$-dimensional $2$-step nilpotent algebras.

####  1.3.1. $2$-dimensional central extensions of ${\mathcal{A}}_{3,1}$

Let us use the notation from subsection 1.2.1. Hence, consider the vector
space in ${\bf T}_{2}(\mathcal{A}_{3,1})$ generated by the following two
cocycles:

$\theta_{1}=\alpha_{1}\nabla_{1}+\alpha_{2}\nabla_{2}+\alpha_{3}\nabla_{3}+\alpha_{4}\nabla_{4};$
---
$\theta_{2}=\beta_{1}\nabla_{1}+\beta_{2}\nabla_{2}+\beta_{3}\nabla_{3}+\beta_{4}\nabla_{4}$.

By the results obtained in the subsection 1.2.1, we can consider

$\theta_{1}\in\\{\nabla_{1},\nabla_{1}+\nabla_{2},\nabla_{1}+\nabla_{4}\\}.$

Therefore, we have:

  1. (1)

$\theta_{1}=\nabla_{1}$. Then, we can suppose $\beta_{1}=0$.

    1. (a)

If $\beta_{4}=0$, we have the following cases:

      1. (i)

If $\beta_{2}=0$, we have the representative
$\langle\nabla_{1},\nabla_{3}\rangle$.

      2. (ii)

If $\beta_{2}\neq 0$, we have the family of representatives
$\langle\nabla_{1},\nabla_{2}+\lambda\nabla_{3}\rangle$.

    2. (b)

If $\beta_{4}\neq 0$, we can suppose $\beta_{4}=1$.

      1. (i)

If $\beta_{2}=\beta_{3}$, consider $x=y=1$ and $t=-\beta_{2}$, then, we obtain
the representative $\langle\nabla_{1},\nabla_{4}\rangle$.

      2. (ii)

If $\beta_{2}\neq\beta_{3}$, consider $x=\frac{1}{\beta_{2}-\beta_{3}}$, $y=1$
and $t=\frac{\beta_{3}}{\beta_{3}-\beta_{2}}$, then, we obtain the
representative $\langle\nabla_{1},\nabla_{2}+\nabla_{4}\rangle$.

  2. (2)

$\theta_{1}=\nabla_{1}+\nabla_{2}.$ Note that we can consider $\beta_{1}=0$.

    1. (a)

Suppose $\beta_{4}=0$.

      1. (i)

If $\beta_{2}=0$ and $\beta_{3}\neq 0$, by choosing $x=y=-r=1$, we obtain the
representative $\langle\nabla_{1},\nabla_{3}\rangle$.

      2. (ii)

If $\beta_{2}\neq 0$ and $\beta_{3}=0$, we have the representative
$\langle\nabla_{1},\nabla_{2}\rangle$.

      3. (iii)

Suppose $\beta_{2}\beta_{3}\neq 0$.

        1. (A)

If $\beta_{2}=-\beta_{3}$, we obtain the representative
$\langle\nabla_{1}+\nabla_{2},\nabla_{2}-\nabla_{3}\rangle$.

        2. (B)

If $\beta_{2}\neq-\beta_{3}$, by choosing $x=y=1$ and
$r=-\frac{\beta_{3}}{\beta_{2}+\beta_{3}}$, we obtain the family of
representatives $\langle\nabla_{1},\nabla_{2}+\lambda\nabla_{3}\rangle$, with
$\lambda\neq 0$.

    2. (b)

If $\beta_{4}\neq 0$, we can suppose $\beta_{4}=1$.

      1. (i)

If $\beta_{3}=\beta_{2}$, by choosing $x=y=1$ and $t=-\beta_{2}$, we obtain
the representative $\langle\nabla_{1}+\nabla_{2},\nabla_{4}\rangle$.

      2. (ii)

If $\beta_{3}\neq\beta_{2}$, by choosing $x=\beta_{3}-\beta_{2}$,
$y=(\beta_{3}-\beta_{2})^{2}$ and $t=-\beta_{2}(\beta_{3}-\beta_{2})$, we
obtain the representative
$\langle\nabla_{1}+\nabla_{2},\nabla_{3}+\nabla_{4}\rangle$.

  3. (3)

$\theta_{1}=\nabla_{1}+\nabla_{4}$. Note that we can consider $\beta_{4}=0$.

    1. (a)

Suppose $\beta_{1}=0$.

      1. (i)

If $\beta_{2}=0$, we have the representative
$\langle\nabla_{1}+\nabla_{4},\nabla_{3}\rangle$.

      2. (ii)

If $\beta_{2}\neq 0$, we have the family of representatives
$\langle\nabla_{1}+\nabla_{4},\nabla_{2}+\lambda\nabla_{3}\rangle$.

    2. (b)

If $\beta_{1}\neq 0$, we can suppose $\beta_{1}=1$.

      1. (i)

Suppose $\beta_{2}=\beta_{3}$.

        1. (A)

If $\beta_{2}=\beta_{3}=0$, we have the representative
$\langle\nabla_{1},\nabla_{4}\rangle$.

        2. (B)

Otherwise, by choosing $x=1$, $y=\frac{1}{2\beta_{2}}$ and $r=\frac{1}{2}$, we
obtain the case 2(b)i.

      2. (ii)

If $\beta_{2}\neq\beta_{3}$, we have:

        1. (A)

If $\beta_{3}=-\beta_{2}\neq 0$, by choosing $x=-1$, $r=-2\beta_{2}^{2}$,
$y=2\beta_{2}$ and $t=\beta_{2}$, we obtain the representative
$\langle\nabla_{1},\nabla_{2}+\nabla_{4}\rangle$.

        2. (B)

If $\beta_{2}\neq-\beta_{3}$, by choosing $x=1$,
$y=\frac{1}{\beta_{2}+\beta_{3}}$ and
$r=\frac{\beta_{3}}{\beta_{2}+\beta_{3}}$, we obtain, again, the case 2(b)ii.

Summarizing, we have the following distinct orbits:

$\langle\nabla_{1},\nabla_{3}\rangle$,
$\langle\nabla_{1},\nabla_{2}+\lambda\nabla_{3}\rangle$,
$\langle\nabla_{1},\nabla_{2}+\nabla_{4}\rangle$,
$\langle\nabla_{1},\nabla_{4}\rangle$ ,
$\langle\nabla_{1}+\nabla_{2},\nabla_{2}-\nabla_{3}\rangle$,
$\langle\nabla_{1}+\nabla_{2},\nabla_{4}\rangle$,
$\langle\nabla_{1}+\nabla_{2},\nabla_{3}+\nabla_{4}\rangle$,
$\langle\nabla_{1}+\nabla_{4},\nabla_{3}\rangle$ and
$\langle\nabla_{1}+\nabla_{4},\nabla_{2}+\lambda\nabla_{3}\rangle$.

A standard computation shows that these orbits are pairwise distinct. Thus,
these orbits give us the following algebras:

${\mathbb{A}}_{5,4}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ |  |  |
---|---|---|---|---|---|---|---|---
${\mathbb{A}}_{5,5}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=\lambda e_{5}$ |  |
${\mathbb{A}}_{5,6}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{5}$ |  |
${\mathbb{A}}_{5,7}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{5}$ |  |  |
${\mathbb{A}}_{5,8}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}+e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=-e_{5}$ |  |
${\mathbb{A}}_{5,9}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{5}$ |  |
${\mathbb{A}}_{5,10}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ | $e_{3}e_{3}=e_{5}$ |
${\mathbb{A}}_{5,11}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ | $e_{3}e_{3}=e_{4}$ |  |
${\mathbb{A}}_{5,12}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=\lambda e_{5}$ | $e_{3}e_{3}=e_{4}$ |

####  1.3.2. $1$-dimensional central extensions of $4$-dimensional $2$-step
nilpotent algebras

Thanks to [26], we have the classification of all $4$-dimensional nilpotent
algebras. From this list we are choosing only $4$-dimensional $2$-step
nilpotent algebras:

$\mathcal{A}_{4,1}$ | $:$ | $e_{1}e_{1}=e_{2}$ |  |  |  |  |
---|---|---|---|---|---|---|---
$\mathcal{A}_{4,2}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{3}e_{3}=e_{4}$ |  |  |  |
$\mathcal{A}_{4,3}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ |  |  |  |
$\mathcal{A}_{4,4}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{2}=\lambda e_{3}$ |  |  |
$\mathcal{A}_{4,5}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{3}$ |  |  |
$\mathcal{A}_{4,6}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$ |  |  |  |
$\mathcal{A}_{4,7}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=-e_{3}$ |  |  |
$\mathcal{A}_{4,8}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{3}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-\lambda e_{3}$ | $e_{2}e_{2}=-e_{4}$ |  |
$\mathcal{A}_{4,9}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=\lambda e_{4}$ | $e_{2}e_{1}=-\lambda e_{4}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{3}=e_{4}$ |
$\mathcal{A}_{4,10}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{2}e_{2}=e_{4}$ | $e_{3}e_{1}=e_{4}$ |
$\mathcal{A}_{4,11}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |
$\mathcal{A}_{4,12}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ |  |  |  |
$\mathcal{A}_{4,13}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ | $e_{2}e_{2}=2e_{3}+e_{4}$ |  |
$\mathcal{A}_{4,14}^{\lambda}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=\lambda e_{4}$ | $e_{2}e_{2}=e_{3}$ |  |  |
$\mathcal{A}_{4,15}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |  |  |

In the following table we give the description of the second cohomology space
of $4$-dimensional $2$-step nilpotent algebras.

${\rm H}^{2}({\mathcal{A}}_{4,1})$ | $=$ | $\Big{\langle}[\Delta_{12}]-[\Delta_{21}],[\Delta_{13}],[\Delta_{14}],[\Delta_{31}],[\Delta_{33}],[\Delta_{34}],[\Delta_{41}],[\Delta_{43}],[\Delta_{44}]\Big{\rangle}$
---|---|---
${\rm H}^{2}({\mathcal{A}}_{4,2})$ | $=$ | $\Big{\langle}[\Delta_{12}]-[\Delta_{21}],[\Delta_{13}],[\Delta_{31}],[\Delta_{34}]-[\Delta_{43}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,3})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{14}],[\Delta_{22}],[\Delta_{24}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}^{\lambda}_{4,4})$ | $=$ | $\Big{\langle}[\Delta_{12}],[\Delta_{14}],[\Delta_{21}],[\Delta_{22}],[\Delta_{24}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,5})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{14}],[\Delta_{22}],[\Delta_{24}],[\Delta_{41}],[\Delta_{42}],[\Delta_{44}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,6})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,7})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{14}]-[\Delta_{24}]-[\Delta_{31}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}^{\lambda}_{4,8})$ | $=$ | $\Big{\langle}[\Delta_{21}],[\Delta_{12}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}^{\lambda}_{4,9})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,10})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{32}],[\Delta_{33}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,11})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,12})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{14}]-[\Delta_{31}],[\Delta_{22}],[\Delta_{23}]-[\Delta_{42}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,13})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}^{\lambda}_{4,14})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{21}],[\Delta_{13}]-\lambda^{2}[\Delta_{31}]+\lambda[\Delta_{24}]-[\Delta_{42}],[\Delta_{23}]-[\Delta_{32}]\Big{\rangle}$
${\rm H}^{2}({\mathcal{A}}_{4,15})$ | $=$ | $\Big{\langle}[\Delta_{11}],[\Delta_{12}],[\Delta_{13}],[\Delta_{21}],[\Delta_{22}],[\Delta_{23}],[\Delta_{31}],[\Delta_{32}]\Big{\rangle}$

Note that we can just consider central extensions of ${\mathcal{A}}_{4,1}$,
${\mathcal{A}}_{4,2}$, ${\mathcal{A}}_{4,7}$, ${\mathcal{A}}_{4,12}$ and
${\mathcal{A}}_{4,14}^{\lambda}$, since ${\bf
T}_{s}({\mathcal{A}}_{4,i})=\emptyset$, in the other cases.

####  1.3.3. Central extensions of ${\mathcal{A}}_{4,1}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{12}]-[\Delta_{21}],$ | $\nabla_{2}=[\Delta_{13}],$ | $\nabla_{3}=[\Delta_{14}],$ | $\nabla_{4}=[\Delta_{31}],$ | $\nabla_{5}=[\Delta_{33}],$
---|---|---|---|---
$\nabla_{6}=[\Delta_{34}],$ | $\nabla_{7}=[\Delta_{41}],$ | $\nabla_{8}=[\Delta_{43}],$ | $\nabla_{9}=[\Delta_{44}].$ |

Take $\theta=\sum\limits_{i=1}^{9}\alpha_{i}\nabla_{i}\in{\rm
H^{2}}({\mathcal{A}}_{4,1})$. The automorphism group of ${\mathcal{A}}_{4,1}$
consists of invertible matrices of the form

$\phi=\begin{pmatrix}x&0&0&0\\\ r&x^{2}&s&t\\\ u&0&y&v\\\
p&0&q&z\end{pmatrix}.$

Since

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

we have that the action of ${\rm Aut}({\mathcal{A}}_{4,1})$ on the subspace
$\Big{\langle}\sum\limits_{i=1}^{9}\alpha_{i}\nabla_{i}\Big{\rangle}$ is given
by $\Big{\langle}\sum\limits_{i=1}^{9}\alpha_{i}^{*}\nabla_{i}\Big{\rangle},$
where

$\alpha_{1}^{*}$ | $=$ | $\alpha_{1}x^{3},$
---|---|---
$\alpha_{2}^{*}$ | $=$ | $\alpha_{1}xs+\alpha_{2}xy+\alpha_{3}xq+\alpha_{5}uy+\alpha_{6}uq+\alpha_{8}yp+\alpha_{9}pq,$
$\alpha_{3}^{*}$ | $=$ | $\alpha_{1}xt+\alpha_{2}xv+\alpha_{3}xz+\alpha_{5}uv+\alpha_{6}uz+\alpha_{8}vp+\alpha_{9}pz,$
$\alpha_{4}^{*}$ | $=$ | $-\alpha_{1}xs+\alpha_{4}xy+\alpha_{5}uy+\alpha_{6}yp+\alpha_{7}xq+\alpha_{8}uq+\alpha_{9}pq,$
$\alpha_{5}^{*}$ | $=$ | $\alpha_{5}y^{2}+(\alpha_{6}+\alpha_{8})yq+\alpha_{9}q^{2},$
$\alpha_{6}^{*}$ | $=$ | $\alpha_{5}yv+\alpha_{6}yz+\alpha_{8}vq+\alpha_{9}qz,$
$\alpha_{7}^{*}$ | $=$ | $-\alpha_{1}xt+\alpha_{4}xv+\alpha_{5}uv+\alpha_{6}pv+\alpha_{7}xz+\alpha_{8}uz+\alpha_{9}pz,$
$\alpha_{8}^{*}$ | $=$ | $\alpha_{5}yv+\alpha_{6}qv+\alpha_{8}yz+\alpha_{9}qz,$
$\alpha_{9}^{*}$ | $=$ | $\alpha_{5}v^{2}+(\alpha_{6}+\alpha_{8})vz+\alpha_{9}z^{2}.$

Since $\operatorname{Ann}(\mathcal{A}_{4,1})=\langle
e_{2},e_{3},e_{4}\rangle$, we can not have $\alpha_{1}=0$. So, we can consider
$\alpha_{1}=1$. Similarly, we can not have
$\alpha_{2}=\alpha_{4}=\alpha_{5}=\alpha_{6}=\alpha_{8}=0$ and, also,
$\alpha_{3}=\alpha_{6}=\alpha_{7}=\alpha_{8}=\alpha_{9}=0$.

######  Remark 4.

By choosing $\phi$ given by $x=y=z=1$, $s=\alpha_{4}$ and $t=\alpha_{7}$, we
obtain

$\phi\theta=\nabla_{1}+(\alpha_{2}+\alpha_{4})\nabla_{2}+(\alpha_{3}+\alpha_{7})\nabla_{3}+\alpha_{5}\nabla_{5}+\alpha_{6}\nabla_{6}+\alpha_{8}\nabla_{8}+\alpha_{9}\nabla_{9}$.

So, we can consider $\alpha_{4}=\alpha_{7}=0$. Moreover, considering $\phi$
given by $x=v=q=1,$ we see that $\phi\theta$ permutes the coefficients
$\alpha_{2}$ with $\alpha_{3}$, $\alpha_{5}$ with $\alpha_{9}$ and also
$\alpha_{6}$ with $\alpha_{8}$. In other words, we have
$(\alpha_{2}^{*},\alpha_{3}^{*},\alpha_{5}^{*},\alpha_{6}^{*},\alpha_{8}^{*},\alpha_{9}^{*})=(\alpha_{3},\alpha_{2},\alpha_{9},\alpha_{8},\alpha_{6},\alpha_{5})$.

Therefore, we can consider only the following cases:

  1. (1)

$\theta=\nabla_{1}+\alpha_{2}\nabla_{2}+\alpha_{3}\nabla_{3}$. Note that
$\alpha_{2}\alpha_{3}\neq 0.$ By choosing $x=y=z=1$,
$v=\frac{1-\alpha_{3}}{\alpha_{2}}$ and $q=-\frac{\alpha_{2}}{\alpha_{3}}$, we
obtain $\langle\nabla_{1}+\nabla_{3}\rangle$, that we do not consider since
the obtained algebra would have $2$-dimensional annihilator.

  2. (2)

$\theta=\nabla_{1}+\alpha_{2}\nabla_{2}+\alpha_{3}\nabla_{3}+\alpha_{6}\nabla_{6}+\alpha_{8}\nabla_{8}$,
with $(\alpha_{6},\alpha_{8})\neq(0,0)$.

    1. (a)

If $\alpha_{8}=-\alpha_{6}$ and $\alpha_{6}\neq 0$, we have:

      1. (i)

If $\alpha_{2}\neq 0$, then, by choosing $x=1$, $y=\frac{1}{\alpha_{2}}$,
$v=-\frac{\alpha_{3}}{\alpha_{6}}$ and $z=\frac{\alpha_{2}}{\alpha_{6}}$, we
obtain the representative
$\langle\nabla_{1}+\nabla_{2}+\nabla_{6}-\nabla_{8}\rangle$.

      2. (ii)

If $\alpha_{2}=\alpha_{3}=0$, by choosing $x=z=1$ and
$y=\frac{1}{\alpha_{6}}$, we obtain the representative
$\langle\nabla_{1}+\nabla_{6}-\nabla_{8}\rangle$.

    2. (b)

If $\alpha_{8}\neq-\alpha_{6}$, note that we can suppose $\alpha_{8}\neq 0$.
By choosing $x=v=1$, $s=u=-\frac{\alpha_{3}}{\alpha_{6}+\alpha_{8}}$,
$t=-\frac{\alpha_{2}\alpha_{6}}{\alpha_{6}+\alpha_{8}}$,
$p=-\frac{\alpha_{2}}{\alpha_{6}+\alpha_{8}}$ and $q=\frac{1}{\alpha_{8}}$, we
have the family of representatives
$\langle\nabla_{1}+\nabla_{6}+\lambda\nabla_{8}\rangle$.

  3. (3)

$\theta_{1}=\nabla_{1}+\alpha_{2}\nabla_{2}+\alpha_{3}\nabla_{3}+\alpha_{5}\nabla_{5}+\alpha_{6}\nabla_{6}+\alpha_{8}\nabla_{8}$,
with $\alpha_{5}\neq 0$.

    1. (a)

Suppose that $\alpha_{6}=\alpha_{8}=0$. Note that $\alpha_{3}\neq 0$. So, by
choosing $x=1$, $y=\frac{1}{\sqrt{\alpha_{5}}}$ ,
$q=-\frac{\alpha_{2}}{\alpha_{3}\sqrt{\alpha_{5}}}$ and
$z=\frac{1}{\alpha_{3}}$, we obtain the representative
$\langle\nabla_{1}+\nabla_{3}+\nabla_{5}\rangle$.

    2. (b)

Suppose $\alpha_{6}\neq 0$ and $\alpha_{8}=0$. By choosing $x=y=1$,
$z=\frac{1}{\alpha_{6}}$, $q=-\frac{\alpha_{5}}{\alpha_{6}}$,
$u=-\frac{\alpha_{3}}{\alpha_{6}}$,
$p=\frac{2\alpha_{3}\alpha_{5}-\alpha_{2}\alpha_{6}}{\alpha_{6}^{2}}$ and
$s=\frac{\alpha_{3}\alpha_{5}-\alpha_{2}\alpha_{6}}{\alpha_{6}}$, we obtain
the representative $\langle\nabla_{1}+\nabla_{6}\rangle$.

    3. (c)

If $\alpha_{6}=0$ and $\alpha_{8}\neq 0$, by choosing $x=q=1$, $v=-\alpha_{8}$
and $z=\alpha_{5}$, we obtain again the case 2b.

    4. (d)

Suppose $\alpha_{6}\alpha_{8}\neq 0$.

      1. (i)

If $\alpha_{8}\neq-\alpha_{6}$, by choosing $x=q=z=1$ and
$v=-\frac{\alpha_{6}+\alpha_{8}}{\alpha_{5}}$, we obtain again the case 2b.

      2. (ii)

If $\alpha_{8}=-\alpha_{6}$, we have two cases:

        1. (A)

If $\alpha_{3}=0$, by choosing $x=1$, $y=\frac{1}{\sqrt{\alpha_{5}}}$,
$z=\frac{\sqrt{\alpha_{5}}}{\alpha_{6}}$, $u=-\frac{\alpha_{2}}{2\alpha_{5}}$
and $s=-t=-\frac{\alpha_{2}}{2\sqrt{\alpha_{5}}}$, we obtain the
representative $\langle\nabla_{1}+\nabla_{5}+\nabla_{6}-\nabla_{8}\rangle$.

        2. (B)

Otherwise, if $\alpha_{3}\neq 0$, by choosing
$x=\frac{\alpha_{3}^{2}\alpha_{5}}{\alpha_{6}^{2}}$,
$y=\frac{\alpha_{3}^{3}\alpha_{5}}{\alpha_{6}^{3}}$,
$z=\frac{\alpha_{3}^{3}\alpha_{5}^{2}}{\alpha_{6}^{4}}$,
$u=-\frac{\alpha_{2}\alpha_{3}^{2}}{2\alpha_{6}^{2}}$ and
$s=-t=-\frac{\alpha_{2}\alpha_{3}^{3}\alpha_{5}}{2\alpha_{6}^{3}}$, we obtain
the representative
$\langle\nabla_{1}+\nabla_{3}+\nabla_{5}+\nabla_{6}-\nabla_{8}\rangle$.

  4. (4)

$\theta=\nabla_{1}+\alpha_{2}\nabla_{2}+\alpha_{3}\nabla_{3}+\alpha_{5}\nabla_{5}+\alpha_{6}\nabla_{6}+\alpha_{8}\nabla_{8}+\alpha_{9}\nabla_{9}$,
with $\alpha_{9}\neq 0$. By choosing $x=2\alpha_{9}$, $y=v=1$,
$q=-\frac{1+\alpha_{6}+\alpha_{8}-\sqrt{(\alpha_{6}+\alpha_{8})^{2}-4\alpha_{5}\alpha_{9}}}{2\alpha_{9}}$
and
$z=-\frac{\alpha_{6}+\alpha_{8}-\sqrt{(\alpha_{6}+\alpha_{8})^{2}-4\alpha_{5}\alpha_{9}}}{2\alpha_{9}}$,
we obtain the case 2 or the case 3, depending on the value of
$\alpha_{5}^{\ast}$.

A standard computation shows that these orbits are pairwise distinct.
Summarizing, we get the distinct orbits

$\langle\nabla_{1}+\nabla_{3}+\nabla_{5}\rangle$,
$\langle\nabla_{1}+\nabla_{6}+\lambda\nabla_{8}\rangle$,
$\langle\nabla_{1}+\nabla_{2}+\nabla_{6}-\nabla_{8}\rangle$,

$\langle\nabla_{1}+\nabla_{3}+\nabla_{5}+\nabla_{6}-\nabla_{8}\rangle$ and
$\langle\nabla_{1}+\nabla_{5}+\nabla_{6}-\nabla_{8}\rangle$,

which give us the following algebras:

${\mathbb{A}}_{5,13}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{3}=e_{5}$ |  |
---|---|---|---|---|---|---|---|---
${\mathbb{A}}_{5,14}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=\lambda e_{5}$ |  |
${\mathbb{A}}_{5,15}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |
${\mathbb{A}}_{5,16}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{3}=e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$
${\mathbb{A}}_{5,17}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{3}=e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |

####  1.3.4. Central extensions of ${\mathcal{A}}_{4,2}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{12}]-[\Delta_{21}],$ | $\nabla_{2}=[\Delta_{13}],$ | $\nabla_{3}=[\Delta_{31}],$ | $\nabla_{4}=[\Delta_{34}]-[\Delta_{43}].$
---|---|---|---

Take $\theta=\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\in{\rm
H^{2}}({\mathcal{A}}_{4,2})$. Since
$\operatorname{Ann}(\mathcal{A}_{4,2})=\langle e_{2},e_{4}\rangle$, we can not
have $\alpha_{1}\alpha_{4}=0.$ The automorphism group of ${\mathcal{A}}_{4,2}$
consists of invertible matrices of the forms

$\phi_{1}=\begin{pmatrix}x&0&0&0\\\ z&x^{2}&r&0\\\ 0&0&y&0\\\ t&0&w&y^{2}\\\
\end{pmatrix}\,\,\,\text{and}\,\,\,\phi_{2}=\begin{pmatrix}0&0&a&0\\\
e&0&f&a^{2}\\\ b&0&0&0\\\ c&b^{2}&d&0\end{pmatrix}.$

Since

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

we have that the action of ${\rm Aut}({\mathcal{A}}_{4,2})$ on the subspace
$\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\Big{\rangle}$ is given
by $\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}^{*}\nabla_{i}\Big{\rangle},$
where:

if $i=1:$ | $\alpha^{*}_{1}=\alpha_{1}x^{3},$ | $\alpha^{*}_{2}=\alpha_{1}xr+\alpha_{2}xy-\alpha_{4}ty,$ | $\alpha^{*}_{3}=-\alpha_{1}xr+\alpha_{3}xy+\alpha_{4}ty,$ | $\alpha_{4}^{*}=\alpha_{4}y^{3};$
---|---|---|---|---
if $i=2:$ | $\alpha_{1}^{*}=\alpha_{4}b^{3},$ | $\alpha_{2}^{*}=-\alpha_{1}ae+\alpha_{3}ab+\alpha_{4}bd,$ | $\alpha_{3}^{*}=\alpha_{1}ae+\alpha_{2}ab-\alpha_{4}bd,$ |  $\alpha_{4}^{*}=\alpha_{1}a^{3}$.

So, we need to consider only the following cases:

  1. (1)

If $\alpha_{2}=-\alpha_{3}$, by choosing $x=\frac{1}{\sqrt[3]{\alpha_{1}}}$,
$r=\frac{\alpha_{3}}{\alpha_{1}\sqrt[3]{\alpha_{4}}}$ and
$y=\frac{1}{\sqrt[3]{\alpha_{4}}}$, we obtain the representative
$\langle\nabla_{1}+\nabla_{4}\rangle$.

  2. (2)

If $\alpha_{2}\neq-\alpha_{3}$, by choosing
$x=\frac{\alpha_{2}+\alpha_{3}}{\sqrt[3]{\alpha_{1}^{2}\alpha_{4}}}$,
$r=\frac{\alpha_{3}(\alpha_{2}+\alpha_{3})}{\alpha_{1}\sqrt[3]{\alpha_{1}\alpha_{4}^{2}}}$
and $y=\frac{\alpha_{2}+\alpha_{3}}{\sqrt[3]{\alpha_{1}\alpha_{4}^{2}}}$, we
obtain the representative $\langle\nabla_{1}+\nabla_{2}+\nabla_{4}\rangle$.

Hence, we get the distinct orbits $\langle\nabla_{1}+\nabla_{4}\rangle$ and
$\langle\nabla_{1}+\nabla_{2}+\nabla_{4}\rangle$, which give us the following
algebras

${\mathbb{A}}_{5,18}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{3}=e_{4}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |
---|---|---|---|---|---|---|---|---
${\mathbb{A}}_{5,19}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{3}=e_{4}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$

####  1.3.5. Central extensions of ${\mathcal{A}}_{4,7}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{11}],$ | $\nabla_{2}=[\Delta_{12}],$ | $\nabla_{3}=[\Delta_{14}]-[\Delta_{24}]-[\Delta_{31}].$
---|---|---

Take $\theta=\sum\limits_{i=1}^{3}\alpha_{i}\nabla_{i}\in{\rm
H^{2}}({\mathcal{A}}_{4,7})$. The automorphism group of ${\mathcal{A}}_{4,7}$
consists of invertible matrices of the form

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

Since

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

we have that the action of ${\rm Aut}({\mathcal{A}}_{4,7})$ on the subspace
$\Big{\langle}\sum\limits_{i=1}^{3}\alpha_{i}\nabla_{i}\Big{\rangle}$ is given
by $\Big{\langle}\sum\limits_{i=1}^{3}\alpha_{i}^{*}\nabla_{i}\Big{\rangle},$
where

$\alpha^{*}_{1}=\alpha_{1}x^{2}-\alpha_{3}yx+\alpha_{3}wx$, |  $\alpha^{*}_{2}=\alpha_{2}x^{2}$, |  $\alpha^{*}_{3}=\alpha_{3}x^{3}$.
---|---|---

Note that, since $\operatorname{Ann}(\mathcal{A}_{4,7})=\langle
e_{3},e_{4}\rangle$, we can not have $\alpha_{3}=0$. Suppose $\alpha_{3}=1$.
Then, we there are two cases to consider:

  1. (1)

If $\alpha_{2}=0$, by choosing $x=1$ and $y=\alpha_{1}$, we get the
representative $\langle\nabla_{3}\rangle$.

  2. (2)

If $\alpha_{2}\neq 0$, by choosing $x=\alpha_{2}$ and
$y=\alpha_{1}\alpha_{2}$, we get the representative
$\langle\nabla_{2}+\nabla_{3}\rangle$.

Therefore, we have the distinct orbits $\langle\nabla_{3}\rangle$ and
$\langle\nabla_{2}+\nabla_{3}\rangle$, which give us the following algebras:

${\mathbb{A}}_{5,20}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=-e_{3}$ | $e_{2}e_{4}=-e_{5}$ | $e_{3}e_{1}=-e_{5}$
---|---|---|---|---|---|---|---
${\mathbb{A}}_{5,21}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=-e_{3}$ | $e_{2}e_{4}=-e_{5}$ | $e_{3}e_{1}=-e_{5}$

####  1.3.6. Central extensions of ${\mathcal{A}}_{4,12}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{11}],$ | $\nabla_{2}=[\Delta_{14}]-[\Delta_{31}],$ | $\nabla_{3}=[\Delta_{22}],$ | $\nabla_{4}=[\Delta_{23}]-[\Delta_{42}].$
---|---|---|---

Take $\theta=\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\in{\rm
H^{2}}({\mathcal{A}}_{4,12})$. The automorphism group of
${\mathcal{A}}_{4,12}$ consists of invertible matrices of the forms

$\phi_{1}=\begin{pmatrix}x&0&0&0\\\ 0&y&0&0\\\ r&s&xy&0\\\
t&z&0&xy\end{pmatrix}\,\,\,\text{and}\,\,\phi_{2}=\begin{pmatrix}0&a&0&0\\\
b&0&0&0\\\ c&d&0&ab\\\ e&f&ab&0\end{pmatrix}.$

Since

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

we have that the action of ${\rm Aut}({\mathcal{A}}_{4,12})$ on the subspace
$\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\Big{\rangle}$ is given
by $\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}^{*}\nabla_{i}\Big{\rangle},$
where

if $i=1:$ | $\alpha_{1}^{*}=(\alpha_{1}x+\alpha_{2}(t-r))x,$ | $\alpha_{2}^{*}=\alpha_{2}x^{2}y,$ | $\alpha_{3}^{*}=(\alpha_{3}y+\alpha_{4}(s-z))y,$ | $\alpha_{4}^{*}=\alpha_{4}xy^{2};$
---|---|---|---|---
if $i=2:$ | $\alpha_{1}^{*}=(\alpha_{3}b+\alpha_{4}(c-e))b,$ | $\alpha_{2}^{*}=\alpha_{4}ab^{2},$ | $\alpha_{3}^{*}=(\alpha_{1}a+\alpha_{2}(f-d))a,$ | $\alpha_{4}^{*}=\alpha_{2}a^{2}b.$

Note that, since $\operatorname{Ann}(\mathcal{A}_{4,12})=\langle
e_{3},e_{4}\rangle$, we can not have $\alpha_{2}=\alpha_{4}=0$. Besides,
considering $\phi_{2}$ given by $a=b=1$, we see that it change the position of
$\alpha_{2}$ and $\alpha_{4}$. Thus, we need to consider only the case when
$\alpha_{2}\neq 0$. Therefore, assume that $\alpha_{2}=1$.

  1. (1)

Suppose $\alpha_{4}=0$.

    1. (a)

If $\alpha_{3}=0$, by choosing $a=b=1$ and $d=\alpha_{1}$, we have the
representative $\langle\nabla_{4}\rangle$.

    2. (b)

If $\alpha_{3}\neq 0$, by choosing $x=y=\alpha_{3}$ and
$r=\alpha_{1}\alpha_{3}$, we obtain the representative
$\langle\nabla_{2}+\nabla_{3}\rangle$.

  2. (2)

If $\alpha_{4}\neq 0$, by choosing $x=\sqrt[3]{\alpha_{4}}$,
$y=\frac{1}{\sqrt[3]{\alpha_{4}^{2}}}$, $r=\alpha_{1}\sqrt[3]{\alpha_{4}}$ and
$s=-\frac{\alpha_{3}}{\alpha_{4}\sqrt[3]{\alpha_{4}^{2}}}$, we obtain the
representative $\langle\nabla_{2}+\nabla_{4}\rangle$.

Thus, we have three representatives of distinct orbits

$\langle\nabla_{2}+\nabla_{3}\rangle$, $\langle\nabla_{2}+\nabla_{4}\rangle$
and $\langle\nabla_{4}\rangle$,

which give us the following algebras:

${\mathbb{A}}_{5,22}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{3}e_{1}=-e_{5}$ |
---|---|---|---|---|---|---|---
${\mathbb{A}}_{5,23}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{3}=e_{5}$ | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$
${\mathbb{A}}_{5,24}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{3}=e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |

####  1.3.7. Central extensions of ${\mathcal{A}}_{4,14}^{\lambda}$

Let us use the following notations:

$\nabla_{1}=[\Delta_{11}],$ | $\nabla_{2}=[\Delta_{21}],$ | $\nabla_{3}=[\Delta_{13}]-\lambda^{2}[\Delta_{31}]+\lambda[\Delta_{24}]-[\Delta_{42}],$ |  $\nabla_{4}=[\Delta_{23}]-[\Delta_{32}]$.
---|---|---|---

Take $\theta=\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\in{\rm
H^{2}}({\mathcal{A}}_{4,14}^{\lambda})$. The automorphism group of
${\mathcal{A}}_{4,14}^{\lambda}$ consists of invertible matrices of the form

$\phi=\begin{pmatrix}x&a&0&0\\\ 0&y&0&0\\\ b&c&y^{2}&0\\\
d&e&(1+\lambda)ay&xy\end{pmatrix}$

Since

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

we have that the action of ${\rm Aut}({\mathcal{A}}_{4,14}^{\lambda})$ on the
subspace $\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}\nabla_{i}\Big{\rangle}$
is given by
$\Big{\langle}\sum\limits_{i=1}^{4}\alpha_{i}^{*}\nabla_{i}\Big{\rangle},$
where

$\alpha_{1}^{*}$ | $=$ | $\alpha_{1}x^{2}+\alpha_{3}bx(1-\lambda^{2}),$
---|---|---
$\alpha_{2}^{*}$ | $=$ | $\alpha_{2}xy+\alpha_{3}ab(1+\lambda^{3})+\alpha_{4}by(1+\lambda)+\alpha_{1}ax(1-\lambda)+2\alpha_{3}dy\lambda-\alpha_{3}cx\lambda(1+\lambda),$
$\alpha_{3}^{*}$ | $=$ | $\alpha_{3}xy^{2},$
$\alpha_{4}^{*}$ | $=$ | $\alpha_{4}y^{3}+\alpha_{3}ay^{2}(1+\lambda+\lambda^{2}).$

Since $\operatorname{Ann}(\mathcal{A}_{4,14}^{\lambda})=\langle
e_{3},e_{4}\rangle$, we can not have $\alpha_{3}=0$. Then, assume
$\alpha_{3}=1$.

  1. (1)

$\lambda=0$.

    1. (a)

Suppose $\alpha_{2}\neq\alpha_{1}\alpha_{4}$. Then, by choosing
$x=y=\alpha_{2}-\alpha_{1}\alpha_{4}$,
$a=-\alpha_{4}(\alpha_{2}-\alpha_{1}\alpha_{4})$ and
$b=-\alpha_{1}(\alpha_{2}-\alpha_{1}\alpha_{4})$, we obtain the representative
$\langle\nabla_{2}+\nabla_{3}\rangle$.

    2. (b)

Suppose $\alpha_{2}=\alpha_{1}\alpha_{4}$. Then, by choosing $x=y=1$,
$a=-\alpha_{4}$ and $b=-\alpha_{1}$, we obtain the representative
$\langle\nabla_{3}\rangle$.

  2. (2)

$\lambda=1$.

    1. (a)

If $\alpha_{1}=0$, by choosing $x=1$, $y=1$, $a=-\frac{\alpha_{4}}{3}$ and
$d=-\frac{\alpha_{2}}{2}$, we obtain the representative
$\langle\nabla_{3}\rangle$.

    2. (b)

If $\alpha_{1}\neq 0$, by choosing $x=\frac{1}{\sqrt{\alpha_{1}}}$,
$y=\sqrt[4]{\alpha_{1}}$, $a=-\frac{\alpha_{4}\sqrt[4]{\alpha_{1}}}{3}$ and
$c=\frac{\alpha_{2}\sqrt[4]{\alpha_{1}}}{2}$, we obtain the representative
$\langle\nabla_{1}+\nabla_{3}\rangle$.

  3. (3)

$\lambda=-1$.

    1. (a)

If $\alpha_{1}=0$, by choosing $x=y=1$, $a=-\alpha_{4}$ and
$d=\frac{\alpha_{2}}{2}$, we obtain the representative
$\langle\nabla_{3}\rangle$.

    2. (b)

If $\alpha_{1}\neq 0$, by choosing $x=\frac{1}{\sqrt{\alpha_{1}}}$,
$y=\sqrt[4]{\alpha_{1}}$, $a=-\alpha_{4}\sqrt[4]{\alpha_{1}}$ and
$d=\frac{\alpha_{2}-2\alpha_{1}\alpha_{4}}{2\sqrt{\alpha_{1}}}$, we obtain the
representative $\langle\nabla_{1}+\nabla_{3}\rangle$.

  4. (4)

$\lambda\notin\\{0,\pm 1\\}$.

    1. (a)

If $1+\lambda+\lambda^{2}\neq 0$, by choosing $x=y=1$,
$a=-\frac{\alpha_{4}}{1+\lambda+\lambda^{2}}$,
$b=\frac{\alpha_{1}}{\lambda^{2}-1}$,
$c=\frac{\alpha_{2}(\lambda^{3}-1)+\alpha_{4}(\alpha_{1}+2\lambda+\alpha_{1}\lambda^{2})}{\lambda(1+\lambda)(\lambda^{3}-1)}$
and $d=\frac{\alpha_{4}}{\lambda^{3}-1}$, we obtain the representative
$\langle\nabla_{3}\rangle$.

    2. (b)

Suppose $1+\lambda+\lambda^{2}=0$.

      1. (i)

If $\alpha_{4}=0$, by choosing $x=y=1$, $b=\frac{\alpha_{1}}{\lambda^{2}-1}$
and $c=d=\frac{\alpha_{2}}{\lambda^{2}-\lambda}$, we obtain the representative
$\langle\nabla_{3}\rangle$.

      2. (ii)

If $\alpha_{4}\neq 0$, by choosing $x=\alpha_{4}^{2}$, $y=\alpha_{4}$,
$b=\frac{\alpha_{1}\alpha_{4}^{2}}{\lambda^{2}-1}$ and
$d=\frac{\alpha_{4}^{2}(\alpha_{1}\alpha_{4}-\alpha_{2}(\lambda-1))}{2\lambda(1-\lambda)}$,
we obtain the representative $\langle\nabla_{3}+\nabla_{4}\rangle$.

Thus, we get the distinct orbits

$\langle\nabla_{2}+\nabla_{3}\rangle_{\lambda=0}$, $\langle\nabla_{3}\rangle$,
$\langle\nabla_{1}+\nabla_{3}\rangle_{\lambda=\pm 1}$ and
$\langle\nabla_{3}+\nabla_{4}\rangle_{\lambda=\omega,\omega^{2}}$,

where $\omega$ is the cubic root of unit, which give us the following
algebras:

${\mathbb{A}}_{5,25}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=e_{5}$ | $e_{2}e_{2}=e_{3}$ | $e_{4}e_{2}=-e_{5}$ |
---|---|---|---|---|---|---|---
${\mathbb{A}}_{5,26}^{\lambda}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=\lambda e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{4}=\lambda e_{5}$ | $e_{3}e_{1}=-\lambda^{2}e_{5}$
|  | $e_{4}e_{2}=-e_{5}$ |  |  |  |  |
${\mathbb{A}}_{5,27}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{4}=e_{5}$
|  | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |  |  |
${\mathbb{A}}_{5,28}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{4}=-e_{5}$
|  | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |  |  |
${\mathbb{A}}_{5,29}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=\omega e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{3}=e_{5}$ | $e_{2}e_{4}=\omega e_{5}$
|  | $e_{3}e_{1}=-\omega^{2}e_{5}$ | $e_{3}e_{2}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |  |
${\mathbb{A}}_{5,30}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=\omega^{2}e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{3}=e_{5}$ | $e_{2}e_{4}=\omega^{2}e_{5}$
|  | $e_{3}e_{1}=-\omega e_{5}$ | $e_{3}e_{2}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |  |

###  1.4. Classification theorem

Now we are ready to summarize all results related to the algebraic
classification of complex antiassociative algebras.

######  Theorem A.

Let ${\bf A}$ be a complex $5$-dimensional non-2-step nilpotent
antiassociative algebra. Then ${\bf A}$ is isomorphic to one algebra from the
following table:

${\mathbb{A}}_{5,1}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=-e_{3}$ |  |
---|---|---|---|---|---|---
${\mathbb{A}}_{5,2}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ |
${\mathbb{A}}_{5,3}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{4}$ |
${\mathbb{A}}_{5,4}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ |
${\mathbb{A}}_{5,5}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=\lambda e_{5}$
${\mathbb{A}}_{5,6}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{5}$
${\mathbb{A}}_{5,7}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{5}$ |
${\mathbb{A}}_{5,8}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}+e_{5}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=-e_{5}$
${\mathbb{A}}_{5,9}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{3}=e_{5}$
${\mathbb{A}}_{5,10}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{4}$ |  |
|  | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ | $e_{3}e_{3}=e_{5}$ |  |
${\mathbb{A}}_{5,11}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=e_{5}$ | $e_{3}e_{3}=e_{4}$
${\mathbb{A}}_{5,12}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ |  |
|  | $e_{2}e_{1}=-e_{4}$ | $e_{3}e_{1}=\lambda e_{5}$ | $e_{3}e_{3}=e_{4}$ |  |
${\mathbb{A}}_{5,13}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{3}=e_{5}$
${\mathbb{A}}_{5,14}^{\lambda}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=\lambda e_{5}$
${\mathbb{A}}_{5,15}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{3}=e_{5}$ |  |
|  | $e_{2}e_{1}=-e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
${\mathbb{A}}_{5,16}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ |
|  | $e_{3}e_{3}=e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
${\mathbb{A}}_{5,17}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ |  |
|  | $e_{3}e_{3}=e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
${\mathbb{A}}_{5,18}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ |  |
|  | $e_{3}e_{3}=e_{4}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
${\mathbb{A}}_{5,19}$ | $:$ | $e_{1}e_{1}=e_{2}$ | $e_{1}e_{2}=e_{5}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{5}$ |
|  | $e_{3}e_{3}=e_{4}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=-e_{5}$ |  |
${\mathbb{A}}_{5,20}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ |  |
|  | $e_{2}e_{2}=-e_{3}$ | $e_{2}e_{4}=-e_{5}$ | $e_{3}e_{1}=-e_{5}$ |  |
${\mathbb{A}}_{5,21}$ | $:$ | $e_{1}e_{2}=e_{3}+e_{5}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ |  |
|  | $e_{2}e_{2}=-e_{3}$ | $e_{2}e_{4}=-e_{5}$ | $e_{3}e_{1}=-e_{5}$ |  |
${\mathbb{A}}_{5,22}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ | $e_{3}e_{1}=-e_{5}$
${\mathbb{A}}_{5,23}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{1}e_{4}=e_{5}$ | $e_{2}e_{1}=e_{4}$ |  |
|  | $e_{2}e_{3}=e_{5}$ | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |
${\mathbb{A}}_{5,24}$ | $:$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{3}=e_{5}$ | $e_{4}e_{2}=-e_{5}$ |
${\mathbb{A}}_{5,25}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=e_{5}$ | $e_{2}e_{2}=e_{3}$ | $e_{4}e_{2}=-e_{5}$
${\mathbb{A}}_{5,26}^{\lambda}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=\lambda e_{4}$ | $e_{2}e_{2}=e_{3}$ |
|  | $e_{2}e_{4}=\lambda e_{5}$ | $e_{3}e_{1}=-\lambda^{2}e_{5}$ | $e_{4}e_{2}=-e_{5}$ |  |
${\mathbb{A}}_{5,27}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=e_{4}$ |
|  | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{4}=e_{5}$ | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |
${\mathbb{A}}_{5,28}$ | $:$ | $e_{1}e_{1}=e_{5}$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=-e_{4}$ |
|  | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{4}=-e_{5}$ | $e_{3}e_{1}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |
${\mathbb{A}}_{5,29}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=\omega e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{3}=e_{5}$
|  | $e_{2}e_{4}=\omega e_{5}$ | $e_{3}e_{1}=-\omega^{2}e_{5}$ | $e_{3}e_{2}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |
${\mathbb{A}}_{5,30}$ | $:$ | $e_{1}e_{2}=e_{4}$ | $e_{1}e_{3}=e_{5}$ | $e_{2}e_{1}=\omega^{2}e_{4}$ | $e_{2}e_{2}=e_{3}$ | $e_{2}e_{3}=e_{5}$
|  | $e_{2}e_{4}=\omega^{2}e_{5}$ | $e_{3}e_{1}=-\omega e_{5}$ | $e_{3}e_{2}=-e_{5}$ | $e_{4}e_{2}=-e_{5}$ |

In the table above, $\omega$ is the primitive cubic root of unit.

##  2\. The geometric classification of nilpotent 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 $\mathcal{O}(\mu)$
denote the orbit of $\mu\in\mathbb{L}(T)$ under the action of ${\rm
GL}(\mathbb{V})$ and $\overline{\mathcal{O}(\mu)}$ denote the Zariski closure
of $\mathcal{O}(\mu)$.

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

Let $\mathcal{A}$ be represented by $\mu\in\mathbb{L}(T)$. Then $\mathcal{A}$
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 $\mathcal{A}$ is rigid
in $\mathbb{L}(T)$ if and only if $\overline{O(\mu)}$ is an irreducible
component of $\mathbb{L}(T)$.

Given the spaces $U$ and $W$, we write simply $U>W$ instead of
$\dim\,U>\dim\,W$.

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

In the present work we use the methods applied to Lie algebras in [6, 18, 19].
First of all, if $\mathcal{A}\to\mathcal{B}$ and
$\mathcal{A}\not\cong\mathcal{B}$, then
$\mathfrak{Der}(\mathcal{A})<\mathfrak{Der}(\mathcal{B})$, where
$\mathfrak{Der}(\mathcal{A})$ is the Lie algebra of derivations of
$\mathcal{A}$. We compute the dimensions of algebras of derivations and check
the assertion $\mathcal{A}\to\mathcal{B}$ only for such $\mathcal{A}$ and
$\mathcal{B}$ that $\mathfrak{Der}(\mathcal{A})<\mathfrak{Der}(\mathcal{B})$.

To prove degenerations, we construct families of matrices parametrized by $t$.
Namely, let $\mathcal{A}$ and $\mathcal{B}$ 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 $\mathcal{A}\to\mathcal{B}$. In this case $E_{1}^{t},\dots,E_{n}^{t}$ is
called a parametrized basis for $\mathcal{A}\to\mathcal{B}$. 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 variety of $5$-dimensional antiassociative algebras contains
infinitely many non-isomorphic algebras, we have to do some additional work.
Let $\mathcal{A}(*):=\\{\mathcal{A}(\alpha)\\}_{\alpha\in I}$ be a series of
algebras, and let $\mathcal{B}$ be another algebra. Suppose that for
$\alpha\in I$, $\mathcal{A}(\alpha)$ 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 $\mathcal{A}(*)\to\mathcal{B}$ if
$\lambda\in\overline{\\{\mathcal{O}(\mu(\alpha))\\}_{\alpha\in I}}$, and
$\mathcal{A}(*)\not\to\mathcal{B}$ if
$\lambda\not\in\overline{\\{\mathcal{O}(\mu(\alpha))\\}_{\alpha\in I}}$.

Let $\mathcal{A}(*)$, $\mathcal{B}$, $\mu(\alpha)$ ($\alpha\in I$) and
$\lambda$ be as above. To prove $\mathcal{A}(*)\to\mathcal{B}$ 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 $\mathcal{A}(*)\to\mathcal{B}$. In this case
$E_{1}^{t},\dots,E_{n}^{t}$ and $f(t)$ are called a parametrized basis and a
parametrized index for $\mathcal{A}(*)\to\mathcal{B}$, respectively.

One can also use the following Lemma, whose proof is the same as the proof of
Lemma 1.5 from [18].

######  Lemma 5.

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
$\mathcal{A}(*)\to\mathcal{B}$ and for any $\alpha\in I$ the algebra
$\mathcal{A}(\alpha)$ can be represented by a structure
$\mu(\alpha)\in\mathcal{R}$, then there is $\lambda\in\mathcal{R}$
representing $\mathcal{B}$.

###  2.3. The geometric classification of antiassociative algebras

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

######  Theorem B.

The variety of $4$-dimensional antiassociative algebras has dimension $12$ and
it has $3$ irreducible components defined by

$\mathcal{C}_{1}=\overline{\\{\mathcal{O}(\mathcal{A}_{4,8}^{\gamma})\\}},\,\mathcal{C}_{2}=\overline{\\{\mathcal{O}(\mathcal{A}_{4,9}^{\alpha})\\}},\,\mathcal{C}_{3}=\overline{\mathcal{O}({\mathbb{A}}_{4,3})}.$

In particular, there is only one rigid $4$-dimensional antiassociative algebra
${\mathbb{A}}_{4,3}$.

The variety of $5$-dimensional antiassociative algebras has dimension $24$ and
it has $8$ irreducible components defined by

$\mathcal{C}_{1}=\overline{\\{\mathcal{O}({\mathfrak{V}}_{3+2})\\}},$
$\mathcal{C}_{2}=\overline{\\{\mathcal{O}({\mathfrak{V}}_{4+1})\\}},$
$\mathcal{C}_{3}=\overline{\mathcal{O}({\mathbb{A}}_{5,10})},$
$\mathcal{C}_{4}=\overline{\\{\mathcal{O}({\mathbb{A}}_{5,14}^{\lambda})\\}},$

$\mathcal{C}_{5}=\overline{\mathcal{O}({\mathbb{A}}_{5,19})},$
$\mathcal{C}_{6}=\overline{\mathcal{O}({\mathbb{A}}_{5,21})},$
$\mathcal{C}_{7}=\overline{\mathcal{O}({\mathbb{A}}_{5,23})},$
$\mathcal{C}_{8}=\overline{\\{O({\mathbb{A}}_{5,26}^{\lambda})\\}},$

In particular, there are only four rigid algebras in this variety:

${\mathbb{A}}_{5,10},$ ${\mathbb{A}}_{5,19},$ ${\mathbb{A}}_{5,21},$ and
${\mathbb{A}}_{5,23}.$

  * Proof.

Thanks to [26], the variety of $4$-dimensional $2$-step nilpotent
antiassociative algebras has dimension $12$ and it has $2$ irreducible
components. All needed degenerations are given in the table below:

${\mathbb{A}}_{4,2}$ | $\to$ | ${\mathbb{A}}_{4,1}$ | $E_{1}^{t}=e_{1}$ | $E_{2}^{t}=e_{2}$ | $E_{3}^{t}=e_{4}$ | $E_{4}^{t}=te_{3}$ |
---|---|---|---|---|---|---|---
${\mathbb{A}}_{4,3}$ | $\to$ | ${\mathbb{A}}_{4,2}$ | $E_{1}^{t}=e_{1}+\frac{1}{2t}e_{2}$ | $E_{2}^{t}=e_{3}+\frac{1}{4t^{2}}e_{4}$ | $E_{3}^{t}=te_{2}+\frac{1}{2}e_{3}$ | $E_{4}^{t}=e_{4}$ |

Since $\overline{\mathcal{O}({\mathbb{A}}_{4,3})}$ has dimension $12$, the
variety of $4$-dimensional antiassociative algebras has $3$ irreducible
components.

Thanks to [22, Theorem A], the variety of $5$-dimensional $2$-step nilpotent
antiassociative algebras has dimension $24$ and it has $3$ irreducible
components.

${\mathfrak{V}}_{4+1}$ | $:$ | $e_{1}e_{2}=e_{5}$ | $e_{2}e_{1}=\lambda e_{5}$ | $e_{3}e_{4}=e_{5}$ | $e_{4}e_{3}=\mu e_{5}$ |
---|---|---|---|---|---|---
${\mathfrak{V}}_{3+2}$ | $:$ | $e_{1}e_{1}=e_{4}$ | $e_{1}e_{2}=\mu_{1}e_{5}$ | $e_{1}e_{3}=\mu_{2}e_{5}$ | $e_{2}e_{1}=\mu_{3}e_{5}$ | $e_{2}e_{2}=\mu_{4}e_{5}$
|  | $e_{2}e_{3}=\mu_{5}e_{5}$ | $e_{3}e_{1}=\mu_{6}e_{5}$ | $e_{3}e_{2}=\lambda e_{4}+\mu_{7}e_{5}$ | $e_{3}e_{3}=e_{5}$
${\mathfrak{V}}_{2+3}$ | $:$ | $e_{1}e_{1}=e_{3}+\lambda e_{5}$ | $e_{1}e_{2}=e_{3}$ | $e_{2}e_{1}=e_{4}$ | $e_{2}e_{2}=e_{5}$ |

Thanks to the first part of this theorem, all $5$-dimensional split
antiassociative algebras are in orbit closure of families
${\mathfrak{V}}_{4+1}$ and ${\mathfrak{V}}_{3+2},$ and
$\overline{\mathcal{O}({\mathbb{A}}_{5,3})}.$

Let us give some important degenerations.

${\mathbb{A}}_{5,10}$ | $\to$ | ${\mathbb{A}}_{5,5}^{\lambda}$ | $E_{1}^{t}=\frac{t}{\lambda-1}e_{1}+\frac{t}{(\lambda-1)^{2}}e_{3}$
---|---|---|---
$E_{2}^{t}=\frac{t^{2}}{(\lambda-1)^{2}}e_{2}+\frac{t^{2}}{(\lambda-1)^{3}}e_{4}+\frac{\lambda t^{2}}{(\lambda-1)^{4}}e_{5}$ | $E_{3}^{t}=\frac{t^{3}}{(\lambda-1)^{2}}e_{3}$
$E_{4}^{t}=\frac{t^{3}}{(\lambda-1)^{3}}e_{4}$ | $E_{5}^{t}=\frac{t^{4}}{(\lambda-1)^{4}}e_{5}$
${\mathbb{A}}_{5,24}$ | $\to$ | ${\mathbb{A}}_{5,8}$ | $E_{1}^{t}=e_{1}+e_{2}$
$E_{2}^{t}=e_{3}+e_{4}$ | $E_{3}^{t}=te_{2}+(t-1)e_{4}$
$E_{4}^{t}=e_{5}$ | $E_{5}^{t}=-te_{4}-e_{5}$
${\mathbb{A}}_{5,10}$ | $\to$ | ${\mathbb{A}}_{5,12}^{\lambda}$ | $E_{1}^{t}=\frac{\lambda^{2}t^{2}}{1-\lambda^{2}}e_{1}-\frac{\lambda^{2}t^{2}}{(\lambda-1)^{2}(1+\lambda)}e_{3}$
$E_{2}^{t}=\frac{\lambda^{3}t^{4}}{(\lambda-1)^{2}(1+\lambda)}e_{3}+\frac{\lambda^{4}t^{4}}{(\lambda-1)^{3}(1+\lambda)^{2}}e_{4}+\frac{\lambda^{5}t^{4}}{(\lambda-1)^{4}(1+\lambda)^{2}}e_{5}$ | $E_{3}^{t}=\frac{\lambda^{4}t^{3}}{(\lambda^{2}-1)^{2}}e_{2}-\frac{\lambda^{3}t^{3}}{(\lambda-1)^{2}(1+\lambda)}e_{3}$
$E_{4}^{t}=\frac{\lambda^{6}t^{6}}{(\lambda-1)^{4}(1+\lambda)^{2}}e_{5}$ | $E_{5}^{t}=\frac{\lambda^{5}t^{5}}{(\lambda^{2}-1)^{3}}e_{4}+\frac{\lambda^{5}t^{5}}{(\lambda-1)^{4}(1+\lambda)^{2}}e_{5}$
${\mathbb{A}}_{5,14}^{t-1}$ | $\to$ | ${\mathbb{A}}_{5,15}$ | $E_{1}^{t}=te_{1}+t^{2}e_{3}$
$E_{2}^{t}=t^{2}e_{2}$ | $E_{3}^{t}=t(t-1)e_{2}+e_{4}$
$E_{4}^{t}=-t^{3}e_{3}$ | $E_{5}^{t}=t^{3}e_{5}$
${\mathbb{A}}_{5,26}^{1+t}$ | $\to$ | ${\mathbb{A}}_{5,16}$ | $E_{1}^{t}=e_{1}+e_{2}$
$E_{2}^{t}=\frac{2t^{2}+3t-3}{2t-1}e_{4}+\frac{(3+3t+t^{2})(2t^{2}+3t-3)^{2}}{18(1-2t)^{2}t^{2}(1+t)(2+t)}e_{5}$
$E_{3}^{t}=3te_{2}+\frac{2t^{2}+3t-3}{3t^{2}(2t^{2}+3t-2)}e_{3}-\frac{2t^{4}+5t^{3}+2t^{2}-3}{6t^{2}(2t^{2}+t-1)}e_{4}$
$E_{4}^{t}=\frac{2t^{2}+3t-3}{3t}e_{3}+\frac{3-3t-2t^{2}}{3t-6t^{2}}e_{4}-\frac{(3+3t+t^{2})(2t^{2}+3t-3)^{3}}{54(1-2t)^{2}t^{3}(1+t)(2+t)}e_{5}$ | $E_{5}^{t}=\frac{2t^{2}+3t-3}{2t-1}e_{5}$
${\mathbb{A}}_{5,19}$ | $\to$ | ${\mathbb{A}}_{5,17}$ | $E_{1}^{t}=\frac{1}{2\sqrt[6]{t^{5}}}e_{1}-\frac{\sqrt[3]{t}}{2}e_{3}$
$E_{2}^{t}=\frac{1}{4\sqrt[3]{t^{5}}}e_{2}-\frac{1}{2\sqrt[12]{t^{5}}}e_{3}-\frac{1}{4\sqrt[12]{t^{19}}}e_{4}$ | $E_{3}^{t}=\frac{1}{2\sqrt[3]{t^{2}}}e_{3}+\frac{1}{4\sqrt[6]{t^{11}}}e_{4}$
$E_{4}^{t}=\frac{1}{4\sqrt[6]{t^{11}}}e_{4}-\frac{1}{8t^{3}}e_{5}$ | $E_{5}^{t}=\frac{1}{8\sqrt{t^{5}}}e_{5}$
${\mathbb{A}}_{5,19}$ | $\to$ | ${\mathbb{A}}_{5,18}$ | $E_{1}^{t}=\frac{1}{2t^{2}}e_{1}+\frac{1}{2t}e_{3}$
$E_{2}^{t}=\frac{1}{4t^{4}}e_{2}+\frac{1}{2t}e_{3}-\frac{1}{4t^{2}}e_{4}$ | $E_{3}^{t}=\frac{1}{2t^{2}}e_{3}-\frac{1}{2t^{3}}e_{4}$
$E_{4}^{t}=\frac{1}{4t^{4}}e_{4}+\frac{1}{8t^{5}}e_{5}$ | $E_{5}^{t}=\frac{1}{8t^{6}}e_{5}$
${\mathbb{A}}_{5,21}$ | $\to$ | ${\mathbb{A}}_{5,25}$ | $E_{1}^{t}=t^{2}(t^{2}-1)e_{2}+t^{2}(t^{2}-1)e_{4}$
$E_{2}^{t}=(t^{5}-t^{3}+t^{2}-1)e_{1}+(t^{2}-1)e_{2}$ | $E_{3}^{t}=(1-t^{2}+t^{3}-t^{5})e_{4}+(1-t^{2})e_{5}$
$E_{4}^{t}=-t^{2}(t^{2}-1)^{2}e_{3}+(t-t^{3}+t^{4}-t^{6})e_{4}+(t-t^{2}-t^{3}+2t^{4}-t^{6})e_{5}$
$E_{5}^{t}=(t-1)^{2}(1+t)^{3}(t^{2}-t^{3}+t^{4})e_{5}$
${\mathbb{A}}_{5,26}^{\frac{1}{1+t}}$ | $\to$ | ${\mathbb{A}}_{5,27}$ | $E_{1}^{t}=(1+t)e_{1}-\frac{t}{2}e_{2}+\frac{1+t}{2}e_{3}$
$E_{2}^{t}=\sqrt{t}e_{2}$ | $E_{3}^{t}=te_{3}+t(1+t)e_{4}$
$E_{4}^{t}=\frac{\sqrt{t}(1+t)(2+t)}{2}e_{4}$ | $E_{5}^{t}=\frac{t(2+t)}{2}e_{5}$
${\mathbb{A}}_{5,26}^{\frac{\sqrt{-t(8+7t)}-4-5t}{\sqrt{-t(8+7t)}+4+3t}}$ | $\to$ | ${\mathbb{A}}_{5,28}$ |
$E_{1}^{t}=e_{1}+\frac{\left(\sqrt{-t(8+7t)}-t\right)\left(4+3t+\sqrt{-t(8+7t)}\right)}{16\sqrt[4]{t}(1+t)}e_{2}+\frac{\left(4+3t+\sqrt{-t(8+7t)}\right)^{3}}{64(1+t)(\sqrt{-t(8+7t)}-t)}e_{3}$
$E_{2}^{t}=\frac{4+3t+\sqrt{-t(8+7t)}}{4}e_{2}$ | $E_{3}^{t}=\frac{\left(4+3t+\sqrt{-t(8+7t)}\right)^{2}}{16}e_{3}+\sqrt[4]{t}(1+t)e_{4}$
$E_{4}^{t}=(1+t)e_{4}$ | $E_{5}^{t}=\frac{4+3t+\sqrt{-t(8+7t)}}{4}e_{5}$
${\mathbb{A}}_{5,26}^{\frac{t+h}{1+t}}$ | $\to$ | ${\mathbb{A}}_{5,29},\,{\mathbb{A}}_{5,30}$ | $E_{1}^{t}=-\frac{3it(1+h)}{\sqrt{1+t}}e_{2}$
$E_{2}^{t}=-\frac{i(1+t(5+2h))}{3t(1+h)\sqrt{(1+t)}}e_{1}+i\sqrt{1+t}e_{2}$ | $E_{3}^{t}=-e_{3}+\frac{1+3t}{3t}e_{4}$
$E_{4}^{t}=e_{4}$ | $E_{5}^{t}=\frac{i}{\sqrt{1+t}}e_{5}$
$h$ is $\omega$ (for ${\mathbb{A}}_{5,29}$) or $\omega^{2}$ (for
${\mathbb{A}}_{5,30}$), where $\omega$ is the primitive cubic root of unit
${\mathbb{A}}_{5,26}^{\frac{(1+L)(t^{2}-1)}{L-1+t^{2}(1+L)}}$ | $\to$ | ${\mathfrak{V}}_{2+3}$ |
$E_{1}^{t}=\frac{1+L}{2L}e_{1}+\frac{(t-1)^{2}(1+t)+2L(t^{3}-1)+L^{2}(1+t+t^{2}+t^{3})}{2L(1+L)}e_{2}-\frac{(L-1+t^{2}(1+L))^{3}}{8L^{2}t(t^{2}(1+L)-1)}e_{3}$
$E_{2}^{t}=\frac{1}{L}e_{1}+\frac{(t-1)(L-1+t^{2}(1+L))}{L(1+L)}e_{2}-\frac{(L-1+t^{2}(1+L))^{3}}{4tL^{2}(1+L)((1+L)t^{2}-1)}e_{3}$
$E_{3}^{t}=\frac{(1-L)(L-1+t^{2}(1+L))^{2}}{2(L+L^{2})^{2}}e_{3}+\frac{(1-L)(2+t)}{2L^{2}}e_{4}-\frac{L-1+t^{2}(1+L)}{2L^{2}t}e_{5}$
$E_{4}^{t}=\frac{(1-L)(L-1+t^{2}(1+L))^{2}}{2(L+L^{2})^{2}}e_{3}+\frac{2+t+L^{2}(2-t)}{2L^{2}(1+L)}e_{4}-\frac{L-1+t^{2}(1+L)}{2L^{2}t}e_{5}$
$E_{5}^{t}=\frac{(L-1+t^{2}(1+L))^{2}}{(L+L^{2})^{2}}e_{3}+\frac{2+t+Lt}{L^{2}+L^{3}}e_{4}-\frac{L-1+t^{2}(1+L)}{tL^{2}(1+L)}e_{5};$ | $\qquad\lambda=\frac{L^{2}-1}{4}$

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

${\mathbb{A}}_{5,2}$ | $\xrightarrow{(e_{1},e_{2},e_{4},te_{3},e_{5})}$ | ${\mathbb{A}}_{5,1}$ | ${\mathbb{A}}_{5,15}$ | $\xrightarrow{(e_{1}+e_{3},e_{2}+e_{5},e_{3},e_{5},te_{4})}$ | ${\mathbb{A}}_{5,2}$
---|---|---|---|---|---
${\mathbb{A}}_{5,11}$ | $\xrightarrow{(e_{1},e_{2}+te_{3},e_{3},e_{4}+e_{5},\frac{1}{t}e_{5})}$ | ${\mathbb{A}}_{5,3}$ | ${\mathbb{A}}_{5,10}$ | $\xrightarrow{(e_{1},e_{2},te_{3},e_{4},te_{5})}$ | ${\mathbb{A}}_{5,4}$
${\mathbb{A}}_{5,10}$ | $\xrightarrow{(-\frac{1}{t}e_{1}+\frac{1}{t}e_{3},\frac{1}{t^{2}}e_{2},\frac{1}{t}e_{3},-\frac{1}{t^{3}}e_{4},\frac{1}{t^{2}}e_{5})}$ | ${\mathbb{A}}_{5,6}$ | ${\mathbb{A}}_{5,9}$ | $\xrightarrow{(e_{1},e_{2},te_{3},e_{4},t^{2}e_{5})}$ | ${\mathbb{A}}_{5,7}$
${\mathbb{A}}_{5,10}$ | $\xrightarrow{(\frac{1}{t}e_{1},\frac{1}{t^{2}}e_{2},\frac{1}{t^{2}}e_{3},\frac{1}{t^{3}}e_{4},\frac{1}{t^{4}}e_{5})}$ | ${\mathbb{A}}_{5,9}$ | ${\mathbb{A}}_{5,10}$ | $\xrightarrow{(-t^{2}e_{1},t^{4}e_{3},t^{3}e_{2}-t^{3}e_{3},t^{6}e_{5},t^{5}e_{4}+t^{5}e_{5})}$ | ${\mathbb{A}}_{5,11}$
${\mathbb{A}}_{5,16}$ | $\xrightarrow{(t^{2}e_{1},t^{4}e_{2},t^{3}e_{3},t^{4}e_{4},t^{6}e_{5})}$ | ${\mathbb{A}}_{5,13}$ | ${\mathbb{A}}_{5,21}$ | $\xrightarrow{(\frac{1+t}{t}e_{1}+e_{2},\frac{1}{t}e_{2},\frac{1}{t^{2}}e_{3},\frac{1+t}{t^{2}}e_{4},\frac{1+t}{t^{3}}e_{5})}$ | ${\mathbb{A}}_{5,20}$
${\mathbb{A}}_{5,21}$ | $\xrightarrow{(te_{1},t^{2}e_{2},t^{3}e_{3}+t^{3}e_{5},t^{3}e_{4},t^{4}e_{5})}$ | ${\mathbb{A}}_{5,22}$ | ${\mathbb{A}}_{5,23}$ | $\xrightarrow{(te_{1},e_{2},te_{3},te_{4},te_{5})}$ | ${\mathbb{A}}_{5,24}$

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

$\dim\mathcal{O}({\mathfrak{V}}_{3+2})=24,$
$\dim\mathcal{O}({\mathfrak{V}}_{4+1})=20,$
$\dim\mathcal{O}({\mathbb{A}}_{5,10})=20,$
$\dim\mathcal{O}({\mathbb{A}}_{5,14}^{\lambda})=20,$
$\dim\mathcal{O}({\mathbb{A}}_{5,19})=20,$
$\dim\mathcal{O}({\mathbb{A}}_{5,21})=20,$
$\dim\mathcal{O}({\mathbb{A}}_{5,23})=20,$
$\dim\mathcal{O}({\mathbb{A}}_{5,26}^{\lambda})=20.$

Hence,
${\mathfrak{V}}_{4+1},{\mathbb{A}}_{5,10},{\mathbb{A}}_{5,14}^{\lambda},{\mathbb{A}}_{5,19},{\mathbb{A}}_{5,21},{\mathbb{A}}_{5,23},{\mathbb{A}}_{5,26}^{\lambda}$
and ${\mathfrak{V}}_{3+2}$ give $8$ irreducible components.

$\Box$

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