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# Open Mathematics

### formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo

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Volume 15, Issue 1

# Solvable Leibniz algebras with NFn⊕ $\begin{array}{}{F}_{m}^{1}\end{array}$ nilradical

L.M. Camacho
/ B.A. Omirov
• Institute of Mathematics. National University of Uzbekistan, 29, Do’rmon yo’li street., 100125, Tashkent, Uzbekistan
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• Other articles by this author:
/ K.K. Masutova
• Institute of Mathematics. National University of Uzbekistan, 29, Do’rmon yo’li street., 100125, Tashkent, Uzbekistan
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• Other articles by this author:
/ I.M. Rikhsiboev
• Corresponding author
• Universiti Kuala Lumpur, Malaysian Institute of Industrial Technology, Bandar Seri Alam, 81750 Johor Bahru, Malaysia
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Published Online: 2017-12-02 | DOI: https://doi.org/10.1515/math-2017-0115

## Abstract

All finite-dimensional solvable Leibniz algebras L, having N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ as the nilradical and the dimension of L equal to n+m+3 (the maximal dimension) are described. NFn and $\begin{array}{}{F}_{m}^{1}\end{array}$ are the null-filiform and naturally graded filiform Leibniz algebras of dimensions n and m, respectively. Moreover, we show that these algebras are rigid.

MSC 2010: 17A32; 17A65; 17B30

## 1 Introduction

Leibniz algebras over 𝕂 were first introduced by A. Bloh [1] and called D-algebras. The term Leibniz algebra was introduced in the study of a non-antisymmetric analogue of Lie algebras by Loday [2], being so the class of Leibniz algebras an extension of the one of Lie algebras. In recent years it has been common theme to extend various results from Lie algebras to Leibniz algebras [3,4]. Many results of the theory of Lie algebras have been extended to Leibniz algebras. For instance, the classical results on Cartan subalgebras [5], variations of Engel’s theorem for Leibniz algebras have been proven by different authors [6, 7] and Barnes has proven Levi’s theorem for Leibniz algebras [8].

In an effort to classify Lie algebras, many authors place various restrictions on the nilradical. The first work which was devoted to description of such Lie algebras is the paper [9]. Later, Mubarakjanov proposed the description of solvable Lie algebras with a given structure of nilradical by means of outer derivations [10]. In the papers [11, 12, 13, 14], the authors apply the Mubarakjanov’s method to classify the solvable Lie algebras with different kinds of nilradicals. Some results of Lie algebra theory generalized to Leibniz algebras in [3] allow us to apply the Mubarakjanov’s method for Leibniz algebras. In this sense, we can see the papers [15, 16, 17, 18].

It is important to study solvable Leibniz algebras because thanks to the Levi’s theorem for Leibniz algebras, a Leibniz algebra is a semidirect sum of the solvable radical and a semisimple Lie algebra. As the semisimple part can be described by simple Lie ideals, the main problem is to understand the solvable radical.

The first aim of the present paper is to classify solvable Leibniz algebras with nilradical N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ where NFn and $\begin{array}{}{F}_{m}^{1}\end{array}$ are the null-filiform and naturally graded filiform Leibniz algebras of dimensions n and m, respectively. To obtain this classification, we use the results obtained in [16, 17, 18].

The arrangement of this work is as follows. In Section 2 we recall some essential notions and properties of Leibniz algebras. We start Section 3 by establishing the maximal dimension of a solvable Leibniz algebra whose nilradical is N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$; thereafter, we present the classification of solvable Leibniz algebras that can be decomposed as a direct sum of their nilradical and a complementary vector space of maximal dimension. Finally, in Section 4 we study the rigidity of the unique solvable Leibniz algebra obtained in the previous section.

Throughout the paper, we consider finite-dimensional vector spaces and algebras over a field of characteristic zero. Moreover, in the multiplication table of an algebra omitted products are assumed to be zero and if it is not noticed we shall consider non-nilpotent solvable algebras.

## 2 Preliminaries

Let us recite some necessary definitions and preliminary results.

A Leibniz algebra over a field 𝔽 is a vector space L equipped with a bilinear map, called bracket, $[−,−]:L×L→L,$

satisfying the Leibniz identity $[x,[y,z]]=[[x,y],z]−[[x,z],y]$

for all x, y, zL.

The set Annr(L) = {xL | [y, x] = 0, yL} is called the right annihilator of the Leibniz algebra L. Note that Annr(L) is an ideal of L and for any x, yL the elements [x, x], [x, y] + [y, x] ∈ Annr(L).

A linear map d:LL of a Leibniz algebra (L,[–,–]) is said to be a derivation if for all x, yL the following condition holds: $d([x,y])=[d(x),y]+[x,d(y)].$

For a given element x of a Leibniz algebra L the operator of right multiplication Rx : LL, defined as Rx(y) = [y, x] for yL, is a derivation. This kind of derivations are called inner derivations.

Any Leibniz algebra L is associated with the algebra of right multiplications R(L) = {Rx|xL}, which is endowed with a structure of Lie algebra by means of the bracket [Rx, Ry] = RxRyRyRx. Thanks to the Leibniz identity the equality [Rx, Ry] = R[y, x] holds true. In addition, the algebra R(L) is antisymmetric isomorphic to the quotient algebra L/Annr(L).

#### Definition 2.1

([10]). Let d1, d2, …, dn be derivations of a Leibniz algebra L. The derivations d1, d2, …, dn are said to be nil-independent if $α1d1+α2d2+⋯+αndn$

is not nilpotent for any scalars α1, α2, …, αn ∈ 𝔽.

In other words, if for any α1, α2, αn ∈ 𝔽 there exists a natural number k such that (α1d1+α2d2+⋯+ αndn)k = 0, then α1 = α2 = ⋯ = αn = 0.

## 2.1 Solvable Leibniz algebras

For a Leibniz algebra L we consider the following lower central and derived series: $L1=LLk+1=[Lk,L1],k≥1;L[1]=L,L[s+1]=[L[s],L[s]],s≥1.$

A Leibniz algebra L is said to be nilpotent (respectively, solvable), if there exists n ∈ ℕ(m ∈ ℕ) such that Ln = 0 (respectively, L[m] = 0).

It should be noted that the sum of any two nilpotent ideals is nilpotent.

The maximal nilpotent ideal of a Leibniz algebra is said to be a nilradical of the algebra.

Obviously, the index of nilpotency of an n-dimensional nilpotent Leibniz algebra is not greater than n+1. The following theorem describes these algebras, algebras with maximal index of nilpotency.

#### Theorem 2.2

([4]). Any n-dimensional null-filiform Leibniz algebra admits a basis {e1, e2,⋯, en} such that the table of multiplication in the algebra has the following form: $NFn:[ei,e1]=ei+1,1≤i≤n−1$

A Leibniz algebra L is said to be filiform if dim Li = ni, where n = dim L and 2 ≤ in.

Due to [4] and [19] it is known that there are three naturally graded filiform Leibniz algebras. In fact, the third type encloses the class of naturally graded filiform Lie algebras.

#### Theorem 2.3

([17]). Any complex n-dimensional naturally graded filiform Leibniz algebra is isomorphic to one of the following pairwise non isomorphic algebras: $Fn1:[ei,e1]=ei+1,2≤i≤n−1,Fn2:[ei,e1]=ei+1,1≤i≤n−2,Fn3(α):[ei,e1]=−[e1,ei]=ei+1,2≤i≤n−1,[ei,en+1−i]=−[en+1−i,ei]=α(−1)i+1en,2≤i≤n−1,$

where α ∈ {0,1} for even n and α = 0 for odd n.

The following theorems describe solvable Leibniz algebras of maximal dimension with N $\begin{array}{}{F}_{n}^{1}\end{array}$ and $\begin{array}{}{F}_{n}^{1}\end{array}$ nilradical.

#### Theorem 2.4

([17]). Let R be a solvable Leibniz algebra whose nilradical is NFn. Then there exists a basis {e1, e2,⋯, en, x} of the algebra R such that the multiplication table of R with respect to this basis has the following form: $[ei,e1]=ei+1,1≤i≤n−1,[ei,x]=iei,1≤i≤n,[x,e1]=−e1.$

#### Theorem 2.5

([18]). An arbitrary (n+2)-dimensional solvable Leibniz algebra with nilradical $\begin{array}{}{F}_{n}^{1}\end{array}$ is isomorphic to the algebra R( $\begin{array}{}{F}_{n}^{1}\end{array}$) with the multiplication table: $[ei,e1]=ei+1,2≤i≤n−1,[e1,x]=e1,[ei,y]=ei,2≤i≤n,[ei,x]=(i−1)ei,2≤i≤n,[x,e1]=−e1.$

Let R be a solvable Leibniz algebra. Then it can be decomposed in the form R = NQ, where N is the nilradical and Q is the complementary vector space. Since the square of a solvable Leibniz algebra is contained into the nilradical [3], we get the nilpotency of the ideal R2 and consequently, Q2N.

Let us consider the restrictions to N of the right multiplication operator on an element xQ (denoted by Rx|N). From [17], we know that for any xQ, the operator Rx|n is a non-nilpotent derivation of N. Let {x1, x2, …, xm} be a basis of Q, then for any scalars {α1, …, αm} ∈ ℂ∖ {0}, the matrix α1Rx1|N+⋯+αmRxm|N is non nilpotent, which means that the elements {x1, …, xm} are nil-independent derivations, [10].

#### Theorem 2.6

([17]). Let R be a solvable Leibniz algebra and N be its nilradical. Then the dimension of the complementary vector space to N is not greater than the maximal number of nil-independent derivations of N.

Moreover, similarly as in Lie algebras, for a solvable Leibniz algebra R, we have dimN$\begin{array}{}\frac{dimR}{2}.\end{array}$

A nilpotent Leibniz algebra is called characteristically nilpotent if all its derivations are nilpotent. If the nilradical N of a Leibniz algebra is characteristically nilpotent then, according to Theorem 2.6, a solvable Leibniz algebra is nilpotent. Therefore, we shall consider solvable Leibniz algebras with non-characteristically nilpotent nilradical. For more details see [17].

## 2.2 The second cohomology group of a Leibniz algebra

For acquaintance with the definition of cohomology group of Leibniz algebras and its applications to the description of the variety of Leibniz algebras (similar to the Lie algebras case) we refer the reader to the papers [2, 20, 21, 22, 23, 24]. Here we just recall that the second cohomology group of a Leibniz algebra L with coefficients in a corepresentation M is the quotient space $HL2(L,M)=ZL2(L,M)/BL2(L,M),$

where the 2-cocycles, φZL2(L, M) and the 2-coboundaries fBL2(L, M) are defined as follows $(d2φ)(a,b,c)=[a,φ(b,c)]−[φ(a,b),c]+[φ(a,c),b]+φ(a,[b,c])−φ([a,b],c)+φ([a,c],b)=0$(1)

and $f(a,b)=[d(a),b]+[a,d(b)]−d([a,b]) for some linear map d.$

The linear reductive group GLn(𝔽) acts on the variety of n-dimensional Leibniz algebras, Leibn as follows: $(g∗λ)(x,y)=g(λ(g−1(x),g−1(y))),g∈GLn(𝔽),λ∈Leibn.$

The orbits (Orb(−)) under this action are the isomorphism classes of algebras. Note that, Leibniz algebras with open orbits are called rigid.

Due the work [20], the nullity of the second cohomology group with coefficients itself gives a sufficient condition for the rigidity of the algebras.

## 3 Main results

Let NFn be an n-dimensional null-filiform Leibniz algebra with a basis {e1, e2, …, en} and $\begin{array}{}{F}_{m}^{1}\end{array}$ an m-dimensional filiform Leibniz algebra from the first class with a basis {f1, f2, … fm}, then we have the following multiplication: $NFn:[ei,e1]=ei+1,1≤i≤n−1;Fm1:[f1,f1]=f3,[fi,f1]=fi+1,2≤i≤m−1.$

Let us consider the direct sum of these algebras N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$. The following proposition describes derivations of the algebra N.

#### Proposition 3.1

Any derivation of the algebra N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ has the following matrix form: $α1α2α3…αn000…0βm02α1α2…αn−1000…00003α1…αn−2000…00⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋮000…nαn000000000…γnδ1δ2δ3…δm−1δm000…τn0δ1+δ2δ3…δm−1σm000…0002δ1+δ2…δm−2δm−1⋮⋮⋮⋱⋮⋮⋮⋮⋮⋮⋮000…000000(m−1)δ1+δ2.$

#### Proof

The proof is going by straightforward calculation of derivation property. □

From this proposition it is easy to see that the number of nil-independent outer derivations of the algebra N is equal to 3.

Now we consider solvable Leibniz algebra R = N+Q, where N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ and the dimension of Q is no more than three. Thus, we study the case dim Q = 3, i.e. dim R = n+m+3. Several papers described solvable Leibniz algebras with a given nilradical [15, 16, 17]. The most interesting cases are when the complementary space of nilradical has the maximum possible. Namely, they have the second group of cohomology trivial. For this reason, we consider the case dimQ = 3.

From the work [17], it follows that any solvable Leibniz algebra whose nilradical is NFn has dimension n+1. It is also known that any solvable Leibniz algebra whose nilradical is $\begin{array}{}{F}_{m}^{1}\end{array}$ has dimension either m+1 or m+2, [16]. In work [18], it was found a unique (m+2)-dimensional solvable Leibniz algebra with nilradical $\begin{array}{}{F}_{m}^{1}\end{array}$. Then in the case of the solvable Leibniz algebras R with nilradical N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ and dim Q = 3, there is only one possible case.

Taking into account Theorems 2.2 and 2.5, we have the following multiplication of the algebra R: $[ei,e1]=ei+1,1≤i≤n−1,[fi,f1]=fi+1,2≤i≤m−1,[ei,x]=iei,1≤i≤n,[f1,y]=f1,[x,e1]=−e1,[fi,y]=(i−1)fi,2≤i≤m,[fi,z]=fi,2≤i≤m,[y,f1]=−f1,$(2)

where {x, y, z} be a basis of the space Q.

In the following theorem, solvable Leibniz algebras with nilradical N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ and dim Q = 3 are described.

#### Theorem 3.2

Any (n+m+3)-dimensional solvable Leibniz algebra with nilradical N = NFn$\begin{array}{}{F}_{m}^{1}\end{array}$ is isomorphic to the following algebra: $L:[ei,e1]=ei+1,1≤i≤n−1,[fi,f1]=fi+1,2≤i≤m−1,[ei,x]=iei,1≤i≤n,[f1,y]=f1,[x,e1]=−e1,[fi,y]=(i−1)fi,2≤i≤m,[fi,z]=fi,2≤i≤m,[y,f1]=−f1.$

#### Proof

From the above argumentations we have the multiplication (2) and we introduce the following denotations for the algebra R (according to the Mubarakzjanov’s method [10]): $[x,f1]=∑i=1nαi1ei+∑i=1mβi1fi,[x,y]=∑i=1nλi1ei+∑i=1mμi1fi,[x,f2]=∑i=1nαi2ei+∑i=1mβi2fi,[y,x]=∑i=1nλi2ei+∑i=1mμi2fi,[y,e1]=∑i=1nγi1ei+∑i=1mδi1fi,[x,z]=∑i=1nλi3ei+∑i=1mμi3fi,[z,e1]=∑i=1nγi2ei+∑i=1mδi2fi,[z,x]=∑i=1nλi4ei+∑i=1mμi4fi.[fi,x]=∑j=1naijej+∑j=1nbijfj,1≤i≤m,[ei,y]=∑j=1ncij1ej+∑j=1mdij1fj,1≤i≤n,[ei,z]=∑j=1ncij2ej+∑j=1mdij2fj,1≤i≤n,$

From Leibniz identity it follows that [y, e2] = [y,[e1, e1]] = 0 and by induction we can easily find that [y, ei] = 0, with 2 ≤ in. Analogously, we have [z, ei] = 0, with 2 ≤ in.

We consider [x, f3] = [x,[f2, f1]] = [[x, f2], f1] = $\begin{array}{}\sum _{i=3}^{m}{\beta }_{i-1}^{2}{f}_{i}.\end{array}$ Similarly, and using the induction method, it is possible to show that $\begin{array}{}\left[x,{f}_{i}\right]=\sum _{j=i}^{m}{\beta }_{j-i+2}^{2}{f}_{j}\end{array}$ for 3 ≤ im.

However, from the equality [x,[f3, y]] = [[x, f3], y], it follows $\begin{array}{}{\beta }_{i}^{2}\end{array}$ = 0, with 2 ≤ im−1, i.e. $[x,fi]=0,3≤i≤m,[x,f2]=∑i=1nαi2ei+β12f1+β22f2+βm2fm.$

Taking the following change $\begin{array}{}{x}^{\prime }=x-{\mu }_{1}^{1}{f}_{1}-\sum _{i=2}^{m}\frac{1}{i-1}{\mu }_{i}^{1}{f}_{i},\end{array}$ we have $\begin{array}{}\left[x{,}^{\prime }y\right]=\sum _{i=1}^{n}{\lambda }_{i}^{1}{e}_{i}.\end{array}$

Next, making $\begin{array}{}{y}^{\prime }=y-\sum _{i=1}^{n}\frac{1}{i}{\lambda }_{i}^{2}{e}_{i},\end{array}$ we obtain $\begin{array}{}\left[y{,}^{\prime }x\right]=\sum _{i=1}^{m}{\mu }_{i}^{2}{f}_{i}.\end{array}$

Finally, the change of basis $\begin{array}{}{z}^{\prime }=z-\sum _{i=1}^{n}\frac{1}{i}{\lambda }_{i}^{4}{e}_{i}\end{array}$ allows to obtain $\begin{array}{}\left[z{,}^{\prime }x\right]=\sum _{i=1}^{m}{\mu }_{i}^{4}{f}_{i}.\end{array}$

Let us apply the Leibniz identity on the following triples of elements: ${x,x,y}⟹λi1=0,2≤i≤n⟹[x,y]=λ11e1;{y,y,x}⟹μi2=0,2≤i≤m⟹[y,x]=μ12f1;{z,y,x}⟹μi4=0,1≤i≤m⟹[z,x]=0.$

We observe that [x, y] + [y, x] ∈ Annr(R), thus [e1, $\begin{array}{}{\lambda }_{1}^{1}\end{array}$ e1+ $\begin{array}{}{\mu }_{1}^{1}\end{array}$ f1] = 0, i.e. $\begin{array}{}{\lambda }_{1}^{1}\end{array}$ = 0. From the identity 0 = [f2, [x, y] + [y, x]] = [f2, $\begin{array}{}{\mu }_{1}^{1}\end{array}$f1] we have $\begin{array}{}{\mu }_{1}^{1}\end{array}$ = 0.

Analogously, [e1, y] + [y, e1] and [e1, z] + [z, e1] ∈ Annr(R), then, $[e1,[e1,y]+[y,e1]]=0 and [f2,[e1,y]+[y,e1]]=0,$

so we have $\begin{array}{}{c}_{11}^{1}=-{\gamma }_{{1}^{},}^{1}{d}_{11}^{1}=-{\delta }_{1}^{1}\text{\hspace{0.17em}and\hspace{0.17em}}{c}_{11}^{2}=-{\gamma }_{1}^{2},{d}_{11}^{2}=-{\delta }_{1}^{2}.\end{array}$

Similarly, from [e1, [x, z] + [z, x]] = 0 and [f2, [x, z] + [z, x]] = 0 we deduce $\begin{array}{}{\lambda }_{1}^{3}={\mu }_{1}^{3}=0.\end{array}$

Considering the following equalities: $[x,f2]=[x,[f2,y]]=−[x,[y,f2]]=0⇒β12=β22=βm2=0,αi2=0,1≤i≤n,[x,[e1,f1]]=−[[x,f1],e1]=0⇒αi1=0,1≤i≤n−1,[y,[e1,f1]]=[z,[e1,f1]]=0⇒δi1=δi2=0,2≤i≤m−1,$

we immediately get $[x,f1]=αn1en+∑i=1mβi1fi,[y,e1]=∑i=1nγi1ei+δ11f1+δm1fm,[z,e1]=∑i=1nγi2ei+δ12f1+δm2fm.$

As a result of [x, f1]+[f1, x] ∈ Annr(R), we observe that a11 = 0 and b11 = − $\begin{array}{}{\beta }_{1}^{1}\end{array}$.

Since, [f1, [x, f1]] = 0 and [f1, [x, e1]] = 0 it follows that a1, i = b1, j = 0, for 1 ≤ in−1,2 ≤ jm−1, i.e. $[f1,x]=a1,nen−β11f1+b1,mfm,$

and from the equalities [fi, [x, e1]] = 0, for 2 ≤ im, it follows that ai, j = 0, for 2 ≤ im and 1 ≤ jn−1, that is $[fi,x]=ai,nen+∑j=1mbijfj,2≤i≤m.$

We know that [x, fi] + [fi, x] ∈ Annr(R) for 2 ≤ im, then 0 = [f2, [x, fi] + [fi, x]] = bi1f3, that is bi1 = 0 for 2 ≤ im.

Now, we proceed by looking at the product of certain elements of Annr(R). Summarizing the following identities $[e1,[y,ei]+[ei,y]]=[f2,[y,ei]+[ei,y]]=[e1,[z,ei]+[ei,z]]=[f2,[z,ei]+[ei,z]]=0,$

we have the following Leibniz brackets of the basic elements: $(3){[ei,e1]=ei+1,1≤i≤n−1,[fi,f1]=fi+1,2≤i≤m−1,[ei,x]=iei,1≤i≤n,[f1,y]=f1,[x,e1]=−e1,[fi,y]=(i−1)fi,2≤i≤m,[fi,z]=fi,2≤i≤m,[y,f1]=−f1,[x,f1]=αn1en+∑i=1mβi1fi,[f1,x]=a1,nen−β11f1+b1,mfm,[y,e1]=∑i=1nγi1ei+δ11f1+δm1fm,[fi,x]=ai,nen+∑j=2mbijfj,2≤i≤m,[z,e1]=∑i=1nγi2ei+δ12f1+δm2fm,[e1,y]=−γ11e1+∑j=2nc1j1ej−δ11f1+∑j=2md1j1fj,[x,z]=∑i=2nλi3ei+∑i=2mμi3fi,[ei,y]=∑j=2ncij1ej+∑j=2mdij1fj,2≤i≤n,[e1,z]=−γ12e1+∑j=2nc1j2ej−δ12f1+∑j=2md1j2fj,[ei,z]=∑j=2ncij2ej+∑j=2mdij2fj,2≤i≤n.$

The Leibniz identity on the following triples imposes futher constraints on (3).

As a result of the above constraints we observe that $[fm,x]=bmmfm,[fm−1,x]=am−1,nen+(β11+bm,m)fm−1+bm−1,mfm.$

By the induction on decrease i(2 ≤ im) and using the Leibniz identity for the elements {fi, x, f1} we get $[fi,x]=((m−i)β11+bmm)fi+∑j=i+1mbi−j+m,mfj,3≤i≤m,[f2,x]=a2,nen+((m−2)β11+bmm)f2+∑i=3mbm−i+2,mfi.$

Considering the Leibniz identity on the triple {e1, y, e1} we get $\begin{array}{}\left[{e}_{2},y\right]=-2{\gamma }_{1}^{1}{e}_{2}+\sum _{i=3}^{n}{c}_{1,i-1}^{1}{e}_{i}.\end{array}$

Also, using the equalities [ei, y] = [[ei−1, e1], y] = [[ei−1, y], e1] − [ei−1, [y, e1]], for 3 ≤ in and by induction method on i, we obtain that $[ei,y]=−iγ11ei+∑j=i+1nc1,j−j+11ej,2≤i≤n.$

Analogously, we can get $\begin{array}{}\left[{e}_{i},z\right]=-i{\gamma }_{1}^{2}{e}_{i}+\sum _{j=i+1}^{n}{c}_{1,j-i+1}^{2}{e}_{j},\text{\hspace{0.17em}for\hspace{0.17em}}2\le i\le n.\end{array}$

The Leibniz identity on the triples {f2, x, y}, {f1, x, y} and {x, f1, y}, (in this order) gives $[f1,x]=a1,nen−β11f1,[f2,x]=a2,nen+((m−2)β11+bmm)f2,[fi,x]=((m−i)β11+bmm)fi,3≤i≤m,[x,f1]=αn1en+β11f1+β21f2.$

Now, applying the Leibniz identity on the triples {x, f1, z}, {y, x, f1}, {fi, x, z}, {fi, x, y}, with 1 ≤ im, {ei, z, x}, {ei, y, x}, with 1 ≤ in, (in this order) it follows $αn1γ12=β21=μi3=0,2≤i≤m−1,a1n=0,c1i1=c1i2=0,2≤i≤n−1,a2n(nγ12+1)=0.$

From the above results we can simplify the family (3) as follows: $(4):[ei,e1]=ei+1,1≤i≤n−1[fi,f1]=fi+1,2≤i≤m−1,[ei,x]=iei,1≤i≤n,[f1,y]=f1,[x,e1]=−e1,[fi,y]=(i−1)fi,2≤i≤m,[fi,z]=fi,2≤i≤m,[y,f1]=−f1,[x,f1]=αn1en+β11f1,[f1,x]=−β11f1,[f2,x]=a2nen+((m−2)β11+bmm)f2,[y,e1]=∑i=1nγi1ei+δ11f1+δm1fm,[fi,x]((m−i)β11+bmm)fi,3≤i≤m,[z,e1]=∑i=1nγi2ei+δ12f1+δm2fm,[e1,y]=−γ11e1+c1n1en−δ11f1+∑j=2md1j1fj,[x,z]=∑i=1nλi3ei+μm3fm,[ei,y]=−iγ11ei,2≤i≤n[e1,z]=−γ12e1+c1n2en−δ12f1+∑j=2md1j2fj,[ei,z]=−iγ12ei,2≤i≤n$

Finally, considering the Leibniz identity on the following triples we obtain: $Leibniz identity Constrant__{x,z,e1}⟹δ12(β11+1)=0,λj3=0,2≤j≤n−2,λn−13=δ12αn1−c1n2,d1i2=0,2≤i≤m,{x,y,e1}⟹c1n1=δ11αn1,δ11(β11+1)=0,d1i1=0,2≤i≤m,{x,z,y}⟹μm3=0,{x,z,x}⟹λn3=0,c1n2=δ12αn1,{e1,y,x}⟹c1n1=0,{e1,z,x}⟹c1n2=0,{y,e1,x}⟹γi1=0,2≤i≤n,δm1(1−bmm)=0,{z,e1,x}⟹γi2=0,2≤i≤n,δm2(1−bmm)=0,$ $Leibniz identity Constrant__{y,e1,y}⟹δm1(γ11+m−1)=0,{z,e1,z}⟹δm2(γ12+1)=0,{f2,e1,y}⟹δ11=0,{f2,e1,z}⟹δ12=0,{y,e1,z}⟹δm1(γ12+1)=0,{z,e1,y}⟹δm2(γ11+m−1)=0,{x,f1,x}⟹αn1(β11+n)=0,{x,f1,y}⟹αn1(nγ11+1)=0,{x,f1,z}⟹αn1γ12=0.$

Summarizing the above results and renaming the following parameters $\begin{array}{}{a}_{2n}=a,\phantom{\rule{thinmathspace}{0ex}}{\alpha }_{n}^{1}=\alpha ,\phantom{\rule{thinmathspace}{0ex}}{\beta }_{1}^{1}=\beta ,\phantom{\rule{thinmathspace}{0ex}}{b}_{mm}=b,\phantom{\rule{thinmathspace}{0ex}}{\gamma }_{1}^{i}={\gamma }_{i},{\delta }_{m}^{i}={\delta }_{i}\end{array}$ we can simplify the family (4) as follows: $(5):[ei,e1]=ei+1,1≤i≤n−1,[fi,f1]=fi+1,2≤i≤m−1,[x,e1]=−e1,[x,f1]=αen+βf1,[y,e1]=γ1e1+δ1fm,[y,f1]=−f1,[z,e1]=γ2e1+δ2fm,[ei,x]=iei,1≤i≤n,[ei,y]=−iγ1ei,1≤i≤n,[f1,x]=−βf1,[f1,y]=f1,[f2,x]=aen+((m−2)β+b)f2,[fi,y]=(i−1)fi,2≤i≤m,[fi,x]=((m−i)β+b)fi,3≤i≤m,[ei,z]=−iγ2ei,1≤i≤n,[fi,z]=fi,2≤i≤m$

with the relations: $δk(b−1)=δk(γ1+m−1)=δk(γ2+1)=0,k=1,2,a(nγ1+1)=a(nγ2+1)=0,α(nγ1+1)=α(β+n)=αγ2=0.$

Now, by performing a change of basis in the family (5) x′ = x + βy −((m − 1)β + b)z we have: $[ei,e1]=ei+1,1≤i≤n−1,[fi,f1]=fi+1,2≤i≤m−1,[x,e1]=−θe1+δfm,[x,f1]=αen,[y,e1]=γ1e1+δ1fm,[y,f1]=−f1,[z,e1]=γ2e1+δ2fm,[ei,x]=iθei1≤i≤n,[ei,y]=−iγ1ei,1≤i≤n,[f2,x]=aen,[f1,y]=f1,[fi,y]=(i−1)fi,2≤i≤m,[ei,z]=−iγ2ei,1≤i≤n,[fi,z]=fi,2≤i≤m$

Since Rx is not nilpotent, θ ≠ 0. Thus we can assume θ′ = 1 (making ${x}^{\prime }=\frac{1}{\theta }x$).

The Leibniz identity on the following triples imposes further constraints on the parameters. $Leibniz identity Constraint__{x,f1,x}⟹α=0,{y,e1,x}⟹δ1=0,{z,e1,x}⟹δ2=0,{x,e1,x}⟹δ=0,{f2,x,z}⟹a(nγ2+1)=0,{f2,x,y}⟹a(nγ1+1)=0.$

We can distinguish the following cases:

• Case a ≠ 0. The restrictions imply that ${\gamma }_{1}={\gamma }_{2}=-\frac{1}{n}.$ By performing a change of basis $f1′=f1,f2′=nf2−aen,fi′=nfi,y′=y−1nx,z′=z−1nx$

we obtain the algebra L.

• Case a = 0. Making the change y′ = y + γ1x, z′ = z + γ2x, we obtain the algebra L.□

## 4 Rigidity of the algebra L

In order to describe the second group of cohomology of the algebra L we need the description of its derivations. The general form of the derivations of L is given in the following proposition.

#### Proposition 4.1

A derivation d of the algebra L has the following form: $d1(ei)=iei,1≤i≤n,d3(f1)=f1,d2(ei)=ei+1,1≤i≤n−1,d3(fi)=(i−2)fi,3≤i≤m,d2(x)=−e1,d4(fi)=fi,2≤i≤m,d5(fi)=fi+1,2≤i≤m−1,d5(y)=−f1.$

#### Proof

The proof is carried out by straightforward calculations of derivation properties.□

From Proposition 4.1 we conclude that dim BL2(L, L) = (m + n + 3)2−5. The general form of an element of the space ZL2(R(L, L)) is presented below.

#### Proposition 4.2

dim ZL2(L, L) = (m + n + 3)2 − 5.

#### Proof

Let φZL2(L, L). We set en+m+1 := x, en+m+2 := y, en+m+3 := z, en+i := fi, 1 ≤ i ≤ m and $(ei,ej)=∑k=1n+m+3ai,jkek,1≤i,j≤n+m+3.$

For φZL2(L, L) we shall verify equation (1). We consider b = cL, then we get [a, φ(b, b)] + φ(a, b2) = 0 for all aL.

If b = e1, then we have $φ(ei,e2)=−ia1,1n+m+1ei−a1,11ei+1,1≤i≤n,φ(f1,e2)=−a1,1n+m+2f1,φ(x,e2)=a1,11e1,φ(y,e2)=a1,1n+1f1,φ(z,e2)=0,φ(fi,e2)=−((i−1)a1,1n+m+2+a1,1n+m+3)fi−a1,1n+1fi+1,2≤i≤m.$

From the multiplication table of the algebra L it is easy to see that φ(b, b) ∈ I1I2 for all bL, be1, where I1 = spane2, en〉 and I2 = spanf2, …, fm〉.

If b, cI1I2, then we obtain [a, φ(b, c)] = 0 for all aL, and, consequently, φ(b, c) ∈ I1I2, i.e. φ(Ii, Ij) ⊆ I1I2, 1 ≤ i, j ≤ 2.

If a, b, cQ, (remember that L = NQ where Q is the complementary vector space of the nilradical N = NFn${F}_{m}^{1}$), then we derive $an+m+1,n+m+21=−an+m+2,n+m+11,an+m+1,n+m+31=−an+m+3,n+m+11,an+m+2,n+m+3n+1=−an+m+3,n+m+2n+1$

and $φ(x,x)∈I1,φ(x,y)∈Fm1⨁〈e1〉,φ(x,z)∈I2⨁〈e1〉,φ(y,x)∈NFn⨁〈f1〉,φ(y,y)∈I2,φ(y,z)∈Fm1,φ(z,x)∈NFn,φ(z,y)∈Fm1,φ(z,z)∈I2.$

From now on, we consider the equation (1) with combinations of the triples {a, b, c}, where a, b, c are the elements of L. $Triples Relations__{x,x,ei},{x,y,ei},{x,ei,x},i≥3⇒φ(x,ei)∈〈e1〉,i≥3{x,x,f1},{x,y,f1},{x,f1,x}⇒φ(x,f1)=−an+1,12e1+an+m+1,n+1n+1f1+∑i=3man+m+1,n+1n+ifi,{x,y,fi},{x,fi,y},2≤j≤m⇒φ(x,fi)∈〈e1〉,2≤i≤m,{y,y,ei},{x,y,ei},{y,x,ei},{y,ei,x},{y,ei,e1}⇒φ(y,ei)=ai−1,1n+1f1,3≤i≤n{z,z,fi},{y,z,fi},{z,x,fi},{z,fi,y}⇒φ(z,fi)=0,3≤i≤m,{y,y,e1},{y,x,e1}⇒φ(y,e1)∈NFn⨁〈f1〉,{y,y,fi},{y,x,fi},{y,fi,y}⇒φ(y,fi)∈〈f1〉,2≤i≤m,{z,z,f1},{z,x,f1},{z,y,f1}⇒φ(z,f1)=an+m+3,n+1n+1f1+∑i=3man+m+3,n+1n+ifi,{z,z,ei},{z,x,ei},{z,ei,x}⇒φ(z,ei)=0,3≤i≤n,{z,x,e1}⇒φ(z,e1)=∑i=1nan+m+3,1iei,{fi,x,z}⇒φ(f1,x)∈NFn⨁〈f1〉;φ(fi,x)∈NFn⨁I2,2≤i≤m,{ei,y,ej},1≤i,j≤n⇒φ(I1,I1)⊆I1;φ(ei,y)∈I1⨁Fm1,2≤i≤n,{ei,ej,e1},1≤i,j≤n,⇒φ(ei,ej)=−iaj−1,1n+m+1ei+(aj−2,1n+m+1aj−1,11)ei+13≤j≤n,1≤i≤n,{x,ei,e1},2≤i≤n−1⇒φ(x,ei)=(ai−1,11−ai−2,1n+m+1)e1,3≤i≤n;{f1,fj,f1}⇒φ(f1,fi)=−an+i−1,n+1m+n+2f1,3≤i≤m;{y,f1,f2},{f1,z,f2}⇒φ(f1,f2)=an+1,n+2n+1f1;{y,e1,f1},{y,f1,e1}⇒φ(f1,e1)=∑s=1nan+1,1ses+an+1,1n+1f1;{y,ei,f1},2≤i≤n⇒φ(f1,ei)=−ai,n+1n+1f1,2≤i≤n;{f1,x,f1},{f1,z,f1}⇒φ(f1,f1)=∑i=2man+1,n+1n+ifi;$

Given the restrictions above and the cocycle property (d2φ)(x, e1, fi) = 0 for 1 ≤ im, (d2φ)(e1, y, f1) = 0 and (d2φ)(fi, x, ej) = 0, 1 ≤ im, 1 ≤ jn, we derive $φ(e1,f1)∈Fm1⨁〈e1,e2〉,φ(e1,f2)∈NFn,φ(e1,fi)∈〈e1,e2〉,3≤i≤m,φ(fi,e1)∈NFn⨁I2,φ(fi,ej)∈I2,2≤i≤m,2≤j≤n.$

Analogously, we get ${fi,x,fj},2≤i,j≤m⇒φ(I2,I2)⊆I2;{ei,fj,x},{ei,e1,fj},2≤i≤n,2≤j≤m⇒φ(I1,I2)⊆I1;{fi,ej,x},2≤i≤m,2≤j≤n⇒φ(I2,I1)⊆I2;{ei,x,f1},2≤i≤n⇒φ(ei,f1)∈I1⨁(Fm1∖〈f2〉),2≤i≤n;{e1,x,f2}⇒φ(e1,f2)∈〈e1,e2〉;{fi,y,e1},2≤i≤m⇒φ(fi,y)∈L∖〈x〉,2≤i≤m;{ei,z,e1},2≤i≤n⇒φ(ei,z)∈NFn⨁I2,2≤i≤n.$

Summarizing the above discussion and from $\left({d}^{2}\phantom{\rule{thinmathspace}{0ex}}\phi \right)\left({f}_{i}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{1}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}i\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m,\phantom{\rule{thinmathspace}{0ex}}\left({d}^{2}\phantom{\rule{thinmathspace}{0ex}}\phi \right)\left({f}_{i}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{j}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}z\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}j\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m,\phantom{\rule{thinmathspace}{0ex}}\left({d}^{2}\phantom{\rule{thinmathspace}{0ex}}\phi \right)\left({f}_{i}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{j}\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}y\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}0,\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}j\phantom{\rule{thinmathspace}{0ex}}\le \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}m\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\left({d}^{2}\phantom{\rule{thinmathspace}{0ex}}\phi \right)\left(y\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}y\phantom{\rule{thinmathspace}{0ex}},\phantom{\rule{thinmathspace}{0ex}}{f}_{1}\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}0$ we conclude:

$φ(fm,e1)=∑s=1nan+m,1ses+an+m,1n+mfm,φ(fi,f1)=−∑s=1n−1an+i+1,1s+1es+an+i,n+1nen+∑s=1man+i,n+1n+sfs+an+i+1,11x+an+i,n+1n+m+2y+an+i,n+1Z,n+m+32≤i≤m−1,(fm,f1)=−an+m,n+1nen+∑j=1man+m,n+1n+sfs+an+m,n+1n+m+2y+an+m,n+1Z,n+m+3φ(f1,f1)=∑i=3man+1,n+1n+ifi,φ(f3,z)=∑s=1nan+2,n+1ses+(an+2,n+1n+1−n+2,n+m+3n+m+2)f1+(an+1,n+m+3n+m+2+an+1,n+m+3n+m+3)f2++(an+2,n+m+3n+2+an+1,n+m+3n+1)f3+∑s=4man+2,n+m+3n+s−1fs+an+2,n+1n+m+1x+an+2,n+1n+m+2y+an+2,n+1Z,n+m+3φ(fi,z)=∑s=1nan+i−1,n+1ses+(an+i−1,n+1n+1−an+i−2,n+1n+m+2)f1+(i−2)(i−12an+1,n+m+3n+m+2+an+1,n+m+3n+m+3)fi−1++(an+2,n+m+3n+2+(i−2)an+1,n+m+3n+1)fi+∑s=i+1man+2,n+m+3n+s−i+2fs+an+i−1,n+1n+m+1x+an+i−1,n+1n+m+2y++an+i−1,n+1Z,n+m+34≤i≤m,φ(fi,f2)=(an+2,n+2n+2+(i−2)an+1Cn+2n+1)fi+an+2,n+2n+3fi+1,2≤i≤m,(fi,f3)=−((i−1)an,nn+m+2+an+2,n+1n+m+3)fi−(an+1,n+2n+1+an+2,n+1n+1)fi+1,2≤i≤m,φ(fi,fj)=−((i−1)an+j−1,n+1n+m+2+an+j−1,n+1n+m+3)fi+(an+j−2,n+1n+m+2−an+j−1,n+1n+1)fi+1++an+i,n+jn+i+j−1fi+j−1,4≤j≤m,2≤i≤m,φ(f1,y)=−∑j=1nan+m+2,n+1ses−an+m+2,n+1n+1f1+∑s=3m−1(s−2)an+1,n+1n+s+1fs+an+1,n+m+2n+mfm−−an+m+2,n+1n+m+1x−an+m+2,n+1n+m+2y−an+m+2,n+1Z,n+m+3φ(f2,y)=∑s=1nan+2,n+m+2ses+∑s=2man+2,n+m+2n+sfs+an+2,n+m+2n+m+1x−an+1,n+2n+1y+(an+1,n+2n+1−an+2,n+2n+2)z,φ(f3,y)=−2∑s=1n−1an+3,1s+1es+2an+2,n+1nen+(an+2,n+1n+1+an+1,n+2n+1)f1++(an+2,n+1n+2+an+1,n+m+2n+m+2+an+1,n+m+2n+m+3)f2+(an+2,n+m+2n+2−an+m+2,n+1n+1)f3++∑s=4m(an+2,n+m+2n+s−1−(s−3)an+2,n+1n+s)fs+2an+3,11x+2an+2,n+1n+m+2y+2an+2,n+1Z,n+m+3φ(fi,y)=−(i−1)∑s=1n−1an+i,1s+1es+(i−1)an+i−1,n+1nen+(i−2)(an+i−1,n+1n+1−an+i−2,n+1n+m+2)f1++∑s=2i−2(i−s)(∑t=1s−1an+i+t−s,n+1n+t+1)fs+(∑t=2i−1an+t,n+1n+t−(i−2)(i−1)2an+m+2,n+1n+m+2−(i−2)an+m+2,n+1n+m+3)fi−1++(an+2,n+m+2n+2−(i−2)an+m+2,n+1n+1)fi+∑s=i+1m(an+2,n+m+2n+s−i+2−(s−i)∑t=1i−2an+i−t,n+1n+s−t+1)fs++(i−1)an+i,11x+(i−1)an+i−1,n+1n+m+2y+(i−1)an+i−1,n+1n+m+3z,4≤i≤m,φ(y,y)=∑i=2m−2(i−1)(an+m+2,n+1n+i+1−an+1,n+1n+i+2)fi+((m−2)an+m+2,n+1n+m−an+1,n+m+2n+m)fm−1+an+m+2,n+m+2n+mfm.$

and the following restrictions $an+1,n+m+3n+i=an+1,n+1n+i+1,2≤i≤m−1,an+2,n+m+3n+m+2=−an+1,n+2n+1,an+2,n+m+3n+1=−an+2,n+2n+3,an+2,n+m+3n+m+3=an+1,n+2n+1−an+2,n+2n+2,an+i,n+2n+i+1=an+2,n+2n+3,3≤i≤m−1,an+i,n+3n+i+2=0,2≤i≤m−2.$

Analogously, considering the equation (1) for the bellow listed triples we obtain the corresponding relations $TriplesRelations__{z,z,f2},{f2,z,f2},{z,f2,a},⇒φ(z,f2)=0,wherea∈{x,z,e1,f1},{z,z,f1},{z,z,y}⇒(z,z)=∑i=2m−1an+m+3,n+1n+i+1fi+1m−1an+m+3,n+m+2n+mfm,φ(z,y)=an+m+3,n+m+2n+1f1+∑i=2m−1(i−1)an+m+3,n+1n+i+1fi+an+m+3,n+m+2n+mfm;{e1,en,e1}⇒an,1n+m+1=0,an,11=an−1,1n+m+1;{f1,e1,y}⇒a1,n+m+2n+m+2=a1,n+m+2n+m+3=0,an+m+2,n+1n+m+1=−an+1,1,1an+m+2,n+1s=an+1,1,s+11≤s≤n−1;{fi,e1,y},2≤i≤m−1⇒an+2,n+m+2n+m+1=an+2,1,1an+2,n+m+2s=−an+2,1,s+11≤s≤n−1,a1,n+m+2n+1=an+2,1n+3,an+i,1n+s=0,s∈[2;i−1]∪[i+2;m],an+i,1n+i+1=an+2,1n+3,2≤i≤m−1;{x,e1,y}⇒a1,n+m+2n+m+1=0,a1,n+m+22=an+m+2,n+m+11,a1,n+m+2s=0,3≤s≤n,a1,n+m+2n+s=(s−1)an+m+1,1n+s,2≤s≤m,an+m+1,1n+1=an+2,1n+3.$

Now, considering the equations $\begin{array}{}\left({d}^{2}\phi \right)\left(x,x,{e}_{1}\right)=\left({d}^{2}\phi \right)\left(x,{e}_{1},{f}_{1}\right)=\left({d}^{2}\phi \right)\left({f}_{1},x,{e}_{1}\right)=\left({d}^{2}\phi \right)\left({e}_{1},{f}_{1},y\right)=\left({d}^{2}\phi \right)\left(y,{f}_{1},x\right)=\left({d}^{2}\phi \right)\left(y,x,{f}_{1}\right)=\left({d}^{2}\phi \right)\left({f}_{2},x,{e}_{1}\right)=\left({d}^{2}\phi \right)\left({f}_{i},x,y\right)=0,\end{array}$ for 2 ≤ im and we can derive $φ(x,x)=∑i=2n−1(ian+m+1,1i+1−a1,n+m+1i+1)ei+an+m+1,n+m+1nen,φ(e1,x)=−an+m+1,11e1+an+m+1,12e2+∑i=3na1,n+m+1iei−an+2,1n+3f1−∑i=2man+m+1,1n+ifi−−an+m+1,1n+m+1x−an+m+1,1n+m+2y−an+m+1,1n+m+3z,φ(e1,f1)=−an+1,11e1+an+1,12e2−an+1,1n+1f1+∑i=3man+m+1,1n+i−1fi,φ(f1,x)=∑i=1n−1ian+1,1i+1ei+nan+m+2,n+1nen−an+m+1,n+1n+1f1,φ(x,e1)=∑i=1nan+m+1,1iei+an+2,1n+3f1+∑i=2man+m+1,1n+ifi+an+m+1,1n+m+1x+an+1,1n+1y+(an+2,1n+2−an+1,1n+1)z,φ(f2,x)=∑i=1n−1ian+2,1i+1ei+an+2,n+m+1nen+an+2,n+m+1n+2f2+an+2,n+m+1n+3f3,φ(fi,x)=∑s=1n−1san+i,1s+1es−nan+i−1,n+1nen+an+i,n+m+1n+ifi+an+2,n+m+1n+3fi+1,3≤i≤m.$

and the following restrictions: $an+m+1,1n+m+2=an=1,1n+1an+2,m+n+2n=−1nan+2,n+m+1n,an+m+1,n+m+2n+1=an+2,n+m+1n+3.$

Analogously, applying the same arguments: ${x,y,f1},{y,x,y}⇒φ(x,y)=−an+m+2,n+m+11e1+an+2,n+m+1n+3f1+∑i=2m−1(i−1)an+m+1,n+1n+i+1fi++an+m+1,n+m+2n+mfm,an+m+2,n+m+1n+1=−an+2,n+m+1n+3;{y,e1,y}⇒an+m+2,1n+1=−an+2,1n+3;{f1,e1,z},{y,z,f1},{f1,z,y}⇒φ(f1,z)=−an+m+3,n+1n+1f1+∑s=2m−1an+1,n+1n+s+1fs+1m−2an+1,n+m+2n+mfm,an+m+2,n+m+3n+i=an+m+2,n+1n+i+1−an+1,n+1n+i+2,2≤i≤m−2,an+m+2,n+m+3n+m−1=an+m+2,n+1n+m−an+1,n+m+3n+m,a1,n+m+3n+m+2=0,an+m+2,n+1n+2=an+1,n+1n+3;{e1,f1,z}⇒a1,n+m+3n+i=an+m+1,1n+i,2≤i≤m−1;{x,z,f1}⇒an+m+1,n+m+3n+i=an+m+1,n+1n+i+1,2≤i≤m−1;{x,z,y}⇒an+m+1,n+m+3n+m=1m−1an+m+1,n+m+2n+m;{f1,ei,e1},2≤i≤n⇒φ(f1,ei)=−ai−1,1n+m+2f1,3≤i≤n,an,1n+m+2=0;{f2,e2,y}⇒a2,n+m+2n+1=−a1,1n+1,{f2,e1,z},{f2,z,x},{f2,z,y}⇒a1,n+m+3n+1=a1,n+m+3n+m+3=0,φ(f2,z)=−∑s=1n−1an+2,1s+1es−1nan+2,n+m+1nen−an+2,n+2n+3f1+an+2,n+m+3n+2f2++an+m+3,n+m+2n+1f3+an+1,11x−an+1,n+2n+1y+(an+1,n+2n+1−an+2,n+2n+2)z.$

To complete the proof, we use the following equations in order to obtain some expressions for 2-cocycle φ and some restrictions. $(d2φ)(fj,e1,f1)=(d2φ)(fj,ei,e1)=0,2≤j≤m,2≤i≤n,⇒φ(fj,ei),2≤j≤m,1≤i≤n,(d2φ)(ei,fj,f1)=(d2φ)(ei,f2,x)=02≤i≤n,2≤j≤m−1,(d2φ)(ei,f2,y)=01≤i≤n,(d2φ)(e1,fi,x)=(d2φ)(x,fi,e1)=03≤i≤m,⇒φ(ei,fj),1≤i≤n,2≤j≤m,(d2φ)(x,fi,x)=(d2φ)(fi,y,fj)=0,2≤i,j≤m⇒φ(x,fi),φ(y,fi),2≤i≤m,an+i,n+jn+i+j−1=0,2≤i≤m,4≤j≤m,(d2φ)(ei,f1,e1)=0,1≤i≤n,⇒φ(ei,f1),2≤i≤n,an,1n+s=0,2≤s≤m−1.$

Similarly, we get ${y,x,e1}⇒an+m+2,n+m+1i=ian+m+2,1i+1,1≤i≤n−1;{z,x,e1}⇒an+m+3,n+m+1i=ian+m+3,1i+1,1≤i≤n−1;{y,z,y}⇒an+m+2,n+m+3n+m=1m−1an+m+2,n+m+2n+m;{ei,z,x},1≤i≤n⇒a1,n+m+3n+m=an+m+1,1n+m,a1,n+m+32=an+m+3,12,a1,n+m+3i=0,3≤i≤n;ai,n+m+1n+s=iai−1,1n+s,2≤s≤m,2≤i≤n;{x,y,e1}⇒a1,n+m+21=−an+m+2,11;{x,z,e1}⇒a1,n+m+31=−an+m+3,11.$

Given the restrictions above, by applying the multiplication of the algebra L and checking the general condition of cocycle for the other basis elements, we get a general form of the 2-cocycle φ. $(6):φ(ei,e1)=∑s=1nai,1ses+∑s=1mai,1n+sfs+ai,1n+m+1x+ai,1n+m+2y+ai,1n+m+3z,1≤i≤n−1,φ(en,e1)=an−1,1n+m+1e1+∑i=2n−1(an−i,1n+m+1−∑s=1i−1an−i+s,1s)ei+an,1nen,φ(f1,e1)=∑s=1nan+1,1ses+an+1,1n+1f1,φ(fi,e1)=∑s=1nan+i,1ses+(an+2,1n+2+(i−2)an+1,1n+1)fi+an+2,1n+3fi+1,2≤i≤m,φ(x,e1)=∑i=1nan+m+1,1iei+an+2,1n+3f1+∑i=2man+m+1,1n+ifi+(2(n+2)(n−1)∑i=1nai,1i)x+an+1,1n+1y++(an+2,1n+2−an+1,1n+1)z,φ(y,e1)=∑i=1nan+m+2,1iei−an+2,1n+3f1,φ(z,e1)=∑i=1nan+m+3,1iei,φ(ei,e2)=−ia1,1n+m+1ei−a1,11ei+1,1≤i≤n,φ(ei,ej)=−iaj−1,1n+m+1ei+(aj−2,1n+m+1−aj−1,11)ei+1,3≤j≤n,1≤i≤n,φ(f1,ei)=−ai−1,1n+m+2f1,2≤i≤n,φ(fi,ej)=−((i−1)aj−1,1n+m+2+aj−1,1n+m+3)fi−aj−1,1n+1fi+1,2≤j≤n,2≤i≤m,φ(x,e2)=a1,11e1,φ(x,ei)=(ai−1,11−ai−2,1n+m+1)e1,3≤i≤n,φ(y,ei)=ai−1,1n+1f1,2≤i≤n,φ(e1,f1)=−an+1,11e1+an+1,12e2−an+1,1n+1f1+∑i=3man+m+1,1n+i−1fi,φ(ei,f1)=−ian+1,11ei+an+1,12ei+1+ai−1,1n+m+2f1−∑s=3mai−1,1n+s−1fs,2≤i≤n,φ(f1,f1)=∑i=3man+1,n+1n+ifi,φ(fi,f1)=−∑s=1n−1an+i+1,1s+1es+an+i,n+1nen+∑s=1man+i,n+1n+sfs+an+i+1,11x+an+i,n+1n+m+2y++an+i,n+1n+m+3z,2≤i≤m−1,φ(fm,f1)=an+m−1,n+1n+m+2f1−∑i=3m−1(∑s=1i−2an+m−s,n+1n+i−s)fi+an+m,n+1n+mfm,φ(x,f1)=−an+1,12e1+an+m+1,n+1n+1f1+∑i=3man+m+1,n+1n+ifi,φ(y,f1)=∑i=1n−1an+1,1i+1ei+an+m+2,n+1nen+an+m+2,n+1n+1f1+an+1,n+1n+3f2+∑i=3man+m+2,n+1n+ifi−an+1,11x+an+m+2,n+1n+m+2y+1m−1(∑i=2man+i,n+1n+i−m(m−1)2an+m+2,n+1n+m+2)z,φ(z,f1)=an+m+3,n+1n+1f1+∑i=3man+m+3,n+1n+ifi,φ(ei,fj)=−ian+j,11ei+an+j,12ei+1,2≤j≤m,1≤i≤n,φ(f1,f2)=an+1,n+2n+1f1,φ(fi,f2)=((i−2)an+1,n+2n+1+an+2,n+2n+2)fi+an+2,n+2n+3fi+1,2≤i≤m,φ(f1,fi)=−an+i−1,n+1m+n+2f1,3≤i≤m,φ(fi,f3)=−((i−1)an+2,n+1n+m+2+an+2,n+1n+m+3)fi−(an+1,n+2n+1+an+2,n+1n+1)fi+1,2≤i≤m,φ(fi,fj)=−((i−1)an+j−1,n+1n+m+2+an+j−1,n+1n+m+3)fi+(an+j−2,n+1n+m+2−an+j−1,n+1n+1)fi+1,4≤j≤m,2≤i≤m,φ(x,fi)=−an+i,12e1,2≤i≤m,φ(y,f2)=−an+2,n+2n+3f1,φ(y,f3)=(an+1,n+2n+1+an+2,n+1n+1)f1,φ(y,fi)=(an+i−1,n+1n+1−an+i−2,n+1n+m+2)f1,4≤i≤m,φ(e1,x)=−an+m+1,11e1+an+m+1,12e2+∑i=3na1,n+m+1iei−an+2,1n+3f1−∑i=2man+m+1,1n+ifi−$

$(7):−(2(n+2)(n−1)∑i=1nai,1i)x−an+1,1n+1y+(an+1,1n+1−an+2,1n+2)z,φ(e2,x)=a1,11e1−2an+m+1,11e2+(an+m+1,12−a1,13)e3+∑s=4n(a1,n+m+1s−1−(s−2)a1,1s)es+2∑s=1ma1,1n+sfs++2a1,1n+m+1x+2a1,1n+m+2y+2a1,1n+m+3z,φ(ei,x)=(i−1)(ai−1,11−ai−2,1n+m+1)e1+∑s=2i−2(i−s)(∑t=1sai−s+t−1,1t−ai−s−1,1n+m+1)es++((n+i)(n−i+1)(n+2)(n−1)∑t=1i−1at,1t−(i+1)(i−2)(n+2)(n−1)∑t=inat,1t)ei−1−ian+m+1,11ei++(an+m+1,12−∑t=1i−1at,1t+2)ei+1+∑s=i+2n(a1,n+m+1s−i+1−(s−i)∑t=1i−1at,1s−i+t+1)es+i∑s=1mai−1,1n+sfs++iai−1,1n+m+1x+iai−1,1n+m+2y+iai−1,1n+m+3z,3≤i≤n,φ(f1,x)=∑i=1n−1ian+1,1i+1ei+nan+m+2,n+1nen−an+m+1,n+1n+1f1,φ(f2,x)=∑i=1n−1ian+2,1i+1ei+an+2,n+m+1nen+an+2,n+m+1n+2f2+an+2,n+m+1n+3f3,φ(fi,x)=∑s=1n−1san+i,1s+1es−nan+i−1,n+1nen+(an+2,n+m+1n+2−(i−2)an+m+1,n+1n+1)fi+an+2,n+m+1n+3fi+1,3≤i≤m,φ(x,x)=∑i=2n−1(ian+m+1,1i+1−a1,n+m+1i+1)ei+an+m+1,n+m+1nen,φ(y,x)=∑i=1n−1ian+m+2,1i+1ei+an+m+2,n+m+1nen−an+2,n+m+1n+3f1,φ(z,x)=∑i=1n−1ian+m+3,1i+1ei+an+m+3,n+m+1nen,φ(e1,y)=−an+m+2,11e1+an+m+2,12e2+an+2,1n+3f1+∑i=2m(i−1)an+m+1,1n+ifi,φ(ei,y)=−ian+m+2,11ei+an+m+2,12ei+1−ai−1,1n+1f1−∑s=2m(s−1)ai−1,1n+sfs,2≤i≤n,φ(f1,y)=−∑i=1n−1an+1,1s+1es−an+m+2,n+1nen−an+m+2,n+1n+1f1+∑s=3m−1(s−2)an+1,n+1n+s+1fs+an+1,n+m+2n+mfm++an+1,11x−an+m+2,n+1n+m+2y−1m−1(∑t=2man+t,n+1n+t−m(m−1)2an+m+2,n+1n+m+2)z,φ(f2,y)=−∑s=1n−1an+2,1s+1es−1nan+2,n+m+1nen+∑s=2man+2,n+m+2n+sfs+an+2,11x−an+1,n+2n+1y++(an+1,n+2n+1−an+2,n+2n+2)z,φ(f3,y)=−2∑s=1n−1an+3,1s+1es+2an+2,n+1nen+(an+2,n+1n+1+an+1,n+2n+1)f1++1m−1((m−2)an+2,n+1n+2−∑t=3man+t,n+1n+t+(m−1)(m−2)2an+m+2,n+1n+m+2)f2++(an+2,n+m+2n+2−an+m+2,n+1n+1)f3+∑s=4m(an+2,n+m+2n+s−1−(s−3)an+2,n+1n+s)fs++2an+3,11x+2an+2,n+1n+m+2y+2an+2,n+1n+m+3z,φ(fi,y)=−(i−1)∑s=1n−1an+i,1s+1es+(i−1)an+i−1,n+1nen+(i−2)(an+i−1,n+1n+1−an+i−2,n+1n+m+2)f1++∑s=2i−2(i−s)(∑t=1s−1an+i+t−s,n+1n+t+1)fs++(m−i+1m−1∑t=2i−1an+t,n+1n+t−i−2m−1∑t=iman+t,n+1n+t+(i−2)(m−i+1)2an+m+2,n+1n+m+2)fi−1++(an+2,n+m+2n+2−(i−2)an+m+2,n+1n+1)fi+∑s=i+1m(an+2,n+m+2n+s−i+2−(s−i)∑t=1i−2an+i−t,n+1n+s−t+1)fs++(i−1)an+i,11x+(i−1)an+i−1,n+1n+m+2y+(i−1)an+i−1,n+1n+m+3z,4≤i≤m,φ(x,y)=−an+m+2,12e1+an+2,n+m+1n+3f1+∑i=2m−1(i−1)an+m+1,n+1n+i+1fi+an+m+1,n+m+2n+mfm,φ(y,y)=∑i=2m−2(i−1)(an+m+2,n+1n+i+1−an+1,n+1n+i+2)fi+((m−2)an+m+2,n+1n+m−an+1,n+m+2n+m)fm−1++an+m+2,n+m+2n+mfm,$

$(8):φ(z,y)=an+m+3,n+m+2n+1f1+∑i=2m−1(i−1)an+m+3,n+1n+i+1fi+an+m+3,n+m+2n+mfm,φ(e1,z)=−an+m+3,11e1+an+m+3,12e2+∑i=2man+m+1,1n+ifi,φ(ei,z)=−ian+m+3,11ei+an+m+3,12ei+1−∑s=2mai−1,1n+sfs,2≤i≤n,φ(f1,z)=−an+m+3,n+1n+1f1+∑s=2m−1an+1,n+1n+s+1fs+1m−2an+1,n+m+2n+mfm,φ(f2,z)=−∑s=1n−1an+2,1s+1es−1nan+2,n+m+1nen−an+2,n+2n+3f1+an+2,n+m+3n+2f2++an+m+3,n+m+2n+1f3+an+1,11x−an+1,n+2n+1y+(an+1,n+2n+1−an+2,n+2n+2)z,φ(f3,z)=−∑s=1n−1an+3,1s+1es+an+2,n+1nen+(an+2,n+1n+1+an+1,n+2n+1)f1++(an+2,n+m+3n+2−an+m+3,n+1n+1)f3+an+m+3,n+m+2n+1f4++an+3,11x+an+2,n+1n+m+2y+an+2,n+1n+m+3z,φ(fi,z)=−∑s=1n−1an+i,1s+1es+an+i−1,n+1nen+(an+i−1,n+1n+1−an+i−2,n+1n+m+2)f1++(an+2,n+m+3n+2−(i−2)an+m+3,n+1n+1)fi+an+m+3,n+m+2n+1fi+1++an+i,11x+an+i−1,n+1n+m+2y+an+i−1,n+1n+m+3z,4≤i≤m,φ(x,z)=−an+m+3,12e1+∑i=2m−1an+m+1,n+1n+i+1fi+1m−1an+m+1,n+m+2n+mfm,φ(y,z)=−an+m+3,n+m+2n+1f1+∑i=2m−2(an+m+2,n+1n+i+1−an+1,n+1n+i+2)fi++(an+m+2,n+1n+m−1m−2an+1,n+m+2n+m)fm−1+1m−1an+m+2,n+m+2n+mfm,φ(z,z)=∑i=2m−1an+m+3,n+1n+i+1fi+1m−1an+m+3,n+m+2n+mfm.$

Taking into account the expressions (6), (7) and (8), we derive that the dimension of ZL2(L, L) is equal to (m + n + 3)2−5. □

Based on Proposition 4.2, we have the following corollary.

#### Corollary 4.3

dim HL2(L, L) = 0.

Thus, according to the results of the paper [20], we derive the following theorem.

#### Theorem 4.4

The algebra L is rigid.

## Acknowledgement

The authors were supported by Ministerio de Economía y Competitividad (Spain), grant MTM2016-79661-P (European FEDER support included) and Ministry of Education and Science of the Republic of Kazakhstan, grant No. 0828/GF4. The last author was supported by Fundamental Research Grant FRGS/1/2016/STG06/UNIKL/02/1 from Malaysia.

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Accepted: 2017-10-04

Published Online: 2017-12-02

Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 1371–1388, ISSN (Online) 2391-5455,

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