# On Some Structural Components of Nilsolitons

Authors: Hulya Kadioglu
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this paper, we study nilpotent Lie algebras that admit nilsoliton metric with simple pre-Einstein derivation. Given a Lie algebraη, we would like to compute as much of its structure as possible. The structural components we consider in this study are the structure constants, the index, and the rank of the nilsoliton derivations. For this purpose, we prove necessary or sufficient conditions for an algebra to admit such metrics. Particularly, we prove theorems for the computation of the Jacobi identity for a given algebra so that we can solve the system of the equation(s) and find the structure constants of the nilsoliton.


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## Body

## 1. Introduction

On any Lie group, it is possible to define several different Riemannian metrics. Considering any Riemannian metrics, Einstein metrics are the most preferable metrics, as the Ricci tensor complies the Einstein metric:Ric=cg, for some constant c∈ℝ. But, it is not possible to define Einstein metrics on nonabelian nilpotent Lie algebras; therefore, we consider the following weaker condition on a left invariant metric g on a nilpotent Lie group G:(1)Ricg=βI+D,for some β∈ℝ and D∈Derη, where Ricg denotes the Ricci operator of η,g, η is the Lie algebra of G, and Derη denotes the Lie algebra of derivations of η. Equation (1) is called the nilsoliton condition, D is called the nilsoliton derivation, and β is called the nilsoliton constant.Nilsoliton metric Lie algebras are unique up to isometry and scaling. This is one of the reasons that makes nilsolitons an important topic. On the other hand, there is a one to one correspondence between Einstein nilradicals with nilsolitons. In [1], Theorem 2.11 states that a nilpotent Lie algebra η is an Einstein nilradical if and only if η admits a nilsoliton metric. Therefore, it indicates that classification of nilsoliton metrics on a nilpotent Lie algebra is equivalent to the same of Einstein nilradicals. Additionally, if δ is Einstein solvmanifold, then the metric restricted to the submanifold can completely be determined by the nilsoliton metric Lie algebra η=δ,δ. On the contrary, any nilsoliton can uniquely extend to an Einstein solvmanifold. Therefore, the study of solvmanifolds is actually the study of nilsolitons. See [1, 2], for a survey on nilsoliton metric Lie algebras.There are three methods to represent a Lie algebra and its related structures: representing a Lie algebra as a linear Lie algebra, i.e., subalgebra ofgln, using table of its structure constants or using generators and relations [3]. In this paper, we use the table of structure constants related to a Lie algebra. The main reason is that this representation helps one to create and classify Lie algebras by the computer software programs as in [4–6]. We vary the Lie algebra structure by finding structure constants. Namely, we determine a Lie algebra η with a fixed basis Xi:1≤i≤n explicitly by given multiplication table, consisting of structure constants αijk which are defined by the relations(2)Xi,Xj=∑αijkXk.In this work, the nonzero structure constants are encoded by using the index setΛ=i,j,k|αijk≠0,i<j<k, ignoring repetitions due to skew-symmetry. While indexing the structure constants, we use triples i,j,k∈Λ such that i<j<k, and if i,j,k,i,j,m∈Λ, then k=m and i,j1,k,i,j2,k∈Λ, then j1=j2. For this purpose, we fix a basis X1,…,Xn for a nilpotent Lie algebra η with Xi,Xj=∑αijkXk≠0 such that for every i,j,#k:αijk≠0≤1, and for every i,k,#j:αijk≠0≤1. Such basis Xj is called nice and defined by Nikolayevsky in [7]. In our paper, we call this basis as “Nikolayevsky basis.” Using this special basis, we do not need to use sum symbol in equation (2).In this paper, we prove theorems regarding to some structural components of nilsolitons for possible algorithmic approach in classifications. In particular, we present some new concepts and theorems regarding to create Jacobi identity equations for a given nilsoliton metric Lie algebra. We also present methods for the computations of the index and the rank of a nilsoliton. We provide several examples to illustrate the newly proposed concepts and methods.This paper consists of four sections. In the second section, we present preliminary background. In the third section, we prove necessary theorems that help us to calculate structural elements of nilsolitons. In the last section, we present concluding statements.


## 2. Preliminaries

Letημ,<,> be a metric algebra, where μ∈Λ2η⊗η∗. Let B=Xii=1n be a ,-orthonormal basis of ημ (we always assume that basis is ordered). The nil-Ricci endomorphism Ricμ is defined as RicμX,Y>=ricμX,Y, where(3)ricμX,Y=−12∑i=1nX,Xi,Y,Xi+14∑i=1nXi,Xj,XXi,Xj,Y,for X,Y∈η. When η is a nilpotent Lie algebra, the nil-Ricci endomorphism is the Ricci endomorphism. If all elements of the basis are eigenvectors for the nil-Ricci endomorphism Ricμ, we call the orthonormal basis a Ricci eigenvector basis.LetDerη denote the derivation algebra of η. A maximal abelian subalgebra of Derη comprised of semisimple elements which is called a maximal torus. The dimension of a maximal torus is called the rank of η.In the following, we define root vectors and root matrix, Gram matrix, Nikolayevsky basis.Definition 1.
Suppose thatΛ=i,j,k|αijk≠0,i<j<k is a finite set which indexes the set of nonzero structure constants corresponding to a Lie algebra η, ignoring repetitions due to skew-symmetry. For 1≤i,j,k≤n, we define 1×n row vector yijk to be ϵiT+ϵjT−ϵkt, where ϵi=1n is the standard orthonormal basis for ℝn. We call the vectors in yijk|i,j,k∈Λ root vectors for Λ. Let y1,y2,…,ym (where m=Λ) be an enumeration of the root vectors in dictionary order. We define root matrix YΛ for Λ to be the m×n matrix whose rows are the root vectors y1,y2,…,ym.
The Gram matrixUΛ for Λ is the m×m matrix defined by UΛ=YΛYΛT; the i,j entry of UΛ is the inner product of the i th and j th root vectors (unless otherwise stated, the matrix U means the Gram matrix corresponding to the index set Λ. Therefore, from now on, we do not use UΛ). From Theorem 5, in [8], we know that U is a symmetric matrix where its all diagonal entries are 3 and its off-diagonal entries are in the set −2,−1,0,1,2.
Nikolayevsky showed that every Lie algebra admitting a derivation with all the eigenvalues of multiplicity one has a nice basis [7]. We use this type of basis in our study. This way our Gram matrices corresponding to metric nilpotent Lie algebras does not have a 2 and −2 as an entree (Lemma 2 in [4]).
Now, suppose thatΛ=m and 1m represents a column vector 111…1T in ℝm.Theorem 1.
(Theorem 1 in [8]). Let η be a nonabelian metric algebra with Ricci eigenvector basis B. Let U and α2 be the Gram matrix and the structure vector for η with respect to B. Then, η satisfies the nilsoliton condition with nilsoliton constant β if and only if Uα2=2β1m.
The above theorem indicates a Lie algebraη which admits a nilsoliton metric iff there exists a solution v∈ℝm of the linear system Uv=1m where all entries are positive real numbers.


## 3. The Structural Elements

In this section, we introduce notions and prove theorems for the computation structural elements of the nilsoliton metric Lie algebras.


### 3.1. The Jacobi Identity(s)

Now, we present the theorems which help to create possible Jacobi identity equations.Theorem 2.
(see [8]). Let η be an n-dimensional vector space; B=Xii=1n be a basis for η. Suppose that a set of nonzero structure constants αi,jk relative to B, indexed by Λ, defines a skew symmetric product on η. Assume that if i,j,k∈Λ, then i<j<k. Then, η is a Lie algebra if and only if whenever there exists m so that the inner product of root vectors yijl,ylkm=−1 for triples i,j,l and l,k,m or k,l,m in Λ, the equation(4)∑s<mαi,jsαs,km+αj,ksαs,im+αk,isαs,jm=0,holds. Furthermore, a term of form αi,jlαl,km is nonzero if and only if yi,jl,yl,km=−1.Lemma 1.
(Lemma 2.8 in [4]). Let η be an n-dimensional nonabelian nilpotent Lie algebra. Suppose that η admits a derivation D having distinct real positive eigenvalues. Let B be a basis consisting of eigenvectors for the derivation D, and let Λ index the nonzero structure constants with respect to B. Let Y be the m×n root matrix for Λ. If rankY=m, then the following hold:(1)
Λ≤n−1.(2)
Ifi1,j1,k1∈Λ and i2,j2,k2∈Λ, then yi1j1k1,yi2j2k2≠−1.Remark 1.
The reverse of Lemma1 is not true.Proof.
Suppose thatη is an algebra which is indexed by the following index set:(5)Λ=1,2,5,1,3,6,1,5,7,1,6,8,2,4,7,2,5,8,3,4,8.
Its corresponding Gram matrix is as follows:(6)3101100131000101311101013011101031100111310101113.
It is a singular matrix withnullity=1. Also, it does not have −1 entry. From Theorem 2, it is a Lie algebra. Also, since the solution space of U.v=1 is(7)v=t+119,619−t,t,519−t,519−t,319,t|0<t<519,from [8] that it is a nilsoliton metric Lie algebra with the magnitudes of the structure constants(8)α1,25=t+119,α1,36=619−t,α1,57=t,α1,68=519−t,α2,47=519−t,α2,58=319α3,48=t,with simple derivation of type 7<10<13<14<17<20<24<27 with singular Gram matrix U and Λ=n−1.The counter example provided in Remark1 illustrates none of the inner products of root vectors being −1 does not imply that the Gram matrix of the nilsoliton is nonsingular. Additionally, if the cardinality of the index set Λ≤n, −1 does not imply that the Gram matrix is nonsingular.


#### 3.1.1. Constructing the System of Jacobi Identities

In order an algebra to be a Lie algebra, one needs to satisfy the Jacobi identity. Using our index setΛ, the corresponding Jacobi identity turns into equation (4). Also, there has to be at least two product couples in the Jacobi identity. Otherwise, if there is one product couple in Jacobi identity, it leads to αi,jsαs,km=0; therefore, it contradicts to the fact that i,j,s,s,k,m∈Λ. On the other hand, the Jacobi identity is created by each vector triples from the given fixed Nikolayevsky basis. For example, the Jacobi identity for Xi,Xj,Xk is(9)αijsαskm+αjksαsim+αkisαsjm=0.Since we use the Nikolayevsky basis, there is a uniquek which appears in equation (9). Therefore, each Jacobi identity is created with fixed Xi,Xj,Xk, and m∈AoM. Also, because there has to be at least one product couple in the Jacobi identity, then there exist 2 or three product couples in the Jacobi identity. In equation (9), the product couple αijsαskm corresponds to the index triples in Λ such that i,j,s,s,k,m∈Λ, or i,j,s,k,s,m∈Λ. In the following definition, we define the set of all product couples Pm related to the Jacobi identity for a given subset Xi,Xj,Xk of the fixed basis. As one can see, there exist at most 3 product couples in the same Jacobi identity.Definition 2.
If there aret product couples in the Jacobi identity for the same m such that i,j,s,s,k,m∈Λ, or i,j,s,k,s,m∈Λ, then the Jacobi identity which was created by the basis vectorsXi,Xj,Xk. Now, we define the set of all nonzero product couples for a given index triple basis vectors Xi,Xj,Xk as follows:(10)Pm=ps=αijsαs,km,1≤s≤n,ps≠0.
It is called set of product couples for#m and for in equation (4):(11)AoM=m∈ℕ:i,j,s,s,k,m∈Λori,j,s,k,s,m∈Λ.
Here, AoM is the matrix of#m that appears in equation (4).Remark 2.
Letη be an algebra that is indexed by Λ, then the Jacobi identity for a given Xi,Xj,Xk with each m∈AoM is given by(12)p1∓p2∓p3=0.
The proof of Remark2 follows from the definition of Jacobi identity for a given algebra.Theorem 3.
Letη be an algebra that is indexed by Λ, and U be the Gram matrix related to Λ. Let vp belong to the solution space to the linear system Uv=1, then the Jacobi identity for each m∈AoM is given by(13)vi1vj1∓vi2vj2∓vi3vj3=0,where 1≤i1,i2,i3,j1,j3≤Λ.Proof.
Letη be an algebra that is indexed by Λ. By Remark 3, the Jacobi identity for each m∈AoM and each basis vector triple Xi,Xj,Xk is given by(14)p1∓p2∓p3=0.
Here,ps=αijsαs,km. By Theorem 1, the solution vectors of the linear system Uv=1 are the squares of the structure vectors αijs for all i,j,s∈Λ. Therefore, for any v that satisfies Uv=1 is of form v=αi,js2. Thus, for each ps∈Pm,(15)ps=vavb,where a,b∈i1,i2,i3,j1,j3, and va=αijs2 and vb=αskm2. Solving the equations for the structure constants, we have(16)αijs=∓va,for all i,j,s∈Λ. Therefore, equation (12) turns into(17)vi1vj1∓vi2vj2∓vi3vj3=0,which finishes the proof.In some cases, there can be more than two square root product couples in equation (13), i.e., there is more than one m∈AoM. In that case, we need to consider all the cases of the signs between the product couples αi,jsαs,km, αj,ksαs,im, and αk,isαs,jm. The following lemma deals with this matter.Lemma 2.
If there arek product couples in the Jacobi identity, then there are 2k−1−1 possible sign choices.Proof.
Suppose thatpi∈Pm for m in equation (4) where Pm is the set of all product couples, as in Definition 2. Therefore, the Jacobi identity turns into p1∓p2∓p3=0. Without loss of generality, we assume that p1>0. For each ps∈Pm, where 2≤s≤n, there are two possible sign choices +,−. Therefore, we have 2t−1 possible sign choices. Since all the product couples are nonzero, then they can not all be +. Therefore, we drop the case +,+,…,+. Thus, there are 2t−1−1 possible sign choices for the set Pm.
As we know that there is at most three, at least two product couples in the Jacobi identity, the there is at most three different sign choices for a Jacobi identity.Definition 3.
We define the matrix of all sign choicesSC=sjs∈M2t−1−1×t−1 matrix where the entries are of the form(18)sjs=1,ps>0,0,ps<0,for 1≤j≤2t−1−1. For example, if Pm=p1,p2,p3, then SC is a 3×2 matrix(19)SC=100100.
The equations are created by(20)p1+−1SC1,1.p2+−1SC1,2.p3=0,p1+−1SC2,1.p2+−1SC2,2.p3=0,p1+−1SC3,1.p2+−1SC3,2.p3=0.
Therefore, we havep1+p2−p3=0, p1−p2+p3=0, and p1−p2−p3=0. If there is more than one array of number m’s in equation (4) in Theorem 2, then one needs to find common solutions of at least one Jacobi identity that was created by SC matrix for each m. To illustrate the above procedure, we have the following example.Example 1.
Suppose that(21)Λ=1,3,4,1,4,5,1,5,6,2,3,6,1,6,7,2,4,7,1,7,8,2,5,8.
The solution of the equationUv=1 is of the form(22)v=x+y−1/51/55+x1/1131/55−x−y32/55−x−yy4/11−xxT,=α1242α1352α1672α1782α2562α2682α3462α4582T,where x,y are parameters of the solution system. Then, there are two m’s in equation (4). Now, let us denote the set of m as AoM. Therefore, AoM=7,8. Each m∈AoM corresponds to a Jacobi identity. For m=7, we have two product couples of the triples i,j,l and l,k,m or k,l,m in Λ. They are 2,3,61,6,7 and 1,3,42,4,7. Therefore, the first Jacobi identity is(23)α236α167±α134α247=0.
Form=8, we have two product couples of the triples i,j,l and l,k,m or k,l,m in Λ. They are 2,4,71,7,8 and 1,4,52,5,8. So, the corresponding Jacobi identity is(24)α247α178±α145α258=0.
It can easily be seen that there is a unique sign choice for the above equations. The common solution is as follows:(25)α134=2/11,α145=1/5,α156=1/11,α236=2/5,α167=1/5,α247=1/5,α178=2/11,α258=2/11.


### 3.2. The Rank and the Index of a Nilsoliton

The following proposition and its following corollary help us to compute rank of a nilsoliton metric Lie algebra.Proposition 1.
(see Proposition 4.7 in [9]). Let η be a nonabelian Lie algebra that admits a simple derivation D. Let B=Xii=1n be an eigenvector basis with index set Λ, and let Y be the root matrix associated to Λ. Then, the rank of the nilsoliton metric Lie algebra η equals to the nullity of the root matrix Y.Corollary 1.
Letη be a nonabelian n−dimensional Lie algebra that admits a simple derivation. Let B be an eigenvector basis with index set Λ, and let Y be the root matrix associated to Λ. Then,(26)rankη=n+nullityYYt−Λ.Proof.
Suppose thatη is an n−dimensional nonabelian Lie algebra, admitting a simple derivation. Let B be an eigenvector basis with index set Λ where Λ=m, and let Y be the root matrix associated to Λ. By Proposition 1, the root matrix Y is an m×n matrix, whose nullity is rankη. Therefore, from rank-nullity theorem, rankη=n−RankY. We also know that rankY=rankYYT; therefore, we have(27)rankη=n−RankYYT.
On the other hand,YYT is an m×m matrix. Therefore, RankYYT=m−NullityYYT. Then, we have(28)rankη=n−m−NullityYYT=n+NullityYYT−Λ.Definition 4.
LetX∈η, adX, denote the adjoint representation and η∗ denote the dual of the Lie algebra η. Then, the skew symmetric bilinear form Ψf, where f∈η∗, is defined by(29)Ψf:η×η⟶ℝX,Y⟶ΨfX,Y=fX,Y.
The index of a Lie algebraη is the integer infdimηf:f∈η∗ where ηf=kerΨf defined by ηf=X∈η:fX,Y=0,∀y∈η.Proposition 2.
(see Proposition 4 in [10]). The index of an n-dimensional Lie algebra η is the integer,(30)indexη=n−RankRηXi,Xj1≤i,j≤n,where Rη is the quotient field of symmetric algebra Sη.Remark 3.
above proposition tells us that the index of a Lie algebra is the nullity of the matrixEη=Xi,Xj.
Now, we present an example regarding to this notion with the use of the index setΛ.Example 2.
Suppose thatΛ=1,2,3,1,3,4,2,3,5 is the index set of an algebra η of dimension 6. Its corresponding Gram matrix is(31)300031013.
Since there is no−1 entry, η is a Lie algebra. The matrix Eη is as follows:(32)0X3X4000−X30X5000−X4−X50000000000000000.
Its rank is 2; therefore,indexη=4.Proposition 3.
For ann−dimensional nilsoliton represented by an index set Λ with a Nikolayevsky basis, the matrix Eη has the following properties:(1)
The matrix is ann×n matrix with zero diagonal entries(2)
The rank ofEη is even(3)
Thenth row and column of Eη are zero matrices(4)
IfΛ=K, then Eη has 2K nonzero entriesProof.
Suppose that the Nikolayevsky basis is represented byB=Xα1≤α≤n. By the definition of Eη, Eηi,j=Xi,Xj, then(33)Eηij=Xi,Xj=−Xj,Xi=−Eηji.
Therefore,Eη is a skew symmetric matrix, which implies the first and second properties. On the other hand, since η is nilpotent Lie algebra with Nikolayevsky basis, then Λ consists of the elements of form i,j,k such that i<j<k. Therefore, n cannot appear in the first or the second component of the triples in Λ. Thus, Xi,Xn=Xn,Xi=0 for all i∈1,2,…,n, which implies that Eηin=Eηni=0, i.e., the last row and the last column are zero matrices.
The entries ofEη are defined by the index set Λ. If Λ=K, then there exists K non-zero entries on the upper triangular part of Eη. Since the matrix is skew symmetric, then there exists other K nonzero entries in the lower triangular part of the matrix. So, in total, there exists 2K nonzero entries in Eη.Corollary 2.
Suppose thatη is an n−dimensional nilsoliton. Then, indexη is even, if n is even, and index is odd if n is odd.Proof.
The index of the nilsoliton isindexη=NullityEη=n−rankEη. Since Eη is a skew symmetric matrix, its rank is always even, which finishes the proof.


## 3.1. The Jacobi Identity(s)

Now, we present the theorems which help to create possible Jacobi identity equations.Theorem 2.
(see [8]). Let η be an n-dimensional vector space; B=Xii=1n be a basis for η. Suppose that a set of nonzero structure constants αi,jk relative to B, indexed by Λ, defines a skew symmetric product on η. Assume that if i,j,k∈Λ, then i<j<k. Then, η is a Lie algebra if and only if whenever there exists m so that the inner product of root vectors yijl,ylkm=−1 for triples i,j,l and l,k,m or k,l,m in Λ, the equation(4)∑s<mαi,jsαs,km+αj,ksαs,im+αk,isαs,jm=0,holds. Furthermore, a term of form αi,jlαl,km is nonzero if and only if yi,jl,yl,km=−1.Lemma 1.
(Lemma 2.8 in [4]). Let η be an n-dimensional nonabelian nilpotent Lie algebra. Suppose that η admits a derivation D having distinct real positive eigenvalues. Let B be a basis consisting of eigenvectors for the derivation D, and let Λ index the nonzero structure constants with respect to B. Let Y be the m×n root matrix for Λ. If rankY=m, then the following hold:(1)
Λ≤n−1.(2)
Ifi1,j1,k1∈Λ and i2,j2,k2∈Λ, then yi1j1k1,yi2j2k2≠−1.Remark 1.
The reverse of Lemma1 is not true.Proof.
Suppose thatη is an algebra which is indexed by the following index set:(5)Λ=1,2,5,1,3,6,1,5,7,1,6,8,2,4,7,2,5,8,3,4,8.
Its corresponding Gram matrix is as follows:(6)3101100131000101311101013011101031100111310101113.
It is a singular matrix withnullity=1. Also, it does not have −1 entry. From Theorem 2, it is a Lie algebra. Also, since the solution space of U.v=1 is(7)v=t+119,619−t,t,519−t,519−t,319,t|0<t<519,from [8] that it is a nilsoliton metric Lie algebra with the magnitudes of the structure constants(8)α1,25=t+119,α1,36=619−t,α1,57=t,α1,68=519−t,α2,47=519−t,α2,58=319α3,48=t,with simple derivation of type 7<10<13<14<17<20<24<27 with singular Gram matrix U and Λ=n−1.The counter example provided in Remark1 illustrates none of the inner products of root vectors being −1 does not imply that the Gram matrix of the nilsoliton is nonsingular. Additionally, if the cardinality of the index set Λ≤n, −1 does not imply that the Gram matrix is nonsingular.


### 3.1.1. Constructing the System of Jacobi Identities

In order an algebra to be a Lie algebra, one needs to satisfy the Jacobi identity. Using our index setΛ, the corresponding Jacobi identity turns into equation (4). Also, there has to be at least two product couples in the Jacobi identity. Otherwise, if there is one product couple in Jacobi identity, it leads to αi,jsαs,km=0; therefore, it contradicts to the fact that i,j,s,s,k,m∈Λ. On the other hand, the Jacobi identity is created by each vector triples from the given fixed Nikolayevsky basis. For example, the Jacobi identity for Xi,Xj,Xk is(9)αijsαskm+αjksαsim+αkisαsjm=0.Since we use the Nikolayevsky basis, there is a uniquek which appears in equation (9). Therefore, each Jacobi identity is created with fixed Xi,Xj,Xk, and m∈AoM. Also, because there has to be at least one product couple in the Jacobi identity, then there exist 2 or three product couples in the Jacobi identity. In equation (9), the product couple αijsαskm corresponds to the index triples in Λ such that i,j,s,s,k,m∈Λ, or i,j,s,k,s,m∈Λ. In the following definition, we define the set of all product couples Pm related to the Jacobi identity for a given subset Xi,Xj,Xk of the fixed basis. As one can see, there exist at most 3 product couples in the same Jacobi identity.Definition 2.
If there aret product couples in the Jacobi identity for the same m such that i,j,s,s,k,m∈Λ, or i,j,s,k,s,m∈Λ, then the Jacobi identity which was created by the basis vectorsXi,Xj,Xk. Now, we define the set of all nonzero product couples for a given index triple basis vectors Xi,Xj,Xk as follows:(10)Pm=ps=αijsαs,km,1≤s≤n,ps≠0.
It is called set of product couples for#m and for in equation (4):(11)AoM=m∈ℕ:i,j,s,s,k,m∈Λori,j,s,k,s,m∈Λ.
Here, AoM is the matrix of#m that appears in equation (4).Remark 2.
Letη be an algebra that is indexed by Λ, then the Jacobi identity for a given Xi,Xj,Xk with each m∈AoM is given by(12)p1∓p2∓p3=0.
The proof of Remark2 follows from the definition of Jacobi identity for a given algebra.Theorem 3.
Letη be an algebra that is indexed by Λ, and U be the Gram matrix related to Λ. Let vp belong to the solution space to the linear system Uv=1, then the Jacobi identity for each m∈AoM is given by(13)vi1vj1∓vi2vj2∓vi3vj3=0,where 1≤i1,i2,i3,j1,j3≤Λ.Proof.
Letη be an algebra that is indexed by Λ. By Remark 3, the Jacobi identity for each m∈AoM and each basis vector triple Xi,Xj,Xk is given by(14)p1∓p2∓p3=0.
Here,ps=αijsαs,km. By Theorem 1, the solution vectors of the linear system Uv=1 are the squares of the structure vectors αijs for all i,j,s∈Λ. Therefore, for any v that satisfies Uv=1 is of form v=αi,js2. Thus, for each ps∈Pm,(15)ps=vavb,where a,b∈i1,i2,i3,j1,j3, and va=αijs2 and vb=αskm2. Solving the equations for the structure constants, we have(16)αijs=∓va,for all i,j,s∈Λ. Therefore, equation (12) turns into(17)vi1vj1∓vi2vj2∓vi3vj3=0,which finishes the proof.In some cases, there can be more than two square root product couples in equation (13), i.e., there is more than one m∈AoM. In that case, we need to consider all the cases of the signs between the product couples αi,jsαs,km, αj,ksαs,im, and αk,isαs,jm. The following lemma deals with this matter.Lemma 2.
If there arek product couples in the Jacobi identity, then there are 2k−1−1 possible sign choices.Proof.
Suppose thatpi∈Pm for m in equation (4) where Pm is the set of all product couples, as in Definition 2. Therefore, the Jacobi identity turns into p1∓p2∓p3=0. Without loss of generality, we assume that p1>0. For each ps∈Pm, where 2≤s≤n, there are two possible sign choices +,−. Therefore, we have 2t−1 possible sign choices. Since all the product couples are nonzero, then they can not all be +. Therefore, we drop the case +,+,…,+. Thus, there are 2t−1−1 possible sign choices for the set Pm.
As we know that there is at most three, at least two product couples in the Jacobi identity, the there is at most three different sign choices for a Jacobi identity.Definition 3.
We define the matrix of all sign choicesSC=sjs∈M2t−1−1×t−1 matrix where the entries are of the form(18)sjs=1,ps>0,0,ps<0,for 1≤j≤2t−1−1. For example, if Pm=p1,p2,p3, then SC is a 3×2 matrix(19)SC=100100.
The equations are created by(20)p1+−1SC1,1.p2+−1SC1,2.p3=0,p1+−1SC2,1.p2+−1SC2,2.p3=0,p1+−1SC3,1.p2+−1SC3,2.p3=0.
Therefore, we havep1+p2−p3=0, p1−p2+p3=0, and p1−p2−p3=0. If there is more than one array of number m’s in equation (4) in Theorem 2, then one needs to find common solutions of at least one Jacobi identity that was created by SC matrix for each m. To illustrate the above procedure, we have the following example.Example 1.
Suppose that(21)Λ=1,3,4,1,4,5,1,5,6,2,3,6,1,6,7,2,4,7,1,7,8,2,5,8.
The solution of the equationUv=1 is of the form(22)v=x+y−1/51/55+x1/1131/55−x−y32/55−x−yy4/11−xxT,=α1242α1352α1672α1782α2562α2682α3462α4582T,where x,y are parameters of the solution system. Then, there are two m’s in equation (4). Now, let us denote the set of m as AoM. Therefore, AoM=7,8. Each m∈AoM corresponds to a Jacobi identity. For m=7, we have two product couples of the triples i,j,l and l,k,m or k,l,m in Λ. They are 2,3,61,6,7 and 1,3,42,4,7. Therefore, the first Jacobi identity is(23)α236α167±α134α247=0.
Form=8, we have two product couples of the triples i,j,l and l,k,m or k,l,m in Λ. They are 2,4,71,7,8 and 1,4,52,5,8. So, the corresponding Jacobi identity is(24)α247α178±α145α258=0.
It can easily be seen that there is a unique sign choice for the above equations. The common solution is as follows:(25)α134=2/11,α145=1/5,α156=1/11,α236=2/5,α167=1/5,α247=1/5,α178=2/11,α258=2/11.


## 3.1.1. Constructing the System of Jacobi Identities

In order an algebra to be a Lie algebra, one needs to satisfy the Jacobi identity. Using our index setΛ, the corresponding Jacobi identity turns into equation (4). Also, there has to be at least two product couples in the Jacobi identity. Otherwise, if there is one product couple in Jacobi identity, it leads to αi,jsαs,km=0; therefore, it contradicts to the fact that i,j,s,s,k,m∈Λ. On the other hand, the Jacobi identity is created by each vector triples from the given fixed Nikolayevsky basis. For example, the Jacobi identity for Xi,Xj,Xk is(9)αijsαskm+αjksαsim+αkisαsjm=0.Since we use the Nikolayevsky basis, there is a uniquek which appears in equation (9). Therefore, each Jacobi identity is created with fixed Xi,Xj,Xk, and m∈AoM. Also, because there has to be at least one product couple in the Jacobi identity, then there exist 2 or three product couples in the Jacobi identity. In equation (9), the product couple αijsαskm corresponds to the index triples in Λ such that i,j,s,s,k,m∈Λ, or i,j,s,k,s,m∈Λ. In the following definition, we define the set of all product couples Pm related to the Jacobi identity for a given subset Xi,Xj,Xk of the fixed basis. As one can see, there exist at most 3 product couples in the same Jacobi identity.Definition 2.
If there aret product couples in the Jacobi identity for the same m such that i,j,s,s,k,m∈Λ, or i,j,s,k,s,m∈Λ, then the Jacobi identity which was created by the basis vectorsXi,Xj,Xk. Now, we define the set of all nonzero product couples for a given index triple basis vectors Xi,Xj,Xk as follows:(10)Pm=ps=αijsαs,km,1≤s≤n,ps≠0.
It is called set of product couples for#m and for in equation (4):(11)AoM=m∈ℕ:i,j,s,s,k,m∈Λori,j,s,k,s,m∈Λ.
Here, AoM is the matrix of#m that appears in equation (4).Remark 2.
Letη be an algebra that is indexed by Λ, then the Jacobi identity for a given Xi,Xj,Xk with each m∈AoM is given by(12)p1∓p2∓p3=0.
The proof of Remark2 follows from the definition of Jacobi identity for a given algebra.Theorem 3.
Letη be an algebra that is indexed by Λ, and U be the Gram matrix related to Λ. Let vp belong to the solution space to the linear system Uv=1, then the Jacobi identity for each m∈AoM is given by(13)vi1vj1∓vi2vj2∓vi3vj3=0,where 1≤i1,i2,i3,j1,j3≤Λ.Proof.
Letη be an algebra that is indexed by Λ. By Remark 3, the Jacobi identity for each m∈AoM and each basis vector triple Xi,Xj,Xk is given by(14)p1∓p2∓p3=0.
Here,ps=αijsαs,km. By Theorem 1, the solution vectors of the linear system Uv=1 are the squares of the structure vectors αijs for all i,j,s∈Λ. Therefore, for any v that satisfies Uv=1 is of form v=αi,js2. Thus, for each ps∈Pm,(15)ps=vavb,where a,b∈i1,i2,i3,j1,j3, and va=αijs2 and vb=αskm2. Solving the equations for the structure constants, we have(16)αijs=∓va,for all i,j,s∈Λ. Therefore, equation (12) turns into(17)vi1vj1∓vi2vj2∓vi3vj3=0,which finishes the proof.In some cases, there can be more than two square root product couples in equation (13), i.e., there is more than one m∈AoM. In that case, we need to consider all the cases of the signs between the product couples αi,jsαs,km, αj,ksαs,im, and αk,isαs,jm. The following lemma deals with this matter.Lemma 2.
If there arek product couples in the Jacobi identity, then there are 2k−1−1 possible sign choices.Proof.
Suppose thatpi∈Pm for m in equation (4) where Pm is the set of all product couples, as in Definition 2. Therefore, the Jacobi identity turns into p1∓p2∓p3=0. Without loss of generality, we assume that p1>0. For each ps∈Pm, where 2≤s≤n, there are two possible sign choices +,−. Therefore, we have 2t−1 possible sign choices. Since all the product couples are nonzero, then they can not all be +. Therefore, we drop the case +,+,…,+. Thus, there are 2t−1−1 possible sign choices for the set Pm.
As we know that there is at most three, at least two product couples in the Jacobi identity, the there is at most three different sign choices for a Jacobi identity.Definition 3.
We define the matrix of all sign choicesSC=sjs∈M2t−1−1×t−1 matrix where the entries are of the form(18)sjs=1,ps>0,0,ps<0,for 1≤j≤2t−1−1. For example, if Pm=p1,p2,p3, then SC is a 3×2 matrix(19)SC=100100.
The equations are created by(20)p1+−1SC1,1.p2+−1SC1,2.p3=0,p1+−1SC2,1.p2+−1SC2,2.p3=0,p1+−1SC3,1.p2+−1SC3,2.p3=0.
Therefore, we havep1+p2−p3=0, p1−p2+p3=0, and p1−p2−p3=0. If there is more than one array of number m’s in equation (4) in Theorem 2, then one needs to find common solutions of at least one Jacobi identity that was created by SC matrix for each m. To illustrate the above procedure, we have the following example.Example 1.
Suppose that(21)Λ=1,3,4,1,4,5,1,5,6,2,3,6,1,6,7,2,4,7,1,7,8,2,5,8.
The solution of the equationUv=1 is of the form(22)v=x+y−1/51/55+x1/1131/55−x−y32/55−x−yy4/11−xxT,=α1242α1352α1672α1782α2562α2682α3462α4582T,where x,y are parameters of the solution system. Then, there are two m’s in equation (4). Now, let us denote the set of m as AoM. Therefore, AoM=7,8. Each m∈AoM corresponds to a Jacobi identity. For m=7, we have two product couples of the triples i,j,l and l,k,m or k,l,m in Λ. They are 2,3,61,6,7 and 1,3,42,4,7. Therefore, the first Jacobi identity is(23)α236α167±α134α247=0.
Form=8, we have two product couples of the triples i,j,l and l,k,m or k,l,m in Λ. They are 2,4,71,7,8 and 1,4,52,5,8. So, the corresponding Jacobi identity is(24)α247α178±α145α258=0.
It can easily be seen that there is a unique sign choice for the above equations. The common solution is as follows:(25)α134=2/11,α145=1/5,α156=1/11,α236=2/5,α167=1/5,α247=1/5,α178=2/11,α258=2/11.


## 3.2. The Rank and the Index of a Nilsoliton

The following proposition and its following corollary help us to compute rank of a nilsoliton metric Lie algebra.Proposition 1.
(see Proposition 4.7 in [9]). Let η be a nonabelian Lie algebra that admits a simple derivation D. Let B=Xii=1n be an eigenvector basis with index set Λ, and let Y be the root matrix associated to Λ. Then, the rank of the nilsoliton metric Lie algebra η equals to the nullity of the root matrix Y.Corollary 1.
Letη be a nonabelian n−dimensional Lie algebra that admits a simple derivation. Let B be an eigenvector basis with index set Λ, and let Y be the root matrix associated to Λ. Then,(26)rankη=n+nullityYYt−Λ.Proof.
Suppose thatη is an n−dimensional nonabelian Lie algebra, admitting a simple derivation. Let B be an eigenvector basis with index set Λ where Λ=m, and let Y be the root matrix associated to Λ. By Proposition 1, the root matrix Y is an m×n matrix, whose nullity is rankη. Therefore, from rank-nullity theorem, rankη=n−RankY. We also know that rankY=rankYYT; therefore, we have(27)rankη=n−RankYYT.
On the other hand,YYT is an m×m matrix. Therefore, RankYYT=m−NullityYYT. Then, we have(28)rankη=n−m−NullityYYT=n+NullityYYT−Λ.Definition 4.
LetX∈η, adX, denote the adjoint representation and η∗ denote the dual of the Lie algebra η. Then, the skew symmetric bilinear form Ψf, where f∈η∗, is defined by(29)Ψf:η×η⟶ℝX,Y⟶ΨfX,Y=fX,Y.
The index of a Lie algebraη is the integer infdimηf:f∈η∗ where ηf=kerΨf defined by ηf=X∈η:fX,Y=0,∀y∈η.Proposition 2.
(see Proposition 4 in [10]). The index of an n-dimensional Lie algebra η is the integer,(30)indexη=n−RankRηXi,Xj1≤i,j≤n,where Rη is the quotient field of symmetric algebra Sη.Remark 3.
above proposition tells us that the index of a Lie algebra is the nullity of the matrixEη=Xi,Xj.
Now, we present an example regarding to this notion with the use of the index setΛ.Example 2.
Suppose thatΛ=1,2,3,1,3,4,2,3,5 is the index set of an algebra η of dimension 6. Its corresponding Gram matrix is(31)300031013.
Since there is no−1 entry, η is a Lie algebra. The matrix Eη is as follows:(32)0X3X4000−X30X5000−X4−X50000000000000000.
Its rank is 2; therefore,indexη=4.Proposition 3.
For ann−dimensional nilsoliton represented by an index set Λ with a Nikolayevsky basis, the matrix Eη has the following properties:(1)
The matrix is ann×n matrix with zero diagonal entries(2)
The rank ofEη is even(3)
Thenth row and column of Eη are zero matrices(4)
IfΛ=K, then Eη has 2K nonzero entriesProof.
Suppose that the Nikolayevsky basis is represented byB=Xα1≤α≤n. By the definition of Eη, Eηi,j=Xi,Xj, then(33)Eηij=Xi,Xj=−Xj,Xi=−Eηji.
Therefore,Eη is a skew symmetric matrix, which implies the first and second properties. On the other hand, since η is nilpotent Lie algebra with Nikolayevsky basis, then Λ consists of the elements of form i,j,k such that i<j<k. Therefore, n cannot appear in the first or the second component of the triples in Λ. Thus, Xi,Xn=Xn,Xi=0 for all i∈1,2,…,n, which implies that Eηin=Eηni=0, i.e., the last row and the last column are zero matrices.
The entries ofEη are defined by the index set Λ. If Λ=K, then there exists K non-zero entries on the upper triangular part of Eη. Since the matrix is skew symmetric, then there exists other K nonzero entries in the lower triangular part of the matrix. So, in total, there exists 2K nonzero entries in Eη.Corollary 2.
Suppose thatη is an n−dimensional nilsoliton. Then, indexη is even, if n is even, and index is odd if n is odd.Proof.
The index of the nilsoliton isindexη=NullityEη=n−rankEη. Since Eη is a skew symmetric matrix, its rank is always even, which finishes the proof.


## 4. Conclusion

In this paper, we prove theorems for the computations of structural elements of ann-dimensional nilsoliton η. We prove theorems regarding to the Jacobi identity or identities that have to be satisfied, rank, and index of the nilsoliton. In the future, we plan to use these theorems to create a computer algorithm for the classifications of nilsolitons for a given dimension. The theorems appearing in this study will allow us to pare down the number of cases to consider in our procedure.



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