Power-associative evolution algebras
Moussa OUATTARA Souleymane SAVADOGO
Département de Mathématiques et Informatique
Université Ouaga I Pr Joseph KI-ZERBO
03 BP 7021 Ouagadougou 03
Burkina Faso
[email protected]@yahoo.fr
Abstract
The paper is devoted to the study of evolution algebras that are power-associative algebras.
We give the Wedderburn decomposition of evolution algebras that are power-associative algebras and we prove that these algebras are Jordan algebras. Finally, we use this decomposition to classify these algebras up to dimension six.
2010 Mathematics Subject Classification : Primary 17D92, 17A05, Secondary 17D99, 17A60
Keywords : Evolution algebras, Power associativity, Wedderburn decomposition, Nilalgebras.
1 Introduction
The notion of evolution algebras was introduced in 2006 by J.P. Tian and P. Vojtěchovský ([13]). In 2008, J.P. Tian laid the foundations of this algebra in his monograph ([14]).
In ([13], Theorem 1.5), the authors show that evolution algebras are commutative (hence flexible) not necessarily power-associative (nor associative).
Let E be n-dimensional algebra over a commutative field F. We say that E is n-dimensional evolution algebra, if it admits a basis B={ei ; 1≤i≤n} such that:
[TABLE]
The basis B is called natural basis of E and the matrix (aij) is the matrix of structural constants of E, relative to the natural basis B.
In ([2], p.127), the authors exhibit a necessary condition for an evolution algebra to be power-associative. They show in particular that the only power-associative evolution algebras are such that aii2=aii (i=1,…,n). However, this condition is not sufficient. It is verified for all nil evolution algebras of the fact that for any i=1,…,n, aii=0 ([5, Proof of Theorem 2.2]).
In [4, Example 4.8], for k=n=4, we find a power-associative algebra which is not associative. Indeed, it is isomorphic to N4,6 (see Table 1). Except the type [1,1,1], all algebras in [7, Table 1] are nil associative evolution algebras. In [7, Table 2], the first line is associative, while the third is power-associative. In [3, Table 1], the fifth, sixth and seventh algebras, of dimension 3, are associative. They are respectively isomorphic to E3,4, N3,3(1) and N3,2 (see Table 1). In [9, Theorem 6.1], the algebras E4,1, E4,2, E4,3, E4,5 and E4,10 are associative.
Since, the evolution algebras are not defined by identities and thus do not form a variety of non-associative algebras, such as that of Lie algebras, alternative algebras or Jordan algebras. Therefore, research on these algebras follows different paths [3, 2, 4, 5, 6, 7, 9, 11, 13].
In section 2, we give an example of an algebra which is power-associative but which is not an evolution algebra. We also recall some definitions and known results about nil evolution algebras.
In section 3, we give Wedderburn decomposition of a power-associative evolution algebra. Since the semisimple component is associative, we deduce that an evolution algebra is power-associative if and only if its nil radical is power-associative. We show that the nil power-associative evolution algebras are Jordan algebras. Thus, power-associative evolution algebras are identified with those which are Jordan algebras.
In section 4, we determine power-associative evolution algebras up to dimension six.
2 Preliminaries
2.1 Example of Jordan algebra that is not evolution algebra
An algebra E is baric if there is a non trivial homomorphism of algebra ω:E⟶K.
**Example
2.1.1. ** In [2, Example 2.2], the authors show that zygotic algebra for simple Mendelian inheritance Z(n,2) (case n=2) is not an evolution algebra. This algebra is a Jordan algebra. In fact, algebra Z(n,2) is the commutative duplicate of the gametic algebra for simple Mendelian G(n,2).
**Example
2.1.2. ** The gametic algebra for simple Mendelian inheritance G(n,2) is not an evolution algebra.
Proof. The multiplication table of G(n,2) in the natural basis B1={ai ; 1≤i≤n} is given by aiaj=21(ai+aj). The linear mapping ω:ai⟼1 is a weight function of G(n,2) and for all x,y∈G(n,2), we have xy=21(ω(x)y+ω(y)x) ([15]). Suppose that G(n,2) is an evolution algebra in the basis B2={ei ; 1≤i≤n}.
Since ω is weight function, there is i0∈{1,…,n} such that ω(ei0)=0. For j=i0, ei0ej=0 leads to ω(ei0)ω(ej)=0, so ω(ej)=0. Then 0=ei0ej=21ω(ei0)ej gives ej=0 impossible. Such a basis B2 does not exist. □
2.2 Results about nil evolution algebras
For an element a∈E, we define principal powers as follows a1=a and ak+1=aka while that of E are defined by
[TABLE]
Definition 2.2.1.
We will say that algebra E is:
i) right nilpotent if there is a nonzero integer n such that En=0 and the minimal such number is called the index of right nilpotency ;
ii) nil if there is a nonzero integer n(a) such that an(a)=0, for all a∈E and the minimal such number is called the nil-index of E.
Theorem 2.2.2** ([5], Theorem 2.7).**
The following statements are equivalent:
i)* there is a natural basis in which the matrix of structural constants is of a form*
[TABLE]
ii)* E is a right nilpotent algebra ;*
iii)* E is a nilalgebra.*
We define the following subspace of E:
∙ The annihilator ann1(E)=ann(E)={x∈E:xE=0} ;
∙ anni(E) is defined by \mathrm{ann}^{i}(\mathcal{E})/\mathrm{ann}^{i-1}(\mathcal{E})=\mathrm{ann}\big{(}\mathcal{E}/\mathrm{ann}^{i-1}(\mathcal{E})\big{)}.
∙ Ui⊕U1={x∈anni(E)∣xanni−1(E)=0}.
In ([6, Lemme 2.7]), the authors show that ann(E)=span{ei∈B∣ei2=0}.
In ([7]), the authors show that anni(E)=span{e∈B∣e2∈anni−1(E)} and that the basis B=B1∪⋯∪Br where Bi={e∈B∣e2∈anni−1(E),e∈anni−1(E)}. Then, for Ui:=span{Bi}, i=1,…,r, we have U1⊕⋯⊕Ui=anni(E) (i=1,…,r). They prove then that Ui⊕U1 is an invariant of nil evolution algebras.
The type of a nil evolution algebra E is the sequence [n1,n2,…,nr] where r and ni are integers defined by annr(E)=E ; ni=dimK(anni(E))−dimK(anni−1(E)) and n1+⋯+ni=dimK(anni(E)) for all i∈{1,…,r}.
3 Power associativity
3.1 Some definitions
Definition 3.1.1. Algebra E is called:
i) associative if for all x, y, z∈E, (x,y,z)=0 where (x,y,z)=(xy)z−x(yz) is the associator of the elements x, y, z of E ;
ii) Jordan if it is commutative and (x2,y,x)=0, for all x,y∈E ;
iii) power-associative if the subalgebra generated by x is associative. In the other words, for all x∈E, xixj=xi+j for all integers i,j≥1.
Definition 3.1.2.
Algebra E is said to be third power-associative if for all x∈E, x2x=xx2.
Remark 3.1.3.
Since any commutative algebra is third power-associative, we deduce that any evolution algebra is third power-associative.
Definition 3.1.4.
Algebra E is said to be fourth power-associative if x2x2=x4 for all x∈E.
Theorem 3.1.5** ([1]).**
Let F be a field of characteristic =2,3,5. Algebra E is power-associative if and only if x2x2=x4, for all x∈E.
3.2 Characterization of associative algebras
If E is a n-dimensional algebra with basis {ei ; 1≤i≤n}, then E is associative if and only if (eiej)ek=ei(ejek) for all 1≤i,j,k≤n ([10, Proposition 1]). We then deduce:
Lemma 3.2.1**.**
An evolution algebra with natural basis B={ei ; 1≤i≤n} is associative if and only if ei2ej=ei(eiej)=0 for all 1≤i=j≤n.
Theorem 3.2.2** ([8], Theorem 1).**
A commutative power-associative nilalgebra A of nilindex 3
and of dimension 4 over a field F of characteristic =2 is associative, and A3=0.
For the evolution algebra, this theorem has the following generalization:
Theorem 3.2.3**.**
Let E be a finite-dimensional evolution algebra. Then E is an associative nilalgebra if and only if x3=0, for all x∈E. In this case E3=0.
Proof.
Let B={ei ; 1≤i≤n} be a natural basis of E.
Let’s suppose that E is an associative nilalgebra. Then for all integer i we have ei3=0. Let’s set x=∑i=1nxiei. We have x3=∑i,j=1nxi2xjei2ej=0 because ei2ej=0 since E is nil and associative.
Conversely the partial linearization of the identity x3=0 give
[TABLE]
For i distinct from j we have ei2ej=0 and E is associative by Lemma 3.2.1.
We have
\mathcal{E}^{3}=\big{<}e_{i}^{2}e_{j} ; 1\leq i,j\leq n\big{>}=0.
□
3.3 Characterization of power-associative algebras
In the following a field F is of characteristic =2.
E is a n-dimensional evolution algebra over a field F and with a natural basis B={ei ; 1≤i≤n} whose multiplication table is given by (1).
Lemma 3.3.1**.**
Algebra E is a fourth power-associative if and only if the following conditions are satisfied:
1)* ei4=ei2ei2 for all 1≤i≤n ;*
2)* 2ei2ej2=(ei2ej)ej+(ej2ei)ei for all 1≤i<j≤n ;*
3)* ei3ej+(ei2ej)ei=0 for all 1≤i=j≤n ;*
4)* (ei2ej)ek+(ei2ek)ej=0 for all 1≤i≤n and 1≤j<k≤n with i=j and i=k.*
Proof.
Suppose that E is a fourth power-associative algebra and x,y,z∈E. We have
[TABLE]
The partial linearization of (3) gives:
[TABLE]
[TABLE]
Since the basis vectors {ei ; 1≤i≤n} check identities (3), (4) and (5), we have
(\refeq2)⟹1) taking x=ei.
(\refeq3)⟹3) taking x=ei and y=ej with i=j.
(\refeq4)⟹2) taking x=ei and y=z=ej with i=j.
(\refeq4)⟹4) taking x=ei, y=ej and z=ek ; i,j,k are pairwise distinct.
Conversely, suppose that conditions 1), 2), 3) and 4) are satisfied. Let x=∑i=1nxiei be an element of E. We have the following equalities :
[TABLE]
We have
[TABLE]
and
[TABLE]
So
[TABLE]
□
Corollary 3.3.2**.**
Let E be a nil evolution algebra. Then E is fourth power-associative algebra if and only if the following conditions are satisfied:
1)* ei2ej2=0 for all 1≤i≤j≤n ;*
2)* (ei2ej)ek=0 for all 1≤i,j,k≤n.*
Proof.
Since E is a nilalgebra, then for all 1≤i,j,k≤n, we have ei3=0 and ajkakj=0 [5, Proof of theorem 2.2]. According to 2) of Lemma 3.3.1, we have
2ei2ej2=(ei2ej)ej+(ej2ei)ei=aijej3+ajiei3=0, hence 1).
Let’s show that (ei2ej)ek=0 for i,j,k two by two distinct and consider 4) of Lemma 3.3.1, i.e.
[TABLE]
Since ajkakj=0, we consider two cases:
If ajk=0, then (ei2ej)ek=0 and (6) leads to (ei2ek)ej=0;
If akj=0, then (ei2ek)ej=0 and (6) gives (ei2ek)ej=0;
thus 4) of Lemma 3.3.1 is equivalent to (ei2ej)ek=0, for all i,j,k pairwise distinct.
□
3.4 Characterization of Jordan algebras
Lemma 3.4.1**.**
E* is Jordan algebra if and only if the following conditions are satisfied:*
1)* ei2ei2=ei4 for all 1≤i≤n ;*
2)* ei3ej=0 for all 1≤i=j≤n ;*
3)* (ei2ej)ei=0 for all 1≤i=j≤n ;*
4)* ei2ej2=(ei2ej)ej=(ej2ei)ei for all 1≤i=j≤n ;*
5)* (ei2ej)ek=0 for all 1≤i,j,k≤n with i,j,k pairwise distinct.*
Proof.
Suppose that E is Jordan algebra and x,y,z∈E. We have
[TABLE]
By linearizing (7), we obtain
[TABLE]
Basis vectors {ei ; 1≤i≤n} verify identities (7), (8).
(\refeq10)⟹1) ; taking x=y=ei ;
(\refeq10)⟹3) ; taking x=ei and y=ej with i=j ;
(\refeq11)⟹2) ; taking x=y=ei and z=ej with i=j ;
(\refeq11)⟹4) ; taking x=ei and y=z=ej ; we also take x=ej and y=z=ei with i=j ;
(\refeq11)⟹5) ; taking x=ei, y=ej and z=ek i,j,k pairwise distinct.
Conversely, suppose that conditions 1), 2), 3), 4) and 5) are satisfied. Let ∑i=1nxiei and y=∑i=1nyiei be elements of E. We have the following equalities:
[TABLE]
□
Corollary 3.4.2**.**
Let E be a nil evolution algebra. Then E is Jordan algebra if and only if the following conditions are satisfied:
1)* ei2ej2=0 for all 1≤i≤j≤n ;*
2)* (ei2ej)ek=0 for all 1≤i,j,k≤n.*
Proof.
Since 2) is exactly 5) of lemma 3.4.1, it remains to show 1).
Since E is nilalgebra, then for all 1≤i≤n, we have ei3=0. According to 4) of Lemma 3.4.1, we have
ei2ej2=(ei2ej)ej=aijej3=0, hence 1).
□
Proposition 3.4.3**.**
Let E be a nil finite-dimensional evolution algebra. The following statements are equivalent:
1)* E is fourth power-associative algebra.*
2)* E is Jordan algebra.*
Proof.
This proposition is a consequence of Corollary 3.3.2 and Corollary 3.4.2.
□
3.5 Wedderburn decomposition
Any finite-dimensional commutative power-associative algebra which is not nilalgebra contains at least one nonzero idempotent ([12], Proposition 3.3 P 33).
Moreover, in ([1]) the Author shows that any commutative power-associative algebra E, which has a nonzero idempotent, admits the following Peirce decomposition: E=E1(e)⊕E0(e)⊕E21(e) where Eλ(e)={x∈E ; ex=λx}.
Definition 3.5.1. ([2], Definition 2.4)
An evolution subalgebra of an evolution algebra E is a subalgebra E′⊆E such that E′ is an evolution algebra i.e. E′ has a natural basis. We say that E′ has the extension property if there exists a natural basis B′ of E′ which can be extended to a natural basis B of E.
E is a n-dimensional evolution algebra over
the field F of characteristic =2 and with a natural basis B={ei ; 1≤i≤n} whose multiplication table is given by (1).
Lemma 3.5.2**.**
If E is fourth power-associative, then for all i=j, ei3ej=(ei2ej)ei=0.
Proof.
Suppose that there exists i0=j0∈{1,…,n} such that ei03ej0=0. Then ai0i0ai0j0ej02=0.
Since E is power-associative, 0=ei03ej0+(ei02ej0)ei0=ai0i0ai0j0ej02+ai0j0aj0i0ei02. And since ai0j0=0, then
[TABLE]
By multiplying (9) by ei0, we get 0=ai0i0ej02ei0+aj0i0ei02ei0=ai0i0aj0i0ei02+aj0i0ai0i0ei02=2ai0i0aj0i0ei02 then aj0i0=0.
(\refeq12) leads to ai0i0ej02=0: impossible. We deduce that for all i=j, ei3ej=0 and (ei2ej)ei=0.
□
Lemma 3.5.3**.**
Let E be a fourth power-associative evolution algebra and not nilalgebra.
Then E admits a nonzero idempotent e such that the evolution subalgebra of E generated by e has extension property.
Proof.
Since E is not nilalgebra, there exists i0∈{1,…,n} such that ei03=0. Without losing the generality, we can assume that e13=0. Thus e=a11−2e12 is an idempotent. Indeed, e2=a11−4e12e12=a11−4e14=a11−2e12=e.
Let’s show that the family {e,e2,e3,…,en} is a natural basis of E. Let α1,…,αn be scalars such that
[TABLE]
By multiplying (10) by e, we get
[TABLE]
For i=1, 0=e13ei=a11e12ei leads to e12ei=0 because a11=0. Thus (\refeq14) leads to α1e=0, then α1=0 and (\refeq13) leads to α2=α3=⋯=αn=0 because {e2,e3,…,en} is linearly independent. It has been shown in passing that eei=0 for i=2,…,n. So the family {e,e2,e3,…,en} is a natural basis of E.
□
Theorem 3.5.4** (of Wedderburn).**
Let E be a not nil power-associative evolution algebra. Then E admits s nonzero pairwise orthogonal idempotents u1,u2,…,us, such that
[TABLE]
direct sum of algebras, where s≥1 is an integer and N is either zero or a nil power-associative evolution algebra.
Furthermore Ess=Ku1⊕Ku2⊕⋯⊕Kus is the semisimple component of E and N=Rad(E) is the nil radical of E.
Proof.
Suppose
that E is a not nilalgebra. Then E admits nonzero idempotent u1 such that the evolution subalgebra Ku1 generated by u1 has extension property and we denote E=E1(u1)⊕E0(u1)⊕E21(u1) the Peirce decomposition of E relative
to u1. Then {u1,e2,e3,…,en} is a natural basis of E.
Let’s v=α1u1+α2e2+α3e3+⋯+αnen∈E21(u1).
We have u1v=α1u12=α1u1=21v=21(α1u1+α2e2+α3e3+⋯+αnen). Then α1=21α1 and αi=0, for i=1. We also have α1=0 and v=0, i.e. E21(u1)=0.
Now let’s v=α1u1+α2e2+α3e3+⋯+αnen∈E1(u1).
We have u1v=α1u1=v=α1u1+α2e2+α3e3+⋯+αnen leads to αi=0 for i=1. Thus, E1(u1)=Ku1, so E=Ku1⊕E0(u1) direct sum of algebras.
If E0(u1) is a nilalgebra, it is finished and s=1. Otherwise we repeat the reasoning with the evolution subalgebra E0(u1) of dimension n−1. And so on, we get result because dimK(E) is finite.
□
Theorem 3.5.5**.**
The following statements are equivalent:
1)* E is power-associative algebra.*
2)* E is Jordan algebra.*
Proof.
2)⟹1) because any Jordan algebra is power-associative.
1)⟹2): Suppose that E is power-associative. Let E=Ess⊕N be the Wedderburn decomposition of E where Ess is the semisimple
component of E and N is the nil radical of E.
Since Ess is associative and that (12) is a direct somme of algebras, then E is a Jordan algebra if and only if N is a Jordan algebra. And since N is a nil fourth power-associative evolution algebra,
then it is Jordan algebra by Proposition 3.4.3
□
4 Classification
According to Theorem 3.5.4 and Theorem 3.5.5, classification problem of power-associative evolution algebras is reduced to that of nil power-associative evolution algebras.
Definition 4.0.1. An algebra E is decomposable if there are nonzero ideals I and J such that E=I⊕J. Otherwise, it is indecomposable.
Lemma 4.0.2** ([7], Corollary 2.6).**
Let E be a finite-dimensional evolution algebra such that dimF(ann(E))≥21dimF(E)≥1. Then E is decomposable.
Moreover for classification, it is sufficient to determine the indecomposable evolution algebras.
We are interested in the classification of indecomposable evolution algebras up to dimension 6. Then dimF(ann(E))=1 or 2.
4.1 Nil indecomposable associative evolution algebra
Suppose that N is an associative algebra. We have ei2ej=0, for all i,j=1,…,n. Then for all i∈{1,…,n} such that ei2=0, we have ei2∈ann(N).
- dim(ann(N))=1.
Without losing the generality, we can suppose that ann(N)=Ken.
Thus, for all i=n, we have ei2=αien with αi=0.
We set vi=ei and vn=α1en with 1≤i≤n−1. We have:
v12=vn, vi2=αiα1−1vn=βivn, vn2=0 with βi=0 and 2≤i≤n−1.
- dim(ann(N))=2.
Without losing the generality, we can suppose that ann(N)=Ken−1⊕Ken.
Thus, for all i=n−1,n, we have ei2=αi,n−1en−1+αi,nen with (αi,n−1,αin)=0.
We can suppose that N2=Ke12⊕Ken−22, otherwise we swap the vectors e1,e2,…,en−2.
We set vi=ei (1≤i≤n), vn−1=v12, vn=vn−22, vi2=αi,n−1vn−1+αi,nvn. Also, there is i0∈{2,…,n−3} such that αi0,n−1αi0n=0, otherwise N would be decomposable.
4.1.1 One-dimensional classification
Theorem 4.1.1**.**
The only 1-dimensional nil evolution algebra, indecomposable and associative, is N1,1: e12=0 of type [1].
4.1.2 Two-dimensional classification
Theorem 4.1.2**.**
The only 2-dimensional nil evolution algebra indecomposable and associative is N2,2: e12=e2, e22=0, of type [1,1].
Proof.
dim(N)=2 leads to dim(ann(N))=1, otherwise N would be decomposable.
So, N is isomorphic to N2,2: e12=e2 and e22=0.
□
4.1.3 Three-dimensional classification
Theorem 4.1.3**.**
The only 3-dimensional nil evolution algebra indecomposable and associative is N3,3(α): e12=e3, e22=αe3, e32=0, of type [1,2] where α∈F∗.
Proof.
dim(N)=3 leads to dim(ann(N))<21dim(N)=1.5, so dim(ann(N))=1. Thus, N is isomorphic to N3,3(α): e12=e3, e22=αe3 and e32=0 with α∈F∗.
□
4.1.4 Four-dimensional classification
Theorem 4.1.4**.**
The only 4-dimensional nil evolution algebra indecomposable and associative is N4,5(α,β): e12=e4, e22=αe4, e32=βe4, e42=0, of type [1,3], where α,β∈F∗.
Proof.
dim(N)=4 leads to dim(ann(N))<21dim(N)=2, so dim(ann(N))=1.
Thus, N is isomorphic to N4,5(α,β): e12=e4, e22=αe4, e32=βe4, e42=0 where α,β∈F∗.
□
4.1.5 Five-dimensional classification
Theorem 4.1.5**.**
Let N be a nil indecomposable associative evolution algebra of dimension five. Then N is isomorphic to one and only one of the algebras N5,8(α,β,γ) and N5,9(α,β) in Table 2.
Proof.
dim(N)=5 leads to dim(ann(N))<21dim(N)=2.5, so dim(ann(N))=1,2.
-
dim(ann(N))=1 ; N is isomorphic to N5,8(α,β,γ): e12=e5, e22=αe5, e32=βe5, e42=γe5, e52=0 with α,β,γ∈F∗.
-
dim(ann(N))=2 ; N is isomorphic to N5,9(α,β): e12=e4, e22=αe4+βe5, e32=e5, e42=e52=0 with α,β∈F∗.
□
4.1.6 Six-dimensional classification
Theorem 4.1.6**.**
Let N be a nil indecomposable associative evolution algebra of dimension six. Then N is isomorphic to one and only one of the algebras N6,16(α,β,γ,δ), N6,17(α,β,γ) and N6,18(α,β,γ,δ) in Table 3.
Proof.
dim(N)=6 leads to dim(ann(N))<21dim(N)=3 so dim(ann(N))=1,2.
-
dim(ann(N))=1 ; N is isomorphic to N6,16(α,β,γ,δ): e12=e6, e22=αe6, e32=βe6, e42=γe6, e52=δe6, e62=0 with α,β,γ,δ∈F∗.
-
dim(ann(N))=2 ; the multiplication table of N is of the form: e12=e5, e22=αe5+βe6, e32=γe5+δe6, e42=e6, e52=e62=0 with αβ=0 and (γ,δ)=0.
If αδ−βγ=0 and γδ=0, then the algebra is isomorphic to N6,18(α,β,γ,δ).
γδ=0 or αδ−γβ=0. We consider three cases :
i)
If γ=0 (δ=0), then N is isomorphic to N6,17(α,β,δ) : e12=e5, e22=αe5+βe6, e32=δe6, e42=e6, e52=e62=0.
ii)
If δ=0 (γ=0), by swapping e5 and e6 we find again N6,17(β,α,γ).
iii)
αδ−βγ=0 and γδ=0. We have δ=α−1βγ and by setting w1=e1, w2=e4, w3=e3, w4=e2, w5=e5 and w6=αe5+βe6, we get algebra N6,17(−αβ−1,β−1,α−1γ).
□
4.2 Nil indecomposable evolution algebra which is not associative
Suppose that N is not associative. There are i0,j0∈{1,…,n} distincts such that ei02ej0=0. We have 0=ei02ej0=ai0j0ej02 leads to ai0j0=0, ei02=0 and ej02=0.
Since N is power-associative, we have 0=(ei02ej0)ek=ai0j0ej02ek then ej02ek=0, for all integers k=j0 and since ej03=0, then ej02ek=0, for all integers k. We deduce that ej02∈ann(N).
Since ei02∈ann(N), necessarily aii0=0 for all 1≤i≤n if not, assuming that ai1i0=0 for some i1, we have , for all 1≤k≤n, 0=(ei12ei0)ek=ai1i0ei02ek, so ei02ek=0 i.e. ei02∈ann(N) : contradiction.
Suppose that for all i=i0,j0, ei02ei=0. Since
[TABLE]
We have
[TABLE]
This is impossible because ai0j0ej02=0.
Then there is j1 distinct from i0 and j0 such that ei02ej1=0. The argument at the beginning of the paragraph tells us that ej12∈ann(N).
Since N is nil, then ann(N)=0, dim(N)>3 and without losing generality, we will set i0=1, j0=2 and j1=3.
Theorem 4.2.1** ([8], Theorem 2).**
Let A be a commutative power-associative nilalgebra of nil-index 4 and dimension 4 over a field F of characteristic = 2, then A4=0 and there is y∈A2 such that yA2=0.
For the evolution algebras, this theorem has the following generalization:
Theorem 4.2.2**.**
Let E be a finite-dimensional nil power-associative evolution algebra which is not associative. Then E is of nil-index 4, E4=0 and there is y∈E2 such that yE2=0.
Proof.
Let’s suppose that E is a power-associative evolution nilalgebra which is not associative and x=∑i=1nxiei. We have x3=∑i,j=1nxi2xjei2ej and x4=∑i,j,k=1xi2xjxk(ei2ej)ek=0 by Corollary 3.3.2. Since E is not associative, there are integers i0=j0 such that ei02ej0=0. In this case ej02ei0=0 and then the element a=ei0+ej0 checks a3=ei02ej0+ej02ei0=ei02ej0=0. Thus the nilindex of E is 4.
We have \mathcal{E}^{3}=\big{<}e_{i}^{2}e_{j} ; 1\leq i,j\leq n\big{>}\neq 0 because E is not associative and \mathcal{E}^{4}=\big{<}(e_{i}^{2}e_{j})e_{k} ; 1\leq i,j,k\leq n\big{>}=0 by Corollary 3.3.2. Since ei02∈annE necessarily aki0=0 and ek2ei0=aki0ei02=0 for all integer k, hence ei0E2=0. In fact ei0∈E2 otherwise we would have e_{i_{0}}^{2}\in\mathcal{E}^{[3]}=\mathcal{E}^{2}\mathcal{E}^{2}=\big{<}e_{i}^{2}e_{j}^{2} ; 1\leq i,j\leq n\big{>}=0: absurd because ei02=0, hence the theorem
□
4.2.1 Four-dimensional classification
Theorem 4.2.3**.**
The only 4-dimensional nil indecomposable power-associative which is not associative evolution algebra is N4,6 in Table 1.
Proof.
Since dim(N)=4, we have dim(ann(N))=1. From the above
[TABLE]
Let’s set v2=a12e2+a14e4, v3=a13e3. We have e12=v2+v3 and 0=e12e12=v22+v32 leads to v32=−v22.
Then we set v4=v22=a122a24e4 ; so N is isomorphic to
[TABLE]
of type [1,2,1].
□
4.2.2 Five-dimensional classification
Theorem 4.2.4**.**
Let N be a nil indecomposable power-associative which is not associative evolution algebra, of dimension five. Then N is isomorphic to one and only one of the algebras N5,10(α), N5,11(α) and N5,12(α,β) in Table 2.
Proof.
We have dim(N)=5 leads to dim(ann(N))<21dim(N)=2.5, so dim(ann(N))=1,2.
- dim(ann(N))=1.
[TABLE]
Since 0=(e12e4)ej=a14a4jej2, we have a14a4j=0 (with j=2,3). we then distinguish two cases.
1.1) a14=0 i.e. e12=a12e2+a13e3+a15e5 with a12a13=0.
Let’s set v2=a12e2+a15e5 and v3=a13e3. We have e12=v2+v3 and 0=e12e12=v22+v32, so v32=−v22.
We set v5=v22=a122a25e5, so v32=−v5 and there are scalars α42, α43 and α45 not all zero such that e42=α42v2+α43v3+α45v5.
We have 0=e12e42=α42v22+α43v32=(α42−α43)v5, so α42=α43 ; this relation ensures the power associativity of N.
The family {e1,v2,v3,e4,v5} is a natural basis of N and its multiplication table is defined by:
[TABLE]
We have \mathrm{ann}(N)=\big{<}v_{5}\big{>}, v2,v3∈ann2(N), e1∈ann3(N) and since e_{4}^{2}\in\big{<}v_{2},v_{3},v_{5}\big{>} then the possible types of N are [1,3,1] or [1,2,2].
1.1.1) The type of N is [1,3,1], then α42=0.
We have
\mathrm{ann}(N)=\big{<}v_{5}\big{>}, \mathrm{ann}^{2}(N)=\big{<}v_{2},v_{3},e_{4},v_{5}\big{>} and ann3(N)=N.
So U3⊕U1={x∈ann3(N) ; x[\mathrm{ann}^{2}(N)]=0\}=\big{<}e_{1},v_{5}\big{>} and (\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}=\big{<}v_{2}+v_{3}\big{>}. N is isomorphic to
N5,10(α):e12=e2+e3,e22=e5,e32=−e5,e42=αe5,e52=0 with α=0.
1.1.2) The type of N is [1,2,2], then α42=0.
We have \mathrm{ann}(N)=\big{<}v_{5}\big{>}, \mathrm{ann}^{2}(N)=\big{<}v_{2},v_{3},v_{5}\big{>} and ann3(N)=N.
So \mathcal{U}_{3}\oplus\mathcal{U}_{1}=\big{<}e_{1},e_{4},v_{5}\big{>} and (\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}=\big{<}v_{2}+v_{3},e_{4}^{2}\big{>}. Thus,
\dim\big{(}\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=1,2.
a) \dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=1 implies α45=0. N is isomorphic to
N5,11(α):e12=e2+e3,e22=e5,e32=−e5,e42=α(e2+e3),e52=0 with α=0.
b) \dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=2 implies α45=0.
N is isomorphic to
N5,12(α,β):e12=e2+e3,e22=e5,e32=−e5,e42=α(e2+e3)+βe5,e52=0 with αβ=0.
1.2) a14=0 i.e. e12=a12e2+a13e3+a14e4+a15e5 and e42=a45e5 with a12a13a14a45=0.
Let’s set v2=a12e2+a15e5, v3=a13e3 and v4=a14e4 ; we have e12=v2+v3+v4 and 0=e12e12=v22+v32+v42 implies v42=−(v22+v32).
We set v5=v22=a122a25e5. There is α∈F∗ such that v32=αv22=αv5.
The family {e1,v2,v3,v4,v5} is a natural basis of N and its multiplication table is defined by: e12=v2+v3+v4,v22=v5,v32=αv5,v42=−(1+α)v5, and v52=0 with α(1+α)=0.
Let’s show that N is isomorphic to N5,10(β) for some nonzero β.
We set w3=v3+v4 and w4=av3+bv4 ; we have: 0=w3w4=av32+bv42=[αa−(1+α)b]v5 then b=1+ααa. For a=1, b=1+αα and w4=v3+1+ααv4.
The family {e1,v2,w3,w4,v5} is a natural basis of N and its multiplication table is defined by:
e12=v2+w3,v22=v5,w32=−v5,w42=[α−1+αα2]v5=1+ααv5,v52=0 with 1+αα=0.
We deduce that N\simeq N_{5,10}\big{(}\beta\big{)} where β=1+αα.
- dim(ann(N))=2. We have
e12=a12e2+a13e3+a14e4+a15e5,e22=a24e4+a25e5,e32=a34e4+a35e5,e42=e52=0 with a12a13=0, (a24,a25)=0 and (a34,a35)=0.
We set v2=a12e2+a15e5 and v3=a13e3+a14e4. We have: e12=v2+v3 and 0=e12e12=v22+v32 implies v32=−v22. Let’s set v4=v22=a122e22, so v32=−v4.
Then N=\big{<}e_{1},v_{2},v_{3},v_{4}\big{>}\oplus\big{<}v_{5}\big{>} direct sum of evolution algebras where ann(N)=Kv4⊕Kv5.
□
4.2.3 Six-dimensional classification
Theorem 4.2.5**.**
Let N be a nil indecomposable power-associative which is not associative evolution algebra of dimension six. Then N is isomorphic to one and only one of the seven algebras N6,19(α,β) to N6,26 in Table 3.
Proof.
Since dim(N)=6, we have dim(ann(N))<21dim(N)=3 and dim(ann(N))=1,2.
- dim(ann(N))=1.
[TABLE]
Since 0=(e12e4)ei=a14a4iei2, then a14a4i=0 (with i=2,3,5) and 0=(e12e5)ej=a15a5jej2 implies a15a5j=0 (with j=2,3,4),
we then distinguish four cases:
1.1) a14=a15=0 i.e. e12=a12e2+a13e3+a16e6.
We set v2=a12e2+a16e6 and v3=a13e3. We have e12=v2+v3 and
0=e12e12=v22+v32
implies v32=−v22.
Let’s set v6=v22=a122a26e6 ; so v32=−v6 and there are scalars α42,α43,α45,α46
not all zero such that e42=α42v2+α43v3+α45e5+α46v6. Similarly, there are scalars
α52,α53,α54,α56 not all zero such that
e52=α52v2+α53v3+α54e4+α56v6. Thus, we have
[TABLE]
These relations ensure power associativity of N.
The family {e1,v2,v3,e4,e5,v6} is a natural basis of N and its multiplication table is given by:
e12=v2+v3,v22=v6,v32=−v6,e42=α42(v2+v3)+α46v6,e52=α52(v2+v3)+α56v6,v62=0 with (α42,α46)=0 and (α52,α56)=0.
We have: \mathrm{ann}(N)=\big{<}v_{6}\big{>}, v2,v3∈ann2(N) and e1∈ann3(N).
Since e_{4}^{2},e_{5}^{2}\in\big{<}v_{2},v_{3},v_{6}\big{>}
then the possible types of N are: [1,4,1], [1,3,2] or [1,2,3].
1.1.1) Type [1,4,1] leads to α42=α52=0. we have \mathrm{ann}^{2}(N)=\big{<}v_{2},v_{3},e_{4},e_{5},v_{6}\big{>} and ann3(N)=N.
So \mathcal{U}_{3}\oplus\mathcal{U}_{1}=\big{<}e_{1},v_{6}\big{>} and (\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}=\big{<}v_{2}+v_{3}\big{>}.
N≃N6,19(α,β):e12=e2+e3,e22=e6,e32=−e6,e42=αe6,e52=βe6,e62=0 with α,β∈F∗.
1.1.2) Type [1,3,2] leads to α42=0 and α52=0 or α42=0 and α52=0.
a) For α42=0 and α52=0, we have: \mathrm{ann}^{2}(N)=\big{<}v_{2},v_{3},e_{5},v_{6}\big{>} and ann3(N)=N.
So \mathcal{U}_{3}\oplus\mathcal{U}_{1}=\big{<}e_{1},e_{4},v_{6}\big{>} and \big{(}\mathcal{U}_{3}\oplus\mathcal{U}_{1}\big{)}^{2}=\big{<}v_{2}+v_{3},e_{4}^{2}\big{>}. Thus
\dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=1,2.
\dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=1\Longrightarrow\alpha_{46}=0.
N≃N6,20(α,β):e12=e2+e3,e22=e6,e32=−e6,e42=α(e2+e3),e52=βe6,e62=0 with α,β∈F∗.
\dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=2\Longrightarrow\alpha_{46}\neq 0.
N≃N6,21(α,β,γ):e12=e2+e3,e22=e6,e32=−e6,e42=α(e2+e3)+βe6,e52=γe6,e62=0 with α,β,γ∈F∗.
b) For α42=0 and α52=0, we find again case a) by permuting vectors e4 and e5.
1.1.3) Type [1,2,3] leads to α42α52=0. We have \mathrm{ann}^{2}(N)=\big{<}v_{2},v_{3},v_{6}\big{>} and ann3(N)=N.
So \mathcal{U}_{3}\oplus\mathcal{U}_{1}=\big{<}e_{1},e_{4},e_{5},v_{6}\big{>} and \big{(}\mathcal{U}_{3}\oplus\mathcal{U}_{1}\big{)}^{2}=\big{<}v_{2}+v_{3},e_{4}^{2},e_{5}^{2}\big{>}.
Thus, \dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=1,2.
\dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=1\Longrightarrow\alpha_{46}=\alpha_{56}=0.
N≃N6,22(α,β):e12=e2+e3,e22=e6,e32=−e6,e42=α(e2+e3),e52=β(e2+e3),e62=0 with α,β∈F∗.
\dim\big{(}(\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}\big{)}=2\Longrightarrow(\alpha_{46},\alpha_{56})\neq 0.
We have N(α,β,γ,δ):e12=e2+e3,e22=e6,e32=−e6,e42=α(e2+e3)+βe6,e52=γ(e2+e3)+δe6,e62=0
with αγ=0 and (β,δ)=0.
We distinguish four cases:
i)
βδ=0 and αδ−βγ=0 gives algebra N6,23(α,β,γ,δ).
ii)
δ=0 and β=0 gives N≃N6,24(α,β,γ).
iii)
β=0 and δ=0. By permuting e4 and e5 we come back to case ii).
iv)
βδ=0 and αδ−βγ=0. Since δ=α−1βγ, we have e42=α(e2+e3+α−1βe6) and
e52=γ(e2+e3+α−1βe6). Let’s do the basis change w1=e4, w2=αe2, w3=αe3+βe6, w4=e1, w5=e5 and w6=α2e6. In the natural basis {w1,w2,w3,w4,w5,w6} the multiplication table leads to
N6,24(α−1,−α−3β,α−1γ).
1.2) a14=0 and a15=0 i.e. e12=a12e2+a13e3+a14e4+a16e6 and e42=a46e6 with a12a13a14a46=0.
Let’s set v2=a12e2+a16e6 ; v3=a13e3 and v4=a14e4 ; we have: e12=v2+v3+v4 and 0=e12e12=v22+v32+v42 then v42=−(v22+v32).
Let’s set v6=v22=a122a26e6. There is α∈F∗ such that v32=αv22=αv6. Thus, v42=−(1+α)v6 with 1+α=0.
There are scalars α52,α53,α54,α56 not all zero such that e52=α52v2+α53v3+α54v4+α56v6.
We have 0=e12e52=[α52+αα53−(1+α)α54]v6 implies α52=−αα53+(1+α)α54.
We have
0=e52e52=[α522+αα532−(1+α)α542]v6 leads to 0=α522+αα532−(1+α)α542. We have:
α522+αα532−(1+α)α542=(1+α)2α542+α2α532−2α(1+α)α53α54+αα532−(1+α)α542=α(1+α)(α54−α53)2.
So, 0=α522+αα532−(1+α)α542=α(1+α)(α54−α53)2⟹α54=α53.
Moreover α52=(1+α)α54−αα53=α54. Thus,
[TABLE]
Let’s show that N is isomorphic to one of the following algebras: N6,18(α′,β′), N6,20(α′,β′), N6,21(α′,β′,γ′) for some (α′,β′,γ′) with α′,β′,γ′∈F∗.
Let’s set w3=v3+v4, w4=e5 and w5=av3+bv4 ; we have: 0=w_{3}w_{5}=\big{(}\alpha a-(1+\alpha)b\big{)}v_{6} implies b=1+ααa.
For a=1, b=1+αα and w5=v3+1+ααv4.
The family {e1,v2,w3,w4,w5,v6} is a natural basis of N and its multiplication table is defined by: e12=v2+w3,v22=v6,w32=−v6,w42=α52(v2+w3)+α56v6,w52=[α−1+αα2]v6=[1+αα]v6,v62=0 with 1+αα=0 and (α52,α56)=0.
For α52=0, then N≃N6,19(α56,1+αα) with α56,1+αα∈F∗.
For α56=0, then N≃N6,20(α52,1+αα) with α56,1+αα∈F∗.
For α52α56=0, then N≃N6,21(α52,α56,1+αα) with α52,α56,1+αα∈F∗.
1.3) a14=0 and a15=0 i.e. e12=a12e2+a13e3+a15e5+a16e6 and e52=a56e6 with a12a13a15a56=0.
By permuting the vectors e4 and e5 of natural basis, we find again case 1.2
1.4) a14a15=0 i.e. e12=a12e2+a13e3+a14e4+a15e5+a16e6, e42=a46e6 and e52=a56e6 with a12a13a14a15a46a56=0. Let’s set v2=a12e2+a16e6, v3=a13e3, v4=a14e4 and v5=a15e5 ; we have: e12=v2+v3+v4+v5 and 0=e12e12=v22+v32+v42+v52 implies v52=−(v22+v32+v42).
Let’s set v6=v22=a122a26e6. There are α,β∈F∗ such that v32=αv6 and v42=βv6.
Thus, v52=−(1+α+β)v6 with 1+α+β=0.
The family {e1,v2,v3,v4,v5,v6} is a natural basis of N and its multiplication table is defined by:
e12=v2+v3+v4+v5,v22=v6,v32=αv6,v42=βv6,v52=−(1+α+β)v6,v62=0 with αβ(1+α+β)=0.
Let’s show that there are vectors x=x3v3+x4v4+x5v5 and y=y3v3+y4v4+y5v5 such that the family {e1,v2,v3+v4+v5,x,y,v6} is a natural basis of N. we set w3=v3+v4+v5.
The family {x,y} is orthogonal to w3 if and only if
[TABLE]
Then x and y are written
[TABLE]
The orthogonality of x and y leads to α(1+β)x3y3+β(1+α)x4y4−αβ(x3y4+x4y3)=0, so
[TABLE]
We distinguish three cases :
i)
if α+1=0, by taking x3=0 and y3=α−1(1+α)y4, it comes that
[TABLE]
ii)
If β+1=0, by taking x4=0 et y4=β−1(1+β)y3, it comes that
[TABLE]
α=β=−1. Identity (15) leads to x3y4+x4y3=0 and we must have x3y4−x4y3=2x3y4=0. Then
[TABLE]
In each of the three cases, there are α′,β′∈F∗ such that x2=α′v6 and y2=β′v6. Thus, the multiplication table of the natural basis {e1,v2,v3+v4+v5,x,y,v6} of N is given by:
e12=v2+w3, v22=v6, w32=−v6, x2=α′v6, y2=β′v6,v62=0 with α′β′=0. So
N is isomorphic to N6,19(α′,β′).
- dim(ann(N))=2. We have:
[TABLE]
We have 0=(e12e4)e2=a14a42e22 and 0=(e12e4)e3=a14a43e32, so a14a42=a14a43=0.
We then distinguish two cases
2.1) a14=0 i.e. e12=a12e2+a13e3+a15e5+a16e6 with a12a13=0.
Let’s set v2=a12e2+a15e5+a16e6 and v3=a13e3, we have e12=v2+v3 and 0=e12e12=v22+v32 leads to v32=−v22.
Let’s set v5=v22=a122e22. So v32=−v5 and there are scalars α, β, γ et δ not all zero such that e42=αv2+βv3+γv5+δe6. By setting w=γv5+δe6, we have e42=αv2+βv3+w.
Moreover 0=e12e42=αv22+βv32=(α−β)v5 leads to α=β. This relation ensure the power associativity of N.
2.1.1) The family {v5,w} is linearly independent.
In that case, the family {e1,v2,v3,e4,v5,w} is a natural basis of N and its multiplication table is defined by
[TABLE]
If α=0, then N=\big{<}e_{1},v_{2},v_{3},v_{5}\big{>}\oplus\big{<}e_{4},w\big{>}, direct sum of evolution algebras.
We deduce that α=0 and N is isomorphic to
N6,25(α):e12=e2+e3, e22=e5
e32=−e5, e42=α(e2+e3)+e6, e52=0, e62=0 with α∈F∗ and its type is [2,2,2].
We have \mathrm{ann}(N)=\big{<}e_{5},e_{6}\big{>} ; \mathrm{ann}^{2}(N)=\big{<}e_{2},e_{3},e_{5},e_{6}\big{>} and ann3(N)=N.
So, \mathcal{U}_{3}\oplus U_{1}=\big{<}e_{1},e_{4},e_{6}\big{>} and (U_{3}\oplus U_{1})^{2}=\big{<}e_{2}+e_{3},e_{6}\big{>}.
2.1.2) The family {v5,w} is linearly dependent. Then, there is γ∈F such that w=γv5. In this case N=\big{<}e_{1},v_{2},v_{3},v_{5},e_{4}\big{>}\oplus\big{<}v_{6}\big{>} direct sum of evolution algebras where ann(N)=Kv5⊕Kv6.
2.2) a14=0 i.e. e12=a12e2+a13e3+a14e4+a15e5+a16e6 and e42=a45e5+a46e6 with a12a13a14=0 and (a45,a46)=0.
We set v2=a12e2+a15e5+a16e6, v3=a13e3 and v4=a14e4.
we have e12=v2+v3+v4 and 0=e12e12=v22+v32+v42 leads to v42=−(v22+v32) and we set v5=v22=a122e22.
2.2.1) The family {v5,v32} is linearly independent. By setting v6=v32=a132e32, we have
N≃N6,26:e12=e2+e3+e4, e22=e5, e32=e6, e42=−(e5+e6), e52=0, e62=0 and its type is [2,3,1].
We have: \mathrm{ann}(N)=\big{<}e_{5},e_{6}\big{>} ; \mathrm{ann}^{2}(N)=\big{<}e_{2},e_{3},e_{4},e_{5},e_{6}\big{>} and ann3(N)=N.
So, \mathcal{U}_{3}\oplus\mathcal{U}_{1}=\big{<}e_{1},e_{6}\big{>} and (\mathcal{U}_{3}\oplus\mathcal{U}_{1})^{2}=\big{<}e_{2}+e_{3}+e_{4}\big{>}.
2.2.2) The family is {v5,v32} is linearly dependent. There is α∈F such that v32=αv5.
We have v42=−(1+α)v5 and N=\big{<}e_{1},v_{2},v_{3},v_{4},v_{5}\big{>}\oplus\big{<}v_{6}\big{>} direct sum of evolution algebras where ann(N)=Kv5⊕Kv6.
□
4.3 General classification
Let Ei,j be the j-th power-associative evolution algebra of dimension i over the commutative fields F. If Ei,j is nilalgebra, we denote Ni,j. According to Definition 4.0.1 and Theorem 3.5.4, we determine Ei,j for 1≤i≤6. Moreover, Ess denotes the semisimple component in the Wedderburn decomposition of Ei,j where s is the number of idempotents pairwise orthogonal.
Proposition 4.3.1**.**
Let E be a power-associative evolution algebra of dimension ≤4. Then E is isomorphic to one and only one of the algebras in Table 1.
Proposition 4.3.2**.**
Let E be a power-associative evolution algebra of dimension 5. Then E is isomorphic to one and only one of the algebras in Table 2.
Proposition 4.3.3**.**
*Let E be a power-associative evolution algebra of dimension 6. Then E is isomorphic to one and only one of the algebras in Table 3.
*
