Chains in evolution algebras
Yolanda Cabrera Casado, Maria Inez Cardoso Gonçalves, Daniel Gonçalves, Dolores Martín Barquero, Cándido Martín González
aa r X i v : . [ m a t h . R A ] A ug CHAINS IN EVOLUTION ALGEBRAS
YOLANDA CABRERA CASADO, MARIA INEZ CARDOSO GONC¸ ALVES, DANIEL GONC¸ ALVES,DOLORES MART´IN BARQUERO, AND C ´ANDIDO MART´IN GONZ ´ALEZ
Abstract.
In this work we approach three-dimensional evolution algebras from certain con-structions performed on two-dimensional algebras. More precisely, we provide four differentconstructions producing three-dimensional evolution algebras from two-dimensional algebras.Also we introduce two parameters, the annihilator stabilizing index and the socle stabilizingindex, which are useful tools in the classification theory of these algebras. Finally, we usemoduli sets as a convenient way to describe isomorphism classes of algebras. Introduction
A type of genetic algebras called evolution algebras appeared in [21]. This type of non-associative algebras arose in order to model Non-Mendelian genetics. For example, the clas-sification into isotopism classes of three dimensional evolution algebras are used to describethe spectrum of genetic patterns of three distinct genotypes during a mitosis process, see[12]. It should also be noted that these algebras have multiple connections with other areasof mathematics, such as graph theory and stochastic processes, see [7, 11, 20].The search for invariants to classify algebras is one of the mainstreams in Mathematics. Forexample, the celebrated Elliot classification program for C*-algebras has attracted the interestof a generation of researches, with outstanding results. In the setting of evolution algebras,the classification of nilpotent algebras is studied in [10, 14, 15]. The four dimensional perfectnon-simple evolution algebras over a field with mild restrictions are classified in [3]. Theclassification of general evolution algebras over finite fields is given in [13]. In the paper [16],the three-dimensional evolution algebras over a real field are classified. For three-dimensionalevolution algebras over a general field, which verify certain restrictions, the classification isdone in [2, 6], resulting in 116 non-isomorphic families. Given the huge number of differenttypes of evolution algebras of dimension 3, it is necessary to classify them according to othercharacteristics, in order to categorize them in a more efficient and convenient manner. Forexample, in [9], the evolution algebras of dimension three are classified according to propertiessuch as irreducibility or degeneracy. Recently, a new branch of study has arisen in [18]. Itsapplication in the setting of evolution algebras debuts in [8]: the determination of evolutionalgebras with a faithful associative and commutative representation. So far, this idea hasbeen pursued only in dimension 2.
Mathematics Subject Classification.
Key words and phrases. evolution algebra, diagonalizable evolution algebra, annihilator, socle.The first and the last two authors are supported by the Junta de Andaluc´ıa through projects FQM-336and UMA18-FEDERJA-119 and by the Spanish Ministerio de Ciencia e Innovaci´on through project PID2019-104236GB-I00, all of them with FEDER funds. The third author is supported by Conselho Nacional deDesenvolvimento Cient´ıfico e Tecnol´ogico (CNPq) grant numbers 304487/2017-1 and 406122/2018-0 andCapes-PrInt grant number 88881.310538/2018-01 - Brazil.
The philosophy of this work is to approach three-dimensional evolution algebras from cer-tain constructions performed on two-dimensional algebras. We give four constructions pro-ducing three-dimensional evolution algebras from two-dimensional algebras. Also we use twoparameters, asi(A) (Definition 3.3) and ssi( A ) (Definition 5.3), which help up to classifythese algebras. We also use moduli sets (see Subsection 2.1) as a convenient way to describeisomorphism classes of algebras.Roughly speaking the study of three-dimensional evolution algebras falls into two disjointclasses: those with nonzero annihilator (sections 3 and 4); and the ones with zero annihilator(sections 5 and 6). The classification of algebras A with nonzero annihilator fall into 5disjoint cases depending on two parameters: (1) the values of the annihilating stabilizing index(abbreviated asi(A)) and (2) the value of dim(ann( A )). In the case asi(A) = dim(ann(A)) = 1all the algebras come from a construction Adj ( B, α ) for a suitable evolution algebra B ofdimension two (see Theorem 4.2). The construction Adj is described in Definition 3.9.The (3-dimensional) algebras A with zero annihilator have nonzero socle, so we can usedim(Soc( A )) as one of the parameters for the classification task. The main results are Propo-sition 6.3, in which dim(Soc( A )) = 3, Theorem 6.6, Theorem 6.13 and Theorem 6.14, for thecase dim(Soc( A )) = 2, and Theorem 6.20 for the case dim(Soc( A )) = 1. The algebras withdim(Soc( A )) = 3 are either simple or direct sum of evolution algebras of dimension less orequal to 2.The evolution algebras A with dim(Soc( A )) = 2 are separated in three large classes: (1)those whose socle has the extension property (Theorem 6.6); (2) those for which Soc( A ) =span( { e , e + e } ) with { e i } natural (Theorem 6.13); and (3) those for which Soc( A ) =span( { e + e , e + e } ) with { e i } natural (Theorem 6.14). In all the cases the constructionAdj defined in Definition 5 . A )) = 1 wedescribe these algebras in terms of two new constructions: Adj and Adj (Subsection 6.3),see Theorem 6.20 and Theorem 6.24 depending on whether Soc( A ) = 0 or Soc( A ) = 0,respectively.At last, notice that a convenient way to describe the isomorphism classes of a collectionof algebras is via a moduli set, that is, via an action of a group in a set such that theisomorphism classes of the algebras are in one-to-one correspondence with the orbits of theaction. We will provide moduli sets for the three dimensional evolution algebras. In fact, theclassification work we develop in this paper can be seen as the description of moduli sets forthree-dimensional evolution algebras.2. Preliminaries An evolution algebra over a field K is a K -algebra A which has a basis B = { e i } i ∈ Λ suchthat e i e j = 0 for every i, j ∈ Λ with i = j . Such a basis is called a natural basis . From nowon, all the evolution algebras we will consider will be finite dimensional and Λ will denote afinite set { , . . . , n } .Let A be an evolution algebra with a natural basis B = { e i } i ∈ Λ . Denote by M B = ( ω ij )the structure matrix of A relative to B , i.e., e i = P j ∈ Λ ω ji e j . Notation 2.1.
We use the following notations in this paper:
HAINS IN EVOLUTION ALGEBRAS 3 (1) We denote by N the natural numbers including 0, by N ∗ the set N \ { } , by Z theintegers and, if K is a field, then we denote by M n ( K ) the algebra of n × n matriceswith coefficientes in K .(2) For n ∈ N ∗ , the notation K n stands for the cartesian product K n × · · · × K , while thenotation K h n i is used for K h n i := { λ n : λ ∈ K } and ( K × ) h n i := { λ n : λ ∈ K × } .(3) For integers 1 ≤ i, j ≤ n , we denote by E ij the n × n matrix obtained by permuting the i th and j th rows of the identity matrix. For example, for n = 2 we have E = (cid:18) (cid:19) , while for n = 3 we have E = ! . To be formal we should specify that the matrix size depends on the chosen n , but wewill refrain from doing so in order to get a lighter notation. The reader should be ableto deduce the value of n from the context.(4) We denote the cyclic group of order two by Z , and if M is a matrix in M n ( K ) thenwe use matrix powers M i , for i ∈ Z , with the meaning M := Id and M := M .(5) Let G be a group acting on a set X . We denote the set of orbits of X under the actionof G by X/G .(6) Let B be a K -algebra. We denote by End K ( B ) the algebra of linear maps from B to B and by End( B ) that of endomorphisms of B as algebra.(7) We denote by Sym ( B ) the K -vector space of all symmetric bilinear forms on B . Definition 2.2.
Let A be an evolution algebra, B = { e i } i ∈ Λ a natural basis and u = P i ∈ Λ α i e i an element of A . The support of u relative to B , denoted supp B ( u ), is defined as the setsupp B ( u ) = { i ∈ Λ | α i = 0 } . If X ⊆ A , we put supp B ( X ) = ∪ x ∈ X supp B ( x ). Following [5],an evolution subalgebra A ′ of an evolution algebra A is said to have the extension property if A ′ has a natural basis which can be extended to a natural basis of A . Remark 2.3.
Assume that { e i } i ∈ I is a natural basis of an evolution algebra A . A usefulcriterion for an element z of A with z = 0 to be natural (i.e. embeddable in a natural basis)is that dim(span( { e i : i ∈ supp( z ) } )) = 1 (see [1, Teorema 3.3]).We recall that an evolution algebra A is non-degenerate if there exists a natural basis { e i } such that e i = 0 for every i . In [5, Corollary 2.19], it is proved that this definition does notdepend on the chosen basis since to be non-degenerated is equivalent to have ann( A ) = 0.With this in mind, we have Proposition 2.4.
Let A be a non-degenerate evolution algebra, then(1) Let B = { e i } i ∈ Λ be a natural basis of A . If z, z ′ are different elements such that zz ′ = 0 , then supp( z ) ∩ supp( z ′ ) has cardinal different from (supports relative to B ).(2) If A is a -dimensional evolution algebra with natural basis { e , e , e } and dim( A ) =2 , then any other natural basis of A is (up to permutations and nonzero multiples) ofthe form { e + ke , e + k ′ e , e } , where k, k ′ are different scalars in K × .Proof. Consider z, z ′ different and zz ′ = 0, if the intersection of their supports has cardinal 1(say i ∈ supp( z ) ∩ supp( z ′ )), then 0 = e i contradicting the fact that A is non-degenerate. Letus prove the second assertion. Since dim( A ) = 2 any natural vector has support of cardinal Y. CABRERA, M. I. GONC¸ ALVES, D. GONC¸ ALVES, D. MART´IN, AND C. MART´IN less or equal to 2. Indeed, suppose there is a natural element with support of cardinal 3.Then, by Remark 2.3, dim(span( { e i : i = 1 , , } )) = 1. This implies that dim( A ) = 1, acontradiction.A priori, the cardinals of the support of the three elements in a natural basis have thepossibilities: (2 , , , (2 , , , (2 , , , (1 , , . Let us check that the first possibility, (2 , , , ,
1) gives that the supports are (up to permutationsand scaling if necessary): { , } , { , } , { } . So the basis is as in the statement of the proposi-tion. The third possibility implies again that the support of cardinal 2 has intersection withsome of the supports of cardinal one, a contradiction. The last possibility, (1 , , (cid:3) Moduli sets.
In this subsection we introduce the moduli sets that we will be used todescribe isomorphism classes of algebras, more concretely in our case to describe isomorphismclasses of three-dimensional evolution algebras.
Definition 2.5.
For a class of K -algebras C , we will say that ( G, M ) is a moduli set for C if: (1) G is a group, M is a G -set and (2) the set of isomorphism classes of algebras in C isin one-to-one correspondence with the orbits of M /G , that is, orbits of M under the actionof G .An easy example from the theory of evolution algebras is the following. Example 2.6.
Consider the class of two-dimensional simple evolution algebras over a fixedfield K . It is easy to check that for any algebra in this class there is a natural basis such thatthe structure matrix of the algebra, relative to this basis, is of one of the following forms:Type I (cid:18) yx (cid:19) , with xy − = 0 , x, y ∈ K × . Type II (cid:18) yx (cid:19) , with xy = 0 . Type III (cid:18) yx (cid:19) , with xy = 0 . (1)Let us consider, for instance, the class of algebras of type I. It can be checked that twoalgebras of type I are isomorphic if and only if either they have the same structure matrixor the structure matrix of one of them is (cid:18) yx (cid:19) and the other one is (cid:18) xy (cid:19) . Thus, we candefine a moduli set ( F , M ), where M = { (cid:18) xy (cid:19) : xy = 1 , x, y ∈ K × } and take the (multiplicative) group F = {± } acting on M in such a way that 1 ∈ F actsas the identity, while ( − · (cid:18) xy (cid:19) := (cid:18) yx (cid:19) . HAINS IN EVOLUTION ALGEBRAS 5
Therefore the isomorphism classes of algebras of type I are in one to one correspondence withthe F -set M / F . In other words, the moduli set ( F , M ) classifies the algebras of type I.Furthermore, we can identify M with the Zarisky open subset { ( x, y ) ∈ K : xy = 1 , x, y ∈ K × } modulo the identification of each ( x, y ) with ( y, x ).If we had K = R , then we could represent this F -set in the real plane, removing the axisand the graphic of the function y = 1 /x , and identifying symmetric points relative to the line y = x : xy y = 1 /x y = x zone Izone IIzone IIIzoneIVzone V so that we would have 5 zones and each point representing an isomorphism class of algebras oftype I. The fact that the set consists of 5 connected components has to do with the existenceof 5 homotopy classes of algebras of type I. However we will not pursue this homotopy ideasfurther.Complementing our comment about moduli sets in the introduction, we observe that oncewe have constructed a moduli set ( M , G ) for a class of algebras C , we have parametrized theisomorphism classes of C by the orbits of M /G and so this is a useful tool for classificationtasks. Next we describe some of the moduli sets that will be used in our work (some ofthe moduli set that will appear require further background to be described, and so we willintroduce then as needed).(D ( K ) ⋊ Z , K \ { } ) : Among the moduli sets used in the classification of three-dimensionalevolution algebras, we will use the group of non uniform scales jointly with the symmetriesdefined below. Roughly speaking, this moduli set consists of the diagonal group of 2 × ( K ) = (cid:26)(cid:18) λ λ (cid:19) : λ i ∈ K × (cid:27) be the group of diagonal matrices. Con-sider the subgroup of GL ( K ) generated by D ( K ) and E = (cid:18) (cid:19) , which we will denoteby D ( K ) ⋊ Z . We haveD ( K ) ⋊ Z = ( λ λ ! : λ i ∈ K × ) ⊔ ( λ λ ! : λ i ∈ K × ) This is a subgroup of the general linear group GL ( K ), therefore there is an induced rep-resentation (D ( K ) ⋊ Z ) × ( K \ { } ) ! ( K \ { } ) such that, for M ∈ D ( K ) ⋊ Z and v ∈ K \ { } (as column vector), the action M v is the usual multiplication. So we have amoduli set (D ( K ) ⋊ Z , K \ { } ). Observe that two vectors v, v ′ ∈ K \ { } are in the same Y. CABRERA, M. I. GONC¸ ALVES, D. GONC¸ ALVES, D. MART´IN, AND C. MART´IN orbit under the action of D ( K ) ⋊ Z if and only if the number of zero entries in v coincidewith the number of zero entries in v ′ . A set of representatives of the orbits are (1 ,
0) and(1 , K × ) h i , K × ) : This moduli set is constructed by considering the group ( K × ) h i := { k : k ∈ K × } and its natural action on the set K × given by(2) ( K × ) h i × K × ! K × , such that for any g ∈ ( K × ) h i and λ ∈ K × we have g · λ = gλ . Note that the cardinal of theset of orbits K × / ( K × ) h i depends greatly of the nature of the ground field K . For instance, if K is algebraically closed it has cardinal one. If K = R , then K × / ( K × ) h i has cardinal 2 andif K = Q , then there are countable many orbits.3. Annihilator chain
In this section we will define the construction of new algebras by the procedure of adjunctionof type one. We will study the isomorphism problem for this class of algebras. In order todo this work we use the upper annihilating series and we define the annihilator stabilizingindex of an algebra. We will apply all this tools and results in the next section to classify thethree-dimensional evolution algebras with non-zero annihilator.We recall that given a nonassociative algebra A , we have the following sequences of sub-spaces: A = 0 , A = A, A k +1 = k X i =1 A i A k +1 − i for k > . If J is an ideal of A , we consider J as an algebra and the powers of J are defined as in theprevious case. An algebra (ideal) A is nilpotent if there exist n ∈ N ∗ such that A n = 0. Definition 3.1.
Let A be an algebra and denote by P ( A ) the power set of A . We define themap ρ A : P ( A ) ! P ( A ), where ρ A ( S ) = SA ∪ AS = { sa : s ∈ S, a ∈ A }∪{ as : s ∈ S, a ∈ A } for any S ∈ P ( A ). We denote ρ A ( x ) := ρ A ( { x } ).In our classification results we will make use of the upper annihilating series, as defined in[10, Definition 3.3]. We recall this definition bellow. Definition 3.2.
Let A be an algebra. We define ann (0) ( A ) := { } and ann ( i ) ( A ) in thefollowing way ann ( i ) ( A ) / ann ( i − ( A ) := ann( A/ (ann ( i − ( A )) . The chain of ideals: { } = ann (0) ( A ) ⊆ ann (1) ( A ) ⊆ · · · ⊆ ann ( i ) ( A ) ⊆ · · · is called the upper annihilating series .Observe that ann (1) ( A ) = ann( A ) = { x ∈ A : xA = Ax = 0 } .Next we define the key classifying parameter in this section, namely the annihilator stabi-lizing index. Definition 3.3.
Let A be an algebra. If there exists k such that k = min { q : ann ( q ) ( A ) =ann ( q +1) ( A ) } , then we call it the annihilator stabilizing index of A , denoted by asi(A). HAINS IN EVOLUTION ALGEBRAS 7
Remark 3.4.
Observe that if A is finite dimensional then asi(A) always exists. If the chainof annihilators does not stabilize, we write asi(A) = ∞ . Using the map ρ A of Definition 3.1 we obtain the following useful description of ann ( i ) ( A ). Lemma 3.5.
Let A be an algebra. Then: (i) ann ( i ) ( A ) = { x ∈ A : xA ∪ Ax ⊂ ann ( i − ( A ) } for i ≥ . (ii) ann ( i ) ( A ) = { x ∈ A : ρ kA ( x ) ⊂ ann ( i − k ) ( A ) for all k ∈ { , , . . . , i }} .Proof. First we prove (i). If a ∈ ann ( i ) ( A ), then aA ∪ Aa ⊂ ann ( i − ( A ) by definition. Forthe other inclusion, notice that if aA ∪ Aa ⊂ ann ( i − ( A ), then ¯ a (cid:16) A/ ann ( i − ( A ) (cid:17) = ¯0 and (cid:16) A/ ann ( i − ( A ) (cid:17) ¯ a = ¯0, so ¯ a ∈ ann (cid:16) A/ ann ( i − ( A ) (cid:17) = ann ( i ) ( A ) / ann ( i − ( A ) . Therefore a ∈ ann ( i ) ( A ). For (ii), first recall that if k = 0, then ρ A is the identity map. For k = 1, if x ∈ ann ( i ) ( A ) then we have by definition that ρ A ( x ) ⊂ ann ( i − ( A ). Hence ρ A ( x ) ⊂ ann ( i − ( A )(by (i)) and iterating k times we obtain that ρ kA ( x ) ⊂ ann ( i − k ) ( A ). For the other inclusion,let x ∈ A be such that ρ kA ( x ) ⊂ ann ( i − k ) ( A ) for all k ∈ { , , . . . , i } . In particular, for k = 0we get x ∈ ann ( i ) ( A ). (cid:3) Before we proceed to the next subsection we study the relation between the absorptionradical of an algebra A and its upper annihilating series. We start with the definition of theabsorption property and absorption radical. Definition 3.6.
Let I be an ideal of an algebra A . We say that I has the absorption property if xA ∪ Ax ⊂ I implies x ∈ I . The absorption radical of A is the intersection of all ideals of A having the absorption property, denoted by rad( A ). Proposition 3.7.
Let A be an algebra. (i) x ∈ ann ( k ) ( A ) if and only if ρ kA ( x ) = 0 . (ii) ann ( k ) ( A ) ⊆ rad( A ) for any k . (iii) If there exists an annihilator stabilizing index k , then ann ( k ) ( A ) is an absorption ideal.Therefore rad( A ) = ann ( k ) ( A ) . (iv) ann ( i ) ( A ) is a nilpotent ideal of A for all i .Proof. For (i) we take k = i in Lemma 3.5 (ii).(ii) We will prove that ann ( k ) ( A ) ⊂ I for any absorbent ideal I of A . This is triviallytrue for k = 0. Assume that ann (0) ( A ) , . . . , ann ( k − ( A ) are contained in I . We prove thatann ( k ) ( A ) ⊂ I . For this, let x ∈ ann ( k ) ( A ). Then, by item (i) in Lemma 3.5, we have that xA and Ax are contained in ann ( k − ( A ) ⊂ I . Since I is absorbent we conclude that x ∈ I .(iii) Let k := asi(A). We prove that rad( A ) ⊂ ann ( k ) ( A ). For this, it suffices to provethat ann ( k ) ( A ) is absorbent. Suppose that xA ∪ Ax ⊆ ann ( k ) ( A ). We have to prove that x ∈ ann ( k ) ( A ). Since xA ∪ Ax ⊆ ann ( k ) ( A ), we have that ¯ x ∈ ann (cid:16) A/ ann ( k ) ( A ) (cid:17) =ann ( k +1) ( A ) / ann ( k ) ( A ) = n ¯0 o , which implies that x ∈ ann ( k ) ( A ). Hence rad( A ) ⊆ ann ( k ) ( A ).By (ii) we obtain that rad( A ) = ann ( k ) ( A ).(iv) Let J = ann ( i ) ( A ) and observe that J ⊇ J ⊇ J ⊇ . . . . Moreover J = J J ⊆ ann ( i − ( A ) by (i). Then ann ( i − ( A ) ⊇ J ⊇ J ⊇ J ⊇ . . . , so ann ( i − ( A ) ⊇ J = J J + J J + J J ⊇ J ⊇ J . . . Reasoning in the same way we get ann ( i − α ) ( A ) ⊇ J α . Inparticular, for α = i we have that J i = 0. (cid:3) Y. CABRERA, M. I. GONC¸ ALVES, D. GONC¸ ALVES, D. MART´IN, AND C. MART´IN
Remark 3.8.
The absorption radical in an evolution algebra is a basic ideal, see [4, Propo-sition 3.1] . Adjunction of type one.
In this subsection we will study how to construct an evolutionalgebra of dimension n + 1 by adjunction of an annihilator element to an n -dimensionalevolution algebra with a symmetric bilinear form. Definitions 3.9.
Let A be an evolution K -algebra endowed with a symmetric bilinear form α : A × A ! K such that α diagonalizes with respect to a natural basis of A . Such a symmetricbilinear form will be called compatible symmetric bilinear form . We will say that ( A, α ) is a diagonalizable evolution algebra if it is endowed with a compatible symmetric bilinear form α . We will drop the bilinear form α if it is clear from the context. If A is a diagonalizableevolution K -algebra we can define a new evolution algebra K × A with product( λ, x )( λ ′ , x ′ ) = ( α ( x, x ′ ) , xx ′ ) . The fact that K × A is an evolution algebra is easily seen considering a natural basis { e i } i ∈ I of A which also diagonalizes α . Then defining f := (1 ,
0) and f i = (0 , e i ) for i ∈ I ,we get a natural basis { f i } i ∈ I ∪{ } of K × A . The algebra K × A will be called the adjunctionof an annihilating element to the diagonalizable evolution algebra ( A, α ). This algebra willbe denoted Adj ( A, α ). Two diagonalizable evolution algebras are isometrically isomorphic if there exists an isomorphism of evolution algebras which preserve the symmetric bilinearforms.
Remark 3.10.
If B is a diagonalizable evolution algebra then dim(ann(Adj ( B, α ))) = 1 +dim(ann( B )) . Remark 3.11.
Let ( A , α ) and ( A , α ) be two diagonalizable evolution algebras. If thereexists an isometric isomorphism f : A ! A , then Adj ( A , α ) ∼ = Adj ( A , α ). Lemma 3.12.
Let A be an evolution algebra with dim(ann( A )) = 1 . Then A ∼ = Adj ( B, α ) ,where B := A/ ann( A ) and α is a compatible bilinear form in B . Furthermore, if asi(A) = 1 then ann( B ) = 0 .Proof. We can write ann( A ) = K z for a certain z ∈ A . We will define a compatible scalarproduct in B , and then we will prove that A ∼ = Adj ( B, α ) for a suitable α .First, we have A = ann( A ) ⊕ C for some subspace C which can be chosen to have a basisof natural vectors. If y , y ∈ C we have y y = θ ( y , y ) z + q ( y , y ), where θ : C × C ! K isa symmetric bilinear form and q : C × C ! C is symmetric and bilinear. Then, for any twoelements a , a ∈ A we have a i = k i z + c i (for i = 1 ,
2) and a a = ( k z + c )( k z + c ) = θ ( c , c ) z + q ( c , c ) . Thus, the only thing needed to prove that A = Adj ( C, θ ) is that C is an evolution algebrarelative to q and that θ diagonalizes in some natural basis of C . In order to do that, take anynatural basis { e i } i ∈ Λ of A and write e i = k i z + y i for any i ∈ Λ. Then, the set { y i } i is a systemof generators of the vector space C , because for any c ∈ C we have c = P i h i e i (for somescalars h i ∈ K ) and hence c = P i h i k i z + P i h i y i . Thus P h i k i = 0 and c = P h i y i . On theother hand the vectors y i ’s are pairwise orthogonal: if i = j we have 0 = e i e j = y i y j . Then,there is a basis B = { y i , . . . , y i q } of C which satisfies y i n y i m = 0 for i n = i m . Consequently0 = y i n y i m = θ ( y i n , y i m ) z + q ( y i n , y i m ) ⇒ θ ( y i n , y i m ) = 0 , q ( y i n , y i m ) = 0 . HAINS IN EVOLUTION ALGEBRAS 9
This proves that C is an evolution algebra for the product q with natural basis B andfurthermore θ diagonalizes in B . Also(3) A = Adj ( C, θ ) . On the other hand, the map f : C ! B = A/ ann( A ) such that f ( y ) = ¯ y (the class of y modulo ann( A )) is an isomorphism of vector spaces and for any c , c ∈ Cf ( c c ) = f ( θ ( c , c ) z + q ( c , c )) = q ( c , c ) = c c = f ( c ) f ( c ) . So C is isomorphic as an evolution algebra to A/ ann( A ) = B . If we define in B the uniquecompatible symmetric bilinear form α that makes of f an isometric isomorphism then, apply-ing Remark 3.11, we have Adj ( C, θ ) ∼ = Adj ( B, α ), from which we conclude that(4) A ∼ = Adj ( A/ ann( A ) , α ) . For the last assertion in the statement of the Lemma, observe also that ann( A/ ann( A )) = 0because asi(A) = 1. (cid:3) Example 3.13.
Consider the evolution algebras A and A with natural bases { e , e , e } and { f , f , f } and product relative to these bases given by the matrices and respectively . These evolution algebras satisfy that asi(A i ) = 1, dim(ann( A i )) = 1, and A / ann( A ) ∼ = A / ann( A ) but they are not isomorphic. To prove this statement, note that A = ann( A ) ⊕ I , where I is the ideal generated by { e , e } , i.e., it is a reducible evolution algebra, butthis is not the case for A . Assume that there exists an evolution ideal J in A such that A = ann( A ) ⊕ J . Denote by { α f + α f + α f , β f + β f + β f } a natural basis of J .Then ( α f + α f + α f )( β f + β f + β f ) = 0 implies α = 0 , β = 0 or α = 0 , β = 0.Assume the first case (the other one is analogue). Since J is a subalgebra, ( α f + α f ) ∈ J ,i.e., α f = x ( α f + α f ) + y ( β f + β f ) . This implies xα = 0 and, since α = 0 (else α f + α f + α f ∈ ann( A )), necessarily x = 0and consequently yβ = 0 and yβ = α . We know that α = 0 hence y = 0 and so β = 0. Useagain that J is a subalgebra to obtain ( β f ) ∈ J , i.e. β ( f + f ) = x ′ ( α f + α f ) + y ′ β f ,that is x ′ α = 0, implying again x ′ = 0 and, consequently, β = 0, a contradiction.3.2. Isomorphisms between adjunction algebras of type one.
In this subsection westudy the relation between isomorphism between evolution algebras and isomorphism of theiradjunctions. We start by showing that for a diagonalizable evolution algebra a scaling of theassociated bilinear form yields isomorphic adjunctions.Let B be a K -algebra with α : B × B ! K and consider a new inner product β : B × B ! K given by β ( x, y ) = kα ( x, y ) for a fixed nonzero k ∈ K . Then denote B α := K × B with theproduct ( λ, x )( λ ′ , x ′ ) = ( α ( x, x ′ ) , xx ′ ) and B β := K × B with the product ( λ, x )( λ ′ , x ′ ) = ( β ( x, x ′ ) , xx ′ ). Observe that the map F : B α ! B β such that F ( λ, x ) = ( kλ, x ) is an isomor-phism of K -algebras. Indeed: F (( λ, x )( µ, y )) = F ( α ( x, y ) , xy ) = ( kα ( x, y ) , xy ) = ( β ( x, y ) , xy )= ( kλ, x )( kµ, y ) = F ( λ, x ) F ( µ, y ) . In particular, when (
B, α ) is a diagonalizable evolution algebra, (
B, β ) is also a diagonal-izable evolution algebra and Adj ( B, α ) and Adj ( B, β ) are isomorphic, via the isomorphism F : Adj ( B, α ) ! Adj ( B, β ) such that F ( λ, x ) = ( kλ, x ). HenceAdj ( B, α ) ∼ = Adj ( B, kα )for k ∈ K × := K \ { } .Next we prove that if two diagonalizable evolution algebras are isometrically isomorphicthen their adjunctions are isomorphic. Proposition 3.14.
Let ( B i , α i ) ( i = 1 , ) be two diagonalizable evolution algebras and let β : B ! B be an algebra isomorphism β : B ! B . Assume also that there is φ ∈ B ∗ :=hom k ( B , K ) (the usual dual space) satisfying φ ( xy ) = α ( β ( x ) , β ( y )) − α ( x, y ) for any x, y ∈ B . Then the map F : Adj ( B , α ) ! Adj ( B , α ) such that F ( λ, x ) :=( λ + φ ( x ) , β ( x )) is an algebra isomorphism.Proof. It is easy to check the linearity of F and its bijective character. Furthermore: F (( λ, x )( µ, y )) = F ( α ( x, y ) , xy ) = ( α ( x, y ) + φ ( xy ) , β ( xy )) = ( α ( β ( x ) , β ( y )) , β ( x ) β ( y )) =( λ + φ ( x ) , β ( x ))( µ + φ ( y ) , β ( y )) = F ( λ, x ) F ( µ, y ) . (cid:3) Remark 3.15.
In particular if β : B ! B is an isometric isomorphism then, taking φ = 0,we get that Adj ( B , α ) ∼ = Adj ( B , α ).In our next result we describe when an isomorphism between adjunctions imply an isor-mophism between the algebras. For this, recall that if B be an evolution algebra with zeroannihilator, and with a compatible inner product α , then ann(Adj ( B, α )) = K × { } (thisfollows from Remark 3.10). Proposition 3.16.
Assume that ( B i , α i ) are two diagonalizable evolution algebras and B has zero annihilator. Assume that F : Adj ( B , α ) ! Adj ( B , α ) is an isomorphism. Then,scaling the inner product of B if necessary, we have that:1) There is an isomorphism of algebras β : B ! B .2) There is an element φ ∈ B ∗ such that φ ( xy ) = α ( β ( x ) , β ( y )) − α ( x, y ) for any x, y ∈ B .Proof. Let F : K × B ! K × B be as in the statement of the proposition. Then, up toscalar multiples, we have that F ((1 , ,
0) (because (1 , ∈ ann( Ad ( B , α )) implies F ((1 , ∈ ann( Ad ( B , α )) = K × F (( λ, λ,
0) for any scalar λ . Now, F ((0 , x )) = ( φ ( x ) , β ( x )) for some linear maps φ : B ! K and β : B ! B . We prove that β is an monomorphism: if β ( x ) = 0 then F ((0 , x )) = ( φ ( x ) ,
0) = F (( φ ( x ) , F is anisomorphism, (0 , x ) = ( φ ( x ) ,
0) implies x = 0. Also β is epimorphism, since for any y ∈ B we have (0 , y ) = F (( λ, x )) for some λ ∈ K and x ∈ B . So y = β ( x ). Next we check that HAINS IN EVOLUTION ALGEBRAS 11 β ( xy ) = β ( x ) β ( y ) for any x, y , and simultaneously we check condition 2) in the proposition.For this, notice that F ((0 , x )(0 , y )) = F (( α ( x, y ) , xy )) = ( a ( x, y ) + φ ( xy ) , β ( xy )) , while F ((0 , x )) F ((0 , y )) = ( φ ( x ) , β ( x ))( φ ( y ) , β ( y )) = ( α ( β ( a ) , β ( y )) , β ( x ) β ( y )) . So we get β ( xy ) = β ( x ) β ( y ) and φ ( xy ) = α ( β ( x ) , β ( y )) − α ( x, y ) for any x, y ∈ B . (cid:3) Corollary 3.17.
Let B i , i = 1 , be diagonalizable perfect evolution algebras. Then we have Adj ( B , α ) ∼ = Adj ( B , α ) if and only if B ∼ = B .Proof. Since the perfection of a finite dimensional evolution algebra implies that its anni-hilator is zero, Proposition 3.16 gives that Adj ( B , α ) ∼ = Adj ( B , α ) implies B ∼ = B .Reciprocally, assume β : B ! B is an isomorphism. Let { u i } be a natural basis of B (nec-essarily it diagonalizes α ) and notice that the (natural) basis { β ( u i ) } of B diagonalizes α .Let ( ω ji ) be the structure matrix of B (so u i = P j ω ji u j for any i ). Let (˜ ω ji ) be the inversematrix of ( ω ji ), that is, P j ω ji ˜ ω kj = δ ki (Kronecker delta) for any i, k . Define φ : B ! K bywriting φ ( u k ) := P i ˜ ω ik ( α ( β ( u i ) , β ( u i )) − α ( u i , u i )) for any k . Then it is easy to check that φ ( u i u j ) = α ( β ( u i ) , β ( u j )) − α ( u i , u j ) for any i, j , whence φ ( xy ) = α ( β ( x ) , β ( y )) − α ( x, y )for any x, y ∈ B , and the result follows from Proposition 3.14. (cid:3) Remark 3.18.
A moduli set for algebras of type Adj ( B, α ) is the following. Fix an evolutionalgebra B and consider the group G = Aut( B ) × B ∗ (where B ∗ is the dual space of B ) endowedwith the product ( η, S )( θ, T ) = ( ηθ, Sθ + T ) , where η, θ ∈ Aut( B ) and S, T ∈ B ∗ . Define U to be the K -space of all compatible symmetricbilinear forms α : B × B ! K . There is an action G × U ! U given by(5) ( θ, T ) α := α ′ , where α ′ ( θ ( x ) , θ ( y )) − α ( x, y ) = T ( xy ) , for any x, y ∈ B . Then ( G , U ) is a moduli set, which will be used to classify algebras of typeAdj ( B, α ) (see Proposition 3.16).4.
Classification in terms of the upper annihilating series
In this section we will study degenerate three-dimensional evolution algebras A in termsof their upper annihilating series. Before we state the classification theorem we prove thefollowing lemma. Lemma 4.1.
Let A be an evolution algebra with natural basis B = { e i } , dim(ann ( A )) = 1 and product e = 0 , e = αe , e = βe + γe + δe with γ = 0 or δ = 0 . Then there existsanother natural basis { f i } such that f = 0 , f = f and f ∈ span( { f , f } ) .Proof. First, scaling e we may consider without lost of generality that α = 1. Now, we takethe natural basis { f i } with f = e , f = xe + e and f = x ′ e + e . Note that we can choose x and x ′ such that f ∈ span( { f , f } ). (cid:3) We have the following classification:
Theorem 4.2.
Let A be a three-dimensional evolution algebra over a field K and assume that ann( A ) = { } . Then we have that one, and only one, of the following possibilities holds: (1) If asi(A) = 3 then A is a nilpotent evolution algebra.(2) If asi(A) = 2 , and ann( A ) has dimension , then A is isomorphic to the evolutionalgebra with structure matrix ! , or with structure matrix ! , or withstructure matrix β ! with β = 0 (notice that the first two algebras are non-isomorphic and also non-isomorphic to any algebra of the third type). Furthermore,the algebras with the previous structure matrix (depending on β ), are classified by themoduli set (2) .(3) If asi(A) = 2 , and ann( A ) has dimension , then A is isomorphic to the nilpotentevolution algebra with structure matrix ! .(4) If asi(A) = 1 , and ann( A ) has dimension , then A is isomorphic to the evolutionalgebra with structure matrix ! .(5) If asi(A) = dim(ann(A)) = 1 then A ∼ = Adj ( B, α ) , where B is a two-dimensionalevolution algebra with zero annihilator provided with a compatible symmetric bilinearform α . The classification of evolution algebras in this class is given by the moduli set (5) .Proof. Notice that the maximum value of asi(A) is 3. In this case A is nilpotent. Sincethe classification of low dimensional nilpotent evolution algebras has been achieved up todimension 5 (see [10]), we will not pursue further this class of algebras. So we will focus onthe cases asi(A) ≤
2. Let us first consider asi(A) = 2 and dim(ann( A )) = 1. Suppose thatdim(ann (2) ( A )) = 2. Take a generator e of ann( A ). We have ann (2) ( A ) = K e ⊕ K e for asuitable element e ∈ A . We know that e A ⊂ K e . At this point we know that e = λe .Choose an a ∈ A which is linearly independent with e and e . So { e , e , a } is a basis of A .If e a = 0 then we have a natural basis { e , e , a } . On the contrary e a = ke for a suitable k ∈ K × . Define e := xe + ye + za (so that a wise choice of x, y, z ∈ K will give a naturalbasis { e , e , e } of A ). To pick x, y, z , notice that we need0 = e e = e ( xe + ye + za ) = yλe + zke , and hence yλ + zk = 0, so that z = − yλ/k , which proves that such a natural basis { e , e , e } exists. We have e = 0, e = λe . As dim(ann (2) ( A )) = 2 then e = αe + βe + δe with δ = 0. By Lemma (4.1) we have the desired basis.If α = 0 then { e , e , β − e } is a natural basis with structure matrix ! . If α = 0 then { α β − e , αβ − e , β − e } is a natural basis with structure matrix ! . Notice that the two algebras above are not isomorphic, since any isomophism between thesetwo algebra preserves the annihilators and the set of nonzero idempotents.
HAINS IN EVOLUTION ALGEBRAS 13
Now suppose that dim(ann (2) ( A )) = 3. It is easy to check the structure matrix of thisalgebra is α β ! , where α and β are nonzero. After a change of basis we can take α = 1. Moreover, if we havetwo evolution algebras with structure matrices β ! and β ′ ! respectively, an easythough tedious computation reveals that both algebras are isomorphic if and only if β ′ = k β for some k ∈ K × . Thus a moduli set for this class of algebras is the one in (2). Clearly thisalgebra is not isomorphic to the two algebras with dim(ann (2) ( A )) = 2 considered above.Next we turn to the case asi(A) = 2 and dim(ann( A )) = 2. Since ann( A ) = h e , e i thereare two zero columns in the structure matrix of A . Since asi(A) = 2, we have that A has anatural basis { e , e , e } with structure matrix α β ! , where α or β is nonzero. Notice that this algebra also has a natural basis given by { αe + βe , e , e } , which has structure matrix ! . Let us now analyze the case asi(A) = 1, in this case we haveann( A ) = ann (2) ( A ) = · · · . Then dim(ann( A )) ∈ { , , } . By hypothesis dim(ann( A )) = 0. The easiest case left is theone in which dim(ann( A )) = 2, because we have a basis of the annihilator { e , e } which canbe completed to a natural basis { e , e , w } of A : define e := xe + ye + zw where z = 0.We have w = ae + be + cw for some c ∈ K × . Observe that if c = 0, then wA ⊂ ann( A )so that w ∈ ann (2) ( A ) = K e + K e a contradiction. Then we compute x, y, z so that e getssimplified: e = z ( ae + be + cw ) = z ae + z be + zc ( e − xe − ye ) = z ( za − cx ) e + z ( zb − cy ) e + zce . Since c = 0, we may take z = c − , x = zac − and y = zbc − which imply e = e . Thus wehave a natural basis { e , e , e } with structure matrix ! .Finally, the case asi(A) = 1 with dim(ann( A )) = 1 follows from Lemma 3.12. Observe alsothat ann( A/ ann( A )) = 0 because asi(A) = 1. A moduli set for this class of algebras is givenin Remark 3.18. (cid:3) Socle chain
In this section we will determine the socle in non-degenerate finite evolution algebras. Weestablish a new type of adjunction algebras and we analyse the conditions for these algebrasto be isomorphic. These results will be useful for the classification in terms of the socle ofnon-degenerate three-dimensional evolution algebras.We start studying minimal ideals of evolution algebras. These ideals are not necessarilysimple algebras when considered as algebras on its own. Even if we assume that a minimalideal has nonzero product, it is not a simple algebra as the following example shows.
Example 5.1.
Consider the three-dimensional evolution algebra with natural basis B = { e , e , e } and product e = e + e , e = − e = e + e + e . Let I be the ideal generated by e , so I = span( e , e + e ). Note that I = 0 and I is not simple because J = span( e + e )is a proper ideal of I . Moreover, I is a minimal ideal of the evolution algebra.Since minimal ideals will play a roll in our study, we next delimit the ground in whichminimal ideals live. Furthermore, The next result gives the key for the effective computationof the socle of an evolution algebra. Proposition 5.2.
Let A be evolution algebra with natural basis B = { e i } . Let I ⊳ A beminimal and i ∈ supp B ( I ) be such that e i = 0 . Then I is generated (as an ideal) by e i .Proof. Let i ∈ supp B ( I ) as in the hypothesis. Then there exists a x ∈ I such that x = λ i e i + θ ,where λ i ∈ K × and θ is in the linear span of the remaining basis elements. Multiplying x by e i we obtain that xe i = λ i e i ∈ I and hence e i ∈ I . Now, by the minimality of I , we havethat I = h e i i . (cid:3) In the case of an associative ring U (not necessarily unital), a (left) U -module M is said tobe irreducible (or simple according to other authors) when U M = 0 and the only submodulesof M are 0 and M itself (see for instance [17, Definition 1, p. 4]). The (left) socle of an U -module M is defined as the sum of all irreducible submodules of M (see [17, Definition1, p. 63]). Then, the socle of a ring U is defined as the socle of the (left) U -module U .Consequently the socle of U is the sum of all minimal left ideals I such that U I = 0. Forinstance, if U = K is a field then its socle is K itself. However, if we consider the ring U = K (a field) endowed with the zero product, then its socle is 0. The socle of a moduleover an associative ring is proved in [17, Corollary 2, p. 61] to be a direct sum of irreducible U -submodules.If A is any K -algebra, we define the left multiplication algebra, M , as the subalgebra ofEnd K ( A ) generated by the left multiplication operators L a ( a ∈ A ). We also define theleft multiplication algebra with unit, M , as the subalgebra of End K ( A ) generated by theleft multiplication operators L a ( a ∈ A ) and the identity map A ! A . Then A is a left M -module, and the simple left M -modules of A are those minimal left ideals I of A suchthat AI = { } . So the (left) socle of A , denoted Soc( A ), can be defined as the socle of the M -module A . Applying the classical socle theory we have Soc( A ) = ⊕ α J α , where { J α } is acertain collection of minimal left ideals of A , each one satisfying AJ α = { } . If A turns outto be commutative then Soc( A ) is a direct sum of minimal ideals of A (not annihilated by A ). In the finite-dimensional case it is clear that the socle is always nonzero. HAINS IN EVOLUTION ALGEBRAS 15
We define the chain of socles of an evolution algebra as usual:Soc( A ) ⊂ Soc (2) ( A ) ⊂ · · · ⊂ Soc ( n ) ( A ) ⊂ Soc ( n +1) ( A ) ⊂ · · · where Soc( A/ Soc ( n ) ( A )) = Soc ( n +1) ( A ) / Soc ( n ) ( A ). This implies that each Soc ( n ) ( A ) is anideal. Definition 5.3.
In the previous setting, suppose that there is an n ∈ N ∗ such that Soc ( n ) ( A ) =Soc ( n + k ) ( A ) for each k >
0. We then define the socle stabilizing index , denoted ssi( A ), as theleast natural n such that Soc ( n ) ( A ) = Soc ( n + k ) ( A ) for any k > Example 5.4.
Consider, for instance, the 4-dimensional algebra A whose structure matrixrelative to the natural basis { e i } i =1 is − − . We apply Proposition 5.2 to compute the socle of A . Then h e i = K e and hence thisideal is minimal. Since h e i = span( { e , e } ) this ideal is not minimal. Finally h e i = h e i =span( { e + e } ), and this ideal is also minimal. So we have only two minimal ideals h e i and h e i . Both ideals are not annihilated by A , hence Soc( A ) = h e i ⊕ h e i .Recall that if A is an evolution algebra which decomposes as a direct sum of (possiblyinfinitely many) ideals A = ⊕ α ∈ Λ J α , then each ideal J α is an evolution algebra (see [5,Lemma 5.2.]).5.1. Adjunction of type two.
Let B be an evolution algebra over K . In this subsection ourgoal is to construct a commutative algebra A containing B as a minimal ideal. Furthermore,we want A/B to be a one-dimensional algebra.As a vector space A = B × K and the multiplication of A must be ( b, λ )( b ′ , λ ′ ) := ( bb ′ + α ( b, λ ′ ) + α ( b ′ , λ ) + Φ( λ, λ ′ ) , λλ ′ k ), where k ∈ K ; α : B × K ! B and Φ : K × K ! B arebilinear maps. Then there is a linear map ϕ : B ! B such that ϕ ( b ) = α ( b,
1) for each b ∈ B (i.e. λϕ ( b ) = α ( b, λ ) with λ ∈ K ). Since we want our construction to give a zero annihilatoralgebra, there must exist an element 0 = b ∈ B such that Φ( λ, λ ′ ) = λλ ′ b . Thus, we canrewrite the product in A in the form:(6) ( b, λ )( b ′ , λ ′ ) := ( bb ′ + λ ′ ϕ ( b ) + λϕ ( b ′ ) + λλ ′ b , λλ ′ k ) , with ϕ : B ! B linear and b ∈ B \ { } . Then B × { } is an ideal of A isomorphic to B sowe will identify them. Under these circumstances we have a short exact sequence B i ֒ ! A π ։ K , where K is endowed with the product λ ∗ λ ′ = k λλ ′ , i is the canonical injection i ( b ) := ( b, π is the canonical projection. Definition 5.5.
Let B be an evolution algebra over K and ϕ : B ! B a linear map. Considermoreover b ∈ B and k ∈ K . We define Adj ( B, ϕ, b , k ) as the algebra B × K definedabove with the product (6). Occasionally, to simplify, we write Adj ( B, ϕ, b ) instead ofAdj ( B, ϕ, b , B × { } to be a minimal ideal of A = Adj ( B, ϕ, b , k ), we must requirethe additional condition that the unique ideals of B which are ϕ -invariant are 0 and B itself.Indeed we have Lemma 5.6.
Assume that A = Adj ( B, ϕ, b , k ) has zero annihilator. Then B × { } is aminimal ideal of A if and only if the unique ideals of B which are ϕ -invariant are and B itself. Furthermore, if we assume that B × { } is minimal, then Soc ( A ) = B × { } or A = Soc ( A ) .Proof. Assume first that B × { } is a minimal ideal of A and take any ideal 0 = I ⊳ B satisfying ϕ ( I ) ⊂ I . Then I × { } is a nonzero ideal of A contained in B × { } and hence I = B . Conversely assume that B has no ϕ -invariant ideals other than 0 and B . Take anideal J of A contained in B . Then J = I × { } , where I ⊳ B and ϕ ( I ) ⊂ I . Hence I = 0 or I = B implying that J = 0 or J = B × { } . Let us prove the second assertion. Assume that B × { } is a minimal ideal of A and J is any other minimal ideal of A . If J ∩ ( B × { } ) = 0,then Soc( A ) = J ⊕ ( B × { } ) = A . So if A does not coincide with its socle then we have0 = J ∩ ( B × { } ) ⊂ B × { } . By minimality of B × { } we have B × { } ⊂ J and so, byminimality of J , we have B × { } = J . Consequently Soc( A ) = B × { } . (cid:3) Isomorphisms between adjunction algebras of type two.
We next explicit theisomorphism conditions among non-semisimple algebras Adj ( B, ϕ, b , k ) with zero annihila-tor and such that Soc(Adj ( B, ϕ, b , k )) = B × { } . In order to do this, we will considera K -algebra homomorphism θ : Adj ( B, ϕ, b , k ) ∼ = Adj ( B ′ , ϕ ′ , b ′ , k ′ ) mapping B × { } to B ′ × { } (a condition which is automatic if θ is an isomorphism, because θ would map thesocle of the first algebra into the socle of the second one, take also into account Lemma5.6). Firstly, for any b ∈ B and λ ∈ K , we can write θ ( b, λ ) = ( θ ( b ) + θ ( λ ) , θ ( λ )), where θ : B ! B ′ , θ : K ! B ′ , and θ : K ! K are linear maps. It is easy to check that θ is ahomomorphism of K -algebras and satisfy(7) θ ϕ = L θ (1) θ + θ (1) ϕ ′ θ ,k θ (1) = k ′ θ (1) ,θ ( b ) + k θ (1) = θ (1) + 2 θ (1) ϕ ′ ( θ (1)) + θ (1) b ′ . If θ turns out to be an isomorphism, then θ : B ! B ′ is an isomorphism. Hence wewill consider the isomorphism problem in the mentioned class of algebras Adj ( B, ϕ, b , k ),but with B fixed. As a consequence of equations (7), the algebras Adj ( B, ϕ, b ,
0) andAdj ( B, ϕ ′ , b ′ ,
0) are isomorphic if, and only if, θ ϕ = L θ (1) θ + θ (1) ϕ ′ θ and θ ( b ) = θ (1) +2 θ (1) ϕ ′ ( θ (1))+ θ (1) b ′ . No algebra Adj ( B, ϕ, b ,
0) is isomorphic to Adj ( B, ϕ ′ , b ′ , k ′ ) with k ′ = 0. Furthermore, two algebras Adj ( B, ϕ, b , k ) and Adj ( B, ϕ ′ , b ′ , k ′ ) are isomorphic if HAINS IN EVOLUTION ALGEBRAS 17 and only if(8) θ ϕθ − = L θ (1) + θ (1) ϕ ′ ,k = k ′ θ (1) ,θ ( b ) + k θ (1) = θ (1) + 2 θ (1) ϕ ′ ( θ (1)) + θ (1) b ′ . As a corollary, any algebra Adj ( B, ϕ, b , k ), with k = 0, is isomorphic to a suitable al-gebra Adj ( B, ϕ ′ , b ′ , ( B, ϕ, b , k ) is reduced to the following two questions:(1) When is there an isomorphism Adj ( B, ϕ, b , ∼ = Adj ( B, ϕ ′ , b ′ ,
0) ?(2) When is there an isomorphism Adj ( B, ϕ, b , ∼ = Adj ( B, ϕ ′ , b ′ ,
1) ?Concerning the second question, by (8) we would have θ (1) = 1. If we take the particularsolution θ = 0, then we get θ ϕθ − = ϕ ′ and θ ( b ) = b ′ . Thus ∀ θ ∈ Aut( B ) , we have Adj ( B, ϕ, b , ∼ = Adj ( B, θϕθ − , θ ( b ) , . In order to address the general answer to the above isomorphism questions, notice that (8)yields the following result:
Proposition 5.7.
Let B be a K -algebra, and ϕ, ϕ ′ : B ! B be linear maps with no non-trivialinvariant ideals. Assume that Adj ( B, ϕ, b , is not semisimple and has zero annihilator.Then Adj ( B, ϕ, b , ∼ = Adj ( B, ϕ ′ , b ′ , if and only if there is an automorphism θ ∈ Aut ( B ) ,and elements k ∈ K × , b ∈ B , such that: (9) ϕ ′ = k − ( θϕθ − − L b ) ,b ′ = k − ( θ ( b ) − b − kϕ ′ ( b )) . In particular, we have
Adj ( B, ϕ, b , ∼ = Adj ( B, θϕθ − , θ ( b ) , for any θ ∈ Aut ( B ) . To get a classifying moduli set for the algebras of type Adj ( B, ϕ, b , G := Aut( B ) × B × K × be the group with product defined by( θ, b , k )( θ ′ , b ′ , k ′ ) := ( θθ ′ , θ ( b ′ ) + k ′ b , kk ′ ) . Consider the set End K ( B ) × B and the action G × [End K ( B ) × B ] ! [End K ( B ) × B ] givenby ( θ, b, k ) · ( ϕ, b ) = ( ϕ ′ , b ′ ), where ϕ ′ = k − ( θϕθ − − L b ) ,b ′ = k − ( θ ( b ) − b − kϕ ′ ( b )) . Then Proposition 5.7 can be re-stated, in terms of the above action, by claiming that theisomorphism classes of algebras Adj ( B, ϕ, b ,
0) are in one-to-one correspondence with theelements of the set (End K ( B ) × B ) / G (or in other words, the algebras of type Adj ( B, ϕ, b , G , End K ( B ) × B )).We now focus on the isomorphism question for algebras of type Adj ( B, ϕ, b , Proposition 5.8.
Let B be a K -algebra. Then Adj ( B, ϕ, b , ∼ = Adj ( B, ϕ ′ , b ′ , if andonly if there is an automorphism θ ∈ Aut ( B ) , and an element b ∈ B , such that: ϕ ′ = θϕθ − − L b ,b ′ = θ ( b ) + b + b − θϕθ − ( b ) . To get a classifying moduli set for the algebras of type Adj ( B, ϕ, b , G := Aut( B ) × B be the group with product defined by( θ, b )( θ ′ , b ′ ) := ( θθ ′ , θ ( b ′ ) + b ) . Consider the set End K ( B ) × B and the action G × [End K ( B ) × B ] ! [End K ( B ) × B ] givenby ( θ, b ) · ( ϕ, b ) = ( ϕ ′ , b ′ ), where(10) ϕ ′ = θϕθ − − L b ,b ′ = θ ( b ) + b + b − θϕθ − ( b ) . Then Proposition 5.8 can be re-stated in terms of the above action by claiming that the iso-morphism classes of algebras Adj ( B, ϕ, b ,
1) (for a fixed B ) are in one-to-one correspondencewith the elements of the set (End K ( B ) × B ) / G . Example 5.9.
Let B be the evolution R -algebra with natural basis B = { e , e } and mul-tiplication e = e and e = e . Then G := Aut( B ) = { , θ } ∼ = Z , where θ swaps e and e . We want to classify the algebras Adj ( B, ϕ, b , b ∈ B we have b = αe + βe with α, β ∈ R , so that the matrix of L b relative to B , by columns, is (cid:18) βα (cid:19) .On the other hand, for any ( ϕ, b ) there is some b such that the matrix of ϕ ′ relative to B is diagonal (see equations in (10)). So the action of the group allow us to find a ( ϕ ′ , b ′ ) inthe orbit of each ( ϕ, b ), with ϕ ′ diagonal relative to the basis B . Consequently we focus inthe problem of classifying the pairs ( x, y ) and ( z, t ), where ϕ is represented by the diagonalmatrix diag( x, y ) relative to B , and b has the coordinates ( z, t ) = (0 ,
0) relative to the samebasis. In these terms, the action is equivalent to G × [ R × ( R \ { } )] ! R × ( R \ { } ) , where the element θ ∈ Aut( B ) acts in the form θ · ( x, y, z, t ) = ( y, x, t, z ). It can be provedthat [ R × ( R \ { } )] / G is in one-to-one correspondence with the subset of R given by { ( x, y, z, t ) ∈ R : z < t } ⊔ { ( x, y, z, z ) ∈ R : x ≤ y, z = 0 } . Now, in order to establish a relation between ann( B ) and Adj ( B, ϕ, b ) we prove thefollowing proposition. Proposition 5.10.
Let A = Adj ( B, ϕ, b ) . We have(1) Ker( ϕ ) = { b ∈ B : ( b, ,
1) = 0 } = ann B ×{ } ( { } × K ) .(2) ann( A ) ∩ ( B × { } ) = (ann( B ) ∩ Ker( ϕ )) × { } .(3) If (ann( B ) ∩ Ker( ϕ )) × { } = { } , then dim(ann( A )) ∈ { , } .(4) If there exists b ∈ B such that b = b and ϕ = − L b , then ( b, ∈ ann( A ) .(5) If (ann( B ) ∩ Ker( ϕ )) × { } = { } and ann( A ) = { } , then there is a b ∈ B such that b = b and ϕ = − L b . In this case ann( A ) = K ( b, . HAINS IN EVOLUTION ALGEBRAS 19 (6) If (ann( B ) ∩ Ker( ϕ )) × { } 6 = 0 , ann( A ) = { } , there is a b ∈ B such that b = b ,and ϕ = − L b , then ann( A ) = K ( b, ⊕ (ann( B ) ∩ Ker( ϕ ) × { } ) .(7) If (ann( B ) ∩ Ker( ϕ )) × { } 6 = 0 and ann( A ) = { } , but there is no b ∈ B such that b = b and ϕ = − L b , then ann( A ) = ann( B ) ∩ Ker( ϕ ) × { } .Proof. To prove item (1) notice that if b ∈ Ker( ϕ ) then we have ( b, , λ ) = ( λϕ ( b ) ,
0) =(0 ,
0) and hence ( b, ∈ ann B ×{ } ( { } × K ). Conversely if ( b, { } × K ) = { } we have ϕ ( b ) = 0.For the assertion (2) take ( b, ∈ ann( A ). Then ( b, x, λ ) = 0 for any x ∈ B and λ ∈ K .Thus bx + λϕ ( b ) = 0, which implies that b ∈ ann( B ) (taking λ = 0) and b ∈ Ker( ϕ ) (taking x = 0). Therefore ( b, ∈ (ann( B ) ∩ Ker( ϕ )) × { } . Now if ( b, ∈ (ann( B ) ∩ Ker( ϕ )) × { } we have bB = { } and ϕ ( b ) = 0. So ( b, x, λ ) = ( bx + λϕ ( b ) ,
0) = (0 , b, ∈ ann( A ) ∩ ( B × { } ).Next we prove the statement (3). We assume (ann( B ) ∩ Ker( ϕ )) × { } = { } . If ann( A ) = { } take an arbitrary nonzero ( b, λ ) ∈ ann( A ). Observe that λ = 0, or else ( b, λ ) ∈ ann( A ) ∩ ( B × { } ) = { } by the previously proved item. Now, for any other nonzero element ( b ′ , λ ′ ) ∈ ann( A ) we also have λ ′ = 0, so that there is a nonzero scalar k with λ ′ = kλ . This impliesthat k ( b, λ ) − ( b ′ , λ ′ ) = ( kb − b ′ , ∈ ann( A ) ∩ ( B × { } ) = { } . So we have proved that anynonzero element in ann( A ) is a multiple of ( b, λ ). Whence dim(ann( A )) = 1.For statement (4), consider ( x, λ ) ∈ A . Then ( b, x, λ ) = ( bx + λϕ ( b ) + ϕ ( x ) + λb ,
0) =(0 , b, ∈ ann( A ).To prove the assertion (5) notice that we already know that ann( A ) has dimension 1.Take a generator of the form ( b,
1) of ann( A ) (this is unique). Then 0 = ( b, x, λ ) =( bx + λϕ ( b ) + ϕ ( x ) + λb ,
0) for any x ∈ B and λ ∈ K . This implies that bx + ϕ ( x ) = 0 forany b ∈ B and ϕ ( b ) = − b . Then ϕ = − L b , so that b = b .Next we prove item (6). Observe that K ( b, ⊂ ann( A ) and, by what was proved in item(2), (ann( B ) ∩ Ker( ϕ ) × { } ) ⊂ ann( A ). So we have K ( b, ⊕ (ann( B ) ∩ Ker( ϕ ) × { } ) ⊂ ann( A ). Finally, if ( b ′ , λ ) ∈ ann( A ) and λ = 0, we apply item (2) and obtain ( b ′ , λ ) ∈ (ann( B ) ∩ Ker( ϕ )) × { } . If λ = 0 we have ( b ′ , λ ) = λ ( b ′′ , b ′′ , ∈ ann( A ). Since( b, ∈ ann( A ), we have ( b ′′ − b, ∈ ann( A ) ∩ ( B × { } ), and by item (2) ( b ′′ − b, ∈ (ann( B ) ∩ Ker( ϕ )) × { } . Thus( b ′ , λ ) = λ ( b ′′ ,
1) = λ ( b + b ′′ − b,
1) = λ ( b, λ ( b ′′ − b, ∈ K ( b, ⊕ (ann( B ) ∩ Ker( ϕ ) × { } ) . To prove the final item (7) notice that, on one hand, it is straightforward to check that(ann( B ) ∩ Ker( ϕ )) × { } ⊂ ann( A ). For the converse inclusion, take ( b, λ ) ∈ ann( A ). Weprove that necessarily λ = 0. Otherwise there is an element of the form ( b, ∈ ann( A ).But then, for any x ∈ B we have 0 = ( b, x,
0) = ( bx + ϕ ( x ) , ϕ = − L b . Also0 = ( b, ,
1) = ( ϕ ( b ) + b , ϕ ( b ) = − b . Since ϕ = − L b we have b = b . Sincesuch a b does not exist, we conclude that λ = 0 and ann( A ) ⊂ (ann( B ) ∩ Ker( ϕ )) × { } . (cid:3) Corollary 5.11.
Let A = Adj ( B, ϕ, b ) . Then the following assertions hold:(1) If ann( B ) = 0 and there is no b ∈ B with b = b and ϕ = − L b , then ann( A ) = 0 .(2) If ann( A ) = 0 , then there is no b ∈ B such that b = b and ϕ = − L b .Proof. The first item of the corollary follows from items (3) and (5) in Proposition 5.10. Thesecond item follows from item (4). (cid:3)
As it happens with Adj , the adjunction Adj ( B, ϕ, b , k ) is also an evolution algebra (fora suitable ϕ ), as we show in the next proposition. Proposition 5.12. If B is an evolution algebra, then A = Adj ( B, ϕ, b , k ) is an evolutionalgebra for a suitable ϕ .Proof. Take a natural basis { u i } of B and x any nonzero element. Define ϕ : B ! B by ϕ ( u i ) = − xu i for all i . Next we prove that the collection B := { ( u i , } ∪ { ( x, } is a basisof A . First notice that B is a generator set. Indeed, since x = P λ i u i for suitable scalars λ i ∈ K , we can write (0 ,
1) = − P λ i ( u i ,
0) + ( x, B is linearlyindependent. If P λ i ( u i ,
0) + µ ( x,
1) = (0 ,
0) then µ = 0, implying that λ i = 0 for all i . Thus B is a basis of A and ( u i , x,
1) = ( u i x + ϕ ( u i ) ,
0) = 0 by the definition of ϕ . Since B is asubalgebra of A we conclude that A is an evolution algebra. (cid:3) Classification in terms of the socle
The degenerate evolution algebras of dimension 3 have been classified in Theorem 4.2. So,our main goal in this section is the classification of non-degenerate 3-dimensional evolutionalgebras in terms of the socle. These have nonzero socle, therefore we must start consideringcases in terms of the dimension of the socle.As a first observation regarding the dimension of the socle, notice that if a commutativealgebra A = 0 has zero annihilator and dim(Soc( A )) > A ) >
1. Indeed, if A = K a for some nonzero a , and we take a nonzero ideal I of A , then IA ⊂ K a and IA = 0.Thus IA = K a ⊂ I . So K a is the unique minimal ideal of A , which implies that Soc( A ) = K a has dimension 1.Before we present our classification result in terms of the socle, wee need two auxiliaryresults, which we prove below. Recall that for a inner product h· , ·i : V × V ! K (i.e.symmetric bilinear form) on a K -vector space V , the radical of h· , ·i is defined as the subspacerad( h· , ·i ) := { v ∈ V : h v, V i = 0 } . Lemma 6.1.
Let A be an evolution K -algebra and = u ∈ A . Fix a subspace B such that A = K u ⊕ B and consider the linear map p : A ! K such that a = p ( a ) u + b according to thedecomposition A = K u ⊕ B . Define an inner product π : A × A ! K by π ( x, y ) := p ( xy ) forany x, y ∈ A . Consider now any natural basis B = { b i } i ∈ Λ of A . Then rad( π ) = span { b i : π ( b i , b i ) = 0 } . In particular, if rad( π ) is one-dimensional, any generator of this radical is (up to nonzeroscalar multiples) contained in every natural basis of A .Proof. Take an arbitrary x ∈ rad( π ) and write x = P i x i b i with x i ∈ K . Then for any j wehave 0 = π ( x, b j ) = P i x i π ( b i , b j ) = x j π ( b j , b j ). So, if x j = 0 then π ( b j , b j ) = 0 and hence x ∈ span { b i : π ( b i , b i ) = 0 } . The other inclusion is clear. (cid:3) Lemma 6.2.
Let A be a three-dimensional evolution algebra with a two-dimensional ideal I .Then, one of the following mutually excluding possibilities holds.(1) I is the linear span of two of the elements in a natural basis of A .(2) There is no natural basis such that I is generated by two of its elements, but thereexists a natural basis B = { e , e , e } of A such that I is the linear span of e i and e j + e k , with i, j, k different. In this case I is an evolution ideal. HAINS IN EVOLUTION ALGEBRAS 21 (3) The ideal I does not verify any of the cases (1) and (2) , but there exists a naturalbasis B = { e , e , e } of A such that I is the linear span of e i + e j and e j + e k , with i, j, k different. If A is perfect, then I is not an evolution ideal.Proof. Let { e i } i =1 be a natural basis of A with structure matrix M = ( ω ij ). Assume that u = xe + ye + ze , u = x ′ e + y ′ e + z ′ e is a basis of the ideal I . For i = 1 , , i := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x y zx ′ y ′ z ′ ω i ω i ω i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . Since e u = xω e + xω e + xω e ∈ I we have thatrank x y zx ′ y ′ z ′ xω xω xω = 2 , which implies x ∆ = 0 . Also from e u ∈ I we deduce that x ′ ∆ = 0. Proceeding in this way we get, not only that0 = x ∆ = x ′ ∆ , but also that 0 = y ∆ = y ′ ∆ and 0 = z ∆ = z ′ ∆ . If some ∆ i = 0then I is the linear span of two elements of the basis { e i } of A . So we assume in the sequelthat ∆ i = 0 for any i = 1 , ,
3. The row reduced echelon form of x y zx ′ y ′ z ′ ! is one of thefollowing: α α ′ ! , α ! , α ! , ( α, α ′ ∈ K ) . So we can find a basis { v , v } of I such that one of the following holds:(a) v = e + αe , v = e + α ′ e .(b) v = e + αe , v = e .(c) v = e + αe , v = e .In cases (b) and (c) we have a basis of I consisting of two vectors of the natural basis. So wefocus on the case (a). If α = α ′ = 0, then again there is a basis of I consisting of elements ofthe natural basis. Next we consider the following possibilities:i) If α = 0 and α ′ = 0, or α ′ = 0 and α = 0, then we can construct a new natural basisof A such that I is in the possibility (2).ii) If α = 0 and α ′ = 0, then we can construct a new natural basis of A such that I is inthe possibility (3).Finally let us prove that if A = A and I is generated by e i + e j and e j + e k , with i, j, k different,then I is not an evolution algebra: assume on the contrary that u = x ( e i + e j ) + y ( e j + e k )and u = x ′ ( e i + e j ) + y ′ ( e j + e k ) is a natural basis of I . Then 0 = xx ′ ( e i + e j ) + yy ′ ( e j + e k ) +( xy ′ + x ′ y ) e j = xx ′ e i + yy ′ e k + ( xx ′ + yy ′ + xy ′ + yx ′ ) e j and hence xx ′ = yy ′ = xy ′ + yx ′ = 0,which is inconsistent with xy ′ − yx ′ = 0. (cid:3) Socle of dimension three.
Under this hypothesis the algebra is either simple or adirect sum of simple evolution algebras of dimension ≤
2. In fact the simple algebras providemoduli sets depending on up to 6 parameters, which is one of the most densely populatedcollection of isomorphism classes of algebras.
Proposition 6.3. If A is a -dimensional evolution K -algebra with zero annihilator, whosesocle is A itself, then A is (isomorphic to) one of the following: (1) A simple evolution algebra.(2) A = B ⊕ K , with B a simple evolution algebra of dimension , and the product in A given by ( b, λ )( b ′ , λ ′ ) = ( bb ′ , λλ ′ ) . (3) K ⊕ K ⊕ K with componentwise product.Proof. If the socle consists of a unique minimal ideal then it is a simple algebra. Otherwisewe have a direct sum of two or three minimal ideals. If A = Soc( A ) = I ⊕ J , with dim( I ) = 2and dim( J ) = 1, then both ideals are evolution algebras (see [5, Lemma 5.2.]) with zeroannihilator. Hence J = K e , where e is an idempotent, and I = 0 since on the contrary IA = 0. Thus I is a simple evolution algebra, because any ideal of I is an ideal of A (giventhat I is a summand of A ). Finally, if the socle has three components then A ∼ = K withcomponentwise operations. (cid:3) Socle of dimension two.
In this section we will study the case dim(Soc( A )) = 2. Westart this subsection describing a number of moduli sets which will appear in the classificationTheorem 6.6. Definition 6.4.
Recall that ( K × ) h i := { k : k ∈ K × } is the group of nonzero squares of K .(1) Consider the group Z := { , } and the product group ( K × ) h i × Z . This group actson the set M ( K ) of 2 × K . The action is defined by[( K × ) h i × Z ] × M ( K ) ! M ( K )( λ, i ) · M := λE i M E i , where E = (cid:18) (cid:19) , E = Id and E = E , see Notation 2.1.(2) Consider the group ( K × ) h i × K × . This group acts on the set M ( K ) of 2 × K . The action is defined in the following way[( K × ) h i × K × ] × M ( K ) ! M ( K )( α, β ) · M := (cid:18) α β (cid:19) M (cid:18) α − β − (cid:19) . (3) The product group K × × Z acts on the set M ( K ) of 3 × K . The action is defined in the following way( K × × Z ) × M ( K ) ! M ( K )( λ, i ) · M vx ω ! := λ E i M E i vx λω ! . (4) The product group K × × K × acts on the set M ( K ) of 3 × K . The action is given by( K × × K × ) × M ( K ) ! M ( K )( α, β ) · M vx ω ! := (cid:18) α β (cid:19) M (cid:18) α − β − (cid:19) vx ω α . HAINS IN EVOLUTION ALGEBRAS 23 (5) Consider the product group ( K × ) h i × Z , which acts on the set K \{ } in the followingway [( K × ) h i × Z ] × ( K \ { } ) ! K \ { } ( λ, i ) · v := λvE i . (6) Let the group K × act on the set K \ { } × K × , via the action defined by K × × [( K \ { } ) × K × ] ! ( K \ { } ) × K × λ · ( x, y, z ) := ( λ x, λ y, λz ) . Notation 6.5.
Let W = { ( ω ij ) ∈ GL ( K ) : ω ω = 0 } . Observe that, applying [5, Corol-lary 4.6]), the matrices of this set correspond to the structure matrices of the two-dimensionalsimple evolution algebras.6.2.1. Soc ( A ) has the extension property. We consider first the case in which the socle isgenerated by two vectors of a natural basis of the algebra.
Theorem 6.6.
Let A be a three-dimensional evolution algebra with zero annihilator. Assumethat dim( Soc ( A )) = 2 and Soc ( A ) has the extension property. Then one of the following threeexcluding possibilities holds:(1) Soc ( A ) is a minimal ideal and ssi( A ) = 1 . Equivalently, the structure matrix of A isof the form (cid:18) M v (cid:19) , where M ∈ W and v = 0 . Furthermore,(a) If | supp B ( e ) | = 2 , then the structure matrix of A relative to a suitable basis is M (cid:18) (cid:19) , where M ∈ W . This class of algebras are classified by the action of the group ( K × ) h i × Z on W . More precisely, two algebras of this kind, whose structurematrices are M (cid:18) (cid:19) and M ′ (cid:18) (cid:19) , are isomorphic if and only if M and M ′ are in the same orbit under the actionof ( K × ) h i × Z described in Definition 6.4(1). A moduli set for this class is (( K × ) h i × Z , W ) .(b) If | supp B ( e ) | = 1 , then the structure matrix of A relative to a suitable basis is M (cid:18) (cid:19) , where M ∈ W . This class of algebras are classified by the group ( K × ) h i × K × .More precisely, two algebras of this kind, whose structure matrices are M (cid:18) (cid:19) and M ′ (cid:18) (cid:19) , are isomorphic if and only if M and M ′ are in the same orbit under the actionof ( K × ) h i × K × described in Definition 6.4(2). A moduli set for this class is (( K × ) h i × K × , W ) .(2) Soc ( A ) is a minimal ideal and ssi( A ) = 2 . Equivalently, e / ∈ Soc ( A ) and the structurematrix of A is of the form (cid:18) M v ω (cid:19) where M ∈ W , ω = 0 and v = 0 . Furthermore,(a) If { , } ⊂ supp B ( e ) , then the structure matrix of A relative to an appropriatebasis is (cid:18) M (cid:0) (cid:1) ω (cid:19) , where M ∈ W . This class of algebras are classified by the group K × × Z . More precisely, two algebras of this kind, whose structure matrices are M (cid:18) (cid:19) ω and M ′ (cid:18) (cid:19) ω ′ , are isomorphic if and only if both matrices are in the same orbit under the actionof K × × Z described in Definition 6.4(3). A moduli set in this case is ( K × × Z , W ) . (b) If / ∈ supp B ( e ) , or / ∈ supp B ( e ) , then the structure matrix of A relative to aappropriate basis is of the form (cid:18) M (cid:0) (cid:1) ω (cid:19) , where M ∈ W . This class of algebrascan be classified by the group K × × K × . Concretely, two algebras whose structurematrices are M (cid:18) (cid:19) ω and M ′ (cid:18) (cid:19) ω ′ are isomorphic if and only if both matrices are in the same orbit under the actionof K × × K × described in Definition 6.4(4). A moduli set in this case is ( K × × K × , W ) .(3) Soc ( A ) is the direct sum of two minimal ideals, say K u and K u . Then, the structurematrix of A relative to a suitable basis, B , is ω ω ω ! . Furthermore,(a) If ssi ( A ) = 1 , then the structure matrix relative to B is (cid:18) Id v (cid:19) , with v = 0 .This class of algebras are classified by the group ( K × ) h i × Z . Concretely, twoalgebras whose structure matrices are (cid:18) Id v (cid:19) and (cid:18) Id v ′ (cid:19) are isomorphic if and only if both matrices are in the same orbit under the ac-tion of ( K × ) h i × Z described in Definition 6.4(5). The corresponding moduli is (( K × ) h i × Z , K \ { } ) .(b) If ssi( A ) = 2 , then the structure matrix of A relative to B is (cid:18) Id v ω (cid:19) , with v = 0 and ω = 0 . This class of algebras are classified by the group K × . Moreprecisely, two algebras of this kind, whose structure matrices are (cid:18) Id v ω (cid:19) and (cid:18) Id v ′ ω ′ (cid:19) HAINS IN EVOLUTION ALGEBRAS 25 are isomorphic if and only if both matrices are in the same orbit under the action K × describes in Definition 6.4(6). The moduli for this class is ( K × , K \{ }× K × ) .Proof. We have Soc( A ) = span ( { e , e } ) with { e i } i =1 a natural basis. Now either Soc( A ) isa minimal ideal or it is the direct sum of two such ideals.(1) Assume that Soc( A ) is a minimal ideal of A and that ssi( A ) = 1. Then it is easyto check that Soc( A ) is a simple evolution algebra. Moreover, as ssi ( A ) = 1, thenSoc( A ) = Soc ( A ). This implies that Soc ( A/ Soc( A )) = 0. On the other hand, A/ Soc( A ) = K e implies that e ∈ Soc( A ). Note that the structure matrix is of theform (cid:18) M v (cid:19) , with M the structure matrix of the evolution algebra Soc( A ). Applying[5, Corollary 4.6]) we get that M ∈ W . Conversely, if we have any matrix of theform (cid:18) M v (cid:19) , with v = (cid:18) ω ω (cid:19) = (cid:18) (cid:19) , M invertible and with nonzero upper right andlower down entries, then Soc( A ) is a minimal ideal and ssi( A ) = 1. Indeed, let A bethe evolution algebra and B = { e , e , e } the natural basis such that the structurematrix relative to B is (cid:18) M v (cid:19) . We note that I = span ( { e , e } ) is a minimal ideal of A , because the structure matrix of I is M and M is invertible and has nonzero upperright and lower down entries (see [5, Corollary 4.6]). Therefore I ⊆ Soc ( A ). Now,Soc( A ) = A , since otherwise A would be a direct sum of simple algebras, but as A isnot perfect. So, Soc( A ) = A and I = Soc( A ). Furthermore, ssi ( A ) = 1 because, asSoc ( A ) = I and Soc ( A/ Soc ( A )) = Soc( K e ) = 0, we have Soc ( A ) / Soc( A ) = 0.(a) If ω ω = 0, v can be chosen to be (scaling the basis if necessary) (cid:16) (cid:17) . Now, wewill see that an evolution algebra with structure matrix (cid:18) M (cid:0) (cid:1) (cid:19) is isomorphicto any algebra with structure matrix (cid:18) M ′ (cid:0) (cid:1) (cid:19) if and only if M ′ = kM or M ′ = k (cid:18) (cid:19) M (cid:18) (cid:19) with k ∈ K × and k is the square of some element. Indeed, let A be an evolution algebra with natural basis B = { e , e , e } and structure matrixrelative to B , given by (cid:18) M (cid:0) (cid:1) (cid:19) , with M ∈ W . We assume that there existsanother evolution algebra A ′ with natural basis B ′ = { e ′ , e ′ , e ′ } and structurematrix relative to B ′ of the form (cid:18) M ′ (cid:0) (cid:1) (cid:19) , with M ∈ W , such that A and A ′ are isomorphic. Let φ be the isomorphism of evolution algebras between A and A ′ . As φ (Soc ( A )) = Soc( A ′ ) then φ ( e ) = α e ′ + α e ′ and φ ( e ) = β e ′ + β e ′ .But φ ( e ) φ ( e ) = 0, because φ is an isomorphism of algebras and e e = 0.Therefore α β ( e ′ ) + α β ( e ′ ) = 0. Applying that M ∈ W (so in particular | M | 6 = 0) we get that α β = 0 and α β = 0. This implies that α = β = 0or α = β = 0. Let φ ( e ) = γ e ′ + γ e ′ + γ e ′ . As { φ ( e ) , φ ( e ) , φ ( e ) } is anatural basis of A ′ then, in any case, γ = γ = 0. Now we deal with the case α = β = 0. We have that e i = ω i e + ω i e for i ∈ { , } and therefore,applying φ , we get φ ( e ) = ω α e ′ + ω β e ′ and φ ( e ) = ω α e ′ + ω β e ′ .But φ ( e ) = ( φ ( e )) = α ( e ′ ) and φ ( e ) = ( φ ( e )) = β ( e ′ ) . Then ( e ′ ) = ω β e ′ + ω α β e ′ and ( e ′ ) = ω β α e ′ + ω α e ′ . Furthermore, as e = e + e , we have that φ ( e ) = α e ′ + β e ′ . But φ ( e ) = ( φ ( e )) = γ ( e ′ ) , and therefore( e ′ ) = β γ e ′ + α γ e ′ . So, as the structure matrix relative to B ′ is of the form (cid:18) M ′ (cid:0) (cid:1) (cid:19) with M ∈ W , necessarily β γ = 1 and α γ = 1. Then, β = α = γ and M ′ = β (cid:18) (cid:19) M (cid:18) (cid:19) . Now, if α = β = 0 we proceed analogously tothe previous case and we obtain that M ′ = a M and α = γ . So, this meansthat M and M ′ are in the same orbit under the action of the group ( K × ) h i × Z described in Definition 6.4(1).(b) If ω = 0 or ω = 0, v can be chosen to be (scaling the basis if necessary) (cid:18) (cid:19) . Now, it can be proved that an algebra with structure matrix (cid:18) M (cid:0) (cid:1) (cid:19) is not isomorphic to any algebra with structure matrix (cid:18) M (cid:0) (cid:1) (cid:19) . Moreover,any algebra with structure matrix (cid:18) M (cid:0) (cid:1) (cid:19) is isomorphic to any algebra withstructure matrix (cid:18) M ′ (cid:0) (cid:1) (cid:19) if and only if M ′ = ω α ω αβ ω βα ω β with α, β ∈ K × and α is the square of some element. In other words, both structure matrices M and M ′ are in the same orbit under the action of the group ( K × ) h i × K × describedin Definition 6.4(2).(2) Suppose that Soc( A ) is a minimal ideal and ssi( A ) = 2. Then e / ∈ Soc( A ) ( ω = 0).Assume that e = s + ω e with s = 0 (if s = 0 then A = Soc ( A ), which is not thecase). So e = ω e + ω e + ω e . This means that the structure matrix of A is (cid:18) M v ω (cid:19) , with M ∈ W and v = (cid:18) ω ω (cid:19) = (cid:18) (cid:19) . Reciprocally, if we have an evolutionalgebra with structure matrix (cid:18) M v ω (cid:19) , where M ∈ W and v = 0, then it is easy tocheck that Soc( A ) is a minimal ideal and ssi( A ) = 2.(a) If ω ω = 0 then, scaling the basis if necessary, we get that the structure matrixof A is (cid:18) M (cid:0) (cid:1) k (cid:19) , with k ∈ K × and M is the structure matrix of the simpleevolution algebra Soc( A ). Reciprocally, if we have an evolution algebra as in thehypothesis of the proposition, with structure matrix (cid:18) M (cid:0) (cid:1) ω (cid:19) , where ω ∈ K × and M ∈ W , then it is easy to check that Soc( A ) is a minimal ideal and ssi( A ) = 2.Furthermore, it can be proved that any evolution algebra with structure matrix (cid:18) M (cid:0) (cid:1) ω (cid:19) is isomorphic to any algebra with structure matrix (cid:18) M ′ (cid:0) (cid:1) ω ′ (cid:19) if and onlyif both are in the same orbit under the action of K × × Z described in Definition6.4(3).(b) If ω = 0 and ω = 0, or ω = 0 and ω = 0, scaling the basis if necessary,then we get that the structure matrix of A is (cid:18) M (cid:0) (cid:1) ω (cid:19) , with ω ∈ K × and M is the structure matrix of Soc( A ) (which is a simple evolution algebra./) Con-versely, let A be an evolution algebra under the hypothesis of the proposition HAINS IN EVOLUTION ALGEBRAS 27 with structure matrix (cid:18) M (cid:0) (cid:1) ω (cid:19) , ω ∈ K × and M ∈ W . It is straightforwardthat ssi( A ) = 2. Now, it is also easy to prove that an evolution algebra withstructure matrix M (cid:18) (cid:19) ω is isomorphic to an evolution algebra with structurematrix M ′ (cid:18) (cid:19) ω ′ if and only if both matrices are in the same orbit under theaction of K × × K × described in Definition 6.4(4).(3) Assume Soc( A ) = K v ⊕ K v . Let k , k ∈ K be such that v = k v and v = k v .Now, if we consider u = k − v and u = k − v , then the basis { u , u , e } is a naturalbasis where the u i ’s are idempotents. Observe that u i e = 0, because u i ∈ Soc( A ) =span( { e , e } ). Therefore { u , u , e } is a new natural basis of A and the structurematrix relative to this basis is ω ω ω ! . Now we discuss two cases:(a) If ssi( A ) = 1, or equivalently ω = 0, then the structure matrix relative to { u , u , e } is of the form (cid:18) Id v (cid:19) with v = 0. Two evolution algebras whosestructure matrices are (cid:18) Id v (cid:19) and (cid:18) Id v ′ (cid:19) (for v, v ′ = 0) are isomorphic ifand only if there is a k ∈ ( K × ) h i such that v ′ = kvE i (where i = 0 , K × ) h i × Z described in Definition 6.4(5).(b) If ssi( A ) = 2 or, equivalently, ω = 0. If ω = 0 then K e is not an ideal of A , because if K e ⊳ A then necessarily Soc( A ) = A , which is not the case. Thisimplies that Soc( A/ Soc( A )) = 0 and hence Soc (2) ( A ) contains strictly Soc( A ).So ssi( A ) = 2. The reciprocal is straightforward. In this case, the structurematrix relative to { u , u , e } is (cid:18) Id v ω (cid:19) , with v = 0 and ω = 0. Moreover,two evolution algebras whose structure matrices are (cid:18) Id v ω (cid:19) and (cid:18) Id v ′ ω ′ (cid:19) (for v = (cid:16) ω ω (cid:17) = 0 , v ′ = (cid:16) ω ′ ω ′ (cid:17) = 0 and ω ω ′ = 0) are isomorphic if and only if thereis an k ∈ K × such that ω ′ = k ω , ω ′ = k ω and ω ′ = kω . In other words,two evolution algebras are isomorphic if and only if both structure matrices arein the same orbit under the action K × describes in Definition 6.4(6). (cid:3) Now we focus in the case in which Soc( A ) = span( { e , e + e } ), with { e , e , e } a naturalbasis of A and there is no natural basis of A such that Soc( A ) is generated by two elementsof this natural basis. Lemma 6.7.
Assume that A is a commutative algebra with zero annihilator with a basis { u , u , u } whose multiplication table is summarized in u = u u = 0 , u u = αu , u = u , u u = βu , u = X γ i u i , with α, β ∈ K . Then: (1) If some of β, γ or γ is nonzero then A is not an evolution algebra.(2) If β = γ = γ = 0 then A is isomorphic to the evolution algebra with structure matrix (11) − − ! . Proof.
Observe that α = 0 because ann( A ) = 0. Write u = P i γ i u i with γ i ∈ K . Considerthe inner products h· , ·i i : A × A ! K ( i = 1 , ,
3) such that xy = P i h x, y i i u i . The matricesof the h· , ·i i are α α γ ! , β β γ ! and γ ! respectively.Then A is an evolution algebra if and only if there is a basis of A orthogonalizing simulta-neously the inner products h· , ·i i for i = 1 , ,
3. Suppose that such a basis B exists. Nextwe prove that, scaling if necessary, we may assume that u is in B . Indeed, if we write B = { f , f , f } then we get that rad( h· , ·i ) = span( u ), and hence, applying Lemma 6.1 to u (taking B = span( { u , u } )), we find that u is in B (up to nonzero scalars). Now:(1) If β = 0 or γ = 0 then rad( h· , ·i ) = span( u ), and arguing as before (applyingLemma 6.1) we conclude that u ∈ B .(2) If β = γ = 0 but γ = 0 we have (taking into account again Lemma 6.1) that K u = rad( h· , ·i ) ∩ rad( h· , ·i ) = span( { f i : h f i , f i i = 0 = h f i , f i i } ) . Hence there is only one i such that h f i , f i i = h f i , f i i = 0 and this f i is u up tononzero scalars. So in this case again u ∈ B .The conclusion is that in case some of the scalars β , γ or γ is nonzero we may assume that u , u ∈ B . Now the third element of B is a vector ξ = xu + yu + zu and we must have h ξ, u i i j = 0 for i = 1 , j = 1 , ,
3. Writing the corresponding equations we get (amongother equations) αz = 0 y + βz = 0and since α = 0 (because the annihilator of A is zero) the only solution is y = z = 0. Butthen ξ is a multiple of u and this is a contradiction. In this case A is not an evolution algebra.But we must analyze the possibility that β = γ = γ = 0. In this case, since u ∈ B , wecan find a natural basis of the form { u , xu + yu , x ′ u + y ′ u } (the corresponding equationsare consistent). Then span( { xu + yu , x ′ u + y ′ u } ) = span( { u , u } ) =: B which is an idealof A . In B (which is an evolution algebra) we have u = 0, u u = αu = 0 and u = γ u .Scaling we get u = 0, u u = u and u = u . Then, putting e = u and f = u − u , wehave ef = u ( u − u ) = u − u = 0 and hence the natural basis { u , e, f } of A has therequired structure matrix. (cid:3) Remark 6.8.
In the case of an indecomposable commutative algebra (see [10, Definition2.1]), Lemma 6.7 implies that the algebra is not an evolution algebra.
Corollary 6.9.
Let A be a three-dimensional evolution algebra with zero annihilator. IfSoc ( A ) is two-dimensional, decomposable, does not have the extension property and Soc ( A ) = HAINS IN EVOLUTION ALGEBRAS 29 , then A is decomposable and isomorphic to the evolution algebra whose structure matrix isgiven in (11) .Proof. Assume Soc( A ) = K u ⊕ K u . Take into account that if some u i = 0 we may rescale itto get u i = u i . We complete to a basis { u , u , u } of A . In this basis we get that u u = αu , u u = βu and u = P i γ i u i for certain α, β, γ i ∈ K . After rescaling if necessary we havetwo cases:(1) u i = u i ( i = 1 ,
2) and u u = 0. In this case { u , u , ξ } is a natural basis where ξ = − αu − βu + u , a contradiction.(2) u = 0, u = u . We apply Lemma 6.7. Since A is an evolution algebra, then it isnecessarily isomorphic to the decomposable evolution algebra with structure matrixas in (11). (cid:3) Socle of A does not have the extension property. We proceed with our argumentation.If the socle does not have the extension property then, applying Theorem 6.2, there are twopossibilities. We study next the case in which Soc( A ) = span( { e , e + e } ), where { e , e , e } is a natural basis of A , and such that Soc( A ) does not have the extension property. Proposition 6.10.
Let A be a tridimensional evolution algebra with dim( A ) = 2 . Let { e i } i =1 be a natural basis of A . We have(1) If Soc ( A ) = span( { e , e + e } ) then Soc ( A ) has the extension property if and only if { e , e } is a linearly dependent set and e + e = 0 .(2) If Soc ( A ) = span( { e + e , e + e } ) then ω i = ω i + ω i for each i , where ω ij arethe structural constants of A . If ann( A ) = 0 and there is no natural basis { e ′ i } i =1 of A with Soc ( A ) = span( { e ′ , e ′ + e ′ } ) , then no vector of Soc ( A ) is a natural vector(embeddable in a natural basis of A ).Proof. We start with the first assertion. If Soc( A ) has the extension property then there isa natural basis u = xe + y ( e + e ), u = x ′ e + y ′ ( e + e ), u = x ′′ e + y ′′ e + z ′′ e with x, y, x ′ , y ′ , x ′′ , y ′′ , z ′′ ∈ K . Since u i (for i = 1 ,
2) is a member of a natural basis, applying[1, Theorem 3.3.] and dim( A ) = 2, we see that the unique possibilities are (up to nonzeroscalars and permutations) u = e and u = e + e . In this case, since u u = 0 we get0 = ( e + e )( x ′′ e + y ′′ e + z ′′ e ) = y ′′ e + z ′′ e and y ′′ , z ′′ are not both zero. Hence e and e arelinearly dependent. Furthermore, we have that y ′′ = z ′′ , since otherwise u = x ′′ e + y ′′ ( e + e )would be a linear combination of u and u . Conversely, if there are scalars α, β ∈ K , α = β ,not simultaneously null with αe + βe = 0, then { e , e + e , αe + βe } is a natural basis.Let us prove now the second assertion. We have ( e + e )( e + e ) ∈ Soc( A ), and hencethere are scalars α and β such that α ( e + e ) + β ( e + e ) = ( e + e )( e + e ) = e . So α = ω , β = ω , α + β = ω and consequently ω = ω + ω . Similarly from the facts that ( e + e ) , ( e + e ) ∈ Soc( A ) we conclude that ω i = ω i + ω i , ( i = 1 , , . Assume now that z = x ( e + e ) + y ( e + e ) is natural. By Proposition 2.4 we know thatsupp( z ) = { , , } . If supp( z ) has cardinal 1, for instance supp( z ) = { } , then x = 0. Also x + y = 0, what implies that y = 0, a contradiction. Thus supp( z ) has cardinal 2. If thenatural basis containing z is { z, z ′ , z ′′ } then, applying Proposition 2.4, we get two possibilities(scaling if necessary):(1) z = e + e , z ′ = e + ke and z ′′ = e (where k = 1 and e + ke = 0).(2) z = e + e , z ′ = e + ke and z ′′ = e (where k = 1 and e + ke = 0).(3) z = e + e , z ′ = e + ke and z ′′ = e (where k = 1 and e + ke = 0).Now, we analyze the first possibility. Since e = z − z ′ − k , then e + e = − k ( z − z ′ ) + z ′′ andthis implies that if λ = k − we haveSoc( A ) = span( { e + e , e + e } ) = span( { z, λz ′ + z ′′ } )contradicting the hypothesis that there is no natural basis { e ′ i } of A for which Soc( A ) =span( { e ′ , e ′ + e ′ } ).Reasoning in a similar way, we get that item (2) implies a contradiction. For the thirdpossibility, since z = xe + ( x + y ) e + ye = e + e , then x = y = 1 and x + y = 0, so2 = 0. Therefore K must have characteristic two. In this case, we can see that Soc( A ) =span( { e + e , e + e } ) = span( { z, λz ′ + z ′′ } ) with λ = k − , a contradiction. (cid:3) Thus we proceed assuming that Soc( A ) is the linear span of { e , e + e } for a naturalbasis { e i } of A and Soc( A ) does not have the extension property. A priory, we have twopossibilities for Soc( A ): either it is a minimal ideal of A or a direct sum of two minimal idealsof A . However the next lemma shows that we must focus only in the case in which Soc( A ) isminimal. Proposition 6.11.
Let A be a non-degenerate three-dimensional evolution algebra, Soc ( A ) =span( { e , e + e } ) and Soc ( A ) does not have the extension property. If Soc ( A ) is not minimal,then A itself is decomposable and isomorphic to the evolution algebra with structure matrix (12) − − ! . If Soc ( A ) is minimal, then A is indecomposable and:(1) Soc ( A ) = A and ssi( A ) = 1 .(2) There exist b ∈ B and ϕ ∈ End K ( B ) such that A ∼ = Adj ( B, ϕ, b , , where B = Soc ( A ) is a two-dimensional evolution algebra.Proof. If Soc( A ) is not minimal we apply Corollary 6.9 and we get that the structure matrixrelative to a natural basis is as (12).Let us consider now the case in which Soc( A ) is minimal. Then, since e ∈ Soc( A ), e = e ( e + e ) and e = e ( e + e ), we have e i ∈ Soc( A ) for i = 1 , ,
3. Hence 0 = A ⊂ Soc( A )and, by minimality of the socle, A = Soc( A ). Then A/A ∼ = K with zero product, soSoc( A/ Soc( A )) = { } , implying ssi( A ) = 1. In this case, by Definition 5.1 we have that A ∼ = Adj ( B, ϕ, b , B = Soc( A ) is a two-dimensional evolution algebra, b = ω e + ω ( e + e ) and the matrix of the linear map ϕ in the basis { e , e + e } is (cid:18) ω ω (cid:19) . Indeed,let Ω : A ! B × K be the isomorphism of algebras where Ω( e ) = ( e , e ) = ( e + e , − e ) = (0 , A is indecomposable. Indeed, if A = I ⊕ J , thenSoc ( A ) = I , a contradiction since Soc( A ) does not have the extension property. (cid:3) HAINS IN EVOLUTION ALGEBRAS 31
Lemma 6.12.
Let A = Adj ( B, ϕ, b ) be a three-dimensional evolution algebra and B :=span( { b , b } ) with { b , b } a natural basis of B . Then, the following statements are equivalent:(1) There exists b ∈ B such that ϕ = − L b .(2) { ( b , , ( b , , ( b , } is a natural basis of A .Proof. First, we write u = ( b ,
0) and u = ( b , u i ( b ,
1) = ( b i b + ϕ ( b i ) ,
0) = (0 ,
0) for i = { , } . Therefore { u , u , ( b , } is a natural basisof A . Conversely, if { u , u , ( b , } is a natural basis, then b i b + ϕ ( b i ) = 0 for i ∈ { , } . As { b , b } is a basis of B then ϕ = − L b . (cid:3) Consider a non-degenerate evolution algebra A of dimension 3, with a socle of dimension2, which is an evolution algebra generated by { e , e + e } , where { e i } i =1 is a natural basis of A . Further assume that the socle does not have the extension property. Then, we know byProposition 6.11 that Soc( A ) = A and if A is indecomposable, then Soc( A ) is not a directsum of two one-dimensional ideals, hence it is a minimal ideal and ssi( A ) = 1. Theorem 6.13.
Let A be a non-degenerate three-dimensional evolution algebra with naturalbasis { e , e , e } and B := Soc ( A ) = span( { e , e + e } ) . Assume that Soc ( A ) does not havethe extension property. If A is indecomposable, then:(1) B is an evolution algebra and there is a linear map ϕ : B ! B and an element b ∈ B such that A ∼ = Adj ( B, ϕ, b , (see Proposition 6.11 (2)).(2) B is a minimal ideal of A , equivalently the unique ideals of B which are ϕ -invariantare and B itself (see Lemma (5.6)).(3) There is no b ∈ B such that ϕ = L b (see Lemma 6.12).The moduli set for this class of algebra is (End K ( B ) × B ) / G where G = Aut ( B ) × B × K × (seeProposition 5.7). Thus the isomorphy classes of algebras Adj ( B, ϕ, b , (for fixed B ) are inone-to-one correspondence with the elements of the set (End K ( B ) × B ) / G (see equation (10) for the definition of the action). Finally, we must analyze the case in which Soc( A ) = span( { e + e , e + e } and there isno natural basis { e ′ , e ′ , e ′ } such that Soc( A ) = span( { e ′ , e ′ + e ′ } . We know that no vectorof the socle is in a natural basis of A by Proposition 6.10. Theorem 6.14.
Let A be a non-degenerate three-dimensional evolution algebra with naturalbasis { e , e , e } . Suppose that Soc ( A ) = span( { e + e , e + e } ) and Soc ( A ) does not havethe extension property. Moreover, there is no natural basis { e ′ , e ′ , e ′ } such that Soc ( A ) =span( { e ′ , e ′ + e ′ } ) . Then(1) Soc ( A ) = A is a minimal ideal and ssi( A ) = 1 .(2) There exist b ∈ B and ϕ ∈ End K ( B ) such that A ∼ = Adj ( B, ϕ, b ) , where B = Soc ( A ) is a two-dimensional algebra.The moduli set for this class of algebra is (End K ( B ) × B ) / G where G = Aut ( B ) × B × K × (see Proposition 5.7). Thus the isomorphism classes of algebras Adj ( B, ϕ, b , (for B fixed)are in one-to-one correspondence with the elements of the set (End K ( B ) × B ) / G (see equation (10) for the definition of the action).Proof. To prove item (1), notice first that it is easy to show that e i ∈ Soc( A ) for i = 1 , , = A ⊂ Soc( A ). This implies A/ Soc( A ) ∼ = K with zero product and therefore Soc( A/ Soc( A )) = { } . Hence, ssi( A ) = 1. Moreover, if A has dimension one, then anynonzero vector is natural (see [1, Theorem 3.3.]), a contradiction. So Soc( A ) = A . If Soc( A )is not minimal, then we apply Corollary 6.9 and obtain that there exists a natural basis { e ′ , e ′ , e ′ } such that the structure matrix is as (12), a contradiction because in this caseSoc( A ) = span( { e ′ , e ′ + e ′ } ).For item (2), by Definition 5.1 we have that A ∼ = Adj ( B, ϕ, b ), where B = Soc( A ) is atwo-dimensional algebra, b = ω ( e + e ) + ω ( e + e ) and the matrix of the linear map ϕ in the basis { e + e , e + e } is (cid:18) ω ω (cid:19) . Indeed, let Ω : A ! B × K be the isomorphism ofalgebras where Ω( e ) = ( e − e , e ) = ( e + e , −
1) and Ω( e ) = (0 , (cid:3) Let us give an example of an algebra of the kind in Theorem 6.14. Consider B = C as an R -algebra and A := Adj ( B, ϕ, b ), where b = i and ϕ : C ! C is the linear map given by ϕ ( x + iy ) = − i ( x + y ) for any x, y ∈ R . Thus ϕ (1) = ϕ ( i ) = − i . Then { (1 , , − ( i, , (0 , } is a natural basis of A and B = C is not an evolution R -algebra since it has no zero divisors.Then B ⊳ A is a minimal ideal since C has no proper nonzero ideals (see Lemma 5.6). Alsoit is impossible to have B = span( { e ′ , e ′ + e ′ } ) for some natural basis { e ′ i } i =1 of A , because e ′ ( e ′ + e ′ ) = 0 which is not possible in C . This example proves that the B in Theorem 6.14is not necessarily an evolution algebra.6.3. Socle of dimension one.
In this final subsection we study the case dim(Soc( A )) = 1.We distinguish two cases: Soc( A ) = 0 and Soc( A ) = 0. For this task we need to introducetwo new adjunctions. The first one will be studied in the following item.6.3.1. Adjunction of type three.
In order to study the case Soc( A ) = 0 we need to define athird type of adjunction of algebras. We construct a new algebra A , based on a K algebra B with a nonzero inner product, such that dim(Soc( A )) = 1 + dim(Soc( B )). Moreover, we studywhich conditions are necessary and sufficient for the adjunction to be an evolution algebra. Definition 6.15.
For a given algebra B with nonzero inner product h· , ·i : B × B ! K , define Adj ( B, h· , ·i ) := K × B with the product (13) ( λ, b )( λ ′ , b ′ ) := ( λλ ′ + h b, b ′ i , bb ′ ) , λ, λ ′ ∈ K , b, b ′ ∈ B. Note that the algebra B is not required to be an evolution algebra in this definition. Remark 6.16.
Let A be a non-degenerate 3-dimensional evolution algebra of the form A = K × B , where B is a 2-dimensional algebra and the product of A is given by ( λ, b )( λ ′ , b ′ ) =( λλ ′ , bb ′ ). We know that in this case B is a non-degenerate evolution algebra so that Soc( B ) =0 by Proposition 5.2. Then dim(Soc( A )) = 1 + dim(Soc( B )) > ( B, h· , ·i ). Lemma 6.17. If A = Adj ( B, h· , ·i ) is a -dimensional algebra then Soc ( A ) = K × { } .Proof. The subspace K × { } is an ideal of A and A ( K × { } ) = 0, hence K × { } ⊂ Soc( A ).We claim that K × { } = Soc( A ): if there is another one-dimensional ideal K ( λ, b ), then( λ, b )(1 ,
0) = ( λ, ∈ K ( λ, b ) implying b = 0, so K ( λ, b ) = K ( λ, A ) = A then thereis a minimal ideal J of dimension 2 in A , but if ( λ, b ) ∈ J we have (1 , λ, b ) = ( λ, ∈ J . Weconclude that either J = { } × B or K (1 , ⊂ J (which is not possible by the minimality of HAINS IN EVOLUTION ALGEBRAS 33 J ). Now we check that { } × B is not an ideal of A . Indeed, since (0 , b )(0 , b ′ ) = ( h b, b ′ i , bb ′ ) ∈{ } × B we get that h · , · i = 0, a contradiction. (cid:3) Next we prove that any non-degenerate three-dimensional evolution algebra with one-dimensional socle (and such that the square of the socle is non-zero) is isomorphic to anadjunction Adj ( B, h · , · i ), for suitable B and h· , ·i . In fact we prove a more general result,valid for finite dimensional evolution algebras. Theorem 6.18.
Let A be a non-degenerate finite-dimensional evolution algebra such that dim( Soc ( A )) = 1 and Soc ( A ) = 0 . Then A ∼ = Adj ( B, h· , ·i ) for a suitable B and h· , ·i .Proof. Take a nonzero generator e of Soc( A ) and let ϕ : A ! K be the linear map such that ex = ϕ ( x ) e for any x ∈ A . Observe that ϕ = 0 because A is a non-degenerate evolutionalgebra. This implies that the kernel B := Ker( ϕ ) has dimension dim( A ) −
1. We know thatSoc( A ) = 0, hence e = 0. Thus e / ∈ B and, rescaling if necessary, e can be taken to bean idempotent with A = K e ⊕ B , where eB = 0 and B is a subspace. So, if we multiply x, y ∈ B , we get a part in the socle and a part in B . We can formalize this by saying thatfor any x, y ∈ B one has xy = h x, y i e + β ( x, y ) , where h · , · i : B × B ! K is a bilinear symmetric form and β : B × B ! B is a bilinear map.Thus the multiplication of two arbitrary elements λe + x and λ ′ e + x ′ of A would be( λe + x )( λ ′ e + x ′ ) = λλ ′ e + h x, x ′ i e + β ( x, x ′ ) . Observe that β endows B with an algebra structure, so that writing xy := β ( x, y ) for x, y ∈ B we get the product defined in (13). Summarizing, A ∼ = Adj ( B, h · , · i ). (cid:3) So far, the algebra B given in Theorem 6.18 has not been proved to be an evolution algebra.We address this task in the following lemma. Lemma 6.19.
Let { ( λ i , b i ) } ni =1 be a basis of A = Adj ( B, h· , ·i ) . Then the following assertionsare equivalent:(1) The set { ( λ i , b i ) } is a natural basis of Adj ( B, h· , ·i ) .(2) B is an evolution algebra endowed with an inner product h· , ·i : B × B ! K such that h b i , b j i = − λ i λ j if i = j ; and the set { b i } ni =1 is pairwise orthogonal.Proof. Let { ( λ i , b i ) } ni =1 be a natural basis of A . If i = j we have0 = ( λ i , b i )( λ j , b j ) = ( λ i λ j + h b i , b j i , b i b j )implying that h b i , b j i = − λ i λ j and the b i ’s are pairwise orthogonal.Conversely, let B be an evolution algebra with an inner product h· , ·i : B × B ! K suchthat h b i , b j i = − λ i λ j if i = j . Then the set { ( λ i , b i ) } ni =1 is orthogonal because ( λ i , b i )( λ j , b j ) =( λ i λ j + h b i , b j i , b i b j ) = (0 , b i b j ) = (0 , (cid:3) Isomorphisms between adjunction algebras of type three.
In this section we deal thestudy of isomorphism between two algebras Adj ( B, h· , ·i ) and Adj ( B ′ , h· , ·i ) ′ for suitable( n − B and B ′ . Theorem 6.20.
Let A be a non-degenerate evolution algebra with dim( A ) = n > and witha -dimensional socle whose square is nonzero. Then A ∼ = Adj ( B, h· , ·i ) for a suitable ( n − dimensional evolution algebra B . Furthermore, Adj ( B, h· , ·i ) ∼ = Adj ( B ′ , h· , ·i ′ ) if and onlyif there is an isometric isomorphism β : B ! B ′ (isometric in the sense that h β ( x ) , β ( y ) i ′ = h x, y i for any x, y ∈ B ).Proof. By Theorem 6.18 we have A ∼ = Adj ( B, h· , ·i ) for a suitable B and h· , ·i . Let { ( λ i , b i ) } ni =1 be a natural basis of A . Applying Lemma 6.19 we get that the set { b i } ni =1 is pairwise orthog-onal. We prove that { b i } ni =1 is a system of generators of B . Indeed, given b ∈ B , noticethat (0 , b ) = P i k i ( λ i , b i ) for some scalars k i and hence b = P i k i b i . So { b i } ni =1 is a systemof generators of B . Therefore, we can select a basis of B by removing one vector of the set { b i } ni =1 . After reordering if necessary, we may assume that { b i } n − i =1 is a natural basis of B .Now we prove the second part of the theorem. Assume that the map θ : Adj ( B, h· , ·i ) ! Adj ( B ′ , h· , ·i ′ ) is an isomorphism. Then θ (1 ,
0) = ( k ,
0) for a nonzero scalar k (because θ fixes the socle). Also θ (0 , b ) = ( α ( b ) , β ( b )), where α : B ! K and β : B ! B ′ are linear. Thus θ ( λ, b ) = ( λk + α ( b ) , β ( b )) for an arbitrary ( λ, b ). Since θ is an isomorphim we deduce that β is an isomorphism. Furthermore, for any λ, λ ′ ∈ K and any β, β ′ ∈ B we have: θ (( λ, b )( λ ′ , b ′ )) = θ ( λλ ′ + h b, b ′ i , bb ′ ) = ( k ( λλ ′ + h b, b ′ i ) + α ( bb ′ ) , β ( bb ′ )) ,θ ( λ, b ) θ ( λ ′ , b ′ ) = ( λk + α ( b ) , β ( b ))( λ ′ k + α ( b ′ ) , β ( b ′ )) =(( λk + α ( b ))( λ ′ k + α ( b ′ )) + h β ( b ) , β ( b ′ ) i ′ , β ( b ) β ( b ′ )) . We also get k ( λλ ′ + h b, b ′ i ) + α ( bb ′ ) = ( λk + α ( b ))( λ ′ k + α ( b ′ )) + h β ( b ) , β ( b ′ ) i ′ ⇒ k λλ ′ + k h b, b ′ i + k α ( bb ′ ) = λλ ′ k + λk α ( b ′ ) + λ ′ k α ( b ) + α ( b ) α ( b ′ ) + h β ( b ) β ( b ′ ) i ′ . Then k = 1, and h b, b ′ i + α ( bb ′ ) = λα ( b ′ ) + λ ′ α ( b ) + α ( b ) α ( b ′ ) + h β ( b ) β ( b ′ ) i ′ . This forces λα ( b ′ ) + λ ′ α ( b ) = 0 and h b, b ′ i + α ( bb ′ ) = α ( b ) α ( b ′ ) + h β ( b ) , β ( b ′ ) i ′ . But the identity λα ( b ′ ) + λ ′ α ( b ) = 0 implies α = 0 so that h b, b ′ i = h β ( b ) , β ( b ′ ) i ′ for any b, b ′ ∈ B .Conversely, let β : B ! B ′ be an isometric isomorphism. Define θ : Adj ( B, h· , ·i ) ! Adj ( B ′ , h· , ·i ′ ) by θ ( λ, b ) := ( λ, β ( b )). It is easy to check that θ is an isomorphism of algebras. (cid:3) Remark 6.21.
By Theorem 6.20 the isomorphism classes of algebras
Adj ( B, h· , ·i ) , where B is fixed, can be described by the moduli set ( G , Sym ( B )) , where G = Aut ( B ) . The action G × Sym ( B ) ! Sym ( B ) is β · h i = h i ′ , where h x, y i ′ := h β ( x ) , β ( x ′ ) i for any x, x ′ ∈ B .Thus the algebras in this class are in one-to-one correspondence with the orbit set Sym ( B ) / G . To end the classification of the three-dimensional evolution algebras, we need to classifythe non-degenerate evolution algebras A with one-dimensional socle such that Soc( A ) = 0.This will be the goal of our next subsection.6.3.3. Adjunction of type four.
The focus of this subsection are the non-degenerate, three-dimensional evolution algebras A with one-dimensional socle verifying Soc( A ) = 0. Wepresent an example of such an algebra below. HAINS IN EVOLUTION ALGEBRAS 35
Example 6.22.
Consider B with natural basis { b , b } such that b = 0 and b = b + b .Let ϕ ∈ B ∗ (the usual dual space) be such that ϕ ( b ) = 1 and ϕ ( b ) = 0. Consider the innerproduct h· , ·i : B × B ! K with h b , b i = − h b , b i . Then, define the algebra A = K × B with product ( λ, b )( λ ′ , b ′ ) = ( h b, b ′ i + λϕ ( b ′ ) + λ ′ ϕ ( b ) , bb ′ ) . The vectors (1 , b ) , (1 , b ) , (0 , b ) form a natural basis and the structure matrix of A relativeto this natural basis is h − −
10 1 0 − − h , where h = h b , b i . It can be proved that Soc( A ) = K ( e − e ), with ( e − e ) = 0.Notice that if A is a non-degenerate, three-dimensional evolution algebra A with one-dimensional socle such that Soc( A ) = 0, then we can write Soc( A ) = K e and e = 0. So wehave a linear map ϕ : A ! K such that ex = ϕ ( x ) e for any x ∈ A . Since A is non-degenerate ϕ = 0, so ker( ϕ ) has dimension (dim( A ) − ϕ ) = { x ∈ A : ex = 0 } , the annihilatorof e in A . This suggests the following definition of adjunction, which is constructed from anarbitrary K algebra B with an inner product and a linear form. Definition 6.23.
Let B be a K -algebra with an inner product h· , ·i : B × B ! K and anelement ϕ ∈ B ∗ , that is, ϕ : B ! K is linear. Then we define in K × B the product ( λ, b )( λ ′ , b ′ ) = ( h b, b ′ i + λϕ ( b ′ ) + λ ′ ϕ ( b ) , bb ′ ) for any scalars λ, λ ′ ∈ K and b, b ′ ∈ B . This algebra will be denoted by Adj ( B, h· , ·i , ϕ ) . If A = Adj ( B, h· , ·i , ϕ ) is an evolution algebra, then B is also an evolution algebra. Indeed,if we assume that the collection { ( λ i , b i ) } ni =1 is a natural basis for A then, when i = j , wehave 0 = ( λ i , b i )( λ j , b j ) = ( h b i , b j i + λ i ϕ ( b j ) + λ j ϕ ( b i ) , b i b j )and hence b i b j = 0, which proves that the set { b i } ni =1 is orthogonal. Next we check thatit is a system of generators of the vector space of B . Take an arbitrary b ∈ B . Then(0 , b ) = P i k i ( λ i , b i ), where k i ∈ K . So b = P i k i b i . This way B is an evolution algebra and anatural basis of B can be obtained by removing some b j from the collection { b i } ni =1 . Theorem 6.24. If A is a non-degenerate evolution algebra of dimension n , with one dimen-sional socle and Soc ( A ) = 0 , then there is a ( n − dimensional evolution algebra B endowedwith an inner product h· , ·i : B × B ! K and ϕ : B ! K linear such that A ∼ = Adj ( B, h· , ·i , ϕ ) .Proof. Let Soc( A ) = K e . Then we know that e = 0. The quotient algebra A/ K e is anevolution algebra, hence pick one of its natural basis { ¯ u , · · · , ¯ u n − } . Then { e, u , · · · , u n − } is a basis of A and we have A = K e ⊕ B , where B is the linear span of u , . . . , u n − . Let p : A ! K be the canonical projection satisfying a = p ( a ) e + b for a ∈ A and b ∈ B . Now, weconsider the canonical projection q : A ! B . This allows us to define h· , ·i : B × B ! K by h x, y i := p ( xy ) and endow B with a K -algebra structure by defining a product B × B ! B such that ( x, y ) x ⊙ y := q ( xy ). We also take into account the linear map ϕ : A ! K such that xe = ϕ ( x ) e for any x ∈ A (note that ϕ = L e ), and consider the restriction ϕ : B ! K . Now, we perform the construction Adj ( B, h· , ·i , ϕ ). Finally, we prove thatthere is an isomorphism f : A ∼ = Adj ( B, h· , ·i , ϕ ). Notice that any x ∈ A can be writen as x = λe + b , with λ ∈ K and b ∈ B . Thus we define f ( λe + b ) = ( λ, b ). Then, if y = λ ′ e + b ′ for λ ′ ∈ K and b ′ ∈ B , we have xy = λeb ′ + λ ′ eb + bb ′ = λϕ ( b ′ ) e + λ ′ ϕ ( b ) e + h b, b ′ i e + b ⊙ b ′ , sothat f ( xy ) = ( λϕ ( b ′ ) + λ ′ ϕ ( b ) + h b, b ′ i , b ⊙ b ′ ). On the other hand f ( x ) f ( y ) = ( λ, b )( λ ′ , b ′ ) =( λϕ ( b ′ ) + λ ′ ϕ ( b ) + h b, b ′ i , b ⊙ b ′ ). The isomorphic character of f is trivial. (cid:3) Note that in the conditions of Theorem 6.24, if A happens to have dimension 3, then thereare no minimal ideals of A of dimension 3 or 2. Indeed: if J ⊳ A is minimal and dim( J ) = 2,then K e ∩ J = 0 and A = K e ⊕ J (direct sum of ideals), In this case eJ = 0, whence eA = 0,which is not possible because A is non-degenerate. If dim( J ) = 3 then J = A and A is nota minimal ideal since it contains K e . Thus the unique possible minimal ideals of A are theone-dimensional.Next we want to determine what conditions on A = Adj ( B, h· , ·i , ϕ ) are needed in orderto have Soc( A ) = K (1 , Lemma 6.25. If A = Adj ( B, h· , ·i , ϕ ) is -dimensional and non-degenerate, then Soc ( A ) = K (1 , if and only if both conditions below are satisfied:(1) There is no nonzero b ∈ B such that b ∈ B ⊥ ∩ ker( ϕ ) and Bb ⊂ K b .(2) There is no nonzero b ∈ ker( ϕ ) such that zb = ( ϕ ( z ) + h z, b i ) b for any z ∈ B .Proof. We prove that if (1) and (2) are satisfied, then Soc( A ) = K (1 , K (0 , b ) = 0. Assume on the contrary that K (0 , b ) ⊳ A . Then0 = (1 , , b ) = ( ϕ ( b ) , . Moreover, for any z ∈ B we have (0 , z )(0 , b ) ∈ K (0 , b ), whichimplies ( h z, b i , zb ) ∈ K (0 , b ). Summarizing:(14) b ∈ ker( ϕ ) Bb ⊂ K b h b, B i = 0 . Then the hypothesis imply b = 0, a contradiction. Next we prove that there is no idealof the form K (1 , b ) with b = 0. Assume on the contrary that such an ideal exists. Then0 = (1 , , b ) = ( ϕ ( b ) , b ∈ ker( ϕ ). Also(0 , z )(1 , b ) = ( ϕ ( z ) + h z, b i , zb ) ∈ K (1 , b ) , and therefore zb = α ( z ) b for some linear map α : B ! K such that ϕ ( z ) + h z, b i = α ( z ).Thus ϕ ( z ) b + h z, b i b = zb for any z ∈ B . Consequently, the hypothesis in (2) imply b = 0, acontradiction. So there are no one dimensional ideals other that K (1 , A ) = K (1 ,
0) then (1) and (2) hold. In order to do that,take into account that:(i) K (1 , b ) (with b = 0) is an ideal of A if and only if b ∈ ker( ϕ ) and h b, b ′ i b + ϕ ( b ′ ) b = bb ′ for every b ′ ∈ B ,(ii) K (0 , b ) ⊳ A (for b = 0) if and only if b ∈ ker( ϕ ) ∩ B ⊥ and bB ⊂ K b .So, as Soc ( A ) = K (1 , b ∈ B satisfying either of the previous conditions,and hence items (1) and (2) are verified. (cid:3) Remark 6.26.
Note that both conditions given in the lemma above are satisfied, for instance,if no nonzero b ∈ B satisfies Bb ⊂ K b . HAINS IN EVOLUTION ALGEBRAS 37
Isomorphisms between adjunction algebras of type four.
So far we know that 3-dimen-sional non-degenerate evolution algebras A , with one dimensional socle such that Soc( A ) = 0,are of the form Adj ( B, h· , ·i , ϕ ) for a 2-dimensional evolution algebra B satisfying both condi-tions (1) and (2) of Lemma 6.25. Next we investigate the isomorphism problem for two suchalgebras Adj ( B, h· , ·i , ϕ ) and Adj ( B ′ , h· , ·i ′ , ϕ ′ ). If θ : Adj ( B, h· , ·i , ϕ ) ∼ = Adj ( B ′ , h· , ·i ′ , ϕ ′ )is an isomorphism, then θ ( K × { } ) = K × { } , so θ (1 ,
0) = ( k ,
0) for some k ∈ K × . More-over, θ (0 , z ) = ( α ( z ) , β ( z )) for linear maps α : B ! K and β : B ! B ′ . It can be checkedthat(1) ϕ ′ β = ϕ .(2) β is an isomorphism from B to B ′ .(3) k h z, z ′ i + α ( zz ′ ) = α ( z ) ϕ ( z ′ ) + α ( z ′ ) ϕ ( z ) + h β ( z ) , β ( z ′ ) i ′ for all z, z ′ ∈ B .Conversely, it is straightforward to see that if α : B ! K and β : B ! B ′ satisfy (1), (2) and(3), then θ is an isomorphism from Adj ( B, h· , ·i , ϕ ) to Adj ( B ′ , h· , ·i ′ , ϕ ′ ). Summarizing weclaim Proposition 6.27.
There is an isomorphism θ : Adj ( B, h· , ·i , ϕ ) ∼ = Adj ( B ′ , h· , ·i ′ , ϕ ′ ) if andonly if there is an isomorphism β : B ! B ′ and a nonzero k ∈ K such that(1) k h z, z ′ i + α ( zz ′ ) = α ( z ) ϕ ( z ′ ) + α ( z ′ ) ϕ ( z ) + h β ( z ) , β ( z ′ ) i ′ for any z, z ′ ∈ B , and(2) ϕ ′ β = ϕ . So fix a 2-dimensional evolution algebra B and consider the problem of classifying thealgebras Adj ( B, h· , ·i , ϕ ) (where the only variables are the inner product and ϕ ). Considerthe action Aut( B ) × B ∗ ! B ∗ such that χ · ϕ := ϕχ − for any χ ∈ Aut( B ) and ϕ ∈ B ∗ .Now, take ϕ, ϕ ′ ∈ B ∗ . If the orbits of ϕ and ϕ ′ (under the previous action) are different,then there is no isomorphism from Adj ( B, h· , ·i , ϕ ) to Adj ( B, h· , ·i ′ , ϕ ′ ). If, on the contrary,the orbits of ϕ and ϕ ′ coincide, then ϕ ′ = ϕβ for some isomorphism β : B ′ ! B . ThusAdj ( B, h· , ·i , ϕ ) = Adj ( B, h· , ·i , ϕβ ) ∼ = Adj ( B, h· , ·i ′ , ϕ ). So we focus on the problem offinding the isomorphism condition forAdj ( B, h· , ·i , ϕ ) ∼ = Adj ( B, h· , ·i ′ , ϕ ) , where the unique variable is the inner product in each case. Then the isomorphism exists ifand only if there is linear map α : B ! K , and a nonzero k ∈ K , such that k h z, z ′ i + α ( zz ′ ) = α ( z ) ϕ ( z ′ ) + α ( z ′ ) ϕ ( z ) + h z, z ′ i ′ for any z, z ′ ∈ B . Define the group G := K × B ∗ with the product( k, α )( k ′ , α ′ ) := ( kk ′ , kα ′ + α ) , where k, k ′ ∈ K × , α, α ′ ∈ B ∗ . The identity element of G is (1 , B × B ! K with linear maps B ⊗ B ! K . Thus h· , ·i ( x ⊗ y ) := h x, y i for x, y ∈ B .(2) µ : B ⊗ B ! B is the product of B , that is, µ ( x ⊗ y ) = xy for x, y ∈ B .(3) If α, α ′ ∈ B ∗ then α ⊗ α ′ is the linear map B ⊗ B ! K such that ( α ⊗ α ′ )( x ⊗ y ) = α ( x ) α ′ ( y ), with x, y ∈ B . We can define • : B ∗ × B ∗ ! K by α • α ′ := α ⊗ α ′ + α ′ ⊗ α .So, there is an action G × Sym ( B ) ! Sym ( B ) such that(15) ( k, α ) · h· , ·i := k h· , ·i + αµ − α • ϕ. Observe that the linear map in (15) is an action since (1 , · h· , ·i = h· , ·i , and defining h· , ·i ′ := k h· , ·i + αµ − α • ϕ , we have( k ′ , α ′ ) · [( k, α ) · h· , ·i ] = ( k ′ , α ′ ) · h· , ·i ′ = k ′ h· , ·i ′ + α ′ µ − α ′ • ϕ = k ′ ( k h· , ·i + αµ − α • ϕ ) + α ′ µ − α ′ • ϕ = kk ′ h· , ·i + ( k ′ α + α ′ ) µ − ( k ′ α + α ′ ) • ϕ =( kk ′ , k ′ α + α ′ ) · h· , ·i = (( k ′ , α ′ )( k, α )) · h· , ·i . So, we can claim the following.
Proposition 6.28.
In the previous setting, we have that the pair ( G , Sym ( B )) is a moduliset for the class of algebras Adj ( B, h· , ·i , ϕ ) with B and ϕ fixed. Acknowledgments
The third author was partially supported by Conselho Nacional de Desenvolvimento Cien-t´ıfico e Tecnol´ogico (CNPq) grant numbers 304487/2017-1 and 406122/2018-0 and Capes-PrInt grant number 88881.310538/2018-01 - Brazil. The first and the last two authors aresupported by the Junta de Andaluc´ıa through projects FQM-336 and UMA18-FEDERJA-119and by the Spanish Ministerio de Ciencia e Innovaci´on through project PID2019-104236GB-I00, all of them with FEDER funds.
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E-mail address : [email protected] M. I. Cardoso Gonc¸alves: Departamento de Matem´atica, Universidade Federal de SantaCatarina, Florian´opolis, SC, 88040-900 - Brazil
E-mail address : [email protected] D. Gonc¸alves: Departamento de Matem´atica, Universidade Federal de Santa Catarina,Florian´opolis, SC, 88040-900 - Brazil
E-mail address : [email protected] D. Mart´ın Barquero: Departamento de Matem´atica Aplicada, Escuela de Ingenier´ıas In-dustriales, Universidad de M´alaga, Campus de Teatinos s/n. 29071 M´alaga. Spain.
E-mail address : [email protected] C. Mart´ın Gonz´alez: Departamento de ´Algebra Geometr´ıa y Topolog´ıa, Facultad de Cien-cias, Universidad de M´alaga, Campus de Teatinos s/n. 29071 M´alaga. Spain.
E-mail address ::