The evolution operator of evolution algebras

ORIGINAL ARTICLE The evolution operator of evolution algebras Desamparados Fern´andez-Ternero, V´ıctor M. G´omez-Sousa and Juan N´u˜nez-Vald´es Universidad de Sevilla, Spain ARTICLE HISTORY Compiled January 12, 2021 ABSTRACT This paper deals with the evolution operator of evolution algebras. The main goal is to obtain novel results of this operator, both in the case in which it is a derivation and when it is a homomorphism of algebras. These new results constitute a criterion for classifying evolution algebras satisfying one of these conditions. An algorithm to obtain all these degenerate evolution algebras starting from those of smaller dimensions is also constructed. KEYWORDS Evolution operator; evolution algebras; algorithm AMS CLASSIFICATION 47C05; 17D92; 17A36; 17A20 1. Introduction At present, finding links between concepts from different branches of Mathematics is an increasingly common object of study for researchers. One of the mathematical concepts which plays an important role in some of these connections are the so-called evolution algebras. This paper deals with them, which were firstly introduced in the Ph.D. Thesis of Jin Jung Paul Tian [1] in 2004; then jointly presented with Petr Vojtechovsky [2] in 2006, and later analyzed in greater depth in a book by Tian [3] in 2008. Indeed, these algebras have connections with different mathematical theories, such as Graphs, Probability, Groups and Knots, for instance. These connections allow not only interpret known results from some fields in others, but also transfer the study of certain concepts from some areas to others. Concrete examples of these mentioned connections can be seen, for example, in [4,5] and [6], which relate evolution algebras to graphs and combinatorial structures, in [7] and [8], which link these algebras with Markov chains and in [9–11] and [12], which use them to study different biological populations. Moreover, it is also convenient to emphasize the direct application that evolution algebras have in Non-Mendelian Genetics, since its origin is due precisely to this branch of Biology. In the field of genetic algebras, to which evolution algebras belong, the product algebra has a clear interpretation: the reproduction among individuals of a certain population. CONTACT D. Fern´andez-Ternero. Email: [email protected] The structure of the paper is as follows. With the aim of the paper is self-contained, some basic concepts and results on evolution algebras are recalled in Section 1. Section 2 deals with the study of the evolution operator of an evolution algebra. Regarding this operator, two cases has been dealt with. The first one, when this operator Lis a derivation of the algebra, that is, when [L(x·y) = L(x)·y+x·L(y)].The second one, when the operator is an homomorphism of algebras, that is, when the condition [L(x·y) = L(x)·L(y)] is satisfied, for all pair of elements xand yof the algebra. In this section, an algorithm to obtain all degenerate evolution algebras starting from evolution algebras of smaller dimensions is constructed and new results on evolution subalgebras and ideals related with the evolution operator are also given. A short section of conclusions ends the paper. 2. Preliminaries on evolution algebras Let E≡(E, +,·) be an algebra over a field K. It is said that Eis an evolution algebra if there exists a basis B={ei, i ∈Λ}of E, where Λ is an index set, such that ei·ej= 0, if i6=j. The basis Bis called a natural basis. The previous definition implies that ej·ej=Pi∈Λaijei, with aij ∈K, where only a finite quantity of aij ,called structure constants are non-zero for each j∈Λ fixed. So, the product on Eis determined by the structure matrix A= (aij). A particular type of evolution algebras are those called graphicable. An evolution algebra is graphicable if constants aij are 1 or 0, for all i, j ∈Λ. In general, evolution algebras are neither associative nor power-associative, but they are flexible [2]. There are two types of trivial evolution algebras. The ones called zero evolution algebras, which are those such that aij = 0, for all i, j ∈Λ and the ones called nonzero trivial evolution algebras, those that aij = 0 if and only if i6=j, that is, e2 j=ajej, for all j∈Λ, where aj∈K\ {0}. Both concepts are intrinsic, that is, independent of the basis. An evolution algebra is said to be non-degenerate if e2 i6= 0, for all i∈Λ. On the contrary, the algebra is called degenerate. Regarding subspaces of evolution algebras, let Ebe an evolution algebra and E0⊆E a subspace of E.E0is a subalgebra of Eif it is closed with respect to the second inner law of Eand it is an evolution subalgebra of Eif it is both subalgebra and of evolution, that is, if there exists a natural basis B0={ei, i ∈Λ0}of E0such that ei·ej= 0, if i6=j. It is said that E0verifies the extension property if a natural basis B0of E0can be extended to a natural basis Bof E. A subspace E0is an ideal of Eif E0·E⊆E0. Then, taking these definitions into account and the fact that every ideal of an evolution algebra is a subalgebra, the concepts of evolution ideal and evolution ideal with the extension property can be defined. In fact, it is known that every subalgebra with the extension property is an ideal, so if E0has the extension property, then the concepts of evolution subalgebra and evolution ideal are equivalent. 2 3. The evolution operator of an evolution algebra Let Ebe an evolution algebra with natural basis B={ei, i ∈Λ}and structure matrix A= (aij). Tian defined in [3] the evolution operator associated with Bas the endomorphism L:E→Ewhich maps each generator into its square, that is, L(ej) = e2 j=Pi∈Λaijei, for all j∈Λ. Note that the matrix representation of the evolution operator with respect to the basis Bis the structure matrix A= (aij). An alternative way of defining this operator is L(x) = θ·x, where θ=Pi∈Λei. This is due to that Lis linear and L(ej) = Pi∈Λeiej=e2 j. So, both definitions are equivalent. 3.1. Evolution algebras whose evolution operator is a derivation In this subsection we study those evolution algebras Eover Cwith natural basis B= {ei, i ∈Λ}and structure matrix A= (aij) such that L∈Der(E), i.e., the evolution operator is a derivation. Tian shows in [3] that a derivation dis characterized by the fact that its associated matrix D= (dij ) satisfies the equations 2ajidii =Pk∈Λakidjk, for all i, j ∈Λ, and akjdji +akidij = 0, for all i, j, k ∈Λ with i6=j. Since the first condition is equivalent to 2ADiag(D) = DA (where Diag(D) is the diagonal matrix whose diagonal is the same as that of D), then these algebras are characterized by the equations 2ADiag(A) = A2,(1) akjaji +akiaij = 0,∀k, ∀i6=j. (2) The first equation is equivalent to A(A−2Diag(A)) = 0, or what is the same, A¯ A= 0, where ¯ Ais the matrix obtained from Aby changing the sign of its diagonal elements. Let us analyze the second equation. If aij = 0, then either aji = 0 or the column j is null. If aij 6= 0, then the entries in column iand those in column jare related by the equalities aki =−aji aij akj. Therefore, the structure matrix must satisfy the conditions A¯ A= 0.(3) If aij = 0,with i6=j, then aji = 0 or akj = 0,∀k. (4) If aij 6= 0,with i6=j, then aki =−aji aij akj,∀k. (5) Proposition 3.1. Let Ebe an evolution algebra with structure matrix A= (aij)such that L∈Der(E). Then, Eis degenerate if and only if there exists jsuch that ajj = 0. Proof. The left-to-right implication is a consequence of the definition of degeneracy. For the right-to-left implication we proceed by contradiction and suppose Eis nondegenerate. Let jbe such that ajj = 0. Since Eis non-degenerate, column jis not null and there exists i6=jsuch that aij 6= 0. Since column iis not null, then aji 6= 0, otherwise the Condition (4) would not be satisfied. Switching indices iand jin 3 Condition (5) and taking k=jwe get 0 = ajj =−aij 6= 0, which is a contradiction. Next we analyze the structure of the matrix Awhen L∈Der(E). For this we do not take into account the permutations of the basis B, since then the evolution operator will be the same. We begin by studying the case in which Eis non-degenerate. Theorem 3.2. Let Ebe a non-degenerate evolution algebra with natural basis B= {ei, i ∈Λ}and structure matrix A= (aij). Then, L∈Der(E)if and only if there exists a rearrangement of the basis Bsuch that the structure matrix is a block diagonal matrix Diag(Ma1, Ma2, . . . ), where Mai=ai−ai −aiai, for some ai6= 0. Proof. The right-to-left implication is easy to check. For the left-to-right implication we have the following properties (i) Since Eis non-degenerate, then ajj 6= 0, for all j. (ii) There is no column (or row) jin which all elements, except ajj, are null. By contradiction, suppose ajj is the only non-null element in column (row) j. Then, the entry of A¯ Awhose position is (j, j) would be −a2 jj 6= 0, which contradicts the Condition (3). (iii) If aij 6= 0, with i6=j, then aij =−ajj. By the Condition (4) we know that aji 6= 0. Switching indices iand jin Condition (5) and taking k=jwe get aij =−ajj. (iv) For all j, there exists a unique i6=jwith aij 6= 0. By contradiction, suppose there exists k6=i, j with akj 6= 0. By Condition (5) and using the previous property we have −ajj =ajj −aii (−aii), which is a contradiction, because ajj 6= 0. (v) For all i, there exists a unique j6=iwith aij 6= 0. This follows from the previous property and the Condition (4), from which we deduce that aij 6= 0 if and only if aji 6= 0. Therefore the matrix Ahas the form         . . .. . . . . . ajj . . . −aii . . . . . .. . . . . . −ajj . . . aii . . . . . .. . .         , where the four indicated elements are the only ones nonzero elements of the ith and jth rows and columns. By the Condition (3) we obtain that aii =ajj. By reordering the 4 basis (putting ejin first position and eiin second position), we can write the structure matrix as Diag(Maii , A0), with A0satisfying the above five properties. By reiterating these arrangements we arrive at the desired result. An immediate consequence of the previous result is the following corollary. Corollary 3.3. There does not exist an odd-dimensional non-degenerate evolution algebra Esuch that L∈Der(E). Since we know the structure matrix of these non-degenerate algebras, we can make explicit their space of derivations. Proposition 3.4. Let Ebe a non-degenerate evolution algebra such that L∈Der(E), i.e., with structure matrix A= (aij) = Diag(Ma1, Ma2, . . . ), for some ak6= 0. Then, the space of derivations of Eis formed by those endomorphisms whose matrix representation is Diag(Mb1, Mb2, . . . ), for some bi∈C. Proof. Let D= (dij) be the matrix representation of a derivation d. Then Equations (1) and (2) are satisfied. If dij = 0 with i6=j, then akjdji = 0, for all k. Since no column of Ais null, there exists ksuch that akj 6= 0 and hence dji = 0. As a consequence, dij is null if and only if dji is null. If dij 6= 0, then dji 6= 0. Since aki =−dji dij akj, then aki is null if and only if akj is null, for all k. Due to the structure of the matrix A, this can only happen when iand jare equal to 2r−1 and 2r, for some r > 0. Therefore, if iand jare not equal to 2r−1 and 2r, then dij = 0. As a consequence we have that D=Diag(D1, D2, . . . ), where Dkis a matrix of size 2 ×2, for all k. Let us see that each Dkis equal to Mb, for some b∈C. From the equality (1), it follows that 2MakDiag(Dk) = DkMak, for all k. These systems are of the form 2ak−ak −akakd11 0 0d22=d11 d12 d21 d22ak−ak −akak. By operating we obtain 2akd11 −d22 −d11 d22 =akd11 −d12 d12 −d11 d21 −d22 d22 −d21. The only solution to this system is d11 =d22 =−d12 =−d21, so each Dkis equal to Mb, for some b∈C. Let us now study the case in which Eis degenerate and has dimension n. Since it is degenerate, then there must be a column of Athat is null. By reordering the basis, we can assume that this column is the last one. Then, this matrix is written as A=A00 vt0,(6) where A0is a square matrix of order n−1 and vis a vector with n−1 entries. Condition 5 (3) is equivalent to A0¯ A0= 0 and vt¯ A0= 0. Since Conditions (4) and (5) can be restricted to A0, then this matrix will be the structure matrix of an evolution algebra E0of dimension n−1 whose evolution operator satisfies that L0∈Der(E0). The above reasoning provides us with the following algorithm to obtain all the structure matrices of these degenerate algebras Step 1: Obtain all matrices A0of dimension n−1, both degenerate and nondegenerate. Step 2: Solve the linear system vt¯ A0= 0, for each matrix A0obtained previously. Step 3: Check if A=A00 vt0satisfies Conditions (4) and (5). Let us use this reasoning to compute all evolution algebras of dimensions 2, 3 and 4 whose evolution operator is a derivation. 3.1.1. 2-dimensional evolution algebras On the one hand, the only non-degenerate evolution algebras whose evolution operator is a derivation are those whose structure matrix is Ma, with a6= 0. On the other hand, since the only one of dimension 1 is the zero evolution algebra, the degenerate cases of dimension 2 are 0 0 a0with a∈C, all of which satisfy Conditions (4) and (5). 3.1.2. 3-dimensional evolution algebras From Corollary 3.3, there does not exist non-degenerate evolution algebras. The degenerate cases are   a−a0 −a a 0 v1v20 and   0 0 0 a0 0 v1v20 , with v= (v1, v2)tsatisfying vt¯ A0= 0. Solving these linear systems, we deduce that these matrices must be of the form   a−a0 −a a 0 b−b0 ,  0 0 0 a0 0 b0 0 or   0 0 0 0 0 0 a b 0 , with a, b ∈C. All these matrices satisfy Conditions (4) and (5). 6 3.1.3. 4-dimensional evolution algebras The non-degenerate cases are those of the form Diag(Ma, Mb), with a, b 6= 0. The degenerate cases are     a−a0 0 −a a 0 0 b−b0 0 v1v2v30     ,    0 0 0 0 a0 0 0 b0 0 0 v1v2v30     and     0 0 0 0 0 0 0 0 a b 0 0 v1v2v30     , with v= (v1, v2, v3)tsatisfying vt¯ A0= 0. Solving these linear systems, we deduce that these matrices must be of the form     a−a0 0 −a a 0 0 b−b0 0 c−c0 0     ,    a−a0 0 −a a 0 0 0 0 0 0 b−b c 0     ,    0 0 0 0 a0 0 0 b0 0 0 c d −ad b0     ,     0 0 0 0 a000 0 0 0 0 b0c0     ,    0 0 0 0 0 0 0 0 a b 0 0 c d 0 0     or     0 0 0 0 0 0 0 0 0 0 0 0 a b c 0     , with a, b, c, d ∈C. All these matrices satisfy Conditions (4) and (5), except the third one, in which it is necessary to impose that d= 0. To see this, we can assume that a6= 0, since otherwise it would be a particular case of the fifth model. So, by imposing Condition (4) on the element at position (1,2), we get that d= 0. 3.2. Evolution algebras whose evolution operator is an endomorphism of algebras We begin by defining some matrix operations that will be used throughout this section. Definition 3.5. Let A= (aij ) and B= (bij) be two n×mmatrices. We define the following matrix operations •The Hadamard product AB= (aij bij ). We define A(2) =AA. •A∗B= (ck ij) a n×m(m−1) 2matrix whose columns are indexed by pairs (i, j) with i < j, considering the lexicographical order. The column (i, j) is the Hadamard product of the ith column of Aand the jth column of B. In this section we study those evolution algebras Eover Cwith natural basis B= {ei, i ∈Λ}and structure matrix A= (aij ) such that L∈End(E), i.e., the evolution operator is an endomorphism of algebras. Tian shows in [3] that an endomorphism of algebras gis characterized by the fact that its associated matrix Gsatisfies the equations AG(2) =GA and A(G∗G) = 0. Then, these algebras are characterized by 7 the equations AA(2) =A2, A(A∗A)=0. On the one hand, the first equation is equivalent to A(A(2) −A) = 0. The elements of A(2) −Aare of the form a2 ij −aij =aij(aij −1). Therefore, the above condition is equivalent to the equality Acj(cj−~ 1 )= 0,∀j, where cjis the jth column of Aand ~ 1 is the vector (1,...,1)t. On the other hand, according to the definition of the operation ∗, the second equation is equivalent to A(cicj)=0,∀i6=j. Therefore, the structure matrix must satisfy the conditions Acj(cj−~ 1 )= 0,∀j, (7) A(cicj)=0,∀i6=j. (8) We begin by studying the case in which Eis degenerate and has dimension n. Then, this matrix is written as (6), where A0is a square matrix of order n−1 and vis a vector with n−1 entries. Condition (7) is equivalent to A0c0 j(c0 j−~ 1 )= 0 and vtc0 j(c0 j−~ 1 )= 0,∀j, where c0 jis the jth column of A0. Condition (8) is equivalent to A0c0 ic0 j= 0 and vtc0 ic0 j= 0,∀i6=j. In this way, A0will be the structure matrix of an evolution algebra E0of dimension n−1 with evolution operator L0∈End(E0) and vwill be orthogonal to each vector c0 j(c0 j−~ 1 ) and c0 ic0 j, for all i, j with i6=j. As a consequence of the previous reasoning we can focus on the study of the nondegenerate evolution algebras, since if these are known we can obtain the structure matrices of all the degenerate ones using the following algorithm Step 1: Obtain all matrices A0of dimension n−1, both degenerate and nondegenerate. Step 2: For each matrix A0obtained, find the vector space orthogonal to {c0 j(c0 j− ~ 1 ), c0 ic0 j;i6=j}. If vbelongs to this space, then the searched matrices will be those of the form (6) Let us now study the case in which Eis non-degenerate. We begin with the case where Lis an automorphism, that is to say, Ahas maximum rank. 8 Proposition 3.6. Let Ebe an evolution algebra with structure matrix A= (aij) such that Lis an automorphism. Then, L∈End(E)if and only if Ais obtained by permuting the columns of the identity matrix. In particular, Eis a graphicable evolution algebra. Proof. The right-to-left implication is easy to check. For the left-to-right implication we proceed as follows. Since Ahas maximum rank, from Condition (7) we obtain that cj(cj−~ 1 ) = 0,∀j. That is to say, aij (aij −1) = 0, for all i, j. Equivalently, aij is zero or one, for all i, j. From Condition (8) we obtain that cicj= 0,∀i6=j. Therefore, there cannot be two columns that have a non-null element in the same position. Since no column is null (since the range of Ais maximum), the only possibility is that each column cjis a basis vector ekjof Cn, with kj6=kj0when j6=j0. If the rank of Ais not maximum we have the following result Proposition 3.7. Let Ebe an evolution algebra of dimension nsuch that L∈ End(E). Let ckbe a non-null column that can be written as a linear combination of others columns. Then, Ack=Ac(2) k= 0. In particular, if Eis non-degenerate and Ahas rank n−k, then there are at least k+ 1 of these columns. Proof. Suppose there is a non-null column ckthat is a linear combination of the other columns. That is, there exist λ1, . . . , λk−1, λk+1, . . . , λnsuch that ck= n X i=1 i6=k λici. Let j6=kbe such that λj6= 0 and cj6= 0. Then, ckcj=   n X i=1 i6=k λici  cj= n X i=1 i6=k λicicj. Using Condition (8) we have 0 = A(ckcj) = n X i=1 i6=k λiA(cicj) = λjA(cjcj). Since λj6= 0, then A(cjcj) = 0. Since Condition (7) assures us that A(cjcj−cj) = 0, we deduce that Acj= 0. To finish, it is only necessary to point out that the roles of cjand ckare interchangeable. 9 In addition, from the above reasoning it follows that {e[n−1] i·e[n−1] i:i∈Ω}={e[n] i: i∈Ω}is a generator set of E[n], so there exists Ω0⊆Ω⊆Λ such that {e[n] i:i∈Ω0} is a basis. Furthermore, this is a natural basis, since if i6=j, e[n] i·e[n] j=Le[n−1] i·Le[n−1] j=Le[n−1] i·e[n−1] j=L(0) = 0. Finally, e[n+1] i=e[n] i·e[n] i=Le[n−1] i·Le[n−1] i=Le[n−1] i·e[n−1] i=Le[n] i, Le[n] i=L(Ln(ei)) = Ln+1(ei), Ln+1(ei) = Ln(L(ei)) = Ln X k∈Λ akiek!=X k∈Λ akiLn(ek) = X k∈Λ akie[n] k. An important observation is that from the equality e[n] i2=X k∈Λ akie[n] k we cannot deduce, in general, that the structure matrix associated to the evolution subalgebra E[n], with respect to the natural basis {e[n] i:i∈Ω}, is Aor a submatrix of A, since Ω is contained in Λ. In fact, Proposition 3.7 assures us that this inclusion is strict as long as Lis not an automorphism. In the case where X k∈Λ\Ω akie[n] k= 0, we will have that the structure matrix is the submatrix of Aformed by the rows and columns whose indexes are in Ω. Moreover, we have that E[n]is an evolution ideal of E[n−1]. Since E[n]⊆E[n−1], if x∈E[n]then x·y∈E[n], for all y∈E[n−1]. In particular, E[1] is an evolution ideal of E. In the following example we can see that, in general, E[n]is not an ideal, for n > 1. We also see that, in general, the evolution subalgebra E[n]does not have the extension property. Example 3.12. Let us consider the evolution algebra Ewith basis {e1, e2, e3}and structure matrix A=  a−a λ a−a λ 0 0 1 , where aand λare non-zero scalars. We have previously seen that in this case L∈ 16 End(E). Since A2=  0 0 λ 0 0 λ 0 0 1  and e[2] i=L2(ei), then e[2] 1=e[2] 2= 0, e[2] 3=λ(e1+e2) + e3and E[2] =< e[2] 3>. Let us see that E[2] is not an ideal. For this we consider x=e1∈E, y=λ(e1+e2) + e3∈E[2]. Then, x·y=λe2 1=λa(e1+e2)/∈E[2]. From where it follows that E[2] is not an ideal. Let us now consider E[1]. Since e[1] 1=−e[1] 2=a(e1+e2) and e[1] 3=λ(e1+e2) + e3, then E[1] =< e[1] 1, e[1] 3>. Let us see that E[1] does not have the extension property. By contradiction, suppose there exists x=P3 i=1 xieibeing linearly independent of e[1] 1 and e[1] 3and such that x·e[1] 1=x·e[1] 3= 0. Since x·e[1] 1= 0, then a(x1e2 1+x2e2 2)=0, from where a2(x1−x2)(e1+e2) = 0, so x1=x2. Since x·e[1] 3= 0, then 0 = λ(x1e2 1+x2e2 2) + x3e2 3=x3e2 3=λx3(e1+e2) + x3e3, so x3= 0. Accordingly, x=x1(e1+e2), which is linearly dependent of e[1] 1, being a contradiction. Corollary 3.13. Let Ebe an evolution algebra with natural basis B={ei:i∈Λ}, structure matrix A= (aij)and evolution operator L∈End(E). Let nbe a natural number and Ω⊆Λa subset such that {e[n] i:i∈Ω}is a natural basis of E[n]. Let L0be the evolution operator of E[n]associated to this basis. Then, L0=L|E[n]and L0∈End(E[n]). Proof. To prove that L0=L|E[n], it is only necessary to see that this equality is satisfied by the elements of the natural basis {e[n] i:i∈Ω}. To do this, L0e[n] i=e[n] i2=e[n+1] i=Le[n] i. 17 Furthermore, since L0(x·y) = L(x·y) = L(x)·L(y) = L0(x)·L0(y), for all x, y ∈E[n], it follows that L0∈End(E[n]). A question that arises naturally is whether the sequence of subalgebras E=E[0] ⊇E[1] ⊇ · · · ⊇ E[n]⊇ · · · stabilizes and, if so, when does it. If Eis a finite dimensional algebra, we obtain the following result Proposition 3.14. Let Ebe an evolution algebra with natural basis B={ei:i= 1, . . . , n}, structure matrix A= (aij)of rank n−kand evolution operator L∈End(E). Then, (1) If E[i]=E[i+1], then E[i+r]=E[i], for all r≥0. (2) k= 0 if and only if E[r]=E, for all r≥0. (3) If k > 0, then E[n−k+1+r]=E[n−k+1], for all r≥0. Proof. (1) Let us prove by induction on r. For r= 0 and r= 1 is trivial. Suppose the result holds for r. Then, E[i+r+1] =E[i+r]·E[i+r]=E[i]·E[i]=E[i+1] =E[i]. That is, the result holds for r+ 1. (2) If k= 0, then, by Proposition 3.6, Ais obtained by permuting the columns of the identity matrix. Since E[1] is generated by {e2 i:i= 1, . . . , n}, then E[1] =E. By the previous case, E[r]=E, for all r≥0. If E[r]=E, for all r≥0, then E[1] =E. Since E[1] =Eis generated by {e2 i:i= 1, . . . , n}, then Amust be of rank n, that is to say, k= 0. (3) Since E[1] =Eis generated by {e2 i:i= 1, . . . , n}, then dim(E[1]) = n−k. Due to dimensionality, the equality E[i]=E[i+1] will be obtained for some i≤n−k+1. The result follows from the case 1. We say that an evolution algebra Eis solvable if there exists a natural number n such that E[n]= 0. The smallest natural number that satisfies this property is called the solvability index. If Eis a finite dimensional algebra, we obtain the following result Proposition 3.15. Let Ebe an evolution algebra with natural basis B={ei:i= 1, . . . , n}, structure matrix A= (aij)and evolution operator L∈End(E). Then, E[n]= 0 if and only if An= 0. As a consequence, Eis solvable if and only if Ais a nilpotent matrix. Moreover, the solvability index of Eand the nilpotency index of A are equal. Proof. It is a direct consequence of the fact that E[n]is generated by {e[n] i:i= 1, . . . , n}and that e[n] i=Ln(ei). 18 The following examples show that this result does not hold if L /∈End(E). Example 3.16. Let us consider the evolution algebra Ewith basis {e1, e2}and structure matrix A=1−1 −1 1 . On the one hand, it can be verified that An= 2n−1A, so Ais not nilpotent. On the other hand, e2 1·e2 1= (e1−e2)·(e1−e2) = e2 1+e2 2= 0, e2 2·e2 2= (−e1+e2)·(−e1+e2) = e2 1+e2 2= 0, e2 1·e2 2= (e1−e2)·(−e1+e2) = −e2 1−e2 2= 0. Since E[1] is generated by {e2 1, e2 2}, then E[2] = 0, so Eis solvable. Example 3.17. Let us consider the evolution algebra Ewith basis {e1, e2}and structure matrix A=6−9 4−6. On the one hand, it can be verified that A2= 0, so Ais nilpotent. On the other hand, e2 1·e2 2= (6e1+ 4e2)·(−9e1−6e2) = −54e2 1−24e2 2= =−54(6e1+ 4e2)−24(−9e1−6e2) = −108e1−72e2= =−18 e2 1= 12 e2 2. Therefore, (e2 1·e2 2)[1] =−18 ·12 e2 1·e2 2= (−18)2·12 e2 1=−18 ·122e2 2. By induction it can be proved that (e2 1·e2 2)[n]= (−18)2n·122n−1e2 1= (−18)2n−1·122ne2 2, so Eis not solvable. 4. Conclusions As authors have been able to observe, there exist few papers in the literature dealing with the evolution operator of evolution algebras, in spite of this concept has showed to be very adequate to deep in the study of those algebras. We think that this paper, in which several novel results related to this operator have been obtained, could increase that number and be interesting and useful for researchers interested in this topic. However, several aspects continue open. With the objective of show them to researchers, we give a short list of them. 19 (1) When the evolution operator is a derivation, it is easy to see that the equality [Ln(x·y) = Pn k=0 n kLn−k(x)·Lk(y)],for all n≥1 is satisfied. An operator is said to be a derivation of order nwhen it satisfies the above equality for n. At this respect, we have studied the case in which Lis a derivation of order 1. A question that could be addressed is to study when the evolution operator is a derivation of order n, for n≥2. Moreover, if Lis a derivation, it is also certain the expression [L(x1· · · xn) = Pn i=1 x1· · · xi−1·L(xi)·xi+1 · · · xn,],for all n≥2, as long as the products on both sides of the equality are performed in the same order. An operator is said to be a n-derivation if it meets the above equality for n. In the same way, it could be studied when Lis an n-derivation, for n≥3. (2) Study when the evolution operator is an endomorphism of algebras of rank r, with 2 ≤r≤n−2, in order to obtain a characterization of these algebras that allows a complete classification for dimensions greater than 3. (3) Transfer the results obtained to other branches of Mathematics that have a direct connection with evolution algebras. For example, each evolution algebra has associated a directed graph such that the edge starting from eito ejhas weight aij . Starting from this, how is the fact that the evolution operator is a derivation or an endomorphism of algebras expressed in the graph? What properties do these graphs have? (4) Finally, in the field of applications, study when the evolution operator satisfies certain equalities, different from those studied, that involve the product and that have a greater biological interest to model certain populations. Funding This work was supported by Junta de Andaluc´ıa Research Group FQM-326. References [1] Tian JP. 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