Classification of three-dimensional evolution algebras

Abstract

We classify three dimensional evolution algebras over a field having characteristic different from 2 and in which there are roots of orders 2, 3 and 7.

Power-Associative Evolution Algebras

“…Since e 2 i0 ∈ annE necessarily a ki0 = 0 and e 2 k e i0 = a ki0 e 2 i0 = 0 for all integer k, hence e i0 E 2 = 0. In fact e i0 ∈ E 2 otherwise we would have e 2 i0 ∈ E [3] = E 2 E 2 = e 2 i e 2 j ; 1 ≤ i, j ≤ n = 0: absurd because e 2 i0 = 0, hence the theorem Table 1.…”

“…We have 0 = e 2 1 e 2 4 = α 42 v 2 2 + α 43 v 2 3 = (α 42 − α 43 )v 5 , so α 42 = α 43 ; this relation ensures the power associativity of N. The family {e 1 , v 2 , v 3 , e 4 , v 5 } is a natural basis of N and its multiplication table is defined by: 2) The type of N is [1, 2, 2], then α 42 = 0. We have ann(N ) = v 5 , ann 2 (N ) = v 2 , v 3 , v 5 and ann 3…”

“…We have: ann(N ) = v 6 , v 2 , v 3 ∈ ann 2 (N ) and e 1 ∈ ann 3 (N ). Since e 2 4 , e 2 5 ∈ v 2 , v 3 , v 6 then the possible types of N are: [1,4,1], [1,3,2] • dim (U 3 ⊕ U 1 ) 2 = 1 =⇒ α 46 = 0. N ≃ N 6,20 (α, β) : e 2 1 = e 2 + e 3 , e 2 2 = e 6 , e 2 3 = −e 6 , e 2 4 = α(e 2 + e 3 ), e 2 5 = βe 6 , e 2 6 = 0 with α, β ∈ F * .…”

“…In [3, Table 1], the fifth, sixth and seventh algebras, of dimension 3, are associative. They are respectively isomorphic to E 3,4 , N 3,3 (1) and N 3,2 (see Table 1). In [9, Theorem 6.1], the algebras E 4,1 , E 4,2 , E 4,3 , E 4,5 and E 4,10 are associative.…”

Sociodemographic Factors of Alcohol Consumption in a Population of Hospitalized Patients in Ouagadougou (Burkina Faso)

“…Ainsi I(N 6,23 (α, β, γ, δ)) ≃ K 4 . • N 6,24 (α, β, γ) : on a R c (e 4 )+[R a , R b ](e 4 ) = αc 4 (e 2 +e 3 )+βc 4 e 6 +α(b4 (a 2 −a 3 )−a 4 (b 2 −b 3 ))e 6 ; R c (e 5 ) + [R a , R b ](e 5 ) = γc 5 (e 2 + e 3 ) + γ(b 5 (a 2 − a 3 ) − a 5 (b 2 − b 3 ))e 6 ; donc R c + [R a , R b ] ∈ I(N 6,24 (α, β, γ)) si et seulement si R c = c 2 (e 62 − e 63 ) et [R a , R b ] = (b 1 (a 2 − a 3 ) − a 1 (b 2 − b 3 ))e 61 + α(b 4 (a 2 − a 3 ) − a 4 (b 2 − b 3 ))e 64 + γ(b 5 (a 2 − a 3 ) − a 5 (b 2 − b 3 ))e 65 . Ainsi I(N 6,24 (α, β, γ)) ≃ K 4 .…”