Four Dimensional Absolute Valued Algebras Containing a Nonzero Central Idempotent or with Left Unit

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1 International Journal of Algebra, Vol. 10, 2016, no. 11, HIKARI Ltd, Four Dimensional Absolute Valued Algebras Containing a Nonzero Central Idempotent or with Left Unit A. Moutassim Centre Régional des Métiers de l Education et de Formation, Farah 2. Settat, Maroc M. Benslimane Département de Mathématiques et Informatiques Faculté des Sciences. Tétouan, Maroc L Equipe de Recherche Algèbres et ses Applications Copyright c 2016 A. Moutassim and M. Benslimane. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract An absolute valued algebra is a nonzero real algebra that is equipped with a multiplicative norm ( xy = x y ). We classify, by an algebraic method, all four-dimensional absolute valued algebras containing a nonzero central idempotent. Moreover, we construct a new absolute valued algebras with left unit of four dimension. Keywords: Absolute valued algebra, central idempotent, left unit 1 Introduction An absolute valued algebra is a nonzero real algebra that is equipped with a multiplicative norm ( xy = x y ). In 1947 Albert proved that the finite dimensional unital absolute valued algebras are classified by R, C, H, O, and that every finite dimensional absolute valued algebra is isotopic to one of the algebras R, C, H, O and so has dimension 1, 2, 4, or 8 [1]. Urbanik and

2 514 A. Moutassim and M. Benslimane Wright proved in 1960 that all unital absolute valued algebras are classified by R, C, H, O [12]. It is easily seen that the one-dimensional absolute valued algebras are classified by R, and it is well-known that the two-dimensional absolute valued algebras are classified by C, C, C, C [11]. The four-dimensional absolute valued algebras have been described by M.I. Ramírez Álvarez in 1997 [9]. The problem of classifying all eight-dimensional absolute valued algebras seems still to be open. In the present paper, we classify, by an algebraic method, all four-dimensional absolute valued algebras containing a nonzero central idempotent, this latter is a generalization of a results given in [6]. On the other hand, the class of fourdimensional absolute valued algebras with left unit studied in [2], [3], [7] and [8] contains a subalgebras of dimension 2. Here we extend this result to more general situation. Indeed, we prove that, if A is a four-dimensional absolute valued algebra with left unit containing a subalgebra of dimension 2, then A is isomorphic to a new absolute valued algebra with left unit of four dimension. We note that there are a four-dimensional absolute valued algebras with left unit do not contain two-dimensional subalgebras [9]. 2 Notations and Preliminary Results Throughout the paper, all absolute valued algebras are assumed to be real and not necessarily associatives. Definition 2.1 A is called a normed algebra (resp, absolute valued algebra) if it is endowed with a space norm:. such that xy x y (resp, xy = x y ), for all x, y A. The most natural examples of absolute valued algebras are R, C, H (the algebra of Hamilton quaternions), O (the algebra of Cayley numbers) with norms equal to their usual absolute values. The reader is referred to [4] for basis facts and intrinsic characterization of these classical absolute valued algebras. The algebra C (resp, C and C) obtained by replacing the product of C with the one defined by x y = x y (resp, x y = xy and x y = x y), where means the standard involution of C. We need the following results: Theorem 2.2 ([5]) The norm of any absolute valued algebra containing a nonzero central idempotent comes from an inner product. Theorem 2.3 ([10]) The norm of any absolute valued algebra A with left unit e comes from an inner product (..) satisfying (xy z) = (y xz) and x(xy) = x 2 y for all x, y, z A with x orthogonal to e.

3 Four dimensional absolute valued algebras Four-dimensional absolute valued algebras containing a nonzero central idempotent In this section, we construct a new absolute valued algebras containing a nonzero central idempotent of four dimension. The following results have great importance in this paragraph Lemma 3.1 Let A be an absolute valued algebra containing a nonzero central idempotent e, then 1) x 2 = x 2 e, for all x (Re). 2) xy + yx = 2(x y)e, for all x, y (Re). Proof. 1) According to Theorem 2.2, A is an inner product space and assuming that x = 1. We have That is (x 2 e) = 1, then x 2 = e. 2) is clear x 2 e = x e x + e = 2. Lemma 3.2 Let A be a finite dimensional absolute valued algebra containing a nonzero central idempotent e, then A contains a subalgebra of dimension 2. Proof. The operators L e, R e are linear isometries fixing e, and induce isometries on the orthogonal space (Re) := E. As E has odd dimension and L e commutes with R e, there exists common norm-one eigenvector u E for both L e and R e associated to eigenvalues α, β {1, 1}. That is, u 2 = e. It follows that Lin{e, u} is a two-dimensional subalgebra of A invariant under both L e and R e. Lemma 3.3 Let A be a four-dimensional absolute valued algebra containing a nonzero central idempotent e and a subalgebra B = A(e, i) (i 2 = e, ie = ei = ±i) of dimension 2. If x, y B, then xy B. Proof. According to Theorem 2.2, A is an inner product space and let {e, i, j, k} be an orthonormal basis of A. Since A is a division algebra, then the operator L j of left multiplication by j on A is bijective. Therefore there exists j 1 A such that j 1 = 1 and i = L j (j 1 ) = jj 1, we have (j 1 e) = (jj 1 je) = (i je) = ±(ie je) = ±(i j) = 0.

4 516 A. Moutassim and M. Benslimane And (j 1 i) = (jj 1 ji) = (i ji) = ±(ei ji) = ±(e j) = 0. Then there exists α, β R such that j 1 = αj + βk, this gives That is i = jj 1 = αj 2 + βjk. βjk = αe + i. Hence jk B (β 0). We put x = γj + δk and y = λj + µk (γ, δ, λ, µ R), we have xy = γλj 2 + γµjk + δλkj + δµk 2 = (γλ δµ)e + (γµ δλ)jk (Lemma 3.1). Since jk B, then xy B. We state now the following important result: Theorem 3.4 Let A be a four-dimensional absolute valued algebra containing a nonzero central idempotent e, then A is isomorphic to a new absolute valued algebra of four dimension. Proof. According to Theorem 2.2, A is an inner product space and contains a subalgebra B = A(e, i) (i 2 = e, ie = ei = ±i) of dimension 2 isomorphic to C or C (Lemma 3.2). Let F = {e, i, j, k} be an orthonormal basis of A, we have (jk e) = (jk j) = (jk k) = 0 (Lemma 3.3), then jk = ±i. We can assume that jk = i and we have the two following cases: First case. B isomorphic to C. We have i 2 = e, ie = ei = i and jk = i. Since (ej e) = (ej i) = (ej ij) = 0, then ej = ±ik ({e, i, ij, ik} is an orthonormal basis of A). Similarly, (ek e) = (ek i) = (ek ik) = 0 then ek = ±ij. We put ej = αj + βk, where α, β R (α 2 + β 2 = 1). Since jk = i, then ej = ik and ek = ij. indeed, if ej = ik or ek = ij then or j(e + k) = je + jk = je + i = ik + ie = i(k + e) (e + j)k = ek + jk = ij + i = ji + ei = (e + j)i. Which gives i = j or i = k (A has no zero divisors), a contradiction. Since (ek e) = (ek i) = 0, then ek = α 1 j + β 1 k, where α 1, β 1 R (α β 2 1 = 1). We have (ej ek) = 0, hence αα 1 + ββ 1 = 0. From which we get (αβ 1 α 1 β) 2 = α 2 β 2 1 2αβ 1 α 1 β + α 2 1β 2 = α 2 (β α 2 1) + α 2 1(α 2 + β 2 ) = α 2 + α 2 1

5 Four dimensional absolute valued algebras 517 and (αβ 1 α 1 β) 2 = α 2 β 2 1 2αβ 1 α 1 β + α 2 1β 2 = β 2 1(α 2 + β 2 ) + β 2 (β α 2 1) = β β 2 = 2 (α 2 + α 2 1) This gives α 2 + α 2 1 = 2 α 2 + α 2 1, that is, α 2 + α 2 1 = 1. Hence αβ 1 α 1 β = ±1 and we have 1) If αβ 1 α 1 β = 1, then And α 1 = α 1 (αβ 1 α 1 β) = β 1 ββ 1 α 2 1β (αα 1 = ββ 1 ) = β(β α 2 1) = β β 1 = β 1 (αβ 1 α 1 β) = β 2 1α + α 2 1α (αα 1 = ββ 1 ) = α(β α 2 1) = α Therefore ej = ik = αe+βi and ek = ij = βj+αk. Applying Lemma 3.1, the multiplication table of the elements of the base F of A is given by: e e i αj + βk βj + αk i i - e βj + αk αj βk j αj + βk βj αk - e i k βj + αk αj + βk - i - e 2) If αβ 1 α 1 β = 1, then α 1 = β and β 1 = α. Therefore ej = ik = αe + βi and ek = ij = βj αk. Applying Lemma 3.1, the multiplication table of the elements of the base F of A is given by: e e i αj + βk βj αk i i - e βj αk αj βk j αj + βk βj + αk - e i k βj αk αj + βk - i - e

6 518 A. Moutassim and M. Benslimane Second case. B isomorphic to C. We have i 2 = e, ie = ei = i and jk = i. If we define a new multiplication on A by x y = x y, we obtain an algebra A which contains a subalgebra isomorphic to C. Therefore A has an orthonormal basis which the multiplication tables are given previously. Consequently, the multiplication tables of the elements of the base F of A are given by: or e e -i αj βk βj αk i -i - e βj + αk αj βk j αj βk βj αk - e i k βj αk αj + βk - i - e e e -i αj βk βj + αk i -i - e βj αk αj βk j αj βk βj + αk - e i k βj + αk αj + βk - i - e 4 Four-dimensional absolute valued algebras with left unit In this section, we construct a new absolute valued algebras with left unit of four dimension. Theorem 4.1 Let A be a four-dimensional absolute valued algebra with left unit e. If A contains a subalgebra B = A(e, i) (i 2 = ±e, ei = i and ie = ±i) of dimension 2, then A is isomorphic to a new absolute valued algebra with left unit of four dimension. Proof. According to Theorem 2.3, A is an inner product space and let {e, i, j, k} be an orthonormal basis of A. We have (ij e) = (j ie) = (j i) = 0, (ij i) = (j i 2 ) = (j e) = 0 and (ij j) = (j ij) = 0 (Theorem 2.3), then ij = ±k. We can assume that ij = k and we have the two following cases: First case. B isomorphic to C. We have i 2 = e, ie = ei = i and ij = k, then ik = i(ij) = i 2 j = j.

7 Four dimensional absolute valued algebras 519 Since (j 2 j) = (j j 2 ) = 0 (j 2 k) = (j 2 ij) = (j i) = 0 (je e) = (je i) = 0 (ji e) = (ji i) = 0 Then we put j 2 = αe + βi (α 2 + β 2 = 1) je = α 1 j + β 1 k (α1 2 + β1 2 = 1) ji = α 2 j + β 2 k (α2 2 + β2 2 = 1) jk = α 3 e + β 3 i (α3 2 + β3 2 = 1) Where α, α 1, α 2, α 3, β, β 1, β 2, β 3 R. Using Theorem 2.3, we have α 1 = (je j) = (e j 2 ) = α (β1 2 = β 2 ) β 1 = (je k) = (e jk) = α 3 (β3 2 = α1) 2 α 2 = (ji j) = (i j 2 ) = β (β2 2 = α 2 ) β 2 = (ji k) = (i jk) = β 3 (α3 2 = α2) 2 Therefore j 2 = αe + βi je = αj + β 1 k ji = βj + β 2 k jk = β 1 e β 2 i Similarly, we have (k 2 j) = (k 2 ik) = (k i) = 0 (k 2 k) = (k k 2 ) = 0 (ke e) = (ke i) = 0 (ki e) = (ki i) = 0 Then we put k 2 = λe + δi (λ 2 + δ 2 = 1) ke = λ 1 j + δ 1 k (λ δ1 2 = 1) ki = λ 2 j + δ 2 k (λ δ2 2 = 1) kj = λ 3 e + δ 3 i (λ δ3 2 = 1)

8 520 A. Moutassim and M. Benslimane Where λ, λ 1, λ 2, λ 3, δ, δ 1, δ 2, δ 3 R, Using Theorem 2.3, we have Therefore λ 1 = (ke j) = (e kj) = λ 3 (δ 2 1 = δ 2 3) δ 1 = (ke k) = (e k 2 ) = λ (λ 2 1 = δ 2 ) λ 2 = (ki j) = (i kj) = δ 3 (δ 2 2 = λ 2 3) δ 2 = (ki k) = (i k 2 ) = δ (λ 2 2 = λ 2 ) k 2 = λe + δi ke = λ 1 j λk ki = λ 2 j δk kj = λ 1 e λ 2 i From the equality k 2 = λe + δi, we obtain Since jk 2 = λje + δji = λ( αj + β 1 k) + δ( βj + β 2 k) = (λα + δβ)j + (λβ 1 + δβ 2 )k 0 = (k j) = (k 2 jk) = (λe + δi β 1 e β 2 i) = (λβ 1 + δβ 2 ) Then jk 2 = (λα + δβ)j, that is j(jk 2 ) = (λα + δβ)j 2 Hence k 2 = ±j 2 and we have k 2 = (λα + δβ)j 2 (T heorem 2.3) 1) If k 2 = j 2, then λ = α and δ = β. Moreover (j + k) 2 2 = 4 2j 2 + jk + kj 2 = 4 jk + kj 2 = 0 Then kj = jk, that is λ 1 = β 1 and λ 2 = β 2. On the other hand, we have (je ji) = (j i) = 0 hence αβ + β 1 β 2 = 0. From which we get (αβ 2 ββ 1 ) 2 = α 2 β 2 2 2αβ 1 ββ 2 + β 2 1β 2 = α 4 + 2α 2 β 2 + β 4 (β 2 2 = α 2 and β 2 1 = β 2 ) = (α 2 + β 2 ) 2 = 1 This gives αβ 2 ββ 1 = ±1 and we have

9 Four dimensional absolute valued algebras 521 i) If αβ 2 ββ 1 = 1, then β 1 = β 1 (αβ 2 ββ 1 ) = α 2 β β1β 2 (αβ = β 1 β 2 ) = β(α 2 + β 2 ) (β1 2 = β 2 ) = β And β 2 = β 2 (αβ 2 ββ 1 ) = β2α 2 + αβ 2 (αβ = β 1 β 2 ) = α(α 2 + β 2 ) (β2 2 = α 2 ) = α Therefore je = ki = αj βk, ji = ke = βj + αk and jk = kj = βe αi. Then the multiplication table of the elements of the base of A is given by: e i i -e k -j j αj βk βj + αk αe + βi βe αi k βj αk αj βk βe + αi αe + βi ii) If αβ 2 ββ 1 = 1, then And β 1 = β 1 (αβ 2 ββ 1 ) = α 2 β + β 2 1β (αβ = β 1 β 2 ) = β(α 2 + β 2 ) (β 2 1 = β 2 ) = β β 2 = β 2 (αβ 2 ββ 1 ) = β 2 2α αβ 2 (αβ = β 1 β 2 ) = α(α 2 + β 2 ) (β 2 2 = α 2 ) = α Therefore je = ki = αj + βk, ji = ke = βj αk and jk = kj = βe + αi. Then the multiplication table of the elements of the base of A is given by:

10 522 A. Moutassim and M. Benslimane e i i -e k -j j αj + βk βj αk αe + βi βe + αi k βj αk αj βk βe αi αe + βi 2) If k 2 = j 2, then λ = α and δ = β. Moreover (j + k) 2 2 = 4 jk + kj 2 = 4 (jk kj) = 1 Then kj = jk, that is λ 1 = β 1 and λ 2 = β 2. In a similar manner, we have i) If β 1 = β and β 2 = α, then je = ki = αj βk, ji = ke = βj + αk and jk = kj = βe αi. Therefore the multiplication table of the elements of the base of A is given by: e i i -e k -j j αj βk βj + αk αe + βi βe αi k βj + αk αj + βk βe αi αe βi ii) If β 1 = β and β 2 = α, then je = ki = αj + βk, ji = ke = βj αk and jk = kj = βe + αi. Therefore the multiplication table of the elements of the base of A is given by: e i i -e k -j j αj + βk βj αk αe + βi βe + αi k βj + αk αj + βk βe + αi αe βi Second case. B isomorphic to C. We have i 2 = e, ie = i, ei = i and ij = k. Similarly, we obtain the multiplication tables of the elements of the base of A by:

11 Four dimensional absolute valued algebras 523 e i -i -e k -j j αj βk βj + αk αe + βi βe αi k βj αk αj βk βe + αi αe + βi e i -i -e k -j j αj + βk βj αk αe + βi βe + αi k βj αk αj βk βe αi αe + βi or e i -i -e k -j j αj βk βj + αk αe + βi βe αi k βj + αk αj + βk βe αi αe βi e i -i -e k -j j αj + βk βj αk αe + βi βe + αi k βj + αk αj + βk βe + αi αe βi Remark 4.2 Assume that ij = k, if we substitute k = t we get ij = t, that is we again get the same multiplication tables previously. Acknowledgements. The authors are very grateful to professor A. Kaidi for his advice and help. References [1] A. A. Albert, Absolute valued real algebras, Ann. Math, 48 (1947), [2] M. Benslimane, A. Moutassim, A. Rochdi, Sur les Algèbres Absolument Valuées Contenant un Élément Central Non Nul, Advances in Applied Clifford Algebras, 20 (2010),

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