Pre-Hilbert Absolute-Valued Algebras Satisfying (x, x 2, x) = (x 2, y, x 2 ) = 0

International Journal of Algebra, Vol. 10, 2016, no. 9, 437-450 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2016.6743 Pre-Hilbert Absolute-Valued Algebras Satisfying (x, x 2, x = (x 2, y, x 2 = 0 K. Diaby and O. Diankha Département de Mathématiques et Informatique Faculté des Sciences et Techniques Université Cheikh Anta Diop, Dakar, Senegal A. Rochdi Département de Mathématiques et Informatique Faculté des Sciences Ben M Sik Université Hassan II, 7955 Casablanca, Morocco Copyright c 2016 K. Diaby, O. Diankha and A. Rochdi. 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 We study Pre-Hilbert absolute-valued algebras A satisfying both identities (x, x 2, x = (x 2, y, x 2 = 0. We show that A possesses nonzero idempotents and has finite dimension in each of the following three cases: 1. A possesses an omnipresent idempotent. In this case A is isomorphic to either R, C, C, H, H, O or O. 2. A contains C. In this case A is isomorphic to either C, H or O. 3. A do not contain C. In this case A is isomorphic to either R, C, H or O. Mathematics Subject Classification: 17A35, 17A36 Keywords: Absolute-valued algebra, Omnipresent idempotent

438 K. Diaby, O. Diankha and A. Rochdi 1 Introduction Absolute-valued algebras are defined as those real algebras A satisfying xy = x. y for a given norm. on A, and all x, y A. Classical results assert that the dimension of every finite-dimensional absolute-valued algebra is either 1, 2, 4 or 8 and the norm of the algebra comes from an inner product [1]. Among absolute-valued algebras the ones having unit are R, C, H (the quaternions and O (octonions [15, Theorem 1], the commutative ones are R, C and C [15, Theorem 3]. We find in the works [13], [2] a fundamental compilation of results in this subject previous in 2014. The discovery of pseudo-octonions P by the physicist Okubo [11] was a blessing to the theory and had considerable consequences. Indeed, it was proved that every Pre-Hilbert absolute-valued algebra satisfying the identity xx 2 = x 2 x (:= x 3 is finite-dimensional [9] and is isomorphic to either R, C, C, H, H, O, O or P [10]. Recently, it was proved that every absolute-valued algebra satisfying x 2 x = xx 2 contains non-zero idempotents [2, Proposition 2.8.84]. This result has been applied to prove that R, C, C, H, and O are the unique absolute-valued algebras satisfying the identity x 2 x = xx 2 and that the idempotents are pairwise commuting [3, Théorème 3.19]. Also, it is proved, independently of [2], that every absolute-valued algebra satisfying both x 2 x = xx 2 and (xx 3 x = x(x 3 x is finite-dimensional and isomorphic to either R, C, C, H, H, O, O or P [3, Théorème 3.26]. The first result in [3] has been significantly improved. It is proved, using [2], that R, C, C, H, H, O and O are the unique absolute-valued algebras satisfying the identity x 2 x = xx 2 and having a non-zero idempotent which commutes with all idempotents [6]. However, the question of whether every absolute-valued algebra satisfying xx 2 = x 2 x is finite-dimensional, remains still an open problem. On the other hand, it is well known that the identity xx 2 = x 2 x is stronger than (xx 2 x = x(x 2 x for any real algebra. However, both identities are equal in a 4-dimensional absolute-valued algebra [4, Theorem 3.10]. In addition, there are examples of infinite-dimensional Pre-Hilbert absolute-valued algebras (A,. satisfying (x 2 yx 2 = x 2 (yx 2 and possessing an idempotent e such that x 2 = x 2 e for all x A [5, Remarks 5.2 (1], [3, Proposition 4.44]. Thus, it is natural to ask whether a Pre-Hilbert absolute-valued algebra satisfying (xx 2 x = x(x 2 x and (x 2 yx 2 = x 2 (yx 2 is finite-dimensional. We obtained partial results. We showed that such an algebra A contains non-zero idempotents (Proposition 4. Moreover, A has finite dimension in the following

Pre-Hilbert absolute-valued algebras 439 cases: 1. A contains a sub-algebra isomorphic to C. Here A has unit and is isomorphic to either C, H or O (Theorem 1. 2. A contain no sub-algebras isomorphic to C. Here A is isomorphic to either R, C, H or O (Lemma 4. 3. A has a non-zero omnipresent idempotent. Here A is isomorphic to either R, C, C, H, H, O or O (Theorem 2. 2 Notations and preliminary results Let B be an algebra over an infinite field K of characteristic 2. For every a B we denote by L a, R a : B B the linear multiplications on the left and right by a. Also, (.,.,., I(B, C B (a, Lin K {a 1,..., a n }, K B [a 1,..., a n ] denote the associator, the set of non-zero idempotents of B, the set of all elements of B commuting with a, the linear hull spanned by a 1,..., a n B, and the sub-algebra of B generated by a 1,..., a n B. An element e B is said to be central if [e, B] = 0 where [.,.] means the commutator. It is said to be omnipresent if: 1. B has dimension 2, 2. e belongs to every sub-algebra of B of dimension 2. We have the following result proved in [3, Proposition 1.18]: Proposition 1 Assume that B satisfies the identity (x, x 2, x = 0. If e B is an idempotent then the equality [e, ex + xe x] = (e, ex + xe x, e holds for all x B. Proof. Linearising (x, x 2, x = 0 we get (x, x 2, y + (x, xy + yx, x + (y, x 2, x = 0 (2.1 For x = e we have (e, e, y + (e, ey + ye, e + (y, e, e = 0. So

440 K. Diaby, O. Diankha and A. Rochdi (e, ey + ye y, e = (e, ey + ye, e (e, y, e = (e, e, y (y, e, e (e, y, e = ey + e(ey (yee + ye (eye + e(ye = [e, ey + ye y]. We state another preliminary result: Lemma 1 Assume that B satisfies (x 2, x, x 2 = (x 2, x 2, x 2 = 0. If e B is an idempotent then the equality (e, ex + xe x, e = 0 holds for all x B. Proof. Linearising (x 2, x, x 2 = 0, (x 2, x 2, x 2 = 0, respectively, we get (x 2, x, xy + yx + (x 2, y, x 2 + (xy + yx, x, x 2 = 0 (2.2 (x 2, x 2, xy + yx + (x 2, xy + yx, x 2 + (xy + yx, x 2, x 2 = 0. (2.3 By putting x = e in (2.2 and (2.3 next subtracting, we get the equality (e, ey + ye y, e = 0. The following consequence will be very useful in Section 4: Corollary 1 Let e B is an idempotent and assume that B satisfies the three identities (x, x 2, x = (x 2, x, x 2 = (x 2, x 2, x 2 = 0. Then the equality [e, ex + xe x] = 0 holds for all x B. (2.4 In particular, C B (e is invartiant under L e and we have: [f, (f g 2 ] = 0 for all f, g I(B. (2.5

Pre-Hilbert absolute-valued algebras 441 3 On absolute-valued algebras satisfying (x, x 2, x = 0 An absolute-valued algebra whose norm comes from an inner product is said to be a Pre-Hilbert absolute-valued algebra. Let A be one of the principal absolute-valued algebras C (complex, H (quaternions or O (octonions. We denote by A the absolute-valued algebra obtained by endowing the normed space of A with the product x y := x y where x x means the standard involution of A. Let e, f I(A being distinct and commute, then e + f + ef = 0 and Lin R {e, f} is a sub-algebra of A isomorphic to C. Also, it is well known that if all the elements of a subset B of an absolute-valued algebra A commute with each other, then the linear hull spanned by B is an inner-product space [15, Lemma 1]. We have the following result: Proposition 2 Let (A,. be an algebra and let x, y be in A such that [x, y] = [x 2, y 2 ] = [x 2, xy] = 0. The equality y 2 x 2 + x 2 y 2 = 2 x y xy holds where.. means the inner product associated to the euclidian space Lin R {x, y}. Proof. We can assume that x, y are linearly independent and distinguish the following two cases: 1. If x, y are orthonormal, then y 2 x 2 = (y x(y + x = y x. y + x = 2. Also, Lin R {x 2, y 2 } is an euclidian space because x 2 y 2 = y 2 x 2. So the result is an immediate consequence of the parallelogram low. 2. If x y 0, we have the orthogonal sum y = x y x 2 x + u. In addition, u 0 commutes with x and a simple calculation gives

442 K. Diaby, O. Diankha and A. Rochdi u 2 = y 2 2 x y x y 2 xy + x 2 x 4 x2, xu = xy x y x 2 x2. So [x 2, u 2 ] = [x 2, xu] = 0 and, taking into account the first case, we have ( 0 = u 2 x 2 x 2 + u 2 x u ( x = u 2 x 2 2 x + u2 2 u 2 = y x y x x 2 ( x 2 + x 2 y x y 2 2 x x 2 ( = ( y 2 2 x y 2 x + x y 2 2 x 4 x 2 x 2 + x 2 y 2 2 x y x y 2 xy + x 2 x 4 x2 = y 2 x 2 + x 2 y 2 2 x y xy. Corollary 2 Let A be an absolute-valued algebra and let a A such that (a 2, a 2, a 2 = ((a 2 2, a 2, (a 2 2 = ((a 2 2, (a 2 2, (a 2 2 = 0. Then ( (a 2 2 2 Lin{(a 2 2, (a 2 3 }. Proof. Follows by putting (x, y = (a, a 2 in Proposition 2. In the rest of this section A will be assumed to satisfy (x, x 2, x = 0 and contains a non-zero idempotent e. Also, its norm. comes from an inner product.,.. Let e be the orthogonal space of e in A, we have the following decomposition of A into a direct sum of sub-spaces: A = Re e. (3.6 We can then state the following: Lemma 2 Let x be in C A (e e {0} then x 2 e R A [x]. = x 2 e. In particular

Pre-Hilbert absolute-valued algebras 443 Proof. We have: ( x 2 x e = x x e x x + e = 2. The parallelogram low shows that that is x 2 = x 2 e. ( x 2 + e = 0 x Corollary 3 Let x be in C A (e {0} then 1. x 2 C A (e. 2. The sub-algebra R A [x] is commutative and isomorphic to either R, C or C. Moreover, e R A [x] if dim R A [x] = 2. Proof. If x C A (e e {0} then [ex, e] = 0, according to the Corollary 3, and x 2 = x 2 e by Lemma 3. In particular x 2 commutes with e and e R A [x]. Now 0 = (x, x 2, x = x 2 (x, e, x = x 2 [ex, x]. So {e, x, ex} is a commutative set. But e, x are linearly independent and, according to [9, Lemme 1.1], ex Lin R {e, x}. Consequently, R A [x] coincides with Lin R {e, x} and is a commutative algebra isomorphic to either R, C or C [15, Theorem 3]. Let now x be in C A (e {0} which can be assumed to be not scalar multiple of an idempotent, that is dim R A [x] 2. Then x is written ( λe + w where (λ, w R e C A (e {0}. Thus R A [x] R A [e, w] = R A [w]. So R A [x] is isomorphic to either C or C, and e R A [x]. Moreover, w commutes with e and we have: x 2 = (λe + w 2 = (λ 2 w 2 e + 2λew commutes with e. We have the following key result: Proposition 3 If A contains both a sub-algebra B isomorphic to C and a subalgebra C isomorphic to C then B and C can not contain common non-zero idempotents.

444 K. Diaby, O. Diankha and A. Rochdi Proof. Let e be in I(A B C. There are a B, f C orthonormal to e such that a 2 = f 2 = e, e a = ae = a and e f = fe = f. As a + f commutes with e, the Corollary 3 (2 show that R A [a + f] is isomorphic to either R, C or C. We distinguish the following two cases: 1. If dim R A [a + f] = 1 then (a + f 2 = λ(a + f for some scalar λ. So g = λ 1 (a + f is an idempotent commuting with e and orthogonal to e. This gives g = g 2 = e absurd. 2. If dim R A [a + f] = 2 then R A [a + f] = Lin R {e, a + f} by virtue of Corollary 3 (2. Consequently, e (a + f = a f R A [a + f] and we have: R A [a + f] = Lin R {e, a, f}. Thereby f Lin R {e, a} giving C B absurd. 4 Pre-Hilbert absolute-valued algebras satisfying (x, x 2, x = (x 2, x, x 2 = (x 2, x 2, x 2 = 0 In what follows A is assumed to be a Pre-Hilbert absolute-valued algebra satisfying (x, x 2, x = (x 2, x, x 2 = (x 2, x 2, x 2 = 0. We have the following result: Proposition 4 For every z A {0} the sub-algebra R A [z 2 ] is isomorphic to either R, C or C. In particular A contains non-zero idempotents. Proof. If (z 2 2 is linearly dependent to z 2 then R A [z 2 ] is isomorphic to R. Assume now that z 2 and (z 2 2 are linearly independent. The four equalities (z 2, z 2, z 2 = ((z 2 2, (z 2 2, (z 2 2 = (z 2, (z 2 2, z 2 = ((z 2 2, z 2, (z 2 2 = 0 are written, respectively: (z 2 2 z 2 = z 2 (z 2 2 := (z 2 3, (4.7 ((z 2 2 2(z 2 2 = (z 2 2 ((z 2 2 2 := ( (z 2 2 3, (4.8 (z 2 3 z 2 = z 2 (z 2 3, (4.9 (z 2 3 (z 2 2 = (z 2 2 (z 2 3. (4.10

Pre-Hilbert absolute-valued algebras 445 The equalities (4.7, (4.9, (4.10 show that {z 2, (z 2 2, (z 2 3 } is a commutative set. As the norm of A come from an inner product and z 2, (z 2 2 linearly independent, we have: (z 2 3 Lin{z 2, (z 2 2 }. (4.11 Let now z be arbitrary in A. The three equalities (z 2, z 2, z 2 = ((z 2 2, z 2, (z 2 2 = ((z 2 2, (z 2 2, (z 2 2 = 0 give ( (z 2 2 2 Lin{(z 2 2, (z 2 3 } by Corollary 4. By taking into account the equalities (4.7, (4.9, we get ( (z 2 2 2z 2 = z 2 ((z 2 2 2. (4.12 The equalities (4.7, (4.8, (4.12 show that is also a commutative set. Thus { 2 } z 2, (z 2 2, ((z 2 2 ( (z 2 2 2 Lin{z 2, (z 2 2 }. (4.13 The equalities (4.11, (4.13 show that Lin{z 2, (z 2 2 } is a commutative sub-algebra of A. It coincides with R A [z 2 ] and is isomorphic to R, C or C. We need, in the sequel, the following preliminary results: Lemma 3 Let e, f I(A such that [e, f] 0. Then (e f 2 is linearly dependent to an idempotent g which is the only one non-zero idempotent of A commuting with both e and f. In particular A contains sub-algebras isomorphic to C. Proof. The equality (2.5 gives [e, (e f 2 ] = [f, (e f 2 ] = 0. Now, the Corollary 3 (2 shows that R A [(e f 2 ] is a commutative sub-algebra. Its dimension is 2 because otherwise if would contain both e and f, which is

446 K. Diaby, O. Diankha and A. Rochdi absurd. So dim R A [(e f 2 ] = 1 and there exists a non-zero idempotent g A linearly dependent to (e f 2 and commuting with both e and f. Let now e be in I(A commuting with both e and f. We have: ee + e + e = 0 = fe + f + e. So (e fe = f e and also (e fg = f e. The absence of divisors of zero in A shows that e = g, and then g is unique. Theorem 1 If A possesses a sub-algebra B isomorphic to C then A has a unique non-zero idempotent, the one of B. Consequently, A has finite dimension and is isomorphic to either C, H or O. Proof. Let e be the non-zero idempotent of algebra B. If A contains a non-zero idempotent different from e then the Lemma 4 shows that A has a non-zero idempotent g e commuting with e. In this case Lin R {e, g} is a sub-algebra isomorphic to C and we will have B Lin{e, g} = Re. This contradicts the Proposition 3. So e is the unique non-zero idempotent of A. In particular, e commutes with x 2 for all x A by virtue of Proposition 4. Now, for every x A we have: 0 = [e, x ex xe] by equality (2.4 = [e, x + (e x 2 e x 2 ] = [e, x]. So e is central and, according to [12, Theorem 2.3], A has finite dimension and is isomorphic to either R, C, C, H, H, O or O. But A cannot be equal to any one of algebras C, H, O. Lemma 4 If A contain no sub-algebras isomorphic to C then A has finite dimension and is isomorphic to either R, C, H or O. Proof. According to the Lemma 3, A contains only one idempotent e 0. The remains of the proof is the same as in Theorem 1. Corollary 4 If R[x 2 ] is isomorphic to R for all x A {0} then A is isomorphic to R.

Pre-Hilbert absolute-valued algebras 447 Proof. Each of algebras C, C, H, O contain an element whose square is not a scalar multiple of an idempotent. Such square generates a sub-algebra of dimension 2. Thus A cannot possesses sub-algebras isomorphic to one of the above four algebras. The Lemma 4 conclude. Theorem 2 If A has a non-zero omnipresent e then A is finite-dimensional and is isomorphic to either C, C, H, H, O or O. Proof. Assume that A contains a non-zero idempotent f A which does not commute with e. The Lemma 3 shows that there exists e I(A such that [e, e] = [e, f] = 0. In particular Lin R {e, f} is a sub-algebra of A isomorphic to C. This one, being 2-dimensional, contains the omnipresent element e. Thus e will commute with f, absurd. Consequently, e commutes with all the idempotents of A : [e, I(A] = 0. (4.14 Let now x be in A {0} then the sub-algebra R A [x 2 ] is isomorphic to either R, C or C by Proposition 4. We distinguish the following three cases: 1. If R A [x 2 ] is isomorphic to R then x 2 is a scalar multiple of some idempotent. So x 2 commutes with e by equality (4.14. 2. If R A [x 2 ] is isomorphic to C then clearly [e, x 2 ] = 0 by Theorem 1. 3. If R A [x 2 ] is isomorphic to C then e R A [x 2 ] and commutes with x 2. Thus e commutes with all the squares in A. A similar reasoning to that used in the end of the proof of Theorem 1 shows that e is central and that A has finite dimension. It is isomorphic to either R, C, C, H, H, O or O. The fact that A contains sub-algebras isomorphic to C, and algebras R, C, H, O do not contain, conclude. We summarize Proposition 4, Lemma 4 and Theorems 1, 2 as follows Corollary 5 Every Pre-Hilbert absolute-valued algebra A satisfying (x, x 2, x = (x 2, x, x 2 = (x 2, x 2, x 2 = 0 contains non-zero idempotents. Moreover, A has finite dimension in the following cases:

448 K. Diaby, O. Diankha and A. Rochdi 1. A contains a non-zero omnipresent idempotent. In this case A is isomorphic to either C, C, H, H, O or O. 2. A contains C. In this case A is isomorphic to either C, H or O. 3. A do not contain C. Here A is isomorphic to either R, C, H or O. Remark 1 The absolute-valued algebra P of pseudo-octonions in which subalgebras isomorphic to C exist in abundance, contains no omnipresent idempotents. In fact P contains no non-zero idempotent that is contained in all 4-dimensional sub-algebras of P [7, Corollary 5]. Problem 1 It is irrelevant whether a prehilbert absolute-valued algebra satisfying (x, x 2, x = (x 2, y, x 2 = 0 and containing sub-algebras isomorphic to C is finite-dimensional. Acknowledgements. This paper has benefited from several remarks and suggestions by Professor Antonio Jesús Calderón. The authors are very grateful to him. References [1] A. A. Albert, Absolute valued real algebras, Ann. Math., 48 (1947, 495-501. http://dx.doi.org/10.2307/1969182 [2] M. Cabrera García and Á. Rodríguez Palacios, Non-associative Normed Algebras: Volume 1 the Vidav-Palmer and Gelfand-Naimark Theorems, Encyclopedia of Mathematics and its Applications, Cambridge University Press, 2014. http://dx.doi.org/10.1017/cbo9781107337763 [3] A. Chandid, Algèbres Absolument Valuées qui Satisfont à (x p, x q, x r = 0, Thèse Doctorale, Université Hassan II-Mohammedia, Faculté des Sciences Ben M Sik, Casablanca, 2009.

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450 K. Diaby, O. Diankha and A. Rochdi [15] K. Urbanik and F. B. Wright, Absolute valued algebras, Proc. Amer. Math. Soc., 11 (1960, 861-866. http://dx.doi.org/10.1090/s0002-9939-1960-0120264-6 Received: July 21, 2016; Published: October 10, 2016