MAPS PRESERVING JORDAN TRIPLE PRODUCT A B + BA ON -ALGEBRAS. Ali Taghavi, Mojtaba Nouri, Mehran Razeghi, and Vahid Darvish

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1 Korean J. Math. 6 (018), No. 1, pp MAPS PRESERVING JORDAN TRIPLE PRODUCT A B + BA ON -ALGEBRAS Ali Taghavi, Mojtaba Nouri, Mehran Razeghi, and Vahid Darvish Abstract. Let A and B be two prime -algebras. Let Φ : A B be a bijective and satisfies Φ(A B A) = Φ(A) Φ(B) Φ(A), for all A, B A where A B = A B + BA. Then, Φ is additive. Moreover, if Φ(I) is idempotent then we show that Φ is R-linear -isomorphism. 1. Introduction Let R and R be rings. We say the map Φ : R R preserves product or is multiplicative if Φ(AB) = Φ(A)Φ(B) for all A, B R, see [9]. Motivated by this, many authors pay more attention to the map on rings (and algebras) preserving different kinds of products to establish characteristics of Φ on rings. A natural problem is to study whether the map Φ preserving the new product on ring or algebra R is a ring or algebraic isomorphism. (for example [1 4, 6 8, 10 1]). Recently, Liu and Ji [5] proved that a bijective map Φ on factor von Neumann algebras preserves, A B + BA if and only if Φ is a - isomorphism. Also, the authors in [14] considered such a bijective map Φ : A B on prime C -algebras which preserves A B + ηba, where Received January 6, 018. Revised March 1, 018. Accepted March 16, Mathematics Subject Classification: 47B48, 46L10. Key words and phrases: Jordan triple product, -isomorphism, Prime -algebras. c The Kangwon-Kyungki Mathematical Society, 018. This is an Open Access article distributed under the terms of the Creative commons Attribution Non-Commercial License ( -nc/3.0/) which permits unrestricted non-commercial use, distribution and reproduction in any medium, provided the original work is properly cited.

2 6 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish η is a non-zero scalar such that η ±1. They proved that Φ is additive. Moreover, if Φ(I) is projection then Φ is -isomorphism. The authors of [13], proved that if the map Φ from a prime -ring A onto a -ring B is bijective and preserves Jordan triple product or -Jordan triple product Φ(ABA) = Φ(A)Φ(B)Φ(A) Φ(AB A) = Φ(A)Φ(B) Φ(A) then it is additive. Also, we show that if Φ : A B, where A and B are two prime rings, preserves Jordan triple product then it is multiplicative or anti-multiplicative. Also, we show that Ψ(A) = Φ(A)Φ(I), for A A, is a C-linear or conjugate C-linear -isomorphism. In this paper, motivated by the above results, we consider a map Φ on two prime -algebras A and B with a nontrivial projection such that Φ is bijective and holds in the following condition Φ(A B A) = Φ(A) Φ(B) Φ(A), for all A, B A where A B = A B + BA. We show that Φ described in the above is additive. Also, if Φ(I) is idempotent then Φ is R-linear -isomorphism. It is well known that C -algebra A is prime, in the sense that AAB = 0 for A, B A implies either A = 0 or B = 0.. Main Results We need the following lemma for proving our theorems. Lemma.1. Let A and B be two -algebras and Φ : A B be a map which satisfies in the following case: (.1) Φ(A B A) = Φ(A) Φ(B) Φ(A). If Φ(T ) = Φ(A) + Φ(B) for T, A, B A then we have for all X, Y A. Φ(X T X) = Φ(X A X) + Φ(X B X) Proof. By assumption we have (.) Φ(T ) = Φ(A) + Φ(B).

3 Maps preserving Jordan triple product A B + BA on -algebras 63 Multiplying the left and right sides of (.) by Φ(X), we obtain (.3) Φ(X)Φ(T ) Φ(X) = Φ(X)Φ(A) Φ(X) + Φ(X)Φ(B) Φ(X). Multiplying the left side of (.) by Φ(X), we obtain (.4) Φ(X) Φ(T ) = Φ(X) Φ(A) + Φ(X) Φ(B). Multiplying the right side of (.) by Φ(X), we obtain (.5) Φ(T ) Φ(X) = Φ(A) Φ(X) + Φ(B) Φ(X). Adding times of (.3), (.4) and (.5) together and making use of (.1) we have Φ(X T X) = Φ(X A X) + Φ(X T X). Our first theorem is as follows: Theorem.. Let A and B be two prime -algebras with unit I A and I B respectively, a nontrivial projection and Φ : A B be a bijective map which satisfies in the following condition (.6) Φ(A B A) = Φ(A) Φ(B) Φ(A) for all A, B A. Then Φ is additive. Proof. Let P 1 be a nontrivial projection in A and P = I A P 1. Denote A ij = P i AP j, i, j = 1,, then A = i,j=1 A ij. For every A A we may write A = A 11 + A 1 + A 1 + A. In all that follow, when we write A ij, it indicates that A ij A ij. For showing additivity of Φ on A, we use above partition of A and give some claims that prove Φ is additive on each A ij, i, j = 1,. We prove the above theorem by several claims. Claim 1. We show that Φ(0) = 0. We know that for all A, B A, the following holds Let B = 0 then Φ(A B A) = Φ(A) Φ(B) Φ(A). Φ(0) = Φ(A) Φ(0) Φ(A) = Φ(A)Φ(0) Φ(A) + Φ(0) Φ(A)Φ(A) +Φ(A) Φ(0) + Φ(A)Φ(0) Φ(A)

4 64 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish for every A A. Since Φ is surjective, we can find A such that Φ(A) = 0, then we have Φ(0) = 0. Claim. For each A 11 A 11 and A A we have Φ(A 11 + A ) = Φ(A 11 ) + Φ(A ). Since Φ is surjective, there exists T = T 11 + T 1 + T 1 + T A such that (.7) Φ(T ) = Φ(A 11 ) + Φ(A ). By applying Lemma (.1) to (.7) for P 1 and P, we have and Φ(P 1 T P 1 ) = Φ(P 1 A 11 P 1 ) + Φ(P 1 A P 1 ) = Φ(4A 11) Φ(P T P ) = Φ(P A 11 P ) + Φ(P A P ) = Φ(4A ). Since Φ is injective, we obtain T P 1 + P 1 T P 1 + P 1 T = 4A 11 and T P + P T P + P T = 4A. Hence, we have A 11 = T 11, A = T and T 1 = T 1 = 0. So, Φ(A 11 + A ) = Φ(A 11 ) + Φ(A ). Claim 3. For each A 1 A 1, A 1 A 1 we have Φ(A 1 + A 1 ) = Φ(A 1 ) + Φ(A 1 ). Since Φ is surjective, we can find T = T 11 + T 1 + T 1 + T A such that (.8) Φ(T ) = Φ(A 1 ) + Φ(A 1 ). By applying Lemma (.1) to (.8) for P 1 P, we have Since Φ is injective, we have So, we obtain Φ ((P 1 P ) T (P 1 P )) = Φ ((P 1 P ) A 1 (P 1 P )) +Φ ((P 1 P ) A 1 (P 1 P )) = 0. (P 1 P ) T (P 1 P ) = 0. T 11 + T = 0

5 Maps preserving Jordan triple product A B + BA on -algebras 65 it follows that T 11 = T = 0. On the other hand, by applying Lemma (.1) to (.8) for X 1 and X 1 we have and Φ(X 1 T X 1 ) = Φ(X 1 A 1 X 1 ) + Φ(X 1 A 1 X 1 ) = Φ(X 1 A 1X 1 ) Φ(X 1 T X 1 ) = Φ(X 1 A 1 X 1 ) + Φ(X 1 A 1 X 1 ) By injection, we have for all X 1 A 1 and = Φ(X 1 A 1X 1 ). X 1 T X 1 = X 1 A 1X 1, X 1 T X 1 = X 1 A 1X 1, for all X 1 A 1. Therefore, by primeness we have T 1 = A 1 = and T 1 = A 1. Claim 4. For each A 11 A 11, A 1 A 1, A 1 A 1 we have Φ(A 11 + A 1 + A 1 ) = Φ(A 11 ) + Φ(A 1 ) + Φ(A 1 ) and Φ(A + A 1 + A 1 ) = Φ(A ) + Φ(A 1 ) + Φ(A 1 ). Since Φ is surjective, there exists T = T 11 + T 1 + T 1 + T such that (.9) Φ(T ) = Φ(A 11 ) + Φ(A 1 ) + Φ(A 1 ). By applying Lemma (.1) to (.9) for P and implying Claim 3, we have Φ(P T P ) = Φ(P A 11 P ) + Φ(P A 1 P ) + Φ(P A 1 P ) = Φ(A 1 + A 1). So, we have T P + P T P + P T = A 1 + A 1 then 4T + T 1 + T 1 = A 1 + A 1.

6 66 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish Therefore, we have T 1 = A 1, T 1 = A 1 and T = 0. For showing that T 11 = A 11 we use the following trick. It is easy to check that Φ(4T 11) = Φ((P 1 P ) T (P 1 P )) = Φ((P 1 P ) A 11 (P 1 P )) + Φ((P 1 P ) A 1 (P 1 P )) +Φ((P 1 P ) A 1 (P 1 P )) = Φ(4A 11). By, injection we have T 11 = A 11. Similarly, one can prove Φ(A + A 1 + A 1 ) = Φ(A ) + Φ(A 1 ) + Φ(A 1 ). Claim 5. For each A 11 A 11, A 1 A 1, A 1 A 1, A A, we have Φ(A 11 + A 1 + A 1 + A ) = Φ(A 11 ) + Φ(A 1 ) + Φ(A 1 ) + Φ(A ). We assume T is an element in A such that (.10) Φ(T ) = Φ(A 11 ) + Φ(A 1 ) + Φ(A 1 ) + Φ(A ). By applying Lemma (.1) to (.10) for P 1 and Claim 4, we have Φ(P 1 T P 1 ) = Φ(P 1 A 11 P 1 ) + Φ(P 1 A 1 P 1 ) + Φ(P 1 A 1 P 1 ) Then, we have +Φ(P 1 A P 1 ) = Φ(4A 11) + Φ(A 1) + Φ(A 1) = Φ(4A 11 + A 1 + A 1). 4T 11 + T 1 + T 1 = 4A 11 + A 1 + A 1 so, T 11 = A 11, T 1 = A 1 and T 1 = A 1. Similarly, by Lemma (.1) to (.10) for P and Claim 4, we have Φ(P T P ) = Φ(4A + A 1 + A 1). So, we obtain T = A. Hence, we obtain Φ(A 11 + A 1 + A 1 + A ) = Φ(A 11 ) + Φ(A 1 ) + Φ(A 1 ) + Φ(A ). Claim 6. For each A ij, B ij A ij such that 1 i, j and i j, we have Φ(A ij + B ij ) = Φ(A ij ) + Φ(B ij ).

7 Maps preserving Jordan triple product A B + BA on -algebras 67 By a simple computation, we can show the following (P i + A ij ) (P j + B ij) (P i + A ij ) = A ij + B ij. By using Claim 5, we have Φ(A ij + B ij ) = Φ((P i + A ij ) (P j + B ij) (P i + A ij )) = Φ(P i + A ij ) Φ(P j + B ij ) Φ(P i + A ij ) = (Φ(P i ) + Φ(A ij )) (Φ(P j ) + Φ(B ij)) (Φ(P i ) + Φ(A ij )) = Φ(P i ) Φ(P j ) Φ(P i ) + Φ(P i ) Φ(B ij) Φ(P i ) +Φ(P i ) Φ(P j ) Φ(A ij ) + Φ(P i ) Φ(B ij) Φ(A ij ) +Φ(A ij ) Φ(P j ) Φ(P i ) + Φ(A ij ) Φ(P j ) Φ(A ij ) +Φ(A ij ) Φ(B ij) Φ(P i ) + Φ(A ij ) Φ(B ij) Φ(A ij ) = Φ(P i P j P i ) + Φ(P i B ij P i ) +Φ(P i P j A ij ) + Φ(P i B ij A ij ) +Φ(A ij P j P i ) + Φ(A ij P j A ij ) +Φ(A ij B ij P i ) + Φ(A ij B ij A ij ) = Φ(A ij ) + Φ(B ij ). Claim 7. For each A ii, B ii A ii such that 1 i, we have Φ(A ii + B ii ) = Φ(A ii ) + Φ(B ii ). Since Φ is surjective, we can find T = T ii + T ij + T ji + T jj A such that (.11) Φ(T ) = Φ(A ii ) + Φ(B ii ). By applying Lemma (.1) to (.11) for P j, we have Φ(P j T P j ) = Φ(P j A ii P j ) + Φ(I P j B ii P j ) = 0. Since Φ is injective, we obtain P j T + T P j + P j T P j = 0. So, T ij = T ji = T jj = 0. Hence, we have T = T ii. On the other hand, for each C ij A ij from Claim 6 and Lemma (.1) for P j + C ij we have Φ(TiiC ij ) = Φ ((P j + C ij ) T (P j + C ij )) = Φ ((P j + C ij ) A ii P j + C ij ) + Φ ((P j + C ij ) B ii (P j + C ij )) = Φ(A iic ij ) + Φ(B iic ij ) = Φ(A iic ij + B iic ij ).

8 68 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish So, we have (T ii A ii B ii)c ij = 0 for all C ij A ij. By primeness, we have T ii = A ii + B ii. Hence, the additivity of Φ comes from the above claims. In the rest of this paper, we show that Φ is -isomorphism. Theorem.3. Let A and B be two prime -algebras with unit I A and I B respectively, a nontrivial projection and Φ : A B be a bijective map which satisfies in the following condition (.1) Φ(A B A) = Φ(A) Φ(B) Φ(A) for all A, B A. If Φ(I A ) is idempotent, then Φ is R-linear -isomorphism. Proof. We prove the above theorem by some claims. Claim 1. Φ is a Q-linear map. By additivity of Φ, it is easy to check that Φ is Q-linear. Claim. We show that Φ is unital. For any A A, by knowing that Φ(I) is idempotent, we have ( I Φ A I ) ( ) ( ) I I = Φ Φ(A) Φ ( ) I ( ) I ( ) ( ) I I = Φ Φ(A) + Φ(A) Φ + Φ Φ(A) Φ. Since Φ is surjective, we can find A such that Φ(A) = I B, then ( I Φ A I ) = Φ(I) by injectivity of Φ we get So, A = I. I A I = I. Claim 3. Φ preserves projections on the both sides.

9 Maps preserving Jordan triple product A B + BA on -algebras 69 Suppose that P A is a projection. From the Claim 1 we have ( I Φ(P ) = Φ P I ) ( ( ) ( ) ) ( ) I I I = Φ Φ(P ) + Φ(P )Φ Φ ( ) I = Φ(P ) Φ = Φ(P ). Then Also, So, Φ(P ) = Φ(P ). Φ(P ) = Φ (P I4 ) P = ( Φ (P ) Φ ( ) I + Φ 4 = 1 Φ(P ) Φ(P ) = Φ(P ) Φ(P ) + Φ(P )Φ(P ) = Φ(P ). Φ(P ) = Φ(P ). ( ) ) I Φ (P ) Φ(P ) 4 Since Φ 1 has the same characteristics of Φ then Φ is the preserver of the projections on the both sides. Remark.4. We note here that if P i and P j = I P i are two Orthogonal projections then Φ(P i ) and Φ(P j ) are so. Φ(P i )Φ(P j ) = Φ(P i )(Φ(I) P i ) = Φ(P i )(Φ(I) Φ(P i )) = 0. Remark.5. From ( I Φ A I ) = Φ ( ) I Φ(A) Φ ( ) I we have Φ(A ) = Φ(A). It means that Φ preserves star.

10 70 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish Claim 4. Φ(A ii ) = B ii. Let X A ii be an arbitrary element, then we obtain Φ(4X) = Φ(P i X P i ) = (Φ(P i ) Φ(X ) + Φ(X )Φ(P i ) ) Φ(P i ) = Φ(P i )Φ(X) + Φ(X)Φ(P i ) + Φ(P i )Φ(X)Φ(P i ) since Φ is Q-linear, so we show that 4Φ(X) = Φ(P i )Φ(X) + Φ(P i )Φ(X)Φ(P i ) + Φ(X)Φ(P i ). From the above equation, we obtain the following relations and So, we have it follows that Φ(X) = Φ(P i )Φ(X)Φ(P j ) = 0, Φ(P j )Φ(X)Φ(P i ) = 0 Φ(P j )Φ(X)Φ(P j ) = 0. Φ(P i )Φ(X)Φ(P j ) = Φ(P i )Φ(X)Φ(P i ) i,j=1 Φ(A ii ) B ii. Since Φ 1 has the properties as Φ then we have Φ(A ii ) = B ii. Claim 5. Φ(A ij )Φ(P j ) = Φ(P i )Φ(A ij )Φ(P j ) for A ij A ij and 1 i, j such that i j. Since Φ is star preserving we have Φ(A ij ) = Φ(P i A ij P i ) So, we showed that = Φ(P i ) Φ(A ij) Φ(P i ) = (Φ(P i ) Φ(A ij) + Φ(A ij)φ(p i )) Φ(P i ) = Φ(P i )Φ(A ij ) + Φ(A ij )Φ(P i ) + Φ(P i )Φ(A ij )Φ(P i ). Φ(A ij ) = Φ(P i )Φ(A ij ) + Φ(A ij )Φ(P i ) + Φ(P i )Φ(A ij )Φ(P i ). We multiply the right side of above equation by Φ(P j ), to obtain (.13) Φ(A ij )Φ(P j ) = Φ(P i )Φ(A ij )Φ(P j ).

11 Maps preserving Jordan triple product A B + BA on -algebras 71 Similarly, one can show that Φ(P j )Φ(A ji ) = Φ(P j )Φ(A ji )Φ(P i ). Claim 6. Suppose that A ii, B ii A ii for 1 i. Then Φ(A ii B ii ) = Φ(A ii )Φ(B ii ). Let C ij A ij be an arbitrary element. Therefore, we have Φ(A ii B ii C ij ) = Φ((P j + C ij ) (BiiA ii) (P j + C ij )) = Φ((P j + C ij ) Φ(BiiA ii) Φ(P j + C ij ) = ((Φ(P j + C ij ) Φ(BiiA ii) +Φ(BiiA ii)φ(p j + C ij ) ) Φ(P j + C ij ) = (Φ(C ij ) Φ(BiiA ii)) Φ(P j ) = Φ(A ii B ii )Φ(C ij ). So, by the above equation, we obtain Φ(A ii B ii )Φ(C ij ) = Φ(A ii (B ii C ij )) = Φ(A ii )Φ(B ii C ij ) = Φ(A ii )Φ(B ii )Φ(C ij ). We have (Φ(A ii B ii ) Φ(A ii )Φ(B ii ))Φ(C ij ) = 0. We multiply the above equation by Φ(P j ) from the left side, then we have By Claim 5, we have (Φ(A ii B ii ) Φ(A ii )Φ(B ii ))Φ(C ij )Φ(P j ) = 0. (Φ(A ii B ii ) Φ(A ii )Φ(B ii ))Φ(P i )Φ(C ij )Φ(P j ) = 0. By primeness, we obtain Φ(A ii B ii ) = Φ(A ii )Φ(B ii ). Claim 7. Suppose that A ij A ii and B ji B ji. Then Φ(A ij B ji ) = Φ(A ij )Φ(B ji ).

12 7 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish Since Φ preserves star, we have Φ(A ij B ji + B ji A ij ) ( = Φ (A ij + B ji ) I ) 4 (A ij + B ji ) ( ) I = Φ(A ij + B ji ) Φ Φ(A ij + B ji ) 4 ( ( ) ( ) ) I I = Φ(A ij + B ij ) Φ + Φ Φ(A ij + B ji ) Φ(A ij + B ij ) 4 4 = 1 (Φ(A ij) + Φ(B ji ) ) Φ(A ij + B ji ) = Φ(A ij )Φ(B ji ) + Φ(B ji )Φ(A ij ). It follows that Φ(A ij B ji ) + Φ(B ji A ij ) = Φ(A ij )Φ(B ji ) + Φ(B ji )Φ(A ij ). Multiplying the left side of the above equation by Φ(P i ) and applying Claims 4 and 5 to obtain Φ(P i )Φ(A ij B ji )+Φ(P i )Φ(B ji A ij ) = Φ(P i )Φ(A ij )Φ(B ji )+Φ(P i )Φ(B ji )Φ(A ij ). So, Φ(A ij B ji ) = Φ(A ij )Φ(B ji ). Claim 8. For A ii A ii and B ij A ij we have Φ(A ii B ij ) = Φ(A ii )Φ(B ij ). Let T ji in A ji such that i j, Claims 6 and 7 imply that Φ(A ii B ij )Φ(T ji ) = Φ(A ii B ij T ji ) Since B is prime and by Claim 5, we have = Φ(A ii )Φ(B ij T ji ) = Φ(A ii )Φ(B ij )Φ(T ji ). Φ(A ii B ij ) = Φ(A ii )Φ(B ij ). Claim 9. For A ij A ij and B jj A jj we have Φ(A ij B jj ) = Φ(A ij )Φ(B jj ).

13 So, Maps preserving Jordan triple product A B + BA on -algebras 73 For each T ji A ji such that i j, we have It should be clear that Φ(T ji )Φ(A ij B jj ) = Φ(T ji A ij B jj ) = Φ(T ji A ij )Φ(B jj ) = Φ(T ji )Φ(A ij )Φ(B jj ). Φ(A ij B jj ) = Φ(A ij )Φ(B jj ). Φ(AB) = Φ((A ii + A ij + A ji + A jj )(B ii + B ij + B ji + B jj )) = Φ(A ii B ii + A ii B ij + A ij B ji + A ij B jj + A ji B ii +A ji B ij + A jj B ji + A jj B jj ) = Φ(A ii )Φ(B ii ) + Φ(A ii )Φ(B ij ) + Φ(A ij )Φ(B ji ) + Φ(A ij )Φ(B jj ) +Φ(A ji )Φ(B ii ) + Φ(A ji )Φ(B ij ) + Φ(A jj )Φ(B ji ) + Φ(A jj )Φ(B jj ) = (Φ(A ii ) + Φ(A ij ) + Φ(A ji ) + Φ(A jj ))(Φ(B ii ) + Φ(B ij ) + Φ(B ji ) +Φ(B jj )) = Φ(A)Φ(B). Claim 10. Φ is R-linear. For every λ R, there exists two rational number sequences {r n }, {s n } such that r n λ s n and lim r n = lim s n = λ when n. It is clear that Ψ preserves positive elements, then Ψ preserves order. So, by the additivity of Φ we have Hence, r n I = Φ(r n I) Φ(λI) Φ(s n I) = s n I. Φ(λI) = λi for λ R. It means that Φ is R-linear. References [1] Z. F. Bai and S.P. Du, Multiplicative Lie isomorphism between prime rings, Comm. Algebra 36 (008), [] J. Cui and C. K. Li, Maps preserving product XY Y X on factor von Neumann algebras, Linear Algebra Appl. 431 (009), [3] P. Ji and Z. Liu, Additivity of Jordan maps on standard Jordan operator algebras, Linear Algebra Appl. 430 (009),

14 74 A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish [4] C. Li, F. Lu, and X. Fang, Nonlinear mappings preserving product XY + Y X on factor von Neumann algebras, Linear Algebra Appl. 438 (013), [5] L. Liu and G. X. Ji, Maps preserving product X Y + Y X on factor von Neumann algebras, Linear and Multilinear Algebra. 59 (011), [6] F. Lu, Additivity of Jordan maps on standard operator algebras, Linear Algebra Appl. 357 (00), [7] F. Lu, Jordan maps on associative algebras, Comm. Algebra 31 (003), [8] F. Lu, Jordan triple maps, Linear Algebra Appl. 375 (003), [9] W.S. Martindale III, When are multiplicative mappings additive? Proc. Amer. Math. Soc. 1 (1969), [10] L. Molnár, On isomorphisms of standard operator algebras, Studia Math. 14 (000), [11] A. Taghavi, H. Rohi, and V. Darvish, Additivity of maps preserving Jordan η - products on C -algebras, Bulletin of the Iranian Mathematical Society. 41 (7) (015) [1] A. Taghavi, V. Darvish, and H. Rohi, Additivity of maps preserving products AP ± P A on C -algebras, Mathematica Slovaca. 67 (1) (017) [13] A. Taghavi, M. Nouri, M. Razeghi, and V. Darvish, Maps preserving Jordan and -Jordan triple product on operator -algebras, submitted. [14] V. Darvish, H. M. Nazari, H. Rohi, and A. Taghavi, Maps preserving η-product A B + ηba on C -algebras, Journal of Korean Mathematical Society. 54 (3) (017) Ali Taghavi Department of Mathematics, Faculty of Mathematical Sciences University of Mazandaran, P. O. Box Babolsar, Iran. taghavi@umz.ac.ir Mojtaba Nouri Department of Mathematics, Faculty of Mathematical Sciences University of Mazandaran, P. O. Box Babolsar, Iran. mojtaba.nori010@gmail.com Mehran Razeghi Department of Mathematics, Faculty of Mathematical Sciences University of Mazandaran, P. O. Box Babolsar, Iran. razeghi.mehran19@yahoo.com Vahid Darvish Department of Mathematics, Faculty of Mathematical Sciences University of Mazandaran, P. O. Box Babolsar, Iran. vahid.darvish@mail.com