Product of derivations on C -algebras

Transcription

1 Int. J. Nonlinear Anal. Appl. 7 (2016) No. 2, ISSN: (electronic) Product of derivations on C -algebras Sayed Khalil Ekrami a,, Madjid Mirzavaziri b, Hamid Reza Ebrahimi Vishki b a Department of Mathematics, Payame Noor University, P.O. Box , Tehran, Iran b Department of Pure Mathematics and Center of Excellence in Analysis on Algebraic Structures (CEAAS), Ferdowsi University of Mashhad, P.O. Box 1159, Mashhad 91775, Iran (Communicated by M. Eshaghi) Abstract Let A be an algebra. A linear mapping δ : A A is called a derivation if δ(ab) = δ(a)b + aδ(b) for each a, b A. Given two derivations δ and δ on a C -algebra A, we prove that there exists a derivation on A such that δδ = 2 if and only if either δ = 0 or δ = sδ for some s C. Keywords: Derivation; C -algebra MSC: Primary 46L57; Secondary 47B47, 16W Introduction Let A be an algebra. A linear mapping δ : A A is called a derivation if it satisfies the Leibniz rule δ(ab) = δ(a)b + aδ(b) for each a, b A. When A is a -algebra, δ is called a -derivation if δ(a ) = δ(a) for each a A. Let δ be a -derivation on a C -algebra A, then δ 2 is a derivation if and only if δ = 0. To see this, note that δ 2 is a derivation if and only if δ 2 (x)y + 2δ(x)δ(y) + xδ 2 (y) = δ 2 (xy) = δ 2 (x)y + xδ 2 (y). The latter is equivalent to the fact that δ(x)δ(y) = 0 for each x, y A. Thus δ(x)δ(x) = δ(x)δ(x ) = 0 for each x A. Hence δ(x) 2 = δ(x)δ(x) = 0. This shows that δ(x) = 0 for each x A. As a typical example of a non-zero derivation in a non-commutative algebra, we can consider the inner derivation δ a implemented by an element a A which is defined as δ a (x) = xa ax for each x A. Even for an inner derivation δ a on an algebra A, it is very probable that δ 2 a is not a derivation. Corresponding author addresses: ekrami@pnu.ac.ir, khalil.ekrami@gmail.com (Sayed Khalil Ekrami), mirzavaziri@um.ac.ir, mirzavaziri@gmail.com (Madjid Mirzavaziri), vishki@um.ac.ir (Hamid Reza Ebrahimi Vishki) Received: July 2016 Revised: December 2016

2 110 Ekrami, Mirzavaziri, Ebrahimi Vishki These considerations show that the set of derivations on an algebra A is not in general closed under product. There are various researches seeking for some conditions under which the product of two derivations will be again a derivation. Posner [9] was the first one who studied the product of two derivations on a prime ring. He showed that if the product of two derivations on a prime ring, with characteristic not equal to 2, is a derivation then one of them must be equal to zero. The same question has been investigated by several authors on various algebras, see for example [1, 2, 3, 5, 6, 7, 8] and references therein. In the realm of C -algebras, Mathieu [5] showed that, if the product of two derivations δ and δ on a C -algebra is a derivation then δδ = 0. The same result was proved by Pedersen [8] for unbounded densely defined derivations on a C -algebra. There are known algebras A such that each derivation on A is inner which is implemented by an element of the algebra A or an algebra B containing A. For example, each derivation on a von Neumann algebra M is inner and is implemented by an element of M. Moreover, each derivation on a C -algebra A acting on a Hilbert space H is inner and implemented by an element of the weak closure M of A in B(H) (See [4, 10]). In the present paper, we are concerned with the following problem: Given two derivations δ and δ on a C -algebra A, find necessary and sufficient condition under which there exists a derivation on A satisfying δδ = 2. We affirm that the condition is: either δ = 0 or δ = sδ for some s C. We do this in two steps; for the matrix algebra M n (C) and for an arbitrary C -algebra. 2. The equation for the case of matrix algebras In this section we are mainly concerned with the structure of derivations on the matrix algebra M n (C). Let A = [a ij ] M n (C). We denote the diagonal matrix whose diagonal entries are a ii by A D. Proposition 2.1. Let A = [a ij ], B = [b ij ] M n (C). Then there exists a C = [c ij ] M n (C) such that δ A δ B = δ C 2 if and only if either δ B = 0 or δ A = sδ B for some s C. Proof. Let {E ij } 1 i,j n be the standard system of matrix units for M n (C). First we show that a ik b lj = b ik a lj for all 1 i, k, l, j n if and only if AXB = BXA for all X M n (C). To see this, suppose that a ik b lj = b ik a lj for all 1 i, k, l, j n then we can write We thus have (E ii AE kl )(E ll BE jj ) = a ik b lj E ij = b ik a lj E ij = (E ii BE kl )(E ll AE jj ). ( E ii )AE kl B( E jj ) = ( E ii )BE kl A( E jj ). i=1 j=1 This shows that AE kl B = BE kl A for each 1 k, l n. We can therefore deduce that AXB = BXA for all X M n (C). On the other hand, if AXB = BXA for all X M n (C), then a ij b kl E il = (E ii AE jk )(E kk BE ll ) = (E ii BE jk )(E kk AE ll ) = b ij a kl E il. We can assume that a 11 = b 11 = c 11 = 0. This is due to the fact that δ A a11 I = δ A, δ B b11 I = δ B and δ C c11 I = δ C. Then δ A δ B = δ C 2 if and only if ABE kl AE kl B BE kl A + E kl BA = C 2 E kl 2CE kl C + E kl C 2, i=1 j=1

3 Product of derivations on C -algebras 7 (2016) No. 2, for each 1 k, l n. This is equivalent to the fact that E ii (ABE kl AE kl B BE kl A + E kl BA)E jj = E ii (C 2 E kl 2CE kl C + E kl C 2 )E jj, for each 1 i, j, k, l n. Now for i k and j l we have For i k and j = l we have ( (0 a ik b lj b ik a lj + 0)E ij = (0 2c ik c lj + 0)E ij. (2.1) a im b mk a ik b ll b ik a ll + 0)E il = ( For i = k and j l we have (0 a kk b lj b kk a lj + c im c mk 2c ik c ll + 0)E il. (2.2) b lm a mj )E kj = (0 2c kk c lj + And finally for i = k and j = l we have ( a km b mk a kk b ll b kk a ll + b lm a ml )E kl = ( c lm c mj )E kj. (2.3) c km c mk 2c kk c ll + c lm c ml )E kl. (2.4) If k l then putting i = l and j = k in the equation (2.1) we have c 2 lk = a lkb lk. Thus for i k and j l we have (a ik b lj + b ik a lj ) 2 = 4c 2 ik c2 lj = 4a ikb ik a lj b lj. This implies that a ik b lj = b ik a lj, for i k, j l. (2.5) Now, if b lj 0 for some 1 l, j n with l j, then the equation a ik = a lj b lj b ik, for i k, implies the existence of some α and β with α + β 0 such that α(a A D ) = β(b B D ). (2.6) If b lj = 0 for all 1 l, j n with l j, then B = B D and so the equation (2.6) holds for α = 0 and any nonzero β C. Interchanging l i, j k and k l in (2.3) we have b im a mk a ll b ik b ll a ik = It follows from (2.2) and (2.7) that a im b mk = c im c mk 2c ll c ik, for i k. (2.7) b im a mk, for i k. Returning to the fact that a im b mk = b im a mk for m i, k, we have a ii b ik + a ik b kk = b ii a ik + b ik a kk, for i k.

4 112 Ekrami, Mirzavaziri, Ebrahimi Vishki This implies that Putting k = l in (2.4) we get a ik (b ii b kk ) = b ik (a ii a kk ). (2.8) a km b mk a kk b kk = Thus it follows from (2.4) that For l = 1 we have c km c mk c kk c kk. a kk b kk a kk b ll b kk a ll + b ll a ll = c kk c kk 2c kk c ll + c ll c ll. c 2 kk = a kk b kk, and then a kk b ll + b kk a ll = 2c kk c ll. Thus for all 1 k, l n we have (a kk b ll + b kk a ll ) 2 = 4c 2 kk c2 ll = 4a kk b kk a ll b ll. This implies that a kk b ll = b kk a ll, for all k, l. A similar argument as about the equation (2.5) implies the existence of some α and β with α + β 0 such that α A D = β B D. Using (2.8) we have b jj a ik (b ii b kk ) = b ik b jj (a ii a kk ) = b ik a jj (b ii b kk ). Now let B D / CI. Then b ii b kk for some i and k. This shows that b jj a ik = a jj b ik. So we have α = α and β = β. By a similar argument we can say that if A D / CI then α = α and β = β. We therefore have and if A D / CI or B D / CI then αa = βb for some α and β with α + β 0. On the other hand, if A D = si and B D = ti for some s, t C then α A D + α(a A D ) = s(α α)i + αa, β B D + β(b B D ) = t(β β)i + βb. Therefore s(α α)i + αa = t(β β)i + βb. Summarizing these we can say that δ A δ B = δ C 2 if and only if αa = βb + ri for some α, β, r C with α + β 0. This is equivalent to the fact that either δ B = 0 or δ A = sδ B for some s C. A natural question is the following: Is it true in general that δδ = 2 on an algebra A is equivalent to either δ = 0 or δ = sδ for some s C? In this case we of course have = sδ. The following example shows that the answer is not affirmative in general. Example 2.2. Let A be the subalgebra of M 2 (C) generated by E 11 and E 12. If δ = δ E12 and δ = δ E11 then for each X = xe 11 + ye 12 A we have δδ (X) = δ(xe 11 + ye 12 xe 11 ) = δ(ye 12 ) = 0. Thus δδ = δ 2 0. But δ 0 and δ is not a multiple of δ.

5 Product of derivations on C -algebras 7 (2016) No. 2, Lemma 2.3. Let A be the subalgebra of M 2 (C) generated by E 11 and E 12. Then each derivation on A is of the form δ = δ ce12 de 11 for some c, d C. Proof. Let δ : A A be a derivation defined by δ(xe 11 + ye 12 ) = f(x, y)e 11 + g(x, y)e 12. Since δ is linear, f(x, y) = f(x, 0) + f(0, y) = xf(1, 0) + yf(0, 1). We therefore have f(x, y) = ax + by and g(x, y) = cx + dy for some a, b, c, d C. Moreover, implies δ((xe 11 + ye 12 )(x E 11 + y E 12 )) = δ(xe 11 + ye 12 )(x E 11 + y E 12 ) + (xe 11 + ye 12 )δ(x E 11 + y E 12 ) f(xx, xy )E 11 + g(xx, xy )E 12 = f(x, y)x E 11 + f(x, y)y E 12 + xf(x, y )E 11 + xg(x, y )E 12. We thus have f(xx, xy ) = f(x, y)x + xf(x, y ), g(xx, xy ) = f(x, y)y + xg(x, y ). By using the fact that f(x, y) = ax + by and g(x, y) = cx + dy, we have f(x, y) = 0. δ = δ ce12 de 11. Whence Proposition 2.4. Let A be the subalgebra of M 2 (C) generated by E 11 and E 12 and δ, δ be two derivations on A. Then δδ = 2 if and only if δ = 0 or δ = δ α E 12 for some α C implies δ = δ αe12 for some α C, or equivalently δ = 0 or δ 2 = 0 implies δ 2 = 0. Proof. Let δ = δ αe12 βe 11, δ = δ α E 12 β E 11 and = δ re12 se 11. Then δδ = 2 if and only if rs = βα and s 2 = ββ. The latter is equivalent to the fact that δ = 0 or δ = δ α E 12 for some α C implies δ = δ αe12 for some α C. On the other hand, a derivation δ on A is of the form δ λe12 for some λ C if and only if δ 2 = Derivations on C -algebras Theorem 3.1. Let A be a C -algebra and δ, δ be two derivations on A. Then there exists a derivation on A such that δδ = 2 if and only if either δ = 0 or δ = sδ for some s C. Proof. Let A act faithfully on the Hilbert space H with the orthonormal basis {ξ i } i I. For a bounded operator T B(H), let t ij = T ξ j, ξ i for i, j I. We thus have T ξ j = i I t ijξ i and we can write T = [t ij ] i,j I. The latter is called the matrix representation of T. For i, j I, let E ij B(H) q I t qpe qp be the operator defined by E ij ξ j = ξ i and E ij ξ k = 0 for k j. Then we have T = p I for every T B(H). By the Kadison-Sakai theorem [4, 10], δ = δ R, δ = δ S and = δ T for some R, S and T in B(H). Thus δδ = 2 if and only if RSE kl RE kl S SE kl R + E kl SR = T 2 E kl 2T E kl T + E kl T 2, for each k, l I. This is equivalent to the fact that E ii (RSE kl RE kl S SE kl R + E kl SR)E jj = E ii (T 2 E kl 2T E kl T + E kl T 2 )E jj,

6 114 Ekrami, Mirzavaziri, Ebrahimi Vishki for each i, j, k, l I. For i k and j l we have Similarly, for i k and j = l we have Also, for i = k and j l we have r ik s lj + s ik r lj = 2t ik t lj. m I r im s mk r ik s ll s ik r ll = m I r kk s lj s kk r lj + m I t im t mk 2t ik t ll. s lm r mj = 2t kk t lj + m I t lm t mj. And finally for i = k and j = l we have m I r km s mk r kk s ll s kk r ll + m I s lm r ml = m I Now a similar verification as in Proposition 2.1 implies the result. t km t mk 2t kk t ll + m I t lm t ml. References [1] M. Barraa and S. Pedersen, On the product of two generalized derivations, Proc. Amer. Math. Soc. 127 (1999) [2] T. Creedon, Products of derivations, Proc. Edinburgh Math. Soc. 41 (1998) [3] C.K. Fong and A.R. Sourour, On the operator identity A k XB k 0, Canad. J. Math. 31 (1979) [4] R.V. Kadison, Derivations of operator algebras, Ann. Math., 83 (1966) [5] M. Mathieu, Properties of the product of two derivations of a C -algebra, Canad. Math. Bull. 32 (1989) [6] M. Mathieu, More properties of the product of two derivations of a C -algebra, Bull. Austral. Math. Soc. 42 (1990) [7] M. Mathieu, Posner s second theorem deduced from the first, Proc. Amer. Math. Soc. 114 (1992) [8] S. Pedersen, The product of two unbounded derivations, Canad. Math. Bull. 33 (1990) [9] E.C. Posner, Derivations in prime rings, Proc. Amer. Math. Soc. 8 (1957) [10] S. Sakai, Derivations of W -algebras, Ann. Math. 83 (1966)