Nonlinear maps preserving the mixed Jordan triple η - ∗ -product between factors
aa r X i v : . [ m a t h . OA ] J u l Nonlinear maps preserving the mixed Jordantriple η - ∗ -product between factors Fangjuan Zhang
School of Science, Xi’an University of Posts and Telecommunications, Xi’an710121, P. R China
Abstract
Let A and B be two factor von Neumann algebras and η be a non-zero complexnumber. A nonlinear bijective map φ : A → B has been demonstrated to satisfy φ ([ A, B ] η ∗ ⋄ η C ) = [ φ ( A ) , φ ( B )] η ∗ ⋄ η φ ( C )for all A, B, C ∈ A . If η = 1 , then φ is a linear ∗ -isomorphism, a conjugate linear ∗ -isomorphism, the negative of a linear ∗ -isomorphism, or the negative of a conjugatelinear ∗ -isomorphism. If η = 1 and satisfies φ ( I ) = 1 , then φ is either a linear ∗ -isomorphism or a conjugate linear ∗ -isomorphism. Key words:
Mixed Jordan triple η - ∗ -product ; Isomorphism; von Neumannalgebras Let A be a ∗ -algebra. For a non-zero scalar η, the Jordan η - ∗ -productof two elements A, B ∈ A is defined by A ⋄ η B = AB + ηBA ∗ . The Jordan( − ∗ -product, customarily called the skew Lie product, has been extensivelystudied because it is naturally seen in the problem of representing quadraticfunctionals with sesquilinear functionals ([12,13,14]) and that of characterizing ⋆ This research was supported by the National Natural Science Foundation of China(No. 11601420) and the Natural Science Basic Research Plan in Shaanxi Provinceof China (Program No. 2018JM1053). ∗ Corresponding author.
Email address: [email protected] (Fangjuan Zhang).
Preprint submitted to Elsevier deals ([10,11]). We often write the Jordan ( − ∗ -product by [ A, B ] ∗ , theJordan (1)- ∗ -product by A • B and the Jordan ( − η )- ∗ -product by [ A, B ] η ∗ , i.e., [ A, B ] ∗ = AB − BA ∗ , A • B = AB + BA ∗ , [ A, B ] η ∗ = AB − ηBA ∗ . A notnecessarily linear map φ between ∗ -algebras A and B is said to preserve theJordan η - ∗ -product if φ ( A ⋄ η B ) = φ ( A ) ⋄ η φ ( B ) for all A, B ∈ A . Recently,researchers have focused on maps preserving the Jordan η - ∗ -product between ∗ -algebra ([1,2,3,8,9]).Huo et al. [1] reported a more general problem, considering the Jordantriple η - ∗ -product of three elements A, B and C in a ∗ -algebra A as definedby A ⋄ η B ⋄ η C = ( A ⋄ η B ) ⋄ η C (note that ⋄ η is not necessarily associative).Moreover, a map φ between ∗ -algebras A and B is said to preserve the Jordantriple η - ∗ -product if φ ( A ⋄ η B ⋄ η C ) = φ ( A ) ⋄ η φ ( B ) ⋄ η φ ( C ) for all A, B, C ∈ A . In[1], let η = − φ be a bijection betweentwo von Neumann algebras, one of which has no central abelian projections,satisfying φ ( I ) = I and preserving the Jordan triple η - ∗ -product. Huo et al.reported that φ is a linear ∗ -isomorphism if η is not real and φ is the sumof a linear ∗ -isomorphism and a conjugate linear ∗ -isomorphism if η is real.Nevertheless, Huo et al. have not considered the case η = − . Li et al. in[2] considered maps that preserve the Jordan triple ( − ∗ -product withoutthe assumption φ ( I ) = I and confirmed that such a map between factors isa linear ∗ -isomorphism, a conjugate linear ∗ -isomorphism, the negative of alinear ∗ -isomorphism, or the negative of a conjugate linear ∗ -isomorphism. In[3], Zhao and Li discussed maps that preserve the Jordan triple (1)- ∗ -productwithout the assumption φ ( I ) = I and the results are similar to [2]. The Lieproduct of two elements A, B ∈ A is defined by [
A, B ] = AB − BA.
A map φ between factor von Neumann algebras A and B is said to preserve Lie productif φ ([ A, B ]) = [ φ ( A ) , φ ( B )] for all A, B ∈ A . In [7], Zhang and Zhang studiednonlinear bijective maps preserving Lie products between factors.The mixed Jordan triple η - ∗ -product of three elements A, B and C in a ∗ -algebra A defined by [ A, B ] η ∗ ⋄ η C. Moreover, a not necessarily linear map φ between factor von Neumann algebras A and B preserves mixed Jordan triple η - ∗ -products if φ ([ A, B ] η ∗ ⋄ η C ) = [ φ ( A ) , φ ( B )] η ∗ ⋄ η φ ( C )for every A, B, C ∈ A . The primary result of this study confirms the following:Let A and B be two factor von Neumann algebras and Let η be a non-zerocomplex number and φ : A → B be a bijection that preserves the mixed Jor-dan triple η - ∗ -product. Then the following statements hold:(1) If η = 1 , then φ is a linear ∗ -isomorphism, a conjugate linear ∗ -isomorphism,the negative of a linear ∗ -isomorphism, or the negative of a conjugate linear ∗ -isomorphism.(2) If η = 1 and satisfies φ ( I ) = 1 , then φ is either a linear ∗ -isomorphism or2 conjugate linear ∗ -isomorphism.As usual, R and C denote respectively the real field and complex field.Throughout, algebras and spaces are over C . A von Neumann algebra A is aweakly closed, self-adjoint algebra of operators on a Hilbert space H containingthe identity operator I. A is a factor means that its center only contains thescalar operators. It is well known that the factor A is prime, in the sense that A A B = { } for A, B ∈ A implies either A = 0 or B = 0 . Let P ( A ) be thespace of all projection operators of A . Lemma 1.1 ([3, Lemma 2.2])
Let A be a factor and A ∈ A . Then AB + BA ∗ = 0 for all B ∈ A implies that A ∈ i R I (i is the imaginary number unit). Lemma 1.2 ([4, Lemma 2.2])
Let A be a factor von Neumann algebra and A ∈ A . If [ A, B ] ∗ ∈ C I for all B ∈ A , then A ∈ C I. Lemma 1.3 ([5, Problem 230])
Let A be a Banach algebra with the identity I. If A, B ∈ A and λ ∈ C are such that [ A, B ] = λI, where [ A, B ] = AB − BA, then λ = 0 . - ∗ -product preserving mapsTheorem 2.1 Let A , B be two factor von Neumann algebras. Suppose that φ is a bijective map from A to B with φ ([ A, B ] ∗ • C ) = [ φ ( A ) , φ ( B )] ∗ • φ ( C ) for all A, B, C ∈ A , then φ is a linear ∗ -isomorphism, a conjugate linear ∗ -isomorphism, the negative of a linear ∗ -isomorphism, or the negative conjugatelinear ∗ -isomorphism. First we give a key technique. Assume that A , A , · · · , A n and T are in A with φ ( T ) = Σ ni =1 φ ( A i ) . Then for S , S , S ∈ A , we obtain φ ([ T, S ] η ∗ ⋄ η S ) = [ φ ( T ) , φ ( S )] η ∗ ⋄ η φ ( S ) = n X i =1 φ ([ A i , S ] η ∗ ⋄ η S ) , (1) φ ([ S , T ] η ∗ ⋄ η S ) = [ φ ( S ) , φ ( T )] η ∗ ⋄ η φ ( S ) = n X i =1 φ ([ S , A i ] η ∗ ⋄ η S ) (2)and φ ([ S , S ] η ∗ ⋄ η T ) = [ φ ( S ) , φ ( S )] η ∗ ⋄ η φ ( T ) = n X i =1 φ ([ S , S ] η ∗ ⋄ η A i ) . (3)Choose an arbitrary nontrivial projection P ∈ A , write P = I − P . Denote A ij = P i A P j , i, j, = 1 , , then A = P i,i =1 A ij . For every A ∈ A , we3an write it as A = P i,i =1 A ij , where A ij denotes an arbitrary element of A ij . We will complete the proof of Theorem 2.1 by proving several lemmas.
Lemma 2.1 φ (0) = 0 . Proof
Since φ is surjective, there exists A ∈ A such that φ ( A ) = 0 . Hence φ (0) = φ ([0 , A ] ∗ • A ) = [ φ (0) , φ ( A )] ∗ • φ ( A ) = 0 . Lemma 2.2 φ ( R I ) = R I, φ ( C I ) = C I and φ preserves self-adjoint ele-ments in both directions. Proof
Let λ ∈ R be arbitrary. It is easily seen that0 = φ ([ λI, B ] ∗ • C ) = [ φ ( λI ) , φ ( B )] ∗ • φ ( C )holds true for any B, C ∈ A . Since φ is surjective, by Lemma 1.1, whichindicates that [ φ ( λI ) , φ ( B )] ∗ ∈ i R I. Then [ φ ( λI ) , B ] ∗ ∈ C I for any B ∈ B . We obtain from Lemma 1.2 that φ ( λI ) ∈ C I, so there exists λ ∈ C such that ( λ − λ ) B ∈ C I for any B ∈ B , then φ ( λI ) ∈ R I. Note that φ − has the same properties as φ. Similarly, if φ ( A ) ∈ R I, then A ∈ R I. Therefore, φ ( R I ) = R I. Since φ ( R I ) = R I, exists λ ∈ R such that φ ( λI ) = I. For any A = A ∗ ∈ A and B ∈ A , we obtain0 = φ ([ A, λI ] ∗ • B ) = [ φ ( A ) , I ] ∗ • φ ( B ) , by the surjectivity of φ and Lemma 1.1, the above equation indicates [ φ ( A ) , I ] ∗ ∈ i R I. Then there exists λ ∈ i R such that φ ( A ) ∗ = φ ( A ) + λI. However,0 = φ ([ A, A ] ∗ • B ) = [ φ ( A ) , φ ( A )] ∗ • φ ( B )for all A = A ∗ ∈ A and B ∈ A . Similarly, [ φ ( A ) , φ ( A )] ∗ ∈ i R I. Then λφ ( A ) ∈ i R I. If λ = 0 , then φ ( A ) ∈ R I. It follows from φ ( R I ) = R I that A = A ∗ ∈ R I, which is contradiction. Thus λ = 0 . Now we get that φ ( A ) = φ ( A ) ∗ . Similarly,if φ ( A ) = φ ( A ) ∗ , then A = A ∗ ∈ A . Therefore φ preserves self-adjoint elementsin both directions.Let λ ∈ C be arbitrary. For every A = A ∗ ∈ A , we obtain that0 = φ ([ A, λI ] ∗ • B ) = [ φ ( A ) , φ ( λI )] ∗ • φ ( B )holds true for any B ∈ A . By the surjectivity of φ and Lemma 1.1 again,the above equation indicates [ φ ( A ) , φ ( λI )] ∗ ∈ i R I. Since A = A ∗ , we have φ ( A ) = φ ( A ) ∗ . Hence [ φ ( A ) , φ ( λI )] ∈ i R I. We obtain from Lemma 1.3 that4 φ ( A ) , φ ( λI )] = 0 , and then Bφ ( λI ) = φ ( λI ) B for any B = B ∗ ∈ B . Thus forany B ∈ B , since B = B + i B with B = B + B ∗ and B = B − B ∗ i , we get Bφ ( λI ) = φ ( λI ) B for any B ∈ B . Hence φ ( λI ) ∈ C I. Similarly, if φ ( A ) ∈ C I, then A ∈ C I. Therefore, φ ( C I ) = C I. Lemma 2.3 φ ( P ( A ) + R I ) = P ( B ) + R I. Proof
Fix a nontrivial projection P ∈ P ( B ) . Based on Lemma 2.2, exists A = A ∗ ∈ A such that φ ( A ) = P + R I. For any B = B ∗ ∈ A and C ∈ A , wethen have the following: φ ([ A, B ] ∗ • C ) = [ φ ( A ) , φ ( B )] ∗ • φ ( C )= [ P, φ ( B )] ∗ • φ ( C ) = [([ P, φ ( B )] ∗ • P ) , P ] ∗ • φ ( C )= [([ φ ( A ) , φ ( B )] ∗ • φ ( A )) , φ ( A )] ∗ • φ ( C ) = φ ([([ A, B ] ∗ • A ) , A ] ∗ • C ) . By the injectivity of φ, we concur that [([ A, B ] ∗ • A ) , A ] ∗ • C = [ A, B ] ∗ • C, which indicates [([ A, B ] ∗ • A ) , A ] ∗ − [ A, B ] ∗ ∈ i R I for all B = B ∗ ∈ A . Forevery X ∈ A , we have X = X + i X , where X = X + X ∗ and X = X − X ∗ i areself-adjoint. We obtain [ A, [ A, [ A, X ]]] − [ A, X ] ∈ i R I, i.e., A X − A XA + 3 AXA − XA − AX + XA ∈ i R I (4)holds true for any X ∈ A . Let U be the group of unitary operators of A and let ϕ be the set of thefunctions U → f ( U ) defined on U with non-negative real values, zero excepton a finite subset of U and such that P U ∈U f ( U ) = 1 . For A ∈ A and f ∈ ϕ, we put f · A = P U ∈U f ( U ) U AU ∗ . For any U ∈ U , by Eq. (4),( A − A ) U − A U A + 3
AU A − U ( A − A ) = αI for certain α ∈ i R I. It follows that A − A − A U AU ∗ + 3 AU A U ∗ − U ( A − A ) U ∗ = αU ∗ , and so A − A − A f · A + 3 Af · A − f · A + f · A = αU ∗ for any f ∈ ϕ. Since A is a factor von Neumann algebra, we obtain from [6, Lemma 5 (PartIII , Chapter 5)] that there exist λ , λ , λ ∈ C such that A − A − λ A + 3 λ A − ( λ − λ ) I = αU ∗ . U ( A − A ) U ∗ − λ U A U ∗ + 3 λ U AU ∗ − ( λ − λ ) I = αU ∗ and then f · A − f · A − λ f · A + 3 λ f · A − ( λ − λ ) I = αU ∗ for any f ∈ ϕ. By[6, Lemma 5 (Part III , Chapter 5)] again, we have αU ∗ = 0 for any U ∈ U . Hence α = 0 . Thus we have( A − A ) U − A U A + 3
AU A − U ( A − A ) = 0 (5)and A − A = 3 λ A − λ A + ( λ − λ ) I (6)for any U ∈ U . By Eqs. (5)–(6), we conclude that( λ A − λ A ) U − A U A + AU A − U ( λ A − λ A ) = 0and( λ A − λ A ) U AU ∗ − A U A U ∗ + AU A U ∗ − U ( λ A − λ A ) AU ∗ = 0for any U ∈ U . Thus( λ A − λ A ) f · A − A f · A + Af · A − λ f · A + λ f · A = 0for any f ∈ ϕ. By applying [6, Lemma 5 (Part III , Chapter 5)] again, weobtain λ ( λ A − λ A ) − λ A + λ A + ( λ − λ λ ) I = 0 , i.e., ( λ − λ ) A + ( λ − λ λ ) A + ( λ − λ λ ) I = 0 . (7)If λ = λ , we obtain from φ ( C I ) = C I and φ ( A ) = P + R I / ∈ C I that A / ∈ C I, then λ = λ λ = λ , so by Eq. (6), we have ( A − λ I ) = A − λ I. Let B = A − λ I, then B = B and [ B, [ B, [ B, X ]]] = [
B, X ]for any X ∈ A . This indicates that B XB − BXB = 0 (8)for any X ∈ A . Let E = ( B + B ) and E = ( B − B ) . It follows from thefact B = B that E and E are idempotents of A and that B = E − E , B = E + E , E E = E E = 0 . This along with Eq. (8) shows us that E XE = 0 for any X ∈ A . Then E = 0 or E = 0 , and so A = λ I + E or A = λ I − E is the sum of a scalarand an idempotent of A . λ = λ , by Eq. (7), we have A = λA + µI for certain λ, µ ∈ C . Thisalong with Eq. (5) indicates that( λ + 4 µ − AU − U A ) = 0 (9)for all U ∈ U . Since
A / ∈ C I, we have AU − U A = 0 for some U ∈ U . By Eq.(9), we obtain λ + 4 µ − . Let E = A + (1 − λ ) I, then E = A + (1 − λ ) A + 14 (1 − λ ) I = λA + µI + (1 − λ ) A + 14 (1 − λ ) I = A + 14 ( λ + 4 µ − λ + 1) I = A + 12 (1 − λ ) I = E. Hence A = ( λ − I + E is the sum of a scalar and an idempotent of A . Since A = A ∗ , then A = αI + E, α ∈ R I, E ∈ P ( A ) . If E = 0 or E = I, from φ ( A ) = P + R I, we obtain φ ( R I ) = P + R I. It follows φ ( R I ) = R I that P = 0or P = I, since P is a nontrivial projection, which is a contradiction. Thus, A is the sum of a real number and a nontrivial projection of A . Applyingthe same argument to φ − , we can obtain the reverse inclusion and equalityfollows. Lemma 2.4
Let i, j ∈ { , } with i = j. Then φ ( A ii + B ij ) = φ ( A ii )+ φ ( B ij ) and φ ( A ii + B ji ) = φ ( A ii ) + φ ( B ji ) for all A ii ∈ A ii , B ij ∈ A ij and B ji ∈ A ji . Proof
Let X = P i,i =1 X ij ∈ A such that φ ( X ) = φ ( A ii )+ φ ( B ij ) . It followsfrom Eq. (2) that φ ( X ij + X ∗ ij ) = φ ([ P i , X ] ∗ • P j ) = φ ([ P i , A ii ] ∗ • P j )+ φ ([ P i , B ij ] ∗ • P j ) = φ ( B ij + B ∗ ij ) . By the injectivity of φ, we have X ij = B ij . We obtain from Eq. (2) that φ ( X ji + X ∗ ji ) = φ ([ P j , X ] ∗ • P i ) = φ ([ P j , A ii ] ∗ • P i ) + φ ([ P j , B ij ] ∗ • P i ) = 0 , which indicates that X ji = 0 . For every T ij ∈ A ij , by applying Eq. (2) again,we obtain φ ( T ij X jj + X ∗ jj T ∗ ij ) = φ ([ T ij , X ] ∗ • P j )= φ ([ T ij , A ii ] ∗ • P j ) + φ ([ T ij , B ij ] ∗ • P j ) = 0 , which implies that T ij X jj = X ∗ jj T ∗ ij = 0 for all T ij ∈ A ij . By the primeness of A , we get X jj = 0 . For every T ij ∈ A ij , by applying Eq. (1), we obtain φ ( X ii T ij + T ∗ ij X ∗ ii ) = φ ([ X, T ij ] ∗ • P j )= φ ([ A ii , T ij ] ∗ • P j ) + φ ([ B ij , T ij ] ∗ • P j ) = φ ( A ii T ij + T ∗ ij A ∗ ii ) , X ii T ij = A ii T ij for all T ij ∈ A ij . By the primeness of A , we obtain X ii = A ii . Thus φ ( A ii + B ij ) = φ ( A ii ) + φ ( B ij ) . In the second case,we can similarly prove that the conclusion is valid.
Lemma 2.5
Let i, j ∈ { , } with i = j. Then φ ( A ij + B ji ) = φ ( A ij ) + φ ( B ji ) for all A ij ∈ A ij and B ji ∈ A ji . Proof
Choose X = P i,i =1 X ij ∈ A such that φ ( X ) = φ ( A ij ) + φ ( B ji ) . Itfollows from Eq. (1) that φ ( X ji + X ∗ ji ) = φ ([ X, P i ] ∗ • P i )= φ ([ A ij , P i ] ∗ • P i ) + φ ([ B ji , P i ] ∗ • P i ) = φ ( B ji + B ∗ ji ) , Thus we have X ji = B ji . Similarly, X ij = A ij . For every T ij ∈ A ij , By applyingEq. (2), we obtain φ ( T ij X jj + X ∗ jj T ∗ ij ) = φ ([ T ij , X ] ∗ • P j )= φ ([ T ij , A ij ] ∗ • P j ) + φ ([ T ij , B ji ] ∗ • P j ) = 0 , from this, we get X jj = 0 . In the same manner, we obtain X ii = 0 . Lemma 2.6 φ (Σ i,j =1 A ij ) = Σ i,j =1 φ ( A ij ) for all A ij ∈ A ij . Proof
Let X = P i,i =1 X ij ∈ A such that φ ( X ) = Σ i,j =1 φ ( A ij ) . It followsfrom Eq. (2) that φ ([ P , X ] ∗ • P ) = Σ i,j =1 φ ([ P , A ij ] ∗ • P ) , i.e., φ ( X + X ∗ ) = φ ( A + A ∗ ) , which implies that X = A . Similarly, X = A . For every T ∈ A , by applying Eq. (2) again, we obtain φ ([ T , X ] ∗ • P ) =Σ i,j =1 φ ([ T , A ij ] ∗ • P ) for all T ∈ A . Thus we have X = A . In the samemanner, we obtain X = A . Lemma 2.7
Let i, j ∈ { , } with i = j. Then φ ( A ij + B ij ) = φ ( A ij ) + φ ( B ij ) for all A ij ∈ A ij and B ij ∈ A ij . Proof
It follows from A ij + B ij + A ∗ ij + B ij A ∗ ij = [ − i I, i P i +i A ij ] ∗ • ( P j + B ij )and Lemmas 2.6, 2.4, 2.5 that φ ( A ij + B ij ) + φ ( A ∗ ij ) + φ ( B ij A ∗ ij )= φ ( A ij + B ij + A ∗ ij + B ij A ∗ ij )= φ ([ − i2 I, i P i + i A ij ] ∗ • ( P j + B ij ))= [ φ ( − i2 I ) , φ (i P i + i A ij )] ∗ • φ ( P j + B ij )= [ φ ( − i2 I ) , φ (i P i ) + φ (i A ij )] ∗ • ( φ ( P j ) + φ ( B ij ))8 φ ([ − i2 I, i P i ] ∗ • P j ) + φ ([ − i2 I, i P i ] ∗ • B ij )+ φ ([ − i2 I, i A ij ] ∗ • P j ) + φ ([ − i2 I, i A ij ] ∗ • B ij )= φ ( B ij ) + φ ( A ij + A ∗ ij ) + φ ( B ij A ∗ ij )= φ ( B ij ) + φ ( A ij ) + φ ( A ∗ ij ) + φ ( B ij A ∗ ij ) , which indicates that φ ( A ij + B ij ) = φ ( A ij ) + φ ( B ij ) . Lemma 2.8
Let i ∈ { , } . Then φ ( A ii + B ii ) = φ ( A ii ) + φ ( B ii ) for all A ii ∈ A ii and B ii ∈ A ii . Proof
Choose X = P i,i =1 X ij ∈ A such that φ ( X ) = φ ( A ii ) + φ ( B ii ) . Itfollows from Eq. (2) that φ ( X ij + X ∗ ij ) = φ ([ P i , X ] ∗ • P j )= φ ([ P i , A ii ] ∗ • P j ) + φ ([ P i , B ii ] ∗ • P j ) = 0 . Thus we have X ij = 0 . Similarly, X ji = 0 . For every T ij ∈ A ij , By applyingEq. (2) again, we have φ ( T ij X jj + X ∗ jj T ∗ ij ) = φ ([ T ij , X ] ∗ • P j )= φ ([ T ij , A ii ] ∗ • P j ) + φ ([ T ij , B ii ] ∗ • P j ) = 0 , which implies that T ij X jj = X ∗ jj T ∗ ij = 0 . By the primeness of A , we obtain X jj = 0 . Therefore, φ ( X ii ) = φ ( A ii ) + φ ( B ii ) . (10)For every T ij ∈ A ij , it follows from Eq. (1) and Lemmas 2.5 and 2.7 that φ ( X ii T ij + T ∗ ij X ∗ ii ) = φ ([ X, T ij ] ∗ • P j )= φ ([ A ii , T ij ] ∗ • P j ) + φ ([ B ii , T ij ] ∗ • P j )= φ ( A ii T ij + T ∗ ij A ∗ ii ) + φ ( B ii T ij + T ∗ ij B ∗ ii )= φ ( A ii T ij ) + φ ( T ∗ ij A ∗ ii ) + φ ( B ii T ij ) + φ ( T ∗ ij B ∗ ii )= φ ( A ii T ij + B ii T ij ) + φ ( T ∗ ij A ∗ ii + T ∗ ij B ∗ ii )= φ ( A ii T ij + B ii T ij + T ∗ ij A ∗ ii + T ∗ ij B ∗ ii ) , which indicates that X ii = A ii + B ii . This together with Eq. (10) shows that φ ( A ii + B ii ) = φ ( A ii ) + φ ( B ii ) . Lemma 2.9 φ is additive and φ ( P ( A )) = P ( B ) . Proof
By Lemmas 2.6–2.8, φ is additive. By Lemmas 2.3 and 2.2, we have φ ( P ( A )) = P ( B ) . emark 2.1 Since [ P , B ] ∗ • C = [ B, P ] ∗ • C for all B = B ∗ ∈ A and C ∈ A , from Lemma 2.9[ Q , φ ( B )] ∗ • φ ( C ) = [ φ ( B ) , Q ] ∗ • φ ( C ) , where Q i ∈ P ( B ) , i = 1 , . The surjectivity of φ indicates that [ Q , φ ( B )] ∗ − [ φ ( B ) , Q ] ∗ ∈ i R I. It follows from Lemma 2.2 that [ Q + Q , B ] ∈ i R I holdstrue for all B = B ∗ ∈ B . By Lemma 1.3, [ Q + Q , B ] = 0 . Thus for every B ∈ B , because B = B + i B with B = B + B ∗ and B = B − B ∗ i , we get[ Q + Q , B ] = 0 for all B ∈ B . From this, exists λ ∈ R such that Q + Q = λI. Multiplying by Q and Q from the left and right respectively in the aboveequation, we obtain Q + Q Q = λQ and Q Q + Q = λQ . Therefore, wecan concur that (1 − λ )( Q − Q ) = 0 by subtracting the above two equations.By the injectivity of φ, exists P = P such that Q = Q . Thus λ = 1 andthen Q = I − Q . Lemma 2.10 φ ( A ij ) = B ij , i, j = 1 , . Proof
Let i, j ∈ { , } with i = j and A ij ∈ A ij . By the fact A ij =[ P i , A ij ] ∗ • I , Lemma 2.9 and Remark 2.1, we have φ ( A ij ) = ( Q i φ ( A ij ) − φ ( A ij ) Q i ) φ ( I − φ ( I φ ( A ij ) ∗ Q i − Q i φ ( A ij ) ∗ ) . From this and Lemma 2.2, we get Q i φ ( A ij ) Q i = Q j φ ( A ij ) Q j = 0 . Thus φ ( A ij ) = Q i φ ( A ij ) Q j + Q j φ ( A ij ) Q i . (11)For every B ∈ A , we obtain from the fact [ A ij , P i ] ∗ • B = 0 , Lemma 2.9 andRemark 2.1 that [ φ ( A ij ) , Q i ] ∗ • φ ( B ) = 0 . Thus [ φ ( A ij ) , Q i ] ∗ ∈ i R I, whichtogether with Eq. (11) indicates that Q j φ ( A ij ) Q i − Q i φ ( A ij ) ∗ Q j ∈ i R I. Multi-plying by Q j and Q i from the left and right respectively in the above equation,we have Q j φ ( A ij ) Q i = 0 . It follows from Eq. (11) that φ ( A ij ) = Q i φ ( A ij ) Q j , and then we obtain φ ( A ij ) ⊆ B ij . Applying the same argument to φ − , weobtain B ij ⊆ φ ( A ij ) . Thus φ ( A ij ) = B ij , i = j. Let A jj ∈ A jj and i = j. It follows from Lemma 2.9 and Remark 2.1 that0 = φ ([ P i , A jj ] ∗ • P j ) = [ Q i , φ ( A jj )] ∗ • Q j = Q i φ ( A jj ) Q j + Q j φ ( A jj ) ∗ Q i and0 = φ ([ P j , A jj ] ∗ • P i ) = [ Q j , φ ( A jj )] ∗ • Q i = Q j φ ( A jj ) Q i + Q i φ ( A jj ) ∗ Q j . Q i φ ( A jj ) Q j = Q j φ ( A jj ) Q i = 0 . Now we obtain φ ( A jj ) = Q i φ ( A jj ) Q i + Q j φ ( A jj ) Q j . (12)For every A ji ∈ A ji and C ∈ A , we have T ji = φ − ( A ji ) ∈ A ji . Therefore0 = φ ([ A ji , A jj ] ∗ • C ) = [ T ji , φ ( A jj )] ∗ • φ ( C ) . Using the surjectivity of φ, the above equation indicates [ T ji , φ ( A jj )] ∗ ∈ i R I. It follows from Eq. (12) that T ji φ ( A jj ) Q i − Q i φ ( A jj ) T ∗ ji ∈ i R I. (13)By Remark 2.1, multiplying by Q j and Q i from the left and right respectivelyin Eq. (13), we can get that T ji φ ( A jj ) Q i = 0 for all T ji ∈ B ji . By the primenessof B , we obtain that Q i φ ( A jj ) Q i = 0 , thus φ ( A jj ) ⊆ B jj . Applying the sameargument to φ − , we can obtain B jj ⊆ φ ( A jj ) . Consequently, φ ( A jj ) = B jj . Lemma 2.11 φ ( AB ) = φ ( A ) φ ( B ) for all A, B ∈ A . Proof
It follows from Remark 2.1 and Lemma 2.10 that φ ([ P i , A ij ] ∗ • B ji ) =[ φ ( P i ) , φ ( A ij )] ∗ • φ ( B ji ) = [ Q i , φ ( A ij )] ∗ • φ ( B ji ) . Thus φ ( A ij B ji ) = φ ( A ij ) φ ( B ji ) . (14)For T ji ∈ B ji , we have X ji = φ − ( T ji ) ∈ A ji by Lemma 2.10. Therefore φ ( A ii B ij ) T ji = φ ( A ii B ij X ji ) = φ ([ A ii , B ij ] ∗ • X ji ) = φ ( A ii ) φ ( B ij ) T ji . By the primeness of B , we obtain φ ( A ii B ij ) = φ ( A ii ) φ ( B ij ) . (15)It follows from Eqs. (14)–(15) that φ ( A ij B jj ) T ji = φ ( A ij B jj X ji ) = φ ( A ij ) φ ( B jj X ji ) = φ ( A ij ) φ ( B jj ) T ji . In the same manner, we obtain φ ( A ij B jj ) = φ ( A ij ) φ ( B jj ) . (16)By Eq. (15), we have φ ( A jj B jj ) T ji = φ ( A jj B jj X ji ) = φ ( A jj ) φ ( B jj X ji ) = φ ( A jj ) φ ( B jj ) T ji . Thus φ ( A jj B jj ) = φ ( A jj ) φ ( B jj ) . (17)From Eqs. (14)-(17) and Lemmas 2.9, 2.10, we obtain φ ( AB ) = φ ( A ) φ ( B ) forall A, B ∈ A . emma 2.12 φ is a linear ∗ -isomorphism, or a conjugate linear ∗ -isomorphism,or the negative of a linear ∗ -isomorphism, or the negative of a conjugate linear ∗ -isomorphism. Proof
It follows from Lemmas 2.9 and 2.11 that φ is a ring isomorphism.By Lemma 2.2, exists λ ∈ R \ { } such that φ ( I ) = λI. By the equality φ ( I ) = φ ( I ) , we concur that φ ( I ) = I or φ ( I ) = − I. In the rest of thissection, we deal with these two cases respectively.
Case 1 If φ ( I ) = I, then φ is either a linear ∗ -isomorphism or a conjugatelinear ∗ -isomorphism.For every rational number q, we have φ ( qI ) = qI. Indeed, since q is rationalnumber, exist two integers r and s such that q = rs . Since φ ( I ) = I and φ isadditive, we get that φ ( qI ) = φ ( rs I ) = rφ ( s I ) = rs φ ( I ) = qI. Let A be a positive element in A . Then A = B for some self-adjointelement B ∈ A . It follows from Lemma 2.11 that φ ( A ) = φ ( B ) . By Lemma2.2, we get that φ ( B ) is self-adjoint. So φ ( A ) is positive. This shows that φ preserves positive elements.Let λ ∈ R . Choose sequence { a n } and { b n } of rational numbers such that a n ≤ λ ≤ b n for all n and lim n →∞ a n = lim n →∞ b n = λ. It follows from a n I ≤ λI ≤ b n I that a n I ≤ φ ( λI ) ≤ b n I. Taking the limit, we obtain that φ ( λI ) = λI. Hence for all A ∈ A , we have φ ( λA ) = φ (( λI ) A ) = φ ( λI ) φ ( A ) = λφ ( A ) . Thus φ is real linear. For every A ∈ A , it follows from − φ ( A ) = φ (i A ) = φ (i I ) φ ( A )that φ (i I ) = − , which implies that φ (i I ) = i I or φ (i I ) = − i I. By Lemma2.11, we obtain that φ (i A ) = i φ ( A ) or φ (i A ) = − i φ ( A ) for all A ∈ A . For all A ∈ A , A = A + i A , where A = A + A ∗ and A = A − A ∗ i areself-adjoint elements. By Lemmas 2.2 and 2.9, if φ (i A ) = i φ ( A ) , then φ ( A ∗ ) = φ ( A − i A ) = φ ( A ) − φ (i A ) = φ ( A ) − i φ ( A )= φ ( A ) ∗ − i φ ( A ) ∗ = φ ( A ) ∗ + (i φ ( A )) ∗ = φ ( A ) ∗ Similarly, if φ (i A ) = − i φ ( A ) , we also obtain φ ( A ∗ ) = φ ( A ) ∗ . Then φ is eithera linear ∗ -isomorphism or a conjugate linear ∗ -isomorphism. Case 2 If φ ( I ) = − I, then φ is either the negative of a linear ∗ -isomorphismor the negative of a conjugate linear ∗ -isomorphism.Consider that the map ψ : A → B defined by ψ ( A ) = − φ ( A ) for all A ∈ A . It is easy to see that ψ satisfies ψ ([[ A, B ] ∗ , C ]) = [[ ψ ( A ) , ψ ( B )] ∗ , ψ ( C )] for all A, B, C ∈ A and ψ ( I ) = I. Then the arguments for Case 1 ensure that ψ is either a linear ∗ -isomorphism or a conjugate linear ∗ -isomorphism. So φ is12ither the negative of a linear ∗ -isomorphism or the negative of a conjugatelinear ∗ -isomorphism.Combining Cases 1–2, the proof of Theorem 2.1 is finished. η - ∗ -product preserving mapsTheorem 3.1 Let A , B be two factor von Neumann algebras and let η ∈ C \{ , } . Suppose that φ is a bijective map from A to B with φ ([ A, B ] η ∗ ⋄ η C ) =[ φ ( A ) , φ ( B )] η ∗ ⋄ η φ ( C ) for all A, B, C ∈ A . Then φ is additive. Theorem 3.2
Let A , B be two factor von Neumann algebras, let η ∈ C \ { , } and let φ : A → B is a bijective map, satisfying φ ( I ) = I andpreserving the mixed Jordan triple η - ∗ -product, Then φ is either a linear ∗ -isomorphism or a conjugate linear ∗ -isomorphism. Proof of Theorem 3.1
In the following, we will complete the proof byproving several claims.
Claim 1 φ (0) = 0 . Since φ is surjective, there exists A ∈ A such that φ ( A ) = 0 . Then weobtain φ (0) = φ ([[0 , A ] η ∗ ⋄ η A ) = [[ φ (0) , φ ( A )] η ∗ ⋄ η φ ( A ) = 0 . Claim 2 φ ( A + A ) = φ ( A ) + φ ( A ) for all A ∈ A and A ∈ A . Let X = P i,i =1 X ij ∈ A such that φ ( X ) = φ ( A ) + φ ( A ) . For any λ ∈ C , [ I, λP − η ] η ∗ ⋄ η A = 0 . By applying Eq. (3) and Claim 1, we obtain φ ([ I, λP − η ] η ∗ ⋄ η X ) = φ ([ I, λP − η ] η ∗ ⋄ η A ) . By the injectivity of φ, we get that[ I, λP − η ] η ∗ ⋄ η X = [ I, λP − η ] η ∗ ⋄ η A , i.e.,( λ + λη ) X + λX + ληX = ( λ + λη ) A . Assume that λ = 0 and λ + λη = 0 , we have X = A , X = X = 0 . Inthe same manner, we obtain X = A . Claim 3 φ ( A + A ) = φ ( A ) + φ ( A ) for all A ∈ A and A ∈ A . Let X = P i,i =1 X ij ∈ A such that φ ( X ) = φ ( A ) + φ ( A ) . For any λ ∈ C , since [ I, λP − λη P − η ] η ∗ ⋄ η A = 0 , Applying Eq. (3) and Claim 1 again, we obtain φ ([ I, λP − λη P − η ] η ∗ ⋄ η X ) = φ ([ I, λP − λη P − η ] η ∗ ⋄ η A ) . The injectivity of φ implies that13 I, λP − λη P − η ] η ∗ ⋄ η X = [ I, λP − λη P − η ] η ∗ ⋄ η A , i.e.,( λ + λη ) X − ( λ + λη ) X + ( λη − λη ) X = ( λη − λη ) A for all λ ∈ C . Thus we get X = X = 0 . Since [ A , λP ] η ∗ ⋄ η I = 0 . It follows from Eq. (1) that φ ([[ X, λP ] η ∗ ⋄ η I ) = φ ([[ A , λP ] η ∗ ⋄ η I ) . Thus we obtain ( λ − λ | η | ) X + η ( λ − λ ) X ∗ = ( λ − λ | η | ) A + η ( λ − λ ) A ∗ for all λ ∈ C , which indicates that X = A . In thesame manner, we obtain X = A . Claim 4
Let i, j, k ∈ { , } with i = j. Then φ ( A kk + A ij ) = φ ( A kk )+ φ ( A ij )for all A kk ∈ A kk and A ij ∈ A ij . We only prove the case i = k = 1 and j = 2 , the proof of the other casesis similar. Now assume that X ∈ A satisfies φ ( X ) = φ ( A ) + φ ( A ) . Forany λ ∈ C , since [ I, λP − η ] η ∗ ⋄ η A = 0 , by applying Eq. (3) and Claim 1 again,we obtain φ ([ I, λP − η ] η ∗ ⋄ η X ) = φ ([ I, λP − η ] η ∗ ⋄ η A ) for any λ ∈ C . Thus we get X = A , X = X = 0 . Since [ I, λP − λη P − η ] η ∗ ⋄ η A = 0 , we have φ ([ I, λP − λη P − η ] η ∗ ⋄ η X ) = φ ([ I, λP − λη P − η ] η ∗ ⋄ η A )for any λ ∈ C , which indicates that X = A . Claim 5 φ ( A + A + A ) = φ ( A ) + φ ( A ) + φ ( A ) and φ ( A + A + A ) = φ ( A ) + φ ( A ) + φ ( A ) for all A ∈ A , A ∈ A , A ∈ A and A ∈ A . Choose X = P i,i =1 X ij ∈ A such that φ ( X ) = φ ( A ) + φ ( A ) + φ ( A ) . For any λ ∈ C , it follows from Claim 4 and Eq. (3) that φ (( λ + ηλ ) X + λX + ληX )= φ ([ I, λP − η ] η ∗ ⋄ η X )= φ ([ I, λP − η ] η ∗ ⋄ η A ) + φ ([ I, λP − η ] η ∗ ⋄ η A ) + φ ([ I, λP − η ] η ∗ ⋄ η A )= φ ( ληA ) + φ ( λA ) = φ ( ληA + λA ) , which indicates that X = A , X = A and X = 0 . Thus we get X = X + A + A . I, − λη P + λP − η ] η ∗ ⋄ η A = 0 , by applying Eq. (3) and Claim 4 again, weobtain φ (( − λη − λ ) X + ( − λη + ηλ ) X )= φ ([ I, − λη P + λP − η ] η ∗ ⋄ η X )= φ ([ I, − λη P + λP − η ] η ∗ ⋄ η A ) + φ ([ I, − λη P + λP − η ] η ∗ ⋄ η A )= φ (( − λη − λ ) A ) + φ (( − λη + ηλ ) A )= φ (( − λη − λ ) A + ( − λη + ηλ ) A ) . This indicates that X = A . In the second case, we can similarly prove thatthe conclusion is valid.
Claim 6
Let i, j ∈ { , } with i = j. Then φ ( A ij + B ij ) = φ ( A ij ) + φ ( B ij )for all A ij , B ij ∈ A ij . Since [[ I, P i + A ij − η ] η ∗ , P j + B ij ] η = A ij + B ij + ηA ∗ ij + ηB ij A ∗ ij , it follows fromClaims 5, 4 and 3 that φ ( A ij + B ij ) + φ ( ηA ∗ ij ) + φ ( ηB ij A ∗ ij ) = φ ([ I, P i + A ij − η ] η ∗ ⋄ η ( P j + B ij ))= [ φ ( I ) , φ ( P i + A ij − η )] η ∗ ⋄ η φ ( P j + B ij )= [ φ ( I ) , φ ( P i − η ) + φ ( A ij − η )] η ∗ ⋄ η ( φ ( P j ) + φ ( B ij )))= [ φ ( I ) , φ ( P i − η )] η ∗ ⋄ η φ ( P j ) + [ φ ( I ) , φ ( P i − η )] η ∗ ⋄ η φ ( B ij )+[ φ ( I ) , φ ( A ij − η )] η ∗ ⋄ η φ ( P j ) + [ φ ( I ) , φ ( A ij − η )] η ∗ ⋄ η φ ( B ij )= φ ( B ij ) + φ ( A ij + ηA ∗ ij ) + φ ( ηB ij A ∗ ij )= φ ( B ij ) + φ ( A ij ) + φ ( ηA ∗ ij ) + φ ( ηB ij A ∗ ij ) . Then φ ( A ij + B ij ) = φ ( A ij ) + φ ( B ij ) . Claim 7 φ ( T A + T B + ηA T ∗ + ηB T ∗ ) = φ ( T A )+ φ ( T B )+ φ ( ηA T ∗ ) + φ ( ηB T ∗ ) for all T , A , B ∈ A and A , B ∈ A . It follows from Claims 3 and 6 that15 ( T A + T B + ηA T ∗ + ηB T ∗ )= φ ([[ I, T − η ] η ∗ ⋄ η ( A + B + A + B ))= [ φ ( I ) , φ ( T − η )] η ∗ ⋄ η ( φ ( A ) + φ ( B ) + φ ( A ) + φ ( B ))= φ ([ I, T − η ] η ∗ ⋄ η A ) + φ ([ I, T − η ] η ∗ ⋄ η B )+ φ ([ I, T − η ] η ∗ ⋄ η A ) + φ ([ I, T − η ] η ∗ ⋄ η B )= φ ( T A ) + φ ( T B ) + φ ( ηA T ∗ ) + φ ( ηB T ∗ ) . Claim 8 φ ([ I, T − η ] η ∗ ⋄ η ( A + B )) = φ ([ I, T − η ] η ∗ ⋄ η A ) + φ ([ I, T − η ] η ∗ ⋄ η B ) forall T ∈ A and A, B ∈ A . Let A = P i,i =1 A ij and B = P i,i =1 B ij , it follows from Claims 5, 6 and 7that φ ([ I, T − η ] η ∗ ⋄ η ( A + B ))= φ ( T A + T B + T A + T B + ηA T ∗ + ηB T ∗ + ηA T ∗ + ηB T ∗ )= φ ( T A + T B ) + φ ( T A + T B + ηA T ∗ + ηB T ∗ )+ φ ( ηA T ∗ + ηB T ∗ )= φ ( T A ) + φ ( T B ) + φ ( T A ) + φ ( T B )+ φ ( ηA T ∗ ) + φ ( ηB T ∗ ) + φ ( ηA T ∗ ) + φ ( ηB T ∗ )= φ ( T A ) + φ ( T A + ηA T ∗ ) + φ ( ηA T ∗ )+ φ ( T B ) + φ ( T B + ηB T ∗ ) + φηB T ∗ )= φ ( T A + T A + ηA T ∗ + ηA T ∗ )+ φ ( T B + T B + ηB T ∗ + ηB T ∗ )= φ ([ I, T − η ] η ∗ ⋄ η A ) + φ ([ I, T − η ] η ∗ ⋄ η B ) . Claim 9 φ is additive.Let A and B be in A . To show that φ ( A + B ) = φ ( A ) + φ ( B ) , we choose X ∈ A such that φ ( X ) = φ ( A ) + φ ( B ) . For any T ∈ A , it follows fromClaim 8 and Eq. (3) that φ ([ I, T − η ] η ∗ ⋄ η X ) = φ ([ I, T − η ] η ∗ ⋄ η A ) + φ ([ I, T − η ] η ∗ ⋄ η B )16 φ ([ I, T − η ] η ∗ ⋄ η ( A + B )) , which indicates that[ I, T − η ] η ∗ ⋄ η X = [ I, T − η ] η ∗ ⋄ η ( A + B ) . Thus we get T ( X − A − B ) − η ( X − A − B ) T ∗ = 0 . By i T replacing T inthe above equation, we get T ( X − A − B ) + η ( X − A − B ) T ∗ = 0 and hence T ( X − A − B ) = 0 for all T ∈ A , this indicates that X − A − B = 0 . Weconcur that φ is additive, and the proof of Theorem 3.1 is completed. Proof of Theorem 3.2
In the rest of this section, we will deal with twocases.
Case 1 | η | = 1 . For all
A, B, C ∈ A , it follows from φ ([ A, B ] η ∗ ⋄ η C ) = [ φ ( A ) , φ ( B )] η ∗ ⋄ η φ ( C )that φ ( ABC − ηBA ∗ C + ηCB ∗ A ∗ − | η | CAB ∗ )= φ ( A ) φ ( B ) φ ( C ) − ηφ ( B ) φ ( A ) ∗ φ ( C )+ ηφ ( C ) φ ( B ) ∗ φ ( A ) ∗ − | η | φ ( C ) φ ( A ) φ ( B ) ∗ . (18)Replacing η with − η in Eq. (18), we obtain φ ( ABC + ηBA ∗ C − ηCB ∗ A ∗ − | η | CAB ∗ )= φ ( A ) φ ( B ) φ ( C ) + ηφ ( B ) φ ( A ) ∗ φ ( C ) − ηφ ( C ) φ ( B ) ∗ φ ( A ) ∗ − | η | φ ( C ) φ ( A ) φ ( B ) ∗ . (19)It follows from Eqs. (18)–(19) and Theorem 3.1 that φ ( ABC − CAB ∗ ) = φ ( A ) φ ( B ) φ ( C ) − φ ( C ) φ ( A ) φ ( B ) ∗ . (20)Taking A = I in Eq. (20), since φ ( I ) = I, we obtain φ ( BC − CB ∗ ) = φ ( B ) φ ( C ) − φ ( C ) φ ( B ) ∗ for all B, C ∈ A . Based on the result of [9], φ is a ∗ -ring isomorphism. Case 2 | η | 6 = 1 . Take α = −| η | . Then we obtain α (1 − | η | ) = 1 and α = 0 . lam 2.1 φ ( αI ) = αI and φ preserves self-adjoint elements in both direc-tions.Since I = φ ([ αI, I ] η ∗ ⋄ η I ) = [ φ ( αI ) , I ] η ∗ ⋄ η I = (1 − | η | ) φ ( αI ) = α φ ( αI ) , we have φ ( αI ) = αI. Let A ∈ A such that A = A ∗ . Then φ ( A ) = φ ([ αI, A ] η ∗ ⋄ η I ) = [ αI, φ ( A )] η ∗ ⋄ η I )= ( α − αη ) φ ( A ) + ( αη − α | η | ) φ ( A ) ∗ . Therefore φ ( A ) ∗ = 1 − α + αηαη − α | η | φ ( A ) = φ ( A ) , which indicates the sufficiency. The necessity can be obtained by considering φ − . Clam 2.2 α φ ( αP ) is a projection in B if and only if P is a projection in A . Since ( αP ) ∗ = αP, it follows from Claim 2.1 that φ ( αP ) ∗ = φ ( αP ) . Thus α φ ( αP ) is self-adjoint. Moreover, φ ( αP ) = φ ([ αP, I ] η ∗ ⋄ η αP )= [ φ ( αP ) , I ] η ∗ ⋄ η φ ( αP ) = (1 − | η | ) φ ( αP ) = 1 α φ ( αP ) and ( α φ ( αP )) = α φ ( αP ) . So α φ ( αP ) is a projection, which shows the suffi-ciency. The necessity can be proved by considering φ − . Clam 2.3
Let i, j ∈ { , } with i = j. Then φ ( A ij ) = B ij . Choose a projection Q i ∈ B , i = 1 , , by Clam 2.2, we have P i = α φ − ( αQ i )is projection in A . It is easy to see that Q i = α φ ( αP i ) . For any A ij ∈ A ij , since φ ( α (1 − η ) A ij ) = φ ([ I, αP i ] η ∗ ⋄ η A ij )= [ I, αQ i ] η ∗ ⋄ η φ ( A ij ) = α (1 − η ) Q i φ ( A ij ) + αη (1 − η ) φ ( A ij ) Q i , we obtain Q j φ ( α (1 − η ) A ij ) Q j = 0 . In the same manner, we obtain Q i φ ( αη ( η − A ij ) Q i = 0 . Since A ij is arbitrary, we obtain φ ( A ij ) = B ij + B ji for some B ij ∈ B ij and B ji ∈ B ji . Because 18 = φ ([ I, A ij ] η ∗ ⋄ η αP i )= [ I, φ ( A ij )] η ∗ ⋄ η αQ i ) = α (1 − η ) B ji + αη (1 − η ) B ∗ ji and α (1 − η ) = 0 , we obtain B ji = 0 , this indicates φ ( A ij ) ∈ B ij . By considering φ − , we get φ ( A ij ) = B ij . Clam 2.4 φ ( A ii ) ⊆ B ii , i = 1 , . Let A ii ∈ A ii and i = j, we have0 = φ ([ I, αP j ] η ∗ ⋄ η A ii )= [ I, αQ j ] η ∗ ⋄ η φ ( A ii ) = α (1 − η ) Q j φ ( A ii ) + αη (1 − η ) φ ( A ii ) Q j , which indicates that Q j φ ( A ii ) Q i = Q i φ ( A ii ) Q j = 0 and φ ( A ii ) = B ii + B jj forsome B ii ∈ B ii and B jj ∈ B jj . Let T ij ∈ B ij and i = j. It follows from Claim 2.3 that φ − ( T ij ) ∈ A ij , thus0 = φ ([ I, φ − ( T ij )] η ∗ ⋄ η A ii )= [ I, T ij ] η ∗ ⋄ η φ ( A ii ) = (1 − η ) T ij B jj + η (1 − η ) B jj T ∗ ij , which indicates that B jj = 0 . So we have φ ( A ii ) = B ii ⊆ B ii . Clam 2.5 φ ( AB ) = φ ( A ) φ ( B ) for all A, B ∈ A . To prove φ ( AB ) = φ ( A ) φ ( B ) , we need to consider that φ ( A ij B kl ) = φ ( A ij ) φ ( B kl ) for any i, j, k, l ∈ { , } . If j = k, it follows from Claims 2.3and 2.4 that φ ( A ij B kl ) = φ ( A ij ) φ ( B kl ) = 0 , then we just need to prove thecases with j = k. It follows from φ ( B ij ) φ ( A ii ) ∗ = 0 that φ ( A ii B ij ) + φ ( ηB ∗ ij A ∗ ii ) = φ ([ A ii , B ij ] η ∗ ⋄ η I )= [ φ ( A ii ) , φ ( B ij )] η ∗ ⋄ η I ) = φ ( A ii ) φ ( B ij ) + ηφ ( B ij ) ∗ φ ( A ii ) ∗ . By Claims 2.3 and 2.4, we obtain φ ( A ii B ij ) = φ ( A ii ) φ ( B ij ) . For any T ij ∈ B ij , i = j, we have C ij = φ − ( T ij ) ∈ A ij by Clam 2.3. So φ ( A ii B ii ) T ij = φ ( A ii B ii C ij ) = φ ( A ii ) φ ( B ii C ij ) = φ ( A ii ) φ ( B ii ) T ij . By the primeness of B , we obtain that φ ( A ii B ii ) = φ ( A ii ) φ ( B ii ) . Since φ ( B ji ) φ ( A ij ) ∗ = 0 , we have 19 ( A ij B ji ) + φ ( ηB ∗ ji A ∗ ij ) = φ ([ A ij , B ji ] η ∗ ⋄ η I )= [ φ ( A ij ) , φ ( B ji )] η ∗ ⋄ η I ) = φ ( A ij ) φ ( B ji ) + ηφ ( B ji ) ∗ φ ( A ij ) ∗ . which indicates that φ ( A ij B ji ) = φ ( A ij ) φ ( B ji ) . For any T ji ∈ B ji , i = j, we have C ji = φ − ( T ji ) ∈ A ji . So φ ( A ij B jj ) T ji = φ ( A ij B jj C ji ) = φ ( A ij ) φ ( B jj C ji ) = φ ( A ij ) φ ( B jj ) T ji . This indicates that φ ( A ij B jj ) = φ ( A ij ) φ ( B jj ) . Combining Cases 1–2, similar to the case 1 of Theorem 2.1, the proof ofTheorem 3.2 is finished.
References [1] Huo DH, Zheng BD, Liu HY. Nonlinear maps preserving Jordan triple η - ∗ -products. J. Math. Anal. Appl. 2015;430:830-844.[2] Li CJ, Chen QY, Wang T. Nonlinear maps preserving the Jordan triple ∗ -product on factor von Neumann algebras. Chin. Ann. Math. Ser. B.2018;39(4):633-642.[3] Zhao FF, Li CJ. Nonlinear maps preserving the Jordan triple ∗ -product betweenfactors. Indagationes Mathematicae. 2018;29:619-627.[4] Liang YX, Zhang JH. Nonlinear mixed Lie triple derivable mappings onfactor von Neumann algebras. Acta Mathematica Sinica, Chinese Series.2019;62(1):13-24.[5] Halmos PR. A Hilbert Space Problem Book, 2nd ed. Springer-Verlag, New York,Heideberg, Berlin. 1982.[6] Dixmier J. Von Neumann Algebras. North-Holland Publishing Company. 1981.[7] Zhang JH, Zhang FJ. Nonlinear maps preserving Lie products on factor vonNeumann algebras. Linear Algebra Appl. 2008;429:18-30.[8] Li CJ, Lu FY, Fang XC. Nonlinear maps preserving product XY + Y X ∗ onfactor von Neumann algebras. Linear Algebra Appl. 2013;438:2339-2345.[9] Zhang FJ. Nonlinear preserving product XY − ξY X ∗ on prime ∗ -rings. ActaMathematica Sinica, Chinese Series. 2014;57(4):775-784.[10] Breˇsar M, Foˇsner A. On ring with involution equipped with some new product.Publ. Math. Debrecen. 2000;57:121-134.[11] Moln´ar L. A condition for a subspace of B ( H ) to be an ideal. Linear AlgebraAppl. 1996;235:229-234.
12] ˇSemrl P. On Jordan ∗ -derivations and an application. Colloq. Math.1990;59:241-251.[13] ˇSemrl P. Quadratic functionals and Jordan ∗ -derivations. Studia Math.1991;97:157-165.[14] ˇSemrl P. Quadratic and quasi-quadratic functionals. Proc. Amer. Math. Soc.1993;119:1105-1113.-derivations. Studia Math.1991;97:157-165.[14] ˇSemrl P. Quadratic and quasi-quadratic functionals. Proc. Amer. Math. Soc.1993;119:1105-1113.