MAPS PRESERVING ZERO JORDAN PRODUCTS ON HERMITIAN OPERATORS

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1 Illinois Journal of Mathematics Volume 49, Number 2, Summer 2005, Pages S MAPS PRESERVING ZERO JORDAN PRODUCTS ON HERMITIAN OPERATORS MIKHAIL A. CHEBOTAR, WEN-FONG KE, AND PJEK-HWEE LEE Abstract. Let H be a separable complex Hilbert space and B s(h) the Jordan algebra of all Hermitian operators on H. Let θ : B s(h) B s(h) be a surjective R-linear map which is continuous in the strong operator topology such that θ(x)θ(y)+θ(y)θ(x) = 0 for all x, y B s(h) with xy + yx = 0. We show that θ(x) = λuxu for all x B s(h), where λ is a nonzero real number and u is a unitary or anti-unitary operator on H.. Introduction Given any (associative) ring R we can render it into a Jordan ring by defining, for any two elements a, b R, the Jordan product a b = ab+ba. An additive map ϕ : R R of rings is called a Jordan homomorphism if ϕ(a b) = ϕ(a) ϕ(b) for all a, b R. In the case that 2 is invertible in R in the sense that 2x = a has a unique solution in R for every a R, this condition is equivalent to ϕ(a 2 ) = ϕ(a) 2 for all a R. Obviously, (associative) homomorphisms and anti-homomorphisms are Jordan homomorphisms. Jordan homomorphisms have been thoroughly investigated in the literature [], [2], [5], [8], [9], [], [22]. The results proved usually read that a Jordan homomorphism must be a homomorphism or an anti-homomorphism. For instance, Herstein [5] showed that a Jordan homomorphism of any ring onto a prime ring is either a homomorphism or an anti-homomorphism. In the case that the ring R is endowed with an involution a a, the set S = S(R) = {a R a = a} of all symmetric elements of R is itself a Jordan ring. In [0] Jacobson and Rickart proved that, given a matrix ring R = M n (A), where A is a ring and n 3, with a canonical involution such that the symmetric elements are trace-valued, any Jordan homomorphism of S can be lifted to a homomorphism of R in a unique way. At his 96 AMS talk [6] Herstein posed the problem of characterizing the Jordan homomorphisms of the symmetric elements of a simple ring with involution. This problem was Received October 4, 2004; received in final form December 9, Mathematics Subject Classification. Primary 47B48. Secondary 6W0, 7C c 2005 University of Illinois

2 446 M.A. CHEBOTAR, W.-F. KE, AND P.-H. LEE solved by Martindale [2] in the presence of orthogonal idempotents. Using Zelmanov s brilliant work on prime Jordan algebras [23], an idempotent-free solution was later obtained by McCrimmon [4] and Martindale [3]. Let H be a complex Hilbert space and B(H) the algebra of all bounded linear operators on H with the adjoint map a a. The Jordan subalgebra B s (H) = S(B(H)) of all Hermitian operators on H (i.e., the set of all bounded observables) plays an important role in the mathematical description of quantum mechanics. As with any algebraic structure the study of the automorphisms of this algebra is of considerable importance. It is well-known (see, for example, [6, p. 99]) that every Jordan automorphism on B s (H) is of the form x uxu, where u is a unitary or anti-unitary operator on H. A unified treatment of the Jordan automorphisms is presented in [4]. In many cases maps on B s (H) defined by certain local properties or preserving certain properties are exactly Jordan automorphisms [6], [7], [8], [9]. In this note we shall show that Jordan automorphisms can be characterized by their action on zero Jordan products. More precisely, we show that a surjective R-linear map θ : B s (H) B s (H) which is continuous in the strong operator topology and preserves zero Jordan products must be of the form x λϕ(x) where λ is a nonzero real number and ϕ is a Jordan automorphism of B s (H). Moreover, θ is a Jordan automorphism if, in addition, θ() =. 2. The results Let C be a commutative ring with 2 (i.e., C is a unital ring in which 2 is invertible), c c an automorphism of order or 2 on C, and R = M n (C), the n by n matrix algebra over C. We denote by e ij the matrix which has in the (i, j)-position and zeros elsewhere. Note that R can be equipped with the involution defined by a = a for a C and e ij = e ji. For simplicity we shall write a ij = ae ij for a C and put M = {e ii i n} {a ij + a ji a C, i j n}. Then the Jordan ring S = S(R) is the linear span of M over the subring F = {c C c = c} of C. Let J be an arbitrary Jordan algebra over F and θ : S J an F -linear map which preserves zero Jordan products, that is, θ(x) θ(y) = 0 whenever x, y S satisfy x y = 0. To begin with, we investigate the product θ(x) θ(y). A relation between θ(x y) and θ(x) θ(y) will be established for arbitrary x, y S. As a consequence, we see that the map θ which preserves zero Jordan products also preserves equal Jordan products, that is, θ(x) θ(y) = θ(u) θ(v) for all x, y, u, v S with x y = u v. Theorem 2.. Let S be the Jordan algebra defined above, J a Jordan algebra over F and θ : S J an F -linear map preserving zero Jordan products.

3 MAPS PRESERVING ZERO JORDAN PRODUCTS 447 Then (2.) θ(x) θ(y) = θ() θ(x y) 2 for all x, y S. In particular, we have θ(x) θ(y) = θ(u) θ(v) for all x, y, u, v S with x y = u v. Proof. Since S is spanned by M over F and θ is F -linear, it suffices to verify (2.) for x, y M. Note that (2.) is obvious if x y = 0. Thus we need only consider the following cases: (a) x = y = e ii. From (e ii ) e ii = 0 it follows that θ(e ii ) θ(e ii ) = 0 and hence (2.2) θ(e ii ) θ(e ii ) = θ() θ(e ii ). Now x y = 2e ii, so we have θ(x) θ(y) = θ() θ(e ii ) = θ() θ(x y). 2 (b) x = e ii, y = a ij + a ji, where i j and a C. From (e ii e jj ) (a ij +a ji ) = 0 it follows that θ(e ii e jj ) θ(a ij +a ji ) = 0, or equivalently, (2.3) θ(e ii ) θ(a ij + a ji ) = θ(e jj ) θ(a ij + a ji ). Since θ(e hh ) θ(a ij + a ji ) = 0 for h / {i, j}, this together with (2.3) yields n θ() θ(a ij + a ji ) = θ(e hh ) θ(a ij + a ji ) and hence h= = θ(e ii ) θ(a ij + a ji ) + θ(e jj ) θ(a ij + a ji ) = 2θ(e ii ) θ(a ij + a ji ), (2.4) θ(e ii ) θ(a ij + a ji ) = 2 θ() θ(a ij + a ji ). Now x y = a ij + a ji, so (2.4) means exactly θ(x) θ(y) = 2 θ() θ(x y). (c) x = a ij + a ji, y = b ij + b ji, where i j and a, b C. Let c = 2 (ab + ab) F. From it follows that (e ii + a ij + a ji e jj ) ( c ii + b ij + b ji + c jj ) = 0 θ(e ii + a ij + a ji e jj ) θ( c ii + b ij + b ji + c jj ) = 0. Expansion of the last identity yields A + B = 0, where A = θ(e ii e jj ) θ( c ii + c jj ) + θ(a ij + a ji ) θ(b ij + b ji ),

4 448 M.A. CHEBOTAR, W.-F. KE, AND P.-H. LEE and B = θ(a ij + a ji ) θ( c ii + c jj ), since θ(e ii e jj ) θ(b ij + b ji ) = 0 by (2.3). Replacing a by 2a, we get 2A + 4B = 0. Therefore, A = B = 0. By (2.2) it follows from A = 0 that θ(a ij + a ji ) θ(b ij + b ji ) = θ(e ii e jj ) θ(c ii c jj ) Now x y = 2c(e ii + e jj ), so we have = cθ(e ii e jj ) θ(e ii e jj ) = c(θ(e ii ) θ(e ii ) + θ(e jj ) θ(e jj )) = cθ() (θ(e ii ) + θ(e jj )). θ(x) θ(y) = cθ() (θ(e ii ) + θ(e jj )) = 2 θ() θ(2c(e ii + e jj )) = θ() θ(x y). 2 (d) x = a ij + a ji, y = b jk + b kj, where i, j, k are distinct and a, b C. From it follows that (a ij + [ab] ik + a ji + [ab] ki ) ( [bb] jj + b jk + b kj e kk ) = 0 θ(a ij + [ab] ik + a ji + [ab] ki ) θ( [bb] jj + b jk + b kj e kk ) = 0. Expansion of the last identity yields A + B = 0, where and A = θ(a ij + a ji ) θ(b jk + b kj ) θ([ab] ik + [ab] ki ) θ(e kk ), B = θ([ab] ik + [ab] ki ) θ(b jk + b kj ) θ(a ij + a ji ) θ([bb] jj ), since θ(a ij + a ji ) θ(e kk ) = 0 and θ([ab] ik + [ab] ki ) θ([bb] jj ) = 0. Replacing b by 2b, we get 2A + 4B = 0. Therefore, A = B = 0. By (2.4) it follows from A = 0 that θ(a ij + a ji ) θ(b jk + b kj ) = θ(e kk ) θ([ab] ik + [ab] ki ) = 2 θ() θ([ab] ik + [ab] ki ). Now x y = [ab] ik + [ab] ki, so the last identity means exactly θ(x) θ(y) = 2θ() θ(x y). Thus we have proved that θ(x) θ(y) = 2θ() θ(x y) for all x, y S. Therefore, if x, y, u, v S satisfy x y = u v, then it is obvious that θ(x) θ(y) = θ(u) θ(v). Using an argument similar to that for [3, Theorem 3.2], we may apply the preceding theorem to the case of Hermitian operators.

5 MAPS PRESERVING ZERO JORDAN PRODUCTS 449 Corollary 2.2. Let F s (H) be the Jordan algebra of all Hermitian operators of finite rank on a complex Hilbert space H, J a Jordan algebra over R and θ : F s (H) J an R-linear map preserving zero Jordan products. Then θ(x) θ(y) = θ(u) θ(v) for all x, y, u, v F s (H) with x y = u v. Proof. Fix x, y, u, v F s (H). There is a projection p F s (H) such that pxp = x, pyp = y, pup = u and pvp = v. Let {e, e 2,..., e n } be an orthonormal basis of the range of p and A the subalgebra of B(H) consisting of all operators a on H defined by a(w) = n i,j= α ij w e j e i for w H, where α ij C and w e j denotes the inner product of w and e j. Note that A is isomorphic to R = M n (C) via the isomorphism a n i,j= α ije ij and we may assume that x, y, u, v S = S(R), the Jordan algebra of all Hermitian matrices. Therefore for the restriction of θ to S we may apply Theorem 2.. Hence θ(x) θ(y) = θ(u) θ(v) for all x, y, u, v F s (H) with x y = u v and the proof is complete. We are now in a position to prove the main theorem of the present paper using some ideas from the proof of [3, Theorem 3.3]. Theorem 2.3. Let B s (H) be the Jordan algebra of all Hermitian operators on a separable complex Hilbert space H and θ : B s (H) B s (H) a surjective R-linear map which is continuous in the strong operator topology. Suppose that θ preserves zero Jordan products. Then λ = θ() is a nonzero real number and there exists a unitary or anti-unitary operator u on H such that θ(x) = λuxu Proof. First we show that for all x B s (H). (2.5) θ() θ(x y) = θ(x) θ(y) 2 for all x, y B s (H). Let x 0, y 0 F s (H). Since H is separable, F s (H) is dense in B s (H) in the strong operator topology. In particular, there exists a sequence {p n } in F s (H) converging to in the strong operator topology. We may assume further that p n (x 0 y 0 ) = (x 0 y 0 )p n = x 0 y 0 for all n. According to Corollary 2.2 the restriction of θ to F s (H) preserves equal Jordan products, so it follows from 2 p n (x 0 y 0 ) = x 0 y 0 that 2 θ(p n) θ(x 0 y 0 ) = θ(x 0 ) θ(y 0 ) for all n. Since θ is continuous in the strong operator topology, we obtain (2.6) 2 θ() θ(x 0 y 0 ) = θ(x 0 ) θ(y 0 ) by passing to the limit in the previous equation.

6 450 M.A. CHEBOTAR, W.-F. KE, AND P.-H. LEE Let x B s (H) and {x n } a sequence in F s (H) converging to x. It follows from (2.6) that 2 θ() θ(x n y 0 ) = θ(x n ) θ(y 0 ) for all n. Passing to the limits in this equation we get (2.7) 2 θ() θ(x y 0) = θ(x) θ(y 0 ). Finally, let y B s (H) and {y n } a sequence in F s (H) converging to y. It follows from (2.7) that 2 θ() θ(x y n) = θ(x) θ(y n ) for all n. Passing to the limits in this equation we obtain (2.5). For any projection p in B s (H) we have (2.8) θ() θ(p) = θ(p) θ(p) = 2θ(p) 2. Thus [θ() θ(p)]θ(p) = θ(p)[θ() θ(p)], and its expansion yields This together with (2.8) gives θ()θ(p) 2 = θ(p) 2 θ(). θ()[θ() θ(p)] = [θ() θ(p)]θ(). Since every element of B s (H) is an R-linear combination of projections [20, Theorem 3], we have for all x B s (H), and, a fortiori, θ()[θ() θ(x)] = [θ() θ(x)]θ() θ()[θ() θ(x y)] = [θ() θ(x y)]θ() for all x, y B s (H). In view of (2.5), we conclude that θ() commutes with θ(x) θ(y) for all x, y B s (H). Since θ : B s (H) B s (H) is surjective, this implies that θ() commutes with x y for all x, y B s (H), and, in particular, commutes with every projection p in B s (H). In other words, θ() is a central element in B s (H), that is, a real number. Note that θ() is nonzero by (2.5). Set λ = θ(). Then (2.5) reduces to λθ(x y) = θ(x) θ(y), and hence ϕ = λ θ is a surjective Jordan homomorphism on B s (H) which is also continuous in the strong operator topology. We claim that ϕ is injective. Suppose on the contrary that ϕ(a) = 0 for some nonzero a B s (H). Let J be the Jordan ideal of B s (H) generated by a. Then the restriction of ϕ to the Jordan ideal J is zero. Let I be a nonzero -ideal of B(H) such that J contains all the symmetric elements of I. (For instance, take I to be the ideal

7 MAPS PRESERVING ZERO JORDAN PRODUCTS 45 generated by all the x 4 for x J; see the proof of [7, Theorem 2.6].) Since the nonzero -ideal I contains F (H), the ideal of all finite-rank operators on H, we have F s (H) = S(F (H)) J and so ϕ is zero on F s (H). Recall that ϕ is continuous and F s (H) is dense in B s (H) in the strong operator topology, so we conclude that ϕ is zero on B s (H), a contradiction. Therefore ϕ is a Jordan automorphism. As we mentioned in the introduction, there exists a unitary or anti-unitary operator u on H such that ϕ(x) = uxu for all x B s (H) and the proof is thereby complete. Acknowledgement. We would like to thank the referee for her/his valuable comments. References [] G. Ancochea, Le théorème de von Staudt en géométrie projective quaternionienne, J. Reine Angew. Math. 84 (942), MR (5,72d) [2] G. Ancochea, On semi-automorphisms of division algebras, Ann. of Math. (2) 48 (947), MR (8,30c) [3] M. Brešar and P. Šemrl, Mappings which preserve idempotents, local automorphisms, and local derivations, Canad. J. Math. 45 (993), MR (94k:47054) [4] G. Cassinelli, E. De Vito, P. J. Lahti, and A. Levrero, Symmetry groups in quantum mechanics and the theorem of Wigner on the symmetry transformations, Rev. Math. Phys. 9 (997), MR (98k:8082) [5] I. N. Herstein, Jordan homomorphisms, Trans. Amer. Math. Soc. 8 (956), MR (7,938f) [6], Lie and Jordan structures in simple, associative rings, Bull. Amer. Math. Soc. 67 (96), MR (25 #3072) [7], Topics in ring theory, The University of Chicago Press, Chicago, Ill.-London, 969. MR (42 #608) [8] L.-K. Hua, On the automorphisms of a field, Proc. Nat. Acad. Sci. U.S.A. 35 (949), MR (0,675d) [9] N. Jacobson and C. E. Rickart, Jordan homomorphisms of rings, Trans. Amer. Math. Soc. 69 (950), MR (2,387h) [0], Homomorphisms of Jordan rings of self-adjoint elements, Trans. Amer. Math. Soc. 72 (952), MR (3,79b) [] I. Kaplansky, Semi-automorphisms of rings, Duke Math. J. 4 (947), MR (9,72e) [2] W. S. Martindale, Jordan homomorphisms of the symmetric elements of a ring with involution, J. Algebra 5 (967), MR (35 #636) [3], Jordan homomorphisms onto nondegenerate Jordan algebras, J. Algebra 33 (990), MR (9m:7045) [4] K. McCrimmon, The Zelmanov approach to Jordan homomorphisms of associative algebras, J. Algebra 23 (989), MR (90j:7053) [5] L. Molnár, The set of automorphisms of B(H) is topologically reflexive in B(B(H)), Studia Math. 22 (997), MR (98e:47068) [6], Local automorphisms of some quantum mechanical structures, Lett. Math. Phys. 58 (200), MR (2002k:47077) [7], Order-automorphisms of the set of bounded observables, J. Math. Phys. 42 (200), MR (2002m:802)

8 452 M.A. CHEBOTAR, W.-F. KE, AND P.-H. LEE [8], Conditionally multiplicative maps on the set of all bounded observables preserving compatibility, Linear Algebra Appl. 349 (2002), MR (2003c:47070) [9] L. Molnár and M. Barczy, Linear maps on the space of all bounded observables preserving maximal deviation, J. Funct. Anal. 205 (2003), MR (2004j:47074) [20] C. Pearcy and D. Topping, Sums of small numbers of idempotents, Michigan Math. J. 4 (967), MR (36 #2006) [2] B. Simon, Quantum dynamics: from automorphism to hamiltonian, Studies in Mathematical Physics, Essays in Honor of Valentine Bargmann, E. H. Lieb, B. Simon, A. S. Wightman, eds., Princeton Series in Physics, Princeton University Press, Princeton, New Jersey 976, pp [22] M. F. Smiley, Jordan homomorphisms onto prime rings, Trans. Amer. Math. Soc. 84 (957), MR (8,75b) [23] E. I. Zelmanov, Prime Jordan algebras. II, Sibirsk. Mat. Zh. 24 (983), 89 04, 92. MR (85d:70) M.A. Chebotar, Department of Mechanics and Mathematics, Tula State University, Tula, Russia address: mchebotar@tula.net Current address: Department of Management and Information Technology, Southern Taiwan University of Technology, Yung-Kang, Taiwan address: mchebotar@yahoo.com W.-F. Ke, Department of Mathematics, National Cheng Kung University, Tainan, Taiwan address: wfke@mail.ncku.edu.tw P.-H. Lee, Department of Mathematics, National Taiwan University, Taipei, Taiwan address: phlee@math.ntu.edu.tw