On symmetry of strong Birkhoff orthogonality in B ( H , K ) and K ( H , K )

In this paper, complete characterizations of left (or right) symmetric points for strong Birkhoff orthogonality in B ( H , K ) and K ( H , K ) are given, where H , K are complex Hilbert spaces and B ( H , K ) ( K ( H , K ) ) is the space of all bounded linear (compact) operators from H into K .


Introduction
This paper is concerned with a strengthened version of Birkhoff orthogonality in Hilbert C * -modules B(H, K) and K (H, K), where H, K are complex Hilbert spaces and B(H, K) (K (H, K)) is the space of all bounded linear (compact) operators from H into K. A Hilbert C * -module M over a C * -algebra A is the norm completion of an inner product A-module which is a complex vector space endowed with the norm x = x, x 1/2 given by an A-valued inner product ·, · which is compatible with a right A-module structure M × A (x, a) → xa ∈ M, where ·, · satisfies the following properties:  The main object of this paper is Birkhoff orthogonality which was first introduced by Birkhoff [8] and given many important properties by James [14,15]. Definition 1 Let X be a Banach space over the scalar field K. Then x ∈ X is said to be Birkhoff orthogonal to y ∈ X , denoted by x ⊥ B y, if x + λy ≥ x for each λ ∈ K. This is viewed as a generalization of the usual orthogonality relation in Hilbert spaces defined by using inner products since they are equivalent to each other in Hilbert spaces; and is known as one of the most important generalized orthogonality relations in Banach spaces because of its geometric meaning which is closely related to the best approximation in norms and support hyperplanes of unit balls. Indeed, if x is a unit vector of a Banach space X and y ∈ X , then x ⊥ B y means that the straight line {x + λy : λ ∈ K} is tangent to the unit ball of X at x.
From the definition, Birkhoff orthogonality is obviously homogeneous; that is, if x ⊥ B y then αx ⊥ B β y for each scalars α, β. However, the relation ⊥ B is not symmetric in general. Indeed, James [15] proved that if x ⊥ B y implies y ⊥ B x in a Banach space X with dim X ≥ 3, then X is a Hilbert space. See also the book of Amir [2] for related topics. The survey by Alonso-Martini-Wu [1] provides a good exposition of basic and important results on Birkhoff orthogonality. Although Birkhoff orthogonality is rarely symmetric in the global sense, it can have some interesting properties for local symmetry. The following definitions (originally for Birkhoff orthogonality) were given by Sain [25] and Sain-Ghosh-Paul [26]. Definition 2 Let X be a Banach space, and let ⊥ be a generalized orthogonality relation in X .
In terms of the preceding definition, Turnšek [29,30] showed that the set of all right symmetric points for ⊥ B in B(H) coincides with that of scalar multiples of isometries or coisometries on H, while there is no nonzero left symmetric point for where H is a complex Hilbert space. Since then, many results on local symmetry of Birkhoff orthogonality have been published, especially, in the fields of Banach (or Hilbert) space operators; see [11,13,21,26] for results on spaces of bounded linear operators between Banach spaces, and [3,7,12,[22][23][24][25]27,28] for important techniques about Birkhoff orthogonality in such spaces. See also [18][19][20] for developments in the setting of operator algebras. Similar investigations have been carried out for approximate symmetry of Birkhoff orthogonality in [10]. The aim of this paper is to advance the study of local symmetry for a strengthened version of Birkhoff orthogonality. The following definition was given by Arambašić and Rajić [4].

Definition 3
Let M be a Hilbert C * -module over a C * -algebra A, and let x, y ∈ M. Then x is said to be A-strongly Birkhoff orthogonal to y, denoted by x ⊥ A y, if x ⊥ B ya for each a ∈ A.
Basics for this generalized orthogonality relation are found in [4][5][6]. In particular, we have x, y = 0 ⇒ x ⊥ A y ⇒ x ⊥ B y in Hilbert C * -modules over A; see [4, page 112]. For local symmetry of strong Birkhoff orthogonality, in the setting of von Neumann algebras R (a very special case of Hilbert C * -modules), it is known that an element A ∈ R is left symmetric for ⊥ R in R if and only if |A| is a scalar multiple of a minimal projection in R; while A is right symmetric for ⊥ R in R if and only if it is right invertible in R; see [18]. However, we have not known complete characterizations of left (or right) symmetric points for strong Birkhoff orthogonality in general Hilbert C * -modules.
In this paper, taking a step forward, we present complete characterizations of left (or right) symmetric points for strong Birkhoff orthogonality in B(H, K) and K (H, K), which may provide some hints for the general case. It is shown that an element S in B(H,

Preliminaries
Throughout this paper, the term "Hilbert space" always means a nontrivial complex Hilbert space. If H is a Hilbert space, its inner product is denoted by ·, · . An element For basics about Hilbert space operators, the readers are referred to, for example, the books of Blackadar [9] and Kadison and Ringrose [16,17]. Let H, K be Hilbert spaces. Let A be a C * -algebra. Then A is said to be unital if it has the multiplicative unit. If A is non-unital, there exists a unital C * -algebra A I , called the unitization of A, such that A is an ideal of A I with dim(A I /A) = 1. Let a ∈ A. Then a is self-adjoint if a * = a; and is positive if it is self-adjoint and has the nonnegative spectrum in in which case, A and B are * -isomorphic. It is well-known that each C * -algebra is * -isomorphic to a C * -subalgebra of B(H) for some Hilbert space H (by the Gelfand-Naimark-Segal theorem); see [16,Theorem 4.5.6]. Another important fact is that if A is an abelian A von Neumann algebra R acting on a Hilbert space H is a C * -subalgebra of B(H) that is closed with respect to the weak-operator topology defined as the weak topology induced by the family of functionals on B(H) of the form A → Ax, y . A bit stronger one, the strong-operator topology, is also important.
with U * U = I H ; and a coisometry is the adjoint of an isometry, that is, an operator U ∈ B(H, K) with UU * = I K . These notions are special cases of partial isometries.
Moreover, V * is also a partial isometry with the initial projection V V * and the final projection V * V .
We recall that each T ∈ B(H, K) has the polar decomposition T = U |T | = |T * |U , where |T | = (T * T ) 1/2 and U is a partial isometry from the range projection of |T | onto the range projection of T , where the range projection of T is defined to be the projection from H onto T (H); see [

Left symmetric points
In this section, we provide characterizations of left symmetric points for ⊥ B(H) (⊥ K (H) ) in B(H, K) (K (H, K) If S ∈ M is left symmetric for ⊥ A in M, then S is rank one.
We are now ready to provide characterizations of left symmetric points for strong Birkhoff orthogonality in B(H, K) and K (H, K). It will turn out that the converse of the preceding lemma holds true. The proof to be given here is essentially due to Arambašić and Rajić

Suppose that S ∈ M is nonzero. Then S is left symmetric for ⊥ A in M if and only if S is rank one.
Proof The "only if" part is completed by Lemma 1. We shall show the "if" part. Suppose that S is rank one. Let T be an element of M such that S ⊥ A T . Then, by [6, Lemma 2.1 (6)], one has S * S ⊥ A S * T in A; and then S * S ⊥ B S * T (T * S) in A. On the other hand, the rank one operator S is represented by Sx = x, x 0 y 0 for some x 0 , y 0 ∈ H. It is easy to check that S * x = x, y 0 x 0 for each x ∈ H, and that S * T T * S = aS * S, where a = y 0 −2 T * y 0 2 . Hence it follows from S * S ⊥ B aS * S and S * S = 0 that a = 0; which implies that S * T T * S = 0 and ( S, T =)S * T = 0. Thus T ⊥ A S; and, S is left symmetric for ⊥ A in A.
The following are immediate consequences of Theorem 2. symmetric in B(H, K).

Corollary 2 Let H, K be Hilbert spaces, and let S ∈ B(H, K) be nonzero. Suppose
Proof Since all S, T ∈ B(H, K) are at most rank one, S ⊥ B(H) T always implies that T ⊥ B(H) S by Theorem 2.
It should be mentioned that Corollary 2 is a special case of [6, Theorem 2.6].

Right symmetric points
Our next aim is to characterize right symmetric points for strong Birkhoff orthogonality in B(H, K) and K (H, K). The following elementary facts are needed in the sequel. The proof can be essentially found, for example, in [16, Proposition 2.5.13].

(ii) S(H) = K if and only if |S * | is injective.
One of the most important arguments in our study of right symmetry is the following.

Suppose that S ∈ M is nonzero. If |S| = S I H and S(H) = K, then S is not right symmetric for ⊥ A in M.
Proof We may assume that S = 1. Let A be the von Neumann algebra generated by |S| and I H . Then there exists a Stonean space K such that A is * -isomorphic to ∈ (0, 1). Let e be the characteristic function of the clopen set O, and let E = ϕ −1 (e). Next, pick an arbitrary unit vector for each x ∈ H; which implies that T = T x 0 = 1. From this, we obtain T ⊥ A S since for each R ∈ A and each λ. On the other hand, since it follows that S ⊥ B T ; and then S ⊥ A T . Thus S is not right symmetric for ⊥ A in M. S of B(H, K) is said to be right invertible if ST = I K for some T ∈ B(K, H). Typical examples are given by coisometries in B(H, K). B(H, K) be nonzero. Suppose that either of the following statements holds:

Then S is right symmetric for ⊥ B(H) in B(H, K) if and only if S is right invertible.
Proof Suppose that S is right symmetric for ⊥ B (H) in B(H, K). We first show that the existence of such an S assures that (i) ⇒ (ii). To see this, it is enough to show that Suppose to the contrary that S(H) = K. Then S is an isometry by Lemma 3; which implies that K is also infinite dimensional. Let (e a ) be an orthonormal basis for H. Then  B(H, K). However, then, one has that S ⊥ B U (U * S). Since UU * (K) = U (H), we derive that UU * S = S; which implies that S = 0, a contradiction. Therefore From what we have shown in above, in either case, dim H ≥ dim K holds. By considering an injection from an orthonormal basis for K into that of H, we can construct an isometry V ∈ B (K, H). We shall show that k 0 = inf{ S * y : y ∈ K , y = 1} > 0. For this purpose, suppose to the contrary that k 0 = 0. Take a sequence (y n ) of unit vectors in K satisfying S * y n → 0. Since V is an isometry, it follows that for each R ∈ B(H) and each λ. Therefore V * ⊥ B(H) S. However, then, S ⊥ B(H) V * since S is right symmetric for ⊥ B (H) in B(H, K). In particular, S ⊥ B V * (V S) with V * S ∈ B(H); which implies that S = 0, a contradiction. Hence we have k 0 > 0. Now we have S * y ≥ k 0 y for each y ∈ K. From this, S * is injective and has the closed range, that is, S * (K) = S * (K). In particular, S * can be viewed as an isomorphism between K and S * (K). Let S 0 ∈ B(S * (K), K) be such that S 0 S * = I K , and let E be a projection from H onto S * (K). Putting T = S 0 E ∈ B(H, K) yields that T S * = S 0 E S * = S 0 S * = I K . Thus ST * = I K holds; and S is right invertible.
Conversely, if ST = I K for some T ∈ B (K, H), then R ∈ B(H, K) and R ⊥ B(H) S imply that R ⊥ B S(T R). This shows that R = 0; and hence, S ⊥ B(H) R. The proof is complete.
Hence the problem remains only in the case that H is finite dimensional and dim H < dim K. In this case, we have a consequence that is natural but completely different from that of Theorem 3. We note that the case dim H = 1 was already completed in Corollary 2. On the other hand, if H is nonzero finite dimensional and dim H < dim K, then each S ∈ B(H, K) always has a nonzero element T ∈ B(H, K) satisfying T ⊥ B(H) S. Indeed, as in the first paragraph of the proof of Theorem 4, we have dim S(H) < dim K. Let F be the projection from K onto S(H), and let y 0 be an element of (I − F)(K) with y 0 = 1. Fix an arbitrary x 0 ∈ H with x 0 = 1. Then T x = x, x 0 y 0 defines a nonzero element of B (H, K) satisfying T x, Sx = 0 for each x ∈ H (that is, S * T = 0). Thus we obtain T ⊥ B(H) S.
We finally consider right symmetric points for ⊥ K (H) in K (H, K). Since K (H, K)  For the converse, suppose that S(H) = K. Then |S * | is injective by Lemma 2; which implies that |S * |y, y > 0 for each nonzero y ∈ K. Let H be a nonzero positive element of K (H). Since 0 ≤ |S * | ≤ I K , at least, we have H − |S * |H ≤ I K − |S * | H ≤ H . Suppose that H − |S * |H = H . Then there exists a sequence (y n ) of unit vectors in K such that (H − |S * |H )y n → H . Since H − |S * |H ∈ K (K), we obtain a subsequence (y n k ) of (y n ) with the property that ((H − |S * |H )y n k ) converges to some z 0 ∈ K. On the other hand, the reflexivity of K generates a subsequence (y n k l ) of (y n k ) that converges weakly to some y 0 ∈ K with y 0 ≤ 1. Since bounded linear operators are weak to weak continuous, we drive that ((H − |S * |H )y n k l ) converges weakly to (H − |S * |H )y 0 . Hence one has that (H − |S * |H )y 0 = z 0 , and that This proves that there is no nonzero element T ∈ K (H, K) satisfying T ⊥ K (H) S; and thus, S is right symmetric for ⊥ K (H) in K (H, K). This completes the proof.  Conversely, we assume that K is separable. Let (e a ) and ( f n ) n∈N be orthonormal bases for H and K, respectively, and let (e a n ) n∈N be a countably infinite subset of (e a ). Define an operator S ∈ K (H, K) by letting Sx = n∈N 2 −n x, e a n f n for each x ∈ H. Then we have S(H) = K since { f n : n ∈ N} ⊂ S(H). Therefore, by Theorem 5, S is a (nonzero) right symmetric point for ⊥ K (H) in K (H, K).