Some linear preserver problems on B(H) concerning rank and corank

arXiv:math/9809102v1 [math.OA] 18 Sep 1998 SOME LINEAR PRESERVER PROBLEMS ON B(H) CONCERNING RANK AND CORANK LAJOS MOLN´ AR Abstract. As a continuation of the work on linear maps between operator algebras which preserve certain subsets of operators with finite rank, or finite corank, here we consider the problem inbetween, that is, we treat the question of preserving operators with infinite rank and infinite corank. Since, as it turns out, in this generality our preservers cannot be written in a nice form what we have got used to when dealing with linear preserver problems, hence we restrict our attention to certain important classes of operators like idempotents, or projections, or partial isometries. We conclude the paper with a result on the form of linear maps which preserve the left ideals in B(H). 1. Introduction Linear preserver problems represent one of the most active research topics in matrix theory (see the survey paper [8]). In the last decade considerable attention has been also paid to similar questions in infinite dimension, that is, to linear preserver problems on operator algebras (see the survey paper [2]). In both cases, the problem is to characterize those linear maps on the algebra in question which leave invariant a given subset, or relation, or function. One of the most important such questions concerns the rank. This is because in many cases preserver problems can be reduced to the problem of rank preservers. Therefore, it is not surprising that a lot of work has been done on such preservers (see, for example, [1, 5] for the finite dimensional case and [7, 11] for the infinite dimensional case as well as the references therein). In our recent paper [6], we considered, among other things, the very similar problem of corank preservers which problem deserves attention, of course, only in the infinite dimensional case. If His a (complex) infinite dimensional Hilbert space, denote by B(H) the algebra of all bounded linear operators acting on H. The result [6, Theorem 3] reads as follows. Let 1991 Mathematics Subject Classification. Primary: 47B49. Key words and phrases. Linear preservers, partial isometries, idempotents, projections, one-sided ideals. This research was supported from the following sources: 1) Joint Hungarian-Slovene research project supported by OMFB in Hungary and the Ministry of Science and Technology in Slovenia, Reg. No. SLO-2/96, 2) Hungarian National Foundation for Scientific Research (OTKA), Grant No. T–016846 F–019322, 3) A grant from the Ministry of Education, Hungary, Reg. No. FKFP 0304/1997. 1 2 LAJOS MOLN´ AR φ:B(H)→B(H) be a bijective linear map which is weakly continuous on norm bounded sets. If φpreserves the corank-koperators in both directions, then there exist invertible operators A, B ∈B(H) such that φis of the form φ(T) = ATB (T∈B(H)). Now, it seems to be a natural question to consider the problem of such preservers which are ”inbetween” rank preservers and corank preservers, that is, to determine those linear maps which preserve the operators with infinite rank and infinite corank. We say that an operator A∈B(H) has infinite rank and infinite corank if the (Hilbert space) dimensions of rng A and rng A⊥are both infinite. Here, rng Astands for the range of A. We consider separable Hilbert spaces since in this case there is only one sort of infinite dimension. Unfortunately, the preservers above do not have such a nice form which we have got used to when dealing with linear preserver problems. Namely, there exist preservers of the above kind which cannot be expressed in terms of multiplications by fixed operators and, possibly, by transposition. To see this, let ψ:B(H)→B(H) be a linear map with norm less than 1 whose range consists of finite rank operators. Then it follows from a basic Banach algebra fact that the linear map φdefined by φ(T) = T−ψ(T) (T∈B(H)) is a bijection of B(H) onto itself, and it is easy to check that φpreserves the operators with infinite rank and infinite corank in both directions (observe that this map preserves the Fredholm index as well which preserver problem might also seem to be natural after discussing corank preservers). So, in order to have one of the desired nice forms for our preservers we should somehow modify the problem by, for example, restricting the set of operators with infinite rank and infinite corank which we want preserve. This is exactly what we are doing here considering the important sets of idempotents, projections and partial isometries, respectively. In the last result of the paper we describe the linear bijections of B(H) which preserve the left ideals in both directions. As it will be clear from the proof, this problem is also connected with the problem of rank preservers. Let us fix the concepts and notation that we shall use throughout. By a projection we mean a self-adjoint idempotent in B(H). An element W∈ B(H) is called a partial isometry if it is an isometry on a closed subspace of H and 0 on its orthogonal complement. Algebraically, Wcan be characterized by the equation W W ∗W=W. We say that the operators A, B ∈B(H) are orthogonal to each other if A∗B=AB∗= 0. This means that the ranges of Aand Bas well as the orthogonal complements of their kernels are orthogonal to each other. If x, y ∈H, then x⊗ydenotes the operator defined by (x⊗y)z=hz, yix(z∈H). In what follows F(H) stands for the ideal of all finite rank operators in B(H). SOME LINEAR PRESERVER PROBLEMS ON B(H) 3 2. Results We begin with the description of all linear bijections φof B(H) which preserve the partial isometries of infinite rank and infinite corank in both directions (this means that Wis a partial isometry with infinite rank and infinite corank if and only if so is φ(A)). Theorem 1. Let Hbe a separable infinite dimensional Hilbert space. Let φ: B(H)→B(H)be a linear bijection which preserves the partial isometries of infinite rank and infinite corank in both directions. Then there exist unitary operators U, V ∈B(H)such that φis either of the form φ(T) = UTV (T∈B(H)) or of the form φ(T) = UTtrV(T∈B(H)) where tr denotes the transpose with respect to an arbitrary but fixed complete orthonormal sequence in H. In the proof we shall use the following two auxiliary results. Lemma 1. Let T, S ∈B(H)be partial isometries with S=ST ∗S. Then we have TT∗S=Sand ST∗T=S. Proof. Denote Q=TS∗. Since SS∗and T∗Tare projections, we compute SS∗=ST∗SS∗T S∗=Q∗(SS∗)Q≤Q∗Q=S(T∗T)S∗≤SS∗. This implies Q∗Q=SS∗. In particular, we obtain kQk ≤ 1 (in fact, the norm of Qis either 0 or 1). But Qis an idempotent. Indeed, we have Q2=TS∗TS∗=T(ST∗S)∗=TS∗=Q. So, Qis a contractive idempotent. It is easy to see that this implies that Qis a self-adjoint idempotent, that is, a projection. To verify this, pick arbitrary elements x∈ker Qand y∈rng Q. Then we have kyk2≤ kµx +yk2(µ∈C). An elementary argument shows that this implies that x⊥y. Hence the kernel and the range of Qare orthogonal to each other and this verifies that Qis a projection. Now, from Q∗Q=SS∗we obtain Q=SS∗. Therefore, TS∗=SS∗and, as SS∗is the projection onto rng S, it follows that the range of Sis included in that of T. Since TT∗is the projection onto rng T, we have T T ∗S=S. Similarly, from the equality S∗=S∗TS∗one can deduce T∗TS∗=S∗which is equivalent to ST∗T=S. Lemma 2. Suppose that T, S ∈B(H)are partial isometries. The operator T+λS is a partial isometry for every λ∈Cwith |λ|= 1 if and only if T and Sare orthogonal to each other. 4 LAJOS MOLN´ AR Proof. Suppose first that (T+λS)(T+λS)∗(T+λS) = T+λS holds for every λ∈Cwith |λ|= 1. Using the fact that T, S are partial isometries, one can conclude that 0 = λ2ST ∗S+λ(TT∗S+ST∗T) + ¯ λTS∗T+SS∗T+TS∗S. Since this is valid for every λ∈Cof modulus 1, choosing the particular values λ= 1,−1, i, −i, it is easy to deduce that ST∗S= 0(1) TT∗S+ST ∗T= 0(2) TS∗T= 0(3) SS∗T+T S∗S= 0.(4) Multiplying (2) by T∗from the left and taking (3) into account, we obtain T∗S= 0. Similarly, multiplying (4) by S∗from the right and taking (1) into account, we have TS∗= 0. So, Tand Sare orthogonal. As for the reverse implication, if T, S are mutually orthogonal partial isometries, then it is just a simple calculation that T+λS is a partial isometry for every λ∈Cof modulus 1. Proof of Theorem 1. Let {x1,... ,xk} ⊆ Hand {y1,... ,yk} ⊆ Hbe two systems of pairwise orthogonal unit vectors. We claim that the image of the finite rank partial isometry R=Pk j=1 xj⊗yjunder φis also a finite rank partial isometry. Let (en) be an orthonormal sequence in the orthogonal complement of {x1,... ,xk}which generates a closed subspace of infinite codimension. Similarly, let (fn) be an orthonormal sequence in {y1,... ,yk}⊥. Denote U=Pnen⊗fnand let V=U+R. Clearly, Uand V are partial isometries of infinite rank and infinite corank. Moreover, for every λ∈Cof modulus 1, the operator R+λU = (V−U)+λU is also a partial isometry of infinite rank and infinite corank. Therefore, φ(V) + (λ−1)φ(U) is a partial isometry for every λ∈Cwith |λ|= 1. This means that with the notation V′=φ(V), U′=φ(U) we have (V′+ (λ−1)U′)(V′+ (λ−1)U′)∗(V′+ (λ−1)U′) = (V′+ (λ−1)U′) for every λ∈Cof modulus 1. Performing the above operations we obtain a polynomial in λ, ¯ λwith operator coefficients which equals 0 on the perimeter of the unit disc in the complex plane. Just as in the proof of Lemma 2, choosing the particular values λ= 1,−1, i, −iwe find that the coefficients of the polynomial in question are all 0. Therefore, we have −U′+U′V′∗U′= 0,(5) 2U′+V′V′∗U′+U′V′∗V′−U′U′∗V′−2U′V′∗U′−V′U′∗U′= 0,(6) SOME LINEAR PRESERVER PROBLEMS ON B(H) 5 U′+V′U′∗V′−U′U′∗V′−V′U′∗U′= 0,(7) and (8) −2U′−V′V′∗U′−V′U′∗V′−U′V′∗V′+ 2U′U′∗V′+U′V′∗U′+ 2V′U′∗U′= 0, where the left hand sides of (5), (6), (7), (8) are the coefficients of λ2,λ,¯ λ and 1, respectively. From (5) and (6) we deduce V′V′∗U′+U′V′∗V′=U′U′∗V′+V′U′∗U′.(9) We prove that φ(R) = V′−U′is a partial isometry. Indeed, we compute (V′−U′)(V′−U′)∗(V′−U′) =(10) V′−U′−V′V′∗U′−V′U′∗V′−U′V′∗V′+U′U′∗V′+U′V′∗U′+V′U′∗U′. From (9) we know that −V′V′∗U′−U′V′∗V′+U′U′∗V′+V′U′∗U′= 0. So, we have to show that V′−U′−V′U′∗V′+U′V′∗U′=V′−U′. By (5) we have U′V′∗U′=U′. It remains to verify that V′U′∗V′=U′. From (7) and (9) we infer that U′+V′U′∗V′−V′V′∗U′−U′V′∗V′= 0.(11) By Lemma 1 it follows that V′V′∗U′=U′and U′V′∗V′=U′. Now, (11) gives V′U′∗V′=U′. Consequently, the right hand side of the equation (10) is equal to V′−U′which verifies that φ(R) is a partial isometry. We next prove that φ(R) has finite rank. By the preserver property of φit is sufficient to prove that φ(R) has infinite corank. We have seen that for every λ∈Cof modulus 1, the operator R+λU is a partial isometry of infinite rank and infinite corank. This implies that for R′=φ(R), the operator R′+λU′is a partial isometry for every λ∈Cwith |λ|= 1. According to Lemma 2, we obtain that R′and U′are orthogonal to each other. Since the range of U′is infinite dimensional, it follows that R′is of infinite corank which implies that R′is a finite rank partial isometry. We next prove that φpreserves the partial isometries in general. To see this, let Wbe a partial isometry. If it is of finite rank, then there is now nothing to prove. So, let Wbe of infinite rank. In that case we have an orthogonal sequence (Wn) of partial isometries of infinite rank and infinite corank whose sum is W. By the preserver property of φit follows that the operators An=φ(W)−Pn+1 k=1 φ(Wk) = φ(W−Pn+1 k=1 Wk) and Bn=Pn k=1 φ(Wk) = φ(Pn k=1 Wk) are partial isometries. Because of the same reason, An+λBnis a partial isometry for every λ∈Cof modulus 1. By Lemma 2 this implies that Anand Bnare orthogonal to each other. The statement [10, Lemma 1.3] tells us that the series of pairwise orthogonal partial isometries is convergent in the strong operator topology and its sum is also a partial isometry. Consider the operators A=φ(W)−Pnφ(Wn) 6 LAJOS MOLN´ AR and B=Pnφ(Wn). By the just mentioned result Pφ(Wn)∗is strongly convergent as well, and Pnφ(Wn)∗=B∗. We then also have φ(W)∗− Pnφ(Wn)∗=A∗. Since Anand Bnare orthogonal for every n∈N, it is now easy to verify that Ais orthogonal to B. The operator Bis a partial isometry. As for A, we know that (An) strongly converges to A and, as we have seen, (A∗ n) strongly converges to A∗. It is well-known that the multiplication is strongly countinuous on the norm-bounded subsets of B(H). Consequently, we infer that (AnA∗ n) strongly converges to AA∗and then that (AnA∗ nAn) strongly converges to AA∗A. Since Anis a partial isometry for every n, we obtain that Ais also a partial isometry. Now, since φ(W) is the sum of the mutually orthogonal partial isometries Aand B, it follows that φ(W) is a partial isometry as well. We have assumed that φ−1has the same preserver properties as φ. Therefore, φpreserves the partial isometries in both directions. Suppose that Wis a maximal partial isometry, that is, suppose that Wis a partial isometry and there is no nonzero partial isometry which is orthogonal to W. If V∈B(H) is a nonzero partial isometry which is orthogonal to φ(W), then V+λφ(W) is a partial isometry for every λ∈Cwith |λ|= 1. This gives us that φ−1(V) + λW is also a partial isometry for every λ∈Cof modulus 1. By Lemma 2 this results in the orthogonality of φ−1(V) and Wwhich is a contradiction. Consequently, we obtain that φpreserves the maximal partial isometries which are precisely the isometries and the coisometries. It is wellknown that the set of all extreme points of the unit ball of B(H) consists of these operators exactly. So, φis a linear map on B(H) which preserves the extreme points of the unit ball. The form of all linear maps with this property acting on a von Neumann factor was determined in [9]. The result [9, Theorem 1] says that there is a unitary operator U∈B(H) such that either there exists a *-homomorphism ψ:B(H)→B(H) such that φ(T) = Uψ(T) (T∈B(H)) or there exists a *-antihomomorphism ψ′:B(H)→B(H) such that φ(T) = Uψ′(T) (T∈B(H)). Since our map φis bijective, the same must hold for the corresponding morphism ψor ψ′above. Now, referring to folk results on the form of *- automorphisms and *-antiautomorphisms of B(H), we conclude the proof. We continue with a result of the same spirit on idempotent preservers. Theorem 2. Let Hbe a separable infinite dimensional Hilbert space. Suppose that φ:B(H)→B(H)is a linear bijection which preserves the idempotents of infinite rank and infinite corank in both directions. Then there is an invertible operator A∈B(H)such that φis either of the form φ(T) = ATA−1(T∈B(H)) SOME LINEAR PRESERVER PROBLEMS ON B(H) 7 or of the form φ(T) = ATtrA−1(T∈B(H)). In the proof we shall use the following lemma which is certainly wellknown and is included here only for the sake of completeness. Lemma 3. If P, Q ∈B(H)are idempotents, then (i)P+Qis an idempotent if and only if PQ =QP = 0; (ii)P−Qis an idempotent if and only if PQ =QP =Q. Proof. It follows from elementary algebraic computations. Proof of Theorem 2. If P, Q ∈B(H) are idempotents, then we write P≤Q if PQ =QP =P. Clearly, this is equivalent to the condition that rng P⊆ rng Qand ker Q⊆ker P. Let us say that an idempotent P∈B(H) is regular if it has infinite rank and infinite corank. We prove that for any two regular idempotents P, Q we have P≤Qif and only if for every regular idempotent R∈B(H), if Q+Ris a regular idempotent, then so is P+R. The necessity is almost evident. To the sufficiency suppose first that rng P*rng Q. Let x∈Hbe such that Px =xand Qx 6=x. Choose a regular idempotent R≤I−Qfor which Q+Ris a regular idempotent and Rx 6= 0 (observe that (I−Q)x6= 0). Since P+Ris an idempotent, we have P R =RP = 0. It follows that 0 = RP x =Rx which is a contradiction. Hence, we have rng P⊆rng Q. The relation ker Q⊆ker Pcan be proved in a similar manner. Using the above characterization and the preserver property of φ, we obtain that φpreserves the relation ≤between regular idempotents. Now, if Ris a finite rank idempotent, then Rcan be written in the form R=Q−Pwith some regular idempotents P≤Q. Since φ(P)≤φ(Q), it follows that φ(R) = φ(Q)−φ(P) is also an idempotent. We prove that φ(R) is of finite rank. Choosing a regular idempotent Pwith R≤P, it follows that P−Ris a regular idempotent and hence φ(P)−φ(R) is also an idempotent. By Lemma 3 (ii) this implies that φ(R)≤φ(P). If φ(R) is not of finite rank, then it is regular which implies that Ris also regular and this is a contradition. Therefore, using the preserver properties of φand φ−1we obtain that φpreserves the finite rank idempotents in both directions. It is now easy to see that φis a linear bijection of F(H) onto itself. By Lemma 3 (i), for any idempotents R, R′∈F(H) we have RR′=R′R= 0 if and only if φ(R)φ(R′) = φ(R′)φ(R) = 0. Using this property it is easy to verify that φpreserves the rank-one idempotents in both directions. By [11, Theorem 4.4] we infer that there is an invertible bounded linear operator A∈B(H) such that φis either of the form φ(T) = ATA−1(T∈F(H)) or of the form φ(T) = ATtrA−1(T∈F(H)). 8 LAJOS MOLN´ AR Without loss of generality we may assume that φis of the first form and then that A=I. We intend to show that φ(T) = T(T∈B(H)). Let P∈B(H) be a regular idempotent. If Ris any finite rank idempotent with R≤P, then just as above, we obtain R=φ(R)≤φ(P). Since R≤Pwas arbitrary, it now follows that P≤φ(P). Since φ−1has the same preserver property as φ, it follows that P≤φ−1(P). But φpreserves the order between the regular idempotents. Hence, we have φ(P)≤P. Therefore, φ(P) = P for every regular idempotent P. Since every idempotent of finite corank is the sum of two regular idempotents, we obtain that φ(P) = Pholds for every idempotent P∈B(H). Since every element of B(H) is a finite linear combination of projections [4, Theorem 2], we conclude that φ(T) = Tis valid for every T∈B(H). This completes the proof. In a similar fashion one can verify the following result concerning projection preservers. Theorem 3. Let Hbe a separable infinite dimensional Hilbert space. Suppose that φ:B(H)→B(H)is a linear bijection which preserves the projections of infinite rank and infinite corank in both directions. Then there is a unitary operator U∈B(H)such that φis either of the form φ(T) = UTU∗(T∈B(H)) or of the form φ(T) = UTtrU∗(T∈B(H)). Our final result describes the linear bijections φof B(H) which preserve the left ideals in both directions (this means that L ⊆ B(H) is a left ideal if and only if φ(L) is a left ideal). As it will be clear from the proof, this problem is also connected with the problem of rank preservers. Theorem 4. Let Hbe a Hilbert space. Suppose that φ:B(H)→B(H)is a linear bijection preserving the left ideals of B(H)in both directions. Then there are invertible operators A, B ∈B(H)such that φis of the form φ(T) = ATB (T∈B(H)). Proof. The minimal left ideals of B(H) are precisely the sets {x⊗y:x∈H} for nonzero y∈H. Since φclearly preserves the minimal left ideals of B(H) in both directions, we easily deduce that φis a linear bijection of F(H) onto itself which preserves the rank-one operators. By [11, Theorem 3.3] (see also [7]) it follows that there are linear bijections A, B :H→Hsuch that φis either of the form φ(x⊗y) = Ax ⊗By (x, y ∈H)(12) or of the form φ(x⊗y) = Ay ⊗Bx (x, y ∈H). Since φis left ideal preserving, the second possibility above obviously cannot occur. SOME LINEAR PRESERVER PROBLEMS ON B(H) 9 We prove that φ(I) is invertible. First we note the following. It is true in any algebra with unit that an element fails to have a left inverse if and only if this element is included in a maximal left ideal (recall that every proper left ideal is included in a maximal left ideal). Therefore, φpreserves the left invertible elements of B(H) in both directions. We recall that an operator Sin B(H) is left invertible if and only if Sis injective and S has closed range. Now, let x, y ∈Hbe arbitrary nonzero vectors. Let λ∈C. By Fredholm alternative x⊗y−λI is injective if and only if it is surjective. This gives us that x⊗y−λI is left invertible if and only if it is invertible. Since the spectrum of any element in B(H) is nonempty, we infer that there is a λ∈Cfor which x⊗y−λI is not left invertible. Suppose that x⊗yis not quasinilpotent, that is, hx, yi 6= 0. Then the scalar λabove can be chosen to be nonzero. It follows that Ax ⊗By −λφ(I) is not left invertible. On the other hand, φ(I) is left invertible and hence it is a left Fredholm operator (see [3, 2.3. Definition, p. 356]). But any compact perturbation of a left Fredholm operator has closed range [3, 2.5. Theorem, p. 356]. So, the operator Ax ⊗By −λφ(I) is not left invertible but it has closed range. Therefore, this operator is not injective, that is, there exists a nonzero vector z∈Hsuch that λφ(I)z=hz, ByiAx. Clearly, this implies that Ax ∈rng φ(I). Since x∈Hwas arbitrary, we conclude that H= rng A⊆φ(I) which means that φ(I) is surjective. This gives us that φ(I) is invertible. We next show that the linear operators A, B in (12) are bounded. Let x, y ∈H. We have seen above that x⊗y−λI is not left invertible if and only if λ∈σ(x⊗y), where σ(.) denotes the spectrum. Similarly, by Fredholm alternative again, φ(I)−1(Ax ⊗By −λφ(I)) is not left invertible if and only if λ∈σ(φ(I)−1Ax ⊗By). Since φpreserves the left invertible operators in both directions, we obtain σ(x⊗y) = σ(φ(I)−1Ax ⊗By). By the spectral radius formula we have |hx, yi| =|hφ(I)−1Ax, Byi| (x, y ∈H). Now, an easy application of the closed graph theorem shows that A, B are continuous. Evidently, we may suppose without any loss of generality that A=B=I. Let S∈B(H) be invertible and write C=φ(S). We claim that C=S. Let x∈Hbe an arbitrary unit vector. Then S(I−λx ⊗x) has a left inverse if and only if λ6= 1. Consequently, the operator C−λSx ⊗xis injective for every λ6= 1 and for every unit vector x∈H. Let z∈Hbe a nonzero vector. Let y=S−1Cz which is also nonzero since C=φ(S) is left invertible. If hz, yi 6= 0, then choosing λ=kyk2/hz, yiwe see that Cz −λ(1/kyk2Sy ⊗y)(z) = Sy −Sy = 0. Since zis nonzero, we deduce λ= 1 which means kyk2=hz, yi. Therefore, for every nonzero vector z∈Hwe have two possibilities. Either