# Maps Preserving Peripheral Spectrum of Generalized Jordan Products of Self-Adjoint Operators

Authors: Wen Zhang; Jinchuan Hou
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

LetA1 and A2  be standard real Jordan algebras of self-adjoint operators on complex Hilbert spaces H1  and  H2, respectively. For  k≥2, let  (i1,…,im) be a fixed sequence with i1,…,im∈{1,…,k} and assume that at least one of the terms in (i1,…,im) appears exactly once. Define the generalized Jordan product  T1∘T2∘⋯∘Tk=Ti1Ti2⋯Tim+Tim⋯Ti2Ti1 on elements in  Ai. Let Φ:A1→A2 be a map with the range containing all rank-one projections and trace zero-rank two self-adjoint operators. We show that Φ satisfies that σπ(Φ(A1)∘⋯∘Φ(Ak))=σπ(A1∘⋯∘Ak) for all A1,…,Ak, where σπ(A) stands for the peripheral spectrum of A, if and only if there exist a scalar c∈{-1,1} and a unitary operator U:H1→H2 such that Φ(A)=cUAU* for all A∈A1, or Φ(A)=cUAtU* for all A∈A1, where At is the transpose of A for an arbitrarily fixed orthonormal basis of H1. Moreover, c=1 whenever m is odd.


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## Body

## 1. Introduction

Recently, the study of maps preserving spectrum of products of operators attracted attentions of researchers. In [1], Molnár characterized surjective maps Φ on bounded linear operators acting on a Hilbert space preserving the spectrum of the product of operators; that is, AB and Φ(A)Φ(B) always have the same spectrum. This similar question was studied by Huang and Hou in [2] by replacing the spectrum by several spectral functions such as the left spectrum and spectral boundary. Hou et al. [3, 4] studied, respectively, further the maps Φ between certain operator algebras preserving the spectrum of a generalized product T1*T2*⋯*Tk and a generalized Jordan product T1∘T2∘⋯∘Tk of low rank operators. Note that the linear maps between Banach algebras which preserve the spectrum are extensively studied in connection with a longstanding open problem due to Kaplansky on invertibility preserving linear maps ([5–10] and the references therein).Moreover, there has been considerable interest in studying peripheral spectrum preserving maps on operator algebras. Recall that the peripheral spectrum of an elementT in a complex Banach algebra A is defined by
(1)σπ(T)={z∈σ(T):|z|=r(T)},
where σ(T) and r(T) stand for the spectrum and the spectral radius of T, respectively. Recall also that a set-valued map Λ:A→2C is said to be a spectral function if ∅≠Λ(T)⊆σ(T) for every T∈A. Since σ(T) is compact, σπ(T) is a well-defined nonempty set and is an important spectral function. Observe that it is always true that σπ(TS)=σπ(ST). In [11], Tonev and Luttman studied maps preserving peripheral spectrum of the usual operator products on standard operator algebras. Recall that a standard operator algebra is a subalgebra of B(X) that contains the identity I and all finite rank operators, where B(X) stands for as usual the Banach algebra of all bounded linear operators on Banach space X. They studied also the corresponding problems in uniform algebras (see [12, 13]). Miura and Honma [14] generalized the result in [13] and characterized surjective maps ϕ and ψ satisfying σπ(ϕ(T)ψ(S))=σπ(TS) on standard operator algebras. Cui and Li studied in [15] the maps preserving peripheral spectrum of Jordan products AB+BA of operators on standard operator algebras. In [16] the maps preserving peripheral spectrum of Jordan semitriple products BAB of operators were characterized. The authors studied in [17, 18], respectively, further the maps between certain operator algebras which preserve peripheral spectrum of a generalized product T1*T2*⋯*Tk and a generalized Jordan product T1∘T2∘⋯∘Tk as defined below.Definition 1.
Fix a positive integerk≥2 and a finite sequence (i1,i2,…,im) such that {i1,i2,…,im}={1,2,…,k} and there is an ip not equal to iq for all other q; that is, ip appears just one time in the sequence. For operators T1,…,Tk, the operators,
(2)T1*T2*⋯*Tk=Ti1Ti2⋯Tim,(3)T1∘T2∘⋯∘Tk=Ti1Ti2⋯Tim+Tim⋯Ti2Ti1
are, respectively, called generalized product and generalized Jordan product of T1,…,Tk, while m is called the width of the products.Evidently, the generalized Jordan productT1∘⋯∘Tk (the generalized product T1*⋯*Tk) covers the Jordan product T1T2+T2T1 and the Jordan triple product T1T2T3+T3T2T1 (the usual product T1T2 and the Jordan semitriple product T1T2T1), and so forth. We also remark that the notations T1*T2*⋯*Tk and T1∘T2∘⋯∘Tk are not unique for T1,T2,…,Tk because they depend on the choice of the integers k≥2, m≥2, and the sequence (i1,i2,…,im). In this paper, we presume that k, m, and the sequence (i1,i2,…im) are arbitrary but fixed throughout the paper.Let us consider the case of Hilbert spaces. Denote byB(H) the set of all bounded linear operators on a complex Hilbert space H and T* the adjoint of T∈B(H). If T=T*, T is self-adjoint. Denote by Bs(H) the real Jordan algebra of all self-adjoint operators in B(H). A real Jordan subalgebra of Bs(H) is said to be standard if it contains the identity I and all finite rank self-adjoint operators. In [14] Miura and Honma characterized the surjective maps between standard operator algebras on Hilbert spaces that preserve the peripheral spectrum of skew products T*S of operators. Cui and Li studied in [15] the maps preserving peripheral spectrum of skew Jordan products AB*+B*A of operators on standard operator algebras on complex Hilbert spaces. A characterization of maps preserving peripheral spectrum of Jordan products of self-adjoint operators AB+BA on standard real Jordan subalgebras of Bs(H) was also given in [15]. In [16] the maps preserving peripheral spectrum of Jordan skew semitriple products BA*B of operators were characterized, and then, the maps preserving peripheral spectrum of the skew generalized products of operators on Hilbert space H were characterized in [17].Products of self-adjoint operators in Hilbert space play a role in several different areas of pure and applied mathematics. In this paper, we characterize the maps preserving the peripheral spectrum of generalized Jordan products of self-adjoint operators between the standard real Jordan algebras of self-adjoint operators on complex Hilbert spaces. LetAi be a standard real Jordan algebra in Bs(Hi), i=1,2, and Φ:A1→A2 a map with range containing all rank-one projections and all rank-two self-adjoint operators with zero trace. We show that Φ satisfies that σπ(Φ(A1)∘⋯∘Φ(Ak))=σπ(A1∘⋯∘Ak) for all A1,…,Ak in A1 if and only if there exist a scalar c∈{-1,1} and a unitary operator U:H1→H2 such that Φ(A)=cUAU* for all A∈A1, or Φ(A)=cUAtU* for all A∈A1, where At is the transpose of A with respect to an arbitrary but fixed orthonormal basis of H1. Moreover, c=1 whenever m is odd. We also characterize the maps from A1 into A2 that preserves the peripheral spectrum of generalized product on Ai.


## 2. Generalized Jordan Products of Self-Adjoint Operators

LetH1 and H2 be two complex Hilbert spaces and Bs(H1) and Bs(H2) the real linear spaces of all self-adjoint operators in B(H1) and B(H2), respectively. Then Bs(H1) and Bs(H2) are real Jordan algebras. Recall that a standard real Jordan algebra on Hi is a Jordan subalgebra of Bs(Hi) which contains all finite rank self-adjoint operators and the identity operator. In this section, we will characterize maps preserving peripheral spectrum of generalized Jordan products of self-adjoint operators.Theorem 2.
LetA1 and A2 be standard real Jordan algebras of self-adjoint operators on complex Hilbert spaces H1 and H2, respectively. Consider the product T1∘⋯∘Tk defined in (3) of Definition 1 with the width m. Assume that Φ:A1→A2 is a map the range of which contains all rank-one projections and all rank-two self-adjoint operators with zero trace. Then Φ satisfies
(4)σπ(Φ(A1)∘⋯∘Φ(Ak))=σπ(A1∘⋯∘Ak)
for all A1,A2,…,Ak∈A1 if and only if there exist a unitary operator U∈B(H1,H2) and a scalar c∈{-1,1} such that either(1)
Φ(A)=cUAU* for every A∈A1, or(2)
Φ(A)=cUAtU* for every A∈A1. Here At is the transpose of A with respect to an arbitrary but fixed orthonormal basis of H1.
Moreover,c=1 whenever m is odd.To prove Theorem2, we consider the special case that Aip=A and Aiq=B for all q≠p. Thus there exist nonnegative integers r, s with r+s=m-1≥1 such that A1∘A2∘⋯∘Ak=BrABs+BsABr. For this special case we have.Theorem 3.
LetA1 and A2 be standard real Jordan algebras of self-adjoint operators on complex Hilbert spaces H1 and H2, respectively. Assume that Φ:A1→A2 is a map the range of which contains all rank-one projections and all rank-two self-adjoint operators with zero trace, and r,s are nonnegative integers with r+s≥1. Then Φ satisfies
(5)σπ(BrABs+BsABr)=σπ(Φ(B)rΦ(A)Φ(B)s+Φ(B)sΦ(A)Φ(B)r)
for all A,B∈A1 if and only if there exist a unitary operator U∈B(H1,H2) and a scalar c∈{-1,1} such that Φ(A)=cUAU* for every A∈A1, or Φ(A)=cUAtU* for every A∈A1. Moreover, c=1 whenever r+s is even. Here At is the transpose of A with respect to an arbitrary but fixed orthonormal basis of H1.IfΦ meets (4), then it also meets (5) for some r, s with r+s=m-1 by taking Aip=A and Aiq=B for q≠p. Hence it is obvious that the truth of Theorem 3 will imply the truth of Theorem 2.Thus we focus our attention to prove Theorem3. We will do it by decomposing the proof in a number of steps and use of technical lemmas.Note that, ifs=r>0, then the question is reduced to the generalized product BrABr of self-adjoint operators, which will be discussed in the next section. So, unless specified otherwise, we always assume in this section that s>r≥0.Lemma 4.
For any unit vectorx∈H and nonzero B∈Bs(H), we have
(6)σπ(Brx⊗xBs+Bsx⊗xBr)={{〈Br+sx,x〉+∥Brx∥∥Bsx∥}if〈Br+sx,x〉>0;{〈Br+sx,x〉-∥Brx∥∥Bsx∥}if〈Br+sx,x〉<0;{±∥Brx∥∥Bsx∥}if〈Br+sx,x〉=0.Proof.
In fact, if there exist nonzeroα,β∈R such that Brx=αx, Bsx=βx, clearly (6) holds. Now assume that Brx and x or Bsx and x are linearly independent. Then there exist nonzero γ∈R and z∈H such that (Brx⊗xBs+Bsx⊗xBr)z=γz; that is,
(7)〈Bsz,x〉Brx+〈Brz,x〉Bsx=γz.
It follows that
(8)〈Bsz,x〉〈Brx,x〉+〈Brz,x〉〈Bsx,x〉=γ〈z,x〉,(9)〈Bsz,x〉〈Br+sx,x〉+〈Brz,x〉〈B2sx,x〉=γ〈Bsz,x〉,(10)〈Brz,x〉〈Br+sx,x〉+〈Bsz,x〉〈B2rx,x〉=γ〈Brz,x〉.
We consider the following two cases.
Case
1  (〈Br+sx,x〉=0). If 〈Brz,x〉≠0, it follows from (10) that 〈Bsz,x〉≠0. Then (9) and (10) imply that γ=±∥Brx∥∥Bsx∥. If 〈Brz,x〉=0, it follows from (9) that 〈Bsz,x〉=0, but this contradicts (7). So σπ(Brx⊗xBs+Bsx⊗xBr)={±∥Brx∥∥Bsx∥}.
Case
2  (〈Br+sx,x〉≠0). In this case, there must be 〈Brz,x〉≠0 and 〈Bsz,x〉≠0. Then it follows from (9) and (10) that
(11)(γ-〈Br+sx,x〉)2=∥Brx∥2∥Bsx∥2,
which implies that γ=〈Br+sx,x〉±∥Brx∥∥Bsx∥. So
(12)σ(Brx⊗xBs+Bsx⊗xBr)={0,〈Br+sx,x〉±∥Brx∥∥Bsx∥}.
Now the result follows immediately.In Lemmas5 and 6, we always assume that Φ:A1→A2 is a map satisfying (5) with range containing all rank-one projections and all rank-two self-adjoint operators of zero trace, and assume that r, s are nonnegative integers with r+s≥1. Recall that a self-adjoint operator A is said to be positive, denote by A≥0, if 〈Ax,x〉≥0 for all x∈H; while A≥B means that A-B≥0.Lemma 5.
Φ
(
I
)
=
I or -I. Φ(I)=-I may occur only if r+s is odd.Proof.
For anyA,B∈A1, since
(13)σπ(BrABs+BsABr)=σπ(Φ(B)rΦ(A)Φ(B)s+Φ(B)sΦ(A)Φ(B)r),
it follows that r(A)=r(Φ(A)) holds for every A∈A1. Let Φ(I)=B. By the assumption on the range of Φ, for any unit vector y∈H2, there exists A∈A1 such that Φ(A)=y⊗y. We consider the following two cases.
Case
1  (s>r=0). It follows from (5) that
(14)σπ(2As)=σπ(Φ(A)sΦ(I)+Φ(I)Φ(A)s)=σπ(y⊗yB+By⊗y),
which implies that ∥By⊗y+y⊗yB∥=2 for all unit vectors y∈H2. Then by Lemma 4, we have
(15)2=∥By⊗y+y⊗yB∥=|〈By,y〉|+∥By∥≤2∥By∥,
and hence ∥By∥≥1 for all unit vectors y∈H2, and ∥B∥≥1. On the other hand, for any unit vector y∈H2, we have 2|〈By,y〉|≤|〈By,y〉|+∥By∥=2. Hence |〈By,y〉|≤1 holds for all unit vectors y∈H2 and consequently, ∥B∥≤1. So we must have ∥B∥=1 and ∥By∥=1 for all unit vectors y∈H2. Now it follows from (15) that B=ɛI with ɛ∈{-1,1}. In particular, if s is even, as As≥0, (14) and (15) imply that 〈By,y〉=1 for all unit vectors y∈H2 and hence B=I.
Case
2  (s>r>0). By (5) we have
(16)σπ(2Ar+s)=σπ(Φ(A)sΦ(I)Φ(A)r+Φ(A)rΦ(I)Φ(A)s)=σπ(2y⊗yBy⊗y),
which implies that ∥y⊗yBy⊗y∥=∥Ar+s∥=1 for all unit vectors y∈H2. Then |〈By,y〉|=1 holds for each unit vector y, and so B=ɛI with ɛ∈{-1,1}. Particularly, if r+s is even, then Ar+s≥0 and it follows from (16) that 〈By,y〉=1 holds for every unit vector y. Hence B=I.IfΦ(I)=-I, then -Φ satisfies the conditions in Theorem 3, so we may as well assume Φ(I)=I in the sequel, and thus σπ(A)=σπ(Φ(A)) holds for every A∈A1.Lemma 6.
Φ preserves rank-one projections in both directions.Proof.
We consider the following two cases.
Case
1  (s>r=0). Consider the following.
Case
1.1 (
s
is even). For any unit vector x∈H1, let Φ(x⊗x)=B and Φ(I-x⊗x)=T. It follows from {0}=σπ((x⊗x)s(I-x⊗x)+(I-x⊗x)(x⊗x)s)=σπ(BsT+TBs) that BsT+TBs=0.
Note that ifA≥0, and AS+SA=0, then AS=SA=0. If fact AS+SA=0 implies that A2S=SA2. Since A≥0, we must have AS=SA and 2AS=AS+SA=0, which forces AS=SA=0.
Now, asBs≥0 and BsT+TBs=0, we see that BsT=TBs=0. It follows from σπ(B)={1} that Bs≠0, which implies that {0}≠ranBs⊆ker⁡T, where ran T stands for the range of T. For any unit vector y∈ker⁡T, pick A∈A1 such that Φ(A)=y⊗y. It follows from σπ((I-x⊗x)As+As(I-x⊗x))=σπ(Ty⊗y+y⊗yT)={0} that (I-x⊗x)As+As(I-x⊗x)=0, which, together with As≥0, implies that (I-x⊗x)As=As(I-x⊗x)=0. So we have A=x⊗x and Φ(x⊗x)=y⊗y is rank-one.
Case
1.2 (
s
is odd). For any unit vector x∈H1, let A=x⊗x and Φ(A)=B. We will prove that B is a rank-one projection.
Claim
1.2.1  (dim⁡ker⁡(B-I)=1). Note that σπ(B)=σπ(A)={1}. Then 1∈σ(B)⊆(-1,1]. It follows that either (i) dim⁡ker⁡(B-I)≥1 or (ii) B-I is injective but not surjective.
Assume that (ii) occurs. Since1∈σπ(B), we have ∥B∥=1 and B≤I. So, according to some suitable space decomposition of H2, B has an operator matrix representation of the form
(17)(a0b000a0c0b0***0c***00***),
where a>1/2 and b,c≥0. To see this, one can first choose three orthonormal vectors x1, x2, x3 such that 1-d<〈Bxj,xj〉<1 for some sufficiently small d∈(0,1/4). Suppose the compression B^ of B onto the span of {x1,x2,x3} has eigenvalues μ1≥μ2≥μ3. Then
(18)μ2≥(μ2+μ3)2≥[(3-3d)-1]2>12.
Let μ2=a. Then B^ is similar to
(19)(a0*0a****).
Thus, there exists a space decomposition such that B has an operator matrix of the form
(20)(aI2B12B12**).
Clearly, there are unitary U, V such that UB12V* has operator matrix of the form
(21)(b000c0),
where b,c≥0. So B has the desired operator matrix form. Under the same decomposition, take S=(0110)⊕0; then σπ(BSs+SsB) has two different points with r(BSs+SsB)≥2a>1 and there exists R∈A1 such that Φ(R)=S. It follows that σπ(R)={-1,1}. So ∥R∥=1 and ∥Rsu∥≤1 for all unit vectors u∈H1. But σπ(ARs+RsA)=σπ(x⊗xRs+Rsx⊗x) is either a singleton or {±∥Rsx∥} with ∥Rsx∥≤1. This contradicts the fact r(ARs+RsA)=r(BSs+SsB)≥2a>1.
Sodim⁡ker⁡(B-I)≥1. Assume that dim⁡ker⁡(B-I)=n≥2. According to the space decomposition H2=ker⁡(B-I)⊕ker⁡(B-I)⊥, B has an operator matrix In⊕N. Under the same space decomposition, take M=(0110)⊕0. Similar to the previous discussion, one gets a contradiction again. So dim⁡ker⁡(B-I)=1.
Claim
1.2.2.  There exists a unit vector y∈H2 such that B=y⊗y.
If it is not true, then, by Claim1.2.1, there exist a unit vector y∈ker⁡(B-I) and a nonzero B2∈A2 with B2y=0 such that B=y⊗y+B2. So there exists a unit vector z∈[y]⊥ such that B2z≠0. Let C1=y⊗y and C2=z⊗z. Then σπ(BC1s+C1sB)=σπ(BC1+C1B)={2}, C1C2s+C2sC1=0, and BC2s+C2sB≠0. Since the range of Φ contains all rank-one projections, there exist D1 and D2 in A1 such that Φ(D1)=C1 and Φ(D2)=C2. Then σπ(D1)=σπ(D2)={1}, σπ(AD1s+D1sA)=σπ(BC1s+C1sB)={2}, D1D2s+D2sD1=0, and AD2s+D2sA≠0.
Since{2}=σπ(AD1s+D1sA)=σπ(x⊗xD1s+D1sx⊗x), it follows from (6) that |〈D1sx,x〉|+∥D1sx∥=2, which, together with ∥D1∥=1, implies that D1sx=x. So, according to the space decomposition H1=[x]⊕[x]⊥, D1s=[1]⊕Z with σ(Z)⊆(-1,1]. If D2 has an operator matrix (v11V12V12*V22) accordingly, then
(22)0=D1sD2+D2D1s=(2v11V12+V12ZZV12*+V12*ZV22+V22Z).
Since I+Z is invertible, we see that V12=0. Clearly, v11=0. So, D2=0⊕V22. But then it contradicts the fact that AD2s+D2sA≠0. So Claim 1.2.2 holds and Φ preserves rank-one projections.
Conversely, assume thatΦ(A) is a rank-one projection; then a similar discussion shows that A is a rank-one projection, too.
Case
2  (s>r>0).Consider the following.
Case
2.1 (
r
+
s
is even). For any unit vector x∈H1, let Φ(x⊗x)=B and Φ(I-x⊗x)=T. It follows from {0}=σπ((I-x⊗x)r(x⊗x)(I-x⊗x)s+(I-x⊗x)s(x⊗x)(I-x⊗x)r)=σπ(TrBTs+TsBTr) that TrBTs+TsBTr=0. Since Ts-r≥0 and TrBTrTs-r+Ts-rTrBTr=TrBTs+TsBTr=0, we see that TsBTr=TrBTs=0. If ker⁡T={0}, then ker⁡Tr={0} and ker⁡Ts={0}, which, together with TrBTs=0, imply that BTs=0, and thus B=0, a contradiction.
So,ker⁡T≠{0}. Take a unit vector y∈ker⁡T and A∈A1 such that Φ(A)=y⊗y. It follows from σπ(Ar(I-x⊗x)As+As(I-x⊗x)Ar)=σπ(2y⊗yTy⊗y)={0} that Ar(I-x⊗x)As+As(I-x⊗x)Ar=0, which, together with As-r≥0, implies that Ar(I-x⊗x)As=As(I-x⊗x)Ar=0. Hence we have A=x⊗x and Φ(x⊗x)=y⊗y.
Case
2.2 (
r
+
s
is odd). For any unit vectors x∈H1, let A=x⊗x and Φ(A)=B. We will prove that B is a rank-one projection.
Claim
2.2.1  (dim⁡ker⁡(B-I)=1). Note that σπ(B)=σπ(A)={1}. Then 1∈σ(B)⊆(-1,1]. It follows that either (i) dim⁡ker⁡(B-I)≥1 or (ii) B-I is injective but not surjective.
Assume that (ii) occurs. Since1∈σπ(B), we have ∥B∥=1 and B≤I. So, like shown in Case 1.2.1, with respect to some suitable space decomposition of H2, B has an operator matrix representation of the form
(23)(a0b000a0c0b0***0c***00***),
where a>1/2 and b,c≥0. Under the same decomposition, take S=(0110)⊕0, and then {±2a}=σπ(SrBSs+SsBSr). As S has rank-two and zero trace, there exists R∈A1 such that Φ(R)=S. It follows that σπ(R)={-1,1}. So ∥R∥=1 and ∥Rsu∥∥Rru∥≤1 for all unit vectors u∈H1. But σπ(RrARs+RsARr)=σπ(Rrx⊗xRs+Rsx⊗xRr)={±∥Rrx∥∥Rsx∥}. This contradicts the fact r(ARs+RsA)=r(BSs+SsB)≥2a>1.
Sodim⁡ker⁡(B-I)≥1. Assume that dim⁡ker⁡(B-I)=n≥2. According to the space decomposition H2=ker⁡(B-I)⊕ker⁡(B-I)⊥, B has an operator matrix In⊕N. Under the same space decomposition, take M=(0110)⊕0. Similar to the previous discussion, one gets a contradiction again. So dim⁡ker⁡(B-I)=1.
Claim
2.2.2. There exists a unit vector y∈H2 such that B=y⊗y.
If it is not true, then, by Claim2.2.1, there exist a unit vector y∈ker⁡(B-I) and a nonzero B2∈A2 with B2y=0 such that B=y⊗y+B2. So there exists a unit vector z∈[y]⊥ such that B2rz≠0 and B2sz≠0. Let C1=y⊗y and C2=z⊗z. Since the range of Φ contains all rank-one projections, there exist D1 and D2 in A1 such that Φ(D1)=C1 and Φ(D2)=C2. Then σπ(D1rAD1s+D1sAD1r)=σπ(C1rBC1s+C1sBC1r)=σπ(2y⊗yBy⊗y)={2}, which, together with (6), implies that 〈D1r+sx,x〉+∥D1rx∥∥D1sx∥=2. It follows from 〈D1r+sx,x〉≤∥D1rx∥∥D1sx∥≤1 that 〈D1r+sx,x〉=1. So D1r+sx=x, and according to the space decomposition H1=[x]⊕[x]⊥, D1r+s=[1]⊕Z with σ(Z)⊆(-1,1]. Thus under the same space decomposition we have D1=[1]⊕Y with σ(Y)⊆(-1,1]. Write D2 in the operator matrix (v11V12V12*V22) accordingly; then
(24)0=D1sD2D1r+D1rD2D1s=(2v11V12(Ys+Yr)(Ys+Yr)V12*YrV22Ys+YsV22Yr).
Clearly, v11=0. So, ArD2As+AsD2Ar=0. But then this contradicts the fact that σπ(ArD2As+AsD2Ar)=σπ(BrC2Bs+BsC2Br)≠{0}. So Claim 2.2.2 holds and Φ preserves rank-one projections.
Conversely, assume thatΦ(A) is a rank-one orthogonal projection; then, a similar discussion implies that A is a rank-one projection.The following lemma was proved in [19].Lemma 7.
LetH be a complex Hilbert space and A,B∈B(H) self-adjoint operators. If |〈Ax,x〉|+∥Ax∥∥x∥=|〈Bx,x〉|+∥Bx∥∥x∥ holds for all x∈H, then A=±B.Now we are in a position to present our proof of Theorem3, except the case r=s.Proof of Theorem3.
The “if” part is obvious. Let us check the “only if” part.
By Lemma6, Φ preserves rank-one projections in both directions. It follows that there exists a bijective map T:H1→H2 such that
(25)Φ(x⊗x)=Tx⊗Tx
for all unit vectors x∈H1, where ∥Tx∥=∥x∥ and T(λx)=λTx for any x∈H1 and λ∈C.
We consider the following two cases.
Case
1  (s>r=0). For any unit vectors x,y∈H1, we have σπ((y⊗y)(x⊗x)s+(x⊗x)s(y⊗y))=σπ((Ty⊗Ty)(Tx⊗Tx)s+(Tx⊗Tx)s(Ty⊗Ty)). By (6), 〈x,y〉=0 if and only if 〈Tx,Ty〉=0, and when 〈x,y〉≠0,
(26)|〈Tx,Ty〉|2+|〈Tx,Ty〉|=|〈x,y〉|2+|〈x,y〉|.
It follows that
(27)|〈Tx,Ty〉|=|〈x,y〉|
holds for all x,y∈H1.
Wigner’s theorem [20] states that every bijective map T between Hilbert spaces H1, H2 satisfying (27) must have the form Tx=ϕ(x)Ux for any x∈H1, where U is a unitary or a conjugate unitary (i.e., antiunitary) operator and ϕ is a so-called phase-function which means that its values are of modulus one. Thus, by Wigner’s theorem, there exists a unitary or conjugate unitary operator U:H1→H2 such that Φ(x⊗x)=Ux⊗Ux for every unit vector x∈H1.
Assume first thatU is unitary. Let A∈A1 be arbitrary. For any unit vector x∈H1,
(28)σπ(A(x⊗x)s+(x⊗x)sA)=σπ(Φ(A)(Ux⊗Ux)s+(Ux⊗Ux)sΦ(A)).
Applying (6), for any unit vector x∈H1, one has
(29)|〈Ax,x〉|+∥Ax∥=|〈U*Φ(A)Ux,x〉|+∥U*Φ(A)Ux∥,
and hence Lemma 7 implies that U*Φ(A)U=±A. Hence, Φ(A)=±UAU* for every A∈A1. We claim that Φ(A)=UAU* for every A∈A1. Otherwise, there exists some nonzero B0 such that Φ(B0)=-UB0U*. Let M+={A∈A1:Φ(A)=UAU*} and M-={B∈A1:B≠0,Φ(B)=-UBU*}. Then M+∩M-=∅, M+∪M-=A1, and B0∈M-. Note that, as Φ(I)=I, I∈M+. It follows that, for any B∈M-, we have σπ(2B)=σπ(IB+BI)=σπ(Φ(I)Φ(B)+Φ(B)Φ(I))=-σπ(2B). Therefore, σπ(B)={-∥B∥,∥B∥} holds for all B∈M-. Let B=∫-∥B∥∥B∥‍λdEλ be the spectral resolution of B. Then the spectral project E=∫(1/2)∥B∥∥B∥‍dEλ≠0. Though we do not know if E∈A1, we can take unit vector x∈EH1 so that 〈Bx,x〉≥(1/2)∥B∥>0. Thus, 〈Bx,x〉+∥Bx∥≥∥B∥>0. By Lemma 4, we have σπ(x⊗xB+Bx⊗x)={〈Bx,x〉+∥Bx∥}. Since x⊗x∈A1 and σπ(x⊗x)={1}, x⊗x∈M+. But then,
(30){〈Bx,x〉+∥Bx∥}=σπ(x⊗xB+Bx⊗x)=σπ(Φ(x⊗x)Φ(B)+Φ(B)Φ(x⊗x))=-σπ(x⊗xB+Bx⊗x)={-〈Bx,x〉-∥Bx∥},
a contradiction. So, Φ(A)=UAU* holds for every A∈A1.
Now assume thatU is conjugate unitary. Take arbitrarily an orthonormal basis {ei}i∈Λ of H and define J by J(∑i∈Λ‍ξiei)=∑i∈Λ‍ξi¯ei. Then J:H1→H1 is conjugate unitary and J2=I. Let V=UJ. Then V is unitary and a similar discussion as above implies that Φ(A)=VAtV* for all A∈A1 and At is the transpose of A for the orthonormal basis {ei}i∈Λ of H1.
Case  2  (s>r>0). For any unit vectors x,y∈H1, we have
(31){2|〈x,y〉|2}=σπ((x⊗x)r(y⊗y)(x⊗x)s+(x⊗x)s(y⊗y)(x⊗x)r)=σπ((Tx⊗Tx)r(Ty⊗Ty)(Tx⊗Tx)s+(Tx⊗Tx)s(Ty⊗Ty)(Tx⊗Tx)r)={2|〈Tx,Ty〉|2}.
Hence
(32)|〈Tx,Ty〉|=|〈x,y〉|
holds for all x,y∈H1. Thus, by Wigner’s theorem, there exists a unitary or conjugate unitary operator U:H1→H2 such that Φ(x⊗x)=Ux⊗Ux for every unit vector x∈H1.
Now assume thatU is unitary. Let A∈A1 be arbitrary. For any unit vector x∈H1, since
(33){2〈Ax,x〉}=σπ((x⊗x)rA(x⊗x)s+(x⊗x)sA(x⊗x)r)=σπ((Ux⊗Ux)rΦ(A)(Ux⊗Ux)s+(Ux⊗Ux)sΦ(A)(Ux⊗Ux)r)={2〈Φ(A)Ux,Ux〉},
we have
(34)〈Ax,x〉=〈Φ(A)Ux,Ux〉∀unitvectorsx∈H1.
So we get Φ(A)=UAU* for every A∈A1.
Similar to the cases>r=0, if U is conjugate unitary, then there exists a unitary operator V such that Φ(A)=VAtV* for all A∈A1.
Hence we have shown that, in the caseΦ(I)=I, there exists a unitary U such that either Φ(A)=UAU* for every A∈A1; or Φ(A)=UAtU* for every A∈A1, where At is the transpose of A with respect to an arbitrarily given orthonormal basis of H1.
IfΦ(I)=-I, considering Ψ=-Φ gives Φ(A)=-UAU* for every A∈A1 or Φ(A)=-UAtU* for every A∈A1. It is clear that this case does not occur if r+s is even.


## 3. Generalized Products of Self-Adjoint Operators on Hilbert Spaces

In this section, we will characterize maps preserving peripheral spectrum of generalized products of self-adjoint operators. Its special case, Theorem10, makes up for the gap for the case s=r in the proof of Theorem 3.LetA be a real Jordan algebra in Bs(H). If a generalized product T1*T2*⋯*Tk defined in (2) satisfies that T1*T2*⋯*Tk∈A for any T1,T2,…,Tk∈A, that is, the general product is closed in A, we say that T1*T2*⋯*Tk is a generalized product on A. The following lemma was proved in [3].Lemma 8.
LetT1*T2*⋯*Tk=Ti1⋯Tip⋯Tim be a generalized product on a standard real Jordan algebra A⊆Bs(H) defined as in (2) of Definition 1. Then there exists a positive integer n with m=2n-1 such that ip=n, and ij=i2n-j for all j=1,…,n.The following is the main result in this section. Observe that we do not need the assumption that the range of the map contains all rank-two self-adjoint operators with zero trace.Theorem 9.
LetA1 and A2 be standard real Jordan algebras of self-adjoint operators on complex Hilbert spaces H1 and H2, respectively. Consider the generalized product T1*⋯*Tk on Bs(Hi) as in Lemma 8 with width m. Assume that Φ:A1→A2 is a map the range of which contains all rank-one projections. Then Φ satisfies
(35)σπ(Φ(A1)*⋯*Φ(Ak))=σπ(A1*⋯*Ak)
for all A1,A2,…,Ak∈A1 if and only if one of the following conditions holds. (1)
There exists a unitary operatorU:H1→H2 such that Φ(A)=UAU* for all A∈A1.(2)
There exists a unitary operatorU:H1→H2 such that Φ(A)=UAtU* for all A∈A1,
whereAt is the transpose of A for an arbitrarily but fixed orthonormal basis of H1.To prove Theorem9, we consider the special case by taking Aip=A and Aiq=B if q≠p. By Lemma 8, there exists positive integer r(=n) with 2r=m-1≥2 such that A1*A2*⋯*Ak=BrABr. It is clear that Theorem 9 is an immediate consequence of the following result.Theorem 10.
LetA1 and A2 be standard real Jordan algebras of delf-adjoint operators on complex Hilbert spaces H1 and H2, respectively. Assume that Φ:A1→A2 is a map the range of which contains all rank-one projections and r is nonnegative integer with r≥1. Then Φ satisfies
(36)σπ(BrABr)=σπ(Φ(B)rΦ(A)Φ(B)r)∀A,B∈A1
if and only if one of the following conditions holds. (1)
There exists a unitary operatorU:H1→H2 such that Φ(A)=UAU* for all A∈A1.(2)
There exists a unitary operatorU:H1→H2 such that Φ(A)=UAtU* for all A∈A1,
whereAt is the transpose of A for an arbitrarily but fixed orthonormal basis of H1.To prove Theorem10, it suffices to check the “only if” part. Assume in the following that Φ:A1→A2 is a map satisfying (36) with range containing all rank-one projections.Lemma 11.
Φ
(
I
)
=
I.Proof.
It follows from (36) that r(A)=r(Φ(A)) holds for every A∈A1. Let Φ(I)=B. For any unit vector y∈H2, there exists A∈A1 such that Φ(A)=y⊗y. Then
(37)σπ(A2r)=σπ(Φ(A)rΦ(I)Φ(A)r)=σπ(y⊗yBy⊗y),
which, together with A2r≥0, implies that 〈By,y〉=1 for all unit vectors y∈H2. So B=I.Lemma 12.
Φ preserves rank-one projections in both directions.Proof.
For any unit vectorx∈H1, let Φ(x⊗x)=B and Φ(I-x⊗x)=T. It follows from {0}=σπ((x⊗x)r(I-x⊗x)(x⊗x)r)=σπ(BrTBr) that BrTBr=0. Similarly, we have TrBTr=0. If ker⁡T={0}, then ker⁡Tr={0}, which, together with TrBTr=0, implies that BTr=0, and thus B=0, a contradiction.
So, there exist a unit vectory∈ker⁡T. Take A∈A1 such that Φ(A)=y⊗y. It follows from σπ(Ar(I-x⊗x)Ar)=σπ(y⊗yTy⊗y)={0} that Ar(I-x⊗x)Ar=0, which implies that A=x⊗x. Thus Φ(x⊗x)=y⊗y and, therefore, Φ preserves rank-one projections.
Similarly one can show thatΦ(A) is a rank-one projection will imply that A is a rank-one projection.Proof of Theorem10.
By Lemma12, Φ preserves rank-one projections in both directions. It follows that there exists a bijective map T:H1→H2 such that
(38)Φ(x⊗x)=Tx⊗Tx
for all unit vectors x∈H1, where ∥Tx∥=∥x∥ and T(λx)=λTx for any x∈H1 and λ∈C.
For any unit vectorsx,y∈H1, we have
(39){|〈x,y〉|2}=σπ((x⊗x)r(y⊗y)(x⊗x)r)=σπ((Tx⊗Tx)r(Ty⊗Ty)(Tx⊗Tx)r)={|〈Tx,Ty〉|2}.
Hence
(40)|〈Tx,Ty〉|=|〈x,y〉|
holds for all x,y∈H1. Thus, by Wigner’s theorem again, there exists a unitary or conjugate unitary operator U:H1→H2 such that Φ(x⊗x)=Ux⊗Ux for every unit vector x∈H1.
Now assume thatU is unitary. Let A∈A1 be arbitrary. For any unit vector x∈H1, since
(41){〈Ax,x〉}=σπ((x⊗x)rA(x⊗x)r)=σπ((Ux⊗Ux)rΦ(A)(Ux⊗Ux)r)={〈Φ(A)Ux,Ux〉},
we have
(42)〈Ax,x〉=〈Φ(A)Ux,Ux〉∀unitvectorsx∈H1.
Hence we get Φ(A)=UAU* for every A∈A1.
Similarly,U is conjugate unitary implies that there exists a unitary operator such that Φ(A)=VAtV* for all A∈A1.Remark 13.
Finally, we remark that if we do not require that the generalized product is closed in the involved standard real Jordan algebrasAi, i=1,2, we can still obtain a characterization of the maps Φ from A1 into A2 with range containing all rank-one projections which preserves the peripheral spectrum of an arbitrarily given generalized product. In fact, such maps have the same form stated in Theorem 2.



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