# Characterizations of Double Commutant Property onBH

Authors: Chaoqun Chen; Fangyan Lu; Cuimei Cui; Ling Wang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

LetH be a complex Hilbert space. Denote by BH the algebra of all bounded linear operators on H. In this paper, we investigate the non-self-adjoint subalgebras of BH of the form T+B, where B is a block-closed bimodule over a masa and T is a subalgebra of the masa. We establish a sufficient and necessary condition such that the subalgebras of the form T+B has the double commutant property in some particular cases.


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## Body

## 1. Introduction

LetH be a complex Hilbert space. We denote by BH the algebra of all bounded linear operators on H. Given a nonempty subset A of BH,the commutant of A is the set A′≔T∈BH:TA=ATforallA∈A. The double commutant of A is A′′≔A′′. Clearly, A⊆A′′. von Neumann’s double commutant theorem states that if A⊆BH is a self-adjoint algebra of operators whose kernel kerA≔⋂A∈AkerA=0, then the closure of A in any of the weak operator, strong operator, and weak∗ topologies is the double commutant A′′. In fact, if A is a WOT-closed, unital C∗-subalgebra of BH, then A=A′′. In this paper, we analyze the settings of non-self-adjoint subalgebras of BH whose double commutant coincides with themselves. We say that such algebras satisfy the double commutant property.In the past several decades, a great deal of effort has been devoted to the study of the subalgebras ofBH with the double commutant property. For a few references, see [1–8]. In recent years, there has been renewed interest in the study of double commutant property [9–15]. For singly generated algebras, Ruston [16] showed that every algebraic operator in BH has the double commutant property. Turner [8] proved that a normal operator satisfies the double commutant property if and only if it is reductive. For nonsingly generated algebras, Davidson and Pitts [2] researched the noncommutative analytic Toeplitz algebra with the double commutant property. Marcoux and Mastnak [12] analyzed the non-self-adjoint subalgebras of BH whose double commutant agrees with themselves; specifically, they considered the class of algebras of the form D+R in finite dimensional space, where R is a bimodule over a masa and D is a unital subalgebras of the masa.In this note, we will investigate the subalgebras ofBH with the double commutant property, which extends the result in [13] extensively.


## 2. Preliminaries

LetH be a Hilbert space, if 0≠x,y∈H, we denote by x⊗y the rank-one operator on H given by x⊗yz≔z,yx. For a subalgebra W of BH, we define the annihilator of W as W⊥=T∈BH:TW=0=WTforallW∈W. Given a collection Pαα of orthogonal projections in BH, we denote by ∨αPα the orthogonal projection onto ∨RanPαα. Note that all projections considered on the manuscript are orthogonal projections. Given projections P and Q in BH, we define the P,Q-block of BH as follows:
(1)BP,Q≔QBHP=QTP:T∈BH.LetM⊆BH be a maximal abelian self-adjoint subalgebra (that is, M is a masa), Pγ,Qγγ∈Γ be a collection of projections in M, then we say that a subspace B of BH is block-generated over M if
(2)B=∨BPγ,Qγ:γ∈Γ.With loss of generality, we assume that eachPγ≠0≠Qγ.Definition 1.
LetM⊆BH be a masa and let B=∨BPγ,Qγ:γ∈Γ be a block-generated bimodule for some family of projections Pγ,Qγγ∈Γ⊆M. We say that B is disconnected if there exist ∅≠Γ1,Γ2⊆Γ, and projections E1,F1,E2,F2∈M so that
(1)
Γ=Γ1∪Γ2(2)
0≠∨BPγ,Qγ:γ∈Γk⊆BEk,Fk,k=1,2(3)
E1∨F1 is orthogonal to E2∨F2
Otherwise, we say thatB is connected.Marcoux and Mastnak proved the following proposition in [12]. Now, we give another simpler proof.Proposition 2 (see [12]).
LetM⊆BH be a masa and let B=∨BPγ,Qγ:γ∈Γ be a block-generated bimodule over M with Pγ≠0≠Qγ for all γ. Then,
(1)
B⊥=BQ0⊥,P0⊥, where P0=∨γPγandQ0=∨γQγ(2)
B is connected if and only if B′=B⊥+ℂIProof.
(1)
It is clear thatBQ0⊥,P0⊥⊆B⊥. We only need to show that B⊥⊆BQ0⊥,P0⊥.
∀T∈B⊥, we have TQγBHPγ=QγBHPγT=0, ∀γ∈Γ. Since BH is prime, we obtain TQγ=0=PγT,∀γ∈Γ. So, TQ0=0=P0T. This implies that T=TQ0⊥ and T=P0⊥T; therefore, T=P0⊥TQ0⊥, so we have T∈BQ0⊥,P0⊥.
(2)
IfB is connected, it is easy to vertify that B⊥+ℂI⊆B′; we will prove that B′⊆B⊥+ℂI.
∀T∈B′, let A=x⊗y∈BH, then TQγx⊗yPγ=Qγx⊗yPγT. This implies that for each γ∈Γ, there exists λγ∈ℂ so that TQγ=λγQγ and PγT=λγPγ. We claim that λγ1=λγ2,∀γ1,γ2∈Γ.
LetΓ1=γ∈Γ:λγ=λγ0, then γ0∈Γ1, so Γ1≠∅. Let Γ2=γ∈Γ:λγ≠λγ0. Suppose that Γ2≠∅. For γ1∈Γ1,γ2∈Γ2, we have
(3)λγ1Qγ1Qγ2=TQγ1Qγ2=TQγ2Qγ1=λγ2Qγ2Qγ1=λγ2Qγ1Qγ2.
Sinceλγ1≠λγ2, we get Qγ1Qγ2=0. Similarily, we have Pγ2Qγ1=0, Pγ1Qγ2=0, and Pγ1Pγ2=0. Let
(4)E1=∨γ∈Γ1Pγ∨Qγ,E2=∨γ∈Γ2Pγ∨Qγ,then E1⊥E2, and ∨BPγ,Qγ:γ∈Γk⊆BEk,Ek,k=1,2. This contradicts the connection of B. Therefore, for all γ∈Γ, there exists λ∈ℂ so that TQγ=λQγ and PγT=λPγ. Thus, T−λIQγ=0=PγT−λI for all γ∈Γ. It ensures that T−λIB=BT−λI=0 for all B∈B. So, we have T−λI∈B⊥, i.e., T∈B⊥+ℂI.
On the other hand, suppose thatB is disconnected, then there exists E1,E2,F1,F2 as Definition 1. Let λ1≠λ2∈ℂ and T=λ1E1∨F1+λ2E2∨F2. ∀B∈B, write B=B1+B2, where Bi∈∨BPγ,Qγ,γ∈Γi. The fact that TB=λ1B1+λ2B2=BT implies that T∈B′. However, if B1,B2≠0,∀δ∈ℂ, we have
(5)T−δIB=TB−δB=λ1−δB1+λ2−δB2≠0.
Thus,T−δI∉B⊥,T∉B⊥+ℂI. This is a contradiction.☐Proposition 3 (see [13]).
LetM⊆BH be a masa and let B=∨BPγ,Qγ:γ∈Γ be a block-generated M-bimodule for some family of projections Pγ,Qγγ∈Γ⊆M. Then, there is a partition Γi:i∈Ω of Γ so that the subspaces Bi=∨λ∈ΓiBPλ,Qλ are connected for each i∈Ω, and i≠j∈Ω implies that Bi∨Bj is disconnected.By the proposition above, we can decomposeB into a direct sum B=⊕iBi, where each Bi is a connected subspace of BH.Definition 4.
LetM⊆BH be a masa and let B be a block-generated bimodule over M. Let B=⊕iBi be the decomposition of B as in Proposition 3 for each i∈Ω; let
(6)Ei≔∨Pγ:γ∈Γi,Fi≔∨Qγ:γ∈Γi.
We defineBEi,Fi as the block closure of Bi and Bc=⊕iBEi,Fi as the block closure of B.By the connection ofBi and Proposition 2, we have Bi′=ℂI+BFi⊥,Ei⊥=BEi,Fi′. Now,
(7)B′=⊕iBi′=∩iBi′=∩iBEi,Fi′=Bc′.LetT⊆M and suppose that A=T+B satisfies A=A′′. Then
(8)A′=T+B′=T′∩B′=T′∩Bc′⊆Bc′,it follows that Bc⊆Bc′′⊆A′′=A. It is obvious that B⊆Bc; we obtain A=T+B=T+Bc. So, we can consider the form B=⊕iBEi,Fi if A=T+B satisfies A=A′′ in the following.


## 3. The Double Commutant Theorem onBH

In this section, we discuss non-self-adjoint subalgebras ofBH which have the double commutant property. We concentrate on the case A=T+B, where T is a subalgebra of M and B∈BH is a block-generated M-bimodule. First, we consider the case B=BE,E⊥.Theorem 5.
LetM⊆BH be a masa. Suppose that A=T+BE,E⊥, where E is a projection in M and T is a subspace of M. If A=A′′, then
(1)
T=T′′(2)
for any nonzero projectionP in T, we have P≰E,P≰E⊥(3)
SpanRanE⊥AE:E⊥AE∈T′,A∈BH is dense in E⊥, where Span denotes the linear expansion of setProof.
SinceA′=T′∩BE,E⊥′, by Proposition 2, BE,E⊥′=ℂI+BE,E⊥, we have
(9)A′=ℂI+E⊥TE:T∈BH,E⊥TE∈T′.(1)
It is clear thatA′⊆T′. So, we have T′′⊆A″=A. Let S∈T′′, then there exists U∈T and T∈BH such that S=U+E⊥TE. Since E∈M,T⊆M, we get E∈T′. It follows that ES=SE,UE=EU. Thus, we obtain(10)E⊥TE=E⊥S−UE=E⊥SE−E⊥UE=E⊥ES−E⊥EU=0.
Therefore,S=U∈T. This implies that T′′⊆T. Hence, T′′=T.
(2)
We assume that there exists nonzero projectionP∈T so that P≤E⊥.
Then,EP=PE=0. Let T∈BH such that E⊥TE∈T′, then PE⊥TE=E⊥TEP=0. It follows from (9) that
(11)A′=ℂI+E⊥−PTEP⊥:E⊥TE∈T′,T∈BH.
So,E⊥−PBHP⊆A′′=T+E⊥BHE. Hence, ∀B∈BH, there exists
U∈T and A∈BH such that E⊥−PBP=U+E⊥AE. Therefore,
(12)E⊥−PBP=E⊥−PU+E⊥AEP=E⊥−PUP+E⊥−PE⊥AEP=0.
It implies thatE⊥−PBHP=0. Since BH is prime, we get P=E⊥. Together with (11), we obtain A′=ℂI. Thus, A=A′′=BH. This is a contradiction.
Similarly, we getP≰E.
(3)
LetP be a projection onto Span¯RanE⊥AE:E⊥AE∈T′,A∈BH.
Then,P≤E⊥. For any E⊥AE∈T′, we have E⊥−PE⊥AE=0 and E⊥AEE⊥−P=0. Hence, E⊥−P∈A′′=A. Therefore, there exists T∈T and B∈BH such that E⊥−P=T+E⊥BE. It follows that E⊥BE=0. So, E⊥−P∈T. From (2), we get E⊥−P=0, i.e., P=E⊥.☐Then, we concern the general caseB=⊕iBEi,Fi,i∈Ω.Theorem 6.
LetM⊆BH be a masa, and B=⊕iBEi,Fi be a block-closed bimodule over M. If A=T+B satifies A=A′′, then the following are equivalent:
(1)
∑iEi=I (convergence being in the SOT)(2)
∑iFi=I (convergence being in the SOT)Proof.
(1)⇒(2). LetT∈A′, then we have T∈B′. Thus, for all S∈BH,
TFiSEi=FiSEiT,i∈Ω. Then, for 0≠x,y∈H, TFix⊗yEi=Fix⊗yEiT,i∈Ω, it implies that there exists ηi∈ℂ so that EiT=ηiEi,i∈Ω. Therefore, T=∑iEiT=⊕iηiEi. It follows that
(13)A′⊆⊕iℂEi.
For anyi∈Ω, since ∑jEj=I, we have Fi=∑jEjFi=EiFi. It implies that Fi≤Ei,i∈Ω. From (13), we get EiBHEi⊆A′′=A=T+B for any i∈Ω. So,
(14)Ei−FiBHFi=Ei−FiEiBHEiFi⊆Ei−FiT+BFi=Ei−FiTFi+Ei−FiFiBHEiFi=0.
Thus,Ei−FiBHFi=0. Therefore, Ei−Fi=0,Ei=Fi,i∈Ω. Hence, ∑iFi=∑iEi=I.
(2)⇒(1). The argument is similar to (1)⇒(2).☐Particularly, ifT=ℂI, then we obtain the necessary and sufficient conditions for A to satisfy the double commutant property.Theorem 7.
LetM⊆BH be a masa and let B=⊕iBEi,Fi be a block-closed bimodule over M. Suppose that A=ℂI+B. Then, A=A′′ if and only if one of the following is true:
(1)
∑iEi=IandB=⊕iBEi,Ei(2)
∑iEi≠I≠∑iFiProof.
First, we prove the necessary part. LetA=A′′. If ∑iEi=I, then from the proof of Theorem 6, we have Fi=Ei,i∈Ω. So, B=⊕iBEi,Ei; if ∑iEi≠I, then ∑iEi≠I≠∑iFi. Otherwise, ∑iFi=I. Thus, we get ∑iEi=I from Theorem 6. This is a contradiction.
Now, we consider the sufficient part. If∑iEi=IandB=⊕iBEi,Ei, then A=ℂI+B=B is the von Neumann algebra with identity element. Hence, from von Neumann’s double commutant theorem, we have A=A′′; if ∑iEi≠I≠∑iFi, then E0⊥≠0≠F0⊥, where E0=∨iEi,F0=∨iFi. By Proposition 2,
(15)A′=B′=∩iBEi,Fi′=∩iℂI+BEi,Fi⊥=∩iℂI+BFi⊥,Ei⊥⊇∩iBFi⊥,Ei⊥=BF0⊥,E0⊥.
So,A′′⊆BF0⊥,E0⊥′=ℂI+BF0⊥,E0⊥⊥=ℂI+BE0,F0.
LetX∈A′′. Then, there exists T∈BH and η∈ℂ so that X=F0TE0+ηI. For any i∈Ω, let Pi be a projection onto RanEi∨Fi and let P0=I−∑iPi. It is clear that Pi∈A′,i∈Ω∪0. Thus, XPi=PiX,i∈Ω∪0. Therefore, we can write X=∑i∈Ω∪0Xi, where Xi=PiXPi. ∀i∈Ω,
(16)Xi=PiXPi=PiF0TE0+ηIPi=FiTEi+ηPi.
SinceX0=P0XP0=P0F0TE0+ηIP0=ηP0, we have
(17)X=ηP0+∑i∈ΩFiTEi+ηPi=ηI+∑i∈ΩFiTEi∈A.
So,A′′⊆A. Hence, A=A′′.☐Now, we concern the class of algebras of the formA=T+B which satisfies A=A′′, where T⊆M spans by projections, B∈BH is a block-closed M bimodule, and T∩B=0.LetM⊆BH be a masa and P∈M be an orthogonal projection. For A⊆BRanP, we denote the relative commutant of A with respect to BRanP as A†=T∈BRanP:AT=TAforallA∈A and the relative annihilator of A with respect to BRanP as A0=T∈BRanP:AT=0=TAforallA∈A.Theorem 8.
LetM⊆BH be a masa, B=⊕iBEi,Fi be a block-closed M-bimodule, and P0 be the maximal projection spanned by the projections over B, i.e., P0=⊕iFiEi. Suppose that T=SpanP1,P2,⋯,Ps is a subalgebra of M, where P0,P1,P2,⋯,Ps are mutually orthogonal projections which ranks are no less than 2 and ∑i=0sPi=I. Suppose that A=T+B, T∩B=0. Then the following are equivalent:
(1)
A=A′′(2)
For each1≤j≤s,Pj≰∑iEi, and Pj≰∑iFi
To prove the theorem, we need several lemmas.Lemma 9.
LetD,E,F∈M be nonzero projections. Let T∈BH and TX=XT for all X∈BD⊥E,DF, then TDF=λDF for some λ∈ℂ.Proof.
SinceTX=XT for all X∈BD⊥E,DF, we have TDFBHD⊥E=DFBHD⊥ET. For x0,y0∈H,TDFx0⊗y0D⊥E=DFx0⊗y0D⊥ET. Thus, TDFx0=λDFx0 and D⊥ET∗y0=λ¯D⊥E∗y0 for some λ∈ℂ. We also have TDFx⊗y0D⊥E=DFx⊗y0D⊥ET, ∀x∈H. So, we get TDFx=λDFx for all x∈H. Thus, TDF=λDF.☐Lemma 10.
LetB=⊕iBEi,Fi, then B′=BF0⊥,E0⊥+Span¯Ei∨Fi:i∈Ω, where E0=∨iEi and F0=∨iFi.Proof.
It is clear thatB′⊇BF0⊥,E0⊥+Span¯Ei∨Fi:i∈Ω.
On the other hand,∀T∈B′, since B′=∩iBEi,Fi′=∩iBEi,Fi⊥+ℂI, there exists Ti∈BEi,Fi⊥ and λi∈ℂ for all i∈Ω so that T=Ti+λiI. Since Ti∈BEi,Fi⊥=BFi⊥,Ei⊥, we have TiFi=0,i∈Ω. It follows that λiFi=T−TiFi=TFi,i∈Ω. Thus, λi≤T for each i∈Ω. Let X=∑iλiEi∨Fi, then X∈Span¯Ei∨Fi:i∈Ω and
(18)T−X=Ti+λiI−Ei∨Fi−∑k≠iλkEk∨Fk=Ti+λiI−Ei−Fi+EiFi−∑k≠iλkEk∨Fk=Ti+λiI−EiI−Fi−∑k≠iλkEk∨Fk.
BecauseI−EiI−Fi∈BEi,Fi⊥ and Ek∨Fk∈BEi,Fi⊥,k≠i, we have T−X∈BEi,Fi⊥ for all i∈Ω. Furthermore, T−X∈∩iBEi,Fi⊥=BF0⊥,E0⊥. So, T∈BF0⊥,E0⊥+Span¯Ei∨Fi:i∈Ω.☐Proof of Theorem8.
(1)⇒(2). With loss of generality, we only need to show thatP1≤∑iFi and P1≰∑iFi. We argue by contradiction. Assume P1≤∑iFi.
∀i∈Ω,P1Ei≤∑iFiEi=FiEi≤P0. So, P1Ei=P0P1Ei=0. Thus, P1⊥Ei=Ei,i∈Ω. In fact, ∀T∈P1′, TP1=P1T. We can write T=P1TP1+P1TP1⊥+P1⊥TP1+P1⊥TP1⊥. But P1TP1⊥+TP1P1⊥=0=P1⊥P1. So, T=P1TP1+P1⊥TP1⊥. Hence, P1′⊆P1BHP1+P1⊥BHP1⊥. It is clear that P1BHP1+P1⊥BHP1⊥⊆P1′. Therefore, P1′=P1BHP1+P1⊥BHP1⊥.
SinceA′⊆P1′=P1BHP1+P1⊥BHP1⊥, we have ∀T∈A′,TP1=P1T.
From above, we can writeT=T1+T4, where T1=P1TP1,T4=P1⊥TP1⊥. Since BP1⊥Ej,P1Fj=P1BEj,FjP1⊥∈A, for all S∈BP1⊥Ej,P1Fj, we have TS=ST. Then, T1S=ST4. For each i∈Ω, let Qi=P1Fi, then ∑iQi=P1∑iFi=P1. Thus, there exists i∈Ω such that Qi≠0. By reindexing if necessary, we assume that Q1≠0.
By Lemma9, for each Q1≠0, there exists αi∈C so that TQi=αiQi. So, with the decomposition of H=RanQ1⊕RanQ2⊕⋯⊕RanP1⊥, we can write
(19)T=α10000α200⋮⋮⋱⋮000T4.
Because of the arbitrary ofT,
(20)A′⊆ℂQ1⊕⋯⊕P1⊥BHP1⊥.
Thus,Q1BHQ1⊆A′′=A. For all 0≠A∈Q1BHQ1, A=Q1AQ1. Since A∈A, we can write A=∑js1⋋jPj+⊕iFiAiEi for some λj∈ℂ,1≤j≤s, and Ai∈BH,i∈Ω. Then, A=AQ1=λ1Q1, it implies that Q1BHQ1⊆ℂQ1 and Q1=P1. P1 is a rank-1 projection. It is a contradiction.
(2)⇒(1). It is clear thatA⊆A′′. We need to prove that A′′⊆A. Let T∈A′′. With the decomposition of H=RanP0⊕RanP1⊕⋯⊕RanPs, we can write T=Ti,j0≤i,j≤s, where Ti,j=PiTPj. Given 0≤j,k≤s, let Aj,k=PjAPk∣RanPk:A∈A⊆BRanPk,RanPj and A′′j,k=PjAPk∣RanPk:A∈A′′.Claim 11.
Ak,k††=Ak,k,0≤k≤s.
Fix0≤k≤s, for each i∈Ω, we define Eik=EiPk,Fik=FiPk and E0k=∑iEik,F0k=∑iFik. By the hypothesis, ∑iEik≠Pk≠∑iFik,1≤k≤s. It is clear that Pk is the identify element of BRanPk. So, by Theorem 7, we have Ak,k††=Ak,k,1≤k≤s.
Fork=0, since P0=⊕iFiEi, we have FiP0=FiEi=P0Ei=EiP0. It implies that Ei0=Fi0, and ∑iEi0=∑iFiEi=P0, by Theorem 7, A0,0††=A0,0.Claim 12.
A′′k,k⊆Ak,k,0≤k≤s.
Since for0≤k≤s,Pk∈A⊆A′′, we have Tk,k=PkTPk∈A′′ and Tk,k∈BRanPk. Let W∈A′, then WPk=PkW, 0≤k≤s. So W=W0⊕W1⊕⋯⊕Ws, where Wk=PkWPk. Thus, WT=TW implies that
(21)WkTk,r=Tk,rWr,0≤k,r≤s.
In particular,WkTk,k=Tk,kWk,0≤k≤s. Therefore, we have
(22)Tk,k∈PkA′Pk′,0≤k≤s.
For1≤k≤s, define E0k0=Pk−∑iEik,F0k0=Pk−∑iFik. Let Ak∈Ak,k†⊆BRanPk, 1≤k≤s. By Lemma 10,
(23)Ak∈BF0k0,E0k0+Span¯Eik∨Fik:i∈Ω.
Hence, there existsBk∈BF0k0,E0k0 and αi∈ℂ,i∈Ω, so that Ak=Bk+∑iαiEik∨Fik. Let Y=Bk+∑iαiEi∨Fi; now, we prove Y∈A′. Since Ei∨Fi∈A′ for each i∈Ω, we need to show that Bk∈A′. In fact,
(24)Bk=E0k0BkF0k0=PkE0k0BkF0k0Pk.
If1≤j≠k≤s,thenBkPj=PjBk=0; if 1≤j=k≤s, then PkBk=Bk=BkPk. Thus, Bk∈T′. Furthermore, for all X∈B,X=∑iFiXEi, then
(25)XBk=∑iXEiPkE0k0Bk=∑iXEikE0k0Bk=0,BkX=BkF0k0Pk∑iFiX=BkF0k0F0kX=0.
Hence,Bk∈B′. Therefore, Bk∈A′. Now, Ak=PkYPk∈PkA′Pk; this implies that Ak,k†⊆PkA0Pk. So, from (22), we have Tk,k∈PkA′Pk†⊆Ak,k††=Ak,k.
Thus,Ak,k′′⊆Ak,k,1≤k≤s.
Now, we considerA0,0=∑iEi0BHEi0. It is easy to verify that A0,0†=⊕Ei0≠0ℂEi0. Given Z∈A0,0†, there exists αi∈ℂ so that Z=∑Ei0≠0αiEi0. Let Y=∑Ei0≠0αiEi∨Fi, then Y∈A′ and Z=P0YP0∈P0A′P0. It implies that A0,0†⊆P0A′P0. So, T0,0∈P0A′P0†⊆A0,0††=A0,0 by (22). Hence, A0,0′′⊆A0,0.Claim 13.
A0,k′′⊆A0,k,Ak,0′′⊆Ak,0,1≤k≤s.
LetB∈BF0k0,E0k0⊆BRanPk, 1≤k≤s, then from Claim 12, we have B=PkBPk and B∈A′. Therefore, B0=P0BP0=0, Bk=PkBPk=B. So, from (21), we get
(26)0=B0T0,k=T0,kBk=T0,kB.
Lety0∈H satisfying F0k0y0≠0. From the above, ∀x∈H,T0,kE0k0x⊗y0F0k0=0. Thus, T0,kE0k0=0. It implies that
(27)T0,k=T0,kE0k.
For eachi∈Ω, since Ei∨Fi∈A′ and T0,k=P0TPk∈A′′, we get Ei∨FiT0,k=T0,kEi∨Fi. Thus,
(28)Ei0T0,k=P0Ei∨FiP0T0,k=T0,kPkEi∨FiPk=T0,kEik∨Fik.
Together with (27) and (28), we have
(29)Fi0T0,k=Ei0T0,k=T0,kE0kEik∨Fik=T0,kEik.
So,T0,k∈⊕iBEik,Fi0∈A0,k.
A similar argument shows thatTk,0∈Ak,0.Claim 14.
A′′m,n⊆Am,n,1≤m≠n≤s.
LetY∈BF0m0,E0m0, then Y=PmYPm and Y∈A′ by Claim 12. So, we have Yn=PnYPn=0, Ym=PmYPm=Y. Together with (21), we have
(30)0=Tm,nYn=YmTm,n=YTm,n.
Thus,Tm,n=F0mTm,n. Similarly, Tm,n=Tm,nE0n. Therefore,
(31)Tm,n=F0mTm,nE0n.
SinceEi∨Fi∈A′, i∈Ω, Tm,n∈A′′, we have Ei∨FiPmTm,n=Tm,nPnEi∨Fi. So, Eim∨FimTm,n=Tm,nEin∨Fin. By (31), FimTm,n=Eim∨FimF0mTm,n=Tm,nE0nEin∨Fin=Tm,nEin. Thus, Tm,n∈⊕iBEin,Fim∈Am,n.
Now from the above, for any0≤i,j≤s, we have A′′i,j⊆Ai,j. So, A′′⊆A.
We complete the proof.



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*Source: 6654100-2021-08-13.xml*