2-LOCAL ISOMETRIES OF SOME NEST ALGEBRAS

Abstract

Let H be a complex separable Hilbert space with $\dim H \geq 2$. Let $\mathcal {N}$ be a nest on H such that $E_+ \neq E$ for any $E \neq H, E \in \mathcal {N}$. We prove that every 2-local isometry of $\operatorname {Alg}\mathcal {N}$ is a surjective linear isometry.

1 Introduction

Let X be a Banach space and $B(X)$ the algebra of all bounded linear operators on X. Suppose that S is a subset of $B(X)$ . Following [4, 6], a map $\phi : X \rightarrow X$ (which is not assumed to be linear) is called a 2-local S-map if for any $a, b \in X$ , there exists $\phi _{a, b} \in S$ , depending on a and b, such that

$$ \begin{align*} \phi_{a, b}(a)=\phi(a) \quad\text{and}\quad \phi_{a, b}(b)=\phi(b). \end{align*} $$

Here, X is said to be 2-S-reflexive if every 2-local S-map belongs to S.

The concept of a 2-local S-map dates back to the paper [13], where Šemrl investigated 2-local automorphisms and 2-local derivations, motivated by Kowalski and Słodkowski [5]. Then in [8], the earliest investigation of 2-local $\operatorname {Iso}(X)$ -maps (also called 2-local isometries in some papers) was carried out by Molnár, where $\operatorname {Iso}(X)$ denotes the set of all surjective linear isometries of X. By an isometry of X, we mean a function $\varphi : X \rightarrow X$ such that $\|\varphi (a)-\varphi (b)\|=\|a-b\|$ for all $a, b \in X$ . In [8], Molnár proved that $B(H)$ is 2- $\operatorname {Iso}(B(H))$ -reflexive, where H is an infinite-dimensional separable Hilbert space. Recently, there has been a growing interest in 2- $\operatorname {Iso}(X)$ -reflexive problems for several operator algebras and function algebras (see, for example, [1, 9, 12]). However, the 2- $\operatorname {Iso}(X)$ -reflexivity in the context of nest algebras has not yet been considered. In this paper, we study 2- $\operatorname {Iso}(X)$ -reflexivity in some nest algebras.

Throughout, H will denote a separable Hilbert space over $\mathbb {C}$ with $\dim H \geq 2$ , along with its dual space $H^*$ . For a subset $S \subseteq H$ , we set $S^\perp := \{f \in H^* : f(S)=0\}$ .

By a subspace lattice on H, we mean a collection $\mathcal {L}$ of closed subspaces of H with $(0)$ and H in $\mathcal {L}$ such that, for every family $\{E_r\}$ of elements of $\mathcal {L}$ , both $\bigvee \{E_r\}$ and $\bigwedge \{E_r\}$ belong to $\mathcal {L}$ , where $\bigvee \{E_r\}$ denotes the closed linear span of $\{E_r\}$ and $\bigwedge \{E_r\}$ denotes the intersection of $\{E_r\}$ . We say a subspace lattice is a nest if it is totally ordered with respect to inclusion. When there is no confusion, we identify the closed subspace and the orthogonal projection on it.

Let $\mathcal {L}$ be a subspace lattice on H and $E \in \mathcal {L}$ . Define

$$ \begin{align*} \begin{aligned} E_{-} & =\bigvee\{F \in \mathcal{L}: F \nsupseteq E\}\quad \text{for } E \neq (0);\quad (0)_-=(0),\\E_{+} & =\bigwedge\{F \in \mathcal{L}: F \nsubseteq E\}\quad \text{for } E \neq H;\quad H_+=H,\\\mathcal{J}(\mathcal{L}) & =\{E \in \mathcal{L}: E \neq(0) \text{ and } E_{-} \neq H\}. \end{aligned} \end{align*} $$

If $\mathcal {N}$ is a nest on H, then it is not difficult to verify that

$$ \begin{align*} H=\bigvee\{E: E \in \mathcal{J}(\mathcal{N})\} \quad\text{and}\quad (0)=\bigwedge \{E_{-}: E \in \mathcal{J}(\mathcal{N})\}. \end{align*} $$

It follows that the subspaces $\bigcup \{E: E \in \mathcal {J}(\mathcal {N})\}$ and $\bigcup \{E_{-}^{\perp }: E \in \mathcal {J}(\mathcal {N})\}$ are both dense in H and $H^*$ , respectively, where $E_{-}^{\perp }=(E_{-})^{\perp }$ .

Denote by $B(H)$ , $K(H)$ and $F(H)$ the algebra of all bounded linear operators on H, the algebra of all compact operators on H and the algebra of all bounded finite rank operators on H, respectively.

By a nest algebra $\operatorname {Alg}\mathcal {N}$ , we mean the set of all operators in $B(H)$ leaving each element in $\mathcal {N}$ invariant, that is, $\operatorname {Alg}\mathcal {N}=\{T \in B(H): T E \subseteq E \text { for all } E \in \mathcal {N}\}$ . Denote $F(\mathcal {N})=\operatorname {Alg}\mathcal {N}\cap F(H)$ and $K(\mathcal {N})=\operatorname {Alg}\mathcal {N}\cap K(H)$ .

For $x \in H$ and $f \in H^*$ , the rank-one operator $x \otimes f$ is defined as the map $z \mapsto f(z)x$ . The following well-known result about rank-one operators will be repeatedly used.

Proposition 1.1 [7].

If $\mathcal {L}$ is a subspace lattice, then $x \otimes y \in \operatorname {Alg}\mathcal {L}$ if and only if there exists an element $E \in \mathcal {L}$ such that $x \in E$ and $y \in E_-^\perp $ .

2 Main result

Our main result is the following theorem.

Theorem 2.1. Let $\mathcal {N}$ be a nest on H such that $E_+ \neq E$ for any $E \neq H, E \in \mathcal {N}$ . If $\phi $ is a 2-local isometry of $\operatorname {Alg}\mathcal {N}$ , then $\phi $ is a surjective linear isometry.

The proof of Theorem 2.1 will be organised in a series of lemmas. In what follows, $\mathcal {N}$ is a nest on H such that $E_+ \neq E$ for any $E \neq H, E \in \mathcal {N}$ and $\phi $ is a 2-local isometry of $\operatorname {Alg}\mathcal {N}$ . For $A, B \in \operatorname {Alg}\mathcal {N}$ , the symbol $\phi _{A, B}$ stands for a surjective linear isometry from $\operatorname {Alg}\mathcal {N}$ to itself such that $\phi _{A, B}(A)=\phi (A)$ and $\phi _{A, B}(B)=\phi (B)$ . For a nest $\mathcal {M}$ , we denote by $\mathcal {M}^{\perp }$ the nest $\{I-E: E \in \mathcal {M}\}$ . A conjugation is a conjugate linear map on H such that $J^2=I$ and $ \langle J x, y\rangle = \langle J y, x\rangle $ for all $x, y \in H$ .

Proposition 2.2 below is cited from the paper by Moore and Trent [11] where they summarise the results in [2, 10] and characterise the surjective linear isometries on nest algebras.

Proposition 2.2. Let $\mathcal {M}$ be a nest on H and $\rho : \operatorname {Alg}\mathcal {M} \rightarrow \operatorname {Alg}\mathcal {M}$ be a surjective linear isometry. Then there are unitary operators U and V in $B(H)$ such that U and $U^*$ lie in $\operatorname {Alg}\mathcal {M}$ . Moreover, one of the following cases holds:

1. (1) $\rho (A)=U V^* A V$ for every $A \in \operatorname {Alg} \mathcal {M}$ and the map $E \mapsto V^* E V$ is an order isomorphism of $\mathcal {M}$ ;

2. (2) $\rho (A)=U V^* J A^* J V$ for every $A \in \operatorname {Alg}\mathcal {M}$ , where J is a conjugation on H such that $JE=EJ$ for each $E \in \mathcal {M}$ and the map $E \mapsto V^* J E J V$ is an order isomorphism from $\mathcal {M}$ onto $\mathcal {M}^{\perp }$ .

Remark 2.3. (1) It can be easily verified that the map $T \mapsto JT^*J$ is a *-anti-isomorphism of $B(H)$ and J maps an orthonormal basis onto another orthonormal basis.

(2) For any $a, b \in H$ ,

$$ \begin{align*}\begin{aligned} \langle (Jf \otimes Jx)a, b \rangle&=\langle \langle a, Jx \rangle Jf, b\rangle= \langle a, Jx \rangle \langle Jf, b \rangle= \langle x, Ja\rangle \langle Jb, f \rangle\\ &= \langle \langle Jb,f \rangle x, Ja \rangle = \langle (x \otimes f)Jb, Ja \rangle = \langle a, J(x \otimes f)Jb\rangle, \end{aligned}\end{align*} $$

so $(Jf \otimes Jx)^*=J(x \otimes f)J$ .

(3) If $\rho $ is a surjective linear isometry of $\operatorname {Alg}\mathcal {M}$ , then according to Proposition 2.2, for any rank-one operator $x \otimes f \in \operatorname {Alg} \mathcal {M}$ , $\rho $ maps $x \otimes f$ to either $UV^*x \otimes V^*f$ or $UV^*Jf \otimes V^*Jx$ , both of which are also rank-one operators. Since every finite rank operator in $\operatorname {Alg}\mathcal {M}$ can be written as a sum of finitely many rank-one operators in $\operatorname {Alg}\mathcal {M}$ and $\rho $ preserves linear independence, it follows that $\rho $ preserves the rank of a finite rank operator. Since $\rho ^{-1}$ is also a surjective linear isometry, $\rho $ preserves the rank in both directions.

Lemma 2.4. $\phi $ is rank preserving and $\phi |_{F(\mathcal {N})}$ is linear.

Proof. It follows from Remark 2.3 that $\phi $ is rank preserving. According to Proposition 2.2, $\phi _{A, B}(X)=U_{A, B}V_{A, B}^*XV_{A, B}$ or $\phi _{A, B}(X)=U_{A, B}V_{A, B}^*JX^*JV_{A, B}$ , where $U_{A, B}$ and $V_{A, B}$ are unitary operators in $B(H)$ depending on $A, B$ and $U_{A, B}, U_{A, B}^*$ lie in $\operatorname {Alg}\mathcal {N}$ .

First, we show that $\phi $ is complex homogeneous. For any $A \in \operatorname {Alg}\mathcal {N}$ and $\lambda \in \mathbb {C}$ , $\phi (\lambda A)=\phi _{A, \lambda A}(\lambda A)=\lambda \phi _{A, \lambda A}(A)=\lambda \phi (A)$ .

Next, we prove that $\phi $ is additive on $F(\mathcal {N})$ . For any $A, B \in F(\mathcal {N})$ , since $\phi $ is rank preserving, $\phi (A)$ and $\phi (B)$ are in $F(\mathcal {N})$ . We claim that $\operatorname {tr}(\phi (A)\phi (B)^*)=\operatorname {tr}(AB^*)$ . Indeed, if $\phi _{A, B}(X)=U_{A, B}V_{A, B}^*XV_{A, B}$ , then

$$ \begin{align*} \operatorname{tr}(\phi(A)\phi(B)^*)=\operatorname{tr}(U_{A, B}V_{A, B}^*AV_{A, B}V^*_{A, B}B^*V_{A, B}U_{A, B}^*)=\operatorname{tr}(AB^*). \end{align*} $$

If $\phi _{A, B}(X)=U_{A, B}V_{A, B}^*JX^*JV_{A, B}$ , then

$$ \begin{align*}\begin{aligned} \operatorname{tr}(\phi(A)\phi(B)^*)&=\operatorname{tr}(U_{A, B}V_{A, B}^*JA^*JV_{A, B}V^*_{A, B}(JB^*J)^*V_{A, B}U_{A, B}^*)\\ &=\operatorname{tr}(U_{A, B}V_{A, B}^*JA^*JV_{A, B}V^*_{A, B}JBJV_{A, B}U_{A, B}^*) =\operatorname{tr}(JA^*BJ) =\operatorname{tr}(AB^*). \end{aligned} \end{align*} $$

Thus, for any $A, A^\prime \in F(\mathcal {N})$ , by the linearity of $\operatorname {tr}$ ,

$$ \begin{align*} \operatorname{tr}((\phi(A+A^{\prime})-\phi(A)-\phi(A^{\prime})) \phi(B)^*)=\operatorname{tr}(((A+A^{\prime})-A-A^{\prime}) B^*)=0. \end{align*} $$

By replacing B with $A+A^\prime $ , A and $A^\prime $ , we obtain

$$ \begin{align*} \operatorname{tr} ((\phi(A+A^{\prime})-\phi(A) - \phi(A^{\prime}))(\phi(A+A^{\prime})-\phi(A)-\phi(A^{\prime}))^*)=0. \end{align*} $$

It follows that $\phi (A+A^{\prime })-\phi (A)-\phi (A^{\prime })=0$ , which means that $\phi $ is additive on $F(\mathcal {N})$ .

By Lemma 2.4 and [3, Corollary 2.2] where Hou and Cui characterise rank-1 preserving linear maps between nest algebras acting on Banach spaces, we can easily prove Lemma 2.5.

Lemma 2.5. One of the following statements holds.

1. (1) There exist injective linear transformations

   $$ \begin{align*} D: \bigcup\{E: E \in \mathcal{J}(\mathcal{N})\} \rightarrow H \quad\text{and}\quad C: \bigcup\{E_{-}^{\perp}: E \in \mathcal{J}(\mathcal{N})\} \rightarrow H^* \end{align*} $$
   such that $\phi (x \otimes f)=D x \otimes C f$ for every $x \otimes f \in F(\mathcal {N})$ .

2. (2) There exist injective linear transformations

   $$ \begin{align*} D: \bigcup\{E_{-}^{\perp}: E \in \mathcal{J}(\mathcal{N})\} \rightarrow H \quad\text{and}\quad C: \bigcup\{E: E \in \mathcal{J}(\mathcal{N})\} \rightarrow H^* \end{align*} $$
   such that $\phi (x \otimes f)=D f \otimes C x$ for every $x \otimes f \in F(\mathcal {N})$ .

By categorising and discussing the two cases in Lemma 2.5, we can obtain the following result.

Lemma 2.6. One of the following statements holds.

1. (1) There exist unitary operators $C, D \in B(H)$ such that $\phi (A)=DAC^*$ for any $A \in K(\mathcal {N})$ .

2. (2) There exist bounded conjugate linear operators $C, D$ such that $CJ, DJ \in B(H)$ are unitary operators and $\phi (A)=(DJ)JA^*J(CJ)^*$ for any $A \in K(\mathcal {N})$ .

Proof. We consider two cases.

Case 1. If Lemma 2.5(1) holds, then based on the assumption on $\mathcal {N}$ , there exist injective linear transformations $D: \bigcup \{E : E \in \mathcal {J}(\mathcal {N})\}\rightarrow H $ and $C: H^* \rightarrow H^*$ such that $\phi (x \otimes f)=D x \otimes C f$ for every $x \otimes f \in F(\mathcal {N})$ . Thus, for any $x \otimes f \in \operatorname {Alg}\mathcal {N}$ ,

$$ \begin{align*} \|Dx\|\, \|Cf\|=\|Dx \otimes Cf\|=\|\phi(x \otimes f)-\phi(0)\|=\|x \otimes f -0\|=\|x\|\, \|f\|.\end{align*} $$

Fix $x_0 \neq 0 \in (0)_+$ . Then $x_0 \otimes f$ is in $\operatorname {Alg}\mathcal {N}$ for any $f \neq 0, f \in ((0)_+)_-^\perp =H^*$ . It follows that $\|Dx_0\|\,\|Cf\| =\|x_0\|\,\|f\|$ . So ${\|Cf\|}/{\|f\|}={\|x_0\|}/{\|Dx_0\|}$ for any $f \neq 0, f \in H^*$ , which means that $C \in B(H^*)$ and $\|C\|={\|x_0\|}/{\|Dx_0\|}$ .

For any $E \in \mathcal {J}(\mathcal {N})$ , fix $f_0 \neq 0, f \in E_-^\perp $ . Then $x \otimes f_0 \in \operatorname {Alg}\mathcal {N}$ for any $x \neq 0, x \in E$ . It follows that $\|Dx\|\,\|Cf_0\| =\|x\|\,\|f_0\|$ . Therefore, ${\|Dx\|}/{\|x\|} ={\|f_0\|}/{\|Cf_0\|} ={\|Dx_0\|}/{\|x_0\|}$ , which means that $\|D|_E\| = {\|Dx_0\|}/{\|x_0\|}$ . Since $\bigcup \{E : E \in \mathcal {J}(\mathcal {N})\}$ is dense in H, we can extend D to an operator in $B(H)$ also denoted by D such that ${\|Dx\|}/{\|x\|}= {\|Dx_0\|}/{\|x_0\|}$ for any $x \neq 0, x \in H$ . So we can assume that $C, D$ are isometries. Since $\phi $ is an isometry, by the linearity of $\phi |_{F(\mathcal {N})}$ and the continuity of $\phi $ , we have $\phi (A)=DAC^*$ for all $A \in K(\mathcal {N})$ .

By the Riesz–Frechet theorem, $H^*$ can be identified with H through a conjugate linear surjective isometry. For any $E \neq H, E \in \mathcal {N}$ , we have $(E_+)_-=E$ by the hypothesis on $\mathcal {N}$ . Thus, x is in $(E_+)_-^{\perp }$ for any $x \in E_+ \ominus E$ , and so $x \otimes x \in \operatorname {Alg}\mathcal {N}$ . Let $\mathcal {N}=\{E_j:j \in \Omega \}$ and $\{e_i^j : i \in \Lambda _j\}$ be an orthonormal basis of $(E_j)_+ \ominus E_j$ . Then $K:=\sum _{i,j} e_i^j \otimes e_i^j/(i\cdot j)$ is a compact operator in $\operatorname {Alg}\mathcal {N}$ . Moreover, K is an injective operator with dense range. We claim that $\phi (K)$ is also an injective operator with dense range.

For the case when $\phi (K)=U_{K, 0}V_{K, 0}^*KV_{K, 0}$ , since $U_{K, 0}, V_{K, 0}$ are unitary operators, $\phi (K)$ is also an injective operator with dense range.

For the case when $\phi (K)=U_{K, 0}V_{K, 0}^*JK^*JV_{K, 0}$ , since $\operatorname {Ker} K= (\operatorname {Ran} K^*)^\perp $ , $K^*$ is an injective operator with dense range. As J is a conjugate linear isometry, it follows that $\phi (K)$ is also an injective operator with dense range.

Therefore, $\phi (K)=\sum _{i,j} De_i^j \otimes Ce_i^j/(i\cdot j)$ is an injective operator with dense range, which implies D and C have dense ranges. Consequently, D and C are surjective isometries (unitary operators).

Case 2. If Lemma 2.5(2) holds, then there exist injective linear transformations $D: H^* \rightarrow H$ and $C: \bigcup \{E \in \mathcal {N} \mid E_{-} \neq H\} \rightarrow H^*$ such that $\phi (x \otimes f)=D f \otimes C x$ for every $x \otimes f \in F(\mathcal {N})$ .

According to the Riesz–Frechet theorem, we can consider D as an injective conjugate linear transformation from H to H, and C as an injective conjugate linear transformation from $\bigcup \{E \in \mathcal {N} \mid E_{-} \neq H\}$ to H. Similarly to Case 1, we can conclude that $DJ$ and $CJ$ are unitary operators. By Remark 2.3,

$$ \begin{align*} \phi(x \otimes f)=Df \otimes Cx & =(DJ)(Jf \otimes Jx)(CJ)^* \\ & =(DJ)(J(x \otimes f)J)^*(CJ)^* =(DJ)(J(x \otimes f)^*J)(CJ)^* \end{align*} $$

for any $x \otimes f \in \operatorname {Alg}\mathcal {N}$ . By the linearity of $\phi |_{F(\mathcal {N})}$ and the continuity of $\phi $ , we have $\phi (A)=(DJ)(JA^*J)(CJ)^*$ for any $A \in K(\mathcal {N})$ .

Lemma 2.7. $\phi (P)\phi (T)^*\phi (P)=\phi (PT^*P)$ for any $T \in \operatorname {Alg}\mathcal {N}$ and any $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ .

Proof. By Lemma 2.2, $\phi _{P, T}(X)=U_{P, T}V_{P, T}^*XV_{P, T}$ or $\phi _{P, T}(X)=U_{P, T}V_{P, T}^*JX^*JV_{P, T}$ . To simplify the notation, denote $U_{P, T}, V_{P, T}$ by $U, V$ , respectively. For $\phi _{P, T}(X)=UV^*XV$ ,

$$ \begin{align*} \begin{aligned} \phi(P)\phi(T)^*\phi(P)&=UV^*PV (UV^*TV)^* UV^*PV =UV^*PT^*PV =UV^*\langle T^*x, f \rangle PV\\ &=\langle T^*x, f \rangle UV^*PV =\langle T^*x, f \rangle \phi(P) =\phi(\langle T^*x, f \rangle P) =\phi(PT^*P). \end{aligned} \end{align*} $$

For $\phi _{P, T}(X)=UV^*JX^*JV$ , using Remark 2.3,

$$ \begin{align*} \begin{aligned} \phi(P)\phi(T)^*\phi(P) &=UV^*JP^*JV (UV^*JT^*JV)^* UV^*JP^*JV =UV^*JP^*TP^*JV\\ &=UV^*J(PT^*P)^*JV =UV^*J(\langle T^*x, f \rangle x \otimes f)^*JV\\ &=\langle T^*x, f \rangle UV^*J(x \otimes f)^*JV\\ &=\langle T^*x, f \rangle\phi(P) =\phi(\langle T^*x, f \rangle P) =\phi(PT^*P). \end{aligned} \end{align*} $$

Furthermore, if $\phi $ is the form in Lemma 2.6(1), then $DPC^* \phi (T)^* DPC^*=DPT^*PC^*$ , which implies that

(2.1) $$ \begin{align} P(C^* \phi(T)^* D-T^*)P=0 \end{align} $$

for any $T \in \operatorname {Alg}\mathcal {N}$ and $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ .

If $\phi $ is the form in Lemma 2.6(2), then it follows that

$$ \begin{align*} \begin{aligned} (DJ)JP^*J(CJ)^* \phi(T)^* (DJ)JP^*J(CJ)^*&=(DJ)J(PT^*P)^*J(CJ)^*\\ &=(DJ)(JP^*J)(JTJ)(JP^*J)(CJ)^*, \end{aligned} \end{align*} $$

which implies that

(2.2) $$ \begin{align} (JP^*J)((CJ)^* \phi(T)^* (DJ)-(JTJ))(JP^*J)=0 \end{align} $$

for any $T \in \operatorname {Alg}\mathcal {N}$ and any $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ .

Under the assumption on $\mathcal {N}$ , Lemmas 2.8 and 2.9 follow from Proposition 2.2.

Lemma 2.8. Let $\rho : \operatorname {Alg}\mathcal {N} \rightarrow \operatorname {Alg}\mathcal {N}$ be a surjective linear isometry. If Case (1) in Proposition 2.2 holds for $\rho $ , then $V, V^*$ are in $\operatorname {Alg}\mathcal {N}$ .

Proof. It is sufficient to show that $V^* E V=E$ for all $E \in \mathcal {N}$ . We prove it by the principle of transfinite induction.

It is evident that $V^*(0)V=(0)$ . Moreover, for any given $F \in \mathcal {N}$ , if the equation $V^* G V=G$ holds for all $G \in \mathcal {N}$ such that $G < F$ , then because $E \mapsto V^* E V$ is an order isomorphism from $\mathcal {N}$ onto $\mathcal {N}$ , it follows that $V^* F V = F$ .

Lemma 2.9. Let $\rho : \operatorname {Alg}\mathcal {N} \rightarrow \operatorname {Alg}\mathcal {N}$ be a surjective linear isometry. If Case (2) in Proposition 2.2 holds for $\rho $ , then the following statements hold.

1. (1) $E_- \neq E$ for any $E \neq (0), E \in \mathcal {N}$ .

2. (2) $\mathcal {N}$ is finite.

3. (3) We denote $\mathcal {N}=\{E_0, E_1, \ldots , E_n\}$ where $(0)=E_0 < E_1 < \cdots <E_n = H$ . Then $V^*$ and V both map $E_i$ onto $I-E_{n-i}$ for $0 \leq i \leq n$ .

Proof. (1) In the nest $\mathcal {N^\perp }$ , we denote $E_+^{\mathcal {N^\perp }}=\bigwedge \{F \in \mathcal {N^\perp }: F \nsubseteq E\}$ for any $E \neq H, E \in \mathcal {N^\perp }$ , and $E_-^{\mathcal {N^\perp }}=\bigvee \{F \in \mathcal {N^\perp }: F \nsupseteq E\}$ for any $E \neq (0), E \in \mathcal {N^\perp }$ .

Since the map $\pi : E \mapsto V^* E V$ is an order isomorphism from $\mathcal {N}$ onto $\mathcal {N}^{\perp }$ , we have $(I-E)_+^{\mathcal {N^\perp }}\neq (I-E)$ for any $I-E \neq H, I-E \in \mathcal {N^\perp }$ . So

$$ \begin{align*} I-E\!\neq\! (I-E)_+^{\mathcal{N^\perp}}\!=\!\!\bigwedge\{I-F \in \mathcal{N^\perp}: I-F> I-E\}\!=\!\!\bigwedge\{I-F \in \mathcal{N^\perp}: F < E\}\!=\!I-E_- \end{align*} $$

for any $I-E \neq H, I-E \in \mathcal {N^\perp }$ . It follows that $E_- \neq E$ for any $E \neq (0) \in \mathcal {N}$ .

(2) Suppose that $\mathcal {N}$ is infinite, then there is a sequence $\{E_i: i \in \mathbb {N}\} \subseteq \mathcal {N}$ such that $E_i \neq (0)$ or H for any $i \in \mathbb {N}$ and $E_i < E_j$ when $i < j$ . Let $G=\bigvee \{E_i: i \in \mathbb {N}\}$ . Then $G_-=\bigvee \{F \in \mathcal {N}: F < G\}\supseteq \bigvee \{E_i: i \in \mathbb {N}\}=G$ which contradicts $G_- \neq G$ . This implies that $\mathcal {N}$ is finite.

(3) Since $E \mapsto V^* J E J V$ is an order isomorphism from $\mathcal {N}$ onto $\mathcal {N}^{\perp }$ and $EJ=JE$ for any $E \in \mathcal {N}$ , we obtain $E_i \mapsto V^*E_iV=I-E_{n-i}$ for $0 \leq i \leq n$ . Since V is a unitary operator, it follows that $V^*$ and V both map $E_i$ onto $I-E_{n-i}$ for $0 \leq i \leq n$ .

Using the characterisation of the $\phi _{A, B}$ provided by Proposition 2.2, we divide the proof of Theorem 2.1 into two lemmas based on whether $\mathcal {N}$ is isomorphic to $\mathcal {N}^\perp $ .

Lemma 2.10. If $\mathcal {N}$ is not order isomorphic to $\mathcal {N}^\perp $ , then $\phi $ is a surjective linear isometry.

Proof. Since $\mathcal {N}$ is not order isomorphic to $\mathcal {N}^\perp $ , every surjective linear isometry of $\operatorname {Alg}\mathcal {N}$ is of the form in Proposition 2.2(1). We distinguish two cases according to Lemma 2.6.

Case 1. Suppose that Lemma 2.6(1) holds, that is, $\phi (A)=DAC^*$ for every $A \in K(\mathcal {N})$ where $C, D$ are unitary operators. We claim that C and D are both in $\operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ .

For any fixed $E \in \mathcal {N}$ , if $x \neq 0, x \in E$ and $f \neq 0, f \in E_-^\perp $ , then it follows from $\phi (x \otimes f)=Dx \otimes Cf=U_{T, x \otimes f}V^*_{T, x \otimes f}(x \otimes f)V_{T, x \otimes f}$ that

$$ \begin{align*}Dx=\lambda_{T, x \otimes f} U_{T, x \otimes f}V_{T, x \otimes f}^*x \quad\text{and}\quad Cf=\frac{1}{{\overline{\lambda}_{T, x \otimes f}}}V_{T, x \otimes f}^*f,\end{align*} $$

where $\lambda _{T, x \otimes f} \in \mathbb {C}$ is on the unit circle.

By Proposition 2.2 and Lemma 2.8, $U_{T, x \otimes f}, V_{T, x \otimes f}$ are both in $\operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ . Fix $x_0 \neq 0, x_0 \in (0)_+$ . Then $x_0 \otimes f$ is in $\operatorname {Alg}\mathcal {N}$ for any $f \neq 0, f \in H$ . Thus, for any $E \neq (0), E \in \mathcal {N}$ , we have $Cf=V_{T, x_0 \otimes f}^*f/{\overline {\lambda }_{T, x_0 \otimes f}} \in E$ for any $f \neq 0, f \in E$ . Also, for any $E \neq H, E \in \mathcal {N}$ , we have $Cf=V_{T, x_0 \otimes f}^*f/{\overline {\lambda }_{T, x_0 \otimes f}} \in E^\perp $ for any $f \neq 0, f \in E^\perp $ . This shows that C is in $\operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ .

For any fixed $E \in \mathcal {J}(\mathcal {N})$ , there exists an $f_0 \neq 0, f_0 \in E_-^\perp $ . It follows that $Dx=\lambda _{T, x \otimes f_0} U_{T, x \otimes f_0}V_{T, x \otimes f_0}^*x \in E$ for any $x \neq 0, x \in E$ , which means that $D \in \operatorname {Alg}\mathcal {N}$ .

Fix $E \in \mathcal {J}(\mathcal {N})$ . Then, for any $y \in E$ and any $x \in E^\perp \cap (\bigcup \{F: F \in \mathcal {J}(\mathcal {N})\})$ ,

$$ \begin{align*} \begin{aligned} \langle x, D^*y \rangle&=\langle Dx, y \rangle=\langle \lambda_{T, x \otimes f} U_{T, x \otimes f}V_{T, x \otimes f}^*x, y \rangle\\ &=\langle x, \lambda_{T, x \otimes f}^*V_{T, x \otimes f}U_{T, x \otimes f}^* y \rangle \in \langle x, E \rangle = \{0\}. \end{aligned} \end{align*} $$

So $D^*E \perp (E^\perp \cap (\bigcup \{F: F \in \mathcal {J}(\mathcal {N})\}))$ . Since $E^\perp \cap (\bigcup \{F: F \in \mathcal {J}(\mathcal {N})\})$ is dense in $E^\perp $ , it follows that $D^* \in \operatorname {Alg}\mathcal {N}$ . This completes the claim.

For any $T \in \operatorname {Alg}\mathcal {N}$ , denote $G:=C^* \phi (T)^* D-T^*$ . By (2.1), $f(Gx)x \otimes f=0$ for any $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ . Thus, G maps $E_+$ into E for any $E \neq H, E \in \mathcal {N}$ . It is clear that G is in $\operatorname {Alg}\mathcal {N}^\perp $ , and hence G maps every $E^\perp \in \mathcal {N}^\perp $ into $E^\perp $ . It follows that G maps $E_+ \ominus E=E_+ \cap E^\perp $ into $E \cap E^\perp $ for any $E \neq H, E \in \mathcal {N}$ which yields $G=0$ and $\phi (T)=DTC^*$ .

Case 2. Suppose that Lemma 2.6(2) holds, that is, $\phi (x \otimes f)=D f \otimes C x$ for every $x \otimes f \in \operatorname {Alg}\mathcal {N}$ where $C, D$ are conjugate linear operators such that $CJ, DJ \in B(H)$ are unitary operators.

Then for $x_0 \neq 0, x_0 \in (0)_+$ and linear independent $f_1, f_2 \in H$ ,

$$ \begin{align*}\phi(x_0 \otimes f_1)=Df_1 \otimes Cx_0=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}x_0 \otimes V^*_{x_0 \otimes f_1, x_0 \otimes f_2}f_1 \end{align*} $$

and

$$ \begin{align*} \phi(x_0 \otimes f_2)=Df_2 \otimes Cx_0=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}x_0 \otimes V^*_{x_0 \otimes f_1, x_0 \otimes f_2}f_2. \end{align*} $$

It follows that $Df_1$ and $Df_2$ are linearly dependent which leads to a contradiction.

In conclusion, $\phi (T)=DTC^*$ for any $T \in \operatorname {Alg}\mathcal {N}$ and it is clear that $\phi $ is a surjective linear isometry of $\operatorname {Alg}\mathcal {N}$ .

Lemma 2.11. If $\mathcal {N}$ is order isomorphic to $\mathcal {N}^\perp $ , then $\phi $ is a surjective linear isometry.

Proof. According to Lemma 2.9, $\mathcal {N}$ is finite; denote $\mathcal {N}=\{E_0, E_1, \ldots , E_n\}$ where $(0)=E_0 < E_1 < \cdots <E_n = H$ . We distinguish two cases according to Lemma 2.6.

Case 1. Suppose that Lemma 2.6(1) holds, that is, $\phi (A)=DAC^*$ for every $A \in K(\mathcal {N})$ where $C, D$ are unitary operators. In this case, for any $E \in \mathcal {J}(\mathcal {N})$ satisfying $\dim E_-^\perp> 1$ , fix $x_0 \neq 0, x_0 \in E$ . For any linearly independent $f_1, f_2 \in E_-^\perp $ , we have $x_0 \otimes f_1, x_0 \otimes f_2 \in \operatorname {Alg}\mathcal {N}$ .

We claim that $\phi _{x_0 \otimes f_1, x_0 \otimes f_2}$ is not of the form in Proposition 2.2(2). Otherwise,

$$ \begin{align*} \phi(x_0 \otimes f_1)=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}J(x_0 \otimes f_1)^*JV_{x_0 \otimes f_1, x_0 \otimes f_2}=Dx_0 \otimes Cf_1 \end{align*} $$

and

$$ \begin{align*} \phi(x_0 \otimes f_2)=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}J(x_0 \otimes f_2)^*JV_{x_0 \otimes f_1, x_0 \otimes f_2}=Dx_0 \otimes Cf_2. \end{align*} $$

It follows that $f_1$ and $f_2$ are linear dependent, leading to a contradiction.

Thus, for any $f_1 \neq 0, f_1 \in H$ , there exist $x_0 \neq 0, x_0 \in (0)_+$ and $f_2 \neq 0, f_2 \in H$ such that

$$ \begin{align*} \phi(x_0 \otimes f_1)=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}(x_0 \otimes f_1)V_{x_0 \otimes f_1, x_0 \otimes f_2}=Dx_0 \otimes Cf_1. \end{align*} $$

Hence, $Dx_0=\lambda _{x_0 \otimes f_1, x_0 \otimes f_2} U_{x_0 \otimes f_1, x_0 \otimes f_2}V_{x_0 \otimes f_1, x_0 \otimes f_2}^*x_0$ and $Cf_1=V_{x_0 \otimes f_1, x_0 \otimes f_2}^*f_1/{\overline {\lambda }_{x_0 \otimes f_1, x_0 \otimes f_2}}$ for some $\lambda _{x_0 \otimes f_1, x_0 \otimes f_2} \in \mathbb {C}$ on the unit circle. By the arbitrariness of $f_1$ and $V_{x_0 \otimes f_1, x_0 \otimes f_2}^* \in \operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ , we obtain $C \in \operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ .

Similarly, for any $E \in \mathcal {N}$ with $\dim E> 1$ , fix $f_0 \in E_-^\perp $ . Let $x_1, x_2 \in E$ be any linearly independent elements. It is impossible for $\phi _{x_1 \otimes f_0, x_2 \otimes f_0}$ to be in the form of Lemma 2.2(2). Thus, for any $x_1 \neq 0, x_1 \in H$ , there exist $f_0 \neq 0, f_0 \in H_-^\perp $ and $x_2 \neq 0, x_2 \in H$ such that

$$ \begin{align*} \phi(x_1 \otimes f_0)=U_{x_1 \otimes f_0, x_2 \otimes f_0}V^*_{x_1 \otimes f_0, x_2 \otimes f_0}(x_1 \otimes f_0)V_{x_1 \otimes f_0, x_2 \otimes f_0}=Dx_1 \otimes Cf_0. \end{align*} $$

It follows that $D \in \operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ .

For any $T \in \operatorname {Alg}\mathcal {N}$ , denote $G:=C^* \phi (T)^* D-T^*$ . Using a similar method to that in Lemma 2.10, we see that G maps $E_+ \ominus E=E_+ \cap E^\perp $ into $E \cap E^\perp $ for any $E \neq H, E \in \mathcal {N}$ , which yields $G=0$ and $\phi (T)=DTC^*$ .

Case 2. Suppose that Lemma 2.6(2) holds, that is, $\phi (x \otimes f)=D f \otimes C x$ for every $x \otimes f \in \operatorname {Alg}\mathcal {N}$ where $C, D$ are conjugate linear operators such that $CJ, DJ \in B(H)$ are unitary operators.

In this case, for any $E \in \mathcal {J}(\mathcal {N})$ with $\dim E_-^\perp> 1$ , fix $x_0 \in E$ . For any linearly independent $f_1, f_2 \in E_-^\perp $ , $x_0 \otimes f_1, x_0 \otimes f_2$ are in $\operatorname {Alg}\mathcal {N}$ . It is impossible for $\phi _{x_0 \otimes f_1, x_0 \otimes f_2}$ to be in the form of Proposition 2.2(1). Otherwise,

$$ \begin{align*} \phi(x_0 \otimes f_1)=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}x_0 \otimes V^*_{x_0 \otimes f_1, x_0 \otimes f_2}f_1=Df_1 \otimes Cx_0 \end{align*} $$

and

$$ \begin{align*} \phi(x_0 \otimes f_2)=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}x_0 \otimes V^*_{x_0 \otimes f_1, x_0 \otimes f_2}f_2=Df_2 \otimes Cx_0, \end{align*} $$

implying that $f_1, f_2$ are linear dependent, which leads to a contradiction.

Thus, for any $f_1 \neq 0, f_1 \in H$ , there exist $x_0 \neq 0, x_0 \in (0)_+$ and $f_2 \neq 0, f_2 \in H$ such that

$$ \begin{align*} \phi(x_0 \otimes f_1)=U_{x_0 \otimes f_1, x_0 \otimes f_2}V^*_{x_0 \otimes f_1, x_0 \otimes f_2}J(x_0 \otimes f_1)^*JV_{x_0 \otimes f_1, x_0 \otimes f_2}=Df_1 \otimes Cx_0. \end{align*} $$

So $Df_1=\lambda _{x_0 \otimes f_1, x_0 \otimes f_2} U_{x_0 \otimes f_1, x_0 \otimes f_2}V_{x_0 \otimes f_1, x_0 \otimes f_2}^*Jf_1$ and $Cx_0=V_{x_0 \otimes f_1, x_0 \otimes f_2}^*Jx_0/{\overline {\lambda }_{x_0 \otimes f_1, x_0 \otimes f_2}}$ for some $\lambda _{x_0 \otimes f_1, x_0 \otimes f_2} \in \mathbb {C}$ on the unit circle. According to Lemma 2.9, $V_{x_0 \otimes f_1, x_0 \otimes f_2}^*$ and $V_{x_0 \otimes f_1, x_0 \otimes f_2}$ both map $E_i$ onto $I-E_{n-i}$ for $0 \leq i \leq n$ . Since $EJ=JE$ for any $E \in \mathcal {N}$ , by the arbitrariness of $f_1$ and $U_{x_0 \otimes f_1, x_0 \otimes f_2} \in \operatorname {Alg}\mathcal {N} \cap \operatorname {Alg}\mathcal {N}^\perp $ , we see that D maps $E_i$ into $I-E_{n-i}$ and $I-E_i$ into $E_{n-i}$ for $0\leq i\leq n$ , respectively.

Similarly, for any $E \in \mathcal {N}$ with $\dim E> 1$ , fix $f_0 \in E_-^\perp $ . For any linearly independent $x_1, x_2 \in E$ , $x_1 \otimes f_0, x_2 \otimes f_0$ are in $\operatorname {Alg}\mathcal {N}$ . It is impossible for $\phi _{x_1 \otimes f_0, x_2 \otimes f_0}$ to be in the form of Proposition 2.2(1). Thus, for any $x_1 \neq 0, x_1 \in H$ , there exist $f_0 \neq 0, f_0 \in H_-^\perp $ and $x_2 \neq 0, x_2 \in H$ such that

$$ \begin{align*} \phi(x_1 \otimes f_0)=U_{x_1 \otimes f_0, x_2 \otimes f_0}V^*_{x_1 \otimes f_0, x_2 \otimes f_0}J(x_1 \otimes f_0)^*JV_{x_1 \otimes f_0, x_2 \otimes f_0}=Df_0 \otimes Cx_1. \end{align*} $$

So $Df_0=\lambda _{x_1 \otimes f_0, x_2 \otimes f_0} U_{x_1 \otimes f_0, x_2 \otimes f_0}V_{x_1 \otimes f_0, x_2 \otimes f_0}^*Jf_0$ and $Cx_1=V_{x_1 \otimes f_0, x_2 \otimes f_0}^*Jx_1/{\overline {\lambda }_{x_1 \otimes f_0, x_2 \otimes f_0}}$ for some $\lambda _{x_1 \otimes f_0, x_2 \otimes f_0} \in \mathbb {C}$ on the unit circle. Since $V_{x_1 \otimes f_0, x_2 \otimes f_0}^*, V_{x_1 \otimes f_0, x_2 \otimes f_0}$ both map $E_i$ onto $I-E_{n-i}$ for any $0 \leq i \leq n$ and $EJ=JE$ for any $E \in \mathcal {N}$ , by the arbitrariness of $x_1$ , we see that C maps $E_i$ into $I-E_{n-i}$ and $I-E_i$ into $E_{n-i}$ for all $0\leq i\leq n$ , respectively.

By (2.2), $ (JP^*J)((CJ)^* \phi (T)^* (DJ)-(JTJ))(JP^*J)=0 $ for any $T \in \operatorname {Alg}\mathcal {N}$ and any $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ . So $\langle ((CJ)^* \phi (T)^* (DJ)-JTJ)Jf , Jx \rangle =0$ for all $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ which means that $((CJ)^* \phi (T)^* (DJ)-JTJ)$ maps $(E_i)_-^\perp $ into $(E_i)^\perp $ .

Moreover, for any $E_i \in \mathcal {N}$ ,

$$ \begin{align*}E_i \xrightarrow{DJ} I-E_{n-i} \xrightarrow{\phi(T)^*} I-E_{n-i} \xrightarrow{(CJ)^*} E_i,\end{align*} $$

and $JTJ$ maps $E_i$ into $E_i$ . It follows that $((CJ)^* \phi (T)^* (DJ)-JTJ)$ maps $E_i \cap (E_i)_-^\perp $ into $E_i \cap E_i^\perp = \{0\}$ . So $((CJ)^* \phi (T)^* (DJ)-JTJ)=0$ , which implies that $\phi (T)=(DJ) J T^* J (CJ)^*$ for any $T \in \operatorname {Alg}\mathcal {N}$ . It is easy to check that $\phi (T)$ is a surjective linear isometry.

Combining Lemmas 2.10 and 2.11 completes the proof of Theorem 2.1.