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 $F(\mathcal {N})$ .

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 $A \in K(\mathcal {N})$ .

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 $P=x \otimes f \in \operatorname {Alg}\mathcal {N}$ .

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 $V^* F V = F$ .

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 $0 \leq i \leq n$ .

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 $\operatorname {Alg}\mathcal {N}$ .

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.