#  Lie ring isomorphisms between nest algebras on Banach spaces

Juan Deng  ,  Jinchuan Hou  Department of Mathematics, Taiyuan University of
Technology, Taiyuan 030024, P. R. China<EMAIL_ADDRESS><EMAIL_ADDRESS>and  Xiaofei Qi  Department of Mathematics, Shanxi
University, Taiyuan, 030006, P. R. China<EMAIL_ADDRESS>
###### Abstract.

Let ${\mathcal{N}}$ and ${\mathcal{M}}$ be nests on Banach spaces $X$ and $Y$
over the (real or complex) field $\mathbb{F}$ and let $\mbox{\rm
Alg}{\mathcal{N}}$ and $\mbox{\rm Alg}{\mathcal{M}}$ be the associated nest
algebras, respectively. It is shown that a map $\Phi:{\rm
Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ is a Lie ring isomorphism
(i.e., $\Phi$ is additive, Lie multiplicative and bijective) if and only if
$\Phi$ has the form $\Phi(A)=TAT^{-1}+h(A)I$ for all $A\in\mbox{\rm
Alg}{\mathcal{N}}$ or $\Phi(A)=-TA^{*}T^{-1}+h(A)I$ for all $A\in\mbox{\rm
Alg}{\mathcal{N}}$, where $h$ is an additive functional vanishing on all
commutators and $T$ is an invertible bounded linear or conjugate linear
operator when $\dim X=\infty$; $T$ is a bijective $\tau$-linear transformation
for some field automorphism $\tau$ of $\mathbb{F}$ when $\dim X<\infty$.

2010 Mathematical Subject Classification. Primary 47L35, 47B49

Key words and phrases. Banach spaces, nest algebras, Lie ring isomorphisms

This work is supported by National Natural Science Foundation of China
(11171249,11101250, 11271217) and Youth Foundation of Shanxi Province
(2012021004).

##  1\. Introduction and main results

Let $\mathcal{R}$ and $\mathcal{R}^{\prime}$ be two associative rings. Recall
that a map $\phi:\mathcal{R}\rightarrow\mathcal{R}^{\prime}$ is called a
multiplicative map if $\phi(AB)=\phi(A)\phi(B)$ for any $A,B\in\mathcal{R}$;
is called a Lie multiplicative map if $\phi([A,B])=[\phi(A),\phi(B)]$ for any
$A,B\in\mathcal{R}$, where $[A,B]=AB-BA$ is the Lie product of $A$ and $B$
which is also called a commutator. In addition, a map
$\phi:\mathcal{R}\rightarrow\mathcal{R}^{\prime}$ is called a Lie
multiplicative isomorphism if $\phi$ is bijective and Lie multiplicative; is
called a Lie ring isomorphism if $\phi$ is bijective, additive and Lie
multiplicative. If $\mathcal{R}$ and $\mathcal{R}^{\prime}$ are algebras over
a field $\mathbb{F}$, $\phi:{\mathcal{R}}\rightarrow{\mathcal{R}}^{\prime}$ is
called a Lie algebraic isomorphism if $\phi$ is bijective, $\mathbb{F}$-linear
and Lie multiplicative. For the study of Lie ring isomorphisms between rings,
see [3, 5, 10] and the references therein. In this paper we focus our
attention on Lie ring isomorphisms between nest algebras on general Banach
spaces.

Let $X$ be a Banach space over the (real or complex) field ${\mathbb{F}}$ with
topological dual $X^{*}$. $\mathcal{B}(X)$ stands for the algebra of all
bounded linear operators on $X$. A nest $\mathcal{N}$ on $X$ is a complete
totally ordered subspace lattice, that is, a chain of closed (under norm
topology) subspaces of $X$ which is closed under the formation of arbitrary
closed linear span (denote by $\bigvee$) and intersection (denote by
$\bigwedge$), and which includes $\\{0\\}$ and $X$. The nest algebra
associated with a nest $\mathcal{N}$, denoted by ${\rm Alg}\mathcal{N}$, is
the weakly closed operator algebra consisting of all operators that leave
every subspace $N\in\mathcal{N}$ invariant. For $N\in\mathcal{N}$, let
$N_{-}=\bigvee\\{M\in\mathcal{N}\mid M\subset N\\}$ and
$N_{-}^{\perp}=(N_{-})^{\perp}$, where $N^{\perp}=\\{f\in X^{*}\mid
N\subseteq\ker(f)\\}$. If $\mathcal{N}$ is a nest on $X$, then
${\mathcal{N}}^{\perp}=\\{N^{\perp}\mid N\in{\mathcal{N}}\\}$ is a nest on
$X^{*}$ and $({\rm Alg}{\mathcal{N}})^{*}\subseteq{\rm
Alg}{\mathcal{N}}^{\perp}$. If ${\mathcal{N}}=\\{(0),X\\}$, we say that
$\mathcal{N}$ is a trivial nest, in this case,
Alg${\mathcal{N}}={\mathcal{B}}(X)$. Non-trivial nest algebras are very
important reflexive operator algebras that are not semi-simple, not semi-prime
and not self-adjoint. If $\dim X<\infty$, a nest algebra on $X$ is isomorphic
to an algebra of upper triangular block matrices. Nest algebras are studied
intensively by a lot of literatures. For more details on basic theory of nest
algebras, the readers can refer to [6, 8].

In [9], Marcoux and Sourour proved that every Lie algebraic isomorphism
between nest algebras on separable complex Hilbert spaces is a sum
$\alpha+\beta$, where $\alpha$ is an algebraic isomorphism or the negative of
an algebraic anti-isomorphism and $\beta:{\rm
Alg}{\mathcal{N}}\rightarrow{\mathbb{C}}I$ is a linear map vanishing on all
commutators, that is, satisfying $\beta([A,B])=0$ for all $A,B\in{\rm
Alg}{\mathcal{N}}$.

Qi and Hou in [11] generalized the result of Marcoux and Sourour by
classifying certain Lie multiplicative isomorphisms. Note that, a Lie
multiplicative isomorphism needs not be additive. Let ${\mathcal{N}}$ and
${\mathcal{M}}$ be nests on Banach spaces $X$ and $Y$ over the (real or
complex) field $\mathbb{F}$, respectively, with the property that if
$M\in{\mathcal{M}}$ such that $M_{-}=M$, then $M$ is complemented in $Y$
(Obviously, this assumption is not needed if $Y$ is a Hilbert space or if
$\dim Y<\infty$). Let $\mbox{\rm Alg}{\mathcal{N}}$ and $\mbox{\rm
Alg}{\mathcal{M}}$ be respectively the associated nest algebras, and let
$\Phi:{\rm Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ be a bijective
map. Qi and Hou in [11] proved that, if $\dim X=\infty$ and if there is a
nontrivial element in $\mathcal{N}$ which is complemented in $X$, then $\Phi$
is a Lie multiplicative isomorphism if and only if there exists a map
$h:\mbox{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ with $h([A,B])=0$ for
all $A,B\in\mbox{\rm Alg}{\mathcal{N}}$ such that $\Phi$ has the form
$\Phi(A)=TAT^{-1}+h(A)$ for all $A\in\mbox{\rm Alg}{\mathcal{N}}$ or
$\Phi(A)=-TA^{*}T^{-1}+h(A)$ for all $A\in\mbox{\rm Alg}{\mathcal{N}}$, where,
in the first form, $T:X\rightarrow Y$ is an invertible bounded linear or
conjugate-linear operator so that $N\mapsto T(N)$ is an order isomorphism from
$\mathcal{N}$ onto $\mathcal{M}$, while in the second form, $X$ and $Y$ are
reflexive, $T:X^{*}\rightarrow Y$ is an invertible bounded linear or
conjugate-linear operator so that $N^{\bot}\mapsto T(N^{\bot})$ is an order
isomorphism from ${\mathcal{N}}^{\bot}$ onto $\mathcal{M}$. If $\dim
X=n<\infty$, identifying nest algebras with upper triangular block matrix
algebras, then $\Phi$ is a Lie multiplicative isomorphism if and only if there
exist a field automorphism $\tau:{\mathbb{F}}\rightarrow{\mathbb{F}}$ and
certain invertible matrix $T$ such that either $\Phi(A)=TA_{\tau}T^{-1}+h(A)$
for all $A$, or $\Phi(A)=-T(A_{\tau})^{\rm tr}T^{-1}+h(A)$ for all $A$, where
$A_{\tau}=(\tau(a_{ij}))$ for $A=(a_{ij})$ and $A^{\rm tr}$ is the transpose
of $A$. Particularly, above results give a characterization of Lie ring
isomorphisms between nest algebras for finite-dimensional case, and for
infinite-dimensional case under the mentioned assumptions on $\mathcal{N}$ and
$\mathcal{M}$.

Recently, Wang and Lu in [12] generalized Marcoux and Sourour’s result from
another direction, and proved that every Lie algebraic isomorphism between
nest algebras ${\rm Alg}{\mathcal{N}}$ and ${\rm Alg}{\mathcal{M}}$ for any
nests $\mathcal{N}$ and $\mathcal{M}$ on Banach spaces $X$ and $Y$
respectively can be decomposed as $\alpha+\beta$, where $\alpha$ is an
algebraic isomorphism or the negative of an algebraic anti-isomorphism and
$\beta:{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ is a linear map
vanishing on each commutator. Because Lie algebraic isomorphisms were
characterized in [4] for the case that the nest $\mathcal{N}$ has a nontrivial
complemented element, Wang and Lu in [12] mainly dealt with the case that all
nontrivial elements of $\mathcal{N}$ are not complemented.

The purpose of the present paper is to characterize all Lie ring isomorphisms
between nest algebras of Banach space operators for any nests. Note that, the
Lie ring isomorphisms are very different from algebraic ones. For example, the
method used in [4] to characterize Lie algebraic isomorphisms for the case
that the nest $\mathcal{N}$ has a nontrivial complemented element is not valid
for characterizing Lie ring isomorphisms. Algebraic isomorphisms between nest
algebras are continuous, however ring isomorphisms are not necessarily
continuous for finite-dimensional case.

The following are the main results of this paper.

Theorem 1.1. Let ${\mathcal{N}}$ and ${\mathcal{M}}$ be nests on Banach spaces
$X$ and $Y$ over the (real or complex) field $\mathbb{F}$, and, ${\rm
Alg}{\mathcal{N}}$ and ${\rm Alg}{\mathcal{M}}$ be the associated nest
algebras, respectively. Then a map $\Phi:{\rm Alg}{\mathcal{N}}\rightarrow{\rm
Alg}{\mathcal{M}}$ is a Lie ring isomorphism, that is, $\Phi$ is additive,
bijective and satisfies $\Phi([A,B])=[\Phi(A),\Phi(B)]$ for all $A,B\in{\rm
Alg}{\mathcal{N}}$, if and only if $\Phi$ has the form $\Phi(A)=\Psi(A)+h(A)I$
for all $A\in{\rm Alg}{\mathcal{N}}$, where $\Psi$ is a ring isomorphism or
the negative of a ring anti-isomorphism between the nest algebras and
$h:\mbox{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}$ is an additive
functional satisfying $h([A,B])=0$ for all $A,B\in\mbox{\rm
Alg}{\mathcal{N}}$.

The ring isomorphisms and the ring anti-isomorphisms between nest algebras of
Banach space operators were characterized in [7, Theorem 2.2, Theorem 2.7 and
Remark 2.6]. Using these results and Theorem 1.1, we can get more concrete
characterization of Lie ring isomorphisms. Recall that a map $S:W\rightarrow
V$ with $W,V$ linear spaces over a field $\mathbb{F}$ is called $\tau$-linear
if $S$ is additive and $S(\lambda x)=\tau(\lambda)Sx$ for all $x\in W$ and
$\lambda\in{\mathbb{F}}$, where $\tau$ is a field automorphism of
$\mathbb{F}$.

Theorem 1.2. Let ${\mathcal{N}}$ and ${\mathcal{M}}$ be nests on Banach spaces
$X$ and $Y$ over the (real or complex) field $\mathbb{F}$, and let ${\rm
Alg}{\mathcal{N}}$ and ${\rm Alg}{\mathcal{M}}$ be the associated nest
algebras, respectively. Then a map $\Phi:{\rm Alg}{\mathcal{N}}\rightarrow{\rm
Alg}{\mathcal{M}}$ is a Lie ring isomorphism if and only if there exist an
additive functional $h:\mbox{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}$
satisfying $h([A,B])=0$ for all $A,B\in\mbox{\rm Alg}{\mathcal{N}}$ and a
field automorphism $\tau:{\mathbb{F}}\rightarrow{\mathbb{F}}$ such that one of
the following holds.

(1) There exists a $\tau$-linear transformation $T:X\rightarrow Y$ such that
the map $N\mapsto T(N)$ is an order isomorphism from $\mathcal{N}$ onto
$\mathcal{M}$ and

$\Phi(A)=TAT^{-1}+h(A)I\ \ {for\ all}\ \ A\in{\mbox{\rm Alg}}\mathcal{N}.$

(2) $X$ and $Y$ are reflexive, there exists a $\tau$-linear transformation
$T:X^{*}\rightarrow Y$ such that the map $N_{-}^{\perp}\mapsto
T(N_{-}^{\perp})$ is an order isomorphism from ${\mathcal{N}}^{\perp}$ onto
${\mathcal{M}}$ and

$\Phi(A)=-TA^{*}T^{-1}+h(A)I\ \ {for\ all}\ \ A\in{\mbox{\rm
Alg}}\mathcal{N}.$

Moreover, if $\dim X=\infty$, the above $T$ is in fact an invertible bounded
linear or conjugate-linear operator; if ${\mathbb{F}}={\mathbb{R}}$, $T$ is
linear.

For the finite dimensional case, it is clear that every nest algebra on a
finite dimensional space is isomorphic to an upper triangular block matrix
algebra. Let ${\mathcal{M}}_{n}({\mathbb{F}})$ denote the algebra of all
$n\times n$ matrices over $\mathbb{F}$. Recall that an upper triangular block
matrix algebra ${\mathcal{T}}={\mathcal{T}}(n_{1},n_{2},\ldots,n_{k})$ is a
subalgebra of ${\mathcal{M}}_{n}({\mathbb{F}})$ consisting of all $n\times n$
matrices of the form

$A=\left(\begin{array}[]{cccc}A_{11}&A_{12}&\ldots&A_{1k}\\\
0&A_{22}&\ldots&A_{2k}\\\ \vdots&\vdots&\ddots&\vdots\\\
0&0&\ldots&A_{kk}\end{array}\right),$

where $n_{1},n_{2},\cdot\cdot\cdot,n_{k}$ are finite sequence of positive
integers satisfying $n_{1}+n_{2}+\cdot\cdot\cdot+n_{k}=n$ and
$A_{ij}\in{\mathcal{M}}_{n_{i}\times n_{j}}({\mathbb{F}})$, the space of all
$n_{i}\times n_{j}$ matrices over $\mathbb{F}$. Thus by Theorem 1.2, we get a
characterization of Lie ring isomorphisms between upper triangular block
matrix algebras.

Corollary 1.3. Let ${\mathbb{F}}$ be the real or complex field, and $m,n$ be
positive integers greater than 1. Let
${\mathcal{T}}={\mathcal{T}}(n_{1},n_{2},\ldots,n_{k})\subseteq{\mathcal{M}}_{n}({\mathbb{F}})$
and
${\mathcal{S}}={\mathcal{T}}(m_{1},m_{2},\ldots,m_{r})\subseteq{\mathcal{M}}_{m}({\mathbb{F}})$
be upper triangular block matrix algebras, and
$\Phi:{\mathcal{T}}\rightarrow{\mathcal{S}}$ be a map. Then $\Phi$ is a Lie
ring isomorphism if and only if $m=n$, and there exist an additive functional
$\phi:{\mathcal{T}}\rightarrow{\mathbb{F}}$ satisfying $\phi([A,B])=0$ for all
$A,B\in{\mathcal{T}}$, a field automorphism
$\tau:{\mathbb{F}}\rightarrow{\mathbb{F}}$ such that either

(1) ${\mathcal{T}}={\mathcal{S}}$, there exists an invertible matrix
$T\in{\mathcal{T}}$ such that

$\Phi(A)=TA_{\tau}T^{-1}+\phi(A)I\ \ for\ \ all\ \ A\in{\mathcal{T}};$

or

(2) $(n_{1},n_{2},\ldots,n_{k})=(m_{r},m_{r-1},\ldots,m_{1})$, there exists an
invertible block matrix $T=(T_{ij})_{k\times k}$ with
$T_{ij}\in{\mathcal{M}}_{n_{i},n_{j}}({\mathbb{F}})$ and $T_{ij}=0$ whenever
$i+j>k+1$, such that

$\Phi(A)=-TA_{\tau}^{\rm tr}T^{-1}+\phi(A)I\ \ for\ \ all\ \
A\in{\mathcal{T}}.$

Where $A_{\tau}=(\tau(a_{ij}))_{n\times n}$ for $A=(a_{ij})_{n\times
n}\in{\mathcal{M}}_{n}({\mathbb{F}})$ and $A^{\rm tr}$ is the transpose of
$A$. If ${\mathbb{F}}={\mathbb{R}}$, then $\Phi$ is a Lie algebraic
isomorphism.

Corollary 1.3 is also a consequence of [11, Corollary 2.2].

Since the Lie ring isomorphisms between nest algebras on finite-dimensional
Banach spaces were already characterized in [11, Corollary 2.2], to give a
classification of all Lie ring isomorphisms between nest algebras of Banach
space operators, it suffices to prove Theorem 1.1 for the infinite-dimensional
cases without any additional assumption on the nests. It is clear that $\dim
X=\infty\Leftrightarrow\dim Y=\infty$.

The remain part of the paper is to prove the main result Theorem 1.1 under the
assumption that both $X$ and $Y$ are infinite-dimensional. Our approach borrow
and combine some ideas developed in [11] and [12]. In Section 2 we give
preliminary lemmas, some of them are also parts of the proof of the main
result. Section 3 deals with the case that both $(0)$ and $X$ are limit points
of the nest $\mathcal{N}$, that is, $(0)=(0)_{+}$ and $X_{-}=X$. The case that
$X_{-}\not=X$ and $X_{-}$ is complemented or $(0)\not=(0)_{+}$ and $(0)_{+}$
is complemented is discussed in Section 4. And finally, the case that
$X_{-}\not=X$ and $X_{-}$ is not complemented or $(0)\not=(0)_{+}$ and
$(0)_{+}$ is not complemented is considered in Section 5.

##  2\. Preliminary lemmas

In this section, we give some preliminary lemmas, definitions and symbols
which are needed in other sections to prove the main result.

Let $X$ and $Y$ be Banach spaces over $\mathbb{F}$, and let $\mathcal{N}$ and
$\mathcal{M}$ be nests on $X$ and $Y$. Let Alg$\mathcal{N}$ and
Alg$\mathcal{M}$ be associated nest algebras, respectively. It is well known
that the commutant of a nest algebra is trivial, i.e., if
$T\in{\mathcal{B}}(X)$ and $TA=AT$ for every operator $A\in$Alg$\mathcal{N}$,
then $T=\lambda I$ for some scalar $\lambda\in\mathbb{F}.$ This fact will be
used in this paper without any specific explanation. In addition, the symbols
${\rm ran}T$, $\ker T$ and ${\rm rank}T$ stand for the range, the kernel and
the rank (i.e., the dimension of ran$T$) of an operator $T$, respectively. For
$x\in X$ and $f\in X^{*}$, $x\otimes f$ stands for the operator on $X$ with
rank not greater than 1 defined by $(x\otimes f)y=f(y)x$ for every $y$. Some
times we use $\langle x,f\rangle$ to present the value $f(x)$ of $f$ at $x$.

The following lemma is also well known which gives a characterization of rank
one operators in nest algebras.

Lemma 2.1. ([8]) Let ${\mathcal{N}}$ be a nest on a (real or complex) Banach
space $X$ and $x\in X$, $f\in X^{*}$. Then $x\otimes f\in{\rm Alg}\mathcal{N}$
if and only if there exists a subspace $N\in{\mathcal{N}}$ such that $x\in N$
and $f\in N_{-}^{\perp}$.

For any non-trivial element $E\in\mathcal{N}$, define

$None$ $\mathcal{J}({\mathcal{N}},E)=\\{A\in{\rm Alg}{\mathcal{N}}:AE=0\ \
{\rm and}\ \ A^{*}E^{\perp}=0\\}.$

In [12], Wang and Lu proved that $\mathcal{L}$ is a proper maximal commutative
Lie algebra ideal in ${\rm Alg}\mathcal{N}$ if and only if
${\mathcal{L}}=\mathbb{F}I+{\mathcal{J}}({\mathcal{N}},E)$ for some unique
$E\in{\mathcal{N}}$. The following lemma shows that any maximal commutative
Lie ring ideal also rises in this way.

Lemma 2.2. ${\mathcal{J}}$ is a proper maximal commutative Lie ring ideal in
${\rm Alg}{\mathcal{N}}$ if and only if it is a proper maximal commutative Lie
algebra ideal.

Proof. Assume that $\mathcal{J}$ is a maximal commutative Lie ring ideal. Then
for any $A\in{\rm Alg}{\mathcal{N}}$, any $C\in{\mathcal{J}}$ and any
$\lambda\in\mathbb{F}$, we have $[A,\lambda
C]=\lambda[A,C]\in\mathbb{F}\mathcal{J}$, which implies that
$\mathbb{F}\mathcal{J}$ is a Lie ring ideal. It is obvious that
$\mathbb{F}\mathcal{J}$ is also commutative. So
$\mathbb{F}\mathcal{J}\subseteq\mathcal{J}$ as $\mathcal{J}$ is maximal. Note
that $\mathbb{F}{\mathcal{J}}\supseteq{\mathcal{J}}$. Thus we get
$\mathbb{F}\mathcal{J}=\mathcal{J}$. It follows that $\mathcal{J}$ is also a
Lie algebra ideal. The converse is obvious. $\Box$

In the rest part of this paper, we assume that $\Phi:{\rm
Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ is a Lie ring isomorphism.
If ${\mathcal{N}}$ contains at least one nontrivial element, by Lemma 2.2, for
any nontrivial element $E\in\mathcal{N}$,
$\Phi(\mathbb{F}I+\mathcal{J}(\mathcal{N},E))$ is a maximal commutative Lie
ring ideal in ${\rm Alg}\mathcal{M}$. Hence there is a unique nontrivial
element $F\in{\mathcal{M}}$ such that
$\Phi(\mathbb{F}I+\mathcal{J}(\mathcal{N},E))=\mathbb{F}I+\mathcal{J}(\mathcal{M},F).$
Define a map

$None$
$\hat{\Phi}:{\mathcal{N}}\setminus{\\{(0),X\\}}\rightarrow{\mathcal{M}}\setminus{\\{(0),Y\\}}$

by
$\Phi(\mathbb{F}I+\mathcal{J}(\mathcal{N},E))=\mathbb{F}I+\mathcal{J}(\mathcal{M},\hat{\Phi}(E))$.

With the symbols introduced above and by an argument similar to [12, Lemmas
4.1, 4.3, 4.4], one can show that the following lemma is still true for the
Lie ring isomorphism $\Phi$.

Lemma 2.3. $\hat{\Phi}$ in Eq.(2.2) is bijective and is either order-
preserving or order-reversing, that is, $\hat{\Phi}$ is an order isomorphism
or a reverse-order isomorphism from ${\mathcal{N}}$ onto ${\mathcal{M}}$ if we
extend the definition of $\hat{\Phi}$ so that $\hat{\Phi}((0))=(0)$ or $Y$ and
$\hat{\Phi}(X)=Y$ or $(0)$ accordingly.

By Lemma 2.3, for any $A\in{\mathcal{J}}({\mathcal{N}},E)$ with nontrivial
$E\in{\mathcal{N}}$, there exists a unique operator
$B\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$ such that
$\Phi(A)-B\in{\mathbb{F}}I$. Thus we can define another map

$None$ $\bar{\Phi}:\bigcup\\{{\mathcal{J}}({\mathcal{N}},E):E\in{\mathcal{N}}\
{\rm is\
nontrivial}\\}\rightarrow\bigcup\\{{\mathcal{J}}({\mathcal{M}},F):F\in{\mathcal{M}}\
{\rm is\ nontrivial}\\}$

with the property that
$\bar{\Phi}(A)\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$ and
$\Phi(A)-\bar{\Phi}(A)\in{\mathbb{F}}I$ for any
$A\in{\mathcal{J}}({\mathcal{N}},E)$.

Similar to [12, Lemma 4.2], we have

Lemma 2.4. $\bar{\Phi}$ is a bijective map and
$\bar{\Phi}({\mathcal{J}}({\mathcal{N}},E))={\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$
for every non-trivial $E\in{\mathcal{N}}$.

Next we discuss the idempotents in nest algebras. Denote by
$\mathcal{E}(\mathcal{N})$ the set of all idempotents in ${\rm
Alg}{\mathcal{N}}$.

Lemma 2.5. ([11, Lemma 2.2]) Let ${\mathcal{N}}$ be a nest on a (real or
complex) Banach space $X$ and $A\in{\rm Alg}\mathcal{N}$.

(1) $A\in{\mathbb{F}}I+\mathcal{E}({\mathcal{N}})$ if and only if
$[A,[A,[A,T]]]=[A,T]$ for all $T\in{\rm Alg}\mathcal{N}$.

(2) $A$ is the sum of a scalar and an idempotent operator with range in
$\mathcal{N}$ if and only if $[A,[A,T]]=[A,T]$ for all $T\in{\rm
Alg}\mathcal{N}$.

By Lemma 2.5, if $P$ is an idempotent operator in ${\rm Alg}{\mathcal{N}}$,
then $\Phi(P)=Q+\lambda_{P}I$, where $\lambda_{P}\in{\mathbb{F}}$ and $Q$ is
an idempotent operator in ${\rm Alg}{\mathcal{M}}$. Furthermore, if ${\rm
ran}P\in{\mathcal{N}}$, then ${\rm ran}Q\in{\mathcal{M}}$. So we can define a
map

$None$
$\tilde{\Phi}:\mathcal{E}(\mathcal{N})\rightarrow\mathcal{E}(\mathcal{M})$

by $\tilde{\Phi}(P)=\Phi(P)-\lambda_{P}I$. It is easily seen that
$\tilde{\Phi}$ is a bijective map from ${\mathcal{E}}({\mathcal{N}})$ onto
${\mathcal{E}}({\mathcal{M}})$; see [11].

Now, for any nontrivial element $E\in{\mathcal{N}}$, define two sets

$None$
$\Omega_{1}(\mathcal{N},E)=\\{P\in\mathcal{E}(\mathcal{N}):PE=0\\}\quad{\rm
and}\quad\Omega_{2}(\mathcal{N},E)=\\{P\in\mathcal{E}(\mathcal{N}):P^{*}E^{\bot}=0\\}.$

For any nontrivial element $F\in{\mathcal{M}}$, the sets
$\Omega_{1}(\mathcal{M},F)$ and $\Omega_{2}(\mathcal{M},F)$ can be analogously
defined. Note that, if $P\in\Omega_{1}({\mathcal{N}},E)$, then one can easily
check $P^{*}E^{\perp}\not=0$; if $E$ is not complemented, then
$PE=0\Rightarrow(I-P)^{*}E^{\perp}\not=0$ and
$P^{*}E^{\perp}=0\Rightarrow(I-P)E\not=0$. These facts are needed in the proof
of Lemma 2.6.

Still, by an argument similar to [12, Lemmas 5.2-5.5], one can show that the
following lemma is true for Lie ring isomorphisms $\Phi$, with $\hat{\Phi}$
and $\tilde{\Phi}$ defined in Eq.(2.2) and Eq.(2.4) respectively.

Lemma 2.6. Assume that $E\in{\mathcal{N}}$ is nontrivial and not complemented
in $X$.

(1) Either
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},E))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(E))$
or
$I-\tilde{\Phi}(\Omega_{1}({\mathcal{N}},E))\subseteq\Omega_{2}({\mathcal{N}},\hat{\Phi}(E))$,
and either
$\tilde{\Phi}(\Omega_{2}({\mathcal{N}},E))\subseteq\Omega_{2}({\mathcal{N}},\hat{\Phi}(E))$
or
$I-\tilde{\Phi}(\Omega_{2}({\mathcal{N}},E))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(E))$.

(2) If $\Omega_{1}({\mathcal{N}},\hat{\Phi}(E))$ and
$\Omega_{2}({\mathcal{N}},\hat{\Phi}(E))$ are not empty, then
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},E))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(E))$
if and only if
$\tilde{\Phi}(\Omega_{2}({\mathcal{N}},E))\subseteq\Omega_{2}({\mathcal{N}},\hat{\Phi}(E))$.

(3) If $F\in{\mathcal{N}}$ is also nontrivial and not complemented in $X$ with
$F<E$, then $\Omega_{1}({\mathcal{N}},E)\not=\emptyset$ and
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},E))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(E))$
together imply that
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},F))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(F))$.

The following lemma gives a characterization of complemented elements
$E\in{\mathcal{N}}$ by the operators in ${\mathcal{J}}({\mathcal{N}},E)$ and
${\mathcal{E}}({\mathcal{N}})$, which is needed to prove that $\hat{\Phi}$
preserves the complementarity.

Lemma 2.7. Assume that $E\in{\mathcal{N}}$ is a nontrivial element. The
following statements are equivalent.

(1) $E\in{\mathcal{N}}$ is complemented in $X$.

(2) There exists some idempotent $P\in{\rm Alg}{\mathcal{N}}$ such that
$[P,A]=A$ for any $A\in{\mathcal{J}}({\mathcal{N}},E)$.

(3) There exists some idempotent $P\in{\rm Alg}{\mathcal{N}}$ such that
$[P,A]=A$ for any rank one operators $A\in{\mathcal{J}}({\mathcal{N}},E)$.

Furthermore, we have $E={\rm ran}P$ for $P$ in (2) and (3).

Proof. (1)$\Rightarrow$(2). If $E\in{\mathcal{N}}$ is complemented in $X$,
there exists an idempotent $P\in{\rm Alg}{\mathcal{N}}$ such that ${\rm
ran}P=E$. For any $A\in{\mathcal{J}}({\mathcal{N}},E)$, we have $PA=A$ and
$AP=0$. Hence $[P,A]=A$.

(2)$\Rightarrow$(3) is obvious.

(3)$\Rightarrow$(1). Assume that there exists some idempotent $P\in{\rm
Alg}{\mathcal{N}}$ such that $[P,A]=A$ for any rank-1 operator
$A\in{\mathcal{J}}({\mathcal{N}},E)$. According to the space decomposition
$X={\rm ran}P+\ker P$, for any $A\in{\mathcal{J}}({\mathcal{N}},E)$, we have

$P=\left(\begin{array}[]{cc}I&0\\\ 0&0\end{array}\right)\quad{\rm and}\quad
A=\left(\begin{array}[]{cc}A_{11}&A_{12}\\\ A_{21}&A_{22}\end{array}\right).$

Then $[P,A]=A$ implies that $A=\left(\begin{array}[]{cc}0&A_{12}\\\
0&0\end{array}\right)$. So $PA=A$ and $AP=0$ hold for all
$A\in{\mathcal{J}}({\mathcal{N}},E)$. Take any $y\in E$ and any $g\in
E^{\perp}$. It is easy to check that $y\otimes
g\in{\mathcal{J}}({\mathcal{N}},E)$. So $Py\otimes g=y\otimes g$ and $y\otimes
gP=0$. It follows that $Py=y$ and $P^{*}g=0$ for all $y\in E$ and all $g\in
E^{\perp}$. Thus we get $E\subseteq{\rm ran}P$ and $E^{\bot}\subseteq\ker
P^{*}$. If $E\not={\rm ran}P$, then there exist $x\in{\rm ran}P$ and $g\in
E^{\bot}$ such that $\langle x,g\rangle=1$. This leads to a contraction that
$0=\langle x,P^{*}g\rangle=\langle Px,g\rangle=\langle x,g\rangle=1$. Hence we
must have ${\rm ran}P=E$ and $E$ is complemented in $X$. $\Box$

Lemma 2.8. Non-trivial element $E\in{\mathcal{N}}$ is complemented in $X$ with
${\rm ran}P=E$ if and only if $\hat{\Phi}(E)$ is complemented in $Y$ with
${\rm ran}\tilde{\Phi}(P)=\hat{\Phi}(E)$. Here $P\in{\rm Alg}{\mathcal{N}}$ is
an idempotent.

Proof. It is clear that $\widehat{\Phi^{-1}}=\hat{\Phi}^{-1}$. So we need only
to show that $\hat{\Phi}(E)$ is complemented in $Y$ whenever
$E\in{\mathcal{N}}$ is complemented in $X$. Indeed, if $E\in{\mathcal{N}}$ is
complemented, by Lemma 2.7, there exists some idempotent $P\in{\rm
Alg}{\mathcal{N}}$ such that $[P,A]=A$ for any
$A\in{\mathcal{J}}({\mathcal{N}},E)$. By definitions of $\hat{\Phi}$,
$\bar{\Phi}$ and $\tilde{\Phi}$ (ref. (2.2)-(2.4)), there exists some scalar
$\gamma$ such that

$\bar{\Phi}(A)+\gamma
I=\Phi(A)=\Phi([P,A])=[\Phi(P),\Phi(A)]=[\tilde{\Phi}(P),\bar{\Phi}(A)].$

Since $\bar{\Phi}$ maps ${\mathcal{J}}({\mathcal{N}},E)$ onto
${\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$ by Lemma 2.4, we see that
$B+\gamma I=[\tilde{\Phi}(P),B]$ holds for any
$B\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$. Assume that $B$ is of
rank-1. If $\gamma\not=0$, then $[\tilde{\Phi}(P),B]=B+\gamma I$ is a sum of
nonzero scalar and a rank-1 operator, which is impossible since a commutator
can not be the sum of a nonzero scalar and a compact operator. Hence
$\gamma=0$, and $B=[\tilde{\Phi}(P),B]$ for all rank-1 operators
$B\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$. By Lemma 2.7 again,
$\hat{\Phi}(E)$ is complemented in $Y$ with ${\rm
ran}\tilde{\Phi}(P)=\hat{\Phi}(E)$. $\Box$

The following lemma is obvious.

Lemma 2.9. $\Phi({\mathbb{F}}I)={\mathbb{F}}I$.

Finally, we give a lemma, which is needed to prove our main result. Let $E$
and $F$ be subspaces of $X$ and $X^{*}$, respectively. Denote by $E\otimes F$
the set $\\{x\otimes f:x\in E,f\in F\\}$.

Lemma 2.10. Let $X_{i}$ be an infinite-dimensional Banach space, $i=1,2$. Let
$E_{i}$ and $F_{i}$ be closed subspaces with dimensions $>2$ of $X_{i}$ and
$X_{i}^{*}$ respectively. Let $\mathcal{A}_{i}$ be a unital subalgebra of
${\mathcal{B}}(X_{i})$ containing $E_{i}\otimes F_{i}$. Suppose that
$\Psi:{\mathcal{A}}_{1}\rightarrow{\mathcal{A}}_{2}$ is an additive bijective
map satisfying $\Psi(\mathbb{F}I)=\mathbb{F}I$ and
$\Phi({\mathbb{F}}I+E_{1}\otimes F_{1})={\mathbb{F}}I+E_{2}\otimes F_{2}$.
Then there is a map $\gamma:E_{1}\otimes F_{1}\rightarrow\mathbb{F}$ and a
field automorphism $\tau:\mathbb{F}\rightarrow\mathbb{F}$ such that either

(1) $\Psi(x\otimes f)=\gamma(x,f)I+Cx\otimes Df$ for all $x\in E_{1}$ and
$f\in F_{1},$ where $C:E_{1}\rightarrow E_{2}$ and $D:F_{1}\rightarrow F_{2}$
are two $\tau$-linear bijective maps; or

(2) $\Psi(x\otimes f)=\gamma(x,f)I+Df\otimes Cx$ for all $x\in E_{1}$ and
$f\in F_{1},$ where $C:E_{1}\rightarrow F_{2}$ and $D:F_{1}\rightarrow E_{2}$
are two $\tau$-linear bijective maps.

Lemma 2.10 can be proved by a similar approach as that in [2] and we omit its
proof here.

Since the “if” part of Theorem 1.1 is obvious, we need only to check the “only
if” part. If $\mathcal{N}$ is a trivial nest, that is,
${\mathcal{N}}=\\{(0),X\\}$, then ${\mathcal{M}}=\\{(0),Y\\}$ by Lemma 2.3. So
$\Phi$ is a Lie ring isomorphism from ${\mathcal{B}}(X)$ onto
${\mathcal{B}}(Y)$. Bai, Du and Hou showed in [1] that every Lie
multiplicative isomorphism between prime rings with a non-trivial idempotent
element is of the form $\psi+\beta$ with $\psi$ a ring isomorphism or the
negative of a ring anti-isomorphism and $\beta$ a central valued map vanishing
on each commutator. Note that ${\mathcal{B}}(X)$ is prime and contains non-
trivial idempotents if $\dim X\geq 2$. Hence, for the case that $\mathcal{N}$
is trivial, Theorem 1.1 follows from [1].

So, in the rest sections we always assume that ${\mathcal{N}}$ is nontrivial.
Thus ${\mathcal{M}}$ is also nontrivial by Lemma 2.3. We shall complete the
proof of Theorem 1.1 by considering several cases according to the situations
of $(0)_{+}$ and $X_{-}$, which will be dealt with separably in Sections 3-5.

##  3\. The case that $(0)_{+}=(0)$ and $X_{-}=X$

In this section, we deal with the case that both $(0)$ and $X$ are limit
points of $\mathcal{N}$, i.e., $(0)_{+}=(0)$ and $X_{-}=X$.

Keep the definitions of $\hat{\Phi}$, $\bar{\Phi}$ in mind; ref. Eqs.(2.2) and
(2.3).

Lemma 3.1. Assume that $(0)_{+}=(0)$ and $X_{-}=X$. Let $E\in{\mathcal{N}}$ be
a nontrivial element. Then for any $x\in E$ and $f\in E^{\bot}$,
$\bar{\Phi}(x\otimes f)$ is a rank one operator.

Proof. Since, by Lemma 2.3, the bijective map
$\hat{\Phi}:{\mathcal{N}}\rightarrow{\mathcal{M}}$ is either order-preserving
or order-reversing, we have $(0)_{+}=(0)$ and $Y_{-}=Y$ in $\mathcal{M}$.
Write $R=\bar{\Phi}(x\otimes f)$ with $\bar{\Phi}$ defined in Eq.(2.3). Then
$R\in\mathcal{J}(\mathcal{M},\hat{\Phi}(E))$. If, on the contrary, ${\rm
rank}R\geq 2$, then, since $\bigcup\\{M:M\in{\mathcal{M}}\\}$ is a dense
linear manifold of $Y$, there exists a nontrivial element
$M_{0}\in{\mathcal{M}}$ and two vectors $u,v\in M$ such that $Ru$ and $Rv$ are
linearly independent. As $\bigcap\\{M:(0)<M\in{\mathcal{M}}\\}=\\{0\\}$, there
exists some nontrivial $L\in{\mathcal{M}}$ such that $Ru,Rv\not\in L$. Let
$Y_{L}={\rm span}\\{Ru,L\\}$. By Hahn-Banach theorem, there exists some $g\in
Y_{L}^{\bot}$ such that $g(Rv)\not=0$. Then $g\in L^{\bot}$ satisfying
$g(Ru)=0$ and $g(Rv)\neq 0.$ It is easily checked that
$L<\hat{\Phi}(E)<M_{0}$. Take $z\in L$, $h\in M_{0}^{\bot}$, and let
$A=\bar{\Phi}^{-1}(z\otimes g)$, $B=\bar{\Phi}^{-1}(u\otimes h)$. By Lemma
2.4, $A\in\mathcal{J}(\mathcal{N},\hat{\Phi}^{-1}(L))$ and
$B\in\mathcal{J}(\mathcal{N},\hat{\Phi}^{-1}(M_{0})).$

If $\hat{\Phi}$ is order-preserving, we have
$\hat{\Phi}^{-1}(L)<E<\hat{\Phi}^{-1}(M_{0})$. So

$\Phi(A(x\otimes f)B)=\Phi([A,[x\otimes f,B]])=[z\otimes g,[R,u\otimes
h]]=(z\otimes g)R(u\otimes h)=0.$

It follows from the injectivity of $\Phi$ that $A(x\otimes f)B=0$, which
implies $A(x\otimes f)=0$ or $(x\otimes f)B=0.$ If $A(x\otimes f)=0,$ then
$0=\Phi([A,x\otimes f])=[z\otimes g,R]=(z\otimes g)R\not=0$, a contradiction;
if $(x\otimes f)B=0,$ then $0=\Phi([x\otimes f,B])=[R,u\otimes h]=R(u\otimes
h)\not=0,$ a contradiction.

If $\hat{\Phi}$ is order-reversing, we have
$\hat{\Phi}^{-1}(M_{0})<E<\hat{\Phi}^{-1}(L).$ So

$\Phi(B(x\otimes f)A)=\Phi([B,[x\otimes f,A]])=[u\otimes h,[R,z\otimes
g]]=(z\otimes g)R(u\otimes h)=0,$

which yields that either $B(x\otimes f)=0$ or $(x\otimes f)A=0$. By a similar
argument to that of the above, one can obtain a contradiction.

Therefore, we must have ${\rm rank}\bar{\Phi}(x\otimes f)={\rm rank}R=1$.
$\Box$

Proof of Theorem 1.1 for the case that $(0)=(0)_{+}$ and $X_{-}=X$.

By Lemma 3.1, Lemma 2.9 and the bijectivity of $\bar{\Phi}$, we have proved
that, for every nontrivial element $E\in{\mathcal{N}}$,
$\Phi(\mathbb{F}I+E\otimes
E^{\bot})=\mathbb{F}I+\hat{\Phi}(E)\otimes\hat{\Phi}(E)^{\bot}$. So, by Lemma
2.10, there exists a ring automorphism
$\tau_{E}:\mathbb{F}\rightarrow\mathbb{F}$ and a map $\gamma_{E}:E\otimes
E^{\bot}\rightarrow\mathbb{F}$ such that either

$None$ $\Phi(x\otimes f)=\gamma_{E}(x,f)I+C_{E}x\otimes D_{E^{\perp}}f$

holds for all $x\in E$ and $f\in E^{\bot}$, where
$C_{E}:E\rightarrow\hat{\Phi}(E)$ and
$D_{E^{\perp}}:E^{\bot}\rightarrow\hat{\Phi}(E)^{\bot}$ are two $\tau$-linear
bijective maps; or

$None$ $\Phi(x\otimes f)=\gamma_{E}(x,f)I+D_{E^{\perp}}f\otimes C_{E}x$

holds for all $x\in E$ and $f\in E^{\bot}$, where
$C_{E}:E\rightarrow\hat{\Phi}(E)^{\perp}$ and
$D_{E^{\perp}}:E^{\bot}\rightarrow\hat{\Phi}(E)$ are two $\tau$-linear
bijective maps.

It is easily checked that, if there is a nontrivial $E_{0}\in{\mathcal{N}}$
such that Eq.(3.1) holds, then Eq.(3.1) holds for any nontrivial
$E\in{\mathcal{N}}$; If there is a nontrivial $E_{0}\in{\mathcal{N}}$ such
that Eq.(3.2) holds, then Eq.(3.1) holds for any nontrivial
$E\in{\mathcal{N}}$.

Assume that Eq.(3.1) holds for a nontrivial $E\in{\mathcal{N}}$. Then, for any
$N\in{\mathcal{N}}$, any $x\in E\cap N$ and any $f\in N^{\perp}\cap
E^{\perp}$, we have

$\Phi(x\otimes f)=\gamma_{E}(x,f)I+C_{E}x\otimes
D_{E^{\perp}}f=\gamma_{N}(x,f)I+C_{N}x\otimes D_{N^{\perp}}f.$

Since ${\rm rank}I=\infty$, the above equation yields
$\gamma_{N}(x,f)=\gamma_{E}(x,f)$ and $C_{E}x\otimes
D_{E^{\perp}}f=C_{N}x\otimes D_{N^{\perp}}f$ for any $x\in N\cap E$. This
entails that there exists an automorphism
$\tau:{\mathbb{F}}\rightarrow{\mathbb{F}}$ so that $\tau_{N}=\tau_{E}=\tau$
for any $N,E\in{\mathcal{N}}$, and if $N\subset E$, then there exists a scalar
$\alpha_{EN}$ such that $C_{E}|_{N}=\alpha_{EN}C_{N}$ and
$D_{N^{\perp}}|_{E^{\perp}}=\alpha_{EN}D_{E^{\perp}}$. Now fix
$E\in{\mathcal{N}}$. For any $N\in{\mathcal{N}}$, we define

$\left\\{\begin{array}[]{lll}\tilde{C}_{N}=C_{N},&\tilde{D}_{N_{-}^{\perp}}=D_{N^{\perp}},&\mbox{\rm
if}\ \ N=E;\\\
\tilde{C}_{N}=\alpha_{EN}C_{N},&\tilde{D}_{N_{-}^{\perp}}=\frac{1}{\alpha_{EN}}D_{N^{\perp}},&\mbox{\rm
if}\ \ N\subset E;\\\
\tilde{C}_{N}=\frac{1}{\alpha_{EN}}C_{N},&\tilde{D}_{N_{-}^{\perp}}=\alpha_{EN}D_{N^{\perp}},&\mbox{\rm
if}\ \ N\supset E.\end{array}\right.$

It is obvious that $\\{\tilde{C_{N}}:N\not=(0),X\\}$ and
$\\{\tilde{D}_{N_{-}^{\perp}}:N\not=(0),X\\}$ are well defined with
$\tilde{C}_{N}|_{E}=\tilde{C}_{E}$ and
$\tilde{D}_{E^{\perp}}|_{N^{\perp}}=\tilde{D}_{N^{\perp}}$ whenever
$E\subseteq N$. Thus there exist bijective $\tau$-linear maps
$C:\bigcup\\{N\in{\mathcal{N}}:N\not=(0),X\\}\rightarrow\bigcup\\{M\in{\mathcal{M}}:M\not=(0),Y\\}$
and
$D:\bigcup\\{N^{\perp}:N\in{\mathcal{N}},N\not=(0),X\\}\rightarrow\bigcup\\{M^{\perp}:M\in{\mathcal{M}},M\not=(0),Y\\}$
such that $C|_{N}=\tilde{C}_{N}$ and
$D|_{N^{\perp}}=\tilde{D}_{N_{-}^{\perp}}$ for any
$N\in{\mathcal{N}}\backslash\\{(0),X\\}$.

By now, we have shown that, there exist bijective $\tau$-linear maps
$C:\bigcup\\{N\in{\mathcal{N}}:N\not=(0),X\\}\rightarrow\bigcup\\{M\in{\mathcal{M}}:M\not=(0),Y\\}$
and
$D:\bigcup\\{N^{\perp}:N\in{\mathcal{N}},N\not=(0),X\\}\rightarrow\bigcup\\{M^{\perp}:M\in{\mathcal{M}},M\not=(0),Y\\}$,
and a map $\gamma:\bigcup\\{E\times
E^{\perp}:E\in{\mathcal{N}}\setminus\\{(0),X\\}\\}\rightarrow{\mathbb{F}}$,
such that for any $x\in N$ and $f\in N^{\perp}$ with
$N\in{\mathcal{N}}\setminus\\{(0),X\\}$, we have

$None$ $\Phi(x\otimes f)=\gamma(x,f)I+Cx\otimes Df.$

Therefore, for any $A\in{\rm Alg}\mathcal{N}$, any $x\in N$ and any $f\in
N^{\perp}$, by Eq.(3.3), we have

$\Phi([A,x\otimes f])=[\Phi(A),\Phi(x\otimes f)]=\Phi(A)Cx\otimes Df-
Cx\otimes\Phi(A)^{*}Df$

and

$\begin{array}[]{rl}\Phi([A,x\otimes f])=&\Phi(Ax\otimes f-x\otimes A^{*}f)\\\
=&(\gamma(Ax,f)-\gamma(x,A^{*}f))I+CAx\otimes Df-Cx\otimes
DA^{*}f.\end{array}$

Combining the above two equations and noting that $I$ is of infinite-rank, one
obtains that

$Cx\otimes\Phi(A)^{*}Df-Cx\otimes DA^{*}f=\Phi(A)Cx\otimes Df-CAx\otimes Df$

holds for any $x\in N$, $f\in N^{\bot}$ and any nontrivial
$N\in{\mathcal{N}}$. Note that $D$ is bijective. So there exists a scalar
$h(A)$ such that

$None$ $\Phi(A)Cx=CAx+h(A)Cx$

for all $x\in\bigcup\\{N\in{\mathcal{N}}:N\not=(0),X\\}$. It is clear that $h$
is additive as a functional of Alg$\mathcal{N}$. Define
$\Psi(A)=\Phi(A)-h(A)I$ for all $A\in{\rm Alg}{\mathcal{N}}$. Then, by
Eq.(3.4), for any $A,B\in{\rm Alg}{\mathcal{N}}$ and any
$x\in\bigcup\\{N\in{\mathcal{N}}:N\not=(0),X\\}$, we have

$\Psi(AB)Cx=CABx=\Psi(A)CBx=\Psi(A)\Psi(B)Cx.$

Since $\bigcup\\{N\in{\mathcal{N}}:N\not=(0),X\\}$ is dense in $X$ and $C$ is
bijective, it follows that $\Psi(AB)=\Psi(A)\Psi(B)$ for all $A,B\in{\rm
Alg}\mathcal{N}$, that is, $\Psi$ is a ring isomorphism and
$\Phi(A)=\Psi(A)+h(A)I$ for all $A\in{\rm Alg}\mathcal{N}$.

Similarly, if Eq.(3.2) holds, one can check that $\Phi(A)=-\Psi(A)+h(A)I$ for
all $A\in{\rm Alg}\mathcal{N}$, where $\Psi$ is a ring anti-isomorphism and
$h$ is an additive functional.

This completes the proof of Theorem 1.1 for the case that $(0)_{+}=(0)$ and
$X_{-}=X$. $\Box$

##  4\. The case $X_{-}\not=X$ and $X_{-}$ is complemented or $(0)_{+}\not=0$
and $(0)_{+}$ is complemented

We give only the proof in detail for the case that $X_{-}\not=X$ and $X_{-}$
is complemented. The case that $(0)\not=(0)_{+}$ and $(0)_{+}$ is complemented
in $X$ can be dealt with similarly.

Proof of Theorem 1.1 for the case that $X_{-}\not=X$ and $X_{-}$ is
complemented in $X$.

Assume that $X_{-}\not=X$ and $X_{-}$ is complemented in $X$.

Since $X_{-}$ is complemented, there exists an idempotent $P_{0}\in{\rm
Alg}\mathcal{N}$ such that ${\rm ran}P_{0}=X_{-}$.

With $\tilde{\Phi}$ as in Eq.(2.4) and by a similar argument to that in the
proof of [11, Theorem 3.1], one can show that there exist an idempotent
operator $Q_{0}$ and a scalar $\lambda_{P_{0}}$ such that
$\Phi(P_{0})=Q_{0}+\lambda_{P_{0}}I$ with $Q_{0}=\tilde{\Phi}(P_{0})$, ${\rm
ran}Q_{0}\in\mathcal{M}$ and the following statements hold:

(a) If there is an idempotent $P_{1}\in{\rm Alg}{\mathcal{N}}$ such that
$P_{1}<P_{0}$ and $\tilde{\Phi}(P_{1})<Q_{0}$ (or $P_{1}>P_{0}$ and
$\tilde{\Phi}(P_{1})>Q_{0}$), then for any $P\in{\rm Alg}{\mathcal{N}}$,
$P<P_{0}\Rightarrow\tilde{\Phi}(P)<Q_{0}$ and
$P>P_{0}\Rightarrow\tilde{\Phi}(P)>Q_{0}$.

(b) If there is an idempotent $P_{1}\in{\rm Alg}{\mathcal{N}}$ such that
$P_{1}<P_{0}$ and $\tilde{\Phi}(P_{1})>Q_{0}$ (or $P_{1}>P_{0}$ and
$\tilde{\Phi}(P_{1})<Q_{0}$), then for any $P\in{\rm Alg}{\mathcal{N}}$,
$P<P_{0}\Rightarrow\tilde{\Phi}(P)>Q_{0}$ and
$P>P_{0}\Rightarrow\tilde{\Phi}(P)<Q_{0}$.

Claim 4.1. If (a) occurs, then $\Phi=\Psi+h$, where $\Psi:{\rm
Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ is a ring isomorphism and
$h:{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ is an additive map
vanishing all commutators.

(a) implies that ${\rm ran}Q_{0}=Y_{-}$ by Lemmas 2.3 and 2.8. For the
convenience, let ${\mathcal{A}}_{11}=P_{0}({\rm Alg}{\mathcal{N}})P_{0}$,
${\mathcal{A}}_{12}=P_{0}({\rm Alg}{\mathcal{N}})(I-P_{0})$,
${\mathcal{A}}_{22}=(I-P_{0})({\rm Alg}{\mathcal{N}})(I-P_{0})$,
${\mathcal{B}}_{11}=Q_{0}({\rm Alg}{\mathcal{M}})Q_{0}$,
${\mathcal{B}}_{12}=Q_{0}({\rm Alg}{\mathcal{M}})(I-Q_{0})$ and
${\mathcal{B}}_{22}=(I-Q_{0})({\rm Alg}{\mathcal{M}})(I-Q_{0})$. Then ${\rm
Alg}{\mathcal{N}}={\mathcal{A}}_{11}\dot{+}{\mathcal{A}}_{12}\dot{+}{\mathcal{A}}_{22}$
and ${\rm
Alg}{\mathcal{M}}={\mathcal{B}}_{11}\dot{+}{\mathcal{B}}_{12}\dot{+}{\mathcal{B}}_{22}$.

We will prove Claim 4.1 by several steps.

Step 1. $\Phi({\mathcal{A}}_{12})={\mathcal{B}}_{12}$.

The proof is the same as that of [11, Lemma 2.8].

Step 2. $\Phi({\mathcal{A}}_{ii})\subseteq{\mathcal{B}}_{ii}+{\mathbb{F}}I$,
$i=1,2$.

For any $A_{11}\in{\mathcal{A}}_{11}$, denote
$\Phi(A_{11})=S_{11}+S_{12}+S_{22}$, where $S_{ij}\in{\mathcal{B}}_{ij}$. Then

$0=\Phi([A_{11},P_{0}])=[\Phi(A_{11}),\Phi(P_{0})]=[\Phi(A_{11}),Q_{0}],$

which implies that $S_{12}=0$. Let $P\in{\mathcal{A}}_{22}$ be any idempotent
with $P\not=I-P_{0}$. It is clear that $P<(I-P_{0})$. Then $I-P>P_{0}$. As
$\Phi$ meets (a), we have $\tilde{\Phi}(I-P)>Q_{0}$, that is,
$\tilde{\Phi}(P)<I-Q_{0}$. It follows from $[A_{11},P]=0$ that
$[\Phi(A_{11}),\tilde{\Phi}(P)]=[S_{22},\tilde{\Phi}(P)]=0$. By the
arbitrariness of $P$ and the bijectivity of $\tilde{\Phi}$, we see that
$S_{22}$ commutes with every idempotent in ${\mathcal{B}}_{22}$. Note that
${\rm ran}Q_{0}=Y_{-}$. So $\mathcal{B}_{22}=\mathcal{B}(\ker Q_{0})$, which
implies $S_{22}\in{\mathbb{F}}(I-Q_{0})$. It follows that
$\Phi(A_{11})=S_{11}+\lambda(I-Q_{0})=(S_{11}-\lambda Q_{0})+\lambda I$ and
hence, $\Phi(\mathcal{A}_{11})\subseteq\mathcal{B}_{11}+\mathbb{F}I$.

Assume that $A_{22}\in\mathcal{A}_{22}$. In the same way as above, one can
show that, $\Phi(A_{22})=T_{11}+T_{22}$ for some
$T_{ii}\in{\mathcal{B}}_{ii}$, $i=1,2$, with $T_{11}$ commuting with every
idempotent in ${\mathcal{B}}_{11}$. For any $B_{11}\in\mathcal{B}_{11}$, by
the surjectivity of $\Phi$, there exists some $A_{0}\in{\rm Alg}\mathcal{N}$
such that $\Phi(A_{0})=B_{11}$. Furthermore, $A_{0}=A_{11}+\lambda I$ for some
$A_{11}\in{\mathcal{A}}_{11}$ and some scalar $\lambda$ by Step 1. Thus we
have

$[B_{11},T_{11}]=[B_{11},T_{22}+T_{11}]=[\Phi(A_{0}),\Phi(A_{22})]=\Phi([A_{0},A_{22}])=\Phi([A_{11}+\lambda
I,A_{22}])=0$

for all $B_{11}\in\mathcal{B}_{11}$, which implies $T_{11}\in\mathbb{F}I$ as
${\mathcal{A}}_{11}$ is a nest algeba. So
$\Phi(\mathcal{A}_{22})\subseteq\mathcal{B}_{22}+\mathbb{F}I$.

By Steps 1-2, for each $A_{12}\in{\mathcal{A}}_{12}$, there exists
$B_{12}\in{\mathcal{B}}_{12}$ such that $\Phi(A_{12})=B_{12}$; for each
$A_{ii}\in{\mathcal{A}}_{ii}$, $i=1,2$, there exist
$B_{ii}\in{\mathcal{B}}_{ii}$ and $\lambda_{ii}\in{\mathbb{F}}$ such that
$\Phi(A_{ii})=B_{ii}+\lambda_{ii}I$. We claim that $B_{ii}$ and $\lambda_{ii}$
are uniquely determined. In fact, if
$\Phi(A_{ii})=B_{ii}+\lambda_{ii}I=B_{ii}^{{}^{\prime}}+\lambda^{{}^{\prime}}_{ii}I$,
then $B_{ii}-B_{ii}^{{}^{\prime}}\in{\mathbb{F}}I$, which implies that
$B_{ii}=B_{ii}^{{}^{\prime}}$ and $\lambda_{ii}=\lambda^{{}^{\prime}}_{ii}$.
Let $\Psi(A_{ij})=B_{ij}$ and
$\Psi(A)=\Psi(A_{11})+\Psi(A_{12})+\Psi(A_{22})$. Then we define a map
$\Psi:{\rm Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ and a map
$h:{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ with
$h(A)=\Phi(A)-\Psi(A)\in{\mathbb{F}}I$. Now, imitating the proof of [11,
Lemmas 2.10-2.13], one can show that $\Psi:{\rm
Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ is a ring isomorphism and
$h:{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ is an additive map
satisfying $h([A,B])=0$ for all $A,B$. Hence Claim 4.1 is true.

Claim 4.2. If $\Phi$ satisfies (b), then $\Phi=-\Psi+h$, where $\Psi:{\rm
Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ is a ring anti-isomorphism
and $h:{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ is an additive map
vanishing all commutators.

If (b) holds, then ${\rm ran}Q_{0}=(0)_{+}\in{\mathcal{M}}$ and $(0)_{+}$ is
complemented in $\mathcal{M}$ by Lemma 2.3 and 2.8. Consider the map
$\Phi^{\prime}:{\rm Alg}{\mathcal{N}}\rightarrow({\rm Alg}{\mathcal{M}})^{*}$
defined by $\Phi^{\prime}(A)=-\Phi(A)^{*}$ for all $A\in{\rm
Alg}{\mathcal{N}}$.

By imitating the proof of [11, Lemmas 2.14], one can show that
$\Phi^{\prime}:{\rm Alg}{\mathcal{N}}\rightarrow({\rm Alg}{\mathcal{M}})^{*}$
is a Lie multiplicative bijective map. Since, for any nontrivial idempotent
operator $P\in{\rm Alg}{\mathcal{N}}$, $\Phi(P)=\tilde{\Phi}(P)+\lambda_{P}I$
for some $\lambda_{P}\in{\mathbb{C}}$, we have
${\Phi}^{\prime}(P)=-\tilde{\Phi}(P)^{*}-\lambda_{P}I$. Now define a map
$\tilde{{\Phi}^{\prime}}:{\mathcal{E}}{(\mathcal{N})}\rightarrow{\mathcal{E}}{(\mathcal{M})}^{*}$
by $\tilde{{\Phi}^{\prime}}(P)=I-\tilde{\Phi}(P)^{*}$ for all idempotents $P$.
Since $\Phi$ satisfies (b), for any nonzero idempotent $P_{1}\in{\rm
Alg}{\mathcal{N}}$, if $P_{1}<P_{0}$, we have
$\tilde{{\Phi}^{\prime}}(P_{1})=I-\tilde{\Phi}(P_{1})^{*}<I-\tilde{\Phi}(P_{0})^{*}=\tilde{{\Phi}^{\prime}}(P_{0})$;
if $P_{1}>P_{0}$, we have
$\tilde{{\Phi}^{\prime}}(P_{1})=I-\tilde{\Phi}(P_{1})^{*}>I-\tilde{\Phi}(P_{0})^{*}=\tilde{{\Phi}^{\prime}}(P_{0})$.
Hence ${\Phi}^{\prime}$ satisfies (a).

Note that ${\mathcal{M}^{\perp}}=\\{M^{\perp}:M\in{\mathcal{M}}\\}$ is a nest
on $Y^{*}$. Since ${\rm ran}Q_{0}\in{\mathcal{M}}$, we have $\ker
Q_{0}^{*}=({\rm ran}Q_{0})^{\perp}\in{\mathcal{M}}^{\perp}$, and so ${\rm
ran}(I-Q_{0}^{*})\in{\mathcal{M}}^{\perp}$. With respect to the decomposition
$Y^{*}={\rm ran}(I-Q_{0}^{*})\dot{+}{\rm ran}Q_{0}^{*}$, we have ${\rm
Alg}{\mathcal{M}}^{\perp}={\mathcal{D}}_{11}\dot{+}{\mathcal{D}}_{12}\dot{+}{\mathcal{D}}_{22}$,
where ${\mathcal{D}}_{11}=(I-Q_{0}^{*})({\rm
Alg}{\mathcal{M}}^{\perp})(I-Q_{0}^{*})$,
${\mathcal{D}}_{12}=(I-Q_{0}^{*})({\rm Alg}{\mathcal{M}}^{\perp})Q_{0}^{*}$,
${\mathcal{D}}_{22}=Q_{0}^{*}({\rm Alg}{\mathcal{M}}^{\perp})Q_{0}^{*}$. Since
${\rm Alg}{\mathcal{M}}=Q_{0}({\rm Alg}{\mathcal{M}})Q_{0}\dot{+}Q_{0}({\rm
Alg}{\mathcal{M}})(I-Q_{0})\dot{+}(I-Q_{0})({\rm
Alg}{\mathcal{M}})(I-Q_{0})={\mathcal{B}}_{11}\dot{+}{\mathcal{B}}_{12}\dot{+}{\mathcal{B}}_{22}$,
we have ${\mathcal{B}}_{11}^{*}\subseteq{\mathcal{D}}_{22}$,
${\mathcal{B}}_{12}^{*}\subseteq{\mathcal{D}}_{12}$ and
${\mathcal{B}}_{22}^{*}\subseteq{\mathcal{D}}_{11}$. Hence $({\rm
Alg}{\mathcal{M}})^{*}={\mathcal{B}}_{22}^{*}\dot{+}{\mathcal{B}}_{12}^{*}\dot{+}{\mathcal{B}}_{11}^{*}\subseteq{\rm
Alg}{\mathcal{M}}^{\perp}$.

Step 1. $\Phi^{\prime}({\mathcal{A}}_{12})={\mathcal{B}}_{12}^{*}$.

The proof is the same as that of [11, Lemma 2.15].

Step 2.
$\Phi^{\prime}({\mathcal{A}}_{ii})\subseteq{\mathcal{B}}_{jj}^{*}+{\mathbb{F}}I$,
$i,j=1,2$ and $i\not=j$.

For any $A_{11}\in{\mathcal{A}}_{11}$, write
$\Phi^{\prime}(A_{11})=S_{22}^{*}+S_{12}^{*}+S_{11}^{*}$, where
$S_{ij}\in{\mathcal{B}}_{ij}$. Then

$0=\Phi^{\prime}([A_{11},P_{0}])=[\Phi^{\prime}(A_{11}),\Phi^{\prime}(P_{0})]=[\Phi^{\prime}(A_{11}),-Q_{0}^{*}]=[Q_{0}^{*},\Phi^{\prime}(A_{11})],$

which implies that $S_{12}^{*}=0$.

Let $P\in{\mathcal{A}}_{22}$ be any idempotent with $P<I-P_{0}$. As
$\Phi^{\prime}$ satisfies (a), we have
$\widetilde{\Phi^{\prime}}(P)<\widetilde{\Phi^{\prime}}(I-P_{0})=I-\tilde{\Phi}(P_{0})^{*}=Q_{0}^{*}$.
It follows from $[A_{11},P]=0$ that
$0=[\Phi^{\prime}(A_{11}),{\Phi}^{\prime}(P)]=[S_{11}^{*},\Phi^{\prime}(P)]$.
Since $P$ is arbitrary, we see that $S_{11}^{*}$ commutes with every
idempotent in ${\mathcal{B}}_{11}^{*}$, which implies that $S_{11}$ commutes
with every idempotent in ${\mathcal{B}}_{11}$. Noting that
$\mathcal{B}_{11}=\mathcal{B}(\ker Q_{0})$ in this case, so we must have
$S_{11}\in{\mathbb{F}}I_{\ker Q_{0}}$. By the arbitrariness of $A_{11}$ we
obtain that
$\Phi^{\prime}(\mathcal{A}_{11})\subseteq\mathcal{B}_{22}^{*}+\mathbb{F}I$.

Similarly, one can show that, for any $A_{22}\in{\mathcal{A}}_{22}$,
$\Phi^{\prime}(A_{22})=T_{22}^{*}+T_{11}^{*}$, where $T_{22}$ commutes with
every idempotent in ${\mathcal{B}}_{22}$. Taking any
$B_{22}\in\mathcal{B}_{22}$, by the surjectivity of $\Phi^{\prime}$, there
exists some $A_{0}\in{\rm Alg}\mathcal{N}$ such that
$\Phi^{\prime}(A_{0})=B_{22}^{*}$. Furthermore, $A_{0}=A_{11}+\lambda I$ for
some $A_{11}\in{\mathcal{A}}_{11}$ and some scalar $\lambda$. Then

$[B_{22}^{*},T_{22}^{*}]=[B_{22}^{*},T_{22}^{*}+T_{11}^{*}]=[\Phi^{\prime}(A_{0}),\Phi^{\prime}(A_{22})]=\Phi^{\prime}([A_{0},A_{22}])=0$

for all $B_{22}\in\mathcal{B}_{22}$. Thus $[B_{22},T_{22}]=0$ for all
$B_{22}\in{\mathcal{B}}_{22}$, which implies $T_{22}\in\mathbb{F}I_{{\rm
ran}Q_{0}}$ as ${\mathcal{B}}_{22}$ is a nest algebra. Therefore
$\Phi^{\prime}({\mathcal{A}}_{22})\subseteq{\mathcal{B}}_{11}^{*}+{\mathbb{F}}I$.

By Steps 1-2, for each $A_{12}\in{\mathcal{A}}_{12}$, there exists
$B_{12}\in{\mathcal{B}}_{12}$ such that $\Phi^{\prime}(A_{12})=B_{12}^{*}$;
for each $A_{ii}\in{\mathcal{A}}_{ii}$ ($i=1,2$), there exist
$B_{jj}\in{\mathcal{B}}_{jj}$ and $\lambda_{jj}\in{\mathbb{F}}$ such that
$\Phi^{\prime}(A_{ii})=B_{jj}^{*}+\lambda_{jj}I$, $i,j=1,2$ and $i\not=j$. It
is easily seen that $B_{jj}$ and $\lambda_{jj}$ are uniquely determined.
Define a map $\Psi^{\prime}:{\rm Alg}{\mathcal{N}}\rightarrow({\rm
Alg}{\mathcal{M}})^{*}$ by $\Psi^{\prime}(A_{12})=B_{12}^{*}$,
$\Psi^{\prime}(A_{11})=B_{22}^{*}$, $\Psi^{\prime}(A_{22})=B_{11}^{*}$ and
$\Psi^{\prime}(A)=\Psi^{\prime}(A_{11})+\Psi^{\prime}(A_{12})+\Psi^{\prime}(A_{22})$
for $A=A_{11}+A_{12}+A_{22}$. Set
$h^{\prime}(A)=\Phi^{\prime}(A)-\Psi^{\prime}(A)$. It is clear that
$h^{\prime}$ maps ${\rm Alg}{\mathcal{N}}$ into ${\mathbb{F}}I$. By [11, Lemma
2.17], one can show that $\Psi^{\prime}$ is a ring isomorphism. Since, for any
$A\in{\rm Alg}{\mathcal{N}}$, there exists a unique element $S\in{\rm
Alg}{\mathcal{M}}$ such that $\Psi^{\prime}(A)=S^{*}$, we can define a map
$\Psi:{\rm Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ by $\Psi(A)=S$.
Thus $\Psi(A)^{*}=\Psi^{\prime}(A)$ for every $A$ and hence $\Psi$ is a ring
anti-isomorphism. Let $h:{\rm Alg}{\mathcal{N}}\rightarrow{\mathbb{F}}I$ be
the map defined by $h^{\prime}(A)=-h(A)^{*}$ for all $A\in{\rm
Alg}{\mathcal{N}}$. Clearly, $h([A,B])=0$ for all $A,B\in{\rm
Alg}{\mathcal{N}}$. Furthermore, we have

$-\Phi(A)^{*}=\Phi^{\prime}(A)=\Psi^{\prime}(A)+h^{\prime}(A)=\Psi(A)^{*}+h^{\prime}(A)=(\Psi(A)-h(A))^{*},$

which yields $\Phi(A)=-\Psi(A)+h(A)$ for every $A\in{\rm Alg}{\mathcal{N}}$.
This completes the proof of Claim 4.2.

By Claim 4.1 and Claim 4.2, Theorem 1.1 holds for the case that $X_{-}\not=X$
and $X_{-}$ is complemented in $X$. $\Box$

##  5\.  The case that $X_{-}\not=X$ and $X_{-}$ is not complemented or
$(0)\not=(0)_{+}$ and $(0)_{+}$ is not complemented

In this section, we deal with the case that $X_{-}\not=X$ and $X_{-}$ is not
complemented or $(0)\not=(0)_{+}$ and $(0)_{+}$ is not complemented. Here we
borrow some ideas developed in [12]. Note that, not like [12], we do not
assume that all non-trivial elements in the nests are not complemented. Also,
we give only the detail of our proof for the case that $X_{-}\not=X$ and
$X_{-}$ is not complemented in $X$. The other case can be checked similarly.

Assume that $X_{-}\not=X$ and $X_{-}$ is not complemented in $X$.

Note that, there are three possible situations that $(0)$ may have, that is,
(1∘) $(0)_{+}\not=(0)$ and $(0)_{+}$ is complemented in $X$, (2∘)
$(0)_{+}=(0)$ and (3∘) $(0)_{+}\not=(0)$ and $(0)_{+}$ is not complemented in
$X$. By Section 4, Theorem 1.1 is true if the situation (1∘) occurs. So what
we need to deal with is either (2∘) or (3∘).

Recall that $\hat{\Phi}$, $\bar{\Phi}$ and $\tilde{\Phi}$ are maps defined in
Eqs.(2.2)-(2.4), and $\Omega_{1}({\mathcal{N}},E)$ and
$\Omega_{2}({\mathcal{N}},E)$ are defined in Eq.(2.5). By Lemma 2.6, either
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(X_{-}))$
or $\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq
I-\Omega_{2}({\mathcal{N}},\hat{\Phi}(X_{-}))$.

The following lemma is crucial for our purpose.

Lemma 5.1. Assume that $(0)<X_{-}<X$ and $X_{-}$ is not complemented. The
following statements are true:

(1) If
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(X_{-}))$,
then $\Phi(\mathbb{F}I+X\otimes X_{-}^{\bot})=\mathbb{F}I+Y\otimes
Y_{-}^{\bot}$, and there exists a ring automorphism
$\tau:\mathbb{F}\rightarrow\mathbb{F}$, a map $\gamma:X\otimes
X_{-}^{\bot}\rightarrow\mathbb{F}$, bijective $\tau$-linear maps
$C:X\rightarrow Y$ and $D:X_{-}^{\bot}\rightarrow Y_{-}^{\bot}$ such that

$\Phi(x\otimes f)=\gamma(x,f)I+Cx\otimes Df$

holds for all $x\in X$ and $f\in X_{-}^{\bot}$.

(2) If $\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq
I-\Omega_{2}({\mathcal{N}},\hat{\Phi}(X_{-}))$, then
$\Phi(\mathbb{F}I+X\otimes X_{-}^{\bot})=\mathbb{F}I+(0)_{+}\otimes Y^{*}$,
and there exists a ring automorphism $\tau:\mathbb{F}\rightarrow\mathbb{F}$, a
map $\gamma:X\otimes X_{-}^{\bot}\rightarrow\mathbb{F}$, bijective
$\tau$-linear maps $C:X\rightarrow Y^{*}$ and
$D:X_{-}^{\bot}\rightarrow(0)_{+}$ such that

$\Phi(x\otimes f)=\gamma(x,f)I+Df\otimes Cx$

holds for all $x\in X$ and $f\in X_{-}^{\bot}$.

To prove Lemma 5.1, we consider two cases, that is, the case that
${\mathcal{N}}$ has at least two nontrivial elements and the case that
${\mathcal{N}}$ has only one nontrivial element. These will be done by Lemma
5.3 and Lemma 5.4, respectively.

We first consider the case that ${\mathcal{N}}$ has at least two nontrivial
elements.

The following lemma is crucial for the proof of Lemma 5.3.

Lemma 5.2. Assume that $\mathcal{N}$ has at least two nontrivial elements and
$X_{-}$ is not complemented in $X$. Then the following statements hold.

(i)
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(X_{-}))$
if and only if $\hat{\Phi}$ is order-preserving.

(ii) $\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq
I-\Omega_{2}({\mathcal{N}},\hat{\Phi}(X_{-}))$ if and only if $\hat{\Phi}$ is
order-reversing.

Proof. By Lemma 2.6, it suffices to show that
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}(X_{-}))$
implies that $\hat{\Phi}$ is order-preserving and
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq
I-\Omega_{2}({\mathcal{N}},\hat{\Phi}(X_{-}))$ implies that$\hat{\Phi}$ is
order-reversing.

Assume that
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq\Omega_{1}({\mathcal{N}},\hat{\Phi}({X_{-}}))$.
Since $\mathcal{N}$ has at least two nontrivial elements, we may take a
nontrivial element $E\in{\mathcal{N}}$ such that $E<X_{-}$. If, on the
contrary, $\hat{\Phi}$ is order-reversing, then we have
$\hat{\Phi}(E)>\hat{\Phi}(X_{-})$. Fix an idempotent
$P\in\Omega_{1}({\mathcal{N}},X_{-})$. Then we have
$\tilde{\Phi}(P)\in\Omega_{1}({\mathcal{M}},\hat{\Phi}(X_{-}))$. For any
$D_{1}\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(X_{-}))$,
$D_{2}\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$, let
$C_{1}=\bar{\Phi}^{-1}(D_{1})$, $C_{2}=\bar{\Phi}^{-1}(D_{2})$. By the
definition of $\bar{\Phi}$, we have
$C_{1}\in{\mathcal{J}}({\mathcal{N}},X_{-})$ and
$C_{2}\in{\mathcal{J}}({\mathcal{N}},E)$. Furthermore,

$\begin{array}[]{rl}0=&\Phi([C_{1},[C_{2},P]])=[D_{1},[D_{2},\tilde{\Phi}(P)]]\\\
=&[D_{1},D_{2}\tilde{\Phi}(P)-\tilde{\Phi}(P)D_{2}]=D_{1}D_{2}\tilde{\Phi}(P)-D_{1}\tilde{\Phi}(P)D_{2}.\end{array}$

For any $y_{1}\in\hat{\Phi}(X_{-})$, $y_{2}\in\hat{\Phi}(E)$ and any
$g_{1}\in\hat{\Phi}(X_{-})^{\bot}$, $g_{2}\in\hat{\Phi}(E)^{\bot}$, it is
obvious that $y_{1}\otimes
g_{1}\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(X_{-}))$ and $y_{2}\otimes
g_{2}\in{\mathcal{J}}({\mathcal{M}},\hat{\Phi}(E))$. Letting
$D_{i}=y_{i}\otimes g_{i}$, the above equation yields

$None$ $\langle y_{2},g_{1}\rangle
y_{1}\otimes\tilde{\Phi}(P)^{*}g_{2}=\langle\tilde{\Phi}(P)y_{2},g_{1}\rangle
y_{1}\otimes g_{2}.$

Choose $y_{2}\in\hat{\Phi}(E)$ and $g_{1}\in\hat{\Phi}(X_{-})^{\bot}$ such
that $\langle y_{2},g_{1}\rangle\neq 0$. Eq.(5.1) implies that there exists
some scalar $\lambda_{g_{2}}$ such that
$\tilde{\Phi}(P)^{*}g_{2}=\lambda_{g_{2}}g_{2}$ for each
$g_{2}\in\hat{\Phi}(E)^{\bot}$. It follows that there is a scalar $\lambda$
such that $\tilde{\Phi}(P)^{*}g_{2}=\lambda g_{2}$ for all
$g_{2}\in\hat{\Phi}(E)^{\bot}$. Since $\tilde{\Phi}(P)^{*}$ is an idempotent,
either $\lambda=0$ or $\lambda=1$.

If $\lambda=0$, then $\tilde{\Phi}(P)^{*}\hat{\Phi}(E)^{\bot}=\\{0\\}$ and
Eq.(5.1) yields $\langle\tilde{\Phi}(P)y_{2},g_{1}\rangle=0$ for all
$y_{2}\in\hat{\Phi}(E)$ and $g_{1}\in\hat{\Phi}(X_{-})^{\bot}$. It follows
that
$\tilde{\Phi}(P)^{*}\hat{\Phi}(X_{-})^{\bot}\subseteq\hat{\Phi}(E)^{\bot}$. So
$\tilde{\Phi}(P)^{*}\hat{\Phi}(X_{-})^{\bot}=\tilde{\Phi}(P)^{*}(\tilde{\Phi}(P)^{*}\hat{\Phi}(X_{-})^{\bot})\subseteq\tilde{\Phi}(P)^{*}\hat{\Phi}(E)^{\bot}=\\{0\\}$,
which is impossible as
$\tilde{\Phi}(P)\in\Omega_{1}({\mathcal{M}},\hat{\Phi}(X_{-}))$.

If $\lambda=1$, then Eq.(5.1) yields
$\langle\tilde{\Phi}(P)y_{2},g_{1}\rangle=\langle y_{2},g_{1}\rangle$ for all
$y_{2}\in\hat{\Phi}(E)$ and $g_{1}\in\hat{\Phi}(X_{-})^{\bot}$. This implies
that
$(I-\tilde{\Phi}(P))^{*}\hat{\Phi}(X_{-})^{\bot}\subseteq\hat{\Phi}(E)^{\bot}$,
and so

$(I-\tilde{\Phi}(P))^{*}\hat{\Phi}(X_{-})^{\bot}=(I-\tilde{\Phi}(P))^{*}((I-\tilde{\Phi}(P))^{*}\hat{\Phi}(X_{-})^{\bot})\subseteq(I-\tilde{\Phi}(P))^{*}\hat{\Phi}(E)^{\bot}=\\{0\\}.$

This, together with the fact $\tilde{\Phi}(P)\hat{\Phi}(X_{-})=\\{0\\}$
entails ${\rm ran}\tilde{\Phi}(P)=\hat{\Phi}(X_{-})$, a contradiction. Hence
$\hat{\Phi}$ is order-preserving.

Similarly, one can show that $\hat{\Phi}$ is order-reversing if
$\tilde{\Phi}(\Omega_{1}({\mathcal{N}},X_{-}))\subseteq
I-\Omega_{2}({\mathcal{N}},\hat{X_{-}})$. $\Box$

Lemma 5.3. Assume that $\mathcal{N}$ has at least two nontrivial elements and
$(0)<X_{-}<X$ is not complemented in $X$. Then for any $x\in X$ and $f\in
X_{-}^{\bot}$, ${\Phi}(x\otimes f)$ is the sum of a scalar and a rank one
operator. Moreover, the statement (1) and (2) of Lemma 5.1 hold.

Proof. We will complete the proof of the lemma by considering three cases.

Case 1. $x\in X_{-}$ and $f\in X_{-}^{\bot}$.

In this case, let $R=\bar{\Phi}(x\otimes f)$. Then
$R\in\mathcal{J}(\mathcal{M},\hat{\Phi}(X_{-}))$ and $\Phi(x\otimes
f)-\bar{\Phi}(x\otimes f)=\Phi(x\otimes f)-R\in{\mathbb{F}}I$. We show that
$R$ is of rank one. Assume on the contrary that ${\rm rank}R\geq 2$. We will
induce contradiction by considering two subcases.

Subcase 1.1.
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},\hat{\Phi}(X_{-}))$.

By Lemma 5.2, $\hat{\Phi}$ is order-preserving. It is clear in this case that
we have $(0)<Y_{-}=\hat{\Phi}(X_{-})<Y$.

Since ${\rm rank}R\geq 2$, there are two vectors $u,v\in Y\setminus Y_{-}$
such that $Ru$ and $Rv$ are linearly independent. Take $h\in Y_{-}^{\bot}$
such that $h(u)=1$. Then $u\otimes
h\in\Omega_{1}(\mathcal{M},\hat{\Phi}(X_{-}))$. Let
$B=\tilde{\Phi}^{-1}(u\otimes h)$. By Lemma 5.2(i),
$B\in\Omega_{1}({\mathcal{N}},X_{-})$.

If $(0)_{+}=(0)\in\mathcal{N}$, then $(0)_{+}=(0)\in\mathcal{M}$ by Lemma 2.3.
Thus $(0)=\cap\\{M:(0)\not=M\in{\mathcal{M}}\\}$ and there exists some
nontrivial element $M\in\mathcal{M}$ and $g\in M^{\bot}$ such that $g(Ru)=0$
and $g(Rv)=1.$ Obviously $M<\hat{\Phi}(X_{-})$, and so
$\hat{\Phi}^{-1}(M)<X_{-}$. Take a nonzero vector $z\in M$ and let
$A=\bar{\Phi}^{-1}(z\otimes g).$ Then
$A\in\mathcal{J}(\mathcal{N},\hat{\Phi}^{-1}(M))$, which implies
$A^{*}\hat{\Phi}^{-1}(M)^{\bot}=\\{0\\}$. Thus we have

$\Phi(A(x\otimes f)B)=\Phi([A,[x\otimes f,B]])=[z\otimes g,[R,u\otimes
h]]=(z\otimes g)R(u\otimes h)=0.$

So either $A(x\otimes f)=0$ or $(x\otimes f)B=0.$ If $A(x\otimes f)=0,$ then
$0=\Phi([A,x\otimes f])=[z\otimes g,R]=(z\otimes g)R\not=0,$ which is
impossible; If $(x\otimes f)B=0,$ then $0=\Phi([x\otimes f,B])=[R,u\otimes
h]=R(u\otimes h)\not=0,$ which is also impossible.

If $(0)<(0)_{+}\in\mathcal{N}$, then $(0)<(0)_{+}\in\mathcal{M}$ by Lemma 2.3.
Let $M=\hat{\Phi}((0)_{+}).$ Then
$(0)<M=\hat{\Phi}((0)_{+})<\hat{\Phi}(X_{-})$ as $\mathcal{M}$ has at least
two nontrivial elements. By Lemma 2.8, $M$ is not complemented and thus
infinite-dimensional. So, there is a vector $z\in M$ and a functional $g\in
Y^{*}$ such that $g(Ru)=0$ and $g(z)=g(Rv)=1.$ Let
$A=\tilde{\Phi}^{-1}(z\otimes g).$ As $z\otimes
g\in\Omega_{2}({\mathcal{M}},M)$, we have
$A\in\Omega_{2}(\mathcal{N},(0)_{+})$ by Lemma 2.6(3). It follows that
$A^{*}\hat{\Phi}^{-1}(M)^{\bot}=\\{0\\}$. Now by calculating $\Phi(A(x\otimes
f)B)$ in the same way as the above, one can get a contradiction.

So in the case that $\hat{\Phi}$ is order-preserving, $R$ is of rank one.

Subcase 1.2. $\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq
I-\Omega_{2}(\mathcal{M},\hat{\Phi}(X_{-}))$.

By Lemma 5.1(ii), in this case $\hat{\Phi}$ is order-reversing. So, we have
$(0)<(0)_{+}=\hat{\Phi}(X_{-})<Y$.

If $Y=Y_{-}$, there is a nontrivial element $M\in{\mathcal{M}}$ and vectors
$u,v\in M$ such that $Ru$ and $Rv$ are linearly independent. Let $h\in
M^{\bot}$ and $B=\bar{\Phi}^{-1}(u\otimes h).$ Then
$B\in\mathcal{J}(\mathcal{N},\hat{\Phi}^{-1}(M))$ by Lemma 2.4. Choose $g$ in
$Y^{*}$ such that $g(Ru)=0$ and $g(Rv)=1.$ Let $z\in\hat{\Phi}(X_{-})$ and
$A=\tilde{\Phi}^{-1}(z\otimes g)$. By Lemma 2.6(2),
$I-A\in\Omega_{1}(\mathcal{N},X_{-}).$ It follows from $\hat{\Phi}(X_{-})\leq
M$ that $\hat{\Phi}^{-1}(M)\leq X_{-}$. So we get

$\Phi(B(x\otimes f)(I-A))=\Phi([B,[x\otimes f,I-A]])=[u\otimes h,[R,-z\otimes
g]]=0,$

which implies that either $B(x\otimes f)=0$ or $(x\otimes f)(I-A)=0.$ If
$Bx\otimes f=0$, then we get $0=\Phi([B,x\otimes f])=[u\otimes h,R]=-Ru\otimes
h\not=0$, a contradiction. If $x\otimes f(I-A)=0$, then $0=\Phi([x\otimes
f,I-A])=[R,-z\otimes g]=z\otimes R^{*}g\not=0$, again a contradiction.

If $Y_{-}<Y$, there are two vectors $u,v\in Y$ such that $Ru$ and $Rv$ are
linearly independent. If $u,v\in Y_{-}$, we take $h\in Y_{-}^{\bot}$ and let
$B=\Phi^{-1}(u\otimes h).$ Then
$B\in\mathcal{J}(\mathcal{N},\hat{\Phi}^{-1}(Y_{-}))$. Choose $g\in Y^{*}$
such that $g(Ru)=0$ and $g(Rv)=1.$ Let $z\in\hat{\Phi}(X_{-})$ and
$A=\tilde{\Phi}^{-1}(z\otimes g)$. By Lemma 2.6(2),
$I-A\in\Omega_{1}(\mathcal{N},X_{-}).$ Still, by assumption, we have
$\hat{\Phi}(X_{-})<Y_{-}$, and so $\hat{\Phi}^{-1}(Y_{-})<X_{-}$. Thus by a
similar argument to that in the preceding paragraph, one can get a
contradiction. So we can assume that $u\not\in Y_{-}.$ Take $h\in
Y_{-}^{\bot}$ such that $h(u)=1$ and let $B=\tilde{\Phi}^{-1}(u\otimes h).$
Then $I-B\in\Omega_{2}(\mathcal{N},\hat{\Phi}^{-1}(Y_{-}))$ by Lemma 5.2(ii).
In addition, there exists some $g\in Y^{*}$ such that $g(Ru)=0$ and $g(Rv)=1.$
Let $z=Rv$ and $A=\tilde{\Phi}^{-1}(z\otimes g).$ Then, as
$z\in\hat{\Phi}(X_{-})=(0)_{+}$, we see that $z\otimes
g\in\Omega_{2}({\mathcal{M}},(0)_{+})$ and, by Lemma 2.6 and Lemma 5.2,
$I-A\in\Omega_{1}(\mathcal{N},X_{-}).$ Since $\hat{\Phi}(X_{-})<Y_{-}$, we
have $\hat{\Phi}^{-1}(Y_{-})<X_{-}$. Calculating $\Phi(B(x\otimes f)(I-A))$,
one can get a contradiction.

Hence ${\rm rank}R=1$ and $\Phi(x\otimes f)$ is the sum of a scalar and a rank
one operator, that is, the lemma is true for the case that $x\in X_{-}$ and
$f\in X_{-}^{\bot}$. Moreover, $\bar{\Phi}(X_{-}\otimes
X_{-}^{\bot})=\hat{\Phi}(X_{-})\otimes\hat{\Phi}(X_{-})^{\bot}$.

Case 2. $x\in X\setminus X_{-}$ and $f\in X_{-}^{\perp}$ with $\langle
x,f\rangle=1$.

Let $P=x\otimes f$. Clearly, $P\in\Omega_{1}(\mathcal{N},X_{-})$. By Case 1,
we have $\bar{\Phi}(X_{-}\otimes
X_{-}^{\bot})=\hat{\Phi}(X_{-})\otimes\hat{\Phi}(X_{-})^{\bot}$ and hence
${\Phi}(X_{-}\otimes
X_{-}^{\bot})\subseteq{\mathbb{F}}I+\hat{\Phi}(X_{-})\otimes\hat{\Phi}(X_{-})^{\bot}$.
For the seek of convenience, write $\tilde{P}=\tilde{\Phi}(P)$ and
$\widehat{X_{-}}=\hat{\Phi}(X_{-})$. Then $\Phi(P)-\tilde{P}\in{\mathbb{F}}I$
and it suffices to show that $\tilde{P}$ is the sum of a scalar and a rank-1
idempotent operator.

Subcase 2.1.
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},\widehat{X_{-}})$.

In this subcase $\hat{\Phi}$ is order-preserving by Lemma 5.2(i), and hence
$\widehat{X_{-}}=Y_{-}$. Note that $\Phi({\mathbb{F}}I)={\mathbb{F}}I$ by
Lemma 2.9. So, applying the fact proved in Case 1, we have
$\Phi(\mathbb{F}I+X_{-}\otimes X_{-}^{\bot})=\mathbb{F}I+Y_{-}\otimes
Y_{-}^{\bot}.$ Thus it follows from Lemma 2.10 that there exists a ring
automorphism $\tau:\mathbb{F}\rightarrow\mathbb{F}$ and a map
$\gamma:X_{-}\times X_{-}^{\bot}\rightarrow\mathbb{F}$ such that either

$None$ $\Phi(y\otimes g)=\gamma(y,g)I+Cy\otimes Dg\ \ {\rm for\ all}\ \ y\in
X_{-}\ \ {\rm and}\ \ g\in X_{-}^{\bot},$

where $C:X_{-}\rightarrow Y_{-}$ and $D:X_{-}^{\bot}\rightarrow Y_{-}^{\bot}$
are two bijective $\tau$-linear maps; or

$None$ $\Phi(y\otimes g)=\gamma(y,g)I+Dg\otimes Cy\ \ {\rm for\ all}\ \ y\in
X_{-}\ \ {\rm and}\ \ g\in X_{-}^{\bot},$

where $C:X_{-}\rightarrow Y_{-}^{\bot}$ and $D:X_{-}^{\bot}\rightarrow Y_{-}$
are two bijective $\tau$-linear maps.

We first show that Eq.(5.3) can not occur. On the contrary, if Eq.(5.3) holds,
for any $y\in X_{-}$ and $g\in X_{-}^{\bot}$, we have

$\Phi([y\otimes g,P])=\Phi(y\otimes P^{*}g)=\gamma(y,P^{*}g)I+DP^{*}g\otimes
Cy$

and

$\Phi([y\otimes g,P])=[Dg\otimes Cy,\tilde{P}]=Dg\otimes\tilde{P}^{*}Cy,$

which imply $Dg\otimes\tilde{P}^{*}Cy=DP^{*}g\otimes Cy$ for all $y\in X_{-}$
and $g\in X_{-}^{\bot}$. Thus there exists some scalar $\lambda$ such that
$D|_{X_{-}^{\bot}}=\lambda DP^{*}|_{X_{-}^{\bot}}$. Since $P$ is of rank one,
we see that $D|_{X_{-}^{\bot}}$ is also of rank one, but this is impossible as
$X_{-}^{\bot}$ is infinite-dimensional.

So Eq.(5.2) holds. Then, for any $y\in X_{-}$ and $g\in X_{-}^{\bot}$, we have

$\Phi([y\otimes g,P])=\Phi(y\otimes P^{*}g)=\gamma(y,P^{*}g)I+Cy\otimes
DP^{*}g$

and

$\Phi([y\otimes g,P])=[Cy\otimes Dg,\tilde{P}]=Cy\otimes\tilde{P}^{*}Dg.$

It follows that $Cy\otimes\tilde{P}^{*}Dg=\gamma(y,P^{*}g)I+Cy\otimes DP^{*}g$
holds for all $y\in X_{-}$ and $g\in X_{-}^{\bot}.$ Since $I$ is of infinite
rank, we have $\gamma(y,P^{*}g)=0$, and $Cy\otimes\tilde{P}^{*}Dg=Cy\otimes
DP^{*}g$. So $\tilde{P}^{*}Dg=DP^{*}g$ for all $g\in X_{-}^{\bot}.$ Since $P$
is of rank one, it follows that the restriction of $\tilde{P}^{*}$ to
$Y_{-}^{\bot}$ is of rank one. Note that $\tilde{P}^{*}Y^{*}\subseteq
Y_{-}^{\bot}$ and
$\tilde{P}^{*}Y^{*}=\tilde{P}^{*}(\tilde{P}^{*}Y^{*})\subseteq\tilde{P}^{*}Y_{-}^{\bot}.$
So $\tilde{P}^{*}$ is of rank one, which implies that $\tilde{P}$ is also of
rank one, as desired.

Subcase 2.2. $\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq
I-\Omega_{2}(\mathcal{M},\widehat{X_{-}})$.

By Lemma 5.2(ii), $\hat{\Phi}$ is order-reversing, and
$\widehat{X_{-}}=(0)_{+}$. Applying the fact proved in Case 1, we have
$\Phi(\mathbb{F}I+X_{-}\otimes
X_{-}^{\bot})=\mathbb{F}I+(0)_{+}\otimes(0)_{+}^{\bot}.$ Thus, by Lemma 2.10
again, there exists a ring automorphism $\tau:\mathbb{F}\rightarrow\mathbb{F}$
and a map $\gamma:X_{-}\times X_{-}^{\bot}\rightarrow\mathbb{F}$ such that
either

$None$ $\Phi(x\otimes f)=\gamma(x,f)I+Cx\otimes Df\ \ {\rm for\ all}\ \ x\in
X_{-}\ \ {\rm and}\ \ f\in X_{-}^{\bot},$

where $C:X_{-}\rightarrow(0)_{+}$ and
$D:X_{-}^{\bot}\rightarrow(0)_{+}^{\bot}$ are two $\tau$-linear bijective
maps; or

$None$ $\Phi(x\otimes f)=\gamma(x,f)I+Df\otimes Cx\ \ {\rm for\ all}\ \ x\in
X_{-}\ \ {\rm and}\ \ f\in X_{-}^{\bot},$

where $C:X_{-}\rightarrow(0)_{+}^{\bot}$ and
$D:X_{-}^{\bot}\rightarrow(0)_{+}$ are two $\tau$-linear bijective maps.

If Eq.(5.4) holds, by calculating $\Phi([y\otimes g,P])$, one obtains
$(I-\tilde{P})Cy\otimes Dg=Cy\otimes DP^{*}g$ for all $y\in X_{-}$ and $g\in
X_{-}^{\bot}.$ Since $P$ is of rank one, we get $D$ is also of rank one, which
is impossible. So we must have that Eq.(5.5) holds. By calculating
$\Phi([y\otimes g,P])$, one can get $(I-\tilde{P})Dg=DP^{*}g$ for all $f\in
X_{-}^{\bot}$, which implies that the restriction of $I-\tilde{P}$ to
$(0)_{+}$ is of rank one. So $I-\tilde{P}$ is of rank one and $\tilde{P}$ is
the sum of a scalar and a rank one operator, as desired.

Summing up, we have proved that $\Phi(x\otimes f)=\Phi(P)$ is the sum of a
scalar and a rank one operator if $x\in X,f\in X_{-}^{\bot}$ with $\langle
x,f\rangle=1$. Moreover,
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},\widehat{X_{-}})$
implies that Eq.(5.2) holds, while
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq
I-\Omega_{2}(\mathcal{M},\widehat{X_{-}})$ implies that Eq.(5.5) holds.

Case 3. $x\in X\setminus X_{-}$ and $f\in X_{-}^{\bot}.$

Now assume that $x\in X\setminus X_{-}$ and $f\in X_{-}^{\bot}.$ We need still
consider two subcases.

Subcase 3.1. $\langle x,f\rangle=\lambda\neq 0$.

Let $P=\lambda^{-1}x\otimes f$. Then the rank-one idempotent
$P\in\Omega_{1}(\mathcal{N},X_{-})$ and $x\otimes f=\lambda P.$ Write
$\tilde{P}=\tilde{\Phi}(P).$ By what proved in Case 2, $\tilde{P}$ is a rank-1
idempotent.

Subcase 3.1.1.
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},\widehat{X_{-}})$.

By Lemma 5.2(i), $\hat{\Phi}$ is order-preserving, and then
$\widehat{X_{-}}=Y_{-}$. So we still have that either Eq.(5.2) or Eq.(5.3)
holds.

If Eq.(5.2) holds, then for any $y\in X_{-}$ and $g\in X_{-}^{\bot}$, we have

$\Phi([y\otimes g,\lambda P])=[\Phi(y\otimes g),\Phi(\lambda
P)]=Cy\otimes\Phi(\lambda P)^{*}Dg-\Phi(\lambda P)Cy\otimes Dg$

and

$\Phi([y\otimes g,\lambda P])=[\tau(\lambda)Cy\otimes
Dg,\tilde{P}]=\tau(\lambda)Cy\otimes\tilde{P}^{*}Dg.$

It follows that $\tau(\lambda)Cy\otimes\tilde{P}^{*}Dg=Cy\otimes\Phi(\lambda
P)^{*}Dg-\Phi(\lambda P)Cy\otimes Dg$, that is,

$Cy\otimes(\overline{\tau(\lambda)}\tilde{P}^{*}Dg-\Phi(\lambda
P)^{*}Dg)=-\Phi(\lambda P)Cy\otimes Dg\ \ {\rm for\ all}\ y\in X_{-}\ {\rm
and}\ g\in X_{-}^{\perp}.$

Since $C$ is bijective, there exists some scalar $\alpha$ such that
$\tau(\lambda)\tilde{P}^{*}Dg-\Phi(\lambda P)^{*}Dg=\alpha Dg$ for all $g\in
X_{-}^{\bot}$, that is, $(\Phi(\lambda P)^{*}+\alpha
I)|_{Y_{-}^{\bot}}=\tau(\lambda)\tilde{P}^{*}|_{Y_{-}^{\bot}}$ as
$D:X_{-}^{\bot}\rightarrow Y_{-}^{\bot}$ is bijective. Let
$\widetilde{\Phi(\lambda P)}=\Phi(\lambda P)+\overline{\alpha}I.$ Note that
$[\lambda P,A]=[\lambda P,[P,[P,A]]]$ holds for all $A\in{\rm
Alg}\mathcal{N}$. So we get

$None$ $[\widetilde{\Phi(\lambda P)},\Phi(A)]=[\widetilde{\Phi(\lambda
P)},[\tilde{P},[\tilde{P},\Phi(A)]]],\ \ \ \forall A\in{\rm Alg}\mathcal{N}.$

By the bijectivity of $\Phi$, for any $y\in Y$ and $g\in Y_{-}^{\bot}$, there
exists some $A\in{\rm Alg}{\mathcal{N}}$ such that $\Phi(A)=y\otimes g$. Thus
Eq.(5.6) entails

$None$ $(\widetilde{\Phi(\lambda P)}-\widetilde{\Phi(\lambda
P)}\tilde{P})y\otimes g=(\tau(\lambda)\tilde{P}+\widetilde{\Phi(\lambda
P)}-2\widetilde{\Phi(\lambda P)}\tilde{P})y\otimes\tilde{P}^{*}g$

for all $y\in Y$ and $g\in Y_{-}^{\bot}$. By Case 2, we can write
$\tilde{P}=u\otimes h$, where $u\in Y$ and $h\in Y_{-}^{\bot}$ with $\langle
u,h\rangle=1.$ Since $\dim(Y_{-}^{\bot})>2,$ there exists some $g_{1}\in
Y_{-}^{\bot}$ such that $g_{1}$ is linearly independent of $h$. If $\langle
u,g_{1}\rangle\neq 0,$ let $g=g_{1}.$ If $\langle u,g_{1}\rangle=0$, let
$g=h+g_{1}$. Then $g$ and $h$ are linearly independent and $\langle
u,g\rangle\neq 0.$ So $g$ and $\tilde{P}^{*}g$ are also linearly independent.
By Eq.(5.7), we get $(\widetilde{\Phi(\lambda P)}-\widetilde{\Phi(\lambda
P)}\tilde{P})y=0$ for all $y\in Y.$ It follows that $\widetilde{\Phi(\lambda
P)}=\widetilde{\Phi(\lambda P)}\tilde{P}$, which implies
$\widetilde{\Phi(\lambda P)}$ is of rank one. So $\Phi(\lambda
P)=\widetilde{\Phi(\lambda P)}-\bar{\alpha}I$ is the sum of a scalar and a
rank one operator.

We claim that Eq.(5.3) can not occur. If, on the contrary, Eq.(5.3) holds,
then for any $y\in X_{-}$ and $g\in X_{-}^{\bot}$, we have

$\Phi([y\otimes g,\lambda P])=[\Phi(y\otimes g),\Phi(\lambda
P)]=Dg\otimes\Phi(\lambda P)^{*}Cy-\Phi(\lambda P)Dg\otimes Cy$

and

$\Phi([y\otimes g,\lambda P])=\Phi(\lambda y\otimes
P^{*}g)=\tau(\lambda)D(P^{*}g)\otimes Cy.$

It follows that $Dg\otimes\Phi(\lambda P)^{*}Cy=(\Phi(\lambda
P)Dg+\tau(\lambda)D(P^{*}g))\otimes Cy$ for all $y\in X_{-}$ and $g\in
X_{-}^{\perp}.$ So there exists some scalar $\gamma$ such that $\Phi(\lambda
P)^{*}Cy=\gamma Cy$, which implies $\Phi(\lambda P)^{*}=\gamma I$ on
$Y_{-}^{\perp}$ as $C$ is bijective. Now, for any $A\in{\rm Alg}\mathcal{N}$,
by the relation $[\lambda P,A]=[\lambda P,[P,[P,A]]]$, we get $[{\Phi(\lambda
P)},\Phi(A)]=[{\Phi(\lambda P)},[\tilde{P},[\tilde{P},\Phi(A)]]]$.
Particularly, for any $y\in Y$ and $g\in Y_{-}^{\bot}$, there exists some
$A\in{\rm Alg}{\mathcal{N}}$ such that $\Phi(A)=y\otimes g$. Thus we have
$[{\Phi(\lambda P)},y\otimes g]=[{\Phi(\lambda
P)},[\tilde{P},[\tilde{P},y\otimes g]]]$, that is,

$({\Phi(\lambda P)}-{\Phi(\lambda P)}\tilde{P}-\gamma
I+\gamma\tilde{P})y\otimes g=({\Phi(\lambda P)}-2{\Phi(\lambda
P)}\tilde{P}+2\gamma\tilde{P}-\gamma I)y\otimes\tilde{P}^{*}g$

holds for all $y\in Y$ and $g\in Y_{-}^{\bot}$. Still, we can choose $g$ such
that $g$ and $\tilde{P}^{*}g$ are linearly independent. The above equation
yields $({\Phi(\lambda P)}-{\Phi(\lambda P)}\tilde{P}-\gamma
I+\gamma\tilde{P})y=({\Phi(\lambda P)}-2{\Phi(\lambda
P)}\tilde{P}+2\gamma\tilde{P}-\gamma I)y=0$ for all $y\in Y$. This implies
${\Phi(\lambda P)}=\gamma I$, which is contradicting to
$\Phi({\mathbb{F}}I)={\mathbb{F}}I$ and the bijectivity of $\Phi$. So Eq.(5.3)
can not occur.

Subcase 3.1.2. $\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq
I-\Omega_{2}(\mathcal{M},\widehat{X_{-}})$

By Lemma 5.2(ii), $\hat{\Phi}$ is order-reversing and thus
$\widehat{X_{-}}=(0)_{+}$. It follows that either Eq.(5.4) or Eq.(5.5) holds.
By a similar argument to the Subcase 3.1.1 above, one can check that, Eq.(5.4)
can not occur and if Eq.(5.5) holds, then $\Phi(x\otimes f)=\Phi(\lambda P)$
is the sum of a scalar and a rank one operator.

Subcase 3.2. $\langle x,f\rangle=0$.

In this case, take $x_{1}\in X$ such that $\langle x_{1},f\rangle=1$ and let
$x_{2}=x-2x_{1}$, $x_{3}=x-x_{1}.$ Then $\langle x_{i},f\rangle\neq 0$ for
$i=1,2,3$. By Subcase 3.1, $\Phi(x_{i}\otimes f)$ is the sum of a scalar and a
rank one operator. So we can assume $\Phi(x_{i}\otimes
f)=\lambda_{i}I+u_{i}\otimes h_{i}$ for some $\lambda_{i}$, $i=1,2,3.$ Note
that

$(\lambda_{1}+\lambda_{2})I+u_{1}\otimes h_{1}+u_{2}\otimes
h_{2}=\Phi((x_{1}+x_{2})\otimes f)=\Phi(x_{3}\otimes
f)=\lambda_{3}I+u_{3}\otimes h_{3}.$

Since $I$ is of infinite rank, we must have $u_{1}\otimes h_{1}+u_{2}\otimes
h_{2}=u_{3}\otimes h_{3}$. This forces that $\\{u_{1},u_{3}\\}$ or
$\\{h_{1},h_{3}\\}$ is a linearly dependent set. So $u_{1}\otimes
h_{1}+u_{3}\otimes h_{3}$ is of rank one. Thus $\Phi(x\otimes
f)=\Phi((x_{1}+x_{3})\otimes f)=(\lambda_{1}+\lambda_{3})I+u_{1}\otimes
h_{1}+u_{3}\otimes h_{3}$ is the sum of a scalar and a rank one operator.

Combining Cases 1-3 and the bijectivity of $\Phi$, we have shown that

(i) $\Phi({\mathbb{F}}I+X\otimes X_{-}^{\bot})={\mathbb{F}}I+Y\otimes
Y_{-}^{\bot}$ if
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},\widehat{X_{-}})$;

(ii) $\Phi({\mathbb{F}}I+X\otimes X_{-}^{\bot})={\mathbb{F}}I+(0)_{+}\otimes
Y^{*}$ if $\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq
I-\Omega_{2}(\mathcal{M},\widehat{X_{-}})$.

So Lemma 2.10 is applicable. Observing the arguments in Case 2 and Case 3, it
is easily seen that (i) implies Lemma 5.1(1) and (ii) implies Lemma 5.1(2).

The proof of Lemma 5.3 is finished. $\Box$

For the case that the nest has only one nontrivial element, we have

Lemma 5.4. Assume that $\mathcal{N}=\\{(0),X_{-},X\\}$ and $X_{-}$ is not
complemented in $X$. Then for any $x\in X$ and $f\in X_{-}^{\bot}$,
${\Phi}(x\otimes f)$ is the sum of a scalar and a rank one operator. Moreover,
the statements (1) and (2) of Lemma 5.1 hold.

Proof. Not that $(0)_{+}=X_{-}$ in this situation. By Lemma 2.3,
${\mathcal{M}}=\\{(0),Y_{-},Y\\}$ and $(0)_{+}=Y_{-}$ is not complemented in
$Y$. Let $R\in\mathcal{J}(\mathcal{M},Y_{-})$ so that ${\Phi}(x\otimes
f)-R\in{\mathbb{F}}I$ (Lemma 2.2). We remak here that, since $\mathcal{N}$
only contains one nontrivial element, Lemma 5.2 and Lemma 2.6(3) are not
applicable. Similar to the proof of Lemma 5.3, we consider three cases.

Case 1. $x\in X_{-}$ and $f\in X_{-}^{\bot}$.

In this case, $x\otimes f\in{\mathcal{J}}({\mathcal{N}},X_{-})$. We’ll prove
${\rm rank}R=1$. Assume on the contrary that ${\rm rank}R\geq 2$. Then there
are two vectors $u,v\in Y\setminus Y_{-}$ such that $Ru$ and $Rv$ are linearly
independent. Take $h\in Y_{-}^{\bot}$ such that $h(u)=1$. Then $u\otimes
h\in\Omega_{1}(\mathcal{M},Y_{-})$. Note that $Y_{-}$ is infinite-dimensional.
There is a vector $z\in Y_{-}$ and a functional $g\in Y^{*}$ such that
$g(Ru)=0$ and $g(z)=g(Rv)=1$. Let $A=\tilde{\Phi}^{-1}(z\otimes g)$ and
$B=\tilde{\Phi}^{-1}(u\otimes h)$.

By Lemma 2.6(1), either
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},Y_{-})$
or
$I-\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{2}(\mathcal{M},Y_{-})$.
If
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},Y_{-})$,
then $B\in\Omega_{1}({\mathcal{N}},X_{-})$, and by Lemma 2.6(2),
$A\in\Omega_{2}(\mathcal{N},X_{-})$. Thus we get $\Phi(A(x\otimes
f)B)=\Phi([A,[x\otimes f,B]])=[z\otimes g,[R,u\otimes h]]=(z\otimes
g)R(u\otimes h)=0$, which implies either $A(x\otimes f)=0$ or $(x\otimes
f)B=0.$ If $A(x\otimes f)=0,$ then $0=\Phi([A,x\otimes f])=[z\otimes
g,R]=(z\otimes g)R,$ which is impossible; if $(x\otimes f)B=0,$ then
$0=\Phi([x\otimes f,B])=[R,u\otimes h]=R(u\otimes h),$ which is also
impossible. If
$I-\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{2}(\mathcal{M},Y_{-})$,
then $I-B\in\Omega_{2}(\mathcal{N},X_{-})$, and by Lemma 2.6(2),
$I-A\in\Omega_{1}(\mathcal{N},X_{-}).$ Thus $\Phi((I-B)(x\otimes
f)(I-A))=\Phi([I-B,[x\otimes f,I-A]])=[-u\otimes h,[R,-z\otimes g]]=0$. So we
get either $(I-B)(x\otimes f)=0$ or $(x\otimes f)(I-A)=0.$ Still, this is
impossible.

Hence ${\rm rank}R=1$.

Case 2. $x\in X\setminus X_{-}$ and $f\in X_{-}^{\perp}$ with $\langle
x,f\rangle=1$.

In this case $P=x\otimes f\in\Omega_{1}(\mathcal{N},X_{-})$. Write
$\tilde{P}=\tilde{\Phi}(P)$. By Case 1 and Lemma 2.9, we have
$\Phi(\mathbb{F}I+X_{-}\otimes X_{-}^{\bot})=\mathbb{F}I+Y_{-}\otimes
Y_{-}^{\bot}.$ So, by Lemma 2.10, there exists a ring automorphism
$\tau:\mathbb{F}\rightarrow\mathbb{F}$ and a map $\gamma:X_{-}\times
X_{-}^{\bot}\rightarrow\mathbb{F}$ such that either

$None$ $\Phi(y\otimes g)=\gamma(y,g)I+Cy\otimes Dg\ \ {\rm for\ all}\ \ y\in
X_{-}\ \ {\rm and}\ \ g\in X_{-}^{\bot},$

where $C:X_{-}\rightarrow Y_{-}$ and $D:X_{-}^{\bot}\rightarrow Y_{-}^{\bot}$
are two $\tau$-linear bijective maps; or

$None$ $\Phi(y\otimes g)=\gamma(y,g)I+Dg\otimes Cy\ \ {\rm for\ all}\ \ y\in
X_{-}\ \ {\rm and}\ \ g\in X_{-}^{\bot},$

where $C:X_{-}\rightarrow Y_{-}^{\bot}$ and $D:X_{-}^{\bot}\rightarrow Y_{-}$
are two $\tau$-linear bijective maps.

Assume first that Eq.(5.8) holds. If
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},Y_{-})$,
then $\tilde{P}\in\Omega_{1}({\mathcal{M}},Y_{-})$. Thus, for any $y\in Y$ and
any $g\in Y_{-}^{\perp}$, we have

$\Phi([y\otimes g,P])=\Phi(y\otimes P^{*}g)=\gamma(y,P^{*}g)I+Cy\otimes
DP^{*}g$

and

$\Phi([y\otimes g,P])=[Cy\otimes Dg,\tilde{P}]=Cy\otimes\tilde{P}^{*}Dg.$

Since $I$ is of infinite rank, the above two equations yields
$Cy\otimes\tilde{P}^{*}Dg=Cy\otimes DP^{*}g$, and so $\tilde{P}^{*}Dg=DP^{*}g$
for all $g\in X_{-}^{\bot}.$ Since $P$ is of rank one, the restriction of
$\tilde{P}^{*}$ to $Y_{-}^{\bot}$ is of rank one. Note that
$\tilde{P}^{*}Y^{*}\subseteq Y_{-}^{\bot}$ and
$\tilde{P}^{*}Y^{*}=\tilde{P}^{*}(\tilde{P}^{*}Y^{*})\subseteq\tilde{P}^{*}Y_{-}^{\bot}.$
So $\tilde{P}^{*}$ is of rank one, which implies that $\tilde{P}$ is also of
rank one. Hence $\Phi(P)$ is the sum of a scalar and a rank one operator.

If
$I-\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{2}(\mathcal{M},Y_{-})$,
then $I-\tilde{P}\in\Omega_{2}(\mathcal{M},Y_{-})$. For any $y\in Y$ and any
$g\in Y_{-}^{\perp}$, we have

$\Phi([y\otimes g,P])=\Phi(y\otimes P^{*}g)=\gamma(y,P^{*}g)I+Cy\otimes
DP^{*}g$

and

$\Phi([y\otimes g,P])=[Cy\otimes Dg,\tilde{P}]=-[Cy\otimes
Dg,I-\tilde{P}]=(I-\tilde{P})Cy\otimes Dg.$

The above two equations yield $(I-\tilde{P})Cy\otimes Dg=Cy\otimes DP^{*}g$
for all $y\in X_{-}$ and $g\in X_{-}^{\bot}$, and hence $D$ and $DP^{*}$ are
linearly dependent. Since $P$ is of rank one, $D$ is also of rank one, which
is impossible as $X_{-}^{\bot}$ is infinite-dimensional. Therefore, this case
can not occur.

Similarly, if Eq.(5.9) holds, one can show that
$I-\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{2}(\mathcal{M},Y_{-})$
and $\Phi(P)$ is the sum of a scalar and a rank one operator.

Case 3. $x\in X\setminus X_{-}$ and $f\in X_{-}^{\bot}.$

Note that, we still have that either Eq.(5.8) or Eq.(5.9) holds.

Subcase 3.1. $\langle x,f\rangle=\lambda\neq 0$.

Then there exists a rank-one idempotent $P\in\Omega_{1}(\mathcal{N},X_{-})$
such that $x\otimes f=\lambda P.$ Write $\tilde{P}=\tilde{\Phi}(P)$. By Case
2, $\tilde{P}$ is a rank-1 idempotent.

Assume that Eq.(5.8) holds. If
$\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{1}(\mathcal{M},Y_{-})$,
then $\tilde{P}\in\Omega_{1}({\mathcal{M}},Y_{-})$. So, for any $y\in X_{-}$
and $g\in X_{-}^{\bot}$, we have

$\Phi([y\otimes g,\lambda P])=[\Phi(y\otimes g),\Phi(\lambda
P)]=Cy\otimes\Phi(\lambda P)^{*}Dg-\Phi(\lambda P)Cy\otimes Dg$

and

$\Phi([y\otimes g,\lambda P])=\Phi([\lambda y\otimes
g,P])=[\tau(\lambda)Cy\otimes
Dg,\tilde{P}]=\tau(\lambda)Cy\otimes\tilde{P}^{*}Dg.$

It follows that

$Cy\otimes(\overline{\tau(\lambda)}\tilde{P}^{*}Dg-\Phi(\lambda
P)^{*}Dg)=-\Phi(\lambda P)Cy\otimes Dg\ {\rm for\ all}\ y\in X_{-}\ {\rm and}\
g\in X_{-}^{\perp}.$

Hence there exists some scalar $\alpha$ such that
$\tau(\lambda)\tilde{P}^{*}Dg-\Phi(\lambda P)^{*}Dg=\alpha Dg$ for all $g\in
X_{-}^{\bot}$, which implies that $(\Phi(\lambda P)^{*}+\alpha
I)|_{Y_{-}^{\bot}}=\tau(\lambda)\tilde{P}^{*}|_{Y_{-}^{\bot}}$ as $D$ is
bijective. Let $\widetilde{\Phi(\lambda P)}=\Phi(\lambda
P)+\overline{\alpha}I.$ For any $A\in{\rm Alg}\mathcal{N}$, by the relation
$[\lambda P,A]=[\lambda P,[P,[P,A]]]$, we get $[\widetilde{\Phi(\lambda
P)},\Phi(A)]=[\widetilde{\Phi(\lambda P)},[\tilde{P},[\tilde{P},\Phi(A)]]]$.
Particularly, for any $y\in Y$ and $g\in Y_{-}^{\bot}$, By the bijectivity of
$\Phi$, there exists some $A\in{\rm Alg}{\mathcal{N}}$ such that
$\Phi(A)=y\otimes g$. So we have $[\widetilde{\Phi(\lambda P)},y\otimes
g]=[\widetilde{\Phi(\lambda P)},[\tilde{P},[\tilde{P},y\otimes g]]]$, that is,

$None$ $(\widetilde{\Phi(\lambda P)}-\widetilde{\Phi(\lambda
P)}\tilde{P})y\otimes g=(\tau(\lambda)\tilde{P}+\widetilde{\Phi(\lambda
P)}-2\widetilde{\Phi(\lambda P)}\tilde{P})y\otimes\tilde{P}^{*}g$

holds for all $y\in Y$ and $g\in Y_{-}^{\bot}$. By Case 2, we can write
$\tilde{P}=u\otimes h$, where $u\in Y$ and $h\in Y_{-}^{\bot}$ with $\langle
u,h\rangle=1.$ Since $\dim(Y_{-}^{\bot})>2,$ there exists $g$ with $\langle
u,g\rangle\neq 0$ such that $g$ and $h$ are linearly independent. So $g$ and
$\tilde{P}^{*}g$ are also linearly independent. By Eq.(5.10), we get
$(\widetilde{\Phi(\lambda P)}-\widetilde{\Phi(\lambda P)}\tilde{P})y=0$ for
all $y\in Y$. So $\widetilde{\Phi(\lambda P)}=\widetilde{\Phi(\lambda
P)}\tilde{P}$, which implies $\widetilde{\Phi(\lambda P)}$ is of rank one. So
$\Phi(\lambda P)=\widetilde{\Phi(\lambda P)}-\alpha I$ is the sum of a scalar
and a rank one operator.

We claim that the case
$I-\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{2}(\mathcal{M},Y_{-})$
does not happen. Otherwise, we have
$I-\tilde{P}\in\Omega_{2}(\mathcal{M},Y_{-})$. Then for any $y\in X_{-}$ and
$g\in X_{-}^{\bot}$, by calculating $\Phi([y\otimes g,\lambda P])$, one can
obtain

$Cy\otimes\Phi(\lambda P)^{*}Dg=(\Phi(\lambda
P)-\tau(\lambda)(I-\tilde{P}))Cy\otimes Dg\ {\rm for\ all}\ y\in X_{-}\ {\rm
and}\ g\in X_{-}^{\perp}.$

It follows that there exists some scalar $\gamma$ such that $\Phi(\lambda
P)^{*}Dg=\gamma Dg$, which implies $\Phi(\lambda P)^{*}=\gamma I$ on
$Y_{-}^{\perp}$ as $D$ is bijective. Now, for any $A\in{\rm Alg}\mathcal{N}$,
by the relation $[\lambda P,A]=[\lambda P,[P,[P,A]]]$, we get $[{\Phi(\lambda
P)},\Phi(A)]=[{\Phi(\lambda P)},[\tilde{P},[\tilde{P},\Phi(A)]]]$.
Particularly, for any $y\in Y$ and $g\in Y_{-}^{\bot}$, there exists some
$A\in{\rm Alg}{\mathcal{N}}$ such that $\Phi(A)=y\otimes g$. So we have
$[{\Phi(\lambda P)},y\otimes g]=[{\Phi(\lambda
P)},[\tilde{P},[\tilde{P},y\otimes g]]]$, that is,

$None$ $({\Phi(\lambda P)}-{\Phi(\lambda P)}\tilde{P}-\gamma
I+\gamma\tilde{P})y\otimes g=({\Phi(\lambda P)}-2{\Phi(\lambda
P)}\tilde{P}+2\gamma\tilde{P}-\gamma I)y\otimes\tilde{P}^{*}g$

holds for all $y\in Y$ and $g\in Y_{-}^{\bot}$. Still, we can choose $g$ such
that $g$ and $\tilde{P}^{*}g$ are linearly independent. By Eq.(5.11), we get
$({\Phi(\lambda P)}-{\Phi(\lambda P)}\tilde{P}-\gamma
I+\gamma\tilde{P})y=({\Phi(\lambda P)}-2{\Phi(\lambda
P)}\tilde{P}+2\gamma\tilde{P}-\gamma I)y=0$ for all $y\in Y$. This leads to a
contradiction ${\Phi(\lambda P)}=\gamma I$.

If Eq.(5.9) holds, by a similar argument to that of the above, one can show
that
$I-\tilde{\Phi}(\Omega_{1}(\mathcal{N},X_{-}))\subseteq\Omega_{2}(\mathcal{M},Y_{-})$
and $\Phi(\lambda P)$ is also the sum of a scalar and a rank one operator.

Subcase 3.2. $\langle x,f\rangle=0$.

The proof is the same to that of Subcase 3.2 in Lemma 5.3. We omit it here.

Combining Cases 1-3, we see that the statements (i) and (ii) in the proof of
Lemma 5.3 still hold, and this completes the proof of Lemma 5.4. $\Box$

Proof of Lemma 5.1. It is immediate from Lemma 5.3 and Lemma 5.4. $\Box$

Proof of Theorem 1.1 for the case $X_{-}\not=X$ and $X_{-}$ is not
complemented.

Now let us show that Theorem 1.1 is true for the case that $X_{-}\not=X$ and
$X_{-}$ is not complemented. By Lemma 5.1, (1) or (2) holds.

If Lemma 5.1(1) holds, then $\Phi(\mathbb{F}I+X\otimes
X_{-}^{\bot})=\mathbb{F}I+Y\otimes Y_{-}^{\bot}$, and there exists a ring
automorphism $\tau:\mathbb{F}\rightarrow\mathbb{F}$ and a map $\gamma:X\otimes
X_{-}^{\bot}\rightarrow\mathbb{F}$ such that

$None$ $\Phi(x\otimes f)=\gamma(x,f)I+Cx\otimes Df$

holds for all $x\in X$ and $f\in X_{-}^{\bot}$, where $C:X\rightarrow Y$ and
$D:X_{-}^{\bot}\rightarrow Y_{-}^{\bot}$ are two $\tau$-linear bijective maps.
Thus for any $A\in{\rm Alg}{\mathcal{N}}$, any $x\in X$ and any $f\in
X_{-}^{\perp}$, as $\Phi([A,x\otimes f])=[\Phi(A),\Phi(x\otimes f)]$ and $I$
is of infinite rank, one obtains that

$Cx\otimes(\Phi(A)^{*}Df-DA^{*}f)=(\Phi(A)Cx-CAx)\otimes Df.$

As $D$ is injective, the above equation entails that there exists a scalar
$h(A)$ such that

$None$ $\Phi(A)C=CA+h(A)C.$

It is clear that $h$ is an additive functional on Alg${\mathcal{N}}$. Define
$\Psi:{\rm Alg}{\mathcal{N}}\rightarrow{\rm Alg}{\mathcal{M}}$ by
$\Psi(A)=\Phi(A)-h(A)I$ for all $A\in{\rm Alg}{\mathcal{N}}$. Then, $\Psi$ is
an additive bijection. By Eq.(5.13), for any $A,B\in{\rm Alg}{\mathcal{N}}$,
we have

$\Psi(AB)C=CAB=\Psi(A)CB=\Psi(A)\Psi(B)C.$

Since ran$C=Y$, we see that $\Psi(AB)=\Psi(A)\Psi(B)$ for all $A,B\in{\rm
Alg}\mathcal{N}$, that is, $\Psi$ is a ring isomorphism.

Assume that Lemma 5.1(2) holds. Then $\Phi(\mathbb{F}I+X\otimes
X_{-}^{\bot})=\mathbb{F}I+(0)_{+}\otimes Y^{*}$ and there exists a ring
automorphism $\tau:\mathbb{F}\rightarrow\mathbb{F}$ and a map $\gamma:X\otimes
X_{-}^{\bot}\rightarrow\mathbb{F}$ such that

$None$ $\Phi(x\otimes f)=\gamma(x,f)I+Df\otimes Cx$

holds for all $x\in X$ and $f\in X_{-}^{\bot}$, where $C:X\rightarrow Y^{*}$
and $D:X_{-}^{\bot}\rightarrow(0)_{+}$ are two $\tau$-linear bijective maps. A
similar argument to the above, one can check that there exists an additive
functional $h$ and a ring anti-isomorphism $\Psi$ such that
$\Phi(A)=-\Psi(A)+h(A)I$ for all $A\in{\rm Alg}\mathcal{N}$.

This completes the proof of Theorem 1.1 for the case that $(0)<X_{-}<X$ and
$X_{-}$ is not complemented. $\Box$

The proof of Theorem 1.1 for the case that $(0)<(0)_{+}<X$ and $(0)_{+}$ is
not complemented is similar and we omit it here.

Now, combining Sections 3-5, the proof of Theorem 1.1 is finished.

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