Maps completely preserving indefinite Jordan 1-†-zero product

Abstract

By characterizing the bijections preserving orthogonality of idempotents in both directions on the infinite dimensional complete indefinite inner product spaces, we obtain the concrete form of surjective maps completely preserving indefinite Jordan 1-†-zero product between †-standard operator algebras. Our results show that such maps are nonzero constant multiple of isomorphisms or conjugate isomorphisms.

Keywords

†-standard operator algebras / complete preserving problem / indefinite inner product spaces / indefinite Jordan 1-†-zero product

1 Introduction and main results

As is known, an indefinite inner product space can be induced by an invertible self-adjoint operator in a Hilbert space, and vice versa. Every invertible self-adjoint operator J on a Hilbert space, by definition, [x,y]=⟨Jx,y⟩, induces an indefinite inner product ⟨\cdot ,\cdot ⟩. Let H_i(i=1,2) be an indefinite inner product space, and T\in \mathcal{B}(H_1,H_2). Then the indefinite conjugate T^{†} of T is defined by the equality [Tx,y{]}_L=[x,T^{†}y{]}_J, where x\in H_1,y\in H_2,J\in \mathcal{B}\left(H_1\right),L\in \mathcal{B}\left(H_2\right). Obviously, T^{†}=J^{-1}{T}^{\ast }L. We call the standard operator algebra on an indefinite inner product space a †-standard operator algebra if it satisfies †-self-adjointness, that is, A\in \mathcal{A}\Rightarrow A^{†}\in \mathcal{A}; the operator U is called a generalized unitary operator if there exists a nonzero number λ such that U^{†}U=U{U}^{†}=\lambda I.

Indefinite inner product spaces are of great significance for the researches of pure mathematical theories and fields such as physics (see [1, 2, 12]). Therefore, it is meaningful to explore the preservation problems in indefinite inner product spaces. However, through a large number of literature searches, up to now, there are relatively few achievements in this regard. For example, in 2004, Cui et al. [4] characterized the additive bijection that preserved indefinite orthogonality in infinite dimensional complete indefinite inner product spaces. Subsequently, in 2006, the first two scholars characterized the linear bijection \mathrm{\Phi }:\mathcal{B}\left(H\right)\to \mathcal{B}\left(K\right) satisfying AB={BA}^{†}\iff \mathrm{\Phi }\left(A\right)\mathrm{\Phi }\left(B\right)=\mathrm{\Phi }\left(B\right)\mathrm{\Phi }{\left(A\right)}^{†} in [3]. In 2013, Dong and Gao [5] characterized the general maps that preserved the operator †-product idempotency on indefinite inner product spaces and obtained their specific forms. In view of this, this paper will use the idea of complete preservation problems to discuss the general surjective maps that completely preserve the indefinite Jordan 1-zero product on †-standard operator algebras and provide the specific forms of the maps.

\mathrm{\forall }A,B\in B\left(H\right), if AB+{BA}^{†}=0, then A and B satisfy indefinite Jordan 1-†-zero product; if the map \mathrm{\Phi } satisfies AB+{BA}^{†}=0\iff \mathrm{\Phi }\left(A\right)\mathrm{\Phi }\left(B\right)+\mathrm{\Phi }\left(B\right)\mathrm{\Phi }{\left(A\right)}^{†}=0, then it is said that \Phi preserves indefinite Jordan 1-†-zero product in both directions.

Complete preservation problem: Define the map {\mathrm{\Phi }}_n:\mathcal{A}\otimes M_n\left(\mathbb{F}\right)\to \mathcal{B}\otimes M_n\left(\mathbb{F}\right) as {\mathrm{\Phi }}_n\left(\right(A_{ij}{)}_{n\times n})=(\mathrm{\Phi }(A_{ij}{\left)\right)}_{n\times n},\mathrm{\forall }n\in \mathbb{N}. If {\mathrm{\Phi }}_n preserves the indefinite Jordan 1-†-zero product, then \Phi is called n-preserving the indefinite Jordan 1-†-zero product; if for each n\in \mathbb{N}, \mathrm{\Phi } is n-preserving the indefinite Jordan 1-†-zero product, then \mathrm{\Phi } is called completely preserving the indefinite Jordan 1-†-zero product.

The complete preservation problem imposes strong constraints on the algebraic structure of standard operator algebras and can more precisely reflect the essence of homomorphic map. However, the existing conclusions are only part of the whole, and further expansion is still needed (For complete preservation of idempotency and square zero elements, invertibility and spectrum, spectral functions, commutativity and Jordan zero-product, trace rank, zero divisors, see [6-11], respectively). Therefore, this paper discusses the indefinite Jordan 1-†-zero product within the framework of complete preservation and present the specific structural form of surjective map that completely preserves the indefinite Jordan 1-†-zero product between †-standard operator algebras.

Rank-one operator x\otimes f is idempotent if and only if [x,y{]}_J=⟨Jx,y⟩=⟨x,Jy⟩=1. I_1\left(H\right) denotes the set of all rank-one idempotent operators on the Hilbert space; I(H) denotes the set of idempotents that contains all rank-one idempotent operators. ∀P, Q∈I(H), if PQ=QP=0, then P, Q satisfy the orthogonality of idempotents. Let S\subseteq H. Then

S^{\perp }=\{f\in H:[x,f{]}_J=0,\mathrm{\forall }x\in S\},

and S^{\perp } is the orthocomplement of S in H.

To prove the main results, the following lemma needs to be presented. It is the main lemma in [7].

Lemma 1 [7]　 Let H, K be infinite dimensional Hilbert spaces, I\left(H\right)\subseteq \mathcal{B}\left(H\right),I\left(K\right)\subseteq \mathcal{B}\left(K\right) be the set of idempotents containing all rank-one idempotents. Let \mathrm{\Phi }:I\left(H\right)\to I\left(K\right) be a bijection. If \Phi preserves orthogonality in both directions, then either there exists an invertible bounded linear operator or (in the complex case) a conjugate linear operator A:H\to K, such that

\mathrm{\Phi }\left(P\right)=AP{A}^{-1},\mathrm{\forall }P\in I\left(H\right);

or there exists an invertible bounded linear operator or a conjugate linear operator such that

\mathrm{\Phi }\left(P\right)=A{P}^{\ast }{A}^{-1},\mathrm{\forall }P\in I\left(H\right).

In the second case, both H and K must be reflexive.

Theorem 1　 Let H, K be infinite dimensional complete indefinite inner product spaces over \mathbb{C}. \mathcal{A},\mathcal{B} are †-standard operator algebras on H and K respectively. Let \mathrm{\Phi }:\mathcal{A}\to \mathcal{B} be a surjective map. Then the following statements are equivalent:

(1) \mathrm{\Phi } completely preserves the indefinite Jordan 1-†-zero product in both directions;

(2) \mathrm{\Phi } preserves the indefinite Jordan 1-†-zero product in both directions;

(3) There exist a nonzero real number μ and a generalized unitary operator or a conjugate generalized unitary operator U:H→K, such that

\mathrm{\Phi }\left(T\right)=\mu UT{U}^{†},\mathrm{\forall }T\in \mathcal{A}.

Proof　A map satisfying \left(3\right) completely preserves the indefinite Jordan 1-†-zero product in both directions, so it is obvious \left(3\right)\Rightarrow \left(1\right)\Rightarrow \left(2\right). Thus, it is only required to prove \left(2\right)\Rightarrow \left(3\right).

Next, it is assumed that \mathrm{\Phi } 2-preserves the indefinite Jordan 1-†-zero product in both directions.

Assertion 1　 \mathrm{\Phi }\left(0\right)=0.

\mathrm{\forall }T\in \mathcal{A}, there exists

(\begin{array}{c}0\\ T\end{array}\begin{array}{c}0\\ 0\end{array})(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array})+(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array})(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array})(\begin{array}{c}0\\ T\end{array}{\begin{array}{c}0\\ 0\end{array})}^{†}=(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array}).

Applying {\mathrm{\Phi }}_2 into the above equality, there is

\begin{array}{rl}& (\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(T\right)\end{array}\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(0\right)\end{array})(\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(0\right)\end{array}\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(0\right)\end{array})\\ & +(\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(0\right)\end{array}\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(0\right)\end{array})(\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(T\right)\end{array}{\begin{array}{c}\mathrm{\Phi }\left(0\right)\\ \mathrm{\Phi }\left(0\right)\end{array})}^{†}=(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array}).\end{array}

Hence, it is obtained that

(1.1)\mathrm{\Phi }\left(T\right)\mathrm{\Phi }\left(0\right)+\mathrm{\Phi }\left(0\right)\mathrm{\Phi }(T{)}^{†}=0.

Due to the surjectivity of \mathrm{\Phi }, there exists certain T_0\in \mathcal{A} such that \mathrm{\Phi }\left(T_0\right)=I. Let T=T_0 in (1.1), then it is obtained 2\mathrm{\Phi }\left(0\right) = 0. Thus, \mathrm{\Phi }\left(0\right)=0.

Assertion 2　 \mathrm{\Phi } is injective and \mathrm{\Phi }\left(I\right)=\alpha I holds for certain nonzero real number \begin{array}{c}\alpha \end{array}.

\mathrm{\forall }T,S\in \mathcal{A}, if \mathrm{\Phi }(Т) = \mathrm{\Phi }(S), there is

\left(\begin{array}{cc}I& I\\ 0& 0\end{array}\right)\left(\begin{array}{cc}T& -T\\ -T& T\end{array}\right)+\left(\begin{array}{cc}T& -T\\ -T& T\end{array}\right){\left(\begin{array}{cc}I& I\\ 0& 0\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),

which is equivalent to

\begin{array}{rl}& \left(\begin{array}{cc}\mathrm{\Phi }\left(I\right)& \mathrm{\Phi }\left(I\right)\\ 0& 0\end{array}\right)\left(\begin{array}{cc}\mathrm{\Phi }\left(T\right)& \mathrm{\Phi }(-T)\\ \mathrm{\Phi }(-T)& \mathrm{\Phi }\left(T\right)\end{array}\right)\\ & +\left(\begin{array}{cc}\mathrm{\Phi }\left(T\right)& \mathrm{\Phi }(-T)\\ \mathrm{\Phi }(-T)& \mathrm{\Phi }\left(T\right)\end{array}\right)\left(\begin{array}{cc}\mathrm{\Phi }(I{)}^{†}& 0\\ \mathrm{\Phi }(I{)}^{†}& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).\end{array}

Substituting \mathrm{\Phi }(Т) with \mathrm{\Phi }(S), there is

\begin{array}{rl}& \left(\begin{array}{cc}\mathrm{\Phi }\left(I\right)& \mathrm{\Phi }\left(I\right)\\ 0& 0\end{array}\right)\left(\begin{array}{cc}\mathrm{\Phi }\left(S\right)& \mathrm{\Phi }(-T)\\ \mathrm{\Phi }(-T)& \mathrm{\Phi }\left(S\right)\end{array}\right)\\ & +\left(\begin{array}{cc}\mathrm{\Phi }\left(S\right)& \mathrm{\Phi }(-T)\\ \mathrm{\Phi }(-T)& \mathrm{\Phi }\left(S\right)\end{array}\right)\left(\begin{array}{cc}\mathrm{\Phi }(I{)}^{†}& 0\\ \mathrm{\Phi }(I{)}^{†}& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).\end{array}

Therefore,

\left(\begin{array}{cc}I& I\\ 0& 0\end{array}\right)\left(\begin{array}{cc}S& -T\\ -T& S\end{array}\right)+\left(\begin{array}{cc}S& -T\\ -T& S\end{array}\right){\left(\begin{array}{cc}I& I\\ 0& 0\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).

This implies that T=S. Therefore, \mathrm{\Phi } is injective, and thus \mathrm{\Phi } is a bijective from \mathcal{A} to \mathcal{B}.

Next, it is required to prove that \mathrm{\Phi }\left(I\right)=\alpha I holds for certain nonzero real number \alpha.

\mathrm{\forall }T\in \mathcal{A}, on the one hand,

\left(\begin{array}{cc}I& I\\ 0& 0\end{array}\right)\left(\begin{array}{cc}T& -T\\ -T& T\end{array}\right)+\left(\begin{array}{cc}T& -T\\ -T& T\end{array}\right){\left(\begin{array}{cc}I& I\\ 0& 0\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),

it can derive that

(1.2)\mathrm{\Phi }\left(I\right)\mathrm{\Phi }(-T)+\mathrm{\Phi }\left(I\right)\mathrm{\Phi }\left(T\right)=0.

On the other hand,

\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right)\left(\begin{array}{cc}0& -T\\ T& 0\end{array}\right)+\left(\begin{array}{cc}0& -T\\ T& 0\end{array}\right){\left(\begin{array}{cc}0& I\\ I& 0\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),

it can derive that

(1.3)\mathrm{\Phi }\left(I\right)\mathrm{\Phi }(-T)+\mathrm{\Phi }\left(T\right)\mathrm{\Phi }(I{)}^{†}=0.

According to (1.2) and (1.3), there is

(1.4)\mathrm{\Phi }\left(I\right)\mathrm{\Phi }\left(T\right)=\mathrm{\Phi }\left(T\right)\mathrm{\Phi }(I{)}^{†}.

According to the surjectivity of \mathrm{\Phi }, there exists T_1\in \mathcal{A} such that {\mathrm{\Phi }(T}_1)=I. In (1.4), let T=T_1, then \mathrm{\Phi }(I)=\mathrm{\Phi }(I{)}^{†}. Thus, (1.4) turns into \mathrm{\Phi }\left(I\right)\mathrm{\Phi }\left(T\right)=\mathrm{\Phi }\left(T\right)\mathrm{\Phi }\left(I\right). This shows that \mathrm{\Phi }\left(I\right) commutes with every operator in \mathcal{B}. Therefore, there exists a nonzero number α, such that \mathrm{\Phi }(I)=\alpha I. Also, since \mathrm{\Phi }(I)\equiv \mathrm{\Phi }(I{)}^{†}, that is \alpha I=\alpha I^{†}, \alpha is a nonzero real number.

If necessary, replace \mathrm{\Phi } with {\alpha }^{-1}\mathrm{\Phi }. Obviously, {\alpha }^{-1}\mathrm{\Phi } is still a 2-preserving indefinite Jordan 1-†-zero-product in both directions, so it can be assumed that \mathrm{\Phi }(I)=I later.

Assertion 3　 \mathrm{\Phi }(-T)=-\mathrm{\Phi }(T), \mathrm{\Phi }\left(T^2\right)=\mathrm{\Phi }(T{)}^2 and \mathrm{\Phi } preserves idempotents in both directions.

Before proving that \mathrm{\Phi } preserves idempotents in both directions, it is required to first prove \mathrm{\Phi }(-T)=-\mathrm{\Phi }\left(T\right), \mathrm{\Phi }\left(T^2\right)={\mathrm{\Phi }\left(T\right)}^2.

There is \mathrm{\Phi }(-T)=-\mathrm{\Phi }\left(T\right) according to (1.2) and\mathrm{\Phi }\left(I\right)=I. On one hand, \mathrm{\forall }T\in \mathcal{A},

(\begin{array}{c}I\\ 0\end{array}\begin{array}{c}-T\\ 0\end{array})\left(\begin{array}{c}{T}^2{T}^{†}\\ {TT}^{†}\end{array}\begin{array}{c}{T}^2\\ T\end{array}\right)+\left(\begin{array}{c}{T}^2{T}^{†}\\ {TT}^{†}\end{array}\begin{array}{c}{T}^2\\ T\end{array}\right)(\begin{array}{c}I\\ 0\end{array}{\begin{array}{c}-T\\ 0\end{array})}^{†}=(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array}).

Applying {\mathrm{\Phi }}_2 into the above equality, there is

\begin{array}{rl}& \left(\begin{array}{cc}I& -\mathrm{\Phi }\left(T\right)\\ 0& 0\end{array}\right)\left(\begin{array}{cc}\mathrm{\Phi }\left(T^2{T}^{†}\right)& \mathrm{\Phi }\left(T^2\right)\\ \mathrm{\Phi }\left(T{T}^{†}\right)& \mathrm{\Phi }\left(T\right)\end{array}\right)\\ & +\left(\begin{array}{cc}\mathrm{\Phi }\left(T^2{T}^{†}\right)& \mathrm{\Phi }\left(T^2\right)\\ \mathrm{\Phi }\left(T{T}^{†}\right)& \mathrm{\Phi }\left(T\right)\end{array}\right){\left(\begin{array}{cc}I& -\mathrm{\Phi }\left(T\right)\\ 0& 0\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ & \Rightarrow \mathrm{\Phi }\left(T^2\right)-\mathrm{\Phi }(T{)}^2=0.\end{array}

That is \mathrm{\Phi }\left(T^2\right)={\mathrm{\Phi }\left(T\right)}^2. Hence,

\begin{array}{rl}{T}^2=T& \iff \left(\begin{array}{cc}I& I\\ I& I\end{array}\right)\left(\begin{array}{cc}T& -T\\ -T^2& T\end{array}\right)+\left(\begin{array}{cc}T& -T\\ -T^2& T\end{array}\right){\left(\begin{array}{cc}I& I\\ I& I\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ & \Rightarrow \left(\begin{array}{cc}I& I\\ I& I\end{array}\right)\left(\begin{array}{cc}\mathrm{\Phi }\left(T\right)& \mathrm{\Phi }(-T)\\ \mathrm{\Phi }(-T^2)& \mathrm{\Phi }\left(T\right)\end{array}\right)\\ & +\left(\begin{array}{cc}\mathrm{\Phi }\left(T\right)& \mathrm{\Phi }(-T)\\ \mathrm{\Phi }(-T^2)& \mathrm{\Phi }\left(T\right)\end{array}\right){\left(\begin{array}{cc}I& I\\ I& I\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ & \Rightarrow \left(\begin{array}{cc}\mathrm{\Phi }\left(T\right)-\mathrm{\Phi }\left(T^2\right)& 0\\ 2\mathrm{\Phi }\left(T\right)-2\mathrm{\Phi }\left(T^2\right)& \mathrm{\Phi }\left(T\right)-\mathrm{\Phi }\left(T^2\right)\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\\ & \Rightarrow \mathrm{\Phi }\left(T\right)=\mathrm{\Phi }\left(T^2\right)=\mathrm{\Phi }(T{)}^2,\end{array}

which implies that \mathrm{\Phi } preserves idempotents, that is T^2=T\Rightarrow \mathrm{\Phi }(T{)}^2=\mathrm{\Phi }(T).

Similarly, we can obtain \mathrm{\Phi }(T{)}^2=\mathrm{\Phi }(T)\Rightarrow T^2=T. Therefore, \mathrm{\Phi } preserves idempotents in both directions.

Assertion 4　There exists a generalized unitary operator or conjugate generalized unitary operator U:H→K, such that \mathrm{\Phi }\left(P\right)=c^{-1}UP{U}^{†},\mathrm{\forall }P\in I_1\left(H\right).

Before that, it is required to first prove \mathrm{\Phi }\left(T^{†}\right)=\mathrm{\Phi }(T{)}^{†},\mathrm{\forall }T\in \mathcal{A}.

\begin{array}{rl}& \left(\begin{array}{cc}T& -I\\ -I& -T^{†}\end{array}\right)\left(\begin{array}{cc}I& T\\ T^{†}& -I\end{array}\right)+\left(\begin{array}{cc}I& T\\ T^{†}& -I\end{array}\right){\left(\begin{array}{cc}T& -I\\ -I& -T^{†}\end{array}\right)}^{†}\\ & =\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\Rightarrow \mathrm{\Phi }\left(T^{†}\right)=\mathrm{\Phi }(T{)}^{†}.\end{array}

\begin{array}{rl}& \mathrm{\forall }P,Q\in I\left(H\right),\\ & PQ=QP=0\iff \left(\begin{array}{cc}{P}^{†}& P\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ Q& 0\end{array}\right)+\left(\begin{array}{cc}0& 0\\ Q& 0\end{array}\right){\left(\begin{array}{cc}{P}^{†}& P\\ 0& 0\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).\end{array}

Since {\mathrm{\Phi }}_2 preserves the indefinite Jordan 1-†-zero product in both directions, there is \mathrm{\Phi }\left(P\right)\mathrm{\Phi }\left(Q\right)={\mathrm{\Phi }\left(Q\right)\mathrm{\Phi }\left(P^{†}\right)}^{†}=0.

Based on \mathrm{\Phi }\left(T^{†}\right)=\mathrm{\Phi }(T{)}^{†}, there is{\mathrm{\Phi }\left(Q\right)\mathrm{\Phi }\left(P^{†}\right)}^{†}=\mathrm{\Phi }\left(Q\right)\mathrm{\Phi }\left(P\right). Thus,

PQ=QP=0\iff \mathrm{\Phi }\left(P\right)\mathrm{\Phi }\left(Q\right)=\mathrm{\Phi }\left(Q\right)\mathrm{\Phi }\left(P\right)=0.

In other words, \mathrm{\Phi } preserves the orthogonality of idempotents in both directions on I(H).

From the lemma, it is known that \mathrm{\Phi } takes the form of \mathrm{\Phi }\left(P\right)={APA}^{-1}, \mathrm{\forall }P\in I\left(H\right) or \mathrm{\Phi }\left(P\right)={A{P}^{\ast }A}^{-1}, \mathrm{\forall }P\in I(H).

Next, it is required to prove that there exists an operator U: H→K (where U is a generalized unitary operator or a conjugate generalized unitary operator), such that \mathrm{\Phi }\left(P\right)=c^{-1}UP{U}^{†},\mathrm{\forall }P\in I_1\left(H\right) or \mathrm{\Phi }\left(P\right)=c^{-1}U{P}^{†}{U}^{†}, \mathrm{\forall }P\in I_1\left(H\right).

Locate the †-Hermitian matrix x\otimes \frac{Jx}{\parallel Jx{\parallel }^2}(\mathrm{\forall }x\in H). \mathrm{\Phi }\left(T^{†}\right)=\mathrm{\Phi }(T{)}^{†}, so

\mathrm{\Phi }\left(x\otimes \frac{Jx}{{\Vert Jx\Vert }^2}\right)=\mathrm{\Phi }{\left(x\otimes \frac{Jx}{{\Vert Jx\Vert }^2}\right)}^{†},\mathrm{\forall }x\in H.

On the other hand,

\left(x\otimes \frac{Jx}{{\Vert Jx\Vert }^2}\right){}_{}{}^2=\frac{⟨Jx,Jx⟩}{{\Vert Jx\Vert }^2}\left(x\otimes \frac{Jx}{{\Vert Jx\Vert }^2}\right)=x\otimes \frac{Jx}{{\Vert Jx\Vert }^2}.

So x\otimes \frac{Jx}{{\Vert Jx\Vert }^2} is a rank-one idempotent. Therefore, if \mathrm{\Phi }\left(P\right)=AP{A}^{-1}, \mathrm{\forall }P\in I\left(H\right), there will be

A\cdot \left(x\otimes \frac{Jx}{{\Vert Jx\Vert }^2}\right)\cdot A^{-1}=(A^{-1}{)}^{†}\cdot P^{†}\cdot A^{†}.

Left-hand side=Ax\otimes \frac{(A^{-1}{)}^{\ast }Jx}{{\Vert Jx\Vert }^2}=Ax\otimes \frac{(A^{\ast }{)}^{-1}Jx}{{\Vert Jx\Vert }^2}；

Right-hand side=(A^{-1}{)}^{†}\cdot P^{†}\cdot A^{†}=L^{-1}(A^{-1}{)}^{\ast }L{L}^{-1}{P}^{\ast }L{L}^{-1}{A}^{\ast }L=L^{-1}(A^{-1}{)}^{\ast }(x\otimes \frac{Jx}{\parallel Jx{\parallel }^2}{)}^{\ast }{A}^{\ast }L=L^{-1}\left(A^{\ast }{)}^{-1}\right(\frac{Jx}{\parallel Jx{\parallel }^2}\otimes x)A^{\ast }L=\frac{L^{-1}(A^{\ast }{)}^{-1}Jx}{\parallel Jx{\parallel }^2}\otimes LAx.

Hence,

Ax\otimes (A^{\ast }{)}^{-1}Jx=L^{-1}(A^{\ast }{)}^{-1}Jx\otimes LAx,\mathrm{\forall }x\in H.

Based on the above equation, it is known that {\left(A^{\ast }\right)}^{-1}Jx and LAx are linearly dependent, that is, LA=c{\left(A^{\ast }\right)}^{-1}J. Let A=U. Then LU=c{\left(U^{\ast }\right)}^{-1}J. Combining U_{†}=J^{-1}{U}^{\ast }L, it is further known that U^{†}U=U{U}^{†}=cI holds for any nonzero real number c. Here,

\mathrm{\Phi }\left(P\right)=AP{A}^{-1}=UP\left(c^{-1}{U}^{†}\right)=c^{-1}UP{U}^{†},\mathrm{\forall }P\in I_1\left(H\right).

Similarly, if \mathrm{\Phi }\left(P\right)=A{P}^{\ast }{A}^{-1},\mathrm{\forall }P\in I\left(H\right), there is

AJx\otimes (A^{\ast }{)}^{-1}x=L^{-1}(A^{\ast }{)}^{-1}x\otimes LAJx,\mathrm{\forall }x\in H.

Then AJ=c{L}^{-1}(A^{\ast }{)}^{-1}. Let U=AJ, and there is U^{†}U=U{U}^{†}=cI, c\in \mathbb{R} in the same way. Here,

\mathrm{\Phi }\left(P\right)=U{J}^{-1}{P}^{\ast }J{U}^{-1}=U{P}^{†}{U}^{-1}=U{P}^{†}\left(c^{-1}{U}^{†}\right)=c^{-1}U{P}^{†}{U}^{†},\mathrm{\forall }P\in I_1\left(H\right).

Finally, it is proved that it is impossible to have \mathrm{\Phi }(P)=c^{-1}U{P}^{†}{U}^{†}, \mathrm{\forall }P\in I_1(H). Instead, assume that \mathrm{\Phi }\left(P\right)=c^{-1}U{P}^{†}{U}^{†} holds. \mathrm{\forall }x,y\in H (x,y are linearly independent), there exists a linear functional f_1,f_2\in H satisfying

⟨x,J{f}_1⟩=1,⟨x,J{f}_2⟩=0,⟨y,J{f}_1⟩=0,⟨y,J{f}_2⟩=1.

Then,

\begin{array}{rl}& \left(\begin{array}{cc}0& 0\\ (y\otimes f_2)& (y\otimes f_2{)}^{†}\end{array}\right)\left(\begin{array}{cc}0& x\otimes f_1\\ 0& 0\end{array}\right)\\ & +\left(\begin{array}{cc}0& x\otimes f_1\\ 0& 0\end{array}\right){\left(\begin{array}{cc}0& 0\\ (y\otimes f_2)& (y\otimes f_2{)}^{†}\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).\end{array}

Applying {\mathrm{\Phi }}_2 into the above equation, we can get

\begin{array}{rl}& \left(\begin{array}{c}0\\ \mathrm{\Phi }\left(y\otimes f_2\right)\end{array}\begin{array}{c}0\\ (\mathrm{\Phi }\left(y\otimes f_2\right){)}^{†}\end{array}\right)\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}\mathrm{\Phi }\left(x\otimes f_1\right)\\ 0\end{array}\right)\\ & +\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}\mathrm{\Phi }\left(x\otimes f_1\right)\\ 0\end{array}\right)(\begin{array}{c}0\\ 0\end{array}{\begin{array}{c}(\mathrm{\Phi }\left(y\otimes f_2\right){)}^{†}\\ \mathrm{\Phi }\left(y\otimes f_2\right)\end{array})}^{†}=\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}\mathrm{\Phi }\left(x\otimes f_1\right)\mathrm{\Phi }\left(y\otimes f_2\right)\\ \mathrm{\Phi }\left(y\otimes f_2\right)\mathrm{\Phi }\left(x\otimes f_1\right)\end{array}\right)\\ & =\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}{c}^{-1}{UJ}^{-1}\left(f_1\otimes x\right){{JU}^{†}c}^{-1}{UJ}^{-1}\left(f_2\otimes y\right){JU}^{†}\\ c^{-1}{UJ}^{-1}\left(f_2\otimes y\right){{JU}^{†}c}^{-1}{UJ}^{-1}\left(f_1\otimes x\right){JU}^{†}\end{array}\right)\\ & =\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}{c}^{-1}{UJ}^{-1}\left(f_1\otimes x\right)\left(f_2\otimes y\right){JU}^{†}\\ c^{-1}{UJ}^{-1}\left(f_2\otimes y\right)\left(f_1\otimes x\right){JU}^{†}\end{array}\right)\end{array}

is not equal to 0, which contradicts with preservation of the indefinite Jordan 1-†-zero product by {\mathrm{\Phi }}_2. Therefore, the assumption doesn’t hold. That is \mathrm{\Phi }(P)=c^{-1}UP{U}^{†}, \mathrm{\forall }P\in I_1(H), so Assertion 4 holds.

Let \mathrm{\Psi }:\mathcal{A}\to U^{†}\mathcal{A}U be a bijection defined by \mathrm{\Psi }(\cdot )=c^{-1}{U}^{†}\mathrm{\Phi }(\cdot )U. Then \mathrm{\Psi }\left(P\right)=c^{-1}{U}^{†}\mathrm{\Phi }\left(P\right)U=P. Next, it is verified that {\mathrm{\Psi }}_2 also preserves the indefinite Jordan 1-†-zero product in both directions. Take T_i\in \mathcal{A}(i=1,2,\dots ,8), such that

\left(\begin{array}{cc}{T}_1& T_2\\ T_3& T_4\end{array}\right),\left(\begin{array}{cc}{T}_5& T_6\\ T_7& T_8\end{array}\right)\in \mathcal{A}\otimes M_2\subseteq \mathcal{B}\left(H^2\right),

and satisfies

\left(\begin{array}{c}{T}_1\\ T_3\end{array}\begin{array}{c}{T}_2\\ T_4\end{array}\right)\left(\begin{array}{c}{T}_5\\ T_7\end{array}\begin{array}{c}{T}_6\\ T_8\end{array}\right)+\left(\begin{array}{c}{T}_5\\ T_7\end{array}\begin{array}{c}{T}_6\\ T_8\end{array}\right)(\begin{array}{c}\begin{array}{c}{T}_1\\ T_3\end{array}\end{array}{\begin{array}{c}{T}_2\\ T_4\end{array})}^{†}=(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array}).

Thus,

\begin{array}{rl}& {\mathrm{\Psi }}_2\left(\begin{array}{c}{T}_1\\ T_3\end{array}\begin{array}{c}{T}_2\\ T_4\end{array}\right){\mathrm{\Psi }}_2\left(\begin{array}{c}{T}_5\\ T_7\end{array}\begin{array}{c}{T}_6\\ T_8\end{array}\right)+{\mathrm{\Psi }}_2\left(\begin{array}{c}{T}_5\\ T_7\end{array}\begin{array}{c}{T}_6\\ T_8\end{array}\right){\mathrm{\Psi }}_2(\begin{array}{c}\begin{array}{c}{T}_1\\ T_3\end{array}\end{array}{\begin{array}{c}{T}_2\\ T_4\end{array})}^{†}\\ & =({c^{-1})}^2\left(\begin{array}{c}{U}^{†}\\ 0\end{array}\begin{array}{c}0\\ U^{†}\end{array}\right)[(\begin{array}{c}\mathrm{\Phi }\left(T_1\right)\\ \mathrm{\Phi }\left(T_3\right)\end{array}\begin{array}{c}\mathrm{\Phi }\left(T_2\right)\\ \mathrm{\Phi }\left(T_4\right)\end{array})(\begin{array}{c}\mathrm{\Phi }\left(T_5\right)\\ \mathrm{\Phi }\left(T_7\right)\end{array}\begin{array}{c}\mathrm{\Phi }\left(T_6\right)\\ \mathrm{\Phi }\left(T_8\right)\end{array})\\ & +(\begin{array}{c}\mathrm{\Phi }\left(T_5\right)\\ \mathrm{\Phi }\left(T_7\right)\end{array}\begin{array}{c}\mathrm{\Phi }\left(T_6\right)\\ \mathrm{\Phi }\left(T_8\right)\end{array})(\begin{array}{c}\mathrm{\Phi }(T_1{)}^{†}\\ \mathrm{\Phi }(T_2{)}^{†}\end{array}\begin{array}{c}\mathrm{\Phi }(T_3{)}^{†}\\ \mathrm{\Phi }(T_4{)}^{†}\end{array})]\left(\begin{array}{c}U\\ 0\end{array}\begin{array}{c}0\\ U\end{array}\right)\\ & =({c^{-1})}^2\left(\begin{array}{c}{U}^{†}\\ 0\end{array}\begin{array}{c}0\\ U^{†}\end{array}\right)[{\mathrm{\Phi }}_2(\begin{array}{c}\left(T_1\right)\\ \left(T_3\right)\end{array}\begin{array}{c}\left(T_2\right)\\ \left(T_4\right)\end{array}){\mathrm{\Phi }}_2(\begin{array}{c}\left(T_5\right)\\ \left(T_7\right)\end{array}\begin{array}{c}\left(T_6\right)\\ \left(T_8\right)\end{array})\\ & +{\mathrm{\Phi }}_2\left(\begin{array}{c}{T}_5\\ T_7\end{array}\begin{array}{c}{T}_6\\ T_8\end{array}\right){\mathrm{\Phi }}_2(\begin{array}{c}\begin{array}{c}{T}_1\\ T_3\end{array}\end{array}{\begin{array}{c}{T}_2\\ T_4\end{array})}^{†}]\left(\begin{array}{c}U\\ 0\end{array}\begin{array}{c}0\\ U\end{array}\right)\\ & =0.\end{array}

So, {\mathrm{\Psi }}_2 also preserves the indefinite Jordan 1-†--zero product in both directions. Therefore, without loss of generality, it is assumed that \mathrm{\Phi }(Р)=P, where P∈I₁(H).

Assertion 5　 \mathrm{\Phi }(x\otimes f)={\lambda }_{x\otimes f}(x\otimes f) holds for any rank-one operator x\otimes f, where 0\ne {\lambda }_{x\otimes f}\in \mathbb{C}.

For any rank-one operator x\otimes f, there is x,y,h\in H satisfying ⟨y,{Jh}_1⟩. Then y\in \mathrm{k}\mathrm{e}\mathrm{r}f,h\in {\left\{x\right\}}^{\perp }\iff \left(\begin{array}{c}0\\ (y\otimes h)\end{array}\begin{array}{c}0\\ {\left(y\otimes h\right)}^{†}\end{array}\right)\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}x\otimes f\\ 0\end{array}\right)+\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}x\otimes f\\ 0\end{array}\right)(\begin{array}{c}\begin{array}{c}0\\ (y\otimes h)\end{array}\end{array}{\begin{array}{c}0\\ {\left(y\otimes h\right)}^{†}\end{array})}^{†}=\left(\begin{array}{c}0\\ 0\end{array}\begin{array}{c}0\\ 0\end{array}\right).

Applying {\mathrm{\Phi }}_2 into the above equation, we can get y\in \mathrm{k}\mathrm{e}\mathrm{r}\left(\mathrm{\Phi }\right(x\otimes f\left)\right), h\in \left(\mathrm{r}\mathrm{a}\mathrm{n}\right(\mathrm{\Phi }(x\otimes f)){)}^{\perp }. That is

y\in \mathrm{k}\mathrm{e}\mathrm{r}f\iff y\in \mathrm{k}\mathrm{e}\mathrm{r}\left(\mathrm{\Phi }\right(x\otimes f\left)\right),h\in {\left\{x\right\}}^{\perp }\iff h\in \left(\mathrm{r}\mathrm{a}\mathrm{n}\right(\mathrm{\Phi }(x\otimes f)){)}^{\perp }.

Hence, for certain {\lambda }_{x\otimes f}\in \mathbb{C}\mathrm{\setminus }\left\{0\right\},\mathrm{\Phi }(x\otimes f)={\lambda }_{x\otimes f}(x\otimes f) holds.

Assertion 6　There exists a map h:\mathcal{A}\to \mathbb{C}\setminus \left\{0\right\}, such that\mathrm{\Phi }\left(T\right)=h\left(T\right)T holds for every T\in \mathcal{A}.

\mathrm{\forall }T,S\in \mathcal{A}, there exists

\left(\begin{array}{cc}0& 0\\ TS& T\end{array}\right)\left(\begin{array}{cc}T& -T{S}^{†}\\ -ST& ST{S}^{†}\end{array}\right)+\left(\begin{array}{cc}T& -T{S}^{†}\\ -ST& ST{S}^{†}\end{array}\right){\left(\begin{array}{cc}0& 0\\ TS& T\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).

Applying {\mathrm{\Phi }}_2 into the above equation, we can get

(1.5)\mathrm{\Phi }\left(TS\right)\mathrm{\Phi }\left(T\right)=\mathrm{\Phi }\left(T\right)\mathrm{\Phi }\left(ST\right).

For any rank-one operator, let S=x\otimes f in (1.5). Then

\mathrm{\Phi }\left(TS\right)\mathrm{\Phi }\left(T\right)={\lambda }_{TS}\left(TS\right)\mathrm{\Phi }\left(T\right)={\lambda }_{TS}(x\otimes f)\mathrm{\Phi }\left(T\right)={\lambda }_{TS}x\otimes \mathrm{\Phi }(T{)}^{†}f;

\mathrm{\Phi }\left(T\right)\mathrm{\Phi }\left(ST\right)=\mathrm{\Phi }\left(T\right){\lambda }_{TS}\left(ST\right)={\lambda }_{TS}\mathrm{\Phi }\left(T\right)(x\otimes f)T={\lambda }_{TS}\mathrm{\Phi }\left(T\right)x\otimes T^{†}f.

Therefore,

{\lambda }_{TS}Tx\otimes \mathrm{\Phi }(T{)}^{†}f={\lambda }_{TS}\mathrm{\Phi }(T)x\otimes T^{†}f.

Obviously, \mathrm{\forall }x\in H, \mathrm{\Phi }\left(T\right)x and Txare linearly dependent. So there exists certain nonzero number {\lambda }_T\in \mathbb{C}, such that \mathrm{\Phi }\left(T\right)={\lambda }_{T}T. For each T\in \mathcal{A}, we define a functionalh(\cdot ):\mathcal{A}\to \mathbb{C}\setminus \left\{0\right\} in the following way: if T\ne 0,h\left(T\right)={\lambda }_T; h\left(0\right)=1. Then there is \mathrm{\Phi }\left(T\right)=h\left(T\right)T,\mathrm{\forall }T\in \mathcal{A}.

Assertion 7　 \mathrm{\forall }T\in \mathcal{A}, there is always h\left(T\right)=1, so \mathrm{\Phi }\left(T\right)=T.

On one hand, since \mathrm{\Phi }2 is injective and \mathrm{\Phi }\left(I\right)=I,h\left(I\right)=1=h\left(0\right); on the other hand, \mathrm{\Phi }(P)=P implies h\left(P\right)=1.

Next, it is required to prove that for any rank-one operator x\otimes f, there is h(x\otimes f)= 1.

\mathrm{\forall }T,S\in A, there exists

\left(\begin{array}{cc}0& 0\\ TS& -T\end{array}\right)\left(\begin{array}{cc}TS& TS{S}^{†}\\ STS& STS{S}^{†}\end{array}\right)+\left(\begin{array}{cc}TS& TS{S}^{†}\\ STS& STS{S}^{†}\end{array}\right){\left(\begin{array}{cc}0& 0\\ TS& -T\end{array}\right)}^{†}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),

and

(1.6)\left(h\right(TS){)}^2=h(T\left)h\right(STS).

\mathrm{\forall }x,f\in H, there must exist y\in H, such that y and x are linearly independent and⟨y,Jf⟩=1, so there exists a nonzero element g_1,g_2\in H, such that

⟨x,J{g}_1⟩=1,⟨y,J{g}_1⟩=0,⟨x,J{g}_2⟩=0,⟨y,J{g}_2⟩=1.

Let g=g_1+g_2, and there are ⟨x,J\mathrm{g}⟩=1, ⟨y,J\mathrm{g}⟩=1. In (1.6), let T=x\otimes f, S=y\otimes g. Hence,

\left(h\right((x\otimes f)(y\otimes g)){)}^2=h(x\otimes f\left)h\right((y\otimes g)(x\otimes f)(y\otimes g)).

Thus,

\left(h\right(⟨y,Jf⟩x\otimes g){)}^2=h(x\otimes f\left)h\right(⟨x,Jg⟩⟨y,Jf⟩y\otimes g)\Rightarrow (h(x\otimes g){)}^2=h(x\otimes f)h(y\otimes g).

h(x\otimes g)=h(y\otimes g)=1, so h(x\otimes f)=1 holds for any rank-one operator x\otimes f.

Next, it is required to prove that \mathrm{\forall }T\in \mathcal{A}\mathrm{\setminus }\left\{0\right\}, there exist x,f\in H such that ⟨Tx,Jf⟩\ne 0. Using (1.6) again, since T(x\otimes f)=Tx\otimes f and h(⟨Tx,Jf⟩x\otimes f) are still rank-one operators, it is obtained that h\left(T\right(x\otimes f\left)\right)=h(⟨Tx,Jf⟩x\otimes f)=1. Thus,

1=\left(h\right(T(x\otimes f)){)}^2=h(T\left)h\right((x\otimes f)T(x\otimes f))=h(T\left)h\right(⟨Tx,Jf⟩x\otimes f)=h(T).

This completes the proof of \left(2\right)\Rightarrow \left(3\right).

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