Abstract. Let

A

be a von Neumann

J

-algebra of type (B) acting in an indeﬁnite metric space. The aim of the paper is to study

J

-projections from

A

In ([1], Chapter XII) the problem of construction of probability theory for quantum mechanics is posed. An analog of boolean algebra of events is quantum logic. An important interpretation of a quantum logic is the set

B {(H)}^{p r}

of all orthogonal projections on a Hilbert space

H

. In construction of measure theory on logics of projections it is important to know the properties and the structure of projections. The problem to construct a quantum ﬁeld theory sometimes leads to an indeﬁnite metric space ([3]). In indeﬁnite case, the set

P

of all

J

-orthogonal projections is an analog of the logics

B {(H)}^{p r}

. In the present paper we study

J

-projections from von Neumann

J

-algebras of type

(B)

. The main results of this paper were announced in [6].

1. Introduction

Let

H

be a complex Hilbert space with an inner product

(., .)

and let

B (H)

be the set of all bounded linear operators in

H

. Fix a self-adjoint symmetry operator

J

(

J = J^{*} = J^{- 1}

J \neq \pm I

). The form

[x, y] : = (J x, y)

is said to be an indeﬁnite metric, and

H

with

[., .]

is said to be the Krein space (

J

-space) (see [2]). Put

P^{+} : = \frac{1}{2} (I + J)

and

P^{-} : = I - P^{+}

. Put also

Γ^{+} \equiv {f \in H : [f, f] = 1}

and

Γ^{-} \equiv {f \in H : [f, f] = - 1}

. It is clear that

J Γ^{\pm} = Γ^{\pm}

. The set

Γ : = Γ^{+} \cup Γ^{-}

is an indeﬁnite analog of the unit sphere

S

H

. Let

A \in B (H)

. The operator

A^{#} : = J A^{*} J

is said to be

J

-adjoint of

A

. Note that

[A x, y] = [x, B y]

for all

x, y \in H

and some

B \in B (H)

if and only if

B = A^{#}

. An operator

A

is said to be

J

-self-adjoint (

J

-positive,

J

-negative) if

[A x, y] = [x, A y]

(

[A x, x] \geq 0

[A x, x] \leq 0

) for all

x, y \in H

. Note that

A

J

-self-adjoint (

J

-positive,

J

-negative) if and only if

J A

is self-adjoint (positive, negative, respectively).

An operator

p \in B (H)

is said to be a projection if

p^{2} = p

. Any one-dimensional projection has the form

(., x) y

where

x

y \in H

with

(y, x) = 1

. Let

P : = {p \in B (H) : p^{2} = p = p^{#}}

. Thus

P

is the set of all

J

-orthogonal (

J

-self-adjoint) projections in

B (H)

. Any

p \in P

is said to be the

J

-projection. Let

P^{+}

(

P^{-}

) be the set of all

J

-positive (

J

-negative, respectively)

J

-projections. It is clear that

P^{+} \cap P^{-} = {0}

. Any one-dimensional

J

-projection has the form

p_{f} : = [f, f] [., f] f

f \in Γ

. A vector

f \in Γ^{+}

(

f \in Γ^{-}

) if and only if

p_{f}

p_{J f} \in P^{+}

(

\in P^{-}

, respectively).

2. Projections of type

(B)

A von Neumann algebra

A

H

is said to be a von Neumann

J

-algebra if

A \in A

implies

A^{#} \in A

. Following [4], a commutative von Neumann

J

-algebra

Z

is said to be a type

(B)

algebra if

Z

contains a pair

P

Q \in B {(H)}^{p r}

such that

P + Q = I

Q^{#} = P

. A von Neumann

J

-algebra

A

is said to be of type

(B)

if its center (=

A \cap A^{'}

) is of type

(B)

Throughout the rest of the paper,

A

is a von Neumann

J

-algebra of type

(B)

, a pair

P, Q

of orthogonal projections in

A \cap A^{'}

satisfying

P + Q = I

Q^{#} = P

is assumed to be ﬁxed. Set

ℬ : = P B (H) P + Q B (H) Q

. It is clear that

ℬ

is a von Neumann

J

-algebra of type

(B)

and

A \subseteq ℬ

. Put

P^{A} = A \cap P

P^{ℬ} = ℬ \cap P

, and

J : = P - Q

. Clearly,

Let

p

q

be projections. Put

p \leq q

p q = q p = p

. Note that if

p

q \in P

then

p q = p

if and only if

q p = p

. With respect to the standard relations, namely, the ordering

\leq

, the orthogonal relation

p ⊥ q

if and only if

p q = q p = 0

, and the orthocomplementation

p \mapsto p^{⊥} : = I - p

, the set

P

is a quantum logic and

P^{A}

is a sublogic. Any

J

-projection from

P^{ℬ}

is said to be a

J

-projection of type

(B)

Proof. Let $R \in P$ . Then ${(σ (R))}^{2} = σ (R)$ and $J {(σ (R))}^{*} J = J J R^{*} J J = σ (R) .$ Thus $σ (R) \in P$ . It is clear that $σ (σ (R)) = R$ . Hence $σ (P) = P$ . Furthermore, $R_{1} \leq R_{2}$ implies $σ (R_{1}) \leq σ (R_{2})$ , $R_{1} ⊥ R_{2}$ implies $σ (R_{1}) ⊥ σ (R_{2})$ , and $σ (R^{⊥}) = {(σ (R))}^{⊥}$ .

Now, let $R \in P^{+}$ . Then $J (J R J) = - J (J R) J$ is a negative operator. Hence $σ (R)$ is a $J$ -negative $J$ -projection. By (1), $σ (P^{+}) = \frac{1}{2} J (I + J) J = \frac{1}{2} (I + J J J) = \frac{1}{2} (I - J) = P^{-}$ . □

We will see below (Theorem 1) that points of

P

being invariant under

σ

form the logic

P^{ℬ}

Proof. Let $A = A^{#}$ . Then $A = A P + A Q = P A P + (J P) (J A J) (P J) = A_{P} + J A_{P}^{*} J .$ Conversely, let $A = A_{P} + J A_{P}^{*} J$ . Then $A^{#} = J A^{*} J = J (A_{P}^{*} + J A_{P} J) J = J A_{P}^{*} J + A_{P} = A .$ □

Proof. Let $R \in P^{A}$ and $q : = P R$ . By Proposition 2, $R = P R P + J {(P R P)}^{*} J = R_{P} + J R_{P}^{*} J$ . Since ${(R_{P})}^{2} + J {(R_{P}^{*})}^{2} J = R^{2} = R = R_{P} + J R_{P}^{*} J$ , ${(R_{P})}^{2} = R_{P}$ . Hence $R_{P} \in ℒ_{P}^{A}$ .

Conversely, let $q \in ℒ_{P}^{A}$ . Then $J q^{*} J = q^{#} \in A$ . Hence $q + J q^{*} J \in A^{J s}$ . In addition, ${(q + J q^{*} J)}^{2} = q + J q^{*} J$ . By Proposition 2, $q + J q^{*} J \in P^{A}$ . □

dim H = \infty

then the logic

P^{ℬ}

is not a

σ

-logic (cf. [5, Proposition 2]).

To prove this fact, we will construct a sequence of mutually orthogonal

J

-projections

{R_{n}}_{1}^{\infty} \subset P^{ℬ}

such that the supremum

\sum R_{n}

does not exist in

P^{ℬ}

. Let

{φ_{n}}_{1}^{\infty}

be an orthonormal family in

P H

. (Note that

{J φ_{n}}_{1}^{\infty}

is an orthonormal family in

Q H

.) Put

f_{k} \equiv {(k + 1)}^{\frac{1}{2}} φ_{2 k} + k^{\frac{1}{2}} φ_{2 k - 1}

and

f_{k}^{-} \equiv {(k + 1)}^{\frac{1}{2}} φ_{2 k} - k^{\frac{1}{2}} φ_{2 k - 1}

. Then

(f_{k}^{-}, f_{k}) = 1

r_{k} : = (., f_{k}^{-}) f_{k} \in ℒ_{P}^{ℬ}

. By the construction,

{R_{k} : = r_{k} + J r_{k}^{*} J}_{1}^{\infty}

is an orthogonal sequence of

J

-projection (by Proposition 3) from

P^{ℬ}

Assume now that there exists the supremum

R : = \sum_{1}^{\infty} R_{n} \in P^{ℬ}

. Put

Then

P_{m} \in P^{ℬ}

P_{m} \geq P_{m + 1}

\forall m

, and

P_{m} \geq R_{n}

\forall m, n

. Hence

P_{m} \geq R = \sum_{1}^{\infty} R_{n} \geq \sum_{1}^{m} R_{k}

. Thus

P_{m} - \sum_{1}^{m} R_{k} \geq R - \sum_{1}^{m} R_{k}

and

P_{m} - \sum_{1}^{m} R_{k} \geq {(R - \sum_{1}^{m} R_{k})}^{*}

. Finally,

We get a contradiction, since the norm of the right hand side expression tends to inﬁnity when

m \to \infty

Proof. Let $R$ have type $(B)$ , i. e., $R \in P^{ℬ}$ . Then $R = P R P + Q R Q$ , hence $R = J R J$ .

Now, let $R = J R J$ . Then $J R = R J$ , hence $P R = \frac{1}{2} (I + J) R = \frac{1}{2} (R + R J) = R P$ , and consequently, $P R Q = R P Q = 0$ . Similarly, $Q R P = 0$ .

Now, let $P R Q = Q R P = 0$ . Then $R = P R P + Q R Q$ , hence $R \in P^{ℬ}$ . □

Proof. 1) Let us consider, for instance, the case $R \in P^{+}$ and $P R Q = 0$ . We have $0 = J P R Q = Q (J R) Q$ . Since $J R$ is a positive operator, the latter implies that $P (J R) P = J R$ . Thus $R = J P (J R) P = (J P J) R P = Q R P$ . Finally, $R = R^{2} = (Q R P) (Q R P) = 0$ .

2) Let $dim H \geq 4$ . Then we can ﬁnd $ϕ_{P} \in P H \cap S$ and $ϕ_{Q} \in Q H \cap S$ such that $(ϕ_{Q}, J ϕ_{P}) = 0$ . Put $f_{P} : = ϕ_{P} = g_{P}$ and $f_{Q} : = \frac{1}{2} J ϕ_{P} + ϕ_{Q}$ , $g_{Q} : = - \frac{1}{2} J ϕ_{P} + ϕ_{Q}$ . Let us deﬁne $f = f_{P} + f_{Q}$ , $g = g_{P} + g_{Q}$ . It is easy to verify that $f \in Γ^{+}$ , $g \in Γ^{-}$ and $[f, g] = 0$ . Thus $p_{f} + p_{g} \in P$ . Finally, $P (p_{f} + p_{g}) Q = P ([., f] f - [., g] g) Q = ((., J f_{P}) f_{P} - (., J g_{P}) g_{P}) = 0$ and $Q (p_{f} + p_{g}) P = (., J f_{Q}) f_{Q} - (., J g_{Q}) g_{Q} \neq 0$ .

3) Now, let $P R Q = 0$ . Then

P R P = (P R P) (P R P) + (P R Q) (Q R P) = (P R P) (P R P)

and

Q R Q = {(P R P)}^{#} = {(P R P)}^{#} {(P R P)}^{#} = (Q R Q) (Q R Q) .

□

Proof. 1) Since $R \in P$ and $R = R^{*}$ , we have $J R = R^{*} J = R J$ . Hence $P^{\pm} R = R P^{\pm}$ .

2) If, in addition, $R = J R J$ then $J P^{+} R J = (J P^{+} J) J R J = P^{-} R$ . This means that $R = (P^{+} + P^{-}) R = P^{+} R + J (P^{+} R) J$ . □

Proof. The equality $p = q p$ is equivalent to

J p J = J q p J = J p J (J q J) = J p J q .

□

In what follows we will use equations (2) (see below). Let

A \in A^{J s}

. This means that

J A = A^{*} J

. Hence

Consider some properties of projections. Let

r

be a bounded projection on

H

. Let us denote by

r_{o r}

the orthogonal projection onto

r H \cap r^{*} H

. By Proposition 4 [5],

r_{o r}

is the greatest orthogonal projection with the property

r_{o r} \leq r

. A projection

r

is said to be properly skew projection if

r_{o r} = 0

. Let

x

y \in H

(x, y) = 1

and

∥ x ∥ = ∥ y ∥

. Then

(., x) y

is the properly skew projection if and only if

∥ x ∥ > 1

The orthoprojection

r_{o r}

is said to be an orthogonal component of

r

, and

r_{s} : = r - r_{o r}

is said to be a properly skew component of

r

. It is clear that

J r_{o r} J

(

J r_{s} J

) is the orthogonal (properly skew, respectively) component of

J r J

Now, let

r

be a properly skew projection. Let us denote by

r_{m}

the orthogonal projection onto

r H

. Then

r = (r_{m} + r_{m}^{⊥}) r (r_{m} + r_{m}^{⊥}) = r_{m} + r_{m} r r_{m}^{⊥} .

Let

r_{m} r r_{m}^{⊥} = v ∣ r_{m} r r_{m}^{⊥} ∣

be the polar decomposition of

r_{m} r r_{m}^{⊥}

, where

∣ B ∣ : = {(B^{*} B)}^{1 ∕ 2}

. Since

r

is a properly skew projection,

v H = r_{m} H

. Note that

v^{*} ∣ r_{m}^{⊥} r^{*} r_{m} ∣

is the polar decomposition of

r_{m}^{⊥} r^{*} r_{m}

and

∣ r_{m}^{⊥} r^{*} r_{m} ∣ = v ∣ r_{m} r r_{m}^{⊥} ∣ v^{*}

. It is clear that

the cover projection of

∣ r_{m}^{⊥} r^{*} r_{m} ∣

(of

∣ r_{m} r r_{m}^{⊥} ∣

) is equal to

v^{*} v

(

v v^{*}

, respectively) and

Put

r_{m}^{'} : = v^{*} r_{m} v

. By deﬁnition,

r_{m}^{'} r_{m} = 0

and

(v - v^{*}) (r_{m} + r_{m}^{'}) {(v - v^{*})}^{*} = r_{m}^{'} + r_{m}

. A straightforward veriﬁcation shows that

Let

A_{i} \in B (H)

i = 1, . ., 4

, be such that

P A_{i} P = A_{i}

, for all

i

. Let us identify

A_{1} + A_{2} J + J A_{3} + J A_{4} J

with the matrix

(\begin{matrix} A_{1} & A_{2} \\ A_{3} & A_{4} \end{matrix})

. Thus

and

P^{-} = \frac{1}{2} (\begin{matrix} P & - P \\ - P & P \end{matrix})

. Put

X : = \frac{1}{4} (\begin{matrix} x & x \\ x & x \end{matrix})

Theorem 2. Let $A$ be a von Neumann $J$ -algebra of type $(B)$ , $R = r + J r^{*} J$ , where $r \leq P$ and $r$ is a properly skew projection from $A$ , and let $P^{-} R P^{+} = U ∣ P^{-} R P^{+} ∣$ be the polar decomposition for $P^{-} R P^{+}$ . Then $X = P^{+} R P^{+}$ , $J U J = - U$ and

R = X + U {(X^{2} - X)}^{1 ∕ 2} - {(X^{2} - X)}^{1 ∕ 2} U^{*} + U (I - X) U^{*} .

(6)

Conversely, let $x \in P A P$ be such that $x = x^{*}$ , $x^{2} - 2 x \geq 0$ and $w \in P A P$ be a partial isometry with the initial subspace $\bar{x H}$ . Then the formula $(6)$ deﬁnes a $J$ -projection. Here $U : = \frac{1}{2} (\begin{matrix} w & w \\ - w & - w \end{matrix})$ . If, in addition, for $x$ , $w$ the equalities

x = w (2 P - x) w^{*} a n d w {(x^{2} - 2 x)}^{1 ∕ 2} = - {(x^{2} - 2 x)}^{1 ∕ 2} w^{*}

(7)

hold, then $R$ is a $J$ -projection from $A$ .

Proof. A simple matrix computation shows that

P^{+} R P^{+} = X a n d P^{-} X P^{-} = \frac{1}{4} (\begin{matrix} x & - x \\ - x & x \end{matrix})

(8)

Using (2) with $R$ in place of $A$ and (5) we obtain

∣ P^{-} R P^{+} ∣ = {[{(P^{-} R P^{+})}^{*} (P^{-} R P^{+})]}^{1 ∕ 2} = {[(- P^{+} R P^{-}) (P^{-} R P^{+})]}^{1 ∕ 2} =

{[(P^{+} R P^{+}) (P^{+} R P^{+}) - P^{+} R P^{+}]}^{1 ∕ 2} = {(X^{2} - X)}^{1 ∕ 2} = \frac{1}{4} (\begin{matrix} ∣ r - r^{*} ∣ & ∣ r - r^{*} ∣ \\ ∣ r - r^{*} ∣ & ∣ r - r^{*} ∣ \end{matrix}) .

Put $V : = \frac{1}{2} (\begin{matrix} (v - v^{*}) & 0 \\ 0 & - (v - v^{*}) \end{matrix}) (\begin{matrix} P & P \\ P & P \end{matrix}) = \frac{1}{2} (\begin{matrix} v - v^{*} & v - v^{*} \\ - (v - v^{*}) & - (v - v^{*}) \end{matrix}) .$ Then

V V^{*} = \frac{1}{2} (\begin{matrix} (v v^{*} + v^{*} v) & - (v v^{*} + v^{*} v) \\ - (v v^{*} + v^{*} v) & (v v^{*} + v^{*} v) \end{matrix})

= \frac{1}{2} (\begin{matrix} (r_{m} + r_{m}^{^{'}}) & - (r_{m} + r_{m}^{^{'}}) \\ - (r_{m} + r_{m}^{^{'}}) & (r_{m} + r_{m}^{^{'}}) \end{matrix}) \leq P^{-},

and $V^{*} V = \frac{1}{2} (\begin{matrix} (v v^{*} + v^{*} v) & (v v^{*} + v^{*} v) \\ (v v^{*} + v^{*} v) & (v v^{*} + v^{*} v) \end{matrix}) \leq P^{+} .$ This means that $V$ is a partial isometry. Since the cover projection of $∣ r - r^{*} ∣$ is equal to $r_{m} + r_{m}^{^{'}}$ , the cover projection of $∣ P^{-} R P^{+} ∣$ is equal to $\frac{1}{2} (\begin{matrix} (r_{m} + r_{m}^{^{'}}) & (r_{m} + r_{m}^{^{'}}) \\ (r_{m} + r_{m}^{^{'}}) & (r_{m} + r_{m}^{^{'}}) \end{matrix}) .$ Hence we have

V {(X^{2} - X)}^{1 ∕ 2} = \frac{1}{4} (\begin{matrix} (v - v^{*}) & 0 \\ 0 & - (v - v^{*}) \end{matrix}) (\begin{matrix} ∣ r - r^{*} ∣ & ∣ r - r^{*} ∣ \\ ∣ r - r^{*} ∣ & ∣ r - r^{*} ∣ \end{matrix}) =

b y (3) = \frac{1}{4} (\begin{matrix} r - r^{*} & r - r^{*} \\ r - r^{*} & r - r^{*} \end{matrix}) = P^{-} R P^{+} .

Thus $V = U$ , $U {(X^{2} - X)}^{1 ∕ 2} = P^{-} R P^{+}$ , and, by (2),

- {(X^{2} - X)}^{1 ∕ 2} U^{*} = P^{+} R P^{-} .

(9)

Moreover,

\begin{matrix} U (I - X) U^{*} = U (P^{+} - X) U^{*} \\ = \frac{1}{4} (\begin{matrix} (v - v^{*}) & 0 \\ 0 & - (v - v^{*}) \end{matrix}) (\begin{matrix} 2 P - x & 2 P - x \\ 2 P - x & 2 P - x \end{matrix}) (\begin{matrix} v^{*} - v & 0 \\ 0 & - (v^{*} - v) \end{matrix}) \\ = by (4) = \frac{1}{4} (\begin{matrix} x & - x \\ - x & x \end{matrix}) = by (8) = P^{-} R P^{-} . & (10) \end{matrix}

Summarizing (8), (9), and (10) we have $(6)$ . By simple calculations we prove the equality $J U J = - U$ .

Conversely, let $x \in P A P$ be such that $x = x^{*}$ , $x^{2} - 2 x \geq 0$ and let $w \in P A P$ be a partial isometry with the initial subspace $\bar{x H}$ . Then one can directly show that $(6)$ deﬁnes a $J$ -projection.

Now, let (7) hold true for $x$ , $w$ . The matrix entry of $R$ at the ﬁrst row and the second column is equal to

\frac{1}{4} (x + w {(x^{2} - 2 x)}^{1 ∕ 2} + {(x^{2} - 2 x)}^{1 ∕ 2} w^{*} - w (2 P - x) w^{*}) = 0,

and the matrix entry of $R$ at the ﬁrst row and the ﬁrst column is equal to

\frac{1}{4} (x + w {(x^{2} - 2 x)}^{1 ∕ 2} - {(x^{2} - 2 x)}^{1 ∕ 2} w^{*} + w (2 P - x) w^{*}) = \frac{1}{2} (x + w {(x^{2} - 2 x)}^{1 ∕ 2}) .

This means that $R = (\begin{matrix} r & 0 \\ 0 & r^{*} \end{matrix})$ is a $J$ -projection from $A$ . Here $r : = \frac{1}{2} (x + w {(x^{2} - x)}^{1 ∕ 2})$ is a projection from $P A P$ . □

Note that for any

J

-projection

p

there exists (non unique!) representation

p = p_{+} + p_{-}

, where

p_{+} \in P^{+}

p_{-} \in P^{-}

. Let us show that for any

ℛ \in P^{ℬ}

there is a unique special representation of such form.

Let us denote by

{(p J)}_{+}

(

{(p J)}_{-}

) the positive (the negative, respectively) part of self-adjoint operator

p J

p \in P

Proof. By the assumption on $p_{+}$ and $p_{-}$ , we have $p_{+} J = {(p J)}_{+}$ and $p_{-} J = - {(p J)}_{-}$ . Hence $p_{+} = (p_{+} J) J = {(p J)}_{+} J$ and $p_{-} = (p_{-} J) J = - {(p J)}_{+} J$ . □

Proof. Let us prove that the projection $R_{+}^{0}$ exists.

1) First, let $R$ be an orthogonal $J$ -projection of type $(B)$ . The $J$ -projection $P^{+} R$ ( $P^{-} R$ ) is $J$ -positive ( $J$ -negative, respectively). By Proposition 5.2), $R = P^{+} R + J P^{+} R J$ . By the construction, the subspaces $P^{+} R H$ and $J P^{+} R J H \subseteq P^{-} H$ are mutually orthogonal.

2) Now, let $R$ be a properly skew $J$ -projection. By Remark 1.3) and Theorem 2, $(6)$ holds. By Theorem 1 and (10),

J X J = J P^{+} R P^{+} J = P^{-} (J R J) P^{-} = U (I - X) U^{*} .

(11)

Hence $J {(X^{2} - X)}^{1 ∕ 2} J = U {(X^{2} - X)}^{1 ∕ 2} U^{*}$ . Thus

{(X^{2} - X)}^{1 ∕ 2} J U = J U {(X^{2} - X)}^{1 ∕ 2} .

(12)

By Theorem 1 and (9),

J (U {(X^{2} - X)}^{1 ∕ 2}) J = J (P^{-} R P^{+}) J = P^{+} (J R J) P^{-} = - {(X^{2} - X)}^{1 ∕ 2} U^{*} .

Hence

- {(X^{2} - X)}^{1 ∕ 2} U^{*} J = J U {(X^{2} - X)}^{1 ∕ 2} .

(13)

By (12) and (13), $J U = - U^{*} J$ , i. e.

- U = J U^{*} J .

(14)

Let us denote by $F_{+}$ ( $F_{-}$ ) the cover projection of positive $(= X_{+})$ (negative $(= X_{-})$ ) part of self-adjoint operator $X$ , respectively. By (11),

J X_{+} J + J (- X_{-}) J = J X J = U (F_{-} + X_{-}) U^{*} + U (F_{+} - X_{+}) U^{*} .

(15)

Since $[U (F_{-} + X_{-}) U^{*}] [U (F_{+} - X_{+}) U^{*}] = 0$ ,

J X_{+} J = U (F_{-} + X_{-}) U^{*} a n d - J X_{-} J = U (F_{+} - X_{+}) U^{*} .

(16)

By (14) and (16),

J U {(X_{+}^{2} - X_{+})}^{1 ∕ 2} J = - U^{*} J {(X_{+}^{2} - X_{+})}^{1 ∕ 2} J =

- U^{*} U {(X_{-}^{2} + X_{-})}^{1 ∕ 2} U^{*} = - {(X_{-}^{2} + X_{-})}^{1 ∕ 2} U^{*} .

(17)

Put

R_{-}^{0} : = - X_{-} + U {(X_{-}^{2} + X_{-})}^{1 ∕ 2} - {(X_{-}^{2} + X_{-})}^{1 ∕ 2} U^{*} + U (F_{-} + X_{-}) U^{*},

R_{+}^{0} : = X_{+} + U {(X_{+}^{2} - X_{+})}^{1 ∕ 2} - {(X_{+}^{2} - X_{+})}^{1 ∕ 2} U^{*} + U (F_{+} - X_{+}) U^{*} .

It is easy to verify that $R_{\pm}^{0} \in P^{\pm}$ , $R_{+}^{0} R_{-}^{0} = 0$ , $R = R_{+}^{0} + R_{-}^{0}$ . Since $R_{+}^{0} J$ is a positive operator, $R_{-}^{0} J$ is a negative operator, and $(R_{+}^{0} J) (R_{-}^{0} J) = 0$ , the subspaces $R_{+}^{0} H$ and $R_{-}^{0} H$ are mutually orthogonal. By (16), (17),

J R_{+}^{0} J = J X_{+} J + J U {(X_{+}^{2} - X_{+})}^{1 ∕ 2} J

- J {(X_{+}^{2} - X_{+})}^{1 ∕ 2} U^{*} J + J U (F_{+} - X_{+}) U^{*} J =

U (F_{-} + X_{-}) U^{*} - {(X_{-}^{2} + X_{-})}^{1 ∕ 2} U^{*} + U {(X_{-}^{2} + X_{-})}^{1 ∕ 2} - X_{-} = R_{-}^{0}

3) Now, let us consider the general case of $R$ . We have $R = R_{o r} + R_{s}$ . We know that $R_{o r}$ is an orthogonal $J$ -projection of type (B), and $R_{s}$ is a properly skew $J$ -projection of type $(B)$ . Let ${(R_{o r})}_{+} : = P^{+} R_{o r}$ and let ${(R_{s})}_{+}$ be the $J$ -projection from step 2), generated by $R_{s}$ . Put $R_{+}^{0} : = {(R_{o r})}_{+} + {(R_{s})}_{+}$ . Thus the projection $R_{+}^{0}$ is that in question.

By Lemma 1, the $J$ -projection $R_{+}^{0}$ is unique. □

J

-projection

R \in P^{+}

is said to be a generator (for a

J

-projection

ℛ

) if

ℛ = R + J R J

and the subspaces

R H

J R H

are mutually orthogonal. Let

F^{+}

(F^{-})

be the cover projection of

P^{+} R P^{+}

(of

P^{-} R P^{-}

, respectively).

Proof. 1) Let $R$ be a generator. Put $ℛ : = R + J R J$ . By Theorem 3, $R = ℛ_{+}^{0}$ . Here $ℛ_{+}^{0}$ is the generator for $ℛ$ from the proof of Theorem 3. From (16) it follows that $U H$ and $J F^{+} H$ are mutually orthogonal.

2) Let us prove some properties. Let $R \in P^{+}$ . Then

P^{+} R P^{+} = P^{+} (J R) P^{+} i s a p o s i t i v e o p e r a t o r a n d

P^{-} R P^{-} = - P^{-} (J R) P^{-} i s a n e g a t i v e o p e r a t o r .

(18)

Furthermore, $P^{+} R^{*} R P^{+} = P^{+} (J R^{*}) R P^{+} = P^{+} R R^{*} J P^{+} = P^{+} R R^{*} P^{+}$ and $P^{-} R^{*} R P^{-} = P^{-} R R^{*} P^{-}$ . We have $(P^{+} R P^{+} x, x) \geq (F^{+} x, x)$ for all $x \in H$ (see Proposition 1, [7]) and

∣ R P^{+} ∣^{2} = P^{+} R^{*} R P^{+} = P^{+} R J R P^{+} =

P^{+} R (2 P^{+} - I) R P^{+} = 2 {(P^{+} R P^{+})}^{2} - P^{+} R P^{+} .

a) Hence follows that the initial projection of $R P^{+}$ is equal to the cover projection of $P^{+} R P^{+}$ (i.e., is equal to $F^{+}$ ).

In the same way we can prove the following assertion.

b) The ﬁnal projection of $P^{-} R$ is equal to the cover projection of $P^{-} R P^{-}$ (i.e. $F^{-}$ ). We have

(P^{-} R P^{+}) (P^{+} R^{*} P^{-}) = - P^{-} R P^{+} J P^{+} R P^{-} = - P^{-} R P^{+} R P^{-} =

- P^{-} R (I - P^{-}) R P^{-} = {(P^{-} R P^{-})}^{2} - P^{-} R P^{-} .

c) From (18) it follows that the ﬁnal projection of $P^{-} R P^{+}$ (and hence of $U$ ) is equal to the cover projection of $P^{-} R P^{-}$ (i.e., is equal to $F^{-}$ )

By b), $P^{-} R = F^{-} P^{-} R$ . By a), $R P^{+} = R P^{+} F^{+}$ . Hence

R P^{+} J R J = R P^{+} J P^{-} R J = R P^{+} (F^{+} J F^{-}) P^{-} R J .

Now, let $R \in P^{+}$ be a $J$ -projection such that $U H$ and $J F^{+} H$ are mutually orthogonal. Hence $0 = J F^{+} J (U U^{*})$ , i.e. by c), $F^{+} J F^{-} = 0$ . Finally, $R P^{+} J R J = R P^{+} (F^{+} J F^{-}) P^{-} R J = 0$ . Similarly, $R P^{-} (J R J) = 0$ . Thus $R (J R J) = R (P^{+} + P^{-}) J R J = 0 .$ Hence

R + J R J \in P^{ℬ}, R J (J R J) = R (P^{+} - P^{-}) (J R J) = 0 .

This means that $R H$ and $J R H$ are mutually orthogonal. Thus $R$ is a generator. □

Let

R

ℛ

J

-projections of type

(B)

and

R_{+}

ℛ_{+}

be their generators. Then

R ℛ = 0

(i.e., the subspaces

R H

ℛ H

are

J

-mutually orthogonal) implies

R + ℛ = (R_{+} + ℛ_{+}) + J (R_{+} + ℛ_{+}) J

. But the

J

-positive

J

-projection

R_{+} + ℛ_{+}

is not a generator for

R + ℛ

, in general.

3. Two-dimensional (minimal)

J

-projections of type

(B)

In what follows we denote by

ℜ

and

ℑ

the real and the imaginary part of a complex number.

Let

f \in H

. Put

f_{P} : = P f

f_{Q} : = Q f

. In terms of

f_{P}

f_{Q}

we can give another simple description for vectors from

Γ

In connection with Corollary 2 we formulate the following elementary proposition.

Proof. Let $α$ , $β$ be real numbers such that $2 β α = 1$ . Put $g_{P} = α y$ , $g_{Q} = β J x$ and $g = g_{P} + g_{Q}$ . Then $1 = 2 β α = 2 [g_{Q}, g_{P}] = 2 ℜ [g_{Q}, g_{P}] .$ By Proposition 7 (2), $g \in Γ^{+}$ . Since $ℑ [g_{Q}, g_{P}] = 0$ , we have $[g, J g] = 0$ , by Proposition 7 (3). Thus $g \in K$ . Since the set ${(α, β) : 2 β \bar{α} = 1}$ is inﬁnite, the set $K$ is inﬁnite too. □

Proof. 1) First, let $Q$ be a two-dimensional $J$ -projection of type $(B)$ . Then $Q = (., x) y + (., J y) J x$ , where $x$ , $y \in P H$ and $(x, y) = 1$ . Let $K$ be from Lemma 2 and let $f \in K$ . Since $[f, J f] = 0$ , we have $p_{f} p_{J f} = 0$ , by Proposition 8. This means that $p_{f} + p_{J f}$ is a two-dimensional $J$ -projection of type $(B)$ . Since $J f, f \in r a n (Q)$ , we have $Q = p_{f} + p_{J f} .$ Since $K$ is inﬁnite and $p_{f}$ is a $J$ -positive $J$ -projection, Corollary 4 is true for the case of two-dimensional $J$ -projection.

2) Let us consider the general case. For any non-zero $J$ -projection $R$ of type $(B)$ we can ﬁnd a two-dimensional $J$ -projection $Q$ of type $(B)$ such that $Q \leq R$ . Then $R - Q$ is also a $J$ -projection of type $(B)$ . Let $R_{+}^{0}$ be the $J$ -positive $J$ -projection such that $R_{+}^{0} H$ , $J R_{+}^{0} H$ are mutually orthogonal (Theorem 3) and $R - Q = R_{+}^{0} + J R_{+}^{0} J$ . The set ${R_{+}^{0} + p_{f} : f \in K}$ is inﬁnite. The operator $R_{0} : = R_{+}^{0} + p_{f}$ , $f \in K$ is a $J$ -positive $J$ -projection and $R = R^{0} + J R^{0} J$ . □

We shall show that there is a unique special

f_{0} \in K

. In order to prove Proposition 11 we need the following proposition.

Proof. Let $f \in N$ . By Proposition 7 (2) and 7 (3),

1 = 2 [f_{Q}, f_{P}] = 2 β \bar{α} (x, y) = 2 β \bar{α} .

(19)

We have $0 = (f, J f) = ∥ f_{P} ∥^{2} - ∥ f_{Q} ∥^{2} = (∣ α ∣^{2} - ∣ β ∣^{2}) ∥ x ∥^{2} .$ Hence $∣ α ∣ = ∣ β ∣$ . By (19), $α = β$ and $∣ α ∣ = \frac{1}{\sqrt{2}}$ . Finally, $f = \frac{λ}{\sqrt{2}} (y + J x)$ , $∣ λ ∣ = 1$ . □

Note the following. Let

x

y \in P H

be such that

(x, y) = 1

and

∥ x ∥ = ∥ y ∥

and let

f_{0}

be from Proposition 9. Then

f_{0} \in P^{+} H

if and only if

∥ x ∥ = 1

if and only if

(., x) y

is an orthogonal projection.

A vector

f \in Γ^{+}

is said to be a generator (for a two-dimensional

J

-projection

R

of type

(B)

) if

R = p_{f} + p_{J f}

and

p_{f} H ⊥ p_{J f} H

. It is clear that a vector

f \in Γ^{+}

is a generator if and only if the

J

-projection

p_{f}

is a generator.

Proof. It is clear that conditions 1) and 2) are equivalent.

2) $\Rightarrow$ 3). Let $(f, J f) = 0 = [f, J f]$ . By Proposition 7 (1), we have $ℜ (J P^{+} f, P^{-} f) = 0$ . By Proposition 8, $ℑ (J P^{+} f, P^{-} f) = 0$ . Hence $(J P^{+} f, P^{-} f) = 0$ .

3) $\Rightarrow$ 4). Since $f \in Γ^{+}$ , we have $2 ℜ [f_{Q}, f_{P}] = 1$ , by Proposition 7 (2). Since $(J P^{+} f, P^{-} f) = 0$ , we have $ℑ [f_{Q}, f_{P}] = 0$ , by Proposition 8. Hence $2 [f_{Q}, f_{P}] = 1$ . Finally, $(J P^{+} f, P^{-} f) = 0$ implies (see Proposition 7 (1)) $∥ f_{P} ∥ = ∥ f_{Q} ∥$ .

4) $\Rightarrow$ 2). The equality $2 [f_{Q}, f_{P}] = 1$ implies $f \in Γ^{+}$ (see Proposition 7 (2), and by Proposition 8, $[f, J f] = 0$ . By Proposition 7 (1), $∥ f_{P} ∥ = ∥ f_{Q} ∥$ implies $(f, J f) = 0$ . □

The following proposition is a two-dimensional analog of Theorem 3 specifying the structure of

J

-projection

R_{+}^{0}

Proof. Let $p \in ℳ$ . By the deﬁnition of $R$ , $p = p_{f}$ , where $f \in Γ^{+}$ . It is clear that $p_{f} (J p_{f} J) = p_{f} p_{J f} = 0$ . Hence $[f, J f] = 0$ . Since ${(J R)}_{+} {(J R)}_{-} = 0$ , $0 = (J p_{f}) (J J p_{f} J)$ . This means $(f, J f) = 0$ . Since $p_{f} < R$ , $f \in R H$ . By the deﬁnition of R again, $f_{P} = α y$ , $f_{Q} = β J x$ . By Proposition 9, $f = λ f_{0}$ , $∣ λ ∣ = 1$ . Hence $p_{f} = p_{f_{0}}$ . □

dim H > 2

, then, for any

J

-projection

p_{f}

f \in Γ

, the set of two-dimensional

J

-projections

q

such that

p_{f} < q

is inﬁnite. Therefore the following proposition seems to be interesting.

Proof. Let us consider, for instance, the case $f \in Γ^{+}$ . Fix $α$ such that $α [f_{Q}, f_{P}] = 1$ . Put $y_{0} : = \bar{α} f_{P}$ , $x_{0} : = J f_{Q}$ . Then $(x_{0}, y_{0}) = 1$ . Thus $(., x_{0}) y_{0}$ is a projection. By the construction, $(., x_{0}) y_{0} \leq P$ . Note that

[f, f_{p}] = [f_{P} + f_{Q}, f_{P}] = [f_{Q}, f_{P}], [f, f_{Q}] = [f_{P}, f_{Q}] .

(20)

Let us deﬁne the two-dimensional $J$ -projection

R : = (., x_{0}) y_{0} + (., J y_{0}) J x_{0}

of type (B). From the equality $α [f_{Q}, f_{P}] = 1$ and (20), it follows that $R p_{f} = p_{f}$ . Thus $p_{f} < R$ . By Proposition 6, $p_{J f} < R$ . This means that $f$ , $J f \in R H$ . Hence the lineal generated by the set ${f, J f}$ is equal to the subspace $R H$ . Any $J$ -projection is uniquely determined by its range space. Therefore $R$ is the unique two-dimensional $J$ -projection with the property $p_{f} < R$ . □

Problem: Does there exist

sup {p, J p J}

P^{ℬ}

for any

p \in P

with respect to the order

\leq

? (cf. Proposition 12).

The author expresses deep gratitude to O.E.Tikhonov for his valuable comments and suggestions.

References

[1] Birkhoff G. Lattice Theory. Providence, Rhode Island, (1967).

[2] Azizov T.Ya. and Iokhvidov I.S. Foundations of the theory of linear operators in spaces with indeﬁnite metric. Nauka, Moscow, (1986), [in Russian]. [English transl., Linear operators in space with an indeﬁnite metric. Wiley, New York, (1989).]

[3] Dadashyan K.Yu. and Khoruzhij S.S.On ﬁeld algebras in quantum theory with indeﬁnite metric. Theoretical and Mathematical Physics, 54 (1983), 57–77. [in Russian]. [English transl., 54 (1983), 35–48.]

[4] Mel’tser M.M. On a classiﬁcation of von Neumann $J$ -algebra. Functional analysis and its applications, 13 (1979) , No.4, 83–84 [in Russian]. [English transl., 13 (1979) , No.4, 305–307.]

[5] Matvejchuk M.S. Probability measures in $W^{*} J$ -algebras in Hilbert spaces with conjugation, Proc. Amer. Math. Soc., 126 (1998), No.4, 1155–1164.

[6] Matvejchuk M.S. and Ionova A.M. The structure of $J$ -projections from logics of type $(B)$ . $8^{t h}$ Bienuial IQSA Meeting Quantum Structures 06, Abstracts, Malta, (2006), July, 9-14, 62–63.

[7] Matvejchuk M.S. A description of indeﬁnite measures in $W^{*} J$ -factors. Dokl. Akad. Nauk SSSR, 319 (1991), No.3, 558–561 [in Russian]. [English transl., Sov. Math. Dokl., 44 (1992), No.1, 161-165.]

Ulyanovsk State Pedagogical University, 432000, Ulyanovsk, Tolstogo str. 87, Apartm. 33, Russia

1. Introduction

2. Projections of type (B)

3. Two-dimensional (minimal) J-projections of type (B)

References

2. Projections of type $(B)$

3. Two-dimensional (minimal) $J$ -projections of type $(B)$