Biordered sets of von Neumann algebras

A projection is a bounded self-adjoint idempotent operator acting on

H

. We denote by

P (M)

(or

P

for brevity if there is no ambiguity) the set of all projections in

M

. For

p, q \in P (M)

deﬁne

This relation is a partial order on

P

. It is known that

P

is a complete lattice with respect to

\leq

[see ?]. Since projections are selfadjoint,

{(p q)}^{*} = q p

for any two projections. Hence the dual relation

≼

deﬁned by

for any two projection

p

and

q

, coincides with the relation

\leq

deﬁned above.

is a closed subspace of

H

uniquely determined by

p

. Subspaces of

H

arising in this way are called subspaces belonging to

M

. Let

Ξ_{M}

[or simply

Ξ

for brevity] denote the set of all subspaces belonging to

M

Ξ

is partially ordered set with respect to inclusion. Furthermore, this gives a mapping

which is an order-isomorphism. For

V \in Ξ

, let

p_{V}

, denote the unique projection with

im p_{V} = V

. Then the map

V \mapsto p_{V}

is also an order-isomorphism of

Ξ

onto

P

and is the inverse of

im : P \to Ξ

8.2. The biordered set

E (M)

In the following we use the notations and terminology from Nambooripad [1979] associated with biordered sets. In particular, the set

E (S)

of all idempotents of a semigroup

S

is a biordered set with respect to the partial binary operation on

E (S)

induced from

S

([see Nambooripad, 1979, Theorem 1]). Therefore given a von Neumann algebra

M

, the set

E (M)

of all idempotents in

M

is the biordered set of idempotents of the multiplicative semigroup of

M

. It is well known that a von Neumann algebra is uniquely determined by the set projections

P (M)

M

[see ?]. Since

P (M) \subseteq E (M)

, it follows that the algebra

M

is uniquely determined by the biordetred set

E (M)

. Hence we shall refer to

E (M)

as the biordered set of

M

. We thus infer that for every von Neumann algebra

M

E (M)

is a nonempty and non-trivial biordered set.

are closed and complementary subspaces in

Ξ

. Therefore

(U_{e}, V_{e})

is a complementary pair of subspaces in

Ξ

associated with the idempotent

e \in E (M)

; for brevity, we will denote by

(U : V)

a complementary pair of subspaces in

M

Proposition 8.1. Every idempotent $e$ in $M$ is associated with a pair $(U_{e} : V_{e}) = (U : V)$ of complementary subspaces in $M$ . Conversely, let $(U : V)$ be any complementary pair of subspaces in $M$ and let $p = p_{U}$ and $q = p_{V}$ are projections determined by the subspaces $U$ and $V$ respectively. Then

\begin{matrix} e (U : V) = (1 - q) p & is the unique idempotent in M such that \\ im e (U : V) = U and N (e (U : V)) = V . \end{matrix}

Therefore the map $e \mapsto (U_{e} : V_{e})$ is a bijection of $E (M)$ onto the set of all complementary pairs of subspaces in $M$ .

Proof. The remarks preceding the statement shows that every idempotent $e \in E (S)$ determines a unique pair of complementary subspaces in $M$ . Conversely, let $(U : V)$ be any pair of complementary subspaces in $M$ . Since $p \mapsto im (p)$ is a bijection there exist unique $p = p_{U}$ and $q = p_{V}$ in $P (M)$ such that $U = im (p)$ and $V = im (q)$ . Then $q' = 1 - q$ is a projection such that $im (q') = V^{⊥}$ and $N (q') = V$ . Then $V^{⊥}$ and $U$ are complements of $V$ . It follows that $p ∣ V^{⊥}$ is a linear homeomorphism of $V^{⊥}$ onto $U$ . Hence

\begin{matrix} im (q' p) = im (p ∣ V^{⊥}) = U . & (*) & Similarly U and V^{⊥} are complements of U^{⊥} and we have \\ im (p q') = im (q' ∣ U) = V^{⊥} . & (**) \end{matrix}

Now $p q' p$ is a projection such that $p q' p \leq p$ . Also by Equation (1) and (2) $im (p q' p) = U = im (p)$ . Since the map $im : p \mapsto im (p)$ is an order-isomorphism we have $p q' p = p$ . Similarly $q' p q' = q'$ . Clearly both $q' p$ and $p q'$ are idempotents. Also by equation (1) $im (q' p) = U$ and since $p ∣ V^{⊥}$ is a linear homeomorphism of $V^{⊥}$ onto $U$ , $N (q' p) = N (q') = V$ . Since $q'$ and $p$ are projections in $M$ , $e (U : V) = q' p$ is an idempotent in $M$ . Thus the idempotent $e (U : V)$ satisﬁes all the requirements. Since $q'$ and $p$ are projections uniquely determined by $V$ and $U$ respectively the idempotent $q' p$ is uniquely determined by the pair $(U : V)$ . It follows that the assignments $e (U : V) \mapsto (U : V)$ is a bijection. $□$

The result above shows that we may identify idempotents in

M

with the corresponding complementary pair of subspaces in

M

by the correspondence

e (U : V) \mapsto (U : V)

so that

E (M)

will also denote the set of all complementary pairs of subspaces in

M

. Here it will be useful to determine explicitly the biorder structure on

E (M)

in terms of complementary pairs of subspaces in

M

. Recall that basis of the structure of a biordered set on a set

E

consists of two quasiorder relations on

E

denoted by

ω r

and

ω l

respectively and a partial binary operation determined by these relations [see Chapter 3 in E. Krishnan, 2000].

Let

e_{i} \in E (M)

i = 1, 2

. Then for each

i = 1, 2

, by Proposition 8.1, there exist complementary pairs of subspaces

(U_{i} : V_{i})

with

Given idempotents

e_{1}, e_{2} \in E (M)

, it follows that [see for example Cliﬀord and Preston, 1961, § 2.2 Exercise 6] that

e_{1} ω l e_{2}

if and only if

V_{1} = im (e_{1}) \subseteq im (e_{2}) = V_{2}

. In this case we can show that the basic product

e_{2} e_{1} = e ()

8. Biordered sets of von Neumann algebras

8.1. Projections

8.2. The biordered set $E (M)$

8. Biordered sets of von Neumann algebras

8.1. Projections

8.2. The biordered set E(M)

8.2. The biordered set $E (M)$