F. Bimodules over von Neumann Algebras 481Theorem F. 7. Let M be a finite van Neumann algebra. Then, there exists
a unique conditional expectation ctr (called the center-valued trace) from M
onto Z(M) such that ctr(xy) = ctr(yx) for every x, y E M. The center-
valued trace ctr is normal and faithful. Any (normal) tracial stater on M
is of the form μ o ctr, where μ is a (normal) state on Z ( M). For projections
e, f EM, one has e,:::) f if and only if ctr(e):::; ctr(!).
Corollary F .8. Let A be a maximal abelian *-subalgebra of M. Then, every
projection in M is equivalent to a projection in A.For the proof, we need a lemma.
Lemma F.9. Let A be a maximal abelian *-subalgebra of M. Then, for
any nonzero projection qo E A with Aqo =I qoM qo, there exists a nonzero
projection qi EA such that qi :::; qo and ctr(qi) :::; ctr(qo)/2.Proof. Replacing ACM with Aqo C qoMqo, we may assume qo = 1. Note
that Z(M) is a proper subalgebra of the maximal abelian subalgebra A.
Choose a nonzero projection p E A which does not belong to Z(M). We
observe that the center trace ctr(p) is not a projection (otherwise, we would
have p = p ctr(p) = ctr(p)). Hence, replacing p with 1 - p if necessary,
we may assume that the spectral projection z = X(o,i; 2 1(ctr(p)) E Z(M) is
nonzero. Letting qi= pz, we are done. D
Proof of Corollary F.8. We first show that for any nonzero projection
p E M, there is a nonzero projection q E A such that q ,:::) p. Truncating
by the nonzero central projection X(c,i] ( ctr(p)) if necessary, we may assume
that ctr(p) ~ c for some c > 0 (e.g., c = II ctr(p)ll/2 does the job). If there
is a nonzero projection q E A such that Aq = qM q, then q is an abelian
projection and q;::) p. (See, e.g., Lemma V.1.25 in [183].) Otherwise, we can
use Lemma F.9 several times and find projections 1 = qo ~qi ~ · · · ~ qn = q
in A such that ctr(q) :::; 2-n :::; c.
Now, let a projection e E M be given and take a maximal projection
f E A such that f ,:::) e. After conjugating by a unitary element, we may
assume that f :::; e. Suppose by contradiction that p = e - f =I 0. Applying
the above result top E (1 - f)M(l - f), we can find a nonzero projection
q E A(l - f) such that q ;::) p. It follows that f + q ;::) f + p = e, which is
impossible, so we have f = e. D
Let 1{ be a right M-module and denote by NC JE(H) the commutant of
the right M-action. By the representation theory of von Neumann algebras,
1{ is isomorphic to a right M-submodule of £^2 Q9 L^2 (M) - i.e., there exists
an isometry V: 1{-+ £^2 Q9 L^2 (M) such that V(ex) = (Ve)x for every e E 1{
and x EM. Since the projection P = VV* E JIB(£^2 @L^2 (M)) commutes with