- First examples 41
Actually, it suffices to. show that we can always find two orthogonal
projections P1,P2 which are Murray-von Neumann equivalent. Suppose we
could prove this and P1,p2 were such projections. Since M has no abelian
projections, the same is true of p2MP2 and hence we would be able to find
equivalent orthogonal projections q1, q2 ~ P2· ·Transporting the qi's back
underneath p1, we would thus have four equivalent orthogonal projections.
Applying the procedure again underneath q2 would give us eight such pro-
jections, and so on.
So, how can we construct two orthogonal equivalent projections? Simply
take a noncentral projection p EM and find some m EM such that pm(l-
p) #- 0. The partial isometry in the polar decomposition of this operator
will have orthogonal support and range projections. 0
Proposition 2.4.9. Let M be a von Neumann algebra. Then the following
are equivalent:
(1) M is a finite direct sum of finite homogeneous van Neumann alge-
bras (that·is,
M ~ Mk(1)(A1) EB··· EB Mk(j)(Aj)where each A is an abelian van Neumann algebra);
(2) M is a nuclear C*-algebra;
(3) M is an exact C* -algebra.Proof. Thanks to Corollary 2.4.4 and the fact that every nuclear C* -algebra
is exact, we only have to prove the implication (3) ::::? (1).
Assume first that M is not type I - i.e., has a summand N without
abelian projections. Since N is infinite dimensional, we can find a pairwise
orthogonal sequence of projections p1,p2, ... in N. None of the corners PiNPi
contains an abelian projection (since N doesn't) and hence the previous
lemma provides an embedding Mi(C) '---t PiNPi· Evidently this yields an
embedding
n nSince exactness passes to subalgebras, and we've seen that ITn Mn(C) is not
exact, it follows that M can't be exact.
Hence we've reduced to the type I case: M = IJiEI A® IIB(Hi)· Since
IIB(H), for an infinite-dimensional Hilbert space H, also contains a copy of
ITn Mn ( C), it follows that each cardinal number i must be a natural number.
Similar reasoning shows that only a finite number of the remaining Ai's can
be nonzero, so the proof is complete. 0