3.8. Nuclearity and tensor products 103
Now let P E IIB(JC) be the orthogonal projection onto the span of the
vectors {v1, ... , vn} C JC and define a u.c.p. map p: A-+ PIIB(JC)P by com-
pression. The matrix of p( a) with respect to { v1, ... , Vn} is
[ (avj, vi) ]i,j.
Define a map 'l/J: PIIB(JC)P-+ M by declaring
'l/J(ei,j) = blbj,
where { ei,j} is a set of matrix units relative to { v1, ... , vn}· According to
Example 1.5.13, 'l/J is a c.p. map (though it may have large norm).
For a E A and e E S we compute
e('l/J o p(a)) = e (.t (avj, Vi)btbj) = .t (avj, Vi)e(btbj)·
i,J=l i,3=1
On the other hand, for a E ~ and e E S we have
e(t,o(a)) =μ(a 0 Ye)
n n
~ (a 0 Ye(~= vi 0 bi), L::>i 0 bi)
i=l i=l
n
= L (avj, vi) (yebj, bi)
i,j=l
n
= L (avj, vi) (7r(blbj )Ye i, i)
i,j=l
n
= L (avj, vi)e(blbj)
i,j=l
= e('l/J 0 p(a)).
Thus 'l/J o p is a factorable map which is close to tp on our prescribed finite
sets and this is all we needed to show. 0
For dessert, we have a few important applications to choose from. The
first, which is an immediate consequence of the bicommutant theorem and
the previous result, shows that semidiscreteness of the commutant does not
depend on the representation.
Corollary 3.8.6. Let M c IIB('h'.) be a von Neumann algebra. Then M is
semidiscrete if and only if M' is semidiscrete.
From the C*-point of view, the next application is probably the most
important.