1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

3.2. Analytic preliminaries 71

Of course 7rB is defined similarly and it is routine to check that we get
a pair of *-homomorphisms with commuting ranges in this way. It is also
evident that 7rA x 7rB = n.
The last thing to check is that both 7r A and 7r B are nondegenerate. But
if {fj} c A is an approximate unit for A, we have that

7rA(fj) (n(x)v) = n(L fjai ® bi)v j~ n(x)v,

i

for all v E H and x E A 0 B, by an argument analogous to that above
(apply Lemma 3.2.5 to the finite set of bi's). Thus 7rA(fj) -+ lrt in the
strong operator topology - i.e., 7rA is nondegenerate and the same holds for
~· D

As usual, the restriction to nondegenerate representations is only for
convenience. For a general representation 7r: A 0 B -+ ll:ll(H) we can still de-
fine restrictions by first cutting to the closure of {n(x)v: x E A0B, v EH},
applying the previous theorem and then extending by zero on the orthogonal
vectors.

Exercises

Exercise 3.2.1. Let {vj}JEJ CH and {wihEI C /(,be orthonormal bases.
Show that
H ®/(,~EB H ~EB JC.
iEJ jEJ

Using this isomorphism make the following statement rigorous: ll:ll(H ® K) is
just the Ix I matrices with entries in ll:ll(H) which define bounded operators
on EBi H. Under this identification what matrix does the operator T ® 1
correspond to? How about 1 ® S?

Exercise 3.2.2. Show that

( ~ 1i;) 0 (~IC;) 9' iE~J 1i; 0 IC;

Exercise 3.2.3. Wouldn't it be more appropriate to use the notation 7rA x
7rB in Corollary 3.2.4? Is this question rhetorical? If you are confused, see
Lemma 3.3.9 in the next section.
Exercise 3.2.4. Show that if 7r: A 0 B -+ C is a -homomorphism and
both A and B are unital, then there exist -homomorphisms 7rA: A-+ C
and 7rB: B -+ C, with commuting ranges, such that 7r = 7rA x 7rB.
Exercise 3.2.5. Show that Theorem 3.2.6 still holds when ll:ll(H) is replaced
by an arbitrary von Neumann algebra. The result fails, in general, if ll:ll(H)