1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

3.2. Analytic preliminaries 69

if we take unit vectors hn E 1-l and kn E JC such that llSll = lim llShnll and

llTll = lim llTknll, then hn ®kn E 1-l 0 Kare still unit vectors and

ll(S ® T)hn ® knll = ll(Shn) ® (Tkn)ll = llShnllllTknll-+ llSllllTll·

This completes the proof. D

It is a simple, though slightly tedious, exercise to show that the ten-
sor product operators above satisfy the same tensor calculus as elementary
tensors in an algebraic tensor product. Moreover, the adjoint and mul-
tiplication are compatible as well. It would be tempting to think that
JB(1-l) 0 JB(JC) and JB(1-l ® K) are isomorphic algebras but sadly this isn't
quite correct (though "up to weak closure" it is). However, we do have
identifications JB(1-l) ~ JB(1-l) ® Cl]( and JB(K) ~ Clh'. ® B(JC) and, since
JB(1-l) ®Cl](, Clh'. ® B(JC) c JB(1-l ® K) are commuting subalgebras, Propo-
sition 3.1.17 provides us with a natural *-homomorphism

JB(1-l) 0 lB(JC) --+ JB(1-l ® K), L Si® Ti f-+ L Si® Ti.

i i

The following corollary is immediate from these remarks.

Corollary 3.2.4 (Tensor product morphisms). Given two -representations
1TA: A--+ JB(1-l) and 1f"E: B--+ JB(K), there is an induced -representation

1fA 0 1fB; A 0 B --+ JB(1-l ®JC)

such that 1fA 01fB(a ® b) = 1fA(a) ® 1fB(b) for all a EA and. b E B.^8

We've noted that if 1fA: A--+ G and 1fB: B--+ Gare -homomorphisms
with commuting ranges, then the product map 1fA x 1fB: A 0 B --+ G is also
a -homomorphism. It is an important fact that in the case G = JB(1-l) every
-homomorphism A0B--+ JB(1-l) arises this way. It turns out that the proof
is completely trivial in the unital case and rather unpleasant in the absence
of units. Hence, if one only cares about unital C -algebras, then spend the
three seconds required to work out Exercise 3.2.4 and skip the rest of this
section.

Lemma 3.2.5. Let 1f: A 0 B --+ JB(1-l) be a *-homomorphism. Then for
each fixed a E A, the linear map
B--+ JB(1-l), b r-+ 7r(a ® b)
is bounded.

8Note that Q9 has different meanings on the left and right hand sides of 1fA 0 7rB(a Q9 b) =
7rA(a)@ 1fB (b). Of course, if you didn't notice this abuse of notation in the paragraph preceding
this corollary, then it probably won't matter here either.