/ 3.1. Algebraic tensor products 65
We complete the *-algebra structure on A0B by defining multiplication.
As with the involution we know how it must go; the proof amounts to
showing it is well-defined.
Proposition 3.1.15 (Multiplication). The tensor product A0B has a mul-
tiplication defined by( ~ ai ® bi) ( ~ Cj ® dj) = ~ aiCj ® bidj.
i J i,J
To prove that this multiplication works, we first consider L(A 0 B), the
vector space of all linear maps A 0 B --+ A 0 B. If Ma : A --+ A is left
multiplication by a E A and Mb: B --+ B is left multiplication by b E B,
then, thanks to Proposition 3.1.4, for every pair (a, b) EA x B we have the
tensor product map Ma® Mb E L(A 0 B). It is routine to check that
AX B--+ L(A 0 B), (a, b) H Ma® Mb
is a bilinear map. By universality there is a linear map M: A0B--+ L(A0B)
such that M(a ® b) =Ma® Mb. Finally one checks that the bilinear map
A0B x A0B--+ A0B, (x,y) H M(x)y
defines our multiplication.
We close this section with two simple consequences of the existence of
product and tensor product maps (Propositions 3.1.4 and 3.1.5). The proofs
are straightforward calculations.
Proposition 3.1.16 (Tensor product morphisms). Given *-homomorphisms
cp: A --+ C and '¢: B --+ D, the tensor product map cp 0 '¢: A 0 B --+ C 0 D
is also a *-homomorphism.
Proposition 3.1.17 (Product morphisms). Given two *-homomorphisms
1fA: A--+ C and 1fB: B--+ C with commuting ranges (i.e., [7rA(a), 7rB(b)] =
0 for all a E A, b E B ), the product map 1fA x 1fB: A 0 B --+ C is also a
*-homomorphism.
Exercises
Exercise 3.1.1. Observe that_if C and Dare both unital, then Proposition
3.1.16 is a special case of Proposition 3.1.17. How about in the nonunital
case?
Exercise 3.1.2. Justify the following identifications (which, by the way, get
used all of the time): A ~ A 0 C ~ A 0 Cl B C A 0 B.
Exercise 3.1.3. Prove that if A is a C*-algebra, then for any n and any
choice of matrix units {ei,j}i,j=l C Mn(C) th~re is a *-algebra isomorphism