1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

3.2. Analytic preliminaries 67

form, it is not complete (unless either 1i or K is finite dimensional) in the

corresponding norm. Of course, this isn't a real problem.

Proposition 3.2.1 (Tensor product of Hilbert spaces). If 1i and K are

Hilbert spaces, then 1i 0 K is a pre-Hilbert space with respect to the inner

product

c2:. hi&! ki, L hj &! kj) = L(hi, hj)(ki, kj).

i j i,j

The Hilbert space completion will be denoted 1i &! K.

Proof. First we must show that there is a sesquilinear map

(-, ·): 1i 0 K x 1i 0 K-+ C

satisfying the formula given above. The proof is similar to the proof of
Proposition 3.1.15 so we only sketch the main points. First, let (7-i 0 K),c
denote the vector space of conjugate linear functionals on 1i 0 K. By Corol-
lary 3.1.7, for each pair (h, k) E 1i x K there is an element f(h,k) E (7-i0K),c
such that

f(h,k) (~Vi &J Wi) = ~ (h, Vi) (k, Wi)

i i

for all 'l:::i Vi&!Wi E 7-i0K. One checks that the mapping 1i x K-+ (7-i0K),c,
(h, k) f-+ f(h·,k), is bilinear and thus induces a linear map M: 1i 0 K -+
(7-i 0 K),c. We then define the desired sesquilinear form by

(v, w) = M(v)w

for all v, w E 1i 0 K.

That we have a positive definite form on 1i 0 K follows from the calcu-
lation

(Lei&!ki,Lei&!ki) = L(ei,ej)(ki,kj) = Lilkill^2 ,

i i i,j i

whenever { ei} c 1i is an orthonormal set of vectors. D

The following fact is a nice exercise. It does not follow from Corollaries
3.1.9 and 3.1.10, but the proof is similar.

Proposition 3.2.2 (Orthonormal bases and vector representations). Sup-
pose that {vihEI C 1i and {wj}jEJ CK are orthonormal bases. Then

{Vi &J Wj }(i,j)EixJ C 1{ &J /(,