# Morita equivalence for operator algebras and tensor products, question about proof

This is a bit of a dumb question I know, but I was reading "Morita equivalence for C*-algebras and W*-algebras" by Rieffel, in this section about Morita equivalences and how they relate to forming tensor products of $W^*$-algebras (page 90) I came across this proof that I just can't seem to grasp and was seeking clarification about a few things.

On page 90 there's this proposition:

8.5. Proposition. Let $M$, $M_1$, $N$, $N_1$ be $W^*$-algebras. Let $X$ be a normal $N$-rigged $M$-module and $Y$ a normal $N_1$-rigged $M_1$-module. Then the algebraic tensor product $X \otimes Y$ over the complex numbers , with $N \otimes N_1$-valued inner product defined by

$\langle x \otimes y,x' \otimes y' \rangle_{N \otimes N_1} = \langle x,x' \rangle_{N} \otimes \langle y,y' \rangle_{N_1}$

and completed for the usual norm, is a normal $N \otimes N_1$-rigged $M \otimes M_1$-bimodule. If $X$ and $Y$ are in fact equivalence bimodules and if an $M \otimes M_1$-valued inned product is defined by

$\langle x \otimes y , x' \otimes y' \rangle_{M \otimes M_1} = \langle x,x' \rangle_{M} \otimes \langle y,y' \rangle_{M_1}$,

then $X \otimes Y$ (completed) becomes and equivalence module.

The proof focuses on proving that the inner products are positive and the "rest of the proof of this proposition is carried out by routine calculations", which, as it were, I'm stuck with, pardon my lack of familiarity with the whole subject but I just have to ask:

• What is this "usual norm" they're referring to? I've seen a tensor product of $W^*$-algebras be referred to as the completion of the algebraic tensor product $X \otimes Y$ with respect to certain norm (don't know if that's Sakai's definition they refer to in the paper).

• How does the range of $\langle - , - \rangle_{M \otimes M_1}$ and $\langle - , - \rangle_{N \otimes N_1}$ span weakly dense subsets on $M \otimes M_1$ and $N \otimes N_1$ respectively?

## 1 Answer

To answer your first question, I think the "usual norm" is the one described on page 63 of Rieffel's paper, i.e. $\lVert z \rVert = \lVert \langle z,z\rangle\rVert^{1/2}$ for $z \in X \otimes Y$. Note that the inner product takes values in a $W^*$-algebra. Depending on your point of view, a $W^*$-algebra is either the same as a von Neumann algebra (i.e. comes along as a weakly closed *-subalgebra of $B(H)$ for some Hilbert space $H$), then the norm on the right hand side of the last equation is the operator norm of $B(H)$. Or $W^*$-algebra refers to the abstract concept of a $C^*$-algebra, which has a predual. In this case the norm is the norm as a $C^*$-algebra. By choosing a faithful representation, you can switch from the latter point of view to the first, so it doesn't really make a difference.

In your case the inner product takes values in the spatial tensor product of $W^*$-algebras $N \otimes N_1$. I will suppose that $N \subset B(H)$ and $N_1 \subset B(K)$ for Hilbert spaces $H$ and $K$ as weakly closed $*$-subalgebras. Then the spatial tensor product is given by forming the tensor product $H \otimes K$ of the two Hilbert spaces. Note that $B(H) \otimes_{alg} B(K) \subset B(H \otimes K)$. In particular, $N \otimes_{alg} N_1 \subset B(H \otimes K)$ and $N \otimes N_1$ is the weak closure of $N \otimes_{alg} N_1$, which agrees with the double commutant $(N \otimes_{alg} N_1)''$. By definition the algebraic tensor product is weakly dense in $N \otimes N_1$.

The commutant of any subset $S \subset B(H)$ is weakly closed, therefore we have $(N \otimes_{alg} N_1)' = (N \otimes N_1)'$. Now look at the tensor product on $X \otimes Y$ defined by $$\langle x \otimes y, x' \otimes y' \rangle_{N \otimes N_1} = \langle x, x'\rangle_N \otimes \langle y, y' \rangle_{N_1}$$ and let $\widetilde{N} = \{ \langle x,x' \rangle_N | x, x' \in X \}$ and $\widetilde{N}_1 = \{ \langle y,y' \rangle_{N_1} | y, y' \in Y \}$. We know that $\widetilde{N} \subset N$ and $\widetilde{N}_1 \subset N_1$ are weakly dense by assumption. Anything that commutes with $\widetilde{N}$, therefore also has to commute with $N$ and likewise for $N_1$. Therefore $$(\widetilde{N} \otimes_{alg} \widetilde{N}_1)' = (N \otimes_{alg} N_1)' = (N \otimes N_1)'$$ which should answer your second question. I hope I didn't mess it up somewhere.

• I thought I had accepted your answer, so rude of me, thanks so much for taking the time to answer. – Samuel M Jun 6 '14 at 18:54