# Commutativity of the spatial tensor product

Let $A, B$ be two arbitrary $C^{*}$-algebras (not neccesary unital).

Let also $A \otimes B$ be an algebraic tensor product, whilst $A \otimes_{\min} B$ stands for the minimal tensor product, which is constructed as follows: if $\pi_{A}$ and $\pi_{B}$ are two representations of $A, B$ on Hilbert spaces $\mathcal{H_{1}}, \mathcal{H_{2}}$ respectively, then one can construct the map $\pi(a \otimes b) = \pi(a) \otimes \pi(b)$.

Furthermore, if one endows $A \otimes B$ with the norm $\| \cdot \|$ defined as $$\| \sum_{i}{a_{i} \otimes b_{i}} \| = \sup \| (\pi_{A} \otimes \pi_{B})(\sum_{i}{a_{i} \otimes b_{i}}) \|$$ where the supremum is taken over the all representations $\pi_{A}$ of $A$ and $\pi_{B}$ of $B$, then the completition of the algebraic tensor product w.r.t to the given normed is called a spatial tensor product of star-algebras.

I would like to show that in fact there exists a $*$-isomorphism $A \otimes_{min} B = B \otimes_{min} A$. In fact, the very first candidate to be such a map is $a \otimes b \mapsto b \otimes a$.

How to rigorously check that the map above is indeed a $*$-isomorphim?

(i can show that for a given non-degenerate representation $\pi$ of $A \otimes B$ on $\mathcal{H}$ there exist $\pi_{A}$, $\pi_{B}$ such that $\pi(a \otimes b) = \pi_{A}(a) \otimes \pi_{B}(b) = \pi_{B}(b) \otimes \pi_{A}(a)$ $-$ in unital case it is clear, in a non-unital one it is clear as well via the construction using approximaing units.

$$\newcommand{\ra}{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} A \otimes B & \ra{a \otimes b \mapsto b \otimes a} & B \otimes A \\ \da{\pi} & & \da{???} \\ B(\mathcal{H}) & \ra{\pi_{A} \otimes \pi_{B} \mapsto \pi_{B} \otimes \pi_{A}} & B(\mathcal{H}) \\ \end{array}$$

Looks as if it possible to obtain $a \otimes b \mapsto b \otimes a$ as a composition of three underlying arrows but i can't figure out an easy way to do this.

As you say, there is a $*$-homomorphism $$\phi: A \otimes B \to B \otimes_{\mathrm{min}} A.$$ This morphism is indeed bounded with respect to the minimal norm on $A \otimes B$:
\begin{align*} \left \lVert \phi \left (\sum_{i=1}^n a_i \otimes b_i \right ) \right \rVert_{\mathrm{min}} & = \left \lVert \sum_{i=1}^n b_i \otimes a_i \right \rVert_{\mathrm{min}} = \left \lVert \sum_{i=1}^n \pi_B(b_i) \otimes \pi_A(a_i) \right \rVert_{\mathbb B(H_B \otimes H_A)} \\ & = \left \lVert \sum_{i=1}^n \pi_A(a_i) \otimes \pi_B(b_i) \right \rVert_{\mathbb B(H_A \otimes H_B)} \\ & = \left \lVert \sum_{i=1}^n a_i \otimes b_i \right \rVert _{\mathrm{min}} \end{align*} So in fact, $\phi$ is isometric. This shows that we get a $*$-homomorphism $$\phi : A \otimes_{\mathrm{min}} B \to B \otimes_{\mathrm{min}} A$$ which is isometric and clearly has dense image. That means that $\phi$ is an isomorphism.
Of course, I cheated a bit by using that $$\mathbb B(H_A \otimes H_B) \cong \mathbb B(H_B \otimes H_A),$$ where $\cong$ means isometric isomorphism of Banach spaces.
• Thanks! Btw, is it OK, to show that the spatial norm endows $A \otimes_{min} B$ with the structure of $C^{*}$-algebra and then to prove that any injective $*$-homomorphism of $C^{*}$ algebras is in fact an isometry? – hyperkahler Jun 5 '17 at 16:32
• Also, as far as i got, it is crucial for $\pi_{A}$ and $\pi_{B}$ to be faitful in order to make the second equality true. – hyperkahler Jun 5 '17 at 17:18
• This are all standard facts, i.e. that the closure of the algebraic tensor product in the minimal norm is a C*-algebra and that injective *-homomorphisms are isometric. And, by definition of the minimal norm, you indeed consider faithful representations of $A$ and $B$. – user42761 Jun 5 '17 at 20:48