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.

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

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.

  • $\begingroup$ 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? $\endgroup$ – hyperkahler Jun 5 '17 at 16:32
  • $\begingroup$ 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. $\endgroup$ – hyperkahler Jun 5 '17 at 17:18
  • 1
    $\begingroup$ 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 $B$. $\endgroup$ – user42761 Jun 5 '17 at 20:48
  • $\begingroup$ Got it now, thanks! $\endgroup$ – hyperkahler Jun 5 '17 at 22:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.