When is the multiplication map of the algebraic tensor product of C*-algebras injective?

Question

A classic result, of Murray and Von Neumann I believe, is that if $\mathcal M\subseteq B(H)$ is a factor then the $*$-homomorphism $\pi : \mathcal M \odot \mathcal M' \rightarrow B(H)$ given by $\pi(x_1\otimes x_2) = x_1x_2$ on simple tensors is injective. On the other hand, if $A \subseteq B(H)$ is a commutative C*-algebra then this same multiplication map from $A\odot A \rightarrow B(H)$ is clearly not injective.

Question: For $A,B\in B(H)$ commuting C*-algebras, when is the multiplication map $A\odot B\rightarrow B(H)$ injective?

Are there any conditions on one or both of the C*-algebras that ensure this? E.g. $A$ simple? I've been looking but cannot find anything.

Answer 1

For a commuting pair of C*-algebras $A,B\subset B(H)$, the multiplication map is injective if and only if there are no nonzero elements $a\in A$ and $b\in B$ with $ab=0$. In particular, if $A$ is simple (and non-degenerate on $H$), then the multiplication map is injective.

This is because the pure states on C*-algebras are "excised"; for any pure state $\phi$ on $A$, there is a net $e_i$ in $A$ with $0\le e_i\le 1$ and $\|e_i\|=1$ that satisfies $\|e_iae_i - \phi(a)e_i^2\|\to0$ for all $a\in A$. It follows that any pure states on $\phi$ on $A$ and $\psi$ on $B$ give rise to a state $\phi\times\psi$ on $C^*(A,B)$ as long as $ab\neq0$ for any nonzero $a\in A_+$ and $b\in B_+$ (which immediately implies $\|ab\|=\|a\| \|b\|$ for every pair $(a,b)\in A_+\times B_+$ by functional calculus). See "Another proof of Proposition 3.4.7" in my book with Brown (p. 82).

Also, at the algebraic level, it is known that if $A$ is a central simple algebra (any simple C*-algebra is central simple), then every ideal in the algebraic tensor product $A \otimes B$ (over the relevant field) is of the form $A \otimes I$ for some ideal $I$ in $B$. See e.g., Drozd & Kirichenko "Finite Dimensional Algebras" Theorem 4.3.2.