A question on $Z^{*}$ algebras

Question

A $Z^{*}$ algebra is a $C^{*}$ algebra which satisfies each of the following equivalent conditions:

1. All elements of $A$ are left zero divisor.

2. All elements are right zero divisor.

3. All elements are two sided zero divisors

4. All positive elements are two sided zero divisor.

The commutative(topological) interpretation for this concept is the following:

For a locally compact Hausdorff space $X$, $C_{0}(X)$ is NOT a $Z^{*}$ algebra if and only if $X$ is an approximately $\sigma$- compact space (Briefly $A\sigma C$ space). that is, there are a sequence of compact subsets $K_{n}$ of $X$ such that $\cup K_{n}$ is dense in $X$. Note that the product of two $A\sigma C$ spaces is a $A\sigma C$ space. Equivalently if $X \times Y$ is not a $A\sigma C$ space then $X$ or $Y$ is not a $A\sigma C$ space. This is a motivation to ask:

Assume that $A$ and $B$ are two $C^{*}$ algebras such that their minimal tensor product is a $Z^{*}$ algebra. Is it true to say that $A$ or $B$ is a $Z^{*}$ algebra?

Note that if the answer to the following question were affirmative, then the answer to the above question would be affirmative, too:

positive element in C* tensor product

Answer 1

Recall that a completely positive map $\phi\colon A\to B$ is said to be faithful if $\phi(a) \neq 0$ for all nonzero $a \in A^+$.

Lemma: If $\phi_i\colon A_i\to B_i$ are faithful cp maps, then $\phi_1\otimes\phi_2\colon A_1\otimes_{\min}A_2\to B_1\otimes_{\min}B_2$ is faithful. Proof: Since $\phi_1\otimes\phi_2=(\mathrm{id}_{B_1}\otimes\phi_2)\circ(\phi_1\otimes\mathrm{id}_{A_2})$, we may assume one of $\phi_i$ (say $\phi_2$) is $\mathrm{id}$. Let a nonzero $a \in (A_1\otimes_{\min}A_2)^+$ be given. By definition of the minimal tensor product, there are states $f_i$ on $A_i$ such that $(f_1\otimes f_2)(a) > 0$. Observe that $(f_1\otimes f_2)(a) = f_1((\mathrm{id}_{A_1}\otimes f_2)(a))$, where $\mathrm{id}_{A_1}\otimes f_2\colon A_1\otimes A_2\to A_1\otimes{\mathbb C}\cong A_1$ is the slice map. Hence, $(\mathrm{id}_{A_1}\otimes f_2)(a) \in A^+\setminus\{0\}$ and $$(\mathrm{id}_{B_1}\otimes f_2)((\phi_1\otimes\mathrm{id}_{A_2})(a)) = \phi_1((\mathrm{id}_{A_1}\otimes f_2)(a)) \neq 0,$$ which implies $(\phi_1\otimes\mathrm{id}_{A_2})(a) \neq 0$. $\Box$

Now, if $A_i$ are non-$Z^*$ $\mathrm{C}^*$-algebras, then there are $a_i\in A_i$ which are not left zero divisors. This means that the completely positive maps $\phi_i\colon A_i\ni x\mapsto a_i^*xa_i \in A_i$ are faithful. By the above lemma, $\phi_i\otimes\phi_2$ is faithful, which means that $a_1\otimes a_2$ is not a left zero divisor.