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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

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3  
The answer is YES, but I'd like to know who is studying $Z^*$-algebras (I haven't heard of this notion) and why. My google search found only one paper dealing with $Z^*$-algebras, and that was yours. –  Narutaka OZAWA Jun 27 at 8:21
    
@NarutakaOZAWA Thank you very much for your comment. As you pointed out, I wrote a note "A note on $Z^{*}$ algebra" in arxive. In that note I essentially used your very valuable answer "mathoverflow.net/questions/136345/…;. After that I presented that note in arxive, I received an email from an specialist who said me this notion is related to subject "Ring of quotient of a ring" which had been investigated by professors A.I Singh, Pere Ara, Martin Matiues, etc. any way thanks again for your comment and I would appreciate if you ... –  Ali Taghavi Jun 27 at 12:31
    
...give your positive answer to this question. In my note on $Z^{*}$ algebra I observed that there is a pathology in extension theory. So I am very interested to find an example of two $Z^{*}$ algebra $A$ and $B$ such that for every extension $0\to A\to C\to B\to 0$, we necessarily have $C$ is a $Z^{*}$ algebra. –  Ali Taghavi Jun 27 at 12:35

1 Answer 1

up vote 4 down vote accepted

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.

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thank you very much for your very interesting answer. In the proof of the lemma where did you used "completely" that is $M_{n}(A)$? –  Ali Taghavi Jun 27 at 22:42
    
One needs complete positivity for well-definedness of $\phi_1\otimes\phi_2$. –  Narutaka OZAWA Jun 27 at 23:18
    
thanks again for your beautiful argument. –  Ali Taghavi Jun 27 at 23:55
    
according to your last comment, what is an example of two bounded linear maps $\phi_{i}:A_{i}\to B_{i}\;\;,i=1,2$ such that $\phi_{1}\otimes \phi_{2}$ is not a bounded operator on minimal tensor product $A_{1} \otimes_{min} A_{2}$? –  Ali Taghavi Jun 29 at 10:24
1  
The most typical example is the transpose map $T$ on $B(\ell_2)$, which is positive but not completely positive and $T\otimes\mathrm{id}_{B(\ell_2)}$ is unbounded. See e.g., mathoverflow.net/questions/86550/… –  Narutaka OZAWA Jun 29 at 10:37

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