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A $Z^*$ algebra is a $C^*$ algebra whose all positive elements are zero divisor.
They form a category with usual structures.

Question. Is this category equivalent to the category of $C^*$ algebras?

The equivalency means in terms of natural transformation between functors.
If the answer would be affirmative then the study of (category of) $C^*$ algebras would be equivalent to study of (category of) $Z^*$ algebras.

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    $\begingroup$ Where does this definition come from? Note that $C(X)$ is not a $Z^*$-algebra for any compact X, and also $C_0({\mathbb R}^n)$ is not a $Z^*$-algebra. Unital $C^*$-algebras are not $Z^*$-algebras. $\endgroup$
    – Yemon Choi
    Commented May 24, 2023 at 3:11
  • $\begingroup$ @YemonChoi Yes algebras you mentioned are not $Z^*$ algebra. But a commutative algebra $C_0(X)$ is a $Z^*$ algebra iff $X$ is not approximately sigma compact $\endgroup$
    – Ali Taghavi
    Commented May 24, 2023 at 6:27
  • $\begingroup$ A space X is approximately sigma compact if it has a dense sigma compact subspace. For exqmple the long line is not approximately sigma compqct $\endgroup$
    – Ali Taghavi
    Commented May 24, 2023 at 6:29
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    $\begingroup$ Which category of $C^*$-algebras do you mean? $\endgroup$
    – David Roberts ♦
    Commented May 24, 2023 at 10:00
  • $\begingroup$ I mean the usual category of all C^* algebra(including unital one $\endgroup$
    – Ali Taghavi
    Commented May 29, 2023 at 17:31

1 Answer 1

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There is at least an easy faithful functor from the category of C${}^*$-algebras into the category of Z${}^*$-algebras. Fix an uncountable set $X$ and map the C${}^*$-algebra $A$ to the Z${}^*$-algebra $C_0(X;A)$. A morphism $f:A \to B$ would go to the morphism $(x_i)_{i \in X} \mapsto (f(x_i))_{i \in X}$. However, this can take non-isomorphic C${}^*$-algebras to isomorphic Z${}^*$-algebras (e.g., any time $A$ and $A \oplus A$ are not isomorphic).

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    $\begingroup$ Thank you and +1 very much for your answer $\endgroup$
    – Ali Taghavi
    Commented May 29, 2023 at 17:30
  • $\begingroup$ Do you mean that $C_0(X,A\oplus A)\sim C_0 (X\times X,A)$ hence isomorphic to $C_0(X,A)$ since $X$ homeomorphic to $X\times X$? Or you mean some things else? $\endgroup$
    – Ali Taghavi
    Commented Oct 8, 2023 at 11:11
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    $\begingroup$ Well, $C_0(X, A\oplus A) \sim C(X,A)$ because $X$ is in bijection with $X \coprod X$ (disjoint union of two copies of $X$). We need $X$ to be uncountable so that anything in $C_0(X,A)$ is zero off of a proper (because countable) subset of $X$, and therefore a zero divisor. $\endgroup$
    – Nik Weaver
    Commented Oct 8, 2023 at 21:28
  • $\begingroup$ So your answer is actually $A\mapsto A\otimes C_0(X)$. The tensor product of a C^* algebra and a Z^* algebra is again Z^* algebra. X being uncountable is not an approximately sigma compact. After all I am really curious about this category, the category of Z^* algebras. Please see my conversation with David Roberts about a precise structure for the category of C^* algebras(what kind of morphisms we are consideringt, etc). Any way it would be interesting that the two categories would be isomorphic in a reasonable sense $\endgroup$
    – Ali Taghavi
    Commented Oct 9, 2023 at 8:22

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