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Is there a name for the following property of a $C^{*}$ algebra $A$?

$$A \simeq A \otimes A\;\;\; \text{The spatial tensor product}$$

Example of this situation is $A=C(X)$ where $X$ is the Cantor set or $A=\mathcal{K}$ where $\mathcal{K}$ is the algebra of compact operators on a separable Hilbert space.

For any such $C^{*}$ algebra, after fixing an isomorphisms between the two algebras, one can consider the following functional equation

$$T(a\otimes b)=T(a) \otimes T(b), \;\;\; T(a^{*})=(T(a))^{*}$$ where $T$ is a linear operator on $A$.

Does this imply that $T$ is a bounded operator? Is there a non scalar example of such $T$ for $A=\mathcal{K}$, the algebra of compact operators?

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    $\begingroup$ Don't you think this could depend on the choice of isomorphism between $A$ and $A \otimes A$? $\endgroup$
    – Will Sawin
    Commented Apr 30, 2016 at 18:25
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    $\begingroup$ The property $A \cong A \otimes A$ is called self-absorbing. So-called strongly self-absorbing $C^*$-algebras have played a cornerstone role in the classification program. $\endgroup$
    – Ulrich Pennig
    Commented Apr 30, 2016 at 20:04
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    $\begingroup$ btw. the only scalars that satisfy your condition are $0$ and $1$, since the first equation gives you $\lambda = \lambda^2$. $\endgroup$
    – Ulrich Pennig
    Commented Apr 30, 2016 at 20:07
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    $\begingroup$ How do you define $T\otimes T$ on $A\otimes A$ without using boundedness in the first place? Remember that $A\otimes A$ is a completion of the algebraic tensor product and therefore you need some form of continuity to extend operators from the algebraic to the spatial tensor product. $\endgroup$
    – Johannes Hahn
    Commented Apr 30, 2016 at 20:11
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    $\begingroup$ @AliTaghavi Well, yes for simple tensors this is a valid definition. But then you have only defined $t\otimes T$ on the algebraic tensor product which is a dense subspace of the/any C*-algebra-tensor product. You need some form of continuity to extend this maps from the subspace to the whole space. $\endgroup$
    – Johannes Hahn
    Commented May 2, 2016 at 19:53

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