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Our following question is motivated by this very interesting answer

Assume that $A$ is a $C^{*}$ algebra. Put $X=\{a\otimes b \mid a,b \in G(A)\}$ where $G(A)$ is the space of all invertible elements of $A$.

Assume that there are two continuous maps $f,g: X \to A$ such that $ x=f(x)\otimes g(x),\;\;\;\forall x \in X$.

Does this imply that $A$ is a finite dimensional commmutative algebra?

Note: After a few modification the above question can be generalized to the category of Banach spaces or other similar categories.

For example on can consider the following:

"Classification of all Banach space $V$ such that for every tensor product norm with $\parallel x\otimes y \parallel= \parallel x \parallel. \parallel y \parallel$ we have two continuous maps $f,g$ from the space of simple tensors to $V$ (or from $V \bar{\otimes} V$ to $V$) with $x=f(x) \otimes g(x)$"

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Well, you don't get "finite dimensional", since this is true for all infinite-dimensional commutative $C^*$-algebras too.

We can suppose $A = C_0(\Omega)$ for some locally compact Hausdorff $\Omega$. Then $X$ is identified with an appropriate subset of $C_0(\Omega^2)$: those $x : \Omega^2 \to \mathbb{C}$ which can be written $x(\omega_1, \omega_2) = a(\omega_1) b(\omega_2)$ where $a,b \in C_0(\Omega)$ are invertible, i.e. nowhere vanishing.

Fix some $\omega_0 \in \Omega$ and put $$f(x)(\omega) = \frac{x(\omega, \omega_0)}{\sqrt{x(\omega_0, \omega_0)}}, \qquad g(x)(\omega) = \frac{x(\omega_0, \omega)}{\sqrt{x(\omega_0, \omega_0)}}.$$ Then $f,g : X \to A$ are continuous, and for any $x \in X$, by writing $x = a \otimes b$, it's easy to check that you have $f(x) \otimes g(x) = x$.

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  • $\begingroup$ Very elegant answer. Thank you very much. Do you think that for $A=B(H)$, such maps f,g do not exist? $\endgroup$ Commented Jan 2, 2017 at 19:13
  • $\begingroup$ @AliTaghavi: I have not thought about it and may not have a chance to do so. $\endgroup$ Commented Jan 2, 2017 at 19:18

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