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Let $A$ and $B$ be Banach algebras. Consider a right Banach $A$-module, $E$, and a right Banach $B$-module, $F$, as well as a Banach algebra morphism $\pi\colon A\to\mathcal L_B(F)$ into the bounded $B$-linear operators on $F$. Then $\pi$ makes $F$ into a left Banach $A$-module, and the actions of $A$ and $B$ commute. Thus, the (completed) tensor product $E\otimes_AF$ has the structure of a right Banach $B$-module.

Question: Is the tensor product of two compact operators again compact?

To be more precise, let $F\in\mathcal K_A(E)$ and $G$ be both $A$- and $B$-linear. Is $F\otimes_AG\in\mathcal K_B(E\otimes_AF)$? As for Hilbert C*-modules, the compact operators on a Banach module are the closure of the linear span of the finite rank operators. Thus, it suffices to consider the case where $F=F_1F_2$ with $F_1\in\mathcal L_A(A,E)$, $F_2\in\mathcal L_A(E,A)$, and where $G=G_1G_2$ with $G_1\in\mathcal L_B(B,F)$, $G_2\in\mathcal L_B(F,B)$. However, the tensor product $F_i\otimes_AG_i$ is not defined since $B$ itself does not carry an $A$-module structure.

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    $\begingroup$ I am not sure what a "compact" or "finite-rank" operator should mean here. For a Hilbert $C^*$-module, the "finite-rank" operators are defined using the $A$-valued inner product: they are very rarely finite-rank when considered just as linear maps. $\endgroup$
    – Matthew Daws
    Commented Mar 25, 2019 at 8:44
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    $\begingroup$ An operator in $\mathcal L_A(E,F)$ is called rank one if it can be written as the composition of an operator in $\mathcal L_A(E,A)$ and an operator in $\mathcal L_A(A,F)$. Finite rank operators are the linear span of the rank one operators, and compact operators are the closure of the finite rank operators. $\endgroup$
    – Benedikt Hunger
    Commented Mar 25, 2019 at 9:16
  • $\begingroup$ Thanks! Sorry, one more question: is $E\otimes_A F$ the projective tensor product of $E$ and $F$, quotiented by the "$A$-balanced" tensors, that is, closed span of things like $x\cdot a\otimes y - x\otimes a\cdot y$? $\endgroup$
    – Matthew Daws
    Commented Mar 25, 2019 at 10:53
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    $\begingroup$ In fact, for Banach algebras one can explicitly write down a semi-norm for the algebraic tensor product $E\otimes_A^{\mathrm{alg}}F$: For $x\in E\otimes_A^{\mathrm{alg}}F$ put $\|x\|=\inf\{\sum_i\|e_i\|\|f_i\|:e_i\in E,f_i\in F,x=\sum_ie_if_i\}$. Then $E\otimes_AF$ is the completion of $E\otimes_A^{\mathrm{alg}}F$ with respect to this semi-norm. I am not sure if this is actually the same thing as what you are suggesting. $\endgroup$
    – Benedikt Hunger
    Commented Mar 25, 2019 at 11:59
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    $\begingroup$ Sorry to ask one more question: how do you define $T\otimes_A S$ in general, for $T\in\mathcal{L}_A(E)$ and $S\in\mathcal{L}_B(F)$? I don't understand why this is well-defined on $E \otimes^{\text{alg}}_A F$: it seems like we would need $S$ to be a right $A$-morphism? $\endgroup$
    – Matthew Daws
    Commented Mar 25, 2019 at 15:28

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