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An operator ideal $\mathfrak J$ is a class of continuous operators. Namely, for every pair of complex Banach spaces, $\mathfrak X,\mathfrak Y$, we have that $\mathfrak J(\mathfrak X,\mathfrak Y) \subseteq \mathfrak L(\mathfrak X,\mathfrak Y)$ is a closed two-sided ideal, which means \begin{align*} 1.& \ \ \ A,B\in \mathfrak J(\mathfrak X,\mathfrak Y) \Rightarrow A+B \in \mathfrak J(\mathfrak X,\mathfrak Y), \\ 2.& \ \ \ \mathfrak L(\mathfrak W,\mathfrak X)\mathfrak J(\mathfrak X,\mathfrak Y)\mathfrak L(\mathfrak Y,\mathfrak Z) \subseteq \mathfrak J(\mathfrak W,\mathfrak Z),\ \textrm{and} \\ 3.& \ \ \ \mathfrak J(\mathfrak X,\mathfrak Y) \supseteq \mathfrak F(\mathfrak X,\mathfrak Y), \ \textrm{the finite-rank operators}. \end{align*}

Now for my question: Let $\mathfrak X$ be any complex Banach space and suppose $J$ is a non-trivial closed two-sided ideal of $\mathfrak L(\mathfrak X)$. Can you always find an operator ideal $\mathfrak J$ such that $\mathfrak J(\mathfrak X,\mathfrak X) = J$?

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    $\begingroup$ Shouldn't the middle term in (2) be the set of bounded linear operators between the spaces? $\endgroup$ Oct 14, 2015 at 18:00
  • $\begingroup$ Yes, it seems you are correct. I was going on my memory of a talk I went to yesterday. I am certainly no expert on these topics. $\endgroup$ Oct 14, 2015 at 18:08
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    $\begingroup$ Can't you just take $\mathfrak{I}(\mathfrak{X},\mathfrak{X})=J$ and $\mathfrak{I}(\mathfrak{Y},\mathfrak{Z})=\emptyset$ for all $\mathfrak{Y},\mathfrak{Z} \neq \mathfrak{X}$? $\endgroup$ Oct 14, 2015 at 18:41
  • $\begingroup$ @ChrisHeunen Yes, it seems you are correct! $\endgroup$ Oct 14, 2015 at 19:18
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    $\begingroup$ An operator ideal is usually assumed in the literature to be non-trivial in the sense that it contains all the (continuous) finite rank operators (it is probably enough to have a single rank 1 operator and then obtain all finite rank operators using the other axioms). But you also want the second condition of your definition to be $\mathfrak{L} (\mathfrak{W} ,\mathfrak{X}) \mathfrak{I} (\mathfrak{X} ,\mathfrak{Y}) \mathfrak{L}(\mathfrak{Y},\mathfrak{Z}) \subseteq \mathfrak{I} (\mathfrak{W} ,\mathfrak{Z}) $. $\endgroup$ Oct 14, 2015 at 20:05

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The answer is yes; take

$$ \mathfrak{I}(\mathfrak{Y},\mathfrak{Z} ) = {\rm span}\{ T \in \mathfrak{L}(\mathfrak{Y},\mathfrak{Z}) \mid \exists U \in \mathfrak{L}(\mathfrak{Y},\mathfrak{X}) , \exists V \in \mathfrak{L}(\mathfrak{X},\mathfrak{Z}) , \exists S \in \mathfrak{I}(\mathfrak{X}) , T= VSU \} $$

More generally, a 2003 survey article by Laustsen and Loy on closed ideals in $\mathfrak{L}(\mathfrak{X})$ defines the operator ideal generated by a set of operators; they comment that their definition has its roots in wotk of Porta.

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