Questions tagged [operator-ideals]
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Concrete example of non-nuclear operator $E \to F$ and isometry $F \hookrightarrow G$ so that the composition $E \to F \hookrightarrow G$ is nuclear
DISCLAIMER: I posted the same question a week ago on Mathematics Stack Exchange.
We know by an abstract argument that there exist Banach spaces $E$, $F$, $G$ and maps $E \to F \hookrightarrow G$ such ...
5
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quasi-weakly compact operators, co-ideals of operator ideals, and Banach spaces $X$ with $X^{**}/X$ separable
Throughout, $X$ and $Y$ will denote Banach spaces with $T\in\mathcal{L}(X,Y)$ (the space of continuous linear operators between $X$ and $Y$). We define the operator $\overline{T}\in\mathcal{L}(X^{**}/...
6
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Is every ideal part of an operator ideal?
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) \...
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1
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Noncommutative analogs of classical Banach geometric properties
The scale of Schatten-von Neumann classes is noncommutatitve analog of classical $\ell_p$-spaces. A lot of researchers devoted their lives to study Banach geometric structure of these spaces. ...
6
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2
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Two-sided ideals of $B(H)$ are hereditary
I seem to recall that (not necessarily closed) two-sided ideals of $B(H)$ are hereditary. Is that true?
If it is, can anyone post a proof/reference?
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Second duals of Grothendieck spaces
The classical example of a Grothendieck space is $\ell_\infty$. It is also known that its even duals $\ell_\infty^{**}$, $\ell_\infty^{(4)}$, $\dots$ are Grothendieck spaces.
(See, e.g., this note ...
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1
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Are these ideals in rings of operators on Hilbert space unique?
Suppose that, for every Hilbert space $H$, we have a subset $I(H) \subseteq B(H)$ of bounded linear operators on $H$, and that together all $I(H)$ form a two-sided ideal, in the sense that whenever $h ...
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2
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Factorization through $\ell_{1}$ and operator ideals
Recently, I bumped into the class of operators that factor through $\ell_{1}(X)$ for some set $X$. For now, $X$ is a set with arbitrary cardinality but if it leads to a more concrete answer to my ...