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I am interested in the Banach space $\mathcal{K}=\mathcal{K}(\ell^2)$ of compact operators on $\ell^2$, however my questions can be stated for any $\mathcal{K}(E)$, where $E$ is an arbitrary Banach space. I think that everyone who tries to study "classical" operator spaces like $\mathcal{K}$, $p$-Schatten class operators etc. immediately discovers the similarity with "commutative" counterparts, i.e. $c_0$ and $\ell^p$. This phenomenon is visible when one uses (generalised) singular numbers for certain classes of operators. Again, I have got plenty of questions concerning this stuff, let me list at least two of them:

1) what are the complemented subspaces of $\mathcal{K}$? Is $\mathcal{K}$ complemented in $\mathcal{B}(\ell^2)$? Recently, Haydon and Argyros constructed an HI-space $E$ such that $\mathcal{K}(E)$ has codimension 1 $\mathcal{B}(E)$, thus complemented.

2) is every bounded operator from $p$-Schatten class to $\mathcal{K}$ compact?

What other properties $\mathcal{K}$ shares with $c_0$?

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I've removed the "operator spaces" tag, since nowadays this usually refers to operator spaces in the sense of Effros, Ruan et al. – Yemon Choi Jun 8 2011 at 19:24
What if you search MathSciNet for "operator spaces" to see what it commonly refers to in recent years? – Gerald Edgar Jun 8 2011 at 20:17
@Gerald: Well, of the top 40 hits, all but 2 (maybe 3) seem to use the term in Yemon's sense. – Matthew Daws Jun 8 2011 at 20:33

2 Answers

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Probably in (2) you meant to ask whether every bounded operator from $K$ into a Schatten $p$ class is compact, since every operator from $c_0$ into $\ell_p$, $p<\infty$, is compact. But no either way: $K$ and any Schatten $p$ class contain complemented subpspaces isometrically isomorphic to $\ell_2$ (e.g. operators whose matrix representation has zeroes except in the first column).

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You might look at Rosenthal's expository paper ma.utexas.edu/users/rosenthl/pdf-papers/93.pdf which discusses properties of $K$ and other $C^*$ algebras as Banach spaces and as operator spaces. – Bill Johnson Jun 9 2011 at 2:54
That is very useful. Thank you! – Tomek Kania Jun 9 2011 at 16:41
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It is easy to see that whenever a space has an unconditional basis then the space of diagonal operators of the basis is equivalent to $\ell_\infty$. If $c_0$ embeds in $K(X,Y)$ then $K(X,Y)$ is not complemented in $B(X,Y)$. One reference for this is: M. FEDER. On subspaces of spaces with an unconditional basis and spaces of operators. Illinois J. Math. 34 (1980), 196-205.

It is also a direct consequence of a result from a Studia paper of Tong and Wilken from 1971. Here they prove that if $Y$ has an unconditional basis then $K(X,Y)$ is uncomplemented in $B(X,Y)$ (assuming the spaces are not equal).

As far as I know the Argyros-Haydon space is the first example of a space for which it is known that $K(X)$ is complemented in $B(X)$.

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