TITLE: Space of compact operators
QUESTION [6 upvotes]: 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 $\mathcal{K}$, Schatten $p$-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 $c_0$?

REPLY [3 votes]: This post is not intended as an answer to a question that was asked & answered 10 years ago, but as an appendix for newcomers to provide background about the DPP in the context, which the question owner and the commentators (and the other experts in the field) already know about.

Every reflexive complemented subspace of a C*-algebra is isomorphic to a Hilbert space.

TFAE for a C*-algebra $A$:



$A^{\ast}$ has DPP (Dunford-Pettis property)
$A$ has DPP
$A$ contains no complemented copy of $\ell^2$.
$A^{\ast\ast}$ is a finite type I von Neumann algebra.

Perhaps the major difference between $c_0$ & $K(\ell^2)$, $\ell^1$ & trace class operators on $\ell^2$, is the lack of DPP for the latter.

On the contrary, if a C*-algebra $A$ does not have DPP, then there exists a complete isometry $K(\ell^2)\to A$.

If $E$ is reflexive, then



a. $K(E)$ is Arens regular. (so is $c_0$)
b. $K(E)$ contains no copy of $\ell^1$. (so does $c_0$)
c. $K(E)$ does not have DPP.

As a consequence of (b) and (c), if $X$ is a subspace of $K(E)$, then both $X$ and $K(E)/X$ cannot have DPP simultaneously.

If $E$ is reflexive and has approximation property, then $K(E)^{\ast\ast}= B(E)$. Thus, $K(E)$ is an ideal in its bidual $B(E)$. Equivalently, every left & right multiplication operator $L_a,R_a:K(E)\to K(E)$, defined by $L_ax = ax$ & $R_ax = xa$ are weakly compact.

On the other hand, the left & right multiplication operators on $c_0$ are compact by DPP.