It is well known that neither
1) $c_0$ is isomorphic to a complemented subspace of $\mathcal{B}(H)$
nor
2) $c_0$ is a quotient of $\mathcal{B}(H)$
for a Hilbert space $H$. Can we replace $H$ above by any Banach space? Reflexive space?
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$\begingroup$ PhotonicCrystal got what he wanted, so I vote to close to keep this from returning to the front page. $\endgroup$– Bill JohnsonCommented Dec 3, 2011 at 15:30
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2 Answers
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I am pretty sure that this is still open, though there are many partial results. Look up, e.g., papers by Giovanni Emmanuele on MathSciNet.
answered Oct 30, 2011 at 23:55
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$c_0$ is a quotient of ${\cal B}(c_0)$, namely for any nonzero $x \in c_0$, the evaluation map $T \to T(x)$ maps ${\cal B}(c_0)$ onto $c_0$.
answered Oct 30, 2011 at 23:58
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$\begingroup$ More generally, any Banach space $X$ is isomorphic to a complemented subspace of ${\cal B}(X)$. Namely for any $x_0 \in X$ and $\phi_0 \in X^*$ with $\phi_0(x_0) = 1$, ${\cal B}(X) = A \oplus B$ where $A = \{T: T(x_0) = 0\}$ and $B = \{x \otimes \phi_0: x \in X\}$. $\endgroup$ Commented Oct 31, 2011 at 0:21
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$\begingroup$ Neat answer. On a tangential point, for $X$ a Hilbert space, reflexive complemented subspaces of $\mathcal{B}(X)$ are isomorphic to Hilbert space(s); I think this was first observed in published form in the early 1990s by Pisier. $\endgroup$ Commented Oct 31, 2011 at 0:50
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$\begingroup$ Isn't $A$ of codimension 1 in $\mathcal{B}(X)$? $\endgroup$ Commented Oct 31, 2011 at 10:25
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$\begingroup$ PhotonicCrystal, it would seem to not be of codimension 1, however as a left-ideal it should be maximal in $\mathcal{B}(X)$. $\endgroup$ Commented Oct 31, 2011 at 12:14
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1$\begingroup$ Fairly good answer is given by G. Emmanuele in: G. Emmanuele, On complemented copies of c0 in spaces of operators II. Comment. Math. Univ. Carolin. 35 (1994), 259-261 $\endgroup$ Commented Nov 1, 2011 at 16:41