# Complemented subspaces isomorphic to $c_0$ in $\mathcal{B}(E)$ [closed]

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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## closed as no longer relevant by Bill Johnson, Felipe Voloch, Henry Cohn, Mark Sapir, Ryan BudneyDec 3 '11 at 18:51

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PhotonicCrystal got what he wanted, so I vote to close to keep this from returning to the front page. –  Bill Johnson Dec 3 '11 at 15:30

## 2 Answers

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.

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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 $c_0$.

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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 $B = \{x \otimes \phi_0: x \in X\}$. –  Robert Israel Oct 31 '11 at 0:21
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. –  Philip Brooker Oct 31 '11 at 0:50
Isn't $A$ of codimension 1 in $\mathcal{B}(X)$? –  PhotonicCrystal Oct 31 '11 at 10:25
PhotonicCrystal, it would seem to not be of codimension 1, however as a left-ideal it should be maximal in $\mathcal{B}(X)$. –  Philip Brooker Oct 31 '11 at 12:14
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 –  PhotonicCrystal Nov 1 '11 at 16:41