MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is well known that neither

1) $c_0$ is isomorphic to a complemented subspace of $\mathcal{B}(H)$


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?

share|cite|improve this question

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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

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.

share|cite|improve this answer

$c_0$ 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 $c_0$.

share|cite|improve this answer
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

Not the answer you're looking for? Browse other questions tagged or ask your own question.