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There are 'canonical' examples of maps on operator spaces which are not completely bounded. Nevertheless, I couldn't produce any examples of bounded projections on relatively easy to understand operator spaces which are not c.b. In particular, I failed this task for $\mathcal{K}(H)$ and $\mathcal{B}(H)$, where $H$ is a Hilbert space (I tried projections onto certain subspaces isomorphic to $c_0$ and $\ell_\infty$, respectively).

I would appreciate any examples.

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Symmetrisation (or anti-symmetrisation).

That is: let $T:K(H)\to K(H)$ be the transpose map and let $P=({\rm id}+T)/2$. Then $P:K(H)\to K(H)$ is a projection onto the subspace of symmetric [NOT self-adjoint] compact operators. Since $T$ is not completely bounded, $P$ is not completely bounded.

This argument also shows that the symmetrisation map $M_n\to M_n$ is a norm-one projection which has cb norm $\geq (n-1)/2$.

The examples you tried fail to work because of a general result: any bounded linear map into a minimal operator space will be automatically completely bounded. This is somewhere in the first third of Effros-Ruan's book, for example, though I don't have my copy here.

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  • $\begingroup$ Thanks. How about (Banach) complemented subspaces (Banach) isomorphic to $\mathcal{K}(H)$ or $\mathcal{B}(H)$? $\endgroup$
    – Olaf Kummers
    Commented Apr 18, 2012 at 6:41
  • $\begingroup$ @Olaf: without thinking about it too hard, I can at least point you to a result of Blower mentioned here mathoverflow.net/questions/12211/… which implies, in particular, that either $P(B(H))$ or $(I-P)(B(H))$ is isomorphic as a Banach space to $B(H)$. I rather suspect both are, but don't see at this moment how one would show this. $\endgroup$
    – Yemon Choi
    Commented Apr 18, 2012 at 7:09
  • $\begingroup$ What do you mean both? One of them can be e.g. $\ell_\infty$. $\endgroup$
    – Olaf Kummers
    Commented Apr 18, 2012 at 8:43
  • $\begingroup$ @Olaf: I meant the particular P in my answer, I thought that was obvious from context $\endgroup$
    – Yemon Choi
    Commented Apr 18, 2012 at 10:03
  • $\begingroup$ Sure, sorry. :) $\endgroup$
    – Olaf Kummers
    Commented Apr 18, 2012 at 10:53

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