The aim of this question is to collect nice maps on $M_n(\mathbb{C})$ with the following property:

$\phi_n:M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ with $||\phi||=1$ and $||\phi_n||_{cb}\rightarrow \infty$ when $n\rightarrow \infty$

Here $||\phi_n||_{cb}$ denotes completely bounded norm of $\phi_n$, i.e. in this special case $||\phi_n||_{cb}=||\phi_{n}^{(n)}||$, where $\phi_{n}^{(n)}:M_n(\mathbb{C})\otimes M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})\otimes M_n(\mathbb{C})$ such that $\phi_{n}^{(n)}([a_{ij}])=[\phi_n(a_{ij})]$ for every $[a_{ij}]\in M_n(M_n(\mathbb{C}))$.

There are many examples of such maps, the simplest one is the transposition of matrix.

Please contribute yours...

  • $\begingroup$ Not an answer but rather a question: If you have a Banach space with two non cb isomorphic operator space structures, can you localize to get an example like you want? If the structures are exact this is clear (I think), but what if not? $\endgroup$ – Bill Johnson Apr 28 '11 at 23:24
  • $\begingroup$ Bill: I think one can manage without assuming that spaces are exact. The proper analog for the question above is to consider spaces which are N-K-exact. Here N-K-exact space means $\phi:E\rightarrow F_n\subseteq M_n$, $\psi:F_n\rightarrow E$ has the N-th bound on the norm: $||\phi^{(N)}||\cdot ||\psi^{(N)}||\leq K$. $\endgroup$ – Kate Juschenko Apr 29 '11 at 7:46
  • $\begingroup$ But every finite-dimensional operator space is N-1-exact and therefore we have maps from the question. I am more interested in direct examples on Mn, but probably it is too broad and I should withdraw my question. $\endgroup$ – Kate Juschenko Apr 29 '11 at 7:53

It is well known that each Schur multiplier satisfies $||.||_{cb}=||.||$. These maps are the $D_n$-bimodule maps on $M_n$. For the right module maps, the situation is more complicated. This gives concretes exemples of maps with your property. See the following paper:



Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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