Let $\phi:\mathcal{A}\longrightarrow\mathcal{B}$ be a bounded map between $C^*$ algebras. $\phi$ is said to be completely bounded if the natural extension map
\begin{eqnarray}
\phi_n:M_n(\mathcal{A})&\longrightarrow & M_n(\mathcal{B})\\
((a_{i,j}))&\longmapsto & ((\phi(a_{ij}))
\end{eqnarray}
is also bounded for all $n$. ($M_n(\mathcal{A})$ denotes $n\times n$ matrices whose entries are elements of $\mathcal{A}$.) This bound defines a norm as well which is known as completely bounded norm on the the set of maps.
The standard example of a 'not' completely bounded bounded map is transpose. I could not construct any other example which does not involve transpose. Unfortunately I could not locate any other example from the literature. Please help.
Best Answer
Let $H$ be a Hilbert space. I will use the following standard notations for quantizations
and spaces of operators
Then we have the following table of spaces of completely bounded operators between different quantizations of $H$ up to isometric ($\simeq_1$) or usual ($\simeq$) isomorphism: $$ \begin{array}{cccccc} CB(\downarrow,\rightarrow) & MIN(H) & C(H) & OH(H) & R(H) & MAX(H)\\ MIN(H) & \simeq_1 B(H) & \simeq_1 S_2(H) & \simeq S_2(H) & \simeq_1 S_2(H) & \simeq S_1(H)\\ C(H) & \simeq_1 B(H) & \simeq_1 B(H) & \simeq_1 S_4(H) & \simeq_1 S_2(H) & \simeq_1 S_2(H)\\ OH(H) & \simeq_1 B(H) & \simeq_1 S_4(H) & \simeq_1 B(H) & \simeq_1 S_4(H) & \simeq S_2(H)\\ R(H) & \simeq_1 B(H) & \simeq_1 S_2(H) & \simeq_1 S_4(H) & \simeq_1 B(H) & \simeq_1 S_2(H)\\ MAX(H) & \simeq_1 B(H) & \simeq_1 B(H) & \simeq_1 B(H) & \simeq_1 B(H) & \simeq_1 B(H)\\ \end{array} $$ This characterization was taken from this page.
From this table we see that for $13$ of $25$ cases the spaces of completery bounded operators are proper subspaces of $B(H)$. Thus we have quite a lot of non-completely bounded operators.