# Examples of not completely bounded maps

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.

• What would count as a map not involving the transpose? For instance, the symmetrisation map that projects $K(H)$ onto the space of complex-symmetric operators is not completely bounded; but arguably the definition involves the transpose...
– user16299
May 1 '12 at 7:11
• @Norbert & @ Yemon Choi thanks for the reply and sorry for delayed response. It seems more complicated than what I thought of.
– RSG
May 10 '12 at 14:53
• Not involving transpose means composition of transpose and some other map, say completely positive maps. That was the thing in my mind when I posted the question...
– RSG
May 10 '12 at 14:58

Let $H$ be a Hilbert space. I will use the following standard notations for quantizations

• $MAX(H)$ - maximal quantization of $H$.
• $MIN(H)$ - minimal quantization of $H$.
• $C(H)$ - column quantization of $H$.
• $R(H)$ - row quantization of $H$.
• $OH(H)$ - operator quatization of $H$.

and spaces of operators

• $B(H)$ - bounded operators on $H$.
• $S_p(H)$ - Schatten $p$-class operators on $H$.
• $CB(X,Y)$ - completely bounded operators between operators spaces $X$ and $Y$.

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.

(The original version of this answer was in response to a slightly more general question.)

I'm not sure if the following is what you are really looking for, but you could always take an arbitrary Banach space $E$ and consider the identity map $E\to E$; this is almost never completely bounded from ${\rm MIN}(E)$ to ${\rm MAX(E)}$.

• Not exactly. Though this is a point which I did not know and trying to learn. What I wanted is that whether given any two arbitrary $C^*$ algebra whether there exists any not cb map other than transpose. Even for a simple case whether there is any not completely bounded map between say $\mathcal{B(H)}$ to itself other than transpose ($\mathcal{H}$ is a Hilbert space) or composition of transpose with completely positive map. \par A side question, if any such map exists, are they also positive (like transpose, but not completely positive)?