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.

  • 1
    $\begingroup$ 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... $\endgroup$
    – user16299
    May 1 '12 at 7:11
  • $\begingroup$ @Norbert & @ Yemon Choi thanks for the reply and sorry for delayed response. It seems more complicated than what I thought of. $\endgroup$
    – RSG
    May 10 '12 at 14:53
  • $\begingroup$ 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... $\endgroup$
    – 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 ${\rm MAX(E)}$.

update: see also Jon Bannon's comment and link here https://mathoverflow.net/questions/86550/positive-but-not-completely-positive/88309#88309

  • $\begingroup$ 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)? $\endgroup$
    – RSG
    May 10 '12 at 15:02
  • $\begingroup$ thanks for informing about the paper of Prof Yu and sorry for the late response. however, i am slightly disturbed about the content of the paper. in the first page second column it is claimed that transpose can be written as a difference of two completely positive maps. this makes the transpose as completely bounded ( by the Wittstock’s decomposition theorem, only there will be no imaginary part in it). But transpose is not completely bounded!! i'll read the remaining paper as it does not seem to depend on the particular point. $\endgroup$
    – RSG
    May 28 '12 at 5:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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