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(I'd be grateful if anyone thinking of putting MathJax in the question title refrains from doing so.)


By consulting various standard sources (Effros-Ruan's book, Pisier's book, the lexicon of Wittstock's old group) I can find some descriptions of the underlying Banach space of $\mathcal{CB}(H,K)$ where $H$ and $K$ are Hilbert spaces equipped with various operator space structures. However, it's not clear from these accounts if there has been systematic work on studying what $CB(H,K)$ looks like as an operator space in these situations.

Let me raise two particular problems for which I'd be happy to get more information, or pointers to the literature. It could be that as a dilettante in the world of operator spaces I've just missed some basic arguments.

Q1. Let ${\bf B}$ denote $B(\ell_2)$ with its usual OSS and let ${\rm COL}$ be $\ell_2$ with column OSS. Then $CB({\rm COL}, {\rm COL})$ is completely isometrically isomorphic to ${\bf B}$. In some sense this characterises ${\rm COL}$: a result of Blecher (Theorem 3.4 in his 1995 Math. Ann. paper) tells us that if $X$ is a Hilbertian operator space and $CB(X)$ is isomorphic as an algebra and an operator space to ${\bf B}$ then $X$ is isomorphic to column Hilbert space. However, when does the formal identity map give a completely bounded map ${\bf B} \to CB(X)$? What about ${\bf B} \to CB(X,Y)$ where $X$ and $Y$ can be different?

Remark. For the last part of Q1: clearly this is possible if the formal identity map $X\to Y$ factors in a cb way through ${\rm COL}$. I'd be more interested to know if this can happen when either $X$ or $Y$ is incomparable with ${\rm COL}$.

Q2. If $X$ is a homogeneous, Hilbertian operator space then by definition the underlying Banach space of $CB(X)$ is $B(\ell_2)$. This happens, for instance, if $X$ is row, column, the self-dual OSS, MAX, or MIN. Of the 5 possibilities just listed, are any of the resulting o.s. structures on $CB(X)$ comparable (i.e. weaker or stronger as o.s. structures) to ${\bf B}$?

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Both your questions can be answered by considering rows and columns in $\mathbf B$, and by noticing that in $CB(X,Y)$ colums are completely isometric to $Y$ and rows are completely isometric to $X^*$.

Therefore if $X_1,X_2$ and $Y_1,Y_2$ are operator spaces structures on $\ell_2$ such that $id\colon CB(X_1,Y_1)\to CB(X_2,Y_2)$ is completely bounded, then $id \colon Y_1 \to Y_2$ and $id \colon X_2 \to X_1$ are both completely bounded. The converse is clear, so this is a characterization when $CB(X_1,Y_1)$ and $CB(X_2,Y_2)$ are comparable.

Therefore the answers to your questions are:

Q1 If either $X$ or $Y$ is incomparable with $\mathrm{COL}$, then $\mathbf B$ and $CB(X,Y)$ are incomparable.

Q2: No. If $X$ and $Y$ are homogeneous hilbertian operator spaces, then $CB(X)$ and $CB(Y)$ are comparable if and only if $X$ and $Y$ are completely isomorphic.

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  • $\begingroup$ Thanks! I see that my original questions were somewhat facile (English or French use of the word). $\endgroup$
    – Yemon Choi
    Apr 18, 2016 at 22:35

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