Let $X$ be an operator space and $CB(X)$ be the set of all completely bounded linear maps $f: X \to X$. Note that $CB(X)$ becomes a Banach algebra for the composition of operators.

Is the multiplication $$CB(X)\odot CB(X)\to CB(X): f\otimes g \mapsto f \circ g$$ completely contractive with respect to the projective tensor product norm on $CB(X)\odot CB(X)$? I.e. is $CB(X)$ a completely contractive Banach algebra? This is claimed in the paper New tensor products of C-algebras and characterization of type I C-algebras as rigidly symmetric C*-algebras .


1 Answer 1


Yes, this is true. More generally:$\newcommand{\CB}{\mathop{\sf{CB}}}\newcommand{\cbnorm}[1]{{\lVert#1\rVert}_{\sf cb}}$

Proposition 1. Let $X$, $Y$, $Z$ be operator spaces. Then composition of operators, viewed as a bilinear map $\CB(Y,Z) \times \CB(X,Y) \to \CB(X,Z)$, is jointly completely contractive.

I can't remember if this is proved explicitly in the book of Effros and Ruan, so I will attempt to give a leisurely explanation, which might not be the quickest or most direct approach, but is the one that I find easiest to reconstruct when needed.

I find it convenient to use a "curried" perspective. Namely: given vector spaces $E,F,G$ and a bilinear map $\beta:E\times F \to G$, define $T_\beta: E \to \mathop{\rm Lin}(F,G)$ by $T_\beta(x):y\mapsto \beta(x,y)$.

Fact 2. If $E$, $F$ and $G$ are operator spaces, $\beta$ is jointly completely contractive if and only if $T_\beta$ is a complete contraction from $E$ to $\CB(F,G)$.

(I will try to supply a reference for this equivalence when I get a chance to look in my copy of the Effros–Ruan book.)

Taking Fact 2 for granted, we see that to prove Proposition 1, it is enough to show that $\CB(Y,Z)\to \CB(\CB(X,Y),\CB(X,Z))$, $f \mapsto (g\mapsto f\circ g)$, is completely contractive. We do this in stages, by a kind of "boot-strap" process.

Remark 3. Let $E$ and $F$ be vector spaces, and note that for each $n\in {\mathbb N}$ there is a natural identification of $M_n \mathop{\rm Lin}(E,F)$ with $\mathop{\rm Lin}(E,M_nF)$. Then if $E$ and $F$ are operator spaces, this identification is a bijection between $M_n\CB(E,F)$ and $\CB(E,M_nF)$, and the definition of the operator space structure on $\CB(E,F)$ is that we define the norm on $M_n\CB(E,F)$ by transporting the norm on $\CB(E,M_nF)$ across this bijection.

Lemma 4. Let $E,F, G$ be operator spaces and let $h: F\to G$ be a complete contraction.

(i) For each $k\in\CB(E,F)$ we have $h\circ k \in \CB(E,G)$

(ii) The map $\CB(E,F) \to \CB(E,G)$ defined by $k\mapsto h\circ k$ is contractive.

Proof. Follows from the definition of the cb-norm (exercise).

Corollary 5. Let $X,Y,W$ be operator spaces and let $f:Y\to W$ be a complete contraction. Then the map $S: \CB(X,Y) \to \CB(X,W)$ defined by $g\mapsto f\circ g$ is completely contractive.

Proof. Let $n\in {\mathbb N}$; we need to show that the amplification $S^{(n)} : M_n\CB(X,Y) \to M_n\CB(Y,W)$ is contractive. This follows from Remark 3, by applying Lemma 4 with $h = f^{(n)} : M_nY \to M_n W$.

Proof of Proposition 1. Let $X, Y, Z$ be operator spaces and let $T:\CB(Y,Z) \to \CB(\CB(X,Y),\CB(X,Z))$ be given by composition of operators. As remarked above, we need to show that $T$ is completely contractive. So let $m\in {\mathbb N}$ and consider the amplification $$ T^{(m)}: M_m \CB(Y,Z) \to M_m\CB(\CB(X,Y),\CB(Y,Z)). $$ Using Remark 3 repeatedly, we may identify $T^{(m)}$ with the linear map $$ \Theta:\CB(Y,M_m Z) \to \CB(\CB(X,Y),\CB(Y,M_mZ) $$ defined by composition of operators. But now if $\cbnorm{f:Y\to M_mZ}\leq 1$, applying Corollary 5 with $W=M_mZ$ shows that $\cbnorm{\Theta(f): \CB(X,Y) \to \CB(Y,M_mZ)}\leq 1$. Thus $\Theta$ is a contraction, and we conclude that $\lVert T^{(m)}\rVert \leq 1$ as required.

  • 4
    $\begingroup$ Remark for any passing categorists: yes, this is a closed symmetric monoidal structure on the category of operator spaces and complete contractions. $\endgroup$
    – Yemon Choi
    Commented Oct 3, 2023 at 22:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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