# Do completely bounded maps on an operator space have a completely contractive Banach algebra structure?

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 .

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.

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