Jointly complete boundedness

Let $X, Y, Z$ be operator spaces. A bilinear mapping $\Phi: X \times Y \rightarrow Z$ is called jointly completely bounded [BP91, Def. 5.3 (jointly completely bounded)] if the norms of the joint amplifications of $\Phi$ are uniformly bounded:

$$\Vert\Phi\Vert _\mathrm{jcb}:= \sup \Vert \Phi^{(p \times q)}(x \otimes y)\Vert < \infty,$$

where $p,q\in{\mathbb{N}}$, $x \in \mathrm{Ball}(M_p(X))$, $y \in \mathrm{Ball}(M_q(Y))$ [BP91, Def. 5.3]. The norm $\Vert\Phi\Vert _\mathrm{jcb}$ equals the norm $\Vert\tilde\Phi\Vert _\mathrm{cb}$ of the linearization

$$\tilde\Phi : X \stackrel{\scriptscriptstyle \wedge}{\otimes}Y \rightarrow Z$$

on the projective operator space tensor product. $\mathit{JCB}(X \times Y; Z)$ denotes the operator space consisting of the jointly completely bounded bilinear maps. One obtains a norm on each matrix level by the identification

$$M_n(\mathit{JCB}(X \times Y;Z)) = \mathit{JCB}(X \times Y; {\mathbb{M}}_n(Z)).$$

We have

$$\mathit{JCB}(X \times Y;Z) \stackrel{\mathrm{cb}}{=}\mathit{CB}(X,\mathit{CB}(Y,Z)) \stackrel{\mathrm{cb}}{=}\mathit{CB}(Y,\mathit{CB}(X,Z)).$$

completely isometrically. By taking the transposition $\Phi^t(y,x) := \Phi(x,y)$ we obtain a complete isometry

$$\mathit{JCB}(X \times Y;Z) \stackrel{\mathrm{cb}}{=}\mathit{JCB}(Y \times X; Z).$$