Completely nuclear mappings

Let $X$ and $Y$ be operator spaces. The completely nuclear mappings from $X$ to $Y$ [ER94, §2], [EJR98, §3] are defined by the projective operator space tensor norm. We consider the extension of the canonical identity $X^* \otimes Y = F(X,Y)$:

$$\Phi :X^*\stackrel{\scriptscriptstyle \wedge}{\otimes}Y\rightarrow X^*\stackrel{\scriptscriptstyle \vee}{\otimes}Y\subset CB(X,Y)$$.$$$$

$\stackrel{\scriptscriptstyle \vee}{\otimes}$ is the injective tensor product and $\stackrel{\scriptscriptstyle \wedge}{\otimes}$ the projective tensor product .

A mapping in the range of $\Phi$ is called completely nuclear. One denotes by

$$\mathit{CN}(X,Y):=(X^*\stackrel{\scriptscriptstyle \wedge}{\otimes}Y)/ {\rm Ker}(\Phi)$$

the space of the completely nuclear mappings and endows it with the quotient operator space structure. The operator space norm is denoted by $\nu(\cdot)$. $M_n(\mathit{CN}(X,Y))$ and $\mathit{CN}(X,M_n(Y))$ are in general not isometric.

Nuclear ⁶⁸ mappings are completely nuclear. [ER94, 3.10]

In general, the projective tensor norm does not respect complete isometries. Hence,even for subspaces $Y_0\subset Y$ the canonical embedding $\mathit{CN}(X,Y_0)\to \mathit{CN}(X,Y)$ is generally only completely contractive and not isometric. Since the projective tensor norm respects quotient mappings, every nuclear map $\varphi:X_0\to Y$ on a subspace $X_0\subset X$ with $\nu(\varphi)<1$ has an extension $\tilde{\varphi}$ to the whole of $X$ satisfying $\nu(\tilde{\varphi})<1$.

The completely nuclear mappings enjoy the $\mathit{CB}$-ideal property . Furthermore, the adjoint $\varphi^*$ is completely nuclear if $\varphi$ is, and the inequality: $\nu(\varphi^*)\leq\nu(\varphi)$ [EJR98, Lemma 3.2] holds.

A mapping $\varphi$ is completely nuclear, if and only if there is a factorization of the form

$$B(\ell_2) \stackrel{M(a,b)}{\rightarrow } T(\ell_2)$$

$$\uparrow r \downarrow s$$

$$X\quad \stackrel{\varphi}{\rightarrow } Y$$

Here $a$, $b$ are Hilbert-Schmidt operators defining the mapping $M(a,b): x \mapsto a x b$. For the completely nuclear norm we have: $\nu (\varphi )=1$ precisely if for all $\epsilon>0$ there exists a factorization with $\Vert r\Vert _{cb}\Vert a\Vert _2\Vert b\Vert _2\Vert s\Vert _{cb}\leq 1+\epsilon$ ⁶⁹ [ER94, Thm. 2.1].

The completely nuclear mappings are not local .

Footnotes

...Nuclear ⁶⁸

The completely nuclear mappings owe their definition to the one of the nuclear mappings of the Banach space theory. There, one considers a corresponding mapping $\Phi_B:E^*\otimes_\gamma F \rightarrow B(E,F)$ for two Banach spaces $E$ and $F$.

... ⁶⁹

In the Banach space theory one has an analogous statement: A mapping $\varphi$ is nuclear, if and only if there's a diagram

$$\ell_\infty \stackrel{d}{\rightarrow } \ell_1$$

$$r\uparrow \downarrow s$$

$$E \stackrel{\varphi}{\rightarrow } F,$$

where $d$ is a diagonal operator, i.e. , there is a $(d_i)\in \ell_1$, such that $d((a_i))=(d_i\cdot a_i)$ for all $(a_i)\in \ell_\infty$. The nuclear norm is then computed as: $\nu_B(\varphi)=\inf\Vert r\Vert\Vert d\Vert _{\ell_1}\Vert s\Vert$, where the infimum runs over all possible factorizations.