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Column Hilbert space factorization

Let $ X$, $ Y$ be operator spaces. We say that a linear map $ T:M_1(X) \rightarrow M_1(Y)$ factors through a column Hilbert space , if there is a Hilbert space $ \H$ and completely bounded maps $ T_2:X \rightarrow {\mathcal{C}}_\H$, $ T_1: {\mathcal{C}}_\H\rightarrow Y$ with $ T=T_1 \circ T_2$. We define

$\displaystyle \gamma_2(T) := \inf\Vert T_1\Vert _{\mathrm{cb}} \Vert T_2\Vert _{\mathrm{cb}}$,

where the infimum runs over all possible factorizations. If no such factorisation exists we say $ \gamma_2(T) := \infty$. $ \Gamma_2(X,Y)$ is the Banach space of all linear maps $ T:X\rightarrow Y$ with $ \gamma_2(T) < \infty$ [ER91, Chap. 5],[Ble92b, p. 83].

Let $ X_1$ and $ Y_1$ be operator spaces and $ T \in \Gamma_2(X,Y)$, $ S \in \mathit{CB}(X_1,X)$, $ R \in \mathit{CB}(Y,Y_1)$. Then we have the $ \mathit{CB}$ ideal property

$\displaystyle \gamma_2(RTS) \leq \Vert R\Vert _\mathrm{cb}\gamma_2(T) \Vert S\Vert _\mathrm{cb}$.

We interpret a matrix $ T=[T_{ij}] \in M_n(\Gamma_2(X,Y))$ as a mapping from $ X$ to $ M_n(Y)$: $ [T_{ij}](x):=[T_{ij}(x)]$. $ T$ has a factorization in completely bounded mappings

$\displaystyle X \stackrel{T_2}{\rightarrow } M_{1,n}({\mathcal{C}}_\H)
\stackrel{T_1}{\rightarrow } M_n(Y).$

Again, we define

$\displaystyle \gamma_2(T) := \inf\Vert T_1\Vert _{\mathrm{cb}} \Vert T_2\Vert _{\mathrm{cb}}$,$\displaystyle $

where the infimum is taken over all factorizations. So we get an operator space structure on $ \Gamma_2(X,Y)$ [ER91, Cor. 5.4].

Let $ X,Y$ be operator spaces and $ Y_0$ an operator subspace of $ Y$. Then the inclusion $ \Gamma_2(X,Y_0) \hookrightarrow \Gamma_2(X,Y)$ is completely isometric [ER91, Prop. 5.2].

Let $ X$, $ Y$ be operator spaces. It is well known that every linear map

$\displaystyle T : X \rightarrow Y^*$

defines a linear functional

$\displaystyle f_T:Y \otimes X \rightarrow {\mathbb{C}}$

via

$\displaystyle \langle f_T, y \otimes x \rangle := \langle T(x),y \rangle$.

This identification determines the complete isometry [ER91, Thm. 5.3] [Ble92b, Thm. 2.11]

$\displaystyle \Gamma_2(X,Y^*) \stackrel{\mathrm{cb}}{=}(Y \otimes_{h}X)^*$   .

Let $ X$, $ Y$, $ Z$ be operator spaces. We get a complete isometry

$\displaystyle \Gamma_2(Y\otimes_{h}X, Z) \stackrel{\mathrm{cb}}{=}\Gamma_2(X,\Gamma_2(Y,Z))$

via the mapping
$\displaystyle T$ $\displaystyle \mapsto$ $\displaystyle \widetilde{T}$  
$\displaystyle \widetilde{T}(x)(y)$ $\displaystyle :=$ $\displaystyle T(y \otimes x)$  

[ER91, Cor. 5.5].


next up previous contents index
Next: Multiplicative Structures Up: Hilbertian Operator Spaces Previous: Characterizations   Contents   Index
Prof. Gerd Wittstock 2001-01-07