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Characterizations

In connection with the column Hilbert space , it is enough to calculate the row norm

$\displaystyle \Vert T\Vert _{\mathrm{row}}
:=\sup_{n \in {\mathbb{N}}}
\sup_{\...
...}(X)}\leq 1}
\left\Vert\left[Tx_1 \ldots Tx_n\right]\right\Vert _{M_{1,n}(Y)}
$

or the column norm

$\displaystyle \Vert T\Vert _{\mathrm{col}}
:=\sup_{n \in {\mathbb{N}}}
\sup_{\...
...rray}{c}Tx_1\\  \vdots \\  Tx_n \end{array}
\right]\right\Vert _{M_{n,1}(X)}
$

of an operator $ T$, instead of the $ \mathrm{cb}$-norm, to ascertain the complete boundedness.

Let $ X$ be an operator space and $ S:{\mathcal{C}}_\H\rightarrow X$ bzw. $ T:X\rightarrow {\mathcal{C}}_\H$. Then we have $ \Vert S\Vert _\mathrm{cb}= \Vert S\Vert _{\mathrm{row}}$ resp. $ \Vert T\Vert _\mathrm{cb}= \Vert T\Vert _{\mathrm{col}}$ ([Mat94, Prop. 4] resp. [Mat94, Prop. 2]).

The column Hilbert space is characterized as follows,

(A)
as a hilbertian operator space [Mat94, Thm. 8]:

For an operator space $ X$ on an Hilbert space $ \H$, we have the following equivalences:

  1. $ X$ is completely isometric to $ {\mathcal{C}}_\H$.
  2. For all operator spaces $ Y$ and all $ T:X\rightarrow Y$ we have $ \Vert T\Vert _{\mathrm{cb}}=\Vert T\Vert _{\mathrm{row}}$, and for all $ S:Y\rightarrow X$ we have $ \Vert S\Vert = \Vert S\Vert _{\mathrm{row}}$. For all operator spaces $ Y$ and all $ T:Y\rightarrow X$ we have $ \Vert T\Vert _\mathrm{cb}= \Vert T\Vert _{\mathrm{col}}$, and for all $ S:X\rightarrow Y$ we have $ \Vert S\Vert = \Vert S\Vert _{\mathrm{col}}$.
  3. $ X$ coincides with the maximal hilbertian operator space on columns and with the minimal hilbertian operator space on rows. That means isometrically
    $\displaystyle M_{n,1}(X)$ $\displaystyle =$ $\displaystyle M_{n,1}(\mathit{MAX}(\H))$  
    $\displaystyle M_{1,n}(X)$ $\displaystyle =$ $\displaystyle M_{1,n}(\mathit{MIN}(\H))$  

(B)
as an operator space: For an operator space $ X$ TFAE:
  1. There is a Hilbert space $ \H$, such that $ X \stackrel{\mathrm{cb}}{=}{\mathcal{C}}_\H$ completely isometrically.
  2. We have

    $\displaystyle M_{n,1}(X) = \oplus_2 M_1(X)$

    and

    $\displaystyle M_{1,n}(X) = M_{1,n}(\mathit{MIN}(M_1(X)))$

    isometrically. [Mat94, Thm. 10].
  3. $ \mathit{CB}(X)$ with the composition as multiplication is an operator algebra [Ble95, Thm. 3.4].

next up previous contents index
Next: Column Hilbert space factorization Up: The column Hilbert space Previous: Tensor products   Contents   Index
Prof. Gerd Wittstock 2001-01-07