Operator space tensor products

An operator space tensor product is the completion of the algebraic tensor product with respect to an operator space tensor norm.

An operator space tensor norm $\Vert\cdot\Vert _\alpha$ is defined for each pair $(X,Y)$ of operator spaces and endows their algebraic tensor product $X \otimes Y$ with the structure of an matrix normed space $(X \otimes Y, \Vert\cdot\Vert _\alpha)$ thuch that the following two properties and [BP91, Def. 5.9]. hold.

The completion is called the $\alpha$-operator space tensor product of $X$ and $Y$ and is denoted by $X \otimes_\alpha Y$.

1.

For the complex numbers holds

$${\mathbb{C}}\otimes_\alpha {\mathbb{C}}= {\mathbb{C}}.$$

2.

For all $S \in \mathit{CB}(X_1,X_2)$ and $T \in \mathit{CB}(Y_1,Y_2)$ the operator $S \otimes T : X_1 \otimes Y_1 \rightarrow X_2 \otimes Y_2$ has a continuous extension

$$S \otimes_\alpha T \in \mathit{CB}(X_1 \otimes_\alpha Y_1, X_2 \otimes_\alpha Y_2).$$

The bilinear mapping

$$\otimes_\alpha : \mathit{CB}(X_1,X_2) \times \mathit{CB}(Y_1,Y_2) \rightarrow \mathit{CB}(X_1 \otimes_\alpha Y_1, X_2 \otimes _\alpha Y_2)$$

$$(S,T) \mapsto S \otimes_\alpha T$$

is jointly completely contractive .³⁰

Property may be replaced by the assumptions and .³¹

3.

An operator space tensor product $\otimes_\alpha$ is functorial: For all $S \in \mathit{CB}(X_1,X_2)$ and $T \in \mathit{CB}(Y_1,Y_2)$ the operator $S \otimes T : X_1 \otimes Y_1 \rightarrow X_2 \otimes Y_2$ has a continuous extension

$$S \otimes_\alpha T \in \mathit{CB}(X_1 \otimes_\alpha Y_1, X_2 \otimes_\alpha Y_2),$$

and

$$\Vert S \otimes_\alpha T \Vert _\mathrm{cb}\leq \Vert S\Vert _\mathrm{cb}\Vert T\Vert _\mathrm{cb}.$$

Remark: This is indeed an equality $\Vert S \otimes_\alpha T \Vert _\mathrm{cb}=\Vert S\Vert _\mathrm{cb}\Vert T\Vert _\mathrm{cb}$.

4.

The algebraic shuffle isomorphism $M_p(X) \otimes M_q(Y) \cong M_{pq}(X \otimes Y)$ has a continuous extension to a complete contraction:

$${\mathbb{M}}_p(X) \otimes_\alpha {\mathbb{M}}_q(Y) \rightarrow {\mathbb{M}}_{pq}(X \otimes_\alpha Y).$$

This complete contraction is called the shuffle map of the $\alpha$-operator space tensor product.

Condition is equivalent to the following two conditions: The shuffle mappings

$${\mathbb{M}}_p(X) \otimes_\alpha Y \rightarrow {\mathbb{M}}_p(X \otimes_\alpha Y),$$

$$X \otimes_\alpha {\mathbb{M}}_q(Y) \rightarrow {\mathbb{M}}_q(X \otimes_\alpha Y)$$

are completely contractive.

Operator space tensor producte may have further special properties:

An operator space tensor product $\otimes_\alpha$ is called

symmetric, if $X \otimes_\alpha Y \stackrel{\mathrm{cb}}{=}Y \otimes_\alpha X$ is a complete isometry;

associative, if $(X \otimes_\alpha Y) \otimes_\alpha Z \stackrel{\mathrm{cb}}{=}X \otimes_\alpha (Y \otimes_\alpha Z)$ is acomplete isometry;

injective, if for all subspaces $X_1 \subset X$, $Y_1 \subset Y$ the map $X_1 \otimes_\alpha Y_1 \hookrightarrow X \otimes_\alpha Y$ is acomplete isometry;

projective, if for all subspaces $X_1 \subset X$, $Y_1 \subset Y$ the map $X \otimes_\alpha Y \rightarrow X/X_1 \otimes_\alpha Y/Y_1$ is a complete quotient map;

self dual, if the algebraic embedding $X^* \otimes Y^* \subset (X \otimes_\alpha Y)^*$ has a completely isometric extension $X^* \otimes_\alpha Y^* \subset (X \otimes_\alpha Y)^*$.

In many applications one finds the Haagerup -tensor product. It is not symmetric, but associative, injective, projective and self dual.

Footnotes

... ³⁰

i.e., let $[S_{ij}] \in M_p(\mathit{CB}(X_1,X_2))$, $[T_{kl}] \in M_q(\mathit{CB}(Y_1,Y_2))$, $p,q\in{\mathbb{N}}$, then the norm of the linear operator

$$[S_{ij} \otimes_\alpha T_{kl}] \in M_{pq}(\mathit{CB}(X_1 \otimes_\alpha Y_1, X_2 \otimes_\alpha Y_2))$$

is estimated by

$$\Vert [S_{ij} \otimes T_{kl}] \Vert _\mathrm{cb} \leq \Vert [S_{ij}] \Vert _\mathrm{cb}\Vert [T_{kl}] \Vert _\mathrm{cb}$$.$$$$

Remark: This is indeed an equality.

...itemnr:ORTPaxiom4.³¹

(2) $\Rightarrow$(3),(4): Condition is a special case of . Let $I \in M_p(\mathit{CB}({\mathbb{M}}_p(X),X))$, $J \in M_q(\mathit{CB}({\mathbb{M}}_q(Y),Y))$ be matrices, which are algebraically the the identical mappings of the vector spaces $M_p(X)$ respectivly $M_q(Y)$. By assumption we have

$${ I \otimes_\alpha J \in M_{pq}(\mathit{CB}({\mathbb{M}}_p(X) \otimes_\alpha {\mathbb{M}}_q(Y), X \otimes_\alpha Y))} \mbox{\quad}$$

$$= M_1(\mathit{CB}({\mathbb{M}}_p(X) \otimes_\alpha {\mathbb{M}}_q(Y), {\mathbb{M}}_{pq}(X \otimes_\alpha Y))),$$

$$\Vert I \otimes_\alpha J\Vert _\mathrm{cb} \leq \Vert I\Vert _\mathrm{cb}\Vert J\Vert _\mathrm{cb}= 1.$$

Now $I \otimes_\alpha J: {\mathbb{M}}_p(X) \otimes_\alpha {\mathbb{M}}_q(Y) \rightarrow {\mathbb{M}}_{pq}(X \otimes _\alpha Y)$ is the shuffle-map in .

(3),(4) $\Rightarrow$(2): Let $[S_{ij}] \in M_p(\mathit{CB}(X_1,X_2))$, $[T_{kl}] \in M_q(\mathit{CB}(Y_1,Y_2))$, $p,q\in{\mathbb{N}}$, and $S \in \mathit{CB}(X_1, {\mathbb{M}}_p(X_2))$, $T \in \mathit{CB}(Y_1, {\mathbb{M}}_q(Y_2))$ the corresponding operators. By holds

$$S \otimes_\alpha T \in \mathit{CB}(X_1 \otimes_\alpha Y_1, {\mathbb{M}}_p(X_2) \otimes_\alpha {\mathbb{M}}_q(Y_2)),$$

$$\Vert S \otimes_\alpha T \Vert _\mathrm{cb} \leq \Vert S\Vert _\mathrm{cb}\Vert T\Vert _\mathrm{cb}= \Vert [S_{ij}] \Vert _\mathrm{cb}\Vert [T_{kl}] \Vert _\mathrm{cb}.$$

We apply the shuffle map $A :{\mathbb{M}}_p(X_2) \otimes_\alpha {\mathbb{M}}_q(Y_2) \rightarrow {\mathbb{M}}_{pq}(X_2 \otimes_\alpha Y_2)$ and obtain from

$$[S_{ij} \otimes_\alpha T_{kl}] = A (S \otimes_\alpha T) \in M_1(\mathit{CB}(X_1 \otimes_\alpha Y_1, {\mathbb{M}}_{pq}(X_2 \ot... ...ha Y_2))) = M_{pq}(\mathit{CB}(X_1 \otimes_\alpha Y_1, X_2 \otimes_\alpha Y_2),$$

$$\Vert [S_{ij} \otimes_\alpha T_{kl}] \Vert _\mathrm{cb} \leq \Vert S \otimes_\alpha T \Vert _\mathrm{cb} \leq \Vert [S_{ij}] \Vert _\mathrm{cb}\Vert [T_{kl}] \Vert _\mathrm{cb}.$$

Hence $\otimes_\alpha$ is jointly completely bounded.