Some formulae for the Haagerup tensor product

1. Let $A$ and $B$ be C$^*$-algebras in $B(\H)$. Then on the algebraic tensor product $A \otimes B$ the Haagerup tensor norm is explicitly given by
   $$\mbox{$ \Vert u\Vert _h := \inf\left\{ \left\Vert \sum_{\nu=1}... ...\Vert ^\frac{1}{2}\ \ :\ u = \sum_{\nu=1}^n a_\nu \otimes b_\nu \right \}$}$$$$,$$
   where $n\in{\mathbb{N}}$, $a_\nu \in A$, $b_\nu \in B$.

   The Haagerup norm of $\sum_{\nu=1}^n a_\nu \otimes b_\nu \in A \otimes B$ equals the cb-norm of the elementary operator $B(\H) \ni x \mapsto \sum_{\nu=1}^n a_\nu x b_\nu$. The Haagerup tensor product $A \otimes_h B$ is the completion of the algebraic tensor product $A \otimes B$ with respect to the above norm. The following more general definition in particular yields a completely isometric embedding $A \otimes_h B \hookrightarrow \mathit{CB}(B(\H))$.

2. We have
   $$\mbox{$ \Vert u\Vert _h = \inf \{ \sum_{\kappa=1}^k \Vert x_\kap... ...t \Vert y_\kappa\Vert \ : \ u = \sum_{\kappa=1}^k x_\kappa \odot y_\kappa \}$}$$$$$$
   where $k,l \in {\mathbb{N}}$, $x_\kappa \in M_{n,l}(X)$, $y_\kappa \in M_{l,n}(Y)$. In fact, one summand suffices [BP91, Lemma 3.2].
3. For elements $u \in M_n(X\otimes Y)$ in the algebraic tensor product there is an $l \in {\mathbb{N}}$ and elements $x \in M_{n,l}(X)$, $y \in M_{l,n}(Y)$ such that
   

   $$u = x \odot y$$

   $$\Vert u\Vert _h = \Vert x\Vert \Vert y\Vert.$$

   
   

   The infimum occuring in the formula describing the norm in this case is actually a minimum [ER91, Prop. 3.5].

4. The Haagerup norm of an element $u \in M_n(X \otimes_h Y)$ can also be expressed using a supremum:⁴⁰
   $$\Vert u\Vert _h = \sup \Vert\langle u, \varphi \odot \psi \rangle\Vert _{M_{n^2}},$$
   where $l \in {\mathbb{N}}$, $\varphi \in M_{n,l}(X^ *)$, $\psi \in M_{l,n}(Y^*)$, $\Vert\varphi\Vert = \Vert\psi\Vert = 1$ [ER91, Prop. 3.4].
5. From the definition of the Haagerup norm one easily deduces that the shuffle -map
   $${\mathbb{M}}_p(X) \otimes_h {\mathbb{M}}_q(Y) \rightarrow {\mathbb{M}}_{pq}(X \otimes_h Y)$$
   is a complete contraction. Hence the Haagerup tensor product enjoys property of an operator space tensor product.

   But the shuffle-map is not an isometry in general as shown by the following example:⁴¹

   $${\mathbb{M}}_n(C_l) \otimes_h {\mathbb{M}}_n(R_l) \stackrel{\math... ...)) \stackrel{\mathrm{cb}}{=}{\mathbb{M}}_n({\mathbb{M}}_n(C_l \otimes_h R_l)).$$
   Since the bilinear mapping
   

   $$\otimes_h : \mathit{CB}(X_1,X_2) \times \mathit{CB}(Y_1,Y_2) \rightarrow \mathit{CB}(X_1 \otimes_h X_2, Y_1 \otimes _h Y_2)$$

   $$(S,T) \mapsto S \otimes_h T$$

   
   
   is contractive, it is jointly completely contractive.⁴²

   In fact, using , we see that it is even completely contractive.

6. For the row and column structure the shuffle -map even is a complete isometry. We have the Lemma of Blecher and Paulsen [BP91, Prop. 3.5]:
   $$C_n(X) \otimes_h R_n(Y) \stackrel{\mathrm{cb}}{=}{\mathbb{M}}_n(X \otimes_h Y).$$
   In many cases it suffices to prove a statement about the Haagerup tensor product on the first matrix level and then to deduce it for all matrix levels using the above formula.⁴³

   Here we list some special cases of the Lemma of Blecher and Paulsen:
   

   $$C_n \otimes_h R_n \stackrel{\mathrm{cb}}{=} {\mathbb{M}}_n,$$

   $$C_n \otimes_h X \stackrel{\mathrm{cb}}{=} C_n(X),$$

   $$X \otimes_h R_n \stackrel{\mathrm{cb}}{=} R_n(X),$$

   $$C_n \otimes_h X \otimes_h R_n \stackrel{\mathrm{cb}}{=} {\mathbb{M}}_n(X)$$

   
   

7. In contrast to , for $R_n \otimes_h C_n$ one obtains the finer operator space structure of the trace class
   $$T_n := {\mathbb{M}}_n^* \stackrel{\mathrm{cb}}{=}R_n \stackrel{\scriptscriptstyle \wedge}{\otimes}C_n \stackrel{\mathrm{cb}}{=}R_n \otimes_h C_n$$   .
   For an operator space $X$ we have [Ble92b, Prop. 2.3]:
   

   $$R_n \otimes_h X \stackrel{\mathrm{cb}}{=} R_n \stackrel{\scriptscriptstyle \wedge}{\otimes}X ,$$

   $$X \otimes_h C_n \stackrel{\mathrm{cb}}{=} C_n \stackrel{\scriptscriptstyle \wedge}{\otimes}X ,$$

   $$R_n \otimes_h X \otimes_h C_n \stackrel{\mathrm{cb}}{=} T_n \stackrel{\scriptscriptstyle \wedge}{\otimes}X ,$$

   $$%%\pu R_n \otimes_h X^* \otimes_h C_n \stackrel{\mathrm{cb}}{=} {\mathbb{M}}_n(X)^*$$

   
   

8. By the very construction the bilinear mapping $X \times Y \rightarrow X \otimes_h Y$, $(x,y) \mapsto x \otimes y$ is a complete contraction.

   Hence its amplification, the tensor matrix product
   

   $$\odot_h: {\mathbb{M}}_{n,l}(X) \times {\mathbb{M}}_{l,n}(Y) \rightarrow {\mathbb{M}}_n(X \otimes_h Y)$$

   $$(x, y) \mapsto x \odot y$$

   
   
   also is a complete contraction . The linearization of the tensor matrix product gives the complete contraction
   $${\mathbb{M}}_{n,l}(X) \otimes_h {\mathbb{M}}_{l,n}(Y) \rightarrow {\mathbb{M}}_n(X \otimes_h Y)$$.$$$$

9. The bilinear mapping⁴⁴
   

   $$\otimes_h : \mathit{CB}(X_1,X_2) \times \mathit{CB}(Y_1,Y_2) \rightarrow \mathit{CB}(X_1 \otimes_h Y_1, X_2 \otimes_h Y_2)$$

   $$(S,T) \mapsto S \otimes_h T$$

   
   
   is completely contractive ⁴⁵and gives rise to a complete contraction
   $$\mathit{CB}(X_1,X_2) \otimes_h \mathit{CB}(Y_1,Y_2) \rightarrow \mathit{CB}(X_1 \otimes_h Y_1, X_2 \otimes_h Y_2).$$

10. Let $\H$ and $\mathcal{K}$ be Hilbert spaces. Taking the Haagerup tensor product of the column Hilbert space ${\mathcal{C}}$ and the row Hilbert space ${\mathcal{R}}$ one obtains completely isometrically the space of compact operators $K$ resp. of trace class⁴⁶operators $T$ [ER91, Cor. 4.4]:

    
    

    $${\mathcal{R}}_{\overline{\H}} \otimes_h {\mathcal{C}}_{\mathcal{K}} \stackrel{\mathrm{cb}}{=} T({\H,\mathcal{K}}),$$

    $${\mathcal{C}}_{\mathcal{K}} \otimes_h {\mathcal{R}}_{\overline{\H}} \stackrel{\mathrm{cb}}{=} K({\H,\mathcal{K}}).$$

    
    
    This example also shows that the Haagerup tensor product is not symmetric.

Footnotes

... supremum:⁴⁰

For $x \in X$, $y \in Y$, we have:

$$(\varphi \odot \psi)(x \otimes y)= \langle x \otimes y, \varphi \... ...y , \varphi_{ij} \otimes \psi_{jk} \rangle\right] = \varphi(x)\psi(y) \in M_n.$$

Here, we used the definitions of two fundamental notions in operator space theory: the tensor matrix multiplication $\varphi \odot \psi$ of mappings $\varphi$, $\psi$ and the joint amplification of the duality of tensor products.

... example:⁴¹

Algebraically, we have on both sides the same spaces of matrices. But on the left side one obtains the finer operator space structure $T_n := {\mathbb{M}}_n^*$ of the trace class:

$${{\mathbb{M}}_n(C_l) \otimes_h {\mathbb{M}}_n(R_l) \stackrel{\mat... ...ackrel{\mathrm{cb}}{=} {\mathbb{M}}_l({\mathbb{M}}_n \otimes_h {\mathbb{M}}_n)}$$

$$= {\mathbb{M}}_l(C_n \otimes_h R_n \otimes_h C_n \otimes_h R_n) = {... ...otimes_h T_n \otimes_h R_n) \stackrel{\mathrm{cb}}{=} {\mathbb{M}}_l(M_n(T_n)).$$

On the right side, we get the coarser operator space structure of the matrices $M_n$:

$${\mathbb{M}}_n({\mathbb{M}}_n(C_l \otimes_h R_l) = {\mathbb{M}}_n... ...}_l)) \stackrel{\mathrm{cb}}{=}{\mathbb{M}}_l({\mathbb{M}}_n({\mathbb{M}}_n)).$$

... contractive.⁴²

This follows from the of the property itemnr:ORTPaxiom2 and the properties and of operator space tensor products.

... formula.⁴³

This method can be applied to obtain this complete isometry itself. It is easy to see that on the first matrix level we have

$$M_1(C_n(X) \otimes_h R_n(Y)) = M_n(X \otimes_h Y)$$

isometrically. From this the complete isometry follows - for all $p \in {\mathbb{N}}$ we have:

$$M_p(C_n(X) \otimes_h R_n(Y)) = M_1(C_p(C_n(X)) \otimes_h R_p(R_n(Y))) = M_1(C_{pn}(X) \otimes_h R_{pn}(Y))$$

$$= M_{pn}(X \otimes_h Y) = M_p({\mathbb{M}}_n(X \otimes_h Y))$$

isometrically .

... mapping⁴⁴

The amplification of $\otimes_h$ is nothing but the tensor matrix multiplication $\odot_h$ of operator matrices.

...sec:cb-bilinear ⁴⁵

Let $S \in M_n(\mathit{CB}(X_1,X_2))$, $T \in M_n(\mathit{CB}(Y_1,Y_2))$. For $x \in M_{p,q}(X_1)$, $y \in M_{q,p}(Y_1)$ we have

$$(S \odot T)^{(p)}(x \odot y) = (S^{(p,q)}(x)) \odot (T^{(q,p)}(y)),$$

$$\Vert (S \odot T)^{(p)}(x \odot y) \Vert _{M_{pn}(X_2 \otimes_h Y_2)} \leq \Vert S^{(p,q)}(x) \Vert \Vert T^{(q,p)}(y) \Vert \leq \Vert S\Vert _\mathrm{cb}\Vert T\Vert _\mathrm{cb}\Vert x\Vert \Vert y\Vert.$$

By the definition of the Haagerup norm (in $X_1 \otimes_h Y_1$) we obtain

$$\Vert (S \odot T)^{(p)}(x \odot y) \Vert _{M_{pn}(X_2 \otimes Y_2)} \leq \Vert S\Vert _\mathrm{cb}\Vert T\Vert _\mathrm{cb}\Vert x \odot y \Vert _{M_{p}(X_1 \otimes_h Y_1)},$$

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

This means that $\otimes_h$ is completely contractive.

... class⁴⁶

$T$ is endowed with its natural operator space structure $T({\H,\mathcal{K}}) :\stackrel{\mathrm{cb}}{=}K({\mathcal{K}, \H})^*$.