Trying to understand Haagerup tensor product $B(H)\otimes_{\rm h}B(K)$

Question

I’m self reading Haagerup tensor product of operator spaces. Understanding it properly, I’m trying some examples. Let $H$ And $K$ be Hilbert space. Let $B(H)$ and $K(H)$ denotes the spaces of bounded and compact operators on $H$?

Can someone explain me what is $B(H)\otimes_{\rm h}B(K)$ and $B(H)\otimes_{\rm h}K(H)$? Are these spaces completely isometric to some well known operator space?

Is there any reference/lecture notes where I can find these kind of stuff?

P.S: The same question was first asked on MSE here.

Answer 1

When $H,K$ are finite dimensional, this is well explained in Pisier's book Introduction to operator space theory in the chapter on Haagerup tensor product, and your space is just the completely bounded maps on $B(K,H)$. When $H$ and $K$ are infinite dimensional, the argument works identically with bounded operators replaced by compact operators. To obtain a similar identification for bounded operators, my guess is that it is better to work with larger completions (extended Haagerup tensor product).

Let me expand a bit. When $H,K$ are finite dimensional, $B(K,H) = H_c \otimes_h K^*_{r}$ (where $H_c$ denotes $H$ with the column operator space structure, and $K^*_r$ denotes $K^*$ with the row Hilbert space structure, the dual of $K_c$). More generally, $H_c\otimes_h \cdot = H_c \otimes_{min} \cdot$ and $\cdot \otimes_{h} H_r = \cdot \otimes_{min} H_r$: when one tensorizes on the left (resp. right) by column (resp. row) operator spaces, Haagerup and minimal tensor product coincide. Taking the dual, we obtain $B(K,H)^* = H^*_r \otimes_h K_c$.

Combining all this with the associativity of the Haagerup tensor product, we obtain the natural identifications $$ B(H) \otimes_h B(K) = H_c \otimes_h (H_r^* \otimes_h K_c)\otimes_h K^*_r = H_c \otimes_{min} K^*_r \otimes_{min} B(H,K)^* = B(H,K) \otimes_{min} (B(K,H))^* = CB(B(K,H),B(K,H)).$$ For example, when $H=K=\mathbf{C}^n$, we obtain $$ M_n \otimes_h M_n = M_n(S^1_n) = CB(M_n,M_n).$$