I’m self reading Haagerup tensor product of operator spaces from Effros Ruan's book on operator spaces. I am unable to digest it properly.
Can someone please explain me the thinking process behind Haagerup tensor product in some easy cases? For instance Haagerup tensor product of two $C^*-$algebras or Haagerup tensor product of operator space with $C^*-$ algebra.
Thank you so much in advance.
Usually you cannot compute the Haagerup tensor product of operator spaces (or even $C^\ast$-algebras) explicitly, but the idea behind it is still quite intuitive in my opinion.
Let's take a step back. If $U$ and $V$ are vector spaces, then $U\otimes V$ is characterized by the fact that $(u,v)\mapsto u\otimes v$ is bilinear and for every bilinear map $\beta\colon U\times V\to\mathbb K$ there exists a unique linear map $\tilde \beta\colon U\otimes V\to \mathbb K$ such that $\tilde \beta(u\otimes v)=\beta(u,v)$. In other words, the map $\beta\mapsto \tilde\beta$ is an isomorphism between the space of bilinear maps from $U\times V\to \mathbb K$ and the space of linear maps from $U\otimes V\to \mathbb K$.
The Haagerup tensor product has a similar property, just that everything should be completely bounded as usual for operator spaces: There is a completely bounded bilinear map from $X\times Y\to X\otimes_h Y,\,(x,y)\mapsto x\otimes y$ and a completely isometric isomorphism $\Phi$ from the space of completely bounded bilinear maps from $X\times Y\to \mathbb C$ to the space of completely bounded linear maps $X\otimes_h Y\to \mathbb C$ such that $\Phi(\beta)(x\otimes y)=\beta(x,y)$.
Now let's make things a little more explicit: First, the Haagerup tensor product $X\otimes_h Y$ is a completion of the algebraic tensor product $X\odot Y$ and $x\otimes y$ has the usual meaning as an element of $X\odot Y$. Moreover, if $A$ and $B$ are $C^\ast$-subalgebras of $B(H)$, then there is a nice formula for the norm of $\sum_j a_j\otimes b_j$ in $A\otimes_h B$: It is the norm of the linear map $$ B(H)\to B(H),\,x\mapsto \sum_j a_j x b_j. $$ This hints at the connection to double operator integrals.
Finally, two explicit examples after all: If $\mathcal R$ is the row Hilbert space and $\mathcal C$ is the column Hilbert space, then $\mathcal C\otimes_h \mathcal R\cong K(\ell_2^\ast,\ell_2)$ completely isometrically and $\mathcal R\otimes_h \mathcal C\cong S_1(\ell_2^\ast,\ell_2)$ completely isometrically. In particular, the Haagerup tensor product is not commutative.