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Before asking my question, let me introduce the relevant terminology.

Throughout, let $(A, \Delta)$ be a compact quantum group.

Definition: A representation $v$ on the Hilbert space $H$ is an element $v\in M(B_0(H)\otimes A)$ such that $(\text{id}\otimes \Delta)(v) = v_{(12)}v_{(13)}$. Here the subscripts with the brackets denote the leg numbering notation.

Definition: An intertwiner from the representation $(H_1, v_1)$ to the representation $(H_2, v_2)$ is an element in $B(H_1,H_2)$ such that $(x\otimes 1)v_1 =v_2(x \otimes 1).$

Question: How should the multiplications $(x \otimes 1) v_1$ and $v_2 (x \otimes 1)$ be interpreted?

One way of making sense of these multiplications is as follows:

Let $A\subseteq B(K)$ be a faithful and non-degenerate representation, say the universal GNS representation of $A$. Then we have a canonical inclusion $M(B_0(H_1) \otimes K) \subseteq B(H_1 \otimes K)$ and we can interpret $x \otimes 1$ as an operator $H_1 \otimes K \to H_2 \otimes K$ and $v_1$ as an operator $H_1 \otimes K \to H_1 \otimes K$ and we can simply form the composition $(x \otimes 1)v_1$ in $B(H_1 \otimes K, H_2 \otimes K)$. Similarly for the other side.

However, can one give a definition that is "space-free", i.e. does not refer to a choice of faithful and non-degenerate representation?

EDIT: Maybe the following works:

Viewing $B(H_1, H_2)$ as a corner of $B(H_1\oplus H_2)$, we have canonical inclusions $$B(H_1,H_2) \otimes A \subseteq B(H_1\oplus H_2) \otimes A\subseteq M(B_0(H_1 \oplus H_2) \otimes A)$$ and we also have a canonical inclusion $$M(B_0(H_1)\otimes A)\subseteq M(B_0(H_1 \oplus H_2)\otimes A)$$ so we can perform the multiplication in $M(B_0(H_1\oplus H_2)\otimes A)$.

However, also the above does not seem quite satisfactory.

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Perhaps what you are after is the language of Hilbert $C^\ast$-modules. Here I follow Lance's book (if someone knows a good online reference, please add a comment!) For a $C^\ast$-algebra $A$ and a Hilbert space $H$ we consider the right $A$-module $H\odot A$ with $A$-valued inner-product $$ \big( \xi\otimes a \big| \eta\otimes b \big) = (\xi|\eta) a^*b. $$ A little work is required to show that this is positive definitive, see Lance page 6. Let $H\otimes A$ be the completion.

It is not so hard to show that the algebra of "compact" adjointable operators $\newcommand{\mc}{\mathcal}\mc K(H\otimes A)$ is $*$-isomorphic to $\mc K(H)\otimes A$, see Lance page 10. Thus the algebra of all adjointable operators satisfies $$ \mc L(H\otimes A) \cong M(\mc K(H\otimes A)) \cong M(\mc K(H)\otimes A). $$ Notice the RHS is exactly the algebra which interests us!

Given a bounded linear map $T:H_1\rightarrow H_2$ it is clear that $T\otimes\iota$ is a bounded linear map $H_1\otimes A\rightarrow H_2\otimes A$ which has adjoint $T^*\otimes\iota$. Thus, just using composition of adjointable maps, for $V_1\in\mc L(H_1\otimes A)$ and $V_2\in\mc L(H_2\otimes A)$ we have that $$ (T\otimes\iota)V_1, \quad V_2(T\otimes\iota) \in \mc L(H_1\otimes A, H_2\otimes A). $$ This gives the precise meaning you are after.


This does agree with the other interpretation. I suppose the way to see this is to consider the isomorphism $$ \mc L((H_1\oplus H_2)\otimes A) \cong \begin{pmatrix} \mc L(H_1\otimes A) & \mc L(H_2\otimes A, H_1\otimes A) \\ \mc L(H_1\otimes A, H_2\otimes A) & \mc L(H_2\otimes A) \end{pmatrix}. $$ This is proved by using the isomorphism $(H_1\oplus H_2)\otimes A \cong (H_1\otimes A) \oplus (H_2\otimes A)$. Now chase all the isomorphisms through.

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