Let $(A, \Delta)$ be a compact quantum group and $\{(H\_\alpha, v\_\alpha)\}$ be a collection of representations of $A$. That is,
$$v\_\alpha \in M(B\_0(H\_\alpha) \otimes A); \quad \quad(\text{id}\otimes \Delta)(v\_\alpha) = (v\_\alpha)\_{(12)}(v\_\alpha)\_{(13)}$$


I want to show that there is a direct sum of these representations. More precisely, put $H = \bigoplus\_\alpha H\_\alpha$. Then I want to show that there is a representation $v$ on $H$ such that the natural inclusion
$$i\_\alpha: H\_\alpha \hookrightarrow H$$ is an intertwiner from $v\_\alpha$ to $v$ for all $\alpha$.


**Attempt**: Let $A \subseteq B(K)$ be the universal GNS representation and view all the relevant multiplier algebras as bounded operators in the canonical way. Then we have $v\_\alpha \in B(H\_\alpha \otimes K)$ and we can form
$$v:= \bigoplus\_\alpha v\_\alpha \in B\left(\bigoplus\_i (H\_i \otimes K)\right) \cong B\left(\left(\bigoplus\_i H\_i\right) \otimes K\right)= B(H \otimes K)$$


I managed to show that $v \in M(B\_0(H) \otimes K)$. However, I am stuck at showing that
$$(\text{id}\otimes \Delta)(v) = v\_{(12)}v\_{(13)}$$


Let $\xi =(\xi\_\alpha)\_\alpha \in H, \eta\_1, \eta\_2$. It suffices to check


$$(\text{id}\otimes \Delta)(v)(\xi \otimes \eta\_1 \otimes \eta\_2) = v\_{(12)}v\_{(13)}(\xi \otimes \eta\_1 \otimes \eta\_2).$$


To prove this, one might 'guess' that
$$(\text{id}\otimes \Delta)(v)(\xi \otimes \eta\_1 \otimes \eta\_2) = ((\text{id}\otimes \Delta)(v\_\alpha)(\xi\_\alpha \otimes \eta\_1 \otimes \eta\_2))\_\alpha$$


but I cannot formally prove this identity. Probably, I'm overseeing some easy trick here.



Direct sum of representations of a compact quantum group
Again, this is a definition chase on what exactly $(\iota\otimes\Delta)$ is. This is defined $\newcommand{\mc}{\mathcal}\mc B\_0(H) \otimes A \rightarrow \mc B\_0(H) \otimes A \otimes A$ and then extended by non-degeneracy to the multiplier algebra. So, for $v \in M(\mc B\_0(H)\otimes A)$ and $t\otimes x\in\mc B\_0(H)\otimes A$ and $s\otimes y\otimes z\in\mc B\_0(H)\otimes A\otimes A$ we have
$$ (\iota\otimes\Delta)(v) \big( (t\otimes\Delta(x))(s\otimes y\otimes z) \big)
= \big[ (\iota\otimes\Delta)(v(t\otimes x)) \big] (s\otimes y\otimes z). $$
The point is that the linear span of elements like $(t\otimes\Delta(x))(s\otimes y\otimes z)$ is dense in $\mc B\_0(H)\otimes A\otimes A$ and so this does define $(\iota\otimes\Delta)(v)$ as a multiplier.


In our setting, $H=\bigoplus H\_\alpha$ and so by linearity and density, it is enough to suppose that we have chosen $\alpha,\beta$ and we have $t\in\mc B\_0(H\_\alpha)$ and $s\in\mc B\_0(H\_\beta, H\_\alpha)$. This follows as then $ts\in\mc B\_0(H\_\beta, H\_\alpha)$ and any member of $\mc B\_0(H\_\beta, H\_\alpha)$ can be approximated by such a product (and then an arbitrary member of $\mc B\_0(H)$ approximated by linear combinations).


But then $v(t\otimes x) = v\_\alpha(t\otimes x)$ and so we can use that $(\iota\otimes\Delta)(v\_\alpha) = v\_{\alpha,12} v\_{\alpha,13}$. In other words, I have verified your "guess".


(This is a bit of a sketch, but to fill in the details is really just an exercise in getting the notation correct.)


