Direct sum of representations of a compact quantum group

Question

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.

Answer 1

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.)