Let $\mathbb{G}$ be a compact quantum group with function algebra $(C(\mathbb{G}), \Delta)$ (in the sense of Woronowicz). Let $X \in M(B\_0(H) \otimes C(\mathbb{G}))$ be a (possibly infinite-dimensional) representation of the quantum group $\mathbb{G}$, and let $K$ be an $X$-invariant subspace of $H$, i.e. if $p\in B(H)$ is the projection on the closed subspace $K$, then $(p\otimes 1)X (p\otimes 1) = X(p\otimes 1).$


Where one encounters an invariant subspace $K$, one hopes to define a subrepresentation $X\_K \in M(B\_0(K)\otimes C(\mathbb{G}))$. It looks like there is an obvious way to do this:


Consider the surjective strict completely positive map
$$\Psi: B\_0(H) \to B\_0(K): x \mapsto pxp^\*$$
The strict completely positive map $$\Psi \otimes \iota: B\_0(H) \otimes C(\mathbb{G}) \to B\_0(K) \otimes C(\mathbb{G})$$ extends uniquely to a bounded linear map
$$\Psi \otimes \iota: M(B\_0(H) \otimes C(\mathbb{G}) \to M(B\_0(K)\otimes C(\mathbb{G}))$$
which is strictly continuous on bounded subsets and we define
$$X\_K:= (\Psi \otimes \iota)(X)$$
as our candidate for a subrepresentation. Everything works out nicely, for example, it is easily verified that
$$(\iota \otimes \Delta)(X\_K) = (X\_K)\_{12}(X\_K)\_{13}.$$
However, what is not clear to me is why $X\_K$ must be invertible (I consider representations to be invertible elements in the multiplier algebra by definition, and representations do not need to be unitary). Does invertibility of $X \in M(B\_0(H)\otimes C(\mathbb{G}))$ imply invertibility of $X\_K \in M(B\_0(K) \otimes C(\mathbb{G}))?$ I have a feeling that the answer might be negative because compressing with a projection can make things non-invertible.



Subrepresentations of C*-algebraic compact quantum groups
The following was my original answer, dealing with the case where $X$ is unitary.


It is a nontrivial fact that the orthogonal complement of an invariant subspace is again an invariant subspace. Thus, the projection $p$ in the question will automatically satisfy the stronger property $(p \otimes 1)X = X(p \otimes 1)$. This was part of Woronowicz' original approach to the representation theory of compact quantum groups. You can also find this result as [Proposition 6.2 in the expository notes of Maes and Van Daele](https://arxiv.org/abs/math/9803122).


When $X$ is merely an invertible representation and the orthogonal projection $p$ of $H$ onto $K$ satisfies $(p \otimes 1) X (p \otimes 1) = X(p \otimes 1)$, it is still true that the restriction of $X$ to the invariant subspace $K$ is an invertible representation of $\mathbb{G}$. The only tricky point is that the equality $X(p \otimes 1) = (p \otimes 1)X$ need not hold. The point is that the complementary invariant subspace $L \subset H$ is not the orthogonal complement of $K$, but another complement of $K$.


For every continuous functional $\omega$ on $C(\mathbb{G})$, denote $X(\omega) = (\text{id} \otimes \omega)(X)$. The assumptions say that $X(\omega) K \subset K$ for every $\omega$. The unitarizability means that we can find an invertible $u \in B(H)$ (not necessarily unitary though) such that $Y := (u \otimes 1) X (u^{-1} \otimes 1)$ is a unitary representation. Writing $K' = u(K)$, we have that $Y(\omega)K' \subset K'$ for all $\omega$. Denoting by $q$ the orthogonal projection onto $K'$, the first paragraph says that $Y (q \otimes 1) = (q \otimes 1) Y$.


We then define $e$ as the, potentially nonorthogonal, projection $e = u^{-1}qu$. Then, $X(e \otimes 1) = (e \otimes 1) X$ is invertible. The projection $e$ is still a projection onto $K$. So, $X(e \otimes 1)$ is the restriction of $X$ to $K$, which thus is invertible. The complementary invariant subspace is $L = (1-e)(H)$.


