Subrepresentations of C*-algebraic compact quantum groups

Question

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.

Answer 1

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.

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 $L = (1-e)(H)$.