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.


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

  • $\begingroup$ Thanks for your answer. It implicitly assumes that $X$ is unitary. Does it still hold if $X$ is only invertible and not unitary? In that case, there is a unitary $u \in B(H)$ such that $Y:=(u\otimes 1)X(u^* \otimes 1)$ is a unitary representation and we could hope to apply the unitary case to this representation. This would boil down to finding a projection $s$ such that $su = q$, i.e. $s =qu^*$ but such a projection almost never exists. $\endgroup$
    – Andromeda
    Commented Nov 25, 2021 at 11:25
  • 1
    $\begingroup$ I have updated my answer so as to also incorporate the case where $X$ is merely invertible. $\endgroup$ Commented Nov 25, 2021 at 15:42
  • $\begingroup$ Thanks for your update! In the case that $X$ is not unitary, how does an explicit inverse of $X_K$ look like (in function of an inverse of $X$)? $\endgroup$
    – Andromeda
    Commented Nov 28, 2021 at 16:10
  • $\begingroup$ Yes, using the notation of my answer, the inverse of $X_K$ is given by $X^{-1}(e \otimes 1)$ $\endgroup$ Commented Nov 29, 2021 at 7:47
  • $\begingroup$ Thanks! It all makes sense now. If you have time/interest, I have some other questions about compact quantum groups as well. $\endgroup$
    – Andromeda
    Commented Nov 30, 2021 at 19:50

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