Before asking my question, let me introduce the relevant terminology.


Throughout, let $(A, \Delta)$ be a compact quantum group.


**Definition:** A representation $v$ on the Hilbert space $H$ is an element $v\in M(B\_0(H)\otimes A)$ such that $(\text{id}\otimes \Delta)(v) = v\_{(12)}v\_{(13)}$. Here the subscripts with the brackets denote the leg numbering notation.


**Definition:** An intertwiner from the representation $(H\_1, v\_1)$ to the representation $(H\_2, v\_2)$ is an element in $B(H\_1,H\_2)$ such that $(x\otimes 1)v\_1 =v\_2(x \otimes 1).$


**Definition:** A closed subspace $K$ of $H$ is called invariant under the representation $(H,v)$ if $(e\otimes 1)v(e\otimes 1) = v(e\otimes 1)$ where $e$ is the orthogonal projection of $H$ onto $K$.


**Question:** Let $x: H\_1 \to H\_2$ be an intertwiner as above. Why Is $\ker(x)$ an invariant subspace of $H\_1$?



Kernel of intertwiner is invariant (compact quantum groups)
Let $e$ be the orthogonal projection onto $\ker(x)$.
If the result is not true, then there is $\xi\otimes\eta \in H\_1\otimes K$ with
$$ (e\otimes 1) v\_1 (e\xi\otimes\eta) \ne v\_1 (e\xi\otimes\eta), $$
because the linear span of such vectors in dense in $H\_1\otimes K$. Here $K$ is some auxiliary Hilbert space such that we can regard $A\subseteq\mathcal B(K)$. Similarly, there is $\xi'\otimes\eta'\in H\_1\otimes K$ with
$$ \bigl((e\otimes 1) v\_1 (e\xi\otimes\eta) \bigm| \xi'\otimes\eta'\bigr) \not=
\bigl( v\_1 (e\xi\otimes\eta) \bigm| \xi'\otimes\eta'\bigr). $$
I write $(\cdot\mid\cdot)$ for the inner-product.


The map $T:\mathbb C \rightarrow K$; $\alpha\mapsto\alpha\eta'$ is bounded, and so there is an adjoint $T^\*:K\rightarrow\mathbb C$ which is simply $K \ni \zeta \mapsto (\zeta\mid\eta')$. Then $1\otimes T^\*:H\_1\otimes K \rightarrow H\_1\otimes\mathbb C \cong H\_1$ is bounded. Consider
$$ \xi'' = (1\otimes T^\*)v\_1(e\xi\otimes\eta) \in H\_1. $$
For any $\xi\_0\in H\_1$ we have that $(\xi''\mid\xi\_0) = (v\_1(e\xi\otimes\eta)\mid\xi\_0\otimes\eta')$ and hence in particular
$$ (\xi''\mid\xi') \ne (\xi''\mid e\xi'). $$


Notice that $(x\otimes 1)v\_1(e\otimes 1) = v\_2(xe\otimes 1)=0$ and so
$(x\otimes 1)v\_1(e\xi\otimes\eta) = 0$. Hence
$$ 0 = \bigl((x\otimes 1)v\_1(e\xi\otimes\eta)\bigm|\xi\_0\otimes\eta'\bigr)
= \bigl(v\_1(e\xi\otimes\eta)\bigm|x^\*\xi\_0\otimes\eta'\bigr)
= (\xi''\mid x^\*\xi\_0) = (x\xi''\mid\xi\_0), $$
for any $\xi\_0\in H\_1$. This shows that $x\xi''=0$ so $\xi''\in\ker(x)$ so $e\xi''=\xi''$, which gives the required contradiction.


