This is taken from Timmermann's Invitation to Quantum Groups and Duality.
Let $(A,\Delta)$ be a *-Hopf algebra and let $\chi:V\rightarrow V\otimes A$ be a corepresentation of $(A,\Delta)$ on a vector space $V$. Call a vector subspace $W\subset V$ invariant with respect to $\chi$ if $\chi(W)\subset W\otimes A$.
Question
If $\chi$ is unitary and $W\subset V$ is an invariant subspace, then the orthogonal complement $W^\perp$ is invariant also.
Timmermann proceeds as follows. Define a map from the dual of $A$, $\pi:A'\rightarrow \hom(V)$ by $$\pi(f)v:=(I\otimes f)\chi(v).$$ Timmerman proves that $W\subset V$ is invariant with respect to $\chi$ if and only if $\pi(A')W\subset W$. Now he simply states "Now the claim follows from a standard argument."
The problem is that I don't see what this standard argument is.
Initially I said suppose that $\tilde{w}\in W^\perp$ so that $\langle\tilde{w},w\rangle=0$. Then I calculated $$\langle \pi(f)\tilde{w},w\rangle=\sum\langle {\tilde{w}}_{(0)}f(\tilde{w}_{(1)}),w\rangle$$ where the Sweedler notation is used. This didn't really go anywhere.
The next idea was to say that $\pi$ was non-degenerate and use Timmermann's result that $\pi$ is an algebra homomorphism of $A'$ with the multiplication/convolution $(f,g)\mapsto (f\otimes g)\Delta$. The next part was what do inverses of $f$ look like here?
$$(f\otimes f^{-1})\Delta(a)=1?$$
If I could do something like this I could certainly carry out a standard argument (familiar to me from the classical Peter Weyl Theorem). Let $\tilde{w}\in W^\perp$ and $w\in W$:
$$\langle \pi(f)\tilde{w},w\rangle=\langle \pi(f^{-1})\pi(f)\tilde{w},\pi(f^{-1})w\rangle$$
but I can't quite justify those calculations.
Can anybody out there help?
Someone pointed me in the right direction.
The question had been put to me: what does it mean to say that $\chi$ is unitary. Timmermann defines an "$A$-valued inner product" on $V\otimes A$ by $$\langle v\otimes a |w\otimes b\rangle_A:=\langle v|w\rangle_V a^*b.$$ Now $\chi$ is said to be unitary if $$\langle\chi(v)|\chi(w)\rangle_A=\langle v|w\rangle_V\cdot 1_A.$$
Timmermann also proves that if $\chi$ is unitary then $\pi$ is a *-homomorphism.
This answers the question.
As before suppose that $\langle \tilde{w},w\rangle_V=0$. Now investigate $\langle \pi(f)\tilde{w},w\rangle_V$. Now take the adjoint of $\pi(f)$: $$\langle \tilde{w},\pi(f)^*w\rangle_V=\langle \tilde{w},\pi(f^*)(w)\rangle_V.$$ By assumption $W$ is invariant for $\pi(A')$ so now we can conclude that $\langle \pi(f)\tilde{w},w\rangle_V=0$ and we are done.