Since i could not find the reference book, below the following earlier version of it is used:
CQGRC.pdf - Compact Quantum Groups and Their Representation Categories - Incomplete preliminary version as of January 17, 2014, by Sergey Neshveyev and Lars Tuset.
In [CQGRC], it is Proposition: 1.7.2, CQGRC, page 21 claiming that
$f_z$ has the property (i) of being a is a homomorphism $\Bbb C[G]\to\Bbb C$.
and the proof of (i) claims the relation
$$
\tag{$\dagger$}
\hat\Delta(\rho^z)=\rho^z\otimes \rho^z\ .
$$
So let us comment first on relation $(\dagger)$ above, which is the question.
For the convenience of the reader, we unpack the notations one by one, and describe
the framework:
Framework:
$G=(A,\Delta)$ is the given $C^*$-algebraic CQG. The underlying algebra is in an alternative notation $A=C(G)$.
(I will use preferably $A$ below, since it is simpler to type.)
We need the space of matrix coefficients of $A$, denoted alternatively by $A_0=\Bbb C[G]$.
The functionals $f_z$ are linear functionals $f_z:A_0\to \Bbb C$, and we write $A_0'$ for this space of functionals.
So for each complex $z$ we have $f_z\in A_0'$. To put the hands on some $f_z$ we need to know the structure of $A_0$.
What is $A_0$? The first proposition in CQGRC, ยง1.6, is:
For a compact quantum group $G=(A,\Delta)$ denote by $A_0=\Bbb C[G]\subset C(G)=A$ the linear span of
matrix coefficients of all finite dimensional (co)representations of $G$, it is a dense subspace of $A$ by CQGRC, Corollary 1.5.6.
So let $(U_\alpha)$ be such a maximal
family of irreducible, unitary, mutually inequivalent representations of $G=(A,\delta)$. The index $\alpha$ runs in some index set
$\Lambda$, that we will often omit.
Here,
$$
U_\alpha = \sum m^\alpha_{ij}\otimes u^\alpha_{ij}\in B(H_\alpha)\otimes A\ .
$$
The matrices $m^\alpha_{ij}$ are the elementary matrices, the canonical basis of the matrix space $B(H_\alpha)$.
For an irreducible, unitary representation $U\in B(H) \otimes A_0$ its contragredient cousin is $U^c$.
CQGRC, Proposition 1.3.13 shows that $U$ and $U^{cc}$ are equivalent.
The space of intertwiners between $U$, and $U^{cc}$, in notation $\operatorname{Mor}(U,U^{cc})\operatorname{Mor}(U,U)$ is thus one-dimensional
(Schur),
and we have even an explicitly constructed intertwiner $j(Q_r)\in \operatorname{Mor}(U, U^{cc})$.
Here $Q_r$ is a positive, invertible operator in $B(\bar H)$, and $j$ is the antimorphism on $B(H)$.
We rescale $Q_r$ by a positive scalar $\lambda_U$, so that the obtained invertible, positive operator $\rho_U:=\lambda_U\; j(Q_r)$
is normed by the condition:
$$
\operatorname{Trace}(\rho_U) =
\operatorname{Trace}(\rho_U^{-1}) \ .
$$
From now on, we can forget about $j(Q_r)$, and $\lambda_U$, and need only the information that the space of intertwiners is one-dimensional,
generated by an invertible, positive operator $\rho_U$.
Since $\rho_U>0$, and for each $z$ the map $a\to a^z:=e^z\log a$ is analytic on $(0,\infty)$, we have an analytic functional calculus
defining the positive operator
$$(\rho_U)^z=:\rho_U^z\in B(H)\ .$$
Now, $f_z$ is introduced in CQGRC, Definition 1.7.1, implicitly. We do not have a formula to evaluate
$f_z$ on an elements from $A_0$, instead, we use the matrix coefficient spaces of individual irreducible representations.
Note that for different (inequivalent) unitary (co)representations $U_\alpha$, $U_\beta$ the corresponding
spaces of matrix coefficients
$\mathscr C(\alpha):=\operatorname{Span}(U_\alpha):=\operatorname{Span}(u_{\alpha,ij})$ and
$\mathscr C(\beta):=\operatorname{Span}(U_\beta):=\operatorname{Span}(u_{\beta,kl})$
generated by their entries are linearly independent (orthogonal). Moreover, this independence holds also for the coefficients
of $U_\alpha$, so that extending $f_z$ is uniquely defined on $\mathscr C(\alpha)$ by the relation:
$$
%\Big(\operatorname{id}_{B(H_{\alpha})}\otimes f_z\Big)\;\Big(U_\alpha\Big) \ =\ \rho_{\alpha}^z\in B(H_{\alpha})_{>0}\ .
(\operatorname{id}\otimes f_z)(U_\alpha)\ =\ \rho_{\alpha}^z\in B(H_{\alpha})\ .
$$
We denoted by $\rho_\alpha$ the positive operator $\rho_{H_\alpha}$ constructed above.
(Unpacking, if $m_{\alpha,ij}$ is the canonical basis of $B(H_\alpha)$, and $U_\alpha =\sum m^\alpha_{ij}\otimes u^\alpha_{ij}$ to obtain
$f_z(u^\alpha_{ij})$ we compute $\rho_\alpha$, then its $z$-power by analytic functional calculus, then we write this result in the $m^\alpha$--basis and take the
coefficient in $m^\alpha_{ij}$.
So far we have the $\rho_U$ objects. Then the $\rho_U^z$ objects. Using them we get $f_z\in A_0'$.
We forget about them, and need in fact only $f_1$.
So we need in fact only $\rho_U$ to know $f_1$ on the vector space spanned by matrix coefficients of $U$.
And now, the text also uses the letter $\rho$ for $f_1$.
(This may be nice a posteriori. But at this point, let us be more careful.)
So i will slightly change notation and use instead:
$$ \varrho:=f_1\in A_0'=\mathscr U(G)\overset{\Phi=\Phi_G}\longrightarrow \prod_\alpha B(H_\alpha)\ .
$$
The text identifies $\varrho$ with its image $\Phi(\varrho)$ in the product of matrix spaces.
Let us compute it explicitly, using $\pi_{U_\alpha}$ as in CQGRC, page 20:
$$
\begin{aligned}
\Phi(\varrho) &=(\ \Phi(\varrho)_\alpha\ )_\alpha \ ,\\
\Phi(\varrho)_\alpha &:= \pi_{U_\alpha}(\varrho)\\
&:=(\operatorname{id}\otimes\varrho)(U_\alpha)\\
&:=(\operatorname{id}\otimes f_1)(U_\alpha)\\
&:=\rho_{U_\alpha}^1=\rho_{U_\alpha}=\rho_\alpha\ .
\end{aligned}
$$
In other words, $\Phi(\varrho)$ is an element in $\prod B(H_\alpha)$ which has as $\alpha$-component the
positive operator $\rho_\alpha\in B(H_\alpha)$. I will write $\rho$ for this family, $\rho=(\rho_\alpha)=\Phi(\varrho)$.
(At any rate, we can now imagine why the notation $\rho$ was kept for $\varrho$.)
What is $\varrho^z\in \mathscr U(G)=A_0'$? It is by definition the functional which is mapped by
$\Phi$ into $\rho^z:=(\rho_\alpha^z)_\alpha$.
So identities involving $\varrho^z$ should be rather moved from the $\mathscr U$-spaces to the matrix spaces.
Let us recall $\hat\Delta$ is as a map $\mathscr{U}(G)\to \mathscr{U}(G\times G)$, which is
determined by its evaluation at some functional $\omega \in A_0' =\mathscr{U}(G)$. The image lies in $\mathscr{U}(G\times G)$.
(It can be bigger than $A_0'\otimes A_0'$.) The functional $\hat \Delta(\omega)\in \mathscr{U}(G\times G)$ is determined
by evaluation on elements of the shape $a\otimes b\in A_0\otimes A_0$. So $\hat\Delta$ is determined by the double evaluation
$$
\hat\Delta(\omega)\ a\otimes b :=\omega(ab)\ .
$$
Our $\omega$ of interest is $\rho^z$.
Recall that we also have a map needed in the sequel (and also displayed vertically),
$$
\require{AMScd}
\begin{CD}
\mathscr U(G) @. \omega\\
@V \Phi_G V V @VVV\\
\prod_{\alpha} B(H_\alpha) @. (\ (\operatorname{id}_{H_\alpha}\otimes\omega)(U_\alpha)\ )_\alpha
\end{CD}
$$
The same $\Phi$-mapping can be written also for $G\times G$, the corresponding irreducible representations are parametrized by tuples $(\alpha,\beta)$,
and are of the shape $U_\alpha\odot U_\beta\in B(H_{(\alpha,\beta)})\otimes A_0\otimes A_0$, where $B(H_{(\alpha,\beta)}):=B(H_\alpha)\otimes B(H_\beta)$
it the algebraic tensor product of the two matrix spaces.
(The composition with the product $A_0\otimes A_0\to A_0$ gives rise to the representation denoted by $U_\alpha\odot U_\beta$ in CQGRC.)
Then consider the diagram, which is for a general $\omega$ not commutative:
$$
\require{AMScd}
\begin{CD}
\omega @>\hat\Delta_G>> \hat\Delta(\omega)\\
\\
\mathscr U(G) @>\hat\Delta_G>> \mathscr U(G\times G)\\
@V \Phi_G V V (??) @VV \Phi_{G\times G} V\\
\prod_\alpha B(H_\alpha) @>>\underline\Delta > \prod_{(\alpha,\beta)} B(H_\alpha) \otimes B(H_\beta)\\
\\
(w_\alpha)_{\alpha\in\Lambda} @>>\underline\Delta > (w_\alpha\otimes w_\beta)_{(\alpha,\beta)\in\Lambda\times\Lambda}
\end{CD}
$$
However for the special value $\omega=\varrho$, which is a "group-like" element,
$\hat\Delta(\varrho)=\varrho\otimes \varrho$ we have the commutativity
$$
\require{AMScd}
\begin{CD}
\varrho @>\hat\Delta >> \hat\Delta(\varrho)\\
@V \Phi_G V V @VV \Phi_{G\times G} V\\
\rho=(\rho_\alpha) @>>\underline\Delta > \rho\otimes \rho =(\rho_\alpha\otimes\rho_\beta)_{(\alpha,\beta)}
\end{CD}
$$
In fact, this group-like property of $\varrho$ is checked in the $\rho$-world,
CQGRC, Theorem 1.4.8,
$$
\rho_{U_\alpha\times U_\beta}=\rho_{U_\alpha}\otimes \rho_{U_\beta}\ .
$$
(This is as stated an equality in the category of representations for $G$.
The part from $A$ involved in what we need below,
when $U_\alpha\times U_\beta$ occurs,
is evaluated to a factor.)
So we compute the component $(\alpha,\beta)$ of $\hat\Delta$ evaluated in $\varrho$.
We write explicitly at some point:
- $U_\alpha =\sum m^\alpha_{ij}\otimes u^\alpha_{ij}\in B(H_\alpha)\otimes A_0$, representation of $G$,
- $U_\beta =\sum m^\beta_{kl}\otimes u^\beta_{kl}\in B(H_\beta)\otimes A_0$, representation of $G$, so that
- $U_\alpha\odot U_\beta= U_{(\alpha,\beta)}=\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes u^\alpha_{ij}\otimes u^\beta_{kl}\in B(H_\alpha)\otimes B(H_\beta)\otimes A_0\otimes A_0$
is the corresponding tensor product representation of $G\times G$. (Not for $G$, when the times notation is used.)
$$
\begin{aligned}
(\ \Phi_{G\times G}\ \hat\Delta(\varrho)\ )_{(\alpha,\beta)}
&:=
\big(\ \operatorname{id}_{B(H_{(\alpha,\beta)})}\otimes\hat\Delta(\varrho)\ \Big)(U_{(\alpha,\beta)})
\\
&=
\big(\
\operatorname{id}_{B(H_{\alpha})}\otimes
\operatorname{id}_{B(H_{\beta})}\otimes
\hat\Delta(\varrho)\ \Big)
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes u^\alpha_{ij}\otimes u^\beta_{kl}
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \hat\Delta(\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \hat\Delta(\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\otimes \underbrace{(\varrho\otimes\varrho)(u^\alpha_{ij}\otimes u^\beta_{kl})}_{\in\Bbb C}
\\
&=
\sum m^\alpha_{ij}\otimes m^\beta_{kl}\cdot \varrho(u^\alpha_{ij})\cdot \varrho(u^\beta_{kl})
\\
&=
\sum m^\alpha_{ij}\cdot \varrho(u^\alpha_{ij})\otimes \sum m^\beta_{kl}\cdot \varrho(u^\beta_{kl})
\\
&=
(\operatorname{id}\otimes\varrho)\left(\sum m^\alpha_{ij}\otimes u^\alpha_{ij}\right)
\otimes
(\operatorname{id}\otimes\varrho)\left(\sum m^\beta_{kl}\otimes u^\beta_{kl}\right)
\\
&=
(\operatorname{id}\otimes\varrho)(U_\alpha)
\otimes
(\operatorname{id}\otimes\varrho)(U_\beta)
\\
&=\rho_\alpha\otimes\rho_\beta
\\
&=\underline\Delta(\rho)_{(\alpha,\beta)}
\\
&=(\ \underline\Delta(\Phi(\varrho)\ )_{(\alpha,\beta)}
\ .\qquad\text{ So:}
\\[2mm]
\Phi_{G\times G}\; \hat\Delta(\varrho)
&=\underline\Delta\; \Phi(\varrho)\ .
\end{aligned}
$$
As a final word, a way of giving a sense to the boxed identity from the question,
$$
\hat\Delta(\varrho^z) =\varrho^z\otimes\varrho^z\ ,
$$
which is related to the upper horizontal arrow in the above diagrams,
is by moving it downwards via the $\Phi$ arrows to the lower horizontal arrow,
which is a map $\underline\Delta$ clearly compatible with the functional calculus,
$$
\underline\Delta(\rho^z)_{(\alpha,\beta)}
=
\rho_\alpha^z\otimes\rho_\beta^z
=
(\rho_\alpha\otimes\rho_\beta)^z
=
(\ \underline\Delta(\rho)\ )^z
\ .
$$
The definition of $\varrho^z$ is by taking $\rho^z$ from the L.-most.H.S. and pushing it via $\Phi^{-1}$
into $\mathscr U(G)$. It may be then useful in the vertical $G\times G$-arrow to write
$\varrho^z\otimes\varrho^z$ as a product of $\varrho^z\otimes\epsilon$ and $\epsilon\otimes\varrho^z$,
then go down via $\Phi$ to get by definition of $\varrho$ the commuting operators
$\rho^z\otimes 1$ and $1\otimes \rho^z$, and here we have
$$
(\rho\otimes\rho)^z =
(\ (\rho\otimes 1)\;(1\otimes\rho)\ )^z =
(\rho\otimes 1)^z\;(1\otimes\rho)^z\ .
$$
The notations in CQGRC were my biggest problems. I hope the above
answer - a general nonsense categorial translation - hits the wound point.
Best Answer
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 $L = (1-e)(H)$.