# Woronowicz characters are multiplicative

I'm reading the book "Compact quantum groups and their representation categories" by Neshveyev-Tuset.

Let $$G$$ be a $$C^*$$-algebraic compact quantum group with function algebra $$(C(G), \Delta)$$. Given a unitary representation $$U \in B(H_U)\otimes C(G)$$ of $$G$$, there is a unique invertible bounded intertwiner $$\rho_U \in \operatorname{Mor}(U,U^{cc})$$ such that $$\operatorname{Tr}(-\rho_U) = \operatorname{Tr}(-\rho_U^{-1})$$ on $$\operatorname{End}(U)\subseteq B(H_U)$$ [Proposition 1.4.4, p16].

On p29, the following definition is given:

Definition: The Woronowicz characters is the family $$\{f_z\}_{z \in \mathbb{C}}$$ of linear functionals on $$\mathbb{C}[G]$$ (= the unique dense Hopf $$^*$$-subalgebra of $$C(G)$$, i.e. the space of matrix coefficients) defined by $$(\iota \otimes f_z)(U) = \rho_U^z$$ for all finite-dimensional representations $$U$$ of $$G$$.

Let us write $$\mathscr{U}(G):= \mathbb{C}[G]^*$$ (the algebraic dual space) which becomes a unital $$*$$-algebra in the obvious way (the $$*$$-algebra structure is induced by the Hopf-structure on $$\mathbb{C}[G]$$). Fixing a complete set of irreducible, pairwise non-equivalent unitary representations $$\{U_\alpha \in B(H_\alpha)\otimes C(G)\}_{\alpha \in I}$$, it is straightforward to see that the mapping $$\Phi: \mathscr{U}(G)\to \prod_{\alpha \in I} B(H_\alpha): \omega \mapsto ((\iota \otimes \omega)(U_\alpha))_{\alpha \in I}$$ is an isomorphism of $$*$$-algebras. Writing $$\mathscr{U}(G \times G) = (\mathbb{C}[G]\otimes \mathbb{C}[G])^*$$, we have a natural map $$\widehat{\Delta}: \mathscr{U}(G)\to \mathscr{U}(G \times G)$$ defined by $$\widehat{\Delta}(\omega)(a\otimes b) = \omega(ab)$$.

On p30, the authors write the following:

Why is the boxed equation true? I understand that $$\widehat{\Delta}(\rho) = \rho \otimes \rho$$.

I'm a bit confused when they say that we can do functional calculus in $$\mathscr{U}(G)$$. I guess this means we can look at the $$*$$-isomorphism $$\Phi$$ and do functional calculus in each $$B(H_\alpha)$$ and then use this to make sense of the functional calculus in $$\mathscr{U}(G)?$$

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, §1.6, 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.

• Thanks In the meantime, I was able to solve the question myself. Your answer is similar to what I did. The point is that we can check an equality in $\mathscr{U}(G^n)$ by checking that it holds in each component of the associated direct sum. Dec 14, 2021 at 20:05
• I had to submit, since in the preview most relations were not correctly compiled in mathjax. Dec 14, 2021 at 20:21