# Compact Quantum Groups and the Existence of the Classical Haar Measure

Before I state my question, let me provide the definition of a compact quantum group.

Definition: An ordered pair $\mathscr{G} = (\mathscr{A},\Phi)$ is called a compact quantum group if

• $\mathscr{A}$ is a separable and unital C*-algebra (whose identity element we denote by $1_{\mathscr{A}}$);
• $\Phi$ is a unital $*$-homomorphism from $\mathscr{A}$ to $\mathscr{A} \otimes_{\sigma} \mathscr{A}$, where ‘$\sigma$’ means ‘minimal’;
• the diagram $$\require{AMScd} \begin{CD} \mathscr{A} @>{\Phi}>> \mathscr{A} \otimes_{\sigma} \mathscr{A} \\ @V{\Phi}VV @VV{\Phi \otimes \text{id}}V \\ \mathscr{A} \otimes_{\sigma} \mathscr{A} @>{\text{id} \otimes \Phi}>> \mathscr{A} \otimes_{\sigma} \mathscr{A} \otimes_{\sigma} \mathscr{A} \end{CD}$$ commutes;
• the sets $$\{ (a \otimes 1_{\mathscr{A}}) \Phi(b) \mid a,b \in \mathscr{A} \} \quad \text{and} \quad \{ (1_{\mathscr{A}} \otimes a) \Phi(b) \mid a,b \in \mathscr{A} \}$$ are linearly dense in $\mathscr{A} \otimes_{\sigma} \mathscr{A}$.

In this set of notes on compact quantum groups by S. Woronowicz, one finds the following remarkable theorem whose proof relies upon nothing more than the representation theory of C*-algebras.

Theorem: Let $\mathscr{G} = (\mathscr{A},\Phi)$ be a compact quantum group. Then there exists a unique state (i.e., a normalized positive linear functional) $h$ on $\mathscr{A}$ such that $$h \star a = a \star h = h(a) \cdot 1_{\mathscr{A}},$$ where $$h \star a \stackrel{\text{def}}{=} (\text{id} \otimes h)(\Phi(a)) \quad \text{and} \quad a \star h \stackrel{\text{def}}{=} (h \otimes \text{id})(\Phi(a)).$$ We call $h$ the Haar state on $\mathscr{G}$.

Out of curiosity, I tried to see how Woronowicz’s Theorem implies the existence of the classical Haar measure on a compact Hausdorff topological group $G$. Despite my best efforts, I only managed to succeed in the case where $G$ is additionally second-countable.

Let $G$ be a compact Hausdorff topological group. Suppose further that it is second-countable (or equivalently, metrizable). Then $C(G)$ ― the space of complex-valued continuous functions on $G$ equipped with the supremum norm ― is a separable and unital C*-algebra.

Consequently, $(C(G),\Delta)$ is a compact quantum group, where the co-multiplication $$\Delta: C(G) \to C(G) ~ \widehat{\otimes} ~ C(G) \cong C(G \times G)$$ is defined by $$\forall x,y \in G: \quad {\Delta f}(x,y) \stackrel{\text{def}}{=} f(x y).$$ According to Woronowicz’s Theorem, there exists a unique state $h$ on $C(G)$ such that $$(\spadesuit) \quad h \star f = f \star h = h(f) \cdot \mathbf{1}.$$ View $\Delta f$ as the limit of a net $\displaystyle \left( \sum_{i \in I_{\lambda}} p_{\lambda,i} \otimes q_{\lambda,i} \right)_{\lambda \in \Lambda}$ of finite sums of elementary tensors. Then \begin{align*} \forall x \in G: \quad (h \star f)(x) & = [(\text{id} \otimes h)(\Delta f)](x) \\ & = \left[ (\text{id} \otimes h) \left( \lim_{\lambda} \sum_{i \in I_{\lambda}} p_{\lambda,i} \otimes q_{\lambda,i} \right) \right](x) \\ & = \left[ \lim_{\lambda} \sum_{i \in I_{\lambda}} p_{\lambda,i} \cdot h \left( q_{\lambda,i} \right) \right](x) \quad (\text{By continuity.}) \\ & = \lim_{\lambda} \sum_{i \in I_{\lambda}} {p_{\lambda,i}}(x) \cdot h \left( q_{\lambda,i} \right) \quad (\text{By continuity.}) \\ & = \lim_{\lambda} h \left( \sum_{i \in I_{\lambda}} {p_{\lambda,i}}(x) \cdot q_{\lambda,i} \right) \quad (\text{By the linearity of $h$.}) \\ & = h \left( \lim_{\lambda} \sum_{i \in I_{\lambda}} {p_{\lambda,i}}(x) \cdot q_{\lambda,i} \right) \quad (\text{By continuity.}) \\ & = h({\Delta f}(x,\cdot)) \\ & = h({L_{x}}(f)), \end{align*} where $L_{x}: C(G) \to C(G)$ denotes left translation by $x$. Similarly, $$\forall x \in G: \quad (f \star h)(x) = h({R_{x}}(f)),$$ where $R_{x}: C(G) \to C(G)$ denotes right translation by $x$. By $(\spadesuit)$, $$\forall x \in G: \quad h({L_{x}}(f)) = h({R_{x}}(f)) = h(f).$$ This implies that $h$ is a left- and right-invariant state on $C(G)$. It follows from the Riesz Representation Theorem that $h$ corresponds to a unique regular Borel probability measure on $G$ that is left- and right-invariant. This proves the existence of the classical Haar measure on $G$.

My question is:

How can I strengthen the argument above so that it yields the existence of the classical Haar measure on a general compact Hausdorff topological group? Is there a way to circumvent the requirement of separability in the definition of a compact quantum group?

Thank you very much for your help!

• I don't know about strengthening Woronowicz' argument, but you could use instead the approoch of Van Daele's proof, see ams.org/journals/proc/1995-123-10/S0002-9939-1995-1277138-0.
– UwF
May 6 '14 at 8:16
• Btw, there is another proof, using a fixed point theorem, but requiring cocommutativity, by Xiu-Chi Quan, see link.springer.com/article/10.1007/BF01002248.
– UwF
May 6 '14 at 8:21
• Thanks, UwF! Van Daele’s paper seems to be exactly what I need.
– user36116
May 7 '14 at 0:04

I don't think there is any direct generalization of this proof. In the language of measure theory, Woronowicz's argument goes by starting with a probability measure $\mu$ on $X$ that has full support, then taking a weak* limit of the Cesaro means of the sequence of convolution powers of $\mu$. If $\mu$ doesn't have full support then you can't expect this limit to be translation invariant. And if $X$ is not metrizable, then it doesn't seem obvious right off that there are any probability measures with full support. Certainly there exist compact Hausdorff spaces for which no finite measure has full support.