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Let $(A, \Delta)$ be a compact quantum group in the sense of Woronowicz. Is it true that the comultiplication $\Delta : A \to A \otimes A$ always injective?

This is true for both the universal (because one has a counit) and the reduced (because the Haar state is faithful) version, but the general case seems more delicate.

In particular, is it true for the dual of a discrete group $\Gamma$ realized as a compact $C^*$-algebraic quantum group using a $C^*$-norm that lies strictly between the universal and the reduced $C^*$-norms? What about the case where $\Gamma$ is the free group on two generators?

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  • $\begingroup$ The result for the universal and reduced versions would be a nice textbook exercise. I presume you are using convolution? Is it the case if something is in the kernel of the comultiplication then it is in the null space of the Haar state (by the quotient to the reduced algebra). Then the question would reduce to injective on the null space of the Haar state. Not sure does that all check out though (or useful). $\endgroup$ Dec 19, 2021 at 19:30
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    $\begingroup$ @JPMcCarthy It is true that $\ker(\Delta) \subset \ker h$ ($h$ being the Haar state), because $(\mathrm{id}_A \otimes h)\Delta(a)=h(a)1$ for all $a \in A$. But if you play around a bit, you would soon find that this doesn't seem to help much except of course the reduced case where $h$ is faithful, as mentioned in the description of the question. $\endgroup$ Dec 19, 2021 at 20:47

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No, the comultiplication need not be injective. When $\Gamma$ is a countable group and $\pi : \Gamma \to \mathcal{U}(H)$ is a faithful unitary representation with the property that $\pi \otimes \pi$ is weakly contained in $\pi$, we write $A = C^*_\pi(\Gamma)$ and there is a unique unital $*$-homomorphism $\Delta : A \to A \otimes A$ satisfying $\Delta(\pi(g)) = \pi(g) \otimes \pi(g)$ for all $g \in \Gamma$. Then, $(A,\Delta)$ is a compact quantum group in the sense of Woronowicz.

Now $\Delta$ is faithful if and only if $\pi$ is weakly contained in $\pi \otimes \pi$. That need not be the case, as the following example with $\Gamma = \mathbb{F}_2$ shows.

For every $0 < \rho < 1$, we consider the function $\varphi_\rho : \mathbb{F}_2 \to \mathbb{R}$ given by $\varphi_\rho(g) = \rho^{|g|}$, where we use the word length $|g|$. By [Haa, Lemma 1.2], $\varphi_\rho$ is a positive definite function and we define $\pi_\rho$ as the associated cyclic representation. Denote by $\lambda$ the regular representation. By [Haa, Theorem 3.1], we have that $\pi_\rho$ is weakly contained in $\lambda$ if and only if $\rho \leq 1/\sqrt{3}$.

We now claim that with $\rho = 2/3$, the representation $\pi = \pi_\rho \oplus \lambda$ provides a counterexample for the faithfulness of $\Delta$. By construction, $\pi \otimes \pi$ is weakly equivalent with $(\pi_\rho \otimes \pi_\rho) \oplus \lambda$. Since $\rho^2 < 1/\sqrt{3}$, it follows from [Haa, Theorem 3.1] that $(\pi \otimes \pi) \sim \lambda \prec \pi$. But since $\rho > 1/\sqrt{3}$, we have $\pi_\rho \not\prec \lambda$, so that $\pi \not\prec \pi \otimes \pi$.

[Haa] U. Haagerup, An example of a nonnuclear C$^*$-algebra, which has the metric approximation property. Invent. Math. 50 (1978/79), 279-293.

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  • $\begingroup$ Unless I am mistaken, in the first paragraph, it should be $A = C^*_\pi(\Gamma)$ instead of $C^*_\pi(G)$ (I tried to correct this trivial typo, but obviously edits should be at least 6 characters long, so I can't do it). Anyway, thanks for this fantastic answer! $\endgroup$ Feb 16, 2022 at 23:56
  • $\begingroup$ Thank you. Yes, that was a typo and it is now corrected. $\endgroup$ Feb 17, 2022 at 7:22
  • $\begingroup$ @StefaanVaes Beautiful example! $\endgroup$
    – J. De Ro
    Mar 18, 2022 at 19:13

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