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Let $\mathbb{G}$ be a locally compact quantum group and let $W$ and $V$ be the left and right fundamental unitaries, i.e., they implement the coproduct on $\mathbb{G}$:
$$\Delta(x) = W^*(1\otimes x)W = V(x\otimes 1)V^*.$$
We can lift $\Delta$ to distinct coproducts $\Delta^l$ and $\Delta^r$ on $\mathcal{B}(L^2(\mathbb{G}))$ by setting $$\Delta^l(T) = W^*(1\otimes T)W, ~ T\in\mathcal{B}(L^2(\mathbb{G}))$$ and $$\Delta^r(T) = V(T\otimes 1)V^*, ~ T\in\mathcal{B}(L^2(\mathbb{G})).$$ Then we can define the left and right convolution products $\triangleleft$ and $\triangleright$ respectively on $\mathcal{T}(L^2(\mathbb{G}))$ by setting $$\omega\triangleleft \tau = (\omega\otimes \tau)\Delta^l ~ \text{and} ~\omega\triangleright \tau = (\omega\otimes \tau)\Delta^r$$ which lifts the convolution product on $L^1(\mathbb{G})$. Let $\hat{\triangleleft}$ and $\hat{\triangleright}$ denote the dual convolution products, i.e, the left and right convolution products on $\widehat{\mathbb{G}}$ respectively. A celebrated result of Kalantar and Neufang (https://doi.org/10.4153/CJM-2012-047-x or arXiv:1110.4933) is that anticommutation between $\triangleright$ and $\hat{\triangleright}$ holds: $$(\omega \triangleright \tau)\mathbin{\hat{\triangleright}} \rho = (\omega \mathbin{\hat{\triangleright}} \rho)\triangleright \tau, ~ \omega,\tau,\rho\in\mathcal{T}(L^2(\mathbb{G})).$$ Questions: Do we have anticommutation for the left convolution products? That is, does $$\omega\mathbin{\hat{\triangleleft}}(\tau\triangleleft \rho) = \tau\triangleleft (\omega\mathbin{\hat{\triangleleft}}\rho), ~\omega,\tau,\rho\in\mathcal{T}(L^2(\mathbb{G}))?$$ More generally, can we say anything about the maps $$({\operatorname{id}}\otimes \Delta^l)\hat{\Delta}^l ~ \text{or}~ (\Delta^l\otimes {\operatorname{id}})\hat{\Delta}^l?$$ What if we restrict the domain to $L^\infty(\mathbb{G})$ or $L^\infty(\widehat{\mathbb{G}})$?
An inspection of the proof by Kalantar and Neufang (https://doi.org/10.4153/CJM-2012-047-x or arXiv:1110.4933) does not make this clear. For right convolution products, the proof makes essential use of the fact $(\Delta^r\otimes {\operatorname{id}})\hat{\Delta}^r = \hat{\Delta}^r_{13}$ which holds because the left leg of $V$ lies in $L^\infty(\widehat{\mathbb{G}})'$. An attempt to replicate the proof with a strategic replacement of $L^\infty(\widehat{\mathbb{G}})$ and $L^\infty(\mathbb{G})$ with their commutants fails because neither leg of $W$ lies in a commutant. I also fail to see how to use the extended unitary antipode to transport the result from right convolution products to left convolution products.
This makes me think anticommutation simply doesn't hold for left convolution products. This seems like something worth noting, however, I have failed to find a note of this one way or the other in the literature. This makes me think I'm missing something obvious. I would be pleased with any help even in the case where $\mathbb{G}$ is compact.
