Let $(A,\Delta)$ be a compact quantum group in the sense of Woronowicz, and let $A_0$ be its dense Hopf subalgebra. We can construct from the Haar state $h:A \to \mathbb{C}$ an inner product $$ \langle \cdot,\cdot\rangle: A_0 \times A_0 \to \mathbb{C}, ~~~~~ (a,b) \mapsto h(a^*b). $$ Let $L^2(A_0)$ be the completion of $A_0$ to a Hilbert space. Let $\pi_L:A_0 \to \mathcal{B}(L^2(A_0))$ denote the representation of $A_0$ on $L^2(A_0)$ given by left multiplication. This gives a norm on $A_0$ defined by $\|a\| := \|\pi_L(a)\|_{op}$, where $\|\cdot\|_{op}$ denotes the operator norm of $\mathcal{B}(L^2(A_0))$. The completion of $A_0$ with respect to $\|\cdot\|_{op}$ is called the reduced completion of $A_0$, and the coproduct of $A_0$ extends to this completion to give a compact quantum group. All this is explained in detail in Timmermann's book on quantum groups.

If we repeat the above construction but replace left multiplication by right multiplication do we get the same norm or something different?

Let $(A,\Delta)$ be a compact quantum group in the sense of Woronowicz, and let $A_0$ be its dense Hopf subalgebra. We can construct from the Haar state $h:A \to \mathbb{C}$ an inner product $$ \langle \cdot,\cdot\rangle: A_0 \times A_0 \to \mathbb{C}, ~~~~~ (a,b) \mapsto h(a^*b). $$ Let $L^2(A_0)$ be the completion of $A_0$ to a Hilbert space. Let $\pi_L:A_0 \to \mathcal{B}(L^2(A_0))$ denote the representation of $A_0$ on $L^2(A_0)$ given by left multiplication. This gives a norm on $A_0$ defined by $\|a\| := \|\pi_L(a)\|_{op}$, where $\|\cdot\|_{op}$ denotes the operator norm of $\mathcal{B}(L^2(A_0))$. The completion of $A_0$ with respect to $\|\cdot\|_{op}$ is called the reduced completion of $A_0$, and the coproduct of $A_0$ extends to this completion to give a compact quantum group. All this is explained in detail in Timmermann's book on quantum groups.

If we repeat the above construction but replace left multiplication by right multiplication do we get the same norm or something different? (note that right multiplication will give an anti-algebra map, but it's not clear that this does anything to the value of the norm.)