and ${\varepsilon}_{kl}=1$ is similar.

Again, it is easy to see that $O_n$ and $O_n^+$ fit into the framework of ${\varepsilon}$-orthogonal quantum groups, using the matrices ${\varepsilon}_{\textnormal{comm}}$ and ${\varepsilon}_{\textnormal{free}}$ respectively. However, let us point out that the commutativity relations $R^{\textnormal{comm}}$ do *not* imply $R^{\varepsilon}$ for general ${\varepsilon}$, i.e. $O_n^{\varepsilon}$ is *no interpolation* between $O_n$ and $O_n^+$. We say that a compact matrix quantum group $G$ is a *quantum subgroup* of $H$ (writing $G\subset H$), if there is a surjective $^*$-homomorphism from $C(H)$ to $C(G)$ mapping generators to generators.

\[PropNoInterpolation\] If ${\varepsilon}\neq {\varepsilon}_{\operatorname{comm}}$ and ${\varepsilon}\neq {\varepsilon}_{\operatorname{free}}$, we have: $$O_n\not\subset O_n^{\varepsilon}\subsetneq O_n^+$$ More general, we have in that case: $$S_n\not\subset O_n^{\varepsilon}\subsetneq O_n^+$$

Since the matrix $u$ in $O_n^{\varepsilon}$ is orthogonal and has self-adjoint entries, we have $O_n^{\varepsilon}\subset O_n^+$. The inclusion is strict, since $S_n\subset O_n\subset  O_n^+$ but $S_n\not\subset O_n^{\varepsilon}$, which we will prove next. We may find $i\neq j$ such that ${\varepsilon}_{ij}=1$ (since ${\varepsilon}\neq {\varepsilon}_{\operatorname{free}}$), and $k\neq l$ such that ${\varepsilon}_{kl}=0$ (since ${\varepsilon}\neq{\varepsilon}_{\operatorname{comm}}$). Let $\sigma\in S_n$ be a permutation with the properties: $$\sigma(k)=i,\qquad\sigma(l)=j$$ The associated permutation matrix $a^\sigma\in M_n({{\mathbb C}})$ is defined by $a^\sigma_{pq}=\delta_{p\sigma(q)}$. The evaluation map $\operatorname{ev}_\sigma: C(S_n)\to{{\mathbb C}}$ is given by $\operatorname{ev}_\sigma(u_{pq})=\delta_{p\sigma(q)}$. Now, assume $S_n\subset O_n^{\varepsilon}$, i.e. there is a surjective 