we have:

-   The relations $\mathring R^{\varepsilon}$ imply the relations $R^{\varepsilon}$.

-   If $u_{ik}u_{jk}=u_{ki}u_{kj}=0$ for all $i\neq j$ and all $k$, then the relations $R^{\varepsilon}$ imply the relations $\mathring R^{\varepsilon}$.

\(a) This follows since ${\varepsilon}$ is symmetric.

\(b) The relations $u_{ik}u_{jk}=0$ imply that the elements $u_{ik}^2$ are projections (and thus, the $u_{ik}$ are partial isometries). Indeed, we have $\sum_ju_{jk}^2=1$ by the orthogonality relations, thus: $$u_{ik}^2=u_{ik}^2\sum_j u_{jk}^2=u_{ik}^4+\sum_{i\neq j}u_{ik}^2u_{jk}^2=u_{ik}^4$$ Recall that for ${\varepsilon}_{ij}=1$ and ${\varepsilon}_{kl}=0$, the relations $R^{\varepsilon}$ imply: $$u_{ik}u_{jl}=u_{jk}u_{il}$$ Multiplying this equation from the right with $u_{jl}^2$, we infer $$u_{ik}u_{jl}=0,$$ since the projections $u_{jl}^2$ and $u_{il}^2$ are orthogonal to each other. The case ${\varepsilon}_{ij}=0$ and ${\varepsilon}_{kl}=1$ is similar.

\[DefGEps\] For any quantum subgroup $G\subset O_n^+$ we define $G^{\varepsilon}$ and $\mathring G^{\varepsilon}$ by: $$C(G^{\varepsilon}):=C(G)/ \langle R^{\varepsilon}\rangle,\qquad C(\mathring G^{\varepsilon}):=C(G)/ \langle \mathring R^{\varepsilon}\rangle$$

For ${\varepsilon}={\varepsilon}_{\operatorname{free}}$, we have $G^{\varepsilon}=G$.

\[LemRZwei\] Let $G$ be a compact matrix quantum group with fundamental unitary $u=(u_{ij})_{1\leq i,j\leq n}$.

-   If the relations ($\mathring R^{\varepsilon}2$) hold for $u_{ij}$ in $C(G)$, then they also hold for $u_{ij}':=\sum_k u_{ik}\otimes u_{kj}\in C(G)\otimes C(G)$.

-   If $G\subset O_n^+$ is a quantum subgroup of $O_n^+$, then also $G^{\varepsilon}$ and $\mathring G^{\varepsilon}$ are compact matrix quantum groups and we have $\mathring G^{\varepsilon}\subset G^{\varepsilon}\subset G$.

\(a) We 