$\det(R_{{\mathcal{G}}})$ and $\det(R)$ have the same sign, i.e. $R$ and $R_{{\mathcal{G}}}$ belong to the same connected component of $O(6)$. Furthermore, since ${\widetilde{\mathcal{G}}}$ is a common subgroup of ${\widetilde{\mathcal{I}}}$ and ${\widetilde{\mathcal{I}}}_{{\mathcal{G}}}$, it is possible to choose $R_{{\mathcal{G}}} \in O(6)$ such that $\hat{{\mathcal{G}}} = \hat{{\mathcal{I}}} \cap \hat{{\mathcal{H}}}_{{\mathcal{G}}}$, i.e. $$\label{rel_g}
R^{-1} G R = R_{{\mathcal{G}}}^{-1} G R_{{\mathcal{G}}}, \; \forall G \in {\widetilde{\mathcal{G}}}\Rightarrow (R_{{\mathcal{G}}}R^{-1})^{-1} G (R_{{\mathcal{G}}}R^{-1}) = G, \; \forall G \in {\widetilde{\mathcal{G}}}.$$ Therefore $R_{{\mathcal{G}}}R^{-1}$ belongs to the centraliser of ${\widetilde{\mathcal{G}}}$ in $GL(6,{\mathbb{R}})$, i.e. the set $${\mathcal{Z}}({\widetilde{\mathcal{G}}}, {\mathbb{R}}) := \{ A \in GL(6,{\mathbb{R}}) : A G = G A, \; \forall G \in {\widetilde{\mathcal{G}}}\}\,.$$ Hence there exists a matrix $M_{{\mathcal{G}}} \in \mathcal{Z}({\widetilde{\mathcal{G}}},{\mathbb{R}}) \cap O(6)$, denoted as the *Schur operator* related to ${\mathcal{G}}$, such that $R_{{\mathcal{G}}} = M_{{\mathcal{G}}} R$. Since $R$ and $R_{{\mathcal{G}}}$ have determinants with equal signs by assumption, we have that $\det(M_{{\mathcal{G}}}) > 0$, hence $M_{{\mathcal{G}}}$ is a rotation in $SO(6)$. Let us consider a path $$\label{schur_rotation}
M_{{\mathcal{G}}}(t) : [0,1] \longrightarrow \mathcal{Z}({\widetilde{\mathcal{G}}},{\mathbb{R}}) \cap SO(6)$$ that connects $M_{{\mathcal{G}}}$ to the identity matrix $I_6$, i.e. $M_{{\mathcal{G}}}(0) = I_6$ and $M_{{\mathcal{G}}}(1) = M_{{\mathcal{G}}}$. Such a path is referred to as the *Schur rotation* associated with ${\widetilde{\mathcal{G}}}$. The name comes from Schur’s Lemma in Representation Theory, that 