three non-equivalent Bravais lattices in ${\mathbb{R}}^6$ left invariant by ${\mathcal{I}}$, classified in [@levitov], namely the simple cubic (SC), body-centered cubic (BCC) and face-centered cubic lattice (FCC): $${\mathcal{L}}_{SC} = \left\{ {\boldsymbol{x}}=(x_1, \ldots, x_6) : x_i \in {\mathbb{Z}}\right\},$$ $${\mathcal{L}}_{BCC} = \left\{ {\boldsymbol{x}}= \frac{1}{2}(x_1, \ldots, x_6) : x_i \in {\mathbb{Z}}, \; x_i = x_j \; mod \; 2, \forall i,j=1,\ldots,6 \right\},$$ $${\mathcal{L}}_{FCC} = \left\{ {\boldsymbol{x}}= \frac{1}{2}(x_1, \ldots, x_6) : x_i \in {\mathbb{Z}}, \; \sum_{i=1}^6 x_i = 0 \; mod \; 2 \right\}.$$ The point group of these three lattices coincide, and it is referred to as the *hyperoctahedral group* in six dimensions, and denoted by $B_6$. ${\widetilde{\mathcal{I}}}$ leaves invariant two three-dimensional spaces, usually denoted as $E^{\parallel}$ (the physical space) and $E^{\perp}$ (the orthogonal space), both totally irrational with respect to any of the hypercubic lattices in ${\mathbb{R}}^6$. The matrix $R \in O(6)$ given by decomposes ${\widetilde{\mathcal{I}}}$ into irreducible representations, i.e. $$\label{R}
\hat{{\mathcal{I}}}:= R^{-1}{\widetilde{\mathcal{I}}}R = \rho_3 \oplus \rho_3',$$ with $\rho_3 \simeq T_1$ and $\rho_3' \simeq T_2$. The explicit forms of $\rho_3$ and $\rho_3'$ are given in Table \[irreps\_ico\]. If $\pi^{\parallel} : {\mathbb{R}}^6 \rightarrow E^{\parallel}$ denotes the projection operator into the parallel space (and $\pi^{\perp}$ the corresponding orthogonal projection), we have, by linear algebra: $$\label{inv}
