The reordered basis $(e_0, e_1, e_2, e_4 \mid e_3, e_5, e_6, e_7)$ and the two $4\times4$ blocks of the flow matrix

Conventions. All three declarations live in the namespace OctonionD8. Indices of $\mathrm{Fin}\,8$ are $0,1,\dots,7$ and indices of $\mathrm{Fin}\,4$ are $0,1,2,3$ (zero-based, exactly as in the code). None of the three definitions refers to anything from the imported files (the octonion table, $\mathrm{omul}$, $R$, $L$, $\mathrm{flowMat}$, the Fano plane); they are written out by hand with explicit literal entries.

blockEquiv. This is a bijection

$$\mathrm{blockEquiv} : \{0,\dots,7\} \;\xrightarrow{\ \sim\ }\; \{0,1,2,3\} \sqcup \{0,1,2,3\},$$

where the target is the disjoint union of two copies of $\mathrm{Fin}\,4$, a "left" copy (written $\iota_L(i)$) and a "right" copy (written $\iota_R(i)$). The forward map is the explicit table

$$0 \mapsto \iota_L(0),\quad 1 \mapsto \iota_L(1),\quad 2 \mapsto \iota_L(2),\quad 3 \mapsto \iota_R(0),$$ $$4 \mapsto \iota_L(3),\quad 5 \mapsto \iota_R(1),\quad 6 \mapsto \iota_R(2),\quad 7 \mapsto \iota_R(3).$$

So the left block consists of the indices $0,1,2,4$ (in that order, becoming left-indices $0,1,2,3$) and the right block consists of the indices $3,5,6,7$ (in that order, becoming right-indices $0,1,2,3$). The inverse map is

$$\iota_L(0),\iota_L(1),\iota_L(2),\iota_L(3) \mapsto 0,1,2,4, \qquad \iota_R(0),\iota_R(1),\iota_R(2),\iota_R(3) \mapsto 3,5,6,7,$$

and the two inverse laws are checked by exhaustive computation. Note that the order within each block is increasing, but the split is not "first four / last four": index $3$ goes to the right block and index $4$ to the left block.

blockA. For arbitrary real numbers $c, s$ (no relation between them is assumed; in particular $c^2+s^2=1$ is not required), $A(c,s)$ is the real $4\times 4$ matrix whose entry in row $i$, column $j$ ($i,j\in\{0,1,2,3\}$) is given by

$$A(c,s) = \begin{pmatrix} 0 & -(c+1) & -s & 0 \\ c+1 & 0 & 0 & s \\ s & 0 & 0 & 1-c \\ 0 & -s & c-1 & 0 \end{pmatrix},$$

rows listed top to bottom as $i=0,1,2,3$ and columns left to right as $j=0,1,2,3$. Entry by entry, the nonzero entries are

$$A_{01} = -c-1,\; A_{02} = -s,\; A_{10} = c+1,\; A_{13} = s,\; A_{20} = s,\; A_{23} = 1-c,\; A_{31} = -s,\; A_{32} = c-1,$$

and all other entries ($A_{00},A_{03},A_{11},A_{12},A_{21},A_{22},A_{30},A_{33}$) are $0$. As written, $A_{ji} = -A_{ij}$ for all $i,j$ (the matrix is skew-symmetric for every $c,s$). Degenerate values: at $c=1,s=0$ the only nonzero entries are $A_{01}=-2$, $A_{10}=2$; at $c=-1,s=0$ the only nonzero entries are $A_{23}=2$, $A_{32}=-2$.

blockB. For arbitrary real numbers $c, s$ (again unconstrained), $B(c,s)$ is the real $4\times 4$ matrix with rows $i=0,1,2,3$ top to bottom and columns $j=0,1,2,3$ left to right:

$$B(c,s) = \begin{pmatrix} 0 & -s & 0 & 1-c \\ s & 0 & 1-c & 0 \\ 0 & c-1 & 0 & -s \\ c-1 & 0 & s & 0 \end{pmatrix}.$$

Entry by entry, the possibly nonzero entries are

$$B_{01} = -s,\; B_{03} = 1-c,\; B_{10} = s,\; B_{12} = 1-c,\; B_{21} = c-1,\; B_{23} = -s,\; B_{30} = c-1,\; B_{32} = s,$$

and all other entries ($B_{00},B_{02},B_{11},B_{13},B_{20},B_{22},B_{31},B_{33}$) are $0$. As written, $B_{ji} = -B_{ij}$ for all $i,j$ (skew-symmetric for every $c,s$). Degenerate values: at $c=1,s=0$, $B(1,0)$ is the zero matrix; at $c=-1,s=0$ the nonzero entries are $B_{03}=B_{12}=2$ and $B_{21}=B_{30}=-2$.