Given a vector space {$V$} over a field {$K$}. The projective space {$P(V)$} is the set of equivalence classes of {$x\in V, x\neq 0$} defined by {$x\sim y$} if there exists {$λ\in K$} such that {$x = λy$}.
A projective line is a one-dimensional projective space. All of the points lie on a single line. The line includes a point at infinity.`
Real vector bundle {$E$} over {$S^n$} defines an equivalence class {$[E]\in KO(S^n)$} generating an additive monoid that yields the Grothendieck group {$KO(S^n)$}. Fixing a point in {$S^n$} and modding out by the dimension of the fiber of {$E$} at that point yields the reduced theory {$\widetilde{KO}(S^n)$}.
They provide a recipe for constructing mutually anticommuting linear complex structures {$J_1, J_2, J_3, J_4, J_5, J_6, J_7,\cdots$}.
{$J_1=\textrm{diag}[\hat{i}]$}
{$J_2=\textrm{diag}[j_4]$}
Suppose {$J_3$} exists. Let {$K=J_1J_2J_3$}. Then {$K^2=I$} and consequently, {$K$} satisfies {$x^2-1$}, thus posseses quaternionic eigenspaces {$V_{\pm}$} where it takes values {$\pm 1$} accordingly. {$Kv_+=v_+$} for {$v_+\in V_+$} and {$Kv_-=-v_-$} for {$v_-\in V_-$}. {$K$} can be used to define {$J_3=J_2^{-1}J_1^{-1}K$}.
To define {$J_4$} we must suppose {$V_+$} and {$V_-$} have the same dimension. Suppose {$J_4$} exists. Let {$L=J_3J_4$}. Then {$LK=-KL, L^2=-I, LJ_1=J_1L, LJ_2=J_2L$}. {$L$} preserves the quaternionic structure and is a quaternionic isometry from {$V_+$} to {$V_-$}. {$L$} can be used to define {$J_4=J_3^{-1}L$}.
Suppose {$J_5$} exists. Let {$M=J_1J_4J_5$}. Then {$M^2=I$} and {$M$} commutes with {$K$} and {$J_1$}. Thus {$M$} acts within an eigenspace of {$K$} (say {$V_+$}) and divides it into two mutually orthogonal eigenspaces {$W_\pm$} with {$W_-=J_2W_+$}. {$M$} can be used to define {$J_5=J_4^{-1}J_1^{-1}M$}.
Suppose {$J_6$} exists. Let {$N=J_2J_4J_6$}. Then {$N^2=I$} and {$N$} commutes with {$K$} and {$M$}. Thus {$N$} acts within an eigenspace of {$M$} (say {$W_+$}) and divides it into two mutually orthogonal eigenspaces {$X_\pm$} with {$X_-=J_1W_+$}. {$N$} can be used to define {$J_6=J_4^{-1}J_2^{-1}N$}.
Suppose {$J_7$} exists. Let {$P=J_1J_6J_7$}. Then {$P^2=I$} and {$P$} commutes with {$K$}, {$M$} and {$N$}. Thus {$P$} acts within an eigenspace of {$X$} (say {$X_+$}) and divides it into two mutually orthogonal eigenspaces {$Y_\pm$}. {$P$} can be used to define {$J_7=J_6^{-1}J_1^{-1}P$}.
To define {$J_8$} we must suppose {$Y_+$} and {$Y_-$} have the same dimension. Suppose {$J_8$} exists. Let {$Q=J_7J_8$}. Then {$Q^2=-I$} and {$Q$} commutes with {$K$}, {$M$} and {$N$} but anticommutes with {$P$}. Thus {$Q$} is an isometry mapping {$Y_+\leftrightarrow Y_-$}. {$Q$} can be used to define {$J_8=J_7^{-1}Q$}.
Can we express {$(a,b)\in M_8(\mathbb{R})\oplus M_8(\mathbb{R})$} in terms of {$2\times 2$} matrices {$\begin{pmatrix} a & \\ & b \\ \end{pmatrix}$} ? What problem do we run into if we want to define {$J_7$}?
Thus the {$2\times 2$} matrices {$M_l$} are mapped by {$L_4$} to an isometry that involves multiplying by {$M_l$} in one direction and {$-M_l$} in the opposite direction.
Octonions take us through the Looking Glass {$L$}.
{$LL=-I$}
{$La=-aL$}
{$(La)(Lb)=Lba$}
{$(La)b=L(ba)$}
{$a(Lb)=L(ba)$}
This yields nonassociativity (based on the "subtractiveness" of the mirror)
{$(Lb)a=L(ab)$} but in general {$L(ab)\neq L(ba)$} because of noncommutativity {$ab\neq ba$}
Hackett, Kauffman described the belt trick for the octonions where {$L$} plays the role of a context switch or a mirror. They imagine it as a sphere around the belt trick. You can rotate the sphere 180 degrees whereas the quaternions turn at the base.
The Octonions form seven copies of the quaternions, one for each copy of the circle in the Fano projective plane.
3 of the form {$j(Li)=Lk, (Li)(Lk)=j, (Lk)j=Li$} outer edges of triangle
3 of the form {$L(jL)=j, (jL)j=L,j(L)=jL$} across the center
1 of the form {$ij=k,jk=i,ki=j$} inner circle
what is missing is a line relating the three corners of the triangle {$Li,Lj,Lk$}. Their product is {$L$}.
Further calculations
Define {$A_i=J_i^{-1}J_{i+1}$}.
{$A_i^2=-I$} because {$J_i^{-1}J_{i+1}J_i^{-1}J_{i+1}=-1$}
{$A_i$} is an orthogonal matrix because {$J_i$} and {$J_{i+1}$} are.
{$A_i$} is a skew-symmetric matrix because it is orthogonal {$A_iA_i^T=I$} and {$AA=-I$} means {$A^T=-A$}.
{$A_i$} anticommutes with {$J_i$} but commutes with {$J_1,J_2,\dots,J_{i-1}$}.
Then for {$\gamma(t)=J_ie^{A_i\pi t}=J_i\cos \pi t + J_{i+1}\sin \pi t$} we have {$\gamma (0)=J_i, \gamma(\frac{1}{2})=J_{i+1},\gamma{1}=-J_i$} and {$\gamma(t)$} is a geodesic in {$G_{i-1}/G_i$}.
Think of {$A_i$} as a shift in perspective. Note that {$A_3=-J_3J_4$} and {$A_7=-J_6J_7$} are isometries between eigenspaces {$V_+$} and {$V_-$}.