Welcome octonions friends !

Long time ago when I travelled through octonion land, I conjectured that every $SO_8$ element can be expressed as product $L_a L_b R_c R_d$ for unit octonions $a$, $b$, $c$, $d$. Is this true ?

It can be seen as generalization of the fact that $SO_4$=$S^3 \otimes S^3$ i.e. every element in $SO_4$ can be seen as product $L_h R_\bar{k}$ for unit quaternions $h$, $k$.

Next step is to see what elements in $SO_{16}$ we obtain by multiplying $L_u$ and $R_v$ for unit sedenions $u, v \in \mathbb S$. We need at least 8 such elements sincce 15*8=120. Sedenions are not so nice as octonions since we have zero divisors there. Therefore we need to check the rank of matrix $L_u$ for unit sedenion $u$. There is work of Moreno showing that set $\{(a,b): ab=0; a,b \in S^{15} \subset \mathbb S\}$ is homeomorphic to $G_2$. Indeed, the subset of zero divisors on sphere $S^{15}$ is 11-dimensional set $\{a+b \iota: a,b \in S^6 \subset \mathbb O; a \perp b; a \perp 1; b \perp 1 \}$ (a,b are perpendicular imaginary unit octonions; $\iota$ is extra element in Cayley-Dickson formula defining sedenion multiplication). (The incorrect statement in wikipedia article on sedenions was corrected on June 28th, 2016 by John Baez). The space of zero-divisors on unit sphere $S^{15}$ is 11-dimensional.

**EDIT 0 2018-04-27**
I am trying to write article about octonions to magazine for high school students in Poland. While doing this I discovered that Moreno was right. There is no such thing as "zero divisor" in sedenions. Zero divisor should have norm equal to zero. In sedenions there is norm defined as square root of $x\bar x$ where conjugation is defined as in octonions from Cayley-Dickson formula $\bar x=\bar a-b\iota$. In this case sedenion norm is just length of the vector. So Moreno correctly considered pairs of elements whose product is zero.

**EDIT 1**: Since some people have justified doubts whether element $L_u$ can be in $SO_{16}$ for unit sedenion $u$ I present following.

**EDIT 2**: I extended 7-dimensional to 9-dimensional set of unitary sedenions u such that $L_u \in SO_{16}$.

$\underline{Fact}$: $L_x$ is in $SO_{16}$ for $x=(u+v\iota)/\sqrt 2$ such that $u,v \in \mathbb C \subset \mathbb O$. $\underline {Proof}:$ From Cayley-Dickson formula: $$L_u=\pmatrix{ L_u & \\ & R_u}$$ $$L_{v\iota}=\pmatrix{ 0 & -R_v S \\L_v S & 0}$$. Calculate the transposed matrix: $$L_x^T=\pmatrix{ L_{\bar u} & R_v S \\ -L_v S & R_\bar u }$$ Now $$L_x L_x^T=\pmatrix{2 & 0 \\ 0 & 2 }$$ because position 1,1 of the matrix is $L_u L_\bar u+R_v R_\bar v$ and position 1,2 of the matrix is $(L_u R_v-R_v L_u)S$. Elements $L_u$ and $R_v$ commutes only when $u,v$ are in the same complex subalgebra of octonions. I am using following facts in calculations: $L_u S=S R_\bar u$ $L_u^T=L_\bar u$ and similar. $\square$

The set defined in the Fact is roughly bundle $S^6 \times \times S^3$, because for each imaginary octonion $a$ there is sphere $S^3$ spanned by $<1,a,\iota,a\iota>$ belonging to this set. I said "roughly", becuase points {1,-1} belong to all fibers $S^3$.

Next, we can continue to trentaduonions $\mathbb T$. I claim that set of zero divisors is 27-dimensional subset of $S^{31}$. We need 16 elements of shape $L_u$ or $R_u$, since 16*31=496.

**EDIT 3**: Regarding trentaduonions I don't have ready theory yet. I tested in GAP, so I present following experimental results. For all 32 base elements $e_k$ the matrices $L_{e_k}$ and $R_{e_k}$ are in $SO_{32}$. Among 496 pairs of 32 base elements there 202 such pairs $k$, $l$ that $L_{e_k+e_l}$/length($e_k+e_l$) is in SO(32). 550 out 4960 triples satisfy it; 1002 out of 35960 fours; 1274 out of 201376 fives; 1218 out of 906192 sixs; 970 out of 3 365 856 sevens; 740 out of 10 518 300 eights.
If I try to prove it then I would see whether similar formulas for conjugation $S$ multiplied by $L_u$ are valid in sedenions.

... and so on ...:)

I have also some generalization proposal for any Lie group, which I post in separate question.

2more comments