Paul Schwahn on Nostr: John Carlos Baez Today I revisited our project on the subgroups of F₄. Recall, we ...

John Carlos Baez (npub1nf4…nqe4)

Today I revisited our project on the subgroups of F₄. Recall, we were hoping to show that for each 𝕂=ℝ,ℂ,ℍ, all subalgebras of 𝔥₃(𝕆) which are isomorphic to 𝔥₃(𝕂) are F₄-related.

We were almost done with this; in fact, I think I have shown that a subalgebra isomorphic to 𝔥₃(ℍ) is F₄-related to either the standard 𝔥₃(ℍ)⊂𝔥₃(𝕆) or a "twisted" one, where two of the off-diagonal elements are in the orthogonal complement of ℍ.

It remained to see if Spin(8), i.e. the subgroup of F₄ preserving each off-diagonal space, has an element mapping ℍ to ℍ in one of its 8-dimensional representations and to its complement in one of the others. There was the issue of how exactly Spin(8) acts on each of these copies of 𝕆. Both the note of Bryant and the book of Harvey have different ways of writing things down, and the result depends delicately on which representation you have on which copy of 𝕆.

I did a back-of-the-envelope calculation, and now it seems to me that there are actually two different F₄-orbits of 𝔥₃(ℍ)-subalgebras...

The idea was to use the "Triality Theorem" (quoted from Harvey), which states that under the right isomorphism Cl(𝕆) ≅ End(𝕆⊕𝕆), Spin(8) gets mapped to the set

{diag(g₁,g₂) | g₁,g₂∈SO(𝕆), ∃g₃∈SO(𝕆), ∀x,y∈𝕆, g₁(xy)=g₂(x)g₃(y)}.

Then g₁,g₂ are the half-spinor representation, and g₃ the vector representation. Together with the multiplicative relations between the off-diagonal elements in 𝔥₃(𝕆), this suffices to determine (up to automorphism) how Spin(8) acts on those copies of 𝕆.

I will try to explain the calculation later; for now I urgently have to get some food.