Exactly, 𝕆⊕𝕆 is the spinor module, and the action of g(x) is by L(x) on the left summand, resp. R(x) on the right summand.
For Spin(9), Bryant writes that it is the subgroup of GL(𝕆⊕𝕆) generated by
\[p(r,x)=\begin{pmatrix}rI_8&CR(x)\\-CL(x)&rI_8\end{pmatrix},\]
where 𝑟∈ℝ and 𝑥∈𝕆 with 𝑟²+|𝑥|²=1, and 𝐶 denotes conjugation in the octonions.
I should also look into Krasnov's calculation (or the ones from Yokota), because I haven't yet fully understood how to arrive at these precise subgroups. But it looks like it's the same as what's going on inside F₄: we fix a point 𝑒∈𝕆ℙ²⊂𝔥₃(𝕆), giving as isotropy group Spin(9), and obtain 𝕆⊕𝕆 as one of the spaces in the Peirce decomposition w.r.t 𝑒. Then we intersect everything with the right 𝔥₃(ℂ)-subalgebra.
What do I mean by "right"? I'm not sure. Is the stabilizer of both a 𝔥₂(𝕆)-subalgebra and a 𝔥₃(ℂ)-subalgebra always conjugate to S(U(2) × U(3))? Or do they have to be "compatible" in some way?
Paul Schwahn
npub1zz…59mhx
2025-12-09 18:27:50 UTC
in reply to nevent1q…khy6
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"content": "Exactly, 𝕆⊕𝕆 is the spinor module, and the action of g(x) is by L(x) on the left summand, resp. R(x) on the right summand.\n\nFor Spin(9), Bryant writes that it is the subgroup of GL(𝕆⊕𝕆) generated by\n\\[p(r,x)=\\begin{pmatrix}rI_8\u0026CR(x)\\\\-CL(x)\u0026rI_8\\end{pmatrix},\\]\nwhere 𝑟∈ℝ and 𝑥∈𝕆 with 𝑟²+|𝑥|²=1, and 𝐶 denotes conjugation in the octonions.\n\nI should also look into Krasnov's calculation (or the ones from Yokota), because I haven't yet fully understood how to arrive at these precise subgroups. But it looks like it's the same as what's going on inside F₄: we fix a point 𝑒∈𝕆ℙ²⊂𝔥₃(𝕆), giving as isotropy group Spin(9), and obtain 𝕆⊕𝕆 as one of the spaces in the Peirce decomposition w.r.t 𝑒. Then we intersect everything with the right 𝔥₃(ℂ)-subalgebra.\n\nWhat do I mean by \"right\"? I'm not sure. Is the stabilizer of both a 𝔥₂(𝕆)-subalgebra and a 𝔥₃(ℂ)-subalgebra always conjugate to S(U(2) × U(3))? Or do they have to be \"compatible\" in some way?",
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