According to Bryant's work on special holonomy groups, $G_2\subset SO(7)$ may be defined as the group preserving the following 3-form:


where $\mathrm{d}x_{ijk}=\mathrm{d}x_i\wedge\mathrm{d}x_j\wedge\mathrm{d}x_k$. In a similar fashion, $Spin(7)\subset SO(8)$ may be defined as the group preserving the following 4-form:


But of course, with this definition, it is not obvious at all that $Spin(7)$ is really a double cover of $SO(7)$. Is there any way to understand this fact? (that is, which is the point of defining $\Omega_0$ like this?) In the book On Quaternions and Octonions there is a definition of $Spin(7)\subset SO(8)$ related to 'octonions automorphisms' (they are not really automorphisms) which reveals some of this double-cover nature. Perhaps these could be a bridge linking these two sides, but at this moment it is not clear for me. Any suggestion will be welcomed.


1 Answer 1


There are at least two sources for this: First, Harvey and Lawson, Calibrated geometries (Acta Math. 1982) proves this (i.e., that the stabilizer of $\Omega_0$ is isomorphic to the nontrivial double cover of $\mathrm{SO}(7)$) using properties of the octonions. Second you can find a proof that doesn't rely on knowledge of the octonions but instead relies on the classification of Lie algebras and groups in Bryant's Metrics with exceptional holonomy (Ann. of Math, 1987, Theorem 4 of Section 2).

There are even more 'direct' (i.e., elementary) proofs available, of course. If those don't satisfy you, then ask again and I'll sketch one here when I have time.

  • $\begingroup$ Thank you very much! I will read these two sources; I am sure they will be most helpful for me. $\endgroup$
    – Jjm
    Nov 30, 2014 at 20:20

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