Timeline for Why is the automorphism group of the octonions $G_2$ instead of $SO(7)$
Current License: CC BY-SA 4.0
6 events
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| Mar 18, 2021 at 23:59 | comment | added | Robert Bryant | @p6majo: A $\mathrm{G}_2$ element can have a 1-dimensional, 3-dimensional, or 7-dimensional $+1$-eigenspace of imaginary octonions. For example, if $x$ and $y$ are orthonormal imaginary octonions, then the set of unit imaginary octonions $z$ that are perpendicular to the $3$-plane spanned by $x$, $y$, and $xy$ is a $3$-sphere, and there is a subgroup of $G_2$ isomorphic to $\mathrm{SU}(2)$ that fixes $x$ and $y$ (and hence $xy$) and acts simply transitively on that 3-sphere. (Be careful not to confuse a subspace of fixed vectors of an automorphism with a subspace that is preserved by it.) | |
| Mar 18, 2021 at 21:50 | comment | added | p6majo | However, a typical generator of $G_2$ has four non-zero entries [twice as many as a generator of $SO(7)$]. It seems to me that it fixes three imaginary units. But perhaps it acts non-trivially on the additional constraint $z\cdot xy=0$? | |
| Mar 18, 2021 at 19:59 | comment | added | p6majo | In three dimensions, a "basic" rotation fixes a $1d$-subspace, in four dimensions there can be a fixed $2d$-plane, and so on. Do I understand correctly, that "basic" rotations that leaves a $3d$-subspace invariant would not be part of the automorphim-group? Basically, only combinations of a few "basic" rotations are allowed that rotate all imaginary units more or less simultaneously? | |
| Mar 18, 2021 at 19:49 | vote | accept | p6majo | ||
| Mar 18, 2021 at 18:08 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Added a remark about nonassociativity.
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| Mar 18, 2021 at 18:02 | history | answered | Robert Bryant | CC BY-SA 4.0 |