Timeline for Why is the automorphism group of the octonions $G_2$ instead of $SO(7)$
Current License: CC BY-SA 4.0
9 events
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| Apr 9 at 19:05 | comment | added | Luke Hutchison | I'm just learning about this, but from the Wikipedia page on G2: "The compact form of G2 can be described as the automorphism group of the octonion algebra or, equivalently, as the subgroup of SO(7) that preserves any chosen particular vector in its 8-dimensional real spinor representation (a spin representation)." en.wikipedia.org/wiki/G2_(mathematics) | |
| Mar 18, 2021 at 19:49 | vote | accept | p6majo | ||
| Mar 18, 2021 at 18:02 | answer | added | Robert Bryant | timeline score: 27 | |
| Mar 18, 2021 at 18:02 | comment | added | Robert Bryant | @TheoJohnson-Freyd: Sure. I'll be happy to do this. | |
| Mar 18, 2021 at 17:08 | comment | added | Theo Johnson-Freyd | @Robert Can I talk you into leaving this answer as an answer? | |
| Mar 18, 2021 at 15:50 | comment | added | Robert Bryant | The quaternions are generated by any two imaginary elements x and y that are orthonormal, i.e., they are spanned by 1, x, y, and xy. Meanwhile, the octonions are generated algebraically by any three imaginary elements, say, x, y, and z that are orthonormal and z is perpendicular to xy. This means that any automorphism of the octonions that fixes three such elements is the identity. Thus, SO(7) is too large to be the automorphism group of the octonions because it acts transitively on the set of oriented orthonormal bases of the imaginary octonions. | |
| Mar 18, 2021 at 15:33 | history | edited | Martin Sleziak |
edited tags
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| Mar 18, 2021 at 15:32 | history | edited | Sam Hopkins | CC BY-SA 4.0 |
added 10 characters in body; edited title
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| Mar 18, 2021 at 15:29 | history | asked | p6majo | CC BY-SA 4.0 |