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Martin Sleziak
Martin Sleziak
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Sam Hopkins
Sam Hopkins
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Why is the automorphism group of the octonions G2$G_2$ instead of SO$SO(7)$

I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of G2$G_2$. However, when I do the same calculation for the quaternions, I end up with the three generators of so(3)$SO(3)$, which basically tells me that I can rotate the set of imaginary units anyway I like.

Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of SO(7)$SO(7)$. Instead, the calculations show, that the possible transformations are restricted to the G2$G_2$ subgroup.

Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions?

Why is the automorphism group of the octonions G2 instead of SO(7)

I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of G2. However, when I do the same calculation for the quaternions, I end up with the three generators of so(3), which basically tells me that I can rotate the set of imaginary units anyway I like.

Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of SO(7). Instead, the calculations show, that the possible transformations are restricted to the G2 subgroup.

Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions?

Why is the automorphism group of the octonions $G_2$ instead of $SO(7)$

I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of $G_2$. However, when I do the same calculation for the quaternions, I end up with the three generators of $SO(3)$, which basically tells me that I can rotate the set of imaginary units anyway I like.

Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of $SO(7)$. Instead, the calculations show, that the possible transformations are restricted to the $G_2$ subgroup.

Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions?

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p6majo
p6majo
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Why is the automorphism group of the octonions G2 instead of SO(7)

I can calculate the derivation of the octonions and I clearly find the 14 generators that form the algebra of G2. However, when I do the same calculation for the quaternions, I end up with the three generators of so(3), which basically tells me that I can rotate the set of imaginary units anyway I like.

Intuitively, I don't understand, why it is not possible for the octonions to be rotated in the same way with an arbitrary rotation of SO(7). Instead, the calculations show, that the possible transformations are restricted to the G2 subgroup.

Is there way to understand this geometrically or algebraically? Is it related to the non-associative property of the octonions?