What are the inner automorphisms of the octonions?
Of course this is an odd question. Since the octonions are nonassociative you might even guess the map
𝑓: 𝕆 → 𝕆
given by
𝑓(𝑥) = 𝑔𝑥𝑔⁻¹
for a nonzero octonion 𝑔 isn't well-defined! After all, maybe we have
(𝑔𝑥)𝑔⁻¹ ≠ 𝑔(𝑥𝑔⁻¹)
But in fact this is a red herring.
(1/n)
(Below we see my friend the physicist Nichol Furey, a big fan of the octonions, standing in front of the Fano plane, which can be used to multiply them.)
Luckily, the octonions are 'alternative': the unital subalgebra generated by any two octonions is associative. Furthermore, the inverse 𝑔⁻¹ of any nonzero octonion is in the unital subalgebra generated by 𝑔. Thus 𝑔, 𝑥 and 𝑔⁻¹ all lie in a unital subalgebra generated by two octonions, so
(𝑔𝑥)𝑔⁻¹ = 𝑔(𝑥𝑔⁻¹)
and we can write either one as 𝑔𝑥𝑔⁻¹ without fear.
So the big question is whether
𝑓(𝑥) = 𝑔𝑥𝑔⁻¹
is an automorphism - that is, whether it obeys 𝑓(𝑥𝑦)=𝑓(𝑥)𝑓(𝑦).
(2/n)
In other words: if we have a nonzero octonion 𝑔, do we have
𝑔(𝑥𝑦)𝑔⁻¹ = (𝑔𝑥𝑔⁻¹) (𝑔𝑦𝑔⁻¹)
for all octonions 𝑥 and 𝑦? Again this is not obvious, because the octonions are nonassociative!
And indeed it's not always true.
This paper:
P. J. C. Lamont, Arithmetics in Cayley's algebra, Glasgow Mathematical Journal, 6 no. 2 (1963), 99-106. https://www.cambridge.org/core/services/aop-cambridge-core/content/view/F911E6DABC9FCABB17A531156AD272AF/S2040618500034808a.pdf/div-class-title-arithmetics-in-cayley-s-algebra-div.pdf
claims to settle when it's true! And the answer given is interesting.
(3/n)
Namely, a nonzero octonion 𝑔 has
𝑔(𝑥𝑦)𝑔⁻¹ = (𝑔𝑥𝑔⁻¹) (𝑔𝑦𝑔⁻¹)
for all octonions 𝑥 and 𝑦 precisely when 𝑔 lies at a 0 degree, 60 degree, 120 degree or 180 degree angle from the positive real line. (We think of the real multiples of 1 ∈ 𝕆 as a copy of the real line in the octonions.)
The proof is a terse calculation - but I haven't checked it carefully yet.
(4/n)
In particular, 𝑔 is a 6th root of 1 in the octonions if and only if |𝑔| = 1 and 𝑔 lies at a 0, 60, 120 or 180 degree angle from the positive real line.
So, if Lamont is correct, 6th roots of 1 give inner automorphisms of the octonions! In fact they give all the inner automorphisms, since rescaling 𝑔 goesn't change 𝑔𝑥𝑔⁻¹.
1 and -1 are 6th roots of 1 that give trivial inner automorphisms - just the identity. But the rest are nontrivial!
(5/n)
What are the nontrivial inner automorphisms of the octonions like? How do they sit in the group \(\mathrm{G}_2\) consisting of all automorphisms of the octonions? What sort of subgroup do they generate? (I doubt the composite of inner automorphisms is inner in this nonassociative world.)
Lots of questions!
I thank Charles Wynn for pointing me toward these mysteries.
(6/n, n = 6)
@undefined @johncarlosbaez they generate the whole G_2. Lie subgroups of G_2 are well understood. as the set of inner automorphisms you describe is 6-dimensional the only possibilities we need to exclude is that it generates a copy of SU(3) which is 8 dimensional or a copy of SO(4) (all other proper subgroups of G_2 have dimensions <6). It should be easy to see that it's not SO(4) which is also 6-dimensional.
SU(3) can be excluded as follows.
All SU(3) in G_2 are known to come about as follows. G_2 acts on S^6=unit imaginary octonians and various SU(3) in G_2 are isotropy subgroups of different points in S^6. if all inner automorphisms were in an isotropy group of some point v in S^6 that v would be in the center of octonions. This is impossible and hence Inner automorphisms generate all of G_2.
@herid - Thanks! I get how every SU(3) subgroup of G₂ fixes an imaginary octonion, so yes, that's ruled out. How should we think of SO(4) subgroups of G₂? By general abstract nonsense they are groups of automorphisms of the octonions preserving *some* extra structure on the octonions, but what structure?
@johncarlosbaez sorry, I don't know the answer to this although I am sure this is known. you may want to ask on mathoverflow. in any even as I said it's easy to rule out that inner automorphisms of octonions are contained in SO(4). they are both closed 6 manifolds, the former being S^6 and and there is no embedding of S^6 into SO(4) for easy topological reasons (it would have to be a homeomorphism).
@herid - okay, that does the job. How do you know SO(4) is a subgroup of G₂? I feel I should learn how to classify the maximal closed subgroups of a compact simple Lie group, but I don't know how:
@johncarlosbaez I am really not an expert on this subject but this is well known and I think there are explicit ways to see it. on the level of lie algebras this is easy from looking at the root system and the Weyl group which is D_6. it contains reelections in two perpendicular lines and that gives the inclusion of the correct lie algebra. not sure what the best what to see that it gives SO(4) and not S^3\times S^3. I think if you look a little more closely you see that the diagonal subalgebra in it sits su(3). the corresponding subgroup is the obvious SO(3) in SU(3). there is no SO(3) in S^3\times S^3 so it is ruled out. But as I said this ought to be well known. G_2/SO(4) is a symmetric space so there should be a nice description of it like you wanted. I think it might be the same as the set of quaternion subalgebras of octonions on which G_2 acts transitively.
@herid - nice! Over on the n-Category Cafe @allenk points out that the conjugacy class of elements of order 2 in G₂ is isomorphic to G₂/SO(4), and notes that this is also the space of quaternion subalgebras of 𝕆.
There should be a direct connection between these two facts. Maybe any quaternion subalgebra of 𝕆 is fixed by a unique automorphism of order 2, or something like that. (I'm not sure that's quite right.)
https://golem.ph.utexas.edu/category/2022/11/inner_automorphisms_of_the_oct.html#c061775
@johncarlosbaez I think my last conjecture is correct. a quaternion algebra in octonions is generated by two unit orthogonal imaginary octonions. that gives dimension of such algebras as 6+5-3=8 (we have to substract 3=dim Sp(1)). and G_2 as the group of automorphisms of octonions obviously acts on the set of such algebras transitively. the stabilizer of a point is 6 dimensional so can only be SO(4) we want. it now should be possible to see it more directly.