The n-Category Café

Toby Bartels: if you read this, could you remind me how to get the spacing around absolute value signs to look less wretched?

(Or anyone else who knows.)

This is a fascinating question!

I can’t answer it but it got me looking at octonions again this afternoon.

I enjoyed trying to visualize this inner automorphism. Without loss of generality, we can take $g$ to belong to the complex subalgebra of $\mathbb{O}$ spanned by $1, i_0$ and to relax the constraint on order or angle, I’ll write $g = \cos(\phi/2) + i_0 \sin(\phi/2)$.

It seems that $\alpha: x \mapsto g x g^{-1}$ defines three intersecting quaternion subalgebras of $\mathbb{O}$. We can cycle the three quaternion subalgebras that $g$ defines using $i_n \mapsto i_{2n}$ (provided that $i_0$ is the imaginary unit used to build $g$).

These three quaternion subalgebras define three planes orthogonal to their intersection (one spanned by $i_1,i_3$, another by $i_2,i_6$, and another by $i_4,i_5$) and $\alpha$ rotates each of these planes by the same angle $\phi$. The constraint that fixes the permissible rotation angle $\phi$ is that multiplication between quaternions in two of these three planes has its product in the third, and that these three planes must rotate together under the action of $\alpha$.

Playing a bit with this today I wondered if these three quaternion subalgebras defined by $g$ (or rather by $i_0$) and the map $i_n \mapsto i_{2n}$ might help with your question somehow.

This isn’t an answer, but I’ll contribute what potential clues I know.

G_2 := Aut(octonions) has exactly two conjugacy classes of order 3; pick elements b,c of them. The big difference is that the centralizer of b is isomorphic to U(2) (or at least has that Lie algebra) and the centralizer of c is SU(3). It is this conjugacy class {c} that you’re producing as “inner automorphisms”.

G_2 acts on the pure imaginary unit octonions, a sphere S^6 inside R^7. Your angle condition finds a couple of other S^6s inside the unit octonions S^7, each invariant under G _2. Conjugation by one of your g preserves 1 and g, so acts on the complementary R^6. On that R^6 we can use the imaginary part of g to define a complex structure, so our stabilizer sits inside the U(3) preserving that. By dimension count it’s only SU(3).

Anyway if you want some kind of “correspondence” between your two different kinds of order 3 automorphisms, we should figure what what the spaces are of each. There’s an S^6 worth of inner auts of O.

Meanwhile, if we fix one specific non-inner automorphism tau of Spin(8) having order 3 (which it’s not obvious we can do, because Out(G) is not a subgroup of Aut(G) but rather a subquotient; anyway we indeed can) then the other ones are of the form tau * h where h is in Spin(8), and (tau * h)^3 = 1.

I don’t know what that space {h} looks like. It obviously contains the order 3 elements of the fixed points of tau, at which point we’re back to order 3 elements of G_2. But maybe there’s more? I guess I’ve at least managed to show one containment, for a specific interpretation of your question.

I’m thinking it’s somewhere in the Laves graph.

I got some help from Vít Tuček on MathOverflow. He pointed out that Yokota gave a proof that all inner automorphisms of $\mathbb{O}$ have order 3 using the Moufang identity and two ideas connected to triality. I’d like to spell it out here.

First, Yokota proves these:

Theorem 1.14.2. For any $\alpha_3 \in SO(8)$ there are $\alpha_1, \alpha_2 \in SO(8)$ such that

$$\alpha_1(x) \, \alpha_2(y) = \alpha_3(x y)$$

for all $x, y \in \mathbb{O}$. Moreover $\alpha_1, \alpha_2$ are unique up to a common choice of sign.

Lemma 1.14.3. If $\alpha_1, \alpha_2, \alpha_3 \in SO(8)$, then

$$\alpha_1(x) \alpha_2(y) = \overline{\alpha_3(\overline{x y})}$$

for all octonions $x,y \in \mathbb{O}$ implies that

$$\alpha_2(x) \alpha_3(y) = \overline{\alpha_1(\overline{x y})}$$

for all $x, y \in \mathbb{O}$.

He calls the theorem the “principle of triality for SO(8)”. It’s a step toward proving the fact that Spin(8) has an outer automorphism of order 3 relating its three 8-dimensional representations. He proves the lemma simply by equation-juggling using basic properties of normed division algebras.

Now suppose conjugation by $a \in \mathbb{O}$ defines an inner automorphism of the octonions. We can normalize $a \in \mathbb{O}$ without changing the inner automorphism it defines, and then we have $a \overline{a} = 1$, so $a^{-1} = \overline{a}$. Thus we have

$$(a x \overline{a})(a y \overline{a}) = a (x y) \overline{a}$$

Next we use a cool fact about the octonions called the Moufang identity:

$$(b c)(d b) = b(c d)b$$

where the right-hand side is actually well-defined regardless of how you parenthesize it. Take $b = \overline{a}$ and get

$$(\overline{a} c)(d \overline{a}) = \overline{a}(c d) \overline{a}$$

We can rewrite this equation if we take

$$\alpha_1(x) = \overline{a} x , \; \alpha_2(x) = x \overline{a} , \; \alpha_3(x) = a x a$$

Namely, we can rewrite it as

$$\alpha_1(c) \alpha_2(d) = \overline{\alpha_3(\overline{c d})}$$

Since $\alpha_1, \alpha_2, \alpha_3 \in \text{SO}(8)$, Lemma 1.14.3 jumps in and tells us

$$\alpha_2(c) \alpha_3(d) = \overline{\alpha_1(\overline{c d})}$$

Concretely this means

$$(c \overline a)(a d a) = \overline{\overline{a} (\overline{c d})}$$

or in other words

$$(c \overline a)(a d a) = (c d) a$$

Now, cleverly taking $c = a x$, $d = y a$, we get

$$(a x \overline{a})(a y a^2) = a (x y) a^2$$

But remember we have assumed that conjugation by $a$ is an inner automorphism and $a^{-1} = \overline{a}$. This gives a very similar equation:

$$(a x \overline{a})(a y \overline{a}) = a (x y) \overline{a}$$

Then the uniqueness part of Theorem 1.14.2 implies we must have

$$\overline{a} = \pm a^2$$

or

$$a^3 = \pm 1$$

so $a^6 = 1$ and $a$ defines an inner automorphism of order 3.

This argument is pretty convoluted, so it’s hard to see how the threeness of triality forces inner automorphisms of the octonions to have order 3. However, we can also turn this argument around and show that $a^6 = 1$ implies that $a$ defines an inner automorphism.

I think the answer (or something close to it) turns out to be in Conway and Smith’s On Quaternions and Octonions in their discussion of companions to an orthogonal transformation.

They define an inverse loop (p. 83) and show that any inverse loop will satisfy $x y z=y z x=z x y=1$. They also introduce the isotopies of an inverse loop, namely triples of invertible maps $(\alpha,\beta,\gamma)$ such that $x y z =1$ implies that $x^\alpha y^\beta z^\gamma = 1$ (p. 84). The principle of triality is related to the fact that $(\alpha,\beta,\gamma)$ being an isotopy also implies that $(\beta,\gamma,\alpha)$ and $(\gamma,\alpha,\beta)$ are isotopies.

The connection that we are looking for seems to come from the concept of companions to any member of an isotopy (p. 86). If we take the map $\gamma$ then there will be a pair of elements $b$ and $a$ (in the orbit of the identity) such that $(x y)^\gamma = (x^\gamma b)(a y^\gamma)$. Suppose that we take the map to be $\gamma: x \mapsto a^{-1} x a$ (shown as $\text{T}_a$ on p. 98). Provided that diassociativity holds (which I suspect it must for a Moufang loop), then the companions of $\gamma$ are $a^{-3}$ and $a^3$. That is, we have $a^{-1}(x y)a = ((a^{-1} x a)a^{-3})(a^3 (a^{-1}y a))$.

I might be missing an assumption somewhere but I think we can conclude something close to this: In any Moufang loop, the map $\gamma: x \mapsto a^{-1} x a$ has companions $a^{-3}$ and $a^3$. Therefore the map $\gamma$ is an automorphism only when these companions cancel out due to (1) the loop being associative or (2) the element $a^3$ belonging to the nucleus of the loop.