## November 22, 2022

### Inner Automorphisms of the Octonions

#### Posted by John Baez

What are the inner automorphisms of the octonions?

Of course this is an odd question. Since the octonions are nonassociative you might fear the map

$f : \mathbb{O} \to \mathbb{O}$

given by

$f(x) = g x g^{-1}$

for some octonion $g \ne 0$ is not even well-defined!

But it is.

The reason is that the octonions are alternative: the unital subalgebra generated by any two octonions is associative. Furthermore, the inverse $g^{-1}$ of $g \ne 0$ is in the unital subalgebra generated by $g$. This follows from

$g^{-1} = \frac{1}{|g|^2} \overline{g}$

and the fact that $\overline{g}$ is in the unital subalgebra generated by $g$, since we can write $g = a + b$ where $a$ is a real multiple of the identity and $b$ is purely imaginary, and then $\overline{g} = a - b = 2a - g$.

It follows that whenever $g$ is a nonzero octonion, we have

$(g x) g^{-1} = g (x g^{-1})$

for all octonions $x$, so we can write either as

$f(x) = g x g^{-1} .$

However, there is no reason a priori to expect $f$ to be an automorphism, meaning

$f(xy) = f(x) f(y)$

for all $x,y \in \mathbb{O}$. For which octonions $g$ does this happen?

Of course it happens when $g$ is real, i.e. a real multiple of $1$. But that’s boring—because then $f$ is the identity. Can we find more interesting inner automorphisms of the octonions?

A correspondent, Charles Wynn, told me that $f$ is an automorphism when

$g = \frac{1}{2} + \frac{\sqrt{3}}{2} i$

and $i \in \mathbb{O}$ is any element with $i^2 = -1$. This kind of element $g$ is a particular sort of 6th root of unity in the octonions—one that lies at a $60^\circ$ angle from the positive real axis.

A bit of digging revealed this paper:

In Theorem 2.1, Lamont claims that $f(x) = g x g^{-1}$ is an automorphism of the octonions iff and only if $g$ is either real or

$(\mathrm{Re}(g))^2 = \frac{1}{4} |g|^2.$

In other words, $f$ is an automorphism iff the octonion $g$ lies at an angle of $0^\circ, 60^\circ, 120^\circ$ or $180^\circ$ from the positive real axis. These cases include all 6th roots of unity in the octonions!

I haven’t fully checked the proof, but it seems to use little more than the Moufang identity.

I wonder what this fact means? How do these inner automorphisms sit inside the group of all automorphisms of the octonions, $\mathrm{G}_2$?

Posted at November 22, 2022 5:03 PM UTC

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### Re: Inner Automorphisms of the Octonions

How many 6th roots of unity in the octonions are there?

Are there any general results about numbers of roots of octonion polynomials?

Posted by: David Corfield on November 23, 2022 9:54 AM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

Just like for the complex numbers and quaternions, you can write any octonion as

$a + b i$

where $i$ is some square root of $-1$. The only difference here is that the complex numbers have just two square roots of $-1$, while the quaternions have a 2-sphere of them, and the octonions have a whole 6-sphere of them! In each case, everything of norm 1 that’s orthogonal to the real axis is a square root of $-1$.

So, as long as we’re playing around with just one octonion and its powers, we might as well be in the complex numbers.

Thus, the conditions for an octonion to be a 6th root of $-1$ look just like those for the complex numbers, except that now you have more choices of what you can call $i$. There’s a 6-sphere of sixth roots of $-1$ that lie at a $60^\circ$ degree angle to the positive real axis. They look like this:

$\frac{1}{2} + \frac{\sqrt{3}}{2} i$

where $i$ is any square root of $-1$. And there’s another 6-sphere of sixth roots of $-1$ that lie at a $120^\circ$ degree angle to the positive real axis. They look like this:

$- \frac{1}{2} + \frac{\sqrt{3}}{2} i$

where $i$ is any square root of $-1$.

Besides these, there are two more sixth roots of $-1$ in the octonions: $1$ and $-1$.

Posted by: John Baez on November 23, 2022 10:21 AM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

I thought this might be a path into

How do these inner automorphisms sit inside the group of all automorphisms of the octonions, $G_2$?

by finding the dimension of $Inn(\mathbb{O})$?

But now there’s the concern of when two distinct octonions of the right kind give the same inner automorphism, and since I really ought to be further through my stack of marking by now, I’d better leave that to someone else.

Posted by: David Corfield on November 23, 2022 11:25 AM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

I think we’ve seen this by now: there are 4 kinds of elements of $\mathbb{O}$ that give inner automorphisms of the octonions: $1, -1$, the 6-sphere of elements like this:

$\frac{1}{2} + \frac{\sqrt{3}}{2} i$

and the 6-sphere of elements like this:

$-\frac{1}{2} + \frac{\sqrt{3}}{2} i$

where in each case $i$ ranges over the 6-sphere of square roots of $-1$.

It’s easy to see that if conjugation $g \in \mathbb{O}$ gives an inner automorphism, so does conjugation by $-g$. They give the same inner automorphism.

So, $1$ and $-1$ both give the identity automorphism, and the two 6-spheres give the same inner automorphisms, since negation maps the first 6-sphere to the second, and vice versa.

Thus, the space of inner automorphisms of $\mathbb{O}$ is $S^6 \sqcup \{1\}$.

This is not a group, since composing inner automorphisms does not usually give an inner automorphism in this nonassociative context!

It still gives an automorphism, though. So we can wonder: which automorphisms of $\mathbb{O}$ do we get by composing inner automorphisms?

And now that my math social media life has moved from Twitter to Mastodon, I got an answer on Mathstodon from Herid Fel. We get a group of automorphisms that’s dense in the group of all automorphisms of $\mathbb{O}$, namely $\mathrm{G}_2$.

Posted by: John Baez on November 24, 2022 12:31 PM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

David wrote:

Are there any general results about numbers of roots of octonion polynomials?

From what I’ve said, the roots of an octonion polynomial with real coefficients are easy: you just find the roots in $\mathbb{C}$ as usual, and then let $i$ range over the whole 6-sphere of square roots of $-1$. This has the effect of taking each individual root in $\mathbb{C}$ and spinning it around to get a 6-sphere of roots in $\mathbb{O}$ — except for real roots, which just sit there, each giving a single root in $\mathbb{O}$.

Seeking roots of an octonion polynomial with octonion coefficients, on the other hand, can easily lead you into gibbering madness. For starters, due to nonassociativity and noncommutativity, there’s the question of what you even mean by a polynomial.

Posted by: John Baez on November 23, 2022 10:54 AM | Permalink | Reply to this

### Conjugacy class

Consider the stabilizer in $\mathrm{G}_2$ of $i$. It acts on the complementary $\mathbb{R}^6$ inside the imaginary octonions, as $SU(3)$. If you let $t$ be a central element in that $SU(3)$, its centralizer in $\mathrm{G}_2$ is just that $SU(3)$, so the conjugacy class is $\mathrm{G}_2/SU(3) = S^6$.

That is to say, the $S^6$ you’re discovering is one of the (two) conjugacy classes of order 3.

I wonder if you can find the conjugacy class of order 2 that looks like $\mathrm{G}_2/SO(4)$, and corresponds to the set of $\mathbb{H}$s inside $\mathbb{O}$ instead of the set of $\mathbb{C}$s.

Posted by: Allen Knutson on November 23, 2022 9:07 PM | Permalink | Reply to this

### Re: Conjugacy class

Cool! By the way, there’s a 6-sphere $S$ of octonions of the form

$g = \frac{1}{2} + \frac{\sqrt{3}}{2} i$

where $i \in \mathbb{O}$ has $i^2 = -1$. Conjugating by any such $g$ is an inner automorphism.

Conjugating by $g^2$ gives the square of this automorphism — less obvious than it seems, but true because the octonions are alternative — so it’s also an automorphism, hence another inner automorphism.

Notice that

$g^2 = -\frac{1}{2} + \frac{\sqrt{3}}{2} i$

live on a different 6-sphere $S'$ in the octonions. However, conjugating by $g^2$ has the same effect as conjugating by

$-g^2 = \frac{1}{2} - \frac{\sqrt{3}}{2} i$

and $-g^2$ lives on the same 6-sphere $S$ that $g$ does: it’s just the antipodal point on that 6-sphere!

Conjugating by $g^3$ gives yet another inner automorphism (for the same reason), but

$g^3 = -1$

so this inner automorphism is the identity.

Summary: we can identify $S \subset \mathbb{O}$ with the 6-sphere of inner automorphisms of $\mathbb{O}$, which are all of order 3. Squaring any of these automorphisms gives the antipodal point on this 6-sphere.

Posted by: John Baez on November 24, 2022 5:00 PM | Permalink | Reply to this

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