## 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

### Re: Conjugacy class

I guess what I’m asking for is this. Given a C inside O plus a choice of i vs. -i in that C, you explained how to use it to define an order 3 automorphism of O, by conjugation by exp(2 pi i/6).

Now, given an H inside O plus choices of i and j in that H, I want an order 2 automorphism of O, something like z |-> (iz)j. Is that an automorphism, by chance?

Posted by: Allen Knutson on November 28, 2022 12:57 AM | Permalink | Reply to this

### Re: Conjugacy class

Charles Wynn, who brought the order-3 inner automorphisms to my attention, also had a description of order-2 automorphisms. Let me look at it. Ah, yes! He described these in terms of two orthogonal imaginary unit octonions. But I won’t reveal the details because he hasn’t authorized me to, and he’s trying to publish a paper about such things.

Posted by: John Baez on November 28, 2022 12:38 PM | Permalink | Reply to this

### Re: Conjugacy class

Does he describe the other conjugacy class of order 3 automorphisms? (There are exactly two, and only one class of order 2s.)

At the far end of the Weyl alcove, i.e. the conjugate points of the exponential map from the Lie algebra to the compact group, there’s an 11-dim family of automorphisms that includes the C-based and H-based ones we were discussing, at the two ends. I don’t have much reason to expect that elements of that family will be more describable than general elements of G2, though.

Posted by: Allen Knutson on November 29, 2022 2:08 PM | Permalink | Reply to this

### Re: Conjugacy class

Allen wrote:

Does he describe the other conjugacy class of order 3 automorphisms?

I don’t think so. By the way, he’s here now. In your terminology, here’s how he defines an automorphism $F \colon \mathbb{O} \to \mathbb{O}$ given two anticommuting square roots of $-1$ in the octonions, say $i$ and $j$:

$F(z) = (i j)^{-1} (i(j z))$

This is clearly the identity on the quaternion subalgebra generated by $i$ and $j$. But if $z$ is orthogonal to this quaternion subalgebra, it will ‘anti-associate’ with $i$ and $j$, so

$F(z) = (i j)^{-1} (i(j z)) = -(i j)^{-1} ((i j) z) = -z$

where in the last step I used the alternativity of the octonions: the subalgebra generated by $i j$ and $z$ is associative.

So, $F$ is the identity on the quaternion subalgebra generated by $i$ and $j$, but $-1$ on the orthogonal complement of that algebra!

Pretty slick.

Posted by: John Baez on November 29, 2022 2:19 PM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

Hi!

Not such a big secret so I’ll just repeat a few things I told Dr. Baez:

Let

$F(S,T,X) = (T S)^{-1}(T(S X))$

in the octonion algebra.

Then if $S$ and $T$ are orthogonal imaginaries of reciprocal norm, $F$ fixes the quaternion algebra generated by $S$ and $T$ and negates the complement.

This is an automorphism that belongs to the conjugacy class $\mathrm{G}_2/SO(4)$ as you mentioned. $F$ is the single-point inversion that characterizes symmetric spaces. Whatever that means, I know Jack about Lie theory.

$F$ is a known automorphism type in what’s called conjugacy-closed loops, I saw it in a paper The structure of conjugacy-closed loops by Kenneth Kunen (2000).

Because $F$ isn’t an automorphism for non-orthogonal $S$ and $T$, the octonions as a whole are not conjugacy closed. Their basis elements are, though and an automorphism of standard basis elements (or any set of elements isomorphic to a set of basis elements) is inherited by the algebra as a whole.

Interesting facts: for valid $S$ and $T$, the quantity $F(S,T,X)$ is equal to

$(S T)(T(S X)) = (T((T X)S))S$

Also let $S$, $T$ and $U$ be an octonion triple such that $S(T U) = -(S T)U$. Then define $V, W$ by

$V = -\frac{1}{2} S + \frac{\sqrt{3}}{2} S U$ $W = -\frac{1}{2} T + \frac{\sqrt{3}}{2} T U$

and let

$P = \frac{1}{2} + \frac{\sqrt{3}}{2} U$

Then

$F(S,T,(F(V,W,X)) = P X P^{-1}$

which is an order-3 automorphism! So the order-2 conjugacy class of the stabilizer of $i$ (or whatever it is) gives the transpositions in various $S_3$ permutation subgroups. The conjugations give the 3-cycles.

Posted by: Charles Wynn on November 28, 2022 1:55 PM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

As for the symmetric space thing – any order two automorphism of a group G results in a “symmetric subgroup” H with the result that G/H is a “symmetric space”. Here G = G2 has no outer automorphisms, and no center, so its order 2 automorphisms come from its conjugacy classes of order 2. There’s exactly one of those, the one you’re dealing with, so the only symmetric space for G2 is this G2/SO(4).

Posted by: Allen Knutson on December 1, 2022 4:40 AM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

Thanks, I think I understand now. I knew of the quotient G2/SO(4) because it is isomorphic to the space of quaternionic subalgebras of the octonions. I suppose there is only 1 nontrivial element of G2 that fixes a given quaternion subalgebra, and that is the one which negates that algebra’s complement.

If so then I guess that would explain why the order 2 conjugacy class of G2 is isomorphic to the space of quaternionic subalgebras of the octonions, inasmuch as F(S,T,X) gives a bijection between them.

Posted by: Charles Wynn on December 5, 2022 1:29 AM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

I need to amend what I said about conjugacy closure: the cited paper only says that $F$ is an automorphism for conjugacy-closed loops, not that it’s a conjugacy-closed loop if $F$ is an automorphism. Even basis tables of the octonions may lack conjugacy closure, this needs to be checked.

My understanding of the property is that:

$((Z Y)X)Y^{-1}$

must equal $Z P$ for some $P$, which seems to fail for octonions? Not sure I understand these loop theory definitions properly.

Posted by: Charles Wynn on November 28, 2022 2:43 PM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

Note that Conway-Smith p.98 prove that the map

$T_a: x \mapsto a^{-1}xa$

is an automorphism if and only if $a^3$ is real, and they credit this to Zorn 1935, so we’ve been calling this “Zorn’s proposition” (in a 2016 paper with R. Paluba).

For us the important thing was to consider such $a$ of the form

$a=(1+i+j+k)/2$

for a quaternionic triple $i,j,k$ of imaginary octonions. Then any line in the Fano plane gives such an $a$ (the three points on the line is a quaternionic triple).

For us, this led to a nice story related to $G_2$ character varieties of $4$-punctured spheres. If you’re interested the basic idea is as follows:

If you take three lines passing through a single point in the Fano plane then the corresponding triple $T_1,T_2,T_3$ all live in the same 6d conjugacy class $C$ in $G=G_2(C)$, and their product $g=T_1.T_2.T_3$ lives in a generic (12d) conjugacy class $C_4 \subset G$. The complex character variety

$\{(a,b,c,d)\in C^3\times C_4 \bigl\vert d=abc\}/G$

then has complex dimension $2=(6.3+12)-2.14$, and one can prove it is the cubic surface:

$xyz+ x^2 + y^2 + z^2 = x + y + z$

that I like to call the “Klein cubic surface”, since it first appeared as the character variety constructed in the same way but starting with the standard triple of generators of the Klein complex reflection group in $GL_3(C)$ (the complex reflection group of order 336 containing Klein’s simple group of order 168).

Thus we get two “representations” of this Klein surface as a character variety, either for $G=G_2(C)$ or for $G=GL_3(C)$.

There is a natural “Hurwitz” action of the 3 string braid group on the character variety, and the braid orbit of our preferred triple (T1,T2,T3) has size seven, corresponding to the 7 points (x,y,z)=000,001, …, 110. The standard triple of generators of the Klein complex reflection group in $GL_3(C)$ braids to give the same 7 points (in the other representation of the surface), and we are still a bit puzzled as to “why” this is true.

There is a third representation of the Klein surface as an $SL_2(C)$ character variety (a Fricke-Klein-Vogt surface), whence the 7 points correspond to certain triples of elements of $SL_2(C)$ that generate the (infinite) 237 triangle group, and then we can relate the braid group orbit to a specific degree 7 algebraic solution of the Painleve 6 differential equation (whose monodromy is the original braid group orbit)…

Posted by: Philip Boalch on November 30, 2022 7:57 PM | Permalink | Reply to this

### Re: Inner Automorphisms of the Octonions

Thanks for the comment about Zorn. I had been looking for a citation of who first discovered this, as I’m writing/revising a paper on the n-dimensional Cayley-Dickson algebras. In these the “new” root of -1 at the “last” iteration of the Cayley-Dickson construction generates a copy of the Eisenstein integers, where conjugation by the norm-1 subset gives automorphisms. I wondered who first discovered this for the octonions (where all roots of -1 are isomorphic).

Posted by: Charles Wynn on December 5, 2022 2:24 AM | Permalink | Reply to this

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