### Octonions and the Standard Model (Part 1)

#### Posted by John Baez

I want to talk about some attempts to connect the Standard Model of particle physics to the octonions. I should start out by saying I don’t have any big agenda here. It’d be great if the octonions — or for that matter, *anything* — led to new insights in particle physics. But I don’t have such insights, and for me particle physics is just a hobby. I’m not trying to come up with a grand unified theory. I just want to explain some patterns linking the Standard Model to the octonions.

Understanding these patterns requires knowing a bit of physics and a bit of math. I’ll focus on the math side of things: mainly, I’ll be polishing up some existing ideas and trying to make them more pretty. I’ll assume you either know the physics or can fake it: either way, it won’t be the main focus.

In writing this first post, my attempt to explain an octonionic description of the strong force led me to a construction of the octonions that makes them look very much like the quaternions. I don’t know if it’s new, but I’d never seen it before. The basic idea is that *octonions are to $\mathbb{C}^3$ as quaternions are to $\mathbb{R}^3$*.

The group of automorphisms of the octonions that preserve any chosen octonion $i$ with $i^2 = -1$ happens to be isomorphic to $\mathrm{SU}(3)$. In physics this is interesting because $\mathrm{SU}(3)$ is the gauge group of the strong nuclear force: the force that holds quarks together. Furthermore, when we look at how $\mathrm{SU}(3)$ acts on the octonions, it matches how $\mathrm{SU}(3)$ acts on quarks and leptons.

This idea has been around for a long time, going back at least to here:

- Murat Günaydin and Feza Gürsey, Quark structure and octonions,
*Journal of Mathematical Physics***14**(1973), 1651–1667.

But if you look at that paper, you’ll see swarms of indices, which in my opinion conceal the beauty of what’s going on.

We could try to simply compute the group of automorphisms preserving some octonion $i$ with $i^2 = -1$. But it turns out there’s a description of the octonions that makes it *obvious* that this group at least *contains* a copy of $\mathrm{SU}(3)$.

First remember how the quaternions work. The quaternions are numbers like

$a + b i + c j + d k$

with $a,b,c,d$ real. In the late 1800s Josiah Willard Gibbs decided to chop quaternions into a scalar and vector part; if we do that we can write a quaternion as

$a + \vec{a}$

where $a \in \mathbb{R}$ and $\vec{a} \in \mathbb{R}^3$ is the vector $b i + c j + d k$. Of course, these days kids are discouraged from adding scalars and vectors, but if you use quaternions it’s okay. Here’s how you multiply quaternions:

$(a + \vec{a})(b + \vec{b}) = a b - \vec{a} \cdot \vec{b} + a \vec{b} + b \vec{a} + \vec{a} \times \vec{b}$

So, all four ways of multiplying scalars and vectors are unified in quaternion multiplication:

- multiplication of scalars (the ordinary multiplication of real numbers)
- multiplication of vectors by scalars (scalar multiplication)
- the dot product of vectors
- the cross product of vectors

This is no coincidence: the dot product and cross product were invented by Gibbs when he chopped quaternions into a scalar and vector part, and in 1901 these operations were popularized in Edwin Wilson’s book *Vector Analysis: A Textbook for the Use of Students of Mathematics*, which was based on some lectures by Gibbs.

Now suppose we want to generalize all this using the complex numbers. We’ll take our scalars to live in $\mathbb{C}$ and our vectors to live in $\mathbb{C}^3$.

*One* thing we can do — not the thing I’m interested in, but I have to mention it — is to copy everything that we just did, word for word, replacing $\mathbb{R}$ by $\mathbb{C}$ everywhere. We don’t get the octonions: we get the **biquaternions**, which is a 4-dimensional complex algebra, or 8-dimensonal real algebra. The biquaternions turn out to be isomorphic to the algebra of $2 \times 2$ complex matrices.

In copying everything ‘word for word’, we defined a dot product on $\mathbb{C}^3$ just as we did for $\mathbb{R}^3$:

$\vec{v} \cdot \vec{w} = \sum_{i=1}^3 v_i w_i$

But another alternative is to use the inner product

$\langle \vec{v}, \vec{w}\rangle = \sum_{i=1}^3 \overline{v}_i w_i$

Just as the dot product on $\mathbb{R}^3$ is invariant under the group $\mathrm{O}(3)$ of orthogonal $3 \times 3$ real matrices, this inner product on $\mathbb{C}^3$ is invariant under the group $\mathrm{U}(3)$ of unitary $3 \times$ complex matrices. Since we’re shooting for a connection to $\mathrm{SU}(3)$, let’s use this!

It turns out that this change forces us to use a modified cross product. The cross product on $\mathbb{R}^3$ is invariant under the group $\mathrm{SO}(3)$ of orthogonal $3 \times 3$ real matrices with determinant 1. (So: not reflections, but rotations.) Unfortunately, the cross product on $\mathbb{C}^3$ is *not* invariant under the group $\mathrm{SU}(3)$ of unitary $3 \times 3$ complex matrices with determinant 1.

What’s invariant under $\mathrm{SU}(3)$ is a mutant version of the cross product, where we first take the usual cross product and then take the *complex conjugate* of each component. Let’s call this

$\overline{\times} \colon \mathbb{C}^3 \times \mathbb{C}^3 \to \mathbb{C}^3$

In coordinates, it’s given by

$(\vec{v} \overline{\times} \vec{w})_i = \overline{(\vec{v} \times \vec{w})_i}$

Later I’ll give a coordinate-free definition which makes it clear why this operation is invariant under $\mathrm{SU}(3)$. Some people would say ‘equivariant’ but that sounds too fancy to me right now. I just mean that if $U \in \mathrm{SU}(3)$ we have

$U \vec{v} \overline{\times} U \vec{w} = U(\vec{v} \overline{\times} \vec{w})$

Now we’ve got all the machinery to describe the octonions in a way that makes them look a lot like quaternions. The quaternions were built from real scalars and real vectors:

$\mathbb{H} \cong \mathbb{R} \oplus \mathbb{R}^3$

So now we’ll build the octonions from complex scalars and complex vectors:

$\mathbb{O} \cong \mathbb{C} \oplus \mathbb{C}^3$

To multiply guys in here we’ll copy the formula for multiplying quaternions, but we’ll make sure the multiplication is preserved by the obvious action of $\mathrm{SU}(3)$ — namely, the trivial action on scalars and the action by matrix multiplication on vectors. This action clearly preserves $i \in \mathbb{C} \subseteq \mathbb{O}$, which is another thing we want. Then, we’ll put the most obvious norm on this space, the one with

$\| a + \vec{a} \|^2 = |a|^2 + \langle \vec{a}, \vec{a} \rangle$

and we’ll show that multiplication gets along with this norm:

$\|(a + \vec{a})(b + \vec{b})\| = \|a + \vec{a}\| \|b + \vec{b}\|$

So, we’ll have an 8-dimensional normed division algebra — which implies that we’ve got the octonions!

Since our formula for multiplication will look almost like the formula for multiplying quaternions, the proof that multiplication gets along with the norm will be almost like the proof that works for the quaternions.

Even better, just as the quaternions have $\mathrm{SO}(3)$ acting as automorphisms since their multiplication was built from $\mathrm{SO}(3)$-invariant operations, the octonions will have $\mathrm{SU}(3)$ acting as automorphisms that preserve $i$, because we’ll build their multiplication from $\mathrm{SU}(3)$-invariant operations that preserve $i$!

Here’s the result:

**Theorem 1.** *If we define multiplication on $\mathbb{C} \oplus \mathbb{C}^3$ by
$(a + \vec{a})(b + \vec{b}) = a b - \langle \vec{a}, \vec{b}\rangle + \overline{a} \vec{b} + b \vec{a} + \vec{a} \overline{\times} \vec{b}$
then we obtain a normed division algebra, since $1 \in \mathbb{C}$ acts as the multiplicative identity and
$\|(a + \vec{a})(b + \vec{b})\| = \|a + \vec{a}\| \|b + \vec{b}\|$
This normed division algebra is isomorphic to the octonions.*

Note the formula for multiplication is *exactly* like the formula for multiplying quaternions… except for three things. First, the dot product has been replaced by the inner product. Second, the cross product has been replaced by the mutant cross product $\overline{\times}$. These changes are not weird, because we want operations that are invariant under $\mathrm{SU}(3)$. Third, the term $a \vec{b}$ has been replaced by $\overline{a} \vec{b}$, while the term $b \vec{a}$ remains unchanged. This *is* weird. Of course complex conjugation of scalars is invariant under $\mathrm{SU}(3)$. But the reason we *need* it is to get a normed division algebra.

I’ll give the proof at the end of this post. In reading this proof, the main thing to keep an eye on is the overlines for complex conjugation. I checked my calculations a few times, but if I got them wrong then maybe the $\overline{a} \vec{b}$ business isn’t needed. That would make me very happy.

Now let me tell you how I came up with this stuff. This part is more fancy than the story so far, but it may appeal to certain people.

Let’s start by *assuming* the octonions $\mathbb{O}$ are an 8-dimensional normed division algebra such that the automorphisms preserving any octonion $i$ with $i^2 = -1$ form a copy of $\mathrm{SU}(3)$, and use this to try to *guess* how multiplication in the octonions works.

What is $\mathbb{O}$ like, as a representation of $\mathrm{SU}(3)$? Initially, at least, it’s a *real* representation. It has a 1-dimensional invariant subspace spanned by the element $1$, since $1$ is fixed by every automorphism. And it has a 1-dimensional invariant subspace spanned by $i$, by assumption. Taking the direct sum of these, we get a 2-dimensional trivial subrepresentation which is also a subalgebra: a copy of $\mathbb{C}$ in $\mathbb{O}$.

Since $\mathrm{SU}(3)$ is compact there must be a complementary 6-dimensional subspace

$V \subseteq \mathbb{O}$

that’s also preserved by $\mathrm{SU}(3)$. We can understand this better if we break it into irreducible representations of $\mathrm{SU}(3)$. The lowest-dimensional real irreps of $\mathrm{SU}(3)$ are the 1-dimensional trivial rep and the 6-dimensional irrep coming from the usual action of $\mathrm{SU}(3)$ on $\mathbb{C}^3$. Note: $\mathrm{SU}(3)$ has two inequivalent *complex* irreps of dimension 3, namely its usual rep on $\mathbb{C}^3$ and the dual of this — but these are equivalent as *real* reps, as always for a unitary rep and its dual.

It follows that $V \subseteq \mathbb{O}$ is the 6-dimensional real irrep of $\mathrm{SU}(3)$. Indeed, the only alternative is that $V$ is the direct sum of 6 copies of the 1-dimensional *trivial* rep! But in that case $\mathrm{SU}(3)$ would act trivially on $\mathbb{O}$, so it would not be a group of automorphisms of $\mathbb{O}$. So that alternative is out.

So $V$ must be isomorphic to the usual rep on $\mathbb{C}^3$, and we get

$\mathbb{O} \cong \mathbb{C} \oplus \mathbb{C}^3$

as real reps of $\mathrm{SU}(3)$.

Now we can try to figure out the options for multiplying octonions. Since the multiplication is invariant under $\mathrm{SU}(3)$, it must be built from $\mathrm{SU}(3)$-invariant bilinear maps involving $\mathbb{C}$ and $\mathbb{C}^3$.

This should instantly make you think of the quaternions! The quaternions

$\mathbb{H} \cong \mathbb{R} \oplus \mathbb{R}^3$

have multiplication built from $\mathrm{SO}(3)$-invariant bilinear maps involving $\mathbb{R}$ and $\mathbb{R}^3$. These maps are familiar:

- multiplication of scalars
- multiplication of vectors by scalars
- the dot product of vectors
- the cross product of vectors

So, we should think about how to adjust these in the complex case, to get $\mathrm{SU}(3)$-invariant maps. Multiplication of scalars remains unchanged, as does the multiplication of vectors by scalars, but the dot product and cross product need to be adjusted.

First remember what’s going on with the dot product and cross product in $\mathbb{R}^3$. The usual inner product on $\mathbb{R}^3$ is invariant under $\mathrm{O}(3)$ just by definition. The cross product is more complicated. To build it, we start with the wedge product.

$\wedge \colon \mathbb{R}^3 \times \mathbb{R}^3 \to \Lambda^2 \mathbb{R}^3$

which is invariant under all of $\mathrm{GL}(3,\mathbb{R})$. Then we apply an isomorphism

$\alpha \colon \Lambda^2 \mathbb{R}^3 \to (\mathbb{R}^3)^\ast$

This is invariant under $\mathrm{SL}(3,\mathbb{R})$, since it’s defined by

$\alpha(\omega)(\vec{v}) \; \mathrm{vol} = \omega \wedge \vec{v}$

where $\mathrm{vol}$ is the usual volume 3-form and $\omega$ is anything in $\Lambda^2 \mathbb{R}^3$. Then we apply the so-called **musical isomorphism**:

$\sharp \colon (\mathbb{R}^3)^* \to \mathbb{R}^3$

which invariant under $\mathrm{O}(3)$, since it’s defined by

$\sharp(f) \cdot \vec{v} = f(\vec{v})$

Putting all this crud together, we get the cross product

$\times \colon \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{R}^3$

defined by

$\vec{v} \times \vec{w} = \sharp \alpha(\vec{v} \wedge \vec{w})$

The cross product is invariant under $\mathrm{SO}(3)$, since that’s the intersection of all the groups I just mentioned!

Whew. It’s amazing kids can learn the cross product.

Now let’s copy all this stuff using complex numbers and get the octonions. We’ll replace the dot product with the inner product on $\mathbb{C}^3$:

$\langle \vec{v}, \vec{w} \rangle = \sum_{i=1}^3 \overline{v}_i w_i$

This is invariant under $\mathrm{U}(3)$, but it’s **sesquilinear**: complex-linear in the second argument but conjugate-linear in the first. To define a mutant version of the wedge product, we’ll start with the complex-bilinear wedge product

$\wedge \colon \mathbb{C}^3 \times \mathbb{C}^3 \to \Lambda^2 \mathbb{C}^3$

This is invariant under all of $\mathrm{GL}(3,\mathbb{C})$. Then we apply an isomorphism

$\alpha \colon \Lambda^2 \mathbb{C}^3 \to (\mathbb{C}^3)^\ast$

This is complex-linear and invariant under $\mathrm{SL}(3,\mathbb{C})$, since it’s defined by

$\alpha(\omega)(\vec{v}) \; \mathrm{vol} = \omega \wedge \vec{v}$

where $\mathrm{vol}$ is the usual volume 3-form. Then we apply an isomorphism

$\sharp \colon (\mathbb{C}^3)^\ast \to \mathbb{C}^3$

This is now *conjugate*-linear and invariant under $\mathrm{U}(3)$, since it’s defined by

$\langle \sharp(f) , \vec{v} \rangle = f(\vec{v})$

Putting all this together, we get

$\overline{\times} \colon \mathbb{C}^3 \times \mathbb{C}^3 \to \mathbb{C}^3$

defined by

$\vec{v} \overline{\times} \vec{w} = \sharp \alpha(\vec{v} \wedge \vec{w})$

This is *conjugate-linear* and invariant under $\mathrm{SU}(3)$. If we work it out in coordinate we see

$(\vec{v} \overline{\times} \vec{w})_i = \overline{(\vec{v} \times \vec{w})_i}$

So, it’s clear that we should define octonion multiplication using these maps, and that we should make it look like quaternion multiplication so we can prove we get a normed division algebra. The one unpredictable twist is that now we can take complex conjugates of scalars: this operation is also $\mathrm{SU}(3)$-invariant. We have to do this in the right places to get a normed division algebra!

So let’s see why, by proving this theorem:

**Theorem 1.** *If we define multiplication on $\mathbb{C} \oplus \mathbb{C}^3$ by
$(a + \vec{a})(b + \vec{b}) = a b - \langle\vec{a}, \vec{b}\rangle + \overline{a} \vec{b} + b \vec{a} + \vec{a} \overline{\times} \vec{b}$
then we obtain a normed division algebra, since $1 \in \mathbb{C}$ acts as the multiplicative identity and
$\|(a + \vec{a})(b + \vec{b})\| = \|a + \vec{a}\| \|b + \vec{b}\|$
This normed division algebra is isomorphic to the octonions.*

**Proof.** It’s enough to show that we have a normed division algebra, since the octonions are the only 8-dimensional normed division algebra over the reals. We’ll use the norm with

$\|a + \vec{a}\|^2 = |a|^2 + \langle \vec{a} , \vec{a} \rangle$

and show that

$\|(a + \vec{a})(b + \vec{b})\| = \|a + \vec{a}\| \|b + \vec{b}\|$

It’s a calculation — and if we take all our complex numbers to be real, this calculation will also show that that $\mathbb{R} \oplus \mathbb{R}^3$ is a normed division algebra with the same multiplication formula: it’s the quaternions.

We start with

$\|(a + \vec{a})(b + \vec{b})\|^2 = \|a b - \langle\vec{a}, \vec{b}\rangle + \overline{a} \vec{b} + b \vec{a} + \vec{a} \overline{\times} \vec{b} \|^2$

and use the definition of the norm to break this up into two terms:

$|a b - \langle \vec{a}, \vec{b} \rangle|^2 + \| \overline{a} \vec{b} + b \vec{a} + \vec{a} \overline{\times} \vec{b} \|^2$

We can expand the first term:

$|a b - \langle \vec{a}, \vec{b} \rangle|^2 = |a b|^2 - 2 \mathrm{Re}(a b \langle \vec{b} , \vec{a} \rangle) + |\langle \vec{a}, \vec{b}\rangle|^2$

We can expand the second term:

$\|\overline{a} \vec{b} + b \vec{a} + \vec{a} \overline{\times} \vec{b} \|^2 =$ $\| \overline{a} \vec{b} \|^2 + \|b \vec{a} \|^2 + \|\vec{a} \overline{\times} \vec{b}\|^2 + 2 \mathrm{Re} \big( a b \langle \vec{b}, \vec{a} \rangle + a \langle \vec{b}, \vec{a} \overline{\times} \vec{b} \rangle + \overline{b} \langle \vec{a}, \vec{a} \overline{\times} \vec{b} \rangle \big)$

But note that

$\langle \vec{a}, \vec{a} \overline{\times} \vec{b} \rangle = \overline{\vec{a} \cdot (\vec{a} \times \vec{b})} = 0$

by a well-known vector identity which works for complex vectors just as for real ones. Similarly $\langle \vec{b}, \vec{a} \overline{\times} \vec{b} \rangle = 0$. So, the expanded second term simplifies, and when we add it to the expanded first term the bits involving $\mathrm{Re}(a b \langle \vec{b} , \vec{a} \rangle)$ cancel out. We’re left with this:

$|a b|^2 + |\langle \vec{a}, \vec{b}\rangle|^2 + \| \overline{a} \vec{b} \|^2 + \|b \vec{a} \|^2 + \|\vec{a} \overline{\times} \vec{b}\|^2$

We need to show this equals

$\begin{array}{ccl} \|a + \vec{a} \|^2 \| b + \vec{b}\|^2 &=& (|a|^2 + \|\vec{a}\|^2) (|b|^2 + \|\vec{b}\|^2) \\ \\ &=& |a|^2 |b|^2 + |a|^2 \|\vec{b}\|^2 + |b|^2 \|\vec{a}\|^2 + \|\vec{a}\|^2 \|\vec{b}\|^2 \end{array}$

And if you look, you’ll see 3 terms match, so it’s enough to show

$|\langle \vec{a}, \vec{b}\rangle|^2 + \|\vec{a} \overline{\times} \vec{b}\|^2 = \|\vec{a}\|^2 \|\vec{b}\|^2$

This would be a familiar vector identity if we were working in $\mathbb{R}^3$ with the usual cross product instead of the mutant one. But notice, the mutant cross product is obtained from the ordinary one by componentwise complex conjugation, so

$\|\vec{a} \overline{\times} \vec{b}\|^2 = \|\vec{a} \times \vec{b}\|^2$

Thus, we just need to show

$|\langle \vec{a}, \vec{b}\rangle|^2 + \|\vec{a} \times \vec{b}\|^2 = \|\vec{a}\|^2 \|\vec{b}\|^2$

This works for $\mathbb{C}^3$ and its inner product exactly as it does for $\mathbb{R}^3$ and its dot product: you can either write out both sides using components and see they agree, or do a geometrical argument using the law of cosines and the law of sines. █

Note: after developing the above approach to the octonions, I found it here:

- Ichiro Yokota, Exceptional Lie groups, Section 1.5.

My approach is more ‘geometrical’, less computational, but it amounts to the same thing.

- Part 1. How to define octonion multiplication using complex scalars and vectors, much as quaternion multiplication can be defined using real scalars and vectors. This description requires singling out a specific unit imaginary octonion, and it shows that octonion multiplication is invariant under $\mathrm{SU}(3)$.
- Part 2. A more polished way to think about octonion multiplication in terms of complex scalars and vectors, and a similar-looking way to describe it using the cross product in 7 dimensions.
- Part 3. How a lepton and a quark fit together into an octonion — at least if we only consider them as representations of $\mathrm{SU}(3)$, the gauge group of the strong force. Proof that the symmetries of the octonions fixing an imaginary octonion form precisely the group $\mathrm{SU}(3)$.
- Part 4. Introducing the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$: the $3 \times 3$ self-adjoint octonionic matrices. A result of Dubois-Violette and Todorov: the symmetries of the exceptional Jordan algebra preserving their splitting into complex scalar and vector parts and preserving a copy of the $2 \times 2$ adjoint octonionic matrices form precisely the Standard Model gauge group.
- Part 5. How to think of $2 \times 2$ self-adjoint octonionic matrices as vectors in 10d Minkowski spacetime, and pairs of octonions as left- or right-handed spinors.
- Part 6. The linear transformations of the exceptional Jordan algebra that preserve the determinant form the exceptional Lie group $\mathrm{E}_6$. How to compute this determinant in terms of 10-dimensional spacetime geometry: that is, scalars, vectors and left-handed spinors in 10d Minkowski spacetime.
- Part 7. How to describe the Lie group $\mathrm{E}_6$ using 10-dimensional spacetime geometry.
- Part 8. A geometrical way to see how $\mathrm{E}_6$ is connected to 10d spacetime, based on the octonionic projective plane.
- Part 9. Duality in projective plane geometry, and how it lets us break the Lie group $\mathrm{E}_6$ into the Lorentz group, left-handed and right-handed spinors, and scalars in 10d Minkowski spacetime.

## Re: Octonions and the Standard Model (Part 1)

I want to point out that quaternionic analysis is believed to be related to some special Feynman diagrams of QED (Quantum Electrodynamics), in “On the work of Igor Frenkel”(Perspectives in Representation Theory, A conference celebrating the 60th birthday of Igor Frenkel).~