### Octonions and the Standard Model (Part 3)

#### Posted by John Baez

Now I’ll finally explain how a quark and a lepton fit together into an octonion — in the very simplified picture where we treat these particles merely as representations of $\mathrm{SU}(3)$, the symmetry group of the strong force. I’ll say just enough about physics for mathematicians to get a sense of what this means. (The most substantial part of this post will be a quick intro to ‘basic triples’, a powerful technique for working with octonions.)

One of curious features of the Standard Model is that fermions — the particles that make up ‘matter’, as opposed to particles that carry ‘forces’ — come in two kinds: leptons and quarks.

Quarks are affected by the strong nuclear force, while leptons are not. Each kind of quark comes in 3 ‘colors’ — three states conventionally called red, green and blue — while leptons do not. These two statements say almost the same thing, since strong force is governed by the group $\mathrm{SU}(3)$. Mathematically, the point is that quarks transform in the usual representation of $\mathrm{SU}(3)$ on $\mathbb{C}^3$, while leptons transform in the trivial representation of this group on $\mathbb{C}$.

An interesting fact is that for each kind of quark there’s a corresponding lepton. For example, ordinary matter is mainly made of two kinds of quarks, called up and down quarks, and two kinds of leptons, called electrons and electron neutrinos. These four form the ‘first generation’ of quarks and leptons, and there are two more.

There are physical reasons to expect this ‘one quark per lepton’ pattern: the GIM mechanism and the need to avoid anomalies. However, it is still tempting to seek a theory that *unifies* quarks and leptons.

Many such theories are known. For example, in the Pati–Salam model, the group $\mathrm{SU}(3)$ is replaced with $\mathrm{SU}(4)$ and each quark together with its partner lepton are considered at a fundamental level to be a single particle transforming in the usual representation of $\mathrm{SU}(4)$ on $\mathbb{C}^4$. Extending the color metaphor, the three colors of quark join with the ‘white’ lepton to form a basis of $\mathbb{C}^4$.

The SU(5) grand unified theory has a different take on this issue: here red, green and blue are joined by two other states called ‘up’ and ‘down’. And the Pati–Salam model and SU(5) theory are themselves reconciled in the larger SO(10) theory.

A different approach, less explored and less successful so far, is to use octonions. Here are two references:

Murat Günaydin and Feza Gürsey, Quark structure and octonions,

*Journal of Mathematical Physics*14 (1973), 1651–1667.Geoffrey Dixon,

*Division Algebras: Octonions, Quaternions, Complex Numbers and the Algebraic Design of Physics*, Springer, Berlin, 2013.

though I could probably list dozens more if I worked at it. I don’t know the whole history of this approach. Now I’m getting curious.

The basic idea is this: if we fix an octonion $i \in \mathbb{O}$ with $i^2 = -1$, the group of automorphisms of the octonions that fix this element is $\mathrm{SU}(3)$. This element $i$ also determines a copy of $\mathbb{C}$ in $\mathbb{O}$. This makes $\mathbb{O}$ into a complex vector space, and $\mathrm{SU}(3)$ acts on this vector space. As a representation of $\mathrm{SU}(3)$, this vector space is isomorphic to

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

where $\mathrm{SU}(3)$ acts trivially on $\mathbb{C}$ and in the usual way on $\mathbb{C}^3$. This is just the representation of $\mathrm{SU}(3)$ on one kind of lepton and its corresponding quark!

So, we can imagine trying to come up with a theory where a quark and a lepton are unified into a single octonion, and then somehow the choice of a square root of $-1$ in $\mathbb{O}$ breaks the symmetry, splitting the octonions into two separate representations of $\mathrm{SU}(3)$: the quark and the lepton. This might be part of a larger story in which the complex numbers, so important in quantum physics, are secretly just one special copy of $\mathbb{C}$ sitting inside $\mathbb{O}$.

I have no good idea for how this would work. Since our usual paradigm for spontaneous symmetry breaking involves the Higgs mechanism, one could try to use that. There is a 6-sphere of square roots of $-1$ sitting in the octonions, each giving a copy of $\mathbb{C}$, so one might imagine some sort of Higgs potential, for example a quartic polynomial, that has minima on this 6-sphere. Someone must have tried this. But I don’t find this idea especially inspiring unless it arises naturally somehow, or explains several things at once.

I’ve already proved all the *mathematical* claims I just made about octonions and representations of $\mathrm{SU}(3)$ — except for one: that the group $\mathrm{SU}(3)$ contains *all* the automorphisms of the octonions preserving $i$. For that it’s handy to use a very powerful tool for working with octonions: ‘basic triples’.

For those who prefer their math free of physics, let me state this in the form of a theorem, and then prove it:

**Theorem 4.** Let $i \in \mathbb{O}$ be any element with $i^2 = -1$. Then $i$ generates a copy of $\mathbb{C}$ in $\mathbb{O}$. Left multiplication by this copy of $\mathbb{C}$ makes $\mathbb{O}$ into a complex vector space. The group of automorphisms of the octonions that fix the element $i$ is isomorphic to $\mathrm{SU}(3)$, and this group acts in a complex-linear way on $\mathbb{O}$. This representation is isomorphic to the direct sum of the trivial representation on $\mathbb{C} \subseteq \mathbb{O}$ and the usual representation of $\mathrm{SU}(3)$ on $\mathbb{C}^3$ by matrix multiplication.

**Proof.** In Part 1 we saw a description of the octonions as $\mathbb{C} \oplus \mathbb{C}^3$ where $\mathbb{C}$ is the subalgebra generated by a chosen square root of $-1$. One can see from this description that left multiplication by $\mathbb{C}$ makes the octonions into a complex vector space with $\mathbb{C}^3$ as a complex subspace. In this description, octonion multiplication is given by operations that are manifestly invariant under $\mathrm{SU}(3)$, which acts trivially on $\mathbb{C}$ and by matrix multiplication on $\mathbb{C}^3$.

Thus, the only thing missing is a proof that no *bigger* group than $\mathrm{SU}(3)$ acts as automorphisms of $\mathbb{O}$ fixing the chosen element $i$. To do this, let’s show the group of such automorphisms is an 8-dimensional connected Lie group. Since its subgroup $\mathrm{SU}(3)$ is also an 8-dimensional and connected, it must be the whole thing.

For this I find it enjoyable to use the concept of ‘basic triple’.

In a normed division algebra the norm always comes from an inner product. Any unit vector orthogonal to $1$ is a square root of $-1$, and vice versa. Further, orthogonal square roots of $-1$ anticommute.

So, if our normed division algebra isn’t the real numbers, we can find a square root of $-1$ in it. Choose one and call it $i$. Then $1$ and $i$ span a copy of $\mathbb{C}$.

If this isn’t our whole normed division algebra, we can find another square root of $-1$ that is orthogonal to this copy of $\mathbb{C}$. Choose one and call it $j$. Then you can show $1, i, j$, and $k = i j$ span a copy of $\mathbb{H}$.

If this isn’t our whole normed division algebra yet, we can find another square root of $-1$ that is orthogonal to this whole copy of $\mathbb{H}$. Choose one and call it $\ell$. Then you can show $1, i, j, k, \ell, \ell i, \ell j$ and $\ell k$ span a copy of $\mathbb{O}$.

We say three elements $i, j, \ell \in \mathbb{O}$ are a **basic triple** if they are orthogonal square roots of $-1$ and $\ell$ is also orthogonal to $k = i j$. The multiplication table of octonions looks the same for any basic triple: you can work out what it must be.

Fix a basic triple $i, j, \ell \in \mathbb{O}$. From what I’ve said, any automorphism of $\mathbb{O}$ maps it to some other basic triple — and conversely, given any other basic triple, there is a unique automorphism mapping $i, j, \ell$ to that one. So, the automorphism group of $\mathbb{O}$ acts freely and transitively on the set of basic triples. We say the basic triples form a torsor for $\mathrm{Aut}(\mathbb{O})$.

The important consequence is that $\mathrm{Aut}(\mathbb{O})$ is homeomorphic to the space of basic triples. This lets us understand the topology of $\mathrm{Aut}(\mathbb{O})$.

Since unit octonions form a 7-sphere with $1$ at the ‘north pole’, the square roots of $-1$ form the ‘equator’, which is a 6-sphere. If we choose one of these, say $i$, then the square roots of $-1$ that are orthogonal to it form a 5-sphere. And if we choose one of these, say $j$, then the square roots of $-1$ that are orthogonal to $i, j$ and $k = i j$ form a 3-sphere.

So, the group $\mathrm{Aut}(\mathbb{O})$ is a bundle of 3-spheres over a bundle of 5-spheres over the 6-sphere. As a spinoff, this implies it is a connected Lie group, and its dimension is $6 + 5 + 3 = 14.$

But we’re actually interested in automorphisms of the octonions that fix a choice of $i$. These correspond to basic triples with this fixed choice of $i$. This is just a bundle of 3-spheres over a 5-sphere. So, this is a connected Lie group with dimension $5 + 3 = 8$. And that’s what we needed to show! █

As another spinoff of all this, it follows that $\mathrm{SU}(3)$ is a bundle of 3-spheres over a 5-sphere. But that’s easy to see directly by letting $\mathrm{SU}(3)$ act on the unit sphere of $\mathbb{C}^3$. This is a 5-sphere, and each point is stabilized by an $\mathrm{SU}(2)$ subgroup, which is a 3-sphere.

## Re: Octonions and the Standard Model (Part 3)

John wrote,

The fact that

trivialandtautologousmean two different things here disquiets me more than I’d like to admit.