### Octonions and the Standard Model (Part 5)

#### Posted by John Baez

Last time I stated a couple of theorems connecting the gauge group of the Standard Model to the exceptional Jordan algebra. To prove them, it helps to become pretty comfortable with the exceptional Jordan algebra and its symmetries. And instead of trying to get the job done quickly, I’d prefer to proceed slowly and gently.

One reason is that while the exceptional Jordan algebra consists of $3 \times 3$ self-adjoint matrices of octonions, we can think of the space of $2 \times 2$ self-adjoint matrices of octonions as 10-dimensional Minkowski spacetime. So, to understand the exceptional Jordan algebra we can use facts about spinors and vectors in 10d spacetime! This is worth thinking about in its own right.

The behavior of spin-1/2 particles depends heavily on the dimension of space, or spacetime, modulo 8. For example, in Minkowski spacetime of any dimension that’s 2 more than a multiple of 8, and *only* in these dimensions, Majorana–Weyl spinors are possible. These are spin-1/2 particles that have an *intrinsic handedness* (that’s the ‘Weyl’ part) and are *their own antiparticles* (that’s the ‘Majorana’ part).

In 10d Minkowski spacetime something special happens: Majorana–Weyl spinors can be described using octonions! Mathematically this originates from the fact that there are two representations of $Spin(9,1)$, called the left-handed and right-handed Majorana–Weyl spinor representations, which can both be identified with $\mathbb{O}^2$, the space of pairs of octonions.

This in turn follows from a simpler fact: 10-dimensional Minkowski spacetime can be identified with $\mathfrak{h}_2(\mathbb{O})$, the space of self-adjoint $2 \times 2$ octonionic matrices:

$\mathfrak{h}_2(\mathbb{O}) = \left\{ \left( \begin{array}{cc} t+x & y \\ y^\ast & t-x \end{array} \right) : \; t,x \in \mathbb{R}, y \in \mathbb{O} \right\}$

Here $t$ is time and there are 9 space coordinates: the real number $x$ and the 8 real numbers describing the octonion $y$. Of course Lorentz transformations and rotations let us change these coordinates. The Lorentz group $SO(9,1)$ is precisely the group of transformations of $\mathfrak{h}_2(\mathbb{O})$ that preserve the determinant

$\mathrm{det} \left( \begin{array}{cc} t+x & y \\ y^\ast & t-x \end{array} \right) = t^2 - x^2 - y y^\ast$

Even though the octonions are noncommutative, we have $y y^\ast = y^\ast y = |y|^2$ where $|y|$ is the norm of the octonion $y$, so there’s no ambiguity here.

All this is very similar to something that happens in 4 dimensions. 4d Minkowski spacetime can be identified with $\mathfrak{h}_2(\mathbb{C})$, the space of self-adjoint $2 \times 2$ complex matrices, and $Spin(3,1)$ has two representations on $\mathbb{C}^2$, the left-handed and right-handed Weyl spinor representations. All this should be familiar to students of particle physics. The 10d case works essentially the same way: we just replace complex numbers with octonions. Of course the octonions are noncommutative and nonassociative, but it doesn’t really cause a problem here.

In fact, something similar also happens in 3d Minkowski spacetime, which can be identified with $\mathfrak{h}_2(\mathbb{R})$, and 6d Minkowski spacetime, which can be identified with $\mathfrak{h}_2(\mathbb{H})$. The special features of the 10d case take a while to manifest themselves.

So, to get the story started we’ll let $\mathbb{K}$ be any normed division algebra — $\mathbb{R}, \mathbb{C}, \mathbb{H}$ or $\mathbb{O}$. We’ll say its dimension is $n$, so $n = 1,2,4$ or $8$. We’ll show how to identify $(n+2)$-dimensional Minkowski spacetime with $\mathfrak{h}_2(\mathbb{K})$, the space of self-adjoint $2 \times 2$ matrices with entries in $\mathbb{K}$. And we’ll describe two kinds of spinors, both of which can be identified with elements of $\mathbb{K}^2$.

The rest of today’s post will be based on this:

- John C. Baez, John Huerta, Division algebras and supersymmetry I, in
*Superstrings, Geometry, Topology, and C${}^\ast$-algebras*, eds. R. Doran, G. Friedman and J. Rosenberg,*Proc. Symp. Pure Math.***81**, AMS, Providence, 2010, pp. 65–80.

## Vectors

We begin by writing an element of $\mathfrak{h}_2(\mathbb{K})$ as

$v = \left( \begin{array}{cc} t + x & y \\ y^\ast & t - x \end{array} \right)$

with $t,x \in \mathbb{R}$ and $x \in \mathbb{K}$.

It is clear that

$-det(v) = -t^2 + x^2 + y y^\ast$

is a quadratic form on $\mathfrak{h}_2(\mathbb{K})$ of signature $(n+1,1)$. This is how $\mathfrak{h}_2(\mathbb{K})$ gets identified with $(n+2)$-dimensional Minkowski spacetime.

But we also want to get our hands on the Minkowski metric — that is, a *bilinear* form on $\mathfrak{h}_2(\mathbb{K})$. For this the operation of **trace reversal** is crucial:

$\tilde{v} = v - tr(v)1$

Concretely, this looks like:

$\tilde{v} = \left( \begin{array}{cc} -t + x& y \\ y^\ast & -t - x \end{array} \right)$

So trace reversal amounts to negating $t$: it’s what physicists call ‘time reversal’. Thanks to this time reversal, if we define the bilinear form

$g \colon \mathfrak{h}_2(\mathbb{K}) \times \mathfrak{h}_2(\mathbb{K}) \to \mathbb{R}$

by

$g(v,w) = \frac{1}{2} Re \, tr(v \tilde{w})$

the inner product of time-like vectors with themselves will become negative, and we get the Minkowski metric. Indeed, a little calculation shows

$v \tilde{v} = \tilde{v} v = -det(v) 1$

so

$g(v,v) = -det(v)$

Thus, $g$ is our Minkowski metric.

In what follows we’ll call elements of $\mathfrak{h}_2(\mathbb{K})$ **vectors**, and write

$V = \mathfrak{h}_2(\mathbb{K})$

## Spinors

Next we introduce two spaces of spinors, which we’ll call **right-handed spinors**, $S_+$, and **left-handed spinors**, $S_-$. As real vector spaces, both will be just $\mathbb{K}^2$. However, the group $Spin(n+1,1)$ will act on them in different ways.

To construct these representations, we begin by defining two ways for vectors to act on spinors. First, a map

$\gamma \colon V \otimes S_+ \to S_-$

given by

$\gamma(v \otimes \psi) = v \psi$

where in $v \psi$ we’re multiplying a matrix and a column vector. Second, a map

$\tilde{\gamma} \colon V \otimes S_- \to S_+$

given by

$\tilde{\gamma}(v \otimes \psi) = \tilde{v} \psi$

We can also think of these as maps that send elements of $V$ to linear operators:

$\begin{array}{cccl} \gamma \colon & V & \to & Hom(S_+, S_-), \\ \tilde{\gamma} \colon & V & \to & Hom(S_-, S_+) \end{array}$

Here a word of caution is needed: 25% of the time, and every time we really care about it, our normed division algebra $\mathbb{K}$ will be nonassociative! So $2 \times 2$ matrices with entries in $\mathbb{K}$ cannot be identified with linear operators on $\mathbb{K}^2$ in the usual way. They certainly induce linear operators via left multiplication:

$L_v(\psi) = v \psi$

Indeed, this is how $\gamma$ and $\tilde{\gamma}$ turn elements of $V$ into linear operators:

$\begin{array}{ccl} \gamma(v) & = & L_v \\ \tilde{\gamma}(v) & = & L_{\tilde{v}} \end{array}$

However, because of nonassociativity, composing such linear operators may be different from multiplying the matrices:

$L_v L_w(\psi) = v( w \psi) \neq (v w) \psi = L_{v w} (\psi)$

Now, back to business. Since vectors act on elements of $S_+$ to give elements of $S_-$ and vice versa, they map the space $S_+ \oplus S_-$ to itself. This gives rise to an action of the Clifford algebra $Cliff(V)$ on $S_+ \oplus S_-$:

**Theorem 7.** The vectors $V = \mathfrak{h}_2(\mathbb{K})$ act on the spinors $S_+ \oplus S_- = \mathbb{K}^2 \oplus \mathbb{K}^2$ via the map

$\Gamma \colon V \to End(S_+ \oplus S_-)$

given by

$\Gamma(v)(\psi, \,\phi) = (\widetilde{v} \phi, \, v \psi)$

Furthermore, $\Gamma(v)$ satisfies the Clifford algebra relation:

$\Gamma(v)^2 = g(v,v) 1$

and so extends to a homomorphism $\Gamma \colon Cliff(V) \to End(S_+ \oplus S_-)$, i.e. a representation of the Clifford algebra $Cliff(V)$ on $S_+ \oplus S_-$.

**Proof.** Suppose $v \in V$ and $\Psi = (\psi, \phi) \in S_+ \oplus S_-$. We need to check that

$\Gamma(v)^2(\Psi) = -\det(v) \Psi$

Here we must be mindful of nonassociativity: we have

$\Gamma(v)^2(\Psi) = ( \tilde{v}(v \psi), \, v(\tilde{v} \phi))$

But it’s easy to check that the expressions $\tilde{v}(v \psi)$ and $v (\tilde{v} \phi)$ involve multiplying at most two different nonreal elements of $\mathbb{K}$. The associative law holds in this case, even for the octonions, so in fact

$\Gamma(v)^2(\Psi) = ( (\tilde{v}v) \psi, \, (v \tilde{v}) \phi)$

To conclude, we use the fact that $v \tilde{v} = \tilde{v} v = -det(v) 1$. █

The action of a vector swaps $S_+$ and $S_-$, so acting by vectors twice sends $S_+$ to itself and $S_-$ to itself. This means that while $S_+$ and $S_-$ are *not* modules for the Clifford algebra $Cliff(V)$, they are both modules for the ‘even part’ of the Clifford algebra, generated by products of pairs of vectors. The group $Spin(n+1,1)$ lives in this even part: as is well-known, it is generated by products of pairs of ‘unit vectors’ in $V$: that is, vectors $v$ with $g(v,v) = \pm 1$. As a result, $S_+$ and $S_-$ are both representations of $Spin(n+1, 1)$.

While we will not need this in what follows, one can check that:

When $\mathbb{K} = \mathbb{R}$, $S_+ \cong S_-$ is the Majorana spinor representation of $Spin(2,1)$.

When $\mathbb{K} = \mathbb{C}$, $S_+$ and $S_-$ are the Weyl spinor representations of $Spin(3,1)$.

When $\mathbb{K} = \mathbb{H}$, $S_+$ and $S_-$ are the Weyl spinor representations of $Spin(5,1)$.

When $\mathbb{K} = \mathbb{O}$, $S_+$ and $S_-$ are the Majorana–Weyl spinor representations of $Spin(9,1)$.

## A taste of $\mathfrak{h}_3(\mathbb{K})$

Now that we’ve gotten vectors and spinors set up using $\mathfrak{h}_2(\mathbb{K})$, let me give you a tiny taste of how they’ll shed light on $\mathfrak{h}_3(\mathbb{K})$. A self-adjoint $3 \times 3$ matrix with entries in $\mathbb{K}$ looks like this:

$a = \left( \begin{array}{ccc} \alpha & z & y^\ast \\ z^\ast & \beta & x \\ y & x^\ast & \gamma \end{array} \right)$

where $x,y,z \in \mathbb{K}$ and $\alpha, \beta, \gamma \in \mathbb{R}$. But we can think of the lower right $2 \times 2$ block as a vector in $V = \mathfrak{h}_2(\mathbb{K})$, and write

$a = \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right)$

where $\alpha \in \mathbb{R}$ is a scalar,

$v = \left( \begin{array}{cc} \beta & x \\ x^\ast & \gamma \end{array} \right) \in V$

is a vector, and

$\psi = \left( \begin{array}{cc} z^\ast \\ y \end{array} \right) \in S_-$

is a left-handed spinor. So, we get an isomorphism of vector spaces

$\mathfrak{h}_3(\mathbb{K}) \cong \mathbb{R} \oplus V \oplus S_-$

Yes: we’ve combined a scalar, a vector and a left-handed spinor into a single entity!

You may wonder why we’re treating $\psi$ as a *left-handed* spinor. I’ll talk about that more next time. We’ll see that right-handed spinors also play an important role.

- 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.

The trace is 2t, so there is just a time reversal