### Octonions and the Standard Model (Part 7)

#### Posted by John Baez

Last time I explained the connection between the exceptional Jordan algebra and 10-dimensional Minkowski spacetime. Today I want to report on some work that Greg Egan, John Huerta and I did in November 2015. We figured out how to describe the big 78-dimensional *symmetry group* of the exceptional Jordan algebra in terms of 10d spacetime geometry!

This group is called $\mathrm{E}_6$, and we saw how it’s built up from:

- the double cover of the Lorentz group of 10d spacetime, $\mathrm{Spin}(9,1)$
- the right-handed spinors in 10d spacetime, $S_+$
- the left-handed spinors in 10d spacetime, $S_-$
- scalars, $\mathbb{R}$.

This lets us chop up the Lie algebra of $\mathrm{E}_6$ as follows:

$\mathfrak{e}_6 \cong \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}$

or counting dimensions,

$78 = 45 + 16 + 16 + 1$

Remember from last time: the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$ consists of $3 \times 3$ self-adjoint matrices of octonions, while 10d Minkowski spacetime can be seen as $\mathfrak{h}_2(\mathbb{O})$ — the $2 \times 2$ self-adjoint matrices of octonions.

So, 10d Minkowski spacetime sits inside the exceptional Jordan algebra! It does so in many ways, and there’s no ‘best’ one. But if we pick any one, then the exceptional Jordan algebra, which is 27-dimensional, splits into three parts, each with a nice interpretation in terms of 10d spacetime geometry:

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

or in terms of dimensions:

$27 = 10 + 16 + 1$

The 10-dimensional part $V$ consists of *vectors* in 10d Minkowski spacetime, the 16-dimensional part $S_+$ consists of *left-handed spinors*, and the 1-dimensional part $\mathbb{R}$ consists of *scalars*. Each part is an irreducible representation of $\mathrm{Spin}(9,1)$ — and the key structure on the exceptional Jordan algebra, the determinant, is invariant under $\mathrm{Spin}(9,1)$.

This is cute, because the three most important kinds of particles in physics are spin-1 (vectors), spin-1/2 (spinors), and spin-0 (scalars). But I’ll again remind everyone that I’m not proposing or advocating a theory of physics here. I’m just doing math.

And here’s what Greg Egan, John Huerta and I showed:

**Theorem 9.** The Lie group $\mathrm{Spin}(9,1)$ and abelian Lie groups isomorphic to $S_+$, $S_-$ and $\mathbb{R}$ show up as Lie subgroups of $\mathrm{E}_6$, in such a way that the direct sum of their Lie algebras is — as a vector space — all of $\mathfrak{e}_6$.

**Proof.** First, remember that we can write any element of $\mathfrak{h}_3(\mathbb{O})$ in this way:

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

where:

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

is a **vector** in 10d Minkowski spacetime: an element of the space $V = \mathfrak{h}_2(\mathbb{O})$ of $2\times 2$ self-adjoint octonionic matrices,

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

is a **left-handed spinor**: an element of $S_- = \mathbb{O}^2$, and $a$ is a **scalar**: an element of $\mathbb{R}$.

When we do this, the determinant takes on $\mathfrak{h}_3(\mathbb{O})$ takes on a simpler form. By definition we have

$\mathrm{det} \left( \begin{array}{ccc} \alpha & z & y^\ast \\ z^\ast & \beta & x \\ y & x^\ast & \gamma \end{array} \right) = \alpha \beta \gamma - (\alpha |x|^2 + \beta |y|^2 + \gamma |z|^2) + 2 Re(x y z)$

but in Theorem 8 we saw that

$\mathrm{det} \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) = - \alpha g(v,v) + g(v, [\psi, \psi])$

Here $g$ is the Minkowski metric and the bracket is a way of turning two left-handed spinors into a vector, explained last time.

Next, remember that whenever I say $\mathrm{E}_6$, I mean the group of linear transformations of $\mathfrak{h}_3(\mathbb{O})$ preserving the determinant. To be precise, this is the noncompact form of $\mathrm{E}_6$ called $\mathrm{E}_{6(-26)}$.

Now, here are four subgroups of $\mathrm{E}_6$:

1) First, $\mathrm{Spin}(9,1)$ can be seen as a Lie subgroup of $\mathrm{E}_6$. If we use our identification

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

then $\mathrm{Spin}(9,1)$ acts on each summand, hence the whole space — and its action preserves the determinant

$\mathrm{det} \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) = - \alpha g(v,v) + g(v, [\psi, \psi])$

since this is defined using operations that we saw last time are $\mathrm{Spin}(9,1)$-equivariant. So, $\mathrm{Spin}(9,1) \subseteq \mathrm{E}_6$.

2) Second, the vector space $S_+ = \mathbb{O}^2$, regarded as an abelian Lie group, can be seen as a Lie subgroup of $\mathrm{E}_6$. For this, let $\phi \in S_+$ act on

$\left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

as follows:

$\begin{array}{ccl} v & \mapsto & v \\ \psi &\mapsto& \psi + \gamma(v \otimes \phi) \\ \alpha & \mapsto & \alpha + g(v, [\phi, \phi]) + 2 \langle \phi, \psi \rangle \end{array}$

The bracket here describes a way to turn two *right*-handed spinors into a vector. The angle brackets denote the dual pairing between $S_+$ and $S_-$, which produces a scalar. And $\gamma$ is a way for any vector to act on a right-handed spinor to give a left-handed spinor. I described all these operations last time — and by the way, all these too are $\mathrm{Spin}(9,1)$-equivariant.

We can check that these formulas define a transformation of $\mathfrak{h}_3(\mathbb{O})$ that preserves the determinant $\alpha \mathrm{det}(v) + g(v, [\psi, \psi])$. By adding $\gamma(v \otimes \phi)$ to $\psi$, $g(v, [\psi, \psi])$ gains these three extra terms:

$\begin{array}{ccl} g(v, [\psi, \gamma(v \otimes \phi)]) &=& -det(v) \langle \phi,\psi \rangle \\ g(v, [\gamma(v \otimes \psi), \psi]) &=& -det(v) \langle \phi,\psi \rangle \\ g(v, [\gamma(v \otimes \phi), \gamma(v \otimes \phi)]) &=& -det(v) g(v, [\phi, \phi]) \end{array}$

To prove these equations, just play around with the explicit formulas given last time. These three extra terms cancel out the extra terms that appear in $\alpha \mathrm{det}(v)$, so the determinant is unchanged.

Also, we can check that this map from $S_+$ to determinant-preserving linear transformations of $\mathfrak{h}_3(\mathbb{O})$ is a Lie group homomorphism, where we make $S_+$ into an abelian Lie group using addition. We can also check that it’s one-to-one. All this is a calculation: the only ‘trick’ we need is the identity

$g(v, [\phi, \phi']) = \langle \phi' , \gamma(v \otimes \phi) \rangle$

for all $v \in V$ and $\phi, \phi' \in S_+$. This follows from the definitions of these operations. So, $S_+$ becomes a Lie subgroup of $\mathrm{E}_6$.

3) Similarly, $S_- = \mathbb{O}^2$ can be seen as an abelian Lie subgroup of $\mathrm{E}_6$. For this, let $\eta \in S_-$ act on

$\left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

as follows:

$\begin{array}{ccl} v &\mapsto& v + \frac{1}{2} \alpha [\eta, \eta] + [\psi, \eta] \\ \psi & \mapsto& \psi + \alpha \eta \\ \alpha &\mapsto& \alpha \end{array}$

Again we can show that this preserves the determinant. And again, we can check that this map from $S_+$ to determinant-preserving linear transformations of $\mathfrak{h}_3(\mathbb{O})$ is a Lie group homomorphism, which is one-to-one. Again these are just calculations, and this time no tricks at all are required. So the additive group of $S_+$ becomes a Lie subgroup of $\mathrm{E}_6$.

4) Similarly, $\mathbb{R}^\ast$, the multiplicative group of nonzero real numbers, can be seen as an abelian Lie subgroup of $\mathrm{E}_6$. For this, let any nonzero real number $t$ act on

$\left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

as follows:

$\begin{array}{ccl} v &\mapsto& t^{-2} v \\ \psi &\mapsto& t \psi \\ \alpha &\mapsto& t^4 \alpha \end{array}$

These rescalings clearly preserve the determinant $\alpha \det(v) + g(v, [\psi, \psi])$.

As a consequence, we get inclusions of Lie algebras

$\begin{array}{ccl} \mathfrak{so}(9,1) &\hookrightarrow& \mathfrak{e}_6 \\ S_+ &\hookrightarrow& \mathfrak{e}_6 \\ S_- &\hookrightarrow& \mathfrak{e}_6 \\ \mathbb{R} &\hookrightarrow& \mathfrak{e}_6 \end{array}$

where the last three are treated as abelian Lie algebras. We thus get a linear map (*not* a Lie algebra homomorphism!)

$f: \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R} \to \mathfrak{e}_6$

I will take it for granted that the dimension of $\mathfrak{e}_6$ is 78: this is well-known. The dimensions of the four summands add up to this number: $45 + 16 + 16 + 1 = 78$. So, to show that $f$ is an isomorphism of vector spaces, we just need to check that it is one-to-one.

When we differentiate the above formulas and get actions of the Lie algebras $\mathfrak{so}(9,1)$, $S_+$, $S_-$, and $\mathbb{R}$ on $\mathfrak{h}_3(\mathbb{O})$, we can direct sum them and get a map from $\mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}$ to linear transformations of $\mathfrak{h}_3(\mathbb{O})$. Here’s what we get. Write $L \in \mathfrak{so}(9,1)$, $\phi \in S_+$, $\eta \in S_-$ and $t \in R$. Then $(L, \phi, \eta, t)$ gives this transformation of $\mathfrak{h}_3(\mathbb{O})$:

$(L, \phi, \eta, t) \colon \left( \begin{array}{cc} \alpha & \psi^\dagger \\ \psi & v \end{array} \right) \mapsto \left( \begin{array}{cc} 2\langle \phi, \psi \rangle + 4t \alpha & (L \psi + \gamma(v \otimes \phi) + t \psi)^\dagger \\ L \psi + \gamma(v \otimes \phi) + t \psi & L v + [\psi,\eta] - 2s v \end{array} \right)$

One can check that this map from $\mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}$ to linear transformations of $\mathfrak{h}_3(\mathbb{O})$ is one-to-one. Thus, we get an isomorphism of vector spaces

$\mathfrak{e}_6 \cong \mathfrak{so}(9,1) \oplus S_+ \oplus S_- \oplus \mathbb{R}$

as desired. █

This isomorphism of vector spaces is not an isomorphism of Lie algebras, since for example the abelian Lie subalgebras $S_+$ and $S_-$ don’t commute with each other in $\mathfrak{e}_6$. However, each summand is a Lie subalgebra of $\mathfrak{e}_6$, and we can use this isomorphism of vector spaces to describe the bracket in $\mathfrak{e}_6$ in terms of operations on $\mathfrak{so}(9,1)$, $S_+$, $S_-$ and $\mathbb{R}$. Maybe I’ll do that next time; I think you’re probably tired by now, if you’re reading this at all.

Also — and perhaps more importantly — we can get a *geometrical* understanding of what’s going on here, using the octonionic projective plane $\mathbb{O}\mathrm{P}^2$. But again, I’ll wait and talk about that later.

- 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. This group is built from the double cover of the Lorentz group, left-handed and right-handed spinors, and scalars in 10d Minkowski spacetime.
- Part 8. A geometrical way to see how $\mathrm{E}_6$ is connected to 10d spacetime, based on the octonionic projective plane.

## Re: Octonions and the Standard Model (Part 7)

typo? Surely $\mathbb{h}_3(\mathbb{O})$ isn’t irreducible if it breaks into irreps.