Octonions and the Standard Model (Part 11)
Posted by John Baez
We can think of the exceptional Jordan algebra as a funny sort of spacetime. This spacetime is 27-dimensional, with light rays through the origin moving on a lightcone given by a cubic equation instead of the usual
$t^2 - x^2 - y^2 - z^2 = 0$
in 4-dimensional Minkowski spacetime. But removing this lightcone still chops spacetime into 3 connected components: the past, the future, and the regions you can’t reach from the origin without exceeding the speed of light. The future is still a convex cone, and so is the past. So causality still makes sense like it does in special relativity.
At some point I got interested in seeing what physics would be like in this funny spacetime. Greg Egan and John Huerta joined me in figuring out the very basics of what quantum field theory would be like in this world. Namely, we figured out a bit about what kinds of particles are possible.
One difference is that we must replace the usual Lorentz group with the 78-dimensional group $\mathrm{E}_6$. But an even bigger difference is this. In 4d Minkowski space, every point in your field of view acts essentially like every other, if you turn your head. But in our 27-dimensional spacetime, the analogous fact fails! There is a ‘sky within the sky’: some particles moving at the speed of light can only be seen in certain directions. Thus, the classification of particles that move at the speed of light is much more baroque.
This is a big digression from my main quest here: explaining how people have tried to relate the octonions to the Standard Model. But it would be a shame not to make our results public, and now is a good time.
To explain why this isn’t a completely wacky thing to do, let me start by pointing out that people have completely classified vector spaces with ‘self-dual homogeneous convex cones’. The most famous example is the future cone
$t^2 - x^2 - y^2 - z^2 > 0, \qquad t > 0$
in 4d Minkowski spacetime. The possibilities are very limited: they correspond to formally real Jordan algebras! There are four infinite families and one exception: the exceptional Jordan agebra $\mathfrak{h}_3(\mathbb{O})$. So, just in the name of exploring possibilities, it seems worth taking a look at physics with $\mathfrak{h}_3(\mathbb{O})$ playing the role of spacetime.
For a quick intro, try:
Faraut and Koryani wrote a whole book on this subject, but they just call them ‘symmetric cones’:
- J. Faraut and A. Korányi, Analysis on Symmetric Cones, Oxford University Press, Oxford, 1994.
Anyway, you don’t need to know about this stuff now: I’m just trying to explain why quantum field theory on $\mathfrak{h}_3(\mathbb{O})$ isn’t a completely arbitrary thing to take a look at. I’m not claiming it’s good for real-world physics!
Let me show you how to classify particles when spacetime is $\mathfrak{h}_3(\mathbb{O})$. But first let me show you how to do it when spacetime is $\mathfrak{h}_2(\mathbb{C})$. Remember from Part 5: this is good old 4-dimensional Minkowski spacetime, where we think of a point $(t,x,y,z)$ as a self-adjoint complex matrix
$a = \left( \begin{array}{cc} t + x & -i y + z \\ i y + z & t - x \end{array} \right)$
so that
$\det(a) = t^2 - x^2 - y^2 - z^2$
and the linear transformations preserving this quadratic form are the Lorentz group $\mathrm{SO}(3,1)$.
Classifying particles in 4d Minkowski spacetime
To classify possible kinds of particles in 4d Minkowski spacetime, we use a procedure called Mackey’s theory of induced representations. But I want to explain this using the vocabulary of physics, not representation theory. First we look at the dual of 4d Minkowski spacetime, called ‘energy-momentum space’ This is canonically isomorphic to Minkowski spacetime, but we call its coordinates $(E,p_x,p_y,p_z)$: energy and the three components of momentum, and we call its quadratic form mass squared:
$m^2 = E^2 - p_x^2 - p_y^2 - p_z^2$
This Lorentz-invariant quantity can actually be any real number, mathematically, but we’ve never seen particles where it’s negative—called ‘tachyons’.
The orbits of the connected Lorentz group $\mathrm{SO}_0(3,1)$ are mainly classified by their mass squared, but not completely. Here are the options:
- Massive particles or tardyons, with $m^2 \gt 0$. Most known particles are this kind. For each choice of $m^2 \gt 0$ there are two orbits, one with $E \gt 0$ and one with $E \lt 0$. But we’ve only seen particles with positive energy (or more precisely, all particles we’ve seen have the same sign of energy: there’s an arbitrary sign convention here).
- Massless particles or luxons, with $m^2 = 0$. Photons are of this kind. In this case there are three orbits, namely $E \gt 0$, $E = 0$ and $E \lt 0$.
- Tachyons, with $m^2 \lt 0$. No known particles are of this kind. For each choice of $m^2 \lt 0$ there’s one orbit.
To continue the classification we need to figure out the stabilizer group of an arbitrarily chosen point in the given orbit. Then different kinds of particles will correspond to different irreducible projective unitary representation of this stabilizer group. For example, the stabilizer group for a massive particle is just $\mathrm{SO}(3)$, the rotations of space. Then choosing an irreducible projective unitary representation amounts to choosing a spin $j = 0, \frac{1}{2}, 1, \dots$. So a massive particle is characterized by its spin.
In the 27-dimensional spacetime corresponding to the exceptional Jordan algebra, the analogous group is $\mathrm{F}_4$. So, massive particles are characterized by representations of $\mathrm{F}_4$.
Orbits of the Lorentz group action on $\mathfrak{h}_2(\mathbb{C})$
To prepare you to figure out orbits for the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$, let’s repeat the classification of orbits we just did, but thinking of 4d Minkowski spacetime as the Jordan algebra $\mathfrak{h}_2(\mathbb{C})$. From this viewpoint we’re classifying orbits of $\mathrm{SL}(2,\mathbb{C})$ on $\mathfrak{h}_2(\mathbb{C})$, where $g \in \mathrm{SL}(2,\mathbb{C})$ acts by
$a \mapsto g a g^\ast$
The determinant of $a$ is clearly invariant under such transformations: that’s what I’d been calling the mass squared! But we can actually diagonalize any matrix $\mathfrak{h}_2(\mathbb{C})$ using transformations in $\mathrm{SL}(2,\mathbb{C})$. Furthermore, it’s not just the product of the eigenvalues—the determinant— that’s invariant under these transformations, but also whether each eigenvalue is positive, negative, or zero. The order of the eigenvalues doesn’t matter. So, here is the classification or orbits:
- $++_\delta$: this is the orbit containing all diagonal matrices $\mathrm{diag}(\alpha, \beta)$ with $\alpha, \beta \gt 0$ and having determinant $\alpha \beta= \delta \gt 0$. These give massive particles with $m^2 = \delta$ and $E \gt 0$.
- $--_{\delta}$: this is the orbit containing all diagonal matrices $\mathrm{diag}(\alpha, \beta)$ with $\alpha, \beta \lt 0$ and $\alpha \beta= \delta \gt 0$. These give massive particles with $m^2 = \delta$ and $E \lt 0$.
- $+-_{\delta}$: this is the orbit containing all diagonal matrices $\mathrm{diag}(\alpha, \beta)$ with $\alpha \gt 0, \beta \lt 0$ and $\alpha \beta= \delta \lt 0$. These give tachyons with $m^2 = \delta$ and $E \lt 0$.
- $+0$: this is the orbit containing all diagonal matrices $\mathrm{diag}(\alpha, 0)$ with $\alpha \gt 0$. These give massless particles with $m^2 = 0$ and $E \gt 0$.
- $-0$: this is the orbit containing all diagonal matrices $\mathrm{diag}(\alpha, 0)$ with $\alpha \lt 0$. These give massless particles with $m^2 = 0$ and $E \lt 0$.
- $00$: this is the orbit containing just the origin $\mathrm{diag}(0, 0)$. This give massless particles with $m^2 = 0$ and $E = 0$.
Now let’s copy this, but using $3 \times 3$ matrices of octonions!
Orbits of the $\mathrm{E}_6$ action on $\mathfrak{h}_3(\mathbb{O})$
Any element in $\mathfrak{h}_3(\mathbb{O})$ can be diagonalized by an element of $\mathrm{F}_4 \subseteq \mathrm{E}_6$. So, when computing the orbit of any element, we may assume without loss of generality that it has the form
$\left( \begin{array}{ccc} \alpha & 0 & 0 \\ 0 & \beta & 0 \\ 0 & 0 & \gamma \end{array} \right)$
with $\alpha, \beta, \gamma \in \mathbb{R}$. Then its determinant is $\alpha \beta \gamma$, and this is $\mathrm{E}_6$-invariant.
We can use transformations in $\mathrm{E}_6$ (or even $\mathrm{SO}(3) \subseteq \mathrm{F}_4$) to permute $\alpha, \beta,$ and $\gamma$, so their order doesn’t matter.
We can use transformations in $\mathrm{Spin}_0(9,1) \subseteq \mathrm{E}_6$ to multiply $\beta$ by any positive constant and divide $\gamma$ by that same constant. Thanks to our ability to permute, the same is true of $\alpha$ and $\beta$, or $\alpha$ and $\gamma$.
Thus, if $\alpha, \beta, \gamma \gt 0$, their product is a complete invariant for the action of $\mathrm{E}_6$. We thus get one $\mathrm{E}_6$ orbit for each value of $\delta \gt 0$:
- $+++{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta, \gamma \gt 0$, $\alpha \beta \gamma = \delta$.
We cannot, it seems, use a transformation in $\mathrm{E}_6$ to multiply two of $\alpha, \beta, \gamma$ by $-1$ and leave the third alone. Thus, there is a separate family of $\mathrm{E}_6$ orbits, one for each $\delta \gt 0$:
- $+--{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta \lt 0$, $\gamma \gt 0$, $\alpha \beta \gamma = \delta$.
By the same reasoning, there are two more one-parameter families of orbits with $\delta \lt 0$:
- $++-{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta \gt 0$, $\gamma \lt 0$, $\alpha \beta \gamma = \delta$.
- $+- -{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha \ge 0$, $\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$.
We’re left with the case $\alpha \beta \gamma = 0$. We can assume without loss of generality that $\gamma = 0$. We can do a transformation to normalize $\alpha$ and $\beta$ to $\pm 1$, and also permute them. So, we get 6 orbits:
- $++0$: The orbit of the matrix $diag(1,1,0)$.
- $+-0$: The orbit of the matrix $diag(1,-1,0)$.
- $--0$: The orbit of the matrix $diag(-1,-1,0)$.
- $+00$: The orbit of the matrix $diag(1,0,0)$.
- $-00$: The orbit of the matrix $diag(-1,0,0)$.
- $000$: The orbit of the matrix $diag(0,0,0)$.
So, there’s a total of 6 orbits and 4 one-parameter families of orbits where the parameter takes values in an open half-line.
Stabilizer groups
Next: what are the stabilizer groups for these orbits? I’ll just tell you the answers, or at least most of them. To understand these answers, you need to remember a bit about the relation between $\mathrm{E}_6$ and 10-dimensional spacetime geometry, which I’ve been explaining in painful detail for the last 5 posts.
For a change of pace, let’s organize orbits by rank: every element of $\mathfrak{h}_3(\mathbb{O})$ has a rank, which is the number of nonzero entries in the matrix after it has been diagonalized. This is an $\mathrm{E}_6$-invariant concept.
Rank 3: a rank-3 element $h \in \mathfrak{h}_3(\mathbb{O})$ is one where $h \times h \in \mathfrak{h}_3(\mathbb{O})^\ast$ doesn’t annihilate any nonzero element of $\mathfrak{h}_3(\mathbb{O})$. There are 4 one-parameter families of rank-3 orbits.
For any value of $\delta \gt 0$, there are 2 orbits of matrices with determinant $\delta$:
- $+++{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta, \gamma \gt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to the compact real form of $\mathrm{F}_4$.
- $+--{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha \gt 0$, $\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to the noncompact real form of $\mathrm{F}_4$ called $F_{4(20)}$, which is diffeomorphic to $\mathrm{Spin}(9) \times \mathbb{R}^{16}$.
For any value of $\delta \lt 0$, there are 2 orbits of matrices with determinant $\delta$:
- $++-{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha, \beta \gt 0$, $\gamma \lt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to the noncompact real form of $\mathrm{F}_4$ called $F_{4(20)}$, which is diffeomorphic to $\mathrm{Spin}(9) \times \mathbb{R}^{16}$.
- $---{}_\delta$: the orbit containing the matrices $diag(\alpha, \beta, \gamma)$ with $\alpha ,\beta, \gamma \lt 0$, $\alpha \beta \gamma = \delta$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to the compact real form of $\mathrm{F}_4$.
Rank 2: a rank-2 element $h \in \mathfrak{h}_3(\mathbb{O})$ is one where $h \times h \in \mathfrak{h}_3(\mathbb{O})^\ast$ annihilates some nonzero element of $\mathfrak{h}_3(\mathbb{O})$ but $h \times h \ne 0$.
There are 3 orbits of rank-2 elements:
- $++0$: The orbit of the matrix $diag(1,1,0)$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to $\mathrm{Spin}(9) \ltimes S_+$.
- $+-0$: The orbit of the matrix $diag(1,-1,0)$. I haven’t yet worked out the dimension or stabilizer for this orbit.
- $--0$: The orbit of the matrix $diag(-1,-1,0)$. This orbit is 26-dimensional. The stabilizer of any point in this orbit is a 52-dimensional group isomorphic to $\mathrm{Spin}(9) \ltimes S_+$.
Rank 1: a rank-1 element $h \in \mathfrak{h}_3(\mathbb{O})$ is one where $h \times h = 0$ but $h \ne 0$. There are 2 orbits of rank-1 elements:
- $+00$: The orbit of the matrix $diag(1,0,0)$. This orbit is 17-dimensional. The stabilizer of any point in this orbit is a 61-dimensional group isomorphic to $\mathrm{Spin}(9,1) \ltimes S_+$.
- $-00$: The orbit of the matrix $diag(-1,0,0)$. This orbit is 17-dimensional. The stabilizer of any point in this orbit is a 61-dimensional group isomorphic to $\mathrm{Spin}(9,1) \ltimes S_+$.
Rank 0: The only rank-0 element of $\mathfrak{h}_3(\mathbb{O})$ is zero, so there is just one orbit:
- $000$: The orbit of the matrix $diag(0,0,0)$. This orbit is 0-dimensional. The stabilizer of the unique point in this orbit is all of $\mathrm{E}_6$.
The large and small lightcone
A novel feature of the exceptional Jordan algebra is that it has two kinds of lightcone! The basic lightcone, which I’ll call the large lightcone, consists of all $a \in \mathfrak{h}_3(\mathbb{O})$ obeying the cubic equation $\det(a) = 0$. However, last time I described a ‘cross product’
$\times \colon \mathfrak{h}_3(\mathbb{O}) \times \mathfrak{h}_3(\mathbb{O}) \to \mathfrak{h}_3(\mathbb{O})^\ast$
such that
$\det(a) = (a \times a)(a)$
If $a \times a = 0$ then clearly $\det(a) = 0$. But the converse is not true! So we also have a small lightcone consisting only of $a \in \mathfrak{h}_3(\mathbb{O})$ with $a \times a = 0$. As I explained at the end last time, projectivizing this gives $\mathbb{O}\mathrm{P}^2$.
So, if you’re a denizen of this 27-dimensional spacetime and you only see light rays moving along the small lightcone, your field of view will be $\mathbb{O}\mathrm{P}^2$. That’s 16-dimensional. But if you can also see light rays moving along the large lightcone, your field of view will be larger! It will be 25-dimensional, since we get it by imposing one equation $\det(a) = 0$ and then projectivizing.
So, in this world there’s a ‘sky inside the sky’. This is actually true of the Jordan algebra $\mathfrak{h}_3(\mathbb{K})$ for any normed division algebra $\mathbb{K}$. And when we go on up to $\mathfrak{h}_3(\mathbb{K})$, which only works for associative normed division algebras, we get a ‘sky inside the sky inside the sky’. And so on.
In terms of the orbits we’ve described so far, the large lightcone is
$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; \det(a) = 0 \} = +\!+0 \; \cup +\!-0 \;\cup \; -\!-0 \; \cup \; +00 \; \cup \; -00 \; \cup\; 000$
while the small lightcone is
$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0 \} = +00 \;\cup\; -00 \;\cup\; 000$
We also have the forwards small lightcone:
$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0, \; \tr(a) \gt 0 \} = +00$
and the backwards small lightcone:
$\{ a \in \mathfrak{h}_3(\mathbb{O}) : \; a \times a = 0, \; \tr(a) \lt 0 \} = -00$
The forwards small lightcone is diffeomorphic to $\mathbb{O}\mathrm{P}^2 \times \mathbb{R}^+$, and so is the backwards small lightcone.
Points in the small lightcone can be explicitly described using the identification $\mathfrak{h}_3(\mathbb{O}) \cong \mathbb{R} \oplus V \oplus S_-$ as follows. It is the union of these two sets:
- (1a): Points of the form $a = (r, \frac{1}{2r} [s,s], s)$ where we can choose $s$ freely, including the origin, and choose any $r \ne 0$. This set is 17-dimensional. It does not include points with $r = 0$.
- (1b): Points of the form $a = (0, v_{LL}, 0)$ where $v_{LL}$ is a nonzero lightlike vector. This set is 9-dimensional. We can think of its elements as the limit points of points in set (1a) where $r \to 0$ and $s \to 0$ together.
In this parametrization we get the forwards small lightcone from (1a) with $r \gt 0$ and from (1b) with $v_{LL}$ on the forwards lightcone in $\mathbb{R}^{9,1}$.
Here is another parametrization of the small lightcone. Again, it is the union of two sets:
- (2a): Points of the form $a = (\frac{1}{2} tr([n_v,n_v]) / tr(v_{LL}), v_{LL}, n_v)$ where $v_{LL}$ is a nonzero lightlike vector and $n_v$ belongs to the 8-dimensional kernel of $\tilde{v}_{LL}$. This set is 17-dimensional. It does not include points with $v_{LL} = 0$.
- (2b): Points of the form $a = (r, 0, 0)$ where $r \ne 0$. This set is 1-dimensional. We can think of its elements as the limit points of points in set (2a) where $v_{LL} \to 0$ and $n_v \to 0$ together.
In this parametrization we get the forwards small lightcone from (2a) with $v_{LL}$ on the forwards lightcone in $\mathbb{R}^{9,1}$, and from (2b) with $r \gt 0$.
Stabilizer computation: +00 orbit
Some of the stabilizer groups are harder to work out than others. Let me do it for a hard one: the +00 orbit.
If we look at the stabilizer of the rank-1 element $h = \mathrm{diag}(1,0,0)$, with pieces $(\alpha,v,\psi_-)=(1,0,0)$, this will be fixed by:
1) any element $g$ of $\mathrm{Spin}(9,1)$, since $v=0$
2) any element $\phi_+$ of $S_+$, since we saw in Part 7 that this acts by:
- $\alpha \mapsto \alpha + g(v, [\phi_+,\phi_+]) + 2\langle \phi_+,\psi_- \rangle$, which takes $\alpha = 1$ to $\alpha = 1$ since $v=0$ and $\psi_- = 0$,
- $v \mapsto v$ which clearly preserves $v = 0$,
- $\psi_- \mapsto \psi_- + v \phi_+$ which takes $\psi_- = 0$ to $\psi_- = 0$ since $v = 0$.
These interact nicely so that the semidirect product $\mathrm{Spin}(9,1) \ltimes S_+$ stabilizes $\mathrm{diag}(1,0,0)$. By dimension counting, this is the whole stabilizer, since
$\mathrm{dim} (\mathrm{Spin}(9,1) \ltimes S_+) = 45 + 16 = 61$
and the orbit of $\mathrm{diag}(1,0,0)$ is 17-dimensional, and $61 + 17 = 78$.
Stabilizer computation: ++- orbit
While we’re heard, let me also compute the stabilizer for the ++- orbit. This is like some sort of ‘tachyon’.
By dimension counting, the stabilizer of the rank-3 element $h = diag(-1,1,1)$, with pieces $(r,v,\psi_-)=(-1,diag(1,1),0)$ is a 52-dimensional group. Page 67 of this book:
- Ichiro Yokota, Exceptional Lie Groups.
describes all 3 real forms of $\mathrm{F}_4$, and I believe this can be used to prove the stabilizer of $h$ is $F_{4(20)}$, the real form whose Killing form has signature 20, meaning that it’s positive definite on a 36-dimensional subspace and negative definite on a 16-dimensional subspace (or the other way around if you use the other sign convention). Among the real forms of $\mathrm{F}_4$, this one is characterized by having $\mathrm{Spin}(9)$ as its maximal compact subgroup, so it is diffeomorphic to $\mathrm{Spin}(9) \times \mathbb{R}^{16}$.
It is easy to see that $\mathrm{Spin}(9)$ appears as a subgroup of the stabilizer of $h$, so there is only a bit left to prove here.
The action of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})^\ast$
Now, I’ve been working in $\mathfrak{h}_3(\mathbb{O})$, but for the classification of particles we should be working in energy-momentum space—that is, the dual of $\mathfrak{h}_3(\mathbb{O})$. This is not isomorphic to $\mathfrak{h}_3(\mathbb{O})$ as a space on which $\mathrm{E}_6$; as I mentioned last time, they’re only isomorphic up to an outer automorphism.
In practice, this means that action of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})^\ast$ is very much like its action on $\mathfrak{h}_3(\mathbb{O})$, but with the roles of $S_+$ and $S_-$ reversed. If you’re a real glutton for detail, you can read these notes for more:
- John Baez, Greg Egan and John Huerta, Geometry of the exceptional Jordan algebra.
I’d enjoy thinking about this more someday… but now we should head back towards the Standard Model.
- 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.
- 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.
- Part 10. Jordan algebras, their symmetry groups, their invariant structures — and how they connect quantum mechanics, special relativity and projective geometry.
- Part 11. Particle physics on the spacetime given by the exceptional Jordan algebra: a summary of work with Greg Egan and John Huerta.
Re: Octonions and the Standard Model (Part 11)
This is great fun to read, maybe more so because I’ve forgotten most of what the three of us did on this together at the time!
Do you mean $\mathfrak{h}_4(\mathbb{K})$ there?