December 19, 2020

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:

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:

• 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.
Posted at December 19, 2020 12:16 PM UTC

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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!

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’.

Do you mean $\mathfrak{h}_4(\mathbb{K})$ there?

Posted by: Greg Egan on December 20, 2020 3:54 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Yes, I meant $\mathfrak{h}_4(\mathbb{K})$.

It was fun to remember this stuff and write it up in a more expository way.

Posted by: John Baez on December 20, 2020 6:33 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Every Lie group has two canonical real forms: the compact form of signature $-dim(G)$ and the split form of signature $+rank(G)$. The other forms are of intermediate (necessarily negative) signature.

In the case of $F_4$, the split form is the one with maximal compact $(Sp(1) \times Sp(3))/\mathbb{Z}_2$. It’s notable that this group does not show up in your classification, but that the intermediate form with signature $-20$ and maximal compact $Spin(9)$ does appear.

Is there a good reason for this? Perhaps the stabilizer group is always a real form of negative signature? No. In the usual $\mathbb{R}^{3+1}$ case, the $m^2 \gt 0$ (tardyon) stabilizers are the the compact form $SU(2)$, with signature $-3$, whereas I think the $m^2 \lt 0$ (tachyon) stabilizers are the split form $SL(2,\mathbb{R}) \cong SU(1,1)$.

Along the light cone, the stabilizer degenerates. (I am reminded of Dyson’s discussion of Maxwell’s failure to discover the cosmological constant: Missed opportunities, Bull. Amer. Math. Soc. Volume 78, Number 5 (1972), 635-652.) In the usual case, both $SU(2) = Spin(3)$ and $SU(1,1) = Spin(2,1)$ can degenerate to $\mathbb{R}^2 \rtimes Spin(2)$.

There are various “natural” degenerations of $F_4$. Your analysis sees the degeneration $\mathbb{R}^{16} \rtimes Spin(9)$, which is “selected” by the nonsplit noncompact form. The split form would have “selected” a degeneration to $\mathbb{H}^{7} \rtimes (Sp(1) \times Sp(3))$. Let me briefly describe this representation. The compact Lie group $Sp(3)$ has three fundamental representations: the defining quaternionic (aka symplectic) one of complex dimension $6$; a real (aka orthogonal) one of complex dimension $14$, which arises as $Alt^2(\mathbf{6}) \ominus \mathbf{1}$; and a quaternionic one of complex dimension $14$, which arises as $Alt^3(\mathbf{6}) \ominus \mathbf{6}$. I want the last of these: it is an action of $Sp(3)$ on $\mathbb{H}^7 = \mathbb{R}^{28}$. It preserves the $\mathbb{H}$-module structure, and so commutes with an $Sp(1)$-action.

So another version of my question is why this degeneration did not appear?

Posted by: Theo on December 20, 2020 6:46 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Thanks for your comment, Theo! When I post about such technical subjects it seems to scare away people, which makes me sad because I’m actually eager to explain what I wrote and answer questions… but on the other hand it seems to bring experts out of the woodwork, experts with superpowers.

In the case of $F_4$, the split form is the one with maximal compact $(Sp(1) \times Sp(3))/\mathbb{Z}_2$. It’s notable that this group does not show up in your classification, but that the intermediate form with signature $-20$ and maximal compact $Spin(9)$ does appear.

Is there a good reason for this?

I’ll make a wild guess: it does appear, and it’s the case I never got around to figuring out. Remember, I wrote this:

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 $\mathrm{Spin}(9) \ltimes S_+$.

So, wanna help us figure out this stabilizer?

In the usual $\mathbb{R}^{3+1}$ case, the $m^2 \gt 0$ (tardyon) stabilizers are the the compact form $SU(2)$, with signature $-3$, whereas I think the $m^2 \lt 0$ (tachyon) stabilizers are the split form $SL(2,\mathbb{R}) \cong SU(1,1)$.

That’s right. My physics intuition kicks in if I think of a tachyon energy-momentum as a spacelike vector like $(t,x,y,z) = (0,1,0,0)$ in Minkowski spacetime, and think about the subgroup of $SO_0(3,1)$ that stabilizes this. We’re basically nailing down one of the “3” in $SO_0(3,1)$, so the stabilizer is $SO_0(2,1)$. But this is $PSL(2,\mathbb{R})$. So this matches what you think, since you’re working with double covers: $SL(2,\mathbb{R})$ double covers $PSL(2,\mathbb{R}) \cong SO_0(2,1)$.

Posted by: John Baez on December 20, 2020 7:14 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

So, the one stabilizer that Greg and John and I didn’t calculate is the stabilizer of

$\left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

under the action of $\mathrm{E}_6$.

We can get a subgroup of this guy’s stabilizer if we look at the stabilizer of

$\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \in \mathfrak{h}_2(\mathbb{O})$

under the action of “$SL(2,\mathbb{O})$”, meaning $Spin(9,1)$. Remember, $\mathfrak{h}_2(\mathbb{O})$ with its determinant is isomorphic to (9+1)-dimensional Minkowski spacetime, and the vector

$\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \in \mathfrak{h}_2(\mathbb{O})$

is a unit vector in one of the 9 spacelike directions. So its stabilizer under $Spin(9,1)$ is $Spin(8,1)$.

So, the stabilizer of

$\left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right)$

under the action of $\mathrm{E}_6$ must contain $Spin(8,1)$.

Posted by: John Baez on December 20, 2020 11:30 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

The maximal compact in $Spin(8,1)$ is virtually $Spin(8)$. (Recall that a group is virtually something if it is that something modulo a sequence of finite-index subgroups and overgroups. For compact groups, virtual equivalence is equivalence of Lie algebras.) The smallest representations of $Spin(8)$ are all 8-dimensional (and real). So there is no way that $Spin(8)$ can map into $Sp(1) \times Sp(3)$, since this has a 6-(complex-)dimensional representation (and if you act trivially in that represenation, then you must land inside $Sp(1)$, which also doesn’t happen).

[This is not the only way to rule out a nontrivial map $Spin(8) \to Sp(1) \times Sp(3)$. Another is to inspect the maximal tori.]

So if your stabilizer contains $Spin(8,1)$, then it definitely does not have virtual maximal compact $Sp(1) \times Sp(3)$.

This makes me suspect that your last case will again produce $Spin(9) \ltimes S_+$, although I haven’t done the calculation.

Posted by: Theo on December 21, 2020 12:23 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

I’m thinking the stabilizer of

$\left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

will be $Spin(8,1) \ltimes S_+$. Maybe that’s what you really meant.

It’s cute how this has dimension 52 just like $\mathrm{F}_4$.

Posted by: John Baez on December 21, 2020 12:29 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

That sure sounds better.

On both “sides” of this $+-0$ orbit are orbits with stabilizer $F_{4(20)}$. That should mean that there is a degeneration $F_{4(20)} \stackrel{?}{\rightsquigarrow} Spin(8,1) \ltimes S_+$. This is plausible. At the level of Lie algebras, $Spin(8,1) = Spin(8) \oplus V$, where $V$ is the vector representation. The Killing form is negative-definite on the $Spin(8)$ part and positive-definite on the $\mathbb{R}^8$ part. For comparison, at the level of Lie algebras, $F_{4(20)} = Spin(9) \oplus S_+$. It is easy for the $Spin(8)$ to lift into $Spin(9)$. If you choose such a lift, then $S_+$ will decompose as $S_+ \oplus S_-$ over $Spin(8)$. Applying the triality automorphism to $Spin(8)$ converts this to $V \oplus S_+$. By the way, this same triality isomorphism converted $Spin(9) = Spin(8) \oplus V$ into $Spin(8) \oplus S_-$.

Off the $+-0$ orbit, there are all sorts of nontrivial maps $V \otimes V \to \dots$, $V \otimes S_+ \to \dots$, and so on. On the orbit, what seems to happen is that the $V \otimes V \to Spin(8)$ component of the bracket survives, and we turn off the components with $S_+$ and $S_-$, where these are the names after triality.

For comparison, to degenerate to $Spin(9) \ltimes S$, we turn off the $S \otimes S \to \dots$ components before applying triality to the $Spin(8) \subset Spin(9)$.

One takeaway is the following. The Lie algebra of $Spin(8,1) \ltimes S$ has a natural symmetric pairing. It is the Killing form (indefinite signature) on $Spin(8,1)$, and it is the $Spin(8,1)$-invariant form on the spinor $S$. [Reality check, literally: the spinor for $Spin(8,1)$ is real/complex/quaternionic the same way as for $Spin(8-1) = Spin(7)$, which has a real spinor.] This form has indefinite signature, as it must, since the action gives a closed embedding $Spin(8,1) \subset SO(S)$, and the domain is noncompact. I think the signature is split: $16 = 8+8$, where the two “$8$“s are the half-spinors over $Spin(8) \subset Spin(8,1)$.

So I think the natural pairing on $Spin(8,1) \ltimes S$ has $Spin(8)$ and $S_+$ of one definiteness (negative, say), and $V$ and $S_-$ of the other definiteness, just like [modulo triality] $Spin(9) \ltimes S$.

Posted by: Theo Johnson-Freyd on December 21, 2020 3:24 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Correction: the other real forms of a group have signature (=$(p-q)$ if the Killing form is positive-definite on a $p$-dimensional subspace and negative definite on a $q$-dimensional subspace) between $-dim(G)$ and $+rank(G)$, but it is not necessarily negative.

For example, $E_6$ has a real form with maximal compact $A_1 \times A_5$ and signature $+2$.

Posted by: Theo Johnson-Freyd on December 21, 2020 3:27 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Here is a dissatisfying answer to my question.

We are working inside the real form of $E_6$ with maximal compact $F_4$. (This is not the split form — that has maximal compact $C_4$ — which may help to explain split forms of other groups are not preferred.) At the level of Lie algebras, this decomposes $E_6 = F_4 \oplus \mathbf{26}$, where $\mathbf{26}$ is the traceless elements in the exceptional Jordan algebra, and Killing form is positive definite on $\mathbf{26}$ and negative definite on $F_4$.

In particular, no subgroup can have a Killing form with more than $26$ positive-definite directions. However, the split form of $F_4$ had $28$ positive-definite (and $24 = 3 + 21$ negative-definite) directions.

The five real forms of $E_6$ have (virtual — meaning I don’t report finite group data) maximal compacts $E_6, F_4, Spin(10) \times U(1), SU(2) \times SU(6), Sp(4)$ of dimensions $78, 52, 46, 38, 36$. [I am using the “topologist’s” convention that $Sp(n)$ means the compact symplectic group of rank $n$, and not the “number theorist’s” convention that $n$ must be even and $Sp(n)$ means the split symplectic group whose defining representation is $n$-dimensional.] In other words, this form of $E_6$ is the “second most compact” form. So closed subgroups are going to prefer having lots of compact directions and not very many noncompact directions.

The usual Lorentz case also was the “second most compact” form $Spin(n-1,1)$. This is just a way of saying that we only want one timelike direction. (Otherwise, you would be able to continuously boost from positive-direction time to negative-direction time. As Egan explores in The Clockwork Rocket and its sequels, this is not automatically a contradiction, but it does lead to some counterintuitive phenomena.) Is there something about Jordan algebras that selects “second most compact” forms?

Posted by: Theo Johnson-Freyd on December 21, 2020 3:48 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Okay, I can now show that the stabilizer of

$a = \left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

under the action of $\mathrm{E}_6$ contains the subgroup $Spin(8,1) \ltimes S_-$.

It’s really easy using the formulas for the action of various subgroups of $\mathrm{E}_6$ on $\mathfrak{h}_3(\mathbb{O})$ given in Part 7. The trick is to write any element of $\mathfrak{h}_3(\mathbb{O})$ as

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

where under the subgroup $Spin(9,1) \subset \mathrm{E}_6$

$v = \left( \begin{array}{cc} \beta & x \\ x^\ast & \gamma \end{array} \right) \in V \; :\!\!= \mathfrak{h}_2(\mathbb{O})$

transforms as vector, and

$\psi = \left( \begin{array}{cc} z^\ast \\ y \end{array} \right) \in S_- \; :\!\!= \mathbb{O}^2$

transforms as left-handed spinor and $\alpha \in \mathbb{R}$ transforms trivially, as a scalar. This gives an isomorphism

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

and using this we write

$a = (v_0, 0, 0)$

where

$v_0 = \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right) \in V$

Since $v_0$ is a unit spacelike vector in $V$, the subgroup of $Spin(9,1) \subset \mathrm{E}_6$ that stabilizes it is exactly $Spin(8,1)$. This group also stabilizes $(v_0,0,0)$.

To go further we recall that the left-handed spinor representation of $Spin(9,1)$, $S_- = \mathbb{O}^2$, can be seen as an abelian Lie subgroup of $\mathrm{E}_6$, which acts on $\mathfrak{h}_3(\mathbb{O})$ as follows: $\eta \in S_-$ acts on any element

$(v, \psi, \alpha) \in V \oplus S_- \oplus \mathbb{R} \; \cong \; \mathfrak{h}_3(\mathbb{O})$

to give

$( v + \frac{1}{2} \alpha [\eta, \eta] + [\psi, \eta] , \; \psi + \alpha \eta , \; \alpha )$

We can see immediately see that any $\eta \in S_-$ stabilizes $a = (v_0, 0, 0)$ because this element has $\psi = 0, \alpha = 0$.

So, the stabilizer contains $Spin(8,1)$ and $S_-$, and thus the whole subgroup $Spin(8,1) \ltimes S_- \subset \mathrm{E}_6$.

The dimension of this subgroup is 52, so if we knew the stabilizer had dimension at most 52 we’d be done — up to discrete transformations, which I’m not getting paid enough to deal with.

I don’t know why the stabilizer should have dimension 52, but I’ll note that $a \in \mathfrak{h}_3(\mathbb{O})$ has rank 2, and the other two kinds of rank-2 elements of the exceptional Jordan algebra have 52-dimensional stabilizers, so it somehow seems plausible.

There’s also an abelian subgroup $S_+ \subset \mathrm{E}_6$, and it’s not instantly obvious from the formulas that nothing in here stabilizes $a = (v_0, 0, 0)$.

Posted by: John Baez on December 21, 2020 7:40 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Okay, I’ve shown that the Lie algebra of the stabilizer of

$a = \left( \begin{array}{ccc} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array} \right) \in \mathfrak{h}_3(\mathbb{O})$

under the action of $\mathrm{E}_6$ is exactly the Lie algebra of $Spin(8,1) \ltimes S_- \subset \mathrm{E}_6$. So, the identity component of the stabilizer is exactly $Spin(8,1) \ltimes S_-$, but the stabilizer might in theory have other connected components, as noted by L. Spice.

Let me record the calculation. I worked out how the Lie algebra $\mathfrak{e}_6$ acts on the exceptional Jordan algebra in Part 7. To describe this we’ll use the vector space isomorphisms

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

and

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

Using these we can write an element of $\mathfrak{e}_6$ as $(L,\phi, \eta, t)$ where $L \in \mathfrak{so}(9,1)$, $\phi \in S_+$, $\eta \in S_-$ and $t \in \mathbb{R}$. We can write an element of $mathfrak{h}_3(\mathbb{O})$ as $(v, \psi, a)$ where $v \in V, \psi \in S_-$ and $\alpha \in \mathbb{R}$.

The action of $\mathfrak{e}_6$ on $\mathfrak{h}_3(\mathbb{O})$ then goes like this:

$(L,\phi, \eta, t) \colon (v, \psi, \alpha) \mapsto$ $\big( L v + [\psi,\eta] - 2t v, \; L \psi + \gamma(v \otimes \phi) + t \psi, \; 2\langle \phi, \psi \rangle + 4t \alpha \big)$

We want to see when this transformation annihilates $(v,\psi, \alpha)$, and we’re interested in the case $(v,\psi, \alpha) = a = (v_0, 0, 0)$ where the vector part is

$v_0 = \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right) \in V$

In this case things simplify a lot. We get

$(L,\phi, \eta, t) \colon (v_0, 0, 0) \mapsto \big( L v - 2t v_0, \; \gamma(v_0\otimes \phi) , \; 0 \big)$

We want the right-hand side to be zero, which is true iff

$L v_0 = 2 t v_0, \qquad \gamma(v_0 \otimes \phi) = 0$

First, note that these equations say nothing about $\eta \in S_+$, so that can be anything. Second, recall from Part 7 that the action of vectors on right-handed spinors appearing in $\gamma(v_0 \otimes \phi)$ is defined using matrix multiplication. The ‘vector’

$v_0 = \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right) \in V$

is really a matrix, the ‘right-handed spinor’

$\phi = \left( \begin{array}{c} \phi\\ \phi_2 \end{array} \right) \in S_+$

is really a column vector consisting of two octonions, and

$\gamma(v_0 \otimes \phi) = \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right) \left( \begin{array}{c} \phi\\ \phi_2 \end{array} \right)$

So this vanishes iff $\phi = 0$!

Third and finally, what about the equation

$L v_0 = 2 t v_0 \; ?$

Clearly this equation is true if $t = 0$ and $L \in \mathfrak{so}(9,1)$ annihilates $v_0$. Since $v_0$ is a spacelike vector the set of $L$ that annihilate it form a copy of $\mathfrak{so}(8,1)$. But are there solutions to the above equation with $t \ne 0$?

No! To say that $L \in \mathfrak{so}(9,1)$ is to say that

$g(L v, w) + g(v, L w)= 0$

for all vectors $v, w \in V$, where $g$ is the Minkowski metric. So in particular we have

$g(L v_0, v_0) = 0$

That is, $L v_0$ is orthogonal to $v_0$. So $L v_0 = t v_0$ would say $t v_0$ is orthogonal to $v_0$. But since $v_0$ is spacelike and nonzero, this is only possible if $t = 0$!

(If $v_0$ were lightlike it would be a whole different ball game.)

So we’re done: the annihilator of $a = (v_0, 0, 0) \in \mathfrak{h}_3(\mathbb{O})$ is exactly the vector space $\mathfrak{so}(8,1) \oplus S_+ \subset \mathfrak{e}_6$.

Posted by: John Baez on December 22, 2020 8:04 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

The real split form $F_{4(4)}$ of the Lie group $F_{4}$ occurs as stabilizer of the rank-3 orbit of the non-transitive action of the split form $E_{6(6)}$ of the Lie group $E_{6}$ on its fundamental module $\mathbf{27\simeq }\mathfrak{h}_{3}(\mathbb{O}_{s})$; see e.g. Th. 1 of [BDFMR], as well as Table I therein. $F_{4(4)}$ is diffeomorphic to $\left( Sp(3)\times Sp(1)\right) \ltimes (\mathbf{14}^{\prime },\mathbf{2})$, where $\mathbf{2}$ is the fundamental module of $Sp(1)$, whereas $\mathbf{14}^{\prime }$ denotes the rank-3 antisymmetric (and skew-traceless) representation of $Sp(3)$ (the one that was indicated as Alt$^{3}\left( \mathbf{6}\right) \ominus \mathbf{6}$). Within the study of extremal black hole solutions in maximal supergravity in $4+1$ space-time dimensions, the stabilizers of such orbits were computed in [FG98] (see also [Krut] for a rigorous treatment).

[BDFMR] : L. Borsten, M.J. Duff, S. Ferrara, A. Marrani, W. Rubens, “Small Orbits”, Phys. Rev. D85 (2012) 086002.

[FG98] : S. Ferrara, M. Gunaydin, “Orbits of exceptional groups, duality and BPS states in string theory”, Int. J. Mod. Phys. A13, 2075 (1998).

[Krut] : S. Krutelevich, “On a canonical form of a $3\times 3$ Hermitian matrix over the ring of integral split octonions”, J. Algebra 253, 276 (2002).

Posted by: Alessio Marrani on January 2, 2021 6:08 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

So I guess it’s the forward small lightcone that’s also the set of pure states for a three-level octonionic quantum system. Now that’s a correspondence I’ll have to think about.

Posted by: Blake Stacey on December 20, 2020 6:50 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Right! I think we can say it for any normed division algebra $\mathbb{K}$ like this. Sitting inside $\mathfrak{h}_3(\mathbb{K})$ there’s a cubic surface

$\{a \in \mathfrak{h}_3(\mathbb{K}) : \; det(a) = 0 \}$

which is a ‘cone’ in the sense that $\det(a) = 0$ implies that $\det(c a) = 0$ whenever $c \in \mathbb{R}$. This is the ‘large lightcone’. Sitting inside this is a smaller cone

$\{a \in \mathfrak{h}_3(\mathbb{K}) : \; a \times a = 0 \}$

where $\times$ is the ‘cross product’

$\times : \mathfrak{h}_3(\mathbb{K}) \times \mathfrak{h}_3(\mathbb{K}) \to \mathfrak{h}_3(\mathbb{K})^\ast$

formed as in Part 10. And projectivizing this smaller cone gives $\mathbb{K}\mathrm{P}^2$, which is the set of states of the three-level quantum system with $\mathfrak{h}_3(\mathbb{K})$ as its algebra of observables.

The stuff I wrote near the end of Part 10 explains what’s up with this cross product business. The key point is that a projection in $\mathfrak{h}_3(\mathbb{K})$ — that is, a guy $p$ with $p^2 = p$ — has trace 1 iff

$p \times p = 0$

Trace-1 projections give all the points of $\mathbb{K}\mathrm{P}^2$. But the concepts of projection and trace, while invariant under automorphisms of $\mathfrak{h}_3(\mathbb{K})$, are not invariant under all the determinant-preserving transformations of $\mathfrak{h}_3(\mathbb{K})$. So Freudenthal (or someone) cleverly noticed that you could describe points in $\mathbb{K}\mathrm{P}^2$ more invariantly just as “nonzero elements $a \in \mathfrak{h}_3(\mathbb{K})$ with $a \times a = 0$, up to multiplying by a nonzero real numbers”.

The point is that we can define a guy $a \in \mathfrak{h}_3(\mathbb{K})$ to have any rank 1 if $a \ne 0$ but $a \times a = 0$. This matches the usual concept of ‘rank’ for a matrix in the real and complex cases. And it turns out that any rank-1 guy with positive trace is a projection up to multiplying by a nonzero real number — even in the funky quaternionic and octonionic cases!

Posted by: John Baez on December 20, 2020 7:44 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

How, I wonder, did the name “cross product” get attached to the operation “for two distinct points, give the line through them”?

(Usually, when I have questions like this, the answer is either some important idea I have failed to appreciate, or “well, we wrote it with a $\times$ and the name kind of stuck”.)

Posted by: Blake Stacey on December 26, 2020 2:24 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Blake wrote:

How, I wonder, did the name “cross product” get attached to the operation “for two distinct points, give the line through them”?

(Usually, when I have questions like this, the answer is either some important idea I have failed to appreciate, or “well, we wrote it with a $\times$ and the name kind of stuck”.)

In this case there is no important idea you have failed to appreciate. Freudenthal just used $\times$ for this operation on page 199 of his paper “Toward plane octave-geometry”:

• Hans Freudenthal, Zur ebenen Oktavengeometrie, Indagationes Mathematicae 15 (1953), 195–200.

For some reason the evil Elsevier has not made this old paper open access like many of their other old math papers. You can buy it for \$31.50.

I don’t know when they swallowed up Indagationes Mathematicae, and I also don’t know what that title means. Indications of Mathematics? Mathematical Indignities?

starting on page 162. The whole paper is amazing.

Posted by: John Baez on December 27, 2020 8:55 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

I don’t know a lot about the history of this (but see [1] and the references therein). But I do know that projective geometry was much more widely known in Freudenthal’s times than today, so one possible way to see this might be to consider the real projective plane and use homogeneous coordinates.

So let $p=(p_1:p_2:p_3)$ and $q=(q_1:q_2:q_3)$ be the homogeneous coordinates of two points in the real projective plane, and let $a=(a_1:a_2:a_3)$ be the homogeneous coordinates corresponding to the line with equation $a_1x_1 + a_2x_2 + a_3x_3=0$. Then $p$ lies on $a$ iff $\langle a,p\rangle {:=} a_1p_1+a_2p_2+a_3p_3=0$.

Conversely, the coordinates $a$ of the line through $p$ and $q$ must satisfy $\langle a,p\rangle = \langle b,p\rangle = 0$, so $a$ must be orthogonal to both $p$ and $q$ as vectors in 3-space — and $a:=p\times q$ fits the bill.

Dually, the point on the intersection of the lines with homogeneous coordinates $a$ and $b$ is just $a\times b$.

This and much more can be found, for example, in

[1] Jürgen Richter-Gebert, Perspectives on projective geometry

Posted by: Julian on January 4, 2021 11:20 AM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Thank you!

Posted by: Blake Stacey on December 28, 2020 8:00 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

When you compute stabilisers, you find a large chunk of stabiliser, and then conclude by dimension counting that you’ve found all of it. But do we somehow know in advance that the point stabilisers are connected, or could there be some finite component group on top of the chunk you’ve found? I ask because it seems that, if you’re then trying to classify representations of these groups, then the component group might matter.

Your stabilisers of the various orbit types do not ‘see’ a global sign flip. Is this just coincidence, or is it explained by some larger symmetry (in $\mathsf{E}_6$?) that flips all signs?

Posted by: L Spice on December 21, 2020 8:21 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

L. Spice wrote:

But do we somehow know in advance that the point stabilisers are connected, or could there be some finite component group on top of the chunk you’ve found?

I don’t known any a priori reason why the stabilizers should be connected. So, our calculations could be off by a discrete group. If someone ever gets serious about understanding quantum field theory on the exceptional Jordan algebra, this will add some subtleties: you could have two kinds of particles that behave almost but not quite the same.

This is already a possibility in our universe, since the group $\mathrm{O}(3)$ is not connected: physicists call a massive particle that transforms trivially under $\mathrm{SO}(3)$ a spin-zero particle, but they come in two kinds: scalars, which transform trivially under reflections, and pseudoscalars, which don’t. The pion is the most famous example of a pseudoscalar meson. And yet, we know a lot just knowing how a particle transforms under the identity component of the relevant group.

Your stabilisers of the various orbit types do not ‘see’ a global sign flip. Is this just coincidence, or is it explained by some larger symmetry (in $\mathsf{E}_6?$) that flips all signs?

That’s an interesting question. There’s a transformation of the exceptional Jordan algebra that multiplies every element by $-1$. This is not in $\mathrm{E}_6$ because transformations in $\mathrm{E}_6$ preserve the determinant. But it’s somehow ‘almost as good’, because it multiplies the determinant by $-1$. It could be connected to the outer automorphism of $\mathrm{E}_6$, which squares to the identity. But I don’t see how.

Posted by: John Baez on December 21, 2020 8:46 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

So, if you’re a denizen of this 27-dimensional spacetime and you only see light rays moving along the small lightcone… But if you can see light rays moving along the large lightcone, your field of view will be larger! It will be 25-dimensional…

Yes, the lightcone little group in D=26+1 M-theory is $SO(25)$, while in D=10+1 it’s $SO(9)$. You’re seeing the Jordan algebraic description of this, i.e., through octonionic quantum mechanics with $\mathbb{O}$-qutrit structure. Hisham Sati has argued that D=10+1 M-theory has hidden $\mathbb{OP}^2$ Cayley plane fibers, as the 11-dimensional supergravity supermultiplet consists of $SO(9)$ representations ($\mathbf{36}$, $\mathbf{44}$, $\mathbf{128}$). Using $\mathfrak{h}_3(\mathbb{O})$ the breaking from $SO(25)\rightarrow SO(9)$ occurs when one of the three orthogonal idempotents (of spectral decomposition) is fixed, i.e., when a logical state of the $\mathbb{O}$-qutrit is fixed.

When studying the Monster group, the larger lightcone symmetry in D=26+1 is needed, as well as 8192 spinor structure. Why?

One can write down a D=26+1 superalgebra using anticommutator of 8192 $\times$ 8192 and find central charges for an M2, M5, M10 and dual M13, M18 and M21 branes. These branes generalize the M2 and M5 of D=10+1 M-theory.

The M2 and M10 branes are very use useful. I’ll explain why.

Take the M2-brane and it breaks Poincare symmetry as $SO(25,1)\rightarrow SO(2,1)\times SO(24)$, where $SO(24)$ is the R-symmetry, which is the isometry group of the transverse space of the M2-brane worldvolume. The M2-brane near horizon geometry is $AdS_4\times S^{23}$. Take the transverse space as discretized via the Leech lattice, then the shortest set of vectors are the 196,560 norm four vectors, which form a discretized 23-sphere. The R-symmetry $SO(24)$ is now broken to a maximal finite subgroup, the Conway group $Co_0$. Next dimensionally reduce to D=25+1 along an $S^1$/$\mathbb{Z}_2$ orbifold, where the M2-brane 3-form has a component in the reduced direction. This turns the M2-brane into a “type IIA-like” string, and the Conway group $Co_0$ is reduced to $Co_1$ under the orbifolding. The near horizon geometry of this string is $AdS_3\times S^{23}$. The Monster (S)CFT then lives on the boundary of this $AdS_3$ giving a Monstrous holography as Witten has argued.

Alternatively, take the M10-brane and break Poincare symmetry as $SO(25,1)\rightarrow SO(10,1)\times SO(16)$. This gives a D=10+1 worldvolume that can be used for a worldvolume realization of M-theory. Stretch a 2-brane between 9D boundaries of the M10-brane and one can recover a Horava-Witten model that reduces to the $E_8\times E_8$ heterotic string as one dimensionally reduces the M10 worldvolume along an $S^1$/$\mathbb{Z}_2$ orbifold. The original D=25+1 bosonic string is an M2-brane reduced outside the M10 worldvolume, while inside one recovers a D=9+1 superstring.

Please send my regards to John Huerta.

Posted by: Metatron on December 27, 2020 5:18 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

In the above, use

$SO(26,1)\rightarrow SO(2,1)\times SO(24)$

$SO(26,1)\rightarrow SO(10,1)\times SO(16)$

for the D=26+1 Poincare symmetry breakings.

Posted by: Metatron on December 27, 2020 5:25 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

The orbits of the non-transitive action of the real form $E_{6(-26)}$ of the Lie group $E_{6}$ on its fundamental module $\mathbf{27\simeq }\mathfrak{h}_{3}(\mathbb{O})$ have been computed in [Shu] (see Th. 1 therein). Within the study of extremal black hole solutions in $\mathcal{N}=2$ Maxwell-Einstein gravity theories in $4+1$ space-time dimensions, the stabilizers of such orbits were computed independently in [FG98] (see also [BDFMR] - and Table II therein- for a collection of results and references). From Th. 2 of [BDFMR] and the stabilizers reported in Table II therein, it seems that you swapped the stabilizers of the $++-$ and $---$ rank-3 orbits, because they respectively are the compact real form $F_{4(-52)}$ and the minimally non-compact real form $F_{4(-20)}$ of the Lie group $F_{4}$. Furthermore, the stabilizer of the $+-0$ rank-2 orbit can be computed to be $Spin(8,1)\ltimes \mathbf{16}$, where $\mathbf{16\equiv }S$ is the (non-chiral) spinor representation $S$ (since there is no chirality in odd dimensions, there is no $S_{+}$ nor $S_{-}$ for $Spin(8,1)$, but simply $S$).

[BDFMR] : L. Borsten, M.J. Duff, S. Ferrara, A. Marrani, W. Rubens, “Small Orbits”, Phys. Rev. D85 (2012) 086002.

[FG98] : S. Ferrara, M. Gunaydin, “Orbits of exceptional groups, duality and BPS states in string theory”, Int. J. Mod. Phys. A13, 2075 (1998).

[Shu] : O. Shukuzawa, “Explicit Classifications of Orbits in Jordan Algebra and Freudenthal Vector Space over the Exceptional Lie Groups”, Commun. Algebra 34, 197 (2006).

Posted by: Alessio Marrani on January 2, 2021 5:42 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

I assume you are aware of the upcoming lecture series / conference https://www.perimeterinstitute.ca/conferences/octonions-and-standard-model, but if not, I thought I’d mention it.

Posted by: Theo on February 6, 2021 1:49 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Thanks—yes, I’m going to speak in that.

Posted by: John Baez on February 6, 2021 8:00 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Looks neat! Will the talks end up on PIRSA later?

Posted by: Blake Stacey on February 7, 2021 3:41 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

I think they will.

Posted by: John Baez on February 7, 2021 11:12 PM | Permalink | Reply to this

Re: Octonions and the Standard Model (Part 11)

Looks like they will indeed!

Posted by: Blake Stacey on February 9, 2021 6:22 PM | Permalink | Reply to this

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