## March 31, 2021

### Can We Understand the Standard Model Using Octonions?

#### Posted by John Baez

I gave two talks in Latham Boyle and Kirill Krasnov’s Perimeter Institute workshop Octonions and the Standard Model.

The first talk was on Monday April 5th at noon Eastern Time. The second was exactly one week later, on Monday April 12th at noon Eastern Time.

Here they are…

Can we understand the Standard Model? (video, slides)

Abstract. 40 years trying to go beyond the Standard Model hasn’t yet led to any clear success. As an alternative, we could try to understand why the Standard Model is the way it is. In this talk we review some lessons from grand unified theories and also from recent work using the octonions. The gauge group of the Standard Model and its representation on one generation of fermions arises naturally from a process that involves splitting 10d Euclidean space into 4+6 dimensions, but also from a process that involves splitting 10d Minkowski spacetime into 4d Minkowski space and 6 spacelike dimensions. We explain both these approaches, and how to reconcile them.

Can we understand the Standard Model using octonions? (video, slides)

Abstract. Dubois-Violette and Todorov have shown that the Standard Model gauge group can be constructed using the exceptional Jordan algebra, consisting of 3×3 self-adjoint matrices of octonions. After an introduction to the physics of Jordan algebras, we ponder the meaning of their construction. For example, it implies that the Standard Model gauge group consists of the symmetries of an octonionic qutrit that restrict to symmetries of an octonionic qubit and preserve all the structure arising from a choice of unit imaginary octonion. It also sheds light on why the Standard Model gauge group acts on 10d Euclidean space, or Minkowski spacetime, while preserving a 4+6 splitting.

You can see all the slides and videos and also some articles with more details here.

Posted at March 31, 2021 10:20 PM UTC

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### Re: Can We Understand the Standard Model Using Octonions?

Typo in the second slide deck: On slide 15, “10” should be “10d” twice.

Posted by: Blake Stacey on April 1, 2021 5:02 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

And on the summary slide of the second deck, the algebra for the complex qubit should be $\mathfrak{h}_2(\mathbb{C})$ instead of $\mathfrak{h}_2(\mathbb{O})$.

Posted by: Blake Stacey on April 1, 2021 7:43 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Thanks very much — I’ll fix these!

I want to improve my second batch of slides.

Right now they don’t explain why the subgroup of automorphisms of $\mathfrak{h}_3(\mathbb{O})$ preserving a copy of $\mathfrak{h}_2(\mathbb{O})$ is $Spin(9)$. I explained this in my series of posts.

The quick thing to say is that you can identify $\mathfrak{h}_2(\mathbb{O})$ with 10d Minkowski spacetime and think of $\mathfrak{h}_3(\mathbb{O})$ as the direct sum of 3 spaces:

• $\mathfrak{h}_2(\mathbb{O})$ (10d vectors)

• $\mathbb{O}^2$ (10d spinors)

• $\mathbb{R}$ (scalars).

Then the group of determinant-preserving transformations of $\mathfrak{h}_3(\mathbb{O})$ preserving $\mathfrak{h}_2(\mathbb{O})$ preserves this splitting and all the invariant operations involving vectors, spinor and scalars, so it’s $Spin(9,1)$. The automorphisms of $\mathfrak{h}_3(\mathbb{O})$ preserving $\mathfrak{h}_2(\mathbb{O})$ also preserve the ‘time axis’ in 10d Minkowski spacetime, so they form the subgroup $Spin(9)$.

I also don’t give any explanation of where $(SU(3) \times SU(3)) / \mathbb{Z}_3$ comes from. It’s not that hard to get an idea of where this comes from.

I don’t like how these important facts show up in my talk ‘as if by magic’, stated but not explained. But maybe explaining all this stuff would overload the talk. There’s a limit to how much information people can absorb in 50 minutes. I guess blogging and papers are better for detailed explanations. I want to continue my blog series and use it to expand on these talks.

Posted by: John Baez on April 2, 2021 4:20 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Okay, I massively revised my second deck of slides:

I now give a rough explanation of why the subgroups

$(SU(3) \times SU(3))/\mathbb{Z}_3, \; Spin(9) \subset \mathrm{F}_4$

show up in this game, and why their intersection is the true gauge group of the Standard Model.

Posted by: John Baez on April 4, 2021 12:16 AM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

The fact of F$_4$ having the wrong dimensionality to act on $\mathbb{O}^3$ is interesting, and these slides seem to make the point more forcefully than I recall seeing before. This ties to something else I’ve thought a little about, the possibility of defining a Wigner function for the octonionic qutrit following the Wootters construction for the complex case. The idea goes like this: For simplicity, assume that $d$ is a power of an odd prime. Then we can construct a full set of $d+1$ mutually unbiased orthonormal bases, i.e., a set such that $\langle \psi | \phi \rangle = 1/d$ whenever $|\psi\rangle$ and $|\phi\rangle$ belong to different bases. Each basis defines a measurement with $d$ possible outcomes, and so a probability distribution over those outcomes has $d-1$ free parameters when we account for normalization. Together, all these probabilities provide $(d-1)(d+1) = d^2 - 1$ real parameters, which is just the dimensionality we need for density matrices on $\mathbb{C}^d$ (self-adjointness for $d^2$, less 1 for trace normalization). A discrete Wigner function can be defined as a thing that is summed over to get the probabilities for the MUB measurements, following a nice geometrical pattern. See Veitch et al. (2014) for some worked examples for the regular qutrit.

The interesting thing is that, per Theorem 8.3 of Cohn, Kumar and Minton (2013), the octonionic qutrit has a set of thirteen mutually unbiased bases. So, counting up the parameters, we get $13 \cdot (3 - 1) = 26$, which matches the dimension of the smallest nontrivial representation of F$_4$ (and, I guess, the dimension of the Albert algebra minus 1 for trace normalization).

Posted by: Blake Stacey on April 5, 2021 10:47 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Whoops, that should be $|\langle \psi | \phi \rangle|^2 = 1/d$.

Posted by: Blake Stacey on April 5, 2021 10:51 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Blake wrote:

The fact of F$_4$ having the wrong dimensionality to act on $\mathbb{O}^3$ is interesting, and these slides seem to make the point more forcefully than I recall seeing before.

I wouldn’t say $\mathrm{F}_4$ has the wrong dimensionality: it has dimension 52. I’d say the problem is that its lowest-dimensional nontrivial representation has dimension 26, while $\mathbb{O}^3$ just has dimension 24.

(By the way, 52 is twice 26.)

I wish I had time to get into these mutually unbiased bases. What do Cohn, Kumar and Minton mean when they say the octonionic qutrit has 13 mutually unbiased bases? What do they mean by an octonionic qutrit?

In my talk I’m mainly discussing the Jordan algebra of observables of the octonionic qutrit, which is the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$. Associated to any Euclidean Jordan algebra we have a set of states and a set of pure states. The set of pure states for $\mathfrak{h}_3(\mathbb{O})$ is $\mathbb{O}\mathrm{P}^2$, so this is another important avatar of the octonionic qutrit. $\mathrm{F}_4$ is

• the automorphism group of the Jordan algebra $\mathfrak{h}_3(\mathbb{O})$

and also

• the isometry group of $\mathbb{O}\mathrm{P}^2$ (which is a Riemannian manifold).

So, when I say ‘octonionic qutrit’, I either mean the Jordan algebra of observables $\mathfrak{h}_3(\mathbb{O})$ or the space of pure states $\mathbb{O}\mathrm{P}^2$.

What’s ‘defective’ in this story is that $\mathrm{F}_4$ doesn’t act in any way (other than trivially) on the vector space $\mathbb{O}^3$. This is true despite the fact that

• elements of $\mathfrak{h}_3(\mathbb{O})$ give linear operators on $\mathbb{O}^3$ in the obvious way

and

• points of $\mathbb{O}\mathrm{P}^2$ correspond to ‘1-dimensional octonionic subspaces’ of $\mathbb{O}^3$ (which need to be defined with great care).
Posted by: John Baez on April 6, 2021 4:33 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

I wouldn’t say $\mathrm{F}_4$ has the wrong dimensionality: it has dimension 52. I’d say the problem is that its lowest-dimensional nontrivial representation has dimension 26, while $\mathbb{O}^3$ just has dimension 24.

Yes, sorry for being sloppy there; it was the mismatch between the dimensions of that representation and of $\mathbb{O}^3$ that I had in mind.

Here’s what Cohn, Kumar and Minton have to say. First, after Eq. (2.1) they define a metric on $K\mathbb{P}^{d-1}$ in terms of the inner product of projection matrices: $\rho(x_1, x_2) = \sqrt{1 - \langle \Pi_1, \Pi_2 \rangle}$. With that in mind, here is their Theorem 8.3:

There exists a tight code $\mathcal{C}$ of 39 points in $\mathbb{O}\mathbb{P}^2$. It consists of 13 orthogonal triples such that, for any two points $x_i, x_j$ in distinct triples, $\rho(x_i,x_j) = \sqrt{2/3}$. In other words, if $\Pi, \Pi'$ are the projection matrices corresponding to two distinct points in $\mathcal{C}$, then $\langle \Pi, \Pi' \rangle$ equals 0 if the two points are in the same triple and otherwise equals 1/3.

The construction is inspired by a 12-point set in $\mathbb{C}\mathbb{P}^2$, which starts with the basis vectors

$(1, 0, 0),\ (0, 1, 0),\ (0, 0, 1) \, ,$

and then includes the 9 vectors

$\frac{1}{\sqrt{3}}(1, \omega^a, \omega^b)$

where $\omega$ is the cube root of unity $e^{2\pi i/3}$ and the indices $a,b = 0, 1, 2$. To extend this into $\mathbb{O}\mathbb{P}^2$, they let $1,i,j,k$ be the standard basis of the quaternions, $l$ any of the four remaining standard basis elements of $\mathbb{O}$, and $n = j l$. Defining $\omega$ as before, they construct the vectors

$\begin{array}{cc}(1,\omega^a,\omega^b)/\sqrt{3}, & (1,\omega^a j, \omega^b l)/\sqrt{3}, \\ (1, \omega^a l, \omega^b n)/\sqrt{3}, & (1, \omega^a n, \omega^b j)/\sqrt{3}. \end{array}$

Hopefully that’s less ambiguous than what I wrote before. So, it’s $\mathbb{O}\mathbb{P}^2$ that they have in mind.

Posted by: Blake Stacey on April 6, 2021 8:48 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Thanks! So I guess they’re thinking of a basis as an ‘orthogonal triple’ of points in $\mathbb{O}\mathrm{P}^2$, where a point in $\mathbb{O}\mathrm{P}^2$ corresponds to a rank-one projection in $\mathfrak{h}_3(\mathbb{O})$, and two of these, say $\Pi_1, \Pi_2$, are orthogonal iff

$\langle \Pi_1, \Pi_2 \rangle = 0$

where the inner product on $\mathfrak{h}_3(\mathbb{O})$ is

$\langle \Pi_1, \Pi_2 \rangle = tr(\Pi_1 \circ \Pi_2)$

where $\circ$ is the Jordan product and $tr$ is the trace on $\mathfrak{h}_3(\mathbb{O})$: just the sum of the diagonal entries.

Posted by: John Baez on April 12, 2021 5:59 AM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Hi, I seem to recall you commenting a while back that you didn’t have a lot of confidence that the apparent links between octonions and the standard model would lead anywhere useful.

Of course, I realise that you have prepared these slides for a particular workshop … but are you perhaps slightly less pessimistic now by any chance?

Thanks!

Posted by: Bertie on April 4, 2021 1:32 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

I think any particular researcher’s optimism or pessimism is uninteresting, especially in fundamental physics where the chance of any one idea working out is very low, and the most optimistic people often come up with the worst ideas. So I’m not going to answer that question.

Posted by: John Baez on April 4, 2021 9:55 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

I might be wrong, but I think what Bertie mean by useful is to get the Lagrangian of the SM, with the “usual physics”. Kind of a first principles derivation.

Posted by: Daniel de França on April 4, 2021 11:04 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

That’s exactly what I’ll start out talking about in my first talk! I don’t feel like summarizing today what I will say tomorrow at 9 am (much too early). A video of the talk will appear eventually, and you can already get the slides.

Posted by: John Baez on April 4, 2021 11:20 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Posted by: Blake Stacey on April 6, 2021 7:38 AM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Posted by: Blake Stacey on April 13, 2021 10:07 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

A lovely talk!🙏 It’s obvious you will be a great loss to your university when the time comes to progress your retirement plans 😃

Posted by: Bertie on April 6, 2021 9:35 AM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Thanks! I look forward to not teaching calculus and serving on university committees — it’ll give me more time to give talks and write.

Posted by: John Baez on April 6, 2021 4:37 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

I have an obnoxious kind of question. While 𝔥2(H) is a six dimensional hermitian matrix, why switching time for a spacial dimension ? It would live in the perpendicular 6d space.

Posted by: Daniel de França on April 6, 2021 6:53 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

I wasn’t able to make it to this week’s talk in real time, but I watched the video, and I should be there on Monday.

Your talk this week started out with a very enticing question: Why does $5 = 2+3$? As far as I can tell, your explanation is “Because $10=4+6$”, which is less than satisfying. Indeed, the whole $S(U(2)\times U(3)) = SU(5) \cap (SO(4) \times SO(6))$ is really just an unpacking of the (a?) definition of the LHS, isn’t it?

One other comment: We do know that $S(U(2)\times U(3))$ is compact and connected. If you know furthermore that it fits in a noncompact group like $SO(9,1)$, then you know that it fits in the maximal compact subgroup $SO(9) = SO(9) \times SO(1)$. The maximal compact is unique up to conjugacy, so there’s no further choice needed to refine $S(U(2)\times U(3)) \subset SO(9,1)$ through $SO(9)$.

I’m having trouble getting markdown to give me the formatting that I want. I have two follow-up questions, but I will post them as replies.

Posted by: Theo Johnson-Freyd on April 9, 2021 2:06 PM | Permalink | Reply to this

### Is the intersection special or generic?

It’s all well and good to realize $K = H_1 \cap_G H_2$ for some subgroups $H_1,H_2 \subset G$. But subgroups come in conjugacy classes. When you said that you had $SU(5) \subset SO(10)$ and $SO(4) \times SO(6) \subset SO(10)$, I think you had in mind some specific subgroups for a specific $SO(10)$, but a topologist or representation theorist would find it more natural just to declare the conjugacy classes of these inclusions. Now there is a problem: up to $SO(10)$-conjugacy, there is a unique $SU(5) \subset SO(10)$, but once you have chosen it, you have broken the $SO(10)$-symmetry. In particular, when you go looking for an $SO(4) \times SO(6) \subset SO(10)$, a priori you should look at all of these up to $SU(5)$-conjugacy, and there might very well be multiple orbits.

A typical thing that happens when there are multiple orbits is that there is a “generic orbit” and some “special” orbits. Actually, the “generic orbit” typically is not a single orbit. What it is is a family of orbits, but they are all qualitatively the same. In particular, you could imagine that, up to common $G$-conjugacy, perhaps there are lots of pairs of subgroups $H_1, H_2 \subset G$, but for most of them, the intersection $H_1 \cap H_2$ is always a copy of $K$.

Let’s do a trivial example. Let’s take $H_1 = H_2 = U(1)$ living inside $G = SU(2)$. Up to $SU(2)$-conjugacy, $H_1$ might as well be the standard $U(1) = diag(\lambda,\lambda^*)$. Then there is a 1-dimensional family of choices for $H_2$. There is a special choice: $H_1 = H_2$, in which case they intersect along a $U(1)$. All the other choices are generic, and the intersection is $\{\pm 1\} \subset SU(2)$.

In the Standard Model = Georgi-Glashow $\cap$ Pati-Salam example, what happens? Is this a generic intersection or a special one?

Posted by: Theo Johnson-Freyd on April 9, 2021 2:07 PM | Permalink | Reply to this

### Re: Is the intersection special or generic?

Theo wrote:

When you said that you had SU(5) ⊂ SO(10) and SO(4) × SO(6) ⊂ SO(10), I think you had in mind some specific subgroups for a specific SO(10), but a topologist or representation theorist would find it more natural just to declare the conjugacy classes of these inclusions.

I’m aware of this issue, but since this was a physics talk I didn’t want to bore and bewilder the audience by dwelling on it.

In both cases the specific subgroups chosen are the “blitheringly obvious” ones: we use the standard inclusion of $SU(5)$ in $SO(10)$ given by the usual isomorphism $\mathbb{C}^5 \cong \mathbb{R}^10$, and the usual inclusion of $SO(4) \times SO(6)$ in $SO(10)$ via block diagonal matrices with a $4 \times 4$ block and a $6 \times 6$ block.

When I wanted to discuss the issue you’re talking about, I did so by taking a 10d real inner product space and stacking structures on it until in the end it was a 5d complex inner product space with a 2+3 splitting. I described some ‘compatibility conditions’ between these structures that secretly pick out which sort of $SO(4) \times SO(6)$ subgroups of $SO(10)$ are allowed after we’ve chosen our $SU(5)$ subgroup. Namely: first we make our 10d real inner product space into a complex inner product space, and then we split it into an orthogonal 4d subspace and 6d subspace that are actually complex subspaces.

I’m not sure how generic or nongeneric this is, but it feels nongeneric, because we have to be ‘lucky’ to get the 4+6 real splitting to actually be a 2+3 complex splitting.

Posted by: John Baez on April 9, 2021 10:36 PM | Permalink | Reply to this

### Is there a quaternionic Pati-Salam?

One could just as well contemplate the embedding $S(U(2) \times U(3)) \subset SU(5) \subset Sp(5)$.

Here my topologist is showing. Topologists tend to use the name $Sp(n)$ for the rank-$n$ compact group of $Sp$ type, with Dynkin diagram $C_n$. It is a thing which acts on $\mathbb{H}^n$. [Namely, the group which commutes with the left $\mathbb{H}^\times$ action is an $\mathbb{R}_{\gt 0}$ central extension of $Sp(n)$.] Representation theorists tend to write “$Sp(n)$” only when $n$ is even, in which case it is the split form of the $Sp$-type group with rank $n/2$, and is a thing which acts on $\mathbb{R}^n$ preserving a symplectic form.

Anyway, $Sp(5)$ has the same dimension as $Spin(11)$, but really it is morally the same size as $Spin(10)$. This is because, if you have an $10$-dimensional complex vector space, then to realize it as $5$ quaternionic dimensional requires choosing an antilinear square root of $-\mathrm{id}$, whereas to realize it as $5$-real-dimensional tensored with $\bC$ requires choosing an antilinear square root of $+\mathrm{id}$.

Do you have any interesting stories to tell about splitting $\mathbb{H}^5 = \mathbb{H}^2 \times\mathbb{H}^3$ analogous to the splitting $\mathbb{R}^{10} = \mathbb{R}^4 \times \mathbb{R}^6$? Is there a “quaternionic Pati-Salam model”?

Posted by: Theo Johnson-Freyd on April 9, 2021 2:08 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Theo wrote:

Your talk this week started out with a very enticing question: Why does 5=2+3? As far as I can tell, your explanation is “Because 10=4+6”, which is less than satisfying.

That wasn’t supposed to be the explanation of anything yet, though string theorists have thought a lot about 10=4+6. I was just reviewing work people have done on massaging the Standard Model and its representation on fermions into a mathematically simple form, which is a prerequisite for understanding it in a more conceptual way. The second part of my talk will describe an octonionic “explanation” of the 10=4+6 split… but I’m not claiming this explanation really works: there are problems with it that I don’t know how to solve. It’s really just a vague sketch of the sort of thing that might eventually count as an explanation: something based on math that’s not just pretty but has some physical content.

Posted by: John Baez on April 9, 2021 10:11 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

The copy of $(SU(3) \times SU(3)) / \mathbb{Z}_3 \subset F_4$ is one of my favourite groups. There are many reasons why it is fun.

A very basic reason is that it is the centralizer of an order-$3$ element in $F_4$. Up to conjugacy, I count three order-$3$ elements of $F_4$. Two of them have nonsemisimple centralizers, i.e. centralizers whose Lie algebras contain a $U(1)$ factor. This one has a semisimple centralizer.

A second reason is that the two $SU(3)$s are very different inside $F_4$, and in particular are not conjugate to each other. One way to see this is that the roots of the two copies have different lengths. (Said another way, the Killing form for $F_4$ restricts differently to the two $\mathfrak{su}(3)$s — off by a factor of $2$, I think.) This is presumably related to your comment about the strong force versus a different one.

A third reason is that $SU(3)$ contains a copy of the finite Heisenberg group $3^{1+2}_+$ (the unique noncommutative group of order $27$ in which all nontrivial elements have order $3$) as what a quantum computer programmer would call the “clock” and “shift” matrices, but if you take the diagonal copy $diag(3^{1+2}_+) \subset SU(3) \times SU(3)$ [or maybe the antidiagonal copy, depending on what coordinates you choose for the two $SU(3)$s], then its image in $(SU(3) \times SU(3)) / \mathbb{Z}_3$ is just a $\mathbb{Z}_3^2$. Together with the residual centre of $(SU(3) \times SU(3)) / \bZ_3$, you get a copy of $\mathbb{Z}_3^3 \subset F_4$ which turns out to be maximal abelian. In particular, it does not fit inside of any torus. Serre (I believe) explained that having a finite abelian $p$-subgroup of a Lie group which does not fit inside any torus exactly corresponds to having nontrivial $p$-local cohomology.

The normalizer of this $\mathbb{Z}_3^3 \subset F_4$ is a nonsplit extension $\mathbb{Z}_3^3 \cdot SL_3(\mathbb{F}_3)$. This explains why the number $26$ shows up so much in the representation theory of $F_4$: any representation must diagonalize over $\mathbb{Z}_3^3$, but the 26 nontrivial eigenspaces must have the same dimension because they are permuted by the $SL_3(\mathbb{F}_3)$. Going the other way, Griess explained how to build the exceptional Jordan algebra, and hence $F_4$, just from this group.

This $\mathbb{Z}_3^3 \subset F_4$ is part of a trilogy. There is are analogous subgroups $\mathbb{Z}_2^3 \cdot SL_3(\mathbb{F}_2) \subset G_2$ and $\mathbb{Z}_5^3 \cdot SL_3(\mathbb{F}_5) \subset E_8$. The former can be used to provide a construction of $\mathbb{O}$ directly from $\mathbb{Z}_2^3 \cdot SL_3(\mathbb{F}_2)$. It is an easier version of Griess’s construction of the exceptional Jordan algebra and hence of $G_2$.

There seems to be no such construction of $E_8$ starting with $\mathbb{Z}_5^3 \cdot SL_2(\mathbb{F}_5)$. The problem seems to be the following. The “Weyl group” $SL_3(\mathbb{F}_p)$ suggests that you look at the determinant form $\det$ on $\mathbb{Z}_p^3$. This is an alternating trilinear form, and so, after including $\mathbb{F}_p \subset U(1)$, you get a 3-cocycle for ordinary $U(1)$-valued cohomology of $\mathbb{Z}_p^3$. It gives you as a class in $H^4(B\mathbb{Z}_p^3; \mathbb{Z})$, which I believe that it is the restriction along $\mathbb{Z}_p^3 \subset G$ of the generator of $H^4(BG; \mathbb{Z}) \cong \mathbb{Z}$, where $G = G_2,F_4,E_8$ for $p=2,3,5$. Anyway, the problem is that for $p \geq 5$, this $U(1)$-valued cocycle is not a coboundary: the class is nontrivial. For $p \leq 3$, it is a coboundary. The reason is that in three dimensions $\det$ is a sum of $3! = 6$ terms, and these terms are themselves cocycles, all of which are cohomologous in $U(1)$-cohomology. So as a cohomology class $\det$ is “divisible by 6”. Indeed, $H^3(\mathbb{Z}_p^3; U(1))$ has a copy of $\mathbb{Z}_p$ inside of it (living as the “essential cohomology”, i.e. the kernel of all restrictions to proper subgroups), and this copy is generated by “$\frac16 \det$”. The reason this is a problem for constructing $E_8$ is because the constructions of $\mathbb{O}$ and the exceptional Jordan algebra both begin by choosing an antiderivative of this specific cocycle.

Posted by: Theo Johnson-Freyd on April 9, 2021 2:46 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Neat! I think I’ve been coming at some of the same concepts from a different direction (and a background of near-total ignorance).

It looks like the Markdown processing glitched your link because it contained parentheses.

Terminological note: The “finite Heisenberg group” was conceived by Hermann Weyl in 1925, in the course of devising a finite-dimensional counterpart to the canonical commutation relation that was originally proposed by Max Born. It’s described in Weyl’s Gruppentheorie und Quantenmechanik (along with other neat things like an early recognition of the importance of entanglement). Accordingly, one will hear it called the “Weyl–Heisenberg group”; less frequently, in my experience, it is also termed the “generalized Pauli group”.

I just learned the other day that the normalizer in $SL_5(\mathbb{C})$ of the Weyl–Heisenberg group for dimension 5 is the symmetry group of the Horrocks–Mumford bundle. See He and McKay (2015), section 4. The Hesse SIC, which I care about for quantum-information reasons, has a symmetry group (the Hessian group) that is a semidirect product of the Weyl–Heisenberg group in dimension 3 with the binary tetrahedral group. It looks like the symmetry group of the Horrocks–Mumford bundle is analogous, except we go up to dimension 5 and use the binary icosahedral group instead.

Posted by: Blake Stacey on April 9, 2021 5:48 PM | Permalink | Reply to this

### Re: Can We Understand the Standard Model Using Octonions?

Theo wrote:

A very basic reason is that it is the centralizer of an order-3 element in $F_4$.

Yes, that’s great! Yokota gives a nice account of this in terms of the exceptional Jordan algebra here:

and that’s the basis of Dubois-Violette and Todorov’s work, which I’ll be discussing on Monday. Yokota constructs the compact exceptional Lie groups using octonions, and then describes their maximal subgroups as centralizers of elements of finite order… all very octonionically.

Thanks for all the other nice information about this $(SU(3) \times SU(3))/\mathbb{Z}_3$ subgroup. It could come in handy! It’s nice to hear that the two $SU(3)$’s are intrinsically different.

Posted by: John Baez on April 10, 2021 8:07 AM | Permalink | Reply to this

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