## August 27, 2018

### Exceptional Quantum Geometry and Particle Physics

#### Posted by John Baez

It would be great if we could make sense of the Standard Model: the 3 generations of quarks and leptons, the 3 colors of quarks vs. colorless leptons, the way only the weak force notices the difference between left and right, the curious gauge group $\mathrm{SU}(3) \times \mathrm{SU}(2)\times \mathrm{U}(1)$, the role of the Higgs boson, and so on. I can’t help but hope that all these facts are clues that we have not yet managed to interpret.

These papers may not be on the right track, but I feel a duty to explain them:

After all, the math is probably right. And they use the exceptional Jordan algebra, which I’ve already wasted a lot of time thinking about — so I’m in a better position than most to summarize what they’ve done.

Don’t get me wrong: I’m not claiming this paper is important for physics! I really have no idea. But it’s making progress on a quirky, quixotic line of thought that has fascinated me for years.

Here’s the main result. The exceptional Jordan algebra contains a lot of copies of 4-dimensional Minkowski spacetime. The symmetries of the exceptional Jordan algebra that preserve any one of these copies form a group…. which happens to be exactly the gauge group of the Standard Model!

Formally real Jordan algebras were invented by Jordan to serve as algebras of observables in quantum theory, but they also turn out to describe spacetimes equipped with a highly symmetrical causal structure. For example, $\mathfrak{h}_2(\mathbb{C})$, the Jordan algebra of $2 \times 2$ self-adjoint complex matrices, is the algebra of observables for a spin-$1/2$ particle — but it can also be identified with 4-dimensional Minkowski spacetime! This dual role of formally real Jordan algebras remains somewhat mysterious, though the connection is understood in this case.

When Jordan, Wigner and von Neumann classified formally real Jordan algebras, they found 4 infinite families and one exception: the exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$, consisting of $3\times 3$ self-adjoint octonion matrices. Ever since then, physicists have wondered what this thing is good for.

Now Todorov and Dubois–Violette claim they’re getting the gauge group of the Standard Model from the symmetry group of the exceptional Jordan algebra by taking the subgroup that

1. preserves a copy of 10d Minkowski spacetime inside this Jordan algebra, and

2. also preserves a copy of the complex numbers inside the octonions — which is just what we need to pick out a copy of 4d Minkowski spacetime inside 10d Minkowski spacetime!

But let me explain this in more detail. First, some old stuff:

If you pick a unit imaginary octonion and call it $i$, you get a copy of the complex numbers inside the octonions $\mathbb{O}$. This lets us split $\mathbb{O}$ into $\mathbb{C} \oplus V$, where $V$ is a 3-dimensional complex Hilbert space. The subgroup of the automorphism group of the octonions that fixes $i$ is $\mathrm{SU}(3)$. This is the gauge group of the strong force. It acts on $\mathbb{C} \oplus V$ in exactly the way you’d need for a lepton and a quark.

The exceptional Jordan algebra $\mathfrak{h}_3(\mathbb{O})$ contains the Jordan algebra $\mathfrak{h}_2(\mathbb{O})$ of $2 \times 2$ self-adjoint octonion matrices in various ways. $\mathfrak{h}_2(\mathbb{O})$ can be identified with 10-dimensional Minkowski spacetime, with the determinant serving as the Minkowski metric. Picking a unit imaginary octonion $i$ then chooses a copy of $\mathfrak{h}_2(\mathbb{C})$ inside $\mathfrak{h}_2(\mathbb{O})$, and $\mathfrak{h}_2(\mathbb{C})$ can be identified with 4-dimensional Minkowski spacetime.

All this is well-known to people who play these games. Now for the new part.

1) First, suppose we take the automorphism group of the exceptional Jordan algebra and look at the subgroup that preserves the splitting of $\mathbb{O}$ into $\mathbb{C} \oplus V$ for each entry of these octonion matrices. This subgroup is

$\displaystyle{ \frac{ \mathrm{SU}(3) \times \mathrm{SU}(3) } {\mathbb{Z}/3} }$

It’s not terribly hard to see why this might be true. We can take any element of $\mathfrak{h}_3(\mathbb{O})$ and spit it into two parts using $\mathbb{O} = \mathbb{C} \oplus V$, getting a decomposition one can write as $\mathfrak{h}_3(\mathbb{O}) = \mathfrak{h}_3(\mathbb{C}) \oplus \mathfrak{h}_3(V)$. One copy of $\mathrm{SU}(3)$ acts by conjugation on $\mathfrak{h}_3(\mathbb{C})$ while another acts by conjugation on $\mathfrak{h}_3(V)$. These two actions commute. The center of $\mathrm{SU}(3)$ is $\mathbb{Z}/3$, consisting of diagonal matrices that are cube roots of the identity matrix. So, we get an inclusion of $\mathbb{Z}/3$ in the diagonal of $\mathrm{SU}(3) \times \mathrm{SU}(3)$ and this subgroup acts trivially on $\mathfrak{h}_3(\mathbb{O})$.

2) Next, take the subgroup of $(\mathrm{SU}(3) \times \mathrm{SU}(3))/\mathbb{Z}/3$ that also preserves a copy of $\mathfrak{h}_2(\mathbb{O})$ inside $\mathfrak{h}_3(\mathbb{O})$. This subgroup, Dubois-Violette and Todorov claim, is

$\displaystyle{ \frac{ \mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1) } {\mathbb{Z}/6} }$

And this is the true gauge group of the Standard Model!

People often say the Standard Model has gauge group is $\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1)$, which is okay, but this group has a $\mathbb{Z}/6$ subgroup that acts trivially on all particles—a fact that arises only because quarks have the exact charges they do! So, the ‘true’ gauge group of the Standard model is the quotient $(\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1))/\mathbb{Z}/6$. And this is fundamental to the $\mathrm{SU}(5)$ grand unified theory—a well-known fact that John Huerta and I explained a while ago here. The point is that while $\mathrm{SU}(3) \times \mathrm{SU}(2)\times \mathrm{U}(1)$ is not a subgroup of $\mathrm{SU}(5)$, its quotient by $\mathbb{Z}/6$ is.

I’ll admit, I don’t fully get how

$\displaystyle{ \frac{ \mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1) } {\mathbb{Z}/6} }$

shows up inside

$\displaystyle{ \frac{ \mathrm{SU}(3) \times \mathrm{SU}(3) } {\mathbb{Z}/3} }$

as the subgroup that preserves an $\mathfrak{h}_2(\mathbb{O})$ inside $\mathfrak{h}_3(\mathbb{O})$.

I think it works like this. I described $\mathrm{SU}(3) \times \mathrm{SU}(3)$ one way, but there should be another essentially equivalent way to get two copies of $\mathrm{SU}(3)$ acting on $\mathfrak{h}_3(\mathbb{O})$. Namely, let the first copy act componentwise on each entry of your $3 \times 3$ octonionic matrix, and let the second act by conjugation on the whole matrix. In this alternative picture the $\mathbb{Z}/3$ subgroup lies wholly in the second copy of $\mathrm{SU}(3)$. Then, figure out those elements of $\mathrm{SU}(3) \times \mathrm{SU}(3)$ that preserve a copy of $\mathfrak{h}_2(\mathbb{O})$ inside $\mathfrak{h}_3(\mathbb{O})$: say, the matrices where the last row and last column vanish. All the elements of the first copy of $\mathrm{SU}(3)$ preserve this $\mathfrak{h}_2(\mathbb{O})$, because they act componentwise. But not all elements of the second copy do: only the block diagonal ones with a $2\times 2$ block and a $1 \times 1$ block. The matrices in $\mathrm{SU}(3)$ with this block diagonal form look like

$\left( \begin{array}{cc} \alpha g & 0 \\ 0 & \alpha^{-2} \end{array} \right)$

where $g \in \mathrm{SU}(2)$ and $\alpha \in \mathrm{U}(1)$. These form a group isomorphic to

$\displaystyle{ \frac{ \mathrm{SU}(2) \times \mathrm{U}(1)}{\mathbb{Z}/2} }$

If all this works out, it’s very pretty: the 2 and the 1 in $\mathrm{SU}(2) \times \mathrm{U}(1)$ arise from the choice of a $2 \times 2$ block and $1 \times 1$ block in $\mathfrak{h}_3(\mathbb{O})$… which is also the choice that lets us find Minkowski spacetime inside $\mathfrak{h}_3(\mathbb{O})$.

But I need to check some things, like how we get the $\mathbb{Z}/6$.

Posted at August 27, 2018 5:26 AM UTC

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### Re: Exceptional Quantum Geometry and Particle Physics

$(SU(3) \times SU(3))/(diag(\mathbb{Z}/3))$ is not isomorphic to $SU(3) \times PSU(3)$, so your “I think it works like this” isn’t quite right. It is true that the standard embedding $SU(2) \subset SU(3)$ has centralizer a $U(1)$ living as $diag(\alpha,\alpha,\alpha^{-2})$, and the combination is an $(SU(2) \times U(1)) / (\mathbb{Z}/2) \subset SU(3)$.

There aren’t that many ways to form a group of shape $(SU(3) \times SU(2) \times U(1))/(\mathbb{Z}/6)$. You need to map $\mathbb{Z}/6 = \mathbb{Z}/3 \times \mathbb{Z}/2$ into the center of $SU(3) \times SU(2) \times U(1)$, which is a $\mathbb{Z}/3 \times \mathbb{Z}/2 \times U(1)$. Assuming you want your final answer not to be a product, your map must be nonzero on all three terms. There are two non-zero maps $\mathbb{Z}/6 \to \mathbb{Z}/3$, but they are related by an outer automorphism of $SU(3)$; there is one non-zero map $\mathbb{Z}/6 \to \mathbb{Z}/2$; and when I said “nonzero” I meant prime-by-prime, so I need the unique (up to outer automorphism) injection $\mathbb{Z}/6 \to U(1)$. So there really is no ambiguity in the expression “$(SU(3) \times SU(2) \times U(1))/(\mathbb{Z}/6)$”.

This discussion of $(SU(3) \times SU(3))/(diag(\mathbb{Z}/3))$ reminds me of a puzzle that I learned from David Treumann. The puzzle involves the three most exceptional groups: $G_2 = Aut(\mathbb{O})$, $F_4 = Aut(\mathfrak{h}_3(\mathbb{O}))$, and $E_8$. Among the ways that they are “most exceptional” is that they are the only groups whose simply-connected forms have no center.

There are many constructions of $\mathbb{O}$, and hence many constructions of $G_2$. In particular, it is easy to construct $G_2$ in such a way as to make manifest a maximal abelian subgroup $2^3 \subset G_2$. (By the group “$p^n$”, I always mean the elementary abelian group of that order: $(\mathbb{Z}/p)^n$.) See, Cartan subgroups are maximal abelian, and are the unique connected maximal abelian subgroups of compact Lie groups, but they are not the unique maximal abelian subgroups. A finite group is maximal abelian iff the corresponding grading of the Lie algebra has nothing in degree zero. Let me investigate this $2^3$ a bit more. Its “Weyl group”, i.e. its normalizer mod centralizer, is an $SL_3(\mathbb{F}_2)$ acting in the obvious way. Choose any $2 \subset 2^3$. Then its centralizer in $G_2$ has shape $(SU(2) \times SU(2)) / diag(2)$. (Note that this group still has center of order $2$.) If you take a complementary $2^2 \subset (SU(2) \times SU(2)) / diag(2)$, it does not lift to $SU(2) \times SU(2)$, but rather its preimage is a quaternion group $Q_8 = 2^{1+2}$. A $2^3 \subset G_2$ grades the Lie algebra, and so you get a system of subalgebras corresponding to a flag in $G_2$, which goes $\mathfrak{c} \subset \mathfrak{su}(2) \times \mathfrak{su}(2) \subset \mathfrak{g}_2$, where $\mathfrak{c}$ is the Cartan.

It turns out that there are similar pictures of $F_4$ and $E_8$. Namely, you’ve already mentioned that $F_4$ has a subgroup $(SU(3) \times SU(3)) / diag(3)$. Similarly, $E_8$ contains a $(SU(5) \times SU(5))/diag(5)$. (Warning: there are two central maps $5 \to SU(5)$ up to outer automorphism; you need to use both of them.) These subgroups can be found by centralziing regular elements of orders $3$ and $5$ respectively. There are interesting copies of the extraspecial groups $3^{1+2}$ and $5^{1+2}$ inside $SU(3)$ and $SU(5)$ respectively. If you take a certain “antidiagonal” copy inside the product, you find a copy of $3^2 \subset (SU(3) \times SU(3)) / diag(3)$ or $5^2 \subset (SU(5) \times SU(5))/diag(5)$, and the results give subgroups $3^3 \subset F_4$ and $5^3 \subset E_8$ that turn out to be maximal abelian. The Weyl groups are $SL_3(\mathbb{F}_3)$ and $SL_3(\mathbb{F}_5)$. The gradings on the Lie algebras give “flags” $\mathfrak{c} \subset \mathfrak{su}(3) \times \mathfrak{su}(3) \subset \mathfrak{f}_4$ and $\mathfrak{c} \subset \mathfrak{su}(5) \times \mathfrak{su}(5) \subset \mathfrak{e}_8$.

Greiss found a construction of $F_4$ that makes its maximal abelian $3^3$ manifest in A Moufang loop, the exceptional Jordan algebra, and a cubic form in 27 variables. The puzzle is to do the same for $E_8$.

Posted by: Theo Johnson-Freyd on August 27, 2018 3:20 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Here is a hint to the puzzle. I say “hint” except that the puzzle is an open problem, so I suspect it is a hint but I don’t know.

Consider an elementary abelian group $p^3$. The “determinant” defines a 3-cocycle with $\mathbb{Z}/p$ coefficients on $p^3$, and so a 3-cocycle with $U(1)$-coefficients.

The well-known construction of $\mathbb{O}$ and also Griess’s construction of $\mathfrak{h}_3(\mathbb{O})$ begin by choosing antiderivates for the determinant, i.e. 2-cochains whose derivative is the determinant.

This is something you can only do when $p=2,3$. Indeed, the determinant represents a nontrivial class in $H^4(B5^3)$.

This is surely related to the fact that, whereas the smallest representations of $G_2$ and $F_4$ are $(p^3-1)$-dimensional (and are in fact the regular representations of the maximal abelian $p^3$ subgroups), the smallest representation of $E_8$ is the adjoint (in all three cases, the adjoint is $2(p^3-1)$-dimensional). It’s as if the determinant is nontrivial in $H^4(B5^3)$, but after somehow doubling the dimensions it becomes trivial. For instance, is (complex Lie algebra) $\mathfrak{e}_8$ somehow 124-dimensional over $\mathbb{H}$?

In my other post, I mentioned my love for levels of Lie groups. I think, but now can’t find my notes, that the generating levels of $G_2$ and $F_4$ restrict trivially to the $2^3$ and $3^3$ subgroups, whereas for $E_8$ the generating level restricts nontrivially. So the better thing is to say that in all cases, the generator of $H^4(BG)$ restricts to the “determinant” as a class in $H^4(Bp^3)$.

I’m pretty sure the extension $5^3.SL_3(5) \subset E_8$ is non-split. I don’t remember about $G_2$ or $F_4$. Compare that the extension $(maximal torus).Weyl \subset E_8$ also does not split.

Posted by: Theo Johnson-Freyd on August 27, 2018 5:43 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

My “I’m pretty sure” is wrong. According to Finite simple groups which projectively embed in an exceptional Lie group are classified! the extension $5^3.SL_3(5) \subset E_8$ is split.

Posted by: Theo Johnson-Freyd on August 29, 2018 7:01 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Robert L. Griess Jr. and A. J. E. Ryba Jr., Finite simple groups which projectively embed in an exceptional Lie group are classified!, Bull. Amer. Math. Soc. 36 (1999), 75-93 https://doi.org/10.1090/S0273-0979-99-00771-5

I guess it’s presented as a musical?

Posted by: David Roberts on August 30, 2018 2:22 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Thanks for that intense blast of erudition, Theo! It will take me a while to recover, but this stuff is all very good to know, and potentially useful for physics games.

your “I think it works like this” isn’t quite right.

Okay, thanks—I’ll have to straighten that out sometime.

There aren’t that many ways to form a group of shape $(SU(3) \times SU(2) \times \mathrm{U}(1))/(\mathbb{Z}/6)$.

That’s good to know. There is exactly one ‘correct’ one for the purposes of Standard Model physics, and that’s $\mathrm{S}(\mathrm{U}(3) \times \mathrm{U}(2))$. This is the subgroup of $\mathrm{SU}(5)$ consisting of block diagonal matrices with a $3\times 3$ block and a $2 \times 2$ block. Equivalently, it’s the group

$\{ (g,h) : \det(g) \det(h) = 1 \} \subset \mathrm{U}(3) \times \mathrm{U}(2).$

The physical reasons why this is the ‘true gauge group of the Standard Model’ are described in loving detail in here, along with the homomorphism from $SU(3) \times SU(2) \times \mathrm{U}(1)$ onto this group, and its kernel.

Assuming you want your final answer not to be a product [….]

Right: the physically relevant quotient group $(SU(3) \times SU(2) \times \mathrm{U}(1))/(\mathbb{Z}/6)$ is not a product. This is not out of some distaste for products, but because the Standard Model compels it!

Similarly, $\mathrm{E}_8$ contains a $(SU(5) \times SU(5))/diag(5)$.

Nice! Kostant mentioned that in his talk on $\mathrm{E}_8$ at U.C. Riverside in 2008. He also pointed out this:

There is an element of order 11 in $SL(2,32) \subset \mathrm{E}_8$. The centralizer of this element in $\mathrm{E}_8$ is a product of two copies of the Standard Model gauge group:

$\mathrm{S}(\mathrm{U}(3) \times \mathrm{U}(2)) \times \mathrm{S}(\mathrm{U}(3) \times \mathrm{U}(2)).$

I am now beginning to hope that this ‘doubled Standard Model gauge group’ is somehow related to this fact:

$\mathrm{F}_4$ is the isometry group of the octonionic projective plane, built using $\mathbb{O}$, while $\mathrm{E}_8$ is the isometry group of the so-called ‘octooctonionic projective plane’, built using $\mathbb{O} \otimes \mathbb{O}$.

We’re seeing a nice way to find the Standard Model gauge group in $\mathrm{F}_4$. Perhaps somehow ‘doubling’ this construction gives the doubled Standard Model gauge group in $\mathrm{E}_8$.

But it’s all quite mysterious to me.

By the way, Kostant also explained how this $SL(2,32)$ acts to permute 31 different ‘pieces of eight’: Cartan subalgebras of $\mathrm{E}_8$, whose direct sum is all of $\mathrm{E}_8$:

$31 \times 8 = 248 .$

Posted by: John Baez on August 28, 2018 7:44 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Right: the physically relevant quotient group ($(SU(3) \times SU(2) \times \mathrm{U}(1))/(\mathbb{Z}/6)$ is not a product. This is not out of some distaste for products, but because the Standard Model compels it!

The Standard Model does not compel it. $Spin(10)$ unification (for instance) compels it, but the Standard Model is perfectly consistent if the gauge group is $SU(3) \times SU(2) \times \mathrm{U}(1))/\Gamma$ for any choice of $\Gamma\in\{\mathbb{1}, \mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_6\}$.

Different choices of $\Gamma$ change the quantization of monopole charges, the spectrum of line-operators and the quantization of the theory on a nontrivial 4-manifold.

I would suggest looking at section 3.4 of 1808.00009 for a recent discussion of (potential) global anomalies, which might restrict the choice of $\Gamma$.

I may be prejudiced, but the results in that paper strike me as vastly more interesting.

Posted by: Jacques Distler on August 31, 2018 10:19 PM | Permalink | PGP Sig | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

John wrote:

Right: the physically relevant quotient group ($(\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1))/(\mathbb{Z}/6)$ is not a product. This is not out of some distaste for products, but because the Standard Model compels it!

Jacques wrote:

The Standard Model does not compel it. $Spin(10)$ unification (for instance) compels it, but the Standard Model is perfectly consistent if the gauge group is

$SU(3) \times SU(2) \times \mathrm{U}(1))/\Gamma$

for any choice of $\Gamma\in\{\mathbb{1}, \mathbb{Z}_2,\mathbb{Z}_3,\mathbb{Z}_6\}$.

Your second sentence is true of course, but your first suggests you misunderstood my comment. Theo and I were talking about groups of the form $(SU(3) \times SU(2) \times \mathrm{U}(1))/(\mathbb{Z}/6)$. He noted that there is more than one group of this form, because $\mathbb{Z}/6$ shows up as a normal subgroup of $SU(3) \times SU(2) \times \mathrm{U}(1)$ in more than one way, but only one group of this form that’s not a product. I concurred and said that this one was the only one compatible with the Standard Model.

My remark was not true unless you fix, ahead of time, the physical meaning of the $\mathrm{U}(1)$ generator as weak hypercharge, e.g. by decreeing that the eigenvalue of the corresponding self-adjoint operator on the electron neutrino is -1. I was implicitly doing that.

For example, $\mathrm{U}(1)/(\mathbb{Z}/6) \cong \mathrm{U}(1)$, so if we stick the $\mathbb{Z}/6$ subgroup entirely in the $\mathrm{U}(1)$ factor we get

$(SU(3) \times SU(2) \times \mathrm{U}(1))/(\mathbb{Z}/6) \quad \cong \quad \mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1)$

which is of course a fine gauge group for the Standard Model!

Posted by: John Baez on September 1, 2018 4:08 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

There is, indeed, a unique maximal subgroup, $\Gamma_{\text{max}}$, of the center of $G= SU(3)\times SU(2)\times U(1)$ which acts trivially on the matter representation, $R$, of $G$. That subgroup is isomorphic to $\mathbb{Z}_6$.

$R$ thus descends to a representation of $G/\Gamma$ for any $\Gamma\subset \Gamma_{\text{max}}$. That suffices classically.

The cool result of Garcia-Etxebarria and Montero is that it suffices in the full quantum theory as well.

Posted by: Jacques Distler on September 1, 2018 8:15 AM | Permalink | PGP Sig | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Just to be clear, $G_{\text{min}}=(SU(3)\times SU(2)\times U(1))/\Gamma_{\text{max}}$ is the smallest Standard Model gauge group compatible with what we know. Modding out by any larger group, $G=(SU(3)\times SU(2)\times U(1))/\Gamma$ for $\Gamma\subset\Gamma_{\text{max}}$ is a bit of an act of hubris, in the sense that it (further) restricts the possible UV completions of the theory.

In particular, it eliminates some of your favourite simple GUT gauge groups. But there are other possibilities. For instance, back in the day, Shelly Glashow wrote a paper where $G_{\text{GUT}}= ({SU(3)}^3)/\mathbb{Z}_3$ (where the quotient is by the diagonal subgroup of the $\mathbb{Z}_3^2$ center).

Posted by: Jacques Distler on September 18, 2018 7:11 AM | Permalink | PGP Sig | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Jacques wrote:

Modding out by any larger group,…

I think you meant to type “Modding out by any smaller group,…”

Posted by: John Baez on September 18, 2018 6:10 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

I think you meant to type “Modding out by any smaller group,…”

No, I meant what I wrote.

The gauge group describes the redundancy of our description of the QFT. It is what we mod out by, to get the true physics.

$SU(3)\times SU(2)\times U(1)$ is a bigger group than $(SU(3)\times SU(2)\times U(1))/\Gamma_{\text{max}}$. Modding out by the larger group is an assertion: that our description of the physics has a larger gauge redundancy.

Whether or not that is true is a question to which we do not currently know the answer.

Posted by: Jacques Distler on September 18, 2018 8:44 PM | Permalink | PGP Sig | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Okay - when you wrote “modding out by any bigger group” I thought you were referring to $\Gamma$, since that’s what you were modding out by in the quotient group you wrote down.

Posted by: John Baez on September 19, 2018 1:20 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

I gather this is distinct from Furey’s ideas, which I also haven’t had time to evaluate?

Posted by: Blake Stacey on August 27, 2018 4:39 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

I believe that the truly exciting fact here — the realization that the Standard Model gauge group is the subgroup of the symmetry group of the exceptional Jordan algebra that preserves a copy of 4d Minkowski spacetime — is brand new, and not to be found in Furey’s work.

Some of the raw ingredients go back a long way and are used both by Furey and Dubois-Violette–Todorov. For example, I believe this paper:

• Murat Günaydin and Feza Gürsey, Quark structure and octonions, J. Math. Phys. 14 (1973), 1651–1667.

looked at how $SU(3) \subset \mathrm{G}_2$ acts on the octonions, breaking them up into a 3d ‘quark’ representation and a 1d ‘colorless’ representation. This idea seems to be used in every attempt to connect the octonions to the Standard Model. But there are lots of ideas that Furey is using that Dubois-Violette and Todorov are not, and also vice versa.

I should add that Cohl Furey is exploring a number of different ideas in different papers, not just one idea.

Posted by: John Baez on August 28, 2018 7:57 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Thanks!

Potential applications of octonions to physics are at least adjacent to a topic that I have an official reason to study (i.e., a topic I have a grant to work on), so I really should learn my way around them.

(Is anyone else having trouble accessing math.ucr.edu? It seems to be down for me.)

Posted by: Blake Stacey on September 3, 2018 3:51 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

I’m at the CQT in Singapore, and I’ve not been paying much attention to math.ucr.edu, but John Huerta also says it’s been down for a while — and not just

http://math.ucr.edu/home/baez

but also the main math department website (on a different server, I think) at

http://math.ucr.edu

I’ve sent an email asking people to fix these. Thanks!

Posted by: John Baez on September 6, 2018 4:02 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Here’s the news:

Hi, Dr. Baez. I sent out an email yesterday and the server is still down from the power outage.

Your website is on the main math server. mathdept.ucr.edu is not on the math server which is why it is up. It is hosted by the university.

Hopefully we will have word of it going up soon.

Posted by: John Baez on September 7, 2018 3:56 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

I haven’t read the papers you linked. It is clear that they have found a copy of the Standard Model Gauge Group inside of $F_4$ as the centralizer of an element of order $6$. (Whence the flavor symmetry?)

If we were talking about 3d rather than 4d gauge theory, I would tell you the following story. A major ingredient in 3d gauge theory is the “level”, which for $G$ simply connected lives in $H^4(BG;\mathbb{Z})$. (When $G$ is not simply connected, the correct thing seems to be to use not ordinary cohomology, but rather generalized cohomology with coefficients in the Anderson dual to the sphere spectrum.) Any map of groups $G' \to G''$ induces a map $H^4(BG'') \to H^4(BG')$ of levels. Recall that if $G$ is simple simply-connected, then $H^4(BG) \cong \mathbb{Z}$; the generator is called “level 1”, and $c_2(adjoint rep)$ is twice the dual coxeter number (or maybe minus that).

I mentioned above an embedding $(SU(2) \times SU(2))/2 \subset G_2$. If I’m calculating correctly, the induced map $H^4(BG_2) \to H^4(BSU(2)) \times H^4(BSU(2))$ takes the generator to the pair $(2,6) \in \mathbb{Z}^2$. I haven’t done the exact $F_4$ computation, but it should’t be too hard. I do know that $H^4(BF_4) \to H^4(BSU(3))^2$ takes the generator to some multiple of $(1,2)$. In the $E_8$ case in my puzzle, the induced map $H^4(BE_8) \to H^4(BSU(5))^2$ takes the generator to $(1,1)$. The ratios $6/2$ and $2/1$ are the ratios of long to short roots in $G_2$ and $F_4$ respectively, and the ratio $1/1$ is the fact that $E_8$ is simply-laced.

If you had handed me $(SU(2) \times SU(2))/2$, I may or may not have handed you back $G_2$. (I mean, I probably would have, personally, because I’m stuck on my puzzle.) For instance, I might have instead said “this is $SO(4)$, what more do you want?”. But if you had handed it to me not just as a group, but as a group equipped with the level $(2,6)$, then I probably would have given you $G_2$ back, because the most natural levels for $SO(4)$ are the multiples of $(1,1)$. Specifically, you could hand me $(SU(2) \times SU(2))/2$ with the level $(2,6)$ and ask me to find the “simplest” overgroup $G \supset (SU(2) \times SU(2))/2$ such that $(2,6)$ is the restriction of a level of $G$. The unique answer to this is $G_2$, if I am not mistaken. Here’s how you do it: build $(SU(2) \times SU(2))/2$ gauge theory with level $(2,6)$, take its low-energy effective TFT, look for an irreducible chiral conformal boundary condition, and work out the symmetry group of the boundary condition.

Do “levels” show up also in the Standard Model? Something about the choice of matter fields, for example? In particular, is there a sense in which the levels of $(SU(3) \times SU(2) \times U(1))/6$ are some prescribed things? The map $SU(2) \times U(1) \to SU(3)$ takes the generator of $H^4(BSU(3))$ to $(1,3) \in H^4(BSU(2)) \times H^4(BU(1))$ if I am not mistaken. And the two $SU(3)$s in $F_4$, of course, have levels differing by a factor of $2$. So if you are handed $(SU(3) \times SU(2) \times U(1))/6$ with levels in the ratio $(1:2:6)$ or $(2:1:3)$, then you should be motivated to build $F_4$. But if you are handed some other levels, then you should reject $F_4$.

Posted by: Theo Johnson-Freyd on August 27, 2018 5:09 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Thanks for this great comment, Theo! It will take me a while to digest it, but I’ll start with something simple:

Do “levels” show up also in the Standard Model?

I haven’t heard people talk about them in the Standard Model. Maybe they do in attempts to obtain the Standard Model as the low-energy limit of a string theory, since that would bring conformal field theories into the game. But I’m not very familiar with that literature, so I can’t say for sure.

What definitely shows up in the Standard Model is a favored representation of the group

$\displaystyle{ \frac{ \mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1) } {\mathbb{Z}/6} }$

namely, the representation on all the fermions and antifermions in one generation. To describe this representation it’s easiest to use the isomorphism

$\displaystyle{ \frac{ \mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1) } {\mathbb{Z}/6} } \cong \mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3))$

where $\mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3))$ is the subgroup of $\mathrm{SU}(5)$ consisting of block diagonal matrices with a $2 \times 2$ block and a $3 \times 3$ block. In these terms, the favored representation comes from taking the obvious representation of $\mathrm{SU}(5)$ on $\Lambda \mathbb{C}^5$ and restricting it to $\mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3))$.

This raises a couple of questions. Can you see any connection between this representation and “levels”? Does this description of the Standard Model gauge group as $\mathrm{S}(\mathrm{U}(2) \times \mathrm{U}(3))$ make you have any ideas?

If the answers are both “no”, we might need to get conformal field theory into the game for levels to show up.

Posted by: John Baez on August 28, 2018 4:27 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Probably my hope to be able to tell the difference between, say, $F_4 \supset G$ and $SU(5) \supset G$ by looking at levels will not go anywhere. There is a way to turn any representation into a level: a (complex, say) representation is a map $G \to U(N)$ for some $N$, and you can pull back $c_2 \in H^4(BU(N))$ along this map. I was going to say that in the case of $\Lambda \mathbb{C}^5$, you got zero, because the even parts cancel the odd parts (if you have a super representation, you should take the difference of $c_2$s), but that was a misunderstanding: the matter fields are all fermions. So at the end of the day, this gives a level which can be pulled back from $SU(5)$, and hence not from $F_4$. This means in particular that the representation you like of the gauge group does not extend to $F_4$, unless I have made a mistake somewhere.

Posted by: Theo Johnson-Freyd on August 28, 2018 3:28 PM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Theo wrote:

This means in particular that the representation you like of the gauge group does not extend to $\mathrm{F}_4$, unless I have made a mistake somewhere.

That sounds correct.

I can ‘cheat’ and look up the dimensions of the irreps of $\mathrm{F}_4$, which go

$1 , 26, 52, \dots,$

and that’s enough to show that no $16$- or $32$-dimensional representation of any subgroup of $\mathrm{F}_4$ extends to all of $\mathrm{F}_4$ unless it has a subspace on which $\mathrm{F}_4$, and thus that subgroup, acts trivially.

So, the representation of $\mathrm{S}(\mathrm{U}(3) \times \mathrm{U}(2))$ on $\Lambda \mathbb{C}^5$ or $\Lambda^{\mathrm{ev}} \mathbb{C}^5$ can’t extend to a representation of $\mathrm{F}_4$.

Luckily this was never a hope of mine. If you’re trying to extend that rep to a larger group, you’re playing the ‘grand unified theory’ game, and the most popular candidates for the larger group are $\mathrm{SU}(5)$, $\mathrm{SU}(4) \times \mathrm{SU}(2) \times \mathrm{SU}(2)$, and their common supergroup $\mathrm{Spin}(10)$. These give rise to nice grand unified theories some of which (contrary to rumor) aren’t quite dead yet — claims that they were killed by proton decay experiments turn out to be exaggerated.

If you’re trying to get the exceptional Jordan algebra into the game, and take advantage of this new result by Dubois-Violette and Todorov, it seems one needs to go down some other road… and what this road might be, nobody knows. That’s why I expressed so much caution at the start of my post.

Posted by: John Baez on August 29, 2018 2:21 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Yes, the main problem with connecting this to physics is that the Standard Model fermions aren’t in the $26$ irrep of $F_4$, (the exceptional Jordan Algebra). One does, as you say, have to go to a larger group, such as $E_6$, for this. If your heart is set on the exceptional Jordan algebra, and gravity, one can put the graviweak force in the $so(4,4)$ subalgebra of $f_4$ and the strong force in a $g_2$, so everything works in $f_4 + g_2 + 26 \otimes 7 = e_8$ but then, as well as one generation of Standard Model fermions, one gets mirror fermions. Fortunately, unless you’re Jacques, you can add another $so(8)$ and get rid of the mirror fermions and also get three generations, possibly in something like $e_12$.

Posted by: Garrett on September 16, 2018 1:15 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Urs Schreiber points out that the subgroup

$\frac{\mathrm{SU}(3) \times \mathrm{SU}(3)}{\mathbb{Z}/3} \subset \mathrm{Aut}(\mathfrak{h}_3(\mathbb{O})) \; \;\colon\!\!= \mathrm{F}_4$

is characterized in Section 2.12.2 of

In Theorem 1.9.4, Yokota looks at a $3 \times 3$ diagonal matrix in $\mathrm{SU}(3)$ that’s a cube root of unity:

$w = \mathrm{diag}(\omega,\omega,\omega) \in \mathrm{SU}(3) \subset \mathrm{Aut}(\mathbb{O}) \;\; \colon\!\! = \mathrm{G}_2$

where

$\omega = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \in \mathbb{O},$

$i$ being our chosen unit imaginary octonion. He shows that the subgroup of $\mathrm{Aut}(\mathbb{O}) = \mathrm{G}_2$ consisting of elements that commute with $w$ is the same as the subgroup that commute with multiplication by $i \in \mathbb{O}$, namely $\mathrm{SU}(3)$.

This should come as no surprise, since an element of $\mathbb{O}$ commutes with $i$ iff it commutes with $\omega$. But still, there’s a bit to check.

In Section 2.12.2 he treats $w$ as an element of $\mathrm{F}_4$ using the standard embedding

$\mathrm{G}_2 \subset \mathrm{F}_4.$

He shows that the subgroup of $\mathrm{F}_4$ consisting of elements that commute with $w$ is

$\frac{\mathrm{SU}(3) \times \mathrm{SU}(3)}{\mathbb{Z}/3}.$

So, there’s just a small argument required to get from here to the claim I called 1) in my post: one has to check that the elements of $\mathrm{F}_4$ commuting with $w$ is the same as the subgroup that preserves the splitting

$\mathfrak{h}_3(\mathbb{O}) = \mathfrak{h}_3(\mathbb{C}) \oplus \mathfrak{h}_3(V)$

arising from the splitting $\mathbb{O} = \mathbb{C} \oplus \mathrm{V}$. But this should be fairly easy, since $\mathbb{C} \subseteq \mathbb{O}$ is precisely the subalgebra of elements that commute with $\omega$.

Posted by: John Baez on August 28, 2018 3:47 AM | Permalink | Reply to this

### Re: Exceptional Quantum Geometry and Particle Physics

Taking the coincident brane perspective, where $n$ branes acquire enhanced $U(n)$ symmetry on the worldvolume, one expects 3+2+1=6 branes behind the SM gauge symmetry. (Picture three worldvolumes, with three, two and one coincident branes, respectively.) From noncommutative geometry, there are three algebras, say $A_1\oplus A_2\oplus A_3$ of “noncommutative functions” over the 3+2+1=6 points. The $\mathbb{Z}_6$ symmetry geometrically implies one can freely permute the branes, between the three worldvolumes.

As the points are usually assigned to pure states, it is sensible to assign primitive idempotents of (formally real) Jordan algebras to them.

Posted by: Metatron on August 31, 2018 3:42 AM | Permalink | Reply to this

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