### 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:

Michel Dubois-Violette, Exceptional quantum geometry and particle physics.

Michel Dubois-Violette and Ivan Todorov, Exceptional quantum geometry and particle physics II.

Michel Dubois-Violette and Ivan Todorov, Deducing the symmetry of the standard model from the automorphism and structure groups of the exceptional Jordan algebra.

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

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

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 $\mathbb{Z}/6$.

## 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

connectedmaximal 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 $E_8$.