# Exceptional Quantum Geometry and Particle Physics

• A
• jeffery_winkler
In summary, the article discusses how they get the gauge group of the Standard Model from the symmetry group of the exceptional Jordan algebra. They are talking about SU(3) x SU(2) x U(1)/Z/6 instead of SU(3) x SU(2) x U(1).

#### jeffery_winkler

What do you think about the article

Exceptional Quantum Geometry and Particle Physics

and the following discussion? It also includes links to the original papers.

Exceptional quantum geometry and particle physics

https://arxiv.org/abs/1604.01247

Exceptional quantum geometry and particle physics II

https://arxiv.org/abs/1808.08110

It talks about how they get the gauge group of the Standard Model from the symmetry group of the exceptional Jordan algebra. They are talking about SU(3) x SU(2) x U(1)/Z/6 instead of SU(3) x SU(2) x U(1).

Baez writes

"The exceptional Jordan algebra contains a lot of copies of 4-dimensional Minkowski spacetime. The symmetries of the exceptional Jordan algebra that preserve anyone of these copies form a group…. which happens to be exactly the gauge group of the Standard Model!"

I think the appropriate response might be - so what? OK, they found that group. But it's not functioning as a gauge group, and I don't see a way to augment the relationship so that it gives rise to gauge fields on one of those "copies of 4-dimensional Minkowski spacetime".

There are numerous relationships between exceptional algebraic objects and groups appearing in physics. For example, SU(5) is a simple group containing the SM gauge group, and E8 contains SU(5) x SU(5). You can probably find some subobject of E8 that will pick out the SM group as its complement.

Urs Schreiber
In more detail, the idea rests on the series of group inclusions F4 (SU(3)xSU(3))/Z3 (SU(3)xSU(2)xU(1))/Z6.

F4 is the automorphism group (group of mappings to itself) of h3(O), the algebra of 3x3 octonion-valued matrices.

(SU(3)xSU(3))/Z3 is the subgroup of F4 which maps the subalgebra h2(O) to itself.

And (SU(3)xSU(2)xU(1))/Z6 is a subgroup of that, which maps a subsubalgebra h2(C) to itself.

((edit: That was wrong, that's not how it works... See this tweet by Baez. Basically,

(SU(3)xSU(3))/Z3 is the symmetry group for "h3(C+C^3)", h3(O) with a preferred imaginary unit for each octonion,

and then (SU(3)xSU(2)xU(1))/Z6 is a subgroup of that which picks out an h2(O) within h3(O), but a preferred imaginary unit makes that "h2(C+C^3)", and so there is a preferred h2(C) within the h2(O).))

Further, h3(O) has 27 dimensions, h2(O) has 10 dimensions and can be given a Minkowski metric, and h2(C) has 4 dimensions and can be given a Minkowski metric.

So you could say this is reminiscent of string theory. Bosonic string theory has 26 dimensions (and the speculated "bosonic M-theory" would have 27 dimensions), superstring theory has 10 dimensions, and the observable world has 4 dimensions.

There is a proposal (Pierre Ramond, Hisham Sati) that there might be a 27-dimensional theory consisting of a 16-dimensional space fibered over 11-dimensional M-theory. The 16-dimensional space is OP2, the octonionic projective plane or Cayley plane. It also has the automorphism group F4, and there are various ways to obtain OP2 as a quotient of h3(O).

So as usual, one way to try to give this idea more content, is to make it into string theory.

Last edited:
Urs Schreiber
mitchell porter said:
I think the appropriate response might be - so what? OK, they found that group. But it's not functioning as a gauge group, and I don't see a way to augment the relationship so that it gives rise to gauge fields on one of those "copies of 4-dimensional Minkowski spacetime".

Right, and in fact it doesn't. I had made that point here.

I wish people were more careful with claims about octonions in fundamental physics. After all, they do underly some good physics, notably the existence of the M2-brane cocycle in 11d supergavity. (neatly reviewed in arxiv.org/abs/1003.3436).

arivero
Is there in the literature some model where the 27-dim exceptional Jordan Algebra can be interpreted as 15 quarks and 12 leptons?

It could be some "five quark model" without the top and three colours, but also simply a 3 generations "uncoloured" model with three extra particles. I am not sure if the model of Dubois Violette interprets their own three extra particles as quarks or leptons.

Last edited:
arivero said:
I am not sure if the model of Dubois Violette interprets their own three extra particles as quarks or leptons.

Ah ok, I see, that there is not such a thing as "the model" but two different models, one we had discussed time ago, https://arxiv.org/abs/1604.01247v2, having three extra particles, and the one trying just to play with triality to justify the three generations, and where the diagonal is somehow related to the intersection of any two generations, but it has not any particular particle associated to it.

It should be interesting to consider some alternative where the Albert algebra only provides part of of the standard model but always three generations. For instance, one could consider to break E6 down to SU(6)xSU(2) so that we have 15 "B-number" and 12 "L-number" particles but all of them uncolored, and then assign supercharges from some sequence of successive breakings, say SU(6) to SU(3)xSU(3)xU(1) and them at least one of the SU(3) down to SU(2)xU(1). This would be different from DV Model 1 because here the extra three particles are "B-number", and should be coloured in the complete schema.

Other breakings are surely possible, depending of our likes, either if we want keep at least one SU(3) to make sure we have generation symmetry or if we just worry about producing the SM hypercharges.

Last edited:
Let be think how we could obtain three generations of a "decolored standard model" from the conjugate irreducible representations of E6, 27 + \bar 27.

First we decompose the 27 of E6 into some 15+12 split, to allow for an -arbitrary- B and L quantum number. The most obvious start is to keep all the stuff together in single irreducibles, so the best bet is $SU(6) \times SU(2)$

$$\begin{array}{|c||c|} \hline & SU(6) \times SU(2) \\ \hline L & (\bar 6,2) \\ \hline B & (15,1) \\ \hline \end{array}$$

Where, inspired by the sBootstrap, I have decided to consider that the 12 are leptons while the 15 are decoloured quarks.

Now let's try some path to assign hypercharges. Say, start with SU(6) branching to su3su3u1

$$\begin{array}{|c |r ||c|} \hline & Y_1 & SU(3) \times SU(3) \times SU(2) \times U(1)_1\\ \hline L & -1 &( \bar 3,1,2) \\ \hline L & 1 &( 1, \bar 3,2) \\ \hline B & 0 & (3,3,1) \\ \hline B & 2 & (\bar 3,1,1) \\ \hline B & -2& (1,\bar 3,1) \\ \hline \end{array}$$

Now one su3 down to su2u1:

$$\begin{array}{|c |r |r ||c|} \hline & Y_1 & Y_2 & SU(3) \times SU(2)_T \times SU(2)_S \times U(1)_1 \times U(1)_2\\ \hline L & -1 &0 &( \bar 3,1,2) \\ \hline L & 1 & -1 &( 1, 2 ,2) \\ \hline L & 1 &2 &( 1, 1 ,2) \\ \hline B & 0 &1& (3,2,1) \\ \hline B & 0 &-2& (3,1,1) \\ \hline B & 2 &0& (\bar 3,1,1) \\ \hline B & -2&-1& (1,2,1) \\ \hline B & -2&2& (1,1,1) \\ \hline \end{array}$$

And here we could for instance do a charge assignment for the quark sector of the standard model, three generations.

$$\begin{array}{|c |r |r ||c|| c | l } \hline & Y_1 & Y_2 & su3su2su2 & \frac 13 Y_1 + \frac 16 Y_2 + T \\ \hline L & -1 &0 &( \bar 3,1,2) &-1/3 \\ \hline L & 1 & -1 &( 1, 2 ,2)& +2/3 \\ & & && -1/3 \\ \hline L & 1 &2 &( 1, 1 ,2) & +2/3 \\ \hline B & 0 &1& (3,2,1) & +2/3 & u, c, t\\ & && & -1/3 & d,s,b\\ \hline B & 0 &-2& (3,1,1) & - 1/3 & d,s,b\\ \hline B & 2 &0& (\bar 3,1,1) & +2/3 & u,c,t\\ \hline B & -2&-1& (1,2,1) & - 1/3\\ & && & - 4/3\\ \hline B & -2&2& (1,1,1) & -1/3\\ \hline \end{array}$$

Which is midly satisfying but 1) it is not so right in the lepton sector, and 2) the three extra "quarks" have different charges; the sBootstrap predicted all the three to be Q=-4/3.

Surely the whole point is to try other assignments, even to keep breaking subgroups to get a finer set of hypercharges. But still let me note that the lepton sector can be arguably fixed by using

$$Y= -\frac 16 L + \frac 13 Y_1 + \frac 16 Y_2 + T$$
$$Q= Y +S$$

as then we get 4 doublets with Y = -1/2 and 2 doublets having Y= +1/2, to be complemented with 4 doublets having Y= +1/2 and 2 doublets having Y = -1/2 coming from the $\bar {27}$ representation of E6. Still one needs to cope with the point of having two different isospin groups, but for a first try is nicely half-realistic.

Comparing with a typical E6 assignment of su3su2u1, this assignment has repurposed the extra coloured down-type quark as three generations of a lepton, and also moved some traditional extra leptons to the quark sector. This was justified by the first step in the branching. On other hand, 15+12 has not a obvious representation as a jordan matrix, or at least I am missing it.

Last edited: