The gauge group of the Standard Model of Particle Physics has Lie rank 4, which suggests that a simple Lie group of rank 4 is what is required to unify them. There are five of those, called A4, B4, C4, D4 and E4 (or F4), and they have dimensions 24, 36, 36, 28 and 52 respectively. Which one you go for depends on what exactly you think you can fit into it. I have argued that 36 is the correct dimension, so that the Lie algebra is either so(7,2) or sp(3,1), but I am not sure that my argument starts from the correct assumptions. If, for example, we assume that the 8 unobservable degrees of freedom of the colour su(3) should not be included, but only the 8 observable degrees of freedom in the CKM and PMNS matrices (also closely related to su(3)), then the conclusion reduces from 36 dimensions to 28, and therefore to D4. In this case, the real form is so(6,2), and as I described in the previous post, it seems to work quite well.
But if, perhaps, we want to treat the electro-weak forces in the same way, then the dimension reduces by another 4 to 24, and the Lie group/algebra is of type A4 instead. This is interesting, because the very first published GUT was of type A4, namely the Georgi-Glashow model with Lie group SU(5). They assumed, for some reason, that the Lie group must be compact, hence SU(5). But if we learn one thing from the Standard Model, it is that when physicists say they are using the compact real form, in fact they often use the complex form instead. My concern is to stick with a real form, and not confuse the issues with extraneous complex forms. The Standard Model needs non-compact real forms in order to have non-zero mass, and we need something of the order of 12 masses in the model. If we also assume that we need SU(3) x SU(2) inside the model, then there is only one possibility, that is SU(3,2).
This immediately deals with the main problem of the Georgi-Glashow model, because only the compact part of the gauge group corresponds to gauge bosons in the Yang-Mills sense, so that there are no more gauge bosons beyond those that are already in the SM. In other words, no new forces, no proton decay, no unobserved “new physics”. The non-compact part of the group therefore needs a different interpretation, but if you want an interpretation as particles, then you are more or less forced to interpret tham as fermions, not bosons. Then you get the SM 12 bosons (no Higgs boson) and 12 fermions, but no “supersymmetry” between them, because the bosons (in the 3×3 and 2×2 diagonal blocks of the 5×5 matrices) split 8+3+1 while the fermions split 2x2x3.
And they actually do split like that in SU(3,2). So you get a factor of two from weak hypercharge, that splits leptons from quarks, and you get a factor of two from weak isospin, that splits electrons from neutrinos, and up quarks from down quarks. And the factor of three that is interpreted as colours in the Georgi-Glashow model, is here interpreted as the three generations. In other words, the theoretical unbroken symmetry of three colours is replaced by the experimental broken symmetry of three generations. This change is of course anathema to those theorists (most of them) who prefer the unbroken but unobservable symmetries of their favourite theory, to the broken but observable symmetries of the real world. But I’m not interested in theories that don’t apply to the real world. I’m not interested in hypothetical properties (like quark colour) that cannot be observed, I’m only interested in real properties (like electron mass) that can be, and are, observed.
So perhaps Georgi and Glashow had the right idea all along. Perhaps an A4 GUT actually is the way the real world fits together. But we’ve thrown away all the gauge group, and reduced the symmetries to nothing at all. Is that good, or is it bad? I propose that it is good, because in the real world, all symmetries are at best approximate, and usually badly broken. We pretend that the physical universe is spherically symmetric, but this is obvious junk. I’m not interested in theories of spherical cows. Especially not three-coloured spherical cows, which is how the SM treats the proton. The proton is certainly a cow (unless perhaps it’s a pig), but spherical? I doubt it.
Well now, consider where we’ve got to – a 24-dimensional model, with all symmetries broken down to nothing. That means we have 24 parameters, that tell us exactly how the symmetries have been broken. And that, if I may say so, is pretty much what the Standard Model looks like.
Conclusion number one: ignore everything I’ve said for the past several weeks – ignore C4 and B4, pass swiftly past D4 to A4, where the story began fifty years ago. Conclusion number two: ignore everything I’ve said for the past several years – except for one crucially important thing, that there are two Lorentz groups, not one. Conclusion number three: start again from scratch.
Now for Act 2. I hope you have taken suitable refreshment during the interval, because Act 2 of the opera is where everything gets turned upside down, and general chaos ensues. I don’t know which is your favourite opera in which the two main protagonists pretend to be each other, but there’s lots of them, and you just have to imagine that they are both going around calling themselves the Lorentz group. This is the situation when the curtain goes up on Act 2. You see, the group SU(3,2) contains a subgroup SO(3,2), which contains a Lorentz group SO(3,1), and it also contains a subgroup SU(2,2), which contains a Lorentz group SL(2,C). In the Standard Model, SO(3,1) and SL(2,C) are considered to be essentially the same group. But in SU(3,2) they are completely different. The stage is set for unlimited confusion, and either a comic or a tragic conclusion, depending on the libretto that has been chosen. My reading is that it is a farce, but others may regard it as a tragedy. And I haven’t seen Act 3, so I really don’t know.
Or perhaps I should just go Wilde, and write about the Importance of Being Lorentz.
No, but, seriously, we really have to talk about the Lorentz group. It is a symmetry group of spacetime, of course, everyone understands that. But it is a broken symmetry group – nobody understands that – whenever I point out this obvious fact, I am excommunicated. Somehow, unbroken Lorentz symmetry is a sine qua non of every theory of physics. The only problem is, it is completely and utterly falsified by experiment, and has been for decades. All the mixing angles fail to be Lorentz-invariant. All of them. Lorentz symmetry is broken. Fact. Experimentally demonstrated at vast cost to millions of taxpayers worldwide.
So, in the middle of Act 2, the Lorentz group insinuates itself into the broken gauge group SU(3,2), and turns out to be as broken as everything else, in fact more so, because there are, as I said, two Lorentz groups. It is not the job of Act 2 to sort out the mess, only to describe it. So let’s get on with the job. The first and most important point is that because the Lorentz symmetries are broken, they are already in SU(3,2) – they do not introduce any more degrees of freedom. And that applies to both of them. There are only 24 degrees of freedom altogether. Apparently.
We need a Dirac equation, I am told, before anyone will take us seriously. A Dirac equation is usually written as 4×4 complex matrices acting on spinors, of 4 complex numbers. This works perfectly well in SU(3,2) acting on 5 complex numbers, by restricting to U(2,2) and 4 complex numbers, where the Dirac gamma matrices live together with a complex scalar, and everything works out fine. But what is the fifth complex number doing? A very good question. It isn’t in the Standard Model, but what it really does, as I am sure you can work out for yourself, is extend the two complex numbers for SU(2) to three for SU(3) and therefore extends from a one-generation model to a three-generation model. What’s not to like?
What about SO(3,2)? Does that have anything to do with the Dirac equation? Apparently it does, because the “imaginary” version of the Dirac matrices, that is the i.gamma^mu, generate the Lie algebra so(3,2). So which one is the real Dirac equation? Is it the one that uses U(2,2), or is it the one that uses SO(3,2)? They are different equations, so they cannot both be correct, can they? Or can they? Are they identical twins separated at birth? That’s a really Wilde idea!
Well, I hope I have given you a good idea of the total confusion that transpires in Act 2. The end of Act 2 takes us up to the present day. Time to take another break. When we reconvene for Act 3, do you think this confusion will be resolved? Do you think the two Lorentz groups will embrace each other as long lost brothers? Or do you think they will murder each other? I don’t know, I haven’t read the libretto. But I do know that there are two Lorentz groups, not one.