I know $E_8$ by itself can't be gauge group because it has no complex representation and so would not be chiral. But assuming the existence of mirror matter which also would have $E_8$ gauge group this seems to be fine.

One other thing that I'm not sure is solved is the symmetry breaking whereby down quarks and charged leptons are given different masses. Because in theories like $SU(5)$ they would have to have the same masses. Would mirror matter solve this? Perhaps by having the charged leptons in one $E_8$ and the down quarks in the other $E_8$?

(BTW, I am talking about a supersymmetric theory where bosons have $E_8$ charges and fermions also have $E_8$ charges. So fermions and bosons are not in the same $E_8$ which I think is impossible.)

  • $\begingroup$ Well, first off, the whole point of a GUT is to enjoy the elegance of a simple gauge group. If you're going to switch to $E_8 \times E_8$ you've just tossed out the main motivation. $\endgroup$ – knzhou Nov 30 '18 at 19:45
  • $\begingroup$ @knzhou The idea is to make it as simple as possible..... but no simpler! $\endgroup$ – zooby Dec 1 '18 at 11:04
  • $\begingroup$ Okay, then how about $E_6$? That's simpler in just about every way. Why insist on going for the biggest exceptional group? $\endgroup$ – knzhou Dec 1 '18 at 11:18
  • $\begingroup$ $E_6$ can't combine all 3 generations. Plus $E_8$ has a natural supersymmetric theory if is in the adjoint representation. $\endgroup$ – zooby Dec 1 '18 at 16:41

This is just a quick informal answer; I skip details and may even miss important cases and loopholes...

The standard model has been obtained many times from E8xE8 heterotic string theory. But only one E8 is used. That E8 is broken by a specifically stringy method (embedding Calabi-Yau spin connection in part of the E8), leaving E6, SO(10), or SU(5) - standard GUT groups - and then that GUT group is broken to the standard model group by flux.

But the gauge group is only half what you need, you also need fermions in the appropriate representation of that group. If E8 is broken to e.g. SU(3) x E6, then the E8 adjoint - which the string theory contains, because it has N=1 supersymmetry - will branch into something that includes a 27 rep, which contains an SM generation... but there will also be a 27bar. So every generation will have a mirror, and the Weyl fermions will pair up into Dirac fermions which you expect to be heavy unless there is finetuning.

However, in the case of the heterotic string, somehow chiral generations can be produced independently of antigenerations. I don't know how exactly, something geometric I guess, but it's probably explained e.g. somewhere in Brian Greene's thesis. Then, so long as you have an unequal number of generations and mirror generations produced, there will be some left over after they have paired up.

Again, all of that pertains to obtaining the standard model from a single stringy E8. As for obtaining it from a single E8 within field theory, the most recent works are by Stephen Adler 1 2. You will find some comments there about the theoretical and empirical difficulties of separating generations and their mirrors.

If we now turn to implementing the standard model in a way that is divided between two E8s... You talk about dividing the fermions between one E8 adjoint and the other. But here's a problem. If you just do the straightforward thing, and think in terms of N=1 E8xE8 gauge theory (which, coupled to supergravity, is the field-theoretic limit of the E8xE8 heterotic string in ten dimensions)... the adjoint of each E8 is a singlet with respect to the other E8. That is, a fermion from one E8 adjoint will not transform under any subgroup of the other E8.

Presumably you intend that one part of the standard model gauge group will be a subgroup of one E8, and another part will be a subgroup of the other E8. Now the quarks have to transform under all parts of the standard model gauge group. But we have just observed that, for N=1 E8xE8 gauge theory, each fermion transforms only under one E8 or the other E8, never both.

So it seems you want fermions that will transform under both E8s at the same time, e.g. a biadjoint representation. But you will still face the problem of prising apart generations and their mirrors, and it will no longer be that intriguing case of automatic supersymmetry that comes from having both fermions and gauge bosons in the 248 representation.

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    $\begingroup$ String theory can produce generations independently of anti-generations. Witten (ICM 1986) sketches one method, where on 10-d ($M^4$ × K) with E8 & its 248 rep, the # of generations happens to be the Euler characteristic of K ÷ 2. Intersecting D-brane models also generate one generation per intersection point of two 3-manifolds. $\endgroup$ – alexchandel Sep 29 '19 at 1:43

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