The conventional approach in GUTs is to put all left-chiral fields $F_L$ of the standard model into one representation of the GUT group. For example, the 16 rep for $SO(10)$ GUT:

$$16_L \rightarrow F_L$$

Now, I've read the remark that the right-chiral fields then go into the $\bar{16}$ rep:

$$\bar{16}_R \rightarrow F_R$$

Does this mean that we are realling starting with

$$16_L \oplus \bar{16}_R \rightarrow F_L \oplus F_R$$

in $SO(10)$ GUTs?

I've never seen this kind of notation and I do understand that our right-chiral fields are here just the charge conjugates of the left-chiral fields. This is why almost all papers are dealing exclusively with $16_L$ and simply charge conjugate later after branching to the standard model. Nevertheless, I think, from a group theoretical point of view, we are really starting with $16_L \oplus \bar{16}_R$, because only then we get all the fields that we need to build an invariant Lagrangian. Is there sth wrong with the notation or writing the GUT group rep we are starting with like this?

• $\uparrow$ Read where? – Qmechanic May 9 '15 at 12:04
• @Qmechanic For example, at the bottom of Page 44 in Group theory for unified model building by Slansky – jak May 9 '15 at 12:14
• Physicists' notations for representation theory are almost always horrible, and yes, you need to take the direct sum of the left and right representations to have the full rep a non-chiral field transforms in. I don't precisely get what you mean with "that we are really starting with"...writing everything in the left rep and just saying it conjugates seems equally valid to me, and so you seem to be asking about personal preference. – ACuriousMind May 9 '15 at 12:22
• @ACuriousMind What do you mean by a non-chiral field? All left-chiral fermion fields are put into one rep and all right-chiral fields into another... With "really starting with" I mean that in order to get all fields we need to build a Lorentz invariant Lagrangian we need the reps in terms of the Standard Model gauge group that branch from $16_L \oplus \bar{16}_R$. If we start only with $16_L$ we get only one half. Of course we can then simply take the charge conjugates of these reps, which yields the same result, but in terms of group theory we get these reps from $\bar{16}_R$ – jak May 9 '15 at 12:29
• Well, yeah, implictly they are are always carrying the ${}_R$ reps with them when they say "get them by conjugating the ${}_L$ reps". I really think it's not wrong on the physical level of rigor to just say "conjugate everything to get the right-chiral fields" and not explicitly write that you always have ${}_L\oplus{}_R.$ – ACuriousMind May 9 '15 at 12:33