EDIT: I found an answer, which I do not understand: In Gürsey - Symmetry breaking patterns in E6 he writes: " Because of Fermi-Dirac statistics of fermions they must occur in the symmetric part of 27x27: $(27x27)_A)= \bar{27} +351'$..

This line of thought is used everywhere in Grand Unified Theories, but I can't figure out why this works and it must be very obvious because no one seems to offer an explanation.

Commonly the left-chiral fermions are assigned to some irrep, like the $16$ of $SO(10)$.

Lorentz invariance demands that we couple complex conjugated right-chiral fields to left-chiral fields $$ \chi_R^\dagger \psi_L = (\chi_L^c)^\dagger \psi_L = ((C \chi_L^\star )^\star)^T \Psi_L = \chi_L^T C \psi_L$$

where I rewrote the term because it is conventional in GUTs to work exclusively with left-chiral fields.

In GUTs like SO(10) all SM fields are part of one big multiplet, for example the $16$ which I will call $\Psi$ and mass terms are of the form

$$ \Psi^T C \Psi $$

Isn't this term already invariant under $SO(10)$? Transforming $\Psi \rightarrow O \Psi$ yields

$$ \Psi^T C \Psi \rightarrow \Psi^T O^T C O \Psi = \Psi^T C \Psi $$

Doesn't this term yield Majorana mass terms? In the Standard model we must be careful to use terms of the form $\chi_R^\dagger \psi_L $ instead of Majorana terms $\psi_R^\dagger \psi_L $. In GUTs all standard model fields are part of the 16 and therefore we don't have two distinct fields we can combine.

In GUTs one argues that this term is of the form $16 \otimes 16 $ and therefore we must consider

$$ 16 \otimes 16 = 10 \oplus 120 \oplus 126$$

and this tells us that the Higgs must be either in the $10$, the $120$ or the $126$ representation of $SO(10)$. An invariant term, including a Higgs field multiplet is then for example,

$$ \Psi^T_{16} C \Psi_{16} \phi_{10} ,$$

where the subscript denotes the corresponding $SO(10)$ representation. Why are terms of this form invariant and why does the decomposition quoted above tell us which Higgs representation we must use in order to get invariant terms?

  • $\begingroup$ $SO(10)$ on Wikipedia. $\endgroup$
    – Qmechanic
    Jul 31, 2018 at 11:51
  • $\begingroup$ I wonder do you know this h physics.stackexchange.com/q/567129/42982? $\endgroup$ Jun 17, 2021 at 21:20

2 Answers 2


Let me attempt to answer your question, since your question is about SO(10) GUT model, so I will assume that you have the knowledge of simpler version of GUT namely SU(5) GUT model and also little of group theory.

You have 4 different questions >>>

01. Isn't this term ($\psi^{T} C \psi$) already invariant under SO(10)?

02. Doesn't this term ($\psi^{T} C \psi$) yield Majorana mass terms?

03. Why are terms ($\psi^{T} C \psi \phi_{10}$) of this form invariant ?

04. why does the decomposition quoted above ($16 \times 16= 10+120+126$) tell us which Higgs representation we must use in order to get invariant terms?

Since all these questions are kind of interrelated so while answering sometimes I may not be able to follow the particular ordering but mix things up. I apologize for that. Anyway,

ANS to QUS#01: yes it is! SO(10) is an orthogonal group so any SO(10) invariant quantity (say X) must remain the same under the group rotation, i.e, $O^{T}XO=X$, so you are right about that.

ANS to QUS#02: the term ($\psi^{T} C \psi$) is neither a mass term nor a Yukawa term, it is just an invariant fermion bilinear. To understand how to form fermion bilinears and also how many of such terms possible, please consult any introductory field theory/particle physics book, as for example: Introduction to Elementary Particles--D. Griffiths; chapter#07,page#224,Equation#7.68 [ http://www.amazon.com/Introduction-Elementary-Particles-David-Griffiths/dp/3527406018 ].

ANS to QUS#03 and #04: now, to generate mass for the fermions, at first you need construct Yukawa coupling terms in the Lagrangian. Always remember two things,

(a) structure of Yukawa coupling $\sim$ fermion * fermion * Higgs scalar ,and

(b) Lagrangian always has to be group invariant.

From (a) we get for SO(10), $16\times 16 \times Higgs$ but from (b) this Higgs representation has to be choosen in such a way that such terms are invariant. In this stage you need a little bit of group theory. Group theory says, $16 \times 16= 10+120+126$, which says, fermion * fermion$\neq 1$ ($1=$ group singlet and singlets are group invariants). So, to form Yukawa coupling which has to be group invariant, the only options you have are 10,120 or 126 representations, choosing any other representation under SO(10) will not give you a singlet. To understand in details why other representations of Higgs will not form singlet, please consult with Group Theory for Unified Model Building--R.Slansky; page#105 [http://inspirehep.net/record/10204?ln=en ]. This answers QUS#04 and QUS#03.

Now, while answering QUS#02 above, I was very brief, as to explain it clearly, I need the elements from the last paragraph. So lets go back to your QUS#02.

Now you know that possible Yukawa coupling terms are:

(i) $16_{F} \times 16_{F} \times 10_{H}$

(ii) $16_{F} \times 16_{F} \times 120_{H}$

(iii) $16_{F} \times 16_{F} \times 126_{H}$

,F and H stand for Fermion and Higgs respectively.

For minimality of SO(10) GUT models, 120 representation of Higgs is not used and so I will not talk about it rather concentrate on 10 and 126 dimensional representations.

Lets assume that you already know SU(5) GUT as your question involves SO(10) GUT. Again from group theory one can write down the Branching Rules of the representations that we are interested in for SO(10)-->SU(5) Group Theory for Unified Model Building--R.Slansky; page#106 as:



(C)$126=1+5+\bar{10}+...$ (dots mean higher dimensional representations that we are not interested in).

Now, if you substitute these in the Yukawa couplings, you will get 10 terms that are invariants, but I will only pick two of them for illustration and to attack your question of Majorana mass that you are interested in.

(i) $1_{F}\times \bar{5}_{F} \times 5_{H}$

(ii) $1_{F}\times 1_{F}\times 1_{H} $

If you know SU(5) GUT you already know that first term (i) is a Dirac mass term where as second term (ii) is Majorana mass term, as their basic forms are:

Dirac mass $\sim$ Right Handed Fermion * Left Handed Fermion * Higgs = R * L * $\langle\phi_{H}\rangle$

Majorana mass $\sim$ Right Handed Fermion * Right Handed Fermion * Higgs = R * R * $\langle\phi_{H}\rangle$

(in $\phi_{H}$ I put $\langle\phi_{H}\rangle$ to represent vacuum expectation value [vev in short], as to give the fermions mass, scalar fields have to get vev [to understand vev, see any introductory field theory book as for example: Quantum Field theory---L.Ryder , Chapter#08] )

because, in (i) $1_{F}$ contains right handed neutrino and $\bar{5}_{F}$ contains left handed neutrino and so Dirac type mass ;on the contrary in (ii) both $1_{F}$ contain right handed neutrino and so Majorana type mass.

Hope that it will help, Thanks!

  • $\begingroup$ I wonder do you know this h physics.stackexchange.com/q/567129/42982? $\endgroup$ Jun 17, 2021 at 21:20

@SAS answered most of the questions, however I believe there's a crucial point which still needs to be addressed: the chirality.

Indeed, it is not obvious a priori why $$\Psi^T C \Psi\,\Phi\,,$$ (where $\Phi$ is some Higgs representation) leads to a Dirac-type masses instead of Majorana masses. Why not the common $\bar\Psi \Psi$?

It turns out to be the case in GUTs because we unify the SM 'right' fermions and 'left' ones into a single representation. To simplify things, we can define a multiplet with only one chirality using charge conjugation (this operation lies outside the group, so it defines two different states that are unrelated by any group tranformation): $$\psi_L^c\equiv P_L \psi^c = (\psi_R)^c.$$

Using only left-handed fields, we can define $\Psi =\Psi_L= (e, e^c,...)_L$. Now, we know that we can form a Dirac mass with $e_L$ and $e^c_L$ as $$m\,(e^{cT}_L C e_L + hc) = m\,(\overline{e_R} e_L +hc).$$ So to obtain this form using $\Psi_L$ all we have to do is: $$\Psi_L^T C \Psi_L<\Phi> =m\, e^{cT}_L C e_L +...= m\, \overline{e_R}e_L+...$$

What is $\Phi$?

The yukawa coupling is of the form $\Psi \Psi \Phi$ (The transpose and $C$ do not conjugate any representation). And we know this has to be invariant. In $SO(10)$, with $\Psi \sim \mathbf{16}$, the only invariants we can construct of this form are: $$\mathbf{16}_i.\mathbf{16}_j.[(\mathbf{10}+\mathbf{126})_s+\mathbf{120}_a]$$ (you can verify that $10+120+126=16^2$.) The subscript $s$ and $a$ stand for symmetric and anti-symmetric contractions. I also included the family indices $i$ and $j$. In principle, any of this scalar representation gives fermion masses. However to get the actual spectrum of the SM a special combination is required.

Usually the SM Higgs is assumed to be in the $\mathbf{10}$, but it could be as well in $\mathbf{126}$ depending on the model and other details; there's no direct answer to this question.

Lastly, about your last edit and the comment on Fermi-Dirac statistics: I don't think this makes any sense at all. We can use an anti-sym representation (like $\mathbf{120}$ above) without any problem.

  • $\begingroup$ I wonder do you know this h physics.stackexchange.com/q/567129/42982? $\endgroup$ Jun 17, 2021 at 21:20

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