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In GUT, why in representation $\bar{5}+10$ of $SU(5)$ we do not consider $\bar{\nu}_L$? One says that there are 15 particles-antiparticles per generation but, for me, there are 16 particles-antiparticles.

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  • $\begingroup$ why do you think that $\nu_L \in 5$ describes a different physical state than $\nu_L \in \bar{5}$? $\endgroup$ – jak May 18 '17 at 8:39
  • $\begingroup$ Of course but I meant why in $\bar{5}+10$ (or $5+\bar{10}$) do we have both $e_L^+$ and $e_L^-$ and not both $\bar{\nu}_L$ and $\nu_L$ ? $\endgroup$ – ketherok May 19 '17 at 22:10
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There is no deep reason for this.

When Georgi and Glashow discovered the first GUT model, they noticed that all standard model particles fit perfectly in the $ \bar{5}\oplus 10$ representation of $SU(5)$. That's why they proposed that it is a good idea to study this kind of model with $SU(5)$ as a GUT group.

However, if you like you can consider an $SU(5)$ model with fermions in the $ \bar{5}\oplus 10 \oplus 1$ representation, i.e. simply add the right-chiral fermion by hand to the model.

This is, in fact, exactly what you get when you consider $SO(10)$ as a GUT group. The $15$ standard model fermions plus the right-chiral neutrino fit perfectly in the $16$ of $SO(10)$. When you break $SO(10)$ to $SU(5)$ you get

$$ 16 \to \bar{5}\oplus 10 \oplus 1 $$

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  • $\begingroup$ is there perhaps something related to anomaly cancellation in play here? $\endgroup$ – AccidentalFourierTransform May 22 '17 at 11:40
  • $\begingroup$ @AccidentalFourierTransform I think the singlet makes no difference for the anomaly cancellation. The anomalies within $\bar{5}\oplus 10 $ cancel and equally withing $\bar{5}\oplus 10 \oplus 1$. Otherwise we would have a strong argument for or against right-chiral neutrinos $\endgroup$ – jak Jun 2 '17 at 6:25

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