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In Zee Chapter VII.5 QFT book:

He showed that the standard model fermions form $5^*$ and $10$ representations of $SU(5)$: $$ 5^* \oplus 10. $$ In particular, the $10$ is the anti-symmetric matrix representation from the tensor product of $5 \otimes 5 $: $$5 \otimes 5 = 10 \oplus 15$$ or $$(5 \otimes 5)_A = 10 $$

Question

  1. If the fermion representation $10$ is from the anti-symmetric part of $5 \otimes 5$, can the fermion $10$ be the composite of fermion $5^*$?

In Zee's notation, can we regard the fermion $10$ (the left-handed up and down quark doublet, the right handed $u$ quark, the right-handed electron) as the bound state of some sort of two $5$ (each $5$ forms the left-handed neutrino and electron doublet, the right handed $d$ quark): $$ \psi^{\mu\nu} = (\psi^{*\mu}\psi^{*\nu}-\psi^{*\nu}\psi^{*\mu})^*? $$

  1. How does the fermion statistics match on the left and right handed sides?
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  • $\begingroup$ The representations may compose as you observe, but the spins and fermion numbers cannot. Observing a representation theory coincidence will not generate a new physics model. $\endgroup$
    – Cosmas Zachos
    Commented Jun 21, 2021 at 14:27
  • $\begingroup$ did people try this: can one add new fermions into the model to match the fermi statistics? $\endgroup$
    – ann marie cœur
    Commented Jun 22, 2021 at 1:13
  • $\begingroup$ I don’t know, alas… $\endgroup$
    – Cosmas Zachos
    Commented Jun 22, 2021 at 2:47
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    $\begingroup$ Someone did it earlier this month. arxiv.org/abs/2106.02464 $\endgroup$
    – Mitchell Porter
    Commented Jun 24, 2021 at 10:25
  • $\begingroup$ thanks so much for this (ref) $\endgroup$
    – ann marie cœur
    Commented Jun 24, 2021 at 14:35

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