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): $$ \psi^{\mu\nu} = (\psi^{*\mu}\psi^{*\nu}-\psi^{*\nu}\psi^{*\mu})^*? $$

2. How does the fermion statistics match on the left and right handed sides?

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): $$ \psi^{\mu\nu} = (\psi^{*\mu}\psi^{*\nu}-\psi^{*\nu}\psi^{*\mu})^*? $$

2. How does the fermion statistics match on the left and right handed sides?

In Zee Chapter VII.5 QFT book:

He showed that the standard model fermions form $5^*$ and $10$ of $SU(5)$: $$ 5^* + 10. $$

In particular, the $10$ is the anti-symmetric matrix representation from the tensor product of $5 \times 5 $: $$5 \times 5 = 10 + 15$$ or $$(5 \times 5)_A = 10 $$

question

1. If the fermion representation $10$ is from the anti-symmetric part of $5 \times 5$, 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): $$ \psi^{\mu\nu} = (\psi^{*\mu}\psi^{*\nu}-\psi^{*nu}\psi^{*\mu})^*? $$

2. How does the fermion statistics match on the left and right handed sides?

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): $$ \psi^{\mu\nu} = (\psi^{*\mu}\psi^{*\nu}-\psi^{*\nu}\psi^{*\mu})^*? $$

2. How does the fermion statistics match on the left and right handed sides?