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If quarks come in three colours $r$, $g$ and $b$ than (neglecting all other quantum numbers and spacial freedom for now) a state of a quark would be a vector in $\mathbb{C}^3$.

If we are now looking for a global symmetry redefining the colours than I would be left with the unitary group $U(3)$. However the correct gauge group is $SU(3)$.

Qmechanic
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Daan
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    The ninth gauge boson that couples to quark number would not interact with the other eight and would get renormalized quite differently, for lack of quark loop insertions. – Bert Barrois Jun 14 '18 at 01:42
  • SU(3) is not the "correct" gauge group in any mathematical sense. It is correct phenomenologically, to the extent nobody has seen any evidence of or theoretical utility for a disjoint independent U(1). There is a very different weak hypercharge U(1) in the SM, but it does a very different job, and is supported by experimental evidence. I am not clear on why you choose to consider something avoidable and useless. – Cosmas Zachos Jun 14 '18 at 18:52
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    Possible duplicate of Why $SU(3)$ and not $U(3)$? – probably_someone Jun 14 '18 at 19:00

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