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I know the mathematical difference that one means $ absolutevalue(det) = 1$ and one means det = 1 (rotation) and that ones the subgroup of the other and so on.

But:

enter image description here

has a local/gauged $SU(3)$ colour gauge symmetry and global $U(2)\times U(2)$ flavour symmetry. This is the Weyl-field lagrangian for $u$ & $d$ quarks.

I'm confused why it should be a $U$ or a $SU$ symmetry, since both are unitary: $1=U^{\dagger} U$ makes bilinears like

enter image description here

invariant. So you should choose $U$ as the symmetry group, since its bigger. Do gauged/local symmetries have to be $SU$ and global can be $U$? Or what's the point here?

EDIT: I think that's the reason, since a local $U(N)$ transformation would need a local absolute value

$a(x)\exp(i\theta^{a}(x) t^{a})$

and then you need to use the product rule for differentiation and that makes things complicated.

Qmechanic
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VN23
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    I don't really understand the question. The $a$ index of the $F^a_{\mu\nu}$ there by definition is an SU(3) index, i.e. runs from 1 to 8. If it were a U(3) gauge theory, the index would have to run from 1 to 9, so $F_{\mu\nu}$ would need to be a different object. Also, unitary means $\lvert \det(A) \rvert = 1$, not $\det(A) \neq 0$ (The latter defines GL, not U). – ACuriousMind May 19 '19 at 09:30
  • its a very general and kinda vague question. not about explicit objects. of course, you need an extra generator for U(3) since its a bigger group and you have one extra parameter. – VN23 May 19 '19 at 09:35
  • Can you gauge the U(3) or why use the SU(3) instead? – VN23 May 19 '19 at 09:37
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    You seem to think that there's some deep reason for "choosing" SU over U, but we use SU(3) for color charges because the predictions of the SU(3) theory match experiment (and that of a U(3) theory do not), not for any theoretical reasons. – ACuriousMind May 19 '19 at 09:40
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    Related: https://physics.stackexchange.com/q/52452/2451 – Qmechanic May 19 '19 at 10:10
  • An other point is that $U(N)$ is not the set of complexe matrices with determinant of norm 1. It is the set of unitary matrices, that is, $M \in U(N)$ if and only if $U \cdot U^* = id$. The condition on the derterminant follows. The other group $SU(N)$ is indeed defined as the subgroup of matrices in $U(N)$ with det 1. – InfiniteLooper Jun 09 '20 at 10:20
  • @ACuriousMind - Theorists should explain experiment from a theoretical point of view, i.e. should give a theoretical reason to account for experimental phenomena. – Shen Dec 06 '21 at 13:03

1 Answers1

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$U(3) \sim SU(3)_{color} * U(1)_{B-L}$ has been considered in some versions (e.g. as components of Pati-Salam's SU(4)) of Standard Model extension. If $SU(3)_{color}$ and $U(1)_{B-L}$ are considered as direct product, then they can have different coupling constants. Otherwise (if embedded in $SU(4)$) they share the same CC.

MadMax
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