I am studying quantum field theory using Srednicki's textbook. Problem 83.1 is:

Suppose that the color group is $G_C=SO(3)$ rather than $SU(3)$, and that each quark flavor is represented by a Dirac field in the 3 representation of $SO(3)$.

a) With $n$ flavors of massless quarks, what is the non-anomalous flavor symmetry group?

The answer is: Each Dirac field equals two left-handed Weyl fields. All $2n$ Weyl fields are in the 3 representation of $G_C=SO(3)$ (because it is real). So there is a $G_F=U(2n)$ flavor symmetry; the $U(1)$ is anomalous, leaving a non-anomalous flavor symmetry group $SU(2n)$.

My question is: Why does the 3 representation of $SO(3)$ contain a $U(2n)$ symmetry, and why does the $U(2n)$ symmetry break down into a $U(1)$ group and an $SU(2n)$ group?

  • $\begingroup$ I believe this question is better suited for physics SE. $\endgroup$ – Fabian Jan 26 '17 at 9:08
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    $\begingroup$ It seems OP is partially conflating color and flavor groups. $\endgroup$ – Qmechanic Jan 26 '17 at 21:57
  • $\begingroup$ I agree with @Qmechanic that the color group is a red herring, simply ensures that it does not impede opposite Weyl components forming singlets in the action: so ignore it--it is not part of a) you are addressing; it is tensor multiplying the rest! Now, in U(2n) the U(1) is always commuting with the SU(2n) subgroup generators, so U(n)~U(1) x SU(2n) automatically: there are no couplings involved to be set equal. If you had baryon/fermion number conservation, your flavor group would be SU(n)xSU(n) instead. $\endgroup$ – Cosmas Zachos Jan 26 '17 at 22:46
  • $\begingroup$ cont'. The reason is , cf 190762, there are no conserved charges for the entire SU(2n). But nobody said there is baryon number conservation. Then the kinetic terms of the left handers and conjugated right handers are fine, may violate fermion number, and allow fun mixing among the 2n components, as per your answer. The U(1) is a twist of the axial U(1) of our world and is anomalous. So this part of the problem is an obvious warmup of what is to follow, not provided here. $\endgroup$ – Cosmas Zachos Jan 26 '17 at 22:50

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