# Why is quark flavor just a SU(N) group?

In the standard model one has U(1) for electromagnetism, SU(2) for the weak sector and SU(3) for the color sector. One could say that in the quark part of the fermions, there are $$\underbrace{6}_\text{flavors} \cdot \underbrace{3}_\text{colors} \cdot \underbrace{(2 + 1)}_\text{left and right}$$ “different” spinor particles. Those states can be grouped into color triplets and weak-isospin doublets and singlets depending on their chirality. For massive quarks, the weak-isospin symmetry is only approximate, of course. The Lagrangian stays invariant under the unitary transformations, therefore those are the gauge symmetries.

The weak-isospin doublets made up from up-down, charm-strange and top-bottom are just the same thing in different generations. Their only difference I think they have is the mass. Neglecting the CKM-matrix and the masses, I would think that there is no way to distinguish a down-quark from a strange-quark or bottom-quark. With masses, there is just this one difference.

One then generalizes the flavor symmetry to include the strange quark. The weak-isospin group is extended from SU(2) to just SU(3). This bugs me: The (approximate) SU(2) weak-isospin symmetry groups the (left-handed) up- and down-quarks together in a doublet. They are two states that differ by their electromagnetic charge. When adding the strange quark I would consider this to be another down-type quark (which is distinct from an up-type quark) from the second generation. Why does one put the strange quark onto equal footing with the other ones by including it with one $\mathrm{SU}(N_\mathrm F = 3)$ group? Shoudn't the strange-quark be treated similarly to the down-quark but different from the up-quark?

For any given force, it only sees part of it. So for the weak-isospin gauge symmetry, all the color states appear to be on equal footing. And the strong force does not care about the electric charge or the weak-isospin $z$-component. Is my problem in understanding that I distinguish up-type and down-type by quantum numbers that belong to other symmetry groups and the flavor would not see at all? Or is there something more to this?

• There are two SU(2) groups at work here, I believe. The one that is relevant to your question is regular isospin, not weak isospin. It is an approximate symmetry of the strong force, not the weak force. – Javier May 27 '16 at 1:22

## 1 Answer

This question appears to cover a fair amount of territory. I have asked a number of these types of questions. This seems to be asking what is the distinction between the say the s quark and the c quark in their doublet. The question of “why $SU(3)$” is another question, which in some ways includes the question of why there are 3 families of quarks.

The isospin doublets are $SU(2)$, but there are $3$ of these and the quarks in the $SU(3)$ representation have additional quantum numbers called flavor. The strange quark was named because of some odd properties with weak decays of hadrons. Gell-Mann proposed a quark that had a flavor that was not conserved in weak decays. This turned out to be the case. The (b, t) doublet similar physics is being probed in the “B-machine.” On a deeper level the quarks in these doublets are transformed into each other by charged weak currents that are flavor changing. These are of course the $W^\pm$ bosons. There is in addition the neutral current that conserves flavor or the $Z$ boson.

The $8\oplus 1$ representation of the $SU(3)$ defines the $8$ gluons. The $3\oplus \bar 3$ representation describes the quark families, and conveniently $8\oplus 1 = 3\oplus \bar 3$. So the quark families are a sort of fermion mirror of the gauge bosons. There is a physical state that can also occur where color and flavor are locked in a superconducting-like phase of QCD matter. This may occur in the interiors of neutron stars, and empirical support for this state of matter in neutron stars would give support for the color-flavor correlation. As yet this has not be completely solved to everyone's satisfaction.

The question of why there is $SU(3)$ and not say $SU(N>3)$ is easier to answer empirically. If there were say four or more doublets of quarks the early universe would have had more degrees of freedom to distribute energy. This would have resulted in some very different cosmological measurements. It appears empirically reasonable to say there is no fourth family of quarks that have masses below 10TeV or so.

A theoretical answer to why $SU(3)$ is something I can only offer somewhat speculatively. The exceptional group $G_2$ is the automorphism of the exceptional $E_8$ or $G_2 \simeq Aut({\mathbb O})$. This means that QCD could easily slip into deeper foundations since $g_2\rightarrow su(3)\oplus 3\oplus \bar 3$ as the maximal subgroup. The exceptional group $E_8$ plays a role in the heterotic string, and quantum gravity. This seems to provide a nice back door for looking at gravitation. It also fits nicely with the work of Bern and Dixon on gauge-gravity correspondence.

I hope this helps some.

• This coincidence is nice for three flavors, but we do have SU(4) with charm and SU(5) with bottom. So those connections are not relevant there. I am not sure what that all means, then. – Martin Ueding May 27 '16 at 11:44
• The occurrence of flavors is with the $3\oplus\bar 3$ representation of $SU(3)$. It is not read off directly from the rank = 3 of the group. – Lawrence B. Crowell May 27 '16 at 13:13