Questions related to the notion of color symmetry in the TGD framework

One of the longstanding open problems of TGD has been which of the following options is the correct one.

1. Quarks and leptons are fundamental fermions having opposite H-chiralities. This predicts separate conservation of baryon and lepton numbers in accordance with observations.
2. Leptons correspond to bound states of 3 quarks in CP₂ scale. This option is simple but an obvious objection is that they should have mass of order CP₂ mass. Baryons could decay to 3 leptons, which is also a problem of GUTs.

I haven't been able to answer this question yet and several arguments supporting the quarks + leptons option have emerged.

Consider first what is known.

1. Color is real and baryons are color singlets like leptons.
2. In QCD, it is assumed that quarks are color triplets and that color does not correlate with electroweak quantum numbers, but this is only an assumption of QCD. Because of quark confinement, we cannot be sure of this.

The TGD picture has two deviations from the QCD picture, which could also cause problems.

1. The fundamental difference is that color and electroweak quantum numbers are correlated for the spinor harmonics of H in both the leptonic and quark sector. In QCD, they are not assumed to be correlated. Both u and d quarks are assumed to be color triplets in QCD, and charged lepton L and ν_L are color singlets.
   1. Could the QCD picture be wrong? If so, the quark confinement model should be generalized. Color confinement would still apply, but now the color singlet baryons would not be made up of color triplet quark states, but would have more general irreducible representations of the color group. This is possible in principle, but I haven't checked the details.
   2. Or can one assume, as I have indeed done, that the accompanying color-Kac Moody algebra allows the construction of "observed" quarks as color triplet states. In the case of leptons, one would get color singlets. I have regarded this as obvious. One should carefully check out which option works or whether both might work.
2. The second problem concerns the identification of leptons. Are they fundamental fermions with opposite H-chirality as compared to quarks or are they composites of three antiquarks in the CP₂ scale (wormhole contact). In this case, the proton would not be completely stable since it could decay into three antileptons.
   1. If leptons are fundamental, color singlet states must be obtained using color-Kac-Moody. It must be admitted that I am not absolutely sure that this is the case.
   2. If leptons are states of three antiquarks, then first of all, other electroweak multiplets than spin and isospin doublets are predicted. There are 2 spin-isospin doublets (spin and isospin 1/2) and 1 spin-isospin quartets (spin and isospin 3/2). This is a potential problem. Only one duplicate has been detected.
   3. Limitations are brought by the antisymmetrization due to Fermi statistics, which drops a large number of states from consideration. In addition, masses are very sensitive to quantum numbers, so it will probably happen that the mass scale is the CP₂ mass scale for the majority of states, perhaps precisely for the unwanted states.

It is good to start by taking a closer look at the tensor product of the irreducible representations (irreps) of the color group (for details see this).

1. The irreps are labeled by two integers (n₁,n₂) by the maximal values of color isospin and hypercharge. The integer pairs (n₁,n₂) are not additive in the tensor product, which splits into a direct sum of irreducible representations. There is however a representation for which the weights are obtained as the sum of the integer pairs (n₁,n₂) for the representations appearing in the tensor product.

   Rotation group presentations simplified example. We get the impulse moment j1+j2,... |j1-j2|. Further, three quarks make a singlet.

2. On basis of the triality symmetry, one expects that, by adding Kac-Moody octet gluons, the states corresponding to (p,p+3)-type and (p,p)-type representations can be converted to each other and even the conversion to color singlet (0,0) is possible. This is the previous assumption that I took for granted and there is no need to give it up.

Let's look at quarks and baryons first.

1. U type spinor harmonics correspond to (p+1,p) type color multiplets, while D type spinor harmonics correspond to (p,p+2) type representations.

   From these, quark triplets can be obtained by adding Kac-Moody gluons and the QCD picture would emerge. But is this necessary? Could one think of using only quark spinor harmonics?

2. The three-quark state UUD corresponds to irreducible representations in the decomposed tensor product. The maximum weight pair is (3p+2,3p+2) if p is the same for all quarks, while UDD with this assumption corresponds to the maximum weights (3p+1,3p+1+3). The value of p may depend on the quark, but even then we get (P,P) and (P,P+3) as maximal weight pairs. UUU and DDD states can also be viewed.

   Besides these, there are other pairs with the same triallity and an interesting question is whether color singlets can be obtained without adding gluons. This would change the QCD picture because the fundamental quarks would no longer be color triplets and the color would depend on the weak isospin.

3. The tensor product of a (p,p+3)-type representation and (possibly more) gluon octets yields also (p,p)-type representations. In particular, it should be possible to get (0,0) type representation.

   Consider next the identification of leptons.

   1. For leptons, neutrino nu_L corresponds to a (p,p)-type representation and charged lepton L to a (p+3,p)-type representation.
   2. Could charged antilepton correspond to a representation of the type UDD and antineutrino to a representation of the type UUD?

      Here comes the cold shower! This assumption is inconsistent with charge additivity! UDD is neutral and corresponds to (p,p+3) rather than (p,p). You would expect the charge to be 1 if the correspondence for color and electroweak quantum numbers is the same as for the lepton + quark option!

      UUD corresponds to (p,p) rather than (p,p+3) and the charge is 1. You would expect it to be zero. Lepton charges cannot be obtained correctly by adding charge +1 or -1 to the system.

      In other words, the 3-quark state does not behave for its quantum numbers like a lepton, i.e. an opposite spinor with H-chirality as a spinor harmonic.

      Therefore bound states of quarks cannot be approximated in terms of spinor modes of H for purely group-theoretic reasons. The reason might be that leptonic and quark spinors correspond to opposite H-chiralities. Of course, it could be argued that since the physical leptons are color singlets, this kind of option could be imagined. Aesthetically it is an unsatisfactory option.

   To sum up, the answers to the questions posed above would therefore be the following:
   1. Quark spinor harmonics can be converted into color triplets by adding gluons to the state (Kac-Moody). Even if this is not done, states built from three non-singlet quarks can be converted into singlets by adding gluons.
   2. The states of the fundamental leptons can be converted into color singlets by adding Kac-Moody gluons. Therefore the original scenario, where the baryon and lepton numbers are preserved separately, is group-theoretically consistent.
   3. Building of analogs of leptonic spinor harmonics from antiquarks is not possible since the correlation between color and electroweak quantum numbers is not correct. I should have noticed this a long time ago, but I didn't. In any case, there are also other arguments that support the lepton + quark option. For example, symplectic resp. conformal symmetry representations could involve only quarks resp. leptons
   For a summary of earlier postings see Latest progress in TGD.

   For the lists of articles (most of them published in journals founded by Huping Hu) and books about TGD see this.