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We have a Trinification Model, $$SU(3)_C \times SU(3)_L \times SU(3)_R.$$ The first group is the colour group, the second is the Left Weak force and the third is the Right Weak force. There are 27 fermions due to the extended weak forces with representations: $(3,3,1)$ Left handed Quarks, $(3,1,3)$ Right handed Quarks and $(1,3,3)$ Leptons. Most theories have the latter 2 $SU(3)$'s broken down into $$SU(2)_L \times SU(2)_R \times U(1)_{B-L}.$$ $B\!-\!L$ is a linear combination of the broken 8th Generators of the $SU(3)$'s, $T_L^8$ and $T_R^8$. My question is what combination of generators forms $B\!-\!L$ and what are the generators for fermions.

Edit: What I meant was what is $T_L^8$ and $T_R^8$ for fermions in the same way that, for example, $T_L^3$ for an electron is $\frac{-1}{2}$ a.k.a. the 3rd component of weak isospin.

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$U(1)_{B-L}$ is non-chiral, i.e. vector-like. Therefore, the $B-L$ linear combination is $T_L^8 + T_R^8$ , while the chiral $T_L^8 - T_R^8$ is broken.

The $B-L$ charge is by definition $1/3$ for quarks and $-1$ for leptons.

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  • $\begingroup$ My question is what the $T_L^8$ and $T_R^8$ is for the fermions including the extra exotic fermions similar to the 3rd component of weak isospin. $\endgroup$
    – Advik
    Commented Oct 23 at 2:04
  • $\begingroup$ What would be the factor of $T_L^8 + T_R^8$ which makes it equal to $B - L$ $\endgroup$
    – Advik
    Commented Oct 23 at 5:08

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