$B$-$L$ global symmetry in the grand unification theories

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Does the precise $$B$$-$$L$$ global symmetry present in:

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart

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In the standard model lagrangian, B and L are separately conserved global charges, and B-L, a vector like symmetry, is anomaly-free. GUTs, like the G-G SU(5) violate B and L, but preserve B-L.

Wikipedia effectively defines the SU(5)-model U(1) symmetry X as $$X = 5(B − L) -2Y_W,$$ introduced by Wilczek & Zee in 1979. It is not a generator of SU(5), of course.

So you may readily compute for the left-chiral $$\bar {\bf{5}}$$, $$\overline{ d_R} : ~~(2Q=)~~~ Y=2/3 , ~~~~ B-L= -1/3 ~~~ \leadsto X=-3 \\ e^-_L, \nu_L : ~~~~~~~ Y= -1 , ~~~~~~ B-L= -1 \qquad \leadsto X=-3.$$ So the entire multiplet possesses a common X-charge: -3.

Proceed to verify for the left-chiral 10, X= 1. $$\overline{ u_R} : ~~(2Q=)~~~~~ Y=-4/3 , ~~~~ B-L= -1/3 ~~~\leadsto X=1 \\ d_L,u_L : ~~~~~~~~~ Y=1/3 , ~~~~ B-L= 1/3 \qquad \leadsto X=1 \\ e^+_L : ~~(2Q=)~~~~~ Y= 2 , ~~~~~~~ B-L= 1 \qquad \leadsto X=1.$$

Thence, for the $$\langle \phi ^* \rangle$$, X=2, so the Yukawa (mass) term is chargeless.

Note that B-L is vectorlike, but Y is not, so, ipso facto, X is not!

Similarly for SO(10), except here X is now a generator of SO(10) and is then gauged (hence SSBroken), hence violated by small amounts.

Responses to comment questions.

1) B-L is a good global symmetry for the SM and SU(5) and local for SO(10). So, e.g., in SU(5) proton decay to a pion and a positron, it is visibly preserved!

2) X is fine in the SM and SU(5) as a global symmetry, as a linear combination of good quantum numbers. As defined, it has a unique eigenvalue for each SU(5) rep, not necessarily the same for all reps, as you observe. Same for the SM which has smaller reps, several of which entered into each SU(5) rep. That means that, even for SU(5) which mixes baryons and leptons, it is straightforward to match X eigenvalues and monitor the symmetry of the coupling terms, like the Yukawas.

3) SO(10) largely follows suit, paralleling SU(5), and likewise lacks exotic particles, but gauges X, so SSBreaks it. But now the above two reps of SU(5) plus an extra singlet (the R-chiral neutrino) fit into a spinor 16 of SO(10). If your wrote down the multiplet, it might be easier to parse out. This review, eqn. (3.3), has X as a traceless diagonal generator, so you can check its eigenvalue on the fermion 16-plet (4.2).

Now, in most models, even though X is SSBroken and its corresponding gauge boson made massive, this happens at a high scale and results in a GUT with SU(5)-like symmetry, with effective B-L violating operators of dimension 6, i.e. suppressed by the square of such heavy scales, so very small. Thus, in such models, for all intents and purposes, proton decay is still approximately respectful of B-L!

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
answered Apr 15, 2020 by (370 points)
Thanks, I vote up. So B, L and B-L are all global symmetry in standard model. Agree?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Otherwise, should we say $G=U(1)_{B-L}$ is gauged in SU(5) g.u.t? Thank you~

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
No! Global means not gauged! In G-G, the U(1) charges in a rep sum to not 0, so not traceless, so not part of the SU(5) gauge group.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
Yes, thanks, but 5¯ and 10 are also gauged under SU(5). Should the best way to describe the
p.s. thus 5¯ and 10 are not the best to describe the
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