# $B$-$L$ 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
But B, L are not global symmetry in SU5 or SO10 models. And B-L is a global symmetry in SU5 or SO10 model. Agree?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Yes, to both, as indicated at the beginning. B,L, and B-L are SM global symmetries, although anomalies violate B and L, preserving B-L. Most GUTs violate B and L, but preserve B-L, and of course Y (in the SM), as in G-G proton decay. Have you clicked the checkmark?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
also can B-L symmetry be adiscrete symmetry in standard models, SU5 or SO10? or some of the Proposed GUTs? en.wikipedia.org/wiki/Grand_Unified_Theory#Proposed_theories

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
In SM, SU(5), and SO(10) it is a fine continuous (Lie) symmetry, not just discrete, so why focus on a subgroup, if you have the full enchilada? I am not sure there is a general theory of GUTs --it is always it or miss. They usually have extra fermions for a change, so I don't know how one assigns B and L numbers to those. But, where there is a will, there is a way; recall Gell-Mann--Ramond--Slansky? The ones you link to, however, are truly bad make-work projects. This might be a separate question.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
Dear Sir, I accept the answer. However I am still puzzled a bit. Are we saying that:

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
1) The B-L is a global symmetry in standard models. But B-L is NOT a global symmetry for SU(5) and SO(10) g.u.t.?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
2) The X is a global symmetry in SU(5). But X is NOT a global symmetry for standard models and SO(10) g.u.t.?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
3) what are the modified "B-L" in SO(10) g.u.t. then?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
p.s. the reason I said 1), is because that the $\bar{5}$ and 10 each of them do not carry the same B-L, but each of them does carry the same X sym

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
"B-L is a symmetry in SU(5), so proton decay to a pion and a positron" "B-L is gauged in for SO(10)", so what happened to proton decay in SO(10)?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
When we say $G$ is a global symmetry with a symmetry group $G$, we mean that we should be able to find gauge invariant charged operator that carry the symmetry charge.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Continued - However, for $G=U(1)_{B-L}$, all the charged operator that carry the symmetry charge given as the 5¯ and 10 above, or other representations like 16, they are all gauge non-invariant fields/operator. If so, how can we say $G=U(1)_{B-L}$ is a global symmetry? Can we identify other gauge invariant fields/operator charged under the $G=U(1)_{B-L}$?

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
$U(1)_{B-L}$ generators/charges commute with SU(5) gauge generators, and are, of course, SU(5) singlets, made up of bilinears $\bar 5 \cdot 5$ and $\bar 10 \cdot 10$, manifestly gauge invariant.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
many thanks, almost clear. But 5¯⋅5 and 10¯⋅10 are in trivial representation of
No, think of the electric charge operator. It generates U(1) rotations and is also an EM singlet, and a color singlet, etc...

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
B-L is a good global symmetry for the SM, agree. But $U(1)_{B-L}$ does not transform SU(5)'s representation $5¯$ and $10¯$ in a uniformed manner, even under each multiplet. How can I get convinced that B-L is a good global symmetry for SU(5) g.u.t?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Let us continue this discussion in chat.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
I see, X is the $U(1)_{B-L}$ for the G-G theory. The original $U(1)_{B-L}$ is not global symmetry for the G-G theory, yes?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Is that $e^+_L=\overline{e^-}_R$ always true in QFT?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
On your right hand side, the R should be inside the overbar, so it turns to an L as it slips past the $\gamma^0$ thereof, and the two sides have the same chirality. Beyond that, yes, of course. It’s not in your text?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
Excuse me - I come back to read you said but got confused "SO(10), except here X is now a generator of SO(10) and is then gauged (hence SSBroken), hence violated by small amounts."

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
--> (1) how can the $X$ which acts on $16= \bar{5}+10$ above differently, say $X=-3$ for $\bar{5}$ and $X=1$ for $10$ ---- can still be a generator of SO(10)?Can you clarify?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
--> (2) If this $X$ is gauged, then how is that SSBroken? (which group is SSBroken? down to which group?)

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
--> (3) why violated by small amounts? can you give a ref?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
You said "1) B-L is a good global symmetry for the SM and SU(5), and local for SO(10)." But I understood differently as

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
--> (4) B-L is a good global symmetry for the SM. But B-L is NOT a global symmetry for the SU(5) GG. Only X is a global symmetry for the SU(5) GG. Do you agree? X is not B-L.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
--> (5) I do not see how the B-L nor the X be a good symmetry generator act on the 16 of SO(10). So I regard that B-L and X are NOT good symmetry of SO(10). Do you agree?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
I think the Wilczek & Zee ref in here and Ozer's thesis linked above should answer all questions. B-L is the difference of two SU(5) and SO(10) generators.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
(1) X is not a diagonal operator for the 16 of SO(10): different components are acted upon differently. (2) SO(10)-->SU(5). (4) (B-L) is a linear combination of two SU(5) GG symmetries, X global and Y local. (5) They are good symmetries as generators, in Ozer's thesis.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
-thanks a lot -!

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

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
thanks Cosmos, btw I do NOT think X (or 5(B - L)- 2 Y) in SO(10) GUT is a continuous U(1) group. It shall be a discrete group, is that true?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
These are Lie group generators generating Noether currents. Why discrete?

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
I think X (or 5(B - L)- 2 Y) in SO(10) GUT is at the $Z_4$ center of Spin(10), which is discrete. I am not sure there is any concrete way to embed the U(1) outside of SU(5) to Spin(10) in the way that people said to have the continuous U(1) for X, at least not for the SO(10) GUT with the Spin(10) group ...

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Ask a separate question, but be explicit.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
thanks - I will <3

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
I propose that
Slansky Table 14, the starting point of any and all such questions, as your instructor in your GUT course must have emphasized to you.

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user Cosmas Zachos
but those Table seem to include only Lie algebra, I am not sure they have Lie group consideration included. Thank you a good ref I just learned from you!

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
btw I have one more basic puzzle learning GUT: physics.stackexchange.com/questions/560676

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
maybe you also know already physics.stackexchange.com/questions/564976

This post imported from StackExchange Physics at 2020-11-28 23:03 (UTC), posted by SE-user annie marie heart
Don't know much about its advantages. Gives you longer life for the proton...

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

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