Recall that

  • the electron carries U(1) gauge charge -1.

  • the U(1) is gauged and mediated by the U(1) gauge boson which is the photon with zero U(1) gauge charge, thus $0$.

Now let us take this problem in the grand unified theory:




  • the X is a symmetry and associated with a X charge.

  • the Y is a symmetry and associated with a hyper Y charge.

  • question 1: there is a X gauge boson, is that a gauge boson associated to the mediator for the gauged X? So X gauge boson also has a X charge 0?

  • question 2: there is a Y gauge boson, is that a gauge boson associated to the mediator for the gauged Y? So Y gauge boson also has a hypercharge Y charge 0?

Wikipedia suggests that X and Y have zero X charges.

But Wikipedia suggests that X and Y have nonzero B - L charges $\pm 2/3$, and zero X, thus I conclude that X and Y gauge bosons have nonzero Y charges, due to $$ Y = (5(B-L)-X)/2 $$ https://en.wikipedia.org/wiki/X_and_Y_bosons

So how come Y gauge boson does not have zero Y gauge charge? (this is in contrast to the photon and X gauge boson example.)

  • $\begingroup$ You are looking at the wrong hypercharge: strong instead of weak! Need to correct to weak. $\endgroup$ – Cosmas Zachos Jun 21 at 2:55
  • $\begingroup$ I am not sure, maybe you can write a comment/answer about different X and Y charges? and meanings for two hyper charges of Y? $\endgroup$ – annie marie heart Jun 21 at 4:54

I think this is a perfect storm of confused notations, where absolutely every misread symbol of WP is thrown on the fire. To alleviate confusion, I'll skip SO(10) completely, stick to SU(5), and use different symbols for the charges and the gauge bosons. The short answer is that the twelve $\mathbb{X}$ and $\mathbb{Y}$ gauge fields and their corresponding charges have absolutely nothing to do with the charges X and Y under consideration, beyond a (here, tragic) synonymy.

  1. In the context of gauge theory, as here, Y is always the weak hypercharge of the SM, shifting weak isospins, and never the strong hypercharge you cite, shifting plain isospin, which you should henceforth ignore with extreme prejudice. Weak hypercharge treats different chiral states of fermions aggressively differently. It is one of the 12 unbroken generators of SU(5). Certainly not one of the 6 charges corresponding to the $\mathbb{Y}$s!

  2. The WP article on $\mathbb{X}$ and $\mathbb{Y}$ bosons is sound, and bears reading and re-reading to identify them. Three of them, $\mathbb{X}^{+4/3}$, carry anticolor charges; three of them, $\mathbb{X}^{-4/3}$, carry color charges; three of them, $\mathbb{Y}^{+1/3}$, carry anticolor charges; and the last three of them, $\mathbb{Y}^{-1/3}$, carry color charges. The positive ones carry B-L= 2/3 and the negative ones -2/3. Consequently, given the weak hypercharges $Y(\mathbb{X}^{+4/3})=5/3$, $Y(\mathbb{X}^{-4/3})=-5/3$, $Y(\mathbb{Y}^{+1/3})=5/3$, $Y(\mathbb{Y}^{-1/3})=-5/3$, they all have $X=0$. It is non-negotiable for you to go to Li & Cheng or your favorite GUT review (e.g. Langacker, but note he is using the half-hypercharge convention!) and see their position and function in the 5×5 traceless matrix of the 24 gauge bosons of SU(5): otherwise all reasonable discussions are bound to end up in shadowboxing sloughs. When confused, just look at an easy bifermion decay mode of these bosons; see how this answer ensures X=0 for all.

  3. The vanishing electric charge of the photon is a red herring here. It just reflects the fact that the EM group is abelian, and so its gauge transformation does not allow nonlinear couplings in the covariant derivatives thereof, by dint of the vanishing structure constant. Trying to elevate the vanishing X here to such an abelian principle is not appropriate. You already see that the 12 SU(5) charges corresponding to the $\mathbb{X}$ and $\mathbb{Y}$s are ferociously non-commuting, and don't even close to a subalgebra, I hope you may check.

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  • $\begingroup$ Many thanks +1, I will accept soon in a few days if no one post any answer. $\endgroup$ – annie marie heart Jun 23 at 21:30
  • $\begingroup$ "Three of them, $\mathbb{X}^{+4/3}$, carry anticolor charges; three of them, $\mathbb{X}^{-4/3}$, carry color charges; three of them, $\mathbb{Y}^{+1/3}$, carry anticolor charges; and the last three of them, $\mathbb{Y}^{-1/3}$, carry color charges." What does the upper indices have anything to do with carrying color or anti-color charge ? I do not see how one with positive upper index get color charge, while the other one with positive upper index get anti-color charge? $\endgroup$ – annie marie heart Jun 23 at 21:33
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    $\begingroup$ The upper indices are electric charges. As I indicated, WP gives you the mnemonic. So the boson with a uu decay has two antisymmetrized color indices, so one anti color index. Likewise for the ud decay boson. So positively charged gauge bosons are anti color triplets. The opposite holds for the negatively charged ones. You may, of course, read this off the 5x5 matrix. $\endgroup$ – Cosmas Zachos Jun 23 at 22:20
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    $\begingroup$ Indeed, WP is spectacularly wrong. You should proceed to remove that sentence with extreme prejudice... Its author was clearly victimized by the very same confusion dispatched here. $\endgroup$ – Cosmas Zachos Jun 26 at 17:37
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    $\begingroup$ In fact, later down in that article, WP takes special pains to contrast the X charge with (the charge of) the X boson! The wrong sentence must have crept in in the dark of night. $\endgroup$ – Cosmas Zachos Jun 26 at 17:41

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