# Why isn't weak hypercharge quantized to integer values?

Page 527 of Srednicki's QFT book lists the matter content of the Standard Model as

left-handed Weyl fields in three copies of the representation $(1, 2, −1/2)⊕(1, 1, +1)⊕(3, 2, +1/6)⊕(\bar{3̄}, 1, −2/3)⊕(\bar{3}, 1, +1/3)$, and a complex scalar field in the representation $(1, 2, −1/2)$.

But the irreducible representations of $\text{U}(1)$ are just the endomorphisms $e^{i \theta} \to e^{i n \theta}$ indexed by the integer $n$, as described here. What does it mean for the matter fields to lie in non-integer representations of the $\text{U}(1)$ gauge group?

Put more simply: in QED, the compactness of the electromagnetic gauge group $\text{U}(1)$ quantizes the electric charge to integer multiples of $e$. Why doesn't the compactness of the weak hypercharge gauge group $\text{U}(1)$ do the same thing to weak hypercharge in the Standard Model?

• ? Aren't they all integer multiples of 1/3, a universal normalization? Note that author uses a hypercharge 1/2 the conventional one. – Cosmas Zachos Jul 11 '17 at 19:09
• Why do you worry about hypercharge but not quark electric charge? And how do you know that electron charge is $e$ and not $3\,e$? Or $36369\,e$? Or any $3n\,e$? – OON Jul 11 '17 at 19:11
• Here. So, define $Y=6(Q-T_3)$ unlike Srednicki's $Y=Q-T_3$ and you have all integer hypercharges. – Cosmas Zachos Jul 11 '17 at 19:18
• @CosmasZachos Thanks, that factor of $2$ difference in normalization conventions clears things up. This is perhaps just definitional quibbling, but the elementary electric charge is actually $e$, not $e/3$. But particles (like quarks) that are confined by a nonabelian gauge field are allowed to "break the rules" and have fractional charge, because the color flux tubes make up the "missing" Aharanov-Bohm phase factor, as discussed here. – tparker Jul 12 '17 at 0:57
• @OON ... $n/(2 \times (\text{electron charge}))$ for some $n>1$, then that would be evidence that the elementary electric charge is actually a fraction of the electron charge (although not a conclusive proof, because it would be possible that the magnetic monopole that we found didn't have unit magnetic charge instead). If we found a magnetic monopole with a magnetic charge that wasn't an integer multiple of $1/(2 \times (\text{electron charge}))$, then we'd really have to go back to the drawing board. – tparker Jul 12 '17 at 1:25