Is there a specific geometric meaning why fractional charges are allowed in SU(N) gauge theories?

So in the standard model of particle physics, there exist particles with fractional charge. What this means geometrically is as follows: We are given a smooth manifold with a principal $U(1)$ bundle $P$. Then, as far as I understand, we can construct the associated vector bundle to the one-dimensional representation $\phi \mapsto e^{i\theta/2}\phi$, where $\theta \in \mathbb{R}$. (For a charge of 1/2, or more generally $\phi \mapsto e^{iq\theta}\phi$).

Now, this isn't actually a representation of $U(1)$, it's a representation of a double-cover $U(1) \rightarrow U(1)$, $z \mapsto z^2$ (a somewhat similar problem arises in constructing bundles associated to spin representations). So to proceed with the geometric construction one needs to construct a double-cover of the bundle $U(1)$.

However, particles with fractional charge generally arise in theories with $SU(N)$ also acting. Explicitly, one situation is where we have a manifold $M$ with a principal $U(1) \times SU(2)$ bundle $P$, and we want to construct an associated vector bundle to the representation $\frac{1}{2} \otimes V$, where $V$ is the defining 2-dimensional complex representation of $SU(2)$ and $\frac{1}{2}$ denotes the 'half-charge representation' of $U(1)$. So we need to construct a bundle corresponding to the double cover $U(1) \times SU(2) \rightarrow U(1) \times SU(2)$ which is the identity on the second factor.

A similar thing occurs with quarks, which transform under an $SU(3)$ symmetry, and also have charges with denominators 3.

My question is: Is there a geometric reason why the factor $SU(2)$ or $SU(3)$ allows for fractional charge? Does having a principal $U(1) \times SU(N)$ bundle canonically give an $n-$fold cover? Page 16 of this paper: http://arxiv.org/abs/hep-th/9701069 seems to suggest it's related to the fact that $SU(N)$ has a center consisting of $N$ elements (more specifically, isomorphic to $\mathbb{Z}/n\mathbb{Z}$), but I can't quite follow the argument.

Thanks, and also please correct me if my understanding of the situation is off.

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The group $Spin^c$ vaguely comes to mind... In any case, spin geometry is probably what you want. –  David Roberts May 22 at 0:59
I don't really understand the question. What do you mean by "allow"? Of course it's the $\text{U}(1)$ factor that allows us to talk about these fractional charge representations. –  Qiaochu Yuan May 22 at 1:32
I mean, can one construct a vector bundle on which sections can be said to have fractional charge. –  Alex Zorn May 22 at 14:36

I'm afraid that this answer will be somewhat physics-y. Apologies if this is deemed inappropriate for MO.

First of all, I think that it is slightly misleading to say that one has "fractional charge in $SU(N)$ gauge theory." A careful reading of the lectures linked in the question reveals that one actually has a Yang-Mills-Higgs theory with group $G$ and then chooses a Higgs configuration which breaks the symmetry from $G$ down to a subgroup $H$. Essentially one looks for finite-energy configurations of the Higgs field, which means that the value of the Higgs field must asymptotically (in space) go to a zero of the potential. The potential is $G$-invariant, but its zeros are only invariant under $H$. In this branch of the theory, the unbroken gauge symmetry is $H$. So in fact one has a gauge theory with gauge group $H$.

Now to talk about charge one must first identify a $U(1)$ subgroup of $H$ as the gauge group of electromagnetism. In the examples considered in that paper, the unbroken gauge group $H$ is such that its Lie algebra $\mathfrak{h} = \mathfrak{u}(1) \oplus \mathfrak{k}$, where $\mathfrak{u}(1)$ is the Lie algebra of the electromagnetic subgroup $U(1) < H$ and $\mathfrak{k}$ is the Lie algebra of a subgroup $K < H$. The subgroups $U(1)$ and $K$ may intersect nontrivially, but do so centrally, that is, $U(1) \cap K \subset Z(K)$, the centre of $K$. In other words, lettting $Z = U(1) \cap K$, then the subgroup $$H \cong (U(1) \times K)/Z$$

Now the allowed charges are those which satisfy the Dirac quantisation condition in the presence of the monopole. Because of the quotient by $Z$, the quantisation condition says that the charge $Q$ (in the right units) satisfies not $\exp(i 2\pi Q) = 1$, which would say it is integral, but rather $\exp(i 2\pi Q) \in Z$. If one takes $K = SU(N)$ and take $Z \cong \mathbb{Z}/N\mathbb{Z}$ to be the full centre, then one finds that $Q$ can be fractional with denominator $N$.

It is interesting that there are two known mechanisms for the "quantisation" of charge: the Dirac quantisation condition in the presence of a magnetic monopole and the embedding of the electromagnetic $U(1)$ in a semisimple unification group $G$, whence the quantisation comes from the fact that in an irreducible representation of $G$, the weights of $U(1)$ will naturally be quantised. In the context of the lectures mentioned by the OP, the two mechanisms are two faces of the same coin.

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It seems the question could have been asked in a more math-friendly way. As is it seems like a physics question to me, but suitable interpreted its a math question in disguise... –  Ryan Budney May 22 at 2:48
That's fair Ryan, it is a physics question but I wanted a mathematical answer so this seemed like the best place to ask. –  Alex Zorn May 22 at 14:28
I think you're probably right, I'm trying to think how precisely to ask this in a mathematical way but I'm not sure how to do so. –  Alex Zorn May 22 at 14:34