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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
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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

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up vote 9 down vote accepted

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

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