# Does magnetic monopole violate $U(1)$ gauge symmetry?

Does a magnetic monopole violate $U(1)$ gauge symmetry? In what sense and why?
Insofar as I know, there are at least two types of magnetic monopoles. One is the Dirac monopole while the other is the monopole in Grand Unified Theories (GUT), e.g., 't Hooft-Polyakov monopole.

In the latter case, some non-Abelian gauge theory breaks down to a (compact) $U(1)$ gauge theory wherein monopoles can be observed from a large distance. The monopole field is not from the $U(1)$ degrees of freedom and nothing is singular.
On the contrary, Dirac monopole is singular (topological defect?) and we must introduce Dirac string or patches of vector potential $\vec{A}(x)$. Are these no more than artifacts arising when one tries to describe the monopole field within the $U(1)$ gauge theory by assuming $\vec{B}=\nabla\times\vec{A}$? It seems true since two Maxwell's equations that magnetic monopole might affect can be cast in the differential form $\mathrm{d}F=0$ and $F=\mathrm{d}A$ in Minkowski spacetime, from which $U(1)$ gauge invariance manifests for $\mathrm{d}(A+\mathrm{d}\chi)=\mathrm{d}A$.

However, I later noticed two seminal papers on Dirac monopole http://dx.doi.org/10.1103/PhysRevD.12.3845 & http://dx.doi.org/10.1016/0550-3213(76)90143-7, in which the authors claimed the following

• electromagnetism without monopole $\rightarrow$ connection on a trivial $U(1)$ bundle
• electromagnetism with monopole $\rightarrow$ connection on a nontrivial $U(1)$ bundle
And the wave function of an electron around a Dirac monopole should be regarded as a section which is free from discontinuities.

I've no idea about fibre bundle. Anyway I guess $U(1)$ theory does not necessarily expel Dirac monopoles. It has something to do with the topology of some manifold in the theory? Which manifold? Can anyone shed light on this intriguing issue? Thanks in advance.

• Maybe Yang's paper "Magnetic monopoles, fiber bundles, and gauge fields" can be of help. – Robin Ekman Feb 25 '15 at 10:37
• Regarding your last question, magnetic monopoles arise if there is a broken symmetry group $G\to H$, and the group $\pi_2(G/H)$ is non-trivial. I'd recommend Weinberg's text on classical solutions in quantum field theory. – JamalS Feb 25 '15 at 11:37

No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the divergence of the magnetic field does no vanish (it has a source/sink there!), and hence the equation that allows us to define the gauge field, namely $$\mathrm{d}F = 0$$, is not fulfilled.
It is valid everywhere else though, and so we consider the gauge theory on space with a single point removed. But $$\mathbb{R}^3 - \{0\}$$ is homotopically equivalent to the sphere $$S^2$$, which is topologically non-trivial - you cannot contract the sphere or $$\mathbb{R}^3-\{0\}$$ smoothly to a point because the monopole is "in the way". This is not a true defect on spacetime, but simply a consequence of us "rescuing" the gauge description although the monopole prohibits us from doing so globally. It is a defect in the gauge theory.
The local description on the sphere $$S^2$$ is most easily achieved by just taking two local gauge fields that are defined on the hemisphere overlapping at the equator, and we get a globally consistent solution of we define a gauge transformation on the overlap that glues the local solutions together1, which is just a circle - $$\mathrm{U}(1)$$ again. So, we must give a smooth map $$\mathrm{U}(1)\to\mathrm{U}(1)$$ from the circle to itself. Such maps are just given by winding the circle $$n$$ times around itself, if you write $$\mathrm{U}(1) = \{\mathrm{e}^{\mathrm{i}\phi} \vert \phi\in[0,2\pi)\}$$, then the transition function is $$\mathrm{e}^{\mathrm{i}\phi}\mapsto\mathrm{e}^{n\mathrm{i}\phi}$$. This transition function fully characterizes the bundle, hence there are now $$\mathbb{Z}$$ different bundle structures that can occur. The $$n\in\mathbb{Z}$$ just characterizes the monopole as having magnetic charge $$\frac{2\pi}{e}n$$.
Now, the "Dirac string" is merely an artifact that occurs if we try to force a global solution out of the local ones. If you take the solution on one of the hemispheres and extend it as far as you can, you find that you can extend it to everywhere but the opposite pole. If we "morph back" the sphere into $$\mathbb{R}^3-\{0\}$$, then the pole (like all points) morphs into a ray starting at the location of the monopole. On this ray which corresponds to the pole, the solution is not defined - this is the famous Dirac string. But recall that we had another local solution - if we glue them again (or switch between them as we need to), we get a description on the whole of $$\mathbb{R}^3 - \{0\}$$, and the Dirac string vanishes. The quantization as $$\frac{2\pi}{e}$$ for the magnetic charge arises from the requirement that this artifact must be undetectable in the local solutions, and hence is not allowed to cause a physical consequence, and only for these magnetic charges, the Aharonov-Bohm effect from such a ray vanishes.
The gluing of the local solutions in fact is a construction of the principal bundle without us explicitly saying so known as the construction of a bundle by cocycles. If we have a global solution, then we need no gluing, and we choose $$n = 0$$, which corresponds to the trivial bundle, and no magnetic monopole, since the solution is then, in principle, extendable to all of $$\mathbb{R}^3$$ and we made a whole fuzz about nothing.
1The gluing technically works like this - take the gluing transformation $$\chi : S^1 \to \mathrm{U}(1)$$ and the two local solutions $$A_1,A_2$$ and set $$A_2 = A_1 + \mathrm{d}\chi$$. Or rather, look at your local solutions and find $$\chi$$ such that this works.
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