# Magnetic monopoles gauge theories

I'm quoting 't Hooft:

"[...] Locally stable field configurations may exist that have some topological twist in them [...].Careful analysis of the existing Lie groups and the way they may be broken spontaneously into one or more subgroups $U(1)$, reveals a general feature: Only if the underlying gauge group is compact, and has a compact covering group, must electric charges in the $U(1)$ gauge groups be quantised (otherwise, it would not be forbidden to add arbitrary real numbers to the $U(1)$ charges), and whenever the covering group of the underlying gauge group is compact, magnetic monopole solutions can be constructed. [...]"

1. What are the covering groups?

2. What did he mean by saying electric charges are quantised only when the gauge group and the covering group are compact?

3. And finally how can magnetic monopoles be constructed out of quantised electric charges?

1) Universal covering groups are groups with the property of being simply connected. Each algebra has a unique covering group. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way $$G=\frac{\tilde G}{Ker(\rho)},$$ where $Ker(\rho)$ is the kernel of the group homomorphism $\rho:\tilde G\rightarrow G$. Once you have defined a particular representation you are able to compute this kernel. For example, you start with an $\mathfrak{su}(2)$ algebra. Then if you choose the adjoint representation you can show that $Ker(\rho)=\mathbb Z_2$ and the group will be $G=SU(2)/\mathbb Z_2=SO(3)$. On the other hand, if you choose the defining representation you get $Ker(\rho)=\mathbb 1$ and $G=SU(2)/\mathbb 1=SU(2)$.
2) A topological magnetic monopole has to satsfy the quantization condition $$e^{ieQ_m}=\mathbb 1,$$ where $Q_m$ is the (non-Abelian) magnetic charge. This is a generalization of the Dirac quantization condition. It can be shown that to satisfy this condition the $U(1)$ has to be compact because the electric charge has to be quantized as well. I am not quite sure about the result he is claiming: "a compact $G$ with compact covering $\tilde G$ implies in $U(1)$ compact". The result I know is that that when you have a spontaneous symmetry breaking $G\rightarrow K\times U(1)$, the $U(1)$ is compact if $G$ and $K$ are both semisimple. Otherwise $U(1)$ may be non compact.