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  No monopoles in the Weinberg-Salam model

+ 2 like - 0 dislike

I'm reading Chapter 10.4 on the 't Hooft-Polyakov monopoles in Ryder's Quantum Field Theory.

On page 412 he explains why magnetic monopoles cannot appear in the Weinberg-Salam model. I'm I right by saying that he shows that the electromagnetic gauge group $U(1)_{em}$ is not compactly embedded into the $U(1)\times U(1)$ subgroup of $SU(2)\times U(1)$?

He then immediately concludes that the first fundamental group of the unbroken symmetry, which is $H=U(1)_{em}$, $\pi_1(H)$ must be trivial or doesn't exists. Could someone refer me why?

Comment: I know that in the $SU(2)\times U(1)$ ones must consider the second homotopy group from $S^2$ to the orbit $G/H=SU(2)\times U(1)/U(1)$, where $H$ is the isotropy group of a vacuum state, after symmetry breaking. But the second homotopy group of a quotient can be related through a exact series to the kernel of the map from $\pi_1(H)$ into $\pi_1(G)$.

What I do not understand is by which theorem for $H$ having a non-compact covering group $\pi_1(H)$ must be trivial or non-existing (???)?

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user Anne O'Nyme
asked May 4, 2014 in Theoretical Physics by Anne O'Nyme (175 points) [ no revision ]
I don't own Ryder so I am not sure what you are asking, but you might be interested in this answer.

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user Hunter
Although it is in the same line, it doesn't answer my question. I've also looked into the references but they don't seem to use the same argument.

This post imported from StackExchange Physics at 2014-05-04 11:10 (UCT), posted by SE-user Anne O'Nyme

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