Left-right topology

Question

Are there non-trivial topological solutions (in particular 't Hooft-Polyakov magnetic monopoles) associated with the (local) breaking

\begin{equation} SU(2)_R \times SU(2)_L \times U(1)_{B-L} \to SU(2)_L \times U(1)_Y \end{equation}

?

Answer 1

In order for a theory to present stable monopole solutions it has to satisfy three requirements:

i) It has to have the topological conditions, generally showed as non trivial second homotopy group of the vacuum manifold.

ii) It has to satisfy a 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.

iii) The monopole has to be a solution of the classical equations of montion.

It can be shown that to satisfy ii) the $U(1)_{em}$ has to be compact (the electric charge has to quantized as well), that is, isomorphic to the circle and not to the reals. It turns out that when you have a SSB $G\rightarrow K\times U(1)$, the $U(1)$ is compact if $G$ and $K$ are both semisimple. Otherwise $U(1)$ may be non compact. In your case, $G$ is not semisimple, it has an Abelian factor.

For chiral symmetry breaking such as in QCD where the symmetry is global there are definitely no 't Hooft-Polyakov monopoles since those appear when you spontaneously break a local gauge symmetry. You are breaking a global one.

There are some studies about something called "semilocal defects" that may appear when you break a local and a global symmetry in a "mixed way".

What characterizes the stability of topological solutions as monopoles and vortices are topological quantities as the winding number. Take a vortex, for simplicity. Roughly speaking the winding number would say how the scalar field rotate as we go around the vortex. This rotation is in the internal space. With a global symmetry you are not able to construct such a rotating scalar field. The topological numbers would be trivial.