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  Are lens spaces classified via a Weinberg angle?

+ 7 like - 0 dislike

I am thinking about Kaluza Klein theory in the 3 dimensional lens spaces. These have an isometry group SU(2)xU(1), generically, and in some way interpolate between the extreme cases of manifolds $S^2 \times S^1$ and $S^3$ (or, arguably depending of your parametrisation, across $S^2 \times S^1 ... S^3 ... S^1 \times S^2$).

My question is, simply, if the application of Kaluza Klein construction forces different coupling constants to the two groups $SU(2)$ and $U(1)$, and then a kind of discrete Weinberg angle, or if they have always the same coupling.

Note that when you build a lens space bundle over CP2 (which is equivalent to a $S^1$ bundle over $CP^2$ x $CP^1$, if I read correctly a remark of Atiyah reported by Kreck and Stolz), you get an isometry group of SU(3) times SU(2) times (1), so the same question could be asked here, in this more widely known setup. Of course that odd dimensional spaces do not carry chirality, but my question is just about having different couplings, that is all. It could be interesting, as a plus, to compare with coupling relationships coming from Non Commutative Geometry...

This post has been migrated from (A51.SE)
asked Oct 1, 2011 in Theoretical Physics by anonymous [ no revision ]

1 Answer

+ 9 like - 0 dislike

If your definition of a lens space means $S^3/\Gamma$, a quotient of the three-sphere, then the $U(1)$, $SU(2)$ couplings may (classically) be derived from the parent theory on the full three-sphere whose isometry is $SO(4) \approx SU(2) \times SU(2)$ (at the level of Lie algebras). The subgroup $\Gamma$ acts only inside one of the $SU(2)$ factors, let's say the second one.

Because the original $S^3$ had a symmetry between both $SU(2)$ factors, their gauge couplings coincided. However, you assumed a $\Gamma={\mathbb Z}_k$ case because you claimed that the second $SU(2)$ was broken to a $U(1)$ subgroup; that wouldn't be true for the non-Abelian groups $\Gamma$ isomorphic to the isometries of the Platonic polyhedra. The orbifolding by ${\mathbb Z}_k$ means that only states with $J_{{\rm second},3}$ being a multiple of $k$ are kept in the spectrum. It's therefore natural to define the new elementary charge $e$ as $k$ times the elementary charge in the parent, unorbifolded theory. So the natural value of the Weinberg angle obeys $\tan \theta_W=g'/g = k$; the primed $g$ refers to the $U(1)$ and is amplified $k$-times.

The charges of charged particles under the first $SU(2)$ are also affected by the orbifolding, however, and they're correlated with the charges under the $U(1)$: for large values of $k$, for example, you won't find any triplet under the first $SU(2)$ in the spectrum (but it's less trivial to say which reps of the first $SU(2)$ appear in the spectrum). For this reason, it's somewhat strange to say that the Weinberg angle becomes extreme. In some moral sense, the natural couplings of both gauge groups are still equal even in the orbifolded theory.

This post has been migrated from (A51.SE)
answered Oct 1, 2011 by Luboš Motl (10,278 points) [ no revision ]
In this case, thanks for the answer! I will wait for some days before accepting, just in case someone wants to propose a more extended or general one.

This post has been migrated from (A51.SE)
Comment: the possibility of redefine the new elementary charge also sounds a bit as the freedom in the normalisation of the hypercharge.

This post has been migrated from (A51.SE)

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