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Gauging discrete symmetries

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I read somewhere what performing an orbifolding (i.e. imposing a discrete symmetry on what would otherwise be a compactification torus) is equivalent to "gauging the discrete symmetry". Can anybody clarify this statement in any way? In particular what does it mean to gauge a discrete symmetry if no gauge bosons and no covariant derivatives are introduced?

This post imported from StackExchange Physics at 2014-03-31 16:01 (UCT), posted by SE-user Heterotic
asked May 14, 2013 in Theoretical Physics by Heterotic (515 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

1 Answer

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I think I got the answer now. The main idea is this: When we gauge continuous symmetries we identify all the states $$A^\mu=A^\mu+\partial^\mu\chi$$ (which are continuously many) as a unique physical state.

When we gauge a discrete symmetry (let's assume it's generated by $\theta$) we identify all the states $$|\Psi\rangle=\theta^n|\Psi\rangle$$ where $n=1,\ldots N-1$ and $N$ is the order of the discrete symmetry group. So we identify a finite number of states as a unique physical state. This is exactly what we do when we are orbifolding.

This post imported from StackExchange Physics at 2014-03-31 16:01 (UCT), posted by SE-user Heterotic
answered May 14, 2013 by Heterotic (515 points) [ no revision ]

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