10
$\begingroup$

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?

$\endgroup$

1 Answer 1

14
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Gauging a discrete symmetry means you only keep the states that are invariant under the discrete group. The ones which are not invariant, you throw them out. $\endgroup$ May 31, 2022 at 1:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.