Understanding the statement "orbifold theories are QFTs with finite gauge group"

Question

I'd like to understand the equivalence of orbifold theories in string theory and (2D worldsheet) QFTs with finite gauge group, using the path integral.

Suppose my action is $$S= \frac{1}{2\pi \alpha'} \int d^2\sigma (\partial_i X^\mu \partial^i X_\mu)$$ and the $X^\mu(\sigma)$ are invariant under some finite group action $\Gamma$. To construct the orbifold theory on a Riemann surface $\Sigma$, I want to take the path integral over the untwisted and twisted sectors, i.e. I want to average over all boundary conditions on $X^\mu$ in which $X^\mu$ is periodic up to $\Gamma$-action.

Now, on the other hand, if I want to compute the path integral of a QFT with a finite gauge group, I would "gauge-fix" and then compute the path integral.

How can I see that the two approaches are the same?

This post imported from StackExchange Physics at 2019-09-16 20:25 (UTC), posted by SE-user Dwagg

Comments

When the gauge group is finite, does computing the path-integral require fixing the gauge? If not, then does the question remain if gauge-fixing isn't used?

This post imported from StackExchange Physics at 2019-09-16 20:25 (UTC), posted by SE-user Chiral Anomaly

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@ChiralAnomaly Good question, I am not sure. Indeed gauge-fixing may not be required for finite groups, or the gauge redundancy may only contribute an overall factor. As for the second question-- yes, regardless, I can't understand how the path integral of a QFT with a finite gauge group relates to the path integral prescription for an orbifold theory, with all its twisted boundary conditions and such.

This post imported from StackExchange Physics at 2019-09-16 20:25 (UTC), posted by SE-user Dwagg