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

+ 2 like - 0 dislike

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
asked Aug 23, 2019 in Theoretical Physics by Dwagg (10 points) [ no revision ]
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
@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

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