Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,791 comments
1,470 users with positive rep
820 active unimported users
More ...

  Gauging discrete symmetries

+ 5 like - 0 dislike
661 views

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 (525 points) [ no revision ]
retagged Apr 19, 2014 by dimension10

1 Answer

+ 5 like - 0 dislike

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 (525 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ys$\varnothing$csOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...