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
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  No local degrees of freedom when connection is flat

+ 5 like - 0 dislike
725 views

I was studying Chern-Simons theory and variation of action gives us the flatness conditions $\mathrm{d} A + A \wedge A = 0$. I am wondering how to see that this implies there are no local degrees of freedom.

And what precisely does it mean that a degree of freedom is local?

This post imported from StackExchange Physics at 2014-09-05 08:27 (UCT), posted by SE-user fales
asked Sep 4, 2014 in Theoretical Physics by fales (25 points) [ no revision ]
Possible duplicate: physics.stackexchange.com/q/98484/2451

This post imported from StackExchange Physics at 2014-09-05 08:27 (UCT), posted by SE-user Qmechanic

2 Answers

+ 5 like - 0 dislike

Recall that $\mathrm{d}A + A\wedge A = F = 0$ means that the field strength is vanishing, i.e. the gauge field is always pure gauge locally.

Local degrees of freedom would mean that the equation of motion ($F = 0$) has more than one local solutions that are not related by a symmetry of the theory. But the field being pure gauge locally means that it can always be locally transformed to be $A = 0$, so the local solutions are uniquely zero, thus implying there are no local degrees of freedom.

Globally, the solutions are given by the finite-dimensional space of flat connections modulo the gauge transformations.

Note that we are talking about 3D Chern-Simons here, the higher dimensional CS theories do exhibit local degrees of freedom, see arXiv/hep-th/9506187.

This post imported from StackExchange Physics at 2014-09-05 08:27 (UCT), posted by SE-user ACuriousMind
answered Sep 4, 2014 by ACuriousMind (910 points) [ no revision ]
+ 2 like - 0 dislike

Here is another perspective. By the state-operator correspondence, the Hilbert space a TQFT assigns to a sphere (of codimension 1) is the space of local (point) operators. We compute this Hilbert space by quantizing the phase space of classical configurations on the sphere. Since the sphere is simply connected, there is a unique classical configuration there for Chern-Simons theory. Thus, the algebra of classical observables is one dimensional. If this is to quantize to an irreducible representation, then this Hilbert space must be one dimensional (a more careful analysis can identify the prequantum vector bundle).

EDIT: Interestingly, as noted by ACuriousMind in the other answer, higher dimensional Chern-Simons theories *do* admit local operators. One can see the difference in this perspective by noting the equations of motion eg. in 5d read F ^ F = 0 and do not imply F = 0 except in the abelian case.

answered Sep 6, 2014 by Ryan Thorngren (1,925 points) [ revision history ]
edited Sep 6, 2014 by Ryan Thorngren

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$\hbar$ysicsOverf$\varnothing$ow
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
...