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  No local degrees of freedom when connection is flat

+ 5 like - 0 dislike

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

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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

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