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One is aware that in the axial gauge (say the light-cone gauge $A_{-}=0$) non-supersymmetric Chern-Simons' theory is a quadratic theory. Hence in this gauge there are no gauge-gauge interactions. Then what is non-Abelian about this theory?

  • Are there effects which exist inspite of the above which make the theory different from the Abelian one?

  • I am aware that there are calculations of exact partition functions for Chern-Simons' theory on compact 3-manifolds. Is that the subtle point that the non-Abelian-ness will somehow manifest itself when on compact space-times? (..or on manifolds with boundary?..) That there is some reason such convenient axial gauges may not exist for compact space-times? (..though i can't see why one can't always choose $A_{-}=0$..because 3-manifolds are parallelizable there aren't any topological restrictions but there could be an issue with Gribov ambiguities..I don't know and would like to know about)

  • I guess that for supersymmetric Chern-Simons' theories this question itself doesn;t arise since I guess there are no gauge choices where the theory will become quadratic in the gauge fields.

(- related i would love to know of any proof/argument as to why there can't be such a gauge choice for YM theory which will make it quadratic - axial gauge can kill the quartic term but I guess thats the best one can do..)

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Non-Abelian simply means "related to non-commuting groups" which non-Abelian CS theory surely is. Your light-cone gauge choice is very bad to study the interesting quantities in CS theory that depend on the spacetime topology or the insertion of defects. It's only valid when the CS theory is pretty much trivial and vacuous. Yes, if you tried to improve the light-cone gauge to be compatible with other topologies, the non-commuting character of the theory would get back. –  Luboš Motl Dec 17 '12 at 17:07
    
@LubošMotl But light-cone gauge is quite simplifying - look at the recent higher-spin theory literature! Without the light cone gauge its hard to see how one could have gotten these exact in the `tHooft limit answers for 2-point functions and free energies. And in these cases I don't know what is non-Abelian. So when one is in such scenarios of not having topological issues, then is non-Abelian CS any different from an Abelian one? –  user6818 Dec 17 '12 at 18:14
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I think what you are missing is the fact that, in the above gauge-fixing it is true that the action looks quadratic and non-interacting. However, with the gauge fixing follows a non-linear constraint which essentially encodes the self-interactions (and thereby, non-abelian structure) of the theory. All allowed gauge-fluctuations MUST satisfy the constraint and you are not allowed to forget it. –  Heidar Aug 20 '13 at 6:37

1 Answer 1

I don't know so much about the Chern-Simons theory and its use in supersymmetric theories, so I can not really help regarding all your questions. But the difference between abelian and non-abelian Chern-Simons term is crystal clear from its definition.

Usually, a Chern-Simons term reads in general (below for a $2+1$-dimensional problem)

$$L_{\text{CS}}=\dfrac{k}{4\pi}\int\text{Tr}\left\{ A\wedge dA+\dfrac{2}{3}A\wedge A\wedge A\right\} =\dfrac{k}{8\pi}\int\varepsilon^{ijk}\text{Tr}\left\{ A_{i}\left(\partial_{j}A_{k}-\partial_{k}A_{j}\right)+\dfrac{2}{3}A_{i}\left[A_{j},A_{k}\right]\right\} $$

such that, for an abelian gauge-field, one as

$$L_{\text{CS}}=\dfrac{k}{8\pi}\int\varepsilon^{ijk}\text{Tr}\left\{ A_{i}\left(\partial_{j}A_{k}-\partial_{k}A_{j}\right)\right\} $$

since $\left[A_{j},A_{k}\right]=0$ from the previous definition. In practise, it means that the gauge-field is in the Lie algebra related to the $\text{U}\left(1\right)$ Lie group.

So in short, the abelian Chern-Simons term is just a restriction of the Chern-Simons term for abelian gauge.

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