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  Why is the semiclassical approximation of the abelian Chern-Simons theory exact?

+ 4 like - 0 dislike

I was told that in abelian Chern-Simons theory (say, with a general level matrix $K$), semiclassical approximation is exact because there is no trivalent vertex, which in non-abelian case makes the perturbative series factorially divergent. Why should the converging perturbative series in this case give us an exact result? For instnace, there might be some terms which vanish to all order of perturbation, which is actually the case for non-abelian complex Chern-Simons theory where we consider trans-series, taking into account all the instanton effects, and then do the resurgent analysis. Why is this not the case for abelian Chern-Simons theory?

I am also curious if the exactness of semiclassical approximation in the abelian Chern-Simons theory is somehow related to that of supersymmetric field theories where the effect of bosons and fermions cancel each other out, making the situation localized near the vacua.

This post imported from StackExchange Physics at 2018-06-28 14:15 (UTC), posted by SE-user Sunghyuk Park
asked Jun 8, 2018 in Theoretical Physics by sparklet (20 points) [ no revision ]
retagged Jun 28, 2018

It is a condensed matter physics observation interpreted in the formalism of the Chern-Simons theory. Why it would be classical ?

Maybe the instanton contributions are reduced to the semi-classical solution, if you mean a WKB-like solution.

1 Answer

+ 1 like - 0 dislike

Unless I am missing something, abelian Chern-Simons theory is a free theory, with a quadratic action and so you can just do the path integral. If the topology of the spacetime 3-manifold is complicated enough (e.g. non-trivial torsion in H_1), you can have different topological sectors and you have to sum over them. If the first Betti number is non-zero, you have a non-trivial moduli space of classical solutions ("zero modes") and you have to integrate over them. Apart from that, what remains is a Gaussian integral over non-zero modes.

answered Jun 28, 2018 by 40227 (5,140 points) [ no revision ]

Thanks for the answer. I knew that it is a typical argument in physics saying that we can do the path integral just because it is a free theory. But what is the definition of path integral after all? As far as I know, there is no consistent mathematical definition, and it is unclear to me what does it mean by a path integral or an infinite dimensional Gaussian integral on manifold. If you define it using Feynman diagrams, then for me it sounds like a tautological argument. What is the definition of the path integral on a manifold for a free theory (not using Feynman diagrams)?

@sunghyukPark : look at the ncatlab synthese on https://ncatlab.org/nlab/show/path+integral

Chern-Simons is definitely not a free field theory. But semiclassical versions of completely integrable models are exact even if they are not free. 

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