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  Why are boundary terms important in Chern-Simons theory?

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I am learning about Chern-Simons theory. I work in Euclidean space. The action is given by
I=R4dω=R4ω


where ω is the usual Chern-Simons form, and I have used stokes' theorem. My first question is, what is the boundary of for dimensional Euclidean space? and why are boundary terms important in this theory? I mean, up till now I have shamelessly ignored boundary terms. Why are they important now?

asked Jul 4, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

1 Answer

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The action of Chern-Simons theory on a 3-manifold M3 can be written as a boundary term of Yang-Mills theory on a 4-manifold M4 of boundary M3. If one is interested in M3=R3 then a natural choice is to take M4 to be a four dimensional half space R+×R3 of boundary {0}×R3. Strictly speaking, R4 has no boundary (at least in the most naive sense) and so writing things like R4 is not really correct if no additional precision is given.

To ignore boundary terms is only reasonable in some cases (when they are none or when their contributions obviously vanish). Chern-Simons theory is a natural theory living on some 3-dimensional boundary of 4-dimensional Yang-Mills theory. One can define and study Chern-Simons theory independently of this fact but to remember this often sheds light on important issues (example: quantization of the Chern-Simons level is related to the quantization of the instanton number).

answered Jul 6, 2016 by 40227 (5,140 points) [ revision history ]

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