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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
where $\omega$ 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 $M_3$ can be written as a boundary term of Yang-Mills theory on a 4-manifold $M_4$ of boundary $M_3$. If one is interested in $M_3=\mathbb{R}^3$ then a natural choice is to take $M_4$ to be a four dimensional half space $\mathbb{R}^+ \times \mathbb{R}^3$ of boundary $\{0\} \times \mathbb{R}^3$. Strictly speaking, $\mathbb{R}^4$ has no boundary (at least in the most naive sense) and so writing things like $\partial \mathbb{R}^4$ 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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