# How does extending a Chern-Simons theory to the bulk fix potential singularities?

According to ref.1 (§A.3), the naive definition of Chern-Simons $$S[A]=k\int_M \mathrm{CS}[A]\tag{A.17}$$ is ill-defined, because $$A$$ may have "Dirac string singularities". The solution is to extend $$M$$ to the bulk, such that $$M=\partial X$$, and define $$S[A]=k\int_X c(F)\tag{A.18}$$ with $$c$$ the Chern form of $$F=\mathrm dA$$. It is argued that, provided $$k$$ is properly quantised, this integral is independent of $$X$$ and of the extension of $$A$$.

But how does this procedure fix potential "Dirac string singularities"? We are integrating over all $$A$$, and therefore we will have singular configurations regardless of whether we use $$\mathrm A.17$$ or $$\mathrm A.18$$. How does extending $$A$$ into the bulk help remove these singular configurations?

References.

1. N. Seiberg, E. Witten, Gapped Boundary Phases of Topological Insulators via Weak Coupling, https://arxiv.org/abs/1602.04251.

## 1 Answer

If the three-manifold $$M$$ contains a two-manifold $$\Sigma$$ such that the field strength $$F$$ is smooth on $$\Sigma$$ but

$$\int_\Sigma F \neq 0 \ ,$$

then we can not find a smooth $$A$$ on $$\Sigma$$ such that $$F = d A$$. the Chern-Simons density

$$CS[A] \sim A \wedge F$$

contains the well-defined $$F$$, but also the ill-defined $$A$$, so its not clear whether its integral makes sense.

On the other hand, the square of the first Chern form

$$c(F) \sim F \wedge F$$

is well-defined even in the presence of monopoles, so we can integrate it. To stress: the singular $$A$$ are still integrated over, but they give a finite result.