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

Question

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.

Answer 1

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.