In the paper "A Duality Web in 2 + 1 Dimensions and Condensed Matter Physics",

https://arxiv.org/abs/1606.01989

on page 34, it talks about the periods of a closed $2$-form. Consider the following path integral,

$$\int\mathcal{D}B\;\exp{\left\{-\frac{i}{8\pi}\int_{\mathbb{R}^{4}}\left(\bar{\tau}^{\prime}F^{\prime +}\wedge\star F^{\prime +}-\tau^{\prime}F^{\prime -}\wedge\star F^{\prime -}\right)-\frac{1}{2\pi}\int_{\mathbb{R}^{4}}F^{\prime}\wedge dB\right\}}$$

where $F^{\prime}$ is an arbitrary $2$-form, $F^{\prime\pm}_{ab}=\frac{1}{2}\left(F_{ab}^{\prime}\pm\frac{i}{2}\epsilon_{abcd}F^{\prime cd}\right)$, and $B$ is a $U(1)$-gauge field. It says that path-integral produces a delta functional $\delta[dF^{\prime}]$.

My first question is that how the path-integral produces a delta-functional. Shouldn't there be an extra factor $i$ in front of the second integral $\int F^{\prime}\wedge dB$, so that

I am thinking that the path-integral can be performed by extending the spacetime integral to a complex plane.

$$\int\mathcal{D}B\;\exp{\left\{\frac{-1}{2\pi}\int_{\mathbb{R}^{4}}F^{\prime}\wedge dB\right\}}$$

$$=\int\mathcal{D}B\;\exp{\left\{\frac{1}{2\pi}\int_{\mathbb{R}^{4}}dF^{\prime}\wedge B\right\}}\rightarrow\int\mathcal{D}B\;\exp{\left\{\frac{1}{2\pi}\oint_{\mathbb{C}\times\mathbb{R}^{3}}dF^{\prime}\wedge B\right\}},$$

where the last step means

$$\oint_{\mathbb{C}\times\mathbb{R}^{3}}dF^{\prime}\wedge B=\int_{\mathbb{R}\times\mathbb{R}^{3}}dF^{\prime}\wedge B+\int_{\mathbb{iR}\times\mathbb{R}^{3}}dF^{\prime}\wedge iB,$$

where the integral over circles in upper and lower half plane at infinity vanishes. If the integrand $dF^{\prime}\wedge B$ has no singularity inside the contour, then one obtains the delta functional.

*New Edition: I forgot that the path integral is integrating $e^{iS}$.*

It then says that the $2$-form $F^{\prime}$ is closed and has periods $2\pi\mathbb{Z}$, and so it is the field strength of some gauge field $B^{\prime}$.

What's the definition of periods of a closed $2$-form? Is that related with the fact that $F^{\prime}$ belongs to the second cohomology class $F^{\prime}\equiv F^{\prime}+d\xi$? Why is it $2\pi\mathbb{Z}$?

The delta functional shows that $F^{\prime}$ belongs to the second de Rham cohomology, $[F^{\prime}]\in H^{2}(\mathbb{R}^{4},\mathbb{R})$.

**Question: How do I show that it actually takes integral values? **

From a physical aspect, that $F^{\prime}$ belongs to integral cohomology means that it is the curvature of a non-trivial $U(1)$-bundle, and there exist magnetic monopoles with integer-valued magnetic charges. Is that correct? The delta functional $\delta[dF^{\prime}]$ implies that $F^{\prime}$ is closed. Why does it also implies that the bundle is non-trivial?