# What are periods of field strength?

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

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

where $F^{\prime}$ is an arbitrary $2$-form, $F^{\prime\pm}_{ab}=\frac{1}{2}\left(F_{ab}^{\prime}\pm\frac{i}{2}\epsilon_{abcd}F^{cd}\right)$, 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

$$\int\mathcal{D}B\;\exp{\left\{\frac{i}{2\pi}\int F^{\prime}\wedge dB\right\}}=\int\mathcal{D}B\;\exp{\left\{\frac{-i}{2\pi}\int dF^{\prime}\wedge B\right\}}=\delta[dF^{\prime}]$$

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 $2\pi\mathbb{Z}$?

1. Don't forget that the path integral is over $\exp(\mathrm{i}S)$, not $\exp(S)$.

2. A period of a differential p-form is simply its value when integrated over a matching p-cycle. "The periods" usually refers to a bunch of such numbers obtained by integrating it over each of a set of p-cycles that forms a basis for the p-th homology.

It is not true that $F'$ necessarily has integral periods if it is truly "arbitrary". We could replace any $F'$ for which this holds by $\frac{1}{2}F',\frac{1}{3}F',\dots$ and it would still have to hold, showing the claim cannot be true unless all periods of $F'$ vanish. Since this obviously does not change closedness, closedness alone is not sufficient to guarantee integrality.

However, in the case where $F'$ is the field strength of a gauge field, $\frac{1}{2\pi}\mathrm{Tr}[F']$ is a first Chern class, which is always in integral cohomology, which is in turn equivalent to having integral periods.

In case you are wondering where the "$\frac{1}{2}F',\frac{1}{3}F',\dots$ argument" fails for the curvature of a gauge field: The gauge field $A'$ with $F' = \mathrm{d}A'$ only exists locally on contractible patches, and there are no non-trivial p-cycles on such contractible patches by definition of contractibility, meaning all of $F$'s periods vanish anyway there. So trying to mimick the argument by using $\frac{1}{2}A,\frac{1}{3}A$ fails to tell you anything about $F$ globally.

A neat thing to note is that the claim "on contractible patches all periods are zero" for the ordinary electromagnetic field strength tensor is simply Gauß' law for magnetism.

• Is that correct to say that the gauge potential is globally defined only when monopoles are absent? – The Last Knight of Silk Road Mar 15 '18 at 21:55
• @NewStudent Yes for a $\mathrm{U}(1)$-gauge theory, i.e. electromagnetism, no for a general gauge theory since it is not clear what you might mean by a monopole there. – ACuriousMind Mar 15 '18 at 22:09