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  Periods of Field Strength

+ 1 like - 0 dislike

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


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? 

asked Mar 12 in Theoretical Physics by New Student (85 points) [ revision history ]
edited 5 days ago by New Student

1 Answer

+ 2 like - 0 dislike

Period means integral over a closed cycle. This is the Gauss law for magnetic field, that any closed surface encloses an integer number of magnetic monopoles. It's the same as saying that the Chern class lives in $H^2(X,\mathbb{Z})$. You can read more about it in Nakahara's book.

answered Mar 12 by Ryan Thorngren (1,835 points) [ no revision ]

Thank you. Could you explain the path integral?

It's analogous to the usual Fourier transform formula for the Dirac delta, right?

There should be an extra factor $i$. I explained it in my post. How do you obtain the delta function?

It's not the usual Fourier transform formula for the Dirac delta. $i$ is missing. 

Could you explain why the period has to be $2\pi\mathbb{Z}$? As far as I can see it, the delta functional implies that $F^{\prime}$ is closed and nothing else. 

Yeah the delta function doesn't really do the integral justice because $B$ is not a global 1-form. Integrating it out also imposes the condition on periods of $F$.

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