# A Simple Question about Gaussian Integral

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The action of electromagnetic field is

$$S=\int\left(-\frac{1}{2e^{2}}F\wedge\ast F+\frac{\theta}{8\pi^{2}}F\wedge F\right),$$

where $F=dA$ is the curvature $2$-form, and $A$ is the connection $1$-form, and $\ast$ is the Hodge star in Lorentzian signature.

Define

$$\tau=\frac{\theta}{2\pi}+\frac{2\pi i}{e^{2}}$$

and

$$F^{\pm}=F\pm i\ast F.$$

The above action can be written as

$$S=\frac{i}{16\pi}\int\left(\tau F^{+}\wedge\ast F^{+}-\bar{\tau}F^{-}\wedge\ast F^{-}\right)=\frac{-1}{8\pi}\int\mathrm{Im}\left(\tau F^{+}\wedge\ast F^{+}\right).$$

One can introduce a dual field $F^{\prime}$, and add

$$\frac{1}{2\pi}\int F^{\prime}\wedge\left(F-dA\right)$$

to the above action.

Then, the path integral

$$\mathcal{Z}=\frac{1}{\mathrm{vol}\mathcal{G}}\int\mathcal{D}Ae^{iS}$$

is equivalent to

$$\int\mathcal{D}F^{\prime}\int\mathcal{D}F\exp\left[\frac{i}{16\pi}\int\left(\tau F^{+}\wedge\ast F^{+}-\bar{\tau}F^{-}\wedge\ast F^{-}\right)+\frac{i}{2\pi}\int F^{\prime}\wedge\left(F-dA\right)\right].$$

The functional integral over $F$ seems to be Gaussian, but is not in the standard form because its action is the imaginary part of $\int\tau F^{+}\wedge\ast F^{+}$.

How to perform the functional integral over $F$? Are there any finite dimensional analog examples?

I also posted my question here.

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