Question about Path-Integral from Particle-Vortex Daulity from 3d Bosonization

Question

I have been studying the paper Particle-Vortex Daulity from 3d Bosonization, 

https://arxiv.org/pdf/1606.01893.pdf

by Andrew Karch and David Tong.

On page 11, it says that integrating out gauge field $a$ produces equation of motion $a=A-\tilde{a}$.

My question is how I should perform the path integral

$$\int\mathcal{D}a \exp\{-i\int(\frac{1}{2}a\wedge da+a\wedge d\tilde{a}-a\wedge dA)\}$$
to obtain the equation of motion.

Answer 1

They had to make some hand-waving assumption they called "absence of holonomy" to get rid of the quadratic term in a, which is of course where all the action happens in this duality to give you fermions and everything.

Once its linearized, the path integral is evaluated by analogy with the 1-dimensional integral

$$\int dx e^{i x y} = \delta(y).$$

EDIT: I was being a bit too cavalier and was going to end up with $\delta(\tilde a - A)$, which is too strong. I was altogether dropping the $ada$ term. A better thing to do is complete the square: redefine $a' = a + (1/2)(\tilde a - A)$. Then the integrand becomes

$$ a' da' - (1/4)(\tilde a - A)d(\tilde a - A).$$

One should worry about the factors of 1/2 and 1/4 that we ended up with. In this form, we're free to do the path integral over $a'$, which on a simply connected manifold can be computed as a Gaussian integral and the saddle point (which is unique on a simply connected manifold) is $a' = 0$, in other words $a = A - \tilde a$.

Comments

Could you elaborate why you can use that formula with the $a\wedge da$ term when the holonomy is absent? 

Why does the delta functional produce $a-A+\tilde{a}=0$?

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I made it a bit more rigorous, and ended up also with a Chern-Simons term for $\tilde a - A$.

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Thank you very much!

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Sorry I made a mistake. The coefficients of CS term and BF term are different. 

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That's good. Then there won't be a 1/4 to worry about. It also makes it a valid Chern-Simons term before integrating out $a$.