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  Faddeev-Popov Determinant of Chern-Simons Theory

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I am asking this question in order to figure out the expression of the Faddeev-Popov determinant given by Edward Witten is his paper "Quantum Field Theory and Jones Polynomial"

https://projecteuclid.org/euclid.cmp/1104178138

Starting from the action

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right)$$

the variation gives equation of motion

$$F=dA+A\wedge A=0$$

Denoting the solutions to the equation of motion by $a$, then one expands a generic connection $A$ around a flat connection

$$A=a+B$$

Then the action splits into three pieces:

$$S[A]=\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(a\wedge da+\frac{2}{3}a\wedge a\wedge a\wedge a\right)+\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(B\wedge D_{a}B\right)+$$

$$+\frac{k}{6\pi}\int_{M}\mathrm{Tr}\left(B\wedge B\wedge B\right)$$

where the covariant derivative in the second term is defined as $D_{a}=d+[a,\,\,\,\,]$.

The gauge transformation $A[U]=U^{-1}dU+U^{-1}AU$ splits into two parts:

$$a[U]=U^{-1}dU+U^{-1}aU,\quad B[U]=U^{-1}BU$$

so that the flat connections remain flat and the perturbation part $B$ lives in the adjoint representation.

Then, the path-integral takes the form

$$Z=\int\mathcal{D}a\,e^{iS[a]}\int\mathcal{D}B\,\exp\left\{\frac{ik}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{ik}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)\right\}$$

It's easy to check that the last two terms

$$S[a;B]=\frac{ik}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{ik}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)$$

are indeed invariant under the gauge transformations $a[U]=U^{-1}dU+U^{-1}aU$ and $B[U]=U^{-1}BU$. Assuming the spacetime manifold $M$ is closed, and $k\in\mathbb{Z}$, then the first part 

$$S[a]==\frac{k}{4\pi}\int_{M}\mathrm{Tr}\left(a\wedge da+\frac{2}{3}a\wedge a\wedge a\right)$$

is also gauge invariant up to a shift by $2\pi\mathbb{Z}$ from the Wess-Zumino-Witten term under large gauge transformations.

For reasons which I still don't understand Assuming that the moduli space of flat connections $\mathcal{M}=\mathrm{Hom}(\pi_{1}(M),G)/G$ is a discrete set, then the partition function really is

$$Z=\sum_{m\in\mathcal{M}}e^{iS[a_{m}]}\frac{1}{\mathrm{Vol}}\int\mathcal{D}B\,\exp\left\{\frac{ik}{4\pi}\int_{M}\mathrm{Tr}(B\wedge D_{a}B)+\frac{ik}{6\pi}\int_{M}\mathrm{Tr}(B\wedge B\wedge B)\right\}$$

To fix the gauge, the author imposes the covariant gauge 

$$\mathcal{F}[a;B]=(D_{a})_{\mu}B^{\mu}=0$$

The Faddeev-Popov determinant is then 

$$\Delta[a;B]=\mathrm{Det}\left(\frac{\delta\mathcal{F}[a[U];B[U]]}{\delta U}\right)\Bigg|_{U=\mathrm{id}}=\mathrm{Det}(M)$$

Using chain rule, one has

$$M(x-y)=\int d^{3}z\left\{\frac{\delta\mathcal{F}(x)}{\delta B(z)}\frac{\delta B(z)}{\delta U(y)}+\frac{\delta\mathcal{F}(x)}{\delta a(z)}\frac{\delta a(z)}{\delta U(y)}\right\}$$

In Witten's paper, the final expression of the Faddeev-Popov is quite simple, which is

$$M=(D_{a})_{\mu}(D_{a})^{\mu}$$

However, carrying on the calculation of functional derivatives, I obtained a quite different result.

To be specific, the functional derivatives are given by

$$\frac{\delta\mathcal{F}(x)}{\delta B(z)}=D_{a}(x)\delta(x-z),\quad \frac{\delta\mathcal{F}(x)}{\delta a(z)}=[\delta(x-z),B(z)]$$

$$\frac{\delta B(z)}{\delta U(y)}=[B(z),\delta(z-y)],\quad \frac{\delta a(z)}{\delta U(y)}=D_{a}(z)\delta(z-y)$$

where $(D_{a})_{\mu}(x)\delta(x-y)=\frac{\partial}{\partial x^{\mu}}\delta(x-y)+[a_{\mu}(x),\delta(x-y)]$, and the commutator here carries Lie-algebra indices and so is symmetric. i.e. $[A,B]=AB+BA$.

Plugging the above functional derivatives back into the determinant, what I obtained in the end is $$M(x)=4(D_{a})_{\mu}B^{\mu}(x)$$

This is obviously incorrect.

What am I mistaken in the above procedure? 

asked Jul 27, 2018 in Theoretical Physics by New Student (270 points) [ revision history ]
edited Jul 27, 2018 by New Student

The fact that the moduli space of flat connections is discrete is a simplifying assumption (see top of page 357). It is not always true.

@40227 Thank you.

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