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?
Q&A (4872)
