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".

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


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.

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


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?

(I also posted this question here)


1 Answer 1


Witten wants to evaluate the path integral for a given background $a$, he doesn't want to vary $a$ inside the $B$ action. If he had wanted to do so he wouldn't have chosen the gauge fixing $(D_{a})_{\mu}B^{\mu}=0$, because this gauge fixing is invariant under the combined gauge transformation of $a$ and $B$ given in the question.

The action for the fluctuation field has the following gauge freedom: $$ a[U] = a$$ $$B[U] = U^{-1}BU + U^{-1}D_aU$$ (Infinitesimally: $\delta B[U] = D_a \delta U + [ B, \delta U] = D_A \delta U$)

Please notice that the Chern-Simons action for $B$ is gauge invariant with respect to the above transformation, since $D_a$ is nilpotent due to the flatness of $a$: $$D_a^2 = F_a = 0$$ Thus we have a gauge theory of the non-Abelian $B$ gauge potential with the exterior derivative $d$ replaced by the twisted differential $D_a$.

Now, we have : $$\delta\mathcal{F}[a;B]=(D_{a})_{\mu}\delta B^{\mu}= (D_{a})_{\mu}(D_A)^{\mu} \delta U$$

Thus the Faddeev-Popov Laplacian is:


Please notice that this is different from Witten's result, the second covariant derivative is with respect to the full gauge potential: background $+$fluctuation. We get Witten's result only if we assume that the fluctuations are small and neglect terms higher than the quadratic term. This fact was noticed by Birmingham, Blau, Rakowski and Thompson, where a careful evaluation is given in equations 6.14 and 6.15 on page 195.

  • $\begingroup$ Thank you so much for your help Mr David Bar Moshe. This problem made me suffer from insomnia. Oh, stupid me. Thank you for illuminating me. $\endgroup$
    – Valac
    Commented Jul 30, 2018 at 20:16
  • $\begingroup$ Hi Sir. I have one more question for you. The fluctuation part of the action under the above gauge transformation then produces a "covariant version" of Wess-Zumino-Witten term, which is $\mathrm{Tr}(U^{-1}D_{a}U\wedge U^{-1}D_{a}U\wedge U^{-1}D_{a}U)$. Is this still integral? $\endgroup$
    – Valac
    Commented Jul 31, 2018 at 15:14
  • $\begingroup$ Yes, it should be. The reason is that the twisted exterior differential $D_a$ acts on the various fields according to their representation. For example $\mathrm{Tr}\left(( U^{-1} dU)^3\right)$ is a scalar, therefore: $D_a\mathrm{Tr}\left(( U^{-1} dU)^3\right) = d\mathrm{Tr}\left(( U^{-1} dU)^3\right) $, thus: $d\mathrm{Tr}\left(( U^{-1} dU)^3\right) = \mathrm{Tr}D_a\left(( U^{-1} dU)^3\right) = 0$. The vanishing of the last term can be proved according the same criteria of the standard WZW term. Thus this form is closed. $\endgroup$ Commented Aug 1, 2018 at 11:49
  • $\begingroup$ Therefore on a compact manifold we can find a normalization which makes it integral. Now, this normalization should not depend on the flat connection $a$ at least when its moduli space is connected. Thus if we start from $a=0$, we get the usual normalization which should persist for all values of $a$ on the connected component of $a=0$. $\endgroup$ Commented Aug 1, 2018 at 11:53
  • $\begingroup$ Hi Mr David Bar Moshe. Do you mean $\mathrm{Tr}D_{a}((D^{-1}D_{a}U)^{3})$? $\endgroup$
    – Valac
    Commented Aug 1, 2018 at 15:55

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