Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Faddeev-Popov Determinant of Chern-Simons Theory

+ 3 like - 0 dislike
1105 views

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 Libertarian Feudalist Bot (270 points) [ revision history ]
edited Jul 27, 2018 by Libertarian Feudalist Bot

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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...