The relationship between 3d Chern-Simons theory on the product of the disk and the real line ($D\times \mathbb{R}$) and the chiral WZW model on $S^1\times \mathbb{R}$ was shown in Elitzur et al Nucl.Phys. B326 (1989) 108 (the main details can be found in the top answer to this question, whose notation we follow).

The essential point is that the Chern-Simons gauge field $\tilde{a}$ on the disk can be shown to obey the flatness constraint $\tilde{f}=d\tilde{a}+\tilde{a}\wedge\tilde{a}=0$, which is solved by $$\tilde{a}=-\tilde{d}UU^{-1}. \tag{1}\label{1}$$ Substituting this into the action leads to the chiral WZW model on $S^1\times \mathbb{R}$.

For the quantum theory, one also ought to show that the path integral measure does not have a nontrivial Jacobian, i.e., $$\int \mathcal{D}\tilde{a}~\delta{(\tilde{f})}=\int \mathcal{D}U, \tag{2}\label{2}$$ where $\int \mathcal{D}U$ comes from the Haar measure. My question is, how do we show this?

My attempt: In 1, one way to show this was explained on page 111. They claim that for the change of variables $$\tilde{a}=-\tilde{d}UU^{-1}+\epsilon \tag{3}\label{3}$$ (where $\epsilon$ is a small variation transverse to the space of flat gauge fields), the path integral Jacobian is proportional to $|\textrm{det}(\partial_z+\partial_zU U^{-1})|^2$, and they further claim that this cancels the factor obtained in converting $\delta(\tilde{f})$ to $\delta(\epsilon)$. Here, $z$ is a complex coordinate on $S^1\times \mathbb{R}$.

Here is my attempt to show this. Firstly, by varying \eqref{3}, I obtain $$\delta \tilde{a}=-\tilde{d}(\delta U) U^{-1}+\tilde{d}UU^{-1}\delta U U^{-1}+\delta\epsilon .$$ If the Jacobian is understood to be $|\textrm{det}\frac{\delta\tilde{a}}{\delta U}|$ this seems to imply that $$\int \mathcal{D}\tilde{a}_z\int \mathcal{D}\tilde{a}_{\bar{z}}=|\textrm{det}(\partial_z-\partial_zU U^{-1})U^{-1}|^2\int\mathcal{D}U,$$ which is not exactly what we want.

Next, to show that $$\delta(\tilde{f})\propto \frac{\delta(\epsilon)}{|\textrm{det}(\partial_z+\partial_zU U^{-1})|^2},$$ it seems that we should use a formula of the form $\delta(f(x)) = \delta(x-x_0)/|f'(x_0)|$, together with the explicit form of $|\textrm{det}\frac{\delta \tilde{f}}{\delta \epsilon_z}|$. I was able to show that $$\tilde{f}=\partial_z\epsilon_{\bar{z}}-[\partial_z U U^{-1},\epsilon_{\bar{z}}]-\partial_{\bar{z}}\epsilon_{{z}}+[\partial_{\bar{z}} U U^{-1},\epsilon_{{z}}],$$ but I am not sure how to generalize $\delta(f(x)) = \delta(x-x_0)/|f'(x_0)|$ appropriately.


The reason you are getting the wrong Jacobian is that you are not landing into a Haar measure at the end. The Jacobian that the paper writes is the one that exchange $\mathcal{D}A$ by the Haar measure of $U$. So first you need to parametrize the gauge group like $U(\alpha)=e^{iT_{i}\alpha^{i}}$ and then write the Haar measure in terms of $\alpha$, something like $d\mu(\alpha)$, and finally compute the Jacobian for $\mathcal{D}A\rightarrow d\mu(\alpha)$.

A faster way to obtain the Jacobian is by noting that the measure $$ (\det (\partial_z - A_z))^{-1}(\det (\partial_\bar z - A_\bar z))^{^{-1}}\mathcal{D}A $$ is invariant under gauge transformation $A_{m}\rightarrow A_{m}+[D_{m},\alpha(x)]$ (the signs might be wrong). If you restrict the path integral over $A$ to be just over pure gauge configurations you get a Haar measure for $U$, since it satisfies all the properties of a Haar measure. Recall that the Haar measure is unique up to an overall scaling. So you can identify

$$ \mathcal{D}U=(\det (\partial_z - A_z))^{-1}(\det (\partial_\bar z - A_{\bar z})^{^{-1}}\mathcal{D}A $$

where $A$ is restricted to be pure gauge. So we get the Jacobian:

$$ \mathcal{D}A=(\det (\partial_z - \partial_zUU^{-1}))(\det (\partial_\bar z - \partial_{\bar z}UU^{-1})\mathcal{D}U $$

You can motivate it in a more rigorous way by noting that you can use the Faddeev-Popov procedure in reverse to obtain the Haar measure in terms of $A$. You define the Faddeev-Popov term by the relation

$$ 1=\Delta_{FP}(A_0)\int \mathcal{D}U \delta(A-A_0^U) $$

where $A_0^{U}$ is the gauge connection obtained by acting the action of $U$ on a fixed connection $A_0$. Inserting this $1$ on a path integral on $A$ we can use the delta to trade any functional of $A$ into a functional of $A_0^{U}$ by integrating on $A$ first. You are left with a path integral over $U$ with the correct Jacobian.

For the $\delta(\tilde{f})$ you can use the formula

$$ \prod_{i=1}^{n}\delta (O^{i}_jb^j)=(det(O))^{-1}\delta^{n}(b) $$

applied to the continuum, $n\rightarrow\infty$, $i\rightarrow x$ and $b_i\rightarrow f(x)$. There is just a small subtlety that you kindly remind me. This is the fact that since $(\varepsilon_{z},\varepsilon_{\bar z})$ is defined to be orthogonal to pure gauge configurations, the delta function $\delta(\tilde f)$ factorizes into one delta just involving $\varepsilon_{z}$-terms times another delta involving just the $\varepsilon_{\bar z}$-terms.

  • $\begingroup$ Thank you for the detailed answer. Could you explain why there is a factorization of the determinant into holomorphic and antiholomorphic part when relating $\delta(\tilde{f})$ and $\delta(\epsilon_z)\delta(\epsilon_{\bar{z}})$? I am unable to see this from the formula you wrote for the finite case. $\endgroup$ – Mtheorist Feb 17 at 5:35
  • $\begingroup$ By assumption $(\epsilon_z ,\epsilon_{\bar z})$ is orthogonal to pure gauge so they must be zero if you constraint $\tilde f$ to zero, i.e. the $\epsilon_z$ terms of $\tilde f$ does not cancel the $\epsilon_{\bar z}$ terms. This imply that the delta function for $\tilde f$ factorizes into two deltas, one for the $\epsilon_z$ terms and the other for the $\epsilon_{\bar z}$ terms. Each delta gives a determinant. $\endgroup$ – Nogueira Feb 18 at 3:18
  • $\begingroup$ Thanks again. Regarding the other part of the question, it seems that gauge invariance of the Haar measure only holds for compact gauge group. I am also interested in the transformation of the measure for the case with noncompact gauge group. In arxiv.org/abs/1311.1853, equations 2.15 to 2.17, this seems to be achievable by comparing metrics on the field spaces, but it is not clear to me why the different metrics give rise to the determinants multiplying the Haar measure. Do you have any idea why? $\endgroup$ – Mtheorist Feb 18 at 3:49
  • $\begingroup$ The way I convince my self about the jacobian in the first part of my answer was not by the argument that I presented. I convince my self by working with the Faddeev-Popov method in reverse. So my ideia would be apply the Faddeev-Popov do address this questions. $\endgroup$ – Nogueira Feb 18 at 11:13
  • $\begingroup$ Sorry, I am a little confused by your use of the Faddeev-Popov method. Usually in textbooks the identity is $1=\Delta_{FP}(A)\int \mathcal{D}U\delta(F(A^U))$, for some functional $F$ of $A^U$. How does this relate to your expression $1=\Delta_{FP}(A)\int \mathcal{D}U\delta(A-A_0^U)$, where the delta functional is a functional of $A$ and $A_0^U$? $\endgroup$ – Mtheorist Apr 13 at 6:09

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