I am, once again, going through an introduction to (bosonic) string theory, following the lecture notes by David Tong on the subject, and once again I am stumbling on technicalities around the Polyakov path integral formulation.

This time it is the claimed gauge invariance of the Faddeev-Popov determinant, defined in Tongs notes in eq. (5.1) on page 110 as:


where, for simplification, $g$ and $g_0$ are lorentzian metrics on the zylinder and the integral is over "the Haar measure" on the group $G$ of diffeomorphisms and Weyl transformations. For $\xi$ the diffeomorphism $f$ and Weyl factor $\phi$, $g^\xi=\phi f^*g$ or something along those lines.

Tong claims that this expression is gauge invariant, that is $\forall \epsilon\in G$: $\Delta[g^\epsilon]=\Delta[g]$, and gives a short uncommented proof of it as:


I guess the third equality uses the translation invariance of the Haar measure, but the second step simply seems wrong to me. I think it should be:

$$\int_G\mathcal{D}\xi\delta(g^\epsilon-g_0^\xi)=\int_G\mathcal{D}\xi\delta(g^\epsilon-g_0^{\epsilon\xi})=\int_G\mathcal{D}\xi\delta([g-g_0^\xi]^\epsilon)=\int_G\mathcal{D}\xi\frac{\delta(g-g_0^\xi)}{|\det\frac{\delta h^\epsilon}{\delta h}\vert_{h=0}|}.$$

If we were talking about a representation of a compact topological group it is clear that this determinant is $1$, but in this case I can't see it.

Moreover, there is indirect evidence that the Fadeev-Popov determinant is not gauge invariant: Apparently it can be written as the partition function of a $c=-26$ CFT, but the partition functions of CFT's are only Weyl-invariant for $c=0$ (or flat background metric which we can't assume since we are integrating over all background metrics).

The question is: am I overlooking something, and if yes, what? To be clear, I am convinced that treating this un-invariance correctly gives the right expression for the gauge fixed pathintegral anyway, but the presentation in Tongs notes seems flawed, even apart from all the assumptions made.

Remark: this would also clear up an earlier question of mine, since the un-invariance of the Faddeev-Popov determinant and that of the string measure would exactly cancel in $26$ dimensions, see my earlier question.

  • $\begingroup$ Clearly (5.1) is poorly defined: $g_1-g_2$ is undefined for $g_{1,2}$ both group elements. I would start over and firstly define the FP determinant through a Gaussian path integral involving appropriate (fermion) variables, which you should determine. In Euclidean signature this should admit a perfectly rigorous zeta-function regularisation, see Vassilevich arxiv.org/abs/hep-th/0306138. (Although I cannot off the top of my head recall a published reference where this FP det is done precisely that way.) Then the infinitesimal gauge transformation is some well defined heat-kernel coeff $\endgroup$
    – user21299
    Apr 24, 2020 at 23:41
  • $\begingroup$ I agree with you that the relevant variation of the FP determinant should probably be zero in the critical dimension (critical central charge) so David Tong's explanation should work at least in that case (the one one cares about). This is all off the top of my head though $\endgroup$
    – user21299
    Apr 24, 2020 at 23:44
  • $\begingroup$ The subtraction is well defined, at least, because the $g$'s form a group under addition (they're not elements of the gauge group, they're metrics). It seems like this definition of the FD determinant differs from the convention of others; I've only seen it where the argument of the FD determinant is subjected to the gauge transformation in the integral, i.e. $\Delta[g]^{-1} = \int \mathcal D \xi \delta(g^\xi - g_0)$ $\endgroup$
    – mthibodeau
    Apr 25, 2020 at 2:47
  • $\begingroup$ The thing is, when writing the Fadeev-Popov determinant as the partition function of the $c=-26$ CFT in the general background it clearly is not gauge invariant: $g^{ab}\frac{\delta}{\delta g^{ab}}\ln Z[g]\sim cR$, where $R$ is the ricci scalar. $\endgroup$
    – Leonard
    Apr 25, 2020 at 6:42
  • 1
    $\begingroup$ @mthibodeau Yeah, normally it is defined in that way and then it is trivially gauge invariant, but then the argument Tong wants to make with it becomes impossible due to other anomalies which vanish only in 0 dimensions. $\endgroup$
    – Leonard
    Apr 25, 2020 at 6:43

3 Answers 3


Let $Z[g]$ be the partition function of a conformal field theory with central charge $c$ on a genus $0$ surface, $F[g]=\ln Z[g]$ the "free energy". It is a standard result that \begin{equation} g^{ab}(p)\frac{\delta}{\delta g^{ab}(p)}F[g]\sim c\sqrt{|g|}R[g](p)\qquad(1) \end{equation} where $R[g]$ is the Ricci curvature and the proportionality constant is not zero and independent of $g$. In particular, eq. (1) implies that the partition function can't be Weyl rescaling invariant whenever $c\neq 0$ and the background is curved.

Firstly, the proof of gauge invariance given by Tong and Polchinski is, almost literally cited, this: \begin{equation} \Delta[g^\epsilon]^{-1}=\int\mathcal{D}\xi\delta(g^\epsilon-g_0^\xi)=\int\mathcal{D}\xi\delta([g-g_0^{\epsilon^{-1}\xi}]^\epsilon)=\int\mathcal{D}\xi'\delta([g-g_0^{\xi'}]^\epsilon)=\int\mathcal{D}\xi'\delta(g-g_0^{\xi'})=\Delta[g]^{-1}\qquad(2) \end{equation}

The point where i don't agree is the second to last equality in eq. (2): as is well known there should be a factor of $|\det({\frac{\delta h^\epsilon}{\delta h}\vert_{h=0}})|^{-1}$ appearing. If we were talking about a representation of a compact group I would agree that this is always $1$, but, since we are including Weyl rescalings, the group we are considering is far from compact. In particular consider the case when $\epsilon$ is a Weyl rescaling $h^\epsilon=\phi h$, then we have to determine $\det('\text{multiplication with }\phi')$, which I highly suspect to not be $1$ for general $\phi$ (even when regularized appropriately).

Secondly, assume that we are on a cylinder such that $\exists \epsilon:g=g_0^\epsilon$. Then following Tong almost word by word we find that \begin{align*} \Delta[g]^{-1}&=\int\mathcal{D}\xi\delta(g_0^\epsilon-g_0^\xi)=\int\mathcal{D}\xi\delta(g_0^\epsilon-(g_0^\epsilon)\xi)\\ &=\int\mathcal{D}\xi\delta(2w(g_0^\epsilon)_{ab}+\nabla_{(a}\nu_{b)})=\ldots\\ &=Z_{\text{bosonic ghosts}}[g_0^\epsilon] \end{align*} so that at the end of the day we can write the Fadeev-Popov determinant as the partition function of the ghost CFT: \begin{equation} \Delta[g]=Z_{\text{gh}}[g]\qquad(3) \end{equation} where the right hand side, as discussed above, is not gauge invariant: Let $\epsilon_\phi$ be the Weyl rescaling by $1+\phi$, gauge invariance must imply that $\frac{\delta \Delta[g^{\epsilon_\phi}]}{\delta \phi(p)}\vert_{\phi=0}=0$, but according to eq. (1) and (3) we have \begin{align*} \frac{\delta \Delta[g^{\epsilon_\phi}]}{\delta \phi(p)}\vert_{\phi=0}&=\frac{\delta Z_{\text{gh}}[g^{\epsilon_\phi}]}{\delta \phi(p)}\vert_{\phi=0}=\frac{\delta Z_{\text{gh}}[g+\phi g]}{\delta \phi(p)}\vert_{\phi=0}\\ &=\int\mathrm{d}q\,\frac{\delta Z_{\text{gh}}[g]}{\delta g^{ab}(q)}\frac{\delta \phi(q) g^{ab}(q)}{\delta \phi (p)}\vert_{\phi=0}=\int\mathrm{d} q\,\frac{\delta Z_{\text{gh}}[g]}{\delta g^{ab}(q)}g^{ab}(q)\delta(p-q)\\ &=Z_{\text{gh}}[g]g^{ab}(p)\frac{\delta}{\delta g^{ab}(p)}F_{\text{gh}}[g]\sim \Delta[g]c\sqrt{|g|}R[g](p) \end{align*}

So, since the ghost CFT in this case has $c=-26\neq0$ and $g$ in general might have non zero curvature we find that the Fadeev-Popov determinant can't be gauge invariant.

\newpage Finally, I want to remark that this is actually not a problem for our considerations, but makes it possible in the first place: \begin{align*} Z_{\text{String}}&=\int\mathcal{D}gZ_{\text{Polyakov}}[g]=\int\mathcal{D}g\Delta[g]\int\mathcal{D}\xi\delta(g-g_0^\xi)Z_{\text{Polyakov}}[g]\\ &=\int\mathcal{D}\xi Z_{\text{gh}}[g_0^\xi]Z_{\text{Polyakov}}[g_0^\xi] \end{align*}

The combination $Z_{\text{gh}}[g_0^\xi]Z_{\text{Polyakov}}[g_0^\xi]$ has a conformal anomaly given by $c=D-26$, so it is gauge invariant if and only if $D=26$! In that case we can drop the integration over the gauge group and the associated infinite but constant factor to get \begin{equation*} Z_{\text{String}}=Z_{\text{gh}}[g_0]Z_{\text{Polyakov}}[g_0] \end{equation*} which is our desired result.

  • $\begingroup$ Yes indeed, the FP determinant is not, in general, invariant. You can then straightforwardly calculate it to produce your (1) --- thereby determining the critical dimension --- using e.g. heat kernel methods. $\endgroup$
    – user21299
    Apr 26, 2020 at 18:19
  • $\begingroup$ Yes, the derivation on Pochinski needs being Weyl-anomaly free from the product of F-P determinant and [dX]. Otherwise, the Vol(Weyl) cannot be canceled directly by the measure integration. In this sense, that derivation should be wrong. $\endgroup$
    – Yuan Yao
    Feb 5, 2021 at 13:32

I think it might be useful to look in Polchinski: in a footnote on page 87 of Vol. 1 he addresses the gauge invariance of the FP determinant. I've attached a screenshot here for convenience:


Essentially Polchinski claims the second line of what you've shown is given by gauge invariance of the delta functional. He doesn't actually prove that statement so you'd have to take his word for it but hopefully that helps!

  • 1
    $\begingroup$ Hello goldmans, for a 'larger' gauge transformation, why can't the leading term be non unity and thus have a non unit determinant? Especially: if the $\epsilon$ is a pure Weyl transformation, then $h^\epsilon=\phi*\epsilon$, and multiplying by phi has certainly non unit determinant. $\endgroup$
    – Leonard
    Apr 25, 2020 at 6:33
  • $\begingroup$ @Leonard sorry yes you're absolutely right. I've actually made a larger error in the Taylor expansion I wrote. For example, if your gauge group has a Lie Algebra structure then the new metric is given by $h' = he^{i\omega_a t_a}$ and so even if $\omega$ is small all the terms are only first order in $h$. I removed my edit since I don't believe I know enough on this to say something definitely but I think possibly the thing to look at is the determinant you've written down. When you make a "change of variables" in the integral, it's not the metrics you integrate over but the gauge group. $\endgroup$ Apr 25, 2020 at 16:53
  • $\begingroup$ @Leonard Therefore you perhaps don't want to consider $\det|\frac{\delta h^\epsilon}{\delta h}|_{h=0}$ but rather something like $\det|\frac{\delta(\zeta\circ \epsilon)}{\delta \zeta}|$, i.e. the change in group element near the identity. I'm certainly not an expert on these things and that statement was very schematic so hopefully someone who knows more can say something else about it but I do think that's potentially where the invariance of the delta function comes from. $\endgroup$ Apr 25, 2020 at 17:17
  • $\begingroup$ I don't think the determinant should be anything else, it is just the general rule that $\delta(f(g(x)))=\sum\frac{\delta(g(x)-g_0)}{|\det df(g_0)|}$. $\endgroup$
    – Leonard
    Apr 26, 2020 at 9:40

This doesn't answer the question, but kind of solves the problem. There is a slightly different definition of the Fadeev-Popov determinant that doesn't require to show that the $\delta$ function is invariant but gets the job equally done. Namely we apply the gauge $\xi$ on the metric integrated over and not on the fiducial metric. $$\Delta[g]^{-1}=\int_G\mathcal{D}\xi\,\delta(g^\xi-g_0)\,.$$ This works because $g^\xi = g_0$ has the same solutions as $g = g_0^{\xi^{-1}}$, so since you integrate over all $\xi$, the $\delta$ function will hit the same zeros.

This definition solves the issue because, assuming that the Haar measure is invariant, one has

$$ \begin{aligned} \Delta[g^\zeta]^{-1}&=\int_G\mathcal{D}\xi\,\delta(g^{\zeta\,\xi}-g_0) \\&= \int_G\mathcal{D}(\zeta^{-1}\xi')\,\delta(g^{\zeta\,\zeta^{-1}\xi'}-g_0) \\ &=\int_G\mathcal{D}\xi'\,\delta(g^{\xi'}-g_0) = \Delta[g]^{-1}\,. \end{aligned} $$

As for the second part: true, the partition function is not well defined if $c$ is non zero. But the reason for that is not an issue of the FP determinant (not its Haar measure, nor the $\delta$ function). It's an issue of the field measure that appears in the rest of the path integral, namely the $\mathcal{D}X^\mu\mathcal{D}g$. When $c\neq 0$ you have an anomaly and that is known to lead to a non-invariance of the field measure under gauge transformations.

  • $\begingroup$ Why the downvote though? $\endgroup$
    – MannyC
    Apr 25, 2020 at 19:44
  • $\begingroup$ I have to remark that with this different definition, while it is obviously gauge invariant, the gauge fixing does not work out due to anomalies of the Polyakov action and measure which only vanish in 0 dimensions. $\endgroup$
    – Leonard
    Apr 26, 2020 at 9:42
  • $\begingroup$ I don't understand what goes wrong and what anomalies you are referring to. $\endgroup$
    – MannyC
    Apr 26, 2020 at 10:03
  • 1
    $\begingroup$ @MannyC, when plugging into the path integral and attempting to perform the integration on $g$, you will get the same factor as before, i.e. roughly speaking because of limited characters : $\int \mathcal{D}g \delta(g^\xi-g_0) f(g) = \frac{f(g_0^{\xi^{-1}})}{|det\frac{\delta h^\epsilon}{\delta h}|}$. So it seems this just pushes the problem further ahead. $\endgroup$
    – Frotaur
    Jun 2, 2020 at 19:31
  • $\begingroup$ Of course, thanks! $\endgroup$
    – MannyC
    Jun 2, 2020 at 23:04

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