Gauge invariance of Faddeev-Popov determinant in bosonic string theory

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.