# 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 $$$$g^{ab}(p)\frac{\delta}{\delta g^{ab}(p)}F[g]\sim c\sqrt{|g|}R[g](p)\qquad(1)$$$$ where $$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: $$$$\Delta[g^\epsilon]^{-1}=\int\mathcal{D}\xi\delta(g^\epsilon-g_0^\xi)=\int\mathcal{D}\xi\delta([g-g_0^{\epsilon^{-1}\xi}]^\epsilon)=\int\mathcal{D}\xi'\delta([g-g_0^{\xi'}]^\epsilon)=\int\mathcal{D}\xi'\delta(g-g_0^{\xi'})=\Delta[g]^{-1}\qquad(2)$$$$

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: $$$$\Delta[g]=Z_{\text{gh}}[g]\qquad(3)$$$$ where the right hand side, as discussed above, is not gauge invariant: Let $$\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.