Energy-momentum tensor of Bosonic Ghost Action in String Theory When quantizing bosonic string theory by means of the path integral, one inverts the Faddeev-Popov determinant by going to Grassmann variables, yielding:
$$
S_{\mathrm{ghosts}} = \frac{-i}{2\pi}\int\sqrt{\hat{g}}b_{\alpha\beta}\hat{\nabla^{\alpha}}c^{\beta}d^2\tau,
$$
where the $-i/2\pi$ just comes from convention/Wick rotated or not. My first problem is the notion of the 'fudicial' metric $\hat{g}$. I find its role in the path integral procedure a bit confusing. What is its relation to the 'normal' metric $g$? Why is it introduced? Related to this confusion is that fact in my lecture notes it is said that the energy momentum tensor is given by:
$$
T_{\alpha\beta} :=\frac{-1}{\sqrt{\hat{g}}}\frac{\delta S_g}{\delta\hat{g}^{\alpha\beta}} = \frac{i}{4\pi}\left( b_{\alpha\gamma}\hat{\nabla}_{\beta}c^{\gamma} + b_{\beta\gamma}\hat{\nabla}_{\alpha}c^{\gamma} - c^{\gamma}\nabla_{\gamma}b_{\alpha\beta} - g_{\alpha\beta}b_{\gamma\delta}\nabla^{\gamma}c^{\delta} \right),
$$
I have trouble deriving this. Varying the $\sqrt{\hat{g}}$ in the action yields the last term I would say:
$$
\delta\sqrt{\hat{g}} = -\frac{1}{2}\sqrt{\hat{g}}\hat{g}_{\alpha\beta}\delta \hat{g}^{\alpha\beta}
$$
 However, this term does not have a 'hat' on the covariant derivative, which I find strange. The first and second term follow easily when writing the action with all indices low (except for a factor of 1/2), but I really don't see where the third term comes from and it also misses a hat on the covariant derivative. It looks like there has been done a partial integration, but I don't see why. I guess I am missing the point of the fiducial metric here. Explanation greatly appreciated!
EDIT: In the discussion below I mentioned that $b_{\alpha\beta}$ is traceless: $b_{\alpha\beta}g^{\alpha\beta} = 0$, I forgot to place that here. It is a consequence of the path integral procedure.
 A: Here's part of my answer to the derivation of the EM tensor for the ghost action. It does not match the expression you gave, but I may have made a mistake. Can you check my work?
We start with the action
\begin{equation}
\begin{split}
S_{gh} &= - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} \nabla_\mu c^\beta \\
\end{split}
\end{equation}
Let us now vary the action w.r.t. metric. We get
\begin{equation}
\begin{split}
\delta S_{gh} &= - \frac{i}{2\pi} \int d^2 \sigma \left( \delta \sqrt{g} \right) g^{\alpha\mu} b_{\alpha\beta} \nabla_\mu c^\beta  \\
&~~~~~~~~~~~~~~~~~- \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} \left( \delta g^{\alpha\mu} \right) b_{\alpha\beta} \nabla_\mu c^\beta  \\
&~~~~~~~~~~~~~~~~~- \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} \delta \left( \nabla_\mu c^\beta \right) \\
&= - \frac{i}{4\pi} \int d^2 \sigma  \sqrt{g} \left[ b_{\alpha\mu} \nabla_\beta c^\mu + b_{\beta\mu} \nabla_\alpha c^\mu   - g_{\alpha\beta}   b_{\rho\sigma} \nabla^\rho c^\sigma \right]  \delta  g^{\alpha\beta}   \\
&~~~~~~~~~~~~~~~~~ - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} c^\lambda    \delta \Gamma^\beta_{\mu\lambda} \\
\end{split}
\end{equation}
We now use
\begin{equation}
\begin{split}
\delta \Gamma^\beta_{\mu\lambda} = \frac{1}{2} g^{\beta\rho} \left[ \nabla_\lambda \delta g_{\rho \mu} + \nabla_\mu \delta g_{\rho \lambda} - \nabla_\rho \delta g_{\mu\lambda}\right]
\end{split}
\end{equation}
Note that in particular, it is a tensor. The last term then becomes
\begin{equation}
\begin{split}
I &=   - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} c^\lambda    \delta \Gamma^\beta_{\mu\lambda}  \\
&=   - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g}  b^{\mu\rho} c^\lambda \left[ \nabla_\lambda \delta g_{\rho \mu} + \nabla_\mu \delta g_{\rho \lambda} - \nabla_\rho \delta g_{\mu\lambda}\right]\\
&=   - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g}  b^{\mu\rho} c^\lambda   \nabla_\lambda \delta g_{\rho \mu}  \\
&= - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g}     \nabla_\lambda \left( b_{\alpha\beta} c^\lambda \right)  \delta g^{\alpha\beta} 
\end{split}
\end{equation}
We then have
\begin{equation}
\begin{split}
\delta S_{gh}  &= - \frac{i}{4\pi} \int d^2 \sigma  \sqrt{g} \left[ b_{\alpha\mu} \nabla_\beta c^\mu + b_{\beta\mu} \nabla_\alpha c^\mu   - g_{\alpha\beta}   b_{\rho\sigma} \nabla^\rho c^\sigma +   \nabla_\lambda \left( b_{\alpha\beta} c^\lambda \right)  \right]  \delta  g^{\alpha\beta}  
\end{split}
\end{equation}
