Energy-momentum tensor of Bosonic Ghost Action in String Theory

Question

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.

Answer 1

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}

Answer 2

The mismatch between Prahas EM tensor and the one in the OP is due to the fact that the $b_{\alpha\beta}$ field is traceless. To correctly take this constraint in to account, we can modify the action by introducing a fermionic Lagrange multiplier $\lambda$, $$ S\to S + \frac{-i}{2\pi}\int d^2\sigma\sqrt{\hat{g}}\lambda h^{\alpha\beta}b_{\alpha\beta} \ . $$ This action is on-shell equivalent to the original one, and has an unconstrained $b$-ghost which makes variations easier.

If we add the Lagrange multiplier part of the action to the analysis in Prahas answer, we obtain the EM tensor, \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) + 2 \lambda b_{\alpha\beta}\right] \delta g^{\alpha\beta} \ . \end{split} \end{equation}

By varying the action with respect to the $b$-ghost we can actually solve for the Lagrange multiplier, obtaining, $$ \lambda = \frac{1}{2} \nabla_\alpha c^\alpha \ . $$ Variation with respect to $\lambda$ of course constrains $b$ to be traceless. Now when we substitute the on-shell value for the Lagrange multiplier we obtained the EM tensor in the OP!

This question and its answer are a decade old, but I thought to answer it anyway since the method of Lagrange multipliers is neat. Its worthwhile retroactively linking to this answer from three years ago, where the covariant derivation of the EM tensor is also presented, albeit with fewer steps (and I don't quite understand it).