# Bosonic closed string effective action

Neil Lambert in his lecture notes https://nms.kcl.ac.uk/neil.lambert/SBQG.pdf in section 3.9 states that imposing conformal invariance at one-loop imposes the following equations on the spacetime fields $$g_{\mu\nu},B_{\mu\nu},\phi$$:

\begin{align} 0&=R_{\mu\nu}+\frac{1}{4}H_{\mu\alpha\beta}H_{\nu}^{\ \alpha\beta}-2D_\mu D_\nu\phi+\mathcal{O}(\alpha'),\\ 0&=D^\lambda H_{\lambda\mu\nu}-2D^\lambda \phi H_{\lambda\mu\nu}+\mathcal{O}(\alpha'),\\ 0&=4D^2\phi+4(D\phi)^2-R-\frac{1}{12}H^2+\mathcal{O}(\alpha'). \end{align}

Then he states that these equations of motion can be derived from a spacetime effective action given by $$S_{\text{eff}}=-\frac{1}{2\alpha'^{12}}\int\text{d}^{26}x\sqrt{-g}\ e^{-2\phi}\left\{R-4(D\phi)^2+\frac{1}{12}H^2\right\},$$ by varying this action with respect to the fields $$g_{\mu\nu},B_{\mu\nu},\phi$$.

However I couldn't quite get the equations of motion right. For example, under a variation of the metric field $$\delta g_{\mu\nu}$$ I obtain the equation $$R_{\mu\nu}+\frac{1}{4}H_{\mu\alpha\beta}H_{\nu}^{\ \alpha\beta}-2D_\mu D_\nu\phi=g_{\mu\nu}\left(\frac{1}{2}R+2(D\phi)^2-2D^2\phi+\frac{1}{24}H^2\right).$$ The left hand side looks alright, however the right hand side is nonzero. I tried taking the trace of this equation and inserting it back to cancel the right hand side but it didn't work.

What is the slickest way of doing this calculation? It is certain that I'm making a simple calculational mistake, therefore if anyone has a reference which does these calculations explicitly, that would be of great help.

• One trick is to use locally flat coordinates and then combine the results into tensor expressions. For an example see page 142 of damtp.cam.ac.uk/user/tong/gr/four.pdf Mar 25, 2022 at 11:34
• @Andrew Your comment is not particularly useful. I know how the Ricci scalar transforms under a variation of the metric. I know how to obtain the e.o.m. by varying the fields, but may answers didn't match the lecture notes. That's why I'm asking the question in the first place.
– user242231
Mar 25, 2022 at 15:44
• I mean, the answer is that you're probably making a mistake in your calculation. It's going to be hard for anyone to identify your mistake without going through your working out, and it's probably going to be hard to find someone who will do that for a complicated calculation like this. So I was just trying to suggest a way to simplify the calculation where it's less likely you'd make an error. But, it's just a suggestion -- if you already are aware of it or don't think it's useful then of course feel free to ignore. Mar 25, 2022 at 16:22
• The other things that come to mind to try (again, just suggestions, and I don't know what you did so maybe you already tried things things) are (a) use Mathematica and xTensor to do the variation, (b) try linearizing the action around a simple background and see if you get the linearized equations right, (c) restrict the action to the minisuperspace (all fields only time dependent and with the metric equal to ${\rm diag}(-N^2(t), a^2(t), a^2(t), a^2(t))$ and see if you get the right equations in this restricted situation when you vary wrt $N$, $a$, and the other fields. Mar 25, 2022 at 16:30
• @123456: In standard GR Einstein's equations from the Einstein-Hilbert action contains the step of dropping boundary terms during the computation. This is more complicated when we have the dilaton term $e^{-2\phi}$. Maybe it's worth checking this step.
– psm
Mar 25, 2022 at 17:39

I once worked this out studying Polchinski's book, to which I refer for notation and conventions. I reproduce my work here verbatim.

We do this in separate steps, starting with the variation of the dilaton field \begin{align} \delta_\Phi {S} =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \delta_\Phi \Bigg\{ \sqrt{-G}\, e^{-2\Phi} \Big[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} + 4 \partial_\mu \Phi \partial^\mu \Phi \Big] \Bigg\}\nonumber\\ =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \Bigg\{ e^{-2\Phi} (-2\delta\Phi) \sqrt{-G}\, \Big[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} + 4 \partial_\mu \Phi \partial^\mu \Phi \Big] \nonumber\\ &\qquad \qquad+ \sqrt{-G}\, e^{-2\Phi} 8 \partial_\mu \Phi \partial^\mu \delta \Phi \Bigg\}\nonumber\\ =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \Bigg\{ -2e^{-2\Phi} \sqrt{-G}\, \Big[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} + 4 \partial_\mu \Phi \partial^\mu \Phi \Big] \nonumber\\ &\qquad \qquad-8 \partial^\mu \Big[ \sqrt{-G}\, e^{-2\Phi} \partial_\mu \Phi\Big]\Bigg\} \delta \Phi \nonumber \end{align} \begin{align} \phantom{\delta_\Phi {S}} =&\, -\frac{1}{\kappa_0^2} \int d^Dx\, e^{-2\Phi} \sqrt{-G}\, \Bigg[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} + 4 \partial_\mu \Phi \partial^\mu \Phi \nonumber\\ &\qquad \qquad+ 4(-G)^{-1/2}\big( \partial^\mu\sqrt{-G}\big) \partial_\mu \Phi +4 \big(-2\partial^\mu \Phi\big) \partial_\mu\Phi +4 \partial^\mu \partial_\mu \Phi \Bigg] \delta \Phi \nonumber\\ =&\, -\frac{1}{\kappa_0^2} \int d^Dx\, e^{-2\Phi} \sqrt{-G}\, \Bigg[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} -4 \partial_\mu \Phi \partial^\mu \Phi \nonumber\\ &\qquad \qquad+2 G^{\nu\sigma}\partial^\mu G_{\nu\sigma} \partial_\mu \Phi +4 \partial_\mu \partial^\mu \Phi \Bigg] \delta \Phi \nonumber\\ =&\, -\frac{1}{2\kappa_0^2 \alpha'} \int d^Dx\, e^{-2\Phi} \sqrt{-G}\, 2\delta\Phi \Bigg\{ -4\Bigg[ \frac{D-26}{6} +\alpha' \partial_\mu \Phi \partial^\mu \Phi -\frac{\alpha'}{2}\partial_\mu \partial^\mu \Phi -\frac{\alpha'}{24} H_{\mu\nu\lambda}H^{\mu\nu\lambda} \Bigg]\nonumber\\ &\qquad \qquad+ \alpha'{R} + 2 \alpha' \partial^\mu\partial_\mu \Phi-\frac{\alpha'}{4} H_{\mu\nu\lambda}H^{\mu\nu\lambda} \Bigg\} \nonumber\\ =&\, -\frac{1}{2\kappa_0^2 \alpha'} \int d^Dx\, e^{-2\Phi} \sqrt{-G}\, 2\delta\Phi \Big( -4 \beta^\Phi + \beta^{G\, \mu}_{\;\;\mu} \Big) \end{align} In the last line we have replaced the ordinary derivative by the covariant derivative, used the fact that the spacetime metric is covariantly constant, $$\nabla_\mu G^{\mu\nu}=0$$ and used the definition in (3.7.14).

Let us now consider \begin{align} \delta_B {S} =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \delta_B \Bigg\{ \sqrt{-G}\, e^{-2\Phi} \Big[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} + 4 \partial_\mu \Phi \partial^\mu \Phi \Big] \Bigg\}\nonumber\\ =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} \left( -\frac{1}{6} H^{\mu\nu\lambda} \delta_B H_{\mu\nu\lambda} \right)=-\frac{1}{4\kappa_0^2} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} H^{\mu\nu\lambda} \delta_B \partial_\mu B_{\nu\lambda} \nonumber\\ =&\, \frac{1}{4\kappa_0^2} \int d^Dx\, \partial_\mu \left(\sqrt{-G}\, e^{-2\Phi} H^{\mu\nu\lambda}\right) \delta B_{\nu\lambda} \nonumber\\ =&\, \frac{1}{4\kappa_0^2} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} \Big(+\frac{1}{2} G^{\nu\sigma} \partial_\mu G_{\nu\sigma} H^{\mu\nu\lambda} -2 \partial_\mu \Phi H^{\mu\nu\lambda} + \partial_\mu H^{\mu\nu\lambda} \Big) \delta B_{\nu\lambda} \nonumber\\ =&\, -\frac{1}{2\kappa_0^2 \alpha'} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} \Big(-\frac{\alpha'}{4} G^{\nu\sigma} \partial_\mu G_{\nu\sigma} H^{\mu\nu\lambda} +\alpha' \partial_\mu \Phi H^{\mu\nu\lambda} -\frac{\alpha'}{2} \partial_\mu H^{\mu\nu\lambda} \Big) \delta B_{\nu\lambda} \nonumber\\ =&\, -\frac{1}{2\kappa_0^2 \alpha'} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} \beta^B_{\mu\nu} \delta B_{\nu\lambda} \end{align} Here also, we have replaced the ordinary derivative by the covariant derivative, used the fact that the spacetime metric is covariantly constant, $$\nabla_\mu G^{\mu\nu}=0$$ and used the definition in (3.7.14).

Finally we consider \begin{align} \delta_G {S} =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \delta_G \Bigg\{ \sqrt{-G}\, e^{-2\Phi} \Big[ -\frac{2(D-26)}{3\alpha'} + {R} - \frac{1}{12} H_{\mu\nu\lambda}H^{\mu\nu\lambda} + 4 \partial_\mu \Phi \partial^\mu \Phi \Big] \Bigg\}\end{align} Recall that the variation of the Einstein-Hilbert action is \begin{align} \delta_G \int d^Dx\, \sqrt{-G}\, {R} = \int d^D x \, \sqrt{-G}\left( {R}_{\mu\nu}-\frac{1}{2} G_{\mu\nu}{R}\right) \delta G^{\mu\nu} \end{align} But we need to be careful as we have an extra factor $$e^{-2\Phi}$$. We split the calculation \begin{align} \delta_G \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} G^{\mu\nu} {R}_{\mu\nu} =& \sum_{a=1}^4 \delta_G I_a \end{align} with \begin{align} \delta_G I_1 =&\, \int d^Dx\, (\delta_G \sqrt{-G}) \, e^{-2\Phi} G^{\mu\nu} {R}_{\mu\nu} = \int d^Dx \, \frac{1}{2} \sqrt{-G} G^{\mu\nu} \delta G_{\mu\nu} e^{-2\Phi} G^{\rho\sigma} {R}_{\rho\sigma} \nonumber\\ =&\, \int d^Dx \sqrt{-G} e^{-2\Phi} \Big[ \frac{1}{2} G^{\mu\nu} {R} \Big]\delta G_{\mu\nu} \end{align} Next, \begin{align} \delta_G I_2 =&\, \int d^Dx\, \sqrt{-G} \, (\delta_G e^{-2\Phi}) G^{\mu\nu} {R}_{\mu\nu} = 0 \end{align} Third, \begin{align} \delta_G I_3 =&\, \int d^Dx\, \sqrt{-G} \, e^{-2\Phi}(\delta_G G^{\mu\nu} ) {R}_{\mu\nu} = \int d^Dx\, \sqrt{-G} \, e^{-2\Phi}(- G^{\mu\rho} G^{\nu\sigma} \delta G_{\rho\sigma } ){R}_{\mu\nu} \nonumber\\ =&\, \int d^Dx \sqrt{-G} e^{-2\Phi} \Big[-{R}^{\mu\nu} \Big]\delta G_{\mu\nu} \end{align} Finally, and this is where the change occurs compared to the Einstein-Hilbert action \begin{align} \delta_G I_4 =&\, \int d^Dx\, \sqrt{-G} \, e^{-2\Phi } G^{\mu\nu}\delta_G {R}_{\mu\nu} = \end{align} The variation of the Ricci tensor is $$\delta {R}_{\mu \nu} = \delta {R}^\rho_{\mu \rho\nu } = \nabla_\rho \delta \Gamma^\rho_{\mu \nu} - \nabla_\nu \delta \Gamma^\rho_{\mu \rho}$$. Thus \begin{align} \delta_G I_4 =&\, \int d^Dx\, \sqrt{-G} \, e^{-2\Phi} G^{\mu\nu }( \nabla_\rho \delta_G\Gamma^\rho_{\mu \nu} - \nabla_\nu \delta_G \Gamma^\rho_{\mu \rho} ) \nonumber\\ =& \, \int d^Dx\, e^{-2\Phi} \big[ \sqrt{-G} \, \nabla_\rho(G^{\mu\nu } \delta_G\Gamma^\rho_{\mu \nu}) - \sqrt{-G} \, \nabla_\nu (G^{\mu\nu } \delta_G \Gamma^\rho_{\mu \rho}) \big] \nonumber\\ =& \, \int d^Dx\, e^{-2\Phi} \big[ \partial_\rho(\sqrt{-G} \, G^{\mu\nu } \delta_G\Gamma^\rho_{\mu \nu}) - \partial_\nu (\sqrt{-G} \, G^{\mu\nu } \delta_G \Gamma^\rho_{\mu \rho}) \big] \end{align} In the case of the Einstein-Hilbert action, i.e. $$\Phi=0$$, this is a total derivative and vanishes. This times this is not the case as we get a contribution from the dilaton field upon partial integration \begin{align} \delta_G I_4=&\, 2 \int d^Dx\, e^{-2\Phi} \sqrt{-G} \, G^{\mu\nu } \big[ \partial_\rho \Phi\delta_G\Gamma^\rho_{\mu \nu} - \partial_\nu \Phi \delta_G \Gamma^\rho_{\mu \rho} \big] \end{align} Unfortunately, this time we need to work out the variations of the connections \begin{align} \delta_G \Gamma^\rho_{\mu\nu} =&\, \delta_G \frac{1}{2} G^{\rho\sigma}(\partial_\mu G_{\sigma\nu} + \partial_\nu G_{\sigma\mu} - \partial_\sigma G_{\mu\nu}) \nonumber\\ =&\, \frac{1}{2}\Big[ - G^{\rho\kappa} G^{\sigma\tau} \delta G_{\kappa\tau} (\partial_\mu G_{\sigma\nu} + \partial_\nu G_{\sigma\mu} - \partial_\sigma G_{\mu\nu}) \nonumber\\ &+ G^{\rho\sigma}(\partial_\mu \delta G_{\sigma\nu} + \partial_\nu \delta G_{\sigma\mu} - \partial_\sigma \delta G_{\mu\nu}) \Big] \end{align} Recall that by covariance we can, to that order, replace all partial derivatives by covariant derivatives. This means that we can ignore the first line. We then split the calculation in two \begin{align} \delta_G I_{4a} = &\, 2 \int d^Dx\, e^{-2\Phi} \sqrt{-G} \, G^{\mu\nu } \partial_\rho \Phi \frac{1}{2} G^{\rho\sigma}(\partial_\mu \delta G_{\sigma\nu} + \partial_\nu \delta G_{\sigma\mu} - \partial_\sigma \delta G_{\mu\nu}) \nonumber\\ =&\, - \int d^Dx\, \sqrt{-G} \, G^{\mu\nu} G^{\rho\sigma} \Big[ \partial_\mu \left( e^{-2\Phi} \partial_\rho\Phi \right) \delta G_{\sigma\nu} + \partial_\nu \left( e^{-2\Phi} \partial_\rho\Phi \right) \delta G_{\sigma\mu}\nonumber\\ & - \partial_\sigma \left( e^{-2\Phi} \partial_\rho \Phi\right) \delta G_{\mu\nu}\Big]\nonumber\\ =&\, - \int d^Dx\, \sqrt{-G} \, G^{\mu\nu} G^{\rho\sigma} \Big[2 \partial_\mu \left( e^{-2\Phi} \partial_\rho \Phi\right) \delta G_{\sigma\nu} - \partial_\sigma \left( e^{-2\Phi} \partial_\rho\Phi \right) \delta G_{\mu\nu}\Big]\nonumber\\ =&\, - \int d^Dx\, \sqrt{-G} \, \partial_\sigma \left( e^{-2\Phi} \partial_\rho\Phi\right) (2G^{\sigma\nu} G^{\rho\mu} - G^{\mu\nu} G^{\rho\sigma}) \delta G_{\mu\nu} \end{align} The second part is \begin{align} \delta_G I_{4b} = -&\, 2 \int d^Dx\, e^{-2\Phi} \sqrt{-G} \, G^{\mu\nu } \partial_\nu \Phi \frac{1}{2} G^{\rho\sigma}(\partial_\mu \delta G_{\sigma\rho} + \partial_\rho \delta G_{\sigma\mu} - \partial_\sigma \delta G_{\rho\mu}) \nonumber\\ =&\, \int d^Dx\, \sqrt{-G} \, G^{\mu\nu } G^{\rho\sigma} \Big[ \partial_\mu \left( e^{-2\Phi} \partial_\nu\Phi \right) \delta G_{\rho \sigma} + \partial_\rho \left( e^{-2\Phi} \partial_\nu\Phi \right) \delta G_{ \sigma \mu} \nonumber\\ &- \partial_\sigma \left( e^{-2\Phi} \partial_\nu\Phi \right) \delta G_{ \rho \mu} \Big]\nonumber\\ =&\, \int d^Dx\, \sqrt{-G} \, G^{\mu\nu } G^{\rho\sigma} \partial_\mu \left( e^{-2\Phi} \partial_\nu\Phi \right) \delta G_{\rho \sigma} \nonumber\\ =&\, \int d^Dx\, \sqrt{-G} \, \partial_\sigma \left( e^{-2\Phi} \partial_\rho \Phi \right)G^{\mu\nu } G^{\rho\sigma} \delta G_{\mu\nu} \end{align} Therefore \begin{align} \delta_G I_{4} =&\, \int d^Dx\, \sqrt{-G} \, \partial_\sigma \left( e^{-2\Phi} \partial_\rho \Phi \right)(- 2G^{\sigma\nu} G^{\rho\mu} + G^{\mu\nu} G^{\rho\sigma} + G^{\mu\nu } G^{\rho\sigma})\delta G_{\mu\nu} \nonumber\\ =&\, \int d^Dx\, \sqrt{-G} \, \partial_\sigma \left( e^{-2\Phi} \partial_\rho \Phi \right)2(G^{\mu\nu} G^{\rho\sigma} - G^{\sigma\nu} G^{\rho\mu} )\delta G_{\mu\nu} \nonumber\\ =&\, \int d^Dx\, \sqrt{-G} \, e^{-2\Phi} (-2 \partial_\sigma \Phi\partial_\rho \Phi +\partial_\rho\partial_\sigma \Phi )\, 2(G^{\mu\nu} G^{\rho\sigma} - G^{\sigma\nu} G^{\rho\mu} )\delta G_{\mu\nu} \nonumber\\ =&\, \int d^Dx\, \sqrt{-G} \, e^{-2\Phi} (-4 \partial_\sigma \Phi\partial^\sigma \Phi G^{\mu\nu} + 4 \partial^\mu \Phi\partial^\nu \Phi +2\partial_\sigma \partial^\sigma \Phi G^{\mu\nu }- 2\partial^\mu \partial^\nu \Phi) \delta G_{\mu\nu} \end{align} Adding the four pieces together we find \begin{align} \delta_G \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} G^{\mu\nu} {R}_{\mu\nu} =&\, \int d^Dx \sqrt{-G} e^{-2\Phi} \Big( \frac{1}{2} G^{\mu\nu} {R} -{R}^{\mu\nu} \nonumber\\ & -4 \partial_\sigma \Phi\partial^\sigma \Phi G^{\mu\nu} + 4 \partial^\mu \Phi\partial^\nu \Phi +2\partial_\sigma \partial^\sigma \Phi G^{\mu\nu }- 2\partial^\mu \partial^\nu \Phi \Big)\delta G_{\mu\nu} \end{align} As expected this gives back the Einstein equations when we have a constant $$\Phi$$, but we see that the dilaton field gives a correction that includes its derivative only.

There is another thing that we need to be careful about as well. Any indices upstairs have been raised via the spacetime metric'' $$G^{\mu\nu}$$ so they also carry a metric dependence. For example \begin{align} \delta_G H_{\mu\nu\lambda}H^{\mu\nu\lambda} =&\, H_{\mu\nu\lambda}\delta_G ( G^{\mu\sigma} G^{\nu\rho} G^{\lambda\kappa} H_{\sigma\rho\kappa}) = 3 H_{\mu\nu\lambda} G^{\mu\sigma} G^{\nu\rho} H_{\sigma\rho\kappa} \delta G^{\lambda\kappa} \nonumber\\ =&\, -3 H_{\mu\nu\lambda} G^{\mu\sigma} G^{\nu\rho} H_{\sigma\rho\kappa}G^{\lambda\tau} G^{\kappa\eta} \delta G_{\tau\eta}\nonumber\\ =&\, -3 H^{\sigma\rho\tau} H^\eta_{\;\;\rho\sigma}\delta G_{\tau\eta}= -3 H^{\mu\rho\sigma} H^\nu_{\;\;\rho\sigma}\delta G_{\mu\nu} \end{align} Similarly \begin{align} \delta_G \partial_\mu\Phi \partial^\mu \Phi = &\, \delta_G G^{\mu\nu} \partial_\mu\Phi \partial_\nu \Phi = - G^{\mu\rho} G^{\nu\sigma} \partial_\mu\Phi \partial_\nu \Phi \delta G_{\rho\sigma} = - \partial^\mu \Phi \partial^\nu \Phi \delta G_{\mu\nu} \end{align} So we have \begin{align} \delta_G {S} =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, e^{-2\Phi} \Bigg\{ \delta_G \big( \sqrt{-G}\big) \Big[ -\frac{2(D-26)}{3\alpha'} - \frac{1}{12} H_{\rho\sigma\lambda}H^{\rho\sigma\lambda} + 4 \partial_\sigma \Phi \partial^\sigma \Phi \Big]\nonumber\\ & + \sqrt{-G}\Big( -{R}^{\mu\nu}+\frac{1}{2} G^{\mu\nu}{R} -4 \partial_\sigma \Phi\partial^\sigma \Phi G^{\mu\nu} + 4 \partial^\mu \Phi\partial^\nu \Phi +2\partial_\sigma \partial^\sigma \Phi G^{\mu\nu }- 2\partial^\mu \partial^\nu \Phi \nonumber\\ & \qquad \qquad+\frac{1}{4} H^{\mu\rho\sigma} H^\nu_{\;\;\rho\sigma} -4 \partial^\mu \Phi \partial^\nu \Phi \Big)\delta G_{\mu\nu} \Bigg\} \nonumber\\ =&\, \frac{1}{2\kappa_0^2} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} \Bigg\{ \frac{1}{2} G^{\mu\nu} \delta G_{\mu\nu} \Big[ -\frac{2(D-26)}{3\alpha'} - \frac{1}{12} H_{\rho\sigma\lambda}H^{\rho\sigma\lambda} + 4 \partial_\sigma \Phi \partial^\sigma \Phi \Big]\nonumber\\ & + \Big( -{R}^{\mu\nu}+\frac{1}{2} G^{\mu\nu}{R} -4 \partial_\sigma \Phi\partial^\sigma \Phi G^{\mu\nu} +2\partial_\sigma \partial^\sigma \Phi G^{\mu\nu }- 2\partial^\mu \partial^\nu \Phi +\frac{1}{4} H^{\mu\rho\sigma} H^\nu_{\;\;\rho\sigma} \Big)\delta G_{\mu\nu} \Bigg\} \nonumber\\ &=\, -\frac{1}{2\kappa_0^2\alpha'} \int d^Dx\, \sqrt{-G}\, e^{-2\Phi} \Big[ \alpha'{R}^{\mu\nu} +2 \alpha' \partial^\mu \partial^\nu \Phi-\frac{\alpha'}{4} H^{\mu\rho\sigma} H^\nu_{\;\;\rho\sigma} \nonumber\\ & -\frac{1}{2} G^{\mu\nu}\Big( \alpha' G^{\mu\nu}{R} +4\alpha'\partial_\sigma \partial^\sigma \Phi-\frac{\alpha'}{12} H_{\rho\sigma\lambda}H^{\rho\sigma\lambda}-\frac{2(D-26)}{3} -4 \alpha' \partial_\sigma \Phi \partial^\sigma \Phi\Big) \Big] \end{align} We can rewrite the term between brackets as \begin{align} \mathfrak{B}^{\mu\nu} =\beta^{G\; \mu\nu} -\frac{1}{2} G^{\mu\nu} (\beta^{G\; \sigma}_{\;\;\;\sigma} -4\beta^\Phi) \end{align} Indeed \begin{align} \mathfrak{B}^{\mu\nu} = &\, \alpha' {R}^{\mu\nu} + 2\alpha' \nabla^\mu \nabla^\nu \Phi - \frac{\alpha'}{4} H^{\mu\rho\sigma}H^\nu_{\;\;\rho\sigma} -\frac{1}{2} G^{\mu\nu}\Big(\alpha' {R} + 2\alpha' \nabla^2\Phi - \frac{\alpha'}{4} H^{\lambda\rho\sigma}H_{\lambda\rho\sigma} \nonumber\\ & - \frac{2(D-26)}{3} +2\alpha' \nabla^2 \Phi -4 \alpha' \nabla_\sigma \Phi\nabla^\sigma \Phi +\frac{\alpha'}{6} H^{\lambda\rho\sigma}H_{\lambda\rho\sigma} \Big) \nonumber\\ =&\, \alpha' {R}^{\mu\nu} + 2\alpha' \nabla^\mu \nabla^\nu \Phi - \frac{\alpha'}{4} H^{\mu\rho\sigma}H^\nu_{\;\;\rho\sigma} -\frac{1}{2} G^{\mu\nu}\Big(\alpha' {R} + 4\alpha' \nabla^2\Phi - \frac{\alpha'}{12} H^{\lambda\rho\sigma}H_{\lambda\rho\sigma} \nonumber\\ & - \frac{2(D-26)}{3} i -4 \alpha' \nabla_\sigma \Phi\nabla^\sigma \Phi \Big) \end{align} and we see that if we replace the covariant derivatives by partial derivatives we do indeed find what we were looking for.

• Wish I could upvote this 2 times! Thank you for your efforts.
– user242231
Mar 27, 2022 at 15:32