I am going through the paper https://arxiv.org/abs/1502.01028 which considers the quadratic gravity with the action \begin{align} S = \int d^4x \sqrt{-g} (R - \alpha C_{\mu\nu\rho\sigma}C^{\mu\nu\rho\sigma} + \beta R^2)\tag{1} \end{align} It was shown in the paper that the corresponding equation of motion is \begin{align} R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu} + 2\beta R (R_{\mu\nu}-\frac{1}{4}R g_{\mu\nu}) + 2\beta (g_{\mu\nu}\nabla^2 R - \nabla_\mu\nabla_\nu R) - 4\alpha B_{\mu\nu} = 0\tag{2} \end{align} where $B_{\mu\nu} = (\nabla^\rho\nabla^\sigma + \frac{1}{2}R^{\rho\sigma})C_{\mu\rho\nu\sigma}$ is the Bach tensor. However, from the variation of the action I got two extra terms quadratic in the Weyl tensor in the equation of motion \begin{align} \frac{1}{2} \alpha g_{\mu\nu} C_{\alpha\beta\gamma\delta} C^{\alpha\beta\gamma\delta} - 4 \alpha C_{\mu\alpha\beta\gamma} {C_{\nu}}^{\alpha\beta\gamma} \end{align} The first comes from the variation of the determinant of the metric \begin{align} \delta\sqrt{-g} = -\frac{1}{2}\sqrt{-g} g_{\mu\nu} \delta g^{\mu\nu} \end{align} and the second comes from \begin{align} \delta(C_{\mu\nu\rho\sigma}C^{\mu\nu\rho\sigma}) = \delta(g^{\mu\mu^{'}}g^{\nu\nu^{'}}g^{\rho\rho^{'}}g^{\sigma\sigma^{'}}C_{\mu\nu\rho\sigma}C_{\mu^{'}\nu^{'}\rho^{'}\sigma{'}}) = 2 C^{\mu\nu\rho\sigma} \delta C_{\mu\nu\rho\sigma} + 4 \delta g^{\mu\nu} C_{\mu\alpha\beta\gamma} {C_{\nu}}^{\alpha\beta\gamma} \end{align} To recover the result in the paper I need the two extra terms to cancel out, but it seems not to be the case, for example if we take the trace. It's likely due to my calculational mistake but I have been struggling to find out. I highly appreciate any help.

  • $\begingroup$ Just to remove ambiguity: in your last term, should $C_{\nu}^{\alpha\beta\gamma}$ be $C_{\nu} {}^{\alpha\beta\gamma}$? (The latter can be written in LaTeX as C_{\nu} {}^{\alpha\beta\gamma}.) $\endgroup$ May 21, 2022 at 13:52
  • $\begingroup$ Thank you for your suggestion, I changed it. $\endgroup$ May 21, 2022 at 14:03
  • $\begingroup$ You should be able to show those two terms cancel after splitting the Weyl tensor into is (anti-)self-dual parts. See e.g. chapter 1.6 of amazon.com/Ideas-Methods-Supersymmetry-Supergravity-Gravitation/… $\endgroup$
    – SigmaAlpha
    May 21, 2022 at 14:29
  • $\begingroup$ I am sorry. I found another contribution to terms quadratic in Weyl tensor from $C^{\mu\nu\rho\sigma}\delta C_{\mu\nu\rho\sigma}$, which makes the coefficient of $C_{\mu\alpha\beta\gamma} {C_{\nu}}^{\alpha\beta\gamma}$ to be $-2\alpha$ rather than $-4\alpha$. The new expression vanishes in dimension 4. $\endgroup$ May 23, 2022 at 14:03


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