Derivative of Riemann tensor respect to Riemann tensor I know that, for example we have
$$\frac{\delta g^{jk}}{\delta g^{lm}}=\delta^{j}_{(l}\delta^{k}_{m)}.$$
This topic was discussed previously e.g. on Physicsforums.com and on Phys.SE.
So I was wondering, how can I compute
$\frac{\delta R^{j}_{klm}}{\delta R^{a}_{bcd}}$?
And in general, how can I compute this quantity for a tensor with some symmetries?
 A: The expression you are looking for is
\begin{equation}
\frac{\partial R_{\alpha\beta\mu\nu}}{\partial R_{\rho\sigma\lambda\gamma}} = \frac{1}{2}\left(\delta_{\alpha\beta}^{\rho\sigma}\delta_{\mu\nu}^{\lambda\gamma} + \delta_{\alpha\beta}^{\lambda\gamma}\delta_{\mu\nu}^{\rho\sigma}\right).
\end{equation}
Where the deltas in the equation are
\begin{equation}
\delta_{\alpha\beta}^{\rho\sigma} = \frac{1}{2}\left(\delta_{\alpha}^{\rho}\delta_{\beta}^{\sigma} - \delta_{\alpha}^{\sigma}\delta_{\beta}^{\rho}\right)
\end{equation}
With this, both the symmetric and antisymmetric properties of the Riemann tensor are taken into account. A more explicit expression is the following
\begin{align}
\frac{\partial R_{\alpha\beta\mu\nu}}{\partial R_{\rho\sigma\lambda\gamma}} &= \frac{1}{8}\left[(\delta_{\alpha}^{\rho}\delta_{\beta}^{\sigma}\delta_{\mu}^{\lambda}\delta_{\nu}^{\gamma} - \delta_{\alpha}^{\rho}\delta_{\beta}^{\sigma}\delta_{\mu}^{\gamma}\delta_{\nu}^{\lambda} - \delta_{\alpha}^{\sigma}\delta_{\beta}^{\rho}\delta_{\mu}^{\lambda}\delta_{\nu}^{\gamma} + \delta_{\alpha}^{\sigma}\delta_{\beta}^{\rho}\delta_{\mu}^{\gamma}\delta_{\nu}^{\lambda})\right. \nonumber \\[5pt]
&\left.+ (\delta_{\alpha}^{\lambda}\delta_{\beta}^{\gamma}\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma} - \delta_{\alpha}^{\lambda}\delta_{\beta}^{\gamma}\delta_{\mu}^{\sigma}\delta_{\nu}^{\rho} - \delta_{\alpha}^{\gamma}\delta_{\beta}^{\lambda}\delta_{\mu}^{\rho}\delta_{\nu}^{\sigma} + \delta_{\alpha}^{\gamma}\delta_{\beta}^{\lambda}\delta_{\mu}^{\sigma}\delta_{\nu}^{\rho})\right]
\end{align}
I've also been told by my doctoral advisor that the following equation also works
\begin{equation}
\frac{\partial R_{\alpha\beta\mu\nu}}{\partial R_{\rho\sigma\lambda\gamma}} = \frac{1}{2}\delta_{\alpha\beta}^{\rho\sigma}\delta_{\mu\nu}^{\lambda\gamma},
\end{equation}
but I fail to see how is the symmetric property of the Riemann tensor is considered.
