I've been reading a paper about a Palatini formulation of $f(R, T)$ gravity theory, and when they vary the gravitational action with respect the connection $\widetilde{\Gamma}$, they obtain that \begin{align} \widetilde{\nabla}_{\lambda}\left[ \sqrt{-g}\left( A^{\mu \nu}\delta_{\alpha}^{\lambda} - A^{\mu \lambda}\delta_{\alpha}^{\nu}\right)\right] = 0 \end{align}\begin{align} \widetilde{\nabla}_{\lambda}\left[ \sqrt{-g}\left( A^{\mu \nu}\delta_{\alpha}^{\lambda} - A^{\mu \lambda}\delta_{\alpha}^{\nu}\right)\right] = 0\tag{23} \end{align} with $A^{\mu \nu} = f'(R)g^{\mu \nu}$.$$A^{\mu \nu} = f'(R)g^{\mu \nu}\tag{21}.$$
Then they say: "This equation can be significantly simplified by taking into account that for $\alpha = \lambda$ the equation is identically zero. Hence for the case $\alpha \neq \lambda$, we find \begin{align} \widetilde{\nabla}_{\lambda}\left[ \sqrt{-g}f'(R)g^{\mu \nu}\right] = 0 \end{align} ".
"This equation can be significantly simplified by taking into account that for $\alpha = \lambda$ the equation is identically zero. Hence for the case $\alpha \neq \lambda$, we find \begin{align} \widetilde{\nabla}_{\lambda}\left[ \sqrt{-g}f'(R)g^{\mu \nu}\right] = 0\tag{24} \end{align} ".
but I do not see where does the last equation come from. I mean, I can see why when $\alpha = \lambda$ the equation is identically zero, but I would expect for the case $\alpha \neq \lambda$ that the equation becomes \begin{align} \widetilde{\nabla}_{\lambda}\left[ \sqrt{-g} A^{\mu \lambda}\delta_{\alpha}^{\nu}\right] = 0 \end{align}\begin{align} \widetilde{\nabla}_{\lambda}\left[ \sqrt{-g} A^{\mu \lambda}\delta_{\alpha}^{\nu}\right] = 0\tag{24'} \end{align} since $\delta_{\alpha}^{\lambda} = 0$ for $\alpha \neq \lambda$.
Does anyone know what's going on? Thanks in advance. Here is the source: https://arxiv.org/pdf/1805.07419.pdfhttps://arxiv.org/abs/1805.07419 page 5.
