Let $(M,g)$ be a smooth manifold and $f:M \rightarrow \mathbb{R}$ a smooth function.
Can we get some expression of: $$ \partial_k f g^{jh} R^k_{hij}$$ in terms of the Ricci tensor? (I use the notations $R^a_{cdb} \partial_a = R(\partial_c, \partial_d) \partial_b$ and $R_{bd}= R^c_{bcd}$)
In a paper, I have seen that $$\partial_k f g^{jh} R^k_{jih}= \partial_j fR^j_i$$
I have the following ansatz: $$\partial_k f g^{jh} R^k_{hij} = \partial_k f g^{jh} g^{lk} R_{hijl} =- \partial_k f g^{jh} g^{lk} R_{ihjl}=-\partial_k f g^{jh} g^{lk} g_{hl}R^h_{ihj} \\ =-\partial_k f g^{jh} \underbrace{g^{lk} g_{hl}}_{= \delta_{kh}}R_{ij} = \partial_h f g^{jh} R_{ij} = \partial_h f R^j_i $$ I pretty sure that the first $=$ is correct but the third $=$ just looks wrong to me, because there are three $h$s.
Can someone help me here? Thanks in advance!
There's actually four $h$'s there. To correct that we use the Bianchi identity, $$ R_{hijl} + R_{hjli} + R_{hlij} = 0.$$ Thus (using $\sigma^{jh} R_{hjli} = 0$)
\begin{align} \sigma^{jh} R^k_{hij} &= \sigma^{jh} \sigma^{lk} R_{hijl}\\ &=-\sigma^{jh} \sigma^{lk} (R_{hjli} + R_{hlij}) \\ &=-\sigma^{jh} \sigma^{lk} R_{hlij} \\ &=\sigma^{jh} \sigma^{lk} R_{lhij}\\ &= \sigma^{lk} R^h_{lhi} = \sigma^{lk} R_{li}. \end{align}