TITLE: Riemann and Ricci tensor
QUESTION [5 upvotes]: 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!

REPLY [3 votes]: 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}