Contracting Bianchi Identity

Question

Bianchi Identity is given as

$$R^n_{ikl;m}+R^n_{imk;l}+R^n_{ilm;k}=0 \tag{1}$$

$R^n_{ikl;m}$ is the Riemann/Curvature Tensor with covariant derivative with respect to m. To contract on $i$,$k$ and $l$,$n$, I do the following

$$g^l_ng^{ik}(R^n_{ikl;m}+R^n_{imk;l}+R^n_{ilm;k})=0 \tag{2}$$

$$R^l_{l;m}+R^n_{m;n}+R^n_{m;n}=0 \tag{3}$$

Since $R^l_{l;m}$ is scalar (right?), so

$$R^l_{l;m}=\frac{\partial R}{\partial x^m} \tag{4}$$

So I get,

$$R^n_{m;n}=-\frac{1}{2}R^l_{l;m}=-\frac{1}{2}\frac{\partial R}{\partial x^m} \tag{5}$$

However, the book doesn't have the negative sign in equation $(5)$. What did I do wrong? Anything wrong when I contract the 3rd term in from $(2)$ to $(3)$? Thanks!

Answer 1

Please see this link for a proof with all the minus signs correct: https://en.m.wikipedia.org/wiki/Proofs_involving_covariant_derivatives

Basically, you pick up some minus signs by permuting the indices of the Riemann tensor, which you need to do in order to write it as the Ricci tensor.