Suppose $(\Sigma,h_{ij})$ is a 3 dimensional Riemannian manifold and $S$ is a 2 dimensional submanifold of $\Sigma$.
Is the following equation true?
$$2(\nabla_i R^{ij})n_j=(\nabla_kR)n^k$$
where $\nabla$ is the Levi-Civita Connection with respect to $h_{ij}$, $R_{ij}$ and $R$ are respectively Ricci and scalar curvatures, and $n_i$ is the unit normal to $S$.
Can someone point me in the right direction?
One can show that $2\nabla_iR^{ij}= \nabla^j R$ using the second Bianchi identity:
$$\nabla^i R_{ij} = g^{ik} g^{mn} \nabla_k R_{imnj} = g^{ik} g^{mn} \big(-\nabla_n R_{imjk}-\nabla_j R_{imkn}\big)$$
$$= g^{ik} g^{mn} \big(-\nabla_n R_{mikj}+\nabla_j R_{mikn}\big)= -g^{mn}\nabla _n R_{mj} + \nabla_j R = -\nabla^i R_{ij} + \nabla_j R\ .$$