Using the Ricci identity for the third covariant derivative of a function - Bochner Weitzenböck identity.

Question

Let $\nabla$ be the Riemannian connection. I was able to proof the following identity for 1-forms $\eta \in \mathscr{T}^1(M)$ (I am following Lee's book "Riemannian Manifolds - An Introduction to Curvature"):

"If $\eta$ is a 1-form, let $$\eta_{i;jk}dx^i \otimes dx^j \otimes dx^k$$ be the local expression for $\nabla^2 \eta$. Prove the Ricci Identity $$ \eta_{i;jk}-\eta_{i;kj} = R_{jki}^{~~~~l} \eta_l"$$

Here $R_{jki}^{~~~~l}$ are the coefficients of the Riemannian curvature Endomorphism $R$, which are defined by $$R_{jki}^{~~~~l} \partial_l = R(\partial_j,\partial_k)\partial_i.$$ At the same time I'm reading R.Schoen's and S.-T. Yau's "Lectures on Differential Geometry". In a Proposition (2.2 p.15.) the Bochner Weitzenböck Identity is stated there in the following way:

"Let M be a Riemannian manifold and $f \in C^3(M)$. If $\{x^i\}$ is a normal coordinate system at a point $p \in M$, then we have at $p$ $$\Delta |\nabla f|^2 = 2 \sum |f_{ij}|^2 + 2 \sum R_{ij}f_if_j + 2 \sum f_i(\Delta f)_i ."$$ Following the arguments it is stated "from $f_{ij} = f_{ji}$ and the Ricci formula $f_{jij} = f_{jji} + R_{ij}f_j$ it follows that[...]". Now after some research i assume $f_{jij} = f_{j;ij}$ as far as my understanding goes. So these denote the coefficients of the 3rd covariant derivative $\nabla^3 f$. It makes sense since $\nabla f$ is a 1-form and $f_{j;ij} = f_{j;ji}$ is indeed true by the symmetry of the Hessian (in the Levi-Cevita case). Now I have the following problem, going after the Ricci identity in Lee's case I get $$f_{i;jj} = f_{j;ij} = f_{j;ji} + R_{ijj}^{~~~~~l} f_l$$, but I unfortunately cannot see how it implies the formula with the Ricci tensor involved as firstly the indices seem to mixed up (Using some symmetry identies of the Riemannian curvature tensor did not help me) and secondly I don't quite understand how the contraction on both (!) sides work. Do I also need in this identity the normal coordinates? I hope my question is stated clear enough. Any help is much appreciated.

Kind regards, Volker

Answer 1

UPDATE: It seems that it's a problem of the sign convention. Using $$ R(X,Y)Z = \nabla_{[X,Y]}Z+\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z$$ rather than the signed definition of the Riemannian curvature endomorphism, we get the identity $$\nabla^2 \eta (Z,Y,X) - \nabla^2 \eta (Z,X,Y) = \langle \eta, R(X,Y)Z \rangle$$, whence in local coordinates that translates to

$$\eta_{i;jk} - \eta_{i;kj} = R^l_{ikj}\eta_l$$ that is for $f\in C^3(M)$ $$f_{j;ij} = f_{j;ji} + R^l_{jji}f_l$$ Now summing up over j instead of l we get the desired identity. Does the last step makes sense? Still a bit unsure about that.