Let $R$ be the Riemann curvature tensor. Let $\text{scal}:=\sum_{k,l} R(e_k,e_l,e_k,e_l)$ be the scalar curvature, where $\{e_1,...,e_n\}$ is an orthonormal basis of $T_xM$. Let $X \in T_xM$. I want to show that $$X(\text{scal})=\sum_{k,l=1}^n ∇_XR(e_k,e_l,e_k,e_l)$$ where $∇$ is the Levi-Civita connection.
My attempt:
$X(\text{scal})=X(\sum_{k,l} R(e_k,e_l,e_k,e_l))=\sum_{k,l} X(R(e_k,e_l,e_k,e_l))$
For $X(R(e_k,e_l,e_k,e_l))$, there are terms like $R(∇_Xe_k,e_l,e_k,e_l)$ and I have no idea why these terms get cancelled out. My first instinct was that $R(e_l,∇_Xe_k,e_l,e_k)$ would cancel out $R(∇_Xe_k,e_l,e_k,e_l)$, but I realize that they are in fact equal instead of in opposite signs.
Thank you for helping!
This follows from two things: The metric is covariantly constant and taking covariant derivatives commutes with taking contractions. In this particular case we see that
\begin{equation} X(scal)=\nabla_{X}(scal)=\nabla_{X}({Rc^{j}}_{ j})=\nabla_{X}({{{R^{i}}_{i}}^j}_j)={{{(\nabla_X R)^{i}}_{i}}^j}_j. \end{equation}
Sorry for the poorly placed index notation. I haven't figured out how I can use the tensor package here. I am new.
The requirement that the covariant derivative commutes with taking contractions is used along with the Leibniz rule in order to define the covariant derivative of forms. Letting $C$ denote contraction we see that \begin{equation} \nabla_{X}(C(\omega\otimes Y))=\nabla_X(\omega(Y))=(\nabla_{X}\omega)(Y)+\omega(\nabla_{X}Y)=C(\nabla_{X}(\omega\otimes Y)). \end{equation} This property then extends to the entire space of tensors.