Thread: Riemann Curvature Tensor View Single Post
 Recognitions: Homework Help Because those derivatives aren't acting on T, they are acting on $\omega_c$, and hence $\frac{1}{6}\nabla_{a} \nabla_b \omega_{c}$ will just be some tensor. In contrast, if you look at the term $\frac{1}{6}\nabla_{a} \omega_c \nabla_b$, only the first operator acts on $\omega_c$; $\nabla_{b}$ will act on whatever tensor comes to the right of this expression.