Does the covariant derivative of a covariant derivative of a covector obey the product rule?

Question

Is this true? $\nabla_a(\nabla_bA_c) = \nabla_a\nabla_bA_c + \nabla_b\nabla_aA_c$ I'm confused because sometimes I see written that $\nabla_a(\nabla_bA_c) = \nabla_a\nabla_bA_c$ which seems to contradict the product rule. I understand that the covariant derivative is not itself a tensor but it obeys a lot of the same manipulation rules that tensors follow.

Answer 1

You say you understand it's not a tensor, yet you try to treat it as if it were.

Ok, think about it like this: the covariant derivative is an operator that, when applied to a tensor $A_a$, it yields a new tensor $(\nabla A)_{ab}$. The notation $\nabla_aA_b:\equiv(\nabla A)_{ab}$ is just that: notation. Hence $$\nabla_a(\nabla_bA_c):\equiv \nabla_a(\nabla A)_{bc}:\equiv (\nabla(\nabla A))_{abc}\equiv (\nabla\nabla A)_{abc}.$$