Question about Derivation of $C$ Tensor in Wald's General Relativity

Question

On page 33 of Wald's General Relativity, I can understand how he says that $\hat\nabla_a-\nabla_a$ defines a map of dual vectors at $p$ to tensors of type (0,2) at $p$ because a dual vector $\omega_b$ (a (0,1) tensor) would become a (0,2) tensor after applying $\hat\nabla_a-\nabla_a$ to it.

I am struggling with how $\hat\nabla_a-\nabla_a$ then "defines a tensor of type (1,2) at $p$, which we will denote as $C^c_{\ \ ab}$". If $(\hat\nabla_a-\nabla_a)\omega_b$ is a (0,2) tensor, it would lead me to write it as something like $(\hat\nabla_a-\nabla_a)\omega_b=C_{ab}$ (whereas Wald says $=C^c_{\ \ ab}\omega_c$. Where does the extra upper index $c$ come from, and why contracted with $\omega_c$?

Answer 1

OP asked: I am struggling with how $\hat\nabla_a-\nabla_a$ then "defines a tensor of type (1,2) at $p$, which we will denote as $C^c_{\ \ ab}$".

In the question, you understand that $(\hat\nabla_a-\nabla_a) \omega_b$ is a (0,2) tensor. Then what tensor $C$ would act as a map that takes a (0,1) tensor to a (0,2) tensor? In other words, if

$$ C : \omega_a \rightarrow C_{ab}$$ then what is $C$ (excuse the use of same symbol for $C_{ab}$ and ${C^c}_{ab}$)?

The answer is that $C:= {C^c}_{ab}$. This is because the top index contracts with the one form $\omega_c$.

I should note that Wald is not calling $(\hat\nabla_a-\nabla_a) \omega_b$ as ${C^c}_{ab}$ (so your understanding is correct there) he is calling $ \hat\nabla_a-\nabla_a$ as ${C^c}_{ab}$.