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

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 $$\omega_c$$?

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 $${C^c}_{ab}$$.