Covariant and partial derivative of a vector field (not component)

Question

Is the covariant derivative of a vector field (not the components of a vector) same as the partial derivative? I am adding a screenshot from page 69 from General Relativity: An introduction for physicists by Hobson where they used the partial derivative operator $\partial_bv$ for vector field derivatives, where in other books the covariant derivative operator $\nabla_bv$ is used.

Relevant: is a vector operating on another like $u(v)$ (as in Lie Bracket) also mean covariant derivative of $v$ in the direction of $u$?

Answer 1

It is indeed a notational issue. The left-hand side represents the tensor that results from the covariant derivative of $\boldsymbol{v} $ along the direction of the coordinate vector $\boldsymbol{e}_b$. Personally I would put it slightly different as: $$ \nabla_{\boldsymbol{e}_b}\boldsymbol{v} = \nabla_b v^a \boldsymbol{e}_a $$ where $\nabla_b v^a$ is the quantity given in your eq. (3.32) of your image.

Notice that $\nabla_b v^a $ is just notation for the components of this covariant derivative (maybe some people prefer something like $(\nabla v)_b{}^a$). What I mean is that $\nabla_b$ is not an operator here, although, in practice, it can be treated like that in component calculations if you understand its rules. On the other hand $\nabla_{\boldsymbol{e}_b}$ is a true operator. In fact, (3.32) comes from the Leibniz rule,

$$\nabla_{\boldsymbol{e}_b}\boldsymbol{v} = \nabla_{\boldsymbol{e}_b} (v^a \boldsymbol{e}_a) = (\nabla_{\boldsymbol{e}_b} v^a )\boldsymbol{e}_a +v^a (\nabla_{\boldsymbol{e}_b} \boldsymbol{e}_a)$$

and the fact that

$$\nabla_{\boldsymbol{e}_b} v^a = \partial_b v^a \qquad \nabla_{\boldsymbol{e}_b} \boldsymbol{e}_a= \Gamma^c{}_{ab} \boldsymbol{e}_c$$