# Relation between covariant differential and covariant derivative

1. Dec 1, 2013

### center o bass

In Theodore Frankel's book, "The Geometry of Physics", he observes at page 248 that the covariant derivative of a vector field can be written as

$$\nabla_X v = e_iX^j (v^i_{,j} + \omega^i_{jk} v^k)= e_i(dv^i(X) + \omega^i_k(X) v^k) = e_i (dv^i + \omega^i_k v^k)(X)$$

where $\omega^i_k = \omega^i_{jk} dx^j$, such that we can write

$\nabla v = e_i\otimes (dv^i + \omega^i_k v^k)$. He then goes on to define the "covariant differential" of a vector valued p-form $\alpha = e_i \otimes \alpha^i$ as

$$\nabla \alpha = e_i \otimes (d\alpha^i + \omega^i_k \wedge \alpha^k)$$

I was then left wondering what relation the "covariant differential" had to the covariant derivative of the vector valued p-form? Since we had that $\nabla v (X) = \nabla_X v$ for a 'vector valued zero form', does one have something like $\nabla \alpha (X) = \nabla_X \alpha$ for a vector valued p-form? Is there any relation here?