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Relation between covariant differential and covariant derivative

  1. Dec 1, 2013 #1
    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?
  2. jcsd
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