Relation between covariant differential and covariant derivative

[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?

 

[jvicens]

Thank you for bringing up this interesting observation from Theodore Frankel's book. The relation between the "covariant differential" of a vector valued p-form and the covariant derivative of the same form is indeed an important one in differential geometry.

To understand this relation, we must first understand the concept of a covariant derivative. The covariant derivative is a generalization of the usual derivative in calculus, which takes into account the curvature of a space. In the context of differential geometry, the covariant derivative of a vector field measures how the vector field changes as we move along a curve in the space. It is defined using the connection on the space, which is represented by the $\omega^i_k$ in the equation you mentioned.

Now, in the equation for the covariant derivative of a vector field, we see that the connection $\omega^i_k$ appears in two places - as a coefficient for the derivative of the vector field, and as a coefficient for the vector field itself. This is because the covariant derivative not only measures the change of a vector field along a curve, but also takes into account how the basis vectors of the space change as we move along the curve. This is where the covariant differential comes in.

The covariant differential of a vector valued p-form, as defined by Frankel, takes into account both the change of the form itself and the change of the basis vectors of the space. This is why we see the wedge product $\wedge$ in the equation, which represents the change of basis vectors. So, in essence, the covariant differential is a generalization of the usual differential in calculus, just like the covariant derivative is a generalization of the usual derivative.

Now, to answer your question, the relation between the "covariant differential" and the covariant derivative is that they are essentially the same concept, just applied to different objects. The notation may be different, but the underlying idea is the same. So, we do have $\nabla \alpha (X) = \nabla_X \alpha$ for a vector valued p-form, just like we have $\nabla v (X) = \nabla_X v$ for a vector valued zero form.

I hope this helps to clarify the relation between the "covariant differential" and the covariant derivative. If you have any further questions, please do not hesitate to ask.

Best