Covariant and directional derivative on scalar fields I am trying to decide whether the covariant derivative and the directional derivative are any different for a scalar field. Perhaps I’m misinterpreting something, but I seem to get a covector field when applying the del operator to a scalar field.
If a scalar field is a (0, 0) tensor, then its covariant derivative will be a (0, 1) tensor. And the del operator is defined $\nabla = \bf{e^i} {\partial \over \partial c^i } $. So then:
$$ \nabla f = \bf{e^i} {\partial f \over \partial c^i } $$ 
Now this seems to make sense, but I get a covector. On the other hand, the gradient is usually defined as:
$$ \nabla f = g^{ij} {\partial f \over \partial c^j} \bf{e_i} $$
So is the covariant derivative of a scalar field supposed to be a vector field, or a covector field?
 A: Without a metric
you are immediately able to take the covariant derivative $\nabla f$ of a scalar field, which coincides with its exterior derivative $\mathrm{d}f$
$$\nabla f = \mathrm{d}f = \sum_i \partial_i f \,\omega^i$$
where $\omega^i$ are the basis covector fields. Obviously this is a covector field.
Then the derivative of $f$ in the direction of a vector $v$ admits the following notations:
$$vf = \nabla_{v}f = (\nabla f)(v) = (\mathrm{d}f)(v) \tag{1}$$
If you have a metric
say $g$, then it induces the so-called musical isomorphisms $\sharp$ (which maps covector fields to vectors fields) and $\flat$ (which maps in the other direction).
So then you can define the gradient vector field of a scalar field as
$$\vec{\nabla}f := (\nabla f)^\sharp$$
In this case, the directional derivative $vf$ can be expressed (apart from the notations in $(1)$) by
$$g(\vec{\nabla}f,v)$$.
In short
"Del operator" may be a bit ambiguous. When applied to functions, in my experience people use it to refer to the grandient vector, but judging by what you wrote

the del operator is defined $\nabla = \bf{e^i} {\partial \over \partial c^i } $

you have found a place where "del operator" is used to talk about the covariant derivative.
