Meaning of the object arising from application of contravariant derivative on a scalar field The contravariant derivative operator can be defined by the action of contravariant metric tensor on the covariant derivative:
$$ Z^{ij} \nabla_i = \nabla^j$$
And, I know that applying the contravariant derivative on an invariant scalar field gives me it's gradient:
$$ \nabla^iF = \frac{\partial F}{\partial Z^j} Z^{ij}$$
Now, applying this to the covariant metric tensor to both sides
$$ \nabla_k F= Z_{ik} \nabla^iF = \frac{\partial  F}{\partial Z^j} Z^{ij} Z_{ik}= \frac{\partial F}{\partial Z^k}$$
Does this new object have any interpretation? I know it's not an invariant but it seems to agree with the regular gradient in cartesian coordinates.
The motivation on why I am asking this question is that I have been going over Pavel Grinfelds tensor calculus book, and in it Grinfeld had stated invariants i.e: Tensors are purely geometric objects (meaning they have some sort of interpetation). So, I'm trying to ponder if non invariants have no geometric meaning or not.
 A: The covariant derivative of an object which is invariant (a scalar, vector, higher order tensor, etc) will just yield the object's partial derivative. Observe that
$$ \nabla_i F = \frac{\partial F}{\partial z^i}  $$
implying that the object in "invariant form" would be given as
$$ \nabla_i F  \vec{Z}^{\:i} = \frac{\partial F}{\partial z^i} \vec{Z}^{\:i} \text{ .}$$
This is the invariant gradient object as written as a differential 1-form. If we wish to express it as a classical gradient, then it must be a "physical (tangent) vector". We can of course transform a 1-form to it's associated tangent vector as
$$ (\nabla F)^k = Z^{k\ell} \: \dfrac{\partial F}{\partial z^\ell} $$
$$ \implies(\nabla F)^k \vec{Z}_k = Z^{k\ell} \: \dfrac{\partial F}{\partial z^\ell}\vec{Z}_k \text{ .} $$
The only thing that we would achieve by using a contravariant derivative instead is that
$$ \nabla^j F = Z^{ij}\nabla_i F = Z^{ij}\frac{\partial F}{\partial z^i}  $$
would already have the form of a once covariant tensor
$$ \nabla^j F \vec{Z}_j = Z^{ij}\frac{\partial F}{\partial z^i} \vec{Z}_j $$
which is exactly the tangent vector which corresponds with the gradient.
