Contravariant derivative of a tensor field in terms of generalized coordinates?

[yucheng]

Homework Statement

N/A

Relevant Equations

N/A

1. The laplacian is defined such that
$$ \vec{\nabla} \cdot \vec{\nabla} V = \nabla_i \nabla^i V = \frac{1}{\sqrt{Z}} \frac{\partial}{\partial Z^{i}} \left(\sqrt{Z} Z^{ij} \frac{\partial V}{\partial Z^{j}}\right)$$

($Z$ is the determinant of the metric tensor, $Z_i$ is a generalized coordinate)

But

$$\vec{\nabla} V = \vec{e}^i\nabla_{i} V = \vec{e}^{i} \partial_i V$$

So we need to express the gradient in terms of contravariant components before using the divergence formula.In Pavel Grinfeld's book, the author claims that $$\nabla^{i} V = Z^{ij}\nabla_{j} V$$ and the author calls this the 'contravariant' derivative.

However, I am concerned because:
https://math.stackexchange.com/ques...erivative-or-why-are-all-derivatives-covarian

and I cannot find the term 'contravariant derivative' anywhere. So...?

Since the contravariant derivative is defined as

$$\nabla_{i} F^{j} = \frac{\partial F^{j}}{\partial Z^{i}} + F^{k} \Gamma^{j}_{ki}$$ ,is there a similar interpretation for $$\nabla^{i} F^{j}$$?

Thanks in advance!

P.S. This might be of interest, but maybe not

https://math.stackexchange.com/ques...-application-of-contravariant-derivative-on-a

 

[ergospherical]

You can raise the index on $\nabla$ with the metric as per usual, i.e. $\nabla^i = Z^{ij} \nabla_j$

 

[yucheng]

You can raise the index on $\nabla$ with the metric as per usual, i.e. $\nabla^i = Z^{ij} \nabla_j$

But is there an interpretation for

$$Z^{mi} \nabla_{i} F^{j} = Z^{mi}\frac{\partial F^{j}}{\partial Z^{i}} + F^{k}Z^{mi} \Gamma^{j}_{ki}$$?