How can I prove that the differential operator with respect to covariant coordinates behaves contravariantly?

Question

Given the Operators

$\partial_i=\frac{\partial}{\partial x^i}$ and $\partial^i=\frac{\partial}{\partial x_i}$

I am supposed to show that they behave co- and contravariantly (as implied by the positioning of the indices) in curvilinear coordinates. Using the chain rule I was able to show

$$\widetilde{\partial_i}=\frac{\partial}{\partial \widetilde{x^i}}=\sum_j{\frac{\partial x^j}{\partial \widetilde{x^i}}\frac{\partial}{\partial x^j}}=\sum_j{J^j_\space i\frac{\partial}{\partial x^j}}=\sum_j{J^j_\space i\space\partial_j} \hspace{0mm},$$ where $J$ is the Jacobi-matrix. According to my Professors Textbook "a very similar calculation demonstrates that $∂^i = \frac{∂} {∂x_i}$ transforms like a contravariant vector."- so I should be expecting to get the inverse Jacobian. Using the chain rule one gets the analogous expression $$\widetilde{\partial^i}=\frac{\partial}{\partial \widetilde{x_i}}=\sum_j{\frac{\partial x_j}{\partial \widetilde{x_i}}\frac{\partial}{\partial x_j}}$$ However, the "supposed" inverse Jacobian contains derivatives of and with respect to covariant coordinates. If that really is the inverse Jacobian, the following equation must be true

$$\sum_k{\frac{\partial x^i}{\partial\widetilde{x^k}}\frac{\partial x_j}{\partial\widetilde{x_k}}}=\delta^i_j$$

as required by $J^{-1}J=I$. I fail to recognize any identity that would indicate that the equation as stated above should equal the Kronecker-Delta. I would really appreciate any help with this issue.

Answer 1

When we have as usual $x_i=g_{ij}x^j$ then one of the definitions $$ \partial_i=\frac{\partial}{\partial x^i}\,,\quad\partial^i=\frac{\partial}{\partial x_i}\, $$ must be sacrificed so that the more standard relationship $$ \partial_i=g_{ij}\,\partial^j $$ holds. (I prefer to write $\partial_i=\frac{\partial}{\partial x_i}$ but that's only a minor convention.) In this Q&A it is shown using the chain rule that

• $x^j$ are components of a contravariant vector;
• the derivative $\partial_i$ is a covariant vector;
• a contravariant derivative does not exist (your Professor's claim?);
• $\partial^j$ should better be called derivative with an upper index;

For further literature please see the given link.