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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.

kimchi lover
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Susp1cious
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  • This Q&A is closely related. BTW what is the difference between $x^i$ and $x_i$ ? What textbook is that exactly ? – Kurt G. Nov 20 '21 at 16:53
  • @KurtG. $x^i$ are the contravariant coordinates (normal/regular coordinates with respect to the basis $e_i$) and $x_i$ are its covariant coordinates which can be obtained by "lowering" its index using the property of the metric tensor $x_i = g_{ij}x^j$. – Susp1cious Nov 20 '21 at 17:17
  • @KurtG. It's a textbook written by my Professor. I and other students believe it's rubbish as many topics are being glossed over without providing further satisfactory explanations. I would also really appreciate textbook recommendations. One book that has helped me a lot is "Manifolds, Tensors and Forms" by Paul Renteln. Unfortunately this book does not provide a direct answer to my question and I was not able to answer this question myself. – Susp1cious Nov 20 '21 at 17:27

1 Answers1

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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.

Kurt G.
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  • Is it possible that it is only true for non-curvilinear coordinates? I was able to show this in an earlier exercise. I can send you my derivation if you're interested. – Susp1cious Nov 20 '21 at 20:12
  • @Susp1cious : Yes, I am interested. – Kurt G. Nov 21 '21 at 12:41
  • Generally we want the differential operators to satisfy the following equation, regardless of their coordinates: $$\frac{\partial}{\partial x^i}x^j = \frac{\partial}{\partial\widetilde{x^i}}\widetilde{x^j} = \delta^i_{\space_j}$$. This is analogous to the invariance condition for regular vectors $$x^ie_i = \widetilde{x^i}\widetilde{e_i}$$. If you simply rewrite both equations for covariant coordinates one can easily show that the operators themselves must behave inversely to their respective denominator. – Susp1cious Nov 21 '21 at 18:26
  • Turns out a contravariant derivative apparently exists. I came to my derivation via a dot product, so there's a good chance that it is equal to the contravariant derivative you provided. – Susp1cious Nov 21 '21 at 20:29
  • @Susp1cious : do you mind to edit your question so that we can see your derivation? My concern is that $\widetilde{\partial^i}=\sum_{j}\frac{\partial{\widetilde{x_i}}}{\partial x_j}\partial^j$ contradicts the chain rule. – Kurt G. Nov 22 '21 at 09:44