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In page-82 of Pavel Grinfeld's Tensor Calculus Book, Exercise 91 is the following question:

Similarly, for a contravariant tensor $T^i$, derive the transformation rule for $\frac{dT^i}{dZ^j}$ and show that it is not a tensor.

OK! Seems easy enough, in a new primed coordinate systems, the tensor transforms as follows:

$$ T^{i'} = T^{i} J_{i}^{i'} \tag{1}$$

Where $$J_i^{i'}= \frac{\partial Z^{i'} }{\partial Z^i}$$ Is the jacobian. I.e: Derivative of new coordinates with respect to the old ones

Now, I am confused with respect to which coordinates that I should take the derivative of equation (1) with, this post suggests that it should be the new coordinates$(Z's)$ but the question asks about the old coordinates $(Z)$

tryst with freedom
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  • What do you mean by "with respect to which coordinates that I should take equation (1)"?

    Can you write what transformation law should $dT^i/dZ^i$ follow in order to say it "transforms like a tensor"? Can you write the transformation law that $dT^i/dZ^î$ actually follows?

     – Jackozee Hakkiuz May 08 '21 at 20:24
  • @JackozeeHakkiuz so say we have the cartesian coordinates in beginning, then $T^i$ may take form $T^i(x,y)$ when write it in a new system (say polar) then would it be $T^{i'}(r,\theta)$ or still $T^{i'} (x,y)$? – tryst with freedom May 08 '21 at 20:27
  • I'm not really sure what law it follows because I don't think so the derivative in itself is a tensor meaning we can't say $ \frac{dT^i}{dZ^j} = A_j^i$ @JackozeeHakkiuz – tryst with freedom May 08 '21 at 20:28
  • The components $T^i$ and $T^{i'}$ do not depend on the coordinates of a given point, instead they depend on the geometric point itself. (I write "geometric" to emphasize that the point does not depend on whatever coordinates you choose to name it with). That's why you can differentiate $T^i$ or $T^{i'}$ with respect to any coordinate system. Anyway, I think that is not relevant for what you are being asked to, which is to obtain the transformation law. Have you obtained the transformation law of these kind of objects before? Look at sections 6.4.1 and 6.4.2. – Jackozee Hakkiuz May 08 '21 at 21:16
  • I mean yeah fair point, but if we have a manifold, to do any calculus, we need to assign $T^i$ to a chart. So all I am asking is which chart are we taking the tensor to be defined on ? Do mean $ \partial_i (T \circ x^{-1} )$ or $ \partial_i ((T \circ y^{-1} )(\circ y \circ x^{-1}) )$ where the second way being in terms of new coordinates – tryst with freedom May 08 '21 at 21:19
  • note: $x^{-1}$ takes in coordinates as input and gives back geometric point – tryst with freedom May 08 '21 at 21:20
  • Ah ha! what I mean is easier to say using manifolds. The $T^i$ are scalar functions defined on the manifold itself, independend of any chart. Then for any chart $Z$ you define $$\frac{\partial T^i}{\partial Z^j} = \left(\partial_j(T^i\circ Z^{-1})\right)\circ Z$$ – Jackozee Hakkiuz May 08 '21 at 21:23
  • Also $T\circ x^{-1}=T\circ y^{-1}\circ y\circ x^{-1}$. – Jackozee Hakkiuz May 08 '21 at 21:25
  • Oh that was dumb of me. However, I'm still having a bit of difficulty following what you are saying. I'll reread your comments in the morning (it's late night here) – tryst with freedom May 08 '21 at 21:27
  • Sure. What I mean is that you don't need to introduce explicit charts. The exercise is an application of the chainrule followed by an application of the Leibniz rule. – Jackozee Hakkiuz May 08 '21 at 21:30
  • Let us continue this discussion in chat. – tryst with freedom May 09 '21 at 07:20

1 Answers1

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I found the answer from this lecture at 1:28:16 by Pavel Grinfeld on this topic. It goes as follows:

$$S_j^i = \frac{\partial T^i}{\partial Z^j}$$

Now, in primed coordinates:

$$ S_{j'}^{i'} = \frac{\partial T^{i'}}{\partial Z^{j'} }$$

So, the derivative must be taken in primed coordinates. Basically the object in the new coordinate system by the new derivative operator.

tryst with freedom
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  • Yes, those are the definitions in each coordinate system. The exercise is asking you to see how these two sets of components are related via the Jacobians. If your $S$ were "a tensor" in the sense Grinfeld talks about, then they would obey a certain law (see equation 6.27 in the book). You must prove that this is not the case. – Jackozee Hakkiuz May 09 '21 at 18:42