Showing that a derivative of a contravariant tensor is not a tensor

Question

If $T_i$ is a covariant tensor of order 1 and $Z$ and $Z'$ denote old and new coordinate systems, such that $T_{i'} = T_{i} \frac{\partial Z^{i}}{\partial Z^{i'}}=T_i J^i_{i'}$ then I show that $\frac{\partial T_i}{\partial Z^j}$ is not a tensor this way:

\begin{align*}\frac{\partial T_{i'}}{\partial Z^{j'}} &= \frac{\partial}{\partial Z^{j'}} \left[ T_i\left(Z\left(Z'\right)\right) J^{i'}_i \right]\\ &=\frac{\partial T_i}{\partial Z^k} \frac{\partial Z^k}{\partial Z^{j'}} J^{i}_{i'} + T_i \frac{\partial J^i_{i'}}{\partial Z^{j'}}\\ &= \frac{\partial T_i}{\partial Z^k} J^k_{j'} J^i_{i'} + T_i \frac{\partial^2 Z^i}{\partial Z^{j'} \partial Z^{i'}} \end{align*} And due to the last term, this does not transform as a covariant tensor and I am done (hopefully). However, when I apply the same procedure to a contravariant tensor $T^i$:

\begin{align*}\frac{\partial T^{i'}}{\partial Z^{j'}} &= \frac{\partial}{\partial Z^{j'}} \left[ T^i\left(Z\left(Z'\right)\right) J^{i'}_{i} \right]\\ &=\frac{\partial T^i}{\partial Z^k} \frac{\partial Z^k}{\partial Z^{j'}} J_{i}^{i'} + T^i \frac{\partial J_i^{i'}}{\partial Z^{j'}}\\ &= \frac{\partial T^i}{\partial Z^k} J^k_{j'} J_i^{i'} + T^i \frac{\partial}{\partial Z^{j'}}\left( \frac{\partial Z^{i'}}{\partial Z^i}\right) \end{align*} However, the last term is different, nothing really prevents me from switching the order of partials, getting:

\begin{align*}T^i \frac{\partial}{\partial Z^{j'}}\left( \frac{\partial Z^{i'}}{\partial Z^i}\right) &= T^i \frac{\partial}{\partial Z^{i}}\left( \frac{\partial Z^{i'}}{\partial Z^{j'}}\right) \\ &= T^i \frac{\partial}{\partial Z^{i}}\left( \delta^{i'}_{j'} \right) = 0\end{align*}

which would imply that the $\frac{\partial T^i}{\partial Z^j}$ transforms as a tensor.

What is my mistake?

Note: this is from page-82 of Pavel Grinfeld's Tensor Calculus Book, Exercise 91 and a relevant question about the same problem is here but without answer to this particular mistake i am making. Another relevant question is here but it stops shy of the switching derivatives which seem to render the expression zero.

Answer 1

I will change you notation a bit, I find $Z$ and $Z'$ confusing. Sorry!

Let's use $x=(x^1,\dots,x^n)$ for the first coordinate system and $y=(y^1,\dots,y^n)$ for the second. The confusion comes from the term $$\partial_{x^i}\left(\partial_{y^j}x^k\right)$$ but on the face of it, this expression does not make much sense: inside the parenthesis we are considering a function $x^k(y)$, and we take its derivative with respect to $y^j$. But then what does it mean to take the derivative along $x^i$?

The answer comes from noticing that the actual expression coming from the change of coordinates gives $$\partial_{x^i}\left(\partial_{y^j}x^k\Bigr|_{y(x)}\right)$$ and we use the chain rule to get $$\partial_{x^i}\left(\partial_{y^j}x^k\Bigr|_{y(x)}\right)=\partial_{y^l}\partial_{y^j}x^k\,\partial_{x^i}y^l$$ which should show that indeed $\partial_{x^i}T^j(x)$ is not a tensor.

An easy one-dimensional example: let $y(x)=(x−1)^2$, so that $x(y)=\sqrt{y}+1$. Near $0$, these are diffeomorphisms (inverse to each other). Now, what happens if we coumpute $\partial_x\left(\partial_yx\Bigr|_{y(x)}\right)$? We get $\partial_yx(y)=\frac{1}{2\sqrt{y}}$, so $(\partial_yx)(y(x))=\frac{1}{2(x−1)}$. When you take the next derivative, we finally arrive at $-\frac{1}{2(x−1)^2}$. If you could switch the derivatives, you would instead get $0$.

Intuitively, the reason why you cannot switch the derivatives in the expression $\partial_x\partial_y$, is that $y$ depends on $x$.