In Schutz, Geometrical Methods of Mathematical Physics, is written a Jacobian coordinate transform $\Lambda$, $$ \Lambda^i_j = \frac{\partial x^i}{\partial y^j} $$ The inverse matrix is written $$ \Lambda^k_j = \frac{\partial y^k}{\partial x^j} $$ Schutz then says (the fact that it is an inverse) "is easily proved using the chain rule for partial derivatives": $$ \frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \frac{\partial x^i}{\partial x^k} = \delta^i_k $$ Question 1: is it correct to regard this as a tensor contraction over "$\partial y^j$" ?

Next, Koks, Explorations in Mathematical Physics, has a similar example, but written out in matrix notation: $$ \left[ \begin{array}{cc} \frac{\partial r}{\partial x} &\quad \frac{\partial r}{\partial y} \\ \frac{\partial\theta}{\partial x} &\quad \frac{\partial\theta}{\partial y} \\ \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{\partial x}{\partial r} &\quad \frac{\partial x}{\partial\theta} \\ \frac{\partial y}{\partial r} &\quad \frac{\partial y}{\partial\theta} \\ \end{array} \right] = \left[\begin{array}{cc} 1 &\quad 0 \\ 0 &\quad 1 \end{array}\right] $$

Rewrite this using $\partial x^1 \equiv \partial r$, $\partial x^2 \equiv \partial\theta$, $\partial y^1 \equiv \partial x$, $\partial y^2 \equiv \partial y$ so that it matches the Shutz notation: $$ \left[ \begin{array}{cc} \frac{\partial x^1}{\partial y^1} &\quad \frac{\partial x^1}{\partial y^2} \\ \frac{\partial x^2}{\partial y^1} &\quad \frac{\partial x^2}{\partial y^2} \\ \end{array} \right] \cdot \left[ \begin{array}{cc} \frac{\partial y^1}{\partial x^1} &\quad \frac{\partial y^1}{\partial x^2} \\ \frac{\partial y^2}{\partial x^1} &\quad \frac{\partial y^2}{\partial x^2} \\ \end{array} \right] = \left[\begin{array}{cc} 1 &\quad 0 \\ 0 &\quad 1 \end{array}\right] $$ Now look at an individual element of the matrix product. Applying the chain rule from Schutz, $$ \frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \frac{\partial x^i}{\partial x^k} = \delta^i_k $$ however if we write this out explicitly for the 1,1 element: $$ \frac{\partial x^1}{\partial y^1}\frac{\partial y^1}{\partial x^1} + \frac{\partial x^1}{\partial y^2}\frac{\partial y^2}{\partial x^1} = 2 \frac{\partial x^1}{\partial x^1} = 2 $$ Question 2: this is incorrect...

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    $\begingroup$ You can't "cancel" $\partial y^1$ in $\frac{\partial x^1}{\partial y^1}\frac{\partial y^1}{\partial x^1}$, that's not how the chain rule works in several variables. $\endgroup$ May 3 '14 at 7:58
  • $\begingroup$ $\frac{\partial y}{\partial x}$ is not a quotient @beginner $\endgroup$
    – janmarqz
    May 4 '14 at 4:08
  • $\begingroup$ Is this not a duplicate of math.stackexchange.com/questions/778149/… ? I think that this answer is more satisfying, since it basically tells you that your method is not wrong but there is just an error in the application. $\endgroup$
    – xpnerd
    Apr 4 '17 at 7:02

There is an implicit sum over all values of $j$ in Schutz's notation; his statement is not that $$\frac{\partial x^1}{\partial y^1} \frac{\partial y^1}{\partial x^1} = 1$$ but rather that $$\frac{\partial x^1}{\partial y^1} \frac{\partial y^1}{\partial x^1} +\frac{\partial x^1}{\partial y^2} \frac{\partial y^2}{\partial x^1}= 1$$ i.e., $$\sum_j \frac{\partial x^1}{\partial y^j} \frac{\partial y^j}{\partial x^1} = 1$$ and his statement that $$\frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \delta^i_k$$ has an implicit summation: $$\sum_j \frac{\partial x^i}{\partial y^j} \frac{\partial y^j}{\partial x^k} = \delta^i_k$$


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