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I wonder if the following proposition holds: $\def\vb{\boldsymbol}$ $\def\pdv#1#2{\frac{\partial #1}{\partial #2}}$

Suppose $D,G\subseteq\mathbb{R}^n$, function $\vb{T}\colon D \to G,(x_1,x_2,\dotsc,x_n)\mapsto(y_1,y_2,\dotsc,y_n)$ is $C^1$ diffeomorphic, then$$ \begin{vmatrix} \pdv{y_1}{x_1} & \pdv{y_1}{x_2} & \dots & \pdv{y_1}{x_n} \\ \pdv{y_2}{x_1} & \pdv{y_2}{x_2} & \dots & \pdv{y_2}{x_n} \\ \vdots & \vdots & & \vdots \\ \pdv{y_n}{x_1} & \pdv{y_n}{x_2} & \dots & \pdv{y_n}{x_n} \\ \end{vmatrix} \begin{vmatrix} \pdv{x_1}{y_1} & \pdv{x_1}{y_2} & \dots & \pdv{x_1}{y_n} \\ \pdv{x_2}{y_1} & \pdv{x_2}{y_2} & \dots & \pdv{x_2}{y_n} \\ \vdots & \vdots & & \vdots \\ \pdv{x_n}{y_1} & \pdv{x_n}{y_2} & \dots & \pdv{x_n}{y_n} \\ \end{vmatrix} = 1. $$

So far I deduced that \begin{align*} &\begin{vmatrix} \pdv{y_1}{x_1} & \pdv{y_1}{x_2} & \dots & \pdv{y_1}{x_n} \\ \pdv{y_2}{x_1} & \pdv{y_2}{x_2} & \dots & \pdv{y_2}{x_n} \\ \vdots & \vdots & & \vdots \\ \pdv{y_n}{x_1} & \pdv{y_n}{x_2} & \dots & \pdv{y_n}{x_n} \\ \end{vmatrix} \begin{vmatrix} \pdv{x_1}{y_1} & \pdv{x_1}{y_2} & \dots & \pdv{x_1}{y_n} \\ \pdv{x_2}{y_1} & \pdv{x_2}{y_2} & \dots & \pdv{x_2}{y_n} \\ \vdots & \vdots & & \vdots \\ \pdv{x_n}{y_1} & \pdv{x_n}{y_2} & \dots & \pdv{x_n}{y_n} \\ \end{vmatrix} \\ &= \det\left( \begin{bmatrix} \pdv{y_1}{x_1} & \pdv{y_1}{x_2} & \dots & \pdv{y_1}{x_n} \\ \pdv{y_2}{x_1} & \pdv{y_2}{x_2} & \dots & \pdv{y_2}{x_n} \\ \vdots & \vdots & & \vdots \\ \pdv{y_n}{x_1} & \pdv{y_n}{x_2} & \dots & \pdv{y_n}{x_n} \\ \end{bmatrix} \begin{bmatrix} \pdv{x_1}{y_1} & \pdv{x_1}{y_2} & \dots & \pdv{x_1}{y_n} \\ \pdv{x_2}{y_1} & \pdv{x_2}{y_2} & \dots & \pdv{x_2}{y_n} \\ \vdots & \vdots & & \vdots \\ \pdv{x_n}{y_1} & \pdv{x_n}{y_2} & \dots & \pdv{x_n}{y_n} \\ \end{bmatrix} \right) \\ &= \begin{vmatrix} \sum_{k=1}^n \pdv{y_1}{x_k} \pdv{x_k}{y_1} & \sum_{k=1}^n \pdv{y_1}{x_k} \pdv{x_k}{y_2} & \dots & \sum_{k=1}^n \pdv{y_1}{x_k} \pdv{x_k}{y_n} \\ \sum_{k=1}^n \pdv{y_2}{x_k} \pdv{x_k}{y_1} & \sum_{k=1}^n \pdv{y_2}{x_k} \pdv{x_k}{y_2} & \dots & \sum_{k=1}^n \pdv{y_2}{x_k} \pdv{x_k}{y_n} \\ \vdots & \vdots & & \vdots \\ \sum_{k=1}^n \pdv{y_n}{x_k} \pdv{x_k}{y_1} & \sum_{k=1}^n \pdv{y_n}{x_k} \pdv{x_k}{y_2} & \dots & \sum_{k=1}^n \pdv{y_n}{x_k} \pdv{x_k}{y_n} \\ \end{vmatrix}. \end{align*}

I've no idea what to do next because it's not possible to eliminate partial derivatives like $$ \pdv{y_i}{x_j} \pdv{x_j}{y_i} = 1. $$

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    $\begingroup$ Have you considered what this looks like from the perspective of the chain rule? $\frac{dx}{dx}=\frac{\partial x}{\partial y_1}\frac{\partial y_1}{\partial x}+\frac{\partial x}{\partial y_2}\frac{\partial y_2}{\partial x}+..$ $\endgroup$ Commented Sep 23, 2024 at 15:15
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    $\begingroup$ Partial derivatives can't cancel like that. You can interpret this as evidence that partial derivatives are not fractions, or that $\partial x_j$ actually hides some information, so it's not the same object on the numerator and denominator. $\endgroup$ Commented Sep 23, 2024 at 15:40

2 Answers 2

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Just an illustration with the Cartesian and polar coordinates:

$$\begin{vmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{vmatrix}=r$$

$$\begin{vmatrix}\dfrac x{\sqrt{x^2+y^2}}&\dfrac y{\sqrt{x^2+y^2}}\\-\dfrac y{x^2+y^2}&\dfrac x{x^2+y^2}\end{vmatrix}=\dfrac1{\sqrt{x^2+y^2}}$$

As

$$\begin{pmatrix}\dfrac x{\sqrt{x^2+y^2}}&\dfrac y{\sqrt{x^2+y^2}}\\-\dfrac y{x^2+y^2}&\dfrac x{x^2+y^2}\end{pmatrix}=\begin{pmatrix}\cos\theta&\sin\theta\\-\dfrac{\sin\theta}r&\dfrac{\cos\theta}r\end{pmatrix}=\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix}^{-1},$$ we verify that the matrices are indeed inverses of each other.

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  • $\begingroup$ The answer is in my answer. Set $r=0$. $\endgroup$ Commented Sep 23, 2024 at 16:24
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$$\begin{align*}\mathrm{d}y_i=&\frac{\partial y_i}{\partial x_j}\mathrm{d}x_j\\=&\frac{\partial y_i}{\partial x_j}\frac{\partial x_j}{\partial y_l}\mathrm{d}y_l\end{align*}$$

Thus

$$\frac{\partial y_i}{\partial x_j}\frac{\partial x_j}{\partial y_l}=\mathrm{\delta}_{il}$$

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