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Let $f: U \to V$ be a diffeomorphism, where $U$ and $V$ are both subsets of $\Bbb R^n$. Show that $\det(J_f(x))\det(J_{f^{-1}}(x))=1$ for all $x \in U.$

So I have that $ \det(J_f(x)) = \det \begin{bmatrix} \frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex] \frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n} \end{bmatrix} $ and similarly $\det(J_{f^{-1}}(x)) = \det \begin{bmatrix} \frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex] \frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n} \end{bmatrix}$

Now from the fact that $\det(AB) = \det(A) \det(B)$ I have that

$\det \begin{bmatrix} \frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex] \frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n} \end{bmatrix} \cdot \det \begin{bmatrix} \frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex] \frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n} \end{bmatrix} = \det (\begin{bmatrix} \frac{\partial f_1}{x_1} & \frac{\partial f_1}{x_2} & \dots & \frac{\partial f_1}{x_n} \\[1ex] \frac{\partial f_2}{x_1} & \frac{\partial f_2}{x_2} & \dots & \frac{\partial f_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f_n}{x_1} & \frac{\partial f_n}{x_2} & \dots & \frac{\partial f_n}{x_n} \end{bmatrix} \begin{bmatrix} \frac{\partial f^{-1}_1}{x_1} & \frac{\partial f^{-1}_1}{x_2} & \dots & \frac{\partial f^{-1}_1}{x_n} \\[1ex] \frac{\partial f^{-1}_2}{x_1} & \frac{\partial f^{-1}_2}{x_2} & \dots & \frac{\partial f^{-1}_2}{x_n} \\ \vdots & \vdots & \ddots & \vdots \\[1ex] \frac{\partial f^{-1}_n}{x_1} & \frac{\partial f^{-1}_n}{x_2} & \dots & \frac{\partial f^{-1}_n}{x_n} \end{bmatrix})$

But this feels already very messy and I'm not sure how the multiplication turns out... Is there another way I should approach this?

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Let $f$ map a neighborhood $U$ of $p$ diffeomorphically onto a neighborhood $V$ of $q:=f(p)$. Then the map $g:=f^{-1}:\>V\to U$ is again differentiable, and we have $$g\bigl(f(x)\bigr)=x\qquad(x\in U)\ .$$ The chain rule then says that $$dg(q)\circ df(p)=d(g\circ f)(p)=d\,{\rm id}_U(p)={\rm id}_{T_p}\ .$$ In terms of matrices this means that $$J_g(q) J_f(p)=I_n\ ,$$ so that $$\det\bigl(J_{f^{-1}}(f(p))\bigr)\det\bigl(J_f(p)\bigr)=1\ .$$ Note that you have a slight typo in the stated formula: The $J_{f^{-1}}$ is taken at $q=f(p)$, not at $p$.

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  • $\begingroup$ Could you elaborate on what do you mean by "in terms of matrices"? Isn't $d(g \circ f)(p)$ a matrix already? $\endgroup$ – user713999 Nov 3 '20 at 19:44
  • $\begingroup$ @Daniel: I denoted the matrix of $df(p)$ with respect to the standard basis by $J_f(p)$, because you had matrices in your question. $\endgroup$ – Christian Blatter Nov 3 '20 at 20:51
  • $\begingroup$ I see! Thank you! $\endgroup$ – user713999 Nov 3 '20 at 21:16

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