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?
Best Answer
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 $p$.