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$$S\subseteq \mathbb{R}^N, U\subseteq \mathbb{R}^k, \phi\in C^1(U,\mathbb{R}^N) \ \text{ s.t.} \ \phi(U)=S.$$

$J_\psi$ is the Jacobian matrix. Let $V\subseteq\mathbb{R}^k$ be open, and let $\psi:V\to S$ a bijection so that $h:=\psi^{-1}\circ\phi\in C^1(U,V)$ and $h^{-1}=\phi^{-1}\circ\psi\in C^1(V,U)$

I'm trying to get to the end statement that $J_\psi(h(x))J_h(x)=J_\phi(x)$ for all $x\in U$.

I'm thinking along the lines of chain rules, but the notation is strange as it is. I think that $J_\phi(x)$ would be the values of $\phi$ with respect to the partial derivatives $x_i$, but it's sending me in circles.


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Don't be misled: the jacobian is a ttrick to express a linear operator as a matrix. Try thinking in terms of derivatives as linear operators, and apply the chain rule. –  Siminore Oct 6 '12 at 12:39
I don't quite follow. If we generalise, say $U\subset \mathbb{R}^n$ is open, $f:U\to \mathbb{R}^m, V\subset \mathbb{R}^m$ open and $g:V\to \mathbb{R}^k$ with $f(U)\subset V$. Letting $x\in U$, the definition of the chain rule is $$d(f\circ g)(x)=dg(f(x))\circ df(x)$$ But I still don't understand how that would fit into this notation. Are $\psi$ and $\phi$ the linear operators?! –  Ronan Oct 6 '12 at 12:51
If so, this leads to $$d(\phi\circ \psi)(x)=d\psi (\phi(x))\circ(d\phi(x))$$ If we can say that $d\psi$ is equal to $J_\psi$, and my brain is still undecided on that front, then $$d(\phi\circ \psi)(x)=J_\psi(\psi\circ h(x))\circ d(\psi\circ h(x))$$ Now, if we can expand, and again, not certain that we can: $$d(\phi\circ \psi)(x)=J_\psi(\psi\circ h(x))\circ J_\psi\circ J_h(x)$$ Someone please come along and tell me I'm wrong, this feels non-rigorous. –  Ronan Oct 6 '12 at 13:07

1 Answer 1

Think first of the differentials. If $\rm can$ denotes the canonical basis, in all cases, we have $\left[{\rm d}\phi(x)\right]_{\rm can} = J_\phi(x)$, and so on. And the matrix of the composition if the product of the matrices.

For starters, for all $x \in U$, since $h = \psi^{-1}\circ \phi$, we have: $${\rm d}h(x) = {\rm d}(\psi^{-1}\circ \phi)(x) \color{red}{=} {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\phi(x), $$where we used the chain rule in $\color{red}{=}$. On the other hand, take $h(x) \in V$. Since $\psi^{-1}\circ \psi = {\rm id}$ and the identity is linear, we have: $$ {\rm d}({\rm id})(h(x)) = {\rm d}(\psi^{-1}\circ \psi)(h(x)) \color{red}{=} {\rm d}(\psi^{-1})(\psi(h(x)))\circ {\rm d}\psi(h(x)) = {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\psi(h(x)),$$where again I used the chain rule in $\color{red}{=}$. By my initial remark: $$\begin{cases} {\rm d}h(x) = {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\phi(x) \\ {\rm id} = {\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\psi(h(x))\end{cases} \implies \begin{cases} \left[{\rm d}h(x)\right]_{\rm can} = \left[d(\psi^{-1})(\phi(x))\circ {\rm d}\phi(x)\right]_{\rm can} \\ \left[{\rm id}\right]_{\rm can} = \left[{\rm d}(\psi^{-1})(\phi(x))\circ {\rm d}\psi(h(x))\right]_{\rm can}\end{cases} $$Proceeding: $$\begin{cases} \left[{\rm d}h(x)\right]_{\rm can} = \left[{\rm d}(\psi^{-1})(\phi(x))\right]_{\rm can}\left[{\rm d}\phi(x)\right]_{\rm can} \\ \left[{\rm id}\right]_{\rm can} = \left[{\rm d}(\psi^{-1})(\phi(x))\right]_{\rm can} \left[{\rm d}\psi(h(x))\right]_{\rm can}\end{cases} \implies \begin{cases} J_h(x) = J_{\psi^{-1}}(\phi(x))J_\phi(x) \\ {\rm id} = J_{\psi^{-1}}(\phi(x)) J_\psi(h(x))\end{cases} $$ From here, we have $(J_{\psi^{-1}}(\phi(x)))^{-1} = J_\psi(h(x))$, so: $$J_h(x) = J_{\psi^{-1}}(\phi(x))J_\phi(x) \implies (J_{\psi^{-1}}(\phi(x)))^{-1}J_h(x) = J_\phi(x) \implies J_\psi(h(x))J_h(x)=J_\phi(x).$$

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