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Let $M$ be a smooth manifold of dimension $m$ and $(U,\phi)$, $(V,\psi)$ be two maps on $M$ such that $U\cap V\neq\emptyset$. I will write $\phi=(\phi^1,\dots,\phi^m)$ and $\psi=(\psi^1,\dots,\psi^m)$.

I want to prove that on $U\cap V$, we can write (using Einstein notation) $$\frac{\partial}{\partial \phi^k}=\frac{\partial \psi^i}{\partial\phi^k} \frac{\partial}{\partial \psi^i}$$ for all $k\in\mathbb N\cap[1,m]$.

My attempt: See my answer below.

My question: Is there a more elegant proof?

  • Would the downvoter care to explain the downvote ? – Maximilian Janisch Jul 31 '20 at 13:05
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    I didn't downvote but will assume that the question was downvoted because you simultaneously submitted an answer to a question that you wrote. Why not write the attempt in the body the question? – Axion004 Aug 01 '20 at 23:29
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    @Axion004 But I mean the feature that you can answer your own question is meant to be used . https://stackoverflow.blog/2011/07/01/its-ok-to-ask-and-answer-your-own-questions/ – Maximilian Janisch Aug 02 '20 at 09:49

2 Answers2

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Let $p\in U\cap V$ and $f\in\mathcal C^\infty(M)$. Then, letting $E_k$ denote the $k$th basis vector of $\mathbb R^m$ and $\mathrm d$ the differential, \begin{split} \frac{\partial f}{\partial \phi^k}(p)&=\left(\frac{\partial (f\circ\phi^{-1})}{\partial x^k}\right)(\phi(p))\\ &=\mathrm d(f\circ\phi^{-1})\vert_{\phi(p)}(E_k)\\ &=\mathrm d(f\circ\psi^{-1}\circ\psi\circ\phi^{-1})\vert_{\phi(p)}(E_k)\\ &\overset{{\text{chain rule}}}=\mathrm d(f\circ\psi^{-1})\vert_{\psi(p)}\big(\mathrm d(\psi\circ\phi^{-1})\vert_{\phi(p)}(E_k)\big)\\ &= \mathrm d(f\circ\psi^{-1})\vert_{\psi(p)}\left(\frac{\partial(\psi\circ\phi^{-1})}{\partial x^k}(\phi(p))\right)\\ &=\frac{\partial(f\circ\psi^{-1})}{\partial x^i}(\psi(p)) \cdot \frac{\partial(\psi^i\circ\phi^{-1})}{\partial x^k}(\phi(p))\\ &=\left.\left(\frac{\partial \psi^i}{\partial\phi^k} \frac{\partial}{\partial \psi^i}\right)\right\vert_p f. \end{split}

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A proof can also be found on page 64 of Lee's Introduction to Smooth Manifolds (second edition).