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In the book Mathematical Methods of Classical Mechanics by V.I. Arnold, the author introduces (p.189) the concept of exterior derivative as "the principal linear part of the increment" of the function $$F(\varepsilon)=\int_{\partial V(\varepsilon)} \omega$$

(where $V(\varepsilon)$ is a "curvilinear parallelepiped" with vertexes $x_0, x_0+\varepsilon \xi_1, ..., x_0+\varepsilon \xi_{n+1}$), $\varepsilon \to 0$, which could be shortly written as $$F(\varepsilon)=(d\omega)(x_0)(\xi_1, ...,\xi_{n+1})\varepsilon^{n+1}+o(\varepsilon^{n+1})$$

Then, in order to show the independence of the exterior derivative from the coordinate system, he states that after a change of coordinates, the difference $$\int_{\partial V(\varepsilon)} \omega - \int_{\partial V'(\varepsilon)} \omega$$ (where $V'$ is the curvilinear parallelepiped expressed in new coordinates) is smaller than $o(\varepsilon^{n+1})$, and asks to prove it. Unfortunately I have no clue how to prove it.

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    $\begingroup$ I think you will have to define the term "curvilinear parallelipiped". If $V$ is just a parallelipiped, then $V'$ isn't. But if you allow arbitrary curvilinearity, it is not clear how it behaves with $\varepsilon$. $\endgroup$ – Ben McKay Jul 9 '19 at 11:53
  • $\begingroup$ I just followed the description given by Arnold, but in fact he treats $V$ as an ordinary parallepiped, and $V'$ as a curvilinear one $\endgroup$ – Lo Scrondo Jul 9 '19 at 12:49
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    $\begingroup$ I am pretty sure implicitly the coordinate systems defining $V$ and $V'$ are the "same" at $x$ (the transition map should have derivative that equals the identity there) (this is given the figure illustrating the situation on pg 191). Then it really should just be a change of variables and application of Taylor's theorem. Following Arnold's notation, for $\partial V'(\epsilon)$, instead of integrating over the four segments $\xi t, \xi t + \eta, \eta t, \eta t + \xi$, you would be integrating over $\xi t + O(\xi^2t^2)$ etc. $\endgroup$ – Willie Wong Jul 9 '19 at 13:22
  • $\begingroup$ Thank you @WillieWong , I think yours is the only possible explanation. Just to verify if I've understood correctly...why your Taylor expansion lacks the linear term, i.e. why it is $\xi t \to \xi t + O(\xi^2t^2)$ and not $\xi t \to \xi t + O(\xi t)$? $\endgroup$ – Lo Scrondo Jul 11 '19 at 11:38
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    $\begingroup$ Because the derivatives agree @LoScrondo: if two functions have the same derivative at one point then there Taylor expansion agrees to first order. $\endgroup$ – Willie Wong Jul 11 '19 at 13:12
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A remark, too long for a comment. To check that the exterior derivative is a geometric operation, coordinate-free, it seems better to define first the Lie derivative of a form $\omega$ with respect to a vector field $X$: you define easily $$ \mathcal L_X(\omega)=\frac{d}{dt}(\Phi_X^t)^*(\omega)_{\vert t=0}, $$ where $\Phi_X^t$ is the flow of the vector field $X$. Then you can define the exterior derivative inductively by taking as a definition the Elie Cartan formula, $$ \mathcal L_X(\omega)=d\omega \lrcorner X+d(\omega \lrcorner X), $$ where $\lrcorner$ stands for the interior product. You know what is $df$ when $f$ is a function (0-form); using the above formula, you get for $\omega_{p+1}$ a $(p+1)$-form, $$ d_{p+1}\omega_{p+1} \lrcorner X=\mathcal L_X(\omega_{p+1}) -d_p(\omega_{p+1} \lrcorner X), \tag{$\ast$}$$ where $d_q\omega_q$ is the exterior differentiation of a $q$ form. Indeed $(\ast)$ gives you directly a geometric definition of $d_{p+1}\omega_{p+1} $ from the knowledge of $d_p$. Let us just recall that for a $q+1$ form $\omega$ $$ \langle\omega\lrcorner X,Y_1\wedge\dots\wedge Y_q\rangle= \langle\omega,X\wedge Y_1\wedge\dots\wedge Y_q\rangle. $$

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  • $\begingroup$ I do prefer your demonstration...unfortunately, I've to follow the one present in the book :-( $\endgroup$ – Lo Scrondo Jul 10 '19 at 12:45
  • $\begingroup$ Look, the Lie derivative of a tensor is a very intuitive and simple geometric operation which relies only on the definition of a local flow for a smooth enough vector field. Then you get for free the exterior differentiation. It is a bit unfair to Elie Cartan, but it is also very easy to check Cartan's formula in coordinates for the $X=\partial/\partial x_j$ with the usual formula for the exterior differentiation, so that you get a full proof. $\endgroup$ – Bazin Jul 10 '19 at 16:24

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