# Exterior derivative independence from coordinate systems

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.

• 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 $\varepsilon$. – Ben McKay Jul 9 '19 at 11:53
• I just followed the description given by Arnold, but in fact he treats $V$ as an ordinary parallepiped, and $V'$ as a curvilinear one – Lo Scrondo Jul 9 '19 at 12:49
• 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. – Willie Wong Jul 9 '19 at 13:22
• 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 $\xi t \to \xi t + O(\xi t)$? – Lo Scrondo Jul 11 '19 at 11:38
• Because the derivatives agree @LoScrondo: if two functions have the same derivative at one point then there Taylor expansion agrees to first order. – Willie Wong Jul 11 '19 at 13:12

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 $$\omega$$ $$\langle\omega\lrcorner X,Y_1\wedge\dots\wedge Y_q\rangle= \langle\omega,X\wedge Y_1\wedge\dots\wedge Y_q\rangle.$$
• 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. – Bazin Jul 10 '19 at 16:24