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I was re-reading this old book of mine; and I noticed that in defining the rules of differential forms, it "makes sense" that we have the rule $dx \wedge dx=0$ because if $dx$ is infinitesimal, then to first order approximations we can ignore powers of $dx$. Similarly, the definition for the exterior derivative $d$, of a differential form $\omega=Adx+Bdy+Cdz$, $d\omega=\frac{dA}{dx}dx + \frac{dB}{dy}dy + \frac{dC}{dz}dz $ "makes sense" because it feels like we are just multiplying the top and bottom by the differentials $dx,dy,$ and $dz$.

But it is practically a miracle that by introducing the simple anti-symmetrical commutation relations for differential forms, and applying very elementary operations, we can arrive at all the results of vector calculus such as gradient and cross product, among a large amount of other well known results.

In this particular book, the authors motivate the anti-symmetry condition by properties of determinants and Jacobian's for change of variables in integration. But I was wondering if there are other ways to think about why differential forms should commute anti-symmetrically which might provide some more intuition on just why this "miracle" works.

Thanks!

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    $\begingroup$ It's the same reason that you have the rule $\int_a^b f(x) dx = -\int_b^a f(x) dx$ in single variable calculus. These are oriented integrals, and to represent oriented integrals you need to keep track of orientations. Once you get past a single variable, the order of your variables matter since if you parametrize the plane by $(x,y) \longmapsto (x,y)$ or $(x,y)\longmapsto (y,x)$ the Jacobians have different signs. I believe there's a nice article on this by Terry Tao in the Princeton Companion to Mathematics.. either that or his blog. But the above is the idea in short. $\endgroup$
    – Ryan Budney
    Nov 22, 2010 at 23:42

2 Answers 2

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One way of looking at the antisymmetric relation is a consequence of $dx∧dx=0$ (which feels intuitive to you). Applied to $(dx+dy)∧(dx+dy)=0$, we get $(dx∧dx)+(dx∧dy)+(dy∧dx)+(dy∧dy)=0$. So, $(dx∧dy)+(dy∧dx)=0$, so $(dx∧dy)=-(dy∧dx)$

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    $\begingroup$ Is it obvious that $(dx + dy) \wedge (dx + dy) = 0$ follows from $dx \wedge dx = 0$? It may just be very early in the morning, but I don't see it. $\endgroup$
    – Gunnar Þór Magnússon
    Nov 23, 2010 at 7:39
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    $\begingroup$ @Gunnar: it follows from the assumption that $\omega \wedge \omega = 0$ for any one-form $\omega$. $\endgroup$
    – Eric O. Korman
    Nov 23, 2010 at 13:09
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I like the motivation given by Jack Lee's book Introduction to Smooth Manifolds. Roughly, we want to capture volume by the exterior algebra: say $\omega$ is a tensor that we want to apply to $n$ vectors to get the $n$-dimensional volume of the parallelogram they form. In the case $n=2$ for example, we should have $\omega(X,X) = 0$ since we get a line and not a 2-d region (so 0 area). Now by linearity, $\omega(X,X) = 0$ forces $X$ to be alternating (as in Timothy's answer).

The algebra of forms is the algebra of alternating tensors.

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