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.


  • 23
    $\begingroup$ 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. $\endgroup$ Nov 22, 2010 at 23:42

2 Answers 2


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 $(dx∧dy)=-(dy∧dx)$

  • 1
    $\begingroup$ 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. $\endgroup$ Nov 23, 2010 at 7:39
  • 6
    $\begingroup$ @Gunnar: it follows from the assumption that $\omega \wedge \omega = 0$ for any one-form $\omega$. $\endgroup$ Nov 23, 2010 at 13:09

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .