When, if ever, can we view a differential form, e.g. like $dx \wedge dy$, as the similar looking expression used in physics to represent the product of "infinitesimals" e.g. $dx$ $dy$? In particular, I'm wondering why differential forms are anti-symmetric, e.g. $dx \wedge dy=-dy \wedge dx$, whereas in physics we often are happy to write $dx$ $dy=dy$ $dx$. Am I misunderstanding something basic?

  • 1
    $\begingroup$ books.google.com/books?id=DUnjs6nEn8wC&pg=PA86 $\endgroup$ – Steve Huntsman May 18 '10 at 4:59
  • 1
    $\begingroup$ This question is probably dead now, but in case someone else wanders into it the way I just did (by searching for something else) I'll add references to two books I feel do a great job of describing the roles of differential forms in physics. 1: John Baez, Javier Muniain, "Gauge fields, knots, and gravity". 2: Mikio Nakahara, "Geometry, topology, and physics". I find the former simpler and more devoted to precisely the question you ask, while the latter is probably more comprehensive in covering other applications of topology to physics as well. $\endgroup$ – gspr Aug 19 '10 at 19:14

In both physics and mathematics, there are times when you want a signed multiple integral $dx \wedge dy$, and there are times when you want its unsigned counterpart $dx\;dy = |dx \wedge dy|$. The difference is that in physics, the notation $dx \wedge dy$ is typically paraphrased either with cross products or with antisymmetric indices. The exterior algebra of differential forms is a brilliant definition due to Elie Cartan. Physicists sometimes need ideas that are equivalent to Cartan's work in this topic, but in most areas of physics they simply didn't adopt his notation. One major exception is string theorists and certain gauge theorists, who by now understand Cartan perfectly well.

For example, the most elegant way to understand a surface integral, as you see it in Ampere's law or Stokes' theorem, is as a signed integral. It is the integral of a differential form $$\omega(x(u,v),y(u,v),z(u,v)) = f(u,v) du \wedge dv$$ over a surface. But you can instead write it as the surface integral of a vector field $\vec{\omega} \cdot (\vec{du} \times \vec{dv})$. Any physicist can tell you that it's a signed integral; the only thing missing is Cartan's notation. A related example is Maxwell's equations. In low energy physics you write them as four equations with 3-vectors. In high-energy physics you write them as one or two equations with 4-vectors and 4-tensors with indices. You can also write the same equation using differential forms, but only gauge theorists and string theorists feel that they need that notation.

On the other hand, a mathematician who wants to use a probability density function or find an unsigned area or volume is perfectly happy to integrate with respect to $dx\;dy = |dx \wedge dy|$. Given other examples such as $ds = \sqrt{|dx|^2+|dy|^2}$ and $|dx \wedge dy|^p$, there is also a shift in emphasis: In more elementary use of Leibniz notation, the differentials are meant more as instructions for what kind of integral you are doing. In Cartan's notation, and in these other unsigned variations, the differentials become objects in their own right, basically what physicists would recognize as tensor fields with special transformation laws.

  • 2
    $\begingroup$ Even beasts like $|dx\wedge dy|^p$ can be seen in the wild! :) $\endgroup$ – Mariano Suárez-Álvarez May 18 '10 at 6:31

Concerning the question if one can literally think of differential forms as being products of "infinitesimals":

Yes. There is a way to make fully and usefully precise the statement that expression of the form $d x \wedge d y$ are products of infinitesimals. This is called synthetic differential reasoning.


A detailed account on how differential forms are functions on infintiesimal edges is given here.


Sophus Lie is famously quoted as having said that he found his theorems by "thinking synthetically" this way, but had to write them up in terms of the language of ordinary analysis in lack of a formal language for this synthetic thinking.

And indeed, in much of the physics literature one sees people implicitly reasoning synethetically, speaking and thinking about infinitesimal quantities. A central message is that this kind of very intuitive reasoning can be justified and does make sense in a precise mathematical way. The link above provides the details.

  • $\begingroup$ Urs, the question is about the exterior vs. usual product, what still stays the same question in synthetic approach. $\endgroup$ – Zoran Skoda May 18 '10 at 13:26
  • 1
    $\begingroup$ To get my head around that, I'd like to hear what it is saying for complex differential geometry. More specifically, what does "synthetic" differential geometry enable one to do that cannot be done with the usual theory of complex-analytic spaces, or the usual way in which tensor fields and differential forms are used in conjunction with linear approximations? $\endgroup$ – BCnrd May 18 '10 at 13:29
  • $\begingroup$ Zoran, yes and my reply was, too: the discussion I pointed to explains how the exterior product with its skew symmetry arises from bilinear functions in genuine infinitesimals. The skew symmetry is a consequence of the infinitesimalness. This is one of the key insights that Anders Kock has contributed to the subject. It is at its base a very simple statement (check out the explicit computation that I linked to!) but is of quite some impact. $\endgroup$ – Urs Schreiber May 18 '10 at 14:33
  • $\begingroup$ BCnrd, what the synthetic approach allows you to do is that it gives you a better conceptual understanding. There is nothing in ordinary differential geometry that you can do syntehtically which you can't do otherwise. But the synthetic language is the more conceptual and more abstract ones. It helps to think about stuff. It provides a way to make precise the wide-spread naive intuition about infinitesimal computations. And if you want to know how to generalize differential geometry, to Lie groupoids, stacks, oo-stacks and so forth, then it helps a lot. $\endgroup$ – Urs Schreiber May 18 '10 at 14:36
  • $\begingroup$ Thanks for the reply. It is a bit over my head, but appropriate for the level of MO so I can't complain. $\endgroup$ – user2028 May 19 '10 at 4:05

This question is long dead, but I think there's one aspect of it that wasn't really addressed, which is how there can be such a direct relationship between the antisymmetric object $\mathrm{d}x \wedge \mathrm{d}y$ and the symmetric object $dxdy$.

The reason is that ever-important proviso when integrating a differential form: the result depends on an arbitrary choice of orientation. So for a manifold $M$ admitting an orientation $o$, the relationship is actually $\int_{(M,o)} dx \wedge dy = \begin{cases} \int_M dxdy = \int_M dydx & \text{if } [\mathrm{d}x \wedge \mathrm{d}y] = o\\ -\int_M dxdy = -\int_M dydx & \text{if } [\mathrm{d}x \wedge \mathrm{d}y] \neq o\end{cases}$

Here the integral on the left is integration of forms and the integral on the right is a measure-theoretic integral. Also, the brackets denote taking the orientation-equivalence classes, where $\omega \sim \omega'$ iff $\omega = f\omega'$ for some everywhere-positive function $f$. Actually, this only makes sense if $\omega$ is nonvanishing; for a general 2-form $\omega = f dx\wedge dy$ we extend by locality of the integral / linearity of the measure and have $\int_{(M,o)} f dx\wedge dy = \int_{M_+} f dxdy - \int_{M_-}fdxdy$ where $M_{\pm} = \{p \in M \mid [f_p \mathrm{d}x \wedge \mathrm{d}y] = \pm o_p\}$ (and $M_0$ gets thrown out).

Now, what is going on when we swap the order of symbols under an integral sign? Well, from the equation above, we see that when we swap $\mathrm{d}x \wedge \mathrm{d}y$ with $\mathrm{d}y \wedge \mathrm{d}x$, we're multiplying our integrals by $-1$ (assuming we intend on keeping the same orientation). On the other hand, when we swap $dxdy$ for $dydx$, we're just using a notational variant as far as the product measure is concerned. But there's an ambiguity in standard integral notation between integration over a product measure and "Fubini'ed" integration, and usually if we swap $dxdy$ for $dydx$, what we mean is that we're changing the order in which we want to Fubini! What is the counterpart in terms of integrals of differential forms? Well, the equations, with $\omega = f \mathrm{d}x\wedge \mathrm{d}y$, will be

$\int(\int fdx)dy = \int fdxdy = \int (\int fdy)dx$

$\int(\int i_{\partial_{y|x}} \omega) \mathrm{d}y = \int \omega = \int(\int i_{\partial_{x|_y}} \omega) \mathrm{d}x$

Here $i_X$ is the insertion operator for the vector field $X$, which feeds $X$ into one of $\omega$'s input slots (first or last, by convention, I'm not sure which), lowering the degree by 1. Also, $\partial_{x|y}$ is short for $\frac{\partial}{\partial x}|_y$. Orientations: suppose that $\int \omega$ is oriented like $\mathrm{d}x \wedge \mathrm{d}y$. Then we can take the other integrals to be oriented like $\mathrm{d}x$ or $\mathrm{d}y$ as appropriate, as long as we take the insertion operator to be inserting in the last slot on the LHS, and the first slot on the RHS...

Independently of what happens when you swap the order of symbols, we can talk about the weird role of orientation in integration of forms. A form like $\mathrm{d}x \wedge \mathrm{d}y$ doesn't integrate to volume exactly, because its integral changes sign with orientation whereas volume doesn't. A quantity like this is usually called (somewhat pejoritively) a pseudoform. Whereas a "true form" requires a choice of orientation on its domain of integration, a pseudoform requires a choice of orientation on the normal bundle of its domain of integration (despite what it sounds like, this does NOT require a metric). One reference for this material is in Theodore Frankel's Geometry of Physics, which Steve Huntsman (cryptically) linked to in his comment above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy