TITLE: Basic question about differential forms and physics
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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.