When evaluating the integrals of differential forms over a parameterized manifold, you often end up with something like:
$$\int _{|\gamma(s)|} dx\wedge dy + y dx \wedge dz = \iint_{[a,b] \times [c,d]} f(s,t)\,ds\,dt$$
However, this led me to a conceptual contradiction: the left hand side involves the wedge product of differential one forms, and is thus an expression involving differential two-forms.
The right hand side, however, has differentials (presumably differential one-forms) just multiplied together. Traditionally, I would interpret $ds dt$ as just representing an "infinitesimally" small square in the s-t plane. However, how do I reconcile this notation/concept with that of the differential form?
Best Answer
The classical area element $ \mathrm d x \, \mathrm d y $ is the absolute value of the $ 2 $-form $ \mathrm d x \wedge \mathrm d y $. That is, $$ \mathrm d x \, \mathrm d y = \lvert \mathrm d x \wedge \mathrm d y \rvert \text . $$
This explains the rules for change of variables. As you probably know, if $ x $ and $ y $ are assumed to be differentiable functions of some other variables $ u $ and $ v $, then the calculus of exterior differential forms calculates $ \mathrm d x \wedge \mathrm d y = J \, \mathrm d u \wedge \mathrm d v $, where $ J = \frac { \partial ( x , y ) } { \partial ( u , v ) } $ is the determinant of the matrix of partial derivatives of $ x $ and $ y $ with respect to $ u $ and $ v $ (the Jacobian). But the classical rule for a change of variables is $ \mathrm d x \, \mathrm d y = \lvert J \rvert \, \mathrm d u \, \mathrm d v $, involving the absolute value of the Jacobian. And this is just what you would expect from my first sentence: $ \mathrm d x \, \mathrm d y = \lvert \mathrm d x \wedge \mathrm d y \rvert = \lvert J \, \mathrm d u \wedge \mathrm d v \rvert = \lvert J \rvert \, \lvert \mathrm d u \wedge \mathrm d v \rvert = \lvert J \rvert \, \mathrm d u \, \mathrm d v $. This is also why $ \mathrm d y \, \mathrm d x = \mathrm d x \, \mathrm d y $ (as in Fubini's Theorem) even though $ \mathrm d x \wedge \mathrm d y = - \mathrm d y \wedge \mathrm d x $.
You can think of $ \mathrm d x $ as an operation that takes a vector and returns its $ x $ component, and similarly for $ \mathrm d y $. (This is a special case of interpreting $ \mathrm d f $ as an operation that takes a tangent vector $ v $ at a given point on a differentiable manifold, thinking of $ v $ as a derivation on the algebra of smooth functions, and returns $ v [ f ] $). Then $ \mathrm d x \otimes \mathrm d y $ takes two vectors as inputs, and multiplies the $ x $-component of the first by the $ y $-component of the second. Next, $ \mathrm d x \wedge \mathrm d y $ is $ \frac 1 2 ( \mathrm d x \otimes \mathrm d y - \mathrm d y \otimes \mathrm d x ) $ (although some people prefer a convention without the factor of $ 1 / 2 $). Finally, $ \mathrm d A : = \lvert \mathrm d x \wedge \mathrm d y \rvert $ is the absolute value of $ \mathrm d x \wedge \mathrm d y $.
Why is this the correct area element? If you take a triangle in the $ ( x , y ) $-plane, pick any one of its three vertices, then take the two vectors from that vertex as the two inputs to $ \mathrm d A $, then the output is the area of the triangle, regardless of the chosen vertex. (The convention without the $ 1 / 2 $ comes from using a parallelogram instead.) Then if you want to evaluate $ f ( x , y ) \, \mathrm d A $ on such a triangle, then you evaluate $ f ( x , y ) $ at the chosen vertex; although this is no longer independent of the vertex chosen, it doesn't affect integrals if $ f $ is continuous. Next, to approximate the integral of $ f ( x , y ) \, \mathrm d A $ on a region $ R $ in the plane with a Riemann sum, triangulate the region (or a triangulable region containing it if it has a weird shape), pick a vertex of each triangle, evaluate $ f ( x , y ) \, \mathrm d A $ on each triangle with the chosen vertex (or use $ 0 $ if the vertex is outside $ R $), and add these up. Finally, to define the integral exactly, take the limit as the length of the largest triangle goes to zero, which will always exist if $ f $ is continuous and $ R $ is compact. (There is also a Henstock–Kurzweil version of this limit that handles more functions and regions.) In particular, the integral of $ \mathrm d A $ itself gives you the area of the region (for any compact region, or any region with an area if you use the Henstock–Kurzweil integral).
Nor is this ad hoc. The procedure above can define the integral of any real-valued operator that takes a point and two vectors as input. And not just in $ \mathbb R ^ 2 $ but on any surface in any manifold (where now the arguments are a point and two tangent vectors at that point); you can think of the integrand as a generalized differential form (including the usual exterior differential forms as a special case). Actually, you generally also have to choose an orientation of the region of integration (telling you which vector is first and which is second); it's just an important property of $ \lvert \mathrm d x \wedge \mathrm d y \rvert $ that (unlike with $ \mathrm d x \wedge \mathrm d y $ without the absolute value) the orientation doesn't matter. Also, for most generalized differential forms that you can write down, the integral either never exists (on a region of positive area) or is always zero (on any region), but we still want to allow these into the calculus, because useful differential forms are obtained from them. (An example where the integral almost never exists is $ \mathrm d x \otimes \mathrm d y $ from above; an example where the integral is always zero is $ \mathrm d x ^ 2 = ( \mathrm d x ) ^ 2 $ from below.)
Another good example is the surface-area element $$ \mathrm d \sigma : = \sqrt { ( \mathrm d x \wedge \mathrm d y ) ^ 2 + ( \mathrm d x \wedge \mathrm d z ) ^ 2 + ( \mathrm d y \wedge \mathrm d z ) ^ 2 } $$ in $ \mathbb R ^ 3 $. This is what you integrate on a surface to find its surface area. Or multiply by $ f ( x , y , z ) $ first to integrate a scalar field on a surface. Unlike with $ \mathrm d A $, you can't do this with exterior differential forms by taking more care with orientation. For that matter, consider the arclength element $ \mathrm d s = \sqrt { \mathrm d x ^ 2 + \mathrm d y ^ 2 } $ in $ \mathbb R ^ 2 $. This is a generalized $ 1 $-form; you integrate it on a curve to get the arclength. Note that the curve itself does not have to be differentiable for this to make sense; $ \mathrm d s $ takes a vector to its Euclidean norm, a Riemann sum approximating the integral of $ \mathrm d s $ is the length of a polygonal approximation of the curve, and the definition of the integral as a limit reduces to the usual definition of the length of a rectifiable curve. (Or even of a non-rectifiable curve; then the integral is infinite. Of course, if the curve is rectifiable, that is if this integral is finite, then it has a continuously differentiable parametrization, but that's a theorem that comes after the definition of the arclength as an integral defined like any other integral.) Of course, we can do generalized differential forms of higher rank as well, such as the volume element $ \mathrm d V = \lvert \mathrm d x \wedge \mathrm d y \wedge \mathrm d z \rvert $, higher-dimensional analogues, the volume element on a hypersurface in a higher-dimensional space, etc.
Special cases of this generalized calculus of differential forms are well known. For those integrals that can be done by assigning an orientation, integrating one exterior differential form or its opposite depending on the orientation chosen, and noting that the final result is independent of orientation, we can use exterior pseudoforms (of rank equal to the dimension of the ambient space). Such pseudoforms can also be interpreted as densities (of weight $ 1 $ and rank equal to the dimension of the manifold). This works for $ \mathrm d A $ or $ \mathrm d V $ (with or without multiplying by a scalar field), but not for $ \mathrm d s $ and $ \mathrm d \sigma $. Each of these four (again with or without multiplying by a scalar field) can be interpreted as a density, still with weight $ 1 $ but now with rank possibly less than the dimension of the manifold. (But note that you are more likely to have heard of densities with different weights than densities with different rank, which don't seem to be as well known.) But non-zero exterior forms are never densities (except in rank $ 0 $), and non-zero exterior pseudoforms are never densities (except in top rank).
So if you want to explain how the density $ \mathrm d s $ is constructed out of the differential forms $ \mathrm d x $ and $ \mathrm d y $, by squaring them (resulting in something that is neither an exterior form, nor a pseudoform, nor a density, of any rank), adding these squares, and then taking a principal square root, you need to have a calculus that allows essentially arbitrary operations. This is easy enough to define once you think of it; simply allow arbitrary functions of a point and any finite number of vectors at that point. Then it's a little trickier to define, in general, how to integrate these; but simple enough once the method is understood. (To differentiate these, we have to deal with higher-order differentials such as $ \mathrm d ^ 2 x $, generalizing the inputs from vectors to jets, and I haven't touched upon that here. The exterior derivative is defined so that higher-order differentials cancel.) Once we have this, however, we can say definitively that the density $ \mathrm d A $, classically written $ \mathrm d x \, \mathrm d y $, is not the ordinary product of $ \mathrm d x $ and $ \mathrm d y $ (which, like $ \mathrm d x ^ 2 $, is a generalized $ 1 $-form whose integral is always zero) but rather the absolute value of the wedge product $ \mathrm d x \wedge \mathrm d y $. And that is how you can reconcile densities and exterior differential forms.