Difference between $dx \wedge dy$ and $dxdy = dA$.

Question

(Beginner in differential forms)

In $\mathbb{R}^2$, consider the differential form $\omega = dx \wedge dy$ and infinitesimal area element $dA = dxdy$. I already know that $$\int_{\mathbb{R}^2} w = \int_{\mathbb{R}^2} dA.$$ So is $dx \wedge dy$ more of a precise way of writing $dA$ or are they just different entities whose integral happen to be same.

Answer 1

The term $dx\,dy$ should only occur in nested double integrals. In reality the "area element" dA of elementary calculus refers to Lebesgue measure in ${\mathbb R}^2$. I use to write ${\rm d}A={\rm d}(x,y)$ when necessary. Note that this area element is unsigned, meaning that there is no question of orientation or ordering of the variables $x$, $y$. One then has, e.g., $$\int_{B(0,1)}{\rm d}(x,y)=\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dy\>dx\ .$$ Contrasting this the exterior product $dx\wedge dy$ is signed: By convention $x$ is considered as the first variable and $y$ as the second variable, leading to $$dx\wedge dy={\rm d}(x,y),\quad dy\wedge dx=-{\rm d}(x,y)\ .$$ It follows that $$\int_{B(0,1)} dx\wedge dy=\pi,\quad \int_{B(0,1)} dy\wedge dx=-\pi\ .$$

Answer 2

In A Course In Mathematics For Students of Physics Bamberg & Sternberg refer to objects like $\sigma\,dA$ as densities. You can think of $\sigma\,dA$ as representing, say, the electrical charge on a tiny patch of a surface. This quantity doesn’t depend on the orientation of the surface, nor does the absolute double integral $\int_R\sigma\,dA$.

By contrast, you can view the two-form $f\,dx\wedge dy$ as representing a flux through that same tiny surface patch. Here, the orientation matters: we need to know which side of the surface is “outside” to be able to say which way the flow is going: the value of the double integral $\int_R f\,dx\wedge dy$ depends on the orientation that you’ve chosen for $R$. With the correct choice of orientation, you can indeed have $\int_R\sigma\,dA=\int_R\sigma\,dx\wedge dy$. Choosing the opposite orientation will change the sign of the integral on the right.

These two types of objects behave differently under pullback. For a two-form, we have $$\alpha^*(f\,dx\wedge dy) = (\alpha^*f)(\det J)\,du\wedge dv,$$ where $\alpha = (\alpha_1,\alpha_2)^T$ and $J$ is the Jacobian matrix $$J=\pmatrix{ {\partial\alpha_1\over\partial u} & {\partial\alpha_1\over\partial v} \\ {\partial\alpha_2\over\partial u} & {\partial\alpha_2\over\partial v} }.$$ When $\alpha$ is orientation-preserving ($\det J\gt0$) and one-to-one, then $\int_R \alpha^*(\tau)=\int_{\alpha(R)}\tau$. On the other hand, the pullback of the density $\sigma\,dx\,dy$ is $$\alpha^*(\sigma\,dx\,dy) = (\alpha^*\sigma)\,|\det J|\,du\,dv.$$ The absolute value of the determinant reflects the fact that we don’t care about orientation and we have $\int_R \alpha^*(\sigma\,dx\,dy)=\int_{\alpha(R)}\sigma\,dx\,dy$ without requiring that $\alpha$ be orientation-preserving as we did for the integral of a two-form.