Since differential forms of odd rank are anti-commutative, what does that mean for multivariable integration?

Question

For differential forms $\alpha\in\Omega^p$ and $\beta\in\Omega^q$, they satisfy the relation $$\alpha\wedge\beta=(-1)^{pq}\beta\wedge\alpha$$ Thus, odd ranked forms are anti-commutative. At first, this made me think that changing the order of definite integration of a multivariable function would change the sign of the result, but this is not the case because $$\int_a^b\int_c^df(x,y)\,dx\,dy = \int_c^d\int_a^bf(x,y)\,dy\,dx$$ What are some results of the anti-commutativity property in (definite or indefinite) integration? Additionally, how is the canceled out in the above example?

Answer 1

I asked this question a few days ago, and I think I have stumbled upon the answer. We integrate forms over an oriented manifold (usually a sub-manifold). Let's say we have the following iterated integral: $$ \int_a^b\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y $$ The manifold in this case is the region $$R=\left\{(x,y)\in\mathbb{R}^2\ \middle|\ a\leq y\leq b,\ g(y)\leq x\leq h(y)\right\}$$ which can be equivalently written as $$R=\left\{(x,y)\in\mathbb{R}^2\ \middle|\ h^{-1}(x)\leq y\leq g^{-1}(x),\ \inf_{[a,b]}{g}\leq x\leq \sup_{[a,b]}{h}\right\}$$ Because of the anti-commutativity of 1-forms, $$ \int\!\!\!\int_R f\ \mathrm{d}x\wedge\mathrm{d}y = -\int\!\!\!\int_R f\ \mathrm{d}y\wedge\mathrm{d}x $$ To avoid picking up a minus sign, we must flip the orientation of the region so that $$ \int\!\!\!\int_R f\ \mathrm{d}x\wedge\mathrm{d}y = \int\!\!\!\int_{R'} f\ \mathrm{d}y\wedge\mathrm{d}x $$ where $R'$ has the opposite orientation of $R$. How do we express orientation in iterated integration? Let's look at a 1-dimensional example. Recall that $$ \int_a^b f\ \mathrm{d}x = -\int_b^a f\ \mathrm{d}x $$ The interval of integration $[a,b]$ is the same on both sides of this equation. The difference is the orientation of that interval. Since we have determined two expressions for the region $R$ earlier, we can say that $$ \int_a^b\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y = \int_{\inf_{[a,b]}\!g}^{\sup_{[a,b]}\!h}\!\!\!\int_{h^{-1}(x)}^{g^{-1}(x)} f\ \mathrm{d}y\,\mathrm{d}x $$ Here, the order of differential forms is flipped which picks up a minus sign, but the orientation of the manifold (region) also flipped which cancelled it out. Let $c=\inf_{[a,b]}g$ and $d=\sup_{[a,b]}h$. Then, $a=\inf_{[c,d]} h^{-1}$ and $b=\sup_{[c,d]} g^{-1}$ and we can write $$ \int_{\inf_{[c,d]} h^{-1}}^{\sup_{[c,d]} g^{-1}}\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y = \int_{\inf_{[a,b]}\!g}^{\sup_{[a,b]}\!h}\!\!\!\int_{h^{-1}(x)}^{g^{-1}(x)} f\ \mathrm{d}y\,\mathrm{d}x $$ By flipping one of the integrals on the right, we recover the original orientation of $R$ and recover the minus sign picked up from flipping our 1-forms: $$ \int_{\inf_{[c,d]} h^{-1}}^{\sup_{[c,d]} g^{-1}}\!\!\int_{g(y)}^{h(y)} f\ \mathrm{d}x\,\mathrm{d}y = -\int_{\inf_{[a,b]}\!g}^{\sup_{[a,b]}\!h}\!\!\!\int_{g^{-1}(x)}^{h^{-1}(x)} f\ \mathrm{d}y\,\mathrm{d}x $$ In short, when we change order of integration, we automatically change the orientation of the manifold we are integrating because the canonical place to put the limits in iterated integration are such that this is the case.