Multivariable Substitution Rule VS Pullback of Volume Forms

Question

Let $d\mathbf{x} = dx^1\wedge \cdots \wedge dx^n$ be the volume form on $U\subset\mathbb{R}^n$ and smooth $c : [a_1,b_1]\times\cdots\times [a_n,b_n] \to U$, $\mathbf{t} = (t_1,\dots,t_n) \mapsto \mathbf{x} = \mathbf{c}(\mathbf{t}) = (c^1(t_1,\dots,t_n),\dots,c^n(t_1,\dots,t_n))$.

I’ve encountered the following two formulae:

Multivariable Substitution Rule: $$d\mathbf{x} = d\mathbf{c} = \vert det\left((D\mathbf{c})_\mathbf{t}\right) \vert d\mathbf{t}$$

Pullback of Volume Forms: $$c^*(d\mathbf{x}) = d\mathbf{c} = det\left((D\mathbf{c})_\mathbf{t}\right) d\mathbf{t} $$

$(D\mathbf{c})_\mathbf{t}$ is the total derivative of $\mathbf{c}$ at $\mathbf{t}$.

My question is, are these two formulae referring to different things? It feels like these two equations are talking about the same thing, but then why does one have an absolute value on the Jacobian determinant?

I’m trying to understand this so I can prove that integrals of differential forms only flip signs under a re-parameterisation that reverses orientation.

Answer 1

Yes, they essentially refer to the same thing. As you can guess, this has something to do with keeping signs consistent. Consider the effect of switching the order of variables in integration. Fubini says this is okay, right?

Let's just simply integrate with the variables being $x,y \in \mathbb{R}^2$. In the case where I do regular Riemann integration with substitution, $dxdy = dydx$ so there's no sign change from the "differential". But if we took only the determinant, then notice that changing the order of $x$ and $y$ multiplies the determinant by $-1$ (because we switch two columns which changes the sign of the determinant). So to keep the value the same, we take the absolute value of the determinant.

How does this work in the case of differential forms? We know that $dx \wedge dy = - dy \wedge dx$ so we have one sign change from here. To "counteract" it, we don't take the absolute value of the determinant. So we have two sign switches per variable switch, so the overall integral remains the same.

Answer 2

If both formulas are correct, but lead to different results, they have to be about different things. The crux is in the notation: In the first formula $d{\bf x}$ and $d{\bf t}$ denote the unsigned "volume element" associated to Lebesgue measure in ${\mathbb R}^n$, whereas in the second formula your $d{\bf x}$ as well as $d{\bf t}=dt_1\wedge\ldots\wedge dt_n$ denote the volume form defined in terms of exterior algebra. The latter comes with a sign, and your formula takes care of that.

For myself I use the letter $d$ for signed things and ${\rm d}$ for unsigned ones. E.g., I write $${\rm vol}(B)=\int_B 1\>{\rm d}({\bf x})\ .$$

Answer 3

This question kind of baffles me too. I think the pullback statement is a well-defined, correct statement of equality between the two forms on the domain space (parameterized by t). This can be integrated over a desired region $R$ of the domain. If we want to work over the range, we use the theorem: $$\int_{R} {\bf c}^*(d{\bf x}) = \int_{{\bf c}(R)} d{\bf x}$$ (This assumes $\bf{c}$ is orientation preserving. Otherwise, you need to negate.)

I think the "multivariate rule" is a mnemonic that can be used as part of calculating an integral using variable substitution. But it does not describe a well-defined, correct statement of equality of two forms. (If it did, where does it live, the domain or the range? If it did, then why do you need to change the limits of integration when you wrap both sides by the integral sign?)

Regardless, what is formally true is that $$ \int_R \det(D{\bf c})\, d{\bf t} = \int_{{\bf c}(R)} d{\bf x} $$ where $D({\bf c})$ is the Jacobian matrix. (This assumes $\bf{c}$ is orientation preserving. Otherwise, you need to negate.)