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There's a nice quote online that says:

"The whole point of differential forms is that integration of forms is designed to work consistently with the change of variables formula. Or another way to think of it, is that the change of variables formula is baked into the definition of differential forms."

So I'm trying to understand the integral$$\intop_{M}\omega=\intop_{D}\omega\left(\frac{\partial\mathbf{X}}{\partial u},\frac{\partial\mathbf{X}}{\partial v}\right)du\,dv$$in terms of the change of variables formula that I've found in Schulz's A Practical Introduction to Differential Forms. Have I got this right?

A 2-form $\omega=dx\wedge dy$ acts on a couple of tangent vectors as$$\left(dx\wedge dy\right)\left(\frac{\partial\mathbf{X}}{\partial u},\frac{\partial\mathbf{X}}{\partial v}\right)=\left|\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right|.$$

And, from the change of variable formula$$dx\wedge dy=\left|\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right|\,du\wedge dv.$$

Is it then OK to say (hopefully, it is. It just seems too simple)?$$\omega=\omega\left(\frac{\partial\mathbf{X}}{\partial u},\frac{\partial\mathbf{X}}{\partial v}\right)\,du\wedge dv,$$ and integrate this to get my integral (but without the wedge for some reason)?

Possible lightbulb moment? Is my integral the general case of the red boxed equation (from https://en.wikipedia.org/wiki/Differential_form#Integration_over_chains) here?

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  • $\begingroup$ If you've done surface integrals in your standard Calculus III course, this should be easy to see. You are allowed to do this. $\endgroup$ – Sean Roberson Aug 16 '19 at 19:58
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On $\mathbb{R}^n$, given an $n$-form $\omega = f dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n$ on $U \subseteq \mathbb{R}^n$ open, one defines $$\int_U \omega = \int_U f,$$ i.e. the integral is defined by integrating "without the wedge," using the standard Riemann integral on $\mathbb{R}^n$. Think of it this way: $dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n$ represents the standard infinitesimal volume on $\mathbb{R}^n$, since it is translation-invariant. So integrating with respect to it is integrating with respect to the usual volume form, which is also what Riemann integration does.

Then given a general $n$-manifold $M$ and an $n$-form $\omega$ supported on a coordinate patch $U \subseteq M$, one takes coordinates $\mathbf{x}: D \to U$ and defines $$ \int_U \omega = \int_D \mathbf{x}^\ast \omega.$$


Change-of-coordinates

Let me address the quote you gave. The magic of this definition is that if $U$ and $V$ are open subsets of $\mathbb{R}^n$ and $\alpha: U \to V$ is an orientation-preserving diffeomorphism, then for an n-form $\omega$ on $V$, $$ \int_V \omega = \int_U \alpha^\ast \omega.$$ This implies the definition I gave above for integrating on the manifold $M$ doesn't depend on coordinates. Let's check this: the pullback of $dx_i$ under $\alpha$ is $$ \alpha^\ast dx_i = \sum_{j=1}^n \frac{\partial \alpha_i}{\partial x_j} dx_j,$$ so $$ \alpha^\ast (dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n) = \left( \sum_{j=1}^n \frac{\partial \alpha_1}{\partial x_j} dx_j\right) \wedge \left( \sum_{j=1}^n \frac{\partial \alpha_2}{\partial x_j} dx_j\right) \wedge \cdots \wedge \left(\sum_{j=1}^n \frac{\partial \alpha_n}{\partial x_j}dx_j\right) \\ = \det(D\alpha) dx_1 \wedge dx_2 \wedge \cdots \wedge dx_n,$$ and our desired formula reduces to the change-of-variables $$ \int_V f= \int_U (f \circ \alpha) \det(D\alpha) $$ since $\alpha$ preserves orientation, so $\det(D\alpha) > 0$.


Red-boxed formula

The red-boxed formula is exactly the pullback of $\omega$ in local coordinates, if $M$ is an $n$-manifold inside $\mathbb{R}^N$ for some large $N$, and we express $\omega$ in terms of differential forms on $\mathbb{R}^N$. So, it agrees with the definition I gave above.

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