Are they the same thing?

So given an example, I could work out by change of coordinates, but how can I apply Change of Variable to replace this process?

Change of Variable in $\mathbb{R^k}$. Assume that $f: V \to U$ is a diffeomorphism of open sets in $\mathbb{R^k}$ and $a$ is an integrable function on $U$. Then $$\int_U a dx_1 \cdots dx_k = \int_V (a \circ f) |\det(df)|dy_1 \cdots dy_k.$$

So I will do $$\int_\Omega 1 dx \wedge dy,$$ where $\Omega$ is a two-manifold with boundary. And consider change of variable function $f(x,y) = (r \cos \theta, r \sin \theta).$ So $$dx = \cos \theta dr - r \sin \theta d \theta, dy = \sin \theta dr + r \cos \theta d\theta$$

Hence, $$\int_\Omega r dx \wedge dy = \int_\Omega 1 (\cos \theta dr - r \sin \theta d \theta) \wedge (\sin \theta dr + r \cos \theta d\theta) = \int_\Omega r dr \wedge d \theta.$$

I see the result of computation $r$ is $|\det(df)|$ in the formula. However, I don't really see where the composition $a \circ f$ happens. Should I see the process that bringing $dx \wedge dy$ to $dr \wedge d \theta$ as a result of the composition?

  • $\begingroup$ As an aside, the theorem you called "change of variable" I would have called "change of coordinates", since you are actually changing the set of points in the plane you are integrating over -- e.g. converting the integral over a half-disc to an integral over a rectangle (where $(r,\theta)$ mean the usual orthogonal coordates). A change of variable, on the other hand, would keep the same domain of integration, but change the differential form; e.g. by substituting $dx \wedge dy = r dr \wedge d\theta$ (this time $(r,\theta)$ mean polar coordinates). Of course, they are eventually the same thing. $\endgroup$ – Hurkyl Aug 4 '13 at 23:54
  • $\begingroup$ That's exactly why I was asking if they are the same thing at the beginning! :):) @Hurkyl Best called "Change of differential form"! $\endgroup$ – Tumbleweed Aug 5 '13 at 0:34
  • $\begingroup$ @Jellybean the whole point of defining differential form in terms of pull-backs to euclidean space it that is done in such a way that the choice of coordinates does not change the answer. In other words, the integration is coordinate free. Of course, we use coordinates to actually calculate it, but the change of variables theorem paired with the structure of oriented manifolds allows us to expect an unambiguous outcome. $\endgroup$ – James S. Cook Aug 5 '13 at 1:29
  • $\begingroup$ So you mean, change of coordinate is not appropriate since it is coordinate free. But how about chance of differential forms! :-P $\endgroup$ – WishingFish Aug 5 '13 at 1:44

It is hard to see composition when you have chosen $a=1$ so yes $a \circ f = 1$, but perhaps we could learn more by tweaking your example a bit. Consider, $\alpha = a(x,y) dx \wedge dy $ and calculate: $$ \alpha = a(x,y) dx \wedge dy = a(x,y) r dr \wedge d\theta$$ Very well, then to express this in polar coordinates we should replace $x$ and $y$ with their polar equivalents: $$ \alpha = a(r\cos(\theta),r\sin(\theta)) r dr \wedge d\theta $$ Integration of a form is coordinate invariant, either of the expressions will yield the same value. The difference in the calculation is that for the $x,y$-expression the definition of the integral just says to drop the wedge and perform a double integral over $\Omega$ since $\Omega$ maps to itself under the cartesian coordinate map (the identity). Whereas the subset $\Omega$ maps to some other domain with respect to the polar chart. For an easy example, if $\Omega = \{ (x,y) \ | \ x^2+y^2 \leq 1 \text{ and } \ $y>0$ \} $ then the $\Omega_{polar} = [0,1] \times [0,\pi] \in \mathbb{R}^2_{r\theta}$. Hence, $$ \int_{\Omega} \alpha = \int_{0}^\pi \int_0^1 a(r\cos(\theta),r\sin(\theta)) r dr\, d\theta = \int_{\Omega} a(x,y) dx \, dy $$

All of this said, the main difference between a change of variable and a change of coordinates is that if we change coordinates the coefficient of a differential form may well change sign. For example, $\gamma(x,y) = dx \wedge dy$ if we just re-order to $(y,x)$ then $\gamma(y,x) = -dy \wedge dx$ hence $1 \mapsto -1$. In contrast, if we swap the order of integration, which could be viewed as a change of variable of a very simple type, then the integral is unaltered: $$ \int_{\Omega} dx \, dy = \int_{\Omega} dy \, dx $$ The integral above is not a two-form integration, it is an area integral. When we change variables, we don't care about disorienting the variables, there is no orientation here. For this reason, the change of variables theorem has the absolute value on the Jacobian determinant. In contrast, when we change coordinates on a differential form, the absolute value is not present. Moreover, integrals of forms are taken over oriented spaces.

This is one of the reason it bothers me when texts omit the wedge for form integrals. It does not help clarify this issue. In any event, think about orientation, this is the difference.

  • $\begingroup$ Thank you so much James! This is very helpful for me as well!! $\endgroup$ – WishingFish Aug 4 '13 at 22:51
  • $\begingroup$ Thanks James - should this better called "change of differential form"...? $\endgroup$ – Tumbleweed Aug 5 '13 at 0:36
  • $\begingroup$ Hi James, thank you very much for your explanation. In the first step, how do you get $$f(x,y) dx \wedge dy = f(x,y) r dr \wedge d\theta?$$ $\endgroup$ – Tumbleweed Aug 5 '13 at 0:44
  • $\begingroup$ @Jellybean same as you.. $dx \wedge dy = (\cos \theta dr-r\sin \theta d\theta) \wedge (\sin \theta dr+r\cos \theta d\theta)$ which leaves just the $dr \wedge d\theta$ and $d\theta \wedge dr$-type terms; $dx \wedge dy = r(\sin^2 \theta + \cos^2 \theta) dr \wedge d\theta = rdr \wedge d\theta$ $\endgroup$ – James S. Cook Aug 5 '13 at 1:25
  • $\begingroup$ oooooo, you are doing nothing but working on $dx \wedge dy$! Got it, and thank yoU*! $\endgroup$ – WishingFish Aug 5 '13 at 1:39

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