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 $$\int_U a dx_1 \cdots dx_k = \int_V (a \circ f) \left|\det(df)\right|dy_1 \cdots dy_k.$$

Is $dx_1 \cdots dx_k$ is a short hand for $dx_1 \wedge \cdots \wedge dx_k$? I just don't see a reason why the author suddenly decide to abbreviate it.

Thank you.

  • 1
    $\begingroup$ In tha case of $\mathbb{R}$ or $\mathbb{R}^n$ we usually do not use wedge between $dx_i$'s. $\endgroup$ – tessellation Aug 3 '13 at 5:17
  • $\begingroup$ I think you can say yes, in that you can consider the wedge $dx\wedge dy$ as a small rectangle, as it is done in the definition of en.wikipedia.org/wiki/Wedge_product , and this is what you're doing when integrating in $R^n$ , of course with the n-th-dimensional version of a rectangle. $\endgroup$ – DBFdalwayse Aug 3 '13 at 5:27

No, $dx_1\dots dx_k$ here is not a short hand for $dx_1\wedge \dots \wedge dx_k$. It is a longhand for the $k$-dimensional Lebesgue measure.

There are two similar kinds of integrals in $\mathbb R^k$:

  1. Integrals of functions with respect to the Lebesgue measure.
  2. Integrals of $k$-forms over $k$-manifold $\mathbb R^k$.

The first one uses the measure space structure. The second uses the manifold structure.

Since a $k$-form $\omega$ is visibly associated with a function via $\omega=f\,dx_1\wedge\dots\wedge dx_k$, integrals 1 and 2 can look alike. The difference between them transpires in the change of variables formula.

For integrals of type 1 it is $$\int_U a\, dx_1 \cdots dx_k = \int_V (a \circ f)\, \left|\det(df)\right|\,dy_1 \cdots dy_k.\tag1$$ Here $\left|\det(df)\right|$ arises as the Radon-Nikodym derivative of the pushforward of the measure $ dx_1 \cdots dx_k$ under $f^{-1}$.

For integrals of type 2 it is $$\int_U a\, dx_1 \wedge\cdots\wedge dx_k = \int_V (a \circ f)\, \det(df) \,dy_1\wedge \cdots\wedge dy_k.\tag2$$ Here $\det(df)$ arises from the computation of $f^*(a\, dx_1 \wedge\cdots\wedge dx_k)$, the pullback of $k$-form $dx_1 \wedge\cdots\wedge dx_k$ under $f$.

It is rare to see an honest explanation of the above difference in a calculus textbook. Usually, the authors find it convenient to use (2) in $\mathbb R^1$ and (1) in higher dimensions, with little to no explanation of the reason. Wikipedia follows suit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.