I am reading through Spivak's Calculus on Manifolds and have come across a technicality in one of the problems that is annoying me. It is Problem 4-25, the statement of which is

Let $c$ be a singular $k$-cube and $p:[0,1]^k\to[0,1]^k$ be a 1-1 function such that $p([0,1]^k)=[0,1]^k$ and $\det p'(x)\geq 0$ for $x\in[0,1]^k$. If $\omega$ is a $k$-form, show that $$ \int_c \omega = \int_{c\circ p} \omega. $$

I think Spivak intended this to be a straightforward exercise in the use of the change of variables theorem (his Theorem 3-13, which I copy below, making the change that Problem 3-39 allows, namely removing the assumption that $\det g'(x)\neq 0$).

3-13 Theorem. Let $A\subset\mathbf{R}^n$ be an open set and $p:A\to\mathbf{R}^n$ a 1-1, continuously differentiable function. If $f:p(A)\to\mathbf{R}$ is integrable, then $$ \int_{p(A)} f = \int_A (f\circ p)\lvert\det p'\rvert. $$

Indeed if we unwind the definitions of integrating forms over cubes, Problem 4-25 is solved as long as we can say that $$ \int_{[0,1]^k} (f\circ p)\lvert \det p' \rvert = \int_{[0,1]^k} f. $$ for a (say smooth) function $f:[0,1]^k\to\mathbf{R}$. There's one problem: $[0,1]^k$ is not open so I can't apply the theorem above directly. What I would like to do is replace the integrals over $[0,1]^k$ by integrals over $(0,1)^k$ but I don't really see how this will work. I don't know that the corresponding integrals would then be equal. I don't know that $p$ restricts to a bijection from $(0,1)^k$ to $(0,1)^k$. I don't know that $p((0,1)^k)$ is open because I can't use the inverse function theorem. Any way I try to approach this I seem to run into technical problems. I am happy to assume that everything is smooth.

Note: What it means in Spivak for a function to be differentiable on a non-open set is that the function extends to a differentiable function on an open set.

  • 1
    $\begingroup$ You don't need $g((0,1)^k)$ to be open in order to use the theorem (though this is indeed true, but it's a hard theorem called the theorem of invariance of domain). The two integrals will be the same because their difference will be over a set of measure $0$ (the boundary of the cube). $\endgroup$ – i like xkcd Aug 31 '13 at 1:34
  • $\begingroup$ I understand that $\int_{[0,1]^k} (f\circ p)\lvert\det p'\rvert=\int_{(0,1)^k} (f\circ p)\lvert\det p'\rvert$ and $\int_{[0,1]^k} f = \int_{(0,1)^k} f$. But then why is it true that $\int_{(0,1)^k} (f\circ p)\lvert\det p'\rvert = \int_{(0,1)^k} f$? $\endgroup$ – frakbak Aug 31 '13 at 1:51

This barely fit in a comment:

You are correct, I should have made that more precise. The thing is that, by the theorem you have $$ \int_{(0,1)^k} (f \circ p )|\det p'| = \int_{g((0,1)^k)} f. $$ But observe that $p((0,1)^k)) = p([0,1]^k \setminus \partial[0,1]^k)$. And since $p$ is $1-1$ and onto, this is $$ p((0,1)^k) = p([0,1]^k) \setminus p(\partial[0,1]^k). $$ What you are subtracting has measure $0$, so you can just write $$ \int_{(0,1)^k} (f \circ p )|\det p'| = \int_{[0,1]^k} f, $$ and now you can complete the proof using that $$ \int_{[0,1]^k} f = \int_{(0,1)^k} f. $$

  • $\begingroup$ It seems you are using the fact that $p((0,1)^k)$ is open, then, to apply the theorem? So we do need invariance of domain. $\endgroup$ – frakbak Aug 31 '13 at 2:45
  • $\begingroup$ @frakbak Where am I using that $p((0,1)^k) is open? I am just applying Theorem 3-13 to the open set $(0,1)^k$... right? $\endgroup$ – i like xkcd Aug 31 '13 at 3:04
  • $\begingroup$ Ah. Yes, you are applying the theorem as I stated it. I included in the theorem the revision made by Problem 3-39, which removed the assumption that $\det p'>0$ everywhere. In the original theorem you got that $p(A)$ was open for free because of the inverse function theorem. But with the possibility that $\det p'=0$ you no longer get that for free. When I was thinking about Problem 3-39 I might have included the openness of $p(A)$ as an extra hypothesis, so I probably should have included it in my statement above. But it follows from invariance of domain...which as you say is a hard theorem. $\endgroup$ – frakbak Aug 31 '13 at 3:12

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.