Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem: Let $\omega\in\Omega^r(M^n)$ suppose that $\int_\sum \omega = 0$ for every oriented smooth manifold $\sum \subseteq M^n$ that is diffeomorphic to $S^r$. Show that $d\omega = 0$.

Proof: Assume $d\omega \neq 0$. Then there exists $v_1, \ldots, v_{r+1}\in T_pM$ such that $d\omega_p(v_1, \ldots, v_{r+1}) \neq 0$.

$D^{r+1}\subseteq \mathbb{R}^{r+1}$ a smooth submanifold of $\mathbb{R}^{n}$ with boundary $S^r$. Let $(h,U)$ be a chart around $p$ such that $D^{r+1}$ (with some radius) is mapped to $N = h^{-1}(D^{r+1})$ around $p$. Then $N$ is a smooth submanifold of $M^n$ with boundary equal to $\partial N = h^{-1}(S^r)$ (diffeomorphic to $S^r$).

By definition of the integral and Stokes' theorem:

$\int_{\partial N} \omega = \int_N d\omega = \int_{D^{r+1}}(h^{-1})^* (d\omega)$.

Now let $\alpha = (h^{-1})^*(d\omega)$. Then $\alpha = f(x)dx_1\wedge\ldots\wedge dx_{r+1}$(topform in $\mathbb{R}^{r+1}$). Since $f(x) \neq 0$, it has to be different from zero on a small domain. Assume that $f(x) > 0$. Then $\int_{D^{r+1}}\alpha = \int_{D^{r+1}}f(x)d\mu_{r+1} > 0$.


-- I feel my idea is correct, but I'm not fully sure this is a full good proof. Could this have been done easier? I'm grateful for any feedback.

share|cite|improve this question
I think you need to assume that $r<n$, or there may be no such oriented smooth submanifold. – Akhil Mathew Nov 29 '10 at 0:05
I think this seems reasonable. The idea is that the integral of $\omega$ will vanish on the boundary of any $r+1$-ball in $M$ by Stokes theorem. Now just shrink the ball so it is really small. Then this integral is essentially the value of $\omega$ at the center. So this value is zero. – Akhil Mathew Nov 29 '10 at 0:07
Yes, $r<n$ is assumed in the proof. – M.B. Nov 29 '10 at 0:12
@M.B.: I know this is old, but perhaps you might consider adding an answer to your question to give it closure? It looks like a correct proof to me. – mixedmath May 9 '12 at 5:13
up vote 3 down vote accepted

As suggested by mixedmath since the proof was correct to begin with.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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