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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 $\int_{D^{r+1}}\alpha = \int_{D^{r+1}}f(x)d\mu_{r+1} > 0$.

Contradiction.

-- 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.

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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
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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
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@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
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up vote 3 down vote accepted

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

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