# show that $\omega$ is exact if and only if the integral of $\omega$ over every p-cycle is 0

Let $M$ be an oriented smooth manifold and $\omega$ a closed $p$-form on $M$. Show that $\omega$ is exact if and only if the integral of $\omega$ over every $p$-cycle is $0$.

In particular, how to prove that if the integral of $\omega$ over every $p$-cycle is $0$, then $\omega$ is exact?

• Interesting question. I've always wondered if one can prove it using only the Stokes theorem. – geodude Apr 26 '14 at 16:21
• Stokes alone is not enough, you need Poincare duality and that de Rham cohomology for smooth manifolds is isomorphic to the usual cohomology with real coefficients. What textbook are you using? Take a look at math.stackexchange.com/questions/81844/… – Moishe Kohan Apr 28 '14 at 17:33

Here's my attempt at this:

We'll say "$\omega$ kills boundaries" to mean $$\int_{\partial N} \omega = 0\quad\text{for all N,}$$ which in turn means this: for all compact $(p+1)$-manifolds-with-boundary $N$ and all $f \in C^\infty(N,M)$, we have $$\int_{\partial N} f^*\omega = 0.$$

Similarly, "$\omega$ kills cycles" means that for all closed ($=$ compact, boundaryless) $p$-manifolds $N$ and all $f \in C^\infty(N,M)$, we have $$\int_N f^*\omega = 0.$$

We want to show that $\omega$ is closed iff $\omega$ kills boundaries, and $\omega$ is exact iff $\omega$ kills cycles.

Claim 1. If $\omega$ is closed, then $\omega$ kills boundaries. If $\omega$ is exact, then $\omega$ kills cycles.

Proof. Stokes' Theorem.

Claim 2. If $\omega$ kills boundaries, then $\omega$ is closed.

Proof. Suppose $d\omega \neq 0$. Then $(d\omega)_q \neq 0$ for some $q \in M$. By continuity, there is a neighborhood $U$ of $x$ in which $d\omega$ is nowhere $0$. It should now be pretty easy to find an embedded $(p+1)$-disk in $U$ such that $$\int_{D^{p+1}} d\omega \neq 0,\quad\text{which implies that}\quad\int_{S^p} \omega \neq 0,$$ so we've found a boundary not killed by $\omega$.

Edit: More explicitly, choose local coordinates centered at $q$ and write $$d\omega = \sum_{i_1 < \cdots < i_{p+1}} f_{i_1\cdots i_{p+1}}\,dx_{i_1} \wedge \cdots \wedge dx_{i_{p+1}}.$$ By rearranging indices, we may assume that $f_{1,2,\ldots,p,p+1}(0) > 0$. Then the integral of $d\omega$ on the $(p+1)$-disk $$\{x_1^2 + \cdots x_{p+1}^2 \leq \varepsilon,\,x_{p+2} = \cdots = x_n = 0 \}$$ will be positive for $\varepsilon$ sufficiently small.

Claim 3. If $\omega$ kills cycles, then $\omega$ is exact.

Proof. (Key ingredient: de Rham's Theorem.) If $\omega$ kills cycles, then certainly $\omega$ kills boundaries, so $\omega$ is closed by Claim 2, hence $[\omega] \in H^p_{dR}(M)$. Now, Stokes' Theorem implies that $$\varphi: H^p_{dR}(M) \to \operatorname{Hom}(H_k(M), \mathbf R): [\omega] \mapsto \left( [\sigma] \mapsto \int_\sigma \omega \right)$$ is well-defined. Saying that $\omega$ kills cycles is the same as saying that $\varphi([\omega]) = 0$. But de Rham's Theorem says that $\varphi$ is an isomorphism. Thus, $[\omega] = 0$, i.e., $\omega$ is exact.

## protected by user26857Sep 24 '15 at 21:51

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