# If the integral of a one form only depends on the homotopy class of paths is it closed?

Let $M$ be a smooth manifold, and let $\omega \in \Omega^1(M)$.

Suppose that for every two paths $\gamma_0,\gamma_1:I \to M$ which are homotopic via a homotopy which fixes the endpoints, $\int_{\gamma_0} \omega=\int_{\gamma_1} \omega$.

Is it true that $\omega$ is closed?

(The converse direction is classic).

By repeating the proof of the converse statement "backwards", one can see that our assumption is equivalent to the the following:

For every homotopy (endpoints fixed) $H:I \times I \to M$, $\int_{I \times I}H^*d\omega=0$.

Yes and it's proved using Stokes theorem.

To check $d\omega =0$, it suffices to check at a fixed point (say, $0$) in a local coordinates $0\in \Omega \subset\mathbb R^n$.

For each $i, j$ fixed and for each $r>0$, let $\gamma_0, \gamma_1$ constitutes the upper and lower semicircle of the circle $C_r$ centered at $0$ in the $i,j$-plane. Then from the condition, for each fixed $r$ we have

$$\int_{C_r} \omega = 0.$$

By Stokes theorem,

$$\int_{B_r} d\omega = 0\Rightarrow \int_{B_r} d\omega_{ij} dx^i dx^j = 0.$$

Divide by $\frac{1}{r^2}$ and take $r\to 0$ gives $d\omega_{ij} (0) = 0$. Do this for each $i, j$ gives $d\omega = 0$ at $0$.