0
$\begingroup$

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) $H:I \times I \to M$, $\int_{I \times I}H^*d\omega=0$.

$\endgroup$
5
$\begingroup$

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 $0$.

$\endgroup$

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.