Is it true that if a differential form is not exact then it’s integral is non zero?

Question

Suppose $M$ is a smooth manifold with or without boundary, $S\subset M$ is an oriented compact $k$- dimensional sub manifold without boundary, and $\omega$ is a closed $k$-form on $M$. By Stoke’s theorem we know that if $\int_{S}\omega \neq 0$, then $\omega$ is not exact. What about the converse? Is it true that if $\omega$ is not exact then $\int_{S}\omega\neq0$? I have no idea how to approach this question. Any help is appreciated.

Answer 1

The question relates to the properties of exact differential forms and their integrals over manifolds. Given a smooth manifold $M$ and a closed $k$-form $\omega$ on $M$, by Stokes' theorem, if $$\int_S \omega \neq 0,$$ then $\omega$ is not exact.

The converse does not necessarily hold. The non-exactness of a differential form does not ensure that its integral over a submanifold $S$ is non-zero. A non-exact form may still yield a zero integral over certain submanifolds. This could occur, for example, if the form is identically zero on $S$ or if the form's values and orientations result in an integral that cancels out over $S$.

An illustrative example is provided by the closed 1-form on $\mathbb{R}^2 \setminus \{0\}$ defined by $$ \omega = \frac{-y\,dx + x\,dy}{x^2 + y^2}, $$ which represents the angular form. This form is not exact, as demonstrated by its integral over a loop encircling the origin being $2\pi$ (a result consistent with the complex analysis principles of argument and residue). However, the integral of this form over a path that does not enclose the origin can be zero, indicating that a form's non-exactness does not imply a non-zero integral universally.

Answer 2

Here is a rather general counterexample.

Consider any $M$, $S$ as in your question and any closed but not exact differential form $\omega$.

The idea is to add a "mirrored" version of $M$ and $S$ such that the two integrals (the original and the mirrored one) cancel out.

Indeed, take two copies of $M$ - $M_1$ and $M_2$. Let $S_1$ and $S_2$ be copies of $S$ in $M_1$ and $M_2$, respectively. Let $M' := M_1 \sqcup M_2$, $S' := S_1 - S_2$ (meaning that $S_2$ is oriented differently from $S_1$). Let $\omega'$ be defined on $M_1$ and $M_2$ as $\omega$. Then we clearly have $\int_{S'} \omega' = \int_S\omega - \int_S\omega = 0$ while $\omega'$ is not exact.