condition for one-form to be exact differential

Question

Simple question I suppose:

Having a (smooth) one-form $$\omega=\omega_i dx^i\in \Lambda(M)$$

Is there a test to find out if $\omega$ is exact differential? Or more precisely that there exists a function $f\in C^{\infty}(M)$, s.t $\omega=df$?

Answer 1

No. A necessary condition is that $d\omega = 0$, that is, the form must be closed, but there is no general method to determine if $\omega$ is exact, unless $M$ is an open ball of $\mathbb{R}^n$, for in that case every closed form is exact, a result known as Poincaré Lemma.

Additional Information

The article in Wikipedia about differential forms can help you understanding the basics about differential forms. A search in this site would be very helpful as well. Now I would limit myself to 1-forms, like the form $\omega$. As I have said in a comment, the exterior derivative of $\omega$ is $$ d\omega = \frac{\partial \omega_i}{\partial x^j}dx^j\wedge dx^i. $$ The wedge product between 1-forms is antisymmetric (for general forms it is skew commutative, see the Wikipedia article), that is, $$ dx^i\wedge dx^j = dx^i\otimes dx^j - dx^j\otimes dx^i. $$ Then the product $dx\wedge dx$ is zero. The skew commutativity property is essential to show that any exact form is a closed form. For example, let us suppose that $\omega = df$ for some function $f$ and, for the sake of simplicity, that we are dealing with $\mathbb{R}^2$ with coordinates $x$ and $y$. Then $$ d\omega = d(df)=d(\frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy)= \frac{\partial^2 f}{\partial x\partial y}dy\wedge dx + \frac{\partial^2 f}{\partial x\partial y}dx\wedge dy, $$ where the terms containing $dx\wedge dx$ and $dy\wedge dy$ were omitted. By the equality of mixed partial derivatives and the fact that $dx\wedge dy = -dy\wedge dx$, you can conclude that $d\omega=d(df)=0$.