I'm trying to figure out some counterexamples of the following: If $F$ is an irrotational vector field then there exists some $U$ such that $F = \text{grad}(U)$ (or more generally, using the language of forms, I'm trying to find a closed form which is not exact).
As @peek-a-boo suggested I Googled a bit and found out the following example. In $\mathbb{R}^2 \setminus \{0\}$ the vector field $$F(x,y) := \left( -\frac{y}{x^2 + y^2},\ \frac{x}{x^2 + y^2} \right)$$ is closed (it has zero rotational) but not exact (it is not the gradient of any other function).
I would really appreciate if
- Someone could give me more examples of such vector fields in $\mathbb{R}^2 \setminus \{0\}$.
- More generally, someone could provide me with some examples of $k$-forms $\omega$ on n-dimensional manifolds $M$ such that its de Rham cohomology class $[\omega]$ is nonzero.
EDITED: I corrected the observations made in the first four comments, and also @Ted Shifrin's comment.