Just a short question.
The well-known Poincaré's Lemma states that every closed k-form, which is defined on a contractible set, is also exact.
In the lecture we have learned the concept of simply connected sets, which are more general sorts of spaces. (any contractible set is also simply connected but not vice versa). Furthermore it was shown that every closed 1-form on a simply connected set is also exact…
Now to my question: Is this statement only true for 1-forms, or is there a generalisation, like "Every closed k-form on a simply connected set is also exact"…
Best Answer
No, the correct generalization involves a "higher-dimensional analogue" of simple connectivity. A closed $2$-form will be exact if the region has no "two-dimensional holes." The correct notion is homology or cohomology. If the $k$th cohomology $H^k(X,\Bbb R)$ (which can be computed topologically or in terms of differential forms) vanishes, then every closed $k$-form is exact. Indeed, if the $1$st cohomology vanishes, then every closed $1$-form is exact; this condition is in fact weaker than simple connectivity. (Rather than needing two paths to be homotopic, it is enough for them to be homologous.)
To answer your question specifically, $\Bbb R^3-\{0\}$ is simply connected, but the $2$-form $$\omega = \frac{x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy}{(x^2+y^2+z^2)^{3/2}}$$ is closed but not exact. (Its integral over any sphere centered at the origin, for example, is $4\pi$.)