TITLE: Interesting question in differential geometry
QUESTION [9 upvotes]: Let $ \alpha $ be a closed $ 3 $-form on $ \mathbb{R}^{4} \setminus \{ 0 \} $. Let $ i: S^{3} \hookrightarrow \mathbb{R}^{4} $ be the canonical embedding of $ S^{3} $, and suppose that $ \Omega := {i^{\star}}(\alpha) $ is an orientation-form on $ S^{3} $. Prove that $ \alpha $ cannot be continued to a smooth form on $ \mathbb{R}^{4} $.
I am new at differential geometry, and I found this problem. It sounds interesting, but I have no idea how to solve it. Any help would be deeply appreciated.

REPLY [6 votes]: Suppose $\alpha$ does extend to a smooth form; in that case the extension is closed and it represents a class $[\alpha]\in H^3(\mathbb R^4)$ in de Rham cohomology. The map $i$ induces a map $i^*:H^3(\mathbb R^n)\to H^3(S^3)$, and the hypothesis that $\alpha$ restricts to a volume form means that $i^*([\alpha])\neq0$.
This is impossible, since $[\alpha]=0$ because, in fact, $H^3(\mathbb R^4)=0$.