Interesting question in differential geometry 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.
 A: Suppose that $ \alpha $ can be extended to a $ 3 $-form $ \tilde{\alpha} $ on $ \mathbb{R}^{4} $. Then by continuity, $ \tilde{\alpha} $ is a closed $ 3 $-form on $ \mathbb{R}^{4} $. As closed forms on $ \mathbb{R}^{4} $ are exact (apply the Poincaré Lemma to $ \mathbb{R}^{4} $, which is a contractible space), we have that $ \tilde{\alpha} = d(\beta) $ for some $ 2 $-form on $ \mathbb{R}^{4} $. As $ \Omega = {i^{\star}}(\alpha) $ is required to be an orientation-form on $ \mathbb{S}^{3} $, integrating it on $ \mathbb{S}^{3} $ should yield a non-zero result. Hence,
\begin{align}
0 &\neq \int_{\mathbb{S}^{3}} \Omega \\
  &=    \int_{\mathbb{S}^{3}} {i^{\star}}(\alpha) \\
  &=    \int_{\mathbb{S}^{3}} {i^{\star}}(\tilde{\alpha}) \\
  &=    \int_{\mathbb{S}^{3}} {i^{\star}}(d(\beta)) \\
  &=    \int_{\mathbb{S}^{3}} d({i^{\star}}(\beta)) \quad (\text{Pullback commutes with exterior derivative.}) \\
  &=    \int_{\partial \mathbb{S}^{3}} {i^{\star}}(\beta) \quad (\text{By Stokes' Theorem.}) \\
  &=    \int_{\varnothing} {i^{\star}}(\beta) \quad (\text{$ \mathbb{S}^{3} $ has no boundary.}) \\
  &=    0,
\end{align}
which is an outright contradiction.
A: 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, $H^3(\mathbb R^4)=0$.
A: Hint:  What's the relationship between closed an exact forms on $\mathbb{R}^n$?  Can an orientation form on $S^3$ be exact?
