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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.

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3 Answers 3

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.

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very nice!!!!!!!!!! –  Bombyx mori Dec 22 '12 at 8:06

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, $H^3(\mathbb R^4)=0$.

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Elegant solution! :) –  Haskell Curry Dec 22 '12 at 10:55

Hint: What's the relationship between closed an exact forms on $\mathbb{R}^n$? Can an orientation form on $S^3$ be exact?

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In $\mathbb{R}^n - \{0\}$ there is no relationship in my opinion. Because there are forms which are closed but not exact. Actually $\mathbb{R}^n - \{0\}$ is the standard example abou closed but not exact forms. –  user53969 Dec 22 '12 at 6:14
    
I agree (well, I wouldn't say no relationship because exact always implies closed), but we're talking about extending it to $\mathbb{R}^n$ where there is a strong relationship... –  Jason DeVito Dec 22 '12 at 6:15
    
Thank you! Now it is clear. –  user53969 Dec 22 '12 at 6:28

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