# 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.

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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