Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer
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, $H^3(\mathbb R^4)=0$.

share|cite|improve this answer
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?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.