Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

We know that a 1-form $\omega$ on a manifold $M$ is exact if and only if $\int_{\gamma}\omega=0$ for any closed loop $\gamma$. How can I prove the following generalization: $\omega$ is an exact n-form on $S^n$ if and only if $\int_{S^n}\omega=0$? One direction follows clearly by Stokes, but I am not sure how to generalize the first fact to prove the remaining direction. Thanks!

share|cite|improve this question
Maybe… can be of any help. – M.B. Jun 3 '11 at 17:50
Do you know how to calculate the de Rham cohomology of $S^n$? – Dactyl Jun 3 '11 at 18:08
up vote 7 down vote accepted

You need to prove somehow that $H^n_{deRham}(S^n)\cong\mathbb{R}$ (and that the isomorphism is given by the integral over $S^n$). One possibility is induction and Mayer-Vietoris sequence. Here is another, somewhat more geometrical way. If $g:S^n\to S^n$ is a rotation and $\beta\in \Omega^n(S^n)$ then $g^*\beta-\beta$ is exact (since if $f_1$ is homotopic to $f_2$ then $f_1^*=f_2^*$ on cohomology). When we average over $SO(n)$, we can see that any $n$-form on $S^n$ is cohomologous to a $SO(n)$-invariant $n$-form. Up to multiple there is only one such form - the volume form $\omega$. Any $n$-form is thus of the form $d\alpha+c\omega$, and your claim follows.

(this argument shows that to find de Rham cohomology of a homogeneous space of a connected compact Lie group, we can restrict ourselves to the sub-complex of invariant forms. If the space is symmetric then all invariant forms are closed, i.e. the cohomology is equal to the space of invariant forms.)

share|cite|improve this answer
How can you conclude from induction and the Mayer-Vietoris that the map that gives the isomorphism is actually integration over $S^n$? – Manuel Jun 3 '11 at 22:39
Ah, ok. From Mayer Vietoris we get $H^n(S^n)=\mathbb{R}$. Since $S^n$ is orientable there is a non vanishing form $\omega$, so $[\omega]$ spans $H^n(S^n)$. In particular, we can take $\omega$ such that its integral over $S^n$ is 1, thus integration over $S^n$ gives us a surjective map $H^n(S^n) \rightarrow \mathbb{R}$, so by dimensional considerations this map must be injective. – Manuel Jun 3 '11 at 23:19

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.