# exact differential n-forms

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!

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Maybe math.stackexchange.com/questions/12268/… 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

## 1 Answer

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.)

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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