Q:

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! 

A:

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.)