Showing de Rham cohomology $H^1(S^n)$ is zero

Question

I'm trying to find an elementary way to see that the 1st de Rham cohomology of the n-sphere is zero for $n>1$, $H^1(S^n) = 0$.

This is part of an attempt to find the de Rham cohomology of the n sphere generally. I have shown that $H^k(S^n) = H^{k-1}(S^{n-1})$ (for $k>1$) but to find the de Rham cohomology generally I need to show that $H^1(S^n) = 0$ and I am struggling to do this.

Is it possible to take a closed one form $\alpha$ on a sphere and explicitly find a function $f$ on $S^n$ that has $df = \alpha$, I've tried writing it out in coordinates and explicitly integrating but I haven't managed to get very far with this!

Answer 1

I think the function you are looking for is $f(x)=\int_{\gamma_x} \alpha$ where $\gamma_x$ is a path from some chosen base point $x_0$ to $x$. But then for this to be well defined you have to say that all paths between $x$ and $x_0$ are homotopic, i.e. that your sphere is simply connected, so the whole argument gets a bit circular...

P.S. Traying to write things out in coordinates will not get you far because a closed form not being exact is a "global" thing, not a local one

Answer 2

You can imitate the argument that shows $\pi_1(S^n)=0$ if $n>1$ by removing a point! Try it!

Answer 3

Write $S^n$ as the union of two open sets $U= S^n-N$, $V=S^n-S $where $N,S$ are the north and south pole. Note that the intersection $U\cap V$ is connected if $n\geq 2$. $U,V$ are contracile (diffeomorphic to $R^,$, and on each of these sets the form $\alpha$ is exact and andmits primitive $f_U, f_V$. On the intersection $d(f_U-f_V)=0$ and by connexity $f_U-f_V$ must be a constant $c$. So $f_V=f_U+c$ one the intersection. As a primitive of $\alpha$ one can take $f_U$ on $U$ and $f_U +c$ on $V$. The main point is that the intersection is connected.