[Math] deRham cohomology of $S^n$ without Mayer-Vietoris

at.algebraic-topologydg.differential-geometry

I've got a few days left at the end of a differential geometry class, and would like to compute the deRham cohomology of $S^n$. We've just proved the Poincare lemma, so I know the cohomology of $R^n$, and homotopy invariance of cohomology is an easy consequence.

The way to compute $H^*(S^n)$ is with Mayer-Vietoris, for example as in Bott and Tu. But I don't have time to fully develop the cohomological algebra for that, just to then apply it in only one case. I'd rather get the computation for $S^n$ directly and use it for a couple of applications.

So, it seems to me that one could trace through the Mayer-Vietoris argument in this specific case. At least I'd like to show that a closed $k$-form on $S^n$ is exact when $0 < k < n$.

Here's how I think the argument could go:

Let $\omega$ be a $k$-form on $S^n$.
Let $U$ and $V$ be $S^n$ minus its north and south pole, respectively. Then homotopy invariance and the Poincare lemma give $k-1$ forms $\alpha$ on $U$ and $\beta$ on $V$ with $d\alpha = \omega$ and $d\beta = \omega$.

Now use a partition of unity $f$ and set $\gamma = f_U \alpha + f_V \beta$, a $k-1$ form on $S^n$. So $\omega – d\gamma$ is now supported on $U \cap V$. Since $U\cap V$ is homotopic to $S^{n-1}$, induction gives that $\omega – d\gamma$ is exact, so $\omega = d\tau + d\gamma$.

The problem is that $\omega – d\gamma$ is exact when restricted to $U \cap V$, and I don't see how why $\tau$ would extend to a form on $S^n$.

Am I missing something? Is there a better approach entirely?

Best Answer

Have you done any integration theory? (I assume you have, otherwise you wouldn't necessarily know what the deRham cohomology does for you.) The fastest proof I know is:

  1. Take a closed $k$-form $\omega$ on $S^n$, note that $g^\ast\omega$ is cohomologous to $\omega$ for all $g\in \mathrm{SO}(n{+}1)$ (since $\mathrm{SO}(n{+}1)$ is connected.

  2. Conclude that $\omega$ is cohomologous to $\bar\omega$, the average over $\mathrm{SO}(n{+}1)$ of $g^\ast\omega$ as $g$ varies over $\mathrm{SO}(n{+}1)$.

  3. But $\bar\omega$ is invariant under the action of $\mathrm{SO}(n{+}1)$, so its value at $x\in S^n$ must be invariant under the subgroup (isomorphic to $\mathrm{SO}(n)$) that stabilizes $x$.

  4. However, $\mathrm{SO}(n)$ acting on $\mathbb{R}^n$ only fixes nonzero forms in degree $0$ and $n$.

  5. Thus, if $\bar\omega$ is not zero, it must be either a constant function ($k=0$) or a multiple of the volume form ($k=n$).

Related Question