# deRham cohomology of $S^n$ without Mayer-Vietoris

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?

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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 ($k=n$).

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Very nice argument. I think that using more representation theory one can maybe obtain results for other compact homogeneous spaces. Step 4 is the crucial one. For instance, in case of $M=\mathbf{C}P^n=SU_{n+1}/U_n$, the identity $\Lambda^2\mathbf{R}^{2n}=\Lambda^2\mathbf{C}^n+\mathfrak{su}_n+\mathbf R$ as $U_n$-representations shows that $H^2_{dR}(M)$ is one dimensional ($H^1_{dR}(M)=0$ is even easier as $U_n$ does not fix any nonzero vector in $\mathbf{R}^{2n}$). But it is not clear how far one can go with this. – Claudio Gorodski Apr 30 '13 at 0:28
@Claudio: This argument is actually a standard one, just adapted to this special case. I think it may have been originally due to Cartan. The general fact is that if $G$ is a connected, compact Lie group acting on a compact manifold $M$, then the ring of $G$-invariant differential forms on $M$ (which is closed under exterior derivative) has its $d$-cohomology, which is isomorphic to the deRham cohomology of the full ring of differential forms on $M$. If $G$ acts transitively on $M=G/H$, then this invariant subring is finite dimensional and can be computed algebraically. See Spivak for details. – Robert Bryant Apr 30 '13 at 9:01
Right. I remember now for instance the cohomology of some Grassmannians being computed using this method in the chapter about symmetric spaces in J. Wolf's book. Still, standard arguments adapted to special cases can be very enlightening. – Claudio Gorodski Apr 30 '13 at 16:02
Very cool method, and I like the generalizations, too. I eventually got my partitions of unity sorted out, and managed to do this the 'fake Mayer-Vietoris' way I was looking for. – Bryan Clair May 1 '13 at 3:52

Suppose $1 < k < n$.

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 $d\beta = \omega$.

On $U \cap V$, $d(\alpha -\beta) = 0$. Since $H^{k-1}(S^{n-1}) = 0$, there is a $k-2$ form $\tau$ on $U \cap V$ with $d\tau = (\alpha - \beta) |_{U\cap V}$.

Now let $f_U, f_V$ be a partition of unity subordinate to $U,V$. Then on $U\cap V$, $d(f_U\tau) + d(f_V\tau) = \alpha - \beta$.

Since $f_U\tau$ extends (by 0) to a form on $V$, $\beta + d(f_U \tau)$ is defined on $V$. Since $f_V\tau$ extends to a form on $U$, $\alpha - d(f_V \tau)$ is defined on $U$.

Put $\sigma = \beta + d(f_U\tau) = \alpha - d(f_V\tau)$. $\sigma$ is defined on all of $S^n$, and $d\sigma = \omega$, so that $\omega$ is exact.

In the $k=1$ case, $\alpha$ and $\beta$ differ by a constant, so they extend to all of $S^n$ and $\omega = d\alpha$.

When $k=n$, show that any $n$ form with integral 0 is exact, so $H^n(S^n)$ is generated by the volume form. The only change from the $k < n$ case is that one needs to check that the integral of $\alpha-\beta$ is zero over $S^{n-1}$. Apply Stokes' theorem to the lower hemisphere (for $\alpha$) and the upper hemisphere (for $\beta$), so that $\int_{S^{n-1}} \alpha - \beta = \int_{S^n}\omega = 0$.

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In the $k=1$ case, $\alpha$ and $\beta$ differ by a constant. Why does this suffice to extend them to all of $S^n$ for $n>1$? This fails for $n=1$, of course. I guess it’s that $\alpha - \beta$ is only locally constant, and only in the $n>1$ case where $S^{n-1}$ is connected do we get an actually constant difference, which allows gluing to a global function. – Joe Hannon Jun 12 '14 at 1:34
Is your final step for the $k=n$ case necessary? Isn’t it sufficient to observe that the space of top forms is rank 1, and the nowhere vanishing volume form must be nontrivial in cohomology since its integral is nonzero, and therefore it must be a generator of the top degree cohomology? – Joe Hannon Jun 12 '14 at 2:08

For any connected and oriented $n$-manifold $M$, the sequence: $$\Omega_c^{n-1}(M) \to \Omega_c^n(M) \to \mathbb{R} \to 0$$ is exact, where the first map is $d$ and the second map is $\int_M$. A detailed proof can be found in Madsen and Tornhave's "From Calculus to Cohomology", but here's a sketch:

1. Prove the result for $M = \mathbb{R}^n$. Since every top degree form on $\mathbb{R}^n$ is a multiple of the volume form, this comes down to proving that any smooth compactly supported function $f$ on $\mathbb{R}^n$ with total integral $0$ can be written as $\sum_{j=1}^n \frac{\partial f_j}{\partial x_j}$ for some smooth functions $f_j$, and this can be done with a bit of calculus.
2. For any $\omega \in \Omega_c^n(M)$ and any coordinate neighborhood $U \subseteq M$ there is some $\eta \in \Omega_c^{n-1}(M)$ such that $\omega - d\eta$ is supported in $U$. Indeed, just pick any $\omega_0$ compactly supported in $U$ with total integral $1$ and deduce from step $1$ that the form $\omega - (\int_M \omega) \omega_0$ is exact.
3. Let $\omega \in \Omega_c^n(M)$ be a form with total integral $0$ and let $U$ be a coordinate neighborhood in $M$. By step 2 there is a form $\eta \in \Omega_c^{n-1}(M)$ such that $\omega - d\eta$ is supported in $U$, and this form has total integral $0$ by Stokes' theorem. But by step 1 $(\omega - d\eta)|_U = d\xi$ for some $\xi \in \Omega_c^{n-1}(U)$; extending $\xi$ by $0$ to all of $M$, we obtain $\omega = d(\eta + \xi)$.

An immediate corollary is that $H^n(M) \cong \mathbb{R}$ (with the isomorphism given by integration) for any closed oriented $n$-manifold $M$.

The simplest proof I can think of that $H^k(S^n)$ vanishes for $0 < k < n$ uses Poincare duality. Any closed $k$-form $\omega$ cohomologous to a closed $k$-form $\omega_0$ supported in a small neighborhood of its Poincare dual. For $k < n$ the Poincare dual is a proper subset of $S^n$ and thus $\omega_0$ can be viewed as a closed (compactly supported) $k$-form on $\mathbb{R}^n$ via stereographic projection. Thus $\omega_0$ - and hence $\omega$ - is exact since you already calculated the de Rham cohomology of $\mathbb{R}^n$. Of course, this argument only helps if you have covered Poincare duality...

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Oops, my argument for $k < n$ is wrong as written; I'll try to fix it later. – Paul Siegel Apr 30 '13 at 1:24