2
$\begingroup$

Can you find mistake in my computation of $H^{k}(S^{n})$.

Sphere is disjoint union of two spaces: $S^{n} = \mathbb{R}^{n}\sqcup\mathbb{R^{0}},$ so $H^{k}(S^n) = H^{k}(\mathbb{R}^{n})\oplus H^{k}(\mathbb{R^{0}}).$ In particular $H^{0}(S^{n}) = \mathbb{R}\oplus\mathbb{R}=\mathbb{R}^{2}$ and $H^{k}(S^{n}) = 0,~~~k>0.$

Where is mistake? Thanks a lot!

  • 0
    @mariano: yeah, yeah, you got me. I was just (wrongly) putting pressure on the fact that $S^1$ is indeed connected (I should've just said that). thanks for pointing it out.2012-03-10

3 Answers 3

5

You are wrong: $S^n$ is not the disjoint union $\mathbb R^n \sqcup \mathbb R^0$ - topologically.

Although $S^n$ is $\mathbb R^n$ with one point at infinity, the topology of this point at infinity is very different from that of $\mathbb R^0$.

  • 0
    for example, the sphere is connected, but the disjoint union is not.2012-03-10
3

This should be a comment but is too long.

If your reasoning were correct, we could also do the following: write $S^1$ as the "disjoint union" of two open intervals and two points (by cutting out the north and south poles, for example) Then your idea would show that $H^0(S^1)=\mathbb R^4$. And you can cut it in more pieces...

  • 0
    Indeed! Thanks for the catch :)2012-03-10
0

de Rham cohomology is not additive, unless you have a partition of the manifold $M$ by two open subsets $U$ and $V$, then the Mayer-Vietoris exact sequence implies that $H^0(M)=H^0(U)\oplus H^0(V)$. Here a point $\{N\}$ is not open in $S^n$. However, the Euler characteristic is additive and $\chi(S^n)=\chi(\mathbb{R}^n)+\chi(\{N\})=(-1)^n+1.$