I am trying to compute the de Rham cohomology of $S^2$ using Mayer Vietoris sequence. I considered the open cover $U$, where $U$ is the whole $S^2$ minus the north pole, and $V$, where $V$ is the whole $S^2$ minus the South Pole. Then, these two spaces are homotopic to a point. Even more, the intersection $U\cap V \cong S^1 \times [0,1] \cong S^1$. Thus, with this information we can start considering the following long exact sequence.

$$ 0 \xrightarrow{p} H^0(S^2) \xrightarrow{\alpha_0} H^0(U) \oplus H^0(V)\cong \mathbb{R} \oplus \mathbb{R} \xrightarrow{\beta_0} H^0(U\cap V)\cong \mathbb{R} \xrightarrow{d_0} H^1(S^2) \xrightarrow{\alpha_1} H^1(U) \oplus H^1(V) \cong 0 \xrightarrow{\beta_1} H^1(U \cap V) \cong \mathbb{R} \xrightarrow{d_1} H^2(S^2) \xrightarrow{\alpha_2} 0 \xrightarrow{\beta_2}0$$

I am struggling to compute $H^k(S^2)$. For instance, to compute $H^0(S^2)$, I was thinking of using the exactness of the sequence. We know that $0 = \text{im } p = \ker \alpha_0$ and so by first isomorphism theorem, we know that $H^0(S^2) = H^0(S^2)/\ker(\alpha_0) \cong \text{im } \alpha_0 = _{\text{(by exactness)}} \ker \beta_0 $. But then, I know how to figure out the kernel of $\beta_0$. I know the answer for the $0$-th de Rham cohomology is $H^0(S^2) \cong \mathbb{R}$, but I really don't know if this approach is correct and what I am missing to reach my desire conclusion.

Thanks for your help!


Since $$ 0 \to H^1(U\cap V)\cong \mathbb{R} \xrightarrow{d_1}H^2(S)\to 0$$ is exact then $d_1$ is an isomorphism, so $H^2(S)\cong \mathbb{R}.$

Now note that if $M$ is any manifold, then $$ H^0(M)=\{f\in C^\infty(M) \text{ | } df=0\}=\{f\in C^{\infty}(M) \text{ | }f \text{ is locally constant}\},$$ so if $M$ has $n$ connected components, then $H^0(M)\cong \mathbb{R}^n.$ Since $S^2$ is connected, this shows that $H^0(S^2)=\mathbb{R}.$

Finally, look at the exact sequence $$ 0\to H^0(S^2)\cong \mathbb{R} \to H^0(U)\oplus H^0(V)\cong \mathbb{R}\oplus \mathbb{R}\to H^0(U\cap V) \cong \mathbb{R}\to H^1(S^2) \to 0$$ and use the following:

Proposition. If $0\to A_1\to A_2\to \cdots \to A_n \to 0$ is an exact sequence of vector spaces then $$ \sum_{i=1}^n(-1)^i\dim A_i=0.$$

Then you find that $1-2+1-\dim H^1(S^2)=0,$ so $H^1(S^2)=0.$

  • $\begingroup$ Awesome! Thanks! I completely missed the fact that $d_1$ was an isomorphism! $\endgroup$ – BOlivianoperuano84 Dec 12 '18 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.