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Am I right that de Rham cohomology $H^k(S^2\setminus \{k~\text{points}\})$ of $2-$dimensional sphere without $k$ points are $H^0 = \mathbb{R}$ $H^2 = \mathbb{R}^{N}$ $H^1 = \mathbb{R}^{N+k-1}?$

I received this using Mayer–Vietoris sequence. And I want only to verify my result.

If you know some elementery methods to compute cohomology of this manifold, I am grateful to you.

Calculation:

Let's $M = S^2$, $U_1$ - set consists of $k$ $2-$dimensional disks without boundary and $U_2 = S^2\setminus \{k~\text{points}\}$. $M = U_1 \cup U_2$ each punctured point of $U_2$ covered by disk (which contain in $U_1$). And $U_1\cap U_2$ is a set consists of $k$ punctured disks (which homotopic to $S^1$). Than collection of dimensions in Mayer–Vietoris sequence $0\to H^0(M)\to\ldots\to H^2(U_1 \cap U_2)\to 0$ is $0~~~~~1~~~~~k+\alpha~~~~~k~~~~~0~~~~~\beta~~~~~k~~~~~1~~~~~\gamma~~~~~0~~~~~0$ whrer $\alpha, \beta, \gamma$ are dimensions of $0-$th, $1-$th and $2-$th cohomolody respectively. $1 - (k+\alpha) + k = 0,$ so $\alpha = 1.$ $\beta - k + 1 - \gamma = 0,$ so $\beta = \gamma + (k-1).$ So $H^0 = \mathbb{R}$ $H^2 = \mathbb{R}^{N}$ $H^1 = \mathbb{R}^{N+k-1}$

Thanks a lot!

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    @Aspirin: also, you probably need to know that $N$ is *finite* for that to work, no?2012-03-12

2 Answers 2

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Your result isn't correct.

I won't tell you the result so that you can compute it yourself, though :) There are very few things more rewarding than getting these things right oneself!

(By the way: do it for $k=1$, $2$, and $3$... you'll catch the pattern soon)

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    @Aspirin: Sure they can.2012-03-11
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It helps to use the fact that DeRahm cohomology is a homotopy invariant, meaning we can reduce the problem to a simpler space with the same homotopy type. I think the method you are trying will work if you can straighten out the details, but if you're still having trouble then try this:

$S^2$ with $1$ point removed is homeomorphic to the disk $D^2$. If we let $S_k$ denote $S^2$ with $k$ points removed, then $S_k$ is homeomorphic to $D_{k-1}$ (where $D_{k-1}$ denotes $D^2$ with $k-1$ interior points removed).

Hint: Find a nicer space which is homotopy equivalent to $D_{k-1}$. Can you, for instance, make it $1$-dimensional? If you could, that would immediately tell you something about $H^2(S_k)$. If you get that far and know Mayer-Vietoris, you should be able to work out the calculation.