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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.


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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Mariano is right, your result is not correct. Could you outline your calculation, so we can help find where you went wrong? – you Mar 11 '12 at 4:11
@you: it may take some time – Aspirin Mar 11 '12 at 4:42
@you: I have done – Aspirin Mar 11 '12 at 6:55
What is $N{}{}$? (Your result is correct for exactly one value of $N$ :) ) – Mariano Suárez-Alvarez Mar 11 '12 at 8:20
@MarianoSuárez-Alvarez: I need $H^{*}(S^{n}\setminus\{k~\text{points}\})$ to calculate the cohomology of sphere with $k$ handles. It is enough to satisfy the equality $\dim H^{1}(S^{n}\setminus\{k~\text{points}\}) - \dim H^{2}(S^{n}\setminus\{k~\text{points}\}) = k-1$ – Aspirin Mar 11 '12 at 18:32

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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is it true, that $U_1$ and $U_2$ from Mayer–Vietoris sequence may be not connecting? – Aspirin Mar 11 '12 at 4:47
What do you mean by not connecting? – Mariano Suárez-Alvarez Mar 11 '12 at 6:16
I corrected my question, can you verify? – Aspirin Mar 11 '12 at 6:46
Mariano, I mean not connected – Aspirin Mar 11 '12 at 7:00
@Aspirin: Sure they can. – Mariano Suárez-Alvarez Mar 11 '12 at 9:05

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.

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