# De Rham cohomology of the plane with $m$ holes

I'm trying to compute the De Rham cohomology of the manifold $$M_d=\mathbb{R}^2\setminus\{ p_1\ldots p_d\},$$ where the points $p_1\ldots p_d$ are all distinguished.

This must be a standard exercise in the use of the Mayer-Vietoris sequence, but I run into a difficulty. Namely, taking a suitable open covering, such as $$M_d=U\cup V,$$ where (up to homeomorphism) $U=M_{d-1}$ and $V=M_1$, and then observing that $M_1$ is homotopically equivalent to the $1$-manifold $\mathbb{S}^1$, I can obtain from the Mayer-Vietoris sequence the following piece of information: $$h^1(M_d)-h^2(M_d)=h^1(M_{d-1})-h^2(M_{d-1})+1,$$ (where $h^i(\ldots)=\dim H^i(\ldots)$).

What is causing trouble is the presence of those $h^2$-terms. I guess that they should vanish, since $h^2(M_1)$ does because of the said homotopical equivalence with $\mathbb{S}^1$. But,

is it true that the plane with $m$ holes is homotopically equivalent to a manifold of dimension $1$?

Intuitively I would say that this is true: $M_2$ is homotopically equivalent to an $8$, $M_3$ to a curve that does three loops and so on. The problem is that those are not manifolds: the figure $8$, for example, has a singularity in its centre. This leaves me puzzled.

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$\mathbb{R}^2\setminus \{p_1,\ldots,p_d\}\simeq \bigvee_{i=1}^d \mathbb{S}^1$. – Sigur Dec 9 '12 at 0:05
@Sigur That is true, but it does not answer his question. – Potato Dec 9 '12 at 0:08
$h^2(M_1) = 0$ therefore $h^2(M_2) = 0$ and so on (since we have the sequence $0 = H^2(M_d)\oplus H^2(M_1) \rightarrow H^2(M_{d+1}) \rightarrow H^3(\mathbb{R}^2)=0$ which is exact in the middle term). – AnonymousCoward Dec 9 '12 at 0:11
@AnonymousCoward: Thank you, this is convincing! – Giuseppe Negro Dec 9 '12 at 0:35

Lets pursue this induction: Suppose that in what follows, at each step, $M_{d-1}\cap M_1$ is $\mathbb{R}^2$ less $d$ distinct points. You have a short exact sequence:

$$0 \rightarrow \Omega^*(M_1\cup M_{d-1}) \rightarrow \Omega^*(M_1)\oplus\Omega^*(M_{d-1}) \rightarrow \Omega^*(M_d) \rightarrow 0$$

When $d = 2$, we have the induced long exact sequence

$$0 \rightarrow H^0(\mathbb{R}^2)=\mathbb{R} \rightarrow H^0(M_1)\oplus H^0(M_1) = \mathbb{R}^2 \rightarrow H^0(M_2)$$

$$\rightarrow H^1(\mathbb{R}^2)=0 \rightarrow H^1(M_1)\oplus H^1(M_1) = \mathbb{R}^2 \rightarrow H^1(M_2)$$

$$\rightarrow H^2(\mathbb{R}^2)=0\rightarrow H^2(M_1)\oplus H^2(M_1) = 0 \rightarrow H^2(M_2)$$

$$\rightarrow 0 \cdots$$

Where I have used the poincare lemma in the left column, and that $S^1\sim M_1$ in the middle column. This will show that $h^2(M_d) = 0$ by inducting.

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Your answer is basically the same as mine! +1 Though! – user38268 Dec 9 '12 at 0:59
sorry, i posted about 5 seconds after you and then was unsure what to do about it. – AnonymousCoward Dec 9 '12 at 1:02
This is a good answer, concise and to-the-point. Thank you! – Giuseppe Negro Dec 9 '12 at 4:15

To answer your only question: No. A 1-manifold is either a (open or closed) line segment or a circle or a disjoint union of such. But you can fatten the figure-$8$ (for example) out to get an honest manifold homotopy-equivalent to what you want (plane minus two points).

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But this does not imply that $h^2(M_2)=0$, it seems to me. Indeed, you can still have nontrivial 2-forms on that fat-figure 8. Am I wrong? – Giuseppe Negro Dec 9 '12 at 0:31

Here's a proof using induction that $H^2(M_d) = 0$ for all $d \geq 1$. Now for $d = 1$ this is clear. Then by Mayer Vietoris we have

$$\ldots \leftarrow H^3\left(M_{d-1} \cup( \Bbb{R}^2 -\{p_d\};\Bbb{R}\right)\leftarrow H^2\left(M_{d-1}\cap ( \Bbb{R}^2 -\{p_d\});\Bbb{R}\right) \leftarrow \\ H^2(M_{d-1};\Bbb{R}) \oplus H^2(\Bbb{R}^2 - \{p_d\};\Bbb{R}) \leftarrow \ldots$$

Now the $H^3$ term is zero simply because the the space in question is just $\Bbb{R}^2$. By induction $H^2(M_{d-1})$ is zero and similarly $H^2(\Bbb{R}^2 - \{p_d\})$ is zero by the basis step. It will now follow that because

$$M_{d-1} \cap \Bbb{R}^2 -\{p_d\} = M_d$$

that $H^2(M_d) = 0$ for all $d$.

Computation of $H^1(M_d)$ for all $d$:

You can do this via induction. For $d = 1$ we have $\Bbb{R}^2 - \{p_1\}$ being homeomorphic (IIRC via some exponential function) to the infinite cylinder $\Bbb{S}^1 \times \Bbb{R}$. Now $\pi_1(S^1 \times \Bbb{R}) = \Bbb{Z}$ and so the Hurewicz Theorem gives that $H_1(S^1 \times \Bbb{R};\Bbb{Z} ) = \Bbb{Z}$. By universal coefficients we get that

$$H^1(S^1 \times \Bbb{R};\Bbb{R}) \cong \textrm{Hom}\left(H_1(S^1 \times \Bbb{R};\Bbb{Z} ),\Bbb{R}\right) \cong \Bbb{R}.$$

Now suppose for the inductive step it has been calculated that the first cohomology of $M_{d-1}$ is $\Bbb{R}^{d-1}$. What I suggest you to do now is suppose you have $d$ points in the plane. Now there is at least one coordinate (wlog say that $x$ - coordinate) such that not all the points have the same $x$ - coordinate. Choose a point with largest $x$ - coordinate. This point of course does not have to be unique.

Can you now find a line that divides the plane into two regions $U$ and $V$ with $U$ containing the $d-1$ points $p_1,\ldots,p_{d-1}$ and $V$ containing just $p_d$?

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