Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

Thank you for reading.

share|cite|improve this question
$\mathbb{R}^2\setminus \{p_1,\ldots,p_d\}\simeq \bigvee_{i=1}^d \mathbb{S}^1$. – 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
up vote 3 down vote accepted

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.

share|cite|improve this answer
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).

share|cite|improve this answer
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 $p_d$?

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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