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This question is motivated by Exercise 1.7 from Differential Forms in Algebraic Topology by Bott & Tu. The original question in the text concerns the de Rham cohomology of $\mathbb{R}^2$ with points $P$ and $Q$ deleted. When I computed $H^1$, I constructed two generators (specified by the integrals along contours around $P$ and $Q$), and used the fact that the space of one forms is two-dimensional to show that the cohomology group is $\mathbb{R}^2$.

If I let $X_k = \mathbb{R}^2 \setminus \{k~\text{points}\}$, then based on the computation, I feel tempting to conjecture that $H^1(X_k) = \mathbb{R}^k$, which coincides with my intuition that $X_k$ is homotopically a wedge sum of $k$ circles. On the other hand, the same argument of mine shows $H^1(X_k) = \mathbb{R}^2$ for all $k \ge 2$, but I have a hard time to understand/visualize this result.

If there's an error, could you point it out? If my computation is correct, could you offer your insight on the result? Thank you!

Ruian Chen
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  • Why do you say "the same argument of mine shows $H^1(X_k) = \mathbb{R}^2$"? You construct two generator if $k=2.$ But for $k>2$ you have $k>2$ generators. So, if you made the same argument you get $H^1(X_k) = \mathbb{R}^k$. – mfl Nov 29 '14 at 11:57
  • @mfl Still the space of one forms is only two-dimensional, so how can $H^1$ have dimension $k$? – Ruian Chen Nov 29 '14 at 12:00
  • @RuianChen Why do you think that? – Alex Youcis Nov 29 '14 at 12:07
  • http://math.stackexchange.com/questions/254123/de-rham-cohomology-of-the-plane-with-m-holes – mfl Nov 29 '14 at 12:16
  • @mfl since this exercise appears in the text before introducing M-V sequence, I prefer an explanation without M-V. – Ruian Chen Nov 29 '14 at 12:22
  • @AlexYoucis Why do I think $H^1$ has dimension at most 2? I think $X_k$ is a two-dimensional manifold, so the space of 1-forms is two-dimensional? $H^1$ consists of the closed 1-form modulo exact 1-forms, so its dimension is at most 2? – Ruian Chen Nov 29 '14 at 12:33

1 Answers1

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One way to think about this is de Rham cohomology gives you the free part of simplicial cohomology. For the case of $\mathbb{R}^{2}-\{p_1,\cdots, p_{n}\}$, you can show that the space is homotpically equivalent to $\bigvee_{i=1}^{n}\mathbb{S}^{1}_{i}$. Thus its first homology group should be $\mathbb{Z}^{n}$ and its de Rham cohomology group should be $\mathbb{R}^{n}$.

To visualize it, you might consider the case $\mathbb{R}^{2}-\{*\}$ first. In this case the de Rham cohomology is represented by a ray $(x,0):x>0$, whose Poincare dual is the angular form $\frac{1}{2\pi}d\theta$. If you have to remove $n$ points, then you are actually working with $n$ angular forms on the punctured plane. Each one of them is dual to a corresponding ray. Thus the picture is kind of "fuzzy".

Bombyx mori
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