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I am trying to calculate the cohomology of $X = \mathbb R^2 \setminus \lbrace \mathbb Z \times \mathbb Z \rbrace = \lbrace (x,y) \in \mathbb R^2 : x \text{ and } y \not \in \mathbb Z \rbrace.$

$H^0(X) = \mathbb R $ since $X$ is connected and $H^k(X) = 0$ for $k > 2$ since $\dim X = 2$.

To determine $H^1(X)$ and $H^2(X)$ take $ U_0 = \mathbb R^2 \setminus (\mathbb R \times \mathbb Z) $ and $U_1 = \mathbb R^2 \setminus (\mathbb Z \times \mathbb R)$. $U_0$, $U_1$

Then $U_0 \stackrel{\text{homotopic}}{\simeq} \mathbb Z \simeq U_1, U_0 \cap U_1 \simeq \mathbb Z^2$ and therefore from the Mayer-Vietoris sequence we have $$ H^1(U_0 \cap U_1) = 0 \rightarrow H^2(X) \rightarrow H^2(U_0) \oplus H^2(U_1) = 0, $$ so $H^2(X) = 0$; \begin{matrix} 0 & \rightarrow & H^0(X) & \rightarrow & H^0(U_0) \oplus H^0(U_1) & \rightarrow & H^0(U_0 \cap U_1) & \rightarrow & H^1(X) & \rightarrow & 0 \\ & & \cong & & \cong & & \cong \\ 0 & \rightarrow & \mathbb R & \rightarrow & \mathbb R ^{\omega} \oplus \mathbb R ^{\omega} & \rightarrow & (\mathbb R ^2) ^{\omega} & \rightarrow & H^1(X) & \rightarrow & 0 \end{matrix} (where $\mathbb R ^{\omega} = \prod_{n \in \mathbb N} \mathbb R).$

What can I say about $H^1(X)$ from this?

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  • $\begingroup$ Actually, $\mathbb R^2 \setminus \mathbb Z^2=\mathbb T^2 \not=\{(x,y)\in\mathbb R^2|(x,y)\not\in \mathbb Z^2\}$ $\endgroup$ – gaoxinge Feb 3 '14 at 14:20
  • $\begingroup$ @gaoxinge I have usually seen $\mathbb{R}^{n}/\mathbb{Z}^{n}$ or $\mathbb{Z}^{n}\backslash\mathbb{R}^{n}$ for tori. $\endgroup$ – M. Luethi Feb 3 '14 at 14:25
  • $\begingroup$ @M.Luethi So does the $\{(x,y)\in\mathbb R^2|(x,y)\not\in\mathbb Z^2\}$ have a differential structure? $\endgroup$ – gaoxinge Feb 3 '14 at 14:27
  • $\begingroup$ @gaoxinge Why should it not? It looks like an open subset of a manifold to me. $\endgroup$ – M. Luethi Feb 3 '14 at 14:29
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    $\begingroup$ Dear User, The maps in your exact sequence are pretty explicit, given your explicit descritions of $U_0$ and $U_1$. Why don't you just try to write them down directly, and then just compute $H^1$ as the indicated cokernel? Regards, $\endgroup$ – Matt E Feb 3 '14 at 15:16
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I would suggest using the Thom-Gysin sequence associated to this situation. This comes from the long-exact sequence in cohomology for the pair $(\mathbb{R}^2,\mathbb{R}^2\setminus\mathbb{Z}^2)$. We replace $H^i(\mathbb{R}^2,\mathbb{R}^2\setminus\mathbb{Z}^2)$ with $H^{i-2}(\mathbb{Z}^2)$ via the Thom isomorphism. So, we obtain a long-exact sequence $$\ldots\rightarrow H^{i-2}(\mathbb{Z}^2)\rightarrow H^i(\mathbb{R}^2)\rightarrow H^i(\mathbb{R}^2\setminus\mathbb{Z}^2)\rightarrow\ldots.$$ Setting $i=1$, we actually obtain $$0\rightarrow H^1(\mathbb{R}^2\setminus\mathbb{Z}^2)\rightarrow H^0(\mathbb{Z}^2)\rightarrow 0.$$ Hence, $$H^1(\mathbb{R}^2\setminus\mathbb{Z}^2)\cong H^0(\mathbb{Z}^2).$$ However, $\mathbb{Z}^2$ is a discrete space, so $H^0(\mathbb{Z}^2)\cong\mathbb{R}^{\mathbb{Z}^2}$.

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  • $\begingroup$ What sphere bundle are you using the Gysin sequence on? $\endgroup$ – Lepanais Feb 6 '14 at 21:26
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Isn't it clear considering the fundamental group?

Every isolated point you delete from $\mathbb{R}^2$ gives you a generator of the fundamental group of your space, which will be the free group with $\mathbb{Z}^2$ generators. Then taking the abelianizated of it, and you find $\mathbb{Z}^{\mathbb{Z}^2}$, take the tensor product with $\mathbb{R}$, and you got it without calculation :)

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You can consider $$\begin{align*} U &= \Bbb{R}^2 - \Bbb{Z}^2 \\ V &= \coprod_{(n,m) \in \Bbb{Z}^2} \text{B}((n,m), 1/3)\end{align*}$$, i.e the disjoint union of balls centered at each point with integer coordinates. Aftewards, it is easy to note that $$\begin{align*} U \cup V &= \Bbb{R}^2 \\ U \cap V &= \coprod_{(n,m) \in \Bbb{Z}^2} \text{B}((n,m), 1/3) \setminus \{(n,m) \} \end{align*}$$

It is straightforward that $H^1(V) \cong 0$ and $H^1(\text{B}((n,m), 1/3) \setminus \{(n,m) \}) \cong H^1(\Bbb{S}^1) \cong \Bbb{R}$. Therefore, $H^1(U \cap V) \cong R^{\Bbb{Z}^2}$. Using Mayer-Vietoris, we obtain the following exact sequence

$$\ldots\rightarrow H^{1}(\mathbb{R}^2)\rightarrow H^1(U) \oplus H^1(V)\rightarrow H^1(U \cap V)\rightarrow H^2(\Bbb{R}^2) \rightarrow\ldots. $$

Since $H^1(\Bbb{R}^2) \cong H^2(\Bbb{R}^2) \cong 0$ and $H^1(V) \cong 0$, we conclude

$$ H^1(\Bbb{R}^2 - \Bbb{Z}^2)\cong H^1(U) \oplus H^1(V) \cong H^1(U \cap V) \cong \Bbb{R}^{\Bbb{Z}^2}$$

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