Problem involving the dimension of the Rhan cohomology of $\mathbb{R}^2 \setminus \mathbb{Z}^2$ I'm starting to study Rham cohomology and I came across the statement that $\dim H_{dR}^1(\mathbb{R}^2 \setminus \mathbb{Z}^2)$ is infinite. However, the Rham cohomology of $H_{dR}^0(\mathbb{R}) = \mathbb{R}$ and $H_{dR}^k(\mathbb{R}) = 0$, for all $k \geq 1$. Why does it happen? What is lost from one case to the next?
Any reference or idea that helps prove that $\dim H_{dR}^1(\mathbb{R}^2 \setminus \mathbb{Z}^2)$ would be most welcome.
 A: Remember that the dimension of $H^1_{dR}(X)$ counts the number of holes in your space. See here, for instance.
Then since $\mathbb{R}^2 \setminus \mathbb{Z}^2$ is a space with infinitely many holes, it makes sense that its first cohomology group should be infinite dimensional.
To see this formally is a bit annoying, but here's a sketch of one concrete way to prove it:
You'll want to look at the forms $d\theta_{m,n}$ where $\theta_{m,n}$ is the angular form around the lattice point $(m,n)$. See here for a super explicit computation of $d\theta_{0,0}$.
By multiplying this by a bump function we can guarantee we only need to worry about it near the point $(0,0)$, then we can translate it to any of our punctures $(m,n)$.
Next, you want to show that all of these $d\theta_{m,n}$s are linearly independent in
$H^1 \left ( \mathbb{R}^2 \setminus \mathbb{Z}^2 \right )$. But say you have a dependence
$$\sum_{i=1}^k \lambda_{m_i,n_i} d \theta_{m_i, n_i} = 0$$
in $H^1(\mathbb{R}^2, \mathbb{Z}^2)$. Well the support of all these functions is contained in some compact region (remember the bump functions?) so that this equation must also be true in
$H^1 \left ( U \setminus (\mathbb{Z}^2 \cap U) \right )$ for some open set $U$ with compact closure (in particular it contains only finitely many lattice points). But we know that these $d \theta_{m_i, n_i}$ are linearly independent in this finite case, so all the $\lambda_{m_i, n_i} = 0$, and the claim follows.

A more conceptual approach would be to use the fact that deRham cohomology sends certain colimits of spaces to limits of their homology groups. See here, for instance.
Let $A_n$ be the open square of side length $2n+1$ centered at the origin with all the integer lattice points removed.
(In symbols, $A_n = \left \{ (x,y) \ \middle | \ -n - \frac{1}{2} < x,y < n + \frac{1}{2}, (x,y) \not \in \mathbb{Z}^2 \right \}$)
Then since
$$
\mathbb{R}^2 \setminus \mathbb{Z}^2 = 
\bigcup_{n \in \mathbb{N}} A_n
$$
is a nice colimit, we see that the de Rham cohomology becomes the direct limit
$$\varprojlim H^1 \left ( A_n \right )$$
which we can quickly compute as
$$
\varprojlim \mathbb{R}^{(2n+1)^2}
$$
and this is easily seen to be infinite dimensional.

I hope this helps ^_^
