Let's consider the torus $\mathbb{T}=\mathbb{S}^1\times\mathbb{S}^1$. Again, the covering $\mathbb{R}^2\longrightarrow\mathbb{T}$, $(\alpha,\beta)\longmapsto(\mathrm{e}^{\mathrm{i}\alpha},\mathrm{e}^{\mathrm{i}\beta})$ provides two global 1-forms $\mathrm{d} \alpha$, $\mathrm{d} \beta$

• 0-forms: $f:\mathbb{R}^2\longrightarrow\mathbb{R}$
• 1-forms: $g\mathrm{d} \alpha+h\mathrm{d} \beta$
• 2-forms: $k\mathrm{d} \alpha\wedge\mathrm{d} \beta$

where every function is differentiable and $2\pi$-periodic in each variable. $\mathrm{d}$ acts as in $\mathbb{R}^2$

To compute the next cohomology groups we will use the Fourier series of these periodic functions: $f(\alpha,\beta)=\sum_{\substack{m,n=-\infty}}^\infty f_{mn}\mathrm{e}^{\mathrm{i}m\alpha}\mathrm{e}^{\mathrm{i}n\beta}$, y lo mismo para $g$, $h$ y $k$. This decomposition will be merely formal, without minding convergence or differentiability, in a totally heuristic fashion which, however, yields completely right results

and equals 0 if $mh_{mn}=ng_{mn}$ for all $m$, $n$; on the other hand

Therefore, given $g_{mn}$, $h_{mn}$ with $mh_{mn}=ng_{mn}$, we're looking for $f_{mn}$ with $g_{mn}=\mathrm{i}mf_{mn}$, $h_{mn}=\mathrm{i}nf_{mn}$

For $m\neq 0$, we may take $f_{mn}=\frac{1}{\mathrm{i}m}g_{mn}$, and

In the same way, if $n\neq 0$, we take $f_{mn}=\frac{1}{\mathrm{i}n}h_{mn}$. The very problem is $m=n=0$: $0h_{00}=0g_{00}$ always holds, but $g_{00}=\mathrm{i}0f_{00}$ and $h_{00}=\mathrm{i}0f_{00}$ only holds if $g_{00}=h_{00}=0$. This way, not every pair of functions $g$, $h$ with $\dfrac{\partial h}{\partial \alpha}-\dfrac{\partial g}{\partial \beta}=0$ follow the pattern $\dfrac{\partial f}{\partial \alpha}$, $\dfrac{\partial f}{\partial \beta}$; previously we have to adjust $g_{00}=h_{00}=0$. Therefore,

because $g$ and $h$ retain two extra degrees of freedom

Two generators of $H^1(\mathbb{T})$ are $\mathrm{d}\alpha$ ($g_{00}=1$) and $\mathrm{d} \beta$ ($h_{00}=1$). As with the circumference, these are closely related to the homology generators of $H_1(\mathbb{T})$

$\mathrm{d}(k\mathrm{d} \alpha\wedge\mathrm{d}\beta)=0$, so the kernel of $\mathrm{d}_2$ is the entire $\Omega^2(\mathbb{T})$. On the other hand,

We need to study if given $k_{mn}$ we may find $g_{mn}$, $h_{mn}$ with $\mathrm{i}mh_{mn}-\mathrm{i}ng_{mn}=k_{mn}$ for all $m$, $n$

If $m\neq 0$ or $n\neq 0$ we may take $g_{mn}=0$, $h_{mn}=\frac{1}{\mathrm{i}m}k_{mn}$ or $g_{mn}=-\frac{1}{\mathrm{i}n}k_{mn}$ and $h_{mn}=0$. As before, the problem resides in $m=n=0$. $h_{00}$ is free, but for $h\in \text{im } \mathrm{d}_1$, it is required that $h_{00}=0$. So in the end,

and a generator of $H^2(\mathbb{T})$ is $\mathrm{d}\alpha\wedge\mathrm{d}\beta$ ($h_{00}=1$)