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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 $\mathbb{R}^2$


$$H^0(\mathbb{T})=\dfrac{\text{ker }\mathrm{d}_0}{\text{im }\mathrm{d}_{-1}}=\dfrac{\left\{f\middle|\dfrac{\partial f}{\partial \alpha}=\dfrac{\partial f}{\partial \beta}=0\right\}}{0}=\dfrac{\{f=cte\in\mathbb{R}\}}{0}\simeq\mathbb{R}$$


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

$$\mathrm{d} (g\mathrm{d} \alpha+h\mathrm{d}\beta)=\left(\dfrac{\partial h}{\partial \alpha}-\dfrac{\partial g}{\partial \beta}\right)\mathrm{d} \alpha\wedge\mathrm{d}\beta=\mathrm{i}\sum_{mn}(mh_{mn}-ng_{mn})\mathrm{e}^{\mathrm{i}m\alpha}\mathrm{e}^{\mathrm{i}n\beta}\mathrm{d} \alpha\wedge\mathrm{d}\beta$$

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

$$\mathrm{d}(f)=\dfrac{\partial f}{\partial \alpha}\mathrm{d} \alpha+\dfrac{\partial f}{\partial \beta}\mathrm{d} \beta=\mathrm{i}\sum_{mn}mf_{mn}\mathrm{e}^{\mathrm{i}m\alpha} \mathrm{e}^{\mathrm{i}n\beta} \mathrm{d}\alpha+ \mathrm{i}\sum_{mn}nf_{mn}\mathrm{e}^{\mathrm{i}m\alpha}\mathrm{e}^{\mathrm{i}n\beta}\mathrm{d}\beta$$

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

$$\mathrm{i}mf_{mn}=g_{mn},\qquad\mathrm{i}nf_{mn}=\frac{\mathrm{i}n}{\mathrm{i}m}g_{mn}=\frac{1}{m}mh_{mn}=h_{mn}$$

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,

$$H^1(\mathbb{T})=\dfrac{\left\{g\mathrm{d}\alpha+h\mathrm{d}\beta\middle|\dfrac{\partial h}{\partial \alpha}-\dfrac{\partial g}{\partial \beta}=0\right\}}{\left\{\dfrac{\partial f}{\partial \alpha}\mathrm{d} \alpha+\dfrac{\partial f}{\partial \beta}\mathrm{d} \beta\right\}}\simeq\mathbb{R}^2$$

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,

$$\mathrm{d} (g\mathrm{d} \alpha+h\mathrm{d}\beta)=\mathrm{i}\sum_{mn}(mh_{mn}-ng_{mn})\mathrm{e}^{\mathrm{i}m\alpha}\mathrm{e}^{\mathrm{i}n\beta}\mathrm{d} \alpha\wedge\mathrm{d}\beta$$

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,

$$H^2(\mathbb{T})=\dfrac{\{k\mathrm{d}\alpha\wedge\mathrm{d}\beta\}}{\left\{\left(\dfrac{\partial h}{\partial \alpha}-\dfrac{\partial g}{\partial \beta}\right)\mathrm{d}\alpha\wedge\mathrm{d}\beta\right\}}\simeq\mathbb{R}$$

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