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The case of the sphere $\mathbb{S}^2$ will be substantially more difficult than the previous ones. The ideas developed, however, will be key when studying Poincaré Lemma and Mayer-Vietoris sequences

Let's consider the sphere $\mathbb{S}^2\subset\mathbb{R}^3$ together with the two charts coming from the stereographic projection from the poles $N=(0,0,1)$ and $S=(0,0,-1)$. The domains of both charts are $\mathbb{R}^2$, and the images are the open sets $U=\mathbb{S}^2-\{N\}$, $V=\mathbb{S}^2-\{S\}$; let $W=U\cap V=\mathbb{S}^2-\{N,S\}$.

Let $f\in C^\infty(\mathbb{S}^2)$ with $\mathrm{d} f=0$. $f$ induce the restrictions $f_1=f|_U$, $f_2=f|_V$, which are identified with functions of domain $\mathbb{R}^2$. But according to the cohomology of $\mathbb{R}^2$, $\mathrm{d} f_1=0$ and $\mathrm{d} f_2=0$ imply that $f_1$ and $f_2$ are constant, and since they coincide in the common domain $W$, $f$ has to be constant

$$H^0(\mathbb{S}^2)=\dfrac{\text{ker }\mathrm{d}_0}{\text{im }\mathrm{d}_{-1}}=\dfrac{\{f=cte\in\mathbb{R}\}}{0}\simeq\mathbb{R}$$

Now let $\alpha$ be a 1-form in $\mathbb{S}^2$ with $\mathrm{d}\alpha=0$. The restriction of $\alpha$ to $U$, $V$ induces the 1-forms $\alpha_1$ and $\alpha_2$, whose exterior derivative is also zero. Since $H^1(\mathbb{R}^2)=0$, closed 1-forms are exact and there exist functions $f_1$ and $f_2$ in $U$, $V$ with $\mathrm{d} f_1=\alpha_1$, $\mathrm{d} f_2=\alpha_2$. But how compatible are $f_1$ and $f_2$ in $W$? Here's the trick: in $W$, $\mathrm{d}(f_1-f_2)=\mathrm{d} f_1-\mathrm{d} f_2=\alpha_1-\alpha_2=\alpha-\alpha=0$. The exterior derivative of $f_1-f_2$ is 0 in $W$ and $W$ is connected, so $f_1-f_2$ is indeed some constant $c\in\mathbb{R}$. We define $\tilde{f}_2=f_2+c$, which stills satisfy $\mathrm{d} \tilde{f}_2=\alpha_2$, and now $f_1=\tilde{f}_2$ in $W$, so we may consider


$$f= \begin{cases} f_1 & \text{en $U$}\\ \tilde{f}_2 & \text{en $V$} \end{cases}$$

This function satisfies $\mathrm{d} f=\alpha$, since it already does in $U$, $V$. Therefore

$$H^1(\mathbb{S}^2)=\dfrac{\text{ker }\mathrm{d}_1}{\text{im }\mathrm{d}_0}=0$$

This is the subtle part. Let $\beta$ a 2-form in $\mathbb{S}^2$. Its domain is a manifold of dimension 2, so its exterior derivative is zero, and the same holds for $\beta_1=\beta|_U$, $\beta_2=\beta|_V$. Again $H^2(\mathbb{R}^2)=0$, and there exist suitable 1-forms $\alpha_1$, $\alpha_2$ defined in $U$, $V$ with $\mathrm{d} \alpha_i=\beta_i$. Now, how do $\alpha_1$ and $\alpha_2$ interact in the common domain $W$? $\mathrm{d}(\alpha_1-\alpha_2)=\mathrm{d}\alpha_1-\mathrm{d}\alpha_2=\beta_1-\beta_2=\beta-\beta=0$. But since we know nothing about the first cohomology group of $W$, we can't say anithing about its closed 1-forms. So let's study the cohomology of $W$ quickly

We may parameterize $W$ with $(\theta,z)\longmapsto(\sqrt{1-z^2}\cos(\theta),\sqrt{1-z^2}\sin(\theta),z)$, $(\theta,z)\in\mathbb{R}\times(-1,1)$. This induces the 1-forms $\mathrm{d} \theta$, $\mathrm{d} z$, and the functions $C^\infty(W)$ match the functions $f(\theta,z)$ which are differentiable and $2\pi$-periodic in $\theta$, which in turn admit the Fourier series representation $f(\theta,z)=\sum_{n}f_{n}(z)\mathrm{e}^{\mathrm{i}n\theta}$.

Let $\alpha=g\mathrm{d} \theta+h\mathrm{d} z$

$$\mathrm{d} \alpha=\left(\dfrac{\partial h}{\partial \theta}-\dfrac{\partial g}{\partial z}\right)\mathrm{d} \theta\wedge\mathrm{d} z=\sum_{n}(\mathrm{i}nh_n(z)-g_n'(z))\mathrm{e}^{\mathrm{i}n\theta}\mathrm{d} \theta\wedge\mathrm{d} z$$

and for $\alpha$ to be closed it is required $\mathrm{i}nh_n(z)=g_n'(z)$ for all $n$. Now we're looking for some $f=\sum_{n}f_{n}(z)\mathrm{e}^{\mathrm{i}n\theta}$ with $\mathrm{d} f=\alpha$, that is, $\dfrac{\partial f}{\partial \theta}=g$, $\dfrac{\partial f}{\partial z}=h$, or equivalently, $\mathrm{i}nf_n(z)=g_n(z)$, $f_n'(z)=h_n(z)$. If $n\neq 0$, there is no problem in taking $f_n(z)=\frac{g_n(z)}{\mathrm{i}n}$ and everything is fine, because $f_n'(z)=\frac{g_n'(z)}{\mathrm{i}n}=\frac{\mathrm{i}nh_n(z)}{\mathrm{i}n}=h_n(z)$. For $n=0$, $\mathrm{i}0h_0(z)=g_0'(z)$ holds when $g_0=c\in\mathbb{R}$ is constant. But if $c\neq 0$, the condition $\mathrm{i}0f_0(z)=g_0(z)=c$ cannot be satisfied. On the other hand, this is the only obstruction: if we correct $g$ with $c=0$, then it is enough to take $f_0(z)=\int_0^{z} h_0(t)\mathrm{d} t$. So $H^1(W)\simeq\mathbb{R}$, depending on the choice of $c$, and a generator of the group is the class of the 1-form $\mathrm{d} \theta$, which we will denote $\gamma$ from now on

We have studied part of the cohomology of $W$, now for the sphere. We had $\mathrm{d}(\alpha_1-\alpha_2)=0$; now we know that $\alpha_1-\alpha_2=\lambda\gamma+\mathrm{d} f$ for some $\lambda\in\mathbb{R}$ and some function $f$ in $W$. Suppose for a moment that $\lambda=0$; in this case we'll find a global $\alpha$ with $\mathrm{d} \alpha=\beta$. But before we have to introduce two functions of compact support

Let $\rho_1$ and $\rho_2$ be two positive differentiable functions defined in $[-1,1]$ such that $\rho_1$ is zero in $[1/3,1]$, $\rho_2$ is zero in $[-1,-1/3]$ and $\rho_1+\rho_2=1$. When applying this functions to the height of the sphere we get another two functions (that will again be denoted $\rho_1$ and $\rho_2$) such that the first has its support contained in $U$ and the second has its support contained in $V$, that, $\{\rho_1,\rho_2\}$ is a differentiable partition of unity subordinate to the open cover $\{U,V\}$

Why is this going to be useful? For extensibility reasons. The function $f$ may be non-extendable to the poles. But: if we define $f_1=f\rho_2$ and $f_2=f\rho_1$ (note the change of indexes!) and extend by zero, then $f_1$ is a differentiable function in $U$, $f_2$ is a differentiable function in $V$ and $f_1+f_2=f$ in $W$

Great! Let $\tilde{\alpha}_1=\alpha_1-\mathrm{d} f_1$, $\tilde{\alpha}_2=\alpha_2+\mathrm{d} f_2$. It is still true that $\mathrm{d} \tilde{\alpha}_1=\beta_1$ and $\mathrm{d} \tilde{\alpha}_2=\beta_2$, but now $\tilde{\alpha}_1-\tilde{\alpha}_2=\alpha_1-\mathrm{d} f_1-\alpha_2-\mathrm{d} f_2=\mathrm{d} f-\mathrm{d}(f_1+f_2)=0$ in $W$. So we end up with a global form


$$\alpha= \begin{cases} \tilde{\alpha}_1 & \text{en $U$}\\ \tilde{\alpha}_2 & \text{en $V$} \end{cases}$$

that satisfies $\mathrm{d} \alpha=\beta$ n $\mathbb{S}^2$. That, $\alpha_1$ and $\alpha_2$ could be non-extensible, but we have managed to correct this non-extensibility

If $\lambda\neq 0$, this process is not possible: if there is some global $\alpha$,


$$ \begin{array}{rl} \mathrm{d}(\alpha-\alpha_1)=\beta-\beta_1=0 \Longrightarrow \alpha-\alpha_1=\mathrm{d} f_1 & \qquad\text{in $U$, for some $f_1$} \\ \mathrm{d}(\alpha-\alpha_2)=\beta-\beta_2=0 \Longrightarrow \alpha-\alpha_2=\mathrm{d} f_2 & \qquad\text{in $V$, for some $f_2$} \\ \alpha_1-\alpha_2=\alpha-\mathrm{d} f_1 -\alpha+\mathrm{d} f_2=\mathrm{d} (f_2-f_1) & \qquad\text{in $W$} \\ \end{array} $$

but we had $\alpha_1-\alpha_2=\lambda\gamma+\mathrm{d} f$, and we know that $\gamma$ is not the exterior derivative of any function

To definitely establish that there exist closed and non-exact 2-forms in the sphere, we have to finde some $\beta$ for which, with the induced $\alpha_i$, $\alpha_1-\alpha_2=\lambda\gamma+\mathrm{d} f$ with $\lambda\neq 0$. Let $\beta_1=\mathrm{d}(\rho_2\gamma)$, $\beta_2=\mathrm{d}(\rho_1\gamma)$. In $W$,

$$\beta_1+\beta_2=\mathrm{d}(\rho_2\gamma+\rho_1\gamma)=\mathrm{d} \gamma=0$$

and we may define


$$\beta= \begin{cases} \beta_1 & \text{en $U$}\\ -\beta_2 & \text{en $V$} \end{cases}$$

The induced 1-forms are, up to the derivative of a function, $\alpha_1=\rho_2\gamma$ and $\alpha_2=-\rho_1\gamma$, and


Now it is sure that

$$H^2(\mathbb{S}^2)=\dfrac{\text{ker }\mathrm{d}_2}{\text{im }\mathrm{d}_1}\simeq\mathbb{R}$$

because the exactness of the closed 2-form $\beta$ depends on the induced coefficient $\lambda$

One explicit generator of $H^2(\mathbb{S}^2)$ could be $\mathrm{d}\theta\wedge\mathrm{d} z$, which happens to match the surface measure of the sphere as subset of $\mathbb{R}^3$