Let $S$ be a compact oriented Riemannian surface. And let $\mathcal{H}^1(S)$ be the space of the harmonic $1$-forms of $S$. I’m trying to prove that there is a linear isomorphism:
$$H^1_{dR}(S)\to \mathcal{H}^1(S).$$
By Hodge decomposition theorem, given a $1$-form $\omega$:
$$[\omega]=[\omega’+d\eta+\delta \varphi]= [\omega’+\delta \varphi]$$
where $\omega’$ is harmomic. The proof will be basically complete if we manage to prove that $\delta \varphi=0$. By Hodge decomposition theorem it suffices to prove that $d \delta \varphi=0$ but I don’t know how to prove this.
Best Answer
For an arbitrary $1$-form $\omega$, there is no de Rham class $[\omega]$; this only makes sense if $\omega$ is closed.
By the Hodge decomposition, we can write $\omega$ as $\omega = \omega' + d\eta + \delta\varphi$ where $\omega'$ is harmonic. Since $d\omega' = 0$ and $dd\eta = 0$, we have $d\omega = d\delta\varphi$. When is $d\delta\varphi = 0$? Note that
$$\|\delta\varphi\|^2 = \langle\delta\varphi, \delta\varphi\rangle = \langle d\delta\varphi, \varphi\rangle$$
so if $d\delta\varphi = 0$, we must have $\|\delta\varphi\|^2 = 0$ and hence $\delta\varphi = 0$. Conversely, if $\delta\varphi = 0$, then $d\delta\varphi = 0$. Therefore, the form $\omega$ is closed if and only if $\delta\varphi = 0$.
So, for closed $\omega$ we have $\omega = \omega' + d\eta$ and hence $[\omega] = [\omega' + d\eta] = [\omega']$. This gives the linear map $H^1_{\text{dR}}(S) \to \mathcal{H}^1(S)$. I'll leave it to you to verify that it is indeed an isomorphism.