[proofplan] We prove that every harmonic form is closed, that the only exact harmonic form is zero, and that every de Rham cohomology class contains a harmonic representative. The first two facts follow from the formal adjoint identity for the codifferential and positivity of the Hodge Laplacian. Surjectivity is the analytic content: we use the [Hodge decomposition theorem](/theorems/3941) for smooth forms on a closed Riemannian manifold, which decomposes every form into an exact part, a coexact part, and a harmonic part. Applying that decomposition to a closed form leaves its cohomology class represented by the harmonic summand. [/proofplan] [step:Equip forms with the $L^2$ inner product and record the Laplacian identity] Let $n=\dim M$, and let $\operatorname{vol}_g$ denote the Riemannian volume measure on $M$. We use the endpoint convention that $\Omega^r(M)=\{0\}$ for every integer $r<0$ or $r>n$, and that $d$ and $\delta_g$ are the corresponding zero maps whenever their source or target lies outside degrees $0,\dots,n$. For each integer $r$ with $0 \leq r \leq n$, let \begin{align*} (\cdot,\cdot)_{L^2}: \Omega^r(M) \times \Omega^r(M) &\to \mathbb{R} \end{align*} be the $L^2$ inner product defined by \begin{align*} (\alpha,\beta)_{L^2} = \int_M \langle \alpha(p),\beta(p)\rangle_{\Lambda^r T_p^*M,g}\,d\operatorname{vol}_g(p), \end{align*} where $\langle \cdot,\cdot\rangle_{\Lambda^r T_p^*M,g}$ is the inner product induced by $g_p$ on $\Lambda^r T_p^*M$. The codifferential $\delta_g: \Omega^{r+1}(M) \to \Omega^r(M)$ is the formal $L^2$ adjoint of $d: \Omega^r(M) \to \Omega^{r+1}(M)$, so for all $\alpha \in \Omega^r(M)$ and $\beta \in \Omega^{r+1}(M)$, \begin{align*} (d\alpha,\beta)_{L^2} = (\alpha,\delta_g\beta)_{L^2}. \end{align*} Since $M$ is closed, there are no boundary terms in this adjoint relation. For any $\omega \in \Omega^k(M)$, using $\Delta_g=d\delta_g+\delta_gd$ and the adjoint relation gives \begin{align*} (\Delta_g\omega,\omega)_{L^2} &= (d\delta_g\omega,\omega)_{L^2}+(\delta_gd\omega,\omega)_{L^2} \\ &= (\delta_g\omega,\delta_g\omega)_{L^2}+(d\omega,d\omega)_{L^2} \\ &= \|\delta_g\omega\|_{L^2}^2+\|d\omega\|_{L^2}^2. \end{align*} [/step] [step:Show that harmonic forms are closed and coclosed] Let $\omega \in \mathcal{H}^k_g(M)$. By definition, $\Delta_g\omega=0$. Substituting into the identity from the previous step gives \begin{align*} 0 = (\Delta_g\omega,\omega)_{L^2} = \|\delta_g\omega\|_{L^2}^2+\|d\omega\|_{L^2}^2. \end{align*} Both terms on the right are nonnegative [real numbers](/page/Real%20Numbers). Hence \begin{align*} \|d\omega\|_{L^2}^2=0 \qquad\text{and}\qquad \|\delta_g\omega\|_{L^2}^2=0. \end{align*} Since $d\omega$ and $\delta_g\omega$ are smooth forms, vanishing of their $L^2$ norms implies \begin{align*} d\omega=0 \qquad\text{and}\qquad \delta_g\omega=0. \end{align*} Thus every harmonic $k$-form is closed, so $\Phi_k(\omega)=[\omega]$ is well-defined. [/step] [step:Prove injectivity by testing an exact harmonic form against itself] Suppose $\omega \in \mathcal{H}^k_g(M)$ and $\Phi_k(\omega)=0$ in $H^k_{dR}(M)$. Then there exists a smooth $(k-1)$-form $\eta \in \Omega^{k-1}(M)$ such that \begin{align*} \omega=d\eta. \end{align*} By the previous step, $\delta_g\omega=0$. Therefore, using the adjoint relation between $d$ and $\delta_g$, \begin{align*} \|\omega\|_{L^2}^2 &= (\omega,\omega)_{L^2} \\ &= (d\eta,\omega)_{L^2} \\ &= (\eta,\delta_g\omega)_{L^2} \\ &= 0. \end{align*} Since $\omega$ is smooth and has zero $L^2$ norm, $\omega=0$. Hence $\ker \Phi_k=\{0\}$, so $\Phi_k$ is injective. [/step] [step:Use Hodge decomposition to produce a harmonic representative] Let $[\alpha]\in H^k_{dR}(M)$ be a de Rham cohomology class, and choose a smooth closed representative $\alpha\in\Omega^k(M)$, so $d\alpha=0$. We use the [Hodge decomposition](/theorems/2745) theorem on closed Riemannian manifolds, understood here as an independently established analytic decomposition theorem for smooth forms. Its hypotheses apply because $M$ is a smooth closed Riemannian manifold and $\alpha\in\Omega^k(M)$ is a smooth $k$-form. Applied to $\alpha$, it gives smooth forms \begin{align*} \eta &\in \Omega^{k-1}(M), & \theta &\in \Omega^{k+1}(M), & h &\in \mathcal{H}^k_g(M) \end{align*} such that \begin{align*} \alpha = d\eta+\delta_g\theta+h. \end{align*} Apply $d$ to both sides. Since $d\alpha=0$, $d^2=0$, and $dh=0$ because $h$ is harmonic, we obtain \begin{align*} 0 = d\alpha = d\delta_g\theta. \end{align*} Now test $\delta_g\theta$ against itself: \begin{align*} \|\delta_g\theta\|_{L^2}^2 &= (\delta_g\theta,\delta_g\theta)_{L^2} \\ &= (\theta,d\delta_g\theta)_{L^2} \\ &= 0. \end{align*} Thus $\delta_g\theta=0$. Consequently \begin{align*} \alpha = d\eta+h, \end{align*} so $\alpha-h=d\eta$ is exact. Hence $[\alpha]=[h]$ in $H^k_{dR}(M)$ with $h\in\mathcal{H}^k_g(M)$. Therefore $\Phi_k$ is surjective. [/step] [step:Conclude that the harmonic representative map is an isomorphism] The map $\Phi_k:\mathcal{H}^k_g(M)\to H^k_{dR}(M)$ is linear because \begin{align*} \Phi_k(a\omega_1+b\omega_2) = [a\omega_1+b\omega_2] = a[\omega_1]+b[\omega_2] \end{align*} for all $a,b\in\mathbb{R}$ and all $\omega_1,\omega_2\in\mathcal{H}^k_g(M)$. The preceding steps show that $\Phi_k$ is well-defined, injective, and surjective. Therefore $\Phi_k$ is an isomorphism of real vector spaces. [/step]