[proofplan] We prove the decomposition by applying the Hodge-Fredholm theorem to the Hodge Laplacian on $k$-forms. This gives an $L^2$-[orthogonal projection](/theorems/437) onto harmonic forms and a Green operator whose Laplacian reconstructs the orthogonal complement of the harmonic space. Expanding the Hodge Laplacian identity \begin{align*} \Delta = d d^* + d^* d \end{align*} produces the exact and coexact pieces. Orthogonality follows from the defining adjointness of $d^*$ and from the identity \begin{align*} (\Delta\omega,\omega)_{L^2} = \|d\omega\|_{L^2}^2 + \|d^*\omega\|_{L^2}^2. \end{align*} [/proofplan] [step:Apply the Hodge-Fredholm theorem to the Hodge Laplacian] Fix $k \in \{0,\dots,\dim M\}$. Let \begin{align*} d: \Omega^k(M) &\to \Omega^{k+1}(M) \end{align*} denote the [exterior derivative](/theorems/1525), let \begin{align*} d^*: \Omega^k(M) &\to \Omega^{k-1}(M) \end{align*} denote its formal adjoint with respect to the $L^2$ inner product induced by $g$ and the Riemannian volume measure, and let \begin{align*} \Delta: \Omega^k(M) &\to \Omega^k(M), \\ \omega &\mapsto d d^*\omega + d^*d\omega \end{align*} be the Hodge Laplacian on $k$-forms. Since $(M,g)$ is closed and Riemannian, the Hodge Laplacian is a self-adjoint elliptic differential operator on the compact manifold $M$. By the Hodge-Fredholm theorem for the Hodge Laplacian, there are an $L^2$-orthogonal projection \begin{align*} P_{\mathcal H}: \Omega^k(M) &\to \mathcal H^k(M) \end{align*} and a Green operator \begin{align*} G: \Omega^k(M) &\to \Omega^k(M) \end{align*} such that, for every $\omega \in \Omega^k(M)$, \begin{align*} \omega = P_{\mathcal H}\omega + \Delta G\omega. \end{align*} Elliptic regularity is part of this theorem, so $G\omega$ is smooth whenever $\omega$ is smooth. [guided] Fix $k \in \{0,\dots,\dim M\}$. The operators used in the decomposition must be declared on $k$-forms. The exterior derivative is the map \begin{align*} d: \Omega^k(M) &\to \Omega^{k+1}(M), \end{align*} and its formal adjoint with respect to the $L^2$ inner product induced by the Riemannian metric $g$ and the Riemannian volume measure is the map \begin{align*} d^*: \Omega^k(M) &\to \Omega^{k-1}(M). \end{align*} The Hodge Laplacian on $k$-forms is the differential operator \begin{align*} \Delta: \Omega^k(M) &\to \Omega^k(M), \\ \omega &\mapsto d d^*\omega + d^*d\omega. \end{align*} The analytic input is the Hodge-Fredholm theorem for this operator. Its hypotheses are met because $M$ is closed, hence compact without boundary, and $g$ makes the Hodge Laplacian a self-adjoint elliptic operator on smooth differential forms. The theorem gives an $L^2$-orthogonal projection \begin{align*} P_{\mathcal H}: \Omega^k(M) &\to \mathcal H^k(M) \end{align*} onto the harmonic forms and a Green operator \begin{align*} G: \Omega^k(M) &\to \Omega^k(M) \end{align*} such that every smooth $k$-form $\omega$ satisfies \begin{align*} \omega = P_{\mathcal H}\omega + \Delta G\omega. \end{align*} The elliptic regularity part of the theorem is what keeps the argument inside $\Omega^k(M)$: if $\omega$ is smooth, then $G\omega$ is smooth, so the expressions $dG\omega$ and $d^*G\omega$ are smooth forms of the required degrees. [/guided] [/step] [step:Expand the Green operator identity into exact and coexact terms] Let $\omega \in \Omega^k(M)$ and define \begin{align*} h &:= P_{\mathcal H}\omega \in \mathcal H^k(M), \\ \alpha &:= d^*G\omega \in \Omega^{k-1}(M), \\ \beta &:= dG\omega \in \Omega^{k+1}(M). \end{align*} Using the Hodge Laplacian identity \begin{align*} \Delta = d d^* + d^*d \end{align*} in the Green operator formula gives \begin{align*} \omega &= h + \Delta G\omega \\ &= h + d d^*G\omega + d^*dG\omega \\ &= h + d\alpha + d^*\beta. \end{align*} Thus \begin{align*} \Omega^k(M) \subset d\Omega^{k-1}(M) + d^*\Omega^{k+1}(M) + \mathcal H^k(M). \end{align*} The reverse inclusion follows from the definitions of the three summands as subspaces of $\Omega^k(M)$. [/step] [step:Prove the exact and coexact summands are orthogonal] Let $\eta \in \Omega^{k-1}(M)$ and $\theta \in \Omega^{k+1}(M)$. By definition of the formal adjoint $d^*$, \begin{align*} (d\eta, d^*\theta)_{L^2} = (d^2\eta, \theta)_{L^2}. \end{align*} Since $d^2 = 0$, this gives \begin{align*} (d\eta, d^*\theta)_{L^2} = 0. \end{align*} Therefore $d\Omega^{k-1}(M)$ is orthogonal to $d^*\Omega^{k+1}(M)$. [/step] [step:Prove harmonic forms are orthogonal to exact and coexact forms] Let $h \in \mathcal H^k(M)$. The identity $\Delta h = 0$ and the formal adjointness of $d$ and $d^*$ give \begin{align*} 0 &= (\Delta h,h)_{L^2} \\ &= (d d^*h,h)_{L^2} + (d^*dh,h)_{L^2} \\ &= (d^*h,d^*h)_{L^2} + (dh,dh)_{L^2} \\ &= \|d^*h\|_{L^2}^2 + \|dh\|_{L^2}^2. \end{align*} Both terms are non-negative, so $dh=0$ and $d^*h=0$. If $\eta \in \Omega^{k-1}(M)$, then \begin{align*} (d\eta,h)_{L^2} = (\eta,d^*h)_{L^2}=0. \end{align*} If $\theta \in \Omega^{k+1}(M)$, then \begin{align*} (d^*\theta,h)_{L^2} = (\theta,dh)_{L^2}=0. \end{align*} Hence $\mathcal H^k(M)$ is orthogonal to both $d\Omega^{k-1}(M)$ and $d^*\Omega^{k+1}(M)$. [guided] Take $h \in \mathcal H^k(M)$, so by definition $\Delta h=0$. To show that $h$ is orthogonal to exact and coexact forms, we first prove the stronger identities $dh=0$ and $d^*h=0$. Using the definition of the Hodge Laplacian and the formal adjointness of $d$ and $d^*$, we compute \begin{align*} 0 &= (\Delta h,h)_{L^2} \\ &= (d d^*h,h)_{L^2} + (d^*dh,h)_{L^2} \\ &= (d^*h,d^*h)_{L^2} + (dh,dh)_{L^2} \\ &= \|d^*h\|_{L^2}^2 + \|dh\|_{L^2}^2. \end{align*} Each summand is non-negative, so both must vanish. Therefore $dh=0$ and $d^*h=0$. Now let $\eta \in \Omega^{k-1}(M)$. The exact form $d\eta$ is orthogonal to $h$ because \begin{align*} (d\eta,h)_{L^2} = (\eta,d^*h)_{L^2}=0. \end{align*} Similarly, for $\theta \in \Omega^{k+1}(M)$, the coexact form $d^*\theta$ is orthogonal to $h$ because \begin{align*} (d^*\theta,h)_{L^2} = (\theta,dh)_{L^2}=0. \end{align*} Thus harmonic forms are orthogonal to both the exact and coexact summands. [/guided] [/step] [step:Use orthogonality to identify the sum as direct] The preceding steps show that the three subspaces $d\Omega^{k-1}(M)$, $d^*\Omega^{k+1}(M)$, and $\mathcal H^k(M)$ are pairwise orthogonal in the $L^2$ inner product. If \begin{align*} \gamma_1 + \gamma_2 + h = 0 \end{align*} with $\gamma_1 \in d\Omega^{k-1}(M)$, $\gamma_2 \in d^*\Omega^{k+1}(M)$, and $h \in \mathcal H^k(M)$, then taking the $L^2$ inner product with each summand gives \begin{align*} \|\gamma_1\|_{L^2}^2 = \|\gamma_2\|_{L^2}^2 = \|h\|_{L^2}^2 = 0. \end{align*} Hence $\gamma_1=0$, $\gamma_2=0$, and $h=0$. Therefore the sum is direct, and the decomposition \begin{align*} \Omega^k(M)=d\Omega^{k-1}(M) \oplus d^*\Omega^{k+1}(M) \oplus \mathcal H^k(M) \end{align*} holds with pairwise orthogonal summands. [/step]