[proofplan] We regard $A^{p,*}(X,E)$ as the Dolbeault complex resolving the sheaf $\Omega_X^p\otimes E$. The metrics $\omega$ and $h$ give an $L^2$ [inner product](/page/Inner%20Product), hence a formal adjoint $\bar\partial_E^*$ and the Dolbeault Laplacian. Elliptic Hodge theory for the Dolbeault complex on a compact manifold gives an [orthogonal decomposition](/theorems/436) into harmonic, exact, and coexact parts. This decomposition immediately proves that each Dolbeault cohomology class has a unique harmonic representative, and the Dolbeault resolution identifies this cohomology with $H^q(X,\Omega_X^p\otimes E)$. [/proofplan]

[step:Identify sheaf cohomology with the cohomology of the Dolbeault complex]Let $\mathcal O(E)$ denote the sheaf of holomorphic sections of $E$, and write $\Omega_X^p\otimes E$ for the sheaf $\Omega_X^p\otimes_{\mathcal O_X}\mathcal O(E)$. For fixed $p$, the $E$-valued Dolbeault complex is \begin{align*} 0\to A^{p,0}(X,E)\xrightarrow{\bar\partial_E}A^{p,1}(X,E)\xrightarrow{\bar\partial_E}\cdots\xrightarrow{\bar\partial_E}A^{p,n}(X,E)\to 0. \end{align*} At the sheaf level, the Dolbeault resolution for the holomorphic vector bundle $E$ identifies the sheaf cohomology group $H^q(X,\Omega_X^p\otimes E)$ with the cohomology group \begin{align*} H^{p,q}_{\bar\partial}(X,E):= \frac{\ker\left(\bar\partial_E:A^{p,q}(X,E)\to A^{p,q+1}(X,E)\right)} {\operatorname{im}\left(\bar\partial_E:A^{p,q-1}(X,E)\to A^{p,q}(X,E)\right)}. \end{align*} For $q=0$, the denominator is interpreted as $0$.[/step]

[guided]The theorem is stated in terms of sheaf cohomology, but [harmonic representatives](/theorems/2747) live naturally in the analytic Dolbeault complex. We therefore first name the bridge between the two. Let $\mathcal O(E)$ be the sheaf of holomorphic sections of $E$, and let $\Omega_X^p\otimes E$ mean $\Omega_X^p\otimes_{\mathcal O_X}\mathcal O(E)$. For fixed $p$, the smooth $E$-valued Dolbeault complex is \begin{align*} 0\to A^{p,0}(X,E)\xrightarrow{\bar\partial_E}A^{p,1}(X,E)\xrightarrow{\bar\partial_E}\cdots\xrightarrow{\bar\partial_E}A^{p,n}(X,E)\to 0. \end{align*} The Dolbeault resolution for a holomorphic vector bundle states that this fine resolution computes the sheaf cohomology of $\Omega_X^p\otimes E$. Thus \begin{align*} H^q(X,\Omega_X^p\otimes E)\cong \frac{\ker\left(\bar\partial_E:A^{p,q}(X,E)\to A^{p,q+1}(X,E)\right)} {\operatorname{im}\left(\bar\partial_E:A^{p,q-1}(X,E)\to A^{p,q}(X,E)\right)}. \end{align*} The point of this step is that it reduces the theorem to a purely analytic statement: every cohomology class of the Dolbeault complex has exactly one representative annihilated by $\Delta_{\bar\partial,E}$.[/guided]

[step:Apply elliptic Hodge decomposition to the $E$-valued Dolbeault complex] The Kähler metric $\omega$ and the Hermitian metric $h$ determine an $L^2$ inner product $(\cdot,\cdot)_{L^2}$ on each $A^{p,q}(X,E)$. Let \begin{align*} \bar\partial_E^*:A^{p,q}(X,E)\to A^{p,q-1}(X,E) \end{align*} be the formal adjoint of $\bar\partial_E$ with respect to this inner product. Since $X$ is compact and $\Delta_{\bar\partial,E}$ is an elliptic self-adjoint second-order differential operator, elliptic Hodge theory gives the orthogonal decomposition \begin{align*} A^{p,q}(X,E)= \mathcal H^{p,q}_{\bar\partial}(X,E) \oplus \operatorname{im}\left(\bar\partial_E:A^{p,q-1}(X,E)\to A^{p,q}(X,E)\right) \oplus \operatorname{im}\left(\bar\partial_E^*:A^{p,q+1}(X,E)\to A^{p,q}(X,E)\right). \end{align*} Moreover, \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X,E)= \ker \bar\partial_E\cap \ker \bar\partial_E^*. \end{align*} Indeed, for $\alpha\in A^{p,q}(X,E)$, \begin{align*} (\Delta_{\bar\partial,E}\alpha,\alpha)_{L^2} = (\bar\partial_E\alpha,\bar\partial_E\alpha)_{L^2} + (\bar\partial_E^*\alpha,\bar\partial_E^*\alpha)_{L^2}, \end{align*} so $\Delta_{\bar\partial,E}\alpha=0$ if and only if $\bar\partial_E\alpha=0$ and $\bar\partial_E^*\alpha=0$. [/step]

[step:Show that every harmonic form defines a Dolbeault cohomology class] Let $\alpha\in\mathcal H^{p,q}_{\bar\partial}(X,E)$. From \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X,E)=\ker\bar\partial_E\cap\ker\bar\partial_E^* \end{align*} we get $\bar\partial_E\alpha=0$. Hence $\alpha$ determines a Dolbeault cohomology class \begin{align*} [\alpha]\in H^{p,q}_{\bar\partial}(X,E). \end{align*} Composing this class with the Dolbeault-resolution identification from the first step defines the stated map \begin{align*} \Phi:\mathcal H^{p,q}_{\bar\partial}(X,E)\to H^q(X,\Omega_X^p\otimes E). \end{align*} [/step]

[step:Prove injectivity by orthogonality to exact forms] Suppose $\alpha\in\mathcal H^{p,q}_{\bar\partial}(X,E)$ and $\Phi(\alpha)=0$. Under the Dolbeault identification, this means that there exists \begin{align*} \beta\in A^{p,q-1}(X,E) \end{align*} such that $\alpha=\bar\partial_E\beta$, with the convention that for $q=0$ this forces $\alpha=0$ immediately. For $q>0$, since $\alpha$ is harmonic, $\bar\partial_E^*\alpha=0$. Therefore \begin{align*} (\alpha,\alpha)_{L^2} = (\bar\partial_E\beta,\alpha)_{L^2} = (\beta,\bar\partial_E^*\alpha)_{L^2} = 0. \end{align*} Thus $\alpha=0$, so $\Phi$ is injective. [/step]

[step:Prove surjectivity by extracting the harmonic summand of a closed form]Let $[\eta]\in H^{p,q}_{\bar\partial}(X,E)$ be a Dolbeault cohomology class, and choose a representative \begin{align*} \eta\in A^{p,q}(X,E) \end{align*} with $\bar\partial_E\eta=0$. By the [Hodge decomposition](/theorems/2745), there exist unique forms \begin{align*} \gamma\in A^{p,q-1}(X,E), \qquad \delta\in A^{p,q+1}(X,E), \qquad \alpha\in\mathcal H^{p,q}_{\bar\partial}(X,E) \end{align*} such that \begin{align*} \eta=\alpha+\bar\partial_E\gamma+\bar\partial_E^*\delta. \end{align*} Apply $\bar\partial_E$ to this equality. Since $\bar\partial_E\eta=0$, $\bar\partial_E\alpha=0$, and $\bar\partial_E^2=0$, we obtain \begin{align*} \bar\partial_E\bar\partial_E^*\delta=0. \end{align*} Taking the $L^2$ inner product with $\delta$ and using the definition of the formal adjoint gives \begin{align*} 0 = (\bar\partial_E\bar\partial_E^*\delta,\delta)_{L^2} = (\bar\partial_E^*\delta,\bar\partial_E^*\delta)_{L^2}. \end{align*} Hence $\bar\partial_E^*\delta=0$, and therefore \begin{align*} \eta=\alpha+\bar\partial_E\gamma. \end{align*} Thus $[\eta]=[\alpha]$, so every Dolbeault cohomology class is represented by a harmonic form. Hence $\Phi$ is surjective.[/step]

[guided]We now show why the harmonic part of a closed form is not merely present, but represents the same cohomology class. Start with an arbitrary Dolbeault class $[\eta]$. By definition of Dolbeault cohomology, we may choose a smooth $E$-valued $(p,q)$-form \begin{align*} \eta\in A^{p,q}(X,E) \end{align*} such that $\bar\partial_E\eta=0$. The [Hodge decomposition](/theorems/3941) splits $\eta$ into three mutually orthogonal pieces: a harmonic piece, an exact piece, and a coexact piece. Thus there are forms \begin{align*} \alpha\in\mathcal H^{p,q}_{\bar\partial}(X,E), \qquad \gamma\in A^{p,q-1}(X,E), \qquad \delta\in A^{p,q+1}(X,E) \end{align*} such that \begin{align*} \eta=\alpha+\bar\partial_E\gamma+\bar\partial_E^*\delta. \end{align*} The exact term $\bar\partial_E\gamma$ does not change the cohomology class. The only possible obstruction is the coexact term $\bar\partial_E^*\delta$. We prove that this term vanishes because $\eta$ is closed. Apply $\bar\partial_E$ to the decomposition. Since $\eta$ is closed, $\bar\partial_E\eta=0$. Since $\alpha$ is harmonic, $\bar\partial_E\alpha=0$. Since $\bar\partial_E$ is a differential, $\bar\partial_E^2=0$. Therefore the only remaining term is \begin{align*} \bar\partial_E\bar\partial_E^*\delta=0. \end{align*} Now pair this equation with $\delta$ using the $L^2$ inner product induced by $\omega$ and $h$. The defining property of the formal adjoint says that \begin{align*} (\bar\partial_E\bar\partial_E^*\delta,\delta)_{L^2} = (\bar\partial_E^*\delta,\bar\partial_E^*\delta)_{L^2}. \end{align*} The left-hand side is $0$, so \begin{align*} (\bar\partial_E^*\delta,\bar\partial_E^*\delta)_{L^2}=0. \end{align*} Positive definiteness of the $L^2$ inner product implies $\bar\partial_E^*\delta=0$. Consequently \begin{align*} \eta=\alpha+\bar\partial_E\gamma. \end{align*} Thus $\eta$ and the harmonic form $\alpha$ differ by a $\bar\partial_E$-exact form, so they determine the same Dolbeault cohomology class: \begin{align*} [\eta]=[\alpha]. \end{align*} This proves that every Dolbeault cohomology class has a harmonic representative.[/guided]

[step:Conclude that the harmonic representative is unique and obtain the sheaf cohomology isomorphism] The injectivity step shows that a Dolbeault class has at most one harmonic representative, and the surjectivity step shows that it has at least one. Therefore the map \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X,E)\to H^{p,q}_{\bar\partial}(X,E) \end{align*} sending $\alpha$ to $[\alpha]$ is an isomorphism. Composing with the Dolbeault-resolution isomorphism \begin{align*} H^{p,q}_{\bar\partial}(X,E)\cong H^q(X,\Omega_X^p\otimes E) \end{align*} gives the stated isomorphism \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X,E)\cong H^q(X,\Omega_X^p\otimes E). \end{align*} This is precisely the claimed natural map from harmonic $E$-valued $(p,q)$-forms to Dolbeault cohomology. [/step]