[proofplan] We use Hodge theory to replace cohomology classes by harmonic forms. On a compact Kähler manifold, harmonicity for the de Rham Laplacian is equivalent to Dolbeault harmonicity, so the type decomposition of a harmonic $k$-form gives a direct sum decomposition indexed by $p+q=k$. Finally, complex conjugation sends forms of type $(p,q)$ to forms of type $(q,p)$ and preserves harmonicity, which is exactly the conjugation condition in the definition of a pure Hodge structure. [/proofplan]

[step:Identify complex cohomology with harmonic forms]Let $g$ be the Riemannian metric underlying the Kähler metric on $X$. For each integer $k\ge 0$, denote by $A^k(X;\mathbb C)$ the complex [vector space](/page/Vector%20Space) of smooth complex-valued $k$-forms on $X$, and define \begin{align*} \mathcal H_d^k(X):=\{\alpha\in A^k(X;\mathbb C):\Delta_d\alpha=0\}. \end{align*} Here $d^*$ denotes the formal adjoint of $d$ with respect to the Riemannian metric $g$, and \begin{align*} \Delta_d &:= dd^* + d^*d \end{align*} is the de Rham Laplacian. We use the [Hodge theorem](/theorems/3942) for compact oriented Riemannian manifolds: for the compact Riemannian manifold $(X,g)$, the map \begin{align*} \mathcal H_d^k(X)&\to H^k(X,\mathbb C) \end{align*} sending a harmonic form to its de Rham cohomology class is an isomorphism. Thus every class in $H^k(X,\mathbb C)$ has a unique $d$-harmonic representative.[/step]

[guided]The first move is to replace cohomology classes, which are equivalence classes of closed forms modulo exact forms, by canonical representatives. Let $A^k(X;\mathbb C)$ be the complex vector space of smooth complex-valued $k$-forms on $X$, and define \begin{align*} \mathcal H_d^k(X):=\{\alpha\in A^k(X;\mathbb C):\Delta_d\alpha=0\}. \end{align*} The Hodge theorem for compact oriented Riemannian manifolds applies because a compact Kähler manifold is in particular a compact oriented Riemannian manifold through its underlying Riemannian metric $g$. It states that the map \begin{align*} \mathcal H_d^k(X)&\to H^k(X,\mathbb C) \end{align*} which sends a harmonic form $\alpha$ to its de Rham cohomology class $[\alpha]$ is an isomorphism. This is the mechanism that makes the decomposition canonical. Instead of choosing arbitrary representatives of cohomology classes, we choose the unique representative lying in $\mathcal H_d^k(X)$.[/guided]

[step:Decompose harmonic forms by type] For $p,q\ge 0$, let $A^{p,q}(X)$ denote the smooth complex-valued forms of type $(p,q)$ on $X$, and define \begin{align*} \mathcal H_{\bar\partial}^{p,q}(X):=\{\alpha\in A^{p,q}(X):\Delta_{\bar\partial}\alpha=0\}. \end{align*} Here $\bar\partial^*$ denotes the formal adjoint of $\bar\partial$ with respect to the same metric $g$, and \begin{align*} \Delta_{\bar\partial} &:= \bar\partial\bar\partial^* + \bar\partial^*\bar\partial \end{align*} is the Dolbeault Laplacian. Every form $\alpha\in A^k(X;\mathbb C)$ has a unique type decomposition \begin{align*} \alpha=\sum_{p+q=k}\alpha_{p,q}, \end{align*} where $\alpha_{p,q}\in A^{p,q}(X)$. Let $\alpha\in\mathcal H_d^k(X)$. Since $X$ is compact Kähler, the [Harmonicity Criteria on Compact Kähler Manifolds](/theorems/8055) give \begin{align*} \Delta_d\alpha=0 \iff \Delta_{\bar\partial}\alpha=0. \end{align*} The operator $\Delta_{\bar\partial}$ preserves bidegree, so \begin{align*} 0=\Delta_{\bar\partial}\alpha=\sum_{p+q=k}\Delta_{\bar\partial}\alpha_{p,q}. \end{align*} By uniqueness of the type decomposition, $\Delta_{\bar\partial}\alpha_{p,q}=0$ for every $p,q$ with $p+q=k$. Applying [citetheorem:8055] again to each $\alpha_{p,q}$, regarded as a complex-valued form on $X$, gives $\Delta_d\alpha_{p,q}=0$. Hence \begin{align*} \mathcal H_d^k(X)=\bigoplus_{p+q=k}\mathcal H_{\bar\partial}^{p,q}(X). \end{align*} [/step]

[step:Define the Hodge summands in de Rham cohomology] Define $H^{p,q}(X)\subset H^{p+q}(X,\mathbb C)$ to be the image of $\mathcal H_{\bar\partial}^{p,q}(X)$ under the harmonic de Rham representative map: \begin{align*} H^{p,q}(X):=\{[\alpha]\in H^{p+q}(X,\mathbb C):\alpha\in\mathcal H_{\bar\partial}^{p,q}(X)\}. \end{align*} By the [Dolbeault Hodge Theorem](/theorems/8059), this space is canonically identified with Dolbeault cohomology in bidegree $(p,q)$. The isomorphism $\mathcal H_d^k(X)\to H^k(X,\mathbb C)$ from the Hodge theorem carries the direct sum decomposition of harmonic forms to \begin{align*} H^k(X,\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X). \end{align*} The sum is direct because a cohomology class has a unique $d$-harmonic representative and the type decomposition of that representative is unique. [/step]

[step:Verify compatibility with complex conjugation] Let \begin{align*} c:A^k(X;\mathbb C)&\to A^k(X;\mathbb C) \end{align*} be the complex conjugation map on complex-valued forms. This conjugation preserves closed and exact forms, so it induces the standard complex conjugation on $H^k(X,\mathbb C)$ coming from the rational structure $H^k(X,\mathbb Q)$, fixing $H^k(X,\mathbb Q)$ pointwise. If $\alpha\in A^{p,q}(X)$, then $c(\alpha)=\overline{\alpha}\in A^{q,p}(X)$. Since $d$ is defined over $\mathbb R$, conjugation commutes with $d$, with the formal adjoint $d^*$ for the real Riemannian metric $g$, and therefore with $\Delta_d=dd^*+d^*d$. Let $[\alpha]\in H^{p,q}(X)$, with $\alpha\in\mathcal H_{\bar\partial}^{p,q}(X)$. By [citetheorem:8055], $\alpha$ is $d$-harmonic. Since $c$ commutes with $\Delta_d$, $\overline{\alpha}$ is also $d$-harmonic. Its type is $(q,p)$, so applying [citetheorem:8055] once more gives $\overline{\alpha}\in\mathcal H_{\bar\partial}^{q,p}(X)$. Hence \begin{align*} \overline{H^{p,q}(X)}\subset H^{q,p}(X). \end{align*} Applying the same argument to a class in $H^{q,p}(X)$ and using $c^2=\operatorname{id}$ gives the reverse inclusion. Therefore \begin{align*} \overline{H^{p,q}(X)}=H^{q,p}(X). \end{align*} [/step]

[step:Conclude that the decomposition is a pure Hodge structure] Let $V_{\mathbb Q}:=H^k(X,\mathbb Q)$. The theorem statement identifies the complexification $V_{\mathbb Q}\otimes_{\mathbb Q}\mathbb C$ with $H^k(X,\mathbb C)$ via extension of scalars. \begin{align*} V_{\mathbb Q}\otimes_{\mathbb Q}\mathbb C \cong H^k(X,\mathbb C). \end{align*} The preceding steps give a direct sum decomposition \begin{align*} V_{\mathbb Q}\otimes_{\mathbb Q}\mathbb C=\bigoplus_{p+q=k}H^{p,q}(X) \end{align*} and complex conjugation with respect to the rational vector space $V_{\mathbb Q}$ satisfies \begin{align*} \overline{H^{p,q}(X)}=H^{q,p}(X). \end{align*} These are precisely the decomposition and conjugation axioms for a pure Hodge structure of weight $k$ on $V_{\mathbb Q}$. Thus $H^k(X,\mathbb Q)$ carries the asserted pure Hodge structure. [/step]