[proofplan] Choose the Kähler metric on $X$ and use its Hermitian inner product to define the formal adjoints and Laplacians on complex-valued differential forms. The elliptic Hodge theorem identifies de Rham cohomology with $d$-harmonic forms, while the [Kähler identities](/theorems/3853) identify $d$-harmonicity with $\bar\partial$-harmonicity up to the factor $2$ on each type component. Since the Laplacians preserve bidegree, the harmonic representative of a de Rham class decomposes uniquely into harmonic $(p,q)$-pieces, and the Dolbeault Hodge theorem identifies those pieces with Dolbeault cohomology. Finally, complex conjugation exchanges bidegrees and carries [harmonic representatives](/theorems/2747) to harmonic representatives, giving the stated conjugation symmetry. [/proofplan]

[step:Define the Laplacians associated to the Kähler metric]Let $g$ be the Riemannian metric associated to the Kähler form on $X$, and let $dV_g$ denote its Riemannian volume measure. For each integer $r$ with $0 \leq r \leq 2n$, let \begin{align*} \Omega^r(X;\mathbb C) := C^\infty(X;\Lambda^r T^*X \otimes_{\mathbb R} \mathbb C) \end{align*} denote the space of smooth complex-valued $r$-forms on $X$. For each pair of integers $(p,q)$ with $0 \leq p,q \leq n$, let \begin{align*} \Omega^{p,q}(X) := C^\infty(X;\Lambda^{p,q}T^*X) \end{align*} denote the space of smooth forms of type $(p,q)$. The complex structure gives the direct-sum decomposition \begin{align*} \Omega^k(X;\mathbb C)=\bigoplus_{p+q=k}\Omega^{p,q}(X). \end{align*} Define the [exterior derivative](/theorems/1525) \begin{align*} d: \Omega^r(X;\mathbb C) &\to \Omega^{r+1}(X;\mathbb C), \end{align*} and define its type components \begin{align*} \partial: \Omega^{p,q}(X) &\to \Omega^{p+1,q}(X), \\ \bar\partial: \Omega^{p,q}(X) &\to \Omega^{p,q+1}(X), \end{align*} so that $d=\partial+\bar\partial$. The metric $g$ and the measure $dV_g$ define the $L^2$ inner product \begin{align*} (\alpha,\beta)_{L^2}:=\int_X h_g(\alpha,\beta)\,dV_g, \end{align*} where $h_g$ is the pointwise Hermitian inner product on complex-valued forms induced by $g$. Since $X$ is compact and has no boundary, the formal adjoints \begin{align*} d^*:\Omega^{r+1}(X;\mathbb C)&\to\Omega^r(X;\mathbb C), \\ \partial^*:\Omega^{p+1,q}(X)&\to\Omega^{p,q}(X), \\ \bar\partial^*:\Omega^{p,q+1}(X)&\to\Omega^{p,q}(X) \end{align*} are defined by this $L^2$ inner product. Define the Laplacians \begin{align*} \Delta_d&:=dd^*+d^*d, \\ \Delta_\partial&:=\partial\partial^*+\partial^*\partial, \\ \Delta_{\bar\partial}&:=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial. \end{align*} A form $\alpha\in\Omega^k(X;\mathbb C)$ is called $d$-harmonic when $\Delta_d\alpha=0$, and a form $\beta\in\Omega^{p,q}(X)$ is called $\bar\partial$-harmonic when $\Delta_{\bar\partial}\beta=0$.[/step]

[guided]The first point is to make the analytic objects precise. The Kähler structure supplies a Hermitian metric, hence a Riemannian metric $g$, and compactness of $X$ supplies a finite Riemannian volume measure $dV_g$. These data give the $L^2$ inner product \begin{align*} (\alpha,\beta)_{L^2}:=\int_X h_g(\alpha,\beta)\,dV_g, \end{align*} where $h_g$ is the pointwise Hermitian inner product on complex-valued differential forms. For each integer $r$ with $0 \leq r \leq 2n$, the space of smooth complex-valued $r$-forms is \begin{align*} \Omega^r(X;\mathbb C) := C^\infty(X;\Lambda^r T^*X \otimes_{\mathbb R} \mathbb C). \end{align*} For each bidegree $(p,q)$ with $0 \leq p,q \leq n$, the space of smooth $(p,q)$-forms is \begin{align*} \Omega^{p,q}(X) := C^\infty(X;\Lambda^{p,q}T^*X). \end{align*} The complex structure decomposes complex-valued $k$-forms by type: \begin{align*} \Omega^k(X;\mathbb C)=\bigoplus_{p+q=k}\Omega^{p,q}(X). \end{align*} The exterior derivative is the map \begin{align*} d: \Omega^r(X;\mathbb C) &\to \Omega^{r+1}(X;\mathbb C). \end{align*} On a complex manifold it splits by type into \begin{align*} \partial: \Omega^{p,q}(X) &\to \Omega^{p+1,q}(X), \\ \bar\partial: \Omega^{p,q}(X) &\to \Omega^{p,q+1}(X), \end{align*} with $d=\partial+\bar\partial$. Because $X$ is compact and has no boundary, [integration by parts](/theorems/2098) defines the formal adjoints \begin{align*} d^*:\Omega^{r+1}(X;\mathbb C)&\to\Omega^r(X;\mathbb C), \\ \partial^*:\Omega^{p+1,q}(X)&\to\Omega^{p,q}(X), \\ \bar\partial^*:\Omega^{p,q+1}(X)&\to\Omega^{p,q}(X). \end{align*} The associated Laplacians are \begin{align*} \Delta_d&:=dd^*+d^*d, \\ \Delta_\partial&:=\partial\partial^*+\partial^*\partial, \\ \Delta_{\bar\partial}&:=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial. \end{align*} Thus $d$-harmonic means $\Delta_d\alpha=0$, and $\bar\partial$-harmonic means $\Delta_{\bar\partial}\beta=0$. These are the representatives that will realize the two cohomology theories.[/guided]

[step:Identify de Rham classes with unique $d$-harmonic forms]Apply the [elliptic Hodge theorem](/theorems/???) to the elliptic complex $(\Omega^\bullet(X;\mathbb C),d)$. Its hypotheses hold because $X$ is compact, smooth, and equipped with the Riemannian metric $g$. Therefore every class in $H^k_{\mathrm{dR}}(X;\mathbb C)$ has a unique representative in \begin{align*} \mathcal H^k_d(X):=\{\alpha\in\Omega^k(X;\mathbb C):\Delta_d\alpha=0\}. \end{align*} Equivalently, the map \begin{align*} \mathcal H^k_d(X)&\to H^k_{\mathrm{dR}}(X;\mathbb C), \\ \alpha&\mapsto [\alpha]_{\mathrm{dR}} \end{align*} is an isomorphism of complex vector spaces.[/step]

[guided]We now use compact elliptic theory. The [elliptic Hodge theorem](/theorems/???) applies to the de Rham elliptic complex because $X$ is a compact smooth manifold and the Kähler metric supplies a smooth Riemannian metric $g$. The theorem says that each de Rham cohomology class has exactly one harmonic representative. Define the $d$-harmonic space in degree $k$ by \begin{align*} \mathcal H^k_d(X):=\{\alpha\in\Omega^k(X;\mathbb C):\Delta_d\alpha=0\}. \end{align*} The Hodge theorem gives the isomorphism \begin{align*} \mathcal H^k_d(X)&\to H^k_{\mathrm{dR}}(X;\mathbb C), \\ \alpha&\mapsto [\alpha]_{\mathrm{dR}}. \end{align*} This proves the first assertion of the theorem: the representative exists by surjectivity of this map and is unique by injectivity of this map.[/guided]

[step:Use the Kähler identities to compare $d$-harmonic and $\bar\partial$-harmonic forms]Since $X$ is Kähler, the [Kähler identities](/theorems/???) imply \begin{align*} \Delta_d=2\Delta_\partial=2\Delta_{\bar\partial}. \end{align*} Moreover $\Delta_{\bar\partial}$ preserves bidegree because $\bar\partial$ has bidegree $(0,1)$ and $\bar\partial^*$ has bidegree $(0,-1)$. Hence $\Delta_d$ also preserves the decomposition by type. If \begin{align*} \alpha=\sum_{p+q=k}\alpha_{p,q},\qquad \alpha_{p,q}\in\Omega^{p,q}(X), \end{align*} then \begin{align*} \Delta_d\alpha=0 &\iff 2\sum_{p+q=k}\Delta_{\bar\partial}\alpha_{p,q}=0 \\ &\iff \Delta_{\bar\partial}\alpha_{p,q}=0\quad\text{for every }(p,q)\text{ with }p+q=k. \end{align*} The last equivalence holds because the decomposition by bidegree is a direct sum. Therefore \begin{align*} \mathcal H^k_d(X)=\bigoplus_{p+q=k}\mathcal H^{p,q}_{\bar\partial}(X), \end{align*} where \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X):=\{\beta\in\Omega^{p,q}(X):\Delta_{\bar\partial}\beta=0\}. \end{align*}[/step]

[guided]This is the point where the Kähler condition is used. On an arbitrary compact complex manifold, the de Rham Laplacian and the Dolbeault Laplacian need not agree. On a Kähler manifold, the [Kähler identities](/theorems/???) give the precise comparison \begin{align*} \Delta_d=2\Delta_\partial=2\Delta_{\bar\partial}. \end{align*} The factor $2$ is harmless for kernels: $\Delta_d\alpha=0$ if and only if $\Delta_{\bar\partial}\alpha=0$. We also need compatibility with type. The operator $\bar\partial$ has bidegree $(0,1)$ and its formal adjoint $\bar\partial^*$ has bidegree $(0,-1)$. Consequently both compositions $\bar\partial\bar\partial^*$ and $\bar\partial^*\bar\partial$ preserve bidegree, so $\Delta_{\bar\partial}$ preserves each space $\Omega^{p,q}(X)$. Since $\Delta_d=2\Delta_{\bar\partial}$, the operator $\Delta_d$ also preserves bidegree. Write a $k$-form $\alpha\in\Omega^k(X;\mathbb C)$ uniquely as \begin{align*} \alpha=\sum_{p+q=k}\alpha_{p,q},\qquad \alpha_{p,q}\in\Omega^{p,q}(X). \end{align*} Then \begin{align*} \Delta_d\alpha=0 &\iff 2\sum_{p+q=k}\Delta_{\bar\partial}\alpha_{p,q}=0 \\ &\iff \Delta_{\bar\partial}\alpha_{p,q}=0\quad\text{for every }(p,q)\text{ with }p+q=k, \end{align*} because the decomposition into bidegrees is a direct sum. Defining \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X):=\{\beta\in\Omega^{p,q}(X):\Delta_{\bar\partial}\beta=0\}, \end{align*} we obtain \begin{align*} \mathcal H^k_d(X)=\bigoplus_{p+q=k}\mathcal H^{p,q}_{\bar\partial}(X). \end{align*}[/guided]

[step:Identify the harmonic type components with Dolbeault cohomology]Apply the [Dolbeault Hodge theorem](/theorems/???) to the elliptic complex $(\Omega^{p,\bullet}(X),\bar\partial)$. Its hypotheses hold because $X$ is compact Hermitian, and every Kähler metric is Hermitian. Therefore the map \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X)&\to H^{p,q}_{\bar\partial}(X), \\ \beta&\mapsto [\beta]_{\bar\partial} \end{align*} is an isomorphism. Combining this with \begin{align*} \mathcal H^k_d(X)=\bigoplus_{p+q=k}\mathcal H^{p,q}_{\bar\partial}(X) \end{align*} and the de Rham Hodge isomorphism gives \begin{align*} H^k_{\mathrm{dR}}(X;\mathbb C) \cong \mathcal H^k_d(X) \cong \bigoplus_{p+q=k}\mathcal H^{p,q}_{\bar\partial}(X) \cong \bigoplus_{p+q=k}H^{p,q}_{\bar\partial}(X). \end{align*} This is precisely the [Hodge decomposition](/theorems/2745) isomorphism induced by taking the type components of the unique harmonic representative.[/step]

[guided]The preceding step decomposed harmonic de Rham representatives into harmonic pieces of type $(p,q)$. We now translate those harmonic pieces into Dolbeault cohomology classes. For fixed $p$, the Dolbeault complex is \begin{align*} \Omega^{p,0}(X)\xrightarrow{\bar\partial}\Omega^{p,1}(X)\xrightarrow{\bar\partial}\cdots\xrightarrow{\bar\partial}\Omega^{p,n}(X). \end{align*} The [Dolbeault Hodge theorem](/theorems/???) applies because $X$ is compact and the Kähler metric is in particular a Hermitian metric. It gives an isomorphism \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X)&\to H^{p,q}_{\bar\partial}(X), \\ \beta&\mapsto [\beta]_{\bar\partial}. \end{align*} Combining the three identifications gives \begin{align*} H^k_{\mathrm{dR}}(X;\mathbb C) \cong \mathcal H^k_d(X) \cong \bigoplus_{p+q=k}\mathcal H^{p,q}_{\bar\partial}(X) \cong \bigoplus_{p+q=k}H^{p,q}_{\bar\partial}(X). \end{align*} The phrase “induced by decomposition of forms by type” means the following concrete map: take a de Rham class, choose its unique $d$-harmonic representative $\alpha$, write $\alpha=\sum_{p+q=k}\alpha_{p,q}$, and send the class to the tuple $([\alpha_{p,q}]_{\bar\partial})_{p+q=k}$. The previous steps prove that each $\alpha_{p,q}$ is $\bar\partial$-harmonic and that this construction is an isomorphism.[/guided]

[step:Show complex conjugation exchanges the Dolbeault summands]Define complex conjugation on forms by the antilinear map \begin{align*} C:\Omega^{p,q}(X)&\to\Omega^{q,p}(X), \\ \beta&\mapsto \overline{\beta}. \end{align*} It satisfies \begin{align*} C\bar\partial=\partial C. \end{align*} Because the Kähler metric is real as a Riemannian metric, $C$ commutes with $\Delta_d$. Using $\Delta_d=2\Delta_{\bar\partial}$ on the source and target bidegrees, if $\beta\in\mathcal H^{p,q}_{\bar\partial}(X)$, then \begin{align*} \Delta_{\bar\partial}(C\beta)=\frac{1}{2}\Delta_d(C\beta)=\frac{1}{2}C(\Delta_d\beta)=C(\Delta_{\bar\partial}\beta)=0. \end{align*} Thus $C$ maps $\mathcal H^{p,q}_{\bar\partial}(X)$ bijectively onto $\mathcal H^{q,p}_{\bar\partial}(X)$, with inverse again given by $C$. Passing through the Dolbeault Hodge isomorphisms gives \begin{align*} \overline{H^{p,q}_{\bar\partial}(X)}=H^{q,p}_{\bar\partial}(X). \end{align*} This completes the proof.[/step]

[guided]Complex conjugation changes a form of type $(p,q)$ into a form of type $(q,p)$. Formally, define the antilinear map \begin{align*} C:\Omega^{p,q}(X)&\to\Omega^{q,p}(X), \\ \beta&\mapsto \overline{\beta}. \end{align*} It satisfies \begin{align*} C\bar\partial=\partial C, \end{align*} because conjugating an antiholomorphic differential produces a holomorphic differential. We need to show that conjugation also preserves harmonic representatives in the correct bidegree. The Riemannian metric underlying the Kähler metric is real, so the de Rham Laplacian commutes with complex conjugation: \begin{align*} \Delta_d(C\gamma)=C(\Delta_d\gamma) \end{align*} for every smooth complex-valued form $\gamma\in\Omega^r(X;\mathbb C)$. Now let $\beta\in\mathcal H^{p,q}_{\bar\partial}(X)$. By the Kähler identity $\Delta_d=2\Delta_{\bar\partial}$, the equation $\Delta_{\bar\partial}\beta=0$ implies $\Delta_d\beta=0$. Therefore \begin{align*} \Delta_{\bar\partial}(C\beta)=\frac{1}{2}\Delta_d(C\beta)=\frac{1}{2}C(\Delta_d\beta)=C(\Delta_{\bar\partial}\beta)=0. \end{align*} Thus $C\beta\in\mathcal H^{q,p}_{\bar\partial}(X)$. Since $C^2$ is the identity map, this correspondence is bijective between $\mathcal H^{p,q}_{\bar\partial}(X)$ and $\mathcal H^{q,p}_{\bar\partial}(X)$. Finally, the Dolbeault Hodge theorem identifies harmonic representatives with Dolbeault cohomology classes. Passing the conjugation bijection through these identifications gives \begin{align*} \overline{H^{p,q}_{\bar\partial}(X)}=H^{q,p}_{\bar\partial}(X). \end{align*} This proves the conjugation symmetry and completes the proof.[/guided]