[proofplan] We prove the equality by constructing an anti-linear bijection between the [harmonic representatives](/theorems/2747) of bidegrees $(p,q)$ and $(q,p)$. Complex conjugation sends smooth $(p,q)$-forms to smooth $(q,p)$-forms and is an involution. On a compact Kähler manifold, Dolbeault harmonicity is equivalent to de Rham harmonicity, and de Rham harmonicity is preserved by complex conjugation. The [Dolbeault Hodge theorem](/theorems/8059) then identifies the dimensions of these harmonic spaces with the Hodge numbers. [/proofplan] [step:Use complex conjugation to exchange the two bidegrees] Let $n=\dim_{\mathbb C}X$. By the theorem statement, $A^{p,q}(X)$ denotes the complex [vector space](/page/Vector%20Space) of smooth complex-valued differential forms of type $(p,q)$ on $X$, with $A^{p,q}(X)=0$ if $p>n$ or $q>n$. If $p>n$ or $q>n$, then both $A^{p,q}(X)$ and $A^{q,p}(X)$ are zero. Therefore $H^{p,q}_{\bar\partial}(X)=0$ and $H^{q,p}_{\bar\partial}(X)=0$, so $h^{p,q}(X)=0=h^{q,p}(X)$. We therefore assume $0\leq p,q\leq n$. Let \begin{align*} C_{p,q}:A^{p,q}(X)&\longrightarrow A^{q,p}(X) \end{align*} denote complex conjugation on smooth differential forms, so $C_{p,q}(\alpha)=\overline{\alpha}$. In a holomorphic coordinate chart $(U,z)$ with $z_j=x_j+iy_j$, let $I=(i_1,\dots,i_p)$ and $J=(j_1,\dots,j_q)$ be strictly increasing multi-indices with entries in $\{1,\dots,n\}$. A local $(p,q)$-form is a finite sum of terms \begin{align*} f_{I,J}\,dz_{i_1}\wedge\cdots\wedge dz_{i_p}\wedge d\overline{z}_{j_1}\wedge\cdots\wedge d\overline{z}_{j_q}, \end{align*} where $f_{I,J}:U\to\mathbb C$ is smooth, and conjugation sends this term to \begin{align*} \overline{f_{I,J}}\,d\overline{z}_{i_1}\wedge\cdots\wedge d\overline{z}_{i_p}\wedge dz_{j_1}\wedge\cdots\wedge dz_{j_q}, \end{align*} which has type $(q,p)$ after reordering the wedge factors. Thus $C_{p,q}$ is an anti-[linear map](/page/Linear%20Map). Since conjugating twice gives the identity, $C_{q,p}\circ C_{p,q}=\operatorname{id}_{A^{p,q}(X)}$, so $C_{p,q}$ is an anti-linear bijection with inverse $C_{q,p}$. [/step] [step:Show conjugation preserves harmonic representatives] By the theorem statement, $\mathcal H^{p,q}_{\bar\partial}(X)\subset A^{p,q}(X)$ denotes the space of $\bar\partial$-harmonic $(p,q)$-forms for the Kähler metric $g$ fixed in the theorem statement. We claim that complex conjugation restricts to an anti-linear bijection \begin{align*} C_{p,q}:\mathcal H^{p,q}_{\bar\partial}(X)&\longrightarrow \mathcal H^{q,p}_{\bar\partial}(X). \end{align*} Let $\alpha\in\mathcal H^{p,q}_{\bar\partial}(X)$. Since $X$ is compact Kähler and all harmonic spaces and adjoints are taken with respect to the Kähler metric $g$ induced by $\omega$, [citetheorem:8055] gives the equivalence between $\bar\partial$-harmonicity and $d$-harmonicity for $\alpha$. Hence $\alpha$ is $d$-harmonic. The underlying real Riemannian manifold of $(X,g)$ is compact and oriented, so [citetheorem:8056] applies. Therefore \begin{align*} d\alpha=0. \end{align*} Also, \begin{align*} d^*\alpha=0. \end{align*} The [exterior derivative](/theorems/1525) is real, so $d\overline{\alpha}=\overline{d\alpha}=0$. Also the formal adjoint $d^*$ commutes with complex conjugation for the real Riemannian metric underlying the Hermitian metric. Indeed, complex conjugation is antiunitary for the $L^2$ [inner product](/page/Inner%20Product) by [citetheorem:8051]. Since $\alpha\in A^{p,q}(X)$ has total degree $p+q$, the defining identity for $d^*$ gives, for every smooth form $\gamma\in A^{p+q-1}(X;\mathbb C)$, \begin{align*} (d^*\overline{\alpha},\gamma)_{L^2}=(\overline{\alpha},d\gamma)_{L^2}. \end{align*} Writing $\gamma=\overline{\eta}$ with $\eta=\overline{\gamma}$, using $d\gamma=\overline{d\eta}$, and applying antiunitarity of conjugation, the right-hand side is zero because $d^*\alpha=0$. Therefore $d^*\overline{\alpha}=0$. Again by [citetheorem:8056], $\overline{\alpha}$ is $d$-harmonic, and by [citetheorem:8055] it is $\bar\partial$-harmonic. Since conjugation changes type from $(p,q)$ to $(q,p)$, we have $\overline{\alpha}\in\mathcal H^{q,p}_{\bar\partial}(X)$. The inverse map is the same construction in the opposite bidegree, because complex conjugation is involutive. Hence $C_{p,q}$ restricts to the required anti-linear bijection. [guided] We want to compare $H^{p,q}(X)$ with $H^{q,p}(X)$, and the natural operation that exchanges the two indices is complex conjugation. The point that needs proof is not merely that conjugation changes type, but that it also preserves the special representatives used by Hodge theory: the harmonic forms. Let $\mathcal H^{p,q}_{\bar\partial}(X)$ be the space of $\bar\partial$-harmonic smooth $(p,q)$-forms. Take an element $\alpha\in\mathcal H^{p,q}_{\bar\partial}(X)$. Because $X$ is compact Kähler and the harmonicity notions are all defined using the Kähler metric $g$ associated to $\omega$, [citetheorem:8055] applies and says that $\bar\partial$-harmonicity and $d$-harmonicity are equivalent. Thus $\alpha$ is $d$-harmonic. The underlying real Riemannian manifold of $(X,g)$ is compact and oriented, so [citetheorem:8056] applies. Hence $d$-harmonicity is equivalent to the two first-order equations \begin{align*} d\alpha=0. \end{align*} Also, \begin{align*} d^*\alpha=0. \end{align*} Now we check these two equations after conjugating. Since the exterior derivative $d$ is defined over the real smooth structure, it commutes with complex conjugation on complex-valued forms. Hence \begin{align*} d\overline{\alpha}=\overline{d\alpha}=0. \end{align*} For the adjoint equation, we use the $L^2$ inner product induced by the Hermitian metric. Since $\alpha\in A^{p,q}(X)$ has total degree $p+q$, let $\gamma\in A^{p+q-1}(X;\mathbb C)$ be arbitrary, and set $\eta=\overline{\gamma}$. The formal adjoint relation gives \begin{align*} (d^*\overline{\alpha},\gamma)_{L^2}=(\overline{\alpha},d\gamma)_{L^2}. \end{align*} Since $\gamma=\overline{\eta}$ and $d$ commutes with conjugation, we have $d\gamma=\overline{d\eta}$. By [citetheorem:8051], complex conjugation is antiunitary for the $L^2$ inner product on forms. Therefore the last inner product is the conjugate of an inner product involving $\alpha$ and $d\eta$. Because $d^*\alpha=0$, the formal adjoint identity gives \begin{align*} (\alpha,d\eta)_{L^2}=(d^*\alpha,\eta)_{L^2}=0. \end{align*} Consequently \begin{align*} (d^*\overline{\alpha},\gamma)_{L^2}=0 \end{align*} for every such test form $\gamma$. Nondegeneracy of the $L^2$ inner product implies $d^*\overline{\alpha}=0$. Thus $\overline{\alpha}$ satisfies \begin{align*} d\overline{\alpha}=0. \end{align*} Also, \begin{align*} d^*\overline{\alpha}=0. \end{align*} By [citetheorem:8056], $\overline{\alpha}$ is $d$-harmonic. Applying the Kähler harmonicity equivalence [citetheorem:8055] once more, $\overline{\alpha}$ is $\bar\partial$-harmonic. Since conjugation sends type $(p,q)$ to type $(q,p)$, this proves \begin{align*} \overline{\alpha}\in\mathcal H^{q,p}_{\bar\partial}(X). \end{align*} Finally, conjugation is anti-linear because $\overline{a\alpha+b\beta}=\overline a\,\overline\alpha+\overline b\,\overline\beta$ for $a,b\in\mathbb C$ and smooth forms $\alpha,\beta$ of the same type. In local holomorphic coordinates it exchanges each factor $dz_j$ with $d\overline z_j$, so it sends forms of type $(p,q)$ to forms of type $(q,p)$. It is also an involution: applying it twice returns the original form. Therefore the same operation gives the inverse map from $\mathcal H^{q,p}_{\bar\partial}(X)$ back to $\mathcal H^{p,q}_{\bar\partial}(X)$. Hence complex conjugation is an anti-linear bijection between the two harmonic spaces. [/guided] [/step] [step:Identify Hodge numbers with dimensions of harmonic spaces] Since $X$ is compact and $g$ is the fixed Hermitian metric on $X$, the Dolbeault [Hodge theorem](/theorems/3942) [citetheorem:8059] applies. Therefore the natural map \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X)\longrightarrow H^{p,q}_{\bar\partial}(X) \end{align*} is an isomorphism. Therefore \begin{align*} h^{p,q}(X)=\dim_{\mathbb C}H^{p,q}_{\bar\partial}(X)=\dim_{\mathbb C}\mathcal H^{p,q}_{\bar\partial}(X). \end{align*} The same argument in bidegree $(q,p)$ gives \begin{align*} h^{q,p}(X)=\dim_{\mathbb C}\mathcal H^{q,p}_{\bar\partial}(X). \end{align*} The anti-linear bijection constructed above preserves complex dimension: if $(e_1,\dots,e_m)$ is a complex basis of $\mathcal H^{p,q}_{\bar\partial}(X)$, then $(\overline{e_1},\dots,\overline{e_m})$ is a complex basis of $\mathcal H^{q,p}_{\bar\partial}(X)$. Hence \begin{align*} \dim_{\mathbb C}\mathcal H^{p,q}_{\bar\partial}(X)=\dim_{\mathbb C}\mathcal H^{q,p}_{\bar\partial}(X). \end{align*} Combining the displayed equalities gives \begin{align*} h^{p,q}(X)=h^{q,p}(X). \end{align*} This proves the theorem. [/step]