[proofplan] We represent primitive Hodge classes by harmonic differential forms of the same Hodge type. The [Kähler identities](/theorems/3853) make the Lefschetz operator compatible with [harmonic representatives](/theorems/2747), so a cohomologically primitive class has a primitive harmonic representative. The pointwise Hodge--Riemann relation for primitive forms then identifies the signed cup-product integral with a positive constant times the $L^2$ norm squared of that representative. Orthogonality of different Hodge types follows from bidegree considerations, and the primitive decomposition follows because $L$ has Hodge type $(1,1)$. [/proofplan] [step:Choose harmonic representatives compatible with Hodge type] Let \begin{align*} A^{p,q}(X) \end{align*} denote the space of smooth complex-valued differential forms of type $(p,q)$ on $X$. Let \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X):=\{\eta\in A^{p,q}(X):\Delta_{\bar\partial}\eta=0\} \end{align*} denote the space of $\bar\partial$-harmonic forms of type $(p,q)$ for the Kähler metric determined by $\omega$, where $\Delta_{\bar\partial}:A^{p,q}(X)\to A^{p,q}(X)$ is the $\bar\partial$-Laplacian associated to that metric. By the [Dolbeault Hodge theorem](/theorems/8059) [citetheorem:8059], every class $\alpha\in H^{p,q}(X)$ has a unique representative \begin{align*} a\in\mathcal H^{p,q}_{\bar\partial}(X) \end{align*} whose Dolbeault cohomology class is $\alpha$. Since $X$ is Kähler, the [harmonicity criteria on compact Kähler manifolds](/theorems/8055) [citetheorem:8055] identify $\bar\partial$-harmonic, $\partial$-harmonic, and $d$-harmonic forms. Thus $a$ is also $d$-harmonic and represents the corresponding de Rham cohomology class of $\alpha$. [guided] We first replace cohomology classes by canonical differential forms. Let \begin{align*} A^{p,q}(X) \end{align*} be the space of smooth complex-valued differential forms of type $(p,q)$ on $X$. Let $\Delta_{\bar\partial}:A^{p,q}(X)\to A^{p,q}(X)$ be the $\bar\partial$-Laplacian associated to the Kähler metric determined by $\omega$, and define \begin{align*} \mathcal H^{p,q}_{\bar\partial}(X):=\{\eta\in A^{p,q}(X):\Delta_{\bar\partial}\eta=0\}. \end{align*} The Dolbeault [Hodge theorem](/theorems/3942) [citetheorem:8059] applies because $X$ is compact Hermitian, hence compact Kähler in particular. It gives a unique $\bar\partial$-harmonic representative \begin{align*} a\in\mathcal H^{p,q}_{\bar\partial}(X) \end{align*} of every Dolbeault class $\alpha\in H^{p,q}(X)$. Why is this the correct representative for the cup-product integral? The integral in the theorem is a de Rham pairing, so we need the same form also to represent the associated de Rham class. On a compact Kähler manifold, the harmonicity criteria [citetheorem:8055] say that $d$-, $\partial$-, and $\bar\partial$-harmonicity are equivalent. Therefore the chosen form $a$ is $d$-harmonic. In particular, $a$ is $d$-closed and represents the de Rham class underlying $\alpha$. [/guided] [/step] [step:Show that primitive classes have primitive harmonic representatives] Let $\alpha\in P^{p,q}(X)$ and let \begin{align*} a\in\mathcal H^{p,q}_{\bar\partial}(X) \end{align*} be its harmonic representative. Let \begin{align*} L:A^m(X)\to A^{m+2}(X) \end{align*} denote exterior multiplication by the Kähler form $\omega$, so that \begin{align*} L(\eta)=\omega\wedge\eta \end{align*} for every smooth $m$-form $\eta$. Since $\omega$ is a Kähler form, $L$ has bidegree $(1,1)$. The Kähler identities, in the form of the Lefschetz commutation relations for Kähler Laplacians, state that $L$ commutes with $\Delta_{\bar\partial}$. Hence $L^{n-k+1}a$ is the harmonic representative of the class \begin{align*} L^{n-k+1}\alpha\in H^{n-q+1,n-p+1}(X). \end{align*} Because $\alpha$ is primitive, this class is zero. The unique harmonic representative of the zero class is the zero form, so \begin{align*} L^{n-k+1}a=0. \end{align*} Thus $a$ is primitive as a differential form. [guided] The primitive condition in the statement is a cohomological condition: \begin{align*} L^{n-k+1}\alpha=0. \end{align*} To use the pointwise Hodge--Riemann relation, we need the corresponding differential form to be primitive. Let \begin{align*} L:A^m(X)\to A^{m+2}(X) \end{align*} be the differential-form Lefschetz operator defined by \begin{align*} L(\eta)=\omega\wedge\eta. \end{align*} Since $\omega$ is a closed $(1,1)$-form, this operator induces on cohomology the cup product operator $\gamma\mapsto[\omega]\smile\gamma$, and it sends forms of type $(u,v)$ to forms of type $(u+1,v+1)$. The Kähler identities, equivalently the Lefschetz commutation relations for Kähler Laplacians, imply that $L$ commutes with $\Delta_{\bar\partial}$. Therefore, if \begin{align*} a\in\mathcal H^{p,q}_{\bar\partial}(X) \end{align*} is the harmonic representative of $\alpha$, then \begin{align*} L^{n-k+1}a \end{align*} is harmonic and represents the cohomology class $L^{n-k+1}\alpha$. But $\alpha\in P^{p,q}(X)$ means precisely that this cohomology class is zero. Hodge uniqueness then forces the harmonic representative of that zero class to be the zero form. Hence \begin{align*} L^{n-k+1}a=0. \end{align*} This is exactly the primitive condition for the representative $a$ as a differential form. [/guided] [/step] [step:Convert the signed top-degree integral into an $L^2$ norm] Let $\alpha\in P^{p,q}(X)$ be nonzero, and let \begin{align*} a\in\mathcal H^{p,q}_{\bar\partial}(X) \end{align*} be its primitive harmonic representative. The pointwise Hodge--Riemann relation for primitive forms [citetheorem:8054], applied with the Kähler metric $g$ determined by $\omega$ and its standard Riemannian volume form, gives the pointwise identity \begin{align*} i^{p-q}(-1)^{k(k-1)/2}a\wedge\overline{a}\wedge\omega^{n-k}=(n-k)! |a|^2_g\,d\operatorname{vol}_g, \end{align*} where $|\cdot|_g$ is the pointwise norm induced by the Kähler metric and $\operatorname{vol}_g$ is its Riemannian volume measure. Pairing the left-hand side with the fundamental class $[X]$ gives \begin{align*} h_k(\alpha,\alpha)=(n-k)!\int_X |a|^2_g\,d\operatorname{vol}_g. \end{align*} The right-hand side is positive because $(n-k)!>0$ and $a\ne 0$. The implication $a\ne 0$ follows from uniqueness of harmonic representatives: if $a=0$, then $\alpha=0$, contrary to the hypothesis. [guided] We now use the pointwise linear algebra. The representative \begin{align*} a\in A^{p,q}(X) \end{align*} is primitive by the previous step, so at every point $x\in X$ the value $a_x$ is a primitive $(p,q)$-form on the Hermitian [vector space](/page/Vector%20Space) $T_xX$. The pointwise Hodge--Riemann relation for primitive forms [citetheorem:8054], with the standard metric normalization induced by $g$, applies to this Hermitian vector space and gives \begin{align*} i^{p-q}(-1)^{k(k-1)/2}a\wedge\overline{a}\wedge\omega^{n-k}=(n-k)! |a|^2_g\,d\operatorname{vol}_g. \end{align*} Here $|\cdot|_g$ denotes the pointwise Hermitian norm on forms induced by the Kähler metric, and $\operatorname{vol}_g$ denotes the corresponding Riemannian volume measure. Because $a$ represents $\alpha$ and $\overline{a}$ represents $\overline{\alpha}$, the cohomological top-degree pairing with the fundamental class $[X]$ equals the integral of the differential form against this volume measure: \begin{align*} i^{p-q}(-1)^{k(k-1)/2}\left\langle \alpha\smile\overline{\alpha}\smile[\omega]^{n-k},[X]\right\rangle=(n-k)!\int_X |a|^2_g\,d\operatorname{vol}_g. \end{align*} The factor $(n-k)!$ is strictly positive. The integral on the right is the squared $L^2$ norm of $a$, multiplied by this positive factor. Since the harmonic representative of a nonzero cohomology class cannot be the zero form, $a\ne 0$, and therefore \begin{align*} \int_X |a|^2_g\,d\operatorname{vol}_g>0. \end{align*} It follows that \begin{align*} h_k(\alpha,\alpha)>0. \end{align*} [/guided] [/step] [step:Prove orthogonality of distinct primitive Hodge types] Let $\alpha\in P^{p,q}(X)$ and $\beta\in P^{r,s}(X)$ with $p+q=r+s=k$. Let \begin{align*} a\in\mathcal H^{p,q}_{\bar\partial}(X),\qquad b\in\mathcal H^{r,s}_{\bar\partial}(X) \end{align*} be their harmonic representatives. The form \begin{align*} a\wedge\overline{b}\wedge\omega^{n-k} \end{align*} has type \begin{align*} (p+s,q+r)+(n-k,n-k). \end{align*} A top-degree form on $X$ has type $(n,n)$. Thus the integral can be nonzero only if \begin{align*} p+s+n-k=n,\qquad q+r+n-k=n. \end{align*} Equivalently, \begin{align*} p+s=k,\qquad q+r=k. \end{align*} Since $r+s=k$ and $p+q=k$, these equalities imply $p=r$ and $q=s$. Therefore, if $(p,q)\ne(r,s)$, the integrand has no $(n,n)$ component, and \begin{align*} h_k(\alpha,\beta)=0. \end{align*} [/step] [step:Decompose the primitive space by Hodge type] By the [Hodge decomposition](/theorems/2745) of compact Kähler cohomology, \begin{align*} H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X). \end{align*} The Lefschetz operator $L$ has Hodge type $(1,1)$, so \begin{align*} L^{n-k+1}H^{p,q}(X)\subset H^{n-q+1,n-p+1}(X) \end{align*} for every $p+q=k$. Since the [Hodge decomposition](/theorems/3941) in degree $2n-k+2$ is direct, a class \begin{align*} \gamma=\sum_{p+q=k}\gamma_{p,q},\qquad \gamma_{p,q}\in H^{p,q}(X), \end{align*} satisfies $L^{n-k+1}\gamma=0$ if and only if \begin{align*} L^{n-k+1}\gamma_{p,q}=0 \end{align*} for every $p,q$ with $p+q=k$. Hence \begin{align*} P^k(X)=\bigoplus_{p+q=k}P^{p,q}(X). \end{align*} The orthogonality proved in the previous step shows that this direct sum is orthogonal with respect to the sesquilinear form $h_k$. Together with the positivity step, this proves the Hodge--Riemann bilinear relations for primitive Hodge classes. [/step]