[proofplan]
The Lefschetz operator has Hodge type $(1,1)$, so its primitive kernel is compatible with the [Hodge decomposition](/theorems/2745) of compact Kähler manifolds. The Hodge-Riemann bilinear relations for primitive Hodge classes then give exactly the orthogonality and positivity conditions for the form $Q_\omega$, with the sign convention appearing in the definition of $Q_\omega$. Finally, when the Kähler class is rational, cup product by $[\omega]$ preserves rational cohomology and the top-degree evaluation pairing is rational-valued, so the same argument descends from real to rational primitive cohomology.
[/proofplan]
[step:Use Hodge type compatibility to decompose primitive cohomology]
By the [[Hodge Decomposition for Compact Kähler Manifolds](/theorems/8066)][citetheorem:8066], the complex cohomology group $H^k(X;\mathbb C)$ admits a direct-sum decomposition
\begin{align*}
H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X).
\end{align*}
Since $[\omega]\in H^{1,1}(X)\cap H^2(X;\mathbb R)$ is the Kähler class, the [[Hodge Type Compatibility of Lefschetz Maps](/theorems/8063)][citetheorem:8063] applies to the Lefschetz map $L_\omega$. Thus, for every pair $p,q$,
\begin{align*}
L_\omega\left(H^{p,q}(X)\right)\subset H^{p+1,q+1}(X).
\end{align*}
Iterating gives
\begin{align*}
L_\omega^{n-k+1}\left(H^{p,q}(X)\right)\subset H^{p+n-k+1,q+n-k+1}(X)
\end{align*}
for every $p+q=k$.
By the equivalent kernel description in the theorem statement, obtained from exactness of scalar extension from $\mathbb R$ to $\mathbb C$, a class of $P^k(X;\mathbb C)$ is precisely a complex degree-$k$ cohomology class killed by the complex-[linear map](/page/Linear%20Map) $L_\omega^{n-k+1}$. Let $\alpha\in P^k(X;\mathbb C)$, and write its [Hodge decomposition](/theorems/3941) as
\begin{align*}
\alpha=\sum_{p+q=k}\alpha_{p,q},
\end{align*}
where $\alpha_{p,q}\in H^{p,q}(X)$ for each pair $p,q$. Since $L_\omega^{n-k+1}$ preserves Hodge type up to the same shift on both indices, the decomposition of
\begin{align*}
L_\omega^{n-k+1}\alpha
\end{align*}
into Hodge summands is
\begin{align*}
L_\omega^{n-k+1}\alpha=\sum_{p+q=k}L_\omega^{n-k+1}\alpha_{p,q}.
\end{align*}
The Hodge decomposition is direct, and the left-hand side is zero because $\alpha$ is primitive. Therefore
\begin{align*}
L_\omega^{n-k+1}\alpha_{p,q}=0
\end{align*}
for every $p,q$ with $p+q=k$. Hence
\begin{align*}
\alpha_{p,q}\in P^k(X;\mathbb C)\cap H^{p,q}(X),
\end{align*}
and so
\begin{align*}
P^k(X;\mathbb C)=\bigoplus_{p+q=k}\left(P^k(X;\mathbb C)\cap H^{p,q}(X)\right).
\end{align*}
Because $L_\omega$ is defined over $\mathbb R$, complex conjugation on $H^k(X;\mathbb C)$ preserves $P^k(X;\mathbb C)$. Since the Hodge decomposition of a compact Kähler manifold satisfies $\overline{H^{p,q}(X)}=H^{q,p}(X)$, conjugation sends
\begin{align*}
P^k(X;\mathbb C)\cap H^{p,q}(X)
\end{align*}
to
\begin{align*}
P^k(X;\mathbb C)\cap H^{q,p}(X).
\end{align*}
Thus this direct-sum decomposition defines a real pure Hodge structure on $P^k(X;\mathbb R)$.
[guided]
The goal of this step is to prove that taking primitive classes does not destroy the Hodge decomposition. The Hodge decomposition itself is supplied by the [Hodge Decomposition for Compact Kähler Manifolds][citetheorem:8066], whose hypotheses are satisfied because $X$ is a compact Kähler manifold. It gives
\begin{align*}
H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X).
\end{align*}
Now we must check that the primitive condition is compatible with this direct sum. The primitive condition is defined using the operator
\begin{align*}
L_\omega^{n-k+1}:H^k(X;\mathbb C)\to H^{2n-k+2}(X;\mathbb C),
\end{align*}
where $L_\omega$ is cup product with $[\omega]$. Since $[\omega]$ is the class of a Kähler form, it has Hodge type $(1,1)$. Therefore the [Hodge Type Compatibility of Lefschetz Maps][citetheorem:8063] applies and gives
\begin{align*}
L_\omega\left(H^{p,q}(X)\right)\subset H^{p+1,q+1}(X).
\end{align*}
Applying this statement $n-k+1$ times gives
\begin{align*}
L_\omega^{n-k+1}\left(H^{p,q}(X)\right)\subset H^{p+n-k+1,q+n-k+1}(X).
\end{align*}
Before using the primitive condition, we identify the complex primitive space with the kernel of this complex-linear operator. The theorem statement defines $P^k(X;\mathbb C)$ as $P^k(X;\mathbb R)\otimes_{\mathbb R}\mathbb C$, and scalar extension from $\mathbb R$ to $\mathbb C$ is exact. Therefore
\begin{align*}
P^k(X;\mathbb C)=\ker\left(L_\omega^{n-k+1}:H^k(X;\mathbb C)\to H^{2n-k+2}(X;\mathbb C)\right).
\end{align*}
Let $\alpha\in P^k(X;\mathbb C)$. By the Hodge decomposition, there are uniquely determined classes $\alpha_{p,q}\in H^{p,q}(X)$ such that
\begin{align*}
\alpha=\sum_{p+q=k}\alpha_{p,q}.
\end{align*}
Because $\alpha$ is primitive, one has
\begin{align*}
L_\omega^{n-k+1}\alpha=0.
\end{align*}
Using linearity of $L_\omega^{n-k+1}$, this becomes
\begin{align*}
0=\sum_{p+q=k}L_\omega^{n-k+1}\alpha_{p,q}.
\end{align*}
Each summand lies in a different Hodge summand after the same shift in type. Since the Hodge decomposition is a direct sum, a sum of components in distinct Hodge summands can vanish only if every component vanishes. Hence
\begin{align*}
L_\omega^{n-k+1}\alpha_{p,q}=0
\end{align*}
for every $p,q$ with $p+q=k$. This says exactly that each $\alpha_{p,q}$ is primitive:
\begin{align*}
\alpha_{p,q}\in P^k(X;\mathbb C)\cap H^{p,q}(X).
\end{align*}
Thus primitive cohomology inherits the Hodge decomposition:
\begin{align*}
P^k(X;\mathbb C)=\bigoplus_{p+q=k}\left(P^k(X;\mathbb C)\cap H^{p,q}(X)\right).
\end{align*}
It remains to check that this is a real Hodge structure, not merely a decomposition of a complex [vector space](/page/Vector%20Space). The operator $L_\omega$ is real-linear on $H^k(X;\mathbb R)$ and its complexification commutes with complex conjugation. Therefore complex conjugation preserves its kernel $P^k(X;\mathbb C)$. The Hodge decomposition on a compact Kähler manifold also satisfies $\overline{H^{p,q}(X)}=H^{q,p}(X)$. Combining these two facts gives
\begin{align*}
\overline{P^k(X;\mathbb C)\cap H^{p,q}(X)}=P^k(X;\mathbb C)\cap H^{q,p}(X).
\end{align*}
This is exactly the conjugation condition for the above decomposition to define a real pure Hodge structure of weight $k$ on $P^k(X;\mathbb R)$.
[/guided]
[/step]
[step:Apply the Hodge-Riemann bilinear relations to the primitive summands]
We apply the [[Hodge--Riemann Bilinear Relations for Primitive Hodge Classes](/theorems/8065)][citetheorem:8065] to the compact Kähler manifold $(X,\omega)$. Its hypotheses are satisfied: $X$ is compact Kähler of complex dimension $n$, $[\omega]\in H^{1,1}(X)\cap H^2(X;\mathbb R)$ is the Kähler class, and $P^k(X;\mathbb C)$ is the primitive subspace defined as the kernel of $L_\omega^{n-k+1}$. The cited theorem uses the same normalization
\begin{align*}
Q_\omega(\alpha,\beta)=(-1)^{k(k-1)/2}\left\langle \alpha\smile\beta\smile[\omega]^{n-k},[X]\right\rangle
\end{align*}
for primitive degree-$k$ classes, so its sign convention agrees with the present definition of $Q_\omega$.
That theorem gives, for all $p,q,r,s$ with $p+q=r+s=k$,
\begin{align*}
Q_\omega\left(P^k(X;\mathbb C)\cap H^{p,q}(X),P^k(X;\mathbb C)\cap H^{r,s}(X)\right)=0
\end{align*}
unless $p=s$ and $q=r$. It also gives, for every nonzero
\begin{align*}
\alpha\in P^k(X;\mathbb C)\cap H^{p,q}(X),
\end{align*}
the positivity relation
\begin{align*}
i^{p-q}Q_\omega(\alpha,\overline{\alpha})>0.
\end{align*}
These are precisely the orthogonality and positivity conditions for $Q_\omega$ to polarize the real Hodge structure on $P^k(X;\mathbb R)$.
[guided]
We now use the result that contains the analytic content of the theorem: the [Hodge--Riemann Bilinear Relations for Primitive Hodge Classes][citetheorem:8065]. Its hypotheses match the objects in this theorem. The manifold $X$ is compact Kähler of complex dimension $n$, the class $[\omega]$ is the real $(1,1)$ Kähler class, and the primitive subspace is exactly
\begin{align*}
P^k(X;\mathbb C)=\ker\left(L_\omega^{n-k+1}:H^k(X;\mathbb C)\to H^{2n-k+2}(X;\mathbb C)\right).
\end{align*}
The normalization also matches: the [bilinear form](/page/Bilinear%20Form) used in the cited theorem is
\begin{align*}
Q_\omega(\alpha,\beta)=(-1)^{k(k-1)/2}\left\langle \alpha\smile\beta\smile[\omega]^{n-k},[X]\right\rangle.
\end{align*}
Thus there is no additional sign conversion between the cited result and the form in this theorem.
The cited Hodge-Riemann theorem gives two assertions. First, if $p+q=r+s=k$, then the primitive Hodge summands are orthogonal unless the types are conjugate:
\begin{align*}
Q_\omega\left(P^k(X;\mathbb C)\cap H^{p,q}(X),P^k(X;\mathbb C)\cap H^{r,s}(X)\right)=0
\end{align*}
unless $p=s$ and $q=r$. Second, for every nonzero class
\begin{align*}
\alpha\in P^k(X;\mathbb C)\cap H^{p,q}(X),
\end{align*}
it gives the positivity inequality
\begin{align*}
i^{p-q}Q_\omega(\alpha,\overline{\alpha})>0.
\end{align*}
These are exactly the orthogonality and positivity axioms for $Q_\omega$ to be a polarization of the real pure Hodge structure already constructed on $P^k(X;\mathbb R)$.
[/guided]
[/step]
[step:Descend the construction to rational primitive cohomology when the Kähler class is rational]
Assume now that $[\omega]$ lies in the image of $H^2(X;\mathbb Q)\to H^2(X;\mathbb R)$, and let $\omega_{\mathbb Q}\in H^2(X;\mathbb Q)$ be the chosen rational lift whose image is $[\omega]$. Then cup product with $\omega_{\mathbb Q}$ defines a $\mathbb Q$-linear map
\begin{align*}
L_{\omega_{\mathbb Q}}:H^m(X;\mathbb Q)\to H^{m+2}(X;\mathbb Q)
\end{align*}
for every integer $m$. Hence its iterate
\begin{align*}
L_{\omega_{\mathbb Q}}^{n-k+1}:H^k(X;\mathbb Q)\to H^{2n-k+2}(X;\mathbb Q)
\end{align*}
is also $\mathbb Q$-linear, so
\begin{align*}
P^k(X;\mathbb Q)=\ker L_{\omega_{\mathbb Q}}^{n-k+1}
\end{align*}
is a $\mathbb Q$-subspace of $H^k(X;\mathbb Q)$.
Complexifying this kernel gives
\begin{align*}
P^k(X;\mathbb Q)\otimes_{\mathbb Q}\mathbb C=P^k(X;\mathbb C),
\end{align*}
because kernels of finite-dimensional linear maps commute with extension of scalars from $\mathbb Q$ to $\mathbb C$. Therefore the Hodge decomposition of $P^k(X;\mathbb C)$ obtained above makes $P^k(X;\mathbb Q)$ a rational pure Hodge structure of weight $k$.
Finally, let $\alpha,\beta\in P^k(X;\mathbb Q)$. Since $\alpha$, $\beta$, and $\omega_{\mathbb Q}$ are rational cohomology classes, the cup product
\begin{align*}
\alpha\smile \beta\smile \omega_{\mathbb Q}^{n-k}\in H^{2n}(X;\mathbb Q)
\end{align*}
is rational. Evaluation on the integral fundamental class $[X]$ gives a rational number, and multiplication by the sign $(-1)^{k(k-1)/2}$ preserves rationality. Thus
\begin{align*}
Q_\omega(\alpha,\beta)\in \mathbb Q.
\end{align*}
The orthogonality and positivity identities are the same after extending scalars to $\mathbb C$, so $Q_\omega$ is a $\mathbb Q$-valued polarization of the rational Hodge structure $P^k(X;\mathbb Q)$.
[guided]
Assume that the real Kähler class $[\omega]$ comes from rational cohomology, and let $\omega_{\mathbb Q}\in H^2(X;\mathbb Q)$ be the chosen rational class whose image in $H^2(X;\mathbb R)$ is $[\omega]$. We use $\omega_{\mathbb Q}$ to define cup product on rational cohomology. Then, for every integer $m$, cup product with $\omega_{\mathbb Q}$ is a $\mathbb Q$-linear map
\begin{align*}
L_{\omega_{\mathbb Q}}:H^m(X;\mathbb Q)\to H^{m+2}(X;\mathbb Q).
\end{align*}
Therefore its iterate
\begin{align*}
L_{\omega_{\mathbb Q}}^{n-k+1}:H^k(X;\mathbb Q)\to H^{2n-k+2}(X;\mathbb Q)
\end{align*}
is also $\mathbb Q$-linear, and its kernel
\begin{align*}
P^k(X;\mathbb Q)=\ker L_{\omega_{\mathbb Q}}^{n-k+1}
\end{align*}
is a rational vector subspace of $H^k(X;\mathbb Q)$.
To compare this rational kernel with the complex primitive space, we use finite-dimensional linear algebra. Extension of scalars from $\mathbb Q$ to $\mathbb C$ is exact, so it commutes with kernels of linear maps between finite-dimensional vector spaces. Hence
\begin{align*}
P^k(X;\mathbb Q)\otimes_{\mathbb Q}\mathbb C=P^k(X;\mathbb C).
\end{align*}
The Hodge decomposition already proved for $P^k(X;\mathbb C)$ therefore equips $P^k(X;\mathbb Q)$ with a rational pure Hodge structure of weight $k$.
It remains to verify that the bilinear form is rational-valued on rational primitive classes. Let $\alpha,\beta\in P^k(X;\mathbb Q)$. Because $\alpha$, $\beta$, and $\omega_{\mathbb Q}$ all lie in rational cohomology, their cup product satisfies
\begin{align*}
\alpha\smile\beta\smile\omega_{\mathbb Q}^{n-k}\in H^{2n}(X;\mathbb Q).
\end{align*}
Pairing this rational top-degree cohomology class with the integral fundamental class $[X]$ gives an element of $\mathbb Q$. Multiplying by $(-1)^{k(k-1)/2}$ does not change rationality, so
\begin{align*}
Q_\omega(\alpha,\beta)\in \mathbb Q.
\end{align*}
After tensoring with $\mathbb C$, the orthogonality and positivity conditions are the same as in the real case. Thus $Q_\omega$ is a $\mathbb Q$-valued polarization of the rational Hodge structure $P^k(X;\mathbb Q)$.
[/guided]
[/step]
