Kähler identities give Lefschetz sl-two and hard Lefschetz

By [F3] the Laplacian commutes with $L$, $\mathrm{\Lambda }$ and $H$; hence $L$ and $\mathrm{\Lambda }$ carry harmonic forms to harmonic forms and $H$ acts on a harmonic form of degree $k$ by the scalar $k-n$ [F1]. By [F4] every real de Rham class has a unique harmonic representative, so applying the corresponding operator to that representative defines $\mathbb{R}$-linear maps $L,\mathrm{\Lambda },H$ on $H^{\ast }(X;\mathbb{R})$; on closed forms $L$ is wedge with the closed $\omega$, so on cohomology it is cup product with $[\omega ]$, and the ${\mathfrak{s}\mathfrak{l}}_2$ relations $[L,\mathrm{\Lambda }]=H$, $[H,L]=2L$, $[H,\mathrm{\Lambda }]=-2\mathrm{\Lambda }$ hold on forms by [F2], hence on harmonic forms by commutation with the Laplacian, hence pass to cohomology. Unique harmonic representatives also transport the pointwise primitive decomposition: a harmonic form $u$ of degree $k$ is uniquely ${\sum }_r{L}^r{u}_r$ with $u_r$ primitive harmonic of degree $k-2r$, $r\ge \max (0,k-n)$, because its primitive-component maps are polynomials in $L,\mathrm{\Lambda }$ by [F2], they commute with $\mathrm{\Delta }$ by [F3] and take a smooth harmonic form to smooth harmonic primitive components; conversely every sum of string elements with primitive harmonic seeds is harmonic. Since $L^{n-t+1}{u}_r=0$ and $L^r{u}_r\mathrm{\ne }0$ for $0\le r\le n-t$ when $t=\mathrm{deg}{u}_r\le n$ by [F2], the map $L^{n-k}$ sends the string of $u_r$ bijectively from its $(k)$-position to its $(2n-k)$-position and zero is the only string element killed in the relevant range, so $L^{n-k}:H^k(X;\mathbb{R})\to H^{2n-k}(X;\mathbb{R})$ is an isomorphism for $0\le k\le n$; the kernel of $L^{n-k+1}$ in degree $k\le n$ consists exactly of the sums ${\sum }_r{L}^r{u}_r$ with $r=0$, because $L^{n-k+1}{L}^r{u}_r=L^{r+n-k+1}{u}_r=0$ if and only if $n-k+1+r>n-t$ with $k=t+2r$, that is $r=0$; this is the displayed $P^k$. Summing the decompositions string by string gives $H^k(X;\mathbb{R})={⨁}_{r\ge \max (0,k-n)}{L}^r{P}^{k-2r}$, and since $L$ raises both Hodge indices by one and $\mathrm{\Lambda }$ lowers both by one, a class of type $(p,q)$ has primitive components of type $(p-r,q-r)$.

Assume $[\omega ]$ is the real image of a rational class $\mathrm{\ell }$, so that cup product with $[\omega ]$ on cohomology is the scalar extension of the rational map $a\mapsto a⌣\mathrm{\ell }$ [F4, F5]. In rational bases the matrix of $L^{n-k}:H^k(X;\mathbb{Q})\to H^{2n-k}(X;\mathbb{Q})$ is rational and becomes invertible over $\mathbb{R}$ by step 1.1, so its determinant is a nonzero rational number and the map is already invertible over $\mathbb{Q}$; under the comparison of [F5] its scalar extension to any characteristic-zero field $F$ is therefore an isomorphism $H^k(X;F)\to H^{2n-k}(X;F)$, compatible with cup products. For $0\le k\le n$, the subspaces $P^k=\ker L^{n-k+1}$ are rational; for $k>n$ they are zero by [F2]. Thus the primitive summands $L^r{P}^{k-2r}$ are then rational, and the rational matrix of the direct-sum map ${⨁}_{r\ge \max (0,k-n)}{L}^r{P}^{k-2r}\to H^k(X;\mathbb{Q})$ becomes the isomorphism of step 1.1 over $\mathbb{R}$, hence is invertible with nonzero rational determinant and stays invertible after scalar extension, which proves that the decomposition is rational and persists over $F$. Finally the lowering operator reconstructed on each string by $\mathrm{\Lambda }{L}^{r}u=r(n-t-r+1)L^{r-1}u$ for primitive $u$ of degree $t=k-2r$ [F2] has rational coefficients and, by uniqueness of the primitive decomposition over $\mathbb{R}$, it coincides there with the transported harmonic operator $\mathrm{\Lambda }$; so the ${\mathfrak{s}\mathfrak{l}}_2$ structure itself is rational.