[proofplan] The proof packages together the standard structural consequences of the [Kähler identities](/theorems/3853). Hard Lefschetz gives the stated isomorphisms for the Lefschetz operator associated to the Kähler class. The compact Kähler [Hodge decomposition](/theorems/2745) and Hodge symmetry give the direct-sum decomposition and conjugation rule. Finally, representatives of pure Hodge type multiply by wedge product to representatives of the summed type, and the de Rham cup product is represented by wedge product of closed forms. [/proofplan]

[step:Apply hard Lefschetz to the Kähler class] Let $\omega\in A^{1,1}(X)$ be a Kähler form representing the given class $[\omega]\in H^2(X;\mathbb R)$. For each integer $k$, the operator $L_k:H^k(X;\mathbb R)\to H^{k+2}(X;\mathbb R)$ is cup product with $[\omega]$. By the [Hard Lefschetz theorem](/theorems/3876) for compact Kähler manifolds applied to the compact Kähler manifold $(X,\omega)$, for every integer $k$ with $0\leq k\leq n$, the iterated map \begin{align*} L^{n-k}:H^k(X;\mathbb R)\to H^{2n-k}(X;\mathbb R) \end{align*} is an isomorphism. Here the hypotheses of the theorem are exactly the compactness of $X$, the Kähler condition for $\omega$, and the fact that $X$ has complex dimension $n$. Thus the first asserted cohomological constraint holds. [/step]

[step:Use the compact Kähler Hodge decomposition]For each integer $k\geq 0$, apply [Hodge Decomposition for Compact Kähler Manifolds]([citetheorem:8066]) to the compact Kähler manifold $X$. The theorem applies because $X$ is compact and Kähler. It gives complex subspaces $H^{p,q}(X)\subset H^k(X;\mathbb C)$, indexed by pairs $p,q\geq 0$ with $p+q=k$, such that \begin{align*} H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X). \end{align*} This is exactly the second asserted condition.[/step]

[guided]We now consume the Kähler hypothesis a second time, this time through Hodge theory rather than through the Lefschetz operator. The input theorem is [Hodge Decomposition for Compact Kähler Manifolds]([citetheorem:8066]). Its hypotheses require that $X$ be a compact Kähler manifold, and this is one of the hypotheses of the present theorem. The conclusion of that theorem is the existence, for each integer $k\geq 0$, of complex vector subspaces $H^{p,q}(X)\subset H^k(X;\mathbb C)$ with $p+q=k$ such that the sum over all such pairs is direct and exhausts $H^k(X;\mathbb C)$. Therefore \begin{align*} H^k(X;\mathbb C)=\bigoplus_{p+q=k}H^{p,q}(X). \end{align*} This is not merely a decomposition of differential forms before passing to cohomology; it is a decomposition of the cohomology group itself. That distinction matters because the theorem is asserting a constraint on the cohomology algebra $H^*(X;\mathbb R)$ after extension of scalars to $\mathbb C$.[/guided]

[step:Identify complex conjugation with interchange of Hodge bidegrees] Complex conjugation on complex-valued differential forms sends a smooth form of type $(p,q)$ to a smooth form of type $(q,p)$, and it commutes with the real [exterior derivative](/theorems/1525) $d$. Hence it induces complex-antilinear maps \begin{align*} H^{p+q}(X;\mathbb C)\to H^{p+q}(X;\mathbb C), \qquad [\alpha]\mapsto [\overline{\alpha}]. \end{align*} By [Hodge Symmetry for Compact Kähler Manifolds]([citetheorem:8081]), applied to the compact Kähler manifold $X$, these maps exchange the Hodge summands: \begin{align*} \overline{H^{p,q}(X)}=H^{q,p}(X). \end{align*} This proves the conjugation symmetry assertion. [/step]

[step:Represent pure Hodge classes by closed forms of pure type] Let $p,q,r,s\geq 0$ be integers. Let \begin{align*} a\in H^{p,q}(X), \qquad b\in H^{r,s}(X) \end{align*} be cohomology classes. By the definition of the Hodge summands in the compact Kähler [Hodge decomposition](/theorems/3941), $H^{p,q}(X)$ consists of de Rham classes represented by closed forms in $A^{p,q}(X)$, where $A^{p,q}(X)$ denotes the complex [vector space](/page/Vector%20Space) of smooth complex-valued differential forms of type $(p,q)$ on $X$. Hence there exist closed smooth complex-valued forms \begin{align*} \alpha\in A^{p,q}(X), \qquad \beta\in A^{r,s}(X) \end{align*} such that \begin{align*} a=[\alpha]\in H^{p+q}(X;\mathbb C), \qquad b=[\beta]\in H^{r+s}(X;\mathbb C). \end{align*} Since $d\alpha=0$ and $d\beta=0$, the graded Leibniz rule for the exterior derivative gives \begin{align*} d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^{p+q}\alpha\wedge d\beta=0. \end{align*} Thus $\alpha\wedge\beta$ is a closed form. [/step]

[step:Compute the type of the wedge product and pass to cup product] The exterior product of a form of type $(p,q)$ with a form of type $(r,s)$ has type $(p+r,q+s)$. Therefore \begin{align*} \alpha\wedge\beta\in A^{p+r,q+s}(X). \end{align*} The de Rham product on cohomology is represented by wedge product of closed differential forms, so \begin{align*} a\smile b=[\alpha]\smile[\beta]=[\alpha\wedge\beta]. \end{align*} Because $\alpha\wedge\beta$ is closed and has type $(p+r,q+s)$, the class $[\alpha\wedge\beta]$ lies in $H^{p+r,q+s}(X)$. Hence \begin{align*} H^{p,q}(X)\smile H^{r,s}(X)\subset H^{p+r,q+s}(X). \end{align*} If $p+r>n$ or $q+s>n$, then $A^{p+r,q+s}(X)=0$ by the type convention for a complex manifold of dimension $n$, so the same inclusion is read as inclusion into the zero summand. Combining this product compatibility with the preceding steps proves all asserted Kähler package constraints. [/step]