[proofplan] The proof uses the [Hard Lefschetz Theorem](/theorems/3876) to split each cohomology group below the middle degree into primitive classes and Lefschetz multiples from two degrees lower. The key linear-algebra step is to prove $H^k(X,\mathbb{R}) = P^k(X) \oplus L H^{k-2}(X,\mathbb{R})$ for $0 \leq k \leq n$. Once this two-term splitting is established, iterating it gives the stated direct sum decomposition. [/proofplan]

[step:Use Hard Lefschetz to isolate the primitive summand]For each integer $m$, write \begin{align*} V^m := H^m(X,\mathbb{R}). \end{align*} For $m < 0$, set $V^m := \{0\}$. The operator $L$ is the degree-$2$ [linear map](/page/Linear%20Map) \begin{align*} L: V^m &\to V^{m+2} \\ \alpha &\mapsto [\omega] \cup \alpha. \end{align*} We use the Hard Lefschetz Theorem, which states that for every integer $0 \leq j \leq n$, the map \begin{align*} L^{n-j}: V^j \to V^{2n-j} \end{align*} is an isomorphism. This is the single external input in the proof; the required cited result is not yet resolved in the wiki: Hard Lefschetz Theorem. Fix an integer $k$ with $0 \leq k \leq n$. We prove \begin{align*} V^k = P^k(X) \oplus L V^{k-2}. \end{align*} If $k=0$ or $k=1$, then $V^{k-2}=\{0\}$ by convention. The Hard Lefschetz map \begin{align*} L^{n-k}: V^k \to V^{2n-k} \end{align*} is an isomorphism. Hence $L^{n-k+1}\alpha = 0$ for $\alpha \in V^k$ would imply \begin{align*} L^{n-k}\alpha \in \ker\left(L: V^{2n-k} \to V^{2n-k+2}\right). \end{align*} Since $2n-k+2>2n$, the target is zero only when applying one further Lefschetz operator past the top degree. In degrees $0$ and $1$, the primitive space is exactly the kernel prescribed by the definition, so the desired splitting reduces to $V^k=P^k(X)$ when $L V^{k-2}=\{0\}$. Now assume $2 \leq k \leq n$. By Hard Lefschetz in degree $k-2$, the map \begin{align*} L^{n-k+2}: V^{k-2} \to V^{2n-k+2} \end{align*} is an isomorphism. In particular, $L:V^{k-2}\to V^k$ is injective, because if $\beta \in V^{k-2}$ and $L\beta=0$, then \begin{align*} L^{n-k+2}\beta = L^{n-k+1}(L\beta)=0, \end{align*} and the isomorphism $L^{n-k+2}$ forces $\beta=0$. The restriction of $L^{n-k+1}$ to $L V^{k-2}$ is an isomorphism onto $V^{2n-k+2}$. Indeed, every element of $L V^{k-2}$ has the form $L\beta$ with $\beta \in V^{k-2}$, and \begin{align*} L^{n-k+1}(L\beta)=L^{n-k+2}\beta. \end{align*} Since $L^{n-k+2}:V^{k-2}\to V^{2n-k+2}$ is an isomorphism, this restriction is bijective. Therefore \begin{align*} P^k(X) \cap L V^{k-2}=\{0\}, \end{align*} because $P^k(X)$ is the kernel of $L^{n-k+1}:V^k\to V^{2n-k+2}$ while this same map is injective on $L V^{k-2}$. Finally, because the restriction to $L V^{k-2}$ is surjective, the rank of \begin{align*} L^{n-k+1}:V^k\to V^{2n-k+2} \end{align*} is $\dim V^{2n-k+2}$. Since $L^{n-k+2}:V^{k-2}\to V^{2n-k+2}$ is an isomorphism, this rank is $\dim V^{k-2}$. The [rank-nullity theorem](/theorems/916) gives \begin{align*} \dim P^k(X) = \dim V^k - \dim V^{k-2}. \end{align*} Because $L:V^{k-2}\to V^k$ is injective, $\dim L V^{k-2}=\dim V^{k-2}$. Hence \begin{align*} \dim P^k(X)+\dim L V^{k-2}=\dim V^k. \end{align*} Together with $P^k(X)\cap L V^{k-2}=\{0\}$, this proves \begin{align*} V^k=P^k(X)\oplus L V^{k-2}. \end{align*}[/step]

[guided]The goal of this step is to separate a class in $H^k(X,\mathbb{R})$ into a primitive part and a part that already comes from degree $k-2$ by multiplication with the Kähler class. We write \begin{align*} V^m := H^m(X,\mathbb{R}) \end{align*} for each integer $m$, and set $V^m:=\{0\}$ for $m<0$. The Lefschetz operator is the linear map \begin{align*} L: V^m &\to V^{m+2} \\ \alpha &\mapsto [\omega]\cup \alpha. \end{align*} We use the Hard Lefschetz Theorem: for every $0\leq j\leq n$, the map \begin{align*} L^{n-j}:V^j\to V^{2n-j} \end{align*} is an isomorphism. This is the structural input that makes the decomposition possible; the required cited result is not yet resolved in the wiki: Hard Lefschetz Theorem. Fix $k$ with $0\leq k\leq n$. We want to prove the two-term decomposition \begin{align*} V^k=P^k(X)\oplus L V^{k-2}. \end{align*} The term $P^k(X)$ is the kernel \begin{align*} P^k(X)=\ker\left(L^{n-k+1}:V^k\to V^{2n-k+2}\right). \end{align*} Thus a class is primitive precisely when one more Lefschetz power than the middle-degree isomorphism kills it. For $k=0$ or $k=1$, the space $V^{k-2}$ is zero by convention, so there is no lower-degree contribution. The desired splitting says simply that the degree-$k$ cohomology is its primitive part. Now suppose $2\leq k\leq n$. Apply Hard Lefschetz in degree $k-2$. It gives an isomorphism \begin{align*} L^{n-k+2}:V^{k-2}\to V^{2n-k+2}. \end{align*} This immediately implies that the first Lefschetz map $L:V^{k-2}\to V^k$ is injective. Indeed, if $\beta\in V^{k-2}$ satisfies $L\beta=0$, then \begin{align*} L^{n-k+2}\beta = L^{n-k+1}(L\beta) = 0. \end{align*} Since $L^{n-k+2}$ is an isomorphism, its kernel is zero, so $\beta=0$. Next we examine what $L^{n-k+1}$ does on the subspace $L V^{k-2}\subset V^k$. Every element of this subspace has the form $L\beta$ for some $\beta\in V^{k-2}$, and \begin{align*} L^{n-k+1}(L\beta)=L^{n-k+2}\beta. \end{align*} Because $L^{n-k+2}:V^{k-2}\to V^{2n-k+2}$ is an isomorphism, the restriction of \begin{align*} L^{n-k+1}:V^k\to V^{2n-k+2} \end{align*} to $L V^{k-2}$ is also an isomorphism onto $V^{2n-k+2}$. This has two consequences. First, the primitive part and the Lefschetz image meet only at zero: \begin{align*} P^k(X)\cap L V^{k-2}=\{0\}. \end{align*} Indeed, $P^k(X)$ is the kernel of $L^{n-k+1}$, while the same map is injective on $L V^{k-2}$. Second, the dimensions add up correctly. Since the restriction to $L V^{k-2}$ is onto $V^{2n-k+2}$, the rank of $L^{n-k+1}:V^k\to V^{2n-k+2}$ is $\dim V^{2n-k+2}$. The Hard Lefschetz isomorphism \begin{align*} L^{n-k+2}:V^{k-2}\to V^{2n-k+2} \end{align*} identifies this dimension with $\dim V^{k-2}$. Hence rank-nullity gives \begin{align*} \dim P^k(X)=\dim V^k-\dim V^{k-2}. \end{align*} Since $L:V^{k-2}\to V^k$ is injective, we also have \begin{align*} \dim L V^{k-2}=\dim V^{k-2}. \end{align*} Therefore \begin{align*} \dim P^k(X)+\dim L V^{k-2}=\dim V^k. \end{align*} Together with the zero intersection, this proves \begin{align*} V^k=P^k(X)\oplus L V^{k-2}. \end{align*}[/guided]

[step:Iterate the two-term splitting to obtain the full decomposition] We prove the stated formula by induction on $k$ for $0\leq k\leq n$. For $k=0$ and $k=1$, the convention $V^m=\{0\}$ for $m<0$ gives \begin{align*} V^k=P^k(X), \end{align*} which is the asserted decomposition. Assume $2\leq k\leq n$, and assume the decomposition has been proved in degree $k-2$. The two-term splitting from the previous step gives \begin{align*} V^k=P^k(X)\oplus L V^{k-2}. \end{align*} By the induction hypothesis, \begin{align*} V^{k-2} = \bigoplus_{r=0}^{\lfloor (k-2)/2\rfloor} L^r P^{k-2-2r}(X). \end{align*} Applying the linear map $L:V^{k-2}\to V^k$ to this direct sum gives \begin{align*} L V^{k-2} = \bigoplus_{r=0}^{\lfloor (k-2)/2\rfloor} L^{r+1}P^{k-2-2r}(X). \end{align*} Set $s:=r+1$. Then $s$ ranges from $1$ to $\lfloor k/2\rfloor$, and $k-2-2r=k-2s$. Therefore \begin{align*} L V^{k-2} = \bigoplus_{s=1}^{\lfloor k/2\rfloor} L^s P^{k-2s}(X). \end{align*} Substituting this into $V^k=P^k(X)\oplus L V^{k-2}$ yields \begin{align*} V^k = P^k(X) \oplus \bigoplus_{s=1}^{\lfloor k/2\rfloor} L^s P^{k-2s}(X) = \bigoplus_{s=0}^{\lfloor k/2\rfloor} L^s P^{k-2s}(X). \end{align*} Since $V^k=H^k(X,\mathbb{R})$, this is exactly \begin{align*} H^k(X,\mathbb{R}) = \bigoplus_{r=0}^{\lfloor k/2\rfloor} L^r P^{k-2r}(X). \end{align*} This completes the proof. [/step]