[proofplan] The Lefschetz operator is cup product with the Kähler class, and the Kähler class has Hodge type $(1,1)$. Therefore its $r$-th power has type $(r,r)$, so multiplication by it shifts a Hodge summand $H^{p,q}(X)$ into $H^{p+r,q+r}(X)$. In total degree $n-r$, Hard Lefschetz gives an isomorphism on the whole cohomology group. The Hodge decompositions in degrees $n-r$ and $n+r$ are direct sums, and the type-shift property forces the total isomorphism to split into isomorphisms on the individual Hodge summands. [/proofplan]

[step:Use the Kähler class to define the type shift]Let \begin{align*} [\omega]\in H^2(X,\mathbb C) \end{align*} be the cohomology class of the Kähler form. Since $\omega$ is a Kähler form, it is a real closed form of type $(1,1)$, so its cohomology class lies in $H^{1,1}(X)$. For every pair of integers $k$ and $r\ge 0$, define the complex-[linear map](/page/Linear%20Map) \begin{align*} L_k^r:H^k(X,\mathbb C)\to H^{k+2r}(X,\mathbb C) \end{align*} by $L_k^0=\operatorname{id}_{H^k(X,\mathbb C)}$ and, for $r\ge 1$, by the composite $L_{k+2r-2}\circ\cdots\circ L_{k+2}\circ L_k$. Equivalently, \begin{align*} L_k^r(\alpha)=[\omega]^r\smile \alpha \end{align*} for every class $\alpha\in H^k(X,\mathbb C)$. When the source degree is clear, we write $L^r$ for $L_k^r$. The cup product respects Hodge type: if $\alpha\in H^{p,q}(X)$, then \begin{align*} [\omega]^r\smile \alpha\in H^{p+r,q+r}(X). \end{align*} Thus $L^r$ restricts to a complex-linear map \begin{align*} L^r:H^{p,q}(X)\to H^{p+r,q+r}(X). \end{align*}[/step]

[guided]The operator $L$ is not an arbitrary linear map. For each integer $k$, it is the complex-linear map \begin{align*} L_k:H^k(X,\mathbb C)\to H^{k+2}(X,\mathbb C) \end{align*} defined by multiplying a cohomology class by the Kähler class: \begin{align*} L_k(\alpha)=[\omega]\smile \alpha \end{align*} for every $\alpha\in H^k(X,\mathbb C)$. Therefore, for every integer $r\ge 0$, \begin{align*} L^r(\alpha)=[\omega]^r\smile \alpha \end{align*} whenever $\alpha$ lies in the appropriate cohomological degree. The point of using a Kähler form is that $\omega$ has type $(1,1)$. Passing to cohomology, this means \begin{align*} [\omega]\in H^{1,1}(X). \end{align*} The cup product of a class of type $(a,b)$ with a class of type $(c,d)$ has type $(a+c,b+d)$. Hence the $r$-fold cup power satisfies \begin{align*} [\omega]^r\in H^{r,r}(X). \end{align*} If $\alpha\in H^{p,q}(X)$, then multiplying by $[\omega]^r$ gives \begin{align*} L^r(\alpha)=[\omega]^r\smile \alpha\in H^{p+r,q+r}(X). \end{align*} This proves that the total cohomological Lefschetz map restricts to a well-defined complex-linear map \begin{align*} L^r:H^{p,q}(X)\to H^{p+r,q+r}(X). \end{align*} If $p+r>n$ or $q+r>n$, the target Hodge group is zero by the stated convention, and the same type calculation says precisely that the product has no nonzero component in an out-of-range bidegree.[/guided]

[step:Apply Hard Lefschetz in total degree $n-r$] Assume now that $p+q=n-r$. Since $(X,\omega)$ is a compact Kähler manifold of complex dimension $n$ and $[\omega]$ is its Kähler class, the Hard Lefschetz part of the Kähler package [citetheorem:8078] applies to the integer $n-r$ and gives that the map \begin{align*} L_{n-r}^r:H^{n-r}(X,\mathbb C)\to H^{n+r}(X,\mathbb C) \end{align*} is an isomorphism. The same compact Kähler hypothesis is exactly the hypothesis of the [Hodge decomposition for compact Kähler manifolds](/theorems/8066), as recorded in the pure Hodge structure on Kähler cohomology [citetheorem:8066], so there is a direct sum decomposition \begin{align*} H^{n-r}(X,\mathbb C)=\bigoplus_{a+b=n-r}H^{a,b}(X). \end{align*} In degree $n+r$, the usual decomposition is \begin{align*} H^{n+r}(X,\mathbb C)=\bigoplus_{u+v=n+r}H^{u,v}(X). \end{align*} Re-index this sum by setting $u=a+r$ and $v=b+r$. No nonzero summand is missed: if $u+v=n+r$ and $u<r$, then $v>n$, so $H^{u,v}(X)=0$ by the stated convention; the case $v<r$ is identical. Hence \begin{align*} H^{n+r}(X,\mathbb C)=\bigoplus_{a+b=n-r}H^{a+r,b+r}(X). \end{align*} [/step]

[step:Deduce injectivity on each Hodge summand] Fix $p,q\ge 0$ with $p+q=n-r$. Suppose $\alpha\in H^{p,q}(X)$ satisfies \begin{align*} L^r(\alpha)=0 \end{align*} in $H^{p+r,q+r}(X)$. Since $H^{p,q}(X)$ is a subspace of $H^{n-r}(X,\mathbb C)$ and the total map \begin{align*} L_{n-r}^r:H^{n-r}(X,\mathbb C)\to H^{n+r}(X,\mathbb C) \end{align*} is injective, it follows that $\alpha=0$. Thus the restricted map \begin{align*} L^r:H^{p,q}(X)\to H^{p+r,q+r}(X) \end{align*} is injective. [/step]

[step:Use the direct Hodge decomposition to prove surjectivity on each summand] Fix $\beta\in H^{p+r,q+r}(X)$. Since the total Lefschetz map is surjective, there exists \begin{align*} \alpha\in H^{n-r}(X,\mathbb C) \end{align*} such that \begin{align*} L^r(\alpha)=\beta. \end{align*} Using the direct [Hodge decomposition](/theorems/2745) in degree $n-r$, write uniquely \begin{align*} \alpha=\sum_{a+b=n-r}\alpha_{a,b}, \end{align*} where $\alpha_{a,b}\in H^{a,b}(X)$. By the type-shift already proved, \begin{align*} L^r(\alpha_{a,b})\in H^{a+r,b+r}(X). \end{align*} Therefore \begin{align*} \beta=L^r(\alpha)=\sum_{a+b=n-r}L^r(\alpha_{a,b}) \end{align*} is the [Hodge decomposition](/theorems/3941) of $\beta$ in degree $n+r$. Since $\beta$ belongs entirely to the summand $H^{p+r,q+r}(X)$ and the Hodge decomposition is direct, all components except the $(p+r,q+r)$ component vanish. Hence \begin{align*} \beta=L^r(\alpha_{p,q}). \end{align*} Thus the restricted map \begin{align*} L^r:H^{p,q}(X)\to H^{p+r,q+r}(X) \end{align*} is surjective. [/step]

[step:Assemble the restricted isomorphisms into Hard Lefschetz] For every pair $(p,q)$ with $p+q=n-r$, the restricted map \begin{align*} L^r:H^{p,q}(X)\to H^{p+r,q+r}(X) \end{align*} is both injective and surjective, hence is an isomorphism. Taking the direct sum of these restricted maps over all pairs satisfying $p+q=n-r$ gives \begin{align*} \bigoplus_{p+q=n-r}H^{p,q}(X)\to \bigoplus_{p+q=n-r}H^{p+r,q+r}(X). \end{align*} Under the Hodge decompositions of $H^{n-r}(X,\mathbb C)$ and $H^{n+r}(X,\mathbb C)$, this direct sum map is exactly \begin{align*} L_{n-r}^r:H^{n-r}(X,\mathbb C)\to H^{n+r}(X,\mathbb C). \end{align*} Therefore the direct sum of the Hodge-type restricted isomorphisms recovers the Hard Lefschetz isomorphism in degree $n-r$. [/step]