[proofplan] We pass from cohomology to harmonic differential forms using the [Hodge theorem for compact Riemannian manifolds](/page/Hodge%20Theorem). On a compact [Kähler manifold](/page/K%C3%A4hler%20Manifold), the [Kähler identities](/page/K%C3%A4hler%20Identities) imply that the [Lefschetz operator](/page/Lefschetz%20Operator) commutes with the [Hodge Laplacian](/page/Hodge%20Laplacian), so it preserves [harmonic forms](/page/Harmonic%20Form) and represents [cup product](/page/Cup%20Product) by the [Kähler class](/page/K%C3%A4hler%20Class) on de Rham cohomology. The remaining argument is finite-dimensional representation theory: the Lefschetz operators form an $\mathfrak{sl}_2(\mathbb R)$-triple on harmonic forms, and the finite-dimensional $\mathfrak{sl}_2$ weight-space theorem gives the required Lefschetz isomorphisms. [/proofplan] [step:Represent cohomology classes by harmonic forms] Let $A^m(X;\mathbb R)$ denote the real [vector space](/page/Vector%20Space) of smooth real-valued differential $m$-forms on $X$. Choose the [Kähler metric](/page/K%C3%A4hler%20Metric) whose Kähler form represents $\omega$, and let \begin{align*} \Delta: A^m(X;\mathbb R) &\to A^m(X;\mathbb R) \end{align*} denote the [Hodge Laplacian](/page/Hodge%20Laplacian) on $m$-forms. Define the harmonic subspace \begin{align*} \mathcal H^m(X) := \{\alpha \in A^m(X;\mathbb R) : \Delta \alpha = 0\}. \end{align*} Since $X$ is compact and the chosen Kähler metric is a Riemannian metric, the [Hodge theorem for compact Riemannian manifolds](/page/Hodge%20Theorem) applies and gives a canonical vector-space isomorphism \begin{align*} \mathcal H^m(X) &\longrightarrow H^m(X,\mathbb R),\\ \alpha &\longmapsto [\alpha]. \end{align*} Thus it is enough to prove that the operator induced by wedging with the Kähler form is an isomorphism \begin{align*} L^{n-k}: \mathcal H^k(X) \longrightarrow \mathcal H^{2n-k}(X). \end{align*} [guided] The cohomology groups in the theorem are finite-dimensional, but the most efficient proof works on differential forms. We choose the Kähler metric whose Kähler form represents the class $\omega$. For each integer $m$, let $A^m(X;\mathbb R)$ be the real vector space of smooth real-valued differential $m$-forms on $X$, and let \begin{align*} \Delta: A^m(X;\mathbb R) &\to A^m(X;\mathbb R) \end{align*} be the Hodge Laplacian determined by this metric. The harmonic $m$-forms are \begin{align*} \mathcal H^m(X) := \{\alpha \in A^m(X;\mathbb R) : \Delta \alpha = 0\}. \end{align*} Because $X$ is compact and the Kähler metric is in particular a Riemannian metric, the [Hodge theorem for compact Riemannian manifolds](/page/Hodge%20Theorem) applies. It identifies each de Rham cohomology class with a unique harmonic representative, giving the isomorphism \begin{align*} \mathcal H^m(X) &\longrightarrow H^m(X,\mathbb R),\\ \alpha &\longmapsto [\alpha]. \end{align*} Therefore the desired cohomological statement follows once we prove the same statement on harmonic forms: \begin{align*} L^{n-k}: \mathcal H^k(X) \longrightarrow \mathcal H^{2n-k}(X). \end{align*} This reduction is where compactness is used: without compactness, [harmonic representatives](/theorems/2747) need not give all cohomology classes in this form. [/guided] [/step] [step:Use the Kähler identities to make the Lefschetz operator act on harmonic forms] Define the [Lefschetz operator](/page/Lefschetz%20Operator) on differential forms by \begin{align*} L: A^m(X;\mathbb R) &\to A^{m+2}(X;\mathbb R),\\ \alpha &\mapsto \omega \wedge \alpha. \end{align*} The [Kähler identities](/page/K%C3%A4hler%20Identities) imply the commutator identity \begin{align*} [\Delta,L] := \Delta L - L\Delta = 0. \end{align*} Hence, if $\alpha \in \mathcal H^m(X)$, then \begin{align*} \Delta(L\alpha) = L(\Delta\alpha) = 0, \end{align*} so $L\alpha \in \mathcal H^{m+2}(X)$. Since $\omega$ is closed, wedging with $\omega$ represents [cup product](/page/Cup%20Product) by $\omega$ under the de Rham product. Therefore the restriction of $L$ to harmonic forms corresponds, under the Hodge isomorphism, to the cohomological [Lefschetz operator](/page/Lefschetz%20Operator). [guided] We now verify that the operator appearing in the theorem is visible on harmonic forms. Define \begin{align*} L: A^m(X;\mathbb R) &\to A^{m+2}(X;\mathbb R),\\ \alpha &\mapsto \omega \wedge \alpha. \end{align*} The Kähler condition is used through the [Kähler identities](/page/K%C3%A4hler%20Identities). One consequence is the commutator relation \begin{align*} [\Delta,L] := \Delta L - L\Delta = 0. \end{align*} This identity says exactly that applying $L$ and then the Hodge Laplacian gives the same result as applying the Hodge Laplacian first and then $L$. Thus, for any harmonic form $\alpha \in \mathcal H^m(X)$, \begin{align*} \Delta(L\alpha) = L(\Delta\alpha) = L(0) = 0. \end{align*} So $L\alpha$ is harmonic and has degree $m+2$. The operator $L$ therefore restricts to a [linear map](/page/Linear%20Map) \begin{align*} L: \mathcal H^m(X) \to \mathcal H^{m+2}(X). \end{align*} Finally, since $\omega$ is closed, the de Rham product satisfies \begin{align*} [L\alpha] = [\omega \wedge \alpha] = \omega \smile [\alpha]. \end{align*} Thus the harmonic-form operator is exactly the cohomological Lefschetz operator after identifying cohomology with harmonic forms. [/guided] [/step] [step:Apply the Lefschetz $\mathfrak{sl}_2(\mathbb R)$ representation on harmonic forms] Let $\Lambda: A^m(X;\mathbb R) \to A^{m-2}(X;\mathbb R)$ be the adjoint of $L$ with respect to the $L^2$ inner product induced by the Kähler metric, and define \begin{align*} H: A^m(X;\mathbb R) &\to A^m(X;\mathbb R),\\ \alpha &\mapsto (m-n)\alpha. \end{align*} The [Kähler identities](/page/K%C3%A4hler%20Identities) imply that $\Lambda$ also commutes with $\Delta$, so $L$, $\Lambda$, and $H$ preserve $\mathcal H^*(X) := \bigoplus_{m=0}^{2n}\mathcal H^m(X)$. The [Lefschetz commutation relations](/page/Lefschetz%20Decomposition) give \begin{align*} [H,L] &= 2L, & [H,\Lambda] &= -2\Lambda, & [L,\Lambda] &= H. \end{align*} These are the defining relations of the Lie algebra $\mathfrak{sl}_2(\mathbb R)$, the real Lie algebra with basis $e,f,h$ and brackets $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. Hence $\mathcal H^*(X)$ is a finite-dimensional $\mathfrak{sl}_2(\mathbb R)$-module with $e$ acting as $L$, $f$ acting as $\Lambda$, and $h$ acting as $H$. [guided] The next point is algebraic. Define $\Lambda: A^m(X;\mathbb R) \to A^{m-2}(X;\mathbb R)$ to be the adjoint of $L$ for the $L^2$ inner product coming from the Kähler metric. Define the degree-counting operator \begin{align*} H: A^m(X;\mathbb R) &\to A^m(X;\mathbb R),\\ \alpha &\mapsto (m-n)\alpha. \end{align*} The [Kähler identities](/page/K%C3%A4hler%20Identities) imply that $\Lambda$ commutes with the Hodge Laplacian, just as $L$ does. Therefore $L$, $\Lambda$, and $H$ all preserve the finite-dimensional vector space \begin{align*} \mathcal H^*(X) := \bigoplus_{m=0}^{2n}\mathcal H^m(X). \end{align*} On this space the [Lefschetz commutation relations](/page/Lefschetz%20Decomposition) give \begin{align*} [H,L] &= 2L, & [H,\Lambda] &= -2\Lambda, & [L,\Lambda] &= H. \end{align*} These are precisely the defining relations of $\mathfrak{sl}_2(\mathbb R)$, which is the real Lie algebra with basis $e,f,h$ and bracket relations $[h,e]=2e$, $[h,f]=-2f$, and $[e,f]=h$. Thus we may regard $\mathcal H^*(X)$ as a finite-dimensional $\mathfrak{sl}_2(\mathbb R)$-module by letting $e$ act as $L$, $f$ act as $\Lambda$, and $h$ act as $H$. The weight of a harmonic $m$-form is $m-n$, because $H\alpha=(m-n)\alpha$. [/guided] [/step] [step:Use finite-dimensional $\mathfrak{sl}_2$ theory to obtain the middle-degree symmetry] By the [finite-dimensional $\mathfrak{sl}_2$ weight-space theorem](/page/Representation%20Theory%20of%20sl2), the raising operator $e$ satisfies the following weight-space isomorphism on every finite-dimensional complex $\mathfrak{sl}_2$-module: if $V_r$ denotes the $h$-weight space of weight $r$, then for every $r \le 0$, \begin{align*} e^{-r}: V_r \longrightarrow V_{-r} \end{align*} is an isomorphism. The same conclusion holds for finite-dimensional real $\mathfrak{sl}_2(\mathbb R)$-modules by complexifying the module and then restricting the resulting complex-linear isomorphism to the real weight spaces. Apply this with $V=\mathcal H^*(X)$ and $r=k-n$. Since \begin{align*} V_{k-n} &= \mathcal H^k(X), & V_{n-k} &= \mathcal H^{2n-k}(X), \end{align*} and $e$ acts as $L$, we obtain an isomorphism \begin{align*} L^{n-k}: \mathcal H^k(X) \longrightarrow \mathcal H^{2n-k}(X). \end{align*} [guided] We now use only finite-dimensional representation theory. In any finite-dimensional $\mathfrak{sl}_2(\mathbb R)$-module $V$, write \begin{align*} V_r := \{v \in V : h v = r v\} \end{align*} for the $h$-weight space of weight $r$. The [finite-dimensional $\mathfrak{sl}_2$ weight-space theorem](/page/Representation%20Theory%20of%20sl2) says that for every integer weight $r \le 0$, the raising operator gives an isomorphism \begin{align*} e^{-r}: V_r \longrightarrow V_{-r}. \end{align*} This theorem is usually stated for complex finite-dimensional $\mathfrak{sl}_2$-modules. It applies here over $\mathbb R$ because we may complexify $V$ to $V \otimes_{\mathbb R} \mathbb C$; the operators $e$, $f$, and $h$ extend complex-linearly, the weight spaces complexify as $(V_r) \otimes_{\mathbb R} \mathbb C$, and the complex isomorphism $e^{-r}: (V_r) \otimes_{\mathbb R} \mathbb C \to (V_{-r}) \otimes_{\mathbb R} \mathbb C$ is induced by the original real map $e^{-r}: V_r \to V_{-r}$. Hence the real map is injective and surjective. In our module $V=\mathcal H^*(X)$, the operator $h$ is $H$, and $H$ acts on $\mathcal H^m(X)$ by multiplication by $m-n$. Therefore the weight space of weight $k-n$ is \begin{align*} V_{k-n} = \mathcal H^k(X), \end{align*} and the weight space of weight $n-k$ is \begin{align*} V_{n-k} = \mathcal H^{2n-k}(X). \end{align*} Because $0 \le k \le n$, the weight $k-n$ is nonpositive. Applying the representation-theoretic isomorphism with $r=k-n$ gives \begin{align*} e^{n-k}: V_{k-n} \longrightarrow V_{n-k}. \end{align*} Since $e$ acts as $L$, this is exactly \begin{align*} L^{n-k}: \mathcal H^k(X) \longrightarrow \mathcal H^{2n-k}(X). \end{align*} [/guided] [/step] [step:Transfer the harmonic isomorphism back to cohomology] Under the Hodge isomorphisms \begin{align*} \mathcal H^k(X) &\cong H^k(X,\mathbb R), & \mathcal H^{2n-k}(X) &\cong H^{2n-k}(X,\mathbb R), \end{align*} the harmonic operator $L^{n-k}$ corresponds to cup product by $\omega^{n-k}$. Since the harmonic operator is an isomorphism, the cohomological map \begin{align*} L^{n-k}: H^k(X,\mathbb R) \longrightarrow H^{2n-k}(X,\mathbb R) \end{align*} is an isomorphism. This proves the Hard Lefschetz theorem. [guided] We have proved that wedging with the Kähler form gives an isomorphism on harmonic forms: \begin{align*} L^{n-k}: \mathcal H^k(X) \longrightarrow \mathcal H^{2n-k}(X). \end{align*} The Hodge isomorphisms from the first step identify these harmonic spaces with de Rham cohomology: \begin{align*} \mathcal H^k(X) &\cong H^k(X,\mathbb R), & \mathcal H^{2n-k}(X) &\cong H^{2n-k}(X,\mathbb R). \end{align*} The compatibility proved above says that the harmonic-form operator $L^{n-k}$ corresponds exactly to [cup product](/page/Cup%20Product) with $\omega^{n-k}$, equivalently to the cohomological [Lefschetz operator](/page/Lefschetz%20Operator). Therefore the induced cohomological map \begin{align*} L^{n-k}: H^k(X,\mathbb R) \longrightarrow H^{2n-k}(X,\mathbb R) \end{align*} is an isomorphism. This is precisely the assertion of the Hard Lefschetz theorem for the chosen integer $k$. [/guided] [/step]