[proofplan] The proof is pointwise linear algebra applied smoothly over the tangent bundle. At each point, the Kähler form gives the exterior algebra of the cotangent space the standard $\mathfrak{sl}_2$ Lefschetz representation generated by $L$, $\Lambda$, and their commutator. Finite-dimensional $\mathfrak{sl}_2$ representation theory gives the direct sum into powers of $L$ applied to primitive vectors, and smoothness follows because the projection operators are universal polynomials in $L$ and $\Lambda$. On a compact Kähler manifold, the [Kähler identities](/theorems/3853) imply that $L$ and $\Lambda$ commute with the Hodge Laplacian, so the same projections preserve harmonic forms; Hodge theory then transfers the decomposition to cohomology. [/proofplan]

[step:Construct the pointwise Lefschetz operators on each exterior algebra]Fix a point $x \in X$. Let \begin{align*} V_x := T_x^*X \otimes_{\mathbb{R}} \mathbb{C} \end{align*} denote the complexified cotangent space, and let \begin{align*} A_x^k := \Lambda^k V_x \end{align*} denote the [vector space](/page/Vector%20Space) of complex $k$-covectors at $x$. Define the pointwise Lefschetz operator \begin{align*} L_x: A_x^k &\to A_x^{k+2} \\ \eta &\mapsto \omega_x \wedge \eta. \end{align*} Let \begin{align*} \Lambda_x: A_x^k &\to A_x^{k-2} \end{align*} be the Hermitian adjoint of $L_x$ with respect to the inner product induced by the Kähler metric on $\Lambda^\bullet V_x$. Define \begin{align*} H_x: A_x^k &\to A_x^k \\ \eta &\mapsto [\Lambda_x,L_x]\eta \end{align*} where $[\Lambda_x,L_x] := \Lambda_x L_x - L_x\Lambda_x$. In a unitary coframe at $x$, the standard computation for a Hermitian vector space gives \begin{align*} [H_x,L_x] = -2L_x, \qquad [H_x,\Lambda_x] = 2\Lambda_x. \end{align*} Thus $(\Lambda_x,L_x,H_x)$ is an $\mathfrak{sl}_2$-triple on the finite-dimensional graded vector space, with $\Lambda_x$ as the raising operator and $L_x$ as the lowering operator for the $H_x$-weights, \begin{align*} A_x^\bullet := \bigoplus_{j=0}^{2n} A_x^j. \end{align*}[/step]

[guided]Fix $x \in X$. The decomposition of forms is first a statement about the exterior algebra of one Hermitian vector space, namely the cotangent space at $x$. We write \begin{align*} V_x := T_x^*X \otimes_{\mathbb{R}} \mathbb{C}, \qquad A_x^k := \Lambda^k V_x. \end{align*} The Kähler form at $x$ defines the map \begin{align*} L_x: A_x^k &\to A_x^{k+2} \\ \eta &\mapsto \omega_x \wedge \eta. \end{align*} The metric gives an inner product on every $\Lambda^k V_x$, hence $L_x$ has a Hermitian adjoint \begin{align*} \Lambda_x: A_x^k &\to A_x^{k-2}. \end{align*} A covector $\eta \in A_x^s$ is primitive exactly when $\Lambda_x\eta=0$. The central algebraic point is that $L_x$ and $\Lambda_x$ do not act independently. Define \begin{align*} H_x: A_x^k &\to A_x^k \\ \eta &\mapsto [\Lambda_x,L_x]\eta, \end{align*} where $[\Lambda_x,L_x]=\Lambda_xL_x-L_x\Lambda_x$. In a unitary coframe for the Hermitian vector space $T_xX$, the Kähler form has the standard normal form, and the creation and annihilation operators given by wedging with the coframe and contracting against the dual frame satisfy the usual commutation rules. With the convention $H_x=[\Lambda_x,L_x]$, this computation gives \begin{align*} [H_x,L_x] = -2L_x, \qquad [H_x,\Lambda_x] = 2\Lambda_x. \end{align*} These are precisely the defining commutation relations of an $\mathfrak{sl}_2$-triple after taking $\Lambda_x$ as the raising operator and $L_x$ as the lowering operator for the $H_x$-weights. Therefore the exterior algebra \begin{align*} A_x^\bullet := \bigoplus_{j=0}^{2n} A_x^j \end{align*} is a finite-dimensional representation of $\mathfrak{sl}_2$, with $H_x$ as the weight operator.[/guided]

[step:Decompose each pointwise exterior power into Lefschetz strings]Let \begin{align*} P_x^s := \ker(\Lambda_x: A_x^s \to A_x^{s-2}) \end{align*} denote the primitive subspace in degree $s$. The finite-dimensional representation theory of $\mathfrak{sl}_2$ implies that every vector in $A_x^k$ has a unique expansion \begin{align*} \eta = \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L_x^r \eta_r, \end{align*} where \begin{align*} \eta_r \in P_x^{k-2r}. \end{align*} Equivalently, \begin{align*} A_x^k = \bigoplus_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L_x^r P_x^{k-2r}. \end{align*} Here the lower bound $r \ge \max\{0,k-n\}$ is forced by primitivity: a nonzero primitive form can have degree at most $n$, so $k-2r \le n$ is necessary. The upper bound $r \le \lfloor k/2 \rfloor$ is forced by the non-negativity of the degree $k-2r$.[/step]

[guided]We now use the standard structure theorem for finite-dimensional $\mathfrak{sl}_2$-representations: every such representation is a direct sum of irreducible strings generated by highest-weight vectors. In the Lefschetz convention used here, the highest-weight vectors are exactly the vectors killed by the raising operator $\Lambda_x$. Thus the primitive subspace in degree $s$ is \begin{align*} P_x^s := \ker(\Lambda_x: A_x^s \to A_x^{s-2}). \end{align*} Applying the $\mathfrak{sl}_2$ decomposition to \begin{align*} A_x^\bullet = \bigoplus_{j=0}^{2n} A_x^j \end{align*} gives that every element $\eta \in A_x^k$ can be written uniquely as a finite sum of vectors obtained by applying powers of $L_x$ to primitive vectors: \begin{align*} \eta = \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L_x^r \eta_r, \qquad \eta_r \in P_x^{k-2r}. \end{align*} The index range is not cosmetic; it encodes the degree constraints. Since $\eta_r$ has degree $k-2r$, we must have $k-2r \ge 0$, which gives \begin{align*} r \le \left\lfloor \frac{k}{2} \right\rfloor. \end{align*} Also, a nonzero primitive form in a Hermitian vector space of complex dimension $n$ has degree at most $n$. Therefore $k-2r \le n$, hence \begin{align*} r \ge k-n. \end{align*} Combining this with $r \ge 0$ gives the lower bound \begin{align*} r \ge \max\{0,k-n\}. \end{align*} The same representation-theoretic decomposition is direct, so the primitive summands are unique.[/guided]

[step:Patch the pointwise decomposition into a smooth decomposition of forms]For fixed $k$ and $r$, let \begin{align*} \Pi_{k,r}: \Lambda^k T^*X \otimes_{\mathbb{R}} \mathbb{C} \to L^r P^{k-2r} \end{align*} denote the pointwise projection onto the summand \begin{align*} L^r P_x^{k-2r} \subset A_x^k. \end{align*} Because the decomposition comes from the finite-dimensional $\mathfrak{sl}_2$ representation, each projection $\Pi_{k,r}$ is a universal polynomial in the smooth bundle endomorphisms $L$, $\Lambda$, and $H=[\Lambda,L]$. Hence $\Pi_{k,r}$ is a smooth vector bundle projection. Let $\alpha \in \Omega^k(X;\mathbb{C})$. Define \begin{align*} \beta_r := \Pi_{k,r}\alpha \in \Omega^k(X;\mathbb{C}) \end{align*} for each integer \begin{align*} \max\{0,k-n\} \le r \le \left\lfloor \frac{k}{2} \right\rfloor. \end{align*} By the image description of $\Pi_{k,r}$, there is a unique smooth primitive form \begin{align*} \alpha_r \in \Omega^{k-2r}(X;\mathbb{C}) \end{align*} such that \begin{align*} \beta_r = L^r \alpha_r. \end{align*} Summing the projections gives \begin{align*} \alpha = \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} \Pi_{k,r}\alpha = \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L^r\alpha_r. \end{align*} The pointwise directness proved above implies uniqueness: if \begin{align*} \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L^r\gamma_r = 0 \end{align*} with each $\gamma_r$ primitive of degree $k-2r$, then evaluating at every $x \in X$ gives a pointwise direct-sum decomposition, so $\gamma_r(x)=0$ for every $x$ and every $r$. Thus $\gamma_r=0$ for every $r$.[/step]

[guided]The pointwise decomposition gives an algebraic splitting at each $x \in X$. We must still check that applying it to a smooth form produces smooth primitive pieces, not merely pointwise-defined covectors. For fixed $k$ and $r$, define \begin{align*} \Pi_{k,r}: \Lambda^k T^*X \otimes_{\mathbb{R}} \mathbb{C} \to L^r P^{k-2r} \end{align*} to be the projection onto the pointwise summand \begin{align*} L^r P_x^{k-2r} \subset \Lambda^k T_x^*X \otimes_{\mathbb{R}} \mathbb{C}. \end{align*} Why is this projection smooth in $x$? The decomposition is produced by the finite-dimensional $\mathfrak{sl}_2$ action, and the spectral projections onto the finite Lefschetz strings are universal algebraic expressions in $L$, $\Lambda$, and $H=[\Lambda,L]$. Since $L$ is wedging with the smooth form $\omega$, since $\Lambda$ is its pointwise adjoint with respect to the smooth Kähler metric, and since $H$ is their commutator, these are smooth bundle endomorphisms. Therefore every $\Pi_{k,r}$ is a smooth bundle projection. Now let $\alpha \in \Omega^k(X;\mathbb{C})$. Define \begin{align*} \beta_r := \Pi_{k,r}\alpha \end{align*} for \begin{align*} \max\{0,k-n\} \le r \le \left\lfloor \frac{k}{2} \right\rfloor. \end{align*} Each $\beta_r$ is a smooth $k$-form because $\Pi_{k,r}$ is a smooth bundle map. Its image lies in $L^rP^{k-2r}$, so there is a unique smooth primitive form \begin{align*} \alpha_r \in \Omega^{k-2r}(X;\mathbb{C}) \end{align*} such that \begin{align*} \beta_r = L^r\alpha_r. \end{align*} Adding the projected pieces gives \begin{align*} \alpha = \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} \Pi_{k,r}\alpha = \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L^r\alpha_r. \end{align*} Finally, suppose another primitive decomposition of zero is given: \begin{align*} \sum_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L^r\gamma_r = 0. \end{align*} Evaluating at a point $x \in X$ gives a relation in the direct sum \begin{align*} A_x^k = \bigoplus_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L_x^rP_x^{k-2r}. \end{align*} Directness forces $\gamma_r(x)=0$ for every $r$. Since this holds for every $x \in X$, each $\gamma_r$ is the zero form. Therefore the decomposition of $\alpha$ is unique.[/guided]

[step:Use the Kähler identities to preserve harmonic forms]Assume now that $X$ is compact. Let \begin{align*} \Delta_d: \Omega^k(X;\mathbb{C}) \to \Omega^k(X;\mathbb{C}) \end{align*} denote the Hodge Laplacian of the Kähler metric. The Kähler identities imply \begin{align*} [\Delta_d,L]=0, \qquad [\Delta_d,\Lambda]=0 \end{align*} (citing a result not yet in the wiki: Kähler identities and commutation of the Lefschetz operators with the Hodge Laplacian). Since $H=[\Lambda,L]$, it follows that \begin{align*} [\Delta_d,H]=0. \end{align*} Each projection $\Pi_{k,r}$ is a universal polynomial in $L$, $\Lambda$, and $H$, so \begin{align*} [\Delta_d,\Pi_{k,r}]=0. \end{align*} Let $\alpha \in \Omega^k(X;\mathbb{C})$ be harmonic, so $\Delta_d\alpha=0$. Then \begin{align*} \Delta_d(\Pi_{k,r}\alpha) = \Pi_{k,r}(\Delta_d\alpha) = 0. \end{align*} Thus every summand $\Pi_{k,r}\alpha=L^r\alpha_r$ is harmonic. It remains to check that the primitive factor $\alpha_r$ itself is harmonic. Put $s:=k-2r$. On a primitive $s$-form $\theta$, the corrected commutation relation $[H,L]=-2L$ and the identity $H\theta=(n-s)\theta$ give, by induction on $m$, \begin{align*} \Lambda L^m\theta = m(n-s-m+1)L^{m-1}\theta \end{align*} for $1\le m\le r$. Applying this identity successively gives \begin{align*} \Lambda^rL^r\theta = c_{s,r}\theta, \qquad c_{s,r}:=r!\prod_{m=0}^{r-1}(n-s-m). \end{align*} The index range in the Lefschetz decomposition gives $r\le n-s$, so $c_{s,r}\ne0$. Since $\Lambda$ commutes with $\Delta_d$, we obtain \begin{align*} \Delta_d\alpha_r = c_{s,r}^{-1}\Delta_d\Lambda^r(L^r\alpha_r) = c_{s,r}^{-1}\Lambda^r\Delta_d(L^r\alpha_r) = 0. \end{align*} Therefore every primitive factor $\alpha_r$ is harmonic, and the primitive Lefschetz decomposition preserves harmonic forms.[/step]

[guided]Now assume $X$ is compact and let \begin{align*} \Delta_d: \Omega^k(X;\mathbb{C}) \to \Omega^k(X;\mathbb{C}) \end{align*} be the Hodge Laplacian of the Kähler metric. A form is harmonic precisely when it lies in the kernel of $\Delta_d$. The compatibility with harmonic forms rests on the Kähler identities. These identities imply the commutation relations \begin{align*} [\Delta_d,L]=0, \qquad [\Delta_d,\Lambda]=0 \end{align*} (citing a result not yet in the wiki: Kähler identities and commutation of the Lefschetz operators with the Hodge Laplacian). Since \begin{align*} H=[\Lambda,L], \end{align*} the Jacobi identity for commutators gives \begin{align*} [\Delta_d,H]=0. \end{align*} The projections $\Pi_{k,r}$ used above are universal polynomials in $L$, $\Lambda$, and $H$. Because $\Delta_d$ commutes with each of these three operators, it commutes with every polynomial built from them: \begin{align*} [\Delta_d,\Pi_{k,r}]=0. \end{align*} Let $\alpha \in \Omega^k(X;\mathbb{C})$ be harmonic. Then $\Delta_d\alpha=0$. Applying the projection $\Pi_{k,r}$ and commuting it past the Laplacian gives \begin{align*} \Delta_d(\Pi_{k,r}\alpha) = \Pi_{k,r}(\Delta_d\alpha) = 0. \end{align*} Therefore each projected Lefschetz summand is harmonic. Since \begin{align*} \Pi_{k,r}\alpha=L^r\alpha_r, \end{align*} we still need to prove that the primitive factor $\alpha_r$ is harmonic. Let $s:=k-2r$. If $\theta$ is a primitive $s$-form, then $\Lambda\theta=0$ and $H\theta=(n-s)\theta$. Using the corrected commutator convention $[H,L]=-2L$, we compute for $1\le m\le r$: \begin{align*} \Lambda L^m\theta &= L^m\Lambda\theta+[\Lambda,L^m]\theta \\ &= \sum_{j=0}^{m-1}L^jHL^{m-1-j}\theta \\ &= \sum_{j=0}^{m-1}(n-s-2(m-1-j))L^{m-1}\theta \\ &= m(n-s-m+1)L^{m-1}\theta. \end{align*} Here the first term vanishes because $\theta$ is primitive. Applying this formula successively gives \begin{align*} \Lambda^rL^r\theta = c_{s,r}\theta, \qquad c_{s,r}:=r!\prod_{m=0}^{r-1}(n-s-m). \end{align*} The Lefschetz index range gives $r\le n-s$, hence $c_{s,r}\ne0$. Applying this to $\theta=\alpha_r$ gives \begin{align*} \alpha_r=c_{s,r}^{-1}\Lambda^r(L^r\alpha_r). \end{align*} Because $\Lambda$ commutes with $\Delta_d$ and $L^r\alpha_r$ is harmonic, \begin{align*} \Delta_d\alpha_r = c_{s,r}^{-1}\Delta_d\Lambda^r(L^r\alpha_r) = c_{s,r}^{-1}\Lambda^r\Delta_d(L^r\alpha_r) = 0. \end{align*} Thus every primitive factor $\alpha_r$ is harmonic, so the decomposition of a harmonic form is a decomposition into harmonic primitive Lefschetz summands.[/guided]

[step:Transfer the harmonic decomposition to cohomology] Because $X$ is compact, the Hodge theorem gives an isomorphism \begin{align*} \mathcal{H}^k(X) &\to H^k(X;\mathbb{C}) \\ \alpha &\mapsto [\alpha], \end{align*} where \begin{align*} \mathcal{H}^k(X) := \ker(\Delta_d:\Omega^k(X;\mathbb{C})\to\Omega^k(X;\mathbb{C})) \end{align*} is the finite-dimensional space of harmonic $k$-forms (citing a result not yet in the wiki: Hodge theorem for compact Riemannian manifolds). Since the primitive Lefschetz projections preserve $\mathcal{H}^k(X)$, the direct sum decomposition \begin{align*} \mathcal{H}^k(X) = \bigoplus_{r=\max\{0,k-n\}}^{\lfloor k/2 \rfloor} L^r\bigl(P^{k-2r}(X)\cap \mathcal{H}^{k-2r}(X)\bigr) \end{align*} passes through this isomorphism to give the corresponding decomposition of $H^k(X;\mathbb{C})$. This proves the compact cohomological statement and completes the proof. [/step]