[proofplan] The proof is a pointwise exterior-algebra calculation at the fixed point $x$. We express wedging and contraction by each coordinate form as creation and annihilation operators. Since the curvature form is diagonal in the chosen unitary coframe, the commutator splits into a sum of independent one-coordinate commutators. Each coordinate contributes $\lambda_j$ exactly when $\overline{\zeta_j}$ is present and contributes $-\lambda_j$ exactly when $\zeta_j$ is absent, giving the stated eigenvalue after summing over $j$. [/proofplan]

[step:Introduce the coordinate creation and annihilation operators] Fix $x\in X$ and suppress the fibre $L_x$, since $L_x$ is one-dimensional and the curvature endomorphism is scalar multiplication by the real $(1,1)$-form $i\Theta(L,h)_x$ in the chosen frame. Let $A_x$ denote the complex exterior algebra of forms at $x$, decomposed by bidegree. For each $j\in\{1,\dots,n\}$, define the creation operators \begin{align*} \varepsilon_j:A_x\to A_x \end{align*} by $\varepsilon_j(\alpha)=\zeta_j\wedge \alpha$, and \begin{align*} \overline{\varepsilon}_j:A_x\to A_x \end{align*} by $\overline{\varepsilon}_j(\alpha)=\overline{\zeta_j}\wedge \alpha$. Define the annihilation operators \begin{align*} \iota_j:A_x\to A_x \end{align*} and \begin{align*} \overline{\iota}_j:A_x\to A_x \end{align*} to be contraction with the unitary dual vectors to $\zeta_j$ and $\overline{\zeta_j}$, respectively. These operators satisfy the canonical anti-commutation relations \begin{align*} \iota_j\varepsilon_j+\varepsilon_j\iota_j=\operatorname{id},\qquad \overline{\iota}_j\overline{\varepsilon}_j+\overline{\varepsilon}_j\overline{\iota}_j=\operatorname{id}. \end{align*} For different indices, the corresponding creation and annihilation operators anti-commute in the usual exterior-algebra sense. Because the coframe is unitary, the Kähler form at $x$ is \begin{align*} \omega_x=i\sum_{j=1}^n \zeta_j\wedge\overline{\zeta_j}. \end{align*} With the convention that $\Lambda_\omega$ is the adjoint Lefschetz contraction by this unitary form, this gives \begin{align*} \Lambda_\omega=-i\sum_{j=1}^n \overline{\iota}_j\iota_j. \end{align*} The curvature hypothesis gives \begin{align*} i\Theta(L,h)_x=i\sum_{j=1}^n \lambda_j\,\varepsilon_j\overline{\varepsilon}_j. \end{align*} [/step]

[step:Reduce the commutator to independent one-coordinate operators]Using the preceding formulas and bilinearity of the commutator, define for each $j$ the operator \begin{align*} C_j:A_x\to A_x \end{align*} by \begin{align*} C_j=\varepsilon_j\overline{\varepsilon}_j\overline{\iota}_j\iota_j-\overline{\iota}_j\iota_j\varepsilon_j\overline{\varepsilon}_j. \end{align*} Then \begin{align*} [i\Theta(L,h),\Lambda_\omega]_x=\sum_{j=1}^n\lambda_j C_j. \end{align*} Indeed, when $j\ne k$, the operator $\varepsilon_j\overline{\varepsilon}_j$ commutes with $\overline{\iota}_k\iota_k$ because moving the two creation operators past the two annihilation operators produces four sign changes. Thus all mixed commutators vanish, and only the terms with the same index survive.[/step]

[guided]We want to understand the commutator \begin{align*} [i\Theta(L,h),\Lambda_\omega]_x=i\Theta(L,h)_x\Lambda_\omega-\Lambda_\omega i\Theta(L,h)_x. \end{align*} The reason for introducing creation and annihilation operators is that the curvature and the Lefschetz contraction are both sums of elementary one-coordinate operations. In the chosen unitary coframe, \begin{align*} i\Theta(L,h)_x=i\sum_{j=1}^n\lambda_j\varepsilon_j\overline{\varepsilon}_j \end{align*} and \begin{align*} \Lambda_\omega=-i\sum_{k=1}^n\overline{\iota}_k\iota_k. \end{align*} Multiplying these two sums gives \begin{align*} i\Theta(L,h)_x\Lambda_\omega=\sum_{j=1}^n\sum_{k=1}^n\lambda_j\varepsilon_j\overline{\varepsilon}_j\overline{\iota}_k\iota_k \end{align*} and \begin{align*} \Lambda_\omega i\Theta(L,h)_x=\sum_{j=1}^n\sum_{k=1}^n\lambda_j\overline{\iota}_k\iota_k\varepsilon_j\overline{\varepsilon}_j. \end{align*} If $j\ne k$, moving $\varepsilon_j\overline{\varepsilon}_j$ past $\overline{\iota}_k\iota_k$ creates four exterior-algebra sign changes: $\varepsilon_j$ crosses $\overline{\iota}_k$ and $\iota_k$, and $\overline{\varepsilon}_j$ crosses $\overline{\iota}_k$ and $\iota_k$. Four sign changes multiply to $+1$, so \begin{align*} \varepsilon_j\overline{\varepsilon}_j\overline{\iota}_k\iota_k=\overline{\iota}_k\iota_k\varepsilon_j\overline{\varepsilon}_j. \end{align*} Therefore every mixed commutator with $j\ne k$ is zero. The only surviving terms are the same-index terms, so the commutator is \begin{align*} [i\Theta(L,h),\Lambda_\omega]_x=\sum_{j=1}^n\lambda_j\left(\varepsilon_j\overline{\varepsilon}_j\overline{\iota}_j\iota_j-\overline{\iota}_j\iota_j\varepsilon_j\overline{\varepsilon}_j\right). \end{align*} This is exactly the asserted decomposition into the operators $C_j$.[/guided]

[step:Compute one coordinate contribution on a basis component] Let $I,J\subset\{1,\dots,n\}$ be increasing index sets and define \begin{align*} \alpha_{I,J}=\zeta_I\wedge\overline{\zeta_J}. \end{align*} Fix $j\in\{1,\dots,n\}$. There are four cases. If $j\in I$ and $j\in J$, then $\overline{\iota}_j\iota_j\alpha_{I,J}$ removes both factors $\zeta_j$ and $\overline{\zeta_j}$, and $\varepsilon_j\overline{\varepsilon}_j$ restores them with the same total exterior sign. To see the sign, let \begin{align*} a=\#\{i\in I:i<j\},\qquad b=\#\{k\in J:k<j\}. \end{align*} The contraction $\iota_j$ contributes $(-1)^a$, the contraction $\overline{\iota}_j$ then contributes $(-1)^{p-1+b}$, and restoring the two factors by $\varepsilon_j\overline{\varepsilon}_j$ contributes the same sign $(-1)^{a+p-1+b}$ when the result is reordered into the standard order with all holomorphic factors before all antiholomorphic factors. The product of these two equal signs is $1$. Meanwhile $\varepsilon_j\overline{\varepsilon}_j\alpha_{I,J}=0$ because both factors are already present. Hence \begin{align*} C_j\alpha_{I,J}=\alpha_{I,J}. \end{align*} If $j\in I$ and $j\notin J$, then $\overline{\iota}_j\iota_j\alpha_{I,J}=0$ because $\overline{\zeta_j}$ is absent, while $\varepsilon_j\overline{\varepsilon}_j\alpha_{I,J}=0$ because $\zeta_j$ is already present. Hence \begin{align*} C_j\alpha_{I,J}=0. \end{align*} If $j\notin I$ and $j\in J$, then $\overline{\iota}_j\iota_j\alpha_{I,J}=0$ because $\zeta_j$ is absent, while $\varepsilon_j\overline{\varepsilon}_j\alpha_{I,J}=0$ because $\overline{\zeta_j}$ is already present. Hence \begin{align*} C_j\alpha_{I,J}=0. \end{align*} If $j\notin I$ and $j\notin J$, then $\overline{\iota}_j\iota_j\alpha_{I,J}=0$, while $\varepsilon_j\overline{\varepsilon}_j\alpha_{I,J}$ inserts both factors and $\overline{\iota}_j\iota_j$ removes them with the same total exterior sign. Hence \begin{align*} C_j\alpha_{I,J}=-\alpha_{I,J}. \end{align*} For a condition $P$, let $\mathbb{1}_P$ denote the indicator number, equal to $1$ when $P$ is true and equal to $0$ when $P$ is false. Thus, in all cases, \begin{align*} C_j\alpha_{I,J}=\left(\mathbb{1}_{j\in J}-\mathbb{1}_{j\notin I}\right)\alpha_{I,J}. \end{align*} [/step]

[step:Sum the coordinate contributions to obtain the eigenvalue] Let $e\in L_x$. Since $C_j$ acts only on the exterior-form factor, the preceding computation gives \begin{align*} C_j(\alpha_{I,J}\otimes e)=\left(\mathbb{1}_{j\in J}-\mathbb{1}_{j\notin I}\right)(\alpha_{I,J}\otimes e). \end{align*} Therefore \begin{align*} [i\Theta(L,h),\Lambda_\omega]_x(\alpha_{I,J}\otimes e)=\sum_{j=1}^n\lambda_j\left(\mathbb{1}_{j\in J}-\mathbb{1}_{j\notin I}\right)(\alpha_{I,J}\otimes e). \end{align*} Rewriting the two indicator sums gives \begin{align*} [i\Theta(L,h),\Lambda_\omega]_x(\alpha_{I,J}\otimes e)=\left(\sum_{j\in J}\lambda_j-\sum_{j\notin I}\lambda_j\right)(\alpha_{I,J}\otimes e). \end{align*} Hence every standard component $\zeta_I\wedge\overline{\zeta_J}\otimes e$ is an eigenvector of the curvature commutator, with eigenvalue \begin{align*} \sum_{j\in J}\lambda_j-\sum_{j\notin I}\lambda_j. \end{align*} This proves the stated diagonal action. [/step]