[proofplan] The proof is a pointwise exterior-algebra computation in a unitary frame. We write the curvature wedge operator and the Kähler contraction $\Lambda$ in terms of creation and annihilation operators for the basis forms $\theta_j$ and $\overline{\theta_j}$. On a basis element of bidegree $(n,q)$, every holomorphic direction is already occupied, so the commutator detects precisely the antiholomorphic directions appearing in $J$. Orthogonality of the standard unitary wedge basis then gives the stated quadratic form. [/proofplan] [step:Introduce the exterior algebra operators at the fixed point] Let \begin{align*} V:=T_x^{1,0}X \end{align*} and let $\Lambda^{p,q}V^*$ denote the space of scalar $(p,q)$-covectors at $x$. For each $j\in\{1,\dots,n\}$, define the exterior multiplication maps \begin{align*} \varepsilon_j:\Lambda^{p,q}V^*&\to \Lambda^{p+1,q}V^*,& \alpha&\mapsto \theta_j\wedge \alpha,\\ \overline{\varepsilon}_j:\Lambda^{p,q}V^*&\to \Lambda^{p,q+1}V^*,& \alpha&\mapsto \overline{\theta_j}\wedge \alpha. \end{align*} Let $\varepsilon_j^*$ and $\overline{\varepsilon}_j^*$ be their adjoints with respect to the Hermitian inner product for which the ordered wedge monomials \begin{align*} \theta_I\wedge \overline{\theta_J} \end{align*} are orthonormal. With the convention \begin{align*} \omega_x=i\sum_{j=1}^n \theta_j\wedge \overline{\theta_j}, \end{align*} the Lefschetz operator is exterior multiplication by $\omega_x$, and its adjoint is \begin{align*} \Lambda=-i\sum_{j=1}^n \overline{\varepsilon}_j^*\varepsilon_j^*. \end{align*} Similarly, exterior multiplication by the curvature form is \begin{align*} i\Theta_h(L)_x = i\sum_{j=1}^n \lambda_j\,\varepsilon_j\overline{\varepsilon}_j. \end{align*} Since $e$ is a unit vector in the Hermitian line $L_x$, these scalar exterior-algebra operators act on $\Lambda^{p,q}V^*\otimes L_x$ by tensoring with $\operatorname{id}_{L_x}$. [guided] We work entirely at the fixed point $x$, so the computation is finite-dimensional linear algebra. The unitary coframe $\theta_1,\dots,\theta_n$ gives an orthonormal basis of wedge monomials, and the unit vector $e\in L_x$ lets us ignore the line-bundle factor except for the scalar coefficients $u_J$. For each index $j$, the operator $\varepsilon_j$ adds the holomorphic factor $\theta_j$, while $\overline{\varepsilon}_j$ adds the antiholomorphic factor $\overline{\theta_j}$: \begin{align*} \varepsilon_j:\Lambda^{p,q}V^*&\to \Lambda^{p+1,q}V^*,& \alpha&\mapsto \theta_j\wedge \alpha,\\ \overline{\varepsilon}_j:\Lambda^{p,q}V^*&\to \Lambda^{p,q+1}V^*,& \alpha&\mapsto \overline{\theta_j}\wedge \alpha. \end{align*} Their adjoints $\varepsilon_j^*$ and $\overline{\varepsilon}_j^*$ remove those factors when present, with the sign dictated by the exterior algebra and with value $0$ when the factor is absent. Because the coframe is unitary and \begin{align*} \omega_x=i\sum_{j=1}^n \theta_j\wedge \overline{\theta_j}, \end{align*} the adjoint of exterior multiplication by $\omega_x$ is \begin{align*} \Lambda=-i\sum_{j=1}^n \overline{\varepsilon}_j^*\varepsilon_j^*. \end{align*} The diagonal curvature hypothesis says that exterior multiplication by $i\Theta_h(L)_x$ is \begin{align*} i\Theta_h(L)_x = i\sum_{j=1}^n \lambda_j\,\varepsilon_j\overline{\varepsilon}_j. \end{align*} Thus both operators are sums of independent one-direction operators, one for each $j$. [/guided] [/step] [step:Compute the one-direction commutator on a top holomorphic basis element] Fix an ordered set $J\subset \{1,\dots,n\}$ with $|J|=q$, and define \begin{align*} \eta_J:=\theta_1\wedge\cdots\wedge\theta_n\wedge \overline{\theta_J}\in \Lambda^{n,q}V^*. \end{align*} For $j\notin J$, the form $\eta_J$ contains $\theta_j$ but does not contain $\overline{\theta_j}$. Hence \begin{align*} \varepsilon_j\overline{\varepsilon}_j\eta_J=0 \end{align*} because $\varepsilon_j$ attempts to wedge in a second copy of $\theta_j$, and \begin{align*} \overline{\varepsilon}_j^*\varepsilon_j^*\varepsilon_j\overline{\varepsilon}_j\eta_J=0. \end{align*} Also $\overline{\varepsilon}_j^*\varepsilon_j^*\eta_J=0$, because $\overline{\theta_j}$ is absent. Thus the $j$-summand of the commutator contributes $0$. For $j\in J$, both $\theta_j$ and $\overline{\theta_j}$ occur in $\eta_J$. The operator $\overline{\varepsilon}_j^*\varepsilon_j^*$ removes these two factors, while $\varepsilon_j\overline{\varepsilon}_j$ restores them. Since $\varepsilon_j$ and $\overline{\varepsilon}_j$ are adjoint to the corresponding removals in the orthonormal wedge basis, the combined remove-and-restore operation fixes $\eta_J$: \begin{align*} \varepsilon_j\overline{\varepsilon}_j\overline{\varepsilon}_j^*\varepsilon_j^*\eta_J=\eta_J. \end{align*} The reverse composition vanishes: \begin{align*} \overline{\varepsilon}_j^*\varepsilon_j^*\varepsilon_j\overline{\varepsilon}_j\eta_J=0, \end{align*} because $\varepsilon_j\overline{\varepsilon}_j\eta_J=0$. Therefore the $j$-summand of $[i\Theta_h(L),\Lambda]$ acts on $\eta_J$ as multiplication by $\lambda_j$. Combining the cases $j\in J$ and $j\notin J$ gives \begin{align*} [i\Theta_h(L),\Lambda]\eta_J = \left(\sum_{j\in J}\lambda_j\right)\eta_J. \end{align*} [guided] The point of using top holomorphic degree is that every $\eta_J$ already contains all holomorphic factors: \begin{align*} \eta_J=\theta_1\wedge\cdots\wedge\theta_n\wedge \overline{\theta_J}. \end{align*} So if a curvature wedge term tries to add another $\theta_j$, it immediately vanishes. First suppose $j\notin J$. Then $\eta_J$ contains $\theta_j$, but it does not contain $\overline{\theta_j}$. Applying $\varepsilon_j\overline{\varepsilon}_j$ to $\eta_J$ gives $0$, because after wedging by $\overline{\theta_j}$ the operator $\varepsilon_j$ wedges by $\theta_j$, and $\theta_j$ is already present. Also \begin{align*} \overline{\varepsilon}_j^*\varepsilon_j^*\eta_J=0, \end{align*} because the contraction $\overline{\varepsilon}_j^*$ has no $\overline{\theta_j}$ factor to remove. Therefore direction $j$ contributes nothing when $j\notin J$. Now suppose $j\in J$. Then $\eta_J$ contains both $\theta_j$ and $\overline{\theta_j}$. Applying $\overline{\varepsilon}_j^*\varepsilon_j^*$ removes exactly these two factors, up to the exterior-algebra sign determined by their positions. Applying $\varepsilon_j\overline{\varepsilon}_j$ then wedges the same two unit factors back in. Because these operators are adjoints with respect to the orthonormal wedge basis, the signs cancel in the remove-and-restore composition: \begin{align*} \varepsilon_j\overline{\varepsilon}_j\overline{\varepsilon}_j^*\varepsilon_j^*\eta_J=\eta_J. \end{align*} The reverse order still vanishes: \begin{align*} \overline{\varepsilon}_j^*\varepsilon_j^*\varepsilon_j\overline{\varepsilon}_j\eta_J=0, \end{align*} since the first operation $\varepsilon_j\overline{\varepsilon}_j\eta_J$ attempts to wedge in a duplicate $\theta_j$. Thus the commutator in the $j$-direction detects exactly whether $\overline{\theta_j}$ appears in $\overline{\theta_J}$. It contributes $\lambda_j$ if $j\in J$ and $0$ otherwise, hence \begin{align*} [i\Theta_h(L),\Lambda]\eta_J = \left(\sum_{j\in J}\lambda_j\right)\eta_J. \end{align*} [/guided] [/step] [step:Take the pointwise inner product and use orthonormality] The forms \begin{align*} \eta_J\otimes e = \theta_1\wedge\cdots\wedge\theta_n\wedge\overline{\theta_J}\otimes e, \qquad |J|=q, \end{align*} form an orthonormal basis of $\Lambda^{n,q}T_x^*X\otimes L_x$ for the pointwise Hermitian inner product induced by $\omega$ and $h$. Since \begin{align*} u=\sum_{|J|=q}u_J\,\eta_J\otimes e \end{align*} and each $\eta_J\otimes e$ is an eigenvector of $[i\Theta_h(L),\Lambda]$ with eigenvalue $\sum_{j\in J}\lambda_j$, we obtain \begin{align*} [i\Theta_h(L),\Lambda]u = \sum_{|J|=q} \left(\sum_{j\in J}\lambda_j\right)u_J\,\eta_J\otimes e. \end{align*} Taking the pointwise Hermitian inner product with $u$ and using orthonormality eliminates all cross terms: \begin{align*} \bigl([i\Theta_h(L),\Lambda]u,u\bigr)_x &= \sum_{|J|=q} \left(\sum_{j\in J}\lambda_j\right)u_J\overline{u_J}\\ &= \sum_{|J|=q} \left(\sum_{j\in J}\lambda_j\right)|u_J|^2. \end{align*} This is the asserted formula. [/step]