[proofplan] The estimate is pointwise. At a fixed point we choose unitary coordinates and a unitary frame of $E$, expand an $E$-valued $(n,q)$-form in the standard [orthonormal basis](/page/Orthonormal%20Basis), and write the curvature commutator as a sum of Nakano curvature quadratic forms indexed by the anti-holomorphic slots. The assumed Nakano lower bound applies to each such quadratic form. Finally, a counting argument shows that each coefficient of the $(n,q)$-form is counted exactly $q$ times, and integration over $X$ gives the global $L^2$ estimate. [/proofplan] [step:Expand the form in a unitary frame at a fixed point] Fix $x\in X$. Choose a unitary holomorphic coframe $(dz_1,\dots,dz_n)$ for $T_x^*X$ and a unitary frame $(e_\alpha)_\alpha$ for $E_x$. Let \begin{align*} dz_{[n]}:=dz_1\wedge\cdots\wedge dz_n. \end{align*} For an ordered subset $J=\{j_1<\cdots<j_q\}\subset\{1,\dots,n\}$, define \begin{align*} d\bar z_J:=d\bar z_{j_1}\wedge\cdots\wedge d\bar z_{j_q}. \end{align*} At $x$, write \begin{align*} \alpha_x=\sum_{|J|=q}\sum_\beta a_{J\beta}\,dz_{[n]}\wedge d\bar z_J\otimes e_\beta, \end{align*} where $a_{J\beta}\in\mathbb C$ are the coefficients of $\alpha_x$ in this unitary basis. Because the chosen coframe and frame are unitary, the pointwise norm satisfies \begin{align*} |\alpha_x|_{\omega,h}^2=\sum_{|J|=q}\sum_\beta |a_{J\beta}|^2. \end{align*} [/step] [step:Rewrite the curvature commutator as Nakano quadratic forms] For an ordered subset $K\subset\{1,\dots,n\}$ with $|K|=q-1$, and for each index $j\in\{1,\dots,n\}$, define coefficients $a_{jK,\alpha}\in\mathbb C$ as follows. If $j\in K$, set $a_{jK,\alpha}:=0$. If $j\notin K$, let $J=\{j\}\cup K$ with increasing order, and let $\varepsilon(j,K)\in\{-1,1\}$ be the sign determined by \begin{align*} d\bar z_j\wedge d\bar z_K=\varepsilon(j,K)\,d\bar z_J. \end{align*} Then set \begin{align*} a_{jK,\alpha}:=\varepsilon(j,K)a_{J\alpha}. \end{align*} In the above unitary frame, the standard pointwise formula for the curvature commutator on $E$-valued $(n,q)$-forms is \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha_x,\alpha_x\bigr)_{\omega,h}=\sum_{|K|=q-1}\sum_{j,k,\alpha,\beta} i\Theta(E,h)_{j\bar k\alpha\bar\beta}(x)\,a_{jK,\alpha}\,\overline{a_{kK,\beta}}. \end{align*} [guided] The reason $(n,q)$-forms are special is that the holomorphic part already contains every $dz_j$. Thus the only remaining indices on which the curvature commutator can act nontrivially are the anti-holomorphic indices. We encode this by removing one anti-holomorphic slot at a time. Fix an ordered subset $K\subset\{1,\dots,n\}$ with $|K|=q-1$. For each $j$, the coefficient $a_{jK,\alpha}$ is defined so that \begin{align*} a_{jK,\alpha}\,dz_{[n]}\wedge d\bar z_j\wedge d\bar z_K\otimes e_\alpha \end{align*} has the same coefficient as the corresponding component of $\alpha_x$. If $j\in K$, then $d\bar z_j\wedge d\bar z_K=0$, so we set $a_{jK,\alpha}=0$. If $j\notin K$, the sign $\varepsilon(j,K)$ records the permutation needed to put $\{j\}\cup K$ in increasing order. With this convention, the local expression for the Lefschetz contraction $\Lambda$ and the curvature operator $i\Theta(E,h)\wedge(\cdot)$ gives \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha_x,\alpha_x\bigr)_{\omega,h}=\sum_{|K|=q-1}\sum_{j,k,\alpha,\beta} i\Theta(E,h)_{j\bar k\alpha\bar\beta}(x)\,a_{jK,\alpha}\,\overline{a_{kK,\beta}}. \end{align*} This is exactly the Nakano curvature form evaluated on the tensors obtained by fixing all but one anti-holomorphic slot. The signs disappear from the final quadratic expression because they are already incorporated into the coefficients $a_{jK,\alpha}$. [/guided] [/step] [step:Apply the Nakano lower bound to each anti-holomorphic slot] For each ordered subset $K\subset\{1,\dots,n\}$ with $|K|=q-1$, define a tensor \begin{align*} u_K:=\sum_{j,\alpha}a_{jK,\alpha}\,\partial_{z_j}\otimes e_\alpha\in T_xX\otimes E_x. \end{align*} The assumed Nakano lower bound applied to $u_K$ gives \begin{align*} \sum_{j,k,\alpha,\beta} i\Theta(E,h)_{j\bar k\alpha\bar\beta}(x)\,a_{jK,\alpha}\,\overline{a_{kK,\beta}}\ge c\sum_{j,\alpha}|a_{jK,\alpha}|^2. \end{align*} Summing this inequality over all $K$ with $|K|=q-1$ yields \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha_x,\alpha_x\bigr)_{\omega,h}\ge c\sum_{|K|=q-1}\sum_{j,\alpha}|a_{jK,\alpha}|^2. \end{align*} [/step] [step:Count how many times each coefficient occurs] Fix an ordered subset $J\subset\{1,\dots,n\}$ with $|J|=q$. For each $j\in J$, there is exactly one ordered subset $K=J\setminus\{j\}$ with $|K|=q-1$ such that $\{j\}\cup K=J$. For $j\notin J$, the coefficient $a_{J\alpha}$ does not occur in $a_{jK,\alpha}$ for any $K$ with $\{j\}\cup K=J$. Hence each number $|a_{J\alpha}|^2$ appears exactly $q$ times in the sum \begin{align*} \sum_{|K|=q-1}\sum_{j,\alpha}|a_{jK,\alpha}|^2. \end{align*} Therefore \begin{align*} \sum_{|K|=q-1}\sum_{j,\alpha}|a_{jK,\alpha}|^2=q\sum_{|J|=q}\sum_\alpha |a_{J\alpha}|^2. \end{align*} Using the pointwise norm identity from the first step, we obtain \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha_x,\alpha_x\bigr)_{\omega,h}\ge qc|\alpha_x|_{\omega,h}^2. \end{align*} [/step] [step:Integrate the pointwise estimate over the compact Kähler manifold] Let $dV_\omega$ denote the Riemannian volume measure induced by the Kähler metric $\omega$, equivalently \begin{align*} dV_\omega=\frac{\omega^n}{n!}. \end{align*} The pointwise estimate holds for every $x\in X$. Since $X$ is compact and $\alpha$ is smooth, both pointwise functions \begin{align*} x\mapsto \bigl([i\Theta(E,h),\Lambda]\alpha_x,\alpha_x\bigr)_{\omega,h} \end{align*} and \begin{align*} x\mapsto |\alpha_x|_{\omega,h}^2 \end{align*} are continuous and therefore integrable with respect to $dV_\omega$. Integrating gives \begin{align*} \int_X \bigl([i\Theta(E,h),\Lambda]\alpha_x,\alpha_x\bigr)_{\omega,h}\,dV_\omega(x)\ge qc\int_X |\alpha_x|_{\omega,h}^2\,dV_\omega(x). \end{align*} By the definitions of the $L^2$ [inner product](/page/Inner%20Product) and $L^2$ norm on $E$-valued forms, this is \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}\ge qc\|\alpha\|_{L^2}^2. \end{align*} This proves the desired coercivity estimate. [/step]