[proofplan] The proof is a coercivity argument. We take an arbitrary $\bar\partial_E$-harmonic $E$-valued $(n,q)$-form and apply the Bochner-Kodaira-Nakano identity with the same curvature-commutator convention as in the hypothesis. Harmonicity makes the Dolbeault Laplacian term vanish, while the curvature lower bound forces the $L^2$ norm of the form to vanish. The Dolbeault-Hodge identification for holomorphic vector bundles then converts the absence of [harmonic representatives](/theorems/2747) into the vanishing of $H^q(X,K_X\otimes E)$. [/proofplan]

[step:Apply the Bochner-Kodaira-Nakano identity to a harmonic representative]Let \begin{align*} \alpha\in A^{n,q}(X,E) \end{align*} be a $\bar\partial_E$-harmonic form. Thus \begin{align*} \bar\partial_E\alpha=0 \end{align*} and \begin{align*} \bar\partial_E^*\alpha=0, \end{align*} where $\bar\partial_E^*$ denotes the $L^2$-adjoint of $\bar\partial_E$ with respect to $\omega$ and $h$. Let $\nabla'$ denote the $(1,0)$-part of the Chern connection on $(E,h)$, and let $(\nabla')^*$ denote its $L^2$-adjoint. The hypotheses for the Bochner-Kodaira-Nakano identity are satisfied because $(X,\omega)$ is compact Kähler, $(E,h)$ is a Hermitian holomorphic vector bundle, and $\alpha$ is a smooth $E$-valued form. By that identity, with curvature convention matching the commutator $[i\Theta(E,h),\Lambda]$ in the hypothesis, we have \begin{align*} \|\bar\partial_E\alpha\|_{L^2}^2+\|\bar\partial_E^*\alpha\|_{L^2}^2 = \|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}. \end{align*} Since $\alpha$ is harmonic, the left-hand side is zero. Hence \begin{align*} 0 = \|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}. \end{align*}[/step]

[guided]We begin with an arbitrary harmonic representative because the theorem asserts that every such representative must vanish. The word harmonic here means harmonic for the Dolbeault Laplacian \begin{align*} \Delta_{\bar\partial,E}=\bar\partial_E\bar\partial_E^*+\bar\partial_E^*\bar\partial_E. \end{align*} Equivalently, on a compact Hermitian manifold, harmonicity gives the two first-order equations \begin{align*} \bar\partial_E\alpha=0 \end{align*} and \begin{align*} \bar\partial_E^*\alpha=0. \end{align*} The identity we need is the Bochner-Kodaira-Nakano identity with the sign convention in which the curvature term is exactly \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}. \end{align*} Here $\nabla'$ is the $(1,0)$-part of the Chern connection on $(E,h)$, and $(\nabla')^*$ is its $L^2$-adjoint. The hypotheses of the Bochner-Kodaira-Nakano identity are met: $(X,\omega)$ is compact Kähler, $(E,h)$ is a Hermitian holomorphic vector bundle, and $\alpha$ is a smooth $E$-valued $(n,q)$-form. Applied to $\alpha$, the identity gives \begin{align*} \|\bar\partial_E\alpha\|_{L^2}^2+\|\bar\partial_E^*\alpha\|_{L^2}^2 = \|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}. \end{align*} The reason this identity is useful is that harmonicity annihilates the left-hand side. Indeed, both summands on the left are squared $L^2$ norms of zero forms, so \begin{align*} \|\bar\partial_E\alpha\|_{L^2}^2+\|\bar\partial_E^*\alpha\|_{L^2}^2=0. \end{align*} Therefore the identity becomes \begin{align*} 0 = \|\nabla'\alpha\|_{L^2}^2+\|(\nabla')^*\alpha\|_{L^2}^2+\bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}. \end{align*} This is the central point: the analytic equation for harmonic forms has been converted into a sum of two nonnegative energy terms plus the curvature term appearing in the hypothesis.[/guided]

[step:Use the curvature lower bound to force the harmonic form to vanish] The squared norms \begin{align*} \|\nabla'\alpha\|_{L^2}^2 \end{align*} and \begin{align*} \|(\nabla')^*\alpha\|_{L^2}^2 \end{align*} are nonnegative. Therefore the preceding identity implies \begin{align*} 0\ge \bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}. \end{align*} The assumed lower bound applies to this smooth $E$-valued $(n,q)$-form $\alpha$, so \begin{align*} \bigl([i\Theta(E,h),\Lambda]\alpha,\alpha\bigr)_{L^2}\ge c\|\alpha\|_{L^2}^2. \end{align*} Combining the two inequalities gives \begin{align*} 0\ge c\|\alpha\|_{L^2}^2. \end{align*} Since $c>0$ and $\|\alpha\|_{L^2}^2\ge 0$, it follows that \begin{align*} \|\alpha\|_{L^2}^2=0. \end{align*} Thus $\alpha=0$ as an $L^2$ form. Since $\alpha$ is smooth, it vanishes identically on $X$. [/step]

[step:Identify Dolbeault cohomology with harmonic representatives] By definition of the canonical bundle, $K_X=\Omega_X^n$. By [citetheorem:9104], applied with $p=n$, the natural map from $\bar\partial_E$-harmonic $E$-valued $(n,q)$-forms to Dolbeault cohomology is an isomorphism: \begin{align*} \ker\left(\Delta_{\bar\partial,E}:A^{n,q}(X,E)\to A^{n,q}(X,E)\right)\cong H^q(X,K_X\otimes E). \end{align*} The hypotheses of that theorem are satisfied because $X$ is compact Kähler and $E\to X$ is a holomorphic Hermitian vector bundle. Since every element of the harmonic space on the left is zero by the preceding step, the [vector space](/page/Vector%20Space) on the right is zero. Hence \begin{align*} H^q(X,K_X\otimes E)=0. \end{align*} [/step]