[proofplan] We identify the sheaf cohomology group $H^q(X,K_X\otimes E)$ with Dolbeault cohomology represented by smooth $E$-valued $(n,q)$-forms, and then use Hodge theory on the compact Kähler manifold to choose the unique harmonic representative of any cohomology class. The Bochner-Kodaira-Nakano identity expresses the Dolbeault Laplacian on that representative as a sum of nonnegative connection terms plus the curvature commutator $[i\Theta_h(E),\Lambda]$. Nakano positivity makes this commutator strictly positive on $E$-valued $(n,q)$-forms when $q>0$, so the harmonic representative must have zero $L^2$ norm. Therefore every cohomology class is zero. [/proofplan]

[step:Represent a cohomology class by a harmonic $E$-valued $(n,q)$-form]Fix an integer $q>0$. Let $\mathcal A^{n,q}(X,E)$ denote the space of smooth $E$-valued $(n,q)$-forms on $X$, and let \begin{align*} \bar\partial_E:\mathcal A^{n,q}(X,E)&\to \mathcal A^{n,q+1}(X,E) \end{align*} be the Dolbeault operator induced by the holomorphic structure on $E$. Let $[\alpha]\in H^q(X,K_X\otimes E)$ be arbitrary. By the Dolbeault isomorphism for holomorphic vector bundles on compact complex manifolds (citing a result not yet in the wiki: Dolbeault Theorem for Vector Bundles), there exists a $\bar\partial_E$-closed form $\alpha\in\mathcal A^{n,q}(X,E)$ representing $[\alpha]$. Equip $\mathcal A^{n,q}(X,E)$ with the $L^2$ inner product induced by $\omega$ and $h$. Let $d\mu_\omega$ be the Riemannian measure associated to the Kähler metric $\omega$, equivalently $d\mu_\omega=\omega^n/n!$. For $u,v\in\mathcal A^{n,q}(X,E)$ define \begin{align*} (u,v)_{L^2}:=\int_X \langle u(x),v(x)\rangle_{\omega,h}\,d\mu_\omega(x), \end{align*} where $\langle\cdot,\cdot\rangle_{\omega,h}$ is the pointwise Hermitian inner product induced by $\omega$ on forms and by $h$ on $E$. Since $X$ is compact Kähler, Hodge theory for the elliptic Dolbeault Laplacian applies (citing a result not yet in the wiki: Hodge Theorem for the Dolbeault Laplacian). Thus the class $[\alpha]$ has a harmonic representative \begin{align*} u\in\mathcal A^{n,q}(X,E) \end{align*} satisfying \begin{align*} \bar\partial_E u=0,\qquad \bar\partial_E^*u=0, \end{align*} where $\bar\partial_E^*:\mathcal A^{n,q}(X,E)\to\mathcal A^{n,q-1}(X,E)$ is the $L^2$ adjoint of $\bar\partial_E$.[/step]

[guided]We start with an arbitrary cohomology class and convert it into an analytic object to which curvature estimates can be applied. The Dolbeault theorem for vector bundles identifies \begin{align*} H^q(X,K_X\otimes E) \end{align*} with the $q$-th cohomology of the complex \begin{align*} \mathcal A^{n,0}(X,E)\xrightarrow{\bar\partial_E}\mathcal A^{n,1}(X,E)\xrightarrow{\bar\partial_E}\cdots . \end{align*} Thus an element $[\alpha]\in H^q(X,K_X\otimes E)$ can be represented by a smooth $E$-valued $(n,q)$-form $\alpha$ with $\bar\partial_E\alpha=0$. The compact Kähler metric supplies an $L^2$ inner product. Explicitly, if $d\mu_\omega=\omega^n/n!$ denotes the measure determined by $\omega$, then for $u,v\in\mathcal A^{n,q}(X,E)$ we set \begin{align*} (u,v)_{L^2}:=\int_X \langle u(x),v(x)\rangle_{\omega,h}\,d\mu_\omega(x). \end{align*} This defines the formal adjoint $\bar\partial_E^*$ of $\bar\partial_E$. Since $X$ is compact, the Dolbeault Laplacian \begin{align*} \Box_{\bar\partial_E}:=\bar\partial_E\bar\partial_E^*+\bar\partial_E^*\bar\partial_E \end{align*} is elliptic, and Hodge theory gives a unique harmonic representative in each Dolbeault cohomology class. Therefore we may choose \begin{align*} u\in\mathcal A^{n,q}(X,E) \end{align*} representing $[\alpha]$ such that \begin{align*} \Box_{\bar\partial_E}u=0. \end{align*} Taking the $L^2$ inner product with $u$ gives \begin{align*} 0=(\Box_{\bar\partial_E}u,u)_{L^2} =\|\bar\partial_Eu\|_{L^2}^2+\|\bar\partial_E^*u\|_{L^2}^2, \end{align*} so both terms vanish: \begin{align*} \bar\partial_Eu=0,\qquad \bar\partial_E^*u=0. \end{align*}[/guided]

[step:Apply the Bochner-Kodaira-Nakano identity to the harmonic representative]Let \begin{align*} \nabla^E:\mathcal A^{p,q}(X,E)&\to \mathcal A^{p+1,q}(X,E)\oplus\mathcal A^{p,q+1}(X,E) \end{align*} be the Chern connection of $(E,h)$, decomposed as $\nabla^E=\partial_E+\bar\partial_E$. Let \begin{align*} \Lambda:\mathcal A^{p,q}(X,E)&\to\mathcal A^{p-1,q-1}(X,E) \end{align*} be the adjoint of wedging with $\omega$. The Bochner-Kodaira-Nakano identity for $E$-valued forms on a Kähler manifold states (citing a result not yet in the wiki: Bochner-Kodaira-Nakano Identity) that \begin{align*} \Box_{\bar\partial_E}=\Box_{\partial_E}+[i\Theta_h(E),\Lambda], \end{align*} where $\Theta_h(E)\in\mathcal A^{1,1}(X,\operatorname{End}(E))$ is the Chern curvature and \begin{align*} \Box_{\partial_E}:=\partial_E\partial_E^*+\partial_E^*\partial_E. \end{align*} Taking the $L^2$ inner product with the harmonic form $u$ gives \begin{align*} 0 &=(\Box_{\bar\partial_E}u,u)_{L^2} \\ &=(\Box_{\partial_E}u,u)_{L^2}+([i\Theta_h(E),\Lambda]u,u)_{L^2} \\ &=\|\partial_Eu\|_{L^2}^2+\|\partial_E^*u\|_{L^2}^2+([i\Theta_h(E),\Lambda]u,u)_{L^2}. \end{align*} The first two terms are nonnegative [real numbers](/page/Real%20Numbers) because they are squared $L^2$ norms.[/step]

[guided]The next step is the analytic heart of the proof. The Chern connection $\nabla^E$ associated to the Hermitian holomorphic bundle $(E,h)$ decomposes into types: \begin{align*} \nabla^E=\partial_E+\bar\partial_E. \end{align*} The Kähler form $\omega$ defines the Lefschetz operator $L$ by wedging with $\omega$, and $\Lambda$ denotes the $L^2$ adjoint of $L$. The Bochner-Kodaira-Nakano identity compares the two Laplacians associated to $\partial_E$ and $\bar\partial_E$: \begin{align*} \Box_{\bar\partial_E}=\Box_{\partial_E}+[i\Theta_h(E),\Lambda]. \end{align*} Here $\Theta_h(E)$ is the curvature of the Chern connection, and $[i\Theta_h(E),\Lambda]$ is the graded commutator acting on $E$-valued forms. This identity applies because $X$ is Kähler and $\nabla^E$ is the Chern connection of a Hermitian holomorphic vector bundle. Since $u$ is harmonic for $\Box_{\bar\partial_E}$, we have $\Box_{\bar\partial_E}u=0$. Taking the $L^2$ inner product of the identity with $u$ gives \begin{align*} 0 &=(\Box_{\bar\partial_E}u,u)_{L^2} \\ &=(\Box_{\partial_E}u,u)_{L^2}+([i\Theta_h(E),\Lambda]u,u)_{L^2}. \end{align*} By definition, \begin{align*} \Box_{\partial_E}=\partial_E\partial_E^*+\partial_E^*\partial_E. \end{align*} Using the definition of an adjoint in the $L^2$ inner product, \begin{align*} (\partial_E\partial_E^*u,u)_{L^2}=\|\partial_E^*u\|_{L^2}^2,\qquad (\partial_E^*\partial_Eu,u)_{L^2}=\|\partial_Eu\|_{L^2}^2. \end{align*} Therefore \begin{align*} 0=\|\partial_Eu\|_{L^2}^2+\|\partial_E^*u\|_{L^2}^2+([i\Theta_h(E),\Lambda]u,u)_{L^2}. \end{align*} The first two summands are nonnegative because they are squared norms, so the only remaining issue is to understand the sign of the curvature commutator.[/guided]

[step:Use Nakano positivity to bound the curvature commutator below]We claim that there is a constant $c>0$ such that every $v\in\mathcal A^{n,q}(X,E)$ satisfies \begin{align*} ([i\Theta_h(E),\Lambda]v,v)_{L^2}\ge c\|v\|_{L^2}^2. \end{align*} Fix $x\in X$. Choose holomorphic coordinates $(z_1,\dots,z_n)$ centred at $x$ which are unitary for $\omega$ at $x$, and choose a local $h$-unitary frame $(e_1,\dots,e_r)$ for $E$ at $x$. Write the curvature tensor at $x$ as \begin{align*} i\Theta_h(E)_x = i\sum_{i,j=1}^n\sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, dz_i\wedge d\bar z_j\otimes e_\beta\otimes e^\alpha, \end{align*} where $(e^\alpha)$ is the dual frame. For $v(x)\in\Lambda^{n,q}T_x^*X\otimes E_x$, write \begin{align*} v(x)=\sum_{|K|=q}\sum_{\alpha=1}^r v_{K,\alpha}(x)\, dz_1\wedge\cdots\wedge dz_n\wedge d\bar z_K\otimes e_\alpha. \end{align*} For an ordered multi-index $J$ of length $q-1$, define coefficients $v_{iJ,\alpha}(x)$ by the rule that $v_{iJ,\alpha}(x)$ is the signed coefficient of the component with anti-holomorphic index set $\{i\}\cup J$, and set $v_{iJ,\alpha}(x)=0$ when $i\in J$. The standard local form of the Nakano curvature commutator on $(n,q)$-forms gives \begin{align*} \langle [i\Theta_h(E),\Lambda]v(x),v(x)\rangle_{\omega,h} = \sum_{|J|=q-1} \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, v_{iJ,\alpha}(x)\overline{v_{jJ,\beta}(x)}. \end{align*} For each fixed $J$, the array $\xi_{i\alpha}=v_{iJ,\alpha}(x)$ defines an element of $T_x^{1,0}X\otimes E_x$. Nakano positivity makes each summand nonnegative, and because $X$ is compact and the smallest Nakano eigenvalue of the continuous curvature tensor is strictly positive at every point, there exists $\lambda>0$ such that \begin{align*} \sum_{i,j=1}^n\sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\xi_{i\alpha}\overline{\xi_{j\beta}} \ge \lambda \sum_{i=1}^n\sum_{\alpha=1}^r |\xi_{i\alpha}|^2 \end{align*} for every $x\in X$ and every $\xi\in T_x^{1,0}X\otimes E_x$. Applying this estimate to $\xi_{i\alpha}=v_{iJ,\alpha}(x)$ and summing over $J$ yields \begin{align*} \langle [i\Theta_h(E),\Lambda]v(x),v(x)\rangle_{\omega,h} &\ge \lambda \sum_{|J|=q-1}\sum_{i=1}^n\sum_{\alpha=1}^r |v_{iJ,\alpha}(x)|^2 \\ &= \lambda q \sum_{|K|=q}\sum_{\alpha=1}^r |v_{K,\alpha}(x)|^2 \\ &= \lambda q\, |v(x)|_{\omega,h}^2. \end{align*} Set $c:=\lambda q>0$. Integrating over $X$ with respect to $d\mu_\omega$ gives \begin{align*} ([i\Theta_h(E),\Lambda]v,v)_{L^2} = \int_X \langle [i\Theta_h(E),\Lambda]v(x),v(x)\rangle_{\omega,h}\,d\mu_\omega(x) \ge c\int_X |v(x)|_{\omega,h}^2\,d\mu_\omega(x) = c\|v\|_{L^2}^2. \end{align*}[/step]

[guided]We now explain why Nakano positivity is exactly the positivity notion needed for $(n,q)$-forms. Fix a point $x\in X$. Choose holomorphic coordinates $(z_1,\dots,z_n)$ centred at $x$ that are unitary for the Kähler metric at $x$, and choose an $h$-unitary local holomorphic frame $(e_1,\dots,e_r)$ for $E$ at $x$. In these coordinates and frame, the curvature has the form \begin{align*} i\Theta_h(E)_x = i\sum_{i,j=1}^n\sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, dz_i\wedge d\bar z_j\otimes e_\beta\otimes e^\alpha. \end{align*} An $E$-valued $(n,q)$-form at $x$ can be written as \begin{align*} v(x)=\sum_{|K|=q}\sum_{\alpha=1}^r v_{K,\alpha}(x)\, dz_1\wedge\cdots\wedge dz_n\wedge d\bar z_K\otimes e_\alpha. \end{align*} The commutator $[i\Theta_h(E),\Lambda]$ contracts one anti-holomorphic index against the curvature tensor. Its local Nakano form on $(n,q)$-forms is \begin{align*} \langle [i\Theta_h(E),\Lambda]v(x),v(x)\rangle_{\omega,h} = \sum_{|J|=q-1} \sum_{i,j=1}^n \sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\, v_{iJ,\alpha}(x)\overline{v_{jJ,\beta}(x)}. \end{align*} Here $J$ ranges over ordered anti-holomorphic multi-indices of length $q-1$, and $v_{iJ,\alpha}(x)$ denotes the signed coefficient obtained by adjoining $i$ to $J$. If $i\in J$, this coefficient is defined to be $0$. For each fixed $J$, the coefficients \begin{align*} \xi_{i\alpha}:=v_{iJ,\alpha}(x) \end{align*} form a tensor $\xi\in T_x^{1,0}X\otimes E_x$. Nakano positivity says exactly that the Hermitian quadratic form \begin{align*} \xi\mapsto \sum_{i,j=1}^n\sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\xi_{i\alpha}\overline{\xi_{j\beta}} \end{align*} is positive definite. Since $X$ is compact and the curvature tensor depends continuously on $x$, the minimum of its smallest Nakano eigenvalue over $X$ is a positive number. Denote this number by $\lambda>0$. Then for every $x\in X$ and every tensor $\xi\in T_x^{1,0}X\otimes E_x$, \begin{align*} \sum_{i,j=1}^n\sum_{\alpha,\beta=1}^r R_{i\bar j\alpha\bar\beta}(x)\xi_{i\alpha}\overline{\xi_{j\beta}} \ge \lambda \sum_{i=1}^n\sum_{\alpha=1}^r|\xi_{i\alpha}|^2. \end{align*} Applying this inequality separately for each $J$ gives \begin{align*} \langle [i\Theta_h(E),\Lambda]v(x),v(x)\rangle_{\omega,h} &\ge \lambda \sum_{|J|=q-1}\sum_{i=1}^n\sum_{\alpha=1}^r |v_{iJ,\alpha}(x)|^2. \end{align*} Each coefficient $v_{K,\alpha}(x)$ with $|K|=q$ appears exactly $q$ times in the sum on the right, once for each choice of the distinguished anti-holomorphic index removed from $K$. Therefore \begin{align*} \sum_{|J|=q-1}\sum_{i=1}^n\sum_{\alpha=1}^r |v_{iJ,\alpha}(x)|^2 = q\sum_{|K|=q}\sum_{\alpha=1}^r |v_{K,\alpha}(x)|^2. \end{align*} Thus \begin{align*} \langle [i\Theta_h(E),\Lambda]v(x),v(x)\rangle_{\omega,h} \ge \lambda q\, |v(x)|_{\omega,h}^2. \end{align*} Because $q>0$, the constant $c:=\lambda q$ is strictly positive. Integrating this pointwise inequality over $X$ with respect to $d\mu_\omega$ gives \begin{align*} ([i\Theta_h(E),\Lambda]v,v)_{L^2} \ge c\|v\|_{L^2}^2. \end{align*} This is the step where the hypothesis $q>0$ is used: when $q=0$, the anti-holomorphic index contraction above has no index to remove, and this argument gives no vanishing.[/guided]

[step:Conclude that the harmonic representative is zero] Apply the curvature estimate from the previous step to the harmonic representative $u$. The Bochner-Kodaira-Nakano identity gives \begin{align*} 0 = \|\partial_Eu\|_{L^2}^2+\|\partial_E^*u\|_{L^2}^2+([i\Theta_h(E),\Lambda]u,u)_{L^2} \ge c\|u\|_{L^2}^2. \end{align*} Since $c>0$, this implies $\|u\|_{L^2}=0$. Hence \begin{align*} \int_X |u(x)|_{\omega,h}^2\,d\mu_\omega(x)=0. \end{align*} The function $x\mapsto |u(x)|_{\omega,h}^2$ is continuous and nonnegative, so it vanishes everywhere on $X$. Therefore $u=0$. Since $u$ represents the arbitrary class $[\alpha]\in H^q(X,K_X\otimes E)$, we have $[\alpha]=0$. Thus \begin{align*} H^q(X,K_X\otimes E)=0 \end{align*} for every integer $q>0$. [/step]