[proofplan] We prove the theorem by passing from sheaf cohomology to Dolbeault cohomology, where the pairing is represented by integration of an $(n,n)$-form. First we verify that the integral is independent of the Dolbeault representatives by using the graded Leibniz rule and [Stokes' theorem](/theorems/1530). Then we choose Hermitian metrics and use compact Hodge theory for the Dolbeault Laplacian to identify cohomology classes with [harmonic representatives](/theorems/2747). The Serre star operator sends harmonic $E$-valued $(0,q)$-forms to harmonic $K_X \otimes E^*$-valued $(0,n-q)$-forms and turns the pairing into the positive $L^2$ norm, which proves nondegeneracy. [/proofplan]

[step:Identify sheaf cohomology with Dolbeault cohomology] Let $\mathcal{O}(E)$ denote the sheaf of holomorphic sections of the holomorphic vector bundle $E \to X$, and let $\mathcal{O}(K_X \otimes E^*)$ denote the sheaf of holomorphic sections of $K_X \otimes E^* \to X$. For each integer $r$, let $A^{0,r}(X,E)$ denote the complex [vector space](/page/Vector%20Space) of smooth $E$-valued $(0,r)$-forms on $X$, with the convention that $A^{0,r}(X,E)=0$ if $r<0$ or $r>n$. Let \begin{align*} \bar{\partial}_E: A^{0,r}(X,E) \to A^{0,r+1}(X,E) \end{align*} denote the Dolbeault operator induced by the holomorphic structure on $E$. Similarly, let \begin{align*} \bar{\partial}_{K_X \otimes E^*}: A^{0,r}(X,K_X \otimes E^*) \to A^{0,r+1}(X,K_X \otimes E^*) \end{align*} denote the Dolbeault operator on $K_X \otimes E^*$, where $A^{0,r}(X,K_X \otimes E^*)$ has the same zero convention outside $0 \leq r \leq n$. Define the Dolbeault cohomology groups by \begin{align*} H^r_{\bar{\partial}}(X,E) &:= \frac{\ker(\bar{\partial}_E: A^{0,r}(X,E) \to A^{0,r+1}(X,E))} {\operatorname{im}(\bar{\partial}_E: A^{0,r-1}(X,E) \to A^{0,r}(X,E))}, \\ H^r_{\bar{\partial}}(X,K_X \otimes E^*) &:= \frac{\ker(\bar{\partial}_{K_X \otimes E^*}: A^{0,r}(X,K_X \otimes E^*) \to A^{0,r+1}(X,K_X \otimes E^*))} {\operatorname{im}(\bar{\partial}_{K_X \otimes E^*}: A^{0,r-1}(X,K_X \otimes E^*) \to A^{0,r}(X,K_X \otimes E^*))}. \end{align*} By the Dolbeault theorem for holomorphic vector bundles (citing a result not yet in the wiki: Dolbeault theorem for holomorphic vector bundles), there are natural isomorphisms \begin{align*} H^q(X,\mathcal{O}(E)) &\cong \frac{\ker(\bar{\partial}_E: A^{0,q}(X,E) \to A^{0,q+1}(X,E))} {\operatorname{im}(\bar{\partial}_E: A^{0,q-1}(X,E) \to A^{0,q}(X,E))}, \\ H^{n-q}(X,\mathcal{O}(K_X \otimes E^*)) &\cong \frac{\ker(\bar{\partial}_{K_X \otimes E^*}: A^{0,n-q}(X,K_X \otimes E^*) \to A^{0,n-q+1}(X,K_X \otimes E^*))} {\operatorname{im}(\bar{\partial}_{K_X \otimes E^*}: A^{0,n-q-1}(X,K_X \otimes E^*) \to A^{0,n-q}(X,K_X \otimes E^*))}. \end{align*} Thus it is enough to construct and prove nondegeneracy of the pairing on Dolbeault cohomology. [/step]

[step:Define the Dolbeault pairing by contracting the coefficient bundles] Let \begin{align*} \operatorname{ev}_{E}: E \otimes E^* &\to X \times \mathbb{C} \end{align*} be the fiberwise evaluation morphism, so that for each $x \in X$, \begin{align*} \operatorname{ev}_{E,x}: E_x \otimes E_x^* &\to \mathbb{C}, \\ v \otimes \lambda &\mapsto \lambda(v). \end{align*} If $\alpha \in A^{0,q}(X,E)$ and $\beta \in A^{0,n-q}(X,K_X \otimes E^*)$, then $\alpha \wedge \beta$ is a smooth section of \begin{align*} \Lambda^{0,n}T^*X \otimes K_X \otimes E \otimes E^*. \end{align*} Applying $\operatorname{ev}_{E}$ to the coefficient factor gives a smooth section of \begin{align*} K_X \otimes \Lambda^{0,n}T^*X = \Lambda^{n,0}T^*X \otimes \Lambda^{0,n}T^*X = \Lambda^{n,n}T^*X. \end{align*} Hence \begin{align*} \operatorname{ev}_{E}(\alpha \wedge \beta) \end{align*} is a smooth top-degree complex-valued form on the compact oriented real manifold underlying $X$. Choose a smooth positive volume form \begin{align*} \mu_X \in A^{n,n}(X) \end{align*} on the underlying oriented real manifold. For each pair $(\alpha,\beta)$, define the smooth function \begin{align*} f_{\alpha,\beta}: X &\to \mathbb{C} \end{align*} by the identity \begin{align*} \operatorname{ev}_{E}(\alpha \wedge \beta)=f_{\alpha,\beta}\,\mu_X. \end{align*} Define \begin{align*} B_q: A^{0,q}(X,E) \times A^{0,n-q}(X,K_X \otimes E^*) &\to \mathbb{C}, \\ (\alpha,\beta) &\mapsto \int_X f_{\alpha,\beta}\,d\mu_X. \end{align*} This integral is finite because $X$ is compact and $f_{\alpha,\beta}$ is smooth. The value is the usual integral of the top-degree form $\operatorname{ev}_{E}(\alpha \wedge \beta)$, so it is independent of the auxiliary volume form used to write that form as a function times a volume form. [/step]

[step:Show that exact changes of representatives do not change the integral]Assume $\alpha \in A^{0,q}(X,E)$ and $\beta \in A^{0,n-q}(X,K_X \otimes E^*)$ satisfy \begin{align*} \bar{\partial}_E \alpha = 0, \qquad \bar{\partial}_{K_X \otimes E^*}\beta = 0. \end{align*} Let $\gamma \in A^{0,q-1}(X,E)$. The graded Leibniz rule for the Dolbeault operator and the holomorphicity of the evaluation pairing give \begin{align*} \bar{\partial}\operatorname{ev}_{E}(\gamma \wedge \beta) = \operatorname{ev}_{E}((\bar{\partial}_E\gamma)\wedge \beta) + (-1)^{q-1}\operatorname{ev}_{E}(\gamma \wedge \bar{\partial}_{K_X \otimes E^*}\beta). \end{align*} Since $\bar{\partial}_{K_X \otimes E^*}\beta = 0$, this becomes \begin{align*} \operatorname{ev}_{E}((\bar{\partial}_E\gamma)\wedge \beta) = \bar{\partial}\operatorname{ev}_{E}(\gamma \wedge \beta). \end{align*} The form $\operatorname{ev}_{E}(\gamma \wedge \beta)$ has bidegree $(n,n-1)$. Since $\partial$ of an $(n,n-1)$-form has type $(n+1,n-1)$ and hence vanishes on an $n$-dimensional complex manifold, we have \begin{align*} d\operatorname{ev}_{E}(\gamma \wedge \beta) = \bar{\partial}\operatorname{ev}_{E}(\gamma \wedge \beta). \end{align*} By Stokes' theorem for compact manifolds without boundary (citing a result not yet in the wiki: Stokes' theorem), the integral of the exact top-degree form $d\operatorname{ev}_{E}(\gamma \wedge \beta)$ over the oriented manifold $X$ is zero. Equivalently, if $f_{\bar{\partial}_E\gamma,\beta}:X\to\mathbb{C}$ is defined by \begin{align*} \operatorname{ev}_{E}((\bar{\partial}_E\gamma)\wedge \beta)=f_{\bar{\partial}_E\gamma,\beta}\,\mu_X, \end{align*} then \begin{align*} \int_X f_{\bar{\partial}_E\gamma,\beta}\,d\mu_X = 0. \end{align*} Now let $\delta \in A^{0,n-q-1}(X,K_X \otimes E^*)$. Again using the graded Leibniz rule, \begin{align*} \bar{\partial}\operatorname{ev}_{E}(\alpha \wedge \delta) = \operatorname{ev}_{E}((\bar{\partial}_E\alpha)\wedge \delta) + (-1)^q\operatorname{ev}_{E}(\alpha \wedge \bar{\partial}_{K_X \otimes E^*}\delta). \end{align*} Since $\bar{\partial}_E\alpha = 0$, we obtain \begin{align*} \operatorname{ev}_{E}(\alpha \wedge \bar{\partial}_{K_X \otimes E^*}\delta) = (-1)^q\bar{\partial}\operatorname{ev}_{E}(\alpha \wedge \delta). \end{align*} The form $\operatorname{ev}_{E}(\alpha \wedge \delta)$ has bidegree $(n,n-1)$, so its [exterior derivative](/theorems/1525) equals its $\bar{\partial}$ derivative. Stokes' theorem gives the vanishing of the integral of the exact top-degree form $d\operatorname{ev}_{E}(\alpha \wedge \delta)$. Equivalently, if $f_{\alpha,\bar{\partial}_{K_X \otimes E^*}\delta}:X\to\mathbb{C}$ is defined by \begin{align*} \operatorname{ev}_{E}(\alpha \wedge \bar{\partial}_{K_X \otimes E^*}\delta)=f_{\alpha,\bar{\partial}_{K_X \otimes E^*}\delta}\,\mu_X, \end{align*} then \begin{align*} \int_X f_{\alpha,\bar{\partial}_{K_X \otimes E^*}\delta}\,d\mu_X = 0. \end{align*} Therefore $B_q$ descends to a well-defined bilinear pairing on Dolbeault cohomology.[/step]

[guided]The only point to check before passing to cohomology is that replacing a representative by a $\bar{\partial}$-exact form does not change the integral. We check this separately in the two variables. First suppose the $E$-valued representative changes by $\bar{\partial}_E\gamma$, where $\gamma \in A^{0,q-1}(X,E)$. Since $\beta$ is a Dolbeault cocycle, $\bar{\partial}_{K_X \otimes E^*}\beta = 0$. The graded Leibniz rule for the Dolbeault operator gives \begin{align*} \bar{\partial}\operatorname{ev}_{E}(\gamma \wedge \beta) = \operatorname{ev}_{E}((\bar{\partial}_E\gamma)\wedge \beta) + (-1)^{q-1}\operatorname{ev}_{E}(\gamma \wedge \bar{\partial}_{K_X \otimes E^*}\beta) = \operatorname{ev}_{E}((\bar{\partial}_E\gamma)\wedge \beta). \end{align*} The form $\operatorname{ev}_{E}(\gamma \wedge \beta)$ has type $(n,n-1)$. On a complex $n$-manifold there are no forms of type $(n+1,n-1)$, so the $\partial$ part of its exterior derivative vanishes. Hence \begin{align*} d\operatorname{ev}_{E}(\gamma \wedge \beta) = \bar{\partial}\operatorname{ev}_{E}(\gamma \wedge \beta). \end{align*} Because $X$ is compact and has no boundary, Stokes' theorem applies and says that the exact top-degree form $d\operatorname{ev}_{E}(\gamma \wedge \beta)$ has integral zero over $X$. In the notation of the previous step, if $f_{\bar{\partial}_E\gamma,\beta}:X\to\mathbb{C}$ is defined by \begin{align*} \operatorname{ev}_{E}((\bar{\partial}_E\gamma)\wedge \beta)=f_{\bar{\partial}_E\gamma,\beta}\,\mu_X, \end{align*} then \begin{align*} \int_X f_{\bar{\partial}_E\gamma,\beta}\,d\mu_X = 0. \end{align*} The same argument works in the second variable, with the sign coming from moving $\bar{\partial}$ past an $E$-valued form of degree $q$. Let $\delta \in A^{0,n-q-1}(X,K_X \otimes E^*)$. Since $\bar{\partial}_E\alpha = 0$, \begin{align*} \bar{\partial}\operatorname{ev}_{E}(\alpha \wedge \delta) = \operatorname{ev}_{E}((\bar{\partial}_E\alpha)\wedge \delta) + (-1)^q\operatorname{ev}_{E}(\alpha \wedge \bar{\partial}_{K_X \otimes E^*}\delta) = (-1)^q\operatorname{ev}_{E}(\alpha \wedge \bar{\partial}_{K_X \otimes E^*}\delta). \end{align*} Again $\operatorname{ev}_{E}(\alpha \wedge \delta)$ has type $(n,n-1)$, so $d=\bar{\partial}$ on this form. Stokes' theorem gives the vanishing of the integral of $d\operatorname{ev}_{E}(\alpha \wedge \delta)$. In the notation of the previous step, if $f_{\alpha,\bar{\partial}_{K_X \otimes E^*}\delta}:X\to\mathbb{C}$ is defined by \begin{align*} \operatorname{ev}_{E}(\alpha \wedge \bar{\partial}_{K_X \otimes E^*}\delta)=f_{\alpha,\bar{\partial}_{K_X \otimes E^*}\delta}\,\mu_X, \end{align*} then \begin{align*} \int_X f_{\alpha,\bar{\partial}_{K_X \otimes E^*}\delta}\,d\mu_X = 0. \end{align*} Thus adding a $\bar{\partial}$-exact form in either argument leaves the integral unchanged, so the pairing is well-defined on cohomology classes.[/guided]

[step:Choose metrics and identify cohomology with harmonic forms] Choose a Hermitian metric $h_E$ on $E$ and a Hermitian metric $g$ on the holomorphic tangent bundle $T^{1,0}X$. Let \begin{align*} \mathcal{H}^{0,r}(X,E) &:= \{\alpha \in A^{0,r}(X,E) : \bar{\partial}_E\alpha = 0 \text{ and } \bar{\partial}_E^*\alpha = 0\} \end{align*} denote the finite-dimensional space of harmonic $E$-valued $(0,r)$-forms, where $\bar{\partial}_E^*$ is the formal adjoint of $\bar{\partial}_E$ with respect to the $L^2$ inner product determined by $g$ and $h_E$. Define $\mathcal{H}^{0,r}(X,K_X \otimes E^*)$ analogously. By the Hodge theorem for the Dolbeault Laplacian on compact Hermitian manifolds (citing a result not yet in the wiki: Hodge theorem for the Dolbeault Laplacian), every Dolbeault cohomology class has a unique harmonic representative. Hence \begin{align*} H^q_{\bar{\partial}}(X,E) \cong \mathcal{H}^{0,q}(X,E), \qquad H^{n-q}_{\bar{\partial}}(X,K_X \otimes E^*) \cong \mathcal{H}^{0,n-q}(X,K_X \otimes E^*). \end{align*} It remains to prove that the pairing is nondegenerate on these harmonic spaces. [/step]

[step:Use the Serre star operator to produce the dual harmonic representative]The metrics $g$ and $h_E$ determine a conjugate-linear Serre star operator \begin{align*} S_E: A^{0,q}(X,E) &\to A^{0,n-q}(X,K_X \otimes E^*) \end{align*} characterized by the identity \begin{align*} \operatorname{ev}_{E}(\alpha \wedge S_E\eta) = (\alpha,\eta)_{g,h_E}\,dV_g \end{align*} for all $\alpha,\eta \in A^{0,q}(X,E)$, where $(\alpha,\eta)_{g,h_E}: X \to \mathbb{C}$ is the pointwise Hermitian inner product, linear in the first argument and conjugate-linear in the second, and $dV_g$ is the smooth volume form determined by $g$. In particular, if $f_{\alpha,S_E\eta}:X\to\mathbb{C}$ is defined by \begin{align*} \operatorname{ev}_{E}(\alpha \wedge S_E\eta)=f_{\alpha,S_E\eta}\,dV_g, \end{align*} then \begin{align*} \int_X f_{\alpha,S_E\eta}\,dV_g = \int_X (\alpha,\eta)_{g,h_E}\,dV_g. \end{align*} The standard Serre-star compatibility theorem applies to every compact Hermitian complex manifold equipped with a Hermitian holomorphic vector bundle. In the present notation it states that there are signs depending only on $n$ and $q$ such that \begin{align*} S_E\bar{\partial}_E &= \pm\,\bar{\partial}_{K_X\otimes E^*}^{\,*}S_E, & S_E\bar{\partial}_E^{\,*} &= \pm\,\bar{\partial}_{K_X\otimes E^*}S_E. \end{align*} The hypotheses of this theorem are satisfied because $X$ is compact complex, $g$ is a Hermitian metric on $T^{1,0}X$, and $h_E$ is a Hermitian metric on $E$. Consequently, if $\alpha$ is harmonic, so $\bar\partial_E\alpha=0$ and $\bar\partial_E^*\alpha=0$, then \begin{align*} \bar\partial_{K_X\otimes E^*}(S_E\alpha)=0, \qquad \bar\partial_{K_X\otimes E^*}^{\,*}(S_E\alpha)=0. \end{align*} Thus \begin{align*} S_E(\mathcal{H}^{0,q}(X,E)) = \mathcal{H}^{0,n-q}(X,K_X \otimes E^*) \end{align*} (citing a result not yet in the wiki: Serre star compatibility with the Dolbeault Laplacian). Let $0 \ne [\alpha] \in H^q_{\bar{\partial}}(X,E)$, and let $\alpha_0 \in \mathcal{H}^{0,q}(X,E)$ be its unique harmonic representative. Since $[\alpha] \ne 0$, uniqueness of harmonic representatives gives $\alpha_0 \ne 0$. Define \begin{align*} \beta_0 := S_E\alpha_0 \in \mathcal{H}^{0,n-q}(X,K_X \otimes E^*). \end{align*} Then $\beta_0$ represents a class in $H^{n-q}_{\bar{\partial}}(X,K_X \otimes E^*)$, and \begin{align*} B_q([\alpha],[\beta_0]) &= \int_X f_{\alpha_0,S_E\alpha_0}\,dV_g \\ &= \int_X (\alpha_0,\alpha_0)_{g,h_E}\,dV_g \\ &= \|\alpha_0\|_{L^2}^2 > 0. \end{align*} Thus no nonzero class in $H^q_{\bar{\partial}}(X,E)$ pairs to zero against every class in $H^{n-q}_{\bar{\partial}}(X,K_X \otimes E^*)$.[/step]

[guided]The nondegeneracy argument is the Hodge-theoretic heart of the proof. The goal is: given a nonzero cohomology class $[\alpha]$, produce a class $[\beta]$ in the complementary group such that the integral pairing is nonzero. The Hermitian metrics on $X$ and $E$ give a pointwise Hermitian inner product, taken linear in the first argument and conjugate-linear in the second, \begin{align*} (\cdot,\cdot)_{g,h_E}: A^{0,q}(X,E) \times A^{0,q}(X,E) \to C^\infty(X,\mathbb{C}) \end{align*} and a volume form $dV_g$ on $X$. They also determine the Serre star operator \begin{align*} S_E: A^{0,q}(X,E) \to A^{0,n-q}(X,K_X \otimes E^*). \end{align*} This operator is defined so that wedging a form with the Serre star of another form reproduces the pointwise Hermitian pairing as a top-degree form: \begin{align*} \operatorname{ev}_{E}(\alpha \wedge S_E\eta) = (\alpha,\eta)_{g,h_E}\,dV_g \end{align*} for all $\alpha,\eta \in A^{0,q}(X,E)$. If $f_{\alpha,S_E\eta}:X\to\mathbb{C}$ is defined by \begin{align*} \operatorname{ev}_{E}(\alpha \wedge S_E\eta)=f_{\alpha,S_E\eta}\,dV_g, \end{align*} then integrating this identity gives \begin{align*} \int_X f_{\alpha,S_E\eta}\,dV_g = \int_X (\alpha,\eta)_{g,h_E}\,dV_g. \end{align*} The compatibility needed from Hodge theory is a precise adjoint identity. On a compact Hermitian complex manifold with a Hermitian holomorphic vector bundle, the Serre star satisfies \begin{align*} S_E\bar{\partial}_E &= \pm\,\bar{\partial}_{K_X\otimes E^*}^{\,*}S_E, & S_E\bar{\partial}_E^{\,*} &= \pm\,\bar{\partial}_{K_X\otimes E^*}S_E. \end{align*} These hypotheses are exactly the metrics chosen above on the compact complex manifold $X$ and on $E$. If $\alpha$ is harmonic, then $\bar\partial_E\alpha=0$ and $\bar\partial_E^*\alpha=0$; applying the two displayed identities gives \begin{align*} \bar\partial_{K_X\otimes E^*}(S_E\alpha)=0, \qquad \bar\partial_{K_X\otimes E^*}^{\,*}(S_E\alpha)=0. \end{align*} Thus the Serre star operator carries harmonic forms to harmonic forms in the complementary coefficient bundle: \begin{align*} S_E(\mathcal{H}^{0,q}(X,E)) = \mathcal{H}^{0,n-q}(X,K_X \otimes E^*). \end{align*} This is the analytic bridge between the algebraic-looking pairing and the $L^2$ norm. Now take a nonzero class $[\alpha] \in H^q_{\bar{\partial}}(X,E)$. By the Dolbeault Hodge theorem, $[\alpha]$ has a unique harmonic representative $\alpha_0 \in \mathcal{H}^{0,q}(X,E)$. Since the class is nonzero, this representative is not the zero form. Define \begin{align*} \beta_0 := S_E\alpha_0. \end{align*} By the harmonic compatibility of $S_E$, the form $\beta_0$ lies in $\mathcal{H}^{0,n-q}(X,K_X \otimes E^*)$, so it is $\bar{\partial}$-closed and represents a Dolbeault cohomology class. The defining property of $S_E$ now turns the Serre pairing into the square of the $L^2$ norm: \begin{align*} B_q([\alpha],[\beta_0]) &= \int_X f_{\alpha_0,S_E\alpha_0}\,dV_g \\ &= \int_X (\alpha_0,\alpha_0)_{g,h_E}\,dV_g \\ &= \|\alpha_0\|_{L^2}^2. \end{align*} Because $\alpha_0 \ne 0$ and the Hermitian metric is positive definite, $\|\alpha_0\|_{L^2}^2 > 0$. Therefore $[\alpha]$ cannot pair to zero with every class in the complementary Dolbeault cohomology group.[/guided]

[step:Prove the complementary kernel vanishes and conclude the duality isomorphism] The preceding step proves that the left kernel of the pairing is zero. We now prove that the right kernel is zero by applying the same left-kernel argument to the complementary coefficient bundle. The preceding Hodge-star argument did not use any special property of $E$: for every holomorphic vector bundle $G\to X$ and every degree $r$, the Dolbeault pairing \begin{align*} B^G_r: H^r_{\bar{\partial}}(X,G) \times H^{n-r}_{\bar{\partial}}(X,K_X\otimes G^*) &\to \mathbb C \end{align*} has zero left kernel. Let \begin{align*} F := K_X \otimes E^* \end{align*} denote the holomorphic vector bundle paired with $E$. The evaluation map $K_X\otimes K_X^*\to X\times\mathbb C$ and the natural double-dual identification $(E^*)^*\cong E$ give a holomorphic bundle isomorphism \begin{align*} \iota: K_X \otimes F^* = K_X \otimes (K_X \otimes E^*)^* &\to E \end{align*} over $X$. Let $0 \ne [\beta] \in H^{n-q}_{\bar{\partial}}(X,F)$. Apply the zero-left-kernel result to the bundle $F$ in degree $n-q$. Since \begin{align*} K_X\otimes F^* \cong E, \end{align*} there is a class \begin{align*} [\gamma]\in H^q_{\bar{\partial}}(X,K_X\otimes F^*) \end{align*} such that \begin{align*} B^F_{n-q}([\beta],[\gamma])\ne 0. \end{align*} Let $[\alpha]\in H^q_{\bar{\partial}}(X,E)$ be the image of $[\gamma]$ under the cohomology isomorphism induced by $\iota$. Choose Dolbeault-closed representatives $\beta$ and $\gamma$, and put $\alpha=\iota(\gamma)$. We compare the top-degree forms defining the two pairings. Locally, write decomposable terms of $\beta$ as a $(0,n-q)$-form with coefficient in $K_X\otimes E^*$ and decomposable terms of $\gamma$ as a $(0,q)$-form with coefficient in $K_X\otimes F^*=K_X\otimes K_X^*\otimes E$. Under $\iota$, the factor $K_X\otimes K_X^*$ is evaluated and the remaining coefficient is the corresponding $E$-coefficient of $\alpha$. The coefficient contraction in $B^F_{n-q}$ is the same contraction as the coefficient contraction in $B^E_q$, after moving a $(0,q)$-form past a $(0,n-q)$-form. Therefore \begin{align*} \operatorname{ev}_{E}(\alpha\wedge\beta) = (-1)^{q(n-q)} \operatorname{ev}_{F}(\beta\wedge\gamma). \end{align*} After writing both top-degree forms against the same smooth positive volume form, integration gives \begin{align*} B^E_q([\alpha],[\beta]) = (-1)^{q(n-q)} B^F_{n-q}([\beta],[\gamma]) \ne 0. \end{align*} Thus no nonzero class in $H^{n-q}_{\bar{\partial}}(X,K_X \otimes E^*)$ annihilates all classes in $H^q_{\bar{\partial}}(X,E)$. Therefore the bilinear pairing \begin{align*} H^q_{\bar{\partial}}(X,E) \times H^{n-q}_{\bar{\partial}}(X,K_X \otimes E^*) \to \mathbb{C} \end{align*} has zero left and right kernels. Since compact Hodge theory gives finite-dimensionality of both Dolbeault cohomology groups, these two kernel statements also imply that the two vector spaces have the same dimension. Hence the map \begin{align*} H^{n-q}_{\bar{\partial}}(X,K_X \otimes E^*) &\to H^q_{\bar{\partial}}(X,E)^*, \\ [\beta] &\mapsto \bigl([\alpha] \mapsto B_q([\alpha],[\beta])\bigr) \end{align*} is an isomorphism. Transporting this isomorphism through the Dolbeault isomorphisms from the first step gives \begin{align*} H^{n-q}(X,\mathcal{O}(K_X \otimes E^*)) \cong H^q(X,\mathcal{O}(E))^*. \end{align*} Equivalently, reversing this canonical isomorphism gives the statement's displayed form \begin{align*} H^q(X,\mathcal O(E))^* \cong H^{n-q}(X,\mathcal O(K_X\otimes E^*)). \end{align*} This is exactly the asserted Serre duality isomorphism, with the convention $H^q(X,E)=H^q(X,\mathcal O(E))$. [/step]