[proofplan] The strategy is to replace cohomology classes by canonical [harmonic representatives](/theorems/2747) and to produce, for each nonzero harmonic form, an explicit partner of complementary bidegree that pairs with it to give its positive $L^2$ norm. First we record that the wedge-evaluation pairing descends to cohomology, using the Leibniz rule for $\bar\partial$ together with [Stokes' theorem](/theorems/1530) on the compact manifold $X$. Next we invoke Hodge theory for the bundle-valued Dolbeault complex: the $\bar\partial_E$-Laplacian is elliptic and self-adjoint, so every class has a unique harmonic representative, and harmonicity is equivalent to being simultaneously $\bar\partial_E$- and $\bar\partial_E^*$-closed. We then build the conjugate-linear Hodge–Serre star $\bar{*}_E$, which uses the Hodge star on forms and the metric conjugation $E \to E^*$; it satisfies $\alpha \wedge \bar{*}_E\gamma = \langle \alpha, \gamma\rangle_h\, dV_g$, and an adjoint identity $\bar\partial_E^* = -\bar{*}_{E^*}\bar\partial_{E^*}\bar{*}_E$ shows it carries harmonic forms to harmonic forms. Pairing a nonzero harmonic $\alpha$ with the class of $\bar{*}_E\alpha$ yields $\|\alpha\|_{L^2}^2 > 0$, which gives nondegeneracy in both variables. [/proofplan] [step:Set up the $E$-valued Dolbeault complex and the wedge-evaluation pairing] Fix the complex dimension $n = \dim_{\mathbb C} X$. The complex structure orients $X$, and $X$ is compact without boundary. The Hermitian metric on $X$ restricts to a Riemannian metric $g$ on the underlying $2n$-real-dimensional manifold; let $dV_g$ denote its Riemannian volume form (equivalently, integration against $dV_g$ is integration against the measure $\mathcal H^{2n}$ on $X$). For $0 \le p,q \le n$ let \begin{align*} A^{p,q}(X,E) &= \Gamma\big(X, \Lambda^{p,q}T^*X \otimes E\big) \end{align*} be the smooth $E$-valued $(p,q)$-forms, and let \begin{align*} \bar\partial_E &: A^{p,q}(X,E) \to A^{p,q+1}(X,E) \end{align*} be the Dolbeault operator determined by the holomorphic structure of $E$: in a local holomorphic frame $(e_1, \dots, e_r)$ of $E$ over an [open set](/page/Open%20Set) $U$, every section is $\alpha = \sum_i \omega_i \otimes e_i$ with $\omega_i \in A^{p,q}(U)$, and $\bar\partial_E \alpha = \sum_i (\bar\partial \omega_i) \otimes e_i$. By [$\bar\partial^2=0$](/theorems/3409) applied componentwise, $\bar\partial_E^2 = 0$, so $\big(A^{p,\bullet}(X,E), \bar\partial_E\big)$ is a cochain complex and $H^{p,q}_{\bar\partial}(X,E)$ is defined as in the statement. The dual bundle $E^*$ carries the dual holomorphic structure with Dolbeault operator $\bar\partial_{E^*}$, defined identically using the dual holomorphic frame. Let $\langle \cdot, \cdot\rangle : E \otimes E^* \to \underline{\mathbb C}$ be the canonical evaluation pairing $\langle e, \lambda\rangle = \lambda(e)$; it is a holomorphic bundle morphism because pairing a holomorphic section of $E$ with a holomorphic section of $E^*$ produces a [holomorphic function](/page/Holomorphic%20Function). For $\alpha = \omega \otimes e \in A^{p,q}(X,E)$ and $\beta = \sigma \otimes \lambda \in A^{p',q'}(X,E^*)$ define \begin{align*} \alpha \wedge \beta &:= \langle e, \lambda\rangle\, \omega \wedge \sigma \in A^{p+p',\,q+q'}(X), \end{align*} extended bilinearly; this is a scalar-valued form. When $p' = n-p$ and $q' = n-q$ the result is an $(n,n)$-form, and $\int_X \alpha \wedge \beta \in \mathbb C$ is the integral of a top-degree form over the compact oriented manifold $X$. [/step] [step:Show the pairing descends to Dolbeault cohomology] We prove that $\int_X \alpha \wedge \beta$ depends only on the cohomology classes $[\alpha] \in H^{p,q}_{\bar\partial}(X,E)$ and $[\beta] \in H^{n-p,n-q}_{\bar\partial}(X,E^*)$. Because the evaluation pairing $\langle \cdot, \cdot\rangle$ is holomorphic, $\bar\partial$ obeys the Leibniz rule across it: for $\zeta \in A^{a,b}(X,E)$ and $\eta \in A^{a',b'}(X,E^*)$, \begin{align*} \bar\partial(\zeta \wedge \eta) &= \bar\partial_E \zeta \wedge \eta + (-1)^{a+b}\, \zeta \wedge \bar\partial_{E^*}\eta. \end{align*} Suppose $\alpha \in A^{p,q}(X,E)$ is $\bar\partial_E$-closed and replace $\beta$ by $\beta + \bar\partial_{E^*}\eta$ with $\eta \in A^{n-p,\,n-q-1}(X,E^*)$. The form $\alpha \wedge \eta$ has bidegree $(p + (n-p),\, q + (n-q-1)) = (n, n-1)$. Applying the Leibniz rule with $a+b = p+q$ and using $\bar\partial_E \alpha = 0$, \begin{align*} \bar\partial(\alpha \wedge \eta) &= (-1)^{p+q}\, \alpha \wedge \bar\partial_{E^*}\eta. \end{align*} On an $(n,n-1)$-form the $\partial$-part would land in bidegree $(n+1, n-1) = 0$, so $d(\alpha \wedge \eta) = \bar\partial(\alpha \wedge \eta)$. Since $X$ is compact and without boundary, the [Generalised Stokes' Theorem](/theorems/3555) gives $\int_X d(\alpha \wedge \eta) = 0$, whence \begin{align*} \int_X \alpha \wedge \bar\partial_{E^*}\eta &= (-1)^{p+q}\int_X d(\alpha \wedge \eta) = 0. \end{align*} By the symmetric computation, if $\beta$ is $\bar\partial_{E^*}$-closed and $\alpha$ is replaced by $\alpha + \bar\partial_E \zeta$ with $\zeta \in A^{p,q-1}(X,E)$, then $\zeta \wedge \beta$ has bidegree $(n,n-1)$, $d(\zeta\wedge\beta) = \bar\partial_E\zeta \wedge \beta$, and $\int_X \bar\partial_E\zeta \wedge \beta = 0$. Hence the pairing is well-defined on cohomology. [guided] We must check the integral is unchanged when a representative is altered by a coboundary, in either variable. The only tool that turns "differs by a coboundary" into "integrates to the same value" is [Stokes' theorem](/theorems/1530), so we need to exhibit the difference as an exact form. Why does $\bar\partial$ satisfy a Leibniz rule across the bundle pairing? The pairing $\langle\cdot,\cdot\rangle : E \otimes E^* \to \underline{\mathbb C}$ is a morphism of holomorphic bundles — in dual holomorphic frames it is the constant pairing $\langle e_i, e_j^*\rangle = \delta_{ij}$, which $\bar\partial$ therefore annihilates. Therefore $\bar\partial$ differentiates $\zeta \wedge \eta$ as if the coefficients were ordinary functions wedged together, producing \begin{align*} \bar\partial(\zeta \wedge \eta) &= \bar\partial_E \zeta \wedge \eta + (-1)^{a+b}\, \zeta \wedge \bar\partial_{E^*}\eta, \end{align*} the sign $(-1)^{a+b}$ being the usual Koszul sign for moving $\bar\partial$ past the form $\zeta$ of total degree $a+b$. Now fix $\bar\partial_E$-closed $\alpha \in A^{p,q}(X,E)$ and change $\beta$ to $\beta + \bar\partial_{E^*}\eta$. The change in the integrand is $\alpha \wedge \bar\partial_{E^*}\eta$. We want this to be exact. Look at $\alpha \wedge \eta$: its bidegree is $(p,q) + (n-p, n-q-1) = (n, n-1)$, total real degree $2n-1$. The Leibniz rule gives \begin{align*} \bar\partial(\alpha \wedge \eta) &= \underbrace{\bar\partial_E \alpha}_{=0} \wedge \eta + (-1)^{p+q}\alpha \wedge \bar\partial_{E^*}\eta = (-1)^{p+q}\alpha \wedge \bar\partial_{E^*}\eta. \end{align*} This is almost what we want, but Stokes is stated for $d$, not $\bar\partial$. Here the bidegree saves us: $d = \partial + \bar\partial$, and $\partial(\alpha\wedge\eta)$ would have bidegree $(n+1, n-1)$, which vanishes since there are no $(n+1)$-forms in $n$ complex dimensions. Thus $d(\alpha\wedge\eta) = \bar\partial(\alpha\wedge\eta)$, an exact $(n,n)$-form. [Stokes' theorem](/theorems/1530) requires a compact oriented manifold without boundary and a smooth top-degree form; $X$ is compact (hypothesis), oriented by its complex structure, has empty boundary, and $\alpha\wedge\eta$ is smooth. Hence $\int_X d(\alpha\wedge\eta) = 0$, so \begin{align*} \int_X \alpha \wedge \bar\partial_{E^*}\eta &= (-1)^{p+q}\int_X d(\alpha \wedge \eta) = 0. \end{align*} Running the identical argument with the roles of the two variables exchanged — now $\beta$ is closed and $\alpha$ is changed by $\bar\partial_E\zeta$, with $\zeta\wedge\beta$ again of bidegree $(n,n-1)$ — shows the value is also insensitive to coboundaries in the first slot. The pairing therefore factors through $H^{p,q}_{\bar\partial}(X,E)\times H^{n-p,n-q}_{\bar\partial}(X,E^*)$. [/guided] [/step] [step:Replace classes by harmonic representatives via Hodge theory for $\bar\partial_E$] Equip $A^{p,q}(X,E)$ with the $L^2$ Hermitian inner product \begin{align*} (\alpha, \gamma)_{L^2} &= \int_X \langle \alpha, \gamma\rangle_h \, dV_g, \end{align*} where $\langle \cdot, \cdot\rangle_h$ is the pointwise Hermitian inner product on $\Lambda^{p,q}T^*X \otimes E$ induced by $g$ and $h$ (conjugate-linear in the second argument). Let $\bar\partial_E^* : A^{p,q}(X,E) \to A^{p,q-1}(X,E)$ be the formal adjoint of $\bar\partial_E$ with respect to $(\cdot,\cdot)_{L^2}$, and define the $\bar\partial_E$-Laplacian \begin{align*} \Box_E &= \bar\partial_E \bar\partial_E^* + \bar\partial_E^* \bar\partial_E : A^{p,q}(X,E) \to A^{p,q}(X,E), \end{align*} a second-order, elliptic, formally self-adjoint, non-negative operator. Set $\mathcal H^{p,q}(X,E) = \ker\big(\Box_E|_{A^{p,q}}\big)$, the space of harmonic $E$-valued $(p,q)$-forms. We invoke Hodge theory for this elliptic complex (citing a result not yet in the wiki in its bundle-valued Dolbeault form, the Hodge Theorem for the $\bar\partial_E$-complex on a compact Hermitian manifold; the real-manifold prototype is the [Hodge Decomposition](/theorems/2745)). Its hypotheses are met: $X$ is compact, so $\Box_E$ is an elliptic operator on a compact manifold and the general theory applies. It yields: 1. $\mathcal H^{p,q}(X,E)$ is finite-dimensional; 2. there is an $L^2$-[orthogonal decomposition](/theorems/436) \begin{align*} A^{p,q}(X,E) &= \mathcal H^{p,q}(X,E) \;\oplus\; \bar\partial_E A^{p,q-1}(X,E) \;\oplus\; \bar\partial_E^* A^{p,q+1}(X,E); \end{align*} 3. the projection onto the harmonic summand induces an isomorphism $\mathcal H^{p,q}(X,E) \xrightarrow{\ \sim\ } H^{p,q}_{\bar\partial}(X,E)$. Moreover, for $\alpha \in A^{p,q}(X,E)$ on the compact $X$, \begin{align*} (\Box_E \alpha, \alpha)_{L^2} &= \|\bar\partial_E \alpha\|_{L^2}^2 + \|\bar\partial_E^* \alpha\|_{L^2}^2, \end{align*} so $\alpha$ is harmonic if and only if $\bar\partial_E \alpha = 0$ and $\bar\partial_E^* \alpha = 0$ simultaneously. The same statements hold for $E^*$, giving $\mathcal H^{n-p,n-q}(X,E^*) \cong H^{n-p,n-q}_{\bar\partial}(X,E^*)$. [/step] [step:Construct the conjugate-linear Hodge–Serre star and identify the pointwise pairing] The Hermitian metric $h$ on $E$ induces a conjugate-linear bundle isomorphism \begin{align*} \tau_h &: E \to E^*, & \tau_h(e) &= h(\,\cdot\,, e), \end{align*} so that $\langle e', \tau_h(e)\rangle = h(e', e) = \langle e', e\rangle_h$ for all $e, e' \in E_x$, $x \in X$. Let $* : \Lambda^{p,q}T^*X \to \Lambda^{n-q,n-p}T^*X$ be the $\mathbb C$-linear extension of the Riemannian Hodge star of $g$, and let $\overline{\,\cdot\,}$ denote complex conjugation of forms. Define the conjugate-linear operator $\bar{*} : A^{p,q}(X) \to A^{n-p,n-q}(X)$ by $\bar{*}\,\omega = *\,\overline{\omega}$, and set \begin{align*} \bar{*}_E &: A^{p,q}(X,E) \to A^{n-p,\,n-q}(X,E^*), & \bar{*}_E(\omega \otimes e) &= \bar{*}\,\omega \otimes \tau_h(e), \end{align*} extended conjugate-linearly. Both $\bar{*}$ and $\tau_h$ are built canonically from the metrics, so $\bar{*}_E$ is independent of the chosen frame and is a conjugate-linear isomorphism of bundles of forms. For decomposables $\alpha = \omega \otimes e$ and $\gamma = \sigma \otimes f$ in $A^{p,q}(X,E)$, using the form-level identity $\omega \wedge \bar{*}\sigma = \langle \omega, \sigma\rangle\, dV_g$ and $\langle e, \tau_h(f)\rangle = \langle e, f\rangle_h$, \begin{align*} \alpha \wedge \bar{*}_E \gamma &= \langle e, \tau_h(f)\rangle\, \omega \wedge \bar{*}\sigma = \langle e, f\rangle_h\, \langle \omega, \sigma\rangle\, dV_g = \langle \alpha, \gamma\rangle_h\, dV_g. \end{align*} By bilinear extension this holds for all $\alpha, \gamma \in A^{p,q}(X,E)$. Integrating, \begin{align*} \int_X \alpha \wedge \bar{*}_E \gamma &= (\alpha, \gamma)_{L^2}, \qquad \text{and in particular} \qquad \int_X \alpha \wedge \bar{*}_E \alpha = \|\alpha\|_{L^2}^2. \end{align*} Thus $\int_X \alpha \wedge \bar{*}_E\alpha \ge 0$, with equality if and only if $\alpha = 0$. [guided] The pairing in the theorem is $\int_X \alpha\wedge\beta$, with $\beta$ valued in $E^*$. To turn the *abstract* nondegeneracy question into a *positivity* statement, we want a canonical way to manufacture, from a given $E$-valued form $\alpha$, an $E^*$-valued partner whose wedge with $\alpha$ recovers the $L^2$ norm. This is exactly the job of the conjugate-linear Hodge–Serre star. Two ingredients are combined. On the bundle side, the Hermitian metric $h$ gives the conjugate-linear identification $\tau_h : E \to E^*$, $\tau_h(e) = h(\cdot, e)$. The defining feature we need is that evaluating $\tau_h(e)$ on a vector reproduces the Hermitian inner product: $\langle e', \tau_h(e)\rangle = h(e', e) = \langle e', e\rangle_h$. On the form side, the standard relation $\omega \wedge \bar*\sigma = \langle\omega,\sigma\rangle\,dV_g$ characterizes the conjugate-linear star $\bar* = *\circ\overline{(\cdot)}$: wedging a form with the conjugate-star of another extracts their pointwise Hermitian inner product times the volume form. (The conjugation is what makes the pairing Hermitian rather than $\mathbb C$-bilinear, hence positive on the diagonal.) Tensoring the two operators gives $\bar{*}_E = \bar{*}\otimes\tau_h$. Why is this well-defined globally? Each factor is constructed purely from the metrics $g$ and $h$ without reference to a frame, so the assignment $\omega\otimes e \mapsto \bar*\omega\otimes\tau_h(e)$ glues to a genuine bundle operator. Computing on a decomposable $\alpha = \omega\otimes e$, $\gamma = \sigma\otimes f$: \begin{align*} \alpha\wedge\bar{*}_E\gamma &= (\omega\otimes e)\wedge(\bar*\sigma\otimes\tau_h(f)) = \langle e,\tau_h(f)\rangle\,(\omega\wedge\bar*\sigma)\\ &= \langle e,f\rangle_h\,\langle\omega,\sigma\rangle\,dV_g = \langle\alpha,\gamma\rangle_h\,dV_g. \end{align*} Setting $\gamma=\alpha$ and integrating over $X$ gives $\int_X\alpha\wedge\bar{*}_E\alpha = \int_X\langle\alpha,\alpha\rangle_h\,dV_g = \|\alpha\|_{L^2}^2$. This is the crucial positivity: the diagonal value of the wedge-evaluation pairing against $\bar{*}_E\alpha$ is the squared $L^2$ norm, strictly positive unless $\alpha\equiv 0$. [/guided] [/step] [step:Prove that $\bar{*}_E$ maps harmonic forms to harmonic forms] [claim:The Hodge–Serre star restricts to a conjugate-linear isomorphism $\bar{*}_E : \mathcal H^{p,q}(X,E) \xrightarrow{\sim} \mathcal H^{n-p,n-q}(X,E^*)$] [/claim] [proof] We use two structural identities. First, the double-star sign: applying [Double Hodge Star](/theorems/2740) (valid on the oriented Riemannian $2n$-manifold $(X,g)$), the $\mathbb C$-linear star satisfies $** = (-1)^{k(2n-k)}\operatorname{id}$ on $k$-forms; with $k = p+q$ and $2n$ even this equals $(-1)^{p+q}$. Since the conjugations in $\bar{*}$ cancel ($\overline{\overline{\omega}} = \omega$) and $\tau_{h^*}\circ\tau_h = \operatorname{id}_E$ under the canonical identification $E^{**} = E$, we obtain on $A^{p,q}(X,E)$ \begin{align*} \bar{*}_{E^*}\,\bar{*}_E &= (-1)^{p+q}\,\operatorname{id}, \end{align*} so $\bar{*}_E$ is a conjugate-linear isomorphism with inverse $(-1)^{p+q}\bar{*}_{E^*}$. Second, we derive a formula for the formal adjoint $\bar\partial_E^*$ in terms of $\bar{*}_E$, using the same [integration by parts](/theorems/2098) as Step 2. Let $\gamma \in A^{p,q-1}(X,E)$ and $\alpha \in A^{p,q}(X,E)$. The form $\gamma \wedge \bar{*}_E\alpha$ has bidegree $(p,q-1)+(n-p,n-q) = (n,n-1)$, so its $\partial$-part vanishes for degree reasons and $d(\gamma\wedge\bar{*}_E\alpha) = \bar\partial(\gamma\wedge\bar{*}_E\alpha)$; [Stokes' theorem](/theorems/1530) on the compact boundaryless $X$ gives $\int_X \bar\partial(\gamma\wedge\bar{*}_E\alpha) = 0$. Expanding by the Leibniz rule of Step 2, \begin{align*} 0 &= \int_X \bar\partial_E\gamma \wedge \bar{*}_E\alpha \;+\; (-1)^{p+q-1}\int_X \gamma \wedge \bar\partial_{E^*}\bar{*}_E\alpha. \end{align*} We rewrite both integrals through the Step 4 identity $\int_X \mu \wedge \bar{*}_E\nu = (\mu,\nu)_{L^2}$: the first integral is exactly $(\bar\partial_E\gamma, \alpha)_{L^2}$, while in the second we write $\bar\partial_{E^*}\bar{*}_E\alpha = \bar{*}_E\mu$ with $\mu = (-1)^{p+q-1}\bar{*}_{E^*}\bar\partial_{E^*}\bar{*}_E\alpha$, using $\bar{*}_{E^*}\bar{*}_E = (-1)^{p+q-1}\operatorname{id}$ on $A^{p,q-1}(X,E)$ (the first double-star computation applied with $k = p+q-1$). Since the scalar $(-1)^{p+q-1}$ is real and the two signs $(-1)^{p+q-1}$ multiply to $+1$, this gives \begin{align*} (\bar\partial_E\gamma, \alpha)_{L^2} &= -\big(\gamma,\ \bar{*}_{E^*}\bar\partial_{E^*}\bar{*}_E\alpha\big)_{L^2}. \end{align*} As this holds for every $\gamma \in A^{p,q-1}(X,E)$, the defining property of the formal adjoint yields \begin{align*} \bar\partial_E^* &= -\,\bar{*}_{E^*}\,\bar\partial_{E^*}\,\bar{*}_E \qquad \text{on } A^{p,q}(X,E), \end{align*} the bundle-valued, conjugate-linear analogue of the codifferential formula $d^* = -\bar{*}\,d\,\bar{*}$ underlying [Hodge Star Commutes with the Laplacian](/theorems/2741). The bidegrees compose correctly: $\bar{*}_E : A^{p,q}(X,E) \to A^{n-p,n-q}(X,E^*)$, then $\bar\partial_{E^*} : A^{n-p,n-q}(X,E^*) \to A^{n-p,n-q+1}(X,E^*)$, then $\bar{*}_{E^*} : A^{n-p,n-q+1}(X,E^*) \to A^{p,q-1}(X,E)$. Exchanging the roles of $E$ and $E^*$ gives likewise $\bar\partial_{E^*}^* = -\,\bar{*}_E\,\bar\partial_E\,\bar{*}_{E^*}$. Now let $\alpha \in \mathcal H^{p,q}(X,E)$. By Step 3, $\bar\partial_E\alpha = 0$ and $\bar\partial_E^*\alpha = 0$. We show $\bar{*}_E\alpha$ is harmonic by checking it is $\bar\partial_{E^*}$-closed and $\bar\partial_{E^*}^*$-closed. From $\bar\partial_E^*\alpha = 0$ and the adjoint identity, $0 = \bar\partial_E^*\alpha = -\bar{*}_{E^*}\bar\partial_{E^*}\bar{*}_E\alpha$. As $\bar{*}_{E^*}$ is an isomorphism, $\bar\partial_{E^*}\,\bar{*}_E\alpha = 0$. From $\bar\partial_E\alpha = 0$ and the second adjoint identity, together with $\bar{*}_{E^*}\bar{*}_E = (-1)^{p+q}\operatorname{id}$, \begin{align*} \bar\partial_{E^*}^*\big(\bar{*}_E\alpha\big) &= -\,\bar{*}_E\,\bar\partial_E\,\bar{*}_{E^*}\,\bar{*}_E\,\alpha = -(-1)^{p+q}\,\bar{*}_E\,\bar\partial_E\,\alpha = 0. \end{align*} Hence $\bar{*}_E\alpha \in \mathcal H^{n-p,n-q}(X,E^*)$. Since $\bar{*}_E$ is an isomorphism of the ambient form spaces and carries $\mathcal H^{p,q}(X,E)$ into $\mathcal H^{n-p,n-q}(X,E^*)$, while its inverse $(-1)^{p+q}\bar{*}_{E^*}$ carries $\mathcal H^{n-p,n-q}(X,E^*)$ back into $\mathcal H^{p,q}(X,E)$ by the identical argument, the restriction is a conjugate-linear isomorphism. [/proof] [guided] We have a clean supply of [harmonic representatives](/theorems/2747) (Step 3) and an operator $\bar{*}_E$ that converts $L^2$ geometry into the wedge pairing (Step 4). The missing link is that $\bar{*}_E$ does not just move forms around — it sends *harmonic* $E$-valued $(p,q)$-forms to *harmonic* $E^*$-valued $(n-p,n-q)$-forms, so that the partner $\bar{*}_E\alpha$ is itself a legitimate representative of a complementary cohomology class. There are two facts to assemble. The first is that $\bar{*}_E$ is invertible: by [Double Hodge Star](/theorems/2740), $** = (-1)^{k(2n-k)}$ on $k$-forms, which for $k=p+q$ on the even-dimensional manifold ($2n$ real dimensions) is $(-1)^{p+q}$; the form conjugations cancel and the metric isomorphisms compose to the identity $\tau_{h^*}\tau_h = \operatorname{id}_E$, so $\bar{*}_{E^*}\bar{*}_E = (-1)^{p+q}\operatorname{id}$. The precise sign is irrelevant — all we need is invertibility. The second fact is the adjoint formula $\bar\partial_E^* = -\bar{*}_{E^*}\bar\partial_{E^*}\bar{*}_E$. This is not an extra axiom: it follows from the same [integration by parts](/theorems/210) as Step 2 — wedge $\gamma \in A^{p,q-1}(X,E)$ against $\bar{*}_E\alpha$, observe the product has bidegree $(n,n-1)$ so $\int_X \bar\partial(\gamma\wedge\bar{*}_E\alpha) = 0$ by Stokes, expand by the Leibniz rule, and read off the adjoint using the Step 4 identity $\int_X\mu\wedge\bar{*}_E\nu = (\mu,\nu)_{L^2}$. It is the engine: it expresses the analytically defined adjoint $\bar\partial_E^*$ purely in terms of $\bar\partial_{E^*}$ conjugated by Hodge stars, which is why $\bar{*}_E$ will swap the two conditions "$\bar\partial$-closed" and "$\bar\partial^*$-closed". One should check the bidegrees compose correctly, and they do: $A^{p,q}(E) \xrightarrow{\bar{*}_E} A^{n-p,n-q}(E^*) \xrightarrow{\bar\partial_{E^*}} A^{n-p,n-q+1}(E^*) \xrightarrow{\bar{*}_{E^*}} A^{p,q-1}(E)$, landing exactly where $\bar\partial_E^*$ lands. Now take a harmonic $\alpha$. By the compactness identity $(\Box_E\alpha,\alpha) = \|\bar\partial_E\alpha\|^2 + \|\bar\partial_E^*\alpha\|^2$ from Step 3, harmonicity is the pair of conditions $\bar\partial_E\alpha = 0$ and $\bar\partial_E^*\alpha = 0$. We test the two harmonicity conditions for $\bar{*}_E\alpha$: - *Is $\bar{*}_E\alpha$ holomorphic-closed?* The condition $\bar\partial_E^*\alpha = 0$ reads $-\bar{*}_{E^*}\bar\partial_{E^*}\bar{*}_E\alpha = 0$; cancel the isomorphism $\bar{*}_{E^*}$ to get $\bar\partial_{E^*}(\bar{*}_E\alpha) = 0$. So the co-closedness of $\alpha$ becomes the closedness of $\bar{*}_E\alpha$. - *Is $\bar{*}_E\alpha$ co-closed?* Using the mirror identity $\bar\partial_{E^*}^* = -\bar{*}_E\bar\partial_E\bar{*}_{E^*}$, \begin{align*} \bar\partial_{E^*}^*(\bar{*}_E\alpha) = -\bar{*}_E\bar\partial_E\bar{*}_{E^*}\bar{*}_E\alpha = -(-1)^{p+q}\bar{*}_E\bar\partial_E\alpha = 0, \end{align*} because $\bar\partial_E\alpha = 0$. So the closedness of $\alpha$ becomes the co-closedness of $\bar{*}_E\alpha$. Both conditions hold, so $\bar{*}_E\alpha$ is harmonic of bidegree $(n-p,n-q)$ with values in $E^*$. Running the same reasoning with $E$ and $E^*$ interchanged shows the inverse $(-1)^{p+q}\bar{*}_{E^*}$ maps back into $\mathcal H^{p,q}(X,E)$, so $\bar{*}_E$ is the claimed isomorphism between harmonic spaces. [/guided] [/step] [step:Conclude nondegeneracy in both variables] By Step 3 the harmonic projections give isomorphisms $\mathcal H^{p,q}(X,E) \cong H^{p,q}_{\bar\partial}(X,E)$ and $\mathcal H^{n-p,n-q}(X,E^*) \cong H^{n-p,n-q}_{\bar\partial}(X,E^*)$; both are therefore finite-dimensional. By Step 2 the pairing is well-defined on cohomology, so we may evaluate it on [harmonic representatives](/theorems/2747). Nondegeneracy in the first variable. Let $c \in H^{p,q}_{\bar\partial}(X,E)$ with $c \ne 0$, and let $\alpha \in \mathcal H^{p,q}(X,E)$ be its (nonzero) harmonic representative. By Step 5, $\beta := \bar{*}_E\alpha$ lies in $\mathcal H^{n-p,n-q}(X,E^*)$ and so defines a class $[\beta] \in H^{n-p,n-q}_{\bar\partial}(X,E^*)$. By Step 4, \begin{align*} \int_X \alpha \wedge \beta &= \int_X \alpha \wedge \bar{*}_E\alpha = \|\alpha\|_{L^2}^2 > 0, \end{align*} since $\alpha \ne 0$. Thus $c$ does not pair to zero with every class, so the map $H^{p,q}_{\bar\partial}(X,E) \to \big(H^{n-p,n-q}_{\bar\partial}(X,E^*)\big)^*$ induced by the pairing is injective. Nondegeneracy in the second variable. Let $d \in H^{n-p,n-q}_{\bar\partial}(X,E^*)$ with $d \ne 0$, and let $\beta \in \mathcal H^{n-p,n-q}(X,E^*)$ be its nonzero harmonic representative. By Step 5 the operator $\bar{*}_E : \mathcal H^{p,q}(X,E) \to \mathcal H^{n-p,n-q}(X,E^*)$ is surjective, so there exists $\alpha \in \mathcal H^{p,q}(X,E)$ with $\bar{*}_E\alpha = \beta$; as $\beta \ne 0$ and $\bar{*}_E$ is injective, $\alpha \ne 0$. Then \begin{align*} \int_X \alpha \wedge \beta &= \int_X \alpha \wedge \bar{*}_E\alpha = \|\alpha\|_{L^2}^2 > 0, \end{align*} so $d$ is not annihilated by all classes in $H^{p,q}_{\bar\partial}(X,E)$. Therefore the pairing is nondegenerate in both arguments. Since both cohomology groups are finite-dimensional and the pairing is nondegenerate, the induced [linear map](/page/Linear%20Map) \begin{align*} H^{p,q}_{\bar\partial}(X,E) &\longrightarrow \big(H^{n-p,n-q}_{\bar\partial}(X,E^*)\big)^* \end{align*} is an injection between finite-dimensional spaces of equal dimension (equality of dimensions follows from nondegeneracy on both sides), hence an isomorphism. This is precisely the assertion of Serre duality, completing the proof. [guided] Everything now converges. Step 3 tells us each cohomology class has a unique harmonic representative, and that the groups are finite-dimensional. Step 2 lets us compute the pairing on those representatives without worrying about the choice. Step 4 says wedging a form against its $\bar{*}_E$-image returns the squared $L^2$ norm, and Step 5 guarantees that $\bar{*}_E$ keeps us inside the harmonic spaces, so the partner is a bona fide complementary class. To show nondegeneracy in the first variable, suppose a nonzero class $c$ were to annihilate everything — we derive a contradiction by producing one class with a nonzero pairing against it. Take its harmonic representative $\alpha \ne 0$ and pair against the class of $\beta = \bar{*}_E\alpha$. The value is exactly $\|\alpha\|_{L^2}^2$, which is strictly positive because the $L^2$ norm vanishes only on the zero form. So $c$ cannot annihilate the complementary group; the map to the [dual space](/page/Dual%20Space) is injective. For the second variable we cannot just repeat verbatim, because the partner construction starts from an $E$-valued form. Instead we use that $\bar{*}_E$ is *surjective* onto the harmonic $E^*$-forms (Step 5): a nonzero harmonic $\beta$ is $\bar{*}_E\alpha$ for some nonzero $\alpha$, and the same computation $\int_X\alpha\wedge\beta = \|\alpha\|_{L^2}^2 > 0$ exhibits a class with nonzero pairing against $d = [\beta]$. Finally, injectivity in both variables between finite-dimensional spaces forces equal dimensions and hence an isomorphism $H^{p,q}_{\bar\partial}(X,E) \cong (H^{n-p,n-q}_{\bar\partial}(X,E^*))^*$. The geometric heart of the argument is the single positivity inequality $\int_X\alpha\wedge\bar{*}_E\alpha = \|\alpha\|_{L^2}^2 > 0$: the Hodge star supplies, for every nonzero harmonic form, an explicit dual partner that detects it. [/guided] [/step]