Harmonic star duality for line bundle valued dolbeault cohomology

By [F2], $\mathrm{\#}$ in the forward direction sends $ud\bar{z}\otimes e$ to $-i\psi \bar{u}dz\otimes e^{\ast }$, is conjugate-linear, and is a smooth bundle isomorphism. In the reverse direction it sends $adz\otimes e^{\ast }$ to $i{\psi }^{-1}\bar{a}d\bar{z}\otimes e^{\ast \ast }$; substituting $a=-i\psi \bar{u}$ gives $-u$, so ${\mathrm{\#}}^2=-1$ and the inverse is minus the reverse star. The wedge identity gives $s\wedge \mathrm{\#}s=\mathrm{\mid }s{\mathrm{\mid }}^{2}d{V}_g$. A nonzero smooth $s$ has positive squared norm on a neighborhood of a point, hence its integral is $\mathrm{\parallel }s{\mathrm{\parallel }}_{L^2}^2>0$; the pointwise top form is nonzero exactly where $s$ is nonzero.

For a smooth section $f$ of $E$ and a holomorphic section $\alpha$ of $F$, evaluation contracts $f\alpha$ to a global smooth $(1,0)$-form. Its exterior derivative has only a $(1,1)$ component on a curve, and locally $\alpha =adz\otimes e^{\ast }$ with ${\mathrm{\partial }}_{\bar{z}}a=0$, so $d(f\alpha )={\bar{\mathrm{\partial }}}_{E}f\wedge \alpha$. On compact $X$ the primitive $f\alpha$ has compact support, and [F6] applied to real and imaginary parts gives ${\int }_X{\bar{\mathrm{\partial }}}_{E}f\wedge \alpha =0$. Consequently the integral defines $B([u],\alpha )$ independently of the smooth representative, and coefficientwise wedge and evaluation make it complex-bilinear.

In the holomorphic frame $dz\otimes e^{\ast }$ of $F=K\otimes E^{\ast }$, the tensor metric has weight $w=2\mathrm{/}(\rho \psi )$ by [F2]. The inverse and canonical cocycles are holomorphic, so this is a holomorphic line bundle by [F1]. For a smooth local coefficient $u$ of an $E$-valued $(0,1)$-form put $a=-i\psi \bar{u}$. Applying [F3] to $F$ gives ${\mathrm{\Delta }}_{F,0}^{\mathrm{\prime }\mathrm{\prime }}a=-\psi {\mathrm{\partial }}_z(2(\rho \psi {)}^{-1}{\mathrm{\partial }}_{\bar{z}}(-i\psi \bar{u}))=2i\psi {\mathrm{\partial }}_z((\rho \psi {)}^{-1}{\mathrm{\partial }}_{\bar{z}}(\psi \bar{u}))$. Applying [F3] to $E$ and conjugating gives $-i\psi \overline{{\mathrm{\Delta }}_{E,1}^{\mathrm{\prime }\mathrm{\prime }}u}=2i\psi {\mathrm{\partial }}_z((\rho \psi {)}^{-1}{\mathrm{\partial }}_{\bar{z}}(\psi \bar{u}))$, since $\rho ,\psi$ are real. These equal coefficients prove ${\mathrm{\Delta }}_{F,0}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\#}=\mathrm{\#}{\mathrm{\Delta }}_{E,1}^{\mathrm{\prime }\mathrm{\prime }}$ on smooth forms; the local equalities are global because the operators and star are globally defined.

By [F4], a Hilbert harmonic degree-one form is smooth and harmonic exactly when ${\mathrm{\partial }}_z(\psi u)=0$. Conjugating this equation gives ${\mathrm{\partial }}_{\bar{z}}(\psi \bar{u})=0$, exactly holomorphy of $a=-i\psi \bar{u}$ in the holomorphic frame of $F$. Conversely, if $a$ is a holomorphic section coefficient, the inverse star gives $u=-i{\psi }^{-1}\bar{a}$, which is smooth, satisfies that first-order equation and therefore belongs to the Hilbert harmonic kernel by [F4]. Thus $\mathrm{\#}$ is a conjugate-linear bijection between ${\mathcal{H}}^{0,1}(E)$ and $H^0(X,F)$; the latter is the degree-zero harmonic kernel for $F$ by [F5]. This uses smooth representatives of maximal-domain kernels, not an identification of a maximal domain with smooth forms.

Replace each Dolbeault class by its unique harmonic representative $u$ using [F5]. If $u\mathrm{\ne }0$, then $\mathrm{\#}u\in H^0(X,F)$ by step 3.1 and $B([u],\mathrm{\#}u)=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2}^2>0$ by step 1.1. Conversely every nonzero $\alpha \in H^0(X,F)$ equals $\mathrm{\#}v$ for a nonzero harmonic $v$, so $B([v],\alpha )>0$. This proves nondegeneracy in both variables. To prove the asserted isomorphisms explicitly, choose an $L^2$-orthonormal basis $e_1,\dots ,e_m$ of the finite-dimensional harmonic space; if it is zero take the empty basis. The $\mathrm{\#}{e}_j$ form a complex basis of $H^0(X,F)$ by the conjugate-linear bijection, and $B([e_j],\mathrm{\#}{e}_k)={\delta }_{jk}$. Thus $[e_j]\mapsto B([e_j],\cdot )$ maps a basis to its dual basis and is a complex-linear isomorphism; likewise $\mathrm{\#}{e}_k\mapsto B(\cdot ,\mathrm{\#}{e}_k)$ gives the complex-linear inverse-side duality $H^0(X,F)\cong H^{0,1}(X,E{)}^{\ast }$.

Integrating [F2] gives $B([u],\mathrm{\#}v)=⟨u,v{⟩}_{L^2}$ for harmonic $u,v$. Since $B$ is complex-bilinear and $\mathrm{\#}$ conjugate-linear, this expression is linear in $u$ and conjugate-linear in $v$; pulling back a linear functional on $H^0(X,F)$ along $\mathrm{\#}$ therefore gives a conjugate-linear functional on the harmonic space. Separately, $R(u)(v)=⟨v,u⟩$ is complex-linear in $v$ and satisfies $R(cu)=\bar{c}R(u)$. The harmonic space is finite-dimensional and hence Hilbert in its inherited pairing, so [F7] identifies this conjugate-linear map with the Riesz bijection to its ordinary complex-linear dual. This proves the stated conjugation conventions and completes all claims, with full AC inherited as in [F8].