One dimensional constant zero mode of dolbeault laplacian

Assume the Axiom of Choice (The Axiom of Choice), inherited through the finite-dimensional-kernel and ellipticity items used below. Let $X$ be a compact Riemann surface, $E=X\times \mathbb{C}$ the trivial holomorphic line bundle with the constant Hermitian metric $h_0(1,1)=1$, and $g$ any compatible Riemannian metric on $X$. Then the constant function $1$ satisfies ${\mathrm{\Delta }}^{\mathrm{\prime }\mathrm{\prime }}1=0$, and the kernel of the Dolbeault Laplacian on functions is exactly $H^{0,0}(E)=\{u\in L^2(X):\bar{\mathrm{\partial }}u=0\}=\mathbb{C}\cdot 1.$ Thus the constant zero mode of the Dolbeault Laplacian is one-dimensional, and $H^{0,0}(X,E)=H^0(X,\mathcal{O})=\mathbb{C}$: every holomorphic function on a compact connected Riemann surface is constant. The same conclusion holds for the trivial bundle with any Hermitian metric.

Here $H^{0,q}(E)$ denotes the harmonic space ${\mathcal{H}}^{0,q}(E)=\ker {\mathrm{\Delta }}_q^{\mathrm{\prime }\mathrm{\prime }}$ with its smooth representatives, while $H^{0,q}(X,E)$ denotes smooth Dolbeault cohomology.

Demailly’s Dolbeault results in §7 apply to a holomorphic Hermitian bundle; the flat-connection de Rham decomposition in §3.3 is contextual and is not used for a varying Hermitian weight. The verification above uses proved local operator interfaces and gives the concrete calculation itself.