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The dbar-solvability criterion and the holomorphic-orthogonality pairing

Statement

Assume the Axiom of Choice (The Axiom of Choice), in particular its countable-choice consequence used to choose compatible Hermitian metrics. Let $X$ be a compact connected Riemann surface, let $Ω^{0, 1} (X)$ be the space of smooth $(0, 1)$ -forms, let $Ω^{0, 0} (X) = C^{\infty} (X, C)$ , and define the global Dolbeault group of the trivial holomorphic line bundle by $H^{0, 1} (X, O_{X}) : = Ω^{0, 1} (X) / \overset{ˉ}{\partial} Ω^{0, 0} (X) .$ Every smooth $(0, 1)$ -form is $\overset{ˉ}{\partial}$ -closed because there are no $(0, 2)$ -forms on a Riemann surface. Let $K = Λ^{1, 0} T^{*} X$ and let $Ω (X) = H^{0} (X, K)$ be the space of holomorphic differentials (Holomorphic line bundles and meromorphic sections on a Riemann surface, Meromorphic differentials, orders and residues). For $θ \in Ω^{0, 1} (X)$ , the following are equivalent:

$θ = \overset{ˉ}{\partial} g$ for some smooth $g : X \to C$ .
Its Dolbeault class $[θ] \in H^{0, 1} (X, O_{X})$ is zero.
$\int_{X} θ \land ω = 0$ for every $ω \in Ω (X)$ .

The pairing $B : H^{0, 1} (X, O_{X}) \times H^{0} (X, K) ⟶ C, B ([θ], ω) = \int_{X} θ \land ω$ is well defined and induces the complex-linear isomorphism $H^{0, 1} (X, O_{X}) ≅ H^{0} (X, K)^{*}$ . The quotient definition gives the equivalence of (1) and (2); Stokes' theorem makes $B$ well defined; harmonic-star duality gives the isomorphism, hence the equivalence of (2) and (3) (Harmonic star duality for line bundle valued dolbeault cohomology, The general Stokes theorem).

Facts & Assumptions

Given: A compact connected Riemann surface $X$ , a smooth $(0, 1)$ -form $θ$ , and the full Axiom of Choice.

[F1]

The Riemann surface atlas supplies a connected smooth oriented real surface with its complex orientation. The trivial holomorphic line bundle has global Dolbeault operator $\overset{ˉ}{\partial}$ , and on a curve $(0, 2)$ -forms vanish (Holomorphic line bundles and meromorphic sections on a Riemann surface, Bigraded complex forms and the Dolbeault operators).

[F2]

Holomorphic differentials are the holomorphic sections of $K = Λ^{1, 0} T^{*} X$ ; locally $ω = f (z) d z$ with $\overset{ˉ}{\partial} f = 0$ , so $d ω = d f \land d z = 0$ (Holomorphic line bundles and meromorphic sections on a Riemann surface, Meromorphic differentials, orders and residues, Bigraded complex forms and the Dolbeault operators).

[F3]

The manifold is boundaryless, and for every smooth $1$ -form $η$ on compact $X$ , under the countable-choice consequence of the assumed full AC, Stokes gives $\int_{X} d η = 0$ ; compactness makes $η$ compactly supported (The general Stokes theorem).

[F4]

Assume full AC (The Axiom of Choice). For a compact Riemann surface, a holomorphic line bundle with a Hermitian metric, and a compatible Riemannian metric, the integration pairing $H^{0, 1} (X, E) \times H^{0} (X, K \otimes E^{*}) \to C$ is well defined and induces an isomorphism $H^{0, 1} (X, E) \to H^{0} (X, K \otimes E^{*})^{*}$ . The metrics required here exist for the trivial line bundle under the countable-choice consequence of AC (Harmonic star duality for line bundle valued dolbeault cohomology, Hermitian metric and $L^{2}$ pairing on a compact Riemann surface).

Proof

technique · quotient definition, Stokes' theorem, and the Dolbeault pairing

1.1F1given

By definition, $[θ] = 0$ in $Ω^{0, 1} (X) / \overset{ˉ}{\partial} Ω^{0, 0} (X)$ exactly when $θ$ belongs to the image of the global operator, which is exactly the existence of a smooth $g$ with $θ = \overset{ˉ}{\partial} g$ . Every $(0, 1)$ -form is closed because the next bidegree is $(0, 2) = 0$ , so this quotient is the Dolbeault group stated above.

1.2F2F3givenalgebra

If $θ$ is replaced by $θ + \overset{ˉ}{\partial} g$ and $ω$ is holomorphic, then $d (g ω) = \overset{ˉ}{\partial} g \land ω$ : the term $\partial g \land ω$ has type $(2, 0)$ and vanishes on a curve, while $d ω = 0$ by [F2]. Thus Stokes [F3] gives $\int_{X} (\overset{ˉ}{\partial} g) \land ω = \int_{X} d (g ω) = 0$ . The integral therefore depends only on $[θ]$ . It is complex-bilinear, since wedge product and integration are complex-linear in each factor. If $θ = \overset{ˉ}{\partial} g$ , the same identity gives $\int_{X} θ \land ω = 0$ for every $ω$ , proving (1) $\Rightarrow$ (3).

2.1F4step 1.1step 1.2given∎

Choose a compatible Hermitian metric on $X$ and a Hermitian metric on the trivial line bundle, as supplied by [F4]; the full Axiom of Choice implies the countable-choice assumption used for this metric existence. Apply [F4] to $E = O_{X}$ , so $E^{*} ≅ O_{X}$ and $K \otimes E^{*} ≅ K$ . Its integration pairing is exactly $B$ , hence the induced map $[θ] \mapsto B ([θ], \cdot)$ is an isomorphism, in particular injective. If (3) holds, this functional is zero, so injectivity gives $[θ] = 0$ and (2) follows; then (1) follows from step 1.1. This also covers the zero class and the case $H^{0} (X, K) = 0$ : in the latter case the isomorphism forces $H^{0, 1} (X, O_{X}) = 0$ , so the vacuous orthogonality condition still implies solvability. Full AC supplies the assumptions of harmonic-star duality and the countable choice required by the metric-existence and Stokes interfaces. The quotient calculation in step 1.1 uses no choice.

Depends on

Used by

Abel's theorem for divisors Theorem

Dependency tree · two levels

88 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Karl Otto Forster, Lectures on Riemann Surfaces (GTM 81) (standard reference, not scraped)
Curtis T. McMullen, Riemann Surfaces, Harvard Math 213b course notes (2026) (standard reference, not scraped)
Jean-Pierre Demailly, Complex Analytic and Differential Geometry (author manuscript, Universite Grenoble Alpes) (standard reference, not scraped)