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Scalar holomorphic sheaf acyclicity on pseudoconvex domains

Statement

Assume full AC. Let m≥1 and let Ω⊆Cm be a Hartogs-pseudoconvex domain. Let OΩ be its sheaf of holomorphic functions. Then the derived sheaf cohomology satisfies Hq(Ω,OΩ)=0 for every integer q>0.

Facts & Assumptions

Given: Full AC; an integer m≥1; and a Hartogs-pseudoconvex domain Ω⊆Cm.

[A1]

Full AC is the set-choice postulate (The Axiom of Choice).

[F1]

Smooth forms of bidegree (0,j) form sheaves A0,j with differential ∂ˉ and ∂ˉ2=0 (Bigraded complex forms and the Dolbeault operators).

[F2]

Under full AC, the local Dolbeault lemma makes the smooth (0,∙)-complex exact in positive degrees on polydiscs (The local Dolbeault lemma on nested polydiscs).

[F3]

A C1 function with all coordinate zˉ-derivatives zero is holomorphic (For C1 functions, holomorphy, complex linearity of the real derivative, and the Cauchy–Riemann system agree).

[F4]

Each sheaf of smooth differential-form modules is acyclic for global sections on every open subset of a smooth manifold (Sheaves of smooth-function modules are cohomologically acyclic).

[F5]

An exact resolution by global-section-acyclic sheaves computes right-derived global sections (The acyclic-resolution theorem for right derived functors, Sheaf cohomology as right derived global sections).

[F6]

On a Hartogs-pseudoconvex domain in Cm, every smooth ∂ˉ-closed (0,q)-form is ∂ˉ-exact for 1≤q≤m (Positive-degree Dolbeault vanishing on pseudoconvex domains).

Proof

1.1F1F2F3given

On each coordinate polydisc in Ω, [F2] says the sequence of sheaves 0→OΩ→A0,0→∂ˉA0,1→∂ˉ⋯ is exact in positive degree. Its degree-zero kernel is exactly OΩ: a smooth function in that kernel has all zˉ-derivatives zero and is holomorphic by [F3]. Thus it is an exact sheaf resolution on Ω.

2.1A1F4F5step 1.1

Each resolution term is a sheaf of modules over smooth functions and is Γ(Ω,−)-acyclic by [F4]. The acyclic-resolution theorem [F5] therefore identifies Hq(Ω,OΩ) with the cohomology in degree q of the global smooth Dolbeault complex.

3.1A1F1F6step 2.1algebra∎

For 1≤q≤m, [F6] makes every closed global smooth (0,q)-form exact, so the degree-q cohomology in step 2.1 is zero. For q>m, the complex has no (0,q)-forms on an m-dimensional complex manifold, so its degree-q cohomology is zero as well. Hence Hq(Ω,OΩ)=0 for every q>0.

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