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Dolbeault cohomology of a domain
Definition
Let be open and let . Write for the smooth complex-valued forms of bidegree . Set and , and define The Dolbeault cohomology vector space is It is well-defined because . If is open, restriction of forms induces a map ; these maps are independent of representatives and compose as restrictions do. No identification with sheaf cohomology is asserted.
Facts & Assumptions
Given: An open , a bidegree , and the complex differential forms defined in Bigraded complex forms and the Dolbeault operators.
The spaces of smooth forms split by bidegree and maps into (Bigraded complex forms and the Dolbeault operators).
The Dolbeault operator satisfies (The d, partial and dbar identities).
Proof
By [F2], every image with is killed by , so and the quotient in the Definition is well-defined. For the image is zero by the stated convention; for the target of is zero.
For an inclusion , restriction commutes with coordinate differentiation: the coefficient formula for gives term by term. Hence closed forms restrict to closed forms and exact forms restrict to exact forms.
Define by for closed . If , then ; step 1.2 gives , so the class is independent of the representative.
Restricting a form to itself is the identity, and for open inclusions , . Therefore the induced cohomology maps satisfy the same identity and composition laws.
Depends on
Used by
Dependency tree · two levels
10 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
- Lebl, Tasty Bits of Several Complex Variables, v4.4, Chapter 4 §4.4 (standard reference, not scraped)
- Guillemin and Campbell, MIT 18.117 Lecture Notes, Lectures 1–4 (standard reference, not scraped)