Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30
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A nonclosed dbar form cannot have a potential

Statement refuted

Let g=zˉ1 dzˉ2 on C2. For every nonempty open U⊆C2, the restricted form g∣U is not of the form ∂ˉu for any smooth function u:U→C.

Facts & Assumptions

Given: The nonempty open set U⊆C2 and the smooth (0,1)-form g∣U=zˉ1 dzˉ2.

[F1]

The coordinate formula for ∂ˉ differentiates form coefficients in each zˉj direction and wedges the result with dzˉj (Bigraded complex forms and the Dolbeault operators).

[F2]

For every smooth complex-valued form, ∂ˉ2=0 (The d, partial and dbar identities).

Counterexample

technique · counterexample
1.1

The proposed witness has a nonzero ∂ˉ derivative at every point. [F1, given, algebra] Applying the coefficient formula [F1] to g=zˉ1 dzˉ2 differentiates its coefficient once in each barred coordinate. Only the j=1 derivative is nonzero, and it equals 1, so ∂ˉg=dzˉ1∧dzˉ2. Because the two coordinate covectors are distinct members of the local wedge basis, this (0,2)-form is nonzero at every point of C2; hence g∣U is not ∂ˉ-closed on any nonempty open U.

2.1

Nonclosedness contradicts the necessary condition for having a potential. [F2, step 1.1, given, algebra] If a smooth u:U→C satisfied ∂ˉu=g∣U, applying ∂ˉ and using [F2] would give 0=∂ˉ2u=∂ˉg∣U=dzˉ1∧dzˉ2. The last form is nonzero at every point of the nonempty set U by step 1.1, a contradiction. Thus no such u exists. ∎

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