Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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A critical point free noncompact band need not be a global product

Statement refuted

The assertion that every critical-point-free closed band is a level-preserving product is false when compactness is omitted. On M=R2∖{(0,0)}, the function f(x,y)=x has no critical point, but f−1([−1,1]) is not a product with its levels as fibers. Its Euclidean normalized ascending gradient trajectory from (−1,0) escapes at time 1.

Facts & Assumptions

[F1]

Closed sublevel and level set of a smooth function: Let f:M→R be smooth on a boundaryless smooth n-manifold. Write Ma=f−1((−∞,a]), Ma=f−1({a}), and f−1([a,b]) for the closed band. Both endpoints are included. A regular value may have empty fiber. The smooth-manifold convention is def-smooth-manifold.

[F2]

Regular interval diffeomorphism: Assume ACω. If a<b and the closed band K=f−1([a,b]) of a smooth function on a boundaryless manifold is compact and critical-point-free, its normalized flow gives a level-preserving diffeomorphism T:Ma×[a,b]→K, T(x,t)=Φt−a(x).

Counterexample

Given: The objects and hypotheses in the statement refuted.

1.1F1algebra

Using the closed-band notation, the differential is df=dx, nonzero at every point of M. The level at −1 is a copy of R, while the level at zero is {0}×(R∖{0}), with two connected components. A level-preserving product would restrict to homeomorphisms from one fixed fiber onto both, which is impossible.

2.1F2step 1.1algebra∎

The band is noncompact, for it contains the unbounded sequence (0,j) for positive integers j. The normalized gradient is ∂x, whose trajectory from (−1,0) is (−1+t,0) for t<1. At t=1 its only possible limit in R2 is the removed point, so it cannot extend as a trajectory in M. This is exactly the missing compact-band hypothesis in the regular-interval theorem, not a counterexample to that theorem.

Depends on

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Dependency tree · two levels

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Sources