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.

1.1

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.

F1algebra
2.1

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.

F2step 1.1algebra∎

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