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LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-05
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Away from fixed critical neighborhoods, sufficiently small C1 perturbations create no new critical points on a compact manifold

Statement

Let M be a compact smooth manifold, let f:M→R be Morse, and let U1,…,Ur be pairwise disjoint critical neighbourhoods as in On a compact manifold, a Morse function has finitely many critical points and a uniform Hessian gap on disjoint critical neighborhoods. Then there is δ>0 such that every smooth function g with ∥dg−df∥C0<δ on M∖⋃iUi has no critical points in M∖⋃iUi.

Facts & Assumptions

Given: A compact smooth manifold M, a Morse function f:M→R, and pairwise disjoint critical neighbourhoods U1,…,Ur for the critical points of f.

[F1]

A critical point of a smooth function is precisely a zero of its differential (Critical points and critical values of a smooth function).

[A1]

On a compact set, the norm of a continuous cotangent vector field attains its minimum.

Proof

technique · direct
1.1F1L1given

By [L1], the compact set K:=M∖⋃iUi contains no critical point of f. Hence [F1] gives dfx≠0 for every x∈K.

2.1A1step 1.1choose

By [A1], the continuous function x↦∥dfx∥ attains a positive minimum m on K. Put δ:=m/2.

3.1F1step 2.1algebra

If g satisfies ∥dg−df∥C0<δ on K, then for every x∈K one has ∥dgx∥≥∥dfx∥−∥dgx−dfx∥>m−δ=δ>0. Therefore dgx≠0 on K, so [F1] shows that g has no critical point in K.

4.1step 3.1∎

Since K=M∖⋃iUi, this says that sufficiently small C1 perturbations create no new critical points outside the chosen critical neighbourhoods.

Depends on

Used by

Dependency tree · two levels

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Sources