CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-04

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A degenerate critical set can be nonisolated

Statement refuted

If every critical point of a smooth function is degenerate, then the critical set is still forced to be discrete.

Facts & Assumptions

Given: The smooth function $f : R^{2} \to R$ , $f (x, y) = x^{2}$ .

[F1]

Critical points are the zeros of the differential, the Hessian is computed at a critical point, and degeneracy means the Hessian has nontrivial kernel (Critical points and critical values of a smooth function, The intrinsic Hessian of a smooth function at a critical point, Nondegenerate critical points, nullity, index, and coindex).

Counterexample

technique · direct computation

1.1

The differential is $d f_{(x, y)} = (2 x, 0)$ in the standard coordinates, so $d f_{(x, y)} = 0$ exactly when $x = 0$ . Thus the whole line ${(0, y) : y \in R}$ is critical.

F1givenalgebra

2.1

The Hessian matrix is constant, namely $Hess (f) = (\begin{array}{cc} 2 & 0 \\ 0 & 0 \end{array})$ . Its kernel contains the $y$ -axis, so every critical point on the line from step 1.1 is degenerate.

F1step 1.1algebra

3.1

Since the critical set contains an entire line, it is not discrete and its points are not isolated. Therefore degenerate critical sets can be nonisolated.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Michele Audin and Mihai Damian, Morse Theory and Floer Homology (standard reference, not scraped)