Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedaudited 2026-09-04
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Nondegenerate critical points are isolated

Statement

Let f:MR be smooth. Every nondegenerate critical point of f has an open neighbourhood containing no other critical point of f.

Facts & Assumptions

Given: A smooth function f:MR and a nondegenerate critical point p of f.

[F1]

A critical point is nondegenerate exactly when its Hessian has trivial kernel (The intrinsic Hessian of a smooth function at a critical point, Nondegenerate critical points, nullity, index, and coindex).

[L1]

In coordinates x=(x1,,xn), dfq=i(fx1)xi(x(q))dxqi. (Coordinate formula for the differential of a function)

[L2]

A C1 map RnRn with invertible derivative at a point is a local diffeomorphism there (The Euclidean inverse function theorem).

Proof

technique · dimension split
1.1

If dimM=0, then {p} is open in M, so it already contains no other point and hence no other critical point.

given
1.2

Assume dimM=n>0. Choose a chart x:URn with x(p)=0, write g:=fx1, and define G(u):=(1g(u),,ng(u)). By [L1], for qU one has dfq=0 exactly when G(x(q))=0. [L1, given, assume-case[ positive-dimension], construct]

2.1

The derivative DG(0) is the Hessian matrix of g at 0, and [F1] makes it invertible because p is nondegenerate.

F1step 1.2
3.1

Applying [L2] to G at 0 gives a neighbourhood W of 0 in which G1(0)={0}. Therefore the corresponding neighbourhood x1[W]U contains no critical point except p.

L2step 2.1
4.1

The zero-dimensional case is step 1.1, and the positive-dimensional case is step 3.1. Hence every nondegenerate critical point is isolated.

step 1.1step 3.1

Depends on

Used by

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Sources