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Morse lemma

Statement

Let f:MR be smooth, let p be a nondegenerate critical point of f, and let λ be the index of p. If n=dimM, then there are local coordinates (x1,,xn) centered at p in which

f=f(p)i=1λ(xi)2+i=λ+1n(xi)2.

For n=0, both sums are empty.

Facts & Assumptions

Given: A smooth function f:MR, a nondegenerate critical point p, and its index λ.

[F1]

Index and nondegeneracy are defined from the critical Hessian (Nondegenerate critical points, nullity, index, and coindex).

[L1]

Sylvester's law gives a linear coordinate change that puts any symmetric Hessian matrix into diagonal normal form with its positive, negative, and zero counts recorded on the diagonal (Sylvester's law of inertia: every real symmetric form is congruent to diag(Ip,Iq,0r), and (p,q,r) is unique).

[L2]

The chartwise inertia counts of the Hessian equal the intrinsic index, coindex, and nullity (Sylvester inertia makes the Morse index intrinsic).

[L3]

A nonzero second derivative in one chosen coordinate splits off a signed square after a local coordinate change (A nonzero second derivative splits off a signed square with a smooth parameter).

[L4]

After splitting one signed square, the remaining Hessian is the restricted residual Hessian (Splitting one Morse coordinate preserves the residual Hessian).

Proof

technique · dimension induction
1.1

If n=0, the manifold is locally a point, so f is locally constant at p. The Hessian acts on the zero vector space, hence λ=0 by [F1], and the displayed formula is exactly f=f(p) with both sums empty.

F1givenbase
1.2

Assume the theorem proved in dimensions <n, where n>0. Choose local coordinates u=(u1,,un) centered at p and write g:=fu1f(p). By [L1], after a linear change of the u-coordinates the Hessian matrix of g at 0 is diagonal with entries in {1,1,0}. Since p is nondegenerate and has index λ, [F1] and [L2] force exactly λ negative diagonal entries, exactly nλ positive diagonal entries, and no zero entry. Reorder the coordinates so the first diagonal entry is negative when λ>0 and positive when λ=0; in particular 2g/(u1)2(0)0. [F1, L1, L2, given, assume-case[ positive-dimension], construct]

2.1

Apply [L3] to the first coordinate u1, taking the remaining variables as parameters. After shrinking the chart there are new coordinates (v1,y) with g(v1,y)=ε(v1)2+H(y), where ε{±1}, yRn1, and 0 is a critical point of H.

L3step 1.2construct
3.1

By [L2], the Hessian of g in the (v1,y) chart still has index λ. By [L4], the Hessian of H at 0 is the restriction to the y-coordinates, and the split v1-direction contributes one negative square exactly when ε=1. Therefore Hess0(H) is nondegenerate, with index λ1 when ε=1 and index λ when ε=1.

L2L4step 2.1algebra
4.1

Apply the induction hypothesis to H on Rn1. It yields local coordinates (v2,,vn) putting H into its Morse normal form, and adjoining v1 contributes one additional negative square exactly when ε=1. Therefore the full expression for g has exactly λ negative squares and nλ positive squares.

ihstep 3.1construct
5.1

Combining steps 2.1 and 4.1 proves the displayed normal form for dimension n, and step 1.1 covers the base case.

step 1.1step 2.1step 4.1discharge-induction

Depends on

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