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Sylvester inertia makes the Morse index intrinsic

Statement

Let $f : M \to R$ be smooth and let $p$ be a critical point of $f$ . Then the numbers of positive, negative, and zero directions of any chart matrix of ${Hess}_{p} (f)$ are independent of the chart. Equivalently, the nullity, index, and coindex of $p$ are intrinsic.

Facts & Assumptions

Given: A smooth function $f : M \to R$ and a critical point $p$ .

[F1]

Hessian matrices in two charts are congruent (Critical-point Hessian matrices transform by congruence under chart changes).

[F2]

Nullity, index, and coindex are defined from the Hessian as kernel dimension and maximal negative- and positive-definite dimensions (Nondegenerate critical points, nullity, index, and coindex).

[L1]

Congruent real symmetric matrices have the same inertia data (Sylvester's law of inertia: every real symmetric form is congruent to $diag (I_{p}, - I_{q}, 0_{r})$ , and $(p, q, r)$ is unique).

[L2]

The inertia counts are exactly the numbers of positive, negative, and zero entries in a diagonal normal form (Positive and negative definiteness, the inertia $(p, q, r)$ , rank $p + q$ , and signature $p - q$ of a real symmetric bilinear or quadratic form).

Proof

technique · congruence invariance

1.1F1given

By [F1], any two chart matrices of ${Hess}_{p} (f)$ are congruent real symmetric matrices.

2.1L1L2step 1.1

Therefore [L1] and [L2] give the same triple $(positive, negative, zero)$ for every chart matrix.

3.1F2L2step 2.1

For one diagonal normal form, the zero count is the kernel dimension, the negative count is the maximal dimension of a negative-definite subspace, and the positive count is the maximal dimension of a positive-definite subspace, by [F2] and [L2]. Hence nullity, index, and coindex are the same in every chart.

Sylvester inertia makes the Morse index intrinsic

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources