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A smooth function is Morse if and only if its differential section is transverse to the zero section

Statement

Let $M$ be a smooth manifold and let $f : M \to R$ be smooth. Then $f$ is a Morse function if and only if the smooth section $d f : M \to T^{*} M$ is transverse to the zero section of the cotangent bundle.

Facts & Assumptions

Given: A smooth manifold $M$ and a smooth function $f : M \to R$ .

[F1]

At a critical point $p$ , the Hessian ${Hess}_{p} (f)$ is the intrinsic symmetric bilinear form defined from the second coordinate derivatives, and $p$ is nondegenerate exactly when that bilinear form has zero nullity (The intrinsic Hessian of a smooth function at a critical point, Nondegenerate critical points, nullity, index, and coindex).

[L1]

A smooth map is transverse to an embedded submanifold exactly when its differential plus the target tangent space spans the ambient tangent space, and the zero section is an embedded submanifold of $T^{*} M$ (A smooth map transverse to an embedded submanifold, The zero section is a smooth embedding).

[A1]

In local coordinates $x = (x^{1}, \dots, x^{n})$ near $p$ , if $g = f \circ x^{- 1}$ and $a = x (p)$ , then the cotangent-bundle coordinates identify $d f$ with $x \mapsto (x, \partial_{1} g (x), \dots, \partial_{n} g (x))$ , and the induced map on the fibre quotient along the zero section is multiplication by the Hessian matrix $(\partial_{i} \partial_{j} g (a))$ .

Proof

technique · direct

1.1

The zero set of the section $d f$ is exactly the critical set of $f$ , because $d f_{p} = 0$ means that the differential of $f$ vanishes at $p$ . Thus transversality to the zero section is vacuous away from the critical points.

givenA1

2.1

Fix a critical point $p$ . By [A1], in cotangent-bundle coordinates centered at $0_{p}$ the derivative of $d f$ at $p$ induces on the fibre quotient exactly the Hessian matrix of $f$ at $p$ . Therefore that quotient map is surjective if and only if ${Hess}_{p} (f)$ is nondegenerate in the sense of [F1].

F1A1step 1.1algebra

3.1

By [L1], $d f$ is transverse to the zero section at $p$ if and only if that quotient map is surjective. Combining this with step 2.1 shows that $d f$ is transverse to the zero section at $p$ if and only if $p$ is a nondegenerate critical point of $f$ .

L1step 2.1

4.1

Since the only points at which transversality needs checking are the critical points from step 1.1, step 3.1 proves that $d f$ is transverse to the zero section exactly when every critical point of $f$ is nondegenerate. By [F1], that is exactly the Morse condition.

F1step 1.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

11 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

Shintaro Fushida-Hardy, Morse theory (standard reference, not scraped)
Marco Gualtieri, Topology I: Smooth Manifolds, Part 10 (standard reference, not scraped)