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For a compact Morse function, disjoint local bump perturbations can separate finitely many equal critical values without changing the Hessians

Statement

Let $M$ be a compact smooth manifold and let $f : M \to R$ be Morse. Then every $C^{\infty}$ neighbourhood $U$ of $f$ contains a smooth function $g$ such that:

$g$ has the same critical points as $f$ ;
the Hessian of $g$ at each critical point equals the Hessian of $f$ there; and
distinct critical points of $g$ have distinct critical values.

Facts & Assumptions

Given: A compact smooth manifold $M$ , a Morse function $f : M \to R$ , and a $C^{\infty}$ neighbourhood $U$ of $f$ .

[L1]

A Morse function on a compact manifold has finitely many critical points (A Morse function on a compact manifold has finitely many critical points).

[L2]

Around a compact set inside an open set there exists a smooth bump that is identically $1$ near the compact set and supported in the open set (A manifold bump for a compact set inside an open set).

[A1]

If a continuous cotangent field is nowhere zero on a compact set, then its norm has a positive minimum there.

[A2]

For finitely many fixed smooth bump functions, the linear combination map from the coefficient space into $C^{\infty} (M)$ is continuous. Hence sufficiently small coefficients place that combination inside any prescribed $C^{\infty}$ neighbourhood of $0$ , and at the same time make its differential uniformly small on a chosen compact set.

Proof

technique · direct

1.1

By [L1], the critical points of $f$ are $p_{1}, \dots, p_{r}$ . Choose pairwise disjoint open neighbourhoods $V_{i} ⋐ U_{i}$ of the $p_{i}$ such that each $U_{i}$ contains no critical point other than $p_{i}$ , and by [L2] choose smooth functions $ρ_{i}$ with $ρ_{i} = 1$ on $V_{i}$ and $supp (ρ_{i}) \subseteq U_{i}$ .

L1L2givenchoose

2.1

Let $K : = M ∖ ⋃_{i = 1}^{r} V_{i}$ . This compact set contains no critical point of $f$ , so [A1] gives a constant $m > 0$ with $∥ d f_{x} ∥ \geq m$ for every $x \in K$ . Using [A2], choose real numbers $λ_{i}$ such that the shifted numbers $f (p_{i}) + λ_{i}$ are pairwise distinct, the finite sum $u : = \sum_{i = 1}^{r} λ_{i} ρ_{i}$ satisfies $f + u \in U$ , and $\sum_{i} ∣ λ_{i} ∣ ∥ d ρ_{i} ∥_{C^{0} (K)} < m / 2.$ Define $g : = f + u$ .

A1A2step 1.1givenchooseconstruct

3.1

On each neighbourhood $V_{i}$ one has $g = f + λ_{i}$ , because $ρ_{i} = 1$ there and every $ρ_{j}$ with $j \neq i$ vanishes there. Therefore $p_{i}$ is still a critical point of $g$ , and the Hessian of $g$ at $p_{i}$ equals the Hessian of $f$ there.

step 1.1step 2.1algebra

4.1

For $x \in K$ , step 2.1 gives $∥ d g_{x} - d f_{x} ∥ \leq \sum_{i} ∣ λ_{i} ∣ ∥ d ρ_{i} ∥_{C^{0} (K)} < m / 2,$ so $∥ d g_{x} ∥ \geq ∥ d f_{x} ∥ - ∥ d g_{x} - d f_{x} ∥ > m / 2 > 0$ . Hence $g$ has no critical point on $K$ . Since every critical point of $f$ lies in some $V_{i}$ , step 3.1 shows that the critical set of $g$ is exactly ${p_{1}, \dots, p_{r}}$ .

step 1.1step 2.1step 3.1algebra

5.1

The critical values of $g$ are $g (p_{i}) = f (p_{i}) + λ_{i}$ , which are pairwise distinct by step 2.1. The same step also gives $g \in U$ . Therefore $g$ has the same critical points and Hessians as $f$ , while all of its critical values are distinct.

step 2.1step 4.1

6.1

Thus one can separate all repeated critical values by disjoint local perturbations without changing any critical Hessian.

step 4.1step 5.1∎

Depends on

Used by

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Sources

Marco Gualtieri, Topology I: Smooth Manifolds, Part 11 (standard reference, not scraped)
Shintaro Fushida-Hardy, Morse theory (standard reference, not scraped)