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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

1.1L1L2givenchoose

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 $supp (ρ_{i}) \subseteq U_{i}$ .

2.1A1A2step 1.1givenchooseconstruct

Let $K : = M ∖ ⋃_{i = 1}^{r} V_{i}$ . This compact set contains no critical point of $f$ . If $K \neq \emptyset$ , [A1] gives $m > 0$ with $∥ d f_{x} ∥ \geq m$ for every $x \in K$ ; if $K = \emptyset$ , choose any $m > 0$ and read the differential bound below as vacuous. 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, when $K \neq \emptyset$ , $\sum_{i} ∣ λ_{i} ∣ ∥ d ρ_{i} ∥_{C^{0} (K)} < m / 2.$ Define $g : = f + u$ .

3.1step 1.1step 2.1algebra

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.

4.1step 1.1step 2.1step 3.1algebra

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 ${p_{1}, \dots, p_{r}}$ .

5.1step 2.1step 4.1

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.

6.1step 4.1step 5.1∎

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

Depends on

Used by

Dependency tree · two levels

15 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

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