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A negative-gradient trajectory on a compact Morse manifold has single critical alpha and omega limits

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $M$ be compact, let $f : M \to R$ be Morse, and let $γ$ be a negative-gradient trajectory. Then $γ$ is full and there are critical points $α (γ)$ and $ω (γ)$ such that

$\lim_{t \to - \infty} γ (t) = α (γ), \lim_{t \to \infty} γ (t) = ω (γ) .$

Facts & Assumptions

Given: ${A C}_{ω}$ , a compact smooth manifold $M$ , a Morse function $f$ , and a negative-gradient trajectory $γ$ .

[F1]

Under the stated ${A C}_{ω}$ premise, a smooth vector field on a compact manifold is complete (Every smooth vector field on a compact manifold is complete). Proof 1.1 is the exact use of this choice assumption.

[F2]

Precompact full tails have nonempty compact connected invariant limit sets (Precompact trajectory tails have nonempty compact connected flow-invariant limit sets).

[F3]

Such limit sets for a negative-gradient trajectory consist of critical points (Every precompact end-limit point of a negative-gradient trajectory is critical).

[F4]

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

Proof

technique · direct

1.1F1given

By [F1], the negative-gradient field is complete, so $γ$ has domain $R$ . Both tails have compact closure because they lie in $M$ .

2.1F2F3step 1.1

By [F2] each of $α (γ)$ and $ω (γ)$ is nonempty and connected, and [F3] places it in $Crit (f)$ .

3.1F4step 2.1

By [F4], $Crit (f)$ is finite and hence discrete. A connected subset of a discrete finite set is one point, so both limit sets are single critical points.

4.1F2step 1.1step 3.1∎

Let $p$ be the single point of $ω (γ)$ and let $U$ be any open neighbourhood of $p$ . If no positive tail lies in $U$ , then every compact set $K_{T} = \overline{γ ([T, \infty))}$ meets the closed set $M ∖ U$ . The nested nonempty compact sets $K_{T} \cap (M ∖ U)$ have nonempty intersection by compactness, giving a point of $ω (γ) ∖ U$ , a contradiction. Thus $γ (t) \to p$ as $t \to \infty$ ; the same argument on negative tails proves convergence to the single point of $α (γ)$ .

Depends on

Used by

Dependency tree · two levels

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Sources

Ralph L. Cohen, Bundles, Manifolds, and Homotopy, Theorem 13.2 (standard reference, not scraped)