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

Statement

Let M be compact, let f:MR 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

limtγ(t)=α(γ),limtγ(t)=ω(γ).

Facts & Assumptions

Given: A compact smooth manifold M, a Morse function f, and a negative-gradient trajectory γ.

[F1]

A smooth vector field on a compact manifold is complete (Every smooth vector field on a compact manifold is complete).

[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
1.1

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

F1given
2.1

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

F2F3step 1.1
3.1

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.

F4step 2.1
4.1

A trajectory with singleton tail-limit set converges to that point: otherwise a sequence of tail times outside a fixed neighbourhood would have a limit point in the same tail-limit set. Thus the two displayed limits hold.

step 2.1step 3.1

Depends on

Used by

Dependency tree · two levels

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Sources