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Converse Poincare-Hopf for nowhere-zero fields

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $M$ be a closed connected smooth $n$ -manifold, $n \geq 1$ . Then $M$ admits a nowhere-zero smooth vector field (A smooth vector field is a smooth section of the tangent bundle) if and only if $χ (M) = 0$ (Euler characteristic of a compact manifold).

Facts & Assumptions

Given: A closed connected smooth $n$ -manifold $M$ , $n \geq 1$ .

[F1]

If $M$ admits a nowhere-zero field then $χ (M) = 0$ (A nowhere-zero vector field forces zero Euler characteristic).

[F2]

For $n = 1$ : if $M \neq \emptyset$ then $M$ is diffeomorphic to the circle $S^{1} = R / Z$ , which carries the standard nowhere-zero rotational field $\partial_{θ}$ ; the empty manifold admits the empty field vacuously (Nonempty closed connected 1-manifolds are circles).

[F3]

For $n \geq 2$ : choose an excellent Morse function $f$ and a Riemannian metric $g$ ; the field $X : = {grad}_{g} f$ has only nondegenerate zeros, namely the critical points, and a critical point of index $λ$ has index $(- 1)^{λ} = \pm 1$ (Every compact smooth manifold admits an excellent Morse function, Every smooth manifold admits a riemannian metric, The Riemannian gradient is the metric dual of the differential, The Riemannian gradient vanishes exactly at the critical points, A Morse gradient zero contributes $(- 1)^{λ}$ to the index, The index of a nondegenerate vector-field zero).

[F4]

Poincare-Hopf: $\sum_{p} {ind}_{p} X = χ (M)$ (Poincare-Hopf for closed manifolds).

[F5]

Given two remaining zeros of opposite index, the ball-selection lemma gives a smooth closed ball containing them in its interior and avoiding every other remaining zero (Two points avoiding a finite set lie in a common embedded ball). They can be cancelled by a modification supported in the interior of that ball, agreeing with the current field near its boundary (Opposite-index nondegenerate zeros cancel in a ball). Subsequent balls may overlap previous ones; they need only avoid the other zeros of the current field.

Proof

1.1F1F2givenconstruct

The forward implication is [F1]. For the converse, the empty manifold has the empty nowhere-zero field. A nonempty closed connected $1$ -manifold is a circle by [F2]; transporting its rotational field gives a nowhere-zero field.

1.2F3F4givenalgebra

Let $n \geq 2$ and $χ (M) = 0$ . Choose the Morse gradient $X$ of [F3]. Its finite zero set has indices $\pm 1$ , and their sum is zero by [F4], so the numbers of positive and negative zeros agree. If there are no zeros, the claim follows immediately.

2.1F5step 1.2constructalgebra∎

Otherwise select one zero of each sign and use [F5] with the finite set of all other current zeros. The resulting ball contains exactly that pair and no boundary zero. Cancel the pair inside it, leaving the field unchanged near the boundary and outside the ball. No new zeros are introduced, and every other zero and its local germ are unchanged. Thus the new field again has equally many positive and negative nondegenerate zeros, with two fewer zeros. Repeating this finite process ends with a smooth nowhere-zero field. This uses neither disjoint supports nor a claim that deleting balls preserves connectedness.

Depends on

Used by

A nowhere-zero vector field on an odd sphere Example

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Sources

Victor Guillemin and Alan Pollack, Differential Topology (Prentice-Hall 1974; complete PDF) (standard reference, not scraped)
John W. Milnor, Topology from the Differentiable Viewpoint (complete 76-page PDF, including the appendix Classifying 1-manifolds) (standard reference, not scraped)