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An inward radial field violates the outward boundary formula

Statement refuted

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the canonical smooth tangent-bundle structure and the cited local-index theorem.

The inward radial field on the closed unit ball shows that the conclusion of Poincare-Hopf with outward-pointing boundary fails if "strictly outward" is weakened to "nonzero on $\partial M$ ": on $D^{n} = {\overline{B}}_{2} (0, 1) \subseteq R^{n}$ with $n \geq 3$ odd, the field $X (u) = - u$ is smooth and nonzero on $\partial D^{n}$ but strictly inward there, its only zero is the centre with index $sign \det (- I_{n}) = (- 1)^{n} = - 1$ (The index of a nondegenerate vector-field zero), while $χ (D^{n}) = 1$ (Contractible nonempty spaces have the homology of a point, Euler characteristic of a compact manifold). Hence $\sum_{p} {ind}_{p} X = - 1 \neq 1 = χ (D^{n})$ , so "nonzero on the boundary" is not enough.

Facts & Assumptions

Given: The closed unit ball $D^{n} \subseteq R^{n}$ , $n \geq 3$ odd (Euclidean spheres and closed balls as subspaces of $R^{n}$ , The Axiom of Countable Choice ( ${A C}_{ω}$ )), and the field $X (u) = - u$ (A smooth vector field is a smooth section of the tangent bundle).

[F1]

At a boundary point $u \in \partial D^{n}$ the outward direction is the radial vector $u$ ; the field $X (u) = - u$ has $⟨ X (u), u ⟩ = - 1 < 0$ , so it is nonzero but strictly inward (Inward, outward, and boundary-tangent vectors).

[F2]

The linear field $X (u) = - u$ has linearization $- I_{n}$ and the only zero $0$ , nondegenerate, with index $sign \det (- I_{n}) = (- 1)^{n}$ (The index of a nondegenerate vector-field zero).

[F3]

$D^{n}$ is contractible, so $χ (D^{n}) = 1$ (Contractible nonempty spaces have the homology of a point, Euler characteristic of a compact manifold).

Counterexample

1.1F1F2algebra

The field $X (u) = - u$ is linear with derivative $- I_{n}$ everywhere, so its only zero is the centre $0$ and it is nondegenerate there with index $(- 1)^{n} = - 1$ because $n$ is odd; on the boundary sphere $⟨ X (u), u ⟩ = - 1 < 0$ , so $X$ is nonzero but strictly inward.

2.1F3step 1.1algebra∎

On the other hand $χ (D^{n}) = 1$ by [F3], so the index sum $- 1$ differs from $χ (D^{n}) = 1$ ; therefore the conclusion of the boundary form of Poincare-Hopf fails for this field even though it is nonzero on the boundary, and the outwardness hypothesis of Poincare-Hopf with outward-pointing boundary is load bearing.

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Sources

John W. Milnor, Topology from the Differentiable Viewpoint (complete 76-page PDF, including the appendix Classifying 1-manifolds) (standard reference, not scraped)
Joel W. Robbin and Dietmar A. Salamon, Introduction to Differential Topology (web draft 2018, complete PDF) (standard reference, not scraped)