LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Negation scales the local index by $(- 1)^{n}$

Statement

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

Let $M$ be a smooth $n$ -manifold, $n \geq 1$ , and let $X$ be a smooth vector field with an isolated zero at $p$ (Isolated zero and local index of a vector field). Then $- X$ has an isolated zero at $p$ and ${ind}_{p} (- X) = (- 1)^{n} {ind}_{p} X .$

Facts & Assumptions

Given: A smooth vector field $X$ on the smooth $n$ -manifold $M$ with an isolated zero at $p$ .

[F1]

The index is computed by the normalized field on a small sphere: for a chart with representative $X_{φ}$ and admissible $ε > 0$ , ${ind}_{p} X = \deg f$ where $f (v) : = X_{φ} (ε v) / ∣ X_{φ} (ε v) ∣$ , the degree being the ordinary one for $n \geq 2$ and the reduced degree for $n = 1$ (Isolated zero and local index of a vector field).

[F2]

The antipodal map $v \mapsto - v$ of $S^{n - 1}$ has degree $(- 1)^{n}$ for $n \geq 2$ ; the reduced degree of the antipodal map of $S^{0}$ is $- 1 = (- 1)^{1}$ (Degree of identity constant reflection and antipodal sphere maps, The reduced degree of a map into the 0-sphere).

[F3]

Degree is multiplicative under composition of maps of $S^{n - 1}$ for $n \geq 2$ , and reduced degree is multiplicative under composition for maps into $S^{0}$ (Degree is multiplicative under composition, Reduced degree into the 0-sphere is homotopy invariant and multiplicative).

Proof

1.1F1algebra

The field $- X$ vanishes exactly where $X$ does, so $p$ is an isolated zero of $- X$ ; in the same chart $(- X)_{φ} = - X_{φ}$ , so its normalized map is $v \mapsto - X_{φ} (ε v) / ∣ X_{φ} (ε v) ∣ = (α \circ f) (v)$ , where $α (v) = - v$ is the antipodal map of $S^{n - 1}$ and $f$ is the normalized map of $X$ .

2.1F1F2F3step 1.1algebra∎

For $n \geq 2$ multiplicativity of the degree under composition gives ${ind}_{p} (- X) = \deg α \cdot \deg f = (- 1)^{n} {ind}_{p} X$ by [F2], and for $n = 1$ the same computation with reduced degrees gives ${ind}_{p} (- X) = (- 1)^{1} {ind}_{p} X$ , since $(- 1)^{n} = (- 1)^{1}$ for $n = 1$ .

Depends on

Used by

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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)

Negation scales the local index by (−1)n(−1)n