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The local index is independent of chart, ball and trivialization

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 without boundary, $n \geq 1$ , let $X$ have an isolated zero $p$ , and let $φ, ψ$ be smooth charts of the given smooth structure centered at $p$ , with admissible radii $ε, δ$ as in Isolated zero and local index of a vector field. Then the maps $f_{φ} (v) = \frac{X_{φ} (ε v)}{∣ X_{φ} (ε v) ∣}, f_{ψ} (v) = \frac{X_{ψ} (δ v)}{∣ X_{ψ} (δ v) ∣}$ have the same degree, using reduced degree when $n = 1$ . Thus the local index is independent of chart and radius. On a coordinate ball it is also unchanged by a smooth fibre trivialization preserving the coordinate orientation. Equivalently, base and fibre orientations must be chosen consistently; reversing only the fibre orientation reverses the degree. No orientation of $M$ is required.

Facts & Assumptions

Given: A smooth vector field $X$ on the smooth $n$ -manifold $M$ with an isolated zero $p$ , smooth charts $(φ, U)$ , $(ψ, V)$ centered at $p$ , and admissible radii $ε, δ > 0$ .

[F1]

The chart representatives are related by the chain rule: with $θ : = ψ \circ φ^{- 1}$ defined near $0$ and $u = φ (x)$ , $X_{ψ} (θ (u)) = d θ_{u} (X_{φ} (u))$ for $u$ near $0$ (The chain rule for differentials of smooth maps, Isolated zero and local index of a vector field). In particular $A : = d θ_{0}$ is an isomorphism with $\det A \neq 0$ (The differential of a diffeomorphism is an isomorphism), and $θ (u) = A u + O (∣ u ∣^{2})$ , $θ^{- 1} (w) = A^{- 1} w + O (∣ w ∣^{2})$ , $d θ_{u} = A + O (∣ u ∣)$ for $u \to 0$ .

[F2]

Degree facts for the normalized sphere maps of $n \geq 2$ : homotopic smooth maps $S^{n - 1} \to S^{n - 1}$ have the same degree (Degree is invariant under proper smooth homotopy), and $\deg (G \circ F) = \deg (G) \deg (F)$ for such maps (Degree is multiplicative under composition, Degree of a map between oriented closed manifolds). A diffeomorphism of $S^{n - 1}$ has degree $+ 1$ or $- 1$ according as it preserves or reverses the orientation (Degree of an orientation-preserving or reversing diffeomorphism).

[F3]

For $n = 1$ reduced degrees are used: the balanced source $S^{0}$ gives $\deg (f) = \frac{f (+ 1) - f (- 1)}{2} \in {- 1, 0, + 1}$ , and $\deg (g \circ f) = \deg (g) \deg (f)$ for maps of $S^{0}$ (The reduced degree of a map into the 0-sphere, Reduced degree into the 0-sphere is homotopy invariant and multiplicative).

[F4]

For an invertible linear $A$ , the normalized linear map $L_{A} (v) : = A v / ∣ A v ∣$ is a diffeomorphism of $S^{n - 1}$ with inverse $L_{A^{- 1}}$ , and its local orientation sign at every $v$ is $sign \det A$ : choosing a positively oriented basis $(v, e_{2}, \dots, e_{n})$ of $R^{n}$ , the outward normal $v$ of the ball is the first basis vector, so the induced map on the tangent space $T_{v} S^{n - 1}$ has the orientation sign of $A$ there, namely $sign \det A$ . Hence $\deg L_{A} = sign \det A$ for $n \geq 2$ by [F2], and the same formula holds for the reduced degree at $n = 1$ , since $L_{A} = sign (A) i d$ on $S^{0}$ by [F3].

Proof

1.1F2F3given

Within either chart, varying the radius through positive admissible radii gives the homotopy $v \mapsto X_{φ} (r (t) v) / ∣ X_{φ} (r (t) v) ∣$ , since its numerator is nonzero. Its degree is constant by [F2] or [F3]. Shrink both radii to a common $ρ > 0$ such that the transition and its inverse are defined and all points used below lie in a zero-free punctured coordinate ball.

1.2F1F4constructalgebra

Put $σ (u) = X_{φ} (u) / ∣ X_{φ} (u) ∣$ , $f (v) = σ (ρ v)$ and $u (v) = θ^{- 1} (ρ v)$ . The normalized $ψ$ -field is $g (v) = L_{d θ_{u (v)}} (σ (u (v)))$ by [F1]. Since $d θ_{u (v)}$ tends uniformly to the invertible matrix $A = d θ_{0}$ , interpolation of this matrix to $A$ stays invertible for small $ρ$ , giving $g ≃ L_{A} \circ σ \circ u$ . The radial homotopy $u_{t} (v) = ((1 - t) ∣ u (v) ∣ + t ρ) u (v) / ∣ u (v) ∣$ stays in the zero-free punctured ball; hence $L_{A} σ (u_{t} (v))$ joins this last map to $L_{A} \circ f \circ \tilde{u}$ , where $\tilde{u} = u / ∣ u ∣$ . Finally $\tilde{u} (v) = L_{A^{- 1}} (v) + O (ρ)$ uniformly, so normalized convex interpolation gives $\tilde{u} ≃ L_{A^{- 1}}$ . Thus $g ≃ L_{A} \circ f \circ L_{A^{- 1}}$ .

2.1F2F3F4step 1.1step 1.2algebra

Multiplicativity and [F4] now give $\deg g = (sign \det A)^{2} \deg f = \deg f$ , also for reduced degree at $n = 1$ . Step 1.1 restores the original radii, proving equality of indices. This does not assert that $f_{φ}$ and $f_{ψ}$ themselves are homotopic: for $X (t) = t^{2}$ and $ψ = - φ$ , they are the distinct constant maps of $S^{0}$ , both of degree zero.

3.1F2F3F4step 2.1algebra∎

On a coordinate ball an orientation-preserving trivialization changes components to $B (u) X_{φ} (u)$ with $B (u) \in G L_{n}^{+} (R)$ . Contracting $B (u)$ to $B (0)$ through $B ((1 - t) u)$ gives a nonzero homotopy on the sphere. The resulting degree is $\deg L_{B (0)} \deg f = \deg f$ by [F4]. A negative determinant instead multiplies it by $- 1$ , which is compensated if the base orientation is also reversed. These are exactly the consistent orientation conventions in the statement.

Depends on

Used by

Cited to discharge well-definedness by Isolated zero and local index of a vector field.

Dependency tree · two levels

38 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)