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The index of a nondegenerate vector-field zero

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, $n \geq 1$ , let $X$ be a smooth vector field and let $p$ be a nondegenerate zero of $X$ (Nondegenerate zero of a vector field). Then ${ind}_{p} X = sign \det (D X_{p} : T_{p} M \to T_{p} M) \in {+ 1, - 1} .$ In particular every nondegenerate zero has index $+ 1$ or $- 1$ .

Facts & Assumptions

Given: A smooth $n$ -manifold $M$ , a smooth vector field $X$ and a nondegenerate zero $p$ of $X$ .

[F1]

In a smooth chart of the given smooth structure $(φ, U)$ with $φ (p) = 0$ the chart representative vanishes at $0$ and its derivative there is the vertical derivative: $X_{φ} (0) = 0$ , $D X_{φ} (0)$ corresponds to $D X_{p}$ under the chart trivialization, and $D X_{p}$ invertible is equivalent to $D X_{φ} (0)$ invertible; moreover $X_{φ} (u) = A u + R (u)$ with $A : = D X_{φ} (0)$ and $R (u) = O (∣ u ∣^{2})$ (Nondegenerate zero of a vector field, The induced tangent bundle chart).

[F2]

The index is computed by the normalized field on a small sphere, ${ind}_{p} X = \deg (v \mapsto X_{φ} (ε v) / ∣ X_{φ} (ε v) ∣)$ with the standard orientations ( $n \geq 2$ ), and for $n = 1$ by the reduced degree of the induced map $S^{0} \to S^{0}$ (Isolated zero and local index of a vector field); the value does not depend on the chart or the admissible radius (The local index is independent of chart, ball and trivialization).

[F3]

The normalized linear map $L_{A} (v) : = A v / ∣ A v ∣$ of an invertible $A$ is a diffeomorphism of $S^{n - 1}$ whose local orientation sign is $sign \det A$ , hence $\deg L_{A} = sign \det A$ ; for $n = 1$ this is the reduced degree of $v \mapsto sign (A) v$ . Degree is invariant under homotopies of maps of $S^{n - 1}$ ( $n \geq 2$ ) and, for $n = 1$ , under homotopies of maps of $S^{0}$ (Degree of a map between oriented closed manifolds, Degree is invariant under proper smooth homotopy, Degree is multiplicative under composition, Reduced degree into the 0-sphere is homotopy invariant and multiplicative).

Proof

1.1F1algebra

Take a smooth chart as in [F1] and put $A : = D X_{φ} (0)$ , so $X_{φ} (u) = A u + R (u)$ with $R (u) = O (∣ u ∣^{2})$ and $A$ invertible; write $c : = ∣ A^{- 1} ∣^{- 1} > 0$ and $∣ R (u) ∣ \leq C ∣ u ∣^{2}$ with $C > 0$ . For $ε < c / (2 C)$ and $v \in S^{n - 1}$ the vector $A v + t ε^{- 1} R (ε v)$ with $t \in [0, 1]$ has norm at least $c - ε^{- 1} C ε^{2} \geq c / 2 > 0$ , so $H_{t} (v) : = (A v + t ε^{- 1} R (ε v)) / ∣ \dots ∣$ is a homotopy from the normalized linear map $L_{A}$ to the normalized chart field $v \mapsto X_{φ} (ε v) / ∣ X_{φ} (ε v) ∣$ .

2.1F1F2F3step 1.1algebra∎

For $n \geq 2$ homotopy invariance gives ${ind}_{p} X = \deg L_{A} = sign \det A$ , and for $n = 1$ the same homotopy is one of maps $S^{0} \to S^{0}$ , so by [F3] the reduced degrees agree and again ${ind}_{p} X = \deg L_{A} = sign \det A$ ; since $A$ corresponds to $D X_{p}$ by [F1], $\det A$ and $D X_{p}$ have the same sign and ${ind}_{p} X = sign \det (D X_{p}) \in {+ 1, - 1}$ , because $D X_{p}$ is invertible.

Depends on

Used by

A Morse gradient zero contributes (-1)^λ to the index Corollary
An interval has nonzero Euler characteristic despite being odd-dimensional Counterexample
An inward radial field violates the outward boundary formula Counterexample
Source, sink and saddle indices on a surface Example
The outward radial field on a disk Example
Euler number of a clutched bundle as the clutching degree Lemma
Finite tangent index count and inward boundary sum Lemma
Opposite-index nondegenerate zeros cancel in a ball Lemma
The index sum of an outward field on an even-dimensional manifold Lemma
The local index is additive under a transverse perturbation Lemma
Small-time flow fixed point indices and vector field zero indices Proposition
The index of a zero is its zero-section intersection number Proposition
The outward boundary hypothesis cannot be replaced by nonzero on the boundary Remark
Converse Poincare-Hopf for nowhere-zero fields Theorem
Poincare-Hopf for closed manifolds Theorem
Poincare-Hopf with outward-pointing boundary Theorem

Dependency tree · two levels

48 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)