Alphabeta Math
Pipeline-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.

Vector Field Index Euler Characteristic and Poincare Hopf

1 · Prerequisites

2 · Summary

The page develops the local index of an isolated zero of a smooth vector field, its chart and ball invariance, the determinant formula for nondegenerate zeros, additivity under transverse perturbation, and the zero-section intersection interpretation that bridges to the Euler class. It then proves the Poincare-Hopf index theorems for closed manifolds and for compact manifolds with an outward-pointing boundary field, the converse existence theorem for nowhere-zero fields by cancellation of opposite zeros, and the corollaries that fix the Morse gradient signs, the odd-dimensional vanishing of the Euler characteristic, and the evaluation of the Euler number of the tangent bundle.

Dimension restrictions are stated in each item; the index and Poincare-Hopf theorems cover every dimension n≥1. The zero-dimensional case of the sphere degree used by the index is supplied on this page so that the n=1 cases of the index, the Gauss-degree lemma and the Poincare-Hopf theorems are covered by the same conventions as the higher-dimensional ones. The Axiom of Choice supplies the excellent Morse functions used to establish homological finiteness and evaluate the index sum. The embedding, tube, Stokes, perturbation and geodesic ball-selection arguments use countable choice. Trivializations for degree computations use matching base and fibre orientations, and the reflected field is doubled using its flow collar.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Euler characteristic of a compact manifold

Definition

Let M be a compact smooth n-manifold, possibly with boundary (Smooth manifolds and their smooth charts in the boundaryless case, Smooth charts, atlases, and structures with boundary in the boundary case). Its Euler characteristic is χ(M):=∑i=0n(−1)idim⁡QHi(M;Q)∈Z, the alternating sum of the rational Betti numbers of the singular homology of M (The singular chain complex and singular homology, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis, The rationals as equivalence classes of pairs of integers). The empty manifold has χ(∅)=0, the empty sum. The sum is finite because dim⁡QHi(M;Q)<∞ for every i and Hi(M;Q)=0 for i>n; that finiteness is proved on this page in Finiteness and additivity of the Euler characteristic ↗, which is why it is recorded as the well-definedness pointer rather than assumed here. The definition itself uses no orientation and no choice of field or coefficient system.

When M has a finite CW model, χ(M) agrees with its cell-count Euler characteristic (Euler characteristic of a finite CW complex): the finite rational cellular complex has one generator per cell, and alternating rank-nullity cancels boundary dimensions. Cellular homology (Cellular homology computes singular homology) therefore identifies that cell count with the rational Betti alternating sum. The integral Euler-Poincare formula (Euler–Poincare formula for finite CW complexes) gives the same count. The relative version is proved in Finiteness and additivity of the Euler characteristic ↗. The cited well-definedness proposition assumes the Axiom of Choice (The Axiom of Choice) to obtain an excellent Morse function. The formula for χ makes no selection of auxiliary data.

PropositionStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Finiteness and additivity of the Euler characteristic

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a compact smooth n-manifold, possibly with boundary.

(i) Each dim⁡QHi(M;Q) is finite and Hi(M;Q)=0 for i>n, so χ(M) of Euler characteristic of a compact manifold is a well-defined integer.

(ii) If A⊆M is a compact smooth submanifold, possibly with boundary, and (M,A) is homotopy equivalent as a pair to a finite relative CW pair with ck relative k-cells, then χ(M)=χ(A)+∑k(−1)kck=χ(A)+∑k(−1)kdim⁡QHk(M,A;Q).

(iii) If M=M1∪NM2 with M1,M2 compact smooth submanifolds, possibly with boundary, N=M1∩M2 a common compact smooth submanifold, and the inclusions N↪Mi cofibrations, then χ(M)=χ(M1)+χ(M2)−χ(N).

Facts & Assumptions

Given: The Axiom of Choice and a compact smooth n-manifold M, possibly with boundary.

[F1]

The double DM is a closed smooth n-manifold. Its continuous folding map q([x,+])=q([x,−])=x is well defined on the quotient and is a retraction onto the first labelled copy, with q∘i=id⁡M (The double of a smooth manifold with boundary, The double has a well-defined smooth structure).

[F2]

For a closed smooth manifold with an excellent Morse function, the handle chain complex of The handle chain complex computes singular homology is a complex of finite-dimensional Q-vector spaces with exactly mk(f) generators in degree k whose homology is Hk(M;Q); hence those homology spaces are finite-dimensional and vanish above the dimension (Every compact smooth manifold admits an excellent Morse function, The singular chain complex and singular homology).

[F3]

Homotopy equivalences induce isomorphisms on singular homology with any coefficients, and the long exact sequence of a pair and the Mayer-Vietoris sequence are long exact sequences of Q-vector spaces (Homotopy equivalences induce isomorphisms on singular homology, Long exact sequence of a pair, Mayer–Vietoris sequence in singular homology).

[F4]

Alternating-dimension lemma: for a long exact sequence ⋯→Ak→Bk→Ck→Ak−1→⋯ of finite-dimensional vector spaces vanishing outside a finite range, one has ∑k(−1)k(dim⁡Ak−dim⁡Bk+dim⁡Ck)=0, and the intermediate spaces are finite-dimensional with the same vanishing range (Rank bookkeeping for a long exact sequence of finite-dimensional vector spaces, Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis, The rationals as equivalence classes of pairs of integers).

[F5]

For a CW pair, relative cellular chains have one generator per relative cell and compute relative singular homology (Relative cellular homology computes relative singular homology). For a finite chain complex, writing Zk=ker⁡dk and Bk=im⁡dk+1 gives dim⁡Ck=dim⁡Hk+dim⁡Bk+dim⁡Bk−1; alternating summation cancels the boundary dimensions. Thus the alternating relative cell count equals the alternating rational relative Betti sum, even when the base subcomplex itself has infinitely many cells.

[F6]

The smooth spaces Mi,N in (iii) are CGWH under the assumed AC (Smooth manifolds have CW homotopy type), so the cofibration interface applies. For a closed cofibration A↪X, the homotopy extension property supplies a retraction X×I→X×{0}∪A×I (Cofibrations are characterized by a retraction of the mapping cylinder strip).

Proof

1.1F1F2givenalgebra

For (i), the empty case is immediate. Otherwise the folding retraction [F1] gives q∗i∗=id⁡ on rational homology, so Hi(M;Q) embeds as a direct summand of Hi(DM;Q). Under the assumed Axiom of Choice, choose an excellent Morse function on the closed double and apply [F2]. Its handle complex is finite-dimensional and concentrated in degrees 0,…,n; its homology, and hence the summand for M, is finite-dimensional and zero above n. This establishes well-definedness without presupposing χ(M).

1.2F6construct

For (iii), replace the glued space by the double mapping cylinder P=M1∪N×{0}(N×I)∪N×{1}M2. We verify that collapsing the cylinder is a homotopy equivalence P→M. Let C=M1∪N(N×I), with collapse f:C→M1. By [F6] extend the track H(n,t)=[n,t] and the initial inclusion of M1 to H:M1×I→C. Set j=H1, so j(n)=[n,1]. Then fH joins id⁡M1 to fj relative to N. On C, use H(x,t) for x∈M1 and [n,s+t(1−s)] for [n,s] in the cylinder; the formulas agree at s=0, define a homotopy from id⁡C to jf, and fix the free end N×{1}. Gluing these maps and homotopies to the identity of M2 proves the asserted equivalence. The compact subspaces Mi are closed in M, so their pushout topology is the topology of M1∪M2=M by finite closed pasting.

2.1F3F4F5step 1.1algebra

For (ii), step 1.1 applies to both M and A. In the pair long exact sequence [F3], Hk(M,A;Q) lies between a quotient of Hk(M;Q) and a subspace of Hk−1(A;Q); it is therefore finite-dimensional and vanishes outside a finite range. Alternating summation of this exact sequence gives χ(M)−χ(A)=∑k(−1)kdim⁡Hk(M,A;Q) by [F4], applied after cyclically relabelling the three terms if necessary. The given equivalence of pairs induces isomorphisms on these relative groups: the absolute homology maps are isomorphisms by [F3], and exactness of the pair sequences gives injectivity and surjectivity of the relative maps by lifting and subtracting successive neighbouring classes. Now [F5] computes the relative alternating sum as ∑k(−1)kck, proving both equalities without requiring a finite absolute CW structure on A.

3.1F3F4step 1.1step 1.2algebra∎

The open subsets U=M1∪(N×[0,2/3)) and V=M2∪(N×(1/3,1]) cover P. They deformation retract to M1,M2, while U∩V=N×(1/3,2/3) retracts to N. Thus [F3] and step 1.2 give a Mayer–Vietoris sequence with homology terms those of N, M1⊔M2 and M. All terms are finite-dimensional and vanish outside a finite range by step 1.1. Applying [F4] at t=−1 gives χ(N)−χ(M1)−χ(M2)+χ(M)=0, as required.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The reduced degree of a map into the 0-sphere

Definition

Write S0={−1,+1}=∂[−1,1] for the 0-sphere (Euclidean spheres and closed balls as subspaces of Rn) and orient it by the boundary orientation of the interval [−1,1] (Induced boundary orientation): the point +1 carries the sign +1 and the point −1 the sign −1, since an orientation of a 0-manifold is a sign at each point (Oriented smooth manifolds and oriented charts). Let (P,s) be a finite oriented 0-manifold, that is, a finite set P together with a sign s(x)∈{+1,−1} for every x∈P (Oriented smooth manifolds and oriented charts); it is balanced when ∑x∈Ps(x)=0. For a balanced (P,s) and a map f:P→S0 — every map is continuous because P is discrete and S0 is discrete — define the reduced degree deg⁡(f):=12∑x∈Ps(x) f(x)∈Z, the signed count of the values divided by two, where the values f(x)∈{−1,+1} are read in R. This is well defined: since f takes only the two values ±1, balance gives ∑f(x)=+1s(x)=−∑f(x)=−1s(x), hence ∑xs(x)f(x)=2∑f(x)=+1s(x) is even, and replacing s by −s changes the sign of deg⁡(f) exactly as changing an orientation changes the ordinary degree (Degree of a map between oriented closed manifolds).

The case P=S0 with the standard orientation above is balanced, and then deg⁡(f)=f(+1)−f(−1)2∈{−1,0,+1}. This is the induced map on the reduced group H~0(S0;Z)≅Z generated by the difference of the two point classes (Reduced homology theory and augmentation): the identity has reduced degree +1, the antipodal map −1, and the two constant maps 0. It is the scalar replacement for the fundamental-class degree of Degree of a map between oriented closed manifolds, whose scalar definition is stated only for nonempty connected closed oriented manifolds and therefore does not apply to a source as disconnected as S0.

The motivating balanced sources are boundaries: if W is a compact oriented smooth 1-manifold and ∂W carries the induced boundary orientation, then ∑p∈∂Wε(p)=0 (Oriented boundary counts of a compact oriented 1-manifold cancel), so ∂W is balanced and every map ∂W→S0 has a reduced degree. No choice principle is used in the definition; the cited boundary-count lemma assumes ACω.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Reduced degree into the 0-sphere is homotopy invariant and multiplicative

Statement

Let (P,s) be a balanced oriented finite 0-manifold, so that every map P→S0 has a reduced degree, and let f:P→S0 be a map (The reduced degree of a map into the 0-sphere).

(i) If F:P×[0,1]→S0 is continuous and Ft(x):=F(x,t), then Ft=F0 for every t∈[0,1]; in particular deg⁡Ft=deg⁡F0.

(ii) If g:S0→S0 is any map and deg⁡g denotes its reduced degree as a self-map of S0, then deg⁡(g∘f)=deg⁡(g)deg⁡(f).

Facts & Assumptions

Given: A balanced oriented finite 0-manifold (P,s) with ∑x∈Ps(x)=0 and a map f:P→S0.

[F1]

S0={−1,+1} is oriented by the boundary orientation of [−1,1]: the point +1 has sign +1 and −1 has sign −1. A reduced degree is defined only for a balanced source: ∑x∈Ps(x)=0, and then deg⁡(h)=12∑x∈Ps(x)h(x)∈Z for a map h:P→S0. For P=S0 itself this reads deg⁡(h)=h(+1)−h(−1)2∈{−1,0,+1} (The reduced degree of a map into the 0-sphere).

[F3]

In S0={−1,+1}⊆R the two singletons are open for the subspace topology: {+1}=S0∩B2(+1,1) and {−1}=S0∩B2(−1,1) (Euclidean spheres and closed balls as subspaces of Rn, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). A continuous map from a connected space into S0 is therefore constant: the preimages of the two disjoint open singletons would otherwise separate the domain.

Proof

Proof technique: direct; split the composition law by the three possible reduced degrees of the self-map g.

1.1F2F3algebra

The product P×[0,1] is the disjoint union of the connected subsets {x}×[0,1], one for each x∈P; a continuous map from a connected space into S0 is constant, so F is constant on each {x}×[0,1], hence Ft(x)=F(x,t)=F(x,0)=F0(x) for all x∈P and t∈[0,1]; equal maps have equal reduced degree, which proves (i).

1.2F1algebra

Suppose g is not constant. A map S0→S0 is determined by the pair (g(+1),g(−1)), so a nonconstant g sends +1 and −1 to different values; hence g is either the identity, with σ:=+1 and g(y)=y for both y, or the antipodal map g(y)=−y, with σ:=−1, so that g(y)=σy for all y∈S0; then g∘f=σf and, by linearity of the defining signed count, deg⁡(g∘f)=12∑xs(x)σf(x)=σdeg⁡(f), while σ=deg⁡(g) by [F1], the identity and the antipodal map having reduced degrees +1 and −1; thus deg⁡(g∘f)=deg⁡(g)deg⁡(f).

2.1F1step 1.2algebra∎

If g is constant with value c∈S0, then g∘f is the constant map c and deg⁡(g∘f)=12c∑x∈Ps(x)=0 by balance, while deg⁡(g)=12(c−c)=0 by [F1]; hence again deg⁡(g∘f)=0=deg⁡(g)deg⁡(f), which completes (ii).

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Isolated zero and local index of a vector field

Definition

Assume ACω (The Axiom of Countable Choice (ACω)) for the canonical smooth tangent-bundle structure.

Let M be a smooth n-manifold without boundary, n≥1, and let X be a smooth vector field on M (A smooth vector field is a smooth section of the tangent bundle, Smooth manifolds and their smooth charts). A point p∈M is an isolated zero of X when X(p)=0 and some chart around p contains no other zero of X; equivalently, the set of zeros of X has p as an isolated point in a chart around p (Manifold charts, coordinate domains, and coordinate functions).

For an isolated zero choose a smooth chart (φ,U) of the smooth structure of M with φ(p)=0 and write Xφ(u):=(dφφ−1(u))X(φ−1(u))∈Rn for the chart representative of X (The induced tangent bundle chart). Since p is an isolated zero there is ε>0 with Xφ≠0 on B‾ε(0)∖{0}⊆φ(U), so the normalized field v↦Xφ(εv)/∣Xφ(εv)∣ is a continuous map Sn−1→Sn−1. The local index of X at p is ind⁡pX:=deg⁡(Sn−1→Sn−1, v↦Xφ(εv)∣Xφ(εv)∣), the degree of Degree of a map between oriented closed manifolds computed with the standard orientations of the two copies of Sn−1, for n≥2.

For n=1 the same formula is read in dimension zero: the sphere S0={±1} parametrizes ∂[−ε,ε] by v↦εv, and the displayed map is a map S0→S0 whose reduced degree in the sense of The reduced degree of a map into the 0-sphere is declared to be the index, ind⁡pX=f(+1)−f(−1)2∈{−1,0,+1},f(v)=Xφ(εv)∣Xφ(εv)∣. This case is well posed for every 0<ε′<ε: the two signs of Xφ on (0,ε) and on (−ε,0) are each constant, because Xφ is continuous and nowhere zero on either half-interval, so the two values f(±1) do not depend on ε′; for n≥2 the value is independent of ε by the homotopy v↦Xφ(((1−t)ε+tε′)v)/∣⋅∣ on the zero-free annulus (Degree is invariant under proper smooth homotopy). The value is also independent of the smooth chart and admissible radius, and of trivializations whose fibre orientation matches the chosen base orientation (The local index is independent of chart, ball and trivialization ↗); in particular no orientation of M is required, because a chart change multiplies both the source and the target orientation by the same sign, and the n=1 case is the same statement read on the 0-sphere. Every statement on this page assumes n≥1; the index is a signed integer, ind⁡pX∈Z.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The local index is independent of chart, ball and trivialization

Statement

Assume ACω (The Axiom of Countable Choice (ACω)) for the canonical smooth tangent-bundle structure.

Let M be a smooth n-manifold without boundary, n≥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 fφ(v)=Xφ(εv)∣Xφ(εv)∣,fψ(v)=Xψ(δv)∣Xψ(δv)∣ 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∘φ−1 defined near 0 and u=φ(x), Xψ(θ(u))=dθu(Xφ(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:=dθ0 is an isomorphism with det⁡A≠0 (The differential of a diffeomorphism is an isomorphism), and θ(u)=Au+O(∣u∣2), θ−1(w)=A−1w+O(∣w∣2), dθu=A+O(∣u∣) for u→0.

[F2]

Degree facts for the normalized sphere maps of n≥2: homotopic smooth maps Sn−1→Sn−1 have the same degree (Degree is invariant under proper smooth homotopy), and deg⁡(G∘F)=deg⁡(G)deg⁡(F) for such maps (Degree is multiplicative under composition, Degree of a map between oriented closed manifolds). A diffeomorphism of Sn−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 S0 gives deg⁡(f)=f(+1)−f(−1)2∈{−1,0,+1}, and deg⁡(g∘f)=deg⁡(g)deg⁡(f) for maps of S0 (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 LA(v):=Av/∣Av∣ is a diffeomorphism of Sn−1 with inverse LA−1, and its local orientation sign at every v is sign⁡det⁡A: choosing a positively oriented basis (v,e2,…,en) of Rn, the outward normal v of the ball is the first basis vector, so the induced map on the tangent space TvSn−1 has the orientation sign of A there, namely sign⁡det⁡A. Hence deg⁡LA=sign⁡det⁡A for n≥2 by [F2], and the same formula holds for the reduced degree at n=1, since LA=sign⁡(A) id on S0 by [F3].

Proof

1.1F2F3given

Within either chart, varying the radius through positive admissible radii gives the homotopy v↦Xφ(r(t)v)/∣Xφ(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)=Xφ(u)/∣Xφ(u)∣, f(v)=σ(ρv) and u(v)=θ−1(ρv). The normalized ψ-field is g(v)=Ldθu(v)(σ(u(v))) by [F1]. Since dθu(v) tends uniformly to the invertible matrix A=dθ0, interpolation of this matrix to A stays invertible for small ρ, giving g≃LA∘σ∘u. The radial homotopy ut(v)=((1−t)∣u(v)∣+tρ)u(v)/∣u(v)∣ stays in the zero-free punctured ball; hence LAσ(ut(v)) joins this last map to LA∘f∘u~, where u~=u/∣u∣. Finally u~(v)=LA−1(v)+O(ρ) uniformly, so normalized convex interpolation gives u~≃LA−1. Thus g≃LA∘f∘LA−1.

2.1F2F3F4step 1.1step 1.2algebra

Multiplicativity and [F4] now give deg⁡g=(sign⁡det⁡A)2deg⁡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 fφ and fψ themselves are homotopic: for X(t)=t2 and ψ=−φ, they are the distinct constant maps of S0, both of degree zero.

3.1F2F3F4step 2.1algebra∎

On a coordinate ball an orientation-preserving trivialization changes components to B(u)Xφ(u) with B(u)∈GLn+(R). Contracting B(u) to B(0) through B((1−t)u) gives a nonzero homotopy on the sphere. The resulting degree is deg⁡LB(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.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Nondegenerate zero of a vector field

Definition

Assume ACω (The Axiom of Countable Choice (ACω)) for the canonical smooth tangent-bundle structure.

Let M be a smooth n-manifold, X a smooth vector field on M and p∈M a zero of X (A smooth vector field is a smooth section of the tangent bundle). View X as a smooth section X:M→TM (Smoothness of a section is equivalent to smooth local components). The zero section 0M is a smooth embedding (The zero section is a smooth embedding), so its differential d(0M)p:TpM→T(p,0)TM is injective (The differential of a smooth map), and the vertical quotient at p is Np:=T(p,0)TM/d(0M)p(TpM). Since X is a section, both dXp and d(0M)p are right inverses of the projection dπ(p,0); hence the class of dXp in Hom⁡(TpM,Np) is the vertical derivative DXp∈End⁡(TpM), read through the canonical identification Np≅TpM: explicitly, the difference dXp−d(0M)p takes values in the vertical space ker⁡dπ(p,0), and DXp is that difference followed by the canonical isomorphism from the vertical space V(p,0)=ker⁡dπ(p,0) to TpM. In an induced tangent-bundle chart φ~ over a chart φ with φ(p)=0 the section reads u↦(u,Xφ(u)) and DXp corresponds to the ordinary derivative DXφ(0) of the chart representative Xφ (The induced tangent bundle chart, Local and global frames of a vector bundle); a change of chart conjugates DXp by an isomorphism, so invertibility and the determinant det⁡DXp are independent of the chart (The tangent bundle as a disjoint union).

The zero p is nondegenerate when DXp is invertible. A nondegenerate zero is isolated: in a chart, Xφ(0)=0 and DXφ(0) is invertible, so Xφ is a diffeomorphism near 0 by the inverse function theorem (The smooth inverse function theorem on manifolds) and its only zero near 0 is 0 itself. In particular a nondegenerate zero is an isolated zero and X has no other zero in some neighbourhood of p. The vertical derivative itself requires no further choice once the smooth tangent-bundle structure is supplied.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The index of a nondegenerate vector-field zero

Statement

Assume ACω (The Axiom of Countable Choice (ACω)) for the canonical smooth tangent-bundle structure.

Let M be a smooth n-manifold, n≥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⁡pX=sign⁡det⁡(DXp:TpM→TpM)∈{+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: Xφ(0)=0, DXφ(0) corresponds to DXp under the chart trivialization, and DXp invertible is equivalent to DXφ(0) invertible; moreover Xφ(u)=Au+R(u) with A:=DXφ(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⁡pX=deg⁡(v↦Xφ(εv)/∣Xφ(εv)∣) with the standard orientations (n≥2), and for n=1 by the reduced degree of the induced map S0→S0 (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 LA(v):=Av/∣Av∣ of an invertible A is a diffeomorphism of Sn−1 whose local orientation sign is sign⁡det⁡A, hence deg⁡LA=sign⁡det⁡A; for n=1 this is the reduced degree of v↦sign⁡(A)v. Degree is invariant under homotopies of maps of Sn−1 (n≥2) and, for n=1, under homotopies of maps of S0 (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:=DXφ(0), so Xφ(u)=Au+R(u) with R(u)=O(∣u∣2) and A invertible; write c:=∣A−1∣−1>0 and ∣R(u)∣≤C∣u∣2 with C>0. For ε<c/(2C) and v∈Sn−1 the vector Av+tε−1R(εv) with t∈[0,1] has norm at least c−ε−1Cε2≥c/2>0, so Ht(v):=(Av+tε−1R(εv))/∣⋯∣ is a homotopy from the normalized linear map LA to the normalized chart field v↦Xφ(εv)/∣Xφ(εv)∣.

2.1F1F2F3step 1.1algebra∎

For n≥2 homotopy invariance gives ind⁡pX=deg⁡LA=sign⁡det⁡A, and for n=1 the same homotopy is one of maps S0→S0, so by [F3] the reduced degrees agree and again ind⁡pX=deg⁡LA=sign⁡det⁡A; since A corresponds to DXp by [F1], det⁡A and DXp have the same sign and ind⁡pX=sign⁡det⁡(DXp)∈{+1,−1}, because DXp is invertible.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The local index is additive under a transverse perturbation

Statement

Assume ACω (The Axiom of Countable Choice (ACω)). Let M be a smooth n-manifold, n≥1, let X be a smooth vector field with an isolated zero at p, and let B⊆M be an embedded closed ball with p∈int⁡B, X≠0 on ∂B, and no zero of X in int⁡B other than p (such a ball exists in a chart around p).

(i) Choose an orientation of B and a smooth trivialization of TM∣B preserving that orientation, and orient ∂B as the boundary of B; then ind⁡pX equals the degree of x↦X(x)/∣X(x)∣ from ∂B to Sn−1 (the reduced degree for n=1).

(ii) If X′ is a smooth vector field on M with X′=X outside int⁡B and only nondegenerate zeros q1,…,qk in int⁡B, then ∑i=1kind⁡qiX′=ind⁡pX.

(iii) Consequently every isolated zero can be perturbed, supported in an arbitrarily small ball around it, to finitely many nondegenerate zeros with the same index sum.

Facts & Assumptions

Given: A smooth vector field X on the smooth n-manifold M with an isolated zero p, and a sufficiently small embedded closed ball B⊆M with p∈int⁡B and no other zero of X in int⁡B.

[F1]

The index of the isolated zero p is the degree of the normalized field on the boundary of a small ball in a chart, with the reduced degree for n=1; it is independent of chart and radius, and of trivializations with matching base and fibre orientations (Isolated zero and local index of a vector field, The local index is independent of chart, ball and trivialization).

[F2]

Hopf's boundary lemma: for a compact smooth m-manifold with boundary N⊂Rm and a smooth field Y on N with only isolated zeros and Y≠0 on ∂N, the sum of the indices over the zeros of Y equals the degree of x↦Y(x)/∣Y(x)∣ from ∂N to Sm−1 with the boundary orientation (the reduced degree for m=1) (The index sum of an outward field is the Gauss degree, Induced boundary orientation).

[F3]

Nondegenerate zeros have index sign⁡det⁡(DY)∈{+1,−1} (The index of a nondegenerate vector-field zero, Nondegenerate zero of a vector field).

[F4]

The set of regular values of a smooth map is dense and has null complement; in particular there are regular values of the chart representative Xφ arbitrarily close to (but different from) 0 (Morse-Sard for smooth manifolds, Regular values have null complement and are dense).

[F5]

There is a smooth bump on the chart that equals 1 on a smaller ball and is supported in a slightly larger one (Explicit compactly supported smooth cutoffs), and a chart around p trivializes TM over B (The induced tangent bundle chart, Embedded smooth submanifolds with boundary).

Proof

1.1F1F2F5algebra

Choose a smooth parametrization b:Dn→B and pull back the field as Y(u)=(dbu)−1X(b(u)) on D:=Dn: this is a smooth field on the compact manifold D with boundary, with the single zero b−1(p) and Y≠0 on ∂D; [F2] applied to N=D gives ind⁡b−1(p)Y=deg⁡(∂D→Sn−1,y↦Y(y)/∣Y(y)∣), and this degree is the degree of x↦X(x)/∣X(x)∣ on ∂B in the induced trivialization; any other orientation-compatible trivialization has the same degree by the matrix-contraction argument of [F1]; hence (i).

2.1F1F2F3step 1.1algebra

For (ii), the same computation applies to the transported field Y′ on D: its zeros in int⁡D are b−1(q1),…,b−1(qk) and it agrees with Y on ∂D, so [F2] gives ∑iind⁡b−1(qi)Y′=deg⁡(∂D, Y/∣Y∣)=ind⁡pX by step 1.1; the indices are transported back by [F1], and each equals ±1 by [F3].

3.1F3F4F5step 1.1step 2.1constructalgebra∎

For (iii), let B be an arbitrarily small embedded closed ball around p with no other zero of X in its interior and let φ be a chart on a neighbourhood of B with φ(p)=0; choose a smooth radial bump λ equal to 1 on a ball B1⊂B around p and supported in a slightly larger ball B2 with B1⊆B2, B2‾⊆int⁡B, and B2 containing no zero of X except p, and by [F4] choose a regular value y of Xφ with ∣y∣ smaller than the (positive) minimum of ∣Xφ∣ on the compact collar supp⁡λ∖B1. Then X′:=X−λy (read in the chart) equals X outside supp⁡λ⊆int⁡B, is nowhere zero on the collar supp⁡λ∖B1, and on B1 has the zeros (Xφ)−1(y), a finite set of nondegenerate points because y is a regular value; all these zeros lie in int⁡B2⊆int⁡B, so (ii) applies and gives ∑iind⁡qiX′=ind⁡pX, with each index ±1 by [F3].

PropositionStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The index of a zero is its zero-section intersection number

Statement

Assume countable choice ACω (The Axiom of Countable Choice (ACω)). Let M be a closed oriented smooth n-manifold, n≥1, and give TM the orientation which along the zero section is the direct sum of the horizontal (tangent) orientation of the base and the vertical (fibre) orientation, first factor first (Product orientations, The tangent bundle as a disjoint union), orient both submanifolds by their projections to M, and let X0⊂TM be the zero section (The zero section is a smooth embedding) and let ΓX be the graph of a smooth vector field X transverse to X0, so that the zeros of X are nondegenerate. Then, with the oriented intersection number of The oriented intersection number, I(X0,ΓX)=∑p:X(p)=0ind⁡pX, and at each zero the local oriented intersection sign of The local oriented intersection sign satisfies ε(X0,ΓX)(p)=ind⁡pX.

Facts & Assumptions

Given: A closed oriented smooth n-manifold M, the tangent bundle with its horizontal-then-vertical orientation along the zero section, and a smooth field X whose graph is transverse to the zero section.

[F1]

X0 and ΓX are closed embedded n-submanifolds of the boundaryless 2n-manifold TM with complementary dimensions, and for a transverse pair with one factor compact the intersection is finite (The zero section is a smooth embedding, Transverse embedded submanifolds, Compact transverse complementary intersections are finite).

[F2]

For a compact oriented x-manifold without boundary A, an oriented n-manifold M and a closed oriented embedded z-submanifold Z with x+z=n and transverse inclusion, the oriented intersection number is the finite sum I(A,Z)=∑p∈A∩Zε(p) of the local signs of The local oriented intersection sign, with the product orientation on the ordered sum of tangent spaces, first factor first (The oriented intersection number).

[F3]

Local model of the intersection: over a chart U of M in which TM is trivialized as U×Rn, the zero section is U×{0} and the graph is {(u,Xφ(u))}; at a point p∈X−1(0) the tangent spaces are TpM⊕{0} and {(v,DXpv)}, and the determinant comparing the ordered product TpX0⊕T(p,0)ΓX with the ambient horizontal-then-vertical orientation is det⁡(DXp): the relevant matrix is block-triangular with identity and DXp blocks, exactly as in the push-off computation of Normal push-off zeros are the self-intersection points (The induced tangent bundle chart, Nondegenerate zero of a vector field).

[F4]

The zeros of X are exactly the intersection points of ΓX with X0, and ΓX is transverse to X0 exactly when DXp is invertible at every zero, equivalently when every zero is nondegenerate; a nondegenerate zero has index sign⁡det⁡(DXp)=±1 (Nondegenerate zero of a vector field, The index of a nondegenerate vector-field zero).

Proof

1.1F1F3F4algebra

Write ΓX={(x,X(x)):x∈M}⊆TM. Since ΓX∩X0={(p,0):X(p)=0}, the intersection points are the zeros of X; at such a point the transversality of ΓX to X0 is equivalent to the surjectivity of DXp (the tangent spaces are the horizontal space and the graph of DXp), so transversality means invertibility of DXp at every zero, i.e. nondegeneracy; by [F1] the intersection is finite, and by [F4] each intersection point carries index sign⁡det⁡(DXp).

2.1F2F3F4step 1.1algebra∎

At a zero p, the local sign ε(X0,ΓX)(p) is computed in the chart of [F3] as the determinant sign of the block-triangular matrix with diagonal blocks In and DXp, hence equals sign⁡det⁡(DXp)=ind⁡pX; this proves the pointwise identity and, summing over the finitely many intersection points with [F2], also I(X0,ΓX)=∑pind⁡pX. The countable choice hypothesis is inherited from the transverse-representative selection in the definition of the oriented intersection number for non-transverse maps (The oriented intersection number, The oriented intersection number is homotopy invariant), although for the transverse pair considered here the sum is choice-free.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Negation scales the local index by (−1)n

Statement

Assume ACω (The Axiom of Countable Choice (ACω)) for the canonical smooth tangent-bundle structure.

Let M be a smooth n-manifold, n≥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)nind⁡pX.

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 Xφ and admissible ε>0, ind⁡pX=deg⁡f where f(v):=Xφ(εv)/∣Xφ(εv)∣, the degree being the ordinary one for n≥2 and the reduced degree for n=1 (Isolated zero and local index of a vector field).

[F2]

The antipodal map v↦−v of Sn−1 has degree (−1)n for n≥2; the reduced degree of the antipodal map of S0 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 Sn−1 for n≥2, and reduced degree is multiplicative under composition for maps into S0 (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=−Xφ, so its normalized map is v↦−Xφ(εv)/∣Xφ(εv)∣=(α∘f)(v), where α(v)=−v is the antipodal map of Sn−1 and f is the normalized map of X.

2.1F1F2F3step 1.1algebra∎

For n≥2 multiplicativity of the degree under composition gives ind⁡p(−X)=deg⁡α⋅deg⁡f=(−1)nind⁡pX by [F2], and for n=1 the same computation with reduced degrees gives ind⁡p(−X)=(−1)1ind⁡pX, since (−1)n=(−1)1 for n=1.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The index sum of an outward field is the Gauss degree

Statement

Assume ACω (The Axiom of Countable Choice (ACω)). Let N⊂Rm be a compact smooth m-dimensional submanifold with boundary, m≥1 (Embedded smooth submanifolds with boundary), and let Y be a smooth vector field on N with only isolated zeros and Y≠0 on ∂N. Then, summing the componentwise degrees over the components of ∂N with the boundary orientation (Induced boundary orientation), ∑x∈N, Y(x)=0ind⁡xY=deg⁡(∂N→Sm−1, x↦Y(x)∣Y(x)∣). For m=1 the right-hand side is read as the reduced degree of the map ∂N→S0: the oriented boundary of a compact oriented 1-manifold is balanced (Oriented boundary counts of a compact oriented 1-manifold cancel), so the reduced degree of The reduced degree of a map into the 0-sphere applies. If in addition Y points strictly outward along ∂N (Inward, outward, and boundary-tangent vectors), the right-hand side equals deg⁡(g) for the Gauss map g:∂N→Sm−1 sending x to the outward unit normal. In particular the index sum is independent of Y.

Facts & Assumptions

Given: A compact smooth m-manifold with boundary N⊂Rm, oriented by the ambient orientation of Rm, and a smooth field Y on N with Y≠0 on ∂N and only isolated zeros.

[F1]

The zeros of Y are finitely many: the zero set is closed, and an infinite closed discrete subset of the compact space N would have an accumulation point x∈N at which continuity gives Y(x)=0 while every neighbourhood of x contains other zeros, contradicting isolatedness. (Isolated zero and local index of a vector field)

[F2]

The index of an isolated zero is the degree of x↦Y(x)/∣Y(x)∣ on a small sphere around the zero, with the standard orientations, and for m=1 the reduced degree of that map S0→S0 (Isolated zero and local index of a vector field, The reduced degree of a map into the 0-sphere).

[F3]

For m≥2, a regular value exists by Morse-Sard for smooth manifolds and Regular values have null complement and are dense. For a proper smooth map F:C→Sm−1 from a nonempty connected closed oriented (m−1)-manifold C and a top form ω on Sm−1, ∫CF∗ω=deg⁡(F)∫Sm−1ω, where the degree is the closed-manifold degree of Degree of a map between oriented closed manifolds, equal to the compact-support cohomological degree of Regular-value formula for degree, and where a normalized volume form with integral one exists (Positive volume form on an oriented manifold, Integral of a compactly supported top form, Degree is well defined and independent of the normalized top form).

[F4]

Under ACω, manifold Stokes holds for a compact oriented manifold with boundary and a smooth (m−1)-form α: ∫∂N′α=∫N′dα, the boundary carrying the induced boundary orientation (The general Stokes theorem, Induced boundary orientation).

[F5]

For m=1: ∂N consists of finitely many points with signs ε(x), and ∑x∈∂Nε(x)=0; the reduced degree of a map h:∂N→S0 is 12∑x∈∂Nε(x)h(x) (Oriented boundary counts of a compact oriented 1-manifold cancel, The reduced degree of a map into the 0-sphere).

[F6]

If Y is strictly outward on ∂N, with outward unit normal g (Inward, outward, and boundary-tangent vectors), then ⟨Y/∣Y∣,g⟩>0 pointwise, so t↦(tY/∣Y∣+(1−t)g)/∣⋯∣ is a homotopy from Y/∣Y∣ to g; homotopic maps have equal degree, and for m=1 a homotopy S0×[0,1]→S0 is constant in the time variable, so the reduced degrees agree (Degree is invariant under proper smooth homotopy, Reduced degree into the 0-sphere is homotopy invariant and multiplicative).

Proof

1.1F1F2algebra

By [F1] the zeros p1,…,pk of Y are finite; choose pairwise disjoint closed coordinate balls D1,…,Dk⊆N around them, so small that Y≠0 on Di‾∖{pi} and ∂Di∩∂N=∅, and let N′:=N∖⋃iint⁡Di, a compact oriented m-manifold with boundary on which the normalized field f:=Y/∣Y∣:N′→Sm−1 is smooth. Its boundary is ∂N′=∂N⊔⨆i∂Di, where each ∂Di carries, as a piece of ∂N′, the orientation opposite to the boundary orientation of the removed ball Di, since the outward normals of N′ and of Di are opposite along ∂Di.

2.1F2F3F4step 1.1algebra

For m≥2 choose a volume form ω on Sm−1 with ∫Sm−1ω=1 and apply [F4] to α=f∗ω: since dω=0, ∫∂N′f∗ω=∫N′f∗dω=0. Evaluating the boundary integral componentwise with [F3] gives 0=deg⁡(∂N→Sm−1)−∑iind⁡piY, because each small sphere ∂Di is mapped by f with degree ind⁡piY in its own boundary orientation by [F2] and therefore contributes −ind⁡piY to ∂N′.

3.1F2F4F5step 1.1algebra

For m=1 use instead the 0-form ω on S0 with ω(±1)=±1, so that ∫S0ω=2 and ∫∂N′f∗ω=∑x∈∂N′ε(x)f(x); Stokes gives ∑x∈∂N′ε(x)f(x)=0, and each removed pair contributes f(pi−δi)−f(pi+δi)=−2ind⁡piY with the orientation of step 1.1 by [F2], so ∑iind⁡piY=12∑x∈∂Nε(x)f(x)=deg⁡(∂N→S0) by [F5], the claimed formula; this and step 2.1 prove the first assertion in both dimensions, and with it the index sum depends only on the boundary values of the normalized field.

4.1F3F6step 3.1algebra∎

If Y is strictly outward, [F6] gives a homotopy from x↦Y(x)/∣Y(x)∣ to the Gauss map g, so their degrees agree and the right-hand side equals deg⁡(g); since deg⁡(g) does not involve Y, the index sum is independent of the choice of the outward field.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Poincare-Hopf for closed manifolds

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a closed smooth n-manifold, n≥1, and let X be a smooth vector field on M with only isolated zeros (Isolated zero and local index of a vector field). Then ∑p:X(p)=0ind⁡pX=χ(M), with χ as in Euler characteristic of a compact manifold. In particular the sum is independent of X and vanishes over the empty zero set; for disconnected M the statement is applied componentwise.

Facts & Assumptions

Given: A closed smooth n-manifold M, n≥1, and a smooth field X on M with only isolated zeros.

[A1]

The Axiom of Choice (The Axiom of Choice) is used only through the existence of an excellent Morse function; the embedding, tube and perturbation arguments use ACω (The Axiom of Countable Choice (ACω)).

[F1]

The zeros of X are finite, and by The local index is additive under a transverse perturbation(iii) each zero can be perturbed, supported in an arbitrarily small ball around it, to finitely many nondegenerate zeros with the same index sum; a nondegenerate zero has index ±1 (The index of a nondegenerate vector-field zero, Nondegenerate zero of a vector field).

[F2]

Hopf's boundary lemma: for a compact smooth k-manifold with boundary N⊂Rk and a smooth field Y on N with only isolated zeros and Y strictly outward on ∂N, ∑Y(x)=0ind⁡xY=deg⁡(g:∂N→Sk−1), the Gauss map of the boundary; the right-hand side is the degree of the normalized field, independent of Y (The index sum of an outward field is the Gauss degree).

[F3]

Every smooth n-manifold admits a proper smooth embedding into R2n+1 (The weak Whitney proper embedding theorem), a closed embedded submanifold of Rk has a tubular neighbourhood given by normal addition with a positive radius function, and a smooth nearest-point retraction onto it (The Euclidean tubular neighbourhood theorem, A closed Euclidean submanifold has a smooth neighborhood retraction).

[F4]

For a Riemannian metric g and an excellent Morse function f on M (Every compact smooth manifold admits an excellent Morse function, Every smooth manifold admits a riemannian metric, Morse functions and excellent Morse functions), the field grad⁡gf vanishes exactly at the critical points (The Riemannian gradient is the metric dual of the differential, The Riemannian gradient vanishes exactly at the critical points), all of which are nondegenerate, and a critical point of index λ contributes ind⁡p(grad⁡gf)=(−1)λ (A Morse gradient zero contributes (−1)λ to the index, Nondegenerate critical points, nullity, index, and coindex).

[F5]

For a closed manifold, the alternating sum of (−1)λ over the critical points of a Morse function equals χ(M), and χ is additive over disjoint unions (Morse Euler characteristic identity, The singular homology of a disjoint union is the direct sum).

Proof

1.1F1algebra

Reduction to nondegenerate zeros: by [F1] the zeros of X are finite, and in pairwise disjoint small balls around them X may be replaced by fields whose zeros in those balls are nondegenerate with the same index sum; the replacements paste smoothly with the unchanged field outside and produce a smooth field X0 on M with only nondegenerate zeros and ∑pind⁡pX0=∑pind⁡pX. Since the right-hand side χ(M) does not involve X, it suffices to prove the identity for fields with nondegenerate zeros.

2.1F2F3step 1.1algebra

An invariant: embed M properly in Rk with k=2n+1 by [F3] and let N be a closed tubular neighbourhood given by normal addition (radius ε>0 uniform by compactness of M) with nearest-point retraction r:N→M. Define w(z):=z−r(z)+X0(r(z)) for z∈N; the two summands are orthogonal, since z−r(z)⊥Tr(z)M and X0(r(z))∈Tr(z)M. Hence w(z)=0 iff z=r(z)∈M and X0(z)=0: the zeros of w are exactly the zeros of X0, viewed in M⊆N. On ∂N we have ∣z−r(z)∣=ε and the outward normal is (z−r(z))/ε, so ⟨w(z),z−r(z)⟩=ε2>0: the field w points strictly outward, and in particular w≠0 on ∂N. At a zero p∈M the derivative of w as a map on N⊆Rk is Dwp=DX0p⊕I in the splitting TpN=TpM⊕TpM⊥ (the map z↦z−r(z) has derivative the orthogonal projection onto the normal space), so det⁡Dwp=det⁡(DX0)p and ind⁡pw=ind⁡pX0 by the determinant sign formula. Hopf's boundary lemma [F2] applied to N therefore gives ∑p:X0(p)=0ind⁡pX0=deg⁡(g:∂N→Sk−1), a number depending only on M (through its embedding and tube), not on X0.

3.1F4F5step 1.1step 2.1algebra

Evaluation: choose an excellent Morse function f:M→R by [A1] and [F4] and a Riemannian metric g, and take X0:=grad⁡gf, a field with only nondegenerate zeros. By [F4] each critical point p contributes (−1)ind⁡(p), so the invariant of step 2.1 equals ∑p(−1)ind⁡(p)=χ(M) by [F5]; combining with steps 1.1 and 2.1 gives ∑pind⁡pX=χ(M) for the original field X, which is therefore independent of X and equal to 0 when X has no zeros.

4.1F5step 3.1algebra∎

If M is disconnected, apply the identity on each component and add: the index sum splits over the components, and χ is additive over disjoint unions by [F5], so the same identity holds; the empty zero set is included (the empty alternating sum is 0).

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

A nowhere-zero vector field forces zero Euler characteristic

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a closed smooth n-manifold, n≥1, and suppose M admits a smooth vector field X with no zeros at all. Then χ(M)=0 (Euler characteristic of a compact manifold).

Facts & Assumptions

Given: A closed smooth n-manifold M with a nowhere-zero smooth vector field X.

[F1]

A vector field with no zeros has only isolated zeros, so Poincare-Hopf applies: ∑p:X(p)=0ind⁡pX=χ(M) (Poincare-Hopf for closed manifolds, Isolated zero and local index of a vector field).

Proof

1.1F1algebra

The zero set of X is empty by hypothesis, so it consists of isolated zeros vacuously and the hypothesis of [F1] is satisfied; the index sum ∑p:X(p)=0ind⁡pX is the empty sum 0.

2.1F1step 1.1algebra∎

Poincare-Hopf [F1] now gives χ(M)=∑p:X(p)=0ind⁡pX=0.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A Morse gradient zero contributes (−1)λ to the index

Statement

Assume ACω (The Axiom of Countable Choice (ACω)) for the canonical smooth tangent-bundle structure.

Let M be a closed smooth n-manifold, n≥1, and let f:M→R be a Morse function (Morse functions and excellent Morse functions), let g be a Riemannian metric on M (Riemannian metric and riemannian manifold) and let p be a critical point of f of index λ=ind⁡(p) (Nondegenerate critical points, nullity, index, and coindex). Then grad⁡gf has a nondegenerate zero at p with ind⁡p(grad⁡gf)=(−1)λ,ind⁡p(−grad⁡gf)=(−1)n−λ.

Facts & Assumptions

Given: A closed smooth manifold M, a Morse function f, a Riemannian metric g and a critical point p of f of Morse index λ.

[F1]

The gradient grad⁡gf is a smooth vector field on M and it vanishes exactly at the critical points of f (The Riemannian gradient is the metric dual of the differential, The Riemannian gradient vanishes exactly at the critical points).

[F2]

In coordinates, differentiating the gradient formula at a critical point gives D(grad⁡gf)p=g(p)−1Hp, because dfp=0 kills the derivatives of the inverse metric. Thus the linearization (vertical derivative) of grad⁡gf is the Hessian bilinear form Hess⁡pf, read through g; the critical Hessian of a Morse function is nondegenerate with λ negative and n−λ positive entries in its inertia normal form, so the linearization is invertible and p is a nondegenerate zero of grad⁡gf (The intrinsic Hessian of a smooth function at a critical point, At a critical point, the intrinsic Hessian agrees with the Levi-Civita Hessian, Nondegenerate critical points, nullity, index, and coindex, Nondegenerate zero of a vector field).

[F3]

A nondegenerate zero has index equal to the sign of the determinant of its linearization (The index of a nondegenerate vector-field zero), the sign of the determinant of a nondegenerate symmetric form with λ negative and n−λ positive entries in its inertia normal form is (−1)λ (Nondegenerate critical points, nullity, index, and coindex), and negating the field multiplies the index by (−1)n (Negation scales the local index by (−1)n); the index of the negated function swaps, ind⁡p(−f)=n−λ (Index and coindex swap under negation).

Proof

1.1F1F2algebra

By [F1] the field grad⁡gf has a zero at p (and only at the critical points), and by [F2] its linearization there is the nondegenerate Hessian with λ negative and n−λ positive entries in its inertia normal form; hence p is a nondegenerate zero and sign⁡det⁡(Dgrad⁡gf)p=(−1)λ.

2.1F2F3step 1.1algebra∎

The determinant formula [F3] gives ind⁡p(grad⁡gf)=(−1)λ, and the negation rule gives ind⁡p(−grad⁡gf)=(−1)n(−1)λ=(−1)n−λ, the second formula; equivalently, −grad⁡gf=grad⁡g(−f) has index (−1)ind⁡p(−f)=(−1)n−λ by the index swap of [F3].

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The Morse critical-point sum is the Euler characteristic

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a closed smooth n-manifold with n≥1, and let f:M→R be a Morse function (Morse functions and excellent Morse functions) and let g be a Riemannian metric on M (Every smooth manifold admits a riemannian metric). Then ∑p∈Crit⁡(f)(−1)ind⁡(p)=χ(M), the sum over the finitely many critical points of f.

Facts & Assumptions

Given: A closed smooth n-manifold M, a Morse function f and a Riemannian metric g on M.

[F1]

The field grad⁡gf vanishes exactly at the critical points of f, which are finitely many, and at a critical point of index λ its index is (−1)λ (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, Morse functions and excellent Morse functions).

[F2]

Poincare-Hopf: the index sum of a smooth field with only isolated zeros on a closed manifold equals χ(M) (Poincare-Hopf for closed manifolds, Euler characteristic of a compact manifold).

Proof

1.1F1F2algebra

The critical points of a Morse function on a closed manifold are finitely many and each is a nondegenerate zero of grad⁡gf; hence the field has only isolated zeros and [F2] gives ∑pind⁡p(grad⁡gf)=χ(M).

2.1F1step 1.1algebra∎

By [F1] each summand is ind⁡p(grad⁡gf)=(−1)ind⁡(p), so substituting into step 1.1 gives ∑p∈Crit⁡(f)(−1)ind⁡(p)=χ(M).

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Closed odd-dimensional manifolds have zero Euler characteristic

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a closed smooth n-manifold with n odd. Then χ(M)=0 (Euler characteristic of a compact manifold).

Facts & Assumptions

Given: A closed smooth n-manifold M with n odd.

[F1]

There is an excellent Morse function f on M, and for any Riemannian metric g the field X:=grad⁡gf vanishes exactly at the (finitely many, nondegenerate) critical points of f (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, Morse functions and excellent Morse functions).

[F2]

The field −X has the same zeros as X, and ind⁡p(−X)=(−1)nind⁡pX; for odd n this is −ind⁡pX (Negation scales the local index by (−1)n, Isolated zero and local index of a vector field).

[F3]

Poincare-Hopf: for a smooth field with only isolated zeros on a closed manifold, the index sum equals χ(M) (Poincare-Hopf for closed manifolds).

Proof

1.1F1F3algebra

Choose an excellent Morse function f and a Riemannian metric g; the field X=grad⁡gf has only isolated (indeed nondegenerate) zeros, so [F3] gives χ(M)=∑pind⁡pX.

2.1F2F3step 1.1algebra∎

The field −X has the same zero set, and since n is odd [F2] gives ind⁡p(−X)=−ind⁡pX; applying [F3] to −X gives χ(M)=∑pind⁡p(−X)=−∑pind⁡pX=−χ(M) by step 1.1. Hence 2χ(M)=0 in Z, so χ(M)=0.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

A degree-zero sphere map extends over the ball without zeros

Statement

Assume countable choice ACω (The Axiom of Countable Choice (ACω)). Let m≥1 and let u:Sm→Sm be a continuous map of degree 0 (Degree of a map between oriented closed manifolds). Then u is homotopic to a constant map, and there is a continuous nowhere-zero map F:Dm+1→Rm+1∖{0} with Dm+1=B‾2(0,1) and F∣Sm=u (Euclidean spheres and closed balls as subspaces of Rn). If u is smooth, F can be chosen smooth with F(x)=u(x/∣x∣) for every x with ∣x∣≥23; in particular F restricts to u on Sm.

Facts & Assumptions

Given: m≥1, a continuous map u:Sm→Sm of degree 0, and ACω for the smooth clause.

[A1]

Countable choice is The Axiom of Countable Choice (ACω); it is used only in [F3].

[F1]

For m≥1 and a continuous self-map f of Sm, the sphere degree of Degree of a self map of an oriented sphere reads the multiplier f∗[Sm]=deg⁡(f)[Sm] on the integral top generator of Hm(Sm;Z)≅Z given by Homology of spheres, while the degree of Degree of a map between oriented closed manifolds reads the multiplier of the fundamental classes of the two standard orientations; for Sm these are the same integer. Homotopic maps have equal degree and deg⁡(g∘f)=deg⁡(g)deg⁡(f); every homotopy equivalence Sm→Sm has degree 1 or −1; the identity has degree 1 and constant maps have degree 0 (Degree is homotopy invariant and multiplicative under composition, Degree of identity constant reflection and antipodal sphere maps, Homotopic maps induce the same map on singular homology).

[F2]

For every r≥1 the degree is an isomorphism πr(Sr,b)→Z; hence two based self-maps of Sr are homotopic through based maps if and only if their degrees agree, and the constant map represents the zero element (Based sphere maps are classified by degree).

[F3]

If two smooth maps between smooth manifolds are continuously homotopic, then they are smoothly homotopic (Continuously homotopic smooth maps are smoothly homotopic).

[F4]

The standard smooth step function σ:R→[0,1] is smooth with σ(t)=0 for t≤0 and σ(t)=1 for t≥1 (The standard smooth step function).

[F5]

Sm=B‾2(0,1)∖B2(0,1)=∂Dm+1 is the unit sphere in Rm+1 and is contained in Rm+1∖{0} (Euclidean spheres and closed balls as subspaces of Rn).

Proof

1.1F1F5algebra

Fix b∈Sm. If u(b)=b put R:=id⁡; otherwise put R(x):=x−2⟨x,w⟩w/∣w∣2 with w:=u(b)−b≠0. A direct computation gives ∣R(x)∣=∣x∣, R2=id⁡ and R(w)=−w, hence R(u(b))=R(b)+R(w)=u(b)−(u(b)−b)=b; thus R restricts to a self-map of Sm and is its own continuous inverse, so g:=R∘u is a based self-map of Sm at b with deg⁡g=deg⁡R⋅deg⁡u=0.

2.1F1F2step 1.1

By [F2] the degree is an isomorphism πm(Sm,b)→Z, so g, having degree 0, is based-homotopic to the constant map at b; applying the continuous map R to that based homotopy exhibits a homotopy from u=R−1∘g to the constant map at R(b), proving that u is homotopic to a constant.

3.1F2F5step 2.1algebra

Let H:Sm×[0,1]→Sm be a homotopy with H0=u and H1≡c, and define F(0):=c and F(x):=H(x/∣x∣, 1−∣x∣) for x∈Dm+1∖{0}. Then F is continuous away from 0 as a composite of continuous maps, and for x→0 continuity at every point of Sm×{1} and a finite subcover of the compact sphere give H(v,t)→c uniformly in v as t→1, so F(x)→c; on Sm we have 1−∣x∣=0 and F(x)=H(x,0)=u(x); and F takes values in Sm⊆Rm+1∖{0}, so F is nowhere zero.

4.1A1F3F4F5step 3.1constructalgebra∎

Suppose now that u is smooth. By [F3] the continuously homotopic smooth maps u and the constant c are smoothly homotopic; let H′ be a smooth homotopy from u to c and put φ(t):=σ(3t−1), so that φ is smooth with φ(t)=0 for t≤1/3 and φ(t)=1 for t≥2/3 by [F4]. Then H′′(x,t):=H′(x,φ(t)) is a smooth homotopy from u to c with H′′(x,t)=u(x) for t≤1/3, and we redefine F(0):=c, F(x):=H′′(x/∣x∣,1−∣x∣) for x≠0. This F is smooth on {0<∣x∣<1} and equals c for 0<∣x∣≤1/3 (where 1−∣x∣≥2/3), so it is smooth at 0 as well; on the collar {∣x∣≥2/3} it equals H′′(x/∣x∣,1−∣x∣)=u(x/∣x∣), a smooth function of x on that collar because ∣x∣≥2/3>0, and at the boundary points ∣x∣=1 it equals u(x), where x/∣x∣=x; so F is smooth on Dm+1 with F(x)=u(x/∣x∣) for ∣x∣≥2/3, and it takes values in Sm, hence is nowhere zero.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Two points avoiding a finite set lie in a common embedded ball

Statement

Assume countable choice ACω (The Axiom of Countable Choice (ACω)). Let M be a connected boundaryless smooth n-manifold, n≥2, let p≠q∈M and let F⊂M∖{p,q} be finite. Then there are a smooth embedded closed arc γ⊆M∖F from p to q and a smoothly embedded closed ball B⊆M with p,q∈int⁡B and B∩F=∅ (Embedded smooth submanifolds with boundary).

Facts & Assumptions

Given: A connected boundaryless smooth n-manifold M, n≥2, distinct p,q, a finite set F disjoint from them, and ACω.

[F2]

Under ACω, the open manifold M∖F admits a proper smooth embedding in Euclidean space (The weak Whitney proper embedding theorem). A closed embedded submanifold of complete Euclidean space is complete in its induced Riemannian metric (Closed embedded submanifolds of complete Riemannian manifolds are complete).

[F3]

Under ACω, every connected complete boundaryless Riemannian manifold is geodesically complete and any two points are joined by a minimizing geodesic (Hopf–Rinow theorem).

[F4]

Levi–Civita parallel transport is a linear isomorphism preserving inner products, and parallel sections along a smooth curve are smooth (Parallel transport is a linear isomorphism, Levi civita parallel transport preserves lengths angles and volume).

[F5]

A closed boundaryless embedded submanifold in a smooth ambient manifold has a tubular neighbourhood under ACω (The tubular neighbourhood theorem in a smooth ambient manifold).

Proof

1.1F1givenalgebra

Put U=M∖F. A punctured coordinate ball in dimension n≥2 is path-connected: join two nonzero points by a broken line through a third point avoiding the two lines through the puncture. To see that U is connected, suppose U=A⊔B were a separation. For each x∈F choose a coordinate ball meeting F only at x; its connected punctured ball lies entirely in A or entirely in B. Add x to that side. The resulting two sets are disjoint nonempty open sets covering M, a contradiction. Hence U is connected and path-connected by [F1].

2.1F2F3step 1.1construct

Embed U properly in R2n+1 by [F2]. Its image is closed: a convergent sequence of image points lies in a compact Euclidean ball; properness gives a compact preimage, and a convergent subsequence shows the limit is in the image. The induced metric is complete by [F2]. By [F3] a nonconstant minimizing geodesic γ:[0,1]→U joins p to q. It has constant positive speed and is injective, since deleting any nonconstant loop would shorten it. A continuous injection from the compact interval into a Hausdorff manifold is an embedding, so this is a smooth embedded arc.

3.1F3F4F5step 2.1construct

Geodesic completeness extends γ beyond both endpoints. Choose a>0 small enough that its restriction to [−a,1+a] remains an embedding: the positive tangent makes it locally injective at each endpoint, and compactness separates these small endpoint continuations from the portions of the original arc outside their coordinate neighbourhoods and from each other. Let W=U∖{γ(−a),γ(1+a)} and S=γ((−a,1+a)). Then S is boundaryless and closed in W, since its closure in U is the extended compact arc and only the two removed endpoints are missing. Apply [F5] to S⊂W. Parallel-transport an orthonormal normal basis along the geodesic by [F4]; its tangent is parallel, so the transported vectors stay normal and give a smooth frame of the normal quotient bundle. The tube is therefore parametrized near its zero section by (t,z)∈(−a,1+a)×Rn−1.

4.1F5step 3.1constructalgebra∎

Choose L with 1/2<L<1/2+a. Compactness of [1/2−L,1/2+L] in the zero section supplies ε>0 such that the ellipsoid E={(t,z):((t−1/2)/L)2+∣z∣2/ε2≤1} is contained in the tube domain: cover that compact segment by finitely many product neighbourhoods in the open domain and take a common positive fibre radius. The ellipsoid is affinely diffeomorphic to Dn, and its tube image is a smooth embedded closed ball B⊂W⊂M∖F. The points (0,0) and (1,0) satisfy the strict ellipsoid inequality because L>1/2, so p,q∈int⁡B. The original arc lies in this ball and avoids F, completing both assertions.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Opposite-index nondegenerate zeros cancel in a ball

Statement

Assume countable choice ACω (The Axiom of Countable Choice (ACω)). Let M be a smooth n-manifold, n≥2, let X be a smooth vector field and let B⊆M be a smoothly embedded closed ball whose interior contains exactly two zeros p,q of X, both nondegenerate and of opposite index, with X≠0 on ∂B (Embedded smooth submanifolds with boundary). Then there is a smooth vector field X′ on M with X′=X outside int⁡B (in particular on a neighbourhood of ∂B) and X′≠0 on B; thus X′ has exactly the zeros of X outside B and none in B.

Facts & Assumptions

Given: A smooth field X on the smooth n-manifold M, n≥2, and a closed ball B containing exactly the two nondegenerate zeros p,q in its interior, with opposite indices and X≠0 on ∂B.

[F1]

Choose a smooth parametrization b:Dn→B and pull back the field as Y(u)=(dbu)−1X(b(u)). This is a smooth vector field on the closed Euclidean ball, with exactly the two corresponding nondegenerate zeros and no boundary zero. The differential of b provides matching base and fibre orientations. The boundary-degree lemma identifies its normalized boundary degree with the sum of local indices, which are preserved under this pullback (The local index is additive under a transverse perturbation, The index sum of an outward field is the Gauss degree, The induced tangent bundle chart).

[F2]

A smooth map u:Sm→Sm of degree 0 is homotopic to a constant map and admits a smooth nowhere-zero extension F:Dm+1→Rm+1∖{0} with F(x)=u(x/∣x∣) for ∣x∣≥23; in particular F=u on the boundary sphere (A degree-zero sphere map extends over the ball without zeros).

[F3]

Each of the two zeros is nondegenerate with index ±1, and the two indices are opposite, so their sum is 0 (The index of a nondegenerate vector-field zero, Nondegenerate zero of a vector field).

[F4]

There are smooth bump functions equal to 1 on a prescribed closed collar of the boundary sphere and supported in a slightly larger collar, and smooth radial interpolations with prescribed values near the two ends of an interval exist (Explicit compactly supported smooth cutoffs).

Proof

1.1F1F3algebra

In the parametrization of [F1] the normalized field u(y):=Y(y)/∣Y(y)∣ on the boundary sphere is smooth and its degree equals ind⁡b−1(p)Y+ind⁡b−1(q)Y=0 by [F1] and [F3].

2.1F1F2F3F4step 1.1construct

Fix 0<r0<1 so that Y≠0 on the collar {r0≤∥y∥≤1}, put u0(v):=Y(r0v)/∣Y(r0v)∣, and note that the ball of radius r0 contains the same two zeros, so [F1] gives deg⁡u0=0 as well; by [F2] applied to u0 there is a smooth nowhere-zero F0 on B‾r0(0) with F0(y)=u0(y/∥y∥) for 23r0≤∥y∥≤r0. Put Ψ(v,s):=Y(sv)/∣Y(sv)∣ on Sn−1×[r0,1], choose by [F4] a smooth function σ:[r0,1]→[r0,1] with σ(s)=s for s near 1 and σ(s)=r0 for s near r0, and define F:=F0 on B‾r0(0) and F(y):=Ψ(y/∥y∥,σ(∥y∥)) for r0≤∥y∥≤1: the two formulas agree on the sphere ∥y∥=r0, where both equal u0(y/∥y∥) and are independent of ∥y∥ in a one-sided neighbourhood of it, so F is smooth and nowhere zero on B‾2(0,1), and F(y)=Y(y)/∣Y(y)∣ on a collar {∥y∥≥1−δ} because σ(s)=s there.

3.1F1F4step 2.1constructalgebra

Choose by [F4] a smooth bump μ equal to 1 on a neighbourhood of the collar {∥y∥≥1−δ} and supported in a slightly larger zero-free collar, extend μ∣Y∣ smoothly by zero from that zero-free collar, and write ψ:=μ ∣Y∣+(1−μ)c with a positive constant c, and put X′′:=ψ F; then ψ is smooth and positive on the ball with ψ=∣Y∣ on {∥y∥≥1−δ}, so X′′ is a smooth nowhere-zero field on the ball, and on that collar X′′=∣Y∣⋅Y/∣Y∣=Y. Therefore the field equal to X′′ on B (transported back by db) and to X outside int⁡B is smooth, agrees with X on a neighbourhood of ∂B and outside int⁡B, and is nowhere zero on B.

4.1step 3.1algebra∎

The resulting smooth field X′ on M therefore has no zero in B and coincides with X off int⁡B, so its zero set is exactly the zero set of X outside B, as claimed.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The reflection of an outward field extends over the double

Statement

Assume countable choice ACω (The Axiom of Countable Choice (ACω)). Let M be a compact smooth n-manifold with boundary, n≥1, and let X be a smooth vector field that is nonzero and points strictly outward along ∂M (Inward, outward, and boundary-tangent vectors). Choose the collar generated by the inward field −X. Let DM be the smooth double defined by this collar, with seam involution τ interchanging the two labelled halves (The double of a smooth manifold with boundary, The double has a well-defined smooth structure). Then X+(y)=X(y)  (y∈M+),X+(τy)=−dτy(X(y))  (y∈M+) defines a smooth vector field on this double (an arbitrary fixed collar need not give a smooth field); its zeros are exactly the two copies of the zeros of X, and for every isolated zero p∈int⁡M ind⁡τp(X+)=(−1)nind⁡pX.

Facts & Assumptions

Given: A compact smooth n-manifold M with boundary, a smooth field X nonzero and strictly outward along ∂M, and the labelled double (DM,τ).

[F1]

DM is the quotient of M+⊔M− identifying the two copies of ∂M, with τ the involution interchanging the labelled halves; smooth collar data near the seam give it the structure of a smooth boundaryless manifold, and two collar choices give structures related by a diffeomorphism fixing the seam pointwise and preserving the halves (The double of a smooth manifold with boundary, The double has a well-defined smooth structure).

[F3]

Because X is strictly outward and nonzero on the compact boundary, there is δ>0 such that X has no zero in the δ-neighbourhood of ∂M; the zeros of X therefore lie in the interior at positive distance from ∂M (Inward, outward, and boundary-tangent vectors).

[F4]

The inward field −X has smooth local forward semiflows at boundary points, using smooth coordinate extensions across the face (Inward-pointing fields have local forward semiflows at the boundary). Their differentials at time zero are invertible because −X is transverse to the boundary. Compactness supplies a uniform short time. Uniqueness and strict inwardness make the map c(x,u)=Φu −X(x) injective: a trajectory cannot return to the boundary, since its boundary defining coordinate has positive derivative at any putative return. Thus c is a global collar for short time, and X=−∂u in its coordinates.

[F5]

The index is independent of the chart and of a trivialization with matching base and fibre orientations, and negation scales it by (−1)n (The local index is independent of chart, ball and trivialization, Negation scales the local index by (−1)n).

Proof

1.1F1F4givenconstruct

If ∂M=∅, the double is the disjoint union of two copies, and the two fields are X and −X. Otherwise use the single global flow collar c(x,u) of [F4] to define the double's smooth structure. Its seam charts have signed coordinate u, with τ(x,u)=(x,−u). This is the collar-defined double of [F1], rather than a replacement of an already fixed smooth structure while keeping X unchanged.

2.1F1F4step 1.1algebra

In these charts X=−∂u on the first half. Reflection followed by negation gives −dτ(−∂u)=−∂u on the second half. The prescriptions therefore agree at the seam and give one smooth nonzero expression there. Off the seam they are X and the push-forward of −X, so they are smooth globally.

3.1F3F5step 1.1step 2.1algebra∎

There are no seam zeros, and off the seam the zero set consists precisely of the two copies of X−1(0). For any isolated zero p, a chart at p transported by τ identifies the second field with −X. Chart invariance and [F5] give ind⁡τpX+=ind⁡p(−X)=(−1)nind⁡pX. The same computation applies to the two disjoint copies when the boundary is empty.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The index sum of an outward field on an even-dimensional manifold

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a compact smooth n-manifold with nonempty boundary, n even, and let X be a smooth vector field with only isolated zeros that is nonzero and strictly outward along ∂M (Inward, outward, and boundary-tangent vectors). Then ∑p:X(p)=0ind⁡pX=χ(M).

Facts & Assumptions

Given: A compact smooth even-dimensional manifold M with nonempty boundary, and a smooth field X on M with only isolated zeros, nonzero and strictly outward along ∂M.

[F1]

Reduction: the zeros of X lie in the interior at positive distance from ∂M, because X≠0 on the compact boundary and there are finitely many zeros; hence, by part (iii) of the index-perturbation lemma, they can be perturbed inside disjoint small balls contained in the interior of M and away from a neighbourhood of ∂M, producing a field X0 with only nondegenerate zeros, the same index sum, and still strictly outward on ∂M (The local index is additive under a transverse perturbation, Isolated zero and local index of a vector field, Inward, outward, and boundary-tangent vectors).

[F2]

Doubling: choose the global flow collar of −X and use it to define DM, a closed smooth n-manifold with seam involution τ, and the reflected field X+ of The reflection of an outward field extends over the double is smooth, has zeros exactly the two copies of the zeros of X, and for every zero p, ind⁡τpX+=ind⁡p(−X)=(−1)nind⁡pX (The double of a smooth manifold with boundary, The double has a well-defined smooth structure, Negation scales the local index by (−1)n).

[F3]

Poincare-Hopf on DM: the index sum of X+ on the closed manifold DM equals χ(DM) (Poincare-Hopf for closed manifolds).

[F4]

Additivity: χ(M∪∂MM)=χ(M)+χ(M)−χ(∂M) by part (iii) of Finiteness and additivity of the Euler characteristic; the inclusion ∂M↪M is a cofibration, because the collar of Collar neighborhood theorem is a neighbourhood deformation retract structure, whose mapping-cylinder retraction characterizes cofibrations (Cofibrations are characterized by a retraction of the mapping cylinder strip); and χ(∂M)=0 because ∂M is a closed manifold of odd dimension n−1 (Closed odd-dimensional manifolds have zero Euler characteristic, Euler characteristic of a compact manifold).

Proof

1.1F1algebra

Apply the reduction of [F1] and replace X by a field X0 with only nondegenerate zeros, the same index sum, and still strictly outward on ∂M; it suffices to prove the identity for X0, and by The index of a nondegenerate vector-field zero each of its zeros has index ±1.

2.1F2F3step 1.1algebra

Form the field-adapted double DM and the reflected field X0+; by [F2] the zeros of X0+ are the two copies of the zeros of X0 and, since n is even, ind⁡τpX0+=(−1)nind⁡pX0=ind⁡pX0, so the index sum over DM is twice the index sum over M; Poincare-Hopf [F3] gives 2∑pind⁡pX0=χ(DM).

3.1F4step 1.1step 2.1algebra∎

By [F4], χ(DM)=χ(M∪∂MM)=2χ(M)−χ(∂M)=2χ(M); substituting into step 2.1 gives 2∑pind⁡pX0=2χ(M), hence ∑pind⁡pX0=χ(M) in Z, and by step 1.1 the same identity holds for the original field X.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Poincare-Hopf with outward-pointing boundary

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let M be a compact smooth n-manifold, n≥1, and let X be a smooth vector field with only isolated zeros that is nonzero and strictly outward along ∂M (Inward, outward, and boundary-tangent vectors; the boundary clause is vacuous when ∂M=∅). Then ∑p:X(p)=0ind⁡pX=χ(M).

Facts & Assumptions

Given: A compact smooth n-manifold M, n≥1, and a smooth field X with only isolated zeros, strictly outward along ∂M.

[F1]

If ∂M=∅, the statement is Poincare-Hopf for closed manifolds; if n is even and ∂M≠∅, it is The index sum of an outward field on an even-dimensional manifold.

[F2]

Products of a boundary chart of M with an endpoint half-interval give normal quadrant charts; on boundaryless interiors the ordinary product theorem applies. For odd n, the product W:=M×[0,1], with its two codimension-two corner strata ∂M×{0} and ∂M×{1} rounded by the standard corner-rounding convention, is a compact smooth (n+1)-manifold with boundary (an even-dimensional one); its boundary is the rounded version of ∂M×[0,1]∪M×{0}∪M×{1}, and the rounding changes only a collar of the corner strata, so W is homotopy equivalent to M (the rounded product is a deformation retract of the original product: in each inward normal quadrant, slide (r,s) along (1,1) to the first point of the retained rounded region. The required nonnegative displacement is continuous because the rounding profile is monotone and transverse to (1,1), and is zero on the retained region. Multiplying that displacement by a homotopy parameter gives a deformation fixing the rounded region, supported in the corner collar. The normal formulas agree along the corner stratum. Thus the rounded product is homotopy equivalent to M, so homology is unchanged and χ(W)=χ(M)) (Products of smooth manifolds have a canonical product smooth structure, Attaching a smooth handle with corner rounding, Smooth handle attachment is independent of corner rounding up to diffeomorphism, Homotopy equivalences induce isomorphisms on singular homology, Euler characteristic of a compact manifold).

[F3]

A product-type zero is nondegenerate with the product index: if X has a nondegenerate zero at p and ψ(t)=t−12 has its simple zero at t0=12, then Z(x,t):=(X(x),ψ(t)∂t) has a nondegenerate zero at (p,t0) with ind⁡(p,t0)Z=ind⁡pX⋅sign⁡ψ′(t0)=ind⁡pX; the linearization is block diagonal with blocks DXp and ψ′(t0) (The index of a nondegenerate vector-field zero, Nondegenerate zero of a vector field).

[F4]

The field Z is strictly outward along ∂W: on ∂M×[0,1] the outward normal of W is the outward normal of ∂M in M and the inward boundary defining coordinate r satisfies dr(Z)=dr(X)<0; on M×{0} the outward normal is −∂t and ⟨Z,−∂t⟩=−ψ(0)=12>0; on M×{1} the outward normal is +∂t and ⟨Z,∂t⟩=ψ(1)=12>0 (Inward, outward, and boundary-tangent vectors). Near a lower corner use inward coordinates r≥0, s=t≥0; near an upper corner use r≥0, s=1−t≥0. In a sufficiently small uniform corner neighbourhood, both dr(Z)<0 and ds(Z)<0. Choose the standard monotone rounding whose outward conormal is −a dr−b ds, where a,b≥0 and a+b>0. Its evaluation on Z is strictly positive, so Z stays strictly outward on every rounded face as well. The rounding is supported away from all zeros and from t=1/2.

[F5]

Reduction to nondegenerate zeros of X by The local index is additive under a transverse perturbation can be performed inside the interior of M, leaving a neighbourhood of ∂M fixed, hence preserving strict outwardness.

Proof

1.1F2F5algebra

If ∂M=∅ or n is even the statement is [F1]; assume therefore that n is odd and ∂M≠∅. Apply [F5] to replace X by a field X0 with only nondegenerate zeros, the same index sum and still strictly outward, and put W:=M×[0,1] and Z(x,t):=(X0(x), (t−12)∂t).

2.1F1F2F3F4step 1.1algebra

The zeros of Z are exactly the points (p,12) with X0(p)=0, all interior, and by [F3] each is nondegenerate with ind⁡(p,1/2)Z=ind⁡pX0; the field Z is strictly outward along ∂W by [F4]. Since dim⁡W=n+1 is even, the even-dimensional boundary lemma [F1] applies to (W,Z) and gives ∑pind⁡pX0=∑(p,1/2)ind⁡(p,1/2)Z=χ(W)=χ(M) by [F2].

3.1F1step 1.1step 2.1algebra∎

By step 1.1 the index sum of X0 equals that of X, so ∑pind⁡pX=χ(M); the remaining cases were handled in step 1.1, completing the proof.

CorollaryStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The Euler number of the tangent bundle is the Euler characteristic

Statement

Assume the Axiom of Choice (The Axiom of Choice).

(i) Let M be a closed oriented smooth n-manifold, n≥1. Then the Euler number of the tangent bundle satisfies ⟨e(TM),[M]⟩=χ(M), where e(TM)∈Hn(M;Z) is the Euler class of Euler class by zero-section pullback of the Thom class and the bracket is the Kronecker evaluation; equivalently the diagonal satisfies ΔM⋅ΔM=χ(M) in the self-intersection number of The self-intersection number of a complementary-dimensional oriented submanifold.

(ii) For any closed smooth n-manifold with n≥1, ⟨wn(TM),[M]⟩2≡χ(M)(mod2), with wn the top Stiefel-Whitney class.

Facts & Assumptions

Given: A closed smooth n-manifold M, oriented in part (i).

[F1]

Choose an excellent Morse function f and a Riemannian metric g. The section s=grad⁡gf of TM vanishes exactly at Crit⁡(f) and is transverse to the zero section there (the linearization is the nondegenerate Hessian) (The Riemannian gradient is the metric dual of the differential, The Riemannian gradient vanishes exactly at the critical points, Every compact smooth manifold admits an excellent Morse function, Every smooth manifold admits a riemannian metric, Morse functions and excellent Morse functions).

[F2]

The signed zero count of a transverse section of a rank-n oriented bundle over a closed oriented n-manifold equals ⟨e(E),[M]⟩; the local sign of a zero of a vector field, read on the zero-section/graph intersection in TM with the horizontal-then-vertical orientation, is the index of the zero (The self-intersection number is the Euler number of the normal bundle, The index of a zero is its zero-section intersection number, Euler class by zero-section pullback of the Thom class).

[F3]

The signed zero count of grad⁡gf is ∑p(−1)ind⁡(p), and this equals χ(M); in particular the Euler number is χ(M). The diagonal form is the self-intersection statement for ΔM, and over F2 the unsigned zero count satisfies #Crit⁡(f)≡⟨wn(TM),[M]⟩2, since the canonical mod 2 Euler class is the top Stiefel-Whitney class (A Morse gradient zero contributes (−1)λ to the index, Morse Euler characteristic identity, The diagonal self-intersection is the Euler number of the tangent bundle, The mod two self-intersection is the top Stiefel-Whitney evaluation, The mod-two Euler class is the top Stiefel–Whitney class).

[F4]

Under the assumed AC, M is paracompact Hausdorff and CGWH with CW homotopy type, and the smooth tangent bundle is numerable (Smooth manifolds have CW homotopy type). These are the base and bundle hypotheses of the cited Thom and Stiefel-Whitney results.

Proof

1.1F1F2F3F4algebra

For (i): by [F1] the section s=grad⁡gf is transverse to the zero section with zero set Crit⁡(f), and by [F2] its signed zero count equals ⟨e(TM),[M]⟩; by [F3] that signed count is ∑p(−1)ind⁡(p)=χ(M), so ⟨e(TM),[M]⟩=χ(M).

2.1F3step 1.1algebra

The diagonal statement of [F3] identifies ΔM⋅ΔM with ⟨e(TM),[M]⟩, so it equals χ(M) as well.

3.1F3F4step 1.1algebra∎

For (ii): the unsigned count #Crit⁡(f) of the transverse section grad⁡gf satisfies #Crit⁡(f)≡⟨wn(TM),[M]⟩2 by [F3], while #Crit⁡(f)≡∑p(−1)ind⁡(p)=χ(M)(mod2) because (−1)λ≡1; hence ⟨wn(TM),[M]⟩2≡χ(M)(mod2).

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

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≥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≥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≠∅ then M is diffeomorphic to the circle S1=R/Z, which carries the standard nowhere-zero rotational field ∂θ; the empty manifold admits the empty field vacuously (Nonempty closed connected 1-manifolds are circles).

[F3]

For n≥2: choose an excellent Morse function f and a Riemannian metric g; the field X:=grad⁡gf has only nondegenerate zeros, namely the critical points, and a critical point of index λ has index (−1)λ=±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: ∑pind⁡pX=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≥2 and χ(M)=0. Choose the Morse gradient X of [F3]. Its finite zero set has indices ±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.

LemmaStatement: Literature-sourcedProof: Literature-sourcedjudge pass (gpt-6.1-sol)Open item page →

Nonempty closed connected 1-manifolds are circles

Statement

Assume ACω. Every nonempty closed connected smooth 1-manifold is diffeomorphic to the circle S1=R/Z with its standard smooth structure (The circle as S1=R/Z with basepoint [0], Diffeomorphisms and local diffeomorphisms of manifolds). Here closed means compact with empty boundary; the empty manifold is excluded because it is connected under Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets but is not diffeomorphic to a circle.

Facts & Assumptions

Given: A nonempty closed connected smooth 1-manifold M, and ACω.

[A1]

ACω: every countable family of nonempty sets has a choice function (The Axiom of Countable Choice (ACω)).

[F1]

Every compact smooth 1-manifold W, possibly with boundary, is diffeomorphic to a finite disjoint union of copies of the circle S1 and of the closed interval [0,1]; a diffeomorphism of manifolds with boundary maps ∂W onto the boundary of the target, and each closed-interval component contributes exactly its two endpoints to that boundary (Boundary of a compact 1-manifold has even cardinality).

[F2]

A closed smooth manifold is by definition a compact smooth manifold with empty boundary (Smooth manifolds and their smooth charts, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right); in particular ∂M=∅.

[F3]

The circle is S1=R/Z with the quotient topology and its standard smooth structure, and a diffeomorphism is a bijective smooth map whose inverse is smooth (The circle as S1=R/Z with basepoint [0], Diffeomorphisms and local diffeomorphisms of manifolds).

Proof

1.1A1F1F2algebra

By [F2] the manifold M is compact with ∂M=∅, so [F1] provides a diffeomorphism from M onto a finite disjoint union ⨆i∈FSi1⊔⨆j∈G[0,1]j; a diffeomorphism of manifolds with boundary carries boundary to boundary, and the boundary of the target is the union of the two endpoints of each interval component, so ∅=∂M corresponds to ⨆j∈G{0,1} and forces G=∅.

2.1F1F3step 1.1algebra∎

Consequently M is diffeomorphic to ⨆i∈FSi1, a disjoint union of ∣F∣ copies of the circle. Each circle is a nonempty connected component of that disjoint union, so the union is connected only when ∣F∣≤1, and M is nonempty, so ∣F∣=1; hence M is diffeomorphic to the standard circle S1, the model R/Z of [F3], as claimed.

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

The outward boundary hypothesis cannot be replaced by nonzero on the boundary

Remarks

Assume the Axiom of Choice (The Axiom of Choice) for the applications of the general index theorems below.

The hypothesis in Poincare-Hopf with outward-pointing boundary is strictly stronger than "X≠0 on ∂M": a field that is nonzero on the boundary but not outward, with only isolated zeros, contributes a boundary correction term. On the closed unit ball Dn⊆Rn, n≥3 odd, the inward radial field X(u)=−u is nonzero on ∂Dn and has the single zero 0, which is nondegenerate with linearization −In (Inward, outward, and boundary-tangent vectors, Isolated zero and local index of a vector field); by The index of a nondegenerate vector-field zero its index is sign⁡det⁡(−In)=(−1)n=−1. On the other hand χ(Dn)=1, because Dn is contractible with the rational homology of a point (Contractible nonempty spaces have the homology of a point, The Axiom of Choice, Euler characteristic of a compact manifold). Hence ∑pind⁡pX=−1≠1=χ(Dn): nonzero on the boundary does not suffice, and the outwardness in the boundary form is a genuine hypothesis rather than a convenience.

In even dimensions the inward radial field on Dn has index +1 and happens to agree with χ(Dn)=1, despite not being outward. Thus equality of the index sum with χ does not imply outwardness. For a compact smooth full-dimensional Euclidean domain N⊂Rn, n≥1, the boundary lemma identifies the index sum of a smooth field with only isolated zeros and nonzero on ∂N with the degree of its normalized boundary map (reduced degree when n=1). Outwardness is sufficient to identify that degree with the Gauss degree, which equals χ(N) by the outward-boundary theorem. This sphere-map description uses the Euclidean tangent trivialization and is not asserted for an arbitrary manifold with a possibly nontrivial tangent bundle.

5 · Examples, counterexamples and false statements

None yet.

Sources