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.

✓ 18 results · all verified · 10 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Characteristic Class Obstructions to Immersions and Embeddings

1 · Prerequisites

2 · Summary

This page turns the characteristic classes of the published algebraic-topology pages into immersion and embedding tests for closed smooth manifolds. The arithmetic is done on the stable normal inverse of the tangent bundle: a smooth bundle $ν$ together with a bundle isomorphism $T M \oplus ν ≅ ε^{N}$ onto a trivial bundle. Such an inverse is supplied by any smooth embedding into a Euclidean space, so every closed manifold has one under countable choice, while an immersion of codimension $k$ supplies an actual rank- $k$ inverse. The page keeps that rank distinction visible throughout: the class of an inverse is independent of the chosen inverse, but the rank of the bundle is exactly the codimension in an immersion problem.

The first block sets up the algebra: the pullback of a trivial bundle is canonically trivial, the compactified Euclidean space has no intermediate cohomology, and the normal total Stiefel–Whitney class is the multiplicative inverse of the tangent class, while the normal total Pontryagin class is the rational inverse — the rational coefficient ring is forced by the two-torsion correction in the integral Whitney product. The normal classes ${\overset{ˉ}{w}}_{i}$ and ${\overset{ˉ}{p}}_{i}$ are then defined once and used everywhere.

The second block contains the obstruction theory. A nonzero ${\overset{ˉ}{w}}_{i}$ with $i > k$ forbids immersion and hence embedding in $R^{m + k}$ ; a nonzero rational ${\overset{ˉ}{p}}_{i}$ with $2 i > k$ forbids the same; and for embeddings the top normal class is additionally forced to vanish, together with the Euler class of an oriented normal bundle. For an even-dimensional immersion into twice its dimension, the signed normal push-off count is a finite sum equal to the Euler number plus twice the signed double-point count, so the Euler number is even and a nonzero signed double-point count obstructs a regular homotopy to an embedding. The real projective spaces illustrate all of this: the truncated polynomial computation gives $w (T {R P}^{m}) = (1 + a)^{m + 1}$ and $\overset{ˉ}{w} ({R P}^{m}) = (1 + a)^{- (m + 1)} = \sum_{i \land m = 0} a^{i}$ , so ${R P}^{2^{p}}$ does not immerse in $R^{2^{p + 1} - 2}$ and does not even embed in the larger $R^{2^{p + 1} - 1}$ . Parallelizable manifolds, conversely, have trivial normal classes and admit Euclidean immersions in every positive codimension, while the parallelizable case says nothing about embeddability.

The final block fixes the boundaries of the page: the characteristic classes themselves are consumed from the published algebraic-topology pages rather than rebuilt here, and the class tests are necessary conditions only — two non-isotopic embeddings can carry identical trivial stable classes.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Stable normal inverse of the tangent bundle

Definition

Let $M$ be a smooth $m$ -manifold (Smooth manifolds and their smooth charts). A stable normal inverse of the tangent bundle is a pair $(ν, φ)$ consisting of a smooth real vector bundle $ν \to M$ of finite rank $k$ (Smooth vector bundles, rank, fibres, and trivial bundles) and a smooth bundle isomorphism $φ : T M \oplus ν ⟶ ε^{m + k}$ onto the trivial real bundle $ε^{m + k} = M \times R^{m + k}$ (The tangent bundle as a disjoint union, Whitney sums of vector bundles, Bundle maps, sections, subbundles, and isomorphisms). A rank- $k$ stable normal inverse is one whose bundle has rank $k$ ; when the rank is not named, $k = rank ν$ . Two stable normal inverses $(ν_{0}, φ_{0})$ , $(ν_{1}, φ_{1})$ are stably equivalent when $ν_{0} \oplus ε^{a} ≅ ν_{1} \oplus ε^{b}$ for some $a, b \geq 0$ . Adding a trivial summand, $(ν, φ) \mapsto (ν \oplus ε^{1}, φ \oplus {id}_{ε^{1}})$ , carries rank- $k$ inverses to rank- $(k + 1)$ inverses and preserves stable equivalence.

This is the inverse-bundle form of the published stable normal bundle Stable normal bundle of a compact smooth manifold: under ${A C}_{ω}$ , for compact $M$ and a smooth embedding $i : M ↪ R^{N}$ with $N \geq m$ , the normal bundle gives a rank- $(N - m)$ example with $T M \oplus ν_{i} ≅ ε^{N}$ (the $N > m$ case is An embedding into Euclidean space gives a rank-(n-m) stable normal inverse ↗; for $N = m$ , $d i$ is a fibrewise isomorphism and the normal quotient is the zero bundle). The dimension qualification is needed for an empty source, whose fibrewise immersion condition alone imposes no dimension inequality. Under ${A C}_{ω}$ every closed smooth $M$ has such an inverse by An embedding into Euclidean space gives a rank-(n-m) stable normal inverse ↗. The converse statement that every stable normal inverse is stably isomorphic to the normal bundle of an embedding is the stable classification statement; it is neither asserted nor used on this page. No choice principle is part of the definition; ${A C}_{ω}$ enters only through the metric and tubular identifications of the cited embedding lemma.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passOpen item page →

The pullback of a trivial smooth vector bundle is canonically trivial

Statement

Let $f : N \to M$ be a smooth map and let $ε_{M}^{r} = M \times R^{r}$ be the trivial smooth real rank- $r$ bundle over $M$ . Then the pullback $f^{*} ε_{M}^{r}$ is canonically isomorphic to the trivial bundle $ε_{N}^{r} = N \times R^{r}$ : the constant frame $e_{1}, \dots, e_{r}$ of $ε_{M}^{r}$ pulls back to the nowhere-zero global frame $x \mapsto (x, e_{j})$ of $f^{*} ε_{M}^{r}$ , and the isomorphism is the one determined by that frame (Local and global frames of a vector bundle, A vector bundle is trivial if and only if it has a global frame). This product-bundle assertion is choice-free.

Facts & Assumptions

Given: A smooth map $f : N \to M$ and the trivial smooth rank- $r$ real bundle $ε_{M}^{r} = M \times R^{r} \to M$ .

[F1]

The pullback set is $f^{*} ε_{M}^{r} = {(q, e) \in N \times (M \times R^{r}) : f (q) = {pr}_{1} (e)} = {(q, (f (q), v)) : q \in N, v \in R^{r}}$ , with projection $(q, (f (q), v)) \mapsto q$ and fibrewise vector-space operations inherited from $ε_{M}^{r}$ (Pullback vector bundles as fibre products, Smooth vector bundles, rank, fibres, and trivial bundles).

[F2]

The pullback carries the smooth rank- $r$ vector-bundle structure constructed in The pullback fibre product is a smooth vector bundle: for a bundle chart $Φ_{α} : E ∣_{U_{α}} \to U_{α} \times R^{r}$ of a bundle $E$ , the map $(q, e) \mapsto (q, v)$ with $Φ_{α} (e) = (f (q), v)$ is a bundle chart over $f^{- 1} (U_{α})$ ; the trivial bundle $ε_{M}^{r}$ has the single global bundle chart $Φ (p, v) = (p, v)$ over $M$ .

[F3]

A smooth rank- $r$ vector bundle is trivial if and only if it has a global frame (Local and global frames of a vector bundle, A vector bundle is trivial if and only if it has a global frame); a global frame $(s_{1}, \dots, s_{r})$ determines the bundle isomorphism $N \times R^{r} \to E$ , $(q, (λ_{1}, \dots, λ_{r})) \mapsto \sum_{j} λ_{j} s_{j} (q)$ .

[F4]

Maps into a product of smooth manifolds are smooth exactly when their components are, and smooth maps are continuous ( $C^{r}$ and smooth maps between smooth manifolds).

Proof

1.1F1F4

Define $Φ : f^{*} ε_{M}^{r} \to N \times R^{r}$ by $Φ (q, (f (q), v)) = (q, v)$ , using the description [F1]. Then $Φ$ is well defined, and its inverse is $(q, v) \mapsto (q, (f (q), v))$ . On each fibre it is the linear isomorphism onto ${q} \times R^{r}$ inverse to $v \mapsto (f (q), v)$ , and $Φ$ lies over ${id}_{N}$ . So $Φ$ is a fibrewise-linear bijection over the base.

2.1F2F4step 1.1

$Φ$ is smooth with smooth inverse. Indeed, in the global pullback chart of [F2] coming from the global chart $Φ (p, v) = (p, v)$ of $ε_{M}^{r}$ , the chart map is exactly $Φ$ , i.e. the identity identification of $N \times R^{r}$ ; and the inverse $(q, v) \mapsto (q, (f (q), v))$ has components $q \mapsto q$ , $q \mapsto f (q)$ and $v \mapsto v$ , which are smooth by [F4] since $f$ is smooth. Hence $Φ$ is a diffeomorphism, and consequently a smooth bundle isomorphism over $N$ .

3.1F2F3step 1.1step 2.1∎

Therefore $f^{*} ε_{M}^{r} ≅ ε_{N}^{r}$ as smooth vector bundles over $N$ , by the isomorphism $Φ$ , which was defined by an explicit formula using only $f$ and hence is canonical. Equivalently, the constant sections $s_{j} (p) = (p, e_{j})$ of $ε_{M}^{r}$ pull back to the sections $f^{*} s_{j} (q) = (q, (f (q), e_{j}))$ of $f^{*} ε_{M}^{r}$ , none of whose values is the zero vector; these pullbacks are smooth because in the global pullback chart of step 2.1 the section $f^{*} s_{j}$ reads as the constant map $q \mapsto e_{j}$ , and they form a global frame of $f^{*} ε_{M}^{r}$ that $Φ$ converts into the standard frame of $N \times R^{r}$ . No selection of local trivializations, complements or representatives has been made: the chart of [F2] used above is the single global chart of the product bundle, so the argument uses no choice principle. The case $r = 0$ gives the zero bundle over $N$ , and the empty or disconnected base is covered verbatim.

CorollaryStatement: AI-generatedProof: AI-generatedprecheck passjudge pass (gpt-6.1-sol)Open item page →

The pullback of the Euclidean tangent bundle is canonically trivial

Statement

Assume countable choice ${A C}_{ω}$ . Let $f : N \to R^{r}$ be a smooth map. Under the standard-coordinate identification $T R^{r} ≅ ε_{R^{r}}^{r}$ given by the induced tangent chart of the identity chart (Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure, The induced tangent bundle chart), the pullback tangent bundle is canonically trivial: $f^{*} T R^{r} ≅ f^{*} ε_{R^{r}}^{r} ≅ ε_{N}^{r},$ where the second isomorphism pulls back the constant frame by The pullback of a trivial smooth vector bundle is canonically trivial.

Facts & Assumptions

Given: A smooth map $f : N \to R^{r}$ and countable choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Under ${A C}_{ω}$ the tangent bundle $T R^{r}$ carries its canonical smooth $2 r$ -manifold structure, for which the induced tangent-bundle charts form a smooth atlas; for a chart $(U, x)$ the induced chart is $v \mapsto (x (p), v^{1}, \dots, v^{r})$ with $v = \sum_{i} v^{i} \partial_{x^{i}} ∣_{p}$ (Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure, The induced tangent bundle chart).

[F2]

The pullback of the trivial rank- $r$ bundle $ε_{R^{r}}^{r}$ along $f$ is canonically isomorphic to $ε_{N}^{r}$ (The pullback of a trivial smooth vector bundle is canonically trivial).

Proof

1.1F1

The identity $id : R^{r} \to R^{r}$ is a smooth chart whose domain is all of $R^{r}$ , so by [F1] its induced tangent-bundle chart $\overset{}{id} : T R^{r} \to R^{r} \times R^{r}$ , $v \mapsto (p, v^{1}, \dots, v^{r})$ , is a diffeomorphism onto $R^{r} \times R^{r}$ ; it is linear on every fibre. Hence it is a smooth bundle isomorphism $T R^{r} ⟶ ε_{R^{r}}^{r} = R^{r} \times R^{r},$ the standard-coordinate identification. It is determined by the identity chart alone, so no choice is made in exhibiting it.

2.1F1F2step 1.1∎

Pulling this identification back along $f$ gives a smooth bundle isomorphism $f^{*} T R^{r} ≅ f^{*} ε_{R^{r}}^{r}$ over $N$ , and [F2] gives a canonical isomorphism $f^{*} ε_{R^{r}}^{r} ≅ ε_{N}^{r}$ carrying the pulled-back constant frame to the standard frame. Composing, $f^{*} T R^{r} ≅ f^{*} ε_{R^{r}}^{r} ≅ ε_{N}^{r}$ canonically. The only choice principle used is ${A C}_{ω}$ , inherited through [F1]; the pullback comparison of [F2] is choice-free, and the empty or disconnected case of $N$ is included since all maps displayed are evaluated fibrewise.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Positive intermediate cohomology of compactified Euclidean space vanishes

Statement

Assume AC. For $N \geq 2$ , $0 < q < N$ , and $R = Z$ or $F_{2}$ , the one-point compactification $(R^{N})^{+}$ of $R^{N}$ (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ ) is homeomorphic to $S^{N}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ ) and has $H^{q} ((R^{N})^{+}; R) = 0$ (Topological universal coefficient short exact sequence for cohomology, The Axiom of Choice).

Facts & Assumptions

Given: integers $N \geq 2$ and $0 < q < N$ , a coefficient ring $R = Z$ or $F_{2}$ , and the one-point compactification $(R^{N})^{+} = (R^{N})^{*}$ with added point $\infty$ (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ ).

[F1]

The topology $T^{*}$ of $X^{*} = X \cup {\infty}$ consists of the open sets of $X$ together with the sets $X^{*} ∖ C$ for $C \subseteq X$ closed in $X$ and a compact subset of $X$ ; $X^{*}$ is compact, $X$ is an open subspace with its original topology, and $X^{*}$ is Hausdorff exactly when $X$ is locally compact and Hausdorff (The one-point (Alexandroff) compactification $X^{*} = X \cup {\infty}$ , whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$ , $X^{*}$ is compact and contains $X$ as an open subspace; $X$ is dense in $X^{*}$ exactly when $X$ is not compact; and $X^{*}$ is Hausdorff exactly when $X$ is locally compact and Hausdorff).

[F2]

$R^{N}$ is locally compact, its topology is the Euclidean metric topology and the product topology, and $S^{N} = {y \in R^{N + 1} : ∥ y ∥_{2} = 1}$ carries the subspace topology, hence is metrizable and Hausdorff ( $R^{n}$ is locally compact and $σ$ -compact, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space, Euclidean spheres and closed balls as subspaces of $R^{n}$ , Every subspace of a metrizable space is metrizable and every subspace of a first countable space is first countable, the metric case being the subspace metric already identified with the subspace topology, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[F3]

A map into a finite product is continuous exactly when its components are; sums, products and quotients of continuous real-valued functions with nowhere-vanishing denominator are continuous, and $x \mapsto x_{i}$ and $x \mapsto ∥ x ∥_{2}^{2}$ are continuous on $R^{N}$ (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, Sums, products, absolute values, finite maxima and minima, and quotients of continuous real-valued maps on a topological space are continuous where defined, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it, Continuity of a map of topological spaces at a point and globally).

[F4]

A closed bounded subset of $R^{N}$ is compact (A subset of $R^{n}$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology).

[F5]

A continuous bijection from a compact space onto a Hausdorff space is a homeomorphism, and a continuous map into a subspace that corestricts to the image is continuous as a map onto that image (A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

[F6]

Under AC, for every space $X$ , abelian group $G$ and $n \geq 0$ the evaluation sequence is natural and exact (Topological universal coefficient short exact sequence for cohomology, The Axiom of Choice).

[F7]

For the sphere $S^{N}$ with $N \geq 2$ one has $H_{0} (S^{N}; Z) ≅ Z$ and $H_{j} (S^{N}; Z) = 0$ for $0 < j < N$ and for $j > N$ ; this is the case $N \geq 1$ of the reduced-homology computation of Homology of spheres. For every abelian group $G$ the explicit free resolution $0 \to 0 \to Z \overset{1}{\to} Z \to 0$ computes ${Ext}_{Z}^{1} (Z, G) = 0$ by the definition of $Ext$ (the same resolution computation performed for $G = Z$ in Integral cohomology detects adjacent homology torsion), and ${Ext}_{Z}^{1} (0, G) = {Hom}_{Z} (0, G) = 0$ holds trivially (Ext via a projective resolution of the first variable).

[F8]

A homeomorphism induces isomorphisms on singular cohomology, contravariantly in the map (Singular cohomology is contravariantly functorial).

Proof

1.1F3algebra

Define $h : (R^{N})^{*} \to S^{N}$ by $h (x) = (\frac{2 x}{1 + ∥ x ∥_{2}^{2}}, \frac{∥ x ∥_{2}^{2} - 1}{∥ x ∥_{2}^{2} + 1}), h (\infty) = (0, \dots, 0, 1),$ writing $∥ x ∥_{2}^{2}$ for the squared Euclidean norm. For $x \in R^{N}$ the identity $(2 ∥ x ∥_{2})^{2} + (∥ x ∥_{2}^{2} - 1)^{2} = (∥ x ∥_{2}^{2} + 1)^{2}$ shows $∥ h (x) ∥_{2} = 1$ , so $h (x) \in S^{N}$ ; and $h (x) \neq (0, \dots, 0, 1)$ since $∥ x ∥_{2}^{2} - 1 = ∥ x ∥_{2}^{2} + 1$ is impossible. For $y = (y_{0}, \dots, y_{N}) \in S^{N}$ with $y_{N} \neq 1$ put $ψ (y) : = (y_{0}, \dots, y_{N - 1}) / (1 - y_{N}) \in R^{N}$ . If $r = ∥ ψ (y) ∥_{2}$ then $r^{2} = (1 - y_{N}^{2}) / (1 - y_{N})^{2} = (1 + y_{N}) / (1 - y_{N})$ , so $1 + r^{2} = 2 / (1 - y_{N})$ and $r^{2} - 1 = 2 y_{N} / (1 - y_{N})$ ; hence $h (ψ (y)) = y$ . Conversely, if $h (x) = (z, t)$ then $1 - t = 2 / (1 + ∥ x ∥_{2}^{2})$ and $ψ (h (x)) = 2 x / (1 + ∥ x ∥_{2}^{2}) \cdot (1 + ∥ x ∥_{2}^{2}) / 2 = x$ . So $h$ is a bijection with inverse $ψ$ on $S^{N} ∖ {(0, \dots, 0, 1)}$ and $\infty \mapsto (0, \dots, 0, 1)$ .

2.1F1F2F3F4F5step 1.1

The map $h$ is continuous. On $R^{N}$ its components $x \mapsto 2 x_{i} / (1 + ∥ x ∥_{2}^{2})$ and $x \mapsto (∥ x ∥_{2}^{2} - 1) / (∥ x ∥_{2}^{2} + 1)$ are quotients with denominator $1 + ∥ x ∥_{2}^{2} \geq 1$ never zero, hence continuous by [F3]; the components are continuous, so $h ∣_{R^{N}}$ is continuous into $R^{N + 1}$ and, since its image lies in $S^{N}$ , continuous into the subspace $S^{N}$ by [F5]. At $\infty$ , let $V \subseteq S^{N}$ be open with $(0, \dots, 0, 1) \in V$ . The subspace topology on $S^{N}$ is the metric topology of the maximum metric, so there is $r > 0$ such that every $y \in S^{N}$ with $∥ y - (0, \dots, 0, 1) ∥_{\infty} < r$ lies in $V$ ; choose $M \geq 1$ with $1 / M < r / 2$ and put $C : = {x \in R^{N} : ∥ x ∥_{2} \leq M}$ , which is closed and bounded, hence compact by [F4]. For $x \notin C$ one has $∥ x ∥_{2} > M \geq 1$ , so $∣ 2 x_{i} / (1 + ∥ x ∥_{2}^{2}) ∣ \leq 2 ∥ x ∥_{2} / ∥ x ∥_{2}^{2} = 2 / ∥ x ∥_{2} < 2 / M < r$ and $∣ (∥ x ∥_{2}^{2} - 1) / (∥ x ∥_{2}^{2} + 1) - 1 ∣ = 2 / (∥ x ∥_{2}^{2} + 1) \leq 2 / ∥ x ∥_{2}^{2} < 2 / M < r$ ; hence $h (x) \in V$ . Therefore $W : = (R^{N})^{*} ∖ C$ is open in $(R^{N})^{*}$ by [F1], contains $\infty$ , and satisfies $h (W) \subseteq V$ .

3.1F1F2F5F8step 1.1step 2.1

Hence $h$ is a homeomorphism. Indeed $R^{N}$ is locally compact and Hausdorff by [F2], so $(R^{N})^{*}$ is compact and Hausdorff by [F1], while $S^{N}$ is Hausdorff by [F2]; a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism by [F5]. Consequently $h^{*} : H^{q} (S^{N}; R) \to H^{q} ((R^{N})^{*}; R)$ is an isomorphism for every $q$ by [F8].

4.1F6F7step 3.1algebra

By the homeomorphism of step 3.1 it suffices to compute $H^{q} (S^{N}; R)$ . By [F7], $H_{q} (S^{N}; Z) = 0$ for $0 < q < N$ , while $H_{0} (S^{N}; Z) ≅ Z$ . Apply the universal coefficient sequence of [F6] in degree $q$ with $X = S^{N}$ and $G = R$ : $0 ⟶ {Ext}_{Z}^{1} (H_{q - 1} (S^{N}; Z), R) ⟶ H^{q} (S^{N}; R) ⟶ {Hom}_{Z} (H_{q} (S^{N}; Z), R) ⟶ 0.$ If $q \geq 2$ then $1 \leq q - 1 < q \leq N - 1$ , so both $H_{q - 1}$ and $H_{q}$ vanish and both outer terms are zero by [F7]. If $q = 1$ (so $N \geq 2$ ) then $H_{1} (S^{N}; Z) = 0$ and $H_{0} (S^{N}; Z) ≅ Z$ , so the right term is ${Hom}_{Z} (0, R) = 0$ and the left term is ${Ext}_{Z}^{1} (Z, R) = 0$ by [F7]. In both cases exactness forces $H^{q} (S^{N}; R) = 0$ , for $R = Z$ and for $R = F_{2}$ alike.

5.1F6step 3.1step 4.1∎

Combining steps 3.1 and 4.1, $H^{q} ((R^{N})^{*}; R) ≅ H^{q} (S^{N}; R) = 0$ for $0 < q < N$ , which is the claimed vanishing for the one-point compactification. The case $N = 2$ , $q = 1$ , both coefficient rings, and both orders of the two outer terms in the universal coefficient sequence are covered by the case distinction of step 4.1; the empty coefficient ring and negative $q$ are excluded by the hypotheses, and no further choice beyond AC, used through [F6] and [F7], enters.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

An embedding into Euclidean space gives a rank-(n-m) stable normal inverse

Statement

Assume countable choice ${A C}_{ω}$ . Let $i : M^{m} ↪ R^{n}$ be a smooth embedding of a closed smooth $m$ -manifold with $n > m$ , and let $ν_{i} = i^{*} T R^{n} / d i (T M)$ be its normal quotient, identified with the orthogonal complement of $d i (T M)$ by a Euclidean metric (Normal and conormal bundles of an embedded submanifold, Assuming countable choice, an ambient metric identifies the two normal bundles). Then $ν_{i}$ is a smooth real bundle of rank $n - m$ , and the orthogonal splitting together with the canonical trivialization $i^{*} T R^{n} ≅ ε^{n}$ gives a smooth bundle isomorphism $φ : T M \oplus ν_{i} \to ε^{n}$ . Hence $(ν_{i}, φ)$ is a rank- $(n - m)$ stable normal inverse of $M$ in the sense of Stable normal inverse of the tangent bundle. Consequently every closed smooth $m$ -manifold admits a stable normal inverse: apply Every smooth manifold embeds in some finite-dimensional Euclidean space to obtain an embedding into some $R^{N}$ . The countable-choice hypothesis is exactly the one inherited from the metric and tubular identifications of the published embedding normal-bundle definition; no further choice is made.

Facts & Assumptions

Given: A smooth embedding $i : M^{m} ↪ R^{n}$ of a closed smooth $m$ -manifold with $n > m$ , and countable choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The normal-bundle set of the embedded submanifold $M \subseteq R^{n}$ is the fibrewise quotient $ν = ∐_{p \in M} T_{p} R^{n} / T_{p} M$ , with the smooth vector-bundle structure supplied for such quotients; the defining quotient of the pullback, $ν_{i} = i^{*} T R^{n} / d i (T M)$ , is the same bundle under the canonical identification of $i^{*} T R^{n}$ with $T R^{n} ∣_{M}$ (Normal and conormal bundles of an embedded submanifold).

[F2]

Assume ${A C}_{ω}$ ; for an embedded submanifold and a Riemannian metric $g$ on the ambient manifold, the quotient map restricts to a smooth bundle isomorphism $T S^{⊥} \to T M ∣_{S} / T S$ ; for $S = M \subseteq R^{n}$ with the Euclidean metric this identifies $ν_{i}$ with the orthogonal complement $d i (T M)^{⊥}$ (Assuming countable choice, an ambient metric identifies the two normal bundles).

[F3]

For a compact (in particular closed) smooth $M$ and a smooth embedding $i : M ↪ R^{N}$ with $N \geq m$ , the published normal-bundle definition gives a smooth real bundle $ν_{i}$ of rank $N - m$ with $T M \oplus ν_{i} ≅ ε^{N}$ ; the only choice used is the inherited ${A C}_{ω}$ of the metric and tubular identifications (Stable normal bundle of a compact smooth manifold, the rank of the quotient).

[F4]

Under ${A C}_{ω}$ the identity chart of $R^{n}$ is a global smooth chart, so its induced tangent-bundle chart trivializes the Euclidean tangent bundle, $T R^{n} ≅ ε_{R^{n}}^{n}$ ; pulling this trivialization back along the smooth map $i$ and applying the choice-free product-pullback lemma gives the canonical trivialization $i^{*} T R^{n} ≅ i^{*} ε_{R^{n}}^{n} ≅ ε^{n}$ (Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure, The induced tangent bundle chart, The pullback of a trivial smooth vector bundle is canonically trivial).

[F5]

Under ${A C}_{ω}$ every smooth $m$ -manifold embeds smoothly into some finite-dimensional Euclidean space (Every smooth manifold embeds in some finite-dimensional Euclidean space).

[F6]

A stable normal inverse of $M$ is a pair $(ν, φ)$ with $ν \to M$ a smooth real bundle of finite rank $k$ and $φ : T M \oplus ν \to ε^{m + k}$ a smooth bundle isomorphism; a rank- $k$ stable normal inverse is one with $rank ν = k$ (Stable normal inverse of the tangent bundle, Smooth vector bundles, rank, fibres, and trivial bundles).

Proof

1.1F1F3F6

Regard $i$ as an embedding of $M$ as an embedded submanifold $i (M) \subseteq R^{n}$ and let $ν_{i}$ be its normal quotient as in [F1]. By [F3] the quotient carries a smooth real vector-bundle structure of rank $n - m$ ; the rank is the difference of the ranks of the ambient tangent bundle of $R^{n}$ and of $d i (T M)$ , computed fibrewise, and equals $n - m$ because $d i$ is fibrewise injective.

1.2F2F3F4

By [F2] the Euclidean metric identifies the quotient $ν_{i}$ with the orthogonal complement $d i (T M)^{⊥}$ , which is a smooth subbundle of $i^{*} T R^{n}$ ; the orthogonal decomposition of the Euclidean bundle gives $i^{*} T R^{n} = d i (T M) \oplus d i (T M)^{⊥} ≅ T M \oplus ν_{i}$ , where the first summand is identified with $T M$ through the isomorphism $d i : T M \to d i (T M)$ . Composing this isomorphism with the canonical trivialization $i^{*} T R^{n} ≅ ε^{n}$ of [F4], which exists because the identity chart of $R^{n}$ trivializes $T R^{n}$ and the product-pullback lemma trivializes its pullback, gives a smooth bundle isomorphism $φ : T M \oplus ν_{i} ⟶ ε^{n} .$

2.1F2F4F6step 1.1step 1.2

Since $ν_{i}$ has rank $n - m$ by step 1.1 and $φ$ is a smooth bundle isomorphism onto $ε^{n} = ε^{m + (n - m)}$ , the pair $(ν_{i}, φ)$ is a rank- $(n - m)$ stable normal inverse of $M$ in the sense of [F6].

3.1F3F4F5F6step 1.1step 1.2step 2.1∎

For existence, let $M$ be any closed smooth $m$ -manifold. By [F5] there is a smooth embedding $j : M ↪ R^{N}$ into some finite-dimensional Euclidean space; the construction above applies to $j$ provided $N > m$ . If $N \leq m$ for the particular embedding produced, compose with the inclusion $R^{N} ↪ R^{N + 1} ↪ \dots ↪ R^{m + 1}$ (each an embedding of a linear subspace as a closed subset, hence a smooth embedding with $d i$ injective) to obtain an embedding into some $R^{n}$ with $n > m$ ; replacing the ambient metric by the standard Euclidean one leaves the argument unchanged. Applying steps 1.1–2.1 to that embedding produces a stable normal inverse of $M$ . The only choice principle used is the ${A C}_{ω}$ inherited from [F2] and [F5]; the trivialization [F4] is canonical, and no embedding, metric or complement is selected beyond the given ones.

LemmaStatement: Literature-sourcedProof: Literature-sourcedprecheck passOpen item page →

An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle

Statement

Assume ${A C}_{ω}$ . Let $f : M^{m} ↬ R^{n}$ be a smooth immersion of a closed smooth $m$ -manifold with $n > m$ , so that $(f, d f)$ is a formal immersion and $ν_{f} = f^{*} T R^{n} / d f (T M)$ is its normal bundle of rank $n - m$ (Formal immersion between smooth manifolds, Normal bundle of a formal immersion). Then the splitting of the tangent-normal sequence of Formal immersion gives the tangent normal-bundle identity gives a smooth bundle isomorphism $T M \oplus ν_{f} \to f^{*} T R^{n}$ , which under the canonical trivialization $f^{*} T R^{n} ≅ ε^{n}$ becomes a smooth isomorphism $φ : T M \oplus ν_{f} \to ε^{n}$ . Hence $(ν_{f}, φ)$ is a rank- $(n - m)$ stable normal inverse of $M$ in the sense of Stable normal inverse of the tangent bundle: an immersion of codimension $n - m$ supplies an actual rank- $(n - m)$ representative of the inverse normal class, not merely a stable one. The choice hypothesis is inherited from the bundle-metric splitting in the normal-bundle construction.

Facts & Assumptions

Given: A smooth immersion $f : M^{m} ↬ R^{n}$ of a closed smooth $m$ -manifold with $n > m$ , and countable choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A smooth map $f$ is an immersion exactly when $(f, d f)$ is a formal immersion; a formal immersion from $M^{m}$ to $N^{n}$ is a smooth map together with a fibrewise injective smooth bundle map over it, and, when $M$ is nonempty, necessarily $m \leq n$ (Immersions, submersions, and constant-rank maps, Formal immersion between smooth manifolds).

[F2]

For a formal immersion $(f, F)$ from $M^{m}$ to $N^{n}$ , the normal bundle $ν_{F} = f^{*} T N / F (T M)$ is a smooth quotient bundle of rank $n - m$ over $M$ when $m \leq n$ ; if $M = \emptyset$ and $m > n$ , it is the empty rank-zero bundle; it is intrinsic up to canonical isomorphism, and for $(f, F) = (f, d f)$ with $f$ a genuine immersion it is the normal bundle of the immersion (Normal bundle of a formal immersion).

[F3]

For every formal immersion $(f, F)$ the quotient map fits into the short exact sequence of smooth bundles $0 \to T M \overset{F}{\to} f^{*} T N \to ν_{F} \to 0$ over $M$ , which splits: a smooth complement of $F (T M)$ restricts to an isomorphism onto $ν_{F}$ and yields a smooth bundle isomorphism $T M \oplus ν_{F} \to f^{*} T N$ restricting to $F$ on the tangent summand. If a smooth bundle metric on $f^{*} T N$ is chosen, the orthogonal complement $F (T M)^{⊥}$ is a canonical complement for that metric (Formal immersion gives the tangent normal-bundle identity). The splitting in the general case uses the metric and inherits ${A C}_{ω}$ .

[F4]

Under ${A C}_{ω}$ the identity chart of $R^{n}$ is a global smooth chart, so its induced tangent-bundle chart trivializes the Euclidean tangent bundle, $T R^{n} ≅ ε_{R^{n}}^{n}$ ; pulling this trivialization back along $f$ and applying the choice-free product-pullback lemma gives the canonical trivialization $f^{*} T R^{n} ≅ f^{*} ε_{R^{n}}^{n} ≅ ε^{n}$ (Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure, The induced tangent bundle chart, The pullback of a trivial smooth vector bundle is canonically trivial).

[F5]

A stable normal inverse of $M$ is a pair $(ν, φ)$ with $ν \to M$ a smooth real vector bundle of finite rank $k$ and $φ : T M \oplus ν \to ε^{m + k}$ a smooth bundle isomorphism; a rank- $k$ stable normal inverse is one with $rank ν = k$ (Stable normal inverse of the tangent bundle).

Proof

1.1F1F2

Since $f$ is an immersion, $(f, d f)$ is a formal immersion from $M^{m}$ to $R^{n}$ by [F1]; in particular $d f : T M \to f^{*} T R^{n}$ is fibrewise injective. By [F2] its normal bundle $ν_{f} = f^{*} T R^{n} / d f (T M)$ is a smooth real vector bundle over $M$ of rank $n - m$ .

1.2F2F3

By [F3] the quotient sequence $0 \to T M \overset{d f}{\to} f^{*} T R^{n} \to ν_{f} \to 0$ splits: the tangent-normal sequence admits a smooth bundle isomorphism $s : T M \oplus ν_{f} ⟶ f^{*} T R^{n}$ restricting to $d f$ on the tangent summand. The splitting uses a smooth bundle metric on the pullback bundle (whose existence is the countable-choice input of that lemma), so this step uses exactly the hypothesis ${A C}_{ω}$ and no more.

2.1F3F4step 1.2

Compose the splitting $s$ with the canonical trivialization $t : f^{*} T R^{n} \to ε^{n}$ supplied by [F4], which exists because the identity chart of $R^{n}$ trivializes $T R^{n}$ and the product-pullback lemma trivializes its pullback along $f$ : $φ : = t \circ s : T M \oplus ν_{f} ⟶ ε^{n}$ is a smooth bundle isomorphism, the composite of two smooth bundle isomorphisms.

3.1F3F4F5step 1.1step 1.2step 2.1∎

By step 1.1 the bundle $ν_{f}$ has rank $n - m$ and by step 2.1 the isomorphism $φ$ maps $T M \oplus ν_{f}$ onto $ε^{n} = ε^{m + (n - m)}$ ; so $(ν_{f}, φ)$ is a rank- $(n - m)$ stable normal inverse of $M$ in the sense of [F5]. Thus an immersion of codimension $n - m$ provides an actual rank- $(n - m)$ inverse bundle, not merely a stable one; nothing beyond this rank and the isomorphism is asserted about $ν_{f}$ . The countable-choice hypothesis is the one inherited from the metric splitting of [F3] and from the canonical trivialization [F4]; no bundle metric, complement or frame is chosen in addition to those data.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Smale-Hirsch makes rank reduction sufficient for Euclidean immersion in positive codimension

Statement

Assume countable choice ${A C}_{ω}$ . Let $M^{m}$ be a closed smooth $m$ -manifold, let $k \geq 1$ , and let $(ν, φ)$ be a rank- $k$ stable normal inverse of $M$ , so that $φ : T M \oplus ν \to ε^{m + k}$ (Stable normal inverse of the tangent bundle). Then there exists an immersion $M ↬ R^{m + k}$ . Together with An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle this says that for closed $M$ and $k \geq 1$ the existence of an immersion $M ↬ R^{m + k}$ is equivalent to the existence of a rank- $k$ stable normal inverse; that equivalence is the precise sense in which the normal problem is complete for immersions on this page. No classification of regular homotopy classes and no statement about the normal bundle of the particular immersion produced is asserted.

Facts & Assumptions

Given: A closed smooth $m$ -manifold $M$ , an integer $k \geq 1$ , a rank- $k$ stable normal inverse $(ν, φ)$ with $φ : T M \oplus ν \to ε^{m + k}$ , and ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A formal immersion from $M$ to $N$ is a pair $(f, F)$ with $f : M \to N$ smooth and $F : T M \to T N$ a smooth bundle map over $f$ that is injective on every fibre; a smooth map $f$ is an immersion exactly when $(f, d f)$ is a formal immersion. The spaces $Imm (M, N)$ and $FImm (M, N)$ carry the weak compact-open topologies, and the derivative map $D : f \mapsto (f, d f)$ maps the former into the latter (Formal immersion between smooth manifolds, Space of immersions and space of formal immersions).

[F2]

Assume ${A C}_{ω}$ ; for smooth boundaryless $M^{m}, N^{n}$ with $m < n$ (positive codimension), the derivative map $D : Imm (M, N) \to FImm (M, N)$ is a weak homotopy equivalence (The Smale–Hirsch immersion theorem).

[F3]

A weak homotopy equivalence induces a bijection on path-component sets $π_{0}$ (Weak homotopy equivalence).

[F4]

The constant map $c : M \to R^{m + k}$ , $x \mapsto 0$ , pulls the trivial bundle back to $ε^{m + k}$ : $c^{*} T R^{m + k} ≅ c^{*} ε^{m + k} ≅ ε^{m + k}$ canonically, by the product-pullback lemma and the standard-coordinate trivialization of the Euclidean tangent bundle (Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure, The induced tangent bundle chart, The pullback of a trivial smooth vector bundle is canonically trivial).

[F5]

Conversely, if $M$ is closed and $f : M ↬ R^{m + k}$ is a smooth immersion with $m + k > m$ , its normal bundle is a rank- $k$ stable normal inverse of $M$ (An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle).

Proof

1.1F1F4

Define $F : = φ ∣_{T M} : T M \to ε^{m + k}$ as the restriction of the bundle isomorphism $φ$ to the first summand; this is a smooth bundle map over $M$ , injective on every fibre. By [F4], it determines a smooth fibrewise injective map $T M \to c^{*} T R^{m + k}$ over ${id}_{M}$ . Composing with the canonical map $c^{*} T R^{m + k} \to T R^{m + k}$ gives a smooth bundle map $\tilde{F}$ over $c$ , so $(c, \tilde{F})$ is a formal immersion by [F1]. Thus $FImm (M, R^{m + k})$ is nonempty.

2.1F1F2F3step 1.1

By [F2] with $N = R^{m + k}$ (boundaryless, positive codimension $k \geq 1$ ) the derivative map $D$ is a weak homotopy equivalence; by [F3] it induces a bijection on path components. Since $FImm (M, R^{m + k})$ is nonempty by step 1.1 and $π_{0} (D)$ is surjective, the target's empty-or-non-empty status matches the source's, so $Imm (M, R^{m + k})$ is nonempty: there exists a smooth immersion $M ↬ R^{m + k}$ .

3.1F2F5step 1.1step 2.1∎

Conversely, every smooth immersion $M ↬ R^{m + k}$ of the closed $M$ has a rank- $k$ normal bundle which is a rank- $k$ stable normal inverse by [F5], under the same ${A C}_{ω}$ . Therefore for closed $M$ and $k \geq 1$ the existence of an immersion into $R^{m + k}$ is equivalent to the existence of a rank- $k$ stable normal inverse: reduction of the structure problem to the normal bundle is sufficient as well as necessary, which is the completeness statement of the design. The argument selects no immersion canonically (it only proves nonemptiness of a space), asserts nothing about the regular homotopy class of the immersion produced, and makes no claim about its normal bundle; the only choice used is the ${A C}_{ω}$ assumed by the Smale-Hirsch theorem and by the normal-bundle splitting.

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

The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class

Statement

Assume AC. Let $M$ be a closed smooth $m$ -manifold, let $(ν, φ)$ be a stable normal inverse of $M$ with $φ : T M \oplus ν \to ε^{N}$ , and let $w (E) = \sum_{i} w_{i} (E)$ denote the total Stiefel-Whitney class in the ring $H^{*} (M; F_{2})$ (Stiefel–Whitney classes from the projective-bundle relation, Singular cohomology ring). Then $w (T M) w (ν) = 1 in H^{*} (M; F_{2}) .$ Hence $w (ν)$ is the unique two-sided inverse of $w (T M)$ , and the classes $w_{i} (ν)$ depend only on $M$ , not on the chosen stable normal inverse $(ν, φ)$ . Equivalently the total normal class is $w (ν) = w (T M)^{- 1} = : \overset{ˉ}{w} (M)$ , the Whitney-duality form of the normal Stiefel-Whitney class.

Facts & Assumptions

Given: A closed smooth $m$ -manifold $M$ , a stable normal inverse $(ν, φ)$ with $φ : T M \oplus ν \to ε^{N}$ a smooth bundle isomorphism, and AC (Stable normal inverse of the tangent bundle, The Axiom of Choice).

[F1]

Stiefel-Whitney classes are defined for numerable real bundles over a paracompact Hausdorff CGWH base of CW homotopy type, with $w_{0} = 1$ , $w_{i} = 0$ for $i > rank$ , and total class $w (E) = \sum_{i} w_{i} (E) \in H^{*} (B; F_{2})$ (Stiefel–Whitney classes from the projective-bundle relation, Singular cohomology ring).

[F2]

The Whitney sum formula $w (E \oplus F) = w (E) w (F)$ holds for numerable bundles over such a base, and adjoining a trivial summand does not change the classes: $w (E \oplus ε^{r}) = w (E)$ , so $w (ε^{r}) = 1$ (Whitney sum formula for Stiefel–Whitney classes).

[F3]

The classes depend only on the isomorphism class of the bundle (Naturality of Stiefel–Whitney classes).

[F4]

A closed smooth manifold is a paracompact Hausdorff CGWH space of CW homotopy type, and every smooth bundle over it, in particular $T M$ , $ν$ and the trivial bundle, is numerable (Smooth manifolds have CW homotopy type); this puts $M$ and these bundles in the scope of [F1]–[F3]. AC is the hypothesis of those suppliers.

[F5]

Singular cohomology is graded commutative; over $F_{2}$ the signs are $1$ , so $H^{*} (M; F_{2})$ is a commutative unital ring (Singular cohomology ring, Singular cohomology is graded commutative). If $u v = 1 = u w$ , then $v = v (u w) = (v u) w = w$ , so inverses are unique.

Proof

1.1F2F3F4F5

By [F3] the isomorphism $φ$ gives $w (T M \oplus ν) = w (ε^{N})$ ; by [F2], $w (T M \oplus ν) = w (T M) w (ν)$ and $w (ε^{N}) = 1$ , the latter because $ε^{N}$ is trivial and adjoining trivial summands does not change the classes. Hence $w (T M) w (ν) = 1 in H^{*} (M; F_{2}) .$ The computation happens in the unital ring of [F5], and the bundles involved are numerable over the closed smooth manifold $M$ by [F4], so the cited Whitney and naturality theorems apply.

2.1F2F5step 1.1∎

Equation $w (T M) w (ν) = 1$ exhibits $w (ν)$ as a two-sided inverse of $w (T M)$ , and by [F5] the inverse of a unit is unique; in particular if $(ν_{0}, φ_{0})$ and $(ν_{1}, φ_{1})$ are two stable normal inverses then $w (ν_{0}) = w (T M)^{- 1} = w (ν_{1})$ , so each $w_{i} (ν)$ depends only on $M$ . This justifies the notation $\overset{ˉ}{w} (M) : = w (T M)^{- 1} = w (ν)$ . The argument uses no property of $φ$ beyond its being a bundle isomorphism, no orientation of $M$ , and only the choice assumed in AC, inherited through the AT suppliers [F1]–[F3].

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

The normal Pontryagin class is the rational inverse of the tangent Pontryagin class

Statement

Assume AC. Let $M$ be a closed connected smooth $m$ -manifold, let $(ν, φ)$ be a stable normal inverse of $M$ with $φ : T M \oplus ν \to ε^{N}$ , and let $p (E) = \sum_{i} p_{i} (E)$ denote the total Pontryagin class in the AT normalization $p_{i} (E) = (- 1)^{i} c_{2 i} (E_{C})$ (Pontryagin classes by complexification). Then $p (T M) p (ν) = 1 in H^{*} (M; Q) .$ Hence $p (ν)$ is the unique inverse of $p (T M)$ in $H^{*} (M; Q)$ , and the classes $p_{i} (ν)$ depend only on $M$ , not on the chosen stable normal inverse. All assertions in this item are over $Q$ . The integral Pontryagin Whitney product holds only modulo elements of order two, so no integral multiplicativity is asserted here. Orientation of $M$ is not needed, since $p_{i}$ is defined for every real bundle by complexification; connectedness is exactly the hypothesis of the AT Whitney-product item.

Facts & Assumptions

Given: A closed connected smooth $m$ -manifold $M$ , a stable normal inverse $(ν, φ)$ with $φ : T M \oplus ν \to ε^{N}$ an isomorphism, and AC (Stable normal inverse of the tangent bundle, The Axiom of Choice).

[F1]

Pontryagin classes are defined by $p_{i} (E) = (- 1)^{i} c_{2 i} (E_{C}) \in H^{4 i} (B; Z)$ , with $p_{0} = 1$ , $p_{i} = 0$ whenever $2 i > rank E$ , and total class $p (E) = \sum_{i} p_{i} (E)$ ; the complexification is determined by $E$ up to canonical isomorphism, so the classes depend only on the isomorphism class of $E$ , and no orientation of $E$ is used (Pontryagin classes by complexification).

[F2]

Over $Z [1 / 2]$ , or any coefficient ring in which $2$ is invertible, the Whitney product $p (E \oplus F) = p (E) p (F)$ holds for numerable real bundles over a path-connected paracompact Hausdorff CW base (Pontryagin Whitney product away from two); over such a ring the two-torsion cross terms drop out. Integrally this multiplicativity is not asserted.

[F3]

For a numerable real bundle over a nonempty path-connected paracompact Hausdorff CW base one has the stability $p_{i} (E \oplus ε^{r}) = p_{i} (E)$ and the rank vanishing $p_{i} (E) = 0$ when $2 i > rank E$ (Naturality, stability, and mod-two reduction of Pontryagin classes, Pontryagin classes by complexification).

[F4]

Under AC, $M$ is paracompact Hausdorff CGWH of CW homotopy type and its smooth bundles are numerable (Smooth manifolds have CW homotopy type). A continuous image of compact $M$ in a CW complex lies in a finite subcomplex (The image of a compact space lies in a finite CW subcomplex). Homotopic maps from a paracompact Hausdorff base give isomorphic pullback bundles (Homotopy invariance of vector-bundle pullback). Chern naturality permits a CW-type source and a CW target (Naturality, normalization, and Whitney sum for Chern classes); since complexification commutes with pullback in bundle charts, the same pullback formula holds for $p_{i} = (- 1)^{i} c_{2 i}$ . These facts allow transfer of [F2] and [F3] from a finite CW model to $M$ , as shown below.

[F5]

Singular cohomology is a graded-commutative unital ring (Singular cohomology ring, Singular cohomology is graded commutative). Pontryagin classes have degrees divisible by four, so their total classes commute. If $u v = 1 = u w$ for commuting classes, then $v = v (u w) = (v u) w = w$ .

Proof

1.1F1F2F3F4construct

If $M$ is empty, its cohomology is the zero ring and the identity and inverse assertions hold with $0 = 1$ . Otherwise choose a CW complex $K$ and maps $a : M \to K$ , $b : K \to M$ with $b a ≃ {id}_{M}$ by [F4]. The compact image $a (M)$ lies in a finite subcomplex; take its connected component $L$ containing $a (M)$ , and restrict $b$ to $L$ . A finite CW complex is compact Hausdorff (it is a finite union of characteristic-disk images), hence paracompact (every open cover has a finite, thus locally finite, subcover); its connected components are path connected. For each bundle $E$ among $T M, ν, T M \oplus ν, ε^{N}$ , put $E_{L} = b^{*} E$ . Pullback numerations make these bundles numerable. Then $a^{*} E_{L} ≅ E$ by homotopy invariance, and Chern naturality in [F4] gives $p_{i} (E) = a^{*} p_{i} (E_{L})$ . Pullback preserves sums and trivial bundles in their charts. Thus the Whitney identity and stability of [F2] and [F3], applied on $L$ and pulled back along $a$ , hold for the given bundles on $M$ . Finally $φ$ gives $p (T M \oplus ν) = p (ε^{N})$ by isomorphism invariance [F1].

2.1F2F3step 1.1

Over $Q$ , where $2$ is invertible, [F2] gives $p (T M \oplus ν) = p (T M) p (ν)$ in $H^{*} (M; Q)$ . For the trivial bundle, apply the stability clause of [F3] with the rank-zero bundle $0_{M}$ : $p_{i} (ε^{N}) = p_{i} (0_{M} \oplus ε^{N}) = p_{i} (0_{M})$ for every $i$ , and the rank convention $p_{i} (0_{M}) = 0$ for $i \geq 1$ together with $p_{0} = 1$ gives $p (ε^{N}) = 1$ . Combining with step 1.1, $p (T M) p (ν) = p (T M \oplus ν) = p (ε^{N}) = 1 in H^{*} (M; Q) .$

3.1F1F2F5step 2.1∎

By [F5] the inverse of the unit $p (T M)$ in the unital ring $H^{*} (M; Q)$ is unique, so $p (ν) = p (T M)^{- 1}$ and the classes $p_{i} (ν)$ do not depend on the chosen stable normal inverse $(ν, φ)$ . This is the rational form of the Pontryagin normal-class identity; no integral multiplicativity is obtained, because [F2] carries the two-torsion caveat and the odd Chern cross terms of a complexified real bundle can be nonzero two-torsion by [F1]. For a disconnected closed $M$ the same computation applies to each component, and the identity then holds componentwise. AC is inherited through [F2] and [F3]; no orientation of $M$ is used anywhere.

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

Normal Stiefel-Whitney and Pontryagin classes of a closed manifold

Definition

Assume AC. Let $M$ be a closed smooth $m$ -manifold and choose a stable normal inverse $(ν, φ)$ of $M$ , which exists by An embedding into Euclidean space gives a rank-(n-m) stable normal inverse ( $A C$ implies the required countable choice by AC implies DC implies countable choice, and the embedding lemma applies to closed $M$ ). Define the total normal Stiefel-Whitney class and the total normal Pontryagin class by $\overset{ˉ}{w} (M) : = w (ν) \in H^{*} (M; F_{2}), \overset{ˉ}{p} (M) : = p (ν) \in H^{*} (M; Q),$ with components ${\overset{ˉ}{w}}_{i} (M) \in H^{i} (M; F_{2})$ and ${\overset{ˉ}{p}}_{i} (M) \in H^{4 i} (M; Q)$ (Stiefel–Whitney classes from the projective-bundle relation, Pontryagin classes by complexification, Singular cohomology ring). By The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class and The normal Pontryagin class is the rational inverse of the tangent Pontryagin class the classes ${\overset{ˉ}{w}}_{i} (M)$ and, for connected $M$ , the classes ${\overset{ˉ}{p}}_{i} (M)$ are independent of the chosen inverse, so the definition is well posed; equivalently $\overset{ˉ}{w} (M) = w (T M)^{- 1}$ and $\overset{ˉ}{p} (M) = p (T M)^{- 1}$ are the explicit inverses realized by any inverse bundle. For a disconnected closed $M$ the Pontryagin definition is applied componentwise. These are the classes also called the normal, dual, or (in Skopenkov's terminology) Stiefel-Whitney and Pontryagin classes of the manifold; they are the classes read by the immersion and embedding tests of this page.

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

High normal Stiefel-Whitney classes obstruct low-codimension immersions

Statement

Assume AC. Let $M$ be a closed smooth $m$ -manifold and let $k \geq 1$ . If there is an index $i > k$ with ${\overset{ˉ}{w}}_{i} (M) \neq 0$ (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold), then $M$ does not immerse in $R^{m + k}$ ; equivalently every immersion of $M$ into a Euclidean space has codimension at least $i$ , so fewer than $i$ dimensions of codimension are impossible. In particular, if ${\overset{ˉ}{w}}_{i} (M) \neq 0$ for some $i \geq 1$ , then $M$ does not immerse in $R^{m + i - 1}$ . This is the standard normal Stiefel-Whitney non-immersion test.

Facts & Assumptions

Given: A closed smooth $m$ -manifold $M$ , an integer $k \geq 1$ , an index $i > k$ with ${\overset{ˉ}{w}}_{i} (M) \neq 0$ , and AC (The Axiom of Choice).

[F1]

AC implies the countable choice ${A C}_{ω}$ used by the normal-bundle splitting (AC implies DC implies countable choice).

[F2]

For a smooth immersion $f : M ↬ R^{n}$ of a closed smooth $m$ -manifold with $n > m$ , its normal bundle $ν_{f}$ of rank $n - m$ is a rank- $(n - m)$ stable normal inverse of $M$ , and $T M \oplus ν_{f} ≅ ε^{n}$ (An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle).

[F3]

The normal classes are $\overset{ˉ}{w} (M) = w (T M)^{- 1} = w (ν)$ for any stable normal inverse $(ν, φ)$ of $M$ , and $w_{i} (ν) = {\overset{ˉ}{w}}_{i} (M)$ for every $i$ (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold, The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class).

[F4]

Stiefel-Whitney classes of a bundle vanish above its rank: if $rank E = r$ then $w_{j} (E) = 0$ for $j > r$ (Stiefel–Whitney classes from the projective-bundle relation).

[F5]

A smooth map with invertible differential is a local diffeomorphism (The smooth inverse function theorem on manifolds).

[F6]

Smooth maps are continuous, continuous images of compact topological spaces are compact, and compact subsets of Hausdorff spaces are closed (Smooth maps are continuous, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones).

Proof

1.1F1F2

Suppose, for contradiction, that there is a smooth immersion $f : M ↬ R^{m + k}$ . By [F1] the hypothesis ${A C}_{ω}$ of [F2] holds with $n = m + k > m$ , so the normal bundle $ν_{f}$ of the immersion is a rank- $k$ stable normal inverse of $M$ .

2.1F3F4step 1.1

By [F3] applied to the rank- $k$ inverse $ν_{f}$ , the class ${\overset{ˉ}{w}}_{i} (M) = w_{i} (ν_{f})$ for the given index $i > k$ . But [F4] gives $w_{i} (ν_{f}) = 0$ because $rank ν_{f} = k < i$ , a contradiction with ${\overset{ˉ}{w}}_{i} (M) \neq 0$ . Hence no immersion into $R^{m + k}$ exists.

3.1F5F6step 2.1construct∎

The final sentence uses $k = i - 1$ . For $i \geq 2$ this satisfies $k \geq 1$ , so step 2.1 excludes immersion in $R^{m + i - 1}$ . For $i = 1$ , a nonzero class ${\overset{ˉ}{w}}_{1} (M)$ forces $M$ to be nonempty and $m \geq 1$ . No nonempty compact positive-dimensional manifold immerses in $R^{m}$ : an equal-dimensional immersion is a local diffeomorphism by [F5], so its image is open; the image is also compact by [F6], hence closed in Hausdorff $R^{m}$ . Euclidean space is connected because any two points are joined by their straight segment, and it is noncompact for $m \geq 1$ because the cover by balls of integer radius has no finite subcover. Thus connectedness of $R^{m}$ makes a nonempty open-and-closed image all of $R^{m}$ , contradicting its noncompactness. Thus the codimension-zero instance is excluded too. Negative codimension is impossible because the derivative could not be injective. Consequently every immersion has codimension at least $i$ under the nonzero-class hypothesis. No converse or classification is asserted.

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High normal Pontryagin classes obstruct low-codimension immersions

Statement

Assume AC. Let $M$ be a closed connected smooth $m$ -manifold and let $k \geq 1$ . If ${\overset{ˉ}{p}}_{i} (M) \neq 0$ in $H^{4 i} (M; Q)$ for some $i \geq 1$ with $2 i > k$ (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold), then $M$ does not immerse in $R^{m + k}$ ; equivalently, if ${\overset{ˉ}{p}}_{i} (M) \neq 0$ then every immersion of $M$ has codimension at least $2 i$ , so $M$ does not immerse in $R^{m + 2 i - 1}$ . All assertions are over $Q$ : the integral Whitney product for Pontryagin classes carries a two-torsion correction and the integral form of this test is not asserted.

Facts & Assumptions

Given: A closed connected smooth $m$ -manifold $M$ , an integer $k \geq 1$ , an index $i \geq 1$ with $2 i > k$ and ${\overset{ˉ}{p}}_{i} (M) \neq 0$ in $H^{4 i} (M; Q)$ , and AC (The Axiom of Choice).

[F1]

AC implies the countable choice used by the normal-bundle splitting (AC implies DC implies countable choice).

[F2]

For a smooth immersion $f : M ↬ R^{n}$ of a closed smooth $m$ -manifold with $n > m$ , the normal bundle $ν_{f}$ of rank $n - m$ is a rank- $(n - m)$ stable normal inverse of $M$ , with $T M \oplus ν_{f} ≅ ε^{n}$ (An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle).

[F3]

The normal Pontryagin class is $\overset{ˉ}{p} (M) = p (T M)^{- 1} = p (ν)$ over $Q$ for any stable normal inverse $(ν, φ)$ of $M$ , so $p_{i} (ν) = {\overset{ˉ}{p}}_{i} (M)$ for every $i$ (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold, The normal Pontryagin class is the rational inverse of the tangent Pontryagin class).

[F4]

Pontryagin classes vanish above the rank: if $rank E = r$ then $p_{i} (E) = 0$ for $2 i > r$ (Naturality, stability, and mod-two reduction of Pontryagin classes, Pontryagin classes by complexification).

Proof

1.1F1F2

Suppose for contradiction that $f : M ↬ R^{m + k}$ is a smooth immersion. Since $k \geq 1$ and $M$ is closed, [F2] applies with $n = m + k > m$ under the countable choice granted by [F1]; therefore the normal bundle $ν_{f}$ , of rank $k$ , is a rank- $k$ stable normal inverse of $M$ .

2.1F3F4step 1.1

By [F3] the class $p_{i} (ν_{f}) = {\overset{ˉ}{p}}_{i} (M)$ for the given index $i$ , and $2 i > k = rank ν_{f}$ , so [F4] gives $p_{i} (ν_{f}) = 0$ , contradicting ${\overset{ˉ}{p}}_{i} (M) \neq 0$ . Hence no immersion of $M$ into $R^{m + k}$ exists.

3.1F3step 2.1∎

The final sentence is the case $k = 2 i - 1$ : if ${\overset{ˉ}{p}}_{i} (M) \neq 0$ then no immersion into $R^{m + 2 i - 1}$ exists, because $2 i > 2 i - 1$ . Equivalently every immersion of $M$ has codimension at least $2 i$ under this hypothesis. The argument is stated over $Q$ because the rank vanishing and the inverse identity for $p_{i}$ are used rationally; the integral form is not asserted, since the integral Whitney product for Pontryagin classes only holds modulo two-torsion. No orientation of $M$ is needed beyond the rational normalization.

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Finite normal push-off count for an even-dimensional Euclidean immersion

Statement

Assume AC. Let $f : M^{2 s} ↬ R^{4 s}$ , $s \geq 1$ , be a self-transverse immersion of a closed oriented manifold with no triple points. Orient its orthogonal normal bundle $ν_{f}$ by $T M \oplus ν_{f}$ in the standard ambient orientation. Its unordered double points are finite and have ordering-independent signs $ε (r)$ . There exists a transverse smooth section $σ$ of $ν_{f}$ , nonzero at all double-point preimages. For every sufficiently small $t > 0$ , the map $f_{t} (x) = f (x) + t σ (x)$ is transverse to $f$ , and $I (f, f_{t}) = ⟨ e (ν_{f}), [M] ⟩ + 2 \sum_{r \in Σ (f)} ε (r) .$

Facts & Assumptions

Given: A self-transverse smooth immersion $f : M^{2 s} ↬ R^{4 s}$ of a closed oriented manifold, $s \geq 1$ , with no triple points; $M$ is closed, so compact without boundary. Write $m = 2 s$ , so $4 s = 2 m$ .

[F1]

The double point locus is $Δ_{2} (f) = {(x, y) \in M \times M ∖ Δ_{M} : f (x) = f (y)}$ and the double point set is $Σ (f) = f ({pr}_{1} Δ_{2} (f))$ ; $f$ is self-transverse when $f \times f$ restricted to $M \times M ∖ Δ_{M}$ is transverse to the diagonal $Δ_{R^{4 s}}$ , equivalently $d f_{x} (T_{x} M) + d f_{y} (T_{y} M) = T_{f (x)} R^{4 s}$ at every ordered pair with $f (x) = f (y)$ , $x \neq y$ . For $m$ even a double point carries an ordering-independent local sign $ε (r) = ε (x, y) = ε (y, x)$ given by the local oriented intersection sign of the two oriented branch disks (Self-transverse immersions and the double point locus, The local oriented intersection sign).

[F2]

A smooth immersion restricts to an embedding on a neighbourhood of every point (Every immersion is locally an embedding).

[F3]

The normal bundle $ν_{f} = f^{*} T R^{4 s} / d f (T M)$ is a smooth real bundle of rank $2 s$ over $M$ , represented by the orthogonal complement $d f (T M)^{⊥}$ for the Euclidean metric, and $T M \oplus ν_{f} ≅ f^{*} T R^{4 s} ≅ ε^{4 s}$ (Normal bundle of a formal immersion, Formal immersion gives the tangent normal-bundle identity). The orientation of $M$ together with the standard orientation of $R^{4 s}$ orients $ν_{f}$ , and $e (ν_{f}) \in H^{2 s} (M; Z)$ is its Euler class.

[F4]

For a closed oriented $X$ and an oriented closed embedded $Z \subseteq M$ of complementary dimension, the oriented intersection number $I (g, Z) = \sum_{p \in g^{- 1} (Z)} ε (p)$ is defined for transverse $g$ and depends only on the homotopy class of $g$ ; for maps $f : X \to M$ , $g : Z \to M$ with compact boundaryless oriented sources of complementary dimension, $I (f, g)$ is the sum of the local signs over $X \times_{M} Z$ , which is finite (The oriented intersection number, The local oriented intersection sign). Here both source manifolds are $M$ , and the ambient manifold for $I (f, f_{t})$ is $R^{4 s}$ . Equivalently this count is the intersection of the product map with the diagonal in $R^{4 s} \times R^{4 s}$ ; the factor-interchange sign is positive because $m$ is even.

[F5]

A transverse section of an oriented rank- $2 s$ real bundle over a closed oriented $2 s$ -manifold has finitely many zeros, and the sum of the local signs of the zero locus equals the evaluation $⟨ e (ν_{f}), [M] ⟩$ of the Euler class (The zero locus of a transverse section represents the Euler dual). Under ${A C}_{ω}$ , parametric transversality for a smooth family $F : M \times S \to N$ transverse to an embedded submanifold $Z$ gives that the parameters whose slice fails to be transverse to $Z$ form a null set (Parametric transversality). Smooth partitions of unity exist on smooth manifolds (Smooth partitions of unity exist on manifolds), and the inverse function theorem holds on manifolds (The smooth inverse function theorem on manifolds). Under countable choice every smooth vector bundle has a smooth bundle metric (Every smooth vector bundle admits a smooth bundle metric), including $T M$ . AC supplies the countable choice used by transversality and these metrics (The Axiom of Choice).

Proof

1.1F1F2F5

The double point locus is finite. Choose for each $p \in M$ a neighbourhood $U_{p}$ on which $f$ is injective, as in [F2]; by compactness finitely many $U_{p_{1}}, \dots, U_{p_{N}}$ cover $M$ , and the open set $⋃_{i} (U_{p_{i}} \times U_{p_{i}})$ is a neighbourhood of the diagonal on which $f (x) = f (y)$ forces $x = y$ . Hence $Δ_{2} (f)$ is contained in the compact set $K = (M \times M) ∖ ⋃_{i} (U_{p_{i}} \times U_{p_{i}})$ and is closed in $M \times M$ . Self-transversality [F1] makes the derivative of $(x, y) \mapsto f (x) - f (y)$ an isomorphism at every point of $Δ_{2} (f)$ : the two source tangent dimensions add to $4 s$ and their image planes span $R^{4 s}$ . The inverse function theorem [F5] therefore isolates each ordered coincidence. Thus $Δ_{2} (f)$ is compact and discrete, hence finite (its singleton cover has a finite subcover). Unordered branch pairs are the elements ${x, y}$ of $D (f)$ obtained by quotienting $Δ_{2} (f)$ by interchange. Since no image has three preimages, $D (f) \to Σ (f)$ is bijective. Thus $Σ (f)$ is finite and each image has the ordering-independent sign of its unique branch pair.

1.2F1F3F5

There is a smooth section $σ$ of $ν_{f}$ that is transverse to the zero section and nonzero at all double-point preimages. Take finitely many trivializing charts of the rank- $2 s$ bundle $ν_{f}$ with relatively compact domains whose interiors cover $M$ , and for each chart and each frame vector a smooth bump function supported in the chart, chosen so that the bump interiors still cover $M$ ; by [F5] such a finite family with nonnegative bumps exists. Extending bump times frame vector by zero gives finitely many smooth global sections $s_{1}, \dots, s_{L}$ of $ν_{f}$ that span $ν_{f} (x)$ at every $x \in M$ . The family $F : M \times R^{L} \to ν_{f}$ , $F (x, b) = \sum_{j} b_{j} s_{j} (x)$ , valued in the total space $ν_{f}$ of the bundle, is smooth and has surjective vertical derivative at every point (the $s_{j} (x)$ span), hence is transverse to the zero section $Z \subseteq ν_{f}$ ; by parametric transversality [F5] the parameters $b$ for which $F_{b}$ is not transverse to $Z$ form a null set of $R^{L}$ . For each of the finitely many points $x \in {pr}_{1} Δ_{2} (f)$ the condition $F_{b} (x) = 0$ is the kernel of the surjective linear map $b \mapsto \sum_{j} b_{j} s_{j} (x)$ , a proper linear subspace whence null; their finite union is null, and we choose $b$ outside it and outside the transversality-exceptional set. Then $σ : = F_{b}$ is a smooth section of $ν_{f}$ , transverse to the zero section; since transversality with a rank- $2 s$ zero section in the $2 s$ -manifold $M$ makes the zero locus discrete and $M$ is compact, $σ^{- 1} (0)$ is finite, and by construction $σ$ is nonzero at every double-point preimage.

2.1F3F5step 1.2

For $t > 0$ put $f_{t} (x) = f (x) + t σ (x)$ , using the embedding $R^{4 s}$ as a vector space, and fix an auxiliary smooth bundle metric on $T M$ (which exists on the closed manifold). For small $t$ , $f_{t}$ is an immersion: unit tangent vectors form a compact set, $d f$ is fibrewise injective with $\min ∥ d f (v) ∥ = c > 0$ over unit vectors, while $d σ$ is bounded over the compact $M$ in any fixed finite family of charts, so $∥ t d σ (v) ∥ < c \leq ∥ d f (v) ∥$ for small $t$ and every unit $v$ , whence $d f_{t} = d f + t d σ$ is fibrewise injective. Near the diagonal, intersections of $f$ and $f_{t}$ are the zeros of $σ$ : define the normal-addition map $E : ν_{f} \to R^{4 s}$ by $E (x, v) = f (x) + v$ , whose derivative at each $(x, 0)$ is the isomorphism $u \oplus w \mapsto d f_{x} (u) + w$ from $T_{x} M \oplus ν_{f} (x)$ onto $R^{4 s}$ ; by the inverse function theorem [F5], for each $p \in M$ there is a bundle neighbourhood over a source neighbourhood $U_{p}$ on which $E$ is injective. Choose finitely many smaller source neighbourhoods $V_{i}$ with $\overline{V_{i}} \subset U_{i}$ that cover $M$ , and choose a uniform normal radius small enough that $E$ is injective on all vectors of that radius based in each $V_{i}$ . The open set $⋃_{i} V_{i} \times V_{i}$ contains the source diagonal. If $(x, y)$ is in this set and $f (x) = f_{t} (y) = f (y) + t σ (y)$ , then for uniformly small $t$ both $E (x, 0)$ and $E (y, t σ (y))$ lie in the same injectivity neighbourhood. Their equality gives $(x, 0) = (y, t σ (y))$ , so $x = y$ and $σ (y) = 0$ . Conversely each zero $y$ of $σ$ gives the intersection $(y, y)$ . At a zero, subtracting the $d f$ columns from the $d f_{t}$ columns leaves the normal block $t d^{v e r t} σ$ , so the intersection determinant has the sign of $\det (d^{v e r t} σ)$ because $t^{m} > 0$ ; this sign, computed in the tangent-first orientation of $T M \oplus ν_{f}$ , is exactly the local contribution of the zero locus of the transverse oriented section $σ$ ; by [F5] the total contribution of these near-diagonal intersections is $⟨ e (ν_{f}), [M] ⟩$ , the Koszul sign of the Euler-duality being $+ 1$ because the zero locus is $0$ -dimensional (equivalently because the rank $2 s$ is even).

2.2F1F4F5step 1.1

Near each ordered pair $(x, y)$ with $f (x) = f (y)$ , $x \neq y$ , and for small $t$ there is exactly one nearby intersection of $f$ with $f_{t}$ , with the original local sign. Indeed $G (x^{'}, y^{'}, t) = f (x^{'}) - f (y^{'}) - t σ (y^{'})$ satisfies $G (x, y, 0) = 0$ , and its derivative in $(x^{'}, y^{'})$ at $(x, y, 0)$ is $(u, w) \mapsto d f_{x} u - d f_{y} w$ ; by self-transversality [F1] the sum $d f_{x} (T_{x} M) + d f_{y} (T_{y} M)$ is all of $R^{4 s}$ and the dimensions add up, so this derivative is an isomorphism. The inverse function theorem with the parameter $t$ (apply the ordinary theorem to $(x^{'}, y^{'}, t) \mapsto (G (x^{'}, y^{'}, t), t)$ ) gives for small $t$ a unique solution $(x^{'} (t), y^{'} (t))$ near $(x, y)$ , and the local sign at $t = 0$ is $ε (x, y)$ ; the sign is locally constant because a nonzero determinant of the ordered derivative pair persists for small $t$ . Applying the same statement to the reversed ordered pair $(y, x)$ produces one further nearby ordered intersection with sign $ε (y, x)$ , and by [F1] for even $m = 2 s$ the two signs agree, $ε (y, x) = ε (x, y) = ε (r)$ , where $r$ is the double point. Hence each unordered double point contributes $2 ε (r)$ .

3.1F1F3F4F5step 2.1step 2.2∎

Choose disjoint sufficiently small neighbourhoods of the diagonal and of the finitely many ordered double pairs, and small $t$ so that all the local statements apply. On the compact complement $K^{'}$ of their union in $M \times M$ the distance $∥ f (x) - f (y) ∥$ has a positive minimum $c^{'} > 0$ because $f (x) \neq f (y)$ for $(x, y) \in K^{'}$ ; since $σ$ is bounded, $∥ f_{t} (y) - f (y) ∥ = t ∥ σ (y) ∥ < c^{'}$ for small $t$ , so no further ordered pair satisfies $f (x) = f_{t} (y)$ . Therefore the ordered intersections of $f$ and $f_{t}$ are exactly the near-diagonal zeros of $σ$ (counted with the signs of step 2.1) together with the two nearby ordered intersections contributed by each double point (step 2.2), and all of them are transverse because the derivatives computed in steps 2.1 and 2.2 are isomorphisms. Hence, summing, $I (f, f_{t}) = ⟨ e (ν_{f}), [M] ⟩ + 2 \sum_{r \in Σ (f)} ε (r)$ for every sufficiently small $t > 0$ . The empty cases are included: if $f$ is an embedding then $Σ (f) = \emptyset$ and the equation reads $I (f, f_{t}) = ⟨ e (ν_{f}), [M] ⟩$ , while a zero-free transverse section gives Euler number zero (characteristic-class vanishing, not an embedding conclusion). AC is used through the parametric transversality, the normal-bundle metric and the Euler dual-class supplier; the signs, the finite sums and the local inverse-function computations are choice-free.

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Top normal classes vanish for Euclidean embeddings

Statement

Assume AC. Let $j : M^{m} ↪ R^{m + k}$ be a smooth embedding of a closed smooth manifold, with $m \geq 1$ and $k \geq 1$ , and let $ν_{j}$ be its rank- $k$ normal bundle. Then ${\overset{ˉ}{w}}_{k} (T M) = w_{k} (ν_{j}) = 0 in H^{k} (M; F_{2}) .$ If $ν_{j}$ is integrally oriented, then also $e (ν_{j}) = 0 in H^{k} (M; Z) .$ Together with rank vanishing, ${\overset{ˉ}{w}}_{i} (T M) = 0$ for every $i \geq k$ . Thus a nonzero top normal class obstructs embedding in codimension $k$ , even though rank alone permits that class for an immersion.

Facts & Assumptions

Given: A smooth embedding $j : M^{m} ↪ R^{m + k}$ of a closed smooth manifold with $m \geq 1$ , $k \geq 1$ , its normal bundle $ν_{j}$ of rank $k$ , and AC (The Axiom of Choice).

[F1]

Under ${A C}_{ω}$ (hence under AC) the embedding gives a rank- $k$ stable normal inverse $(ν_{j}, φ)$ of $M$ , where $ν_{j} = j^{*} T R^{m + k} / d j (T M)$ is the normal quotient (An embedding into Euclidean space gives a rank-(n-m) stable normal inverse); if $ν_{j}$ is integrally oriented it is an oriented rank- $k$ bundle in the sense of the Thom interface.

[F2]

Let $Φ$ be a compatible tubular chart for $ν_{j}$ with a metric $h$ and radius $ρ$ (existing under countable choice, hence under AC); its collapse is a based continuous map $c : (R^{m + k})^{+} \to {Th}_{h} (ν_{j})$ sending the tube to the disk-sphere quotient model and every other point to the basepoint. The zero section $s : M \to D (ν_{j})$ followed by the quotient map $q : D (ν_{j}) \to {Th}_{h} (ν_{j})$ is the based zero section $z = q \circ s$ , and on $M$ the collapse satisfies $c \circ j = z$ (Pontryagin–Thom collapse with specified normal data).

[F3]

For either coefficient ring $R = F_{2}$ or $R = Z$ with a supplied integral orientation, the normalized Thom class $u \in H^{k} (D (ν_{j}), S (ν_{j}); R)$ corresponds under the quotient identification to a class $u \in H^{k} ({Th}_{h} (ν_{j}); R)$ of positive degree, and the quotient-map pullback $q^{*} (u)$ is the relative-to-absolute image of the relative Thom class; hence $z^{*} (u) = s^{*} q^{*} (u) = e (ν_{j})$ , the Euler class of Euler class by zero-section pullback of the Thom class, which by The mod-two Euler class is the top Stiefel–Whitney class equals $w_{k} (ν_{j})$ for $R = F_{2}$ , while for $R = Z$ it is the oriented Euler class (Thom class and Thom isomorphism: the AT interface, Euler class by zero-section pullback of the Thom class, The mod-two Euler class is the top Stiefel–Whitney class).

[F4]

Since $m \geq 1$ and $k \geq 1$ , one has $0 < k < N$ for $N = m + k$ , so the one-point compactification has $H^{k} ((R^{N})^{+}; R) = 0$ for $R = Z$ and $R = F_{2}$ (Positive intermediate cohomology of compactified Euclidean space vanishes). Cohomology is contravariantly functorial, so $f^{*} (0) = 0$ and $(c \circ j)^{*} = j^{*} c^{*}$ (Singular cohomology is contravariantly functorial).

[F5]

A closed smooth manifold is a paracompact Hausdorff CGWH space of CW homotopy type over which every smooth bundle, in particular $ν_{j}$ , is numerable (Smooth manifolds have CW homotopy type); this places $M$ and $ν_{j}$ in the scope of the Thom interface and of the mod-two Euler class theorem.

[F6]

For the stable normal inverse $(ν_{j}, φ)$ one has $w (ν_{j}) = w (T M)^{- 1}$ , so $w_{k} (ν_{j}) = {\overset{ˉ}{w}}_{k} (T M)$ ; also $w_{i} (ν_{j}) = 0$ for $i > k$ because $ν_{j}$ has rank $k$ (The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class, Stiefel–Whitney classes from the projective-bundle relation).

Proof

1.1F1F2F3F4F5

Let $u$ denote the normalized Thom class of $ν_{j}$ in degree $k$ , in the relative model, and also its image in $H^{k} ({Th}_{h} (ν_{j}); R)$ under the quotient identification of [F3]; the degree $k$ is positive and the Thom space is based, so the reduced and ordinary descriptions agree in this degree. Pulling back along the collapse gives $c^{*} u \in H^{k} ((R^{m + k})^{+}; R)$ , which is zero by [F4] since $0 < k < m + k$ ; the interface and the mod-two Euler theorem apply to $ν_{j}$ because the closed manifold is a suitable base by [F5].

2.1F2F3F4step 1.1

By [F2] the collapse satisfies $c \circ j = z$ ; functoriality [F4] gives $z^{*} u = (c \circ j)^{*} u = j^{*} c^{*} u = j^{*} 0 = 0 in H^{k} (M; R) .$ The composite $z = q \circ s$ is the zero section followed by the quotient map, so by [F3] $z^{*} u = s^{*} q^{*} u = e (ν_{j})$ . Hence $e (ν_{j}) = 0$ in $H^{k} (M; R)$ for the chosen coefficient ring: for $R = Z$ this is the oriented Euler class, and for $R = F_{2}$ the mod-two Euler class.

3.1F1F3F6step 2.1

For $R = F_{2}$ , the published identification $e_{2} (ν_{j}) = w_{k} (ν_{j})$ of [F3] gives $w_{k} (ν_{j}) = 0$ in $H^{k} (M; F_{2})$ , and by [F1] and [F6] this class equals ${\overset{ˉ}{w}}_{k} (T M)$ . For $R = Z$ with $ν_{j}$ integrally oriented, step 2.1 gives the oriented Euler vanishing $e (ν_{j}) = 0$ .

4.1F3F6step 3.1∎

Finally, for every $i > k$ the rank convention gives $w_{i} (ν_{j}) = 0$ because $rank ν_{j} = k$ , so by [F6] ${\overset{ˉ}{w}}_{i} (T M) = w_{i} (ν_{j}) = 0$ for all $i > k$ ; together with the degree- $k$ vanishing of step 3.1 this gives ${\overset{ˉ}{w}}_{i} (T M) = 0$ for every $i \geq k$ . In particular a nonzero top normal class in degree $k$ is an obstruction to embedding in codimension exactly $k$ , in contrast with the rank test, which only sees the classes of degree $> k$ for immersions. The argument uses AC through the Thom interface and the embedding normal-bundle lemma; orientation is needed only for the integral Euler clause, and no Poincaré duality or ambient fundamental class is used.

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The Euler class of an oriented even-rank normal bundle controls self-intersection

Statement

Assume AC. Let $X$ be an oriented smooth $2 m$ -manifold and $A^{m} \subseteq X$ a closed oriented embedded submanifold whose normal bundle $ν_{A}$ is oriented compatibly, with $m$ even. Then the self-intersection number of The self-intersection number of a complementary-dimensional oriented submanifold satisfies $A \cdot A = ⟨ e (ν_{A}), [A] ⟩,$ and $e (ν_{A})$ is Poincare dual to the zero locus of any smooth section of $ν_{A}$ transverse to the zero section (The self-intersection number is the Euler number of the normal bundle, The zero locus of a transverse section represents the Euler dual). For odd rank the Euler class of an oriented bundle is two-torsion (The Euler class of an oriented odd-rank bundle is two-torsion), so its integral evaluation on a closed oriented odd-dimensional source is zero, whereas a nonzero pairing $⟨ e (ν_{A}), [A] ⟩$ in even rank makes $e (ν_{A})$ non-torsion and prevents $ν_{A}$ from admitting a nowhere-zero section (A nowhere-zero section forces the Euler class to vanish). Over $F_{2}$ , with no orientation hypotheses, the same identities hold with $w_{m} (ν_{A}) = e_{2} (ν_{A})$ (The mod two self-intersection is the top Stiefel-Whitney evaluation, The mod-two Euler class is the top Stiefel–Whitney class).

Give $R^{2 m}$ its standard orientation. Let $M^{m}$ be a closed oriented smooth manifold with $m$ even, and let $f : M ↬ R^{2 m}$ be a smooth immersion. Its normal bundle $ν_{f}$ , the quotient in Normal bundle of a formal immersion, is oriented by the orientations of $M$ and $R^{2 m}$ . For $m > 0$ , An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle gives $T M \oplus ν_{f} ≅ ε^{2 m}$ ; for $m = 0$ both summands are zero bundles, so the identity holds directly. Hence $⟨ e (ν_{f}), [M] ⟩$ is the signed zero count of any transverse section of $ν_{f}$ ; it is invariant under regular homotopy (Regular homotopy of immersions) and vanishes if $ν_{f}$ admits a nowhere-zero section. If $m = 2 s \geq 2$ , $f$ is self-transverse and has no triple points, set $D (f) = \sum_{r \in Σ (f)} ε (r)$ with the ordering-independent branch signs. Then $⟨ e (ν_{f}), [M] ⟩ + 2 D (f) = 0.$ In particular the Euler number is even, $D (f) = - ⟨ e (ν_{f}), [M] ⟩ / 2$ , and $# Σ (f) m o d 2 = - ⟨ e (ν_{f}), [M] ⟩ / 2 m o d 2$ . A nonzero $D (f)$ obstructs regular homotopy to an embedding; in particular odd double-point parity does. No sufficiency criterion or odd-dimensional Whitney identification is asserted.

Facts & Assumptions

Given: An oriented smooth $2 m$ -manifold $X$ and a closed oriented embedded $A^{m} \subseteq X$ with $m$ even and compatibly oriented normal bundle (first paragraph); a closed oriented $M^{m}$ with $m$ even and a smooth immersion $f : M ↬ R^{2 m}$ (second paragraph); AC.

[F1]

For a closed oriented embedded $A^{m} \subseteq X$ of an oriented boundaryless $X$ , the self-intersection number equals the evaluation of the Euler class of the normal bundle, $A \cdot A = ⟨ e (ν_{A}), [A] ⟩$ , and the Euler class is Poincaré dual to the zero locus of a transverse section (The self-intersection number of a complementary-dimensional oriented submanifold, The self-intersection number is the Euler number of the normal bundle, The zero locus of a transverse section represents the Euler dual); mod two the same holds without orientations with $w_{m} (ν_{A}) = e_{2} (ν_{A})$ (The mod two self-intersection is the top Stiefel-Whitney evaluation, The mod-two Euler class is the top Stiefel–Whitney class).

[F2]

The Euler class of an oriented odd-rank bundle is two-torsion, so it pairs to zero against the fundamental class of a closed oriented source of odd dimension; a nowhere-zero section forces the Euler class to vanish, with no converse (The Euler class of an oriented odd-rank bundle is two-torsion, A nowhere-zero section forces the Euler class to vanish, Euler class by zero-section pullback of the Thom class).

[F3]

For $m > 0$ and an immersion $f : M^{m} ↬ R^{2 m}$ of a closed oriented $M$ with the normal bundle $ν_{f} = f^{*} T R^{2 m} / d f (T M)$ oriented by the orientations of $M$ and $R^{2 m}$ , one has $T M \oplus ν_{f} ≅ ε^{2 m}$ , so $ν_{f}$ is a rank- $m$ stable normal inverse of $M$ (Normal bundle of a formal immersion, An immersion into R^n gives a rank-(n-m) representative of the stable normal bundle, Stable normal inverse of the tangent bundle). For $m = 0$ , both $T M$ and the normal quotient are zero bundles, so the same bundle identity holds directly.

[F4]

A regular homotopy of immersions is a smooth family $H : M \times I \to R^{2 m}$ restricting to an immersion at each time (Regular homotopy of immersions). For a smooth map $H$ the vertical derivative $d_{M} H : {pr}_{M}^{*} T M \to H^{*} T R^{2 m}$ is a smooth bundle map over $M \times I$ (Vector bundle maps over a smooth base map); if it is fibrewise injective, its image is a subbundle and the quotient is a smooth vector bundle (Constant-rank kernels and images of bundle maps over one base are subbundles, Quotient vector bundles by a subbundle, A vector bundle quotient by a subbundle is a smooth vector bundle). The local minor formulas for the image and quotient bundles also apply in boundary charts. Under AC the base $M \times I$ is paracompact Hausdorff CGWH of CW type and these bundles are numerable (Smooth manifolds have CW homotopy type). The Euler class is natural for bundle maps over a base map preserving orientation, so a bundle isomorphism over the identity gives equal Euler classes (Naturality, orientation sign, and Whitney product for Euler classes); homotopic maps induce the same map in singular cohomology (Homotopic maps induce equal maps in singular cohomology).

[F5]

The diagonal $Δ_{R^{2 m}} \subseteq R^{2 m} \times R^{2 m}$ is an embedded submanifold of complementary dimension to the product $M \times M$ (The diagonal is an embedded submanifold); oriented intersection numbers of complementary-dimensional maps are homotopy invariant (The oriented intersection number is homotopy invariant) and satisfy the factor-interchange formula with sign $(- 1)^{m^{2}} = 1$ for $m$ even (Intersection number under factor interchange).

[F6]

For a self-transverse immersion $f : M^{2 s} ↬ R^{4 s}$ with $s \geq 1$ of a closed oriented manifold with no triple points, a transverse section $σ$ and all sufficiently small $t > 0$ satisfy $I (f, f_{t}) = ⟨ e (ν_{f}), [M] ⟩ + 2 \sum_{r \in Σ (f)} ε (r)$ (Finite normal push-off count for an even-dimensional Euclidean immersion); if $m \geq 1$ and $f$ is an embedding into $R^{2 m}$ then $e (ν_{f}) = 0$ and ${\overset{ˉ}{w}}_{m} (T M) = 0$ (Top normal classes vanish for Euclidean embeddings).

[F7]

AC implies the countable choice used throughout the normal-bundle and Euler-class suppliers (AC implies DC implies countable choice, The Axiom of Choice).

Proof

1.1F1F2

First consider the embedded case. By [F1] one has $A \cdot A = ⟨ e (ν_{A}), [A] ⟩$ for the closed oriented embedded $A^{m} \subseteq X$ , computed with the tangent-first sign convention of the self-intersection number, and $e (ν_{A})$ is Poincaré dual to the zero locus of any section transverse to the zero section: the Koszul sign of that duality is $(- 1)^{m (n - m)} = (- 1)^{m^{2}} = + 1$ because $m$ is even. [F2] gives that an odd-rank oriented bundle has two-torsion Euler class, so on a closed oriented odd-dimensional source its evaluation vanishes, whereas no such vanishing is available in even rank, and $⟨ e (ν_{A}), [A] ⟩ \neq 0$ excludes a nowhere-zero section of $ν_{A}$ by [F2]. Over $F_{2}$ no orientation is needed and the identity reads $A \cdot_{2} A = ⟨ w_{m} (ν_{A}), [A] ⟩_{2}$ by [F1].

1.2F1F2F3F7

Now let $f : M^{m} ↬ R^{2 m}$ with $m$ even. The normal bundle $ν_{f}$ of the formal immersion $(f, d f)$ is oriented by the orientation of $M$ and the standard orientation of $R^{2 m}$ , and by [F3] the orthogonal decomposition gives $T M \oplus ν_{f} ≅ ε^{2 m}$ ; hence $e (ν_{f})$ is defined integrally and, by the section form of [F1] applied to the immersion normal bundle, $⟨ e (ν_{f}), [M] ⟩$ is the signed count of the zeros of any transverse section of $ν_{f}$ . If $ν_{f}$ admits a nowhere-zero section then $⟨ e (ν_{f}), [M] ⟩ = 0$ by [F2]. We prove the asserted regular-homotopy invariance in the next step.

2.1F3F4step 1.2

If $m = 0$ , every map $M \to R^{0}$ is the same map, so its normal data and Euler number are unchanged by any regular homotopy. For $m > 0$ , let $H : M \times I \to R^{2 m}$ be a regular homotopy from $H_{0}$ to $H_{1}$ , and write $N_{H} : = H^{*} T R^{2 m} / d_{M} H ({pr}_{M}^{*} T M)$ for the vertical quotient. The vertical derivative $d_{M} H$ is a smooth bundle map over $M \times I$ and is fibrewise injective because every $H_{t}$ is an immersion, so its image is a rank- $m$ subbundle and $N_{H}$ is a smooth rank- $m$ vector bundle over $M \times I$ by [F4]. At each time $t$ the restriction of $N_{H}$ to $M \times {t}$ is the normal bundle of $H_{t}$ under the canonical identification, and the orientations of $M$ and $R^{2 m}$ induce an orientation of $N_{H}$ whose restrictions are the orientations of the endpoint normal bundles. With the endpoint inclusions $j_{0}, j_{1} : M \to M \times I$ , naturality of the Euler class [F4] gives $e (ν_{H_{t}}) = j_{t}^{*} e (N_{H})$ for $t = 0, 1$ , and $j_{0}, j_{1}$ are homotopic through $(x, t) \mapsto (x, t)$ ; by [F4] they induce the same map in cohomology, so $e (ν_{H_{0}}) = e (ν_{H_{1}})$ and hence $⟨ e (ν_{H_{0}}), [M] ⟩ = ⟨ e (ν_{H_{1}}), [M] ⟩$ . This is invariance under every regular homotopy, with no genericity assumption on the intermediate slices.

2.2F5F6step 1.2

Suppose now $m = 2 s \geq 2$ and $f$ is self-transverse with no triple points, and put $D (f) = \sum_{r \in Σ (f)} ε (r)$ with the ordering-independent signs of the self-transverse case. By [F6] there are a transverse section $σ$ and small $t > 0$ with $I (f, f_{t}) = ⟨ e (ν_{f}), [M] ⟩ + 2 D (f)$ . We show $I (f, f_{t}) = 0$ . The pair $(f, f_{t})$ is transverse to the diagonal $Δ \subseteq R^{2 m} \times R^{2 m}$ at these intersections, and by [F5] the oriented intersection number of the pair equals $I (f \times f_{t}, Δ)$ . Here orient $Δ$ by $v \mapsto (v, v)$ and identify its normal quotient by $(u, v) \mapsto v - u$ . At a coincidence the normal derivative of the product is $(- d f, d f_{t})$ , so its determinant differs from that of the ordered pair $(d f, d f_{t})$ by $(- 1)^{m} = 1$ , since $m$ is even. Translate the second factor: for $s \in [0, 1]$ put $f_{t, s} (y) = f_{t} (y) + s c v$ with a fixed nonzero vector $v$ and $c$ so large that $im (f_{t}) + c v$ is disjoint from $im f$ ; both images are compact, so such $c$ exists. This is a smooth homotopy of the product map $(f, f_{t, s})$ through the target of [F5], and the endpoint has no intersections with the diagonal, so its intersection number is $0$ ; by homotopy invariance [F5] the number $I (f \times f_{t}, Δ)$ is $0$ as well. Hence $⟨ e (ν_{f}), [M] ⟩ + 2 D (f) = 0$ .

3.1F2F4F5F6step 2.1step 2.2∎

It follows that $2 D (f) = - ⟨ e (ν_{f}), [M] ⟩$ , so the Euler number $⟨ e (ν_{f}), [M] ⟩$ is even, $D (f) = - ⟨ e (ν_{f}), [M] ⟩ / 2$ , and reducing the integer equality $D (f) = - ⟨ e (ν_{f}), [M] ⟩ / 2$ modulo $2$ gives $# Σ (f) \equiv D (f) \equiv - ⟨ e (ν_{f}), [M] ⟩ / 2 (m o d 2)$ , since every sign is $\pm 1 \equiv 1$ mod two. Finally, if $f$ were regularly homotopic to an embedding $g : M ↪ R^{2 m}$ , then step 2.1 would give $⟨ e (ν_{f}), [M] ⟩ = ⟨ e (ν_{g}), [M] ⟩$ , while for the embedding [F6] forces $e (ν_{g}) = 0$ ; hence $D (f) = - ⟨ e (ν_{f}), [M] ⟩ / 2 = 0$ . Contrapositively, a nonzero $D (f)$ , and in particular an odd double-point parity, obstructs a regular homotopy to an embedding. The statement asserts no converse: vanishing of $D$ does not produce a regular homotopy to an embedding, and no odd-dimensional Whitney identification is claimed. AC is used through the normal-bundle and Euler-class suppliers and the oriented intersection theory; the regular-homotopy invariance of step 2.1 is valid for all $m$ , while the push-off count of step 2.2 uses the even-dimensional hypotheses.

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The inverse of one plus the generator in the truncated mod-two polynomial ring

Statement

Let $m \geq 1$ , let $F_{2} [t]$ be the polynomial ring over $F_{2}$ and let $R_{m} = F_{2} [t] / (t^{m + 1})$ be the truncated polynomial ring, so that $t^{m + 1} = 0$ and $1, t, \dots, t^{m}$ is an $F_{2}$ -basis. Write the binary expansion $m = \sum_{j \in B} 2^{j}$ and let $S_{m} = {i \in Z : 0 \leq i \leq m, i \land m = 0}$ , where $\land$ is digitwise AND of binary expansions; set $d (m) = \max S_{m}$ . Then $1 + t$ is a unit of $R_{m}$ and $(1 + t)^{- (m + 1)} = \sum_{i \in S_{m}} t^{i} .$ Consequently the coefficient of $t^{i}$ in $(1 + t)^{- (m + 1)}$ is $1$ exactly for $i \in S_{m}$ , this coefficient equals $(\binom{m + i}{i}) m o d 2$ , and the highest power occurring with nonzero coefficient is $t^{d (m)}$ . If $m + 1$ is a power of two then $S_{m} = {0}$ and $(1 + t)^{- (m + 1)} = 1$ .

Facts & Assumptions

Given: An integer $m \geq 1$ , the ring $F_{2} = Z / 2$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ , For every prime $p$ , the two operations on $Z / p$ make it a field), the polynomial ring $F_{2} [t]$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution), and the quotient ring $R_{m} = F_{2} [t] / (t^{m + 1})$ with the monic polynomial $t^{m + 1}$ (The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ , Multiplication of additive cosets is well defined if and only if the additive subgroup is a two-sided ideal, Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[F1]

Division by the monic polynomial $t^{m + 1}$ gives every class of $R_{m}$ a unique representative of degree at most $m$ ; hence $1, t, \dots, t^{m}$ is an $F_{2}$ -basis of $R_{m}$ and $t^{m + 1} = 0$ in $R_{m}$ (Division by a monic polynomial over a commutative ring, The quotient ring $R / I$ with $(r + I) (s + I) = r s + I$ ). In $F_{2}$ one has $1 + 1 = 0$ (The congruence class $[a]_{n}$ and the quotient set $Z / n$ ).

[F2]

For every commutative ring $R$ , every $λ \in R$ and every integer $j \geq 1$ the formal identity $\frac{1}{(1 - λ x)^{j}} = \sum_{n \geq 0} (\binom{n + j - 1}{j - 1}) λ^{n} x^{n}$ holds in $R ⟦ x ⟧$ , the binomial coefficient acting by repeated addition (Repeated poles expand formally as $(1 - λ x)^{- j} = \sum_{n \geq 0} (\binom{n + j - 1}{j - 1}) λ^{n} x^{n}$ , The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ , Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ); moreover $(\binom{N}{k}) = (\binom{N}{N - k})$ for $0 \leq k \leq N$ (The binomial coefficients are symmetric and increase to the middle level before decreasing).

[F3]

Coefficients of sums and Cauchy products in $R ⟦ x ⟧$ in degree $k$ depend only on the coefficients of degree at most $k$ (Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ).

Proof

1.1F1algebra

Every element of $R_{m}$ has a unique representative of degree at most $m$ by [F1], and $t^{m + 1} = 0$ ; in particular $1, t, \dots, t^{m}$ form a basis and no class has two such representatives. The element $1 + t$ is a unit with the displayed finite inverse: since $m \geq 1$ , in characteristic two $(1 + t) (1 + t + \dots + t^{m}) = 1 + t^{m + 1} = 1 + 0 = 1$ in $R_{m}$ , so $1 + t + \dots + t^{m}$ is an inverse of $1 + t$ and hence $1 + t$ is a unit.

1.2F1givenalgebra

Write $m = \sum_{j \in B} 2^{j}$ with $B$ the finite set of binary digit positions, so $B \neq \emptyset$ because $m \geq 1$ . Choose $M \geq \max B$ with $2^{M + 1} > m$ . In $F_{2} [t]$ the identity $(1 + t)^{2^{j}} = 1 + t^{2^{j}}$ holds for every $j \geq 0$ : it is trivial for $j = 0$ , and from $(u + v)^{2} = u^{2} + 2 u v + v^{2} = u^{2} + v^{2}$ in characteristic two, $(1 + t)^{2^{j + 1}} = ((1 + t)^{2^{j}})^{2} = (1 + t^{2^{j}})^{2} = 1 + t^{2^{j + 1}}$ . Multiplying the identities for $j \in B$ gives $(1 + t)^{m} = \prod_{j \in B} (1 + t^{2^{j}})$ , and iterating $\prod_{j = 0}^{N} (1 + t^{2^{j}}) = 1 + t + \dots + t^{2^{N + 1} - 1}$ gives, in $R_{m}$ , $(1 + t)^{m + 1} = (1 + t) \prod_{j \in B} (1 + t^{2^{j}}), \prod_{j = 0}^{M} (1 + t^{2^{j}}) = 1 + t + \dots + t^{2^{M + 1} - 1} .$

2.1F1step 1.2algebra

Put $P_{m} : = \prod_{j \notin B, 0 \leq j \leq M} (1 + t^{2^{j}})$ , a finite product in $R_{m}$ . Expanding the product over all subsets $A \subseteq {0, \dots, M} ∖ B$ , each subset contributes $t^{σ (A)}$ with $σ (A) = \sum_{j \in A} 2^{j}$ , and distinct subsets have distinct sums by uniqueness of binary expansion; all other coefficients are $0$ . Since $t^{k} = 0$ in $R_{m}$ for every $k \geq m + 1$ by [F1], only the subsets with $σ (A) \leq m$ contribute, and such a sum has binary support inside ${0, \dots, M}$ and disjoint from $B$ , i.e. $σ (A) \land m = 0$ , so $σ (A) \in S_{m}$ . Conversely every $i \in S_{m}$ satisfies $i \leq m < 2^{M + 1}$ , so its binary expansion involves only digits $j \leq M$ and, since $i \land m = 0$ , no digit of $B$ ; the subset $A = {j : 2^{j} occurs in i}$ is admissible and $i = σ (A)$ . Therefore $P_{m} = \sum_{i \in S_{m}} t^{i} in R_{m} .$

3.1F1step 1.2step 2.1algebra

In $R_{m}$ one computes, using step 1.2 and the Frobenius identities, $(1 + t)^{m + 1} P_{m} = (1 + t) \prod_{j = 0}^{M} (1 + t^{2^{j}}) = (1 + t) (1 + t + \dots + t^{2^{M + 1} - 1}) = 1 + t^{2^{M + 1}} = 1,$ the last equality because $2^{M + 1} > m$ forces $t^{2^{M + 1}} = 0$ in $R_{m}$ by [F1]. Hence $P_{m}$ is a two-sided inverse of $(1 + t)^{m + 1}$ in the commutative ring $R_{m}$ , so $(1 + t)^{- (m + 1)} = P_{m} = \sum_{i \in S_{m}} t^{i}$ by step 2.1.

4.1F2F3step 3.1algebra

For the binomial-coefficient description apply [F2] over the commutative ring $F_{2}$ with $λ = 1$ and $j = m + 1$ : in $F_{2} ⟦ x ⟧$ one has $(1 - x)^{- (m + 1)} = \sum_{i \geq 0} ((\binom{i + m}{m}) m o d 2) x^{i} = \sum_{i \geq 0} ((\binom{m + i}{i}) m o d 2) x^{i}$ , the second equality by the symmetry clause of [F2], where $1 - x = 1 + x$ because $1 + 1 = 0$ in $F_{2}$ . By [F3] the coefficientwise truncation map $φ : F_{2} ⟦ x ⟧ \to R_{m}$ , $\sum_{i} a_{i} x^{i} \mapsto \sum_{i = 0}^{m} a_{i} t^{i}$ , is a surjective ring homomorphism: addition is coefficientwise, and in the Cauchy product the coefficient of $x^{k}$ depends only on the coefficients of degree at most $k$ , so truncation at degree $m$ commutes with products in $R_{m}$ , where $t^{m + 1} = 0$ . Since $φ (1 - x) = 1 + t$ and ring homomorphisms carry inverses of units to inverses of units, $φ ((1 - x)^{- (m + 1)}) = (1 + t)^{- (m + 1)} = \sum_{i = 0}^{m} ((\binom{m + i}{i}) m o d 2) t^{i} .$ Comparing coefficients with step 3.1 gives: the coefficient of $t^{i}$ in $(1 + t)^{- (m + 1)}$ equals $(\binom{m + i}{i}) m o d 2$ for every $0 \leq i \leq m$ , and it equals $1$ exactly for the $i \in S_{m}$ by the formula of step 3.1.

5.1step 1.2step 3.1step 4.1algebra∎

The set $S_{m}$ contains $0$ and is finite, so $d (m) = \max S_{m}$ is defined; by steps 3.1 and 4.1 the coefficient of $t^{d (m)}$ is $1$ , while every coefficient of degree $i > d (m)$ with $i \leq m$ is $0$ because such $i \notin S_{m}$ , and degrees above $m$ vanish in $R_{m}$ . Hence the highest power occurring with nonzero coefficient is exactly $t^{d (m)}$ . If $m + 1 = 2^{q}$ is a power of two, then $q \geq 1$ , $m = 2^{q} - 1 = \sum_{j = 0}^{q - 1} 2^{j}$ , and every $i$ with $1 \leq i \leq m$ has some binary digit at a position $j \leq q - 1$ , hence satisfies $i \land m \neq 0$ ; therefore $S_{m} = {0}$ and the inverse is $1$ , consistently with $(1 + t)^{m + 1} = (1 + t)^{2^{q}} = 1 + t^{2^{q}} = 1$ in $R_{m}$ .

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Stiefel-Whitney classes of the tangent bundle of real projective space

Statement

Assume AC. Let $m \geq 1$ , let $L \to {R P}^{m}$ be the tautological line bundle, and let $a \in H^{1} ({R P}^{m}; F_{2})$ be the nonzero degree-one class. Then there is a smooth real bundle isomorphism $T {R P}^{m} \oplus ε^{1} ≅ (m + 1) L,$ and consequently, in $H^{*} ({R P}^{m}; F_{2}) = F_{2} [a] / (a^{m + 1})$ , $w (T {R P}^{m}) = w (L)^{m + 1} = (1 + a)^{m + 1},$ where $w (L) = 1 + x_{L} = 1 + a$ is the rank-one case of Stiefel–Whitney classes from the projective-bundle relation (Real projective bundle and tautological line, Tautological degree-one class on a real projective bundle, Mod-two real projective bundle theorem).

Facts & Assumptions

Given: An integer $m \geq 1$ , real projective space ${R P}^{m}$ with its smooth structure from the affine charts, the tautological line $L = γ^{1} \subseteq ε^{m + 1} = {R P}^{m} \times R^{m + 1}$ , the tangent bundle $T {R P}^{m}$ , and the class $a$ .

[F1]

The affine charts $U_{i} = {[x] : x_{i} \neq 0}$ with coordinates $x_{j} / x_{i}$ form a smooth atlas of ${R P}^{m}$ (Real projective space from affine charts); the tautological line bundle is $γ_{ε^{m + 1}}$ , the subbundle $L = {([x], v) : v \in R x} \subseteq ε^{m + 1}$ (the case $E = ε^{m + 1}$ of Real projective bundle and tautological line).

[F2]

A smooth chart $(U, x)$ produces the induced tangent-bundle chart $v \mapsto (x (p), v^{1}, \dots, v^{m})$ with $v = \sum_{i} v^{i} \partial_{x^{i}} ∣_{p}$ (The induced tangent bundle chart, Coordinate derivations form a basis of the tangent space), and every tangent vector is the velocity $\dot{γ} (0)$ of a smooth curve (Every tangent vector is the velocity of a smooth curve, The velocity derivation of a smooth curve). Smoothness of maps between smooth manifolds is checked in charts ( $C^{r}$ and smooth maps between smooth manifolds, Immersions, submersions, and constant-rank maps).

[F3]

For bundles over the same base one has the Whitney sum, tensor product, dual and Hom bundles, with $Hom (E, F) ≅ E^{*} \otimes F$ (Whitney sum, tensor, dual, Hom, and exterior-power bundles).

[F4]

The standard Euclidean inner product restricts to a smooth metric on the tautological line $L \subseteq ε^{m + 1}$ . Its orthogonal complement $L^{⊥}$ is smooth, and the quotient map identifies $L^{⊥}$ smoothly with $ε^{m + 1} / L$ (Orthogonal complements of subbundles are smooth subbundles, A vector bundle quotient by a subbundle is a smooth vector bundle): in a smooth local frame the inverse is obtained by orthogonal projection. Thus $ε^{m + 1} = L \oplus L^{⊥}$ smoothly, and the metric gives a smooth isomorphism $L ≅ L^{*}$ by $v \mapsto ⟨ v, - ⟩$ .

[F5]

Closed bounded Euclidean subsets are compact, and continuous images of compact spaces are compact (A subset of $R^{n}$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism). A closed smooth manifold is a paracompact Hausdorff CGWH space of CW homotopy type over which every smooth bundle is numerable (Smooth manifolds have CW homotopy type).

[F6]

SW classes are defined by the projective-bundle relation, $w (L) = 1 + x_{L}$ for a line bundle, they are natural under bundle isomorphisms, satisfy the Whitney product formula, and satisfy $w (E \oplus ε^{r}) = w (E)$ (Stiefel–Whitney classes from the projective-bundle relation, Whitney sum formula for Stiefel–Whitney classes, Naturality of Stiefel–Whitney classes, Tautological degree-one class on a real projective bundle).

[F7]

For the trivial rank- $(m + 1)$ bundle $ε^{m + 1}$ over the one-point base, the projective bundle is $P (ε^{m + 1}) = {R P}^{m}$ with tautological line $L$ , so the projective-bundle theorem applies with $B = p t$ , $n = m + 1$ and $x = x_{L}$ : $H^{*} ({R P}^{m}; F_{2})$ is a free $F_{2}$ -module with basis $1, x_{L}, \dots, x_{L}^{m}$ ; the classes $c_{i} \in H^{i} (p t; F_{2})$ of its relation vanish for $i \geq 1$ by the dimension axiom for singular cohomology, so the kernel of the algebra map $F_{2} [x] \to H^{*} ({R P}^{m}; F_{2})$ , $x \mapsto x_{L}$ , is exactly the ideal $(x_{L}^{m + 1})$ ; hence $x_{L}^{m + 1} = 0$ , $x_{L}^{i} \neq 0$ for $0 \leq i \leq m$ , $H^{*} ({R P}^{m}; F_{2}) = F_{2} [x_{L}] / (x_{L}^{m + 1})$ , and $H^{1} ({R P}^{m}; F_{2}) = F_{2} x_{L}$ is one-dimensional, so $x_{L}$ is the unique nonzero degree-one class $a$ of the statement (Mod-two real projective bundle theorem, Singular cohomology satisfies the Eilenberg Steenrod cohomology axioms, Real projective bundle and tautological line). AC is the hypothesis of the projective-bundle theorem (The Axiom of Choice).

Proof

1.1F1F2F5given

The affine charts make ${R P}^{m}$ a boundaryless smooth manifold. It is compact: every line has a unit representative, and the quotient projection $S^{m} \to {R P}^{m}$ is continuous and surjective; $S^{m}$ is closed and bounded, hence compact, and its image is compact by [F5]. Fix a line $ℓ$ and a complement $H$ , and let $p : R^{m + 1} \to ℓ$ be projection along $H$ . The lines transverse to $H$ form an open set $U_{ℓ}$ : on every affine chart, transversality is the nonvanishing of a linear coordinate expression. Each such line is uniquely the graph of $f \in Hom (ℓ, H)$ . In affine coordinates this graph chart and its inverse are ratios of linear expressions with nonzero denominators, so are smooth by [F1] and [F2]. Differentiating at the graph of $0$ and composing $H ≅ R^{m + 1} / ℓ$ gives an isomorphism $Θ_{ℓ} : T_{ℓ} {R P}^{m} \to Hom (ℓ, R^{m + 1} / ℓ)$ .

2.1F1F2F3F4step 1.1

This tangent identification is independent of $H$ . For a second complement $H^{'}$ , write $p^{'}, q^{'}$ for the projections onto $ℓ, H^{'}$ . Near $f = 0$ the graph transition is $f^{'} = q^{'} f ({id}_{ℓ} + p^{'} f)^{- 1} .$ Indeed, a graph vector $u + f (u)$ has $ℓ$ -coordinate $({id}_{ℓ} + p^{'} f) u$ and $H^{'}$ -coordinate $q^{'} f (u)$ . The derivative of this transition at $0$ is $g \mapsto q^{'} g$ , since $f = 0$ there; modulo $ℓ$ , $q^{'} g (u)$ and $g (u)$ agree. Thus both differentials give the same $Θ_{ℓ}$ . These maps define a fibrewise isomorphism $Θ : T {R P}^{m} \to Hom (L, ε^{m + 1} / L)$ , with the quotient and Hom bundles supplied by [F3] and [F4].

3.1F2F3step 1.1step 2.1

The map $Θ$ is a smooth bundle isomorphism. Over $U_{ℓ}$ , the chart differential of $φ_{ℓ}$ trivializes $T {R P}^{m} ∣_{U_{ℓ}} ≅ U_{ℓ} \times Hom (ℓ, H)$ , and the same graph data trivialize $Hom (L, ε^{m + 1} / L) ∣_{U_{ℓ}}$ : at $ℓ^{'} = Γ (f)$ the projection $R^{m + 1} \to ℓ$ along $H$ restricts to a linear isomorphism $L_{ℓ^{'}} \to ℓ$ , while $u + g \mapsto g - f (u)$ ( $u \in ℓ$ , $g \in H$ ) is a linear map killing $L_{ℓ^{'}}$ and inducing an isomorphism $R^{m + 1} / L_{ℓ^{'}} \to H$ . Both depend polynomially on $f = φ_{ℓ} (ℓ^{'})$ , hence smoothly on $ℓ^{'}$ , and relative to these two trivializations $Θ$ is the identity map of $U_{ℓ} \times Hom (ℓ, H)$ : the derivative of the straight slope curve $f + t g$ is $g$ , whose image under the graph trivialization of the Hom-bundle is again $g$ by the formula just displayed. A map that is the identity in local trivializations is smooth, and $Θ$ is bijective with fibrewise-linear inverse, so it is a smooth bundle isomorphism $T {R P}^{m} ≅ L^{*} \otimes (ε^{m + 1} / L)$ .

4.1F3F4step 3.1

By [F4] the Euclidean metric gives smooth isomorphisms $ε^{m + 1} = L \oplus L^{⊥}$ , $L^{⊥} ≅ ε^{m + 1} / L$ and $L^{*} ≅ L$ . Also $Hom (L, L)$ is canonically trivial, with the identity as a nowhere-zero section. Tensoring the splitting with $L^{*}$ and using step 3.1 yields $T {R P}^{m} \oplus ε^{1} ≅ L^{*} \otimes (ε^{m + 1} / L) \oplus L^{*} \otimes L ≅ L^{*} \otimes ε^{m + 1} ≅ (m + 1) L^{*} ≅ (m + 1) L .$ All these isomorphisms are smooth; no continuous metric is substituted for a smooth one.

5.1F5F6F7step 3.1step 4.1∎

Finally $Θ$ and the splitting are used to compute the classes. The bundle $ε^{m + 1}$ over the one-point base has $P (ε^{m + 1}) = {R P}^{m}$ and tautological line $γ_{ε^{m + 1}} = L$ , so by [F7] its tautological class is $x_{L} = x_{ε^{m + 1}} \in H^{1} ({R P}^{m}; F_{2})$ and the projective-bundle relation is $x_{L}^{m + 1} = 0$ (all $c_{i} \in H^{> 0} (p t; F_{2})$ vanish), while $1, x_{L}, \dots, x_{L}^{m}$ is a basis; since $H^{1} ({R P}^{m}; F_{2}) = F_{2} a$ has the unique nonzero class $a$ and a basis element cannot be zero, $x_{L} = a$ . Hence $w (L) = 1 + x_{L} = 1 + a$ by the rank-one case of [F6]. Applying [F6] to the stable isomorphism of step 4.1, $w (T {R P}^{m}) = w (T {R P}^{m} \oplus ε^{1}) = w ((m + 1) L) = w (L)^{m + 1} = (1 + a)^{m + 1}$ in $H^{*} ({R P}^{m}; F_{2}) = F_{2} [a] / (a^{m + 1})$ : the first equality is the stability clause for trivial summands, the second is invariance of the classes under bundle isomorphisms, and the third is the Whitney product formula iterated over the $(m + 1)$ summands.

TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Real projective space Stiefel-Whitney non-immersion obstruction

Statement

Assume AC. Let $m \geq 1$ , let $a$ denote the nonzero degree-one class of $H^{*} ({R P}^{m}; F_{2}) = F_{2} [a] / (a^{m + 1})$ (Mod-two real projective bundle theorem, Real projective bundle and tautological line), and let $\overset{ˉ}{w} ({R P}^{m})$ be the total normal Stiefel-Whitney class (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold). Then $w (T {R P}^{m}) = (1 + a)^{m + 1}, \overset{ˉ}{w} ({R P}^{m}) = (1 + a)^{- (m + 1)} = \sum_{i \in S_{m}} a^{i},$ with $S_{m} = {0 \leq i \leq m : i \land m = 0}$ and digitwise AND, and the highest nonzero term is ${\overset{ˉ}{w}}_{d (m)} ({R P}^{m}) = a^{d (m)}$ , where $d (m) = \max S_{m}$ . Consequently:

(i) ${R P}^{m}$ does not immerse in $R^{m + k}$ for any $k < d (m)$ ; that is, no immersion of ${R P}^{m}$ has codimension less than $d (m)$ (High normal Stiefel-Whitney classes obstruct low-codimension immersions);

(ii) if $m = 2^{p}$ with $p \geq 1$ , then $d (m) = m - 1$ and ${\overset{ˉ}{w}}_{m - 1} ({R P}^{m}) = a^{m - 1} \neq 0$ , so ${R P}^{2^{p}}$ does not immerse in $R^{2^{p + 1} - 2}$ ;

(iii) $w (T {R P}^{m}) = 1$ exactly when $m + 1$ is a power of two. This is a characteristic-class criterion only; it does not assert that these projective spaces are parallelizable.

Facts & Assumptions

Given: An integer $m \geq 1$ , real projective space ${R P}^{m}$ with $H^{*} ({R P}^{m}; F_{2}) = F_{2} [a] / (a^{m + 1})$ , and AC (The Axiom of Choice).

[F1]

The tangent class is $w (T {R P}^{m}) = (1 + a)^{m + 1}$ and the normal class is its inverse, $\overset{ˉ}{w} ({R P}^{m}) = w (T {R P}^{m})^{- 1} = (1 + a)^{- (m + 1)}$ , the latter by the normal Stiefel-Whitney inverse identity (Stiefel-Whitney classes of the tangent bundle of real projective space, The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class, Normal Stiefel-Whitney and Pontryagin classes of a closed manifold).

[F2]

In $F_{2} [t] / (t^{m + 1})$ one has the unit $1 + t$ and the identity $(1 + t)^{- (m + 1)} = \sum_{i \in S_{m}} t^{i}$ with $S_{m} = {0 \leq i \leq m : i \land m = 0}$ , $d (m) = \max S_{m}$ ; if $m + 1$ is a power of two then $S_{m} = {0}$ (The inverse of one plus the generator in the truncated mod-two polynomial ring).

[F3]

If a closed smooth $M^{m}$ has ${\overset{ˉ}{w}}_{i} (M) \neq 0$ for some $i > k$ , then $M$ does not immerse in $R^{m + k}$ (High normal Stiefel-Whitney classes obstruct low-codimension immersions).

[F4]

For the trivial rank- $(m + 1)$ bundle $ε^{m + 1}$ over the one-point base, the projective bundle is $P (ε^{m + 1}) = {R P}^{m}$ , and the projective-bundle theorem gives the ring $H^{*} ({R P}^{m}; F_{2}) = F_{2} [x_{L}] / (x_{L}^{m + 1})$ , free on $1, x_{L}, \dots, x_{L}^{m}$ , because the relation classes $c_{i} \in H^{i} (p t; F_{2})$ vanish for $i \geq 1$ by the dimension axiom for singular cohomology (Mod-two real projective bundle theorem, Singular cohomology satisfies the Eilenberg Steenrod cohomology axioms, Real projective bundle and tautological line); the group $H^{1} ({R P}^{m}; F_{2})$ is then one-dimensional, so the nonzero class $a$ of the statement equals $x_{L}$ and $a^{i} \neq 0$ for $0 \leq i \leq m$ because $1, a, \dots, a^{m}$ is a basis. The tangent-bundle computation is that of the local lemma based on the affine charts of ${R P}^{m}$ (Real projective space from affine charts, Stiefel-Whitney classes of the tangent bundle of real projective space). AC is the hypothesis of these suppliers (The Axiom of Choice).

Proof

1.1F1F2F4

The two displayed computations are [F1] and [F2] read in the ring $F_{2} [a] / (a^{m + 1})$ : $w (T {R P}^{m}) = (1 + a)^{m + 1}$ , and $\overset{ˉ}{w} ({R P}^{m}) = (1 + a)^{- (m + 1)} = \sum_{i \in S_{m}} a^{i}$ with $d (m) = \max S_{m}$ the highest index with nonzero coefficient, so the highest nonzero normal class is ${\overset{ˉ}{w}}_{d (m)} ({R P}^{m}) = a^{d (m)} \neq 0$ by [F4]. By definition of $d (m)$ , every ${\overset{ˉ}{w}}_{i}$ with $i > d (m)$ vanishes.

2.1F2F3F4step 1.1

For clause (i): if $k < d (m)$ then $i : = d (m) > k$ and ${\overset{ˉ}{w}}_{i} ({R P}^{m}) \neq 0$ , so [F3] forbids an immersion of ${R P}^{m}$ into $R^{m + k}$ ; equivalently every immersion has codimension at least $d (m)$ . For clause (ii): if $m = 2^{p}$ with $p \geq 1$ , then the binary expansion of $m$ has the single nonzero digit $2^{p}$ , so $i \land m = 0$ for $0 \leq i \leq m$ exactly when $i < 2^{p} = m$ , that is $S_{m} = {0, 1, \dots, m - 1}$ and $d (m) = m - 1$ ; hence ${\overset{ˉ}{w}}_{m - 1} = a^{m - 1} \neq 0$ and, applying clause (i) with $k = m - 2 < d (m)$ , there is no immersion into $R^{m + (m - 2)} = R^{2 m - 2} = R^{2^{p + 1} - 2}$ . For $m = 2$ this is $k = 0$ : the final clause of [F3] with the nonzero degree-one normal class excludes this equal-dimensional immersion; its proof treats that instance by the local-diffeomorphism and compact-image argument.

3.1F1F2F4∎

For clause (iii): write $m + 1 = \sum_{j \in B} 2^{j}$ with $B$ the set of binary digit positions. In characteristic two, $(1 + a)^{m + 1} = \prod_{j \in B} (1 + a^{2^{j}})$ : this follows by iterating $(1 + a)^{2^{j + 1}} = ((1 + a)^{2^{j}})^{2} = 1 + a^{2^{j + 1}}$ , as in [F2]. If $B = {j_{0}}$ has one element then $m + 1 = 2^{j_{0}}$ is a power of two and $(1 + a)^{m + 1} = 1 + a^{m + 1} = 1$ in $F_{2} [a] / (a^{m + 1})$ . If $B$ has at least two elements and $j_{0} = \min B$ , then $2^{j_{0}} < m + 1$ and the product contains the monomial $a^{2^{j_{0}}}$ with coefficient $1$ (choose the factor $j_{0}$ and the constant term from every other factor); no other selection of factors contributes to degree $2^{j_{0}}$ , and $2^{j_{0}} \leq m$ so this term is nonzero in the truncation. Hence $w (T {R P}^{m}) \neq 1$ in that case, establishing the equivalence. The criterion concerns the tangent class only and says nothing about parallelizability or about Massey-type improvements of the non-immersion bound for general $m$ .

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)Open item page →

Parallelizable manifolds have no stable characteristic-class obstruction to Euclidean immersion

Statement

Assume AC. Let $M$ be a closed smooth $m$ -manifold whose tangent bundle is trivial, $T M ≅ ε^{m}$ . Then for every $k \geq 1$ the trivial bundle $ε^{k}$ , together with the composite isomorphism $T M \oplus ε^{k} ≅ ε^{m} \oplus ε^{k} ≅ ε^{m + k}$ , is a rank- $k$ stable normal inverse of $M$ ; consequently $M$ admits an immersion into $R^{m + k}$ for every $k \geq 1$ , in particular into $R^{m + 1}$ (Smale-Hirsch makes rank reduction sufficient for Euclidean immersion in positive codimension). Moreover $\overset{ˉ}{w} (M) = 1$ and $\overset{ˉ}{p} (M) = 1$ (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold): the inverse total classes of the trivial tangent class are trivial, so every immersion and embedding test on this page based on ${\overset{ˉ}{w}}_{i}$ or ${\overset{ˉ}{p}}_{i}$ returns zero for $M$ . The proposition asserts nothing about embeddability, about the minimal immersion dimension below $m + 1$ , or about other obstructions; dimension, rank and embedding-theoretic issues are unaffected.

Facts & Assumptions

Given: A closed smooth $m$ -manifold $M$ with a trivialization $T M ≅ ε^{m}$ , and AC (The Axiom of Choice).

[F1]

A rank- $k$ stable normal inverse of $M$ is a smooth real bundle $ν$ of rank $k$ together with a smooth bundle isomorphism $φ : T M \oplus ν \to ε^{m + k}$ (Stable normal inverse of the tangent bundle, Smooth vector bundles, rank, fibres, and trivial bundles, Whitney sums of vector bundles).

[F2]

Under the countable choice ${A C}_{ω}$ (implied by AC by AC implies DC implies countable choice), a rank- $k$ stable normal inverse of a closed $M$ with $k \geq 1$ produces an immersion into $R^{m + k}$ (Smale-Hirsch makes rank reduction sufficient for Euclidean immersion in positive codimension, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

The normal classes of a closed $M$ are $\overset{ˉ}{w} (M) = w (T M)^{- 1}$ in $H^{*} (M; F_{2})$ and $\overset{ˉ}{p} (M) = p (T M)^{- 1}$ in $H^{*} (M; Q)$ for connected $M$ , realized as $w (ν)$ , $p (ν)$ for any stable normal inverse $(ν, φ)$ (Normal Stiefel-Whitney and Pontryagin classes of a closed manifold, The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class, The normal Pontryagin class is the rational inverse of the tangent Pontryagin class).

[F4]

The trivial bundle has trivial characteristic classes: $w (ε^{r}) = 1$ and $p (ε^{r}) = 1$ (The normal Stiefel-Whitney class is the multiplicative inverse of the tangent class, The normal Pontryagin class is the rational inverse of the tangent Pontryagin class, Smooth vector bundles, rank, fibres, and trivial bundles).

Proof

1.1F1

Fix a trivialization $τ : T M \to ε^{m}$ and let $k \geq 1$ . Let $φ : T M \oplus ε^{k} \to ε^{m + k}$ be the composite of $τ \oplus {id}_{ε^{k}} : T M \oplus ε^{k} \to ε^{m} \oplus ε^{k}$ with the canonical associativity identification $ε^{m} \oplus ε^{k} = ε^{m + k}$ . This is a smooth bundle isomorphism, so by [F1] the pair $(ε^{k}, φ)$ is a rank- $k$ stable normal inverse of $M$ .

2.1F2step 1.1

By [F2] and the triviality of AC implies ${A C}_{ω}$ , the rank- $k$ stable normal inverse of step 1.1 produces an immersion $M ↬ R^{m + k}$ for every $k \geq 1$ ; taking $k = 1$ gives an immersion into $R^{m + 1}$ , and any larger codimension is obtained by stabilizing or by the same construction.

2.2F3F4step 1.1

The characteristic classes vanish in the normal direction. Since $T M ≅ ε^{m}$ , naturality of the characteristic classes gives $w (T M) = w (ε^{m}) = 1$ and $p (T M) = p (ε^{m}) = 1$ by [F4]. By [F3] the normal classes are the inverses of these units, so $\overset{ˉ}{w} (M) = 1^{- 1} = 1$ and $\overset{ˉ}{p} (M) = 1^{- 1} = 1$ ; equivalently, the inverse bundle of step 1.1 is trivial, $w (ε^{k}) = 1$ , $p (ε^{k}) = 1$ , and the inverse lemmas identify these with the normal classes. Hence every normal Stiefel-Whitney class ${\overset{ˉ}{w}}_{i} (M)$ with $i \geq 1$ and every normal Pontryagin class ${\overset{ˉ}{p}}_{i} (M)$ with $i \geq 1$ vanishes, so no immersion or embedding test of this page based on $\overset{ˉ}{w}$ or $\overset{ˉ}{p}$ obstructs anything for $M$ .

3.1F1F2F3step 1.1step 2.2∎

Parallelizability supplies a rank- $k$ inverse with trivial normal bundle for every $k \geq 1$ , hence the stated Euclidean immersions by [F2], while the positive normal characteristic classes vanish by step 2.2. AC is inherited from the characteristic-class suppliers and implies the countable choice used by the immersion-existence supplier. No embedding or minimal-dimension conclusion is asserted.

CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Embedding obstructions include all immersion normal-class obstructions

Statement

Assume AC. Every smooth embedding is a smooth immersion (Smooth embeddings, Immersions, submersions, and constant-rank maps). Hence, for $k \geq 1$ , the normal-class tests of this page also obstruct embeddings: if $M$ is a closed smooth $m$ -manifold and ${\overset{ˉ}{w}}_{i} (M) \neq 0$ for some $i > k$ , then $M$ admits neither an immersion nor an embedding into $R^{m + k}$ (High normal Stiefel-Whitney classes obstruct low-codimension immersions); if $M$ is connected and ${\overset{ˉ}{p}}_{i} (M) \neq 0$ for some $i$ with $2 i > k$ , then $M$ admits neither an immersion nor an embedding into $R^{m + k}$ (High normal Pontryagin classes obstruct low-codimension immersions). Moreover, for $n \geq m$ , the normal bundle of an embedding into $R^{n}$ is a genuine rank- $(n - m)$ stable normal inverse (for $n > m$ by An embedding into Euclidean space gives a rank-(n-m) stable normal inverse; for $n = m$ , the differential is a fibrewise isomorphism and the normal quotient is the zero bundle), so the rank-vanishing that drives the tests applies to the embedded normal bundle as well. In addition, for $m \geq 1$ and $k \geq 1$ embedding imposes the stronger condition ${\overset{ˉ}{w}}_{k} (T M) = 0$ , and an oriented embedded normal bundle must have $e (ν) = 0$ (Top normal classes vanish for Euclidean embeddings). Thus ${\overset{ˉ}{w}}_{i} (T M) \neq 0$ for any $i \geq k$ obstructs embedding in $R^{m + k}$ ; only $i > k$ is the rank obstruction for immersion. No converse is asserted: these tests are necessary conditions only.

Facts & Assumptions

Given: A closed smooth $m$ -manifold $M$ with $m \geq 1$ , an integer $k \geq 1$ , and AC (The Axiom of Choice).

[F1]

Every smooth embedding is a smooth immersion (Smooth embeddings, Immersions, submersions, and constant-rank maps).

[F2]

For a closed smooth $M$ : if ${\overset{ˉ}{w}}_{i} (M) \neq 0$ for some $i > k$ , then $M$ does not immerse in $R^{m + k}$ (High normal Stiefel-Whitney classes obstruct low-codimension immersions); if $M$ is connected and ${\overset{ˉ}{p}}_{i} (M) \neq 0$ for some $i$ with $2 i > k$ , then $M$ does not immerse in $R^{m + k}$ (High normal Pontryagin classes obstruct low-codimension immersions). These assertions inherit AC from the characteristic-class suppliers; AC also supplies the countable choice required for the normal-bundle splitting (AC implies DC implies countable choice).

[F3]

For $n \geq m$ , the normal bundle of an embedding $M ↪ R^{n}$ is a rank- $(n - m)$ stable normal inverse (by the embedding lemma for $n > m$ , and directly from the fibrewise-isomorphic differential for $n = m$ ) and realizes the normal classes $\overset{ˉ}{w} (M)$ , $\overset{ˉ}{p} (M)$ (An embedding into Euclidean space gives a rank-(n-m) stable normal inverse, Normal Stiefel-Whitney and Pontryagin classes of a closed manifold).

[F4]

If a closed $M^{m}$ with $m \geq 1$ embeds in $R^{m + k}$ with $k \geq 1$ , then ${\overset{ˉ}{w}}_{k} (T M) = 0$ , and if the embedded normal bundle is integrally oriented then its Euler class vanishes (Top normal classes vanish for Euclidean embeddings).

Proof

1.1F1F2

Let $g : M ↪ R^{m + k}$ be a smooth embedding. By [F1] the map $g$ is a smooth immersion, so every non-immersion statement applies to it: in particular, if ${\overset{ˉ}{w}}_{i} (M) \neq 0$ for some $i > k$ , then $M$ has no immersion, hence no embedding, into $R^{m + k}$ ; and if $M$ is connected and ${\overset{ˉ}{p}}_{i} (M) \neq 0$ for some $i$ with $2 i > k$ , then again no immersion and no embedding exists.

1.2F4

The embedding case is in fact stronger in top degree. Let $g : M^{m} ↪ R^{m + k}$ with $m \geq 1$ , $k \geq 1$ . Its normal bundle $ν$ has rank $k$ and, by [F4], satisfies ${\overset{ˉ}{w}}_{k} (T M) = w_{k} (ν) = 0$ , while an integrally oriented embedded normal bundle has $e (ν) = 0$ . Hence if ${\overset{ˉ}{w}}_{i} (T M) \neq 0$ for some $i \geq k$ , then in particular no embedding into $R^{m + k}$ exists, whereas for immersion the rank-vanishing argument only excludes the degrees $i > k$ : the top degree $i = k$ is the embedding-specific condition.

2.1F3step 1.1

The rank-vanishing that drives these tests is realized concretely by the embedding normal bundle: for $n \geq m$ , by [F3] the bundle $ν$ of an embedding $M ↪ R^{n}$ is a genuine rank- $(n - m)$ stable normal inverse, so $w_{i} (ν) = {\overset{ˉ}{w}}_{i} (M)$ and $p_{i} (ν) = {\overset{ˉ}{p}}_{i} (M)$ for every $i$ , and the classes of degree above the rank vanish by the rank convention. Thus the same normal classes carry both the transferred immersion tests and the top-degree embedding test.

3.1F2F3F4step 1.2step 2.1∎

Summarizing, the necessary conditions for an embedding of a closed smooth $M^{m}$ into $R^{m + k}$ with $m \geq 1$ , $k \geq 1$ are: ${\overset{ˉ}{w}}_{i} (M) = 0$ for all $i \geq k$ ; ${\overset{ˉ}{p}}_{i} (M) = 0$ for $2 i > k$ when the normal bundle is rationally considered; and $e (ν) = 0$ for an oriented embedded normal bundle. The first of these contains all the rank obstructions to immersion (degrees $i > k$ ) and adds the embedding-specific top class ${\overset{ˉ}{w}}_{k}$ . No converse is asserted: vanishing of all these classes is far from sufficient for embeddability, as the following boundary remark explains; the companion page separately shows that identical stable normal data do not classify isotopy. AC is inherited from the class suppliers; no additional choice is used.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

A closed three-manifold with H_1 = Z/2 and H_2 = 0 does not embed in S^4

Statement

Assume AC. Let $M$ be a closed connected oriented smooth three-manifold with integral homology $H_{1} (M) = Z / 2$ , $H_{2} (M) = 0$ and $H_{3} (M) = Z$ . Then $M$ admits no smooth embedding into $S^{4}$ , and hence none into $R^{4}$ .

Facts & Assumptions

Given: AC and $M$ as stated; all homology and cohomology below use integral coefficients.

[F1]

For a nonempty proper compact locally contractible $K \subset S^{4}$ , Alexander duality gives ${\tilde{H}}_{i} (S^{4} ∖ K) ≅ {\tilde{H}}^{3 - i} (K)$ (Alexander duality for compact locally contractible subsets of a sphere).

[F2]

UCT gives $0 \to {Ext}_{Z}^{1} (H_{r - 1} (X), Z) \to H^{r} (X) \to Hom (H_{r} (X), Z) \to 0$ (Topological universal coefficient short exact sequence for cohomology).

[F3]

Mayer–Vietoris applies to open covers, sphere homology is zero in degrees one and two, and deformation retractions induce homology isomorphisms (Mayer–Vietoris sequence in singular homology, Homology of spheres, Homotopic maps induce the same map on singular homology).

[F4]

A closed smooth submanifold has a tubular neighbourhood under countable choice; AC supplies countable choice (The tubular neighbourhood theorem in a smooth ambient manifold, AC implies DC implies countable choice, The Axiom of Choice).

Proof

1.1F1F2F4givenconstruct

Suppose $M \subset S^{4}$ is smoothly embedded. Its normal line is oriented by the orientations of $M$ and $S^{4}$ , and has a global positive unit section: in an oriented local line frame the positive unit vector is independent of the frame, so these sections glue. Compactness and [F4] give a product tube $M \times (- ϵ, ϵ)$ . By [F2], $H^{3} (M) = Z$ , since $H_{2} (M) = 0$ and $H_{3} (M) = Z$ . Thus [F1] gives ${\tilde{H}}_{0} (S^{4} ∖ M) = Z$ . The complement is an open manifold and is locally path connected; $H_{0}$ is free on its path components, so it has exactly two components $U, V$ . Every component has nonempty frontier in $M$ , since otherwise it is both open and closed in connected $S^{4}$ . Near any frontier point a hypersurface chart has exactly two connected local sides. Each of the two global halves of the product tube is connected because $M$ is connected. Every complementary component meets one of them, by the local side chart at a frontier point. Therefore the two tube halves lie in distinct components, one in $U$ and one in $V$ . Their closures $A = \overline{U}$ and $B = \overline{V}$ are compact smooth manifolds with common boundary $M$ , are locally contractible, and satisfy $S^{4} = A \cup B$ and $A \cap B = M$ .

2.1F3step 1.1construct

Enlarge $A$ and $B$ by a small portion of the opposite tube half to obtain an open cover of $S^{4}$ . The two open sets retract onto $A, B$ , and their intersection retracts onto $M$ , by moving the collar coordinate linearly to zero on the added halves. Also $U ↪ A$ and $V ↪ B$ are homotopy equivalences: a collar map which moves coordinate $s \geq 0$ slightly into $s > 0$ , and equals $s$ outside a smaller collar, is homotopic to the identity by linear interpolation and supplies homotopy inverses for the interior inclusions. Applying [F3] to the open cover, the segments $H_{2} (M) \to H_{2} (A) \oplus H_{2} (B) \to H_{2} (S^{4})$ and $H_{2} (S^{4}) \to H_{1} (M) \to H_{1} (A) \oplus H_{1} (B) \to H_{1} (S^{4})$ show $H_{2} (A) = H_{2} (B) = 0$ and $H_{1} (A) \oplus H_{1} (B) ≅ Z / 2$ . Hence one of $H_{1} (A), H_{1} (B)$ is zero and the other is $Z / 2$ .

3.1F1F2step 1.1step 2.1algebra∎

Since $S^{4} ∖ A = V$ , applying [F1] to $A$ and then the interior equivalence in step 2.1 gives $H_{1} (B) ≅ H_{1} (V) ≅ H^{2} (A)$ . By [F2] and $H_{2} (A) = 0$ , this equals ${Ext}_{Z}^{1} (H_{1} (A), Z)$ . The latter is zero if $H_{1} (A) = 0$ and is $Z / 2$ if $H_{1} (A) = Z / 2$ : for the second calculation use the free resolution $0 \to Z \overset{2}{\to} Z \to Z / 2 \to 0$ , whose dual has cokernel $Z / 2$ . Thus $H_{1} (A)$ and $H_{1} (B)$ are simultaneously zero or simultaneously $Z / 2$ , contradicting step 2.1. No smooth embedding in $S^{4}$ exists. Composing a putative embedding in $R^{4}$ with inverse stereographic projection would give one in $S^{4}$ , proving the last assertion.

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

Characteristic-class vanishing is only necessary for embedding

The boundary recorded

Assume AC. Remark. For a closed smooth $m$ -manifold every embedding into $R^{n}$ with $n \geq m$ has a rank- $(n - m)$ normal bundle, so the vanishing of all normal Stiefel-Whitney classes in degrees $i \geq n - m$ (when $m \geq 1$ and $n - m \geq 1$ ), of the Euler class of an oriented embedded normal bundle (when $m \geq 1$ and $n - m \geq 1$ ), and of all normal Pontryagin classes with $2 i > n - m$ is necessary for embeddability; the top Stiefel–Whitney and oriented Euler vanishings are embedding-specific here (Embedding obstructions include all immersion normal-class obstructions). It is not sufficient: ${R P}^{3}$ has a rank-one trivial stable normal inverse. Indeed, for $q = (q_{0}, q_{1}, q_{2}, q_{3}) \in S^{3}$ , the three fields $(- q_{1}, q_{0}, - q_{3}, q_{2})$ , $(- q_{2}, q_{3}, q_{0}, - q_{1})$ and $(- q_{3}, - q_{2}, q_{1}, q_{0})$ are perpendicular to $q$ and pairwise orthonormal, by direct dot products, so frame $T_{q} S^{3}$ (The tangent space of a regular level set is the kernel). Each satisfies $X (- q) = - X (q)$ , so their differentials descend through $S^{3} \to {R P}^{3}$ . This map is a local diffeomorphism: on an affine chart its two local inverses are the representatives with the chosen coordinate $1$ , normalized to unit length with the two signs (Real projective space from affine charts). The descended fields therefore give a smooth tangent frame. Thus the parallelizable-manifold proposition gives $\overset{ˉ}{w} = \overset{ˉ}{p} = 1$ and a trivial rank-one inverse; its Euler class is zero by its constant nonzero section (Parallelizable manifolds have no stable characteristic-class obstruction to Euclidean immersion, A nowhere-zero section forces the Euler class to vanish). Nevertheless ${R P}^{3}$ does not smoothly embed in $R^{4}$ : $H_{1} ({R P}^{3}; Z) = Z / 2$ . To compute this, use one cell in each dimension $0, 1, 2, 3$ in the standard quotient cellulation. The 1-cell has coinciding endpoints, so $d_{1} = 0$ , and the boundary of the 2-cell maps twice around ${R P}^{1}$ , so $d_{2} = 2$ with consistent orientations; hence $\ker d_{1} / im d_{2} = Z / 2$ by the cellular incidence and homology theorems (Cellular boundary is the incidence degree matrix, Cellular homology computes singular homology; Hatcher, Algebraic Topology, Example 2.42, printed p.144). For the 3-cell, the two hemispheres of its attaching sphere cover the open 2-cell once each. Their signed incidence contributions are $1$ and $- 1$ : the antipodal identification on $S^{2}$ has degree $(- 1)^{3} = - 1$ (Degree of identity constant reflection and antipodal sphere maps). Thus $d_{3} = 1 - 1 = 0$ by the cellular incidence theorem, so $H_{2} ({R P}^{3}; Z) = 0$ and $H_{3} ({R P}^{3}; Z) = Z$ by the same cellular homology theorem. The descended tangent frame orients this closed connected three-manifold. The proved local obstruction A closed three-manifold with H_1 = Z/2 and H_2 = 0 does not embed in S^4 therefore rules out a smooth embedding in $S^{4}$ , and hence one in $R^{4}$ . Its proof uses the two complementary sides, Mayer–Vietoris, Alexander duality and UCT; no classification of embeddings is required.

Separately, characteristic classes do not classify embeddings up to isotopy. Let $i : S^{2} ↪ R^{3}$ be the unit sphere inclusion and $r (x_{1}, x_{2}, x_{3}) = (x_{1}, x_{2}, - x_{3})$ . Their radial fields $n_{i} (x) = x$ and $n_{r} (x) = r (x)$ trivialize their normal lines: reflection preserves inner products, so $⟨ r (x), d r_{x} (v) ⟩ = ⟨ x, v ⟩ = 0$ for tangent vectors $v$ . Thus the two embeddings have identical stable characteristic classes. If they were isotopic, flattening time near the endpoints and applying The isotopy extension theorem would give an ambient isotopy with time-one diffeomorphism $H$ satisfying $H \circ i = r$ . Continuity of the nonzero differential determinant along the isotopy makes $H$ orientation preserving. Since $H (S^{2}) = S^{2}$ , it permutes the two complementary components; it preserves the bounded ball component because the closure of its image is $H (B^{3})$ , which is compact, while the exterior has unbounded closure. Therefore $H (B^{3}) = B^{3}$ . An orientation-preserving diffeomorphism of the ball carries outward-pointing boundary vectors to outward-pointing vectors and preserves its induced boundary orientation (Induced boundary orientation). Its restriction to $S^{2}$ has degree $1$ (Degree of an orientation-preserving or reversing diffeomorphism), whereas $H ∣_{S^{2}} = r$ has degree $- 1$ (Degree of identity constant reflection and antipodal sphere maps), a contradiction. This proves the failure of isotopy classification by these classes. Isotopy extension applies to an isotopy already given and supplies no criterion for its existence.

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

The characteristic-class construction is cited, not rebuilt

The interface used

Assume AC. Remark. This page computes with the characteristic classes but does not construct them. The exact interfaces are: the Stiefel-Whitney classes $w_{i}$ and the conventions $w_{0} = 1$ , $w_{i} = 0$ for $i > rank$ of Stiefel–Whitney classes from the projective-bundle relation together with the Whitney product and naturality of Whitney sum formula for Stiefel–Whitney classes and Naturality of Stiefel–Whitney classes; the Euler class as the zero-section pullback of the Thom class of Euler class by zero-section pullback of the Thom class; and the Pontryagin classes $p_{i} (E) = (- 1)^{i} c_{2 i} (E_{C})$ of Pontryagin classes by complexification with the Whitney product away from two of Pontryagin Whitney product away from two and the odd-Chern two-torsion fact of Odd Chern classes of a complexified real bundle are two-torsion. Authoring must substitute these exact item ids and the two-torsion qualification before performing the calculations of The normal Pontryagin class is the rational inverse of the tangent Pontryagin class and High normal Pontryagin classes obstruct low-codimension immersions; no construction, normalization or product formula is minted here.

5 · Examples, counterexamples and false statements

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